ISM for
Problem 1-1
Determine the resultant internal normal force acting on the cross section through point A in each
column. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the column
has a mass of 200 kg/m.
( a) Given:
2
kg
wBC := 300
m
L BC := 3m
kg
wCA := 400
m
FB := 5kN
L CA := 1.2m
g := 9.81
m
s
FC := 3kN
Solution:
+
↑Σ Fy = 0;
(
)
(
)
FA − wBC⋅ g ⋅ L BC − wCA⋅ g ⋅ L CA − FB − 2FC = 0
(
)
(
)
FA := wBC⋅ g ⋅ LBC + wCA⋅ g ⋅ LCA + FB + 2FC
FA = 24.5 kN
( b) Given:
g := 9.81
L := 3m
m
2
s
FB := 8kN
w := 200
Ans
kg
m
F1 := 6kN
F2 := 4.5kN
Solution:
+
↑Σ Fy = 0;
FA − ( w⋅ L) ⋅ g − FB − 2F1 − 2F2 = 0
FA := ( w⋅ L ) ⋅ g + FB + 2F1 + 2F2
FA = 34.89 kN
Ans
Problem 1-2
Determine the resultant internal torque acting on the cross sections through points C and D of the
shaft. The shaft is fixed at B.
Given:
T A := 250N⋅ m
T CD := 400N⋅ m
T DB := 300N⋅ m
Solution:
Equations of equilibrium:
+
TA − TC = 0
T C := TA
T C = 250 N⋅ m
+
Ans
T A − TCD + TD = 0
T D := T CD − T A
T D = 150 N⋅ m
Ans
Problem 1-3
Determine the resultant internal torque acting on the cross sections through points B and C.
Given: T D := 500N⋅ m
T BC := 350N⋅ m
T AB := 600N⋅ m
Solution:
Equations of equilibrium:
Σ Mx = 0;
T B + T BC − T D = 0
T B := −TBC + TD
T B = 150 N⋅ m
Σ Mx = 0;
Ans
TC − TD = 0
T C := TD
T C = 500 N⋅ m
Ans
Problem 1-4
A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting
on the section through point A.
P := 80N
Given:
θ := 30deg
φ := 45deg
a := 0.3m
b := 0.1m
Solution:
Equations of equilibrium:
+
ΣF x'=0;
NA − P⋅ cos ( φ − θ ) = 0
NA := P⋅ cos ( φ − θ )
NA = 77.27 N
+ ΣF y'=0;
Ans
VA − P⋅ sin ( φ − θ ) = 0
VA := P⋅ sin ( φ − θ )
VA = 20.71 N
+ ΣΜA=0;
Ans
MA + P⋅ cos ( φ ) ⋅ a cos ( θ ) − P⋅ sin ( φ ) ⋅ ( b + a sin ( θ ) ) = 0
MA := −P⋅ cos ( φ ) ⋅ a cos ( θ ) + P⋅ sin ( φ ) ⋅ ( b + a sin ( θ ) )
MA = −0.555 N⋅ m
Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Problem 1-5
Determine the resultant internal loadings acting on the cross section through point D of member AB.
ME := 70N⋅ m
a := 0.05m
b := 0.3m
Given:
Solution:
Segment AB: Support Reactions
+ ΣΜA=0;
−ME − By⋅ ( 2⋅ a + b) = 0
−ME
By :=
By = −175 N
2a + b
⎛ 150 ⎞
Bx := By⋅ ⎜
200
Bx = −131.25 N
Segment DB:
NB := −Bx
VB := −By
+
ND + NB = 0
At B:
ΣF x=0;
⎝
⎠
ND := −NB
+
ΣF y=0;
ND = −131.25 N
Ans
VD = −175 N
Ans
VD + VB = 0
VD := −VB
+ ΣΜD=0;
+
−MD − ME − By⋅ ( a + b) = 0
MD := −ME − By⋅ ( a + b)
MD = −8.75 N⋅ m
Ans
Problem 1-6
The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal
loadings acting on the cross section at point D.
P := 5000N
Given:
a := 0.8m b := 1.2m
c := 0.6m d := 1.6m
e := 0.6m
Solution:
⎛ b⎞
⎝ d⎠
θ = 36.87 deg
⎛ a + b⎞ − θ
⎝ d ⎠
φ = 14.47 deg
θ := atan ⎜
φ := atan ⎜
Member AB:
+ ΣΜA=0;
FBC⋅ sin ( φ ) ⋅ ( a + b) − P⋅ ( b) = 0
P⋅ ( b)
FBC :=
sin ( φ ) ⋅ ( a + b)
FBC = 12.01 kN
Segment BD:
+
ΣF x=0;
−ND − FBC⋅ cos ( φ ) − P⋅ cos ( θ ) = 0
ND := −FBC⋅ cos ( φ ) − P⋅ cos ( θ )
ND = −15.63 kN
+ ΣF y=0;
Ans
VD + FBC⋅ sin ( φ ) − P⋅ sin ( θ ) = 0
VD := −FBC⋅ sin ( φ ) + P⋅ sin ( θ )
VD = 0 kN
+ ΣΜD=0;
Ans
d−c
(FBC⋅ sin(φ) − P⋅ sin(θ))⋅ sin(θ) − MD = 0
(
MD = 0 kN⋅ m
d−c
) sin(θ)
MD := FBC⋅ sin ( φ ) − P⋅ sin ( θ ) ⋅
Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-7
Solve Prob. 1-6 for the resultant internal loadings acting at point E.
P := 5000N
Given:
a := 0.8m b := 1.2m
c := 0.6m
d := 1.6m
e := 0.6m
Solution:
⎛ b⎞
⎝ d⎠
θ = 36.87 deg
⎛ a + b⎞ − θ
⎝ d ⎠
φ = 14.47 deg
θ := atan ⎜
φ := atan ⎜
Member AB:
+ ΣΜA=0;
FBC⋅ sin ( φ ) ⋅ ( a + b) − P⋅ ( b) = 0
FBC :=
P⋅ ( b)
sin ( φ ) ⋅ ( a + b)
FBC = 12.01 kN
Segment BE:
+
ΣF x=0;
−NE − FBC⋅ cos ( φ ) − P⋅ cos ( θ ) = 0
NE := −FBC⋅ cos ( φ ) − P⋅ cos ( θ )
NE = −15.63 kN
+ ΣF y=0;
Ans
VE + FBC⋅ sin ( φ ) − P⋅ sin ( θ ) = 0
VE := −FBC⋅ sin ( φ ) + P⋅ sin ( θ )
VE = 0 kN
Ans
+ ΣΜE=0;
(FBC⋅ sin(φ) − P⋅ sin(θ))⋅ e − ME = 0
(
)
ME := FBC⋅ sin ( φ ) − P⋅ sin ( θ ) ⋅ e
ME = 0 kN⋅ m
Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-8
The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist and
load weigh 1500 N, determine the resultant internal loadings in the crane on cross sections through
points A, B, and C.
P := 1500N
Given:
w := 750
a := 2.1m
b := 1.5m
c := 0.6m
d := 2.4m
N
m
e := 0.9m
Solution:
Equations of Equilibrium: For point A
+
ΣF x=0;
NA := 0
+
ΣF y=0;
VA − w⋅ e − P = 0
VA := w⋅ e + P
+ ΣΜA=0;
Ans
VA = 2.17 kN
Ans
MA = −1.654 kN⋅ m
Ans
−MA − ( w⋅ e) ⋅ ( 0.5 ⋅ e) − P⋅ ( e) = 0
MA := −( w⋅ e) ⋅ ( 0.5 ⋅ e) − P⋅ ( e)
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
+
ΣF x=0;
NB := 0
+
ΣF y=0;
VB − w⋅ ( d + e) − P = 0
VB := w⋅ ( d + e) + P
+ ΣΜB=0;
Ans
VB = 3.98 kN
Ans
−MB − [ w⋅ ( d + e) ] ⋅ [ 0.5 ⋅ ( d + e) ] − P⋅ ( d + e) = 0
MB := −[ w⋅ ( d + e) ] ⋅ [ 0.5 ⋅ ( d + e) ] − P⋅ ( d + e)
MB = −9.034 kN⋅ m
Ans
Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
+
ΣF x=0;
VC := 0
+
ΣF y=0;
−NC − w⋅ ( b + c + d + e) − P = 0
+ ΣΜB=0;
Ans
NC := −w⋅ ( b + c + d + e) − P
Ans
NC = −5.55 kN
−MC − [ w⋅ ( c + d + e) ] ⋅ [ 0.5 ⋅ ( c + d + e) ] − P⋅ ( c + d + e) = 0
MC := −[ w⋅ ( c + d + e) ] ⋅ [ 0.5 ⋅ ( c + d + e) ] − P⋅ ( c + d + e)
MC = −11.554 kN⋅ m
Ans
Note: Negative sign indicates that NC and MC acts in the opposite direction to that shown
on FBD.
Problem 1-9
The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the
tooth, i.e., at the centroid point A of section a-a.
P := 400N
Given:
Solution:
θ := 30deg
φ := 45deg
a := 4mm
b := 5.75mm
α := φ − θ
Equations of equilibrium: For section a -a
+
ΣF x'=0;
VA − P⋅ cos ( α ) = 0
VA := P⋅ cos ( α )
VA = 386.37 N
+ ΣF y'=0;
Ans
NA − sin ( α ) = 0
NA := P⋅ sin ( α )
NA = 103.53 N
+ ΣΜA=0;
Ans
−MA − P⋅ sin ( α ) ⋅ a + P⋅ cos ( α ) ⋅ b = 0
MA := −P⋅ sin ( α ) ⋅ a + P⋅ cos ( α ) ⋅ b
MA = 1.808 N⋅ m
Ans
Problem 1-10
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
section through point C. Assume the reactions at the supports A and B are vertical.
kN
w1 := 4.5
m
Given:
kN
w2 := 6.0
m
a := 1.8m
b := 1.8m
d := 1.35m
e := 1.35m
Solution:
c := 2.4m
L 1 := a + b + c
L 2 := d + e
Support Reactions:
+ ΣΜA=0;
L2 ⎞
⎛
By⋅ L1 − w1⋅ L 1 0.5 ⋅ L1 − 0.5w2⋅ L2 ⋅ ⎜ L 1 +
= 0
3 ⎠
⎝
L2 ⎞
⎛
By := w1⋅ L 1 ⋅ ( 0.5 ) + 0.5w2⋅ L2 ⋅ ⎜ 1 +
By = 22.82 kN
3⋅ L1
⎝
⎠
(
(
+
ΣF y=0;
)(
)
) (
(
)
)
Ay + By − w1⋅ L1 − 0.5w2⋅ L2 = 0
Ay := −By + w1⋅ L 1 + 0.5w2⋅ L 2
Ay = 12.29 kN
Equations of Equilibrium: For point C
+
ΣF x=0;
NC := 0
+
ΣF y=0;
⎡⎣Ay − w1⋅ ( a + b)⎤⎦ − VC = 0
Ans
VC := −⎡⎣Ay − w1⋅ ( a + b)⎤⎦
+ ΣΜC=0;
VC = 3.92 kN
Ans
MC = 15.07 kN⋅ m
Ans
MC + ⎡⎣w1⋅ ( a + b)⎤⎦ ⋅ 0.5 ⋅ ( a + b) − Ay⋅ ( a + b) = 0
MC := −⎡⎣w1⋅ ( a + b)⎤⎦ ⋅ 0.5 ⋅ ( a + b) + Ay⋅ ( a + b)
Note: Negative sign indicates that VC acts in the opposite direction to that shown on FBD.
Problem 1-11
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
sections through points D and E. Assume the reactions at the supports A and B are vertical.
kN
kN
w1 := 4.5
w2 := 6.0
m
m
a := 1.8m
b := 1.8m
c := 2.4m
Given:
d := 1.35m
Solution:
e := 1.35m
L 1 := a + b + c
L 2 := d + e
Support Reactions:
+ ΣΜA=0;
L2 ⎞
⎛
By⋅ L1 − w1⋅ L 1 0.5 ⋅ L1 − 0.5w2⋅ L2 ⋅ ⎜ L 1 +
= 0
3 ⎠
⎝
L2 ⎞
⎛
By := w1⋅ L 1 ⋅ ( 0.5 ) + 0.5w2⋅ L2 ⋅ ⎜ 1 +
By = 22.82 kN
3⋅ L1
⎝
⎠
(
)(
(
+
ΣF y=0;
)
) (
(
)
)
Ay + By − w1⋅ L1 − 0.5w2⋅ L2 = 0
Ay := −By + w1⋅ L 1 + 0.5w2⋅ L 2
Ay = 12.29 kN
Equations of Equilibrium: For point D
+
ΣF x=0;
ND := 0
+
ΣF y=0;
⎡⎣Ay − w1⋅ ( a)⎤⎦ − VD = 0
Ans
VD := Ay − w1⋅ ( a)
+ ΣΜD=0;
VD = 4.18 kN
Ans
MD = 14.823 kN⋅ m
Ans
VE = 2.03 kN
Ans
ME = −0.911 kN⋅ m
Ans
MD + ⎡⎣w1⋅ ( a)⎤⎦ ⋅ 0.5 ⋅ ( a) − Ay⋅ ( a) = 0
MD := −⎡⎣w1⋅ ( a)⎤⎦ ⋅ 0.5 ⋅ ( a) + Ay⋅ ( a)
Equations of Equilibrium: For point E
+
ΣF x=0;
NE := 0
+
ΣF y=0;
VE − 0.5w2⋅ ( 0.5 ⋅ e) = 0
Ans
VE := 0.5w2⋅ ( 0.5 ⋅ e)
+ ΣΜD=0;
⎛e⎞ = 0
⎝ 3⎠
⎛e⎞
ME := ⎡⎣−0.5 w2⋅ ( 0.5 ⋅ e)⎤⎦ ⋅ ⎜
⎝ 3⎠
−ME − ⎡⎣0.5w2⋅ ( 0.5 ⋅ e)⎤⎦ ⋅ ⎜
Note: Negative sign indicates that ME acts in the opposite direction to that shown on FBD.
Problem 1-12
Determine the resultant internal loadings acting on (a) section a-a and (b) section b-b. Each section is
located through the centroid, point C.
kN
m
θ := 45deg
a := 1.2m
b := 2.4m
w := 9
Given:
L := a + b
Solution:
Support Reactions:
+ ΣΜA=0;
−Bx⋅ L⋅ sin ( θ ) + ( w⋅ L) ( 0.5 ⋅ L) = 0
Bx :=
ΣF y=0;
+
( w⋅ L ) ⋅ ( 0.5 ⋅ L)
L⋅ sin ( θ )
Ay − w⋅ L ⋅ sin ( θ ) = 0
Ay := w⋅ L⋅ sin ( θ )
+
ΣF x=0;
Bx = 22.91 kN
Ay = 22.91 kN
Bx − w⋅ L⋅ cos ( θ ) + Ax = 0
Ax := w⋅ L⋅ cos ( θ ) − Bx
Ax = −0 kN
(a) Equations of equilibrium: For Section a - a :
+ ΣF x=0;
NC + Ay⋅ sin ( θ ) = 0
NC := −Ay⋅ sin ( θ )
+
ΣF y=0;
Ans
VC = −5.4 kN
Ans
VC + Ay⋅ cos ( θ ) − w⋅ a = 0
VC := −Ay⋅ cos ( θ ) + w⋅ a
+ ΣΜA=0;
NC = −16.2 kN
−MC − ( w⋅ a) ⋅ ( 0.5 ⋅ a) + Ay cos ( θ ) ⋅ a = 0
MC := ( w⋅ a) ⋅ ( 0.5 ⋅ a) − Ay cos ( θ ) ⋅ a
MC = −12.96 kN⋅ m
Ans
NC = −7.64 kN
Ans
VC = −15.27 kN
Ans
MC = −12.96 kN⋅ m
Ans
(b) Equations of equilibrium: For Section b - b :
+
ΣF x=0;
NC + w⋅ a⋅ cos ( θ ) = 0
NC := −w⋅ a⋅ cos ( θ )
+
ΣF y=0;
VC − w⋅ a⋅ sin ( θ ) + Ay = 0
VC := w⋅ a⋅ sin ( θ ) − Ay
+ ΣΜA=0;
−MC − ( w⋅ a) ⋅ ( 0.5 ⋅ a) + Ay cos ( θ ) ⋅ a = 0
MC := ( w⋅ a) ⋅ ( 0.5 ⋅ a) − Ay cos ( θ ) ⋅ a
Problem 1-13
Determine the resultant internal normal and shear forces in the member at (a) section a-a and (b)
section b-b, each of which passes through point A. Take θ = 60 degree. The 650-N load is applied
along the centroidal axis of the member.
P := 650N
Given:
θ := 60deg
(a) Equations of equilibrium: For Section a - a :
ΣF y=0;
+
P − Na_a = 0
Na_a := P
+
ΣF x=0;
Na_a = 650 N
Ans
Va_a := 0
Ans
(b) Equations of equilibrium: For Section b - b :
+
ΣF y=0;
−Vb_b + P⋅ cos ( 90deg − θ ) = 0
Vb_b := P⋅ cos ( 90deg − θ )
Vb_b = 562.92 N
+ ΣF x=0;
Ans
Nb_b − P⋅ sin ( 90deg − θ ) = 0
Nb_b := P⋅ sin ( 90deg − θ )
Nb_b = 325 N
Ans
Problem 1-14
Determine the resultant internal normal and shear forces in the member at section b-b, each as a
o
o
function of θ. Plot these results for 0 ≤ θ ≤ 90 . The 650-N load is applied along the centroidal axis
of the member.
Given:
P := 650N
θ := 0
Equations of equilibrium: For Section b - b :
+
ΣF x0;
Nb_b − P⋅ cos ( θ ) = 0
Nb_b := P⋅ cos ( θ )
+
ΣF y=0;
Ans
−Vb_b + P⋅ cos ( θ ) = 0
Vb_b := −P⋅ cos ( θ )
Ans
Problem 1-15
The 4000-N load is being hoisted at a constant speed using the motor M, which has a weight of 450 N.
Determine the resultant internal loadings acting on the cross section through point B in the beam. The
beam has a weight of 600 N/m and is fixed to the wall at A.
Given:
W1 := 4000N
w := 600
N
m
W2 := 450N
a := 1.2m
b := 1.2m
c := 0.9m
d := 0.9m
e := 1.2m
f := 0.45m
r := 0.075m
Solution:
Tension in rope:
T :=
W1
2
T = 2.00 kN
Equations of Equilibrium: For point B
+
ΣF x=0;
(−NB − T) = 0
NB := −T
+
ΣF y=0;
Ans
VB = 4.72 kN
Ans
VB − w⋅ ( e) − W1 = 0
VB := w⋅ ( e) + W1
+ ΣΜB=0;
NB = −2 kN
−MB − [ w⋅ ( e) ] ⋅ 0.5 ⋅ ( e) − W1⋅ ( e + r) + T ⋅ ( f) = 0
MB := −[ w⋅ ( e) ] ⋅ 0.5 ⋅ ( e) − W1⋅ ( e + r) + T⋅ ( f)
MB = −4.632 kN⋅ m
Ans
Problem 1-16
Determine the resultant internal loadings acting on the cross section through points C and D of the
beam in Prob. 1-15.
Given: W1 := 4000N
W2 := 450N
w := 600
a := 1.2m
b := 1.2m
d := 0.9m
e := 1.2m
f := 0.45m
r := 0.075m
N
m
c := 0.9m
Solution:
W1
T = 2.00 kN
2
Equations of Equilibrium: For point C
L C := d + e
Tension in rope:
+
+
ΣF x=0;
ΣF y=0;
+ ΣΜC=0;
T :=
(−NC − T) = 0
NC := −T
NC = −2 kN
Ans
( )
VC := w⋅ ( L C) + W1
VC = 5.26 kN
Ans
VC − w⋅ LC − W1 = 0
( )
( )
( )
( )
(
)
−MC − ⎡⎣w⋅ L C ⎤⎦ ⋅ 0.5 ⋅ LC − W1⋅ L C + r + T ⋅ ( f) = 0
(
)
MC := −⎡⎣w⋅ LC ⎤⎦ ⋅ 0.5 ⋅ L C − W1⋅ LC + r + T⋅ ( f)
MC = −9.123 kN⋅ m
Ans
Equations of Equilibrium: For point D
L D := b + c + d + e
+
ΣF x=0;
ND := 0
ND = 0 kN
Ans
+
ΣF y=0;
VD − w⋅ L D − W1 − W2 = 0
( )
VD := w⋅ ( LD) + W1 + W2
VD = 6.97 kN
Ans
+ ΣΜC=0;
( ) ( )
(
)
MD := −⎡⎣w⋅ ( L D)⎤⎦ ⋅ 0.5 ⋅ ( L D) − W1⋅ ( L D + r) − W2⋅ ( b)
−MD − ⎡⎣w⋅ LD ⎤⎦ ⋅ 0.5 ⋅ LD − W1⋅ LD + r − W2⋅ ( b) = 0
MD = −22.932 kN⋅ m Ans
Problem 1-17
Determine the resultant internal loadings acting on the cross section at point B.
Given:
Solution:
w := 900
kN
m
a := 1m
b := 4m
L := a + b
Equations of Equilibrium: For point B
+
ΣF x=0;
NB := 0
NB = 0 kN
+
ΣF y=0;
Ans
⎡ ⎛ b ⎞⎤
VB − 0.5 ⋅ ⎢w⋅ ⎜ ⎥ ⋅ ( b) = 0
L
⎣ ⎝ ⎠⎦
⎡ ⎛ b ⎞⎤
VB := 0.5 ⋅ ⎢w⋅ ⎜ ⎥ ⋅ ( b)
L
⎣ ⎝ ⎠⎦
VB = 1440 kN
+ ΣΜB=0;
Ans
⎡ ⎛ b ⎞⎤ ⋅ ( b) ⋅ ⎛ b ⎞ = 0
⎥
⎜
⎣ ⎝ L ⎠⎦
⎝ 3⎠
−MB − 0.5 ⋅ ⎢w⋅ ⎜
b
⎡ ⎛ b ⎞⎤
MB := −0.5 ⋅ ⎢w⋅ ⎜ ⎥ ⋅ ( b) ⋅
L
3
⎣ ⎝ ⎠⎦
MB = −1920 kN⋅ m
Ans
Problem 1-18
The beam supports the distributed load shown. Determine the resultant internal loadings acting on the
cross section through point C. Assume the reactions at the supports A and B are vertical.
kN
w1 := 0.5
m
Given:
a := 3m
kN
w2 := 1.5
m
Solution:
L := 3⋅ a
w := w2 − w1
Support Reactions:
+ ΣΜA=0;
⎛ 2L ⎞ = 0
By⋅ L − w1⋅ L ( 0.5 ⋅ L ) − [ 0.5 ( w) ⋅ L ] ⋅ ⎜
3
(
)
⎝
2⎞
⎛
By := ( w1⋅ L ) ⋅ ( 0.5 ) + [ 0.5 ( w) ⋅ L ] ⋅ ⎜
⎝ 3⎠
⎠
By = 5.25 kN
+
ΣF y=0;
Ay + By − w1⋅ L − 0.5 ( w) ⋅ L = 0
Ay := −By + w1⋅ L + 0.5 ( w) ⋅ L
Ay = 3.75 kN
Equations of Equilibrium: For point C
+
ΣF x=0;
NC := 0
+
ΣF y=0;
⎡ ⎛ a ⎞⎤
VC + w1⋅ a + 0.5 ⋅ ⎢w⋅ ⎜ ⎥ ⋅ ( a) − Ay = 0
L
NC = 0 kN
Ans
⎣ ⎝ ⎠⎦
⎡ ⎛ a ⎞⎤
VC := −w1⋅ a − 0.5 ⋅ ⎢w⋅ ⎜ ⎥ ⋅ ( a) + Ay
L
⎣ ⎝ ⎠⎦
VC = 1.75 kN
+ ΣΜC=0;
Ans
⎡ ⎛ a ⎞⎤
⎛ a ⎞ − A ⋅a = 0
MC + w1⋅ a ( 0.5 ⋅ a) + 0.5 ⋅ ⎢w⋅ ⎜ ⎥ ⋅ ( a) ⋅ ⎜
y
L
3
(
)
⎣ ⎝ ⎠⎦
⎝ ⎠
⎡ ⎛ a ⎞⎤
⎛ a ⎞ + A ⋅a
MC := −w1⋅ a ( 0.5 ⋅ a) − 0.5 ⋅ ⎢w⋅ ⎜ ⎥ ⋅ ( a) ⋅ ⎜
y
L
3
(
)
MC = 8.5 kN⋅ m
⎣ ⎝ ⎠⎦
Ans
⎝ ⎠
Problem 1-19
Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18.
kN
w1 := 0.5
m
Given:
a := 3m
kN
w2 := 1.5
m
L := 3⋅ a
Solution:
w := w2 − w1
Support Reactions:
+ ΣΜA=0;
⎛ 2L ⎞ = 0
By⋅ L − w1⋅ L ( 0.5 ⋅ L ) − [ 0.5 ( w) ⋅ L ] ⋅ ⎜
3
(
)
⎝
2⎞
⎛
By := ( w1⋅ L ) ⋅ ( 0.5 ) + [ 0.5 ( w) ⋅ L ] ⋅ ⎜
⎝ 3⎠
⎠
By = 5.25 kN
ΣF y=0;
+
Ay + By − w1⋅ L − 0.5 ( w) ⋅ L = 0
Ay := −By + w1⋅ L + 0.5 ( w) ⋅ L
Ay = 3.75 kN
Equations of Equilibrium: For point D
+
ΣF x=0;
ND := 0
+
ΣF y=0;
⎡ ⎛ 2a ⎞⎤ ⋅ ( 2a) − A = 0
VD + w1⋅ ( 2a) + 0.5 ⋅ ⎢w⋅ ⎜
⎥
y
L
ND = 0 kN
Ans
⎣ ⎝ ⎠⎦
⎡ ⎛ 2⋅ a ⎞⎤ ⋅ ( 2a) + A
VD := −w1⋅ ( 2a) − 0.5 ⋅ ⎢w⋅ ⎜
⎥
y
L
⎣ ⎝
VD = −1.25 kN
+ ΣΜD=0;
⎠⎦
Ans
⎡ ⎛ 2⋅ a ⎞⎤ ⋅ ( 2a) ⋅ ⎛ 2a ⎞ − A ⋅ ( 2a) = 0
MD + ⎡⎣w1⋅ ( 2a)⎤⎦ ( a) + 0.5 ⋅ ⎢w⋅ ⎜
⎥
⎜
y
L
3
⎣ ⎝
⎠⎦
⎝ ⎠
⎡ ⎛ 2⋅ a ⎞⎤ ⋅ ( 2a) ⋅ ⎛ 2a ⎞ + A ⋅ ( 2a)
MD := ⎡⎣−w1⋅ ( 2a)⎤⎦ ( a) − 0.5 ⋅ ⎢w⋅ ⎜
⎥
⎜
y
L
3
⎣ ⎝
MD = 9.5 kN⋅ m
Ans
⎠⎦
⎝ ⎠
Problem 1-20
The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kN
on the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internal
loadings at cross sections through points D, E, and F.
P := 4kN
Given:
a := 1.2m
b := 1.8m
Solution:
Support Reactions: FBD (a) and (b).
Given
+ ΣΜA=0;
By⋅ ( a) + Bx⋅ ( 0.5 ⋅ b) − P⋅ ( a) = 0
[1]
+ ΣΜC=0;
Bx⋅ ( 0.5 ⋅ b) + P⋅ ( a) − By⋅ ( a) − P⋅ ( a) = 0
[2]
Bx := 1kN
Solving [1] and [2]: Initial guess:
⎛⎜ Bx ⎞
:= Find ( Bx , By)
⎜ By
⎝ ⎠
By := 2kN
⎛⎜ Bx ⎞ ⎛ 2.67 ⎞
=⎜
kN
⎜ By
2
⎝
⎠
⎝ ⎠
From FBD (a):
+
ΣF x=0;
ΣF y=0;
+
Bx − Ax = 0
Ax := Bx
Ax = 2.67 kN
Ay − P − By = 0
Ay := P + By
Ay = 6 kN
From FBD (b):
+
+
ΣF x=0;
ΣF y=0;
Cx − Bx = 0
Cx := Bx
Cx = 2.67 kN
Cy + By − P − P = 0
Cy := 2P − By
Cy = 6 kN
Equations of Equilibrium: For point D [FBD (c)].
+
ΣF x=0;
VD := 0
VD = 0 kN
Ans
+
ΣF y=0;
ND := 0
ND = 0 kN
Ans
+ ΣΜD=0;
MD := 0
MD = 0 kN⋅ m
Ans
For point E [FBD (d)].
+
ΣF x=0;
Ax − VE = 0
VE := Ax
VE = 2.67 kN
Ans
+
ΣF y=0;
NE − Ay = 0
NE := Ay
NE = 6 kN
Ans
+ ΣΜE=0;
ME − Ax⋅ ( 0.5 ⋅ b) = 0
ME := Ax⋅ ( 0.5 ⋅ b)
ME = 2.4 kN⋅ m
Ans
For point F [FBD (e)].
+
ΣF x=0;
VF + Ax − Cx = 0
VF := −Ax + Cx
VF = 0 kN
Ans
+
ΣF y=0;
NF − Ay ⋅ −Cy = 0
NF := Ay + Cy
NF = 12 kN
Ans
+ ΣΜF=0;
(
)
MF − Ax + Cx ⋅ ( 0.5 ⋅ b) = 0
(
)
MF := Ax + Cx ⋅ ( 0.5 ⋅ b)
MF = 4.8 kN⋅ m Ans
Problem 1-21
The drum lifter suspends the 2.5-kN drum. The linkage is pin connected to the plate at A and B. The
gripping action on the drum chime is such that only horizontal and vertical forces are exerted on the
drum at G and H. Determine the resultant internal loadings on the cross section through point I.
Given:
P := 2.5 kN
θ := 60deg
a := 200mm
b := 125mm
c := 75mm
d := 125mm
e := 125mm
f := 50mm
Solution:
Equations of Equilibrium: Memeber Ac and BD are
two-force members.
ΣF y=0;
P − 2⋅ F sin ( θ ) = 0
F :=
P
2⋅ sin ( θ )
[1]
[2]
F = 1.443 kN
Equations of Equilibrium: For point I.
+
ΣF x=0;
VI − F⋅ cos ( θ ) = 0
VI := F⋅ cos ( θ )
VI = 0.722 kN
Ans
+
ΣF y=0;
−NI + F⋅ sin ( θ ) = 0
NI := F⋅ sin ( θ )
NI = 1.25 kN
Ans
+ ΣΜI=0;
−MI + F⋅ cos ( θ ) ⋅ ( a) = 0
MI := F⋅ cos ( θ ) ⋅ ( a)
MI = 0.144 kN⋅ m
Ans
Problem 1-22
Determine the resultant internal loadings on the cross sections through points K and J on the drum lifter
in Prob. 1-21.
Given:
P := 2.5 kN
θ := 60deg
a := 200mm
b := 125mm
c := 75mm
d := 125mm
e := 125mm
f := 50mm
Solution:
Equations of Equilibrium: Memeber Ac and BD are
two-force members.
ΣF y=0;
P − 2⋅ F sin ( θ ) = 0
F :=
P
2⋅ sin ( θ )
[1]
[2]
F = 1.443 kN
Equations of Equilibrium: For point J.
+ ΣF y'=0;
+
ΣF x'=0;
NI + F = 0
+ ΣΜJ=0;
VI := 0
VI = 0 kN
Ans
NI := −F
NI = −1.443 kN
Ans
MJ := 0
MJ = 0 kN⋅ m
Ans
Note: Negative sign indicates that NJ acts in the opposite direction to that shown on FBD.
Support Reactions: For Member DFH :
+ ΣΜH=0;
FEF⋅ ( c) − F cos ( θ ) ⋅ ( a + b + c) + F⋅ sin ( θ ) ⋅ ( f) = 0
⎛ a + b + c ⎞ − F⋅ sin ( θ ) ⋅ ⎛ f ⎞
⎜
⎝ c ⎠
⎝ c⎠
FEF := F⋅ cos ( θ ) ⋅ ⎜
FEF = 3.016 kN
Equations of Equilibrium: For point K.
+
ΣF x=0;
+
NK + FEF = 0
NK := FEF
NK = 3.016 kN
Ans
ΣF y=0;
VK := 0
VK = 0 kN
Ans
+ ΣΜK=0;
MK := 0
MK = 0 kN⋅ m
Ans
Problem 1-23
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section at B. Neglect the weight of the wrench CD.
Given:
g := 9.81
m
2
ρ := 12
s
kg
m
a := 0.150m
b := 0.400m
c := 0.200m
d := 0.300m
Solution:
w := ρ ⋅ g
ΣF x=0;
NBx := 0N
Ans
ΣF y=0;
VBy := 0N
Ans
ΣF z=0;
VBz − P + P − w⋅ ( b + c) = 0
P := 60N
VBz := P − P + w⋅ ( b + c)
VBz = 70.6 N
ΣΜx=0;
Ans
T Bx + P⋅ ( b) − P⋅ ( b) − ( w⋅ b) ⋅ ( 0.5 ⋅ b) = 0
T Bx := −P⋅ ( b) + P⋅ ( b) + ( w⋅ b) ⋅ ( 0.5 ⋅ b)
ΣΜy=0;
Ans
MBy = 6.23 N⋅ m
Ans
MBy − P⋅ ( 2a) + ( w⋅ b) ⋅ ( c) + ( w⋅ c) ⋅ ( 0.5 ⋅ c) = 0
MBy := P⋅ ( 2⋅ a) − ( w⋅ b) ⋅ ( c) − ( w⋅ c) ⋅ ( 0.5 ⋅ c)
ΣΜz=0;
T Bx = 9.42 N⋅ m
MBz := 0N⋅ m
Ans
Problem 1-24
The main beam AB supports the load on the wing of the airplane. The loads consist of the wheel
reaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity at
D, and the 2-kN weight of the wing, having a center of gravity at E. If it is fixed to the fuselage at A,
determine the resultant internal loadings on the beam at this point. Assume that the wing does not
transfer any of the loads to the fuselage, except through the beam.
Given:
PC := 175kN
PE := 2kN
a := 1.8m
b := 1.2m
c := 0.6m
d := 0.45m
PD := 6kN
e := 0.3m
Solution:
ΣF x=0;
VAx := 0kN
Ans
ΣF y=0;
NAy := 0kN
Ans
ΣF z=0;
VAz − PD − PE + PC = 0
VAz := PD + PE − PC
VAz = −167 kN
ΣΜx=0;
Ans
MAx − PD⋅ ( a) − PE ⋅ ( a + b + c) + PC⋅ ( a + b) = 0
MAx := PD⋅ ( a) + PE⋅ ( a + b + c) − PC⋅ ( a + b)
ΣΜy=0;
Ans
T Ay = −2.1 kN⋅ m
Ans
T Ay + PD⋅ ( d) − PE ⋅ ( e) = 0
T Ay := −PD⋅ ( d) + PE⋅ ( e)
ΣΜz=0;
MAx = −507 kN⋅ m
MAz := 0kN⋅ m
Ans
Problem 1-25
Determine the resultant internal loadings acting on the cross section through point B of the signpost.
The post is fixed to the ground and a uniform pressure of 50 N/m2 acts perpendicular to the face of the
sign.
Given:
a := 4m
d := 2m
b := 6m
e := 3m
p := 50
m
c := 3m
Solution:
ΣF x=0;
P := p⋅ ( c) ⋅ ( d + e)
VBx − P = 0
VBx := P
VBx = 750 N
Ans
ΣF y=0;
VBy := 0N
Ans
ΣF z=0;
NBz := 0N
Ans
ΣΜx=0;
MBx := 0N⋅ m
Ans
ΣΜy=0;
MBy − P⋅ ( b + 0.5 ⋅ c) = 0
MBy := P⋅ ( b + 0.5 ⋅ c)
MBy = 5625 N⋅ m
ΣΜz=0;
Ans
T Bz − P⋅ [ e − 0.5 ⋅ ( d + e) ] = 0
T Bz := P⋅ [ e − 0.5 ⋅ ( d + e) ]
T Bz = 375 N⋅ m
N
Ans
2
Problem 1-26
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given:
Solution:
P1z := 400N
P2y := 200N
P3y := 80N
a := 0.3m
b := 0.4m
c := 0.3m
d := 0.4m
L := a + b + c + d
Support Reactions:
ΣΜz=0;
2P3y⋅ ( d) + 2P2y⋅ ( c + d) − Ay⋅ ( L ) = 0
d
c+d
Ay := 2P3y⋅ + 2P2y⋅
L
L
Ay = 245.71 N
ΣF y=0;
−Ay − By + 2⋅ P2y + 2⋅ P3y = 0
By := −Ay + 2⋅ P2y + 2⋅ P3y
ΣΜy=0;
2P1z⋅ ( b + c + d) − Az⋅ ( L ) = 0
b+c+d
L
Az = 628.57 N
Bz + Az − 2⋅ P1z = 0
Bz := −Az + 2⋅ P1z
Bz = 171.43 N
Az := 2P1z⋅
ΣF z=0;
By = 314.29 N
Equations of Equilibrium: For point D.
ΣF x=0;
NDx := 0N
ΣF y=0;
VDy − By + 2⋅ P3y = 0
Ans
VDy := By − 2⋅ P3y
VDy = 154.3 N
ΣF z=0;
Ans
VDz + Bz = 0
VDz := −Bz
VDz = −171.4 N
Ans
ΣΜx=0;
T Dx := 0N⋅ m
Ans
ΣΜy=0;
MDy + Bz⋅ ( d + 0.5 ⋅ c) = 0
MDy := −⎡⎣Bz⋅ ( d + 0.5 ⋅ c)⎤⎦
ΣΜz=0;
MDy = −94.29 N⋅ m
Ans
MDz = −148.86 N⋅ m
Ans
MDz + By⋅ ( d + 0.5 ⋅ c) − 2⋅ P3y⋅ ( 0.5 ⋅ c) = 0
MDz := −By⋅ ( d + 0.5 ⋅ c) + 2⋅ P3y⋅ ( 0.5 ⋅ c)
Problem 1-27
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given:
Solution:
P1z := 400N
P2y := 200N
a := 0.3m
b := 0.4m
P3y := 80N
c := 0.3m
d := 0.4m
L := a + b + c + d
Support Reactions:
ΣΜz=0;
2P3y⋅ ( d) + 2P2y⋅ ( c + d) − Ay⋅ ( L ) = 0
d
c+d
Ay := 2P3y⋅ + 2P2y⋅
L
L
ΣF y=0;
−Ay − By + 2⋅ P2y + 2⋅ P3y = 0
By := −Ay + 2⋅ P2y + 2⋅ P3y
ΣΜy=0;
By = 314.29 N
2P1z⋅ ( b + c + d) − Az⋅ ( L ) = 0
b+c+d
L
Az = 628.57 N
Bz + Az − 2⋅ P1z = 0
Bz := −Az + 2⋅ P1z
Bz = 171.43 N
Az := 2P1z⋅
ΣF z=0;
Ay = 245.71 N
Equations of Equilibrium: For point C.
ΣF x=0;
NCx := 0N
ΣF y=0;
VCy − Ay = 0
Ans
VCy := Ay
VCy = 245.7 N
ΣF z=0;
Ans
VCz + Az − 2P1z = 0
VCz := −Az + 2P1z
Ans
VCz = 171.4 N
ΣΜx=0;
T Cx := 0N⋅ m
ΣΜy=0;
MCy − Az⋅ ( a + 0.5 ⋅ b) + 2⋅ P1z⋅ ( 0.5 ⋅ b) = 0
Ans
MCy := Az⋅ ( a + 0.5 ⋅ b) − 2⋅ P1z⋅ ( 0.5 ⋅ b)
ΣΜz=0;
MCy = 154.29 N⋅ m
Ans
MCz = −122.86 N⋅ m
Ans
MCz + Ay⋅ ( a + 0.5 ⋅ b) = 0
MCz := −Ay⋅ ( a + 0.5 ⋅ b)
Problem 1-28
Determine the resultant internal loadings acting on the cross section of the frame at points F and G.
The contact at E is smooth.
Given: a := 1.2m
b := 1.5m
c := 0.9m
d := 0.9m
e := 1.2m
θ := 30deg
Solution:
L :=
2
P := 400N
2
d +e
Member DEF :
+ ΣΜD=0;
Member BCE :
+ ΣΜB=0;
NE ⋅ ( b) − P⋅ ( a + b) = 0
a+b
NE := P⋅
NE = 720 N
b
⎛ e ⎞ ⋅ ( d) − N ⋅ sin ( θ ) ⋅ ( c + d) = 0
E
⎝ L⎠
⎛ L ⎞ ⋅ N ⋅ sin ( θ ) ⋅ ( c + d)
FAC := ⎜
⎡
⎤⎦
⎝ e⋅ d ⎠ ⎣ E
FAC⋅ ⎜
FAC = 900 N
+
ΣF x=0;
⎛ d ⎞ − N ⋅ cos ( θ ) = 0
Bx + FAC⋅ ⎜
E
L
⎝ ⎠
⎛ d ⎞ + N ⋅ cos ( θ )
Bx := −FAC⋅ ⎜
E
⎝L⎠
Bx = 83.54 N
+
ΣF y=0;
⎛ e ⎞ − N ⋅ sin ( θ ) = 0
E
⎝ L⎠
⎛ e ⎞ − N ⋅ sin ( θ )
By := FAC⋅ ⎜
E
⎝L⎠
−By + FAC⋅ ⎜
By = 360 N
Equations of Equilibrium: For point F.
+ ΣF y'=0;
+
ΣF x'=0;
+ ΣΜF=0;
NF := 0
NF = 0 N
Ans
VF − P = 0
VF := P
VF = 400 N
Ans
MF − P⋅ ( 0.5 ⋅ a) = 0
MF := P⋅ ( 0.5 ⋅ a)
MF = 240 N⋅ m
Ans
Equations of Equilibrium: For point G.
+
+
ΣF x=0;
Bx − NG = 0
NG := Bx
NG = 83.54 N
Ans
ΣF y=0;
VG − By = 0
VG := By
VG = 360 N
Ans
MG = 162 N⋅ m
Ans
+ ΣΜG=0;
−MG + By⋅ ( 0.5 ⋅ d) = 0 MG := By⋅ ( 0.5 ⋅ d)
Problem 1-29
The bolt shank is subjected to a tension of 400 N. Determine the resultant internal loadings acting on
the cross section at point C.
Given:
P := 400N
r := 150mm
θ := 90deg
Solution:
Equations of Equilibrium: For segment AC.
+
+
ΣF x=0;
NC + P = 0
ΣF y=0;
+ ΣΜG=0;
MC + P⋅ ( r) = 0
NC := P
NC = 400 N
Ans
VC := 0
VC = 0 N
Ans
MC := −P⋅ ( r)
MC = −60 N⋅ m
Ans
Problem 1-30
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section through B.
Given:
P := 750N
ρ := 12
kg
m
a := 1m
Solution:
MC := 800N⋅ m
g := 9.81
m
2
s
b := 2m
⎛ 4⎞⋅ P
⎝ 5⎠
Py := ⎜
c := 2m
3
Pz := − ⋅ P
5
Equations of Equilibrium: For point B.
ΣF x=0;
VBx := 0kip
ΣF y=0;
NBy + Py = 0
ΣF z=0;
Ans
NBy := −Py
Ans
NBy = −600 N
Ans
VBz + Pz − ρ ⋅ g⋅ c − ρ ⋅ g⋅ b = 0
VBz := −Pz + ρ ⋅ g⋅ c + ρ ⋅ g⋅ b
VBz = 920.9 N
ΣΜx=0;
Ans
MBx + Pz⋅ ( b) − ρ ⋅ g⋅ c⋅ ( b) − ρ ⋅ g⋅ b⋅ ( 0.5 ⋅ b) = 0
MBx := −Pz⋅ ( b) + ρ ⋅ g⋅ c⋅ ( b) + ρ ⋅ g⋅ b⋅ ( 0.5 ⋅ b)
MBx = 1606.3 N⋅ m Ans
ΣΜy=0;
T By := 0N⋅ m
ΣΜz=0;
MBy + Mc = 0
MBy := −MC
MBy = −800 N⋅ m
Ans
Ans
Problem 1-31
The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings
acting on the cross section through A which is located at an angle θ from the horizontal.
Solution:
P := kN
θ := deg
Equations of Equilibrium: For point A.
+ ΣF x=0;
−NA + P⋅ cos ( θ ) = 0
NA := P⋅ cos ( θ )
+
ΣF y=0;
VA − P⋅ sin ( θ ) = 0
VA := P⋅ sin ( θ )
+ ΣΜA=0;
Ans
Ans
MA − P⋅ r⋅ ( 1 − cos ( θ ) ) = 0
MA := P⋅ r⋅ ( 1 − cos ( θ ) )
Ans
Problem 1-32
The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determine
the resultant internal loadings acting on the cross section through point B. Hint: The distance from the
centroid C of segment AB to point O is CO = 0.9745r.
Given:
θ := 22.5deg
r := m
a := 0.9745r
w :=
kN
m
2
Solution:
Equations of Equilibrium: For point B.
ΣF z=0;
π
VB − ⋅ r⋅ w = 0
4
ΣF x=0;
ΣΜx=0;
ΣΜy=0;
π
⋅ r⋅ w⋅ ( 0.09968r) = 0
4
π
MB + ⋅ r⋅ w⋅ ( 0.37293r) = 0
4
TB −
VB := 0.785 ⋅ w⋅ r
Ans
NB := 0
Ans
T B := 0.0783w⋅ r
2
MB := −0.293 w⋅ r
2
Ans
Ans
Problem 1-33
A differential element taken from a curved bar is shown in the figure. Show that dN/dθ = V,
dV/dθ = -N, dM/dθ = -T , and dT/dθ = M,
Solution:
Problem 1-34
The column is subjected to an axial force of 8 kN, which is applied through the centroid of the
cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution
of stress acting over the area’s cross section.
Given:
P := 8kN
b := 150mm
d := 140mm
t := 10mm
Solution:
A := 2( b⋅ t) + d⋅ t
σ :=
P
A
2
A = 4400.00 mm
σ = 1.82 MPa
Ans
Problem 1-35
The anchor shackle supports a cable force of 3.0 kN. If the pin has a diameter of 6 mm, determine the
average shear stress in the pin.
Given:
P := 3.0kN
d := 6mm
Solution:
+
ΣF y=0;
2⋅ V − P = 0
V := 0.5P
V = 1500 N
2
A :=
π⋅d
4
A = 28.2743 mm
V
A
τ avg = 53.05 MPa
τ avg :=
2
Ans
Problem 1-36
While running the foot of a 75-kg man is momentarily subjected to a force which is 5 times his
weight. Determine the average normal stress developed in the tibia T of his leg at the mid section a-a.
The cross section can be assumed circular, having an outer diameter of 45 mm and an inner diameter
of 25 mm. Assume the fibula F does not support a load.
Given:
g = 9.81
m
2
s
M := 75kg
do := 45mm
di := 25mm
A :=
π ⎛ 2
2
⋅ ⎝ do − di ⎞⎠
4
A = 1099.5574 mm
σ :=
5M⋅ g
A
σ = 3.345 MPa
Solution:
2
Ans
Problem 1-37
The thrust bearing is subjected to the loads shown. Determine the average normal stress developed on
cross sections through points B, C, and D. Sketch the results on a differential volume element located
at each section.
3
kPa := 10 Pa
Units Used:
Given:
P := 500N
Q := 200N
dB := 65mm
dC := 140mm
Solution:
2
AB :=
π ⋅ dB
2
AB = 3318.3 mm
4
P
σ B :=
AB
σ B = 150.7 kPa
Ans
2
AC :=
π ⋅ dC
2
AC = 15393.8 mm
4
P
σ C :=
AC
σ C = 32.5 kPa
Ans
2
AD :=
π ⋅ dD
4
Q
σ D :=
AD
2
AD = 7854.0 mm
σ D = 25.5 kPa
Ans
dD := 100mm
Problem 1-38
The small block has a thickness of 5 mm. If the stress distribution at the support developed by the load
varies as shown, determine the force F applied to the block, and the distance d to where it is applied.
Given:
a := 60mm
b := 120mm
t := 5mm
σ 1 := 0MPa
σ 2 := 40MPa
σ 3 := 60MPa
Solution:
⌠
F = ⎮ σ dA
⌡
(
)
F := 0.5 ⋅ σ 2⋅ ( a⋅ t) + σ 2⋅ ( b⋅ t) + 0.5 ⋅ σ 3 − σ 2 ⋅ ( b⋅ t)
F = 36.00 kN
Ans
Require:
⌠
F ⋅ d = ⎮ x⋅ σ dA
⌡
d :=
2a
⎛ 2⋅ b ⎞
⎡⎣0.5⋅ σ 2⋅ ( a⋅ t)⎤⎦ ⋅ + ⎡⎣σ 2⋅ ( b⋅ t)⎤⎦ ⋅ ( a + 0.5⋅ b) + ⎡⎣0.5⋅ ( σ 3 − σ 2) ⋅ ( b⋅ t)⎤⎦ ⋅ ⎜ a +
3 ⎠
3
⎝
d = 110 mm
F
Ans
Problem 1-39
The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a
couple is applied to the lever, determine the average shear stress in the pin between the pin and lever.
Given:
a := 250mm
b := 12mm
d := 6mm
P := 20N
Solution:
+ ΣΜO=0;
V⋅ b − P⋅ ( 2a) = 0
⎛ 2a ⎞
⎝b⎠
V := P⋅ ⎜
V = 833.33 N
2
π⋅d
A :=
4
A = 28.2743 mm
V
A
τ avg = 29.47 MPa
τ avg :=
2
Ans
Problem 1-40
The cinder block has the dimensions shown. If the material fails when the average normal stress
reaches 0.840 MPa, determine the largest centrally applied vertical load P it can support.
Given:
σ allow := 0.840MPa
ao := 150mm
ai := 100mm
bo := [ 2⋅ ( 1 + 2 + 3) + 2] ⋅ mm
bi := [ 2⋅ ( 1 + 3) ] ⋅ mm
Solution:
A := ao⋅ bo − ai⋅ bi
2
A = 1300 mm
Pallow := σ allow⋅ ( A)
Pallow = 1.092 kN
Ans
Problem 1-41
The cinder block has the dimensions shown. If it is subjected to a centrally applied force of P = 4 kN,
determine the average normal stress in the material. Show the result acting on a differential volume
element of the material.
Given:
P := 4kN
ao := 150mm
ai := 100mm
bo := [ 2⋅ ( 1 + 2 + 3) + 2] ⋅ mm
bi := [ 2⋅ ( 1 + 3) ] ⋅ mm
Solution:
A := ao⋅ bo − ai⋅ bi
σ :=
P
A
σ = 3.08 MPa
Ans
2
A = 1300 mm
Problem 1-42
The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod is
subjected to the greater average normal stress and compute its value. Take θ = 30°. The diameter of
each rod is given in the figure.
Given:
W := 250N
θ := 30deg
φ := 45deg
dB := 9mm
dC := 6mm
dD := 7.5mm
Initial guess: FAC := 1N
Given
Solution:
FAD := 1N
+
ΣF x=0;
FAC⋅ cos ( θ ) − FAD⋅ cos ( φ ) = 0
[1]
+
ΣF y=0;
FAC⋅ sin ( θ ) + FAD⋅ sin ( φ ) − W = 0
[2]
⎛⎜ FAC ⎞
:= Find ( FAC , FAD)
⎜ FAD
⎝
⎠
Solving [1] and [2]:
⎛⎜ FAC ⎞ ⎛ 183.01 ⎞
=⎜
N
⎜ FAD
224.14
⎝
⎠
⎝
⎠
Rod AB:
2
AAB :=
π ⋅ dB
2
AAB = 63.61725 mm
4
W
σ AB :=
AAB
σ AB = 3.93 MPa
Rod AD :
2
AAD :=
σ AD :=
π ⋅ dD
2
AAD = 44.17865 mm
4
FAD
σ AD = 5.074 MPa
AAD
Rod AC:
2
AAC :=
σ AC :=
π ⋅ dC
4
FAC
AAC
2
AAC = 28.27433 mm
σ AC = 6.473 MPa
Ans
Problem 1-43
Solve Prob. 1-42 for θ = 45°.
Given:
W := 250N
θ := 45deg
φ := 45deg
dB := 9mm
dC := 6mm
dD := 7.5mm
Initial guess: FAC := 1N
Given
Solution:
FAD := 1N
+
ΣF x=0;
FAC⋅ cos ( θ ) − FAD⋅ cos ( φ ) = 0
[1]
+
ΣF y=0;
FAC⋅ sin ( θ ) + FAD⋅ sin ( φ ) − W = 0
[2]
⎛⎜ FAC ⎞
:= Find ( FAC , FAD)
⎜ FAD
⎝
⎠
Solving [1] and [2]:
⎛⎜ FAC ⎞ ⎛ 176.78 ⎞
=⎜
N
⎜ FAD
⎝
⎠ ⎝ 176.78 ⎠
Rod AB:
2
AAB :=
π ⋅ dB
2
AAB = 63.61725 mm
4
W
σ AB :=
AAB
σ AB = 3.93 MPa
Rod AD :
2
AAD :=
σ AD :=
π ⋅ dD
2
AAD = 44.17865 mm
4
FAD
σ AD = 4.001 MPa
AAD
Rod AC:
2
AAC :=
σ AC :=
π ⋅ dC
4
FAC
AAC
2
AAC = 28.27433 mm
σ AC = 6.252 MPa
Ans
Problem 1-44
The 250-N lamp is supported by three steel rods connected by a ring at A. Determine the angle of
orientation θ of AC such that the average normal stress in rod AC is twice the average normal stress in
rod AD. What is the magnitude of stress in each rod? The diameter of each rod is given in the figure.
Given:
W := 250N
φ := 45deg
dB := 9mm
dC := 6mm
Solution:
dD := 7.5mm
2
AAB :=
Rod AB:
AAD :=
Rod AD :
π ⋅ dB
4
2
AAB = 63.61725 mm
2
π ⋅ dD
4
2
π ⋅ dC
Rod AC:
AAC :=
Since
σ AC = 2σ AD
4
2
AAD = 44.17865 mm
2
AAC = 28.27433 mm
Therefore
Initial guess: FAC := 1N FAD := 2N
Given
FAC
FAD
= 2
AAC
AAD
FAC
AAC
= 2
FAD
AAD
θ := 30deg
[1]
+
ΣF x=0;
FAC⋅ cos ( θ ) − FAD⋅ cos ( φ ) = 0
[2]
+
ΣF y=0;
FAC⋅ sin ( θ ) + FAD⋅ sin ( φ ) − W = 0
[3]
Solving [1], [2] and [3]:
W
σ AB :=
AAB
σ AD :=
σ AC :=
FAD
AAD
FAC
AAC
⎛ FAC ⎞
⎜
⎜ FAD ⎟ := Find ( FAC , FAD , θ )
⎜
⎝ θ ⎠
σ AB = 3.93 MPa
Ans
σ AD = 3.19 MPa
Ans
σ AC = 6.38 MPa
Ans
⎛⎜ FAC ⎞ ⎛ 180.38 ⎞
=⎜
N
⎜ FAD
⎝
⎠ ⎝ 140.92 ⎠
θ = 56.47 deg
Problem 1-45
The shaft is subjected to the axial force of 30 kN. If the shaft passes through the 53-mm diameter hole
in the fixed support A, determine the bearing stress acting on the collar C. Also, what is the average
shear stress acting along the inside surface of the collar where it is fixed connected to
the 52-mm diameter shaft?
Given:
P := 30kN
dhole := 53mm
dshaft := 52mm
dcollar := 60mm
hcollar := 10mm
Solution:
Bearing Stress:
π
2
2
Ab := ⋅ ⎛⎝ dcollar − dhole ⎞⎠
4
σ b :=
P
Ab
σ b = 48.3 MPa
Ans
Average Shear Stress:
(
)(
As := π ⋅ dshaft ⋅ hcollar
τ avg :=
P
As
)
τ avg = 18.4 MPa
Ans
Problem 1-46
The two steel members are joined together using a 60° scarf weld. Determine the average normal and
average shear stress resisted in the plane of the weld.
Given:
P := 8kN
θ := 60deg
b := 25mm
h := 30mm
Solution:
Equations of Equilibrium:
+
ΣF x=0;
N − P⋅ sin ( θ ) = 0
N := P⋅ sin ( θ )
+ ΣF y=0;
A :=
σ :=
N = 6.928 kN
V − P⋅ cos ( θ ) = 0
V := P⋅ cos ( θ )
V = 4 kN
σ = 8 MPa
Ans
τ avg = 4.62 MPa
Ans
h⋅ b
sin ( θ )
N
A
τ avg :=
V
A
Problem 1-47
The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N. Determine the
average normal stress developed in the bolt BC if the bolt has a diameter of 8 mm. Assume A is a pin.
P := 775N
Given:
a := 40mm
b := 30mm
d := 8mm
θ := 20deg
Solution:
Support Reaction:
P⋅ ( a) − FBC⋅ cos ( θ ) ⋅ ( a + b) = 0
ΣF A=0;
FBC :=
P⋅ a
( a + b) ⋅ cos ( θ )
FBC = 471.28 N
Average Normal Stress:
ABC :=
σ :=
π⋅d
4
2
FBC
ABC
σ = 9.38 MPa
Ans
Problem 1-48
The board is subjected to a tensile force of 425 N. Determine the average normal and average shear
stress developed in the wood fibers that are oriented along section a-a at 15° with the axis of the board.
Given:
P := 425N
θ := 15deg
b := 25mm
h := 75mm
Solution:
Equations of Equilibrium:
+
ΣF x=0;
V − P⋅ cos ( θ ) = 0
V := P⋅ cos ( θ )
+ ΣF y=0;
V = 410.518 N
N − P⋅ sin ( θ ) = 0
N := P⋅ sin ( θ )
N = 1N
Average Normal Stress:
A :=
σ :=
h⋅ b
sin ( θ )
N
A
τ avg :=
V
A
σ = 0.0152 MPa
Ans
τ avg = 0.0567 MPa
Ans
Problem 1-49
The open square butt joint is used to transmit a force of 250 kN from one plate to the other. Determine
the average normal and average shear stress components that this loading creates on the face of the
weld, section AB.
Given:
P := 250kN
θ := 30deg
b := 150mm
h := 50mm
Solution:
Equations of Equilibrium:
+
ΣF x=0;
−V + P⋅ sin ( θ ) = 0
V := P⋅ sin ( θ )
+ ΣF y=0;
V = 125 kN
N − P⋅ cos ( θ ) = 0
N := P⋅ cos ( θ )
N = 216.506 kN
Average Normal and Shear Stress:
A :=
σ :=
h⋅ b
sin ( 2θ )
N
A
τ avg :=
V
A
σ = 25 MPa
Ans
τ avg = 14.434 MPa
Ans
Problem 1-50
The specimen failed in a tension test at an angle of 52° when the axial load was 100 kN. If the diameter
of the specimen is 12 mm, determine the average normal and average shear stress acting on the area of
the inclined failure plane. Also, what is the average normal stress acting on the cross section when
failure occurs?
P := 100kN
Given:
d := 12mm
θ := 52deg
Solution:
Equations of Equilibrium:
V − P⋅ cos ( θ ) = 0
+ ΣF x=0;
V := P⋅ cos ( θ )
V = 61.566 kN
N − P⋅ sin ( θ ) = 0
+ ΣF y=0;
N := P⋅ sin ( θ )
N = 78.801 kN
Inclined plane:
π⎛ d ⎞
A := ⎜
4 ⎝ sin ( θ ) ⎠
2
σ :=
N
A
τ avg :=
V
A
σ = 549.05 MPa
Ans
τ avg = 428.96 MPa
Ans
Cross section:
2
A :=
πd
4
σ :=
P
A
σ = 884.19 MPa
Ans
τ avg := 0
τ avg = 0 MPa
Ans
Problem 1-51
A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine the
maximum average shear stress in the specimen and indicate the orientation θ of a section on which it
occurs.
Solution:
Equations of Equilibrium:
V − P⋅ cos ( θ ) = 0
+ ΣF y=0;
V = P⋅ cos ( θ )
Inclined plane:
Aincl =
τ=
dτ
dθ
A
sin ( θ )
V
Aincl
=
τ=
P⋅ cos ( 2θ )
A
P⋅ cos ( θ ) ⋅ sin ( θ )
A
dτ
dθ
τ=
= 0
cos ( 2θ ) = 0
2θ = 90deg
τ max =
P⋅ sin ( 90°)
2A
θ := 45deg
Ans
P
2A
Ans
τ max =
P⋅ sin ( 2θ )
2A
Problem 1-52
The joint is subjected to the axial member force of 5 kN. Determine the average normal stress acting
on sections AB and BC. Assume the member is smooth and is 50-mm thick.
Given:
Solution:
P := 5kN
θ := 45deg
φ := 60deg
dAB := 40mm
dBC := 50mm
t := 50mm
α := 90deg − φ
α = 30.00 deg
AAB := t⋅ dAB
ABC := t⋅ dBC
+
ΣF x=0;
NAB⋅ cos ( α ) − P⋅ cos ( θ ) = 0
NAB :=
P⋅ cos ( θ )
cos ( α )
NAB = 4.082 kN
+
ΣF y=0;
−NAB⋅ sin ( α ) + P⋅ sin ( θ ) − NBC = 0
NBC := −NAB⋅ sin ( α ) + P⋅ sin ( θ )
NBC = 1.494 kN
σ AB :=
NAB
AAB
σ AB = 2.041 MPa
Ans
σ BC :=
NBC
ABC
σ BC = 0.598 MPa
Ans
Problem 1-53
The yoke is subjected to the force and couple moment. Determine the average shear stress in the bolt
acting on the cross sections through A and B. The bolt has a diameter of 6 mm. Hint: The couple
moment is resisted by a set of couple forces developed in the shank of the bolt.
Given:
P := 2.5kN
M := 120N⋅ m
ho := 62mm
hi := 50mm
d := 6mm
θ := 60deg
Solution:
As a force on bolt shank is zero, then
τ A := 0
Ans
Equations od Equilibrium:
ΣF z=0;
P − 2Fz = 0
Fz := 0.5P
ΣMz=0;
Fz = 1.25 kN
( )
M − Fx⋅ hi = 0
Fx :=
M
hi
Average Shear Stress:
Fx = 2.4 kN
πd
A :=
4
2
The bolt shank subjected to a shear force of
τ B :=
VB
A
τ B = 95.71 MPa
VB :=
Ans
2
2
Fx + Fz
Problem 1-54
The two members used in the construction of an aircraft fuselage are joined together using a 30°
fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld.
Assume each inclined plane supports a horizontal force of 2 kN.
Given:
P := 4.0kN
b := 37.5mm
hhalf := 25m
θ := 30deg
Solution:
Equations of Equilibrium:
+
−V + 0.5P⋅ cos ( θ ) = 0
ΣF x=0;
V := 0.5P⋅ cos ( θ )
V = 1.732 kN
N − 0.5P⋅ sin ( θ ) = 0
+ ΣF y=0;
N := 0.5P⋅ sin ( θ )
N = 1 kN
Average Normal and Shear Stress:
A :=
σ :=
(hhalf )⋅ b
sin ( θ )
N
A
τ avg :=
V
A
σ = 533.33 Pa
Ans
τ avg = 923.76 Pa
Ans
Problem 1-55
The row of staples AB contained in the stapler is glued together so that the maximum shear stress the
glue can withstand is τ max = 84 kPa. Determine the minimum force F that must be placed on the
plunger in order to shear off a staple from its row and allow it to exit undeformed through the groove
at C. The outer dimensions of the staple are shown in the figure. It has a thickness of 1.25 mm
Assume all the other parts are rigid and neglect friction.
Given:
τ max := 0.084MPa
a := 12.5mm
b := 7.5mm
t := 1.25mm
Solution:
Average Shear Stress:
A := a⋅ b − [ ( a − 2t) ⋅ ( b − t) ]
τ max =
V
A
(
V = 2.63 N
Fmin := V
Fmin = 2.63 N
)
V := τ max ⋅ A
Ans
Problem 1-56
Rods AB and BC have diameters of 4mm and 6 mm, respectively. If the load of 8 kN is applied to the
ring at B, determine the average normal stress in each rod if θ = 60°.
Given:
W := 8kN
θ := 60deg
dA := 4mm
dC := 6mm
Solution:
2
AAB :=
Rod AB:
Rod BC :
ABC :=
ΣF y=0;
+
π ⋅ dA
4
2
π ⋅ dC
4
FBC⋅ sin ( θ ) − W = 0
FBC :=
W
sin ( θ )
FBC = 9.238 kN
+
ΣF x=0;
FBC⋅ cos ( θ ) − FAB = 0
FAB := FBC⋅ cos ( θ )
FAB = 4.619 kN
σ AB :=
σ BC :=
FAB
AAB
FBC
ABC
σ AB = 367.6 MPa
Ans
σ BC = 326.7 MPa
Ans
Problem 1-57
Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the vertical load of 8 kN is
applied to the ring at B, determine the angle θ of rod BC so that the average normal stress in each rod
is equivalent. What is this stress?
Given:
W := 8kN
dA := 4mm
dC := 6mm
Solution:
2
Rod AB:
Rod BC :
AAB :=
ABC :=
π ⋅ dA
4
2
π ⋅ dC
4
+
ΣF y=0;
FBC⋅ sin ( θ ) − W = 0
+
ΣF x=0;
FBC⋅ cos ( θ ) − FAB = 0
Since
FAB = σ ⋅ AAB
FBC = σ ⋅ ABC
Initial guess:
Given
σ := 100MPa
θ := 50deg
σ ⋅ ABC⋅ sin ( θ ) − W = 0
[1]
σ ⋅ ABC⋅ cos ( θ ) − σ ⋅ AAB = 0
[2]
Solving [1] and [2]:
⎛σ⎞
⎜ := Find ( σ , θ )
⎝θ ⎠
θ = 63.61 deg
Ans
σ = 315.85 MPa
Ans
Problem 1-58
The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normal
stress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.
P := 40kN
Given:
a := 0.9m
Solution:
c :=
2
b := 1.2m
2
2
a +b
h :=
b
c
A := 780mm
c = 1.5 m
v :=
a
c
Joint A:
+
ΣF y=0; ( v) ⋅ FAB − P = 0
FAB :=
P
v
FAB = 66.667 kN
+
ΣF x=0; ( h) ⋅ FAB − FAE = 0
FAE := ( h) ⋅ FAB
FAE = 53.333 kN
σ AB :=
σ AE :=
FAB
A
FAE
A
σ AB = 85.47 MPa
(T)
Ans
σ AE = 68.376 MPa
(C)
Ans
Joint E:
+
ΣF y=0; FEB − 0.75P = 0
FEB := 0.75P
FEB = 30 kN
+
ΣF x=0; FED − FAE = 0
FED := FAE
FED = 53.333 kN
σ EB :=
σ ED :=
FEB
A
FED
A
σ EB = 38.462 MPa
(T)
Ans
σ ED = 68.376 MPa
(C)
Ans
Joint B:
⎛ FEB ⎞
⎝ v ⎠
+
ΣF y=0; ( v) ⋅ FBD − ( v) ⋅ FAB − FEB = 0
FBD := FAB + ⎜
FBD = 116.667 kN
+
ΣF x=0;
FBC := ( h)FAB + ( h)FBD
FBC = 146.667 kN
FBC − ( h)FAB − ( h)FBD = 0
σ BC :=
σ BD :=
FBC
A
FBD
A
σ BC = 188.034 MPa
(T)
Ans
σ BD = 149.573 MPa
(C)
Ans
Problem 1-59
The bars of the truss each have a cross-sectional area of 780 mm2. If the maximum average normal
stress in any bar is not to exceed 140 MPa, determine the maximum magnitude P of the loads that can
be applied to the truss.
σ allow := 140MPa
Given:
a := 0.9m
2
c :=
Solution:
b := 1.2m
2
a +b
2
A := 780mm
c = 1.5 m
b
a
v :=
c
c
For comparison purpose, set P := 1kN
h :=
Joint A:
+
ΣF y=0; ( v) ⋅ FAB − P = 0
FAB :=
P
v
FAB = 1.667 kN
+
ΣF x=0; ( h) ⋅ FAB − FAE = 0
FAE := ( h) ⋅ FAB
FAE = 1.333 kN
σ AB :=
σ AE :=
FAB
A
FAE
A
σ AB = 2.137 MPa
(T)
σ AE = 1.709 MPa
(C)
Joint E:
+
ΣF y=0; FEB − 0.75P = 0
FEB := 0.75P
FEB = 0.75 kN
+
ΣF x=0; FED − FAE = 0
FED := FAE
FED = 1.333 kN
σ EB :=
σ ED :=
FEB
A
FED
A
σ EB = 0.962 MPa
(T)
σ ED = 1.709 MPa
(C)
Joint B:
⎛ FEB ⎞
⎝ v ⎠
+
ΣF y=0; ( v) ⋅ FBD − ( v) ⋅ FAB − FEB = 0
FBD := FAB + ⎜
+
ΣF x=0;
FBC := ( h)FAB + ( h)FBD
FBC − ( h)FAB − ( h)FBD = 0
σ BC :=
σ BD :=
FBC
A
FBD
A
σ BC = 4.701 MPa
(T)
σ BD = 3.739 MPa
(C)
FBD = 2.917 kN
FBC = 3.667 kN
Since the cross-sectional areas are the same, the highest stress occurs in the member BC,
which has the greatest force
(
Fmax := max FAB , FAE , FEB , FED , FBD , FBC
⎛ P ⎞⋅ σ
( allow⋅ A)
⎝ Fmax ⎠
Pallow := ⎜
)
Fmax = 3.667 kN
Pallow = 29.78 kN
Ans
Problem 1-60
The plug is used to close the end of the cylindrical tube that is subjected to an internal pressure of p =
650 Pa. Determine the average shear stress which the glue exerts on the sides of the tube needed to
hold the cap in place.
Given:
p := 650Pa
a := 25mm
di := 35mm
do := 40mm
Solution:
Ap :=
π ⋅ di
2
4
(
)
As := π ⋅ do ( a)
P⋅ ( a) − FBC⋅ cos ( θ ) ⋅ ( a + b) = 0
( )
P := p⋅ Ap
P = 0.625 N
Average Shear Stress:
τ avg :=
P
As
τ avg = 199.1 Pa
Ans
Problem 1-61
The crimping tool is used to crimp the end of the wire E. If a force of 100 N is applied to the handles,
determine the average shear stress in the pin at A. The pin is subjected to double shear and has a
diameter of 5 mm. Only a vertical force is exerted on the wire.
P := 100N
Given:
a := 37.5mm
b := 50mm
c := 25mm
d := 125mm dpin := 5mm
Solution:
From FBD (a):
+
ΣF x=0;
Bx := 0
Bx = 0 N
+ ΣΜD=0;
P⋅ ( d) − By⋅ ( c) = 0
d
By := P⋅
c
By = 500 N
From FBD (b):
+
ΣF x=0;
Ax := 0
Ax = 0 N
+ ΣΜE=0;
Ay⋅ ( a) − By⋅ ( a + b) = 0
Ay := By⋅
a+b
a
Ay = 1166.67 N
Average Shear Stress:
Apin :=
π ⋅ dpin
2
4
( )
VA := 0.5 ⋅ Ay
VA
τ avg :=
Apin
VA = 583.333 N
τ avg = 29.709 MPa
Ans
Problem 1-62
Solve Prob. 1-61 for pin B. The pin is subjected to double shear and has a diameter of 5 mm.
Given:
a := 37.5mm
b := 50mm
c := 25mm
d := 125mm dpin := 5mm P := 100N
Solution:
From FBD (a):
+
ΣF x=0;
Bx := 0
Bx = 0 N
+ ΣΜD=0;
P⋅ ( d) − By⋅ ( c) = 0
d
By := P⋅
c
By = 500 N
Average Shear Stress: Pin B is subjected to doule shear
Apin :=
π ⋅ dpin
2
4
( )
VB := 0.5 ⋅ By
VB
τ avg :=
Apin
VB = 250 N
τ avg = 12.732 MPa
Ans
Problem 1-63
The railcar docklight is supported by the 3-mm-diameter pin at A. If the lamp weighs 20 N, and the
extension arm AB has a weight of 8 N/m, determine the average shear stress in the pin needed to
support the lamp. Hint: The shear force in the pin is caused by the couple moment required for
equilibrium at A.
w := 8
Given:
N
m
P := 20N
a := 900mm
h := 32mm
dpin := 3mm
Solution:
From FBD (a):
+
ΣF x=0;
Bx := 0
+ ΣΜA=0;
V⋅ ( h) − ( w⋅ a) ⋅ ( 0.5a) − P⋅ ( a) = 0
⎛
⎝
V := ( w⋅ a) ⋅ ⎜ 0.5
a⎞
a
+ P⋅
h⎠
h
V = 663.75 N
Average Shear Stress:
Apin :=
τ avg :=
π ⋅ dpin
2
4
V
Apin
τ avg = 93.901 MPa
Ans
Problem 1-64
The two-member frame is subjected to the distributed loading shown. Determine the average normal
stress and average shear stress acting at sections a-a and b-b. Member CB has a square cross section
of 35 mm on each side. Take w = 8 kN/m.
w := 8
Given:
kN
m
a := 3m
Solution:
(
b := 4m
2
c :=
2
a +b
h :=
2
) 2
A := 0.035 m
c = 5m
a
c
v :=
b
c
Member AB:
ΣMA=0;
By⋅ ( a) − ( w⋅ a) ⋅ ( 0.5a) = 0
By := 0.5w⋅ a
+
ΣF y=0;
By = 12 kN
( v) ⋅ FAB − By = 0
FAB :=
By
v
FAB = 15 kN
Section a-a:
σ a_a :=
FAB
A
τ a_a := 0
σ a_a = 12.24 MPa
Ans
τ a_a = 0 MPa
Ans
Section b-b:
+
ΣF x=0;
N − FAB⋅ ( h) = 0
N := FAB⋅ ( h)
N = 9 kN
+
ΣF y=0;
V − FAB⋅ ( v) = 0
V := FAB⋅ ( v)
V = 12 kN
A
Ab_b :=
h
σ b_b :=
N
Ab_b
σ b_b = 4.41 MPa
Ans
τ b_b :=
V
Ab_b
τ b_b = 5.88 MPa
Ans
Problem 1-65
Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN. Determine
the average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member B
which has a thickness of 30 mm.
Given:
Solution:
+
+
P := 5kN
θ := 60deg
φ := 30deg
drod := 10mm
h := 40mm
t := 30mm
AB := t⋅ h
π
2
Arod := ⋅ drod
4
ΣF x=0;
ΣF y=0;
P⋅ cos ( θ ) − FB = 0
FB := P⋅ cos ( θ )
FB = 2.5 kN
Fc − P⋅ sin ( θ ) = 0
FC := P⋅ sin ( θ )
FC = 4.33 kN
Average Normal Stress:
σ B :=
σ C :=
FB
AB
FC
Arod
σ B = 2.083 MPa
Ans
σ C = 55.133 MPa
Ans
Problem 1-66
Consider the general problem of a bar made from m segments, each having a constant cross-sectional
area Am and length Lm. If there are n loads on the bar as shown, write a computer program that can be
used to determine the average normal stress at any specified location x. Show an application of the
program using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m,
P 2 = -1.5 kN, A 2 = 625 mm2.
Problem 1-67
The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear
stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a
diameter of 18 mm.
Given:
P := 15kN
a := 0.5m
b := 1m
c := 1.5m
d := 1.5m
e := 0.5m
θ := 30deg
dpin := 18mm
L := a + b + c + d + e
Solution:
Support Reactions:
ΣΜA=0; −By⋅ ( L) + P⋅ ( L − a) + 4P⋅ ( c + d + e) + 4P⋅ ( d + e) + 2P⋅ ( e) = 0
L−a
c+d+e
d+e
e
+ 4⋅ P⋅
+ 4⋅ P⋅
+ 2⋅ P⋅
L
L
L
L
By := P⋅
By = 82.5 kN
+
ΣF y=0;
−By + P + 4⋅ P + 4⋅ P + 2⋅ P − Ay = 0
Ay := −By + P + 4⋅ P + 4P + 2⋅ P
Ay = 82.5 kN
By
FBC :=
sin ( θ )
Ax := FBC⋅ cos ( θ )
FBC = 165 kN
Ax = 142.89 kN
Average Shear Stress:
Apin :=
π ⋅ dpin
2
4
For Pins B and C:
τ B_and_C :=
0.5FBC
Apin
τ B_and_C = 324.2 MPa
For Pin A:
FA :=
τ A :=
2
Ax + Ay
0.5FA
Apin
2
FA = 165 kN
τ A = 324.2 MPa
Ans
Ans
Problem 1-68
The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the
loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are
in double shear as shown, and each has a diameter of 18 mm.
Given:
τ allow := 80MPa
a := 0.5m
b := 1m
c := 1.5m
d := 1.5m
e := 0.5m
θ := 30deg
dpin := 18mm
L := a + b + c + d + e
Solution:
For comparison purpose, set
P := 1kN
Support Reactions:
ΣΜA=0; −By⋅ ( L) + P⋅ ( L − a) + 4P⋅ ( c + d + e) + 4P⋅ ( d + e) + 2P⋅ ( e) = 0
By := P⋅
+
ΣF y=0;
L−a
c+d+e
d+e
e
+ 4⋅ P⋅
+ 4⋅ P⋅
+ 2⋅ P⋅
L
L
L
L
By = 5.5 kN
−By + P + 4⋅ P + 4⋅ P + 2⋅ P − Ay = 0
Ay := −By + P + 4⋅ P + 4P + 2⋅ P
FBC :=
Ay = 5.5 kN
By
FBC = 11 kN
sin ( θ )
Ax := FBC⋅ cos ( θ )
FA :=
2
Ax + Ay
2
Ax = 9.53 kN
FA = 11 kN
Require:
(
)
Fmax := max FBC , FA
Apin :=
π ⋅ dpin
Fmax = 11 kN
2
4
⎛ P ⎞⋅ τ
⎡⎣ allow⋅ ( 2Apin)⎤⎦
⎝ Fmax ⎠
Pallow := ⎜
Pallow = 3.70 kN
Ans
Problem 1-69
The frame is subjected to the load of 1 kN. Determine the average shear stress in the bolt at A as a
o
function of the bar angle θ. Plot this function, 0 ≤ θ ≤ 90 , and indicate the values of θ for which this
stress is a minimum. The bolt has a diameter of 6 mm and is subjected to single shear.
Given: P := 1kN
dbolt := 6mm
a := 0.6m
b := 0.45m
c := 0.15m
Solution:
Support Reactions:
ΣΜC=0;
FAB⋅ cos ( θ ) ⋅ ( c) + FAB⋅ sin ( θ ) ⋅ ( a) − P⋅ ( a + b) = 0
FAB =
P ⋅ ( a + b)
cos ( θ ) ⋅ ( c) + sin ( θ ) ⋅ ( a)
Average Shear Stress: Pin B is subjected to doule shear
τ=
τ=
dτ
dθ
dτ
dθ
FAB
Abolt :=
Abolt
π ⋅ dbolt
2
4
4P⋅ ( a + b)
2
⎡⎣cos ( θ ) ⋅ ( c) + sin ( θ ) ⋅ ( a)⎤⎦ ⋅ ⎛⎝π ⋅ dbolt ⎞⎠
=
4P⋅ ( a + b)
π ⋅ dbolt
= 0
2
⋅
sin ( θ ) ⋅ ( c) − cos ( θ ) ⋅ ( a)
⎣⎡cos ( θ ) ⋅ ( c) + sin ( θ ) ⋅ ( a)⎤⎦
sin ( θ ) ⋅ ( c) − cos ( θ ) ⋅ ( a) = 0
2
a
c
⎛ a⎞
θ := atan ⎜
⎝ c⎠
tan ( θ ) =
θ = 75.96 deg
Ans
Problem 1-70
The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of the
beam, 1ft ≤ x ≤ 12ft. If the hoist is rated to support a maximum of 7.5 kN, determine the maximum
average normal stress in the 18-mm-diameter tie rod BC and the maximum average shear stress in the
16-mm-diameter pin at B.
Given:
P := 7.5kN
xmax := 3.6m
a := 3m
θ := 30deg
drod := 18mm
dpin := 16mm
Solution:
Support Reactions:
FBC⋅ sin ( θ ) ⋅ ( a) − P⋅ ( x) = 0
ΣΜC=0;
FBC =
P⋅ ( x)
sin ( θ ) ⋅ ( a)
Maximum F BC occurs when x= xmax . Therefore,
FBC :=
Arod :=
τ pin :=
σ rod :=
(
)
P⋅ xmax
sin ( θ ) ⋅ ( a)
FBC = 18.00 kN
2
π ⋅ dpin
π ⋅ drod
4
0.5 ⋅ FBC
Apin
FBC
Arod
Apin :=
2
4
τ pin = 44.762 MPa
Ans
σ rod = 70.736 MPa Ans
Problem 1-71
The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal
and average shear stresses acting over the shaded section, which is oriented at θ from the horizontal.
o
o
Plot the variation of these stresses as a function of θ (0 ≤ θ ≤ 90 ).
Solution:
Equations of Equilibrium:
V − P⋅ cos ( θ ) = 0
+ ΣF x=0;
V = P⋅ cos ( θ )
N − P⋅ sin ( θ ) = 0
+ ΣF y=0;
N = P⋅ sin ( θ )
Inclined plane:
A
Aθ =
sin ( θ )
σ=
N
A
τ avg =
V
A
σ=
P
2
⋅ sin ( θ )
A
Ans
τ avg =
P
⋅ sin ( 2θ )
2A
Ans
Problem 1-72
The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC. If the cable has
a diameter of 15 mm, plot the average normal stress in the cable as a function of the boom position θ for
o
o
Given:
W := 3kN
0 ≤ θ ≤ 90 .
a := 1m
do := 15mm
Solution:
Angle B:
φ B = 0.5 ( 90deg + θ )
φ B = 45deg + 0.5θ
Support Reactions:
ΣΜA=0;
( )
FBC⋅ sin φ B ⋅ ( a) − W⋅ ( 0.5a) cos ( θ ) = 0
FBC =
0.5W⋅ cos ( θ )
sin ( 45deg + 0.5θ )
Average Normal Stress:
σ BC =
σ BC =
FAB
ABC
2
ABC :=
π ⋅ do
4
cos ( θ )
⎛ 2W ⎞ ⋅
⎜
2 sin ( 45deg + 0.5θ )
⎝ π ⋅ do ⎠
Ans
Problem 1-73
The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed
loading along its length and to two concentrated loads as shown, determine the average normal stress in
the bar as a function of for 0 < x ≤ 0.5m.
P1 := 3kN
Given:
w := 8
P2 := 6kN
( − 6 ) m2
kN
m
A := 400⋅ 10
a := 0.5m
L := a + b
Solution:
+
b := 0.75m
ΣF x=0;
−N + P1 + P2 + w⋅ ( L − x) = 0
N = P1 + P2 + w⋅ ( L − x)
Average Normal Stress:
σ=
σ=
N
A
P1 + P2 + w⋅ ( L − x)
A
Ans
Problem 1-74
The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed
loading along its length and to two concentrated loads as shown, determine the average normal stress in
the bar as a function of for 0.5m < x ≤ 1.25m.
P1 := 3kN
Given:
w := 8
kN
m
a := 0.5m
( − 6 ) m2
A := 400⋅ 10
b := 0.75m
L := a + b
Solution:
+
P2 := 6kN
ΣF x=0;
−N + P1 + w⋅ ( L − x) = 0
N = P1 + w⋅ ( L − x)
Average Normal Stress:
σ=
σ=
N
A
P1 + w⋅ ( L − x)
A
Ans
Problem 1-75
The column is made of concrete having a density of 2.30 Mg/m3. At its top B it is subjected to an axial
compressive force of 15 kN. Determine the average normal stress in the column as a function of the
distance z measured from its base. Note: The result will be useful only for finding the average normal
stress at a section removed from the ends of the column, because of localized deformation at the ends.
m
3 kg
Given:
P := 3kN
ρ := 2.3 10
g := 9.81
3
2
s
m
r := 180mm
h := 0.75m
( )
Solution:
A := π ⋅ r
+
ΣF z=0;
2
w := ρ ⋅ g⋅ A
N − P − w⋅ ( h − z) = 0
N = P + w⋅ ( h − z)
Average Normal Stress:
σ=
N
A
σ=
P + w⋅ ( h − z)
A
Ans
Problem 1-76
The two-member frame is subjected to the distributed loading shown. Determine the largest
intensity of the uniform loading that can be applied to the frame without causing either the average
normal stress or the average shear stress at section b-b to exceed σ = 15 MPa, and τ = 16 MPa
respectively. Member CB has a square cross-section of 30 mm on each side.
Given:
Solution:
σ allow := 15MPa
τ allow := 16MPa
a := 4m
A := 0.030 m
2
c :=
2
a +b
2
) 2
c = 5m
a
c
h :=
(
b := 3m
v :=
b
c
kN
wo := 1
m
Set
Member AB:
ΣMA=0;
(
)
−By⋅ ( b) − wo⋅ b ⋅ ( 0.5b) = 0
By := 0.5wo⋅ b
By = 1.5 kN
Section b-b:
By = FBC⋅ ( h)
FBC :=
By
h
FBC = 1.88 kN
+
ΣF x=0;
FBC⋅ ( h) − Vb_b = 0
Vb_b := FBC⋅ ( h)
Vb_b = 1.5 kN
+
ΣF y=0;
−Nb_b + FBC⋅ ( v) = 0
Nb_b := FBC⋅ ( v)
Nb_b = 1.125 kN
A
Ab_b :=
v
σ b_b :=
Nb_b
Ab_b
σ b_b = 0.75 MPa
τ b_b :=
Vb_b
Ab_b
τ b_b = 1 MPa
⎛ σ allow ⎞
Assume failure due to normal stress:
wallow := wo⋅ ⎜
Assume failure due to shear stress:
wallow := wo⋅ ⎜
⎝ σ b_b ⎠
⎛ τ allow ⎞
⎝ τ b_b ⎠
kN
wallow = 20.00
m
kN
wallow = 16.00
m
Controls !
Ans
Problem 1-77
The pedestal supports a load P at its center. If the material has a mass density ρ, determine the radial
dimension r as a function of z so that the average normal stress in the pedestal remains constant. The
cross section is circular.
Solution:
Require:
σ=
P + w1
σ=
A
P + w1 + dw
A + dA
P⋅ dA + w1⋅ dA = A⋅ dw
P + w1
dw
=
dA
A
dw
= σ
dA
2
dA = π ( r + dr) − πr
[1]
2
dA = 2πr⋅ dr
dw = πr ⋅ ( ρ ⋅ g) ⋅ dz
2
From Eq.[1],
πr ⋅ ( ρ ⋅ g) ⋅ dz
2
2πr⋅ dr
= σ
r⋅ ( ρ ⋅ g) ⋅ dz
= σ
2dr
r
z
⌠ 1
ρ⋅g ⌠
⎮ ( 1) dz = ⎮
dr
⎮ r
2σ ⌡0
⌡r
1
⎛r⎞
= ln ⎜
2σ
⎝ r1 ⎠
ρ ⋅ g⋅ z
However,
σ=
⎛ ρ⋅g ⎞ ⋅z
⎜
2σ ⎠
r = r ⋅ e⎝
1
P
π ⋅ r1
2
⎛⎜ π ⋅ r 2⋅ ρ ⋅ g ⎞
1
⋅z
⎜ 2P
⎠
r = r ⋅ e⎝
1
Ans
Problem 1-78
The radius of the pedestal is defined by r = (0.5e-0.08y2) m, where y is given in meters. If the material
has a density of 2.5 Mg/m3, determine the average normal stress at the support.
Given:
ro := 0.5m
h := 3m
g = 9.81
m
2
s
− 0.08y
r = r o⋅ e
( 3) kg3
2
ρ := 2.5 ⋅ 10
m
m
yunit := 1m
Solution:
⎛ − 0.08y2⎞
⎠ ⋅ dy
dr = π ⎝ e
Ao := πro
2
Ao = 0.7854 m
2
( 2)
dV = π r ⋅ dy
2 ⎛ − 0.08y ⎞
dV = πro ⎝ e
⌠
⎮
V := ⎮
⌡
3
0
2
2
⎠ ⋅ dy
2
⎡
⎤
2
⎢πr 2 ⎛ e− 0.08y ⎞ ⋅ y ⎥ dy
⎠ ( unit)⎦
⎣ o ⎝
V = 1.584 m
3
W := ρ ⋅ g⋅ V
W = 38.835 kN
σ :=
W
Ao
σ = 0.04945 MPa
Ans
Problem 1-79
The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at its
center. If it is rotating in the horizontal plane at a constant angular rate of ω, determine the average
normal stress in the bar as a function of x.
Solution:
Equation of Motion :
+
ΣF x=MaN=Mω r2 ;
⎛ L − x⎞ ω⋅ ⎡x + 1 ⎛ L − x⎞⎤
⎢
⎜
⎥
⎝ 2 ⎠ ⎣ 2 ⎝ 2 ⎠⎦
N = m⋅ ⎜
N=
(
m⋅ ω
2
2
⋅ L − 4⋅ x
8
)
Average Normal Stress:
σ=
N
A
σ=
m⋅ ω
2
2
⋅ L − 4⋅ x
8A
(
)
Ans
Problem 1-80
Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are
10mm. thick, determine to the nearest multiples of 5mm the smallest dimension h of the support so that
the average shear stress does not exceed τallow = 2.1 MPa.
Given:
Solution:
P := 4kN
t := 10mm
τ allow := 2.1MPa
a := 300mm
b := 125mm
c :=
2
2
a +b
h :=
a
c
c = 325 mm
v :=
V := P⋅ ( v)
τ allow =
h :=
Use
b
c
V = 1.54 kN
V
t⋅ h
V
t⋅ τ allow
h := 75mm
h = 73.26 mm
h = 75 mm
Ans
Problem 1-81
The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure
shear stress for the bolts is τfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.
Given:
P := 80kN
τ fail := 350MPa
γ := 2.5
Solution:
τ allow :=
τ fail
γ
⎛ P⎞
Vbolt := 0.5 ⋅ ⎜
2
⎝ ⎠
Abolt =
Vbolt
τ allow
τ allow = 140 MPa
Vbolt = 20 kN
⎛ π ⎞ ⋅ d2 = Vbolt
⎜
τ allow
⎝ 4⎠
d :=
4 ⎛ Vbolt ⎞
⋅⎜
π ⎝ τ allow ⎠
d = 13.49 mm
Ans
Problem 1-82
The rods AB and CD are made of steel having a failure tensile stress of σfail = 510 MPa. Using a factor
of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load
shown. The beam is assumed to be pin connected at A and C.
P1 := 4kN
Given:
a := 2m
P2 := 6kN
b := 2m
γ := 1.75
Solution:
P3 := 5kN
c := 3m
d := 3m
σ fail := 510MPa
L := a + b + c + d
Support Reactions:
ΣΜA=0;
FCD⋅ ( L ) − P1⋅ ( a) − P2⋅ ( a + b) − P3⋅ ( a + b + c) = 0
⎛a⎞ + P ⋅a+ b + P ⋅a+ b+ c
3
L
⎝ L⎠ 2 L
FCD := P1⋅ ⎜
ΣΜC=0;
FCD = 6.70 kN
−FAB⋅ ( L ) + P1⋅ ( b + c + d) + P2⋅ ( c + d) + P3⋅ ( d) = 0
FAB := P1⋅
b+c+d
c+d
⎛ d⎞
+ P 2⋅
+ P3⋅ ⎜
L
L
⎝L⎠
FAB = 8.30 kN
Average Normal Stress: Design of rod sizes
σ allow :=
σ fail
γ
σ allow = 291.43 MPa
For Rod AB
Abolt =
FAB
σ allow
⎛ π ⎞ ⋅ d 2 = FAB
⎜
σ allow
⎝ 4 ⎠ AB
dAB :=
4 ⎛ FAB ⎞
⋅⎜
π ⎝ σ allow ⎠
dAB = 6.02 mm
Ans
dCD = 5.41 mm
Ans
For Rod CD
Abolt =
FCD
σ allow
⎛ π ⎞ ⋅ d 2 = FCD
⎜
σ allow
⎝ 4 ⎠ CD
dCD :=
4 ⎛ FCD ⎞
⋅⎜
π ⎝ σ allow ⎠
Problem 1-83
The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is
fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if
the allowable shear stress for the key is τallow = 35 MPa.
Given:
P := 200N
τ allow := 35MPa
L := 500mm
a := 20mm
b := 25mm
Solution:
ΣΜA=0;
Fa_a ⋅ ( a) − P⋅ ( L ) = 0
Fa_a := P⋅
L
a
Fa_a = 5000 N
For the key
Aa_a =
Fa_a
τ allow
b⋅ d =
d :=
Fa_a
τ allow
1 ⎛ Fa_a ⎞
⋅⎜
b τ allow
⎝
d = 5.71 mm
⎠
Ans
Problem 1-84
The fillet weld size a is determined by computing the average shear stress along the shaded plane,
which has the smallest cross section. Determine the smallest size a of the two welds if the force
applied to the plate is P = 100 kN. The allowable shear stress for the weld material is τallow = 100 MPa.
Given:
P := 100kN
τ allow := 100MPa
L := 100mm
θ := 45deg
Solution:
Shear Plane in the Weld:
0.5P
Aweld =
τ allow
L ⋅ a⋅ sin ( θ ) =
a :=
Aweld = L⋅ a⋅ sin ( θ )
1
0.5P
τ allow
⎛ 0.5P ⎞
⋅⎜
L⋅ sin ( θ ) ⎝ τ allow ⎠
a = 7.071 mm
Ans
Problem 1-85
The fillet weld size a = 8 mm. If the joint is assumed to fail by shear on both sides of the block along
the shaded plane, which is the smallest cross section, determine the largest force P that can be applied
to the plate. The allowable shear stress for the weld material is τallow = 100 MPa.
Given:
a := 8mm
L := 100mm
θ := 45deg
τ allow := 100MPa
Solution:
Shear Plane in the Weld:
(
)
P = τ allow⋅ 2Aweld
P := τ allow⋅ ⎡⎣2( L⋅ a⋅ sin ( θ ) )⎤⎦
P = 113.14 kN
Ans
Aweld = L⋅ a⋅ sin ( θ )
Problem 1-86
The eye bolt is used to support the load of 25 kN. Determine its diameter d to the nearest multiples of
5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washer
will not penetrate or shear through it. The allowable normal stress for the bolt is σallow = 150 MPa and
the allowable shear stress for the supporting material is τallow = 35 MPa.
Given:
P := 25kN
dwasher := 25mm
σ allow := 150MPa
τ allow := 35MPa
Solution:
Allowable Normal Stress : Design of bolt size
Abolt =
⎛ π ⎞ ⋅ d2 = P
⎜
σ allow
⎝ 4⎠
P
σ allow
d :=
4 ⎛ P ⎞
⋅⎜
π ⎝ σ allow ⎠
d = 14.567 mm
Use d := 15mm
d = 15 mm
Ans
Allowable Shear Stress: Design of support thickness
Asupport =
P
τ allow
(
)
π ⋅ dwasher ⋅ h =
h :=
(
P
τ allow
⎛
1
π ⋅ dwasher
⋅⎜
d := 10mm
d = 10 mm
⎞
) ⎝ τ allow ⎠
h = 9.095 mm
Use
P
Ans
Problem 1-87
The frame is subjected to the load of 8 kN. Determine the required diameter of the pins at A and B if
the allowable shear stress for the material is τallow = 42 MPa. Pin A is subjected to double shear,
whereas pin B is subjected to single shear.
P := 8kN
Given:
a := 1.5m
Solution:
τ allow := 42MPa
b := 1.5m
c := 1.5m d := 0.6m
θ BC := 45deg
Support Reactions: From FBD (a),
(
ΣΜD=0;
)
FBC⋅ sin θ BC ⋅ ( c) − P⋅ ( c + d) = 0
FBC := P⋅
(
c+d
)
sin θ BC ⋅ ( c)
FBC = 15.839 kN
From FBD (b),
ΣΜA=0;
Dy⋅ ( a + b) − P⋅ ( c + d) = 0
c+d
Dy := P⋅
a+b
Dy = 5.6 kN
+
ΣF x=0;
Ax − P = 0
Ax := P
+
ΣF y=0;
Dy − Ay = 0
Ay := Dy
Ax = 8 kN
FA :=
Apin =
0.5FA
τ allow
4 ⎛ 0.5FA ⎞
⋅⎜
π ⎝ τ allow ⎠
Pin A is subjected to single shear, and
Apin =
FB
τ allow
2
FA = 9.77 kN
⎛ π ⎞ ⋅ d2 = 0.5FA
⎜
τ allow
⎝ 4⎠
d :=
For pin B:
2
Ax + Ay
d = 12.166 mm
Ans
FB := FBC
⎛ π ⎞ ⋅ d2 = F B
⎜
τ allow
⎝ 4⎠
d :=
4 ⎛ FB ⎞
⋅⎜
π ⎝ τ allow ⎠
d = 21.913 mm
Ans
Problem 1-88
The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile
stress of σallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5
kN.
P := 5kN
Given:
σ allow := 200MPa
a := 4m
b := 3m
Solution:
c :=
2
2
a +b
θ := 60deg
h :=
a
c
v :=
b
c
At joint A:
Initial guess:
FAB := 1kN
FAC := 2kN
Given
+
ΣF x=0;
FAC⋅ ( h) − FAB⋅ sin ( θ ) = 0
[1]
+
ΣF y=0;
FAC⋅ ( v) + FAB⋅ cos ( θ ) − P = 0
[2]
Solving [1] and [2]:
⎛⎜ FAB ⎞
:= Find ( FAB , FAC)
⎜ FAC
⎝
⎠
⎛⎜ FAB ⎞ ⎛ 4.3496 ⎞
=⎜
kN
⎜ FAC
4.7086
⎝
⎠
⎝
⎠
For wire AB
AAB =
FAB
σ allow
⎛ π ⎞ ⋅ d 2 = FAB
⎜
σ allow
⎝ 4 ⎠ AB
dAB :=
4 ⎛ FAB ⎞
⋅⎜
π ⎝ σ allow ⎠
dAB = 5.26 mm
Ans
dAC = 5.48 mm
Ans
For wire AC
AAC =
FAC
σ allow
⎛ π ⎞ ⋅ d 2 = FAC
⎜
σ allow
⎝ 4 ⎠ AC
dAC :=
4 ⎛ FAC ⎞
⋅⎜
π ⎝ σ allow ⎠
Problem 1-89
The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile
stress of σallow = 180 MPa, and wire AB has a diameter of 6 mm and AC has a diameter of 4 mm,
determine the greatest force P that can be applied to the chain before one of the wires fails.
σ allow := 180MPa
Given:
a := 4m
b := 3m
dAB := 6mm
Solution:
c :=
2
θ := 60deg
dAC := 4mm
2
a +b
h :=
(
a
c
b
c
v :=
)
FAB = AAB ⋅ σ allow
Assume failure of AB:
⎛ π ⎞ ⋅ d 2⋅ σ
⎝ 4 ⎠ AB allow
FAB := ⎜
FAB = 5.09 kN
At joint A:
Initial guess:
P1 := 1kN
FAC := 2kN
Given
+
ΣF x=0;
FAC⋅ ( h) − FAB⋅ sin ( θ ) = 0
[1]
+
ΣF y=0;
FAC⋅ ( v) + FAB⋅ cos ( θ ) − P1 = 0
[2]
⎛⎜ P1 ⎞
:= Find ( P1 , FAC)
⎜ FAC
⎝
⎠
Solving [1] and [2]:
Assume failure of AC:
(
⎛⎜ P1 ⎞ ⎛ 5.8503 ⎞
=⎜
kN
⎜ FAC
5.5094
⎝
⎠
⎝
⎠
)
FAC = AAC ⋅ σ allow
⎛ π ⎞ ⋅ d 2⋅ σ
⎝ 4 ⎠ AC allow
FAC := ⎜
FAC = 2.26 kN
At joint A:
Initial guess:
P2 := 1kN
FAB := 2kN
Given
+
ΣF x=0;
FAC⋅ ( h) − FAB⋅ sin ( θ ) = 0
[1]
+
ΣF y=0;
FAC⋅ ( v) + FAB⋅ cos ( θ ) − P2 = 0
[2]
Solving [1] and [2]:
Chosoe the smallest value:
⎛⎜ FAB ⎞
:= Find ( FAB , P2)
⎜ P2
⎝
⎠
(
P := min P1 , P2
)
⎛⎜ FAB ⎞ ⎛ 2.0895 ⎞
=⎜
kN
⎜ P2
2.4019
⎝
⎠
⎝
⎠
P = 2.40 kN
Ans
Problem 1-90
The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stress
of σallow = 168 MPa. Determine the greatest load that can be supported without causing the cable to fail
when θ = 30° and φ = 45°. Neglect the size of the winch.
Given:
σ allow := 168MPa
do := 6mm
θ := 30deg
φ := 45deg
Solution:
For the cable:
(
)
T cable = Acable ⋅ σ allow
⎛ π ⎞ ⋅ d 2⋅ σ
⎝ 4 ⎠ o allow
T cable := ⎜
T cable = 4.7501 kN
At joint B:
Initial guess:
FAB := 1 kN
W := 2 kN
Given
+
ΣF x=0;
−T cable cos ( θ ) + FAB⋅ cos ( φ ) = 0
[1]
+
ΣF y=0;
−W + FAB⋅ sin ( φ ) − T cable ⋅ sin ( θ ) = 0
[2]
Solving [1] and [2]:
⎛ FAB ⎞
:= Find ( FAB , W)
⎜
⎝ W ⎠
⎛ FAB ⎞ ⎛ 5.818 ⎞
=⎜
kN
⎜
⎝ W ⎠ ⎝ 1.739 ⎠
Ans
Problem 1-91
The boom is supported by the winch cable that has an allowable normal stress of σallow = 168 MPa. If
it is required that it be able to slowly lift 25 kN, from θ = 20° to θ = 50°, determine the smallest
diameter of the cable to the nearest multiples of 5mm. The boom AB has a length of 6 m. Neglect the
size of the winch. Set d = 3.6 m.
Given:
σ allow := 168MPa
W := 25 kN
d := 3.6m
a := 6m
Solution:
θ := 20deg
Maximum tension in canle occurs when
sin ( θ )
sin ( ψ)
=
a
d
⎡⎛ d ⎞ ⋅ sin ( θ )⎤
⎥
⎣⎝ a ⎠
⎦
ψ := asin ⎢⎜
ψ = 11.842 deg
At joint B:
Initial guess:
FAB := 1 kN
Given
φ := θ + ψ
+
ΣF x=0;
−T cable cos ( θ ) + FAB⋅ cos ( φ ) = 0
[1]
+
ΣF y=0;
−W + FAB⋅ sin ( φ ) − T cable ⋅ sin ( θ ) = 0
[2]
Solving [1] and [2]:
T cable := 2 kN
⎛⎜ FAB ⎞
:= Find ( FAB , T cable )
⎜ Tcable
⎝
⎠
⎛⎜ FAB ⎞ ⎛ 114.478 ⎞
=⎜
kN
⎜ Tcable
⎝
⎠ ⎝ 103.491 ⎠
For the cable:
Acable =
P
σ allow
⎛ π ⎞ ⋅ d 2 = Tcable
⎜
σ allow
⎝ 4⎠ o
do :=
4 ⎛ Tcable ⎞
⋅⎜
π ⎝ σ allow ⎠
do = 28.006 mm
Use
do := 30mm
do = 30 mm
Ans
Problem 1-92
The frame is subjected to the distributed loading of 2 kN/m. Determine the required diameter of the
pins at A and B if the allowable shear stress for the material is τallow = 100 MPa. Both pins are
subjected to double shear.
Given:
w := 2
kN
m
τ allow := 100MPa
r := 3m
Solution:
Member AB is atwo-force member
θ := 45deg
Support Reactions:
ΣΜA=0;
FBC⋅ sin ( θ ) ⋅ ( r) − ( w⋅ r) ⋅ ( 0.5r) = 0
FBC :=
+ ΣF y=0;
0.5w⋅ r
FBC = 4.243 kN
sin ( θ )
Ay + FBC⋅ sin ( θ ) − w⋅ r = 0
Ay := −FBC⋅ sin ( θ ) + w⋅ r
+ ΣF x=0;
Ax − FBC⋅ cos ( θ ) = 0
Ax := FBC⋅ cos ( θ )
2
Ax + Ay
FB := FBC
Ax = 3 kN
Pin A and pin B are subjected to double shear
Average Shear Stress:
FA :=
Ay = 3 kN
2
FA = 4.243 kN
FB = 4.243 kN
Since both subjected to the same shear force V = 0.5 FA and
Apin =
V
τ allow
⎛ π ⎞⋅ d 2 = V
⎜
τ allow
⎝ 4 ⎠ pin
dpin :=
4 ⎛ V ⎞
⋅⎜
π ⎝ τ allow ⎠
dpin = 5.20 mm
Ans
V := 0.5FB
Problem 1-93
Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required to
support is P = 150 kN. The allowable tensile stress, bearing stress, and shear stress is (σt)allow = 175
MPa, (σb)allow = 275 MPa, and τallow = 115 MPa.
Given:
P := 150kN
σ t_allow := 175MPa
τ allow := 115MPa
σ b_allow := 275MPa
d2 := 30mm
Solution:
Allowable Normal Stress: Design of end cap outer diameter
A=
P
σ t_allow
P
⎛ π ⎞ ⋅ ⎛ d 2 − d 2⎞ =
⎜ ⎝ 1
2
⎠
σ t_allow
⎝ 4⎠
d1 :=
4 ⎛
P
⎞+d 2
⋅⎜
2
π ⎝ σ t_allow ⎠
d1 = 44.62 mm
Ans
d3 = 26.35 mm
Ans
t = 15.75 mm
Ans
Allowable Bearing Stress: Design of circular shaft diameter
A=
P
σ b_allow
P
⎛ π ⎞ ⋅ ⎛ d 2⎞ =
⎜ ⎝ 3⎠
σ b_allow
⎝ 4⎠
d3 :=
Allowable Shear Stress:
A=
P
τ allow
4 ⎛
P
⎞
⋅⎜
π ⎝ σ b_allow ⎠
Design of end cap thickness
(π ⋅ d3)⋅ t = τ
t :=
1
⎛
⋅⎜
P
allow
P
⎞
π ⋅ d3 ⎝ τ allow ⎠
Problem 1-94
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the size of square bearing plates A' and B' required to support the loading. Take P = 7.5 kN.
Dimension the plates to the nearest multiples of 10mm. The reactions at the supports are vertical.
Given:
σ b_allow := 2.8MPa
P := 7.5 kN
P1 := 10 kN
P2 := 10 kN
P3 := 15 kN
P4 := 10 kN
a := 1.5m
b := 2.5m
Solution:
L := 3⋅ a + b
Support Reactions:
ΣΜA=0;
By ( 3a) − P2⋅ ( a) − P3⋅ ( 2a) − P4⋅ ( 3a) − P⋅ ( L) = 0
⎛ a ⎞ + P ⋅ ⎛ 2a ⎞ + P ⋅ ⎛ 3a ⎞ + P⋅ ⎛ L ⎞
By := P2⋅ ⎜
⎜
3 ⎜ 3a
4 ⎜ 3a
3a
3a
⎝ ⎠
ΣΜB=0;
⎝ ⎠
⎝ ⎠
⎝ ⎠
By = 35 kN
−Ay⋅ ( 3⋅ a) + P1⋅ ( 3⋅ a) + P2⋅ ( 2⋅ a) + P3⋅ ( a) − P⋅ ( b) = 0
⎛ 3a ⎞ + P ⋅ ⎛ 2a ⎞ + P ⋅ ⎛ a ⎞ − P⋅ ⎛ b ⎞
Ay := P1⋅ ⎜
⎜
2 ⎜ 3a
3 ⎜ 3a
3a
3a
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
Ay = 17.5 kN
For Plate A:
Aplate_A =
Ay
Ay
2
aA =
σ b_allow
aA :=
σ b_allow
Ay
σ b_allow
aA = 79.057 mm
aA = 80mm
Use aA x aA plate:
Ans
For Plate B
Aplate_B =
By
By
2
σ b_allow
aB =
aB :=
σ b_allow
By
σ b_allow
aB = 111.803 mm
Use aB x aB plate:
aB = 120mm
Ans
Rb = 35kN
(b is in subscript)
Problem 1-95
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have
square cross sections of 50mm x 50mm and 100mm x 100mm, respectively.
Given:
σ b_allow := 2.8MPa
P1 := 10 kN
P2 := 10 kN
P3 := 15 kN
P4 := 10 kN
a := 1.5m
b := 2.5m
aA := 50mm
aB := 100mm
Solution:
L := 3⋅ a + b
Support Reactions:
ΣΜA=0;
By ( 3a) − P2⋅ ( a) − P3⋅ ( 2a) − P4⋅ ( 3⋅ a) − P⋅ ( L ) = 0
⎛ a ⎞ + P ⋅ ⎛2⋅ a ⎞ + P ⋅ ⎛ 3⋅ a ⎞ + P⋅ ⎛ L ⎞
By = P2⋅ ⎜
⎜
3 ⎜ 3a
4 ⎜ 3a
3a
3a
⎝ ⎠
ΣΜB=0;
⎝
⎠
⎝
⎠
⎝ ⎠
−Ay⋅ ( 3⋅ a) + P1⋅ ( 3⋅ a) + P2⋅ ( 2⋅ a) + P3⋅ ( a) − P⋅ ( b) = 0
⎛ 3a ⎞ + P ⋅ ⎛ 2a ⎞ + P ⋅ ⎛ a ⎞ − P⋅ ⎛ b ⎞
Ay = P1⋅ ⎜
⎜
2 ⎜ 3a
3 ⎜ 3a
3a
3a
⎝ ⎠
For Plate A:
⎝ ⎠
⎝ ⎠
⎝ ⎠
Ay := ⎛⎝ aA ⎞⎠ ⋅ σ b_allow
2
⎛ 3a ⎞
⎛ 2a ⎞
⎛a⎞
⎛b⎞
⎛ a 2⎞ ⋅ σ
⎝ A ⎠ b_allow = P1⋅ ⎜⎝ 3a ⎠ + P2⋅ ⎜⎝ 3a ⎠ + P3⋅ ⎜⎝ 3a ⎠ − P⋅ ⎜⎝ 3a ⎠
⎛ 3a ⎞ + P ⋅ ⎛ 2a ⎞ + P ⋅ ⎛ a ⎞ − ⎛ a 2⎞ ⋅ σ
⎛ 3a ⎞
⋅⎜
⎜
⎜
2
3
A
b_allow
⎝
⎠
⎝b⎠
⎝b⎠
⎝ b⎠
⎝b⎠
P := P1⋅ ⎜
For Plate B:
P = 26.400 kN
Pcase_1 := P
2
By := ⎛⎝ aB ⎞⎠ ⋅ σ b_allow
⎛a⎞
⎛ a⎞
⎛ 3⋅ a ⎞
⎛L⎞
⎛ a 2⎞ ⋅ σ
⎝ B ⎠ b_allow = P2⋅ ⎜⎝ 3a ⎠ + P3⋅ ⎜⎝ 2⋅ 3a ⎠ + P4⋅ ⎜⎝ 3a ⎠ + P⋅ ⎜⎝ 3a ⎠
3a
⎛ a ⎞ − P ⋅ ⎛ 2⋅ a ⎞ − P ⋅ ⎛ 3⋅ a ⎞ + ⎛a 2⎞ ⋅ σ
⋅
⎜
⎜
3
4
B
b_allow
⎝
⎠
L
⎝L⎠
⎝ L⎠
⎝L ⎠
P := −P2⋅ ⎜
(
)
Pallow := min Pcase_1 , Pcase_2
P = 3.000 kN
Pcase_2 := P
Pallow = 3 kN
Ans
Problem 1-96
Determine the required cross-sectional area of member BC and the diameter of the pins at A and B if
the allowable normal stress is σallow = 21 MPa and the allowable shear stress is τallow = 28 MPa.
Given:
σ allow := 21MPa
τ allow := 28MPa
P := 7.5kip
θ := 60deg
a := 0.6m
b := 1.2m
c := 0.6m
Solution: L := a + b + c
Support Reactions:
ΣΜA=0; By⋅ ( L) − P⋅ ( a) − P⋅ ( a + b) = 0
a
a+b
By := P⋅ + P⋅
L
L
FBC :=
By = 33.362 kN
+
ΣF y=0;
+ ΣF x=0;
FBC = 38.523 kN
Bx = 19.261 kN
−By + P + P − Ay = 0
Ay := −By + P + P
Ay = 33.362 kN
Bx − Ax = 0
Ax := Bx
Ax = 19.261 kN
Member BC:
ABC :=
2
AA =
Ax + Ay
FBC
σ allow
FA
τ allow
2
FA = 38.523 kN
2
ABC = 1834.416 mm
AB =
0.5FBC
τ allow
Ans
⎛ π ⎞ ⋅ d 2 = FA
⎜
τ allow
⎝ 4⎠ A
dA :=
Pin B:
sin ( θ )
Bx := FBC⋅ cos ( θ )
FA :=
Pin A:
By
4 ⎛ FA ⎞
⋅⎜
π ⎝ τ allow ⎠
dA = 41.854 mm
Ans
dB = 29.595 mm
Ans
⎛ π ⎞ ⋅ d 2 = 0.5FBC
⎜
τ allow
⎝ 4⎠ B
dB :=
4 ⎛ 0.5FBC ⎞
⋅⎜
π ⎝ τ allow ⎠
Problem 1-97
The assembly consists of three disks A, B, and C that are used to support the load of 140 kN.
Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the
diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (τallow)b =
350 MPa and allowable shear stress is τallow = 125 MPa.
P := 140kN
Given:
τ allow := 125MPa
σ b_allow := 350MPa
hB := 20mm
hC := 10mm
Solution:
Allowable Shear Stress:
A=
P
Assume shear failure dor disk C
(π ⋅ d2)⋅ hC = τ
τ allow
d2 :=
1
P
allow
⎛
P
⋅⎜
⎞
π ⋅ hC ⎝ τ allow ⎠
d2 = 35.65 mm
Ans
d3 = 27.60 mm
Ans
Allowable Bearing Stress: Assume bearing failure dor disk C
A=
P
⎛ π ⎞ ⋅ ⎛ d 2 − d 2⎞ =
⎜ ⎝ 2
3
⎠
σ b_allow
⎝ 4⎠
P
σ b_allow
d3 :=
2
d2 −
4 ⎛
P
⎞
⋅⎜
π ⎝ σ b_allow ⎠
Allowable Bearing Stress: Assume bearing failure dor disk B
A=
P
σ b_allow
P
⎛ π ⎞ ⋅ ⎛ d 2⎞ =
⎜ ⎝ 1⎠
σ b_allow
⎝ 4⎠
d1 :=
4 ⎛
P
⎞
⋅⎜
π ⎝ σ b_allow ⎠
d1 = 22.57 mm
Since d3 > d1, disk B might fail due to shear.
τ=
P
A
τ :=
P
π ⋅ d1⋅ hB
τ = 98.73 MPa
Therefore
< τallow (O.K.!)
d1 = 22.57 mm
Ans
Problem 1-98
Strips A and B are to be glued together using the two strips C and D. Determine the required thickness
t of C and D so that all strips will fail simultaneously. The width of strips A and B is 1.5 times that of
strips C and D.
Given:
P := 40N
t := 30mm
bA := 1.5m
bB := 1.5m
bC := 1m
bD := 1m
Solution:
Average Normal Stress:
σA = σB
N
Requires,
σB = σC
0.5N
bC ⋅ tC
(bA)⋅ t ( ) ( )
0.5 ( bA) ⋅ t
t :=
C
=
σC = σD
bC
tC = 22.5 mm
Ans
Problem 1-99
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the size of square bearing plates A' and B' required to support the loading. Dimension the
plates to the nearest multiples of 10mm. The reactions at the supports are vertical. Take P = 7.5 kN.
σ b_allow := 2.8MPa
Given:
w := 10
kN
m
P := 7.5kN
a := 4.5m
b := 2.25m
L := a + b
Solution:
Support Reactions:
ΣΜA=0;
By ( a) − w⋅ ( a) ⋅ ( 0.5 ⋅ a) − P⋅ ( L) = 0
⎛ L⎞
By := w⋅ ( 0.5 ⋅ a) + P⋅ ⎜
a
⎝ ⎠
ΣΜB=0;
By = 33.75 kN
−Ay⋅ ( a) + w⋅ ( a) ⋅ ( 0.5 ⋅ a) − P⋅ ( L ) = 0
⎛ b⎞
Ay := w⋅ ( 0.5 ⋅ a) − P⋅ ⎜
a
⎝ ⎠
Ay = 18.75 kN
Allowable Bearing Stress: Design of bearing plates
For Plate A:
Area =
Ay
σ b_allow
Ay
2
aA =
aA :=
σ b_allow
Ay
σ b_allow
aA = 81.832 mm
Use aA x aA plate:
aA := 90mm
aA = 90 mm
Ans
For Plate B
Area =
By
σ b_allow
By
2
aB =
aB :=
σ b_allow
By
σ b_allow
aB = 109.789 mm
Use aB x aB plate:
aB := 110mm
aB = 110.00 mm
Ans
Problem 1-100
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have
square cross sections of 50mm x 50mm and 100mm x 100mm, respectively.
σ b_allow := 2.8MPa
Given:
w := 10
kN
m
aA := 50mm
Solution:
a := 4.5m
b := 2.25m
aB := 100mm
L := a + b
Support Reactions:
By⋅ ( a) − w⋅ ( a) ⋅ ( 0.5 ⋅ a) − P⋅ ( L) = 0
ΣΜA=0;
⎛ a ⎞ − w⋅ ⎛ a ⎞ ⋅ ( 0.5⋅ a)
⎜
⎝L⎠
⎝L⎠
P = By⋅ ⎜
ΣΜB=0;
−Ay⋅ ( a) + w⋅ ( a) ⋅ ( 0.5 ⋅ a) − P⋅ ( b) = 0
⎛ a ⎞ + w⋅ ⎛ a ⎞ ⋅ ( 0.5⋅ a)
⎜
⎝ b⎠
⎝ b⎠
P = −Ay⋅ ⎜
Allowable Bearing Stress:
Assume failure of material occurs under plate A.
⎛ a ⎞ + ( w⋅ a) ⋅ 0.5a
b
⎝ b⎠
P := −⎡⎣⎛⎝ aA ⎞⎠ ⋅ σ b_allow⎤⎦ ⋅ ⎜
2
Ay := ⎛⎝ aA ⎞⎠ ⋅ σ b_allow
2
P = 31 kN
Pcase_1 := P
Assume failure of material occurs under plate B.
⎛ a ⎞ − w⋅ ⎛ a ⎞ ⋅ ( 0.5⋅ a)
⎜
⎝ L⎠
⎝ L⎠
P := By⋅ ⎜
By := ⎛⎝ aB ⎞⎠ ⋅ σ b_allow
2
P = 3.67 kN
Pcase_2 := P
(
)
Pallow := min Pcase_1 , Pcase_2
Pallow = 3.67 kN
Ans
Problem 1-101
The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the average
shear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has a
diameter of 12 mm. If the yield shear stress for the bolt is τy = 175 MPa, and the yield tensile stress for
the rod is σy = 266 MPa, determine the factor of safety with respect to yielding in each case.
τ y := 175MPa w := 12
Given:
kN
m
σ y := 266MPa
a := 1.2m
b := 0.6m
do := 10mm
drod := 12mm
Solution:
c :=
2
2
a +e
Support Reactions:
ΣΜC=0;
e := 0.9m
h :=
a
c
v :=
e
c
L := a + b
⎡⎣FAB⋅ ( v)⎤⎦ ⋅ ( a) − w⋅ ( L) ⋅ ( 0.5⋅ L) = 0
⎛ L ⎞ ⋅ ( 0.5⋅ L)
⎝ a⋅ v ⎠
FAB := w⋅ ⎜
FAB = 27 kN
For bolt A:
Bolt A is subjected to double shear, and
⎛ π ⎞⋅ d 2
⎝ 4⎠ o
A := ⎜
V
A
τ :=
τy
FS :=
For rod AB:
N := FAB
⎛ π ⎞⋅ d 2
⎝ 4 ⎠ rod
A := ⎜
τ
V := 0.5FAB
τ = 171.89 MPa
Ans
FS = 1.02
Ans
N = 27 kN
σ :=
FS :=
N
A
σy
σ
σ = 238.73 MPa
Ans
FS = 1.11
Ans
V = 13.5 kN
Problem 1-102
Determine the intensity w of the maximum distributed load that can be supported by the hanger
assembly so that an allowable shear stress of τallow = 95 MPa is not exceeded in the 10-mm-diameter
bolts at A and B, and an allowable tensile stress of σallow = 155 MPa is not exceeded in the
12-mm-diameter rod AB.
Given:
τ allow := 95MPa
a := 1.2m
b := 0.6m
do := 10mm
2
Solution: c :=
2
Support Reactions:
e := 0.9m
drod := 12mm
a +e
ΣΜC=0;
σ allow := 155MPa
h :=
a
c
v :=
e
c
L := a + b
⎡⎣FAB⋅ ( v)⎤⎦ ⋅ ( a) − w⋅ ( L) ⋅ ( 0.5⋅ L) = 0
⎛ L ⎞ ⋅ ( 0.5⋅ L)
⎝ a⋅ v ⎠
FAB = w⋅ ⎜
Assume failure of pin A or B:
⎛ π ⎞⋅ d 2
⎝ 4⎠ o
A := ⎜
V = τ allow⋅ A
V = 0.5FAB
⎡⎛ π ⎞ 2⎤
⎛ L ⎞ ⋅ ( 0.5⋅ L) = τ
⋅ ⎢⎜ ⋅ do ⎥
allow
⎝ a⋅ v ⎠
⎣⎝ 4 ⎠ ⎦
0.5 ⋅ w⋅ ⎜
a⋅ v
w :=
( 0.5L)
w = 6.632
2
⎡⎛ π ⎞ ⋅ d 2⎤
⎥
⎣⎝ 4 ⎠ o ⎦
⋅ τ allow⋅ ⎢⎜
kN
m
(controls!)
Ans
Assuming failure of rod AB:
N = σ allow⋅ A
N = FAB
⎛ π ⎞⋅ d 2
⎝ 4 ⎠ rod
A := ⎜
⎡⎛ π ⎞
2⎤
⎛ L ⎞ ⋅ ( 0.5⋅ L) = σ
⋅ ⎢⎜ ⋅ drod ⎥
allow
⎝ a⋅ v ⎠
⎣⎝ 4 ⎠
⎦
w⋅ ⎜
w :=
a⋅ v
0.5L
w = 7.791
2
⎡⎛ π ⎞ ⋅ d 2⎤
⎥
⎣⎝ 4 ⎠ rod ⎦
⋅ σ allow⋅ ⎢⎜
kN
m
Problem 1-103
The bar is supported by the pin. If the allowable tensile stress for the bar is (σt)allow = 150 MPa, and
the allowable shear stress for the pin is τallow = 85 MPa, determine the diameter of the pin for which
the load P will be a maximum. What is this maximum load? Assume the hole in the bar has the same
diameter d as the pin. Take t =6 mm and w = 50 mm.
Given:
τ allow := 85MPa
σ t_allow := 150MPa
t := 6mm
Solution:
w := 50mm
Given
Allowable Normal Stress : The effective cross-sectional
area Ae for the bar must be considered here by taking
into account the reduction in cross-sectional area
introduced by the hole. Here, effective area Ae is equal
to (w - d) t, and σallow equals to P /Ae .
σ t_allow =
P
( w − d) ⋅ t
[1]
Allowable Shear Stress: The pin is subjected to double
shear and therefore the allowable τ equals to 0.5P /Apin,
and the area A pin is equal to ( π/4) d2.
⎛ 2 ⎞⋅ ⎛ P ⎞
⎝ π ⎠ ⎜⎝ d2 ⎠
τ allow = ⎜
[2]
Solving [1] and [2]: Initial guess: d := 20mm
⎛P⎞
⎜ := Find ( P , d)
⎝d ⎠
P = 31.23 kN
Ans
d = 15.29 mm
Ans
P := 10kN
Problem 1-104
The bar is connected to the support using a pin having a diameter of d = 25 mm. If the allowable
tensile stress for the bar is (σt)allow = 140 MPa, and the allowable bearing stress between the pin and
the bar is (σb)allow =210 MPA, determine the dimensions w and t such that the gross area of the cross
section is wt = 1250 mm2 and the load P is a maximum. What is this maximum load? Assume the hole
in the bar has the same diameter as the pin.
Given:
σ t_allow := 140 MPa
2
A := 1250mm
Solution:
A = w⋅ t
σ b_allow := 210 MPa
d := 25mm
Given
Allowable Normal Stress : The effective cross-sectional
area Ae for the bar must be considered here by taking
into account the reduction in cross-sectional area
introduced by the hole. Here, effective area Ae is equal
to (w - d) t, that is (A- d t) and σallow equals to P /Ae .
σ t_allow =
P
A − d⋅ t
[1]
Allowable Bearing Stress : The projected area Ab is equal
to (d t), and σallow equals to P /Ab .
⎛P⎞
⎝ d⋅ t ⎠
σ b_allow = ⎜
[2]
Solving [1] and [2]: Initial guess:
⎛P⎞
⎜ := Find ( P , t)
⎝t⎠
And :
w :=
A
t
t := 0.5in
P = 105.00 kN
Ans
t = 20.00 mm
Ans
w = 62.50 mm
Ans
P := 1kip
Problem 1-105
The compound wooden beam is connected together by a bolt at B. Assuming that the connections at
A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B
and the required outer diameter of its washers if the allowable tensile stress for the bolt is (σt)allow =
150 MPa. and the allowable bearing stress for the wood is (σb)allow = 28 MPa. Assume that the hole in
the washers has the same diameter as the bolt.
Given: P1 := 3kN
P2 := 1.5kN
P3 := 2kN
σ t_allow := 150MPa
a := 2m
σ b_allow := 28MPa
b := 1.5m
Solution:
Given
From FBD (a):
ΣΜD=0;
−Cy⋅ ( 4⋅ b) + By⋅ ( 3b) + P2⋅ ( 2⋅ b) + P3⋅ ( b) = 0
[1]
⎡⎣By⋅ ( 2⋅ a + b) − Cy⋅ ( 2⋅ a)⎤⎦ − P1⋅ ( a) = 0
[2]
From FBD (b):
ΣΜA=0;
Solving [1] and [2]: Initial guess:
⎛⎜ By ⎞
:= Find ( By , Cy)
⎜ Cy
⎝ ⎠
By := 1kN
Cy := 2kN
⎛⎜ By ⎞ ⎛ 4.4 ⎞
=⎜
kN
⎜ Cy
⎝ ⎠ ⎝ 4.55 ⎠
For bolt:
Arae =
By
σ t_allow
⎛ π ⎞ ⋅ ⎛ d 2⎞ = By
⎜ ⎝ B⎠
σ t_allow
⎝ 4⎠
dB :=
4 ⎛ By ⎞
⋅⎜
π ⎝ σ t_allow ⎠
dB = 6.11 mm
Ans
For washer:
Area =
By
σ b_allow
By
⎛ π ⎞ ⋅ ⎛ d 2 − d 2⎞ =
⎜ ⎝ w
B ⎠
σ b_allow
⎝ 4⎠
dw :=
2
dB +
By ⎞
4 ⎛
⋅⎜
π ⎝ σ b_allow ⎠
dw = 15.41 mm
Ans
Problem 1-106
The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a single
leaf and therefore it involves single shear in the pin, and the support at B has a double leaf and therefore
it involves double shear. The allowable shear stress for both pins is τallow = 150 MPa. If a uniform
distributed load of w = 8 kN/m is placed on the bar, determine its minimum allowable position x from
B. Pins A and B each have a diameter of 8 mm. Neglect any axial force in the bar.
Given: τ allow := 150MPa
do := 8mm
a := 2m
w := 8
b := 2m
kN
m
Solution:
ΣΜA=0;
By⋅ ( a) − w⋅ ( b − x) ⋅ [ a + x + 0.5 ⋅ ( b − x) ] = 0
[1]
ΣΜB=0;
Ay⋅ ( a) − w⋅ ( b − x) ⋅ [ x + 0.5 ⋅ ( b − x) ] = 0
[2]
Assume failure of pin A:
Arae =
Ay
⎛ π ⎞ ⋅ ⎛ d 2⎞ = Ay
⎜ ⎝ o⎠
τ allow
⎝ 4⎠
τ allow
π
2
Ay := ⋅ ⎛⎝ do ⎞⎠ ⋅ τ allow
4
(
)
Ay = 7.5398 kN
Substitute value of force A into Eq [2],
Given
(
)
(
)
Ay⋅ ( a) − w⋅ b − x1 ⋅ ⎡⎣x1 + 0.5 ⋅ b − x1 ⎤⎦ = 0
Initial guess:
( )
x1 := 0.3m
x1 := Find x1
0.5By
τ allow
x1 = 0.480 m
xcase_1 := x1
Assume failure of pin B:
Arae =
[2]
⎛ π ⎞ ⋅ ⎛ d 2⎞ = 0.5By
⎜ ⎝ o⎠
τ allow
⎝ 4⎠
⎛π⎞
2
By := 2⋅ ⎜ ⋅ ⎛⎝ do ⎞⎠ ⋅ τ allow
4
(
⎝ ⎠
)
By = 15.0796 kN
Substitute value of force A into Eq [1],
Given
(
)
(
)
x2 := Find ( x2)
By⋅ ( a) − w⋅ b − x2 ⋅ ⎡⎣a + x2 + 0.5 ⋅ b − x2 ⎤⎦ = 0
Initial guess:
x2 := 0.3m
[1]
x2 = 0.909 m
xcase_2 := x2
Choose the larger x value:
(
)
x := max xcase_1 , xcase_2
x = 0.909 m
Ans
Problem 1-107
The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a single
leaf and therefore it involves single shear in the pin, and the support at B has a double leaf and therefore
it involves double shear. The allowable shear stress for both pins is τallow = 125 MPa. If x = 1 m,
determine the maximum distributed load w the bar will support. Pins A and B each have a diameter of 8
mm. Neglect any axial force in the bar.
Given:
τ allow := 125MPa
a := 2m
Solution:
x := 1m
do := 8mm
b := 2m
kN
wo :=
m
Given
ΣΜA=0;
Bw⋅ ( a) − wo⋅ ( b − x) ⋅ [ a + x + 0.5 ⋅ ( b − x) ] = 0
[1]
ΣΜB=0;
Aw⋅ ( a) − wo⋅ ( b − x) ⋅ [ x + 0.5 ⋅ ( b − x) ] = 0
[2]
Initial guess:
Bw := 1kN
Solving [1] and [2]:
For pin A:
Aw := 1kN
⎛⎜ Bw ⎞
:= Find ( Bw , Aw)
⎜ Aw
⎝ ⎠
⎛⎜ Bw ⎞ ⎛ 1.75 ⎞
=⎜
kN
⎜ Aw
0.75
⎝
⎠
⎝ ⎠
⎛ Aw ⎞
Ay = w1⋅ ⎜
⎝ wo ⎠
Arae =
Ay
τ allow
⎛ π ⎞ ⋅ ⎛ d 2⎞ = ⎛ w1 ⎞ ⋅ ⎛ Aw ⎞
⎜ ⎝ o⎠ ⎜
⎜
⎝ 4⎠
⎝ τ allow ⎠ ⎝ wo ⎠
⎛ wo ⎞
π
2
w1 := ⋅ ⎛⎝ do ⎞⎠ ⋅ τ allow ⋅ ⎜
4
⎝ Aw ⎠
(
For pin B
)
kN
w1 = 8.378
m
⎛ Bw ⎞
By = w ⋅ ⎜
⎝ wo ⎠
Arae =
0.5By
τ allow
⎛ π ⎞ ⋅ ⎛ d 2⎞ = ⎛ w2 ⎞ ⋅ ⎛ Bw ⎞
⎜ ⎝ o⎠ ⎜
⎜
⎝ 2⎠
⎝ τ allow ⎠ ⎝ wo ⎠
⎛ wo ⎞
π
2
w2 := ⋅ ⎛⎝ do ⎞⎠ ⋅ τ allow ⋅ ⎜
2
⎝ Bw ⎠
(
The smalleer w controls !
(
w := min w1 , w2
w = 7.181
kN
m
)
Ans
)
kN
w2 = 7.181
m
Problem 1-108
The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a single
leaf and therefore it involves single shear in the pin, and the support at B has a double leaf and therefore
it involves double shear. The allowable shear stress for both pins is τallow = 125 MPa. If x = 1 m and w
= 12 kN/m, determine the smallest required diameter of pins A and B. Neglect any axial force in the
bar.
Given: τ allow := 125MPa
x := 1m
do := 8mm
a := 2m
Solution:
b := 2m
w := 12
kN
m
Given
ΣΜA=0;
By⋅ ( a) − w⋅ ( b − x) ⋅ [ a + x + 0.5 ⋅ ( b − x) ] = 0
[1]
ΣΜB=0;
Ay⋅ ( a) − w⋅ ( b − x) ⋅ [ x + 0.5 ⋅ ( b − x) ] = 0
[2]
Initial guess:
By := 1kN
Solving [1] and [2]:
⎛⎜ By ⎞
:= Find ( By , Ay)
⎜ Ay
⎝ ⎠
⎛⎜ By ⎞ ⎛ 21 ⎞
=⎜
kN
⎜ Ay
9
⎝
⎠
⎝ ⎠
For pin A:
Arae =
Ay := 1kN
Ay
τ allow
⎛ π ⎞ ⋅ ⎛ d 2⎞ = ⎛ Ay ⎞
⎜ ⎝ A⎠ ⎜
⎝ 4⎠
⎝ τ allow ⎠
dA :=
4 ⎛ Ay ⎞
⋅⎜
π ⎝ τ allow ⎠
dA = 9.57 mm
Ans
For pin B
Arae =
0.5By
τ allow
⎛ π ⎞ ⋅ ⎛ d 2⎞ = ⎛ By ⎞
⎜ ⎝ B⎠ ⎜
⎝ 2⎠
⎝ τ allow ⎠
dB :=
2 ⎛ By ⎞
⋅⎜
π ⎝ τ allow ⎠
dB = 10.34 mm
Ans
Problem 1-109
The pin is subjected to double shear since it is used to connect the three links together. Due to wear,
the load is distributed over the top and bottom of the pin as shown on the free-body diagram.
Determine the diameter d of the pin if the allowable shear stress is τallow = 70 MPa and the load P = 40
kN. Also, determine the load intensities w1 and w2 .
Given:
τ allow := 70 MPa
P := 40kN
a := 37.5mm
b := 25mm
+ ΣF y=0;
P − w1 ( a) = 0
Solution:
Pin:
P
w1 :=
a
kN
w1 = 1066.67
m
Link:
+ ΣF y=0;
(
Ans
)
P − 2 0.5w2 ⋅ ( b) = 0
P
w2 :=
b
kN
w2 = 1600.00
m
Ans
Shear Stress
Area =
0.5P
τ allow
⎛ π ⎞ ⋅ d2 = P
⎜
τ allow
⎝ 2⎠
d :=
2 ⎛ P ⎞
⋅⎜
π ⎝ τ allow ⎠
d = 19.073 mm
Ans
Problem 1-110
The pin is subjected to double shear since it is used to connect the three links together. Due to wear,
the load is distributed over the top and bottom of the pin as shown on the free-body diagram.
Determine the maximum load P the connection can support if the allowable shear stress for the
material is τallow = 56 MPa and the diameter of the pin is 12.5 mm. Also, determine the load intensities
w1 and w2 .
Given:
τ allow := 56 MPa
d := 12.5mm
a := 37.5mm
b := 25mm
Solution:
Shear Stress
Area =
0.5P
τ allow
⎛ π ⎞ ⋅ d2 = P
⎜
τ allow
⎝ 2⎠
⎛ π ⎞ ⋅ d2 ⋅ τ
( allow)
⎝ 2⎠
P := ⎜
P = 13.7445 kN
Pin:
+ ΣF y=0;
P − w1 ( a) = 0
P
w1 :=
a
kN
w1 = 366.52
m
Ans
Link:
+ ΣF y=0;
(
)
P − 2 0.5w2 ⋅ ( b) = 0
P
w2 :=
b
kN
w2 = 549.78
m
Ans
Ans
Problem 1-111
The cotter is used to hold the two rods together. Determine the smallest thickness t of the cotter and
the smallest diameter d of the rods. All parts are made of steel for which the failure tensile stress is σfail
= 500 MPa and the failure shear stress is τfail = 375 MPa. Use a factor of safety of (F.S.)t = 2.50 in
tension and (F.S.)s = 1.75 in shear.
Given:
σ fail := 500MPa
τ fail := 375MPa
d2 := 40mm
h := 10mm
FS t := 2.50
FS s := 1.75
P := 30kN
Solution:
Allowable Normal Stress : Design of rod size
σ allow :=
σ fail
FS t
Area =
σ allow = 200 MPa
P
σ allow
⎛ π ⎞ ⋅ d2 = ⎛ P ⎞
⎜
⎜σ
⎝ 4⎠
⎝ allow ⎠
4
d :=
⋅
P
π σ allow
d = 13.82 mm
Ans
Allowable Shear Stress : Design of cotter size
τ allow :=
τ fail
FS s
Area =
τ allow = 214.29 MPa
0.5P
τ allow
⎛ 0.5P ⎞
h⋅ t = ⎜
⎝ τ allow ⎠
t :=
1 0.5P
⋅
h τ allow
t = 7 mm
Ans
Problem 1-112
The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine
the average normal stress in the shank, the average shear stress along the cylindrical area of the plate
defined by the section lines a-a, and the average shear stress in the bolt head along the cylindrical area
defined by the section lines b-b.
Given:
P := 8kN
dshank := 7mm
da_a := 18mm
db_b := 7mm
hhead := 8mm
hplate := 30mm
Solution:
Average Normal Stress:
⎛π⎞
2
Ashank := ⎜ ⎛⎝ dshank ⎞⎠
4
⎝ ⎠
σ t :=
P
Ashank
σ t = 207.9 MPa
Ans
τ a_avg = 4.72 MPa
Ans
τ b_avg = 45.47 MPa
Ans
Average Shear Stresses:
(
)
Aa_a := π da_a hplate
τ a_avg :=
P
Aa_a
(
)
Ab_b := π db_b hhead
τ b_avg :=
P
Ab_b
Problem 1-113
The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive
load of 6 kN. Determine the average normal and shear stress acting on the plane through section aa.
Show the results on a ifferential volume element located on the plane.
P := 6kN
Given:
d := 150mm
Solution:
θ := 30deg
φ := 90deg − θ
Equations of Equilibrium:
Va_a − P⋅ cos ( φ ) = 0
+ ΣF x=0;
Va_a := P⋅ cos ( φ )
Va_a = 3 kN
Na_a − P⋅ sin ( φ ) = 0
+ ΣF y=0;
Na_a := P⋅ sin ( φ )
Na_a = 5.196 kN
At inclined plane:
2
A :=
d
sin ( φ )
σ a_a :=
Na_a
A
σ a_a = 0.2 MPa
Ans
τ a_a :=
Va_a
A
τ a_a = 0.115 MPa
Ans
Problem 1-114
Determine the resultant internal loadings acting on the cross sections located through points D and E of
the frame.
kN
Given:
w := 2.5
a := 1.2m b := 0.9m
m
c := 1.5m e := 0.45m
Solution:
d :=
2
a +b
v :=
Support Reactions:
+ ΣΜA=0;
a
d
2
d = 1.5 m
h :=
b
d
L := b + c
By⋅ ( b) − ( w⋅ L) ( 0.5 ⋅ L) = 0
⎛ L⎞
By := ( w⋅ L ) ⋅ ⎜ 0.5
⎝ b⎠
By = 8 kN
+
ΣF y=0;
Ay − By + w⋅ L = 0
Ay := By − w⋅ L
Ay = 2 kN
By = FBC⋅ ( v)
+
FBC :=
By
v
FBC = 10 kN
ΣF x=0;
FBC⋅ ( h) − Ax = 0
Ax := FBC⋅ ( h)
Ax = 6 kN
Segment AD:
+
ΣF x=0;
ND − Ax = 0
ND := Ax
ND = 6 kN
Ans
+
ΣF y=0;
Ay + w⋅ ( e) + VD = 0
VD := −Ay − w⋅ ( e)
VD = −3.13 kN
Ans
MD = −1.153 kN⋅ m
Ans
+ ΣΜD=0;
MD + [ w⋅ ( e) ] ⋅ ( 0.5 ⋅ e) + Ay⋅ ( e) = 0
MD := −[ w⋅ ( e) ] ⋅ ( 0.5 ⋅ e) − Ay⋅ ( e)
Segment CE:
+
ΣF x=0;
NE + FBC = 0
ND := −FBC
ND = −10 kN
Ans
+ ΣF y=0;
VD := 0
VD = 0 kN
Ans
+ ΣΜE=0;
MD := 0
MD = 0 kN⋅ m
Ans
Problem 1-115
The circular punch B exerts a force of 2 kN on the top of the plate A. Determine the average shear
stress in the plate due to this loading.
Given:
P := 2kN
dpunch := 4mm
hplate := 2mm
Solution:
Average Shear Stresses:
(
)
Aa_a := π dpunch hplate
τ a_avg :=
P
Aa_a
τ a_avg = 79.58 MPa
Ans
Problem 1-116
The cable has a specific weight γ (weight/volume) and cross-sectional area A. If the sag s is small, so
that its length is approximately L and its weight can be distributed uniformly along the horizontal axis,
determine the average normal stress in the cable at its lowest point C.
Solution:
Equations of Equilibrium:
ΣMA=0;
⎛ γAL ⎞ ⋅ L = 0
⎝ 2 ⎠ 4
T⋅ s − ⎜
γ=
γ ⋅ A⋅ L
8⋅ s
Average Normal Stresses:
σ=
T
A
σ=
γ⋅L
8⋅ s
Ans
Problem 1-117
The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold
up the frame. If AB weighs 2.0 kN/m and the column FC has a weight of 3.0 kN/m, determine the
resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of
both the beam and column in the calculation.
kN
Given: wb := 2.0
L := 3.6m d := 1.8m
m
kN
wc := 3.0
m
H := 4.8m e := 1.2m
a := 3.6m
b := 3.6m
c := 1.2m
Solution:
Beam AB: L c :=
+ ΣΜA=0;
+ ΣF y=0;
2
2
L +c
c
vb :=
hb :=
Lc
By⋅ ( L) − wAB⋅ ( L ) ( 0.5 ⋅ L ) = 0
L
By := wb⋅ ( L) ⋅ ⎛⎜ 0.5 ⎞
⎝ L⎠
m
By = 7936.64
lb
2
s
−Ay − By + wb⋅ ( L) = 0
( )
ΣF x=0;
Lc
Ay := −By + wb⋅ ( L )
By
FBC :=
vb
Ax := FBC⋅ hb
By = FBC⋅ vb
+
L
( )
−FBC⋅ ( h) + Ax = 0
Ay = 3.6 kN
FBC = 11.38 kN
Ax = 10.8 kN
Segment AD:
+
ΣF x=0;
ND + Ax = 0
ND := −Ax
ND = −10.8 kN
Ans
+
ΣF y=0;
−Ay + wb⋅ ( d) + VD = 0
VD := Ay − wb⋅ ( d)
VD = 0 kN
Ans
+ ΣΜD=0;
MD + ⎡wb⋅ ( d)⎤ ⋅ ( 0.5 ⋅ d) − Ay⋅ ( d) = 0
⎣
⎦
MD := −⎡wb⋅ ( d)⎤ ⋅ ( 0.5 ⋅ d) + Ay⋅ ( d)
⎣
⎦
Member CG:
Hb :=
2
H +b
2
Column FC:
+ ΣΜC=0;
H
vc :=
Fx⋅ ( H) − Ax⋅ c = 0
hc :=
Hb
Fx := Ax⋅ ⎛⎜
MD = 3.24 kN⋅ m
Ans
b
Hb
c⎞
⎝ H⎠
Fx = 2.7 kN
( )
( )
+
ΣF x=0;
FBC⋅ hb − Ax + Fx − FCG⋅ hc = 0
+
ΣF y=0;
FCG :=
( )
FBC⋅ hb − Ax + Fx
FCG = 4.5 kN
hc
−Fy + By + wc⋅ ( H) + FBC⋅ vb + FCG⋅ vc = 0
( )
( )
Fy := By + wc⋅ ( H) + FBC⋅ ( vb) + FCG⋅ ( vc)
Fy = 25.2 kN
Segment FE:
+
ΣF x=0;
VE − Fx = 0
VE := Fx
VE = 2.7 kN
Ans
+
ΣF y=0;
NE + wc⋅ ( e) − Fy = 0
NE := −wc⋅ ( e) + Fy
NE = 21.6 kN
Ans
−ME + Fy⋅ ( e) = 0
ME := Fy⋅ ( e)
ME = 30.24 kN⋅ m
Ans
+ ΣΜE=0;
Problem 1-118
The 3-Mg concrete pipe is suspended by the three wires. If BD and CD have a diameter of 10 mm and
AD has a diameter of 7 mm, determine the average normal stress in each wire.
M := 3000kg
Given:
g := 9.81
m
2
s
h := 2m
Solution:
r := 1m
α := 120deg
dBD := 10mm
dCD := 10mm
W := M⋅ g
W = 29.43 kN
θ := 0.5α
θ = 60 deg
L :=
2
2
r +h
v :=
dAD := 7mm
h
L
Equations of Equilibrium:
ΣΜx=0;
2⋅ F⋅ ( r⋅ cos ( θ ) ) − FAD⋅ ( r) = 0
FBD = FCD = F
ΣΜy=0;
FBD⋅ ( r⋅ sin ( θ ) ) − FCD⋅ ( r⋅ sin ( θ ) ) = 0
FAD = F
ΣF z=0;
3⋅ [ F⋅ ( v) ] − W = 0
F :=
W
3v
F = 10.97 kN
Allowable Normal Stress :
σBD = σCD = σ1
σ1 =
F
Area
σ 1 :=
σAD = σ2
σ2 =
F
Area
σ 2 :=
F
π ⎛
2⎞
⋅ d
4 ⎝ BD ⎠
σ 1 = 139.65 MPa
Ans
σ 2 = 285 MPa
Ans
F
π ⎛
2
⋅ ⎝ dAD ⎞⎠
4
Problem 1-119
The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal
stress in each rod and the average shear stress in the pin A between the members.
P := 5kN
dpin := 25mm
d30 := 30mm
d40 := 40mm
Given:
Solution:
Average Normal Stress:
⎛π⎞
2
A40 := ⎜ ⎛⎝ d40 ⎞⎠
4
⎝ ⎠
σ 40 :=
P
A40
σ 40 = 3.979 MPa
Ans
σ 30 = 7.074 MPa
Ans
τ avg = 5.093 MPa
Ans
⎛π⎞
2
A30 := ⎜ ⎛⎝ d30 ⎞⎠
4
⎝ ⎠
σ 30 :=
P
A30
Average Shear Stresses:
⎛π⎞
2
Apin := ⎜ ⎛⎝ dpin ⎞⎠
4
⎝ ⎠
τ avg :=
0.5P
Apin
Problem 2-1
An air-filled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until the
ball's diameter becomes 175 mm, determine the average normal strain in the rubber.
Given:
d0 := 150mm
d := 175mm
Solution:
ε :=
πd − πd0
πd0
ε = 0.1667
mm
mm
Ans
Problem 2-2
A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having an
outer diameter of 125 mm, determine the average normal strain in the strip.
L 0 := 375mm
Given:
Solution:
L := π ⋅ ( 125) mm
ε :=
πL − πL0
πL 0
ε = 0.0472
mm
mm
Ans
Problem 2-3
The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the
end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
Given:
a := 3m
L CE := 4m
b := 4m
L BD := 4m
∆LCE := 10mm
Solution:
⎛ a ⎞ ⋅ ∆L
⎝ a + b ⎠ CE
∆LBD := ⎜
ε CE :=
ε BD :=
∆LCE
L CE
∆L BD
LBD
∆LBD = 4.2857 mm
ε CE = 0.00250
mm
mm
Ans
ε BD = 0.00107
mm
mm
Ans
Problem 2-4
The center portion of the rubber balloon has a diameter of d = 100 mm. If the air pressure within it
causes the balloon's diameter to become d = 125 mm, determine the average normal strain in the
rubber.
Given:
d0 := 100mm
d := 125mm
Solution:
ε :=
πd − πd0
πd0
ε = 0.2500
mm
mm
Ans
Problem 2-5
The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displace
10 mm downward, determine the normal strain developed in wires CE and BD.
Given:
a := 3m
b := 2m
c := 2m
L CE := 4m
L BD := 3m
∆ tip := 10mm
Solution:
∆LBD
a
=
∆L CE
∆LCE
a+b
a+b
=
∆ tip
a+b+c
⎛ a + b ⎞⋅ ∆
⎝ a + b + c ⎠ tip
∆LCE = 7.1429 mm
⎛ a ⎞⋅ ∆
⎝ a + b + c ⎠ tip
∆LBD = 4.2857 mm
∆LCE := ⎜
∆LBD := ⎜
Average Normal Strain:
ε CE :=
ε BD :=
∆LCE
L CE
∆L BD
LBD
ε CE = 0.00179
mm
mm
Ans
ε BD = 0.00143
mm
mm
Ans
Problem 2-6
The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normal
strain in each wire is εmax = 0.002 mm/mm, determine the maximum vertical displacement of the load
P.
Given:
a := 3m
b := 2m
L CE := 4m
L BD := 3m
ε allow := 0.002
mm
mm
c := 2m
Solution:
∆LBD
a
=
∆L CE
∆LCE
a+b
a+b
=
∆ tip
a+b+c
Average Elongation/Vertical Displacement:
∆LBD := L BD⋅ ε allow
∆LBD = 6.00 mm
⎛ a + b + c ⎞ ⋅ ∆L
⎝ a ⎠ BD
∆ tip := ⎜
∆ tip = 14.00 mm
∆LCE := LCE⋅ ε allow
∆LCE = 8.00 mm
⎛ a + b + c ⎞ ⋅ ∆L
⎝ a + b ⎠ CE
∆ tip := ⎜
∆ tip = 11.20 mm
(Controls !)
Ans
Problem 2-7
The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2
mm, determine the normal strain developed in each wire.
Given:
a := 300mm
θ := 30deg
∆ A := 2mm
Solution:
Consider the triangle CAA':
φ A := 180deg − θ
L CA' :=
φ A = 150 deg
2
2
( )
( )
a + ∆ A − 2⋅ a⋅ ∆ A ⋅ cos φ A
L CA' = 301.734 mm
∆ CA :=
L CA' − a
a
∆ CA = 0.00578
mm
mm
Ans
Problem 2-8
Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a
force is applied to the end D of the member and causes it to rotate by θ = 0.3°, determine the normal
strain in the cable. Originally the cable is unstretched.
Given:
a := 400mm
b := 300mm
c := 300mm
θ := 0.3deg
Solution:
L AB :=
2
2
a +b
L AB = 500 mm
Consider the triangle ACB':
φ C := 90deg + θ
L AB' :=
2
φ C = 90.3 deg
( )
2
a + b − 2⋅ a⋅ b⋅ cos φ C
L AB' = 501.255 mm
ε AB :=
LAB' − L AB
L AB
ε AB = 0.00251
mm
mm
Ans
Problem 2-9
Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a
force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm/mm
determine the displacement of point D. Originally the cable is unstretched.
Given:
a := 400mm
b := 300mm
ε AB := 0.0035
mm
mm
c := 300mm
Solution:
L AB :=
2
2
a +b
L AB = 500 mm
(
)
L AB' := L AB⋅ 1 + ε AB
L AB' = 501.750 mm
Consider the triangle ACB':
φ C = 90deg + θ
L AB' =
2
2
( )
a + b − 2⋅ a⋅ b⋅ cos φ C
⎡ ( a2 + b2 ) − L 2⎤
AB' ⎥
φ C := acos ⎢
2⋅ a⋅ b
⎣
⎦
φ C = 90.419 deg
θ := φ C − 90deg
θ = 0.41852 deg
θ = 0.00730 rad
∆ D := ( b + c) ⋅ θ
∆ D = 4.383 mm
Ans
Problem 2-10
The wire AB is unstretched when θ = 45°. If a vertical load is applied to bar AC, which causes θ =
47°, determine the normal strain in the wire.
Given:
θ := 45deg
∆θ := 2deg
Solution:
2
2
LAB =
L +L
LCB =
( 2L) + L
2
2
LAB =
2L
LCB =
5L
From the triangle CAB:
φ A := 180deg − θ
φ A = 135.00 deg
( ) = sin(φA)
sin φ B
L
LCB
⎛ L⋅ sin ( φ A) ⎞
φ B := asin ⎜
⎝
⎠
5L
φ B = 18.435 deg
From the triangle CA'B:
φ' B := φ B + ∆θ
φ' B = 20.435 deg
( ) = sin(180deg − φA')
sin φ' B
L
LCB
⎛ 5⋅ L⋅ sin ( φ' B) ⎞
φ A' := 180deg − asin ⎜
⎝
L
⎠
φ A' = 128.674 deg
φ' C := 180deg − φ' B − φ A'
φ' C = 30.891 deg
( ) = sin(φ'C)
sin φ' B
L
L A'B
LA'B :=
ε AB :=
( ) ⋅L
sin ( φ' B)
sin φ' C
L A'B − LAB
LAB
LAB :=
2L
ε AB = 0.03977
Ans
Problem 2-11
If a load applied to bar AC causes point A to be displaced to the left by an amount ∆L, determine the
normal strain in wire AB. Originally, θ = 45°.
Given:
θ := 45deg
∆L
L
ε AC =
Solution:
2
2
L AB =
L +L
L CB =
( 2L ) + L
2
2
L AB =
2L
L CB =
5L
From the triangle A'AB:
φ A := 180deg − θ
φ A = 135.00 deg
2
(
( )
)
L A'B =
∆L + LAB − 2⋅ ( ∆L ) ⋅ L AB ⋅ cos φ A
L A'B =
∆L + 2⋅ L + 2⋅ ( ∆L) ⋅ L
ε AB =
2
2
2
L A'B − LAB
LAB
2
ε AB =
∆L + 2⋅ L + 2⋅ ( ∆L ) ⋅ L −
2
2L
2L
2
ε AB =
⎡1 ⋅ ⎛ ∆L ⎞⎤ + 1 + ⎛ ∆L ⎞ − 1
⎢ ⎜
⎥
⎜
⎣2 ⎝ L ⎠⎦
⎝L ⎠
Neglecting the higher-order terms,
⎛ ∆L ⎞
ε AB = ⎜ 1 +
L ⎠
⎝
⎡
⎣
−1
1 ⎛ ∆L ⎞
⎤
⋅⎜
+ .....⎥ − 1
2 ⎝ L ⎠
⎦
ε AB = ⎢1 +
ε AB =
0.5
1 ⎛ ∆L ⎞
⋅⎜
2 ⎝ L ⎠
(Binomial expansion)
Ans
Alternatively,
ε AB =
L A'B − LAB
LAB
ε AB =
ε AB =
∆L⋅ sin ( θ )
2L
1 ⎛ ∆L ⎞
⋅⎜
2 ⎝ L ⎠
Ans
Problem 2-12
The piece of plastic is originally rectangular. Determine the shear strain γxy at corners A and B if the
plastic distorts as shown by the dashed lines.
Given:
a := 400mm
b := 300mm
∆Ax := 3mm
∆Ay := 2mm
∆Cx := 2mm
∆Cy := 2mm
∆Bx := 5mm
∆By := 4mm
Solution:
Geometry : For small angles,
α :=
β :=
ψ :=
θ :=
∆Cx
α = 0.00662252 rad
b + ∆Cy
∆By − ∆Cy
(
a + ∆Bx − ∆Cx
∆Bx − ∆Ax
(
)
β = 0.00496278 rad
)
ψ = 0.00662252 rad
b + ∆By − ∆Ay
∆Ay
a + ∆Ax
θ = 0.00496278 rad
Shear Strain :
−3
γ xy_B := β + ψ
γ xy_B = 11.585 × 10
γ xy_A := −( θ + ψ)
γ xy_A = −11.585 × 10
rad
−3
rad
Ans
Ans
Problem 2-13
The piece of plastic is originally rectangular. Determine the shear strain γxy at corners D and C if the
plastic distorts as shown by the dashed lines.
Given:
a := 400mm
b := 300mm
∆Ax := 3mm
∆Ay := 2mm
∆Cx := 2mm
∆Cy := 2mm
∆Bx := 5mm
∆By := 4mm
Solution:
Geometry : For small angles,
α :=
β :=
ψ :=
θ :=
∆Cx
α = 0.00662252 rad
b + ∆Cy
∆By − ∆Cy
(
a + ∆Bx − ∆Cx
∆Bx − ∆Ax
(
)
β = 0.00496278 rad
)
ψ = 0.00662252 rad
b + ∆By − ∆Ay
∆Ay
a + ∆Ax
θ = 0.00496278 rad
Shear Strain :
−3
γ xy_D := α + θ
γ xy_D = 11.585 × 10
γ xy_C := −( α + β )
γ xy_C = −11.585 × 10
rad
−3
rad
Ans
Ans
Problem 2-14
The piece of plastic is originally rectangular. Determine the average normal strain that occurs along th
diagonals AC and DB.
Given:
a := 400mm
b := 300mm
∆Ax := 3mm
∆Ay := 2mm
∆Cx := 2mm
∆Cy := 2mm
∆Bx := 5mm
∆By := 4mm
Solution:
Geometry :
L AC :=
a +b
2
2
L AC = 500 mm
L DB :=
a +b
2
2
L DB = 500 mm
L A'C' :=
(a + ∆Ax − ∆Cx)2 + (b + ∆Cy − ∆Ay)2
L A'C' = 500.8 mm
L DB' :=
(a + ∆Bx)2 + (b + ∆By)2
L DB' = 506.4 mm
Average Normal Strain :
ε AC :=
ε BD :=
LA'C' − L AC
LAC
LDB' − L DB
L DB
− 3 mm
ε AC = 1.601 × 10
ε BD = 12.800 × 10
mm
− 3 mm
mm
Ans
Ans
Problem 2-15
The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns
of the frame tilt θ = 2°. Determine the approximate normal strain in the wire when the frame is in thi
position. Assume the columns are rigid and rotate about their lower supports.
Given:
a := 4m
b := 3m
c := 1m
θ := 2deg
Solution:
⎛ θ ⎞⋅ π
⎝ 180 ⎠
θ= ⎜
θ = 0.03490659 rad
Geometry : The vertical dosplacement is negligible.
∆Ax := c⋅ θ
∆Ax = 34.907 mm
∆Bx := ( b + c) ⋅ θ
∆Bx = 139.626 mm
2
2
L AB :=
a +b
L AB = 5000 mm
L A'B' :=
(a + ∆Bx − ∆Ax)2 + b2
L A'B' = 5084.16 mm
Average Normal Strain :
ε AB :=
LA'B' − L AB
LAB
ε AB = 16.833 × 10
− 3 mm
mm
Ans
Problem 2-16
The corners of the square plate are given the displacements indicated. Determine the shear strain along
the edges of the plate at A and B.
Given: ax := 250mm
∆v := 5mm
ay := 250mm
∆h := 7.5mm
Solution:
At A :
⎛ θ' A ⎞
tan ⎜
⎝ 2 ⎠
=
ax − ∆h
ay + ∆v
⎛ ax − ∆h ⎞
θ' A := 2⋅ atan ⎜
⎝ ay + ∆v ⎠
⎛ π ⎞ − θ'
A
⎝ 2⎠
γ nt_A := ⎜
θ' A = 1.52056 rad
γ nt_A = 0.05024 rad
Ans
At B :
⎛ φ' B ⎞
tan ⎜
⎝ 2 ⎠
=
ay + ∆v
ax − ∆h
⎛ ay + ∆v ⎞
φ' B := 2⋅ atan ⎜
⎝ ax − ∆h ⎠
γ nt_B := φ' B −
π
2
φ' B = 1.62104 rad
γ nt_B = 0.05024 rad
Ans
Problem 2-17
The corners of the square plate are given the displacements indicated. Determine the average normal
strains along side AB and diagonals AC and DB.
Given:
ax := 250mm
ay := 250mm
∆v := 5mm
∆h := 7.5mm
Solution:
For AB :
2
2
L AB :=
ax + ay
L A'B' :=
(ax − ∆h)2 + (ay + ∆v)2
ε AB :=
LA'B' − L AB
LAB
L AB = 353.55339 mm
L A'B' = 351.89665 mm
ε AB = −4.686 × 10
− 3 mm
mm
Ans
For AC :
( )
L A'C' := 2⋅ ( ay + ∆v)
L AC := 2 ay
ε AC :=
LA'C' − L AC
LAC
L AC = 500 mm
L A'C' = 510 mm
ε AC = 20.000 × 10
− 3 mm
mm
Ans
For DB :
( )
L D'B' := 2⋅ ( ax − ∆h)
L DB := 2 ax
ε DB :=
LD'B' − L DB
LDB
L DB = 500 mm
L D'B' = 485 mm
− 3 mm
ε DB = −30.000 × 10
mm
Ans
Problem 2-18
The square deforms into the position shown by the dashed lines. Determine the average normal strain
along each diagonal, AB and CD. Side D'B' remains horizontal.
Given:
a := 50mm
b := 50mm
∆Bx := −3mm
∆Cx := 8mm
∆Cy := 0mm
L AD' := 53mm
θ' A := 91.5deg
Solution:
For AB :
(
)
∆By := L AD'⋅ cos θ' A − 90deg − b
2
2
∆By = 2.9818 mm
L AB :=
a +b
L AB = 70.7107 mm
L AB' :=
(a + ∆Bx)2 + (b + ∆By)2
L AB' = 70.8243 mm
ε AB :=
LAB' − L AB
L AB
− 3 mm
ε AB = 1.606 × 10
mm
Ans
For CD :
∆Dy := ∆By
2
∆Dy = 2.9818 mm
2
L CD :=
a +b
L CD = 70.7107 mm
L C'D' :=
(a + ∆Cx)2 + (b + ∆Dy)2 − 2⋅ (a + ∆Cx)⋅ (b + ∆Dy)⋅ cos (θ'A)
L C'D' = 79.5736 mm
ε CD :=
LC'D' − L CD
LCD
− 3 mm
ε CD = 125.340 × 10
mm
Ans
Problem 2-19
The square deforms into the position shown by the dashed lines. Determine the shear strain at each of
its corners, A, B, C, and D. Side D'B' remains horizontal.
a := 50mm
Given:
b := 50mm
∆Bx := −3mm
∆Cx := 8mm
θ' A := 91.5deg
Solution:
∆Cy := 0mm
LAD' := 53mm
θ' A = 1.597 rad
Geometry :
(
)
∆By := LAD'⋅ cos θ' A − 90deg − b
∆By = 2.9818 mm
∆Dy := ∆By
∆Dy = 2.9818 mm
(
)
∆Dx := −L AD'⋅ sin θ' A − 90deg
∆Dx = −1.3874 mm
In triangle C'B'D' :
LC'B' :=
(∆Cx − ∆Bx)2 + (b + ∆By)2
LC'B' = 54.1117 mm
LD'B' := a + ∆Bx − ∆Dx
LC'D' :=
LD'B' = 48.3874 mm
(a + ∆Cx − ∆Dx)2 + (b + ∆Dy)2
LC'D' = 79.586 mm
2
2
2
LC'B' + LD'B' − LC'D'
cos θ B' =
2⋅ L C'B' ⋅ L D'B'
( )
(
)(
)
⎡ L 2 + L 2 − L 2⎤
⎢ C'B'
D'B'
C'D' ⎥
θ B' := acos ⎢
⎥
⎣ 2⋅ ( LC'B') ⋅ ( LD'B') ⎦
θ B' = 101.729 deg
θ B' = 1.7755 rad
θ D' := 180deg − θ' A
θ D' = 88.500 deg
θ D' = 1.5446 rad
θ C' := 180deg − θ B'
θ C' = 78.271 deg
θ C' = 1.3661 rad
Shear Strain :
( )
γ xy_B := 0.5π − ( θ B')
γ xy_C := 0.5π − ( θ C')
γ xy_D := 0.5π − ( θ D')
γ xy_A := 0.5π − θ' A
−3
γ xy_A = −26.180 × 10
γ xy_B = −204.710 × 10
γ xy_C = 204.710 × 10
γ xy_D = 26.180 × 10
rad
−3
rad
Ans
rad
Ans
rad
Ans
−3
−3
Ans
Problem 2-20
The block is deformed into the position shown by the dashed lines. Determine the average normal
strain along line AB.
Given:
∆xBA := ( 70 − 30)mm
∆yBA := 100mm
∆xB'A := ( 55 − 30)mm
∆yB'A :=
( 1102 − 152)mm
Solution:
For AB :
2
2
L AB :=
∆xBA + ∆yBA
L AB' :=
∆xB'A + ∆yB'A
ε AB :=
2
L AB = 107.7033 mm
2
L AB' = 111.8034 mm
LAB' − L AB
L AB
ε AB = 38.068 × 10
− 3 mm
mm
Ans
Problem 2-21
A thin wire, lying along the x axis, is strained such that each point on the wire is displaced ∆x = k x2
along the x axis. If k is constant, what is the normal strain at any point P along the wire?
∆x = k⋅ x
Given:
2
Solution:
ε=
d
∆x
dx
ε = 2⋅ k⋅ x
Ans
Problem 2-22
The rectangular plate is subjected to the deformation shown by the dashed line. Determine the averag
shear strain γxy of the plate.
Given:
a := 150mm
b := 200mm
∆a := 0mm
∆b := −3mm
Solution:
⎛ ∆b ⎞
⎝ a⎠
∆θ := atan ⎜
∆θ = −1.146 deg
∆θ = −19.9973 × 10
−3
rad
Shear Strain :
γ xy := ∆θ
−3
γ xy = −19.997 × 10
rad
Ans
Problem 2-23
The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the averag
shear strain γxy of the plate.
Given:
a := 200mm
∆a := 3mm
b := 150mm
∆b := 0mm
Solution:
⎛ ∆a ⎞
⎝ b ⎠
∆θ := atan ⎜
∆θ = 1.146 deg
−3
∆θ = 19.9973 × 10
rad
Shear Strain :
γ xy := ∆θ
γ xy = 19.997 × 10
−3
rad
Ans
Problem 2-24
The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the
average normal strains along the diagonal AC and side AB.
Given:
a := 200mm
b := 150mm
∆Ax := −3mm
∆Bx := 0mm
∆Ay := 0mm
∆By := 0mm
∆Cx := 0mm
∆Dx := −3mm
∆Cy := 0mm
∆Dy := 0mm
Solution:
Geometry :
2
L AC :=
2
a +b
L AC = 250 mm
L AB := b
L AB = 150 mm
L A'C :=
(a + ∆Cx − ∆Ax)2 + (b + ∆Cy − ∆Ay)2
L A'C = 252.41 mm
L A'B :=
2
∆Ax + b
2
L A'B = 150.03 mm
Average Normal Strain :
ε AC :=
ε AB :=
LA'C − L AC
L AC
LA'B − L AB
L AB
− 3 mm
ε AC = 9.626 × 10
mm
− 6 mm
ε AB = 199.980 × 10
mm
Ans
Ans
Problem 2-25
The piece of rubber is originally rectangular. Determine the average shear strain γxy if the corners B
and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines
Given:
a := 300mm
b := 400mm
∆Ax := 0mm
∆Bx := 0mm
∆Ay := 0mm
∆By := 2mm
∆Dx := 3mm
∆Dy := 0mm
Solution:
⎛ ∆By ⎞
∆θ AB := atan ⎜
⎝ a ⎠
∆θ AB = 0.38197 deg
∆θ AB = 6.6666 × 10
−3
rad
⎛ ∆Dx ⎞
∆θ AD := atan ⎜
⎝ b ⎠
∆θ AD = 0.42971 deg
∆θ AD = 7.4999 × 10
−3
rad
Shear Strain :
γ xy_A := ∆θ AB + ∆θ AD
γ xy_A = 14.166 × 10
−3
rad
Ans
Problem 2-26
The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed
lines. Determine the average normal strain along the diagonal DB and side AD.
Given:
a := 300mm
b := 400mm
∆Ax := 0mm
∆Bx := 0mm
∆Ay := 0mm
∆By := 2mm
∆Dx := 3mm
∆Dy := 0mm
Solution:
Geometry :
2
L DB :=
2
a +b
L DB = 500 mm
L AD := b
L AD = 400 mm
L D'B' :=
(a + ∆Bx − ∆Dx)2 + (b + ∆Dy − ∆By)2
L AD' :=
∆Dx + b + ∆Dy
2
(
)2
L D'B' = 496.6 mm
L AD' = 400.01 mm
Average Normal Strain :
ε BD :=
ε AD :=
LD'B' − L DB
LDB
L AD' − L AD
L AD
ε BD = −6.797 × 10
ε AD = 28.125 × 10
− 3 mm
mm
− 6 mm
mm
Ans
Ans
Problem 2-27
The material distorts into the dashed position shown. Determine (a) the average normal strains εx
and εy , the shear strain γxy at A, and (b) the average normal strain along line BE.
Given:
a := 80mm
E x := 80mm
b := 125mm
Bx := 0mm
By := 100mm
∆Ax := 0mm
∆Ay := 0mm
∆Cx := 10mm
∆Cy := 0mm
∆Dx := 15mm
∆Dy := 0mm
E y := 50mm
Solution:
∆xAC := ∆Cx − ∆Ax
∆yAC := ∆Cy − ∆Ay
∆xAC = 10.00 mm
∆yAC = 0.00 mm
⎛ ∆Cx ⎞
∆θ AC := atan ⎜
⎝
∆θ AC = 79.8300 × 10
⎠
b
−3
rad
Since there is no deformation occuring along the y- and x-axis,
ε x_A := ∆yAC
ε x_A = 0
2
Ans
2
∆xAC + b − b
ε y_A :=
b
ε y_A = 0.00319
Ans
γ xy_A := ∆θ AC
γ xy_A = 79.830 × 10
−3
rad
Ans
Geometry :
∆Bx
∆Cx
∆Ex
∆Dx
=
By
b
⎛ By ⎞
∆Bx := ⎜
⋅ ∆Cx
b
⎝
⎠
∆Bx = 8 mm
∆By := 0mm
=
Ey
b
⎛ Ey ⎞
⋅ ∆D
⎝b⎠ x
∆Ex := ⎜
∆Ex = 6 mm
∆Ey := 0mm
L BE :=
(Ex − Bx)2 + (Ey − By)2
L BE = 94.34 mm
L B'E' :=
(a + ∆Ex − ∆Bx)2 + (Ey + ∆Ey − ∆By)2
L B'E' = 92.65 mm
ε BE :=
L B'E' − L BE
L BE
− 3 mm
ε BE = −17.913 × 10
mm
Ans
Note: Negative sign indicates shortening of BE.
Problem 2-28
The material distorts into the dashed position shown. Determine the average normal strain that occurs
along the diagonals AD and CF.
Given: a := 80mm
E x := 80mm
b := 125mm
Bx := 0mm
By := 100mm
∆Ax := 0mm
∆Ay := 0mm
∆Cx := 10mm
∆Cy := 0mm
∆Dx := 15mm
∆Dy := 0mm
E y := 50mm
Solution:
∆xAC := ∆Cx − ∆Ax
∆yAC := ∆Cy − ∆Ay
∆xAC = 10.00 mm
∆yAC = 0.00 mm
⎛ ∆Cx ⎞
∆θ AC := atan ⎜
∆θ AC = 79.8300 × 10
⎝
b
⎠
L AD :=
a +b
2
2
L AD = 148.41 mm
L CF :=
a +b
2
2
L CF = 148.41 mm
L A'D' :=
(a + ∆Dx − ∆Ax)2 + (b + ∆Dy − ∆Ay)2
−3
rad
Geometry :
L A'D' = 157.00 mm
L C'F :=
(a − ∆Cx)2 + (b − ∆Cy)2
L C'F = 143.27 mm
Average Normal Strain :
ε AD :=
ε CF :=
L A'D' − L AD
LAD
LC'F − L CF
LCF
ε AD = 57.914 × 10
− 3 mm
mm
− 3 mm
ε CF = −34.653 × 10
mm
Ans
Ans
Problem 2-29
The block is deformed into the position shown by the dashed lines. Determine the shear strain at
corners C and D.
Given:
a := 100mm
∆Ax := −15mm
∆Bx := −15mm
b := 100mm
L CA' := 110mm
Solution:
Geometry :
⎛ ∆Ax ⎞
∆θ C := asin ⎜
⎝ LCA' ⎠
⎛ ∆Bx ⎞
∆θ D := asin ⎜
⎝
LCA'
⎠
∆θ C = −7.84 deg
∆θ C = −0.1368 rad
∆θ D = −7.84 deg
∆θ D = −0.1368 rad
Shear Strain :
γ xy_C := ∆θ C
−3
γ xy_C = −136.790 × 10
rad
Ans
rad
Ans
γ xy_D := −∆θ D
γ xy_D = 136.790 × 10
−3
Problem 2-30
The bar is originally 30 mm long when it is flat. If it is subjected to a shear strain defined by γxy = 0.0
x, where x is in millimeters, determine the displacement ∆y at the end of its bottom edge. It is distorte
into the shape shown, where no elongation of the bar occurs in the x direction.
Given:
L := 300mm
γ xy = 0.02 ⋅ x
unit := 1mm
Solution:
dy
= tan γ xy
dx
dy
= tan ( 0.02 ⋅ x)
dx
( )
∆y
⌠
⎮
⌡0
L
⌠
1 dy = ⎮ tan ( 0.02 ⋅ x) ⋅ ( unit) dx
⌡0
30
⌠
∆y := ⎮
⌡0
tan ( 0.02x) ⋅ ( unit) dx
∆y = 9.60 mm
Ans
Problem 2-31
The curved pipe has an original radius of 0.6 m. If it is heated nonuniformly, so that the normal strain
along its length is ε = 0.05 cos θ, determine the increase in length of the pipe.
Given:
r := 0.6m
ε = 0.05 ⋅ cos ( θ )
Solution:
⌠
∆L = ⎮ ε dL
⌡
90deg
⌠
∆L = ⎮
⌡0
90deg
⌠
∆L := ⎮
⌡0
0.05 ⋅ cos ( θ ) d( rθ )
0.05 ⋅ r⋅ cos ( θ ) dθ
∆L = 30.00 mm
Ans
Problem 2-32
Solve Prob. 2-31 if ε = 0.08 sin θ .
Given:
r := 0.6m
ε = 0.08 ⋅ sin ( θ )
Solution:
⌠
∆L = ⎮ ε dL
⌡
90deg
⌠
∆L = ⎮
⌡0
90deg
⌠
∆L := ⎮
⌡0
0.08 ⋅ sin ( θ ) d( rθ )
0.08 ⋅ r⋅ sin ( θ ) dθ
∆L = 0.0480 m
Ans
Problem 2-33
A thin wire is wrapped along a surface having the form y = 0.02 x2, where x and y are in mm. Origina
the end B is at x = 250 mm. If the wire undergoes a normal strain along its length of ε = 0.0002x,
2
⎛ dy ⎞ ⋅ dx .
determine the change in length of the wire. Hint: For the curve, y = f (x), ds = 1 + ⎜
⎝ dx ⎠
Given:
Bx := 250mm
ε = 0.0002 ⋅ x
unit := 1mm
Solution:
y = 0.02x
2
dy
= 0.04x
dx
2
⌠
∆L = ⎮ ε ds
⌡
⎛ dy ⎞ ⋅ dx
⎝ dx ⎠
ds =
1+⎜
ds =
1 + ( 0.04x) ⋅ dx
2
B
⌠ x
2
∆L = ⎮
( 0.0002x) ⋅ 1 + ( 0.04x) dx
⌡0
250
⌠
∆L := ( unit) ⋅ ⎮
⌡0
∆L = 42.252 mm
2
( 0.0002x) ⋅ 1 + ( 0.04x) dx
Ans
Problem 2-34
The fiber AB has a length L and orientation If its ends A and B undergo very small displacements uA
and vB, respectively, determine the normal strain in the fiber when it is in position A'B'.
Solution:
Geometry:
L A'B' =
(L⋅ cos (θ) − uA)2 + (L⋅ sin(θ) − vB)2
L A'B' =
L + uA + vB + 2L ⋅ vB⋅ sin ( θ ) − uA⋅ cos ( θ )
2
2
(
2
)
Average Normal Strain:
ε AB =
L A'B' ⋅ −L
L
2
ε AB =
1+
2
uA + vB
L
2
+
(
2⋅ vB⋅ sin ( θ ) − uA⋅ cos ( θ )
L
Neglecting higher-order terms uA2 and vB2 ,
ε AB =
1+
(
2⋅ vB⋅ sin ( θ ) − uA⋅ cos ( θ )
L
) −1
Using the binomial theorem:
ε AB = 1 +
ε AB =
(
)
1 ⎡ 2⋅ vB⋅ sin ( θ ) − uA⋅ cos ( θ ) ⎤
⎢
⎥ + .. − 1
2⎣
L
⎦
vB⋅ sin ( θ )
L
−
uA⋅ cos ( θ )
L
Ans
) −1
Problem 2-35
If the normal strain is defined in reference to the final length, that is,
ε' =
n
⎛ ∆s' − ∆s ⎞
⎜
p → p' ⎝ ∆ s' ⎠
lim
instead of in reference to the original length, Eq.2-2, show that the difference in these strains is
represented as a second-order term, namely, εn - ε'n = εn ε'n.
Solution:
εn =
∆ S' − ∆ S
∆S
⎛ ∆S' − ∆S ⎞ − ⎛ ∆S' − ∆S ⎞
⎜
⎝ ∆S ⎠ ⎝ ∆S' ⎠
ε n − ε'n = ⎜
2
∆S' − 2⋅ ( ∆S ) ⋅ ( ∆S' ) + ∆ S
ε n − ε'n =
( ∆S ) ⋅ ( ∆S' )
2
(∆S' − ∆S )2
ε n − ε'n =
(∆S )⋅ (∆S' )
⎛ ∆S' − ∆S ⎞ ⋅ ⎛ ∆S' − ∆S ⎞
⎜
⎝ ∆S ⎠ ⎝ ∆S' ⎠
ε n − ε'n = ⎜
( )( )
ε n − ε'n = ε n ⋅ ε'n
(Q.E.D.)
Problem 3-1
A concrete cylinder having a diameter of 150 mm. and gauge length of 300 mm is tested in
compression. The results of the test are reported in the table as load versus contraction. Draw the
stress-strain diagram using scales of 10 mm = 2 MPa and 10 mm = 0.1(10-3) mm/mm. From the
diagram, determine approximately the modulus of elasticity.
Contraction
Load
(mm)
(kN)
P :=
δ :=
Given:
d := 150
L := 300
Solution:
10 P
δ
⎛π⎞ 2
A := ⎜ ⋅ d * σ :=
ε :=
L
A
⎝ 4⎠
3
σ=P/A
(MPa)
ε=δ/L
(mm/mm)
⎛ 0.00 ⎞
⎜
⎜ 1.41 ⎟
⎜ 2.69 ⎟
⎜ 4.67 ⎟
⎜
⎟
⎜ 5.80 ⎟
⎜ 7.22 ⎟
σ=⎜
⎟
⎜ 8.49 ⎟
⎜ 9.76 ⎟
⎜ 10.89 ⎟
⎜
⎟
⎜ 13.16 ⎟
⎜ 14.15 ⎟
⎜
⎝ 15.00 ⎠
⎛ 0.000000 ⎞
⎜
⎜ 0.000050 ⎟
⎜ 0.000100 ⎟
⎜ 0.000167 ⎟
⎜
⎟
⎜ 0.000217 ⎟
⎜ 0.000283 ⎟
ε =⎜
⎟
⎜ 0.000333 ⎟
⎜ 0.000375 ⎟
⎜ 0.000417 ⎟
⎜
⎟
⎜ 0.000517 ⎟
⎜ 0.000583 ⎟
⎜
⎝ 0.000625 ⎠
Regression
curve:
0.0
25.0
0.0000
0.0150
47.5
0.0300
82.5
0.0500
102.5
0.0650
127.5
0.0850
150.0
0.1000
172.5
0.1125
192.5
0.1250
232.5
0.1550
250.0
0.1750
265.0
0.1875
x := min ( ε ) , 0.00005 .. max ( ε )
Coeff := loess ( ε , σ , 1.5 )
Y ( x) := interp ( Coeff , ε , σ , x)
16
14
12
10
Modulus of Elasticity:
σ
From the stress-strain diagram,
∆ε := ( 0.0003 − 0)
Y( x )
mm
mm *
6
∆σ := ( 8.0 − 0)MPa *
E approx :=
8
4
∆σ
∆ε
E approx = 26.67 GPa
2
Ans
0 1 .10
4 .
4
4
4
4
4
4
2 10 3 .10 4 .10 5 .10 6 .10 7 .10
ε ,x
Problem 3-2
Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between the
origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of
resilience.
( 6)
σ=P/A
(MPa)
MJ := 10 J
Unit used:
σ :=
Solution:
ε :=
0.0
232.4
Regression curve:
x := min ( ε ) , 0.00005 .. max ( ε )
Coeff := loess ( ε , σ , 0.9 )
Y ( x) := interp ( Coeff , ε , σ , x)
ε=δ/L
(mm/mm)
0.0000
0.0006
318.5
0.0010
345.8
0.0014
360.5
0.0018
373.8
0.0022
400
Modulus of Elasticity:
From the stress-strain diagram,
mm
mm
360
∆σ := ( 232.4 − 0)MPa
320
∆ε := ( 0.0006 − 0)
E approx :=
∆σ
∆ε
280
E approx = 387.3 GPa
Ans
240
σ
Y( x)
Modulus of Resilience:
The modulus of resilience is equal to the area
under the initial linear portion of the curve.
∆ε := ( 0.0006 − 0)
160
mm
mm
120
3 kN
2
80
∆σ := ( 232.4 − 0)10
1
⋅ ∆ε ⋅ ∆σ
2
MJ
ur = 0.0697
3
m
200
m
ur :=
40
Ans
0
5 .10
4
0.001
0.0015
ε,x
0.002
0.0025
Problem 3-3
Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between the
origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The
rupture stress is σr = 373.8 MPa.
Unit used:
Solution:
( 6)
MJ := 10 J
σ :=
Regression curve:
x := min ( ε ) , 0.00005 .. max ( ε )
Coeff := loess ( ε , σ , 0.9 )
Y ( x) := interp ( Coeff , ε , σ , x)
σ=P/A
(MPA)
0.0
232.4
ε :=
ε=δ/L
(mm/mm)
0.0000
0.0006
318.5
0.0010
345.8
0.0014
360.5
0.0018
373.8
0.0022
400
Modulus of Resilience:
The modulus of resilience is equal
to the area under the curve.
360
1
A1 := ⋅ ( 232.4 ) ⋅ ( 0.0004 + 0.0010 )
2
320
A2 := 318.5 ⋅ ( 0.0022 − 0.0010 )
280
1
A3 := ⋅ ( 373.8 − 318.5 ) ⋅ ( 0.0022 − 0.0010 )
2
240
σ
Y( x )
1
A4 := ⋅ ( 318.5 − 232.4 ) ⋅ ( 0.0010 − 0.0006 )
2
Atotal := A1 + A2 + A3 + A4
6 J
ut := Atotal⋅ 10 ⋅
3 *
m
ut = 0.595
MJ
3*
m
Ans
200
160
120
80
40
0
5 .10
4
0.001
0.0015
ε ,x
0.002
0.0025
Problem 3-4
A tension test was performed on a steel specimen having an original diameter of 13 mm and gauge
length of 50 mm. The data is listed in the table. Plot the stress-strain diagram and determine
approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use
a scale of 10 mm = 209 MPa and 10 mm = 0.05 mm/mm. Redraw the elastic region, using the same
stress scale but a strain scale of 10 mm = 0.001 mm/mm.
Load
Elongation
(kN)
(mm)
Given:
d := 12.5*
L := 50*
P :=
δ :=
0.0
0.0000
3
10 ⋅ P
δ
⎛π⎞ 2
A := ⎜ ⋅ d * σ :=
ε := *
*
L
A
⎝ 4⎠
Solution:
σ=P/A
(MPa)
ε=δ/L
(mm/mm.)
⎛ 0.00 ⎞
⎛ 0.00000 ⎞
⎜ 61.12
⎜ 0.00025
⎜
⎜
⎟
⎟
⎜ 187.42 ⎟
⎜ 0.00075 ⎟
⎜ 325.95 ⎟
⎜ 0.00125 ⎟
⎜
⎜
⎟
⎟
448.18
0.00175
⎜
⎜
⎟
⎟
⎜ 480.78 ⎟
⎜ 0.00250 ⎟
⎜
⎜
⎟
⎟
σ = 480.78 * ε = 0.00400 *
⎜
⎜
⎟
⎟
⎜ 488.92 ⎟
⎜ 0.01000 ⎟
⎜ 676.34 ⎟
⎜ 0.02000 ⎟
⎜
⎜
⎟
⎟
⎜ 814.87 ⎟
⎜ 0.05000 ⎟
⎜ 875.99 ⎟
⎜ 0.14000 ⎟
⎜ 794.50 ⎟
⎜ 0.20000 ⎟
⎜
⎜
753.76
⎝
⎝ 0.23000 ⎠
⎠
7.5
0.0125
23.0
0.0375
40.0
0.0625
55.0
0.0875
59.0
0.1250
59.0
0.2000
60.0
0.5000
83.0
1.0000
100.0
2.5000
107.5
7.0000
97.5
10.0000
92.5
11.5000
Curve Fit:
Use:
⎛ x ⎞
⎜ 0.3
F ( x) := ⎜ x ⎟
⎜ 0.6
⎝x ⎠
Fit := linfit ( ε , σ , F)
⎛ −4189.696 ⎞
⎜
Fit = 2285.571
⎜
⎝ 596.538 ⎠
x := min ( ε ) , 0.00005 .. max ( ε )
Y ( x) := F ( x) ⋅ Fit
1000
1000
800
800
600
σ
400
Y( x ) 400
200
200
σ
0
0.001
0.002
0.003
ε
Modulus of Elasticity:
0.004
0.005
600
0
0.05
0.1
0.15
ε,x
From the stress-strain diagram,
0.2
0.25
From the stress-strain diagram,
mm
∆ε := ( 0.00125 − 0)
mm *
∆σ := ( 326 − 0)MPa
*
∆σ
Eapprox :=
∆ε *
Eapprox = 260.8 GPa
*
Ans
σ Y = 448MPa
Ans
σ ult = 890MPa
Ans
σ R = 753.8MPa
Ans
Problem 3-5
The stress-strain diagram for a steel alloy having an original diameter of 12mm and a gauge length of
50 mm is given in the figure. Determine approximately the modulus of elasticity for the material, the
load on the specimen that causes yielding, and the ultimate load the specimen will support.
Given:
d := 12mm
Solution:
A := ⎜
L := 50mm
⎛ π ⎞ ⋅ d2
⎝ 4⎠
Modulus of Elasticity:
From the stress-strain diagram,
∆ε := ( 0.001 − 0)
mm
mm
∆σ := ( 290 − 0)MPa
E :=
∆σ
∆ε *
E = 290 GPa*
Ans
From the stress-strain diagram, σ Y := 290MPa
Yield Load:
( )
PY := σ Y ⋅ A
PY = 32.80 kN
From th stress-strain diagram,
Ultimate Load:
Ans
σ u := 550MPa
( )
Pu := σ u ⋅ A
Pu = 62.20 kN
Ans
Problem 3-6
The stress-strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of
50 mm is given in the figure. If the specimen is loaded until it is stressed to 500MPa, determine the
approximate amount of elastic recovery and the increase in the gauge length after it is unloaded.
d := 12mm
Given:
L := 50mm
σ max := 500MPa
Solution:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
Modulus of Elasticity:
From the stress-strain diagram,
∆ε := ( 0.001 − 0)
mm
mm
∆σ := ( 290 − 0)MPa
E :=
∆σ
∆ε *
E = 290 GPa*
Ans
Elastic Recovery:
rRe :=
σ max
E
AmountRe := rRe ⋅ ( L)
rRe = 0.00172
mm
mm
AmountRe = 0.08621 mm
Ans
mm
From the stress-strain diagram, ε max := 0.08
mm
Permanent set:
(Permanent elongation)
rps := ε max − rRe
Amountps := rps⋅ ( L)
rps = 0.07828
mm
mm
Amountps = 3.91379 mm
Ans
Problem 3-7
The stress-strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of
50 mm is given in the figure. Determine approximately the modulus of resilience and the modulus of
toughness for the material.
( 6)
MJ := 10 J
Unit used:
d := 12mm
Given:
L := 50mm
Solution:
Modulus of Resilience:
The modulus of resilience is equal to the area
under the initial linear portion of the curve.
∆ε := ( 0.001 − 0)
mm
mm
∆σ := ( 290 − 0)MPa
ur :=
1
⋅ ∆ε ⋅ ∆σ
2
ur = 0.145 MPa
Ans
Modulus of Toughness:
The modulus of toughness is equal to the area under the curve, and could be approximated by
counting the number of sqaures. the total number of squares is:
n := 33
∆ε sq := ( 0.04 )
mm
mm
∆σ sq := 100MPa
(
)(
)
ut := n⋅ ∆ε sq ⋅ ∆σ sq
ut = 132 MPa
Ans
Problem 3-8
The stress-strain diagram for a steel bar is shown in the figure. Determine approximately the modulus
of elasticity, the proportional limit, the ultimate stress, and the modulus of resilience. If the bar is
loaded until it is stressed to 450 MPa, determine the amount of elastic strain recovery and the
permanent set or strain in the bar when it is unloaded.
3
Unit used:
kJ := 10 J
Given:
σ max := 450MPa
Solution:
Modulus of Elasticity:
From th stress-strain diagram,
∆ε := ( 0.0015 − 0)
mm
mm
∆σ := ( 325 − 0) ⋅ MPa
E :=
∆σ
∆ε
E = 216.67 GPa
Ans
Modulus of Resilience:
The modulus of resilience is equal to the area under the
initial linear portion of the curve.
∆ε := ( 0.0015 − 0)
mm
mm
∆σ := ( 325 − 0) ⋅ MPa
1
⋅ ∆ε ⋅ ∆σ
2
kJ
ur = 243.75
3
m
ur :=
Elastic Recovery:
Ans
rRe :=
From th stress-strain diagram,
Permanent set:
σ max
E
rRe = 0.00208
ε max := 0.0750
mm
mm
Ans
mm
mm
Ans
mm
mm
(Permanent elongation)
rps := ε max − rRe
rps = 0.07292
Problem 3-9
The σ-ε diagram for elastic fibers that make up human skin and muscle is shown. Determine the
modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.
Modulus of Elasticity:
From the stress-strain diagram,
∆ε := ( 2.00 − 0)
mm
mm
∆σ := ( 77 − 0)MPa
E :=
∆σ
E = 38.5 MPa
∆ε
Ans
Modulus of Resilience:
The modulus of resilience is equal to the area under the initial linear portion of the curve.
∆ε := ( 2.00 − 0)
mm
mm
∆σ := ( 77 − 0)MPa
ur :=
1
⋅ ∆ε ⋅ ∆σ
2
ur = 77.00 MPa
Ans
Modulus of Toughness:
The modulus of toughness is equal to the area under the curve.
A1 := ur
A2 := 0.5 σ 1 + σ 2 ⋅ ε 2 − ε 1
(
)(
)
ut := A1 + A2
ut = 134.75 MPa
Ans
ε 0 := 0
ε 1 := 2.00
ε 2 := 2.25
σ 0 := 0
σ 1 := 77MPa σ 2 := 385MPa
Problem 3-10
An A-36 steel bar has a length of 1250 mm and cross-sectional area of 430 mm2. Determine the length
of the bar if it is subjected to an axial tension of 25 kN. The material has linear-elastic behavior.
2
A := 430mm
Given:
L 0 := 1250mm
σ Y := 250MPa
P := 25kN
E st := 200GPa
Solution:
Normal Stress:
P
σ = 58.140 MPa
A
Hence Hook's law is still valid.
σ :=
[less than yield stress σy]
Normal Strain:
ε :=
σ
E st
ε = 290.6977 × 10
− 6 mm
mm
Thus,
δL := ε ⋅ L0
δL = 0.36337 mm
L := L 0 + δL
L = 1250.363 mm
Ans
Problem 3-11
The stress-strain diagram for polyethylene, which is used to sheath coaxial cables, is determined from
testing a specimen that has a gauge length of 250 mm. If a load P on the specimen develops a strain of
ε = 0.024 mm/mm, determine the approximate length of the specimen, measured between the gauge
points, when the load is removed. Assume the specimen recovers elastically.
Given:
L 0 := 250mm
Solution:
Modulus of Elasticity:
From th stress-strain diagram,
∆ε := ( 0.004 − 0)
mm
mm
∆σ := ( 14.0 − 0) ⋅ MPa
E :=
∆σ
∆ε
E = 3500.00 MPa
Elastic Recovery:
mm
From the stress-strain diagram, ε max := 0.024
mm
σ max
mm
rRe :=
rRe = 0.00743
E
mm
Permanent set:
rps := ε max − rRe
rps = 0.01657
mm
mm
Permanent elongation:
∆L := rps⋅ L0
∆L = 4.14286 mm
L := L 0 + ∆L
L = 254.143 mm
Ans
σ max := 26MPa
Problem 3-12
Fiberglass has a stress-strain diagram as shown. If a 50-mm-diameter bar of length 2 m made from
this material is subjected to an axial tensile load of 60 kN, determine its elongation.
Given:
Solution:
L 0 := 2m
d := 50mm
P := 60kN
unit := 1Pa
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
σ :=
P
A
Given:
6
σ = 30.558 × 10 Pa
6 0.5
σ' = 300( 10 ) ⋅ ε
ε :=
σ' :=
σ
unit
⎡⎢ σ' 2 ⎤⎥
⎢ 9( 1016)⎥
⎣
⎦
ε = 0.010375
( )
∆L := ε ⋅ L 0
mm
mm
∆L = 20.75 mm
Ans
Problem 3-13
The change in weight of an airplane is determined from reading the strain gauge A mounted in the
plane's aluminum wheel strut. Before the plane is loaded, the strain gauge reading in a strut is ε1 =
0.00100 mm/mm, whereas after loading ε2 = 0.00243 mm/mm. Determine the change in the force on
the strut if the cross-sectional area of the strut is 2200 mm2. E al = 70 GPa.
Given:
2
A := 2200mm
E al := 70⋅ GPa
ε 1 := 0.00100
mm
mm
ε 2 := 0.00243
mm
mm
Solution:
Stress-strain Relationship:
Applying Hook's law σ = E ε .
σ 1 := Eal⋅ ε 1
σ 1 = 70.00 MPa
σ 2 := Eal⋅ ε 2
σ 2 = 170.10 MPa
Normal Force:
Applying equation σ = P / A .
P1 := A⋅ σ 1
P1 = 154.00 kN
P2 := A⋅ σ 2
P2 = 374.22 kN
∆P := P2 − P1
∆P = 220.22 kN
Thus,
Ans
Problem 3-14
A specimen is originally 300 mm long, has a diameter of 12 mm, and is subjected to a force of 2.5 kN.
When the force is increased to 9 kN, the specimen elongates 22.5 mm. Determine the modulus of
elasticity for the material if it remains elastic.
Given:
Solution:
d := 12mm
L 0 := 300mm
P1 := 2.5kN
P2 := 9kN
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
Applying equation σ = P / A .
Normal Force:
σ 1 :=
Thus,
∆L := 22.5mm
P1
A
P2
σ 2 :=
A
∆σ := σ 2 − σ 1
∆ε :=
E :=
σ 1 = 22.105 MPa
σ 2 = 79.577 MPa
∆σ = 57.473 MPa
∆L
L0
∆σ
∆ε
E = 766.3 MPa
Ans
Problem 3-15
A structural member in a nuclear reactor is made from a zirconium alloy. If an axial load of 20 kN is to
be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3
with respect to yielding. What is the load on the member if it is 1-m long and its elongation is 0.5 mm?
E zr = 100 GPa, σY = 400 MPa. The material has elastic behavior.
Given:
P := 20kN
L 0 := 1m
E zr := 100⋅ GPa
∆L := 0.5mm
σ Y := 400⋅ MPa
FoS := 3
Solution:
σ allow :=
Areq :=
σY
σ allow = 133.33 MPa
FoS
P
2
σ allow
A := Areq
Areq = 150 mm
∆L
L0
ε = 0.0005
σ := E zr⋅ ε
σ = 50 MPa
P := σ ⋅ A
P = 7.5 kN
ε :=
Ans
mm
mm
Ans
Problem 3-16
The pole is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5 mm,
determine how much it stretches when a horizontal force of 15 kN acts on the pole.
Given:
Solution:
P := 15kN
E st := 200⋅ GPa
a := 1.2m
b := 1m
d := 5mm
θ := 30deg
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
L := a + b
Support Reactions:
ΣΜB=0;
Cx⋅ ( L) + P⋅ ( b) = 0
b
Cx := −P⋅
L
Cx = −6.82 kN
+ ΣF x=0;
Cx + P + Bx = 0
Bx := −Cx − P
Bx = −8.18 kN
FAB :=
L AB :=
σ AB :=
ε AB :=
FAB
A
σ AB
E st
δ AB := ε AB⋅ LAB
−Bx
FAB = 16.36 kN
sin ( θ )
L
L AB = 2.54 m
cos ( θ )
σ AB = 833.393 MPa
ε AB = 0.0041670
δ AB = 10.586 mm
mm
mm
Ans
Problem 3-17
By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness. The stress-strain
diagrams for three types of this material showing this effect are given below. Specify the type that
should be used in the manufacture of a rod having a length of 125 mm and a diameter of 50 mm, that
is required to support at least an axial load of 100 kN and also be able to stretch at most 6 mm.
Given:
d := 50mm
L 0 := 125mm
P := 100kN
δL := 6mm
Solution:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
Normal Stress:
σ :=
P
A
σ = 50.930 MPa
Normal Strain:
ε :=
δL
L0
ε = 0.048000
mm
mm
From the stress-strain diagram,the copolymer will satisfy both
stress and strain requirements.
Ans
Problem 3-18
The steel wires AB and AC support the 200-kg mass. If the allowable axial stress for the wires is σallow
= 130 MPa, determine the required diameter of each wire. Also, what is the new length of wire AB
after the load is applied? Take the unstretched length of AB to be 750 mm. E st = 200 GPa.
g := 9.81
Given:
m
m := 200kg
2
s
θ := 60deg
v :=
4
5
h :=
3
5
L 0 := 750mm
E st := 200GPa
Solution:
σ allow := 130MPa
W := m⋅ g
Axial force in steel wires AB and AC:
Initial guess: FAC := 1N
Given
+
ΣF x=0;
−FAB⋅ cos ( θ ) + FAC⋅ ( h) = 0
[1]
+
ΣF y=0;
FAB⋅ sin ( θ ) + FAC⋅ ( v) − W = 0
[2]
⎛⎜ FAC ⎞
:= Find ( FAC , FAB)
⎜ FAB
⎝
⎠
Solving [1] and [2]:
Wire AB :
Wire AC:
⎛ π ⎞⋅ d 2
⎝ 4 ⎠ AB
4 ⎛ FAB ⎞
⋅⎜
π ⎝ σ allow ⎠
⎛⎜ FAC ⎞ ⎛ 1066.75 ⎞
=⎜
N
⎜ FAB
⎝
⎠ ⎝ 1280.10 ⎠
FAB = σ allow⋅ ⎜
dAB :=
⎛ π ⎞⋅ d 2
⎝ 4 ⎠ AC
4 ⎛ FAC ⎞
⋅⎜
π ⎝ σ allow ⎠
dAB = 3.54 mm
Ans
dAC = 3.23 mm
Ans
FAC = σ allow⋅ ⎜
dAC :=
Applying Hook's law σ = E ε .
Stress-strain Relationship:
ε AB :=
Thus,
FAB := 2N
σ allow
E st
(
ε AB = 0.000650
)
L AB := L0⋅ 1 + ε AB
mm
mm
L AB = 750.487 mm
Ans
Problem 3-19
The two bars are made of polystyrene, which has the stress-strain diagram shown. If the
cross-sectional area of bar AB is 950 mm2 and BC is 2500 mm2, determine the largest force P that can
be supported before any member ruptures. Assume that buckling does not occur.
Given:
a := 1.2m
b := 0.9m
2
AAB := 950mm
Solution:
c :=
2
2
a +b
2
ABC := 2500mm
h :=
a
c
v :=
b
c
+ ΣF y=0;
FAB⋅ ( v) − P = 0
+ ΣF x=0;
FAB⋅ ( h) − FBC = 0 [2]
[1]
Solving Eqs.[1] and [2]:
5
P
3
4
FBC = P
3
FAB =
[1a]
[2a]
Assume tension failure of BC:
From the stress-strain diagram,
(
)(
)
σ R_t := 35MPa
FBC := ABC ⋅ σ R_t
FBC = 87.50 kN
From Eq.[2a],
P := 0.75 ⋅ FBC
P = 65.63 kN
Pcase_1 := P
Assume compression failure of AB:
From the stress-strain diagram,
(
)(
FAB := AAB ⋅ σ R_c
)
From Eq.[1a],
σ R_c := 175MPa
FAB = 166.25 kN
P := 0.60 ⋅ FAB
P = 99.75 kN
Pcase_2 := P
Chosoe the smallest value:
(
)
P := min Pcase_1 , Pcase_2
P = 65.63 kN
Ans
Problem 3-20
The two bars are made of polystyrene, which has the stress-strain diagram shown. Determine the
cross-sectional area of each bar so that the bars rupture simultaneously when the load P = 15 kN.
Assume that buckling does not occur.
Given:
a := 1.2m
Solution:
c :=
2
b := 0.9m
2
a +b
P := 15kN
h :=
a
c
v :=
b
c
+ ΣF y=0;
FAB⋅ ( v) − P = 0
+ ΣF x=0;
FAB⋅ ( h) − FBC = 0 [2]
[1]
Solving Eqs.[1] and [2]:
5
P
3
4
FBC := P
3
FAB :=
FAB = 25.00 kN
FBC = 20.00 kN
For member BC:
From the stress-strain diagram,
⎛ FBC ⎞
ABC := ⎜
⎝ σ R_t ⎠
σ R_t := 35MPa
2
ABC = 571.43 mm
Ans
For member AB:
From the stress-strain diagram,
⎛ FAB ⎞
AAB := ⎜
⎝ σ R_c ⎠
σ R_c := 175MPa
2
AAB = 142.86 mm
Ans
Problem 3-21
The stress-strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a
strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine
the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the
diameter of the post is 80 mm.
P := 80kN
Given:
L AB := 2m
L CD := 0.5m
dAB := 40mm
dCD := 80mm
Solution:
L AC := 1.5m
P
Ay :=
2
P
Cy :=
2
FAB := Ay
FCD := Cy
Support Reactions:
⎛π⎞
2
AreaAB := ⎜ ⋅ dAB
4
⎛π⎞
2
AreaCD := ⎜ ⋅ dCD
4
⎝ ⎠
⎝ ⎠
From the stress-strain diagram,
E :=
32.2MPa
0.01
E = 3220.00 MPa
σ AB :=
ε AB :=
FAB
σ AB = 31.831 MPa
AreaAB
σ AB
E
ε AB = 0.0098854
δ AB := ε AB⋅ LAB
δ AB = 19.771 mm
σ CD :=
ε CD :=
FCD
AreaCD
σ CD
mm
mm
σ CD = 7.958 MPa
E
ε CD = 0.0024714
δ CD := ε CD⋅ LCD
δ CD = 1.236 mm
Angle of tilt α:
tan ( α ) :=
mm
mm
δ AB − δ CD
L AC
⎛ δ AB − δ CD ⎞
α := atan ⎜
⎝
LAC
⎠
α = 0.708 deg
Ans
Problem 3-22
The stress-strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a
strut AB and post CD made from this material, determine the largest load P that can be applied to the
beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm.
Given:
L AB := 2m
L CD := 0.5m
dAB := 12mm
dCD := 40mm
Solution:
L AC := 1.5m
P
Ay =
2
P
Cy =
2
FAB = Ay
FCD = Cy
Support Reactions:
⎛π⎞
2
AreaAB := ⎜ ⋅ dAB
4
⎛π⎞
2
AreaCD := ⎜ ⋅ dCD
4
⎝ ⎠
⎝ ⎠
For rupture of strut AB:
From the stress-strain diagram,
σ R_t := 50.0MPa
( )
P := 2AreaAB⋅ ( σ R_t)
FAB = AreaAB⋅ σ R_t
P = 11.31 kN
Ans
(Controls!)
For rupture of post CD:
From the stress-strain diagram,
σ R_c := 95.0MPa
( )
P := 2AreaCD⋅ ( σ R_c)
FCD = AreaCD⋅ σ R_c
P = 238.76 kN
Problem 3-23
The beam is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5
mm, determine how much it stretches when a distributed load of w = 1.5 kN/m acts on the pipe. The
material remains elastic.
kN
m
L := 3m
w := 1.5
dAB := 5mm
θ := 30deg
Given:
E st := 200⋅ GPa
Solution:
Support Reactions:
ΣΜC=0;
FAB⋅ sin ( θ ) ⋅ ( L ) − w⋅ ( L ) ⋅ ( 0.5 ⋅ L) = 0
FAB :=
⎛π⎞
2
AreaAB := ⎜ ⋅ dAB
4
⎝ ⎠
σ AB :=
ε AB :=
FAB
AreaAB
σ AB
E st
δ AB := ε AB⋅ LAB
w⋅ L
2 sin ( θ )
L AB :=
L
cos ( θ )
σ AB = 229.183 MPa
ε AB = 0.0011459
mm
mm
δ AB = 3.970 mm
Ans
Problem 3-24
The beam is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5
mm, determine the distributed load w if the end B is displaced 18 mm downward.
Given:
L := 3m
δBy := 18mm
dAB := 5mm
θ B := 30deg
E st := 200⋅ GPa
Solution:
( )
Consider triangle BB'C:
L ⋅ sin θ C = δBy
⎛ δBy ⎞
θ C := asin ⎜
⎝ L ⎠
Consider triangle AB'C:
( )
θ' C := 90deg + θ C
2
L'AB :=
L AC := L⋅ tan θ B
(
2
)
( )
L AC + L − 2⋅ L AC ⋅ L ⋅ cos θ' C
L'AB = 3.4731 m
⎛π⎞
2
AreaAB := ⎜ ⋅ dAB
⎝ 4⎠
L AB :=
L
( )
cos θ B
δ AB := L'AB − L AB
ε AB :=
δ AB = 8.9883 mm
δ AB
ε AB = 0.0025947
LAB
σ AB := ε AB⋅ E st
(
mm
mm
σ AB = 518.942 MPa
)
FAB := σ AB⋅ AreaAB
FAB = 10.19 kN
Support Reactions:
ΣΜC=0;
( )
FAB⋅ sin θ B ⋅ ( L) − w⋅ ( L) ⋅ ( 0.5 ⋅ L ) = 0
⎛ 2 sin ( θ B) ⎞
w := ⎜
⎝
L
⎠
⋅ FAB
w = 3.40
kN
m
Ans
Problem 3-25
Direct tension indicators are sometimes used instead of torque wrenches to insure that a bolt has a
prescribed tension when used for connections. If a nut on the bolt is tightened so that the six heads of
the indicator that were originally 3 mm high are crushed 0.3 mm, leaving a contact area on each head
of 1.5 mm2, determine the tension in the bolt shank. The material has the stressstrain diagram shown.
Given:
h := 3mm
δh := 0.3mm
2
Area := 1.5mm
number := 6
unit := 1MPa
Solution:
Stress-strain Relationship:
ε :=
δh
h
ε = 0.1000
mm
mm
From the stress-strain diagram,
σ − 450
ε − 0.0015
=
600 − 450
0.3 − 0.0015
⎡
⎣
⎛ ε − 0.0015 ⎞⎤ ⋅ unit
⎥
⎝ 0.3 − 0.0015 ⎠⎦
σ := ⎢450 + ( 600 − 450) ⎜
σ = 499.50 MPa
Axial Force:
For each head
P := σ ⋅ ( Area)
P = 0.749 kN
Thus, the tension in the bolt is
T := ( number) ⋅ P
T = 4.50 kN
Ans
Problem 3-26
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it,
determine the change in its length and the change in its diameter. E p = 2.70 GPa, νp = 0.4.
Given:
Solution:
P := 300N
L := 200mm
d := 15mm
E P := 2.70 ⋅ GPa
ν := 0.4
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
σ :=
P
A
ε long :=
σ = 1.698 MPa
σ
EP
ε long = 0.0006288
mm
mm
δ := ε long⋅ L
δ = 0.126 mm
ε lat := −ν ⋅ ε long
ε lat = −0.0002515
∆d := ε lat⋅ ( d)
∆d = −0.003773 mm
Ans
mm
mm
Ans
Problem 3-27
The block is made of titanium Ti-6A1-4V and is subjected to a compression of 1.5 mm along the y
axis, and its shape is given a tilt of θ = 89.7°.Determine εy, εx, and γxy .
Given:
L x := 125mm
L y := 100mm
δ y := −1.5 mm
θ := 89.7deg
ν := 0.36
Solution:
Normal Strain:
ε y :=
Poisson's Ratio:
δy
ε y = −0.01500
Ly
mm
mm
Ans
The lateral and longitudinal strain can be related using Poisson's ratio.
( )
ε x := −ν ⋅ ε y
ε x = 0.00540
mm
mm
Ans
Shear Strain:
β := 180deg − θ
β = 90.30 deg
β = 1.58 rad
Thus,
γ xy :=
π
−β
2
γ xy = −0.00524 rad
Ans
Problem 3-28
A short cylindrical block of bronze C86100, having an original diameter of 38 mm and a length of 75
mm, is placed in a compression machine and squeezed until its length becomes 74.5 mm. Deterinme
the new diameter of the block.
Given:
Solution:
L := 75mm
L' := 74.5mm
d := 38mm
ν := 0.34
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
ε long :=
L' − L
L
ε long = −0.0066667
mm
mm
mm
mm
ε lat := −ν ⋅ ε long
ε lat = 0.0022667
∆d := ε lat⋅ ( d)
∆d = 0.086133 mm
d' := d + ∆d
d' = 38.0861 mm
Ans
Problem 3-29
The elastic portion of the stress-strain diagram for a steel alloy is shown in the figure. The specimen
from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When the
applied load on the specimen is 50 kN, the diameter is 12.99265 mm. Determine Poisson's ratio for the
material.
Given:
d := 13mm
d' := 12.99265mm
L := 50mm
P := 50kN
Solution:
Normal Stress:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
σ :=
P
A
Normal Strain:
σ = 376.698 MPa
From the stress-strain diagram, the modulus of elasticity is
E :=
400MPa
0.002
E = 200 GPa
Applying Hook's law σ = E ε .
σ
E
ε long = 0.0018835
mm
mm
d' − d
d
ε lat = −0.0005654
mm
mm
ε long :=
ε lat :=
Poisson's Ratio:
The lateral and longitudinal strain can be related using Poisson's ratio.
ν :=
−ε lat
ε long
ν = 0.30018
Ans
Problem 3-30
The elastic portion of the stress-strain diagram for a steel alloy is shown in the figure. The specimen
from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a load
of P = 20 kN is applied to the specimen, determine its diameter and gauge length. Take ν = 0.4.
Given:
L := 50mm
d := 13mm
P := 20kN
ν := 0.4
Solution:
Normal Stress:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
σ :=
P
A
Normal Strain:
σ = 150.679 MPa
From the stress-strain diagram, the modulus of elasticity is
E :=
400MPa
0.002
E = 200 GPa
Applying Hook's law σ = E ε .
ε long :=
σ
E
ε long = 0.0007534
mm
mm
Thus,
Poisson's Ratio:
δL := ε long⋅ ( L )
δL = 0.037670 mm
L' := L + δL
L' = 50.0377 mm
Ans
The lateral and longitudinal strain can be related using Poisson's ratio.
(
ε lat := −ν ⋅ ε long
)
ε lat = −0.00030136
mm
mm
∆d := ε lat⋅ ( d)
∆d = −0.003918 mm
d' := d + ∆d
d' = 12.99608 mm
Ans
Problem 3-31
The shear stress-strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 6
mm is made of this material and used in the lap joint, determine the modulus of elasticity E and the
force P required to cause the material to yield. Take ν = 0.3.
Given:
d := 6mm
ν := 0.3
Solution:
Modulus of Rigidity:
From the stress-strain diagram,
G :=
350MPa
0.004
Modulus of Elasticity:
G = 87500 MPa
G=
E
2⋅ ( 1 + ν )
E := 2G⋅ ( 1 + ν )
E = 227500 MPa
Ans
Yielding Stress:
τ Y := 350MPa
From the stress-strain diagram,
⎛ π ⎞ ⋅ d2
⎝ 4⎠
τY =
( )
P = 9.896 kN
A := ⎜
P := τ Y ⋅ A
P
A
Ans
Problem 3-32
The brake pads for a bicycle tire are made of rubber. If a frictional force of 50 N is applied to each
side of the tires, determine the average shear strain in the rubber. Each pad has cross-sectional
dimensions of 20 mm and 50 mm. Gr = 0.20 MPa.
Given:
a := 20mm
b := 50mm
V := 50N
G := 0.20MPa
Solution:
A := a⋅ b
2
A = 1000.00 mm
Average Shear Stress:
The shear force is V.
τ :=
V
A
τ = 0.050 MPa
Shear Stress-strain Relationship:
Applying Hooke's law for shear: τ = G⋅ γ
γ :=
τ
G
γ = 0.250 rad
Ans
Problem 3-33
The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm.
Both the plug and the sleeve are 50 mm long. Determine the axial pressure p that must be applied to the
top of the plug to cause it to contact the sides of the sleeve. Also, how far must the plug be
compressed downward in order to do this? The plug is made from a material for which E = 5 MPa, ν
= 0.45.
Given:
d := 30mm
d' := 32mm
E := 5MPa
ν := 0.45
L := 50mm
Solution:
ε lat :=
ν=
d' − d
d
−ε lat
ε long
ε lat = 0.0666667
−ε lat
ε long :=
mm
mm
ε long = −0.14815
ν
mm
mm
Applying Hook's law σ = E ε .
(
)
p = −0.741 MPa
Ans
δL := ε long ⋅ L
δL = −7.41 mm
Ans
p := E ⋅ ε long
(
)
Problem 3-34
The rubber block is subjected to an elongation of 0.75 mm. along the x axis, and its vertical faces are
given a tilt so that θ = 89.3°. Deterinme the strains εx, εy and γxy . Take νr = 0.5.
Given:
L x := 100mm
L y := 75mm
δ x := 0.75mm
θ := 89.3deg
ν := 0.5
Solution:
Normal Strain:
ε x :=
Poisson's Ratio:
δx
ε x = 0.00750
Lx
mm
mm
Ans
The lateral and longitudinal strain can be related using Poisson's ratio.
ε y := −ν ⋅ ε x
ε y = −0.00375
mm
mm
Ans
Shear Strain:
θ = 89.30 deg
θ = 1.56 rad
Thus,
γ xy :=
π
−θ
2
γ xy = 0.01222 rad
Ans
Problem 3-35
The elastic portion of the tension stress-strain diagram for an aluinmum alloy is shown in the figure.
The specimen used for the test has a gauge length of 50 mm. and a diameter of 12.5 mm. When the
applied load is 45 kN, the new diameter of the specimen is 12.48375 mm. Compute shear modulus Gal
for the aluinmum.
Given:
d := 12.5mm
d' := 12.48375mm
L := 50mm
P := 45kN
Solution:
Normal Stress:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
σ :=
P
A
Normal Strain:
σ = 366.693 MPa
From the stress-strain diagram, the modulus of elasticity is
E :=
500MPa
0.00614
E = 81433.22 MPa
Applying Hook's law σ = E ε .
σ
E
ε long = 0.0045030
mm
mm
d' − d
d
ε lat = −0.0013000
mm
mm
ε long :=
ε lat :=
Poisson's Ratio:
The lateral and longitudinal strain can be related using Poisson's ratio.
ν :=
G :=
−ε lat
ε long
E
2⋅ ( 1 + ν )
ν = 0.28870
G = 31.60 GPa
Ans
Problem 3-36
The elastic portion of the tension stress-strain diagram for an aluinmum alloy is shown in the figure. The
specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. If the applied load is
50 kN deterinme the new diameter of the specimen. The shear modulus is Gal = 28 GPa.
Given:
d := 12.5mm
L := 50mm
P := 50kN
G := 28⋅ GPa
Solution:
Normal Stress:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
σ :=
P
A
Normal Strain:
σ = 407.44 MPa
From the stress-strain diagram, the modulus of elasticity is
E :=
500MPa
0.00614
E = 81433.22 MPa
Applying Hook's law σ = E ε .
ε long :=
Poisson's Ratio:
σ
E
ε long = 0.0050033
mm
mm
E
G=
2⋅ ( 1 + ν )
Thus,
ν := ⎜
⎛ E ⎞−1
⎝ 2⋅ G ⎠
ν = 0.454
The lateral and longitudinal strain can be related using Poisson's ratio.
(
ε lat := −ν ⋅ ε long
)
ε lat = −0.00227233
mm
mm
∆d := ε lat⋅ ( d)
∆d = −0.028404 mm
d' := d + ∆d
d' = 12.4716 mm
Ans
Problem 3-37
The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in
each bolt is 4 kN, deterinme the normal strain in the bolts. Each bolt has a diameter of 5 mm. If σY =
280 MPa and Est = 200 GPa, what is the strain in each bolt when the nut is unscrewed so that the
clamping force is released?
Given:
d := 5mm
σ Y := 28MPa
P := 4kN
E := 200⋅ GPa
Solution:
Normal Stress:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
σ :=
Normal Strain:
P
A
σ = 203.7183 MPa ( < σY = 280 MPa )
Since σ < σ Y, Hook's law is still valid.
ε :=
σ
E
ε = 0.0010186
mm
mm
Ans
If the nut is unscrewed, the load is zero. Therefore, the strain ε = 0.
Ans
Problem 3-38
The rigid pipe is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5
mm., determine how much it stretches when a load of P = 1.5 kN acts on the pipe. The material
remains elastic.
Given:
P := 1.5kN
E st := 200⋅ GPa
L BC := 2.4m
d := 5mm
θ := 60deg
Solution:
Support Reactions:
ΣΜC=0;
(
)
(
)
−FAB⋅ cos ( θ ) ⋅ L AB + P⋅ LAB = 0
FAB :=
Normal Stress:
P
cos ( θ )
⎛ π ⎞ ⋅ d2
⎝ 4⎠
Area := ⎜
σ AB :=
Normal Strain:
FAB = 3 kN
FAB
Area
L AB :=
L BC
sin ( θ )
σ AB = 152.789 MPa
L AB = 2.771 m
Applying Hook's law σ = E ε .
ε AB :=
Thus,
σ AB
E st
ε AB = 0.0007639
mm
mm
(
)
δLAB := ε AB⋅ L AB
δLAB = 2.1171 mm
Ans
Problem 3-39
The rigid pipe is supported by a pin at C and an A-36 guy wire AB. If the wire has a diameter of 5
mm., determine the load P if the end B is displaced 2.5 mm. to the right. E st = 200 GPa.
Given:
L := 2.4m
d := 5mm
θ := 60deg
δBx := 2.5mm
E st := 200⋅ GPa
Solution:
Consider triangle BB'C:
( )
L ⋅ sin θ C = δBx
⎛ δBx ⎞
θ C := asin ⎜
⎝ L ⎠
Consider triangle AB'C:
L AC := L⋅ cot ( θ )
θ' C := 90deg + θ C
2
L'AB :=
2
(
)
( )
L AC + L − 2⋅ L AC ⋅ L ⋅ cos θ' C
L'AB = 2.7725 m
Normal Strain:
Normal Stress:
L
L AB :=
sin ( θ )
ε AB :=
L'AB − LAB
L AB
ε AB = 0.0004510
mm
mm
⎛ π ⎞ ⋅ d2
⎝ 4⎠
Area := ⎜
Applying Hook's law σ = E ε .
Thus,
σ AB := ε AB⋅ E st
σ AB = 90.191 MPa
FAB := σ AB⋅ ( Area)
FAB = 1.771 kN
Support Reactions:
ΣΜC=0;
FAB⋅ cos ( θ ) ⋅ ( L) − P⋅ L = 0
P := FAB⋅ cos ( θ )
P = 0.885 kN
Ans
Problem 3-40
While undergoing a tension test, a copper-alloy specimen having a gauge length of 50 mm. is subjected
to a strain of 0.40 mm./mm. when the stress is 490 MPa. If σY = 315 MPa when εY = 0.0025
mm./mm., determine the distance between the gauge points when the load is released.
Given:
L 0 := 50mm
Solution:
mm
mm
mm
ε Y := 0.0025
mm
ε 1 := 0.40
Modulus of Elasticity:
E :=
σ 1 := 490⋅ MPa
σ Y := 315MPa
σY
εY
E = 126.00 GPa
Elastic Recovery:
rRe :=
σ1
E
rRe = 0.0038889
Permanent set:
rps := ε 1 − rRe
rps = 0.39611
mm
mm
mm
mm
Permanent elongation:
∆L := rps⋅ L0
∆L = 19.806 mm
L := L 0 + ∆L
L = 69.806 mm
Ans
Problem 3-41
The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an
inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve
are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the
bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. E al = 70 GPa,
E mg = 45 GPa.
Given:
L b := 80mm
db := 8mm
L s := 30mm
ds_o := 20mm
ds_i := 12mm
P := 8kN
E al := 70GPa
E mg := 45GPa
Solution:
⎛π⎞ 2
Ab := ⎜ ⋅ db
4
⎛π⎞
2
2
As := ⎜ ⋅ ⎛⎝ ds_o − ds_i ⎞⎠
4
⎝ ⎠
Normal Stress:
⎝ ⎠
σ b :=
P
Ab
σ b = 159.15 MPa
σ s :=
P
As
σ s = 39.79 MPa
Normal Strain:
ε b :=
ε s :=
σb
Eal
σs
Emg
ε b = 0.002274
mm
mm
Ans
ε s = 0.000884
mm
mm
Ans
Problem 3-42
A tension test was performed on a steel specimen having an original diameter of 12.5mm and a gauge
length of 50 mm. The data is listed in the table. Plot the stress-strain diagram and determine
approximately the modulus of elasticity, the ultimate stress, and the rupture stress. Use a scale of 20
mm = 50 MPa and 20 mm = 0.05 mm/mm. Redraw the linear-elastic region, using the same stress
scale but a strain scale of 20 mm = 0.001 mm/mm.
Load
Elongation
(kN)
(mm)
Given:
d := 12.5* L := 50*
P :=
δ :=
0.0
0.0000
3
Solution:
P⋅ 10
δ
⎛π⎞ 2
A := ⎜ ⋅ d * σ :=
ε := *
*
L
A
⎝ 4⎠
11.1
0.0175
31.9
0.0600
ε=δ/L
(mm/mm)
37.8
0.1020
40.9
0.1650
43.6
0.2490
53.4
1.0160
62.3
3.0480
64.5
6.3500
62.3
8.8900
58.8
11.9380
σ=P/A
(MPa)
⎛⎜ 0.0000 ⎞
⎛⎜ 0.000000 ⎞
⎜ 90.4509 ⎟
⎜ 0.000350 ⎟
⎜ 259.9446 ⎟
⎜ 0.001200 ⎟
⎟
⎟
⎜
⎜
⎜ 308.0221 ⎟
⎜ 0.002040 ⎟
⎜ 333.2832 ⎟
⎜ 0.003300 ⎟
⎟
⎟
⎜
⎜
σ = 355.2848 * ε = 0.004980 *
⎟
⎟
⎜
⎜
⎜ 435.1423 ⎟
⎜ 0.020320 ⎟
⎜ 507.6661 ⎟
⎜ 0.060960 ⎟
⎟
⎟
⎜
⎜
⎜ 525.5933 ⎟
⎜ 0.127000 ⎟
⎜ 507.6661 ⎟
⎜ 0.177800 ⎟
⎜ 479.1455
⎜ 0.238760
⎠
⎠
⎝
⎝
Use
⎛ x ⎞
⎜ 0.3
F ( x) := ⎜ x ⎟
⎜ 0.6
⎝x ⎠
⎛ 107.325 ⎞
⎜
Fit = 2123.113
⎜
⎝ −2190.474 ⎠
x := min ( ε ) , 0.005 .. max ( ε )
Y ( x) := F ( x) ⋅ Fit
550
550
500
500
450
450
400
400
350
σ
Fit := linfit ( ε , σ , F)
350
300
σ
250
Y( x ) 250
200
200
150
150
100
100
50
50
0
0.001 0.002 0.003 0.004 0.005
ε
300
0
0.05
0.1
0.15
ε ,x
0.2
0.25
From th stress-strain diagram,
Modulus of Elasticity:
From th stress-strain diagram,
mm
∆ε := ( 0.0005 − 0)
mm *
∆σ := ( 125 − 0)MPa *
E approx :=
∆σ
∆ε *
E approx = 250 GPa*
Ans
σ ult = 530MPa
Ans
σ R = 479MPa
Ans
Problem 3-43
A tension test was performed on a steel specimen having an original diameter of 12.5 mm and a gauge
length of 50 mm. Using the data listed in the table, plot the stress-strain diagram and determine
approximately the modulus of toughness. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm/mm.
d := 12.5*
Given:
L := 50*
Load
(kN)
P :=
3
P⋅ 10
δ
⎛π⎞ 2
A := ⎜ ⋅ d * σ :=
ε := *
*
L
A
⎝ 4⎠
Solution:
ε=δ/L
(mm/mm)
σ=P/A
(MPa)
⎛⎜ 0.0000 ⎞
⎛⎜ 0.000000 ⎞
⎜ 90.4509 ⎟
⎜ 0.000350 ⎟
⎜ 259.9446 ⎟
⎜ 0.001200 ⎟
⎟
⎟
⎜
⎜
⎜ 308.0221 ⎟
⎜ 0.002040 ⎟
⎜ 333.2832 ⎟
⎜ 0.003300 ⎟
⎟
⎟
⎜
⎜
σ = 355.2848 * ε = 0.004980 *
⎟
⎟
⎜
⎜
⎜ 435.1423 ⎟
⎜ 0.020320 ⎟
⎜ 507.6661 ⎟
⎜ 0.060960 ⎟
⎟
⎟
⎜
⎜
⎜ 525.5933 ⎟
⎜ 0.127000 ⎟
⎜ 507.6661 ⎟
⎜ 0.177800 ⎟
⎜ 479.1455
⎜ 0.238760
⎠
⎠
⎝
⎝
Use
⎛ x ⎞
⎜ 0.3
F ( x) := ⎜ x ⎟
⎜ 0.6
⎝x ⎠
δ :=
0.0
11.1
Elongation
(mm)
0.0000
0.0175
31.9
0.0600
37.8
0.1020
40.9
0.1650
43.6
0.2490
53.4
1.0160
62.3
3.0480
64.5
6.3500
62.3
8.8900
58.8
11.9380
Fit := linfit ( ε , σ , F)
⎛ 107.325 ⎞
⎜
Fit = 2123.113
⎜
⎝ −2190.474 ⎠
x := min ( ε ) , 0.005 .. max ( ε )
Y ( x) := F ( x) ⋅ Fit
Modulus of Toughness:
550
The modulus of toughness is equal to the area under the
curve, and could be approximated by counting the
number of sqaures. the total number of squares is:
500
450
400
n := 188.5
350
mm
∆ε sq := ( 0.025 )
mm
σ
300
Y( x ) 250
∆σ sq := 25MPa
200
(
150
)(
)
ut := n⋅ ∆ε sq ⋅ ∆σ sq
ut = 117.81 MPa
100
Ans
50
0 0.05 0.1 0.15 0.2 0.25
ε ,x
Problem 3-44
An 8-mm-diameter brass rod has a modulus of elasticity of E br = 100 GPa. If it is 3 m long and
subjected to an axial load of 2 kN, determine its elongation. What is its elongation under the same load
if its diameter is 6 mm?
Given:
P := 2kN
L := 3m
d1 := 8mm
d2 := 6mm
E br := 100⋅ GPa
Solution:
Case 1:
⎛π⎞ 2
A1 := ⎜ ⋅ d1
4
⎝ ⎠
P
A1
σ 1 :=
ε 1 :=
σ1
Ebr
δ 1 := ε 1⋅ L
Case 2:
σ 1 = 39.789 MPa
ε 1 = 0.0003979
mm
mm
δ 1 = 1.194 mm
Ans
⎛π⎞ 2
A2 := ⎜ ⋅ d2
4
⎝ ⎠
σ 2 :=
ε 2 :=
P
A2
σ2
Ebr
δ 2 := ε 2⋅ L
σ 2 = 70.736 MPa
ε 2 = 0.0007074
mm
mm
δ 2 = 2.122 mm
Ans
Problem 4-1
The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measured
from the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a
wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller
exerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings.
Given:
F := 5kN
L := 8m
do := 400mm
t := 50mm
E := 200⋅ GPa
Solution:
di := do − 2t
⎛ π ⎞ ⋅ ⎛d 2 − d 2⎞
i ⎠
⎝ 4⎠ ⎝ o
A := ⎜
Internal Force:
As shown on FBD.
P := −F
Displacement:
δ A :=
P⋅ L
A⋅ E
δ A = −0.00364 mm
Ans
Note: Negative sign indicates that end A moves towards end D.
Problem 4-2
The A-36 steel column is used to support the symmetric loads from the two floors of a building.
Determine the vertical displacement of its top, A, if P 1 = 200 kN, P2 = 310 kN, and the column has a
cross-sectional area of 14625 mm2.
Given:
2
L := 3.6m
A := 14625mm
P1 := 200kN
P2 := 310kN
E := 200⋅ GPa
Solution:
As shown on FBD.
Internal Force:
Displacement:
δ AB :=
δ BC :=
−2P1⋅ L
A⋅ E
(
)
− 2 P1 + P2 ⋅ L
A⋅ E
δ A := δ AB + δ BC
δ A = −1.74769 mm
Ans
Note: Negative sign indicates that end A moves towards end C.
Problem 4-3
The A-36 steel column is used to support the symmetric loads from the two floors of a building.
Determine the loads P 1 and P 2 if A moves downward 3 mm and B moves downward 2.25 mm when
the loads are applied. The column has a cross-sectional area of 14625 mm2.
Given:
2
L := 3.6m
A := 14625mm
δ A := −3mm
δ B := −2.25 mm
E := 200⋅ GPa
Solution:
Internal Force:
As shown on FBD.
Displacement:
Initial guess:
P1 := 1kN
Given
For AB:
δA − δB =
For BC:
δB =
Solving [1] and [2]:
P2 := 2kN
−2P1⋅ L
A⋅ E
(
)
− 2 P1 + P2 ⋅ L
A⋅ E
[1]
[2]
⎛⎜ P1 ⎞
:= Find ( P1 , P2)
⎜ P2
⎝ ⎠
⎛⎜ P1 ⎞ ⎛ 304.69 ⎞
=⎜
kN
⎜ P2
⎝ ⎠ ⎝ 609.38 ⎠
Ans
Problem 4-4
The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with
respect to end D if the diameters of each segment are dAB = 20 mm, dBC = 25 mm, and dCD = 12 mm.
Take E cu = 126 GPa.
Given:
L AB := 2m
dAB := 20mm
L BC := 3.75m dBC := 25mm
L CD := 2.5m
dCD := 12mm
E := 126⋅ GPa
PA := −40kN
PB := 25kN
PC := 10kN
PD := −30kN
Solution:
Internal Force:
Displacement:
As shown on FBD.
⎛π⎞
2
AAB := ⎜ ⋅ dAB
4
δ AB :=
⎛π⎞
2
ABC := ⎜ ⋅ dBC
4
δ BC :=
⎝ ⎠
⎝ ⎠
⎛π⎞
2
ACD := ⎜ ⋅ dCD
4
⎝ ⎠
(
)
E ⋅ ( AAB)
PA⋅ L AB
(PA + 2PB)⋅ (LBC)
E⋅ ( ABC)
(PA + 2PB + 2PC)⋅ (LCD)
δ CD :=
E⋅ ( ACD)
δ A_D := δ AB + δ BC + δ CD
δ A_D = 3.8483 mm
Ans
Note: The positive sign indicates that end A moves away from end D.
Problem 4-5
The A-36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 60 mm2,
Determine the displacement of B and A. Neglect the size of the couplings at B, C, and D.
Given:
L AB := 0.50m
L BC := 1.50m
L CD := 0.75m
PA := 8kN
PB := 2kN
PC := 3.3kN
hB :=
4
5
vB :=
3
5
θ C := 60deg
2
A := 60mm
E := 200⋅ GPa
Solution:
As shown on FBD.
Internal Force:
Displacement:
δ AB :=
δ BC :=
δ CD :=
(
)
PA⋅ L AB
E⋅ A
⎡⎣PA + 2PB⋅ ( vB)⎤⎦ ⋅ ( LBC)
E⋅ A
⎡⎣PA + 2PB⋅ ( vB) + 2PC⋅ sin ( θ C)⎤⎦ ⋅ ( LCD)
E⋅ A
δ B := δ BC + δ CD
δ B = 2.307 mm
Ans
δ A := δ AB + δ BC + δ CD
δ A = 2.641 mm
Ans
Problem 4-6
The assembly consists of an A-36 steel rod CB and a 6061-T6 aluminum rod BA, each having a
diameter of 25 mm. Determine the applied loads P1 and P2 if A is displaced 2 mm to the right and B is
displaced 0.5 mm to the left when the loads are applied. The unstretched length of each segment is
shown in the figure. Neglect the size of the connections at B and C, and assume that they are rigid.
Given:
L BA := 1.2m
L CB := 0.6m
δ A := 2mm
δ B := −0.5 mm
d := 25mm
E CB := 200⋅ GPa
E BA := 68.9 ⋅ GPa
Solution:
Internal Force:
Displacement:
Initial guess:
As shown on FBD.
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
P1 := 1kN
Given
For BA:
δA − δB =
For CB:
δB =
Solving [1] and [2]:
P2 := 2kN
P1⋅ LBA
A⋅ E BA
(P1 − P2)⋅ LCB
A⋅ ECB
⎛⎜ P1 ⎞
:= Find ( P1 , P2)
⎜ P2
⎝ ⎠
[1]
[2]
⎛⎜ P1 ⎞ ⎛ 70.46 ⎞
=⎜
kN
⎜ P2
152.27
⎝
⎠
⎝ ⎠
Ans
Problem 4-7
The 15-mm-diameter A-36 steel shaft AC is supported by a rigid collar, which is fixed to the shaft at
B. If it is subjected to an axial load of 80 kN at its end, determine the uniform pressure distribution p
on the collar required for equilibrium.Also, what is the elongation on segment BC and segment BA?
Given:
L AB := 200mm
L BC := 500mm
d := 15mm
PC := 80kN
E := 200⋅ GPa
rc := 35mm
Solution:
⎛π⎞
2
2
Acollar := ⎜ ⋅ ⎡⎣ 2rc − d ⎤⎦
4
⎝ ⎠
( )
Equations of equilibrium:
+
(
ΣF y=0;
)
p⋅ Acollar − PC = 0
p :=
Internal Force:
Displacement:
PC
Acollar
p = 21.79 MPa
Ans
As shown on FBD.
⎛ π ⎞ ⋅ d2
⎝ 4⎠
PBC := PC
A := ⎜
δ BC :=
δ BA :=
(
PBC⋅ LBC
)
E⋅ A
(
)
PBA⋅ LAB
E⋅ A
PBA := 0
δ BC = 1.132 mm
Ans
δ BA = 0 mm
Ans
Problem 4-8
The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB
and DC. Determine the vertical displacement of the 2.5-kN load if the members were horizontal when
the load was originally applied. Each wire has a cross-sectional area of 16 mm2.
Given:
L DE := 0.9m L CF := 0.9m
P := 2.5kN
2
L DH := 0.3m L HC := 0.6m
A := 16mm
L AH := 0.54m L BG := 1.5m
E := 193⋅ GPa
L AI := 0.9m
L IB := 0.3m
L DC := LDH + L HC
Solution:
L AB := LAI + LIB
Internal Forces in the wires :
From FBD (b):
+
(
)
(
)
ΣΜA=0;
FBG⋅ LAB − P⋅ L AI = 0
ΣF y=0;
FAH + FBG − P = 0
From FBD (a):
⎛ LAI ⎞
FBG := P⋅ ⎜
⎝ LAB ⎠
FBG = 1.875 kN
FAH := P − FBG
FAH = 0.625 kN
+
(
)
(
)
ΣΜD=0;
FCF⋅ LDC − FAH⋅ L DH = 0
ΣF y=0;
FCF + FDE − FAH = 0
⎛ LDH ⎞
FCF := FAH⋅ ⎜
Displacement:
(
FDE⋅ L DE
δ D :=
δ C :=
E⋅ A
(
)
)
FDE := FAH − FCF
FDE = 0.4167 kN
δ C = 0.06071891 mm
E⋅ A
⎛ LHC ⎞
⎝
FCF = 0.2083 kN
δ D = 0.12143782 mm
FCF⋅ LCF
δ H := δ C + ⎜
⎝ LDC ⎠
LDC
⎠
(
)
⋅ δD − δC
δ H = 0.10119819 mm
⎡FAH⋅ (LAH)⎤
⎥
⎣ E⋅ A ⎦
δ A := δ H + ⎢
δ B :=
(
)
FBG⋅ LBG
E⋅ A
⎛ LAI ⎞
δ I := δ A + ⎜
⎝
LAB
⎠
(
)
⋅ δB − δA
δ I = 0.736 mm
Ans
Problem 4-9
The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB
and DC. Determine the angle of tilt of each member after the 2.5-kN load is applied. The members
were originally horizontal, and each wire has a cross-sectional area of 16 mm2.
L DE := 0.9m L CF := 0.9m
Given: P := 2.5kN
2
L DH := 0.3m L HC := 0.6m
A := 16mm
L AH := 0.54m L BG := 1.5m
E := 193⋅ GPa
L AI := 0.9m
L IB := 0.3m
L DC := LDH + L HC
Solution:
L AB := LAI + LIB
Internal Forces in the wires :
From FBD (b):
+
(
)
(
)
ΣΜA=0;
FBG⋅ LAB − P⋅ L AI = 0
ΣF y=0;
FAH + FBG − P = 0
From FBD (a):
⎛ LAI ⎞
FBG := P⋅ ⎜
⎝ LAB ⎠
FBG = 1.875 kN
FAH := P − FBG
FAH = 0.625 kN
+
(
)
(
)
ΣΜD=0;
FCF⋅ LDC − FAH⋅ L DH = 0
ΣF y=0;
FCF + FDE − FAH = 0
⎛ LDH ⎞
FCF := FAH⋅ ⎜
Displacement:
FDE⋅ L DE
δ D :=
E⋅ A
(
δ C :=
(
)
⎝ LDC ⎠
(
δD − δC
LDC
⎡FAH⋅ (LAH)⎤
⎥
⎣ E⋅ A ⎦
(
)
FBG⋅ LBG
(
)
δ H = 0.10119819 mm
⎛ δD − δC⎞
α DC := atan ⎜
⎝
L DC
⎠
α DC = 0.0039 deg
Ans
δ A = 0.21049223 mm
δ B = 0.91078368 mm
E⋅ A
tan β AB :=
)
⋅ δD − δC
δ A := δ H + ⎢
δ B :=
FDE = 0.4167 kN
δ C = 0.06071891 mm
⎛ LHC ⎞
δ H := δ C + ⎜
)
FDE := FAH − FCF
δ D = 0.12143782 mm
E⋅ A
(
FCF = 0.2083 kN
)
FCF⋅ LCF
tan α DC :=
⎝ LDC ⎠
δB − δA
L AB
⎛ δB − δA⎞
β AB := atan ⎜
⎝
LAB
⎠
β AB = 0.0334 deg
Ans
Problem 4-10
The bar has a cross-sectional area of 1800 mm2, and E = 250 GPa. Determine the displacement of its
end A when it is subjected to the distributed loading.
Given:
L := 1.5m
E := 250⋅ GPa
1
2
A := 1800mm
w = 500⋅ x
3 N
m
Solution:
Internal Force:
As shown on FBD.
x
x
⌠
Px = ⎮ w⋅ x dx
⌡0
⌠
⎛ 1⎞ ⎤
⎮ ⎡⎢
⎜ 3 ⎥
Px = ⎮ ⎣500⋅ ⎝ x ⎠ ⋅ x⎦ dx
⌡0
⎛ 4⎞
1500 ⎜ 3
Px =
⋅ ⎝x ⎠
4
Displacement:
unit := 1N⋅ m
⌠
⎮
δA = ⎮
⌡
L
Px
E⋅ A
dx
L = 1.50 m
0
⎡⌠48
⎤
4⎞ ⎥
⎢⎮ ⎛⎜
unit ⎢⎮
1500 3
δ A :=
⋅x
dx⎥
⎜
A⋅ E ⎢⎮ ⎝ 4
⎠ ⎥
⌡
⎣0
⎦
δ A = 2.990 mm
Ans
Problem 4-11
The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area
of each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the horizontal
displacement of point F.
Given:
L CD := 1.2m
L EC := 0.6m
L AB := 1.8m
L AE := 0.3m
P := 30kN
E := 120⋅ GPa
2
L EF := 0.3m
2
ACD := 600mm
AAB := 900mm
2
AEF := 1200mm
Solution:
L AC := LAE + LEC
Internal Forces in the rods :
From FBD :
+
(
)
(
⎛ LAE ⎞
)
ΣΜA=0;
FCD⋅ LAC − P⋅ L AE = 0
ΣF y=0;
FCD + FAB − P = 0
Displacement:
δ C :=
δ A :=
(
)
FCD⋅ LCD
E⋅ ACD
(
)
⎛ LEC ⎞
⎝
(
FCD = 10.00 kN
FAB := P − FCD
FAB = 20.00 kN
δ A = 0.3333 mm
δ E := δ C + ⎜
δ F_E :=
⎝ LAC ⎠
δ C = 0.1667 mm
FAB⋅ L AB
E ⋅ AAB
FCD := P⋅ ⎜
L AC
)
P⋅ L EF
E ⋅ AEF
δ F := δ E + δ F_E
⎠
(
)
⋅ δA − δC
δ E = 0.2778 mm
δ F_E = 0.0625 mm
δ F = 0.340278 mm
Ans
Problem 4-12
The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC.The cross-sectional area
of each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the angle of tilt
of bar AC.
Given:
L CD := 1.2m
L EC := 0.6m
L AB := 1.8m
L AE := 0.3m
P := 30kN
E := 120⋅ GPa
2
L EF := 0.3m
2
ACD := 600mm
AAB := 900mm
2
AEF := 1200mm
Solution:
L AC := LAE + LEC
Internal Forces in the rods :
From FBD :
+
(
(
)
ΣΜA=0;
FCD⋅ LAC − P⋅ L AE = 0
ΣF y=0;
FCD + FAB − P = 0
Displacement:
δ C :=
δ A :=
(
)
⎛ LAE ⎞
)
tan α AC :=
(
E⋅ ACD
(
⎝ LAC ⎠
FCD = 10.00 kN
FAB := P − FCD
FAB = 20.00 kN
δ C = 0.1667 mm
)
FAB⋅ L AB
δA − δC
LAC
)
FCD⋅ LCD
FCD := P⋅ ⎜
δ A = 0.3333 mm
E ⋅ AAB
⎛ δA − δC⎞
α AC := atan ⎜
⎝
L AC
⎠
α AC = 0.01061 deg
Ans
Problem 4-13
A spring-supported pipe hanger consists of two springs which are originally unstretched and have a
stiffness of k = 60 kN/m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm,
and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries have
a total weight of 4 kN, determine the displacement of the pipe when it is attached to the support.
Given:
Solution:
L AB := 0.75m
L CD := 0.75m
L EF := 0.75m
dAB := 5mm
dCD := 5mm
dEF := 12mm
L GE := 0.25m
L EH := 0.25m
k := 60
P := 4kN
E := 193⋅ GPa
kN
m
L GH := L GE + L EH
Internal Forces in the rods :
From FBD (a) :
+
(
)
(
)
ΣΜA=0;
FCD⋅ LGH − P⋅ LGE = 0
ΣF y=0;
FCD + FAB − P = 0
⎛ LGE ⎞
FCD := P⋅ ⎜
FCD = 2.00 kN
FAB := P − FCD
FAB = 2.00 kN
⎝ LGH ⎠
From FBD (b) :
+
ΣF y=0;
(
)
FEF ⋅ − FCD + FAB = 0
FEF := FCD + FAB
Displacement:
FEF = 4.00 kN
⎛π⎞
2
⎛π⎞
2
AAB := ⎜ ⋅ dAB ACD := ⎜ ⋅ dCD
4
4
⎝ ⎠
δ E :=
(
⎝ ⎠
)
FEF⋅ L EF
(
)
⎛ FCD ⎞
⎝ k ⎠
δ D := δ E + ⎜
δ C_D :=
)
FCD⋅ L CD
(
⎝ ⎠
δ E = 0.13744 mm
E⋅ AEF
(
⎛π⎞
2
AEF := ⎜ ⋅ dEF
4
)
E⋅ ACD
δ total := δ D + δ C_D
δ D = 33.47077 mm
Due to symmetry,
δ B := δ D
δ C_D = 0.39583 mm
Due to symmetry,
δ total = 33.8666 mm
δ B_A := δ C_D
Ans
Problem 4-14
A spring-supported pipe hanger consists of two springs, which are originally unstretched and have a
stiffness of k = 60 kN/m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm,
and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe is displaced 82 mm when it
is filled with fluid, determine the weight of the fluid.
Given:
L AB := 0.75m L CD := 0.75m
L EF := 0.75m
Solution:
dAB := 5mm
dCD := 5mm
dEF := 12mm
L GE := 0.25m
L EH := 0.25m
δ total := 82mm
E := 193⋅ GPa
k := 60
kN
m
L GH := L GE + L EH
Iniatlly set: P := 1kN
Internal Forces in the rods :
From FBD (a) :
+
(
)
(
)
ΣΜA=0;
FCD⋅ LGH − P⋅ LGE = 0
ΣF y=0;
FCD + FAB − P = 0
⎛ LGE ⎞
FCD := P⋅ ⎜
FCD = 0.50 kN
FAB := P − FCD
FAB = 0.50 kN
⎝ LGH ⎠
From FBD (b) :
+
ΣF y=0;
(
)
FEF ⋅ − FCD + FAB = 0
FEF := FCD + FAB
Displacement:
FEF = 1.00 kN
⎛π⎞
2
⎛π⎞
2
AAB := ⎜ ⋅ dAB ACD := ⎜ ⋅ dCD
4
4
⎝ ⎠
δ E :=
(
⎝ ⎠
)
FEF⋅ L EF
(
)
⎛ FCD ⎞
⎝ k ⎠
δ D := δ E + ⎜
δ C_D :=
)
FCD⋅ L CD
(
)
E⋅ ACD
δ B := δ D
δ C_D = 0.09896 mm
⎛ δ total ⎞
W := P⋅ ⎜
δ D = 8.36769 mm
Due to symmetry,
δ' total := δ D + δ C_D
δ total
W
=
P
δ' total
⎝ ⎠
δ E = 0.03436 mm
E⋅ AEF
(
⎛π⎞
2
AEF := ⎜ ⋅ dEF
4
⎝ δ'total ⎠
Due to symmetry,
δ B_A := δ C_D
δ' total = 8.4666 mm
W = 9.69 kN
Ans
Problem 4-15
The assembly consists of three titanium rods and a rigid bar AC.The cross-sectional area of each rod is
given in the figure. If a vertical force P = 20 kN is applied to the ring F, determine the vertical
displacement of point F. Eti = 350 GPa.
Given:
L BA := 2m
L DC := 2m
L AE := 0.5m
L EC := 0.75m
2
Solution:
L EF := 1.5m
2
ABA := 60mm
ADC := 45mm
P := 20kN
E := 350⋅ GPa
2
AEF := 75mm
L AC := LAE + LEC
Internal Forces in the rods :
From FBD :
+
(
)
(
⎛ LAE ⎞
)
ΣΜA=0;
FDC⋅ LAC − P⋅ L AE = 0
ΣF y=0;
FDC + FBA − P = 0
Displacement:
δ C :=
δ A :=
(
)
FDC⋅ LDC
E⋅ ADC
(
)
⎛ LEC ⎞
δ F_E :=
(
⎝
)
P⋅ L EF
E ⋅ AEF
δ F := δ E + δ F_E
FDC = 8.00 kN
FBA := P − FDC
FBA = 12.00 kN
δ A = 1.1429 mm
E ⋅ ABA
L AC
⎝ LAC ⎠
δ C = 1.0159 mm
FBA⋅ L BA
δ E := δ C + ⎜
FDC := P⋅ ⎜
⎠
(
)
⋅ δA − δC
δ E = 1.0921 mm
δ F_E = 1.1429 mm
δ F = 2.2349 mm
Ans
Problem 4-16
The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of
500 mm2. If a vertical force of P = 250 kN is applied to the end B of member AB, determine the
vertical displacement of point B.
Given:
a := 1.5m
b := 2m
2
2
L AD = 2.5 m
2
2
L AC = 2.5 m
L AD :=
a +b
L AC :=
a +b
2
E := 200⋅ GPa
Solution:
A := 500mm
2
c :=
L AB := 3m
2
a +b
h :=
⎛a⎞
⎝ b⎠
θ := atan ⎜
P := 250kN
a
c
v :=
b
c
θ = 36.869898 deg
Equation of equilibrium :
For AB : +
(FAD + FAC)⋅ v − P = 0
ΣF y=0;
Since FAD = FAC
FAC :=
P
2v
Displacement:
FAC⋅ L AC
(
)
E⋅ A
*
δ A_C :=
δ B_A :=
Consider triangle AA'C:
(
FAC = 156.25 kN
δ A_C = 3.9063 mm
)
P⋅ LAB
δ B_A = 7.5 mm
E⋅ A
L A'C := L AC + δ A_C
L AA' = δ A
θ' A := 180deg − θ
( ) = sin(θ'A)
sin φ A'
LAC
LA'C
( )
⎛
sin θ' A ⎞
⎝
LA'C
φ A' := asin ⎜ L AC⋅
⎠
θ' C := 180deg − θ' A − φ A'
( ) = sin(θ'A)
sin θ' C
δA
L A'C
δ A := LAC⋅
δ B := δ A + δ B_A
φ A' = 36.80289 deg
θ' C = 0.0670094 deg
( )
sin ( θ' A)
δ A = 4.8731 mm
δ B = 12.37 mm
Ans
sin θ' C
Problem 4-17
The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of
500 mm2. Determine the magnitude of the force P needed to displace point B 2.5 mm downward.
a := 1.5m
Given:
b := 2m
2
2
L AD = 2.5 m
2
2
L AC = 2.5 m
L AD :=
a +b
L AC :=
a +b
2
E := 200⋅ GPa
A := 500mm
2
c :=
Solution:
L AB := 3m
2
a +b
a
c
h :=
⎛a⎞
⎝ b⎠
θ := atan ⎜
δ B := 2.5mm
b
c
v :=
θ = 36.869898 deg
Equation of equilibrium :
For AB : +
(FAD + FAC)⋅ v − P = 0
ΣF y=0;
P
2v
Since FAD = FAC
FAC =
Displacement:
FAC⋅ LAC
δ A_C =
FAD =
(
)
(
)
P⋅ L AC
2v( E ⋅ A)
)
P⋅ L AB
δ B_A =
E⋅ A
δ B = δ A + δ B_A
Consider triangle AA'C:
δ A_C =
E⋅ A
(
P
2v
δA = δB −
(
)
P⋅ LAB
E⋅ A
L A'C = LAC + δ A_C
L AA' = δ A
θ' A := 180deg − θ
2
2
2
θ' A = 143.130102 deg
( )(
)
( )
L A'C = δ A + LAC − 2 δ A ⋅ LAC ⋅ cos θ' A
Initial guess:
P := 1kN
Given
P⋅ ( LAC)⎤
P⋅ ( L AB)⎤
P⋅ ( LAB)⎤
⎡
⎡
⎡
⎢LAC +
⎥ = ⎢δ B −
⎥ + LAC2 − 2⎢δ B −
⎥ ⋅ ( L ) ⋅ cos ( θ'A)
2v( E⋅ A) ⎦
E⋅ A ⎦
E ⋅ A ⎦ AC
⎣
⎣
⎣
Solving :
2
2
P := Find ( P)
P = 50.47 kN
Ans
Problem 4-18
Consider the general problem of a bar made from m segments, each having a constant cross-sectional
area Am and length Lm. If there are n loads on the bar as shown, write a computer program that can be
used to determine the displacement of the bar at any specified location x. Show an application of the
program using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m,
P 2 = -1.5 kN, A 2 = 625 mm2.
Problem 4-19
The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm2 and
is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed
load is applied.
Given:
L := 4m
w := 300
a := 1.5m
b := 2m
2
A := 14mm
Solution:
L BC :=
N
m
E := 68.9 ⋅ GPa
2
2
a +b
v :=
a
h :=
L BC
b
L BC
Support Reactions:
ΣΜA=0;
2
w⋅ L
FBC :=
2⋅ v⋅ b
FBC⋅ ( v) ⋅ ( b) − w⋅ ( L ) ⋅ ( 0.5 ⋅ L) = 0
FBC = 2 kN
Displacement:
δ B_C :=
Consider triangle AB'C:
(
FBC⋅ L BC
)
E⋅ A
L B'C := LBC + δ B_C
2
2
δ B_C = 5.183 mm
L B'C = 2.505183 m
( )
2
L B'C = a + b − 2( a) ⋅ ( b) ⋅ cos θ' A
⎛ a2 + b2 − L 2 ⎞
B'C
θ' A := acos ⎜
2⋅ a⋅ b
⎝
⎠
φ := θ' A − 90deg
θ' A = 90.24775 deg
φ = 0.24775 deg
φ = 0.0043241 rad
δ D := φ ⋅ L
δ D = 17.30 mm
Ans
Problem 4-20
The rigid beam is supported at its ends by two A-36 steel tie rods. If the allowable stress for the steel is
σallow = 115 MPa, the load w = 50 kN/m, and x = 1.2 m, determine the diameter of each rod so that the
beam remains in the horizontal position when it is loaded.
Given:
L CD := 1.8m
L AC := 2.4m
L AB := 1.8m
x := 1.2m
w := 50
kN
m
σ allow := 115⋅ MPa
Solution:
Internal Forces in the rods :
From FBD :
+
2
ΣΜA=0;
FCD⋅ LAC − ( w⋅ x) ⋅ ( 0.5x) = 0
w⋅ x
FCD :=
2⋅ L AC
ΣF y=0;
FCD + FAB − w⋅ x = 0
FAB := w⋅ x − FCD
Displacement:
(
)
FAB = 45.00 kN
To maintain the rigid beam in the horizontal position, the elongation of both
rods AB and CD must be the same.
δA = δC
⎛π⎞
2
AAB = ⎜ ⋅ dAB
4
⎝ ⎠
δA =
Thus,
FCD = 15.00 kN
(
)
FAB⋅ LAB
E⋅ AAB
(
⎛π⎞
2
ACD = ⎜ ⋅ dCD
4
⎝ ⎠
δC =
(
E ⋅ ACD
) = FCD⋅ (LCD)
FAB⋅ L AB
⎡⎛ π ⎞
2⎤
E ⋅ ⎢⎜ ⋅ dAB ⎥
4
⎣⎝ ⎠
⎦
)
FCD⋅ L CD
⎡⎛ π ⎞
2⎤
E⋅ ⎢⎜ ⋅ dCD ⎥
4
⎣⎝ ⎠
⎦
dCD
dAB
=
dAB =
( )
FAB⋅ ( L AB)
FCD⋅ L CD
3⋅ dCD
Allowable Normal Stress:
Assume failure of rods AB.
σ allow =
FAB
⎛ π ⎞⋅ d 2
⎜
⎝ 4 ⎠ AB
dAB :=
4⋅ FAB
( π ) ⋅ σ allow
dCD :=
dAB
3
dAB = 22.321 mm
dCD = 12.887 mm
Assume failure of rods CD.
σ allow =
FCD
⎛ π ⎞⋅ d 2
⎜
⎝ 4 ⎠ CD
dCD :=
4⋅ FCD
( π ) ⋅ σ allow
dAB :=
3⋅ dCD
dCD = 12.887 mm
Ans
dAB = 22.321 mm
Ans
Problem 4-21
The rigid beam is supported at its ends by two A-36 steel tie rods. The rods have diameters dAB = 12
mm and dCD = 7.5 mm. If the allowable stress for the steel is σallow = 115 MPa, determine the intensity
of the distributed load w and its length x on the beam so that the beam remains in the horizontal position
when it is loaded.
Given:
L CD := 1.8m
L AC := 2.4m
L AB := 1.8m
dAB := 12mm dCD := 7.5mm
σ allow := 115⋅ MPa
Solution:
⎛π⎞
2
AAB = ⎜ ⋅ dAB
4
⎛π⎞
2
ACD = ⎜ ⋅ dCD
4
⎝ ⎠
⎝ ⎠
Allowable Normal Stress:
Assume failure of rods AB.
σ allow =
Displacement:
FAB
⎡⎛ π ⎞
2⎤
FAB := σ allow ⋅ ⎢⎜ ⋅ dAB ⎥
4
(
AAB
FAB = 13.006 kN
⎦
To maintain the rigid beam in the horizontal position, the elongation of both
rods AB and CD must be the same.
δA = δC
δA =
Thus,
) ⎣⎝ ⎠
(
)
FAB⋅ LAB
δC =
E⋅ AAB
(
(
)
FCD⋅ L CD
E ⋅ ACD
) = FCD⋅ (LCD)
FAB⋅ L AB
⎡⎛ π ⎞ ⋅ d 2⎤
⎥
⎣⎝ 4 ⎠ AB ⎦
E ⋅ ⎢⎜
FCD
FAB
+
⎛ d 2⎞
CD
FCD := ⎜
⋅F
⎜ d 2 AB
⎝ AB ⎠
2
=
dCD
2
dAB
Internal Forces in the rods :
From FBD :
⎡⎛ π ⎞ ⋅ d 2⎤
⎥
⎣⎝ 4 ⎠ CD ⎦
E⋅ ⎢⎜
FCD = 5.081 kN
Given
(
)
ΣΜA=0;
FCD⋅ LAC − ( w⋅ x) ⋅ ( 0.5x) = 0
ΣF y=0;
FCD + FAB − w⋅ x = 0
Initial guess:
Solving :
w := 1
⎛w⎞
:= Find ( w , x)
⎜
⎝x⎠
kN
m
x := 1m
w = 13.41
x = 1.35 m
kN
m
Ans
Allowable Normal Stress:
Assume failure of rods CD.
σ allow =
Displacement:
FCD
⎡⎛ π ⎞
2⎤
FCD := σ allow ⋅ ⎢⎜ ⋅ dCD ⎥
4
(
ACD
⎦
FCD = 5.081 kN
To maintain the rigid beam in the horizontal position, the elongation of both
rods AB and CD must be the same.
δA = δC
δA =
Thus,
) ⎣⎝ ⎠
(
)
FAB⋅ LAB
δC =
E⋅ AAB
(
(
)
FCD⋅ L CD
E ⋅ ACD
) = FCD⋅ (LCD)
FAB⋅ L AB
⎡⎛ π ⎞ ⋅ d 2⎤
⎥
⎣⎝ 4 ⎠ AB ⎦
E ⋅ ⎢⎜
FCD
FAB
2
=
dCD
2
dAB
⎡⎛ π ⎞ ⋅ d 2⎤
⎥
⎣⎝ 4 ⎠ CD ⎦
E⋅ ⎢⎜
⎛ d 2⎞
AB
FAB := ⎜
⋅F
⎜ d 2 CD
⎝ CD ⎠
FAB = 13.006 kN
The results are the same as assuming failure of rod AB.
Therefore, rods AB and CD fail simultaneously.
Ans
Problem 4-22
The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN
and the soil provides a frictional resistance that is uniformly distributed along its sides of w = 4 kN/m,
determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of
the post A with respect to its bottom B? Neglect the weight of the post.
Given:
L := 2m
d := 60mm
P := 20kN
w := 4
kN
m
E := 13.1 ⋅ GPa
Solution:
For entire post [FBD (a)]
Equation od Equilibrium :
+
ΣF y=0;
−P + w⋅ L + F = 0
F := P − w⋅ L
Internal Force:
+
ΣF y=0;
F = 12 kN
FBD (b).
−P + w⋅ y − Fy = 0
Fy = −P + w⋅ y
Displacement:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
unit := 1kN
⌠
⎮
δ A_B = ⎮
⌡
L
Fy
A⋅ E
dy
0
⎡ L
⎤
⎣0
⎦
unit ⎢⌠
⎮ ( −P + w⋅ y) ⋅ 1 dy⎥
δ A_B :=
A⋅ E ⎢⎮
kN ⎥
⌡
δ A_B = −0.864 mm
Ans
Note: Negative sign indicates that ens A moves towards end B.
Problem 4-23
The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN
and the soil provides a frictional resistance that is distributed along its length and varies linearly from w
= 0 at y = 0 to w = 3 kN/m at y = 2 m, determine the force F at its bottom needed for equilibrium.
Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the
weight of the post.
Given:
L := 2m
d := 60mm
P := 20kN
kN
wo := 3
m
E := 13.1 ⋅ GPa
Solution:
For entire post [FBD (a)]
Equation od Equilibrium :
+
ΣF y=0;
−P + 0.5wo⋅ L + F = 0
F := P − 0.5wo⋅ L
Internal Force:
+
ΣF y=0;
FBD (b).
⎛
⎝
−P + 0.5 ⋅ ⎜ wo⋅
F = 17 kN
y⎞
⋅ y − Fy = 0
L⎠
⎛ wo ⎞ 2
⋅y
⎝ 2⋅ L ⎠
Fy = − P + ⎜
Displacement:
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
unit := 1kN
⌠
⎮
δ A_B = ⎮
⌡
L
Fy
A⋅ E
dy
0
⎡⎢⌠L
⎤
⎛ wo ⎞ 2⎤ 1 ⎥
unit ⎮ ⎡
dy⎥
δ A_B :=
⋅ y ⎥⋅
⎢ ⎢−P + ⎜
A⋅ E ⎮ ⎣
⎝ 2⋅ L ⎠ ⎦ kN ⎥
⎢⌡0
⎣
⎦
δ A_B = −1.026 mm
Ans
Note: Negative sign indicates that ens A moves towards end B.
Problem 4-24
The rod has a slight taper and length L. It is suspended from the ceiling and supports a load P at its
end. Show that the displacement of its end due to this load is δ = PL/(πEr2r1). Neglect the weight of
the material. The modulus of elasticity is E.
Solution:
L + xo
Geometry :
Thus,
xo
r1
=
r2
rx = r1 +
xo =
r2 − r1
L
⋅x
r2 − r1
rx =
Ax = π ⋅ ⎛⎝ rx ⎞⎠
2
Displacement:
L ⋅ r1
Ax =
π
L
2
(
)
r1⋅ L + r2 − r1 ⋅ x
L
(
)
⋅ ⎡⎣r1⋅ L + r2 − r1 ⋅ x⎤⎦
L
⌠
P
dx
δ= ⎮
⎮ Ax⋅ E
⌡0
2 ⎡⌠ L
⎤
1
⎮
⎢
⎥
δ=
⋅
dx
π ⋅ E ⎢⎮ r ⋅ L + r − r ⋅ x 2 ⎥
⎡1
( 2 1) ⎤⎦ ⎥
⎢⎮
⌡ ⎣
⎣0
⎦
P⋅ L
δ=
P⋅ L
2
π⋅E
⋅
(
L
1
r2 − r1 ⋅ ⎡r1⋅ L + r2 − r1 ⋅ x⎤ 0
)⎣
(
) ⎦
2
δ=
1
1 ⎤
⎡
−
⎥
π ⋅ E ⋅ ( r2 − r1) ⎣ r1⋅ L + ( r2 − r1) ⋅ L r1⋅ L⎦
δ=
⎛ 1 − 1 ⎞
π ⋅ E ⋅ ( r2 − r1) ⎝ r2⋅ L r1⋅ L ⎠
δ=
δ=
− P⋅ L
⋅⎢
− P⋅ L
2
− P⋅ L
2
⋅⎜
(
π ⋅ E ⋅ r2 − r1
− P⋅ L
π ⋅ E ⋅ r1⋅ r2
⎛ r2 − r1 ⎞
⋅⎜
) ⎝ r1⋅ r2⋅ L ⎠
Ans
2
Problem 4-25
Solve Prob. 4-24 by including both P and the weight of the material, considering its specific weight to
be γ (weight per volume).
Solution:
Internal Force:
+
ΣF y=0;
FBD (b).
Px − wx − P = 0
L + xo
Geometry :
r2
rx = r1 +
Thus,
=
Px = Wx + P
xo
r1
xo =
r2 − r1
L
⋅x
L ⋅ r1
r2 − r1
rx =
Ax = π ⋅ ⎛⎝ rx ⎞⎠
2
π
Ax =
(
)
r1⋅ L + r2 − r1 ⋅ x
L
(
L
γ
⎛ ⎞
⎛ γ ⋅ π ⎞ ⋅ ⎛r 2⎞ ⋅ x
Wx = ⎜ ⋅ Ax⋅ x0 + x − ⎜
⎝ 3⎠
⎝ 3 ⎠⎝1⎠ 0
(
)
⋅ ⎡r ⋅ L + r2 − r1 ⋅ x⎤⎦
2 ⎣1
)
2
( )
⎛ γ ⋅ π ⎞ ⋅ r ⋅ L + r − r ⋅ x 2⋅ ⎛ L⋅ r1 + x⎞ − ⎛ γ ⋅ π ⎞ ⋅ ⎛ r 2⎞ ⋅ ⎛ L⋅ r1 ⎞
⎜
⎡
( 2 1) ⎤⎦ ⎜ r − r
⎝ 1 ⎠ ⎜r − r
2 ⎣1
⎝ 2 1 ⎠ ⎝ 3 ⎠
⎝ 2 1⎠
⎝ 3⋅ L ⎠
γ⋅π
3
3 3
Wx =
⋅ ⎡⎣⎣⎡r1⋅ L + ( r2 − r1) ⋅ x⎦⎤ − ⎛⎝ r1 ⋅ L ⎞⎠⎤⎦
2
Wx = ⎜
(
)
3⋅ L ⋅ r 2 − r 1
Displacement:
⎡⌠L
⎤
3 ⎛ 3 3⎞ ⎥
⎢
r
⋅
L
r
−
r
+
⋅
x
−
r
⋅
L
⎮ ⎡⎣ 1
⌠
( 2 1) ⎤⎦ ⎝ 1 ⎠ x⎥
γ
⎮ Wx
δ1 =
⋅ ⎢⎮
d
δ1 = ⎮
dx
2
3⋅ E⋅ ( r2 − r1) ⎢⎮
⎥
r1⋅ L + ( r2 − r1) ⋅ x⎤⎦
⎡
⎣
⎮ Ax⋅ E
⎢⌡
⎥
⌡0
⎣0
⎦
L
⎡⌠L
⎤
⌠
γ
1
3 3 ⎮
⎢
⎥
δ1 =
⋅ ⎮ ⎣⎡r1⋅ L + ( r2 − r1) ⋅ x⎦⎤ dx − ⎛⎝ r1 ⋅ L ⎞⎠ ⋅
dx
⎮
2 ⎥
3⋅ E⋅ ( r2 − r1) ⎢⌡
⎮ ⎡⎣r1⋅ L + ( r2 − r1) ⋅ x⎤⎦
⎢0
⎥
⌡
0
⎣
⎦
L
Using the result of Prob.(4-24) for the 2nd Integral, we have
δ1 =
⎡⎡ 2
γ
L ⎤
1 ⎞⎤
3 3 ⎛
⋅ ⎢⎢r1⋅ L + r2 − r1 ⋅ ⎥ − ⎛⎝ r1 ⋅ L ⎞⎠ ⋅ ⎜
⎥
3⋅ E⋅ r2 − r1 ⎣
L ⋅ r 1⋅ r 2
2⎦
(
)⎣
(
)
2
⎝
⎠⎦
2 2
γ ⋅ L ⋅ r1
(
)
δ1 =
⋅−
6⋅ E ⋅ ( r 2 − r 1)
3⋅ E⋅ ( r2 − r1) ⋅ r2
2
γ ⋅ L ⋅ r2 + r1
L
⌠
P
dx
δ = δ1 + ⎮
⎮ Ax⋅ E
⌡0
Using the result of Prob.(4-24) for the 2nd Integral, we have
2 2
γ ⋅ L ⋅ r1
(
)
P⋅ L
δ=
⋅−
+
6⋅ E⋅ ( r2 − r1)
3⋅ E ⋅ ( r 2 − r 1) ⋅ r 2 π ⋅ E ⋅ r 1⋅ r 2
2
γ ⋅ L ⋅ r2 + r1
Ans
Problem 4-26
The support is made by cutting off the two opposite sides of a sphere that has a radius r0. If the
original height of the support is r0 /2, determine how far it shortens when it supports a load P .The
modulus of elasticity is E.
Solution:
Geometry :
r = ro⋅ cos ( θ )
Ay = π ⋅ r
y = ro⋅ sin ( θ )
dy = ro⋅ cos ( θ ) ⋅ dθ
2
Ay = π ⋅ ro ⋅ cos ( θ )
2
2
h = 0.25 ⋅ ro
( )
0.25 ⋅ ro = ro⋅ sin ( θ o)
h = ro⋅ sin θ o
Displacement:
θ
⎛⎜ ⌠h
⌠ o 2P⋅ r ⋅ cos ( θ )
⎞
P
o
⎮
dy⎟ δ = ⎮
δ = 2⋅ ⎜ ⎮
dθ
2
2
⎮ Ay⋅ E
(
)
⋅
⋅
cos
θ
⋅
E
π
r
⎮
⎜ ⌡0
o
⌡0
⎝
⎠
θ o := asin ( 0.25 )
θ o = 14.48 deg
⎛⌠14.48 o
⎞
⎜
2P
1
⎛
⎞ ⋅ ⎜⎮
dθ ⎟
δ= ⎜
⎮
(
)
π
⋅
r
⋅
E
cos
θ
o
⎜
⎝
⎠ ⌡0
⎝
⎠
⎛ 2⋅ P ⎞ ⋅ ( ln ( sec ( θ ) + tan ( θ ) ) ) 14.48
δ= ⎜
0
⎝ π ⋅ r o⋅ E ⎠
δ=
0.511 ⋅ P
o
Ans
π ⋅ ro⋅ E
Alternatively,
Geometry :
2
Ay = π ⋅ x
Ay = π ⋅ ⎛⎝ ro − y ⎞⎠
2
2
Displacement:
⎛⎜ ⌠h
⎞
P
⎮
dy
δ = 2⋅ ⎜
⎮ Ay⋅ E ⎟
⎜ ⌡0
⎝
⎠
⌠
δ= ⎮
⎮
⎮
⌡
0.25⋅ ro
0
2⋅ P
2
2
π ⋅ ⎛⎝ ro − y ⎞⎠ ⋅ E
dθ
0.25 ro
⎛ 2⋅ P ⎞ ⋅ ⎛ 1 ⎞ ⋅ ln ⎛ ro + y ⎞
⎜
⎜
⎝ π ⋅ E ⎠ ⎝ 2⋅ ro ⎠ ⎝ ro − y ⎠ 0
δ= ⎜
δ=
0.511 ⋅ P
π ⋅ ro⋅ E
Ans
Problem 4-27
The ball is truncated at its ends and is used to support the bearing load P. If the modulus of elasticity
for the material is E, determine the decrease in its height when the load is applied.
Solution:
Geometry :
2
2
x = r −y
2
(2
2
Ay = π ⋅ x
Ay = π ⋅ r − y
2
)
Displacement:
⎛⎜ ⌠h
⎞
P
dy⎟
δ = 2⋅ ⎜ ⎮
⎮ A ⋅E
⎜ ⌡0 y
⎝
⎠
0.866 ⋅ r
⌠
δ= ⎮
⎮
⌡0
2
⎛ r ⎞ = r 2 − h2
⎜
⎝ 2⎠
2⋅ P
(2
2
)
dθ
π⋅ r − y ⋅E
⎛ 2⋅ P ⎞ ⋅ ⎛ 1 ⎞ ⋅ ln ⎛ r + y ⎞
⎜
⎜
⎝ π ⋅ E ⎠ ⎝ 2⋅ r ⎠ ⎝ r − y ⎠
δ= ⎜
δ=
2.63 ⋅ P
π ⋅ ro⋅ E
h=
Ans
0.866 r
0
3
⋅r
2
h = 0.866 ⋅ r
Problem 4-28
Determine the elongation of the aluminum strap when it is subjected to an axial force of 30 kN. Eal =
70 GPa.
Given:
L 1 := 250mm
L 2 := 800mm
d1 := 15mm
d2 := 50mm
t := 6mm
P := 30kN
E := 70⋅ GPa
Solution:
Displacement:
δ := 2⋅
P⋅ L 1
(
E⋅ t⋅ d2 − d1
δ = 2.371 mm
)
⎛ d2 ⎞
⋅ ln ⎜
⎝ d1 ⎠
+
P⋅ L 2
( )
E⋅ t⋅ d2
Ans
Problem 4-29
The casting is made of a material that has a specific weight γ and modulus of elasticity E. If it is
formed into a pyramid having the dimensions shown, determine how far its end is displaced due to
gravity when it is suspended in the vertical position.
Solution:
FBD (b).
1
Pz − γ ⋅ A⋅ z = 0
3
Internal Force:
+
ΣF y=0;
Pz =
1
γ ⋅ A⋅ z
3
Displacement:
⎛⎜ ⌠L
⎞
Pz
⎮
dy⎟
δ = ⎜⎮
⎜ ⌡0 A⋅ E
⎝
⎠
⌠
δ= ⎮
⎮
⌡
L
γ ⋅ A⋅ z
dy
3A⋅ E
0
⎞
γ ⎛⎜ ⌠
⎮ z dy
3⋅ E ⎜ ⌡0
L
δ=
⎝
⎠
2
γ⋅L
δ=
6⋅ E
Ans
Problem 4-30
The pedestal is made in a shape that has a radius defined by the function r = 2/(2 + y1/2) m, where y is
in meter. If the modulus of elasticity for the material is E = 100 MPa, determine the displacement of its
top when it supports the 5-kN load.
Given:
H := 4m
do := 1m
P := 5kN
E := 100⋅ 10 Pa
( 6)
d1 := 0.5m
2
r=
2+y
0.5
Solution:
Displacement:
4⋅ π
2
Ay =
unit := 1m
H := 4
Ay = π ⋅ r
(2 + y0.5)2
H
⌠
P
dy
δ= ⎮
⎮ Ay⋅ E
⌡0
⎤
⎡⌠H
2 ⎥
⎢⎮ (
0.5)
⎛ P ⎞ ⎛ 1 ⎞ ⎢⎮ 2 + y
δ := ⎜ ⋅ ⎜
dy⎥
4⋅ π
⎝ E ⎠ ⎝ unit ⎠ ⎢⎮
⎥
⌡
⎣0
⎦
δ = 0.1804 mm
Ans
Problem 4-31
The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is
subjected to an axial force of 150 kN, determine the average normal stress in the concrete and in each
rod. Each rod has a diameter of 20 mm.
L := 1.2m
Given:
rconc := 100mm
P := 150kN E st := 200⋅ GPa
dst := 20mm
E conc := 29⋅ GPa
Solution:
⎛π⎞
2
Ast := 6⎜ ⋅ ⎛⎝ dst ⎞⎠
4
Aconc := π ⋅ ⎛⎝ rconc ⎞⎠ − Ast
2
⎝ ⎠
Compatibility:
δ st = δ conc
(Pst)⋅ L = (Pconc)⋅ L
Given
Ast⋅ E st
Aconc ⋅ E conc
[1]
Pst + Pconc − P = 0
[2]
Equations of equilibrium:
ΣF
+ y=0;
Initial guess:
Pconc := 1kN
Solving [1] and [2]:
Pst := 2kN
⎛⎜ Pconc ⎞
:= Find ( Pconc , Pst)
⎜ Pst
⎝
⎠
⎛⎜ Pconc ⎞ ⎛ 104.152 ⎞
=⎜
kN
⎜ Pst
45.848
⎝
⎠
⎝
⎠
Average Normal Stress:
σ st :=
Pst
Ast
σ conc :=
Pconc
Aconc
σ st = 24.323 MPa
Ans
σ conc = 3.527 MPa
Ans
Problem 4-32
The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is
subjected to an axial force of 150 kN, determine the required diameter of each rod so that one-fourth
of the load is carried by the concrete and three-fourths by the steel.
Given:
L := 1.2m
rconc := 100mm
P := 150kN
Pst := 0.75P
E st := 200⋅ GPa
Pconc := 0.25P
E conc := 29⋅ GPa
Solution:
Pst = 112.50 kN
Pconc = 37.50 kN
δ st = δ conc
Compatibility:
(Pst)⋅ L = (Pconc)⋅ L
Ast⋅ E st
Aconc ⋅ E conc
⎛π⎞
2
Ast = 6⎜ ⋅ ⎛⎝ dst ⎞⎠
4
Aconc = π ⋅ ⎛⎝ rconc ⎞⎠ − Ast
2
⎝ ⎠
Thus,
π ⋅ ⎛ rconc ⎞ − Ast
2
⎝
⎠
Ast
E st Pconc
⋅
Pst Econc
=
π ⋅ ⎛ rconc ⎞
⎝
2
⎠ = ⎛ Est ⋅ Pconc + 1⎞
⎜
Ast
⎝ Pst Econc ⎠
Ast :=
E
π ⋅ ⎛ rconc ⎞
⎝
2
⎠
st Pconc
⋅
+1
Pst Econc
2
Ast = 9523.2948 mm
2
dst =
dst :=
2
3⋅ π
Ast
2
3⋅ π
Ast
dst = 44.95 mm
Ans
Problem 4-33
The A-36 steel pipe has a 6061-T6 aluminum core. It is subjected to a tensile force of 200 kN.
Determine the average normal stress in the aluminum and the steel due to this loading. The pipe has an
outer diameter of 80 mm and an inner diameter of 70 mm.
Given:
L := 400mm
do := 80mm
P := 200kN
di := 70mm
E st := 200GPa
Solution:
E al := 68.9GPa
π
2
Aal := ⋅ ⎛⎝ di ⎞⎠
4
π
2
2
Ast := ⋅ ⎛⎝ do − di ⎞⎠
4
δ st = δ al
Compatibility:
Given
(Pst)⋅ L = (Pal)⋅ L
Aal⋅ Eal
[1]
Pst + Pal − P = 0
[2]
Ast⋅ E st
Equations of equilibrium:
+
ΣF x=0;
Initial guess:
Solving [1] and [2]:
Pal := 1kN
Pst := 2kN
⎛⎜ Pal ⎞
:= Find ( Pal , Pst)
⎜ Pst
⎝ ⎠
⎛⎜ Pal ⎞ ⎛ 105.899 ⎞
=⎜
kN
⎜ Pst
⎝ ⎠ ⎝ 94.101 ⎠
Average Normal Stress:
σ st :=
σ al :=
Pst
Ast
Pal
Aal
σ st = 79.88 MPa
Ans
σ al = 27.52 MPa
Ans
Problem 4-34
The concrete column is reinforced using four steel reinforcing rods,each having a diameter of 18 mm.
Determine the stress in the concrete and the steel if the column is subjected to an axial load of 800 kN.
E st = 200 GPa, E c = 25 GPa.
P := 800kN
Given:
Solution:
bc := 300mm
dst := 18mm
E st := 200⋅ GPa
E c := 25⋅ GPa
⎛π⎞
2
Ast := 4⎜ ⋅ ⎛⎝ dst ⎞⎠
4
Ac := ⎛⎝ bc ⎞⎠ − Ast
2
⎝ ⎠
Set:
L := 1m
δ st = δ conc
Compatibility:
(Pst)⋅ L (Pc)⋅ L
Given
Ast⋅ E st
=
Ac⋅ E c
[1]
Pst + Pc − P = 0
[2]
Equations of equilibrium:
+
ΣF y=0;
Initial guess:
Solving [1] and [2]:
Pc := 1kN
Pst := 2kN
⎛⎜ Pc ⎞
:= Find ( Pc , Pst)
⎜ Pst
⎝ ⎠
⎛⎜ Pc ⎞ ⎛ 732.928 ⎞
=⎜
kN
⎜ Pst
67.072
⎝
⎠
⎝ ⎠
Average Normal Stress:
σ st :=
σ c :=
Pst
Ast
Pc
Ac
σ st = 65.89 MPa
Ans
σ c = 8.24 MPa
Ans
Problem 4-35
The column is constructed from high-strength concrete and four A-36 steel reinforcing rods. If it is
subjected to an axial force of 800 kN, determine the required diameter of each rod so that one-fourth
of the load is carried by the steel and three-fourths by the concrete. E st = 200 GPa, E c = 25 GPa.
Given:
P := 800kN
E c := 25⋅ GPa
Pc := 0.75P
E st := 200⋅ GPa
bc := 300mm
Pst := 0.25P
Solution:
Pst = 200 kN
Pc = 600 kN
δ st = δ c
Compatibility:
(Pst)⋅ L = (Pc)⋅ L
Ast⋅ E st
Ac⋅ E c
⎛π⎞ 2
Ast = 4⎜ ⋅ dst
4
Ac = ⎛⎝ bc ⎞⎠ − Ast
2
⎝ ⎠
2
Thus,
bc − Ast
Ast
2
bc
Ast
=
Est Pc
⋅
Pst E c
⎛ Est Pc
= ⎜
⋅
⎝ Pst Ec
⎞
+1
⎠
2
Ast :=
E
bc
st Pc
⋅
+1
Pst Ec
2
Ast = 3600 mm
2
dst =
dst :=
1
π
Ast
1
π
Ast
dst = 33.85 mm
Ans
Problem 4-36
The A-36 steel pipe has an outer radius of 20 mm and an inner radius of 15 mm. If it fits snugly
between the fixed walls before it is loaded, determine the reaction at the walls when it is subjected to
the load shown.
Given:
Solution:
L AB := 300mm
ro := 20mm
L BC := 700mm
ri := 15mm
P := 8kN
E st := 200GPa
A := π ⋅ ⎛⎝ ro − ri ⎞⎠
2
2
L := L AB + L BC
By superposition :
+
−∆ C + δ C = 0
( −2P) ⋅ L AB
A⋅ E
+
(FC)⋅ L = 0
A⋅ E
⎛ LAB ⎞
⎝ L ⎠
FC := 2⋅ P⋅ ⎜
FC = 4.80 kN
Ans
FA = 11.20 kN
Ans
Equations of equilibrium:
+
ΣF x=0;
FA + FC − 2P = 0
FA := 2P − FC
Problem 4-37
The 304 stainless steel post A has a diameter of d = 50 mm and is surrounded by a red brass C83400
tube B. Both rest on the rigid surface. If a force of 25 kN is applied to the rigid cap, determine the
average normal stress developed in the post and the tube.
Given: L := 200mm dst := 50mm
ro := 75mm t := 12mm
P := 25kN E st := 193⋅ GPa
Solution:
E br := 101⋅ GPa
ri := ro − t
π
2
Ast := ⋅ ⎛⎝ dst ⎞⎠
4
Compatibility:
Abr := π ⋅ ⎛⎝ ro − ri ⎞⎠
2
2
δ st = δ br
Given
(Pst)⋅ L = (Pbr)⋅ L
[1]
Pst + Pbr − P = 0
[2]
Ast⋅ E st
Abr⋅ Ebr
Equations of equilibrium:
+
ΣF y=0;
Initial guess:
Solving [1] and [2]:
Pbr := 1kN
Pst := 2kN
⎛⎜ Pbr ⎞
:= Find ( Pbr , Pst)
⎜ Pst
⎝ ⎠
⎛⎜ Pbr ⎞ ⎛ 14.525 ⎞
=⎜
kN
⎜ Pst
10.475
⎝
⎠
⎝ ⎠
Average Normal Stress:
σ st :=
σ br :=
Pst
Ast
Pbr
Abr
σ st = 5.335 MPa
Ans
σ br = 2.792 MPa
Ans
Problem 4-38
The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigid
surface. If a force of 25 kN is applied to the rigid cap, determine the required diameter d of the steel
post so that the load is shared equally between the post and tube.
Given:
L := 200mm
ro := 75mm
t := 12mm
P := 25kN
E st := 193⋅ GPa
E br := 101⋅ GPa
Pst := 0.5P
Pbr := 0.5P
Solution:
ri := ro − t
Pst = 12.5 kN
Compatibility:
Pbr = 12.5 kN
δ st = δ br
(Pst)⋅ L = (Pbr)⋅ L
Ast⋅ E st
Abr⋅ Ebr
⎛π⎞ 2
Ast = ⎜ ⋅ dst
4
Abr := π ⋅ ⎛⎝ ro − ri ⎞⎠
2
⎝ ⎠
π ⋅ ⎛ ro − ri ⎞
⎝
⎠ = Est ⋅ Pbr
Ast
Pst E br
2
Thus,
2
2
π ⋅ ⎛ ro − ri ⎞
⎝
⎠
Ast :=
Est Pbr
⋅
Pst E br
2
2
2
Ast = 2722.54 mm
2
dst =
dst :=
4
π
Ast
4
π
Ast
dst = 58.88 mm
Ans
Problem 4-39
The load of 7.5 kN is to be supported by the two vertical steel wires for which σY = 500 MPa. If,
originally, wire AB is 1250 mm long and wire AC is 1252.5 mm long, determine the force developed in
each wire after the load is suspended. Each wire has a cross-sectional area of 12.5 mm2.
Given:
2
L AB := 1250mm
L AC := 1252.5mm
A := 12.5mm
W := 7.5kN
E := 200⋅ GPa
E Y := 70500⋅ MPa
∆L := LAC − LAB
Solution:
∆L = 2.50 mm
Compatibility:
δ AC + ∆L = δ AB
+
Given
(TAC)⋅ LAC + ∆L = (TAB)⋅ LAB [1]
A⋅ E
A⋅ E
Equations of equilibrium:
+
ΣF y=0;
Initial guess:
T AC + T AB − W = 0
T AC := 1kN
Solving [1] and [2]:
[2]
T AB := 2kN
⎛⎜ TAC ⎞
:= Find ( T AC , T AB)
⎜ TAB
⎝
⎠
⎛⎜ TAC ⎞ ⎛ 1.249 ⎞
=⎜
kN
⎜ TAB
6.251
⎝
⎠
⎝
⎠
Check Average Normal Stress:
σ AC :=
σ AB :=
TAC
A
TAB
A
σ AC = 99.9 MPa
[ < σY = 500 MPa]
σ AB = 500.1 MPa
[ < σY = 500 MPa]
Ans
Problem 4-40
The load of 4 kN is to be supported by the two vertical steel wires for which σY = 560 MPa. If,
originally, wire AB is 1250 mm long and wire AC is 1252.5 mm long, determine the cross-sectional
area of AB if the load is to be shared equally between both wires. Wire AC has a cross-sectional area
of 13 mm2.
Given:
2
W := 4kN
L AB := 1250mm
AAC := 13mm
T AC := 0.5W
L AC := 1252.5mm
E Y := 560⋅ MPa
T AB := 0.5W
E := 200⋅ GPa
Solution:
∆L := LAC − LAB
∆L = 2.50 mm
T AC = 2 kN
T AB = 2 kN
Compatibility:
+
δ AC + ∆L = δ AB
(TAC)⋅ LAC + ∆L = (TAB)⋅ LAB
(AAC)⋅ E
(AAB)⋅ E
Thus,
AAB :=
(TAB)⋅ LAB
(TAC)⋅ LAC + (∆L)⋅ E
AAC
2
AAB = 3.60911 mm
Ans
Check Average Normal Stress:
σ AC :=
σ AB :=
TAC
AAC
TAB
AAB
σ AC = 153.846 MPa [ < σY = 560 MPa]
σ AB = 554.15 MPa
[ < σY = 560 MPa]
Problem 4-41
The support consists of a solid red brass C83400 post surrounded by a 304 stainless steel tube. Before
the load is applied, the gap between these two parts is 1 mm. Given the dimensions shown, determine
the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the
materials.
Given: L br := 0.251m
E br := 101⋅ GPa
L st := 0.250m
E st := 193⋅ GPa
dbr := 60mm
di := 80mm
Solution:
σ Y_br := 70⋅ MPa
t := 10mm
∆L := Lbr − L st
∆L = 1 mm
π
2
Abr := ⋅ ⎛⎝ dbr ⎞⎠
4
π
2
2
Ast := ⋅ ⎛⎝ do − di ⎞⎠
4
+
Compatibility:
do := di + 2t
δ br + ∆L = δ st
(Fbr)⋅ Lbr + ∆L = (Fst)⋅ Lst
Abr⋅ Ebr
Ast⋅ E st
[1]
Equations of equilibrium:
+
ΣF y=0;
Fst + Fbr − P = 0
[2]
Assume brass yields, then
(
)
Fbr := σ Y_br Abr
σ Y_br
ε br :=
E br
( )
δ br := ε br L br
Thus, only the brass is loaded.
Fbr = 197.92 kN
ε br = 0.00069307
δ br = 0.17 mm
mm
mm
[ < ∆L = 1 mm ]
P := Fbr
P = 197.92 kN
Ans
Problem 4-42
Two A-36 steel wires are used to support the 3.25-kN (~325-kg) engine. Originally, AB is 800 mm
long and A'B' is 800.2 mm long. Determine the force supported by each wire when the engine is
suspended from them. Each wire has a cross-sectional area of 6.25 mm2.
Given:
2
L AB := 800mm
A := 6.25mm
L A'B' := 800.2mm
W := 3.25kN
( 3)
E := 29⋅ 10 ⋅ ksi
∆L := LA'B' − L AB
Solution:
∆L = 0.200 mm
Compatibility:
δ AB + ∆L = δ A'B'
+
Given
(TAB)⋅ LAB + ∆L = (TA'B')⋅ LA'B'
A⋅ E
[1]
A⋅ E
Equations of equilibrium:
+
ΣF y=0;
Initial guess:
Solving [1] and [2]:
T AB + T A'B' − W = 0
T AB := 1kN
[2]
T A'B' := 2kN
⎛⎜ TAB ⎞
:= Find ( TAB , TA'B')
⎜ TA'B'
⎝
⎠
⎛⎜ TAB ⎞ ⎛ 1.469 ⎞
=⎜
kN
⎜ TA'B'
1.781
⎝
⎠
⎝
⎠
Ans
Problem 4-43
The bolt AB has a diameter of 20 mm and passes through a sleeve that has an inner diameter of 40 mm
andan outer diameter of 50 mm. The bolt and sleeve are made of A-36 steel and are secured to the rigid
brackets as shown. If the bolt length is 220 mm and the sleeve length is 200 mm, determine the tension
in the bolt when a force of 50 kN is applied to the brackets.
Given:
L b := 220mm
db := 20mm
do := 50mm
L s := 200mm
P := 25kN
di := 40mm
E := 200GPa
Solution:
π
2
Ab := ⋅ ⎛⎝ db ⎞⎠
4
π
2
2
As := ⋅ ⎛⎝ do − di ⎞⎠
4
δb = δs
Compatibility:
Given
(Pb)⋅ Lb (Ps)⋅ Ls
Ab⋅ E
=
[1]
As⋅ E
Equations of equilibrium:
+
ΣF x=0;
Initial guess:
Solving [1] and [2]:
Pb + Ps − 2P = 0
Pb := 1kN
[2]
Ps := 2kN
⎛⎜ Pb ⎞
:= Find ( Pb , Ps)
⎜ Ps
⎝ ⎠
⎛⎜ Pb ⎞ ⎛ 14.388 ⎞
=⎜
kN
⎜ Ps
35.612
⎝
⎠
⎝ ⎠
Problem 4-44
The specimen represents a filament-reinforced matrix system made from plastic (matrix) and glass
(fiber). If there are n fibers, each having a cross-sectional area of Af and modulus of Ef , embedded in
a matrix having a cross-sectional area of Am and modulus of E m, determine the stress in the matrix and
each fiber when the force P is imposed on the specimen.
Solution:
Compatibility:
δm = δf
( P m) ⋅ L = ( P f ) ⋅ L
Am⋅ E m
n⋅ Af⋅ E f
⎛ Am⋅ Em ⎞
Pm = ⎜
⎝
⋅ Pf
[1]
P − Pm − Pf = 0
[2]
n⋅ Af⋅ E f
⎠
Equations of equilibrium:
+ ΣF y=0;
Solving [1] and [2]:
⎛
Pm = ⎜
⎝
Am⋅ Em
⎞
⋅P
n⋅ Af⋅ E f + Am⋅ E m
⎠
n⋅ Af⋅ Ef
⎛
Pf = ⎜
⎞
⎝ n⋅ Af⋅ Ef + Am⋅ Em ⎠
⋅P
Average Normal Stress:
σm =
σf =
Pm
Am
Pf
n⋅ Af
⎛
Em
⎞
⎝
n⋅ Af⋅ Ef + Am⋅ Em
⎠
⎛
Ef
⎞
⎝
n⋅ Af⋅ Ef + Am⋅ Em
⎠
σm = ⎜
σf = ⎜
⋅P
Ans
⋅P
Ans
Problem 4-45
The distributed loading is supported by the three suspender bars. AB and EF are made from aluminum
andCD is made from steel. If each bar has a cross-sectional area of 450 mm2, determine the maximum
intensity w of the distributed loading so that an allowable stress of (σallow)st = 180 MPa in the steel and
(σallow)al = 94 MPa in the aluminum is not exceeded. Est = 200 GPa, E al = 70 GPa.
L := 2m
Given:
2
b := 1.5m
A := 450mm
E st := 200⋅ GPa
E al := 70⋅ GPa
σ al_allow := 94MPa
σ st_allow := 180MPa
Solution:
(FAB)⋅ L (FCD)⋅ L
=
(Eal)⋅ A (Est)⋅ A
Compatibility:
δA = δC
+
⎛ Eal ⎞
FAB = ⎜
⋅ FCD
[1]
2⋅ FAB = w⋅ ( 2b) − FCD
[2]
⎝
Equations of equilibrium:
+
ΣΜC=0;
FEF⋅ ( b) − FAB⋅ ( b) = 0
ΣF y=0;
FAB + FCD + FEF − w⋅ ( 2b) = 0
E st
⎠
FAB = FEF
Assume failure of AB and EF:
(
)
FAB = 42.30 kN
⋅ FAB
FCD = 120.86 kN
FAB := σ al_allow ⋅ A
⎛ Est ⎞
From [1]:
FCD := ⎜
From [2]:
w :=
⎝ Eal ⎠
FAB
b
+
FCD
2⋅ b
w = 68.49
kN
m
Assume failure of CD:
(
)
FCD = 81.00 kN
⋅ FCD
FAB = 28.35 kN
FCD := σ st_allow ⋅ A
⎛ Eal ⎞
From [1]:
FAB := ⎜
From [2]:
w :=
⎝
Est
FAB
b
+
⎠
FCD
2⋅ b
w = 45.90
kN
m
[Controls !]
Ans
Problem 4-46
The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and
cross-sectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm
and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the
average normal stress in the wire and the block. E st = 200 GPa, Eal = 70 GPa.
Given:
2
L k := 200mm
Ak := 22.5mm
eD := 50mm
AD := 40mm
P := 450N
a := 100mm
b := 150mm
c := 150mm
E st := 200GPa
E al := 70GPa
2
Solution:
Compatibility:
θ BC = θ AD
Given
δ BC
b
=
δD
c
(FBC)⋅ Lk = (FD)⋅ eD
Ak⋅ Est⋅ b
[1]
AD⋅ Eal⋅ c
Equations of equilibrium:
ΣΜA=0;
−FBC⋅ ( b) − FD⋅ ( c) + P⋅ ( a + b) = 0
Initial guess:
Solving [1] and [2]:
FBC := 1kN
[2]
FD := 2kN
⎛⎜ FBC ⎞
:= Find ( FBC , FD)
⎜ FD
⎝
⎠
⎛⎜ FBC ⎞ ⎛ 214.968 ⎞
=⎜
N
⎜ FD
535.032
⎝
⎠
⎝
⎠
Average Normal Stress:
σ BC :=
σ D :=
FBC
Ak
FD
AD
σ BC = 9.554 MPa
Ans
σ D = 13.376 MPa
Ans
Problem 4-47
The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and
cross-sectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm
and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the
rotation of the link about the pin A. Report the answer in radians. Est = 200 GPa, E al = 70 GPa.
Given:
2
L k := 200mm
Ak := 22.5mm
eD := 50mm
AD := 40mm
P := 450N
a := 100mm
b := 150mm
c := 150mm
E st := 200GPa
E al := 70GPa
2
Solution:
Compatibility:
θ BC = θ AD
Given
δ BC
b
δD
=
c
(FBC)⋅ Lk = (FD)⋅ eD
Ak⋅ Est⋅ b
[1]
AD⋅ Eal⋅ c
Equations of equilibrium:
ΣΜA=0;
−FBC⋅ ( b) − FD⋅ ( c) + P⋅ ( a + b) = 0
Initial guess:
Solving [1] and [2]:
FBC := 1kN
[2]
FD := 2kN
⎛⎜ FBC ⎞
:= Find ( FBC , FD)
⎜ FD
⎝
⎠
Displacement:
δ D :=
(FD)⋅ eD
AD⋅ E al
θ AD :=
δD
c
⎛⎜ FBC ⎞ ⎛ 214.968 ⎞
=⎜
N
⎜ FD
⎝
⎠ ⎝ 535.032 ⎠
δ D = 0.009554 mm
θ AD = 63.69 × 10
−6
rad
Ans
Problem 4-48
The three A-36 steel wires each have a diameter of 2 mm and unloaded lengths of LAC = 1.60 m and
LAB = LAD = 2.00 m. Determine the force in each wire after the 150-kg mass is suspended from the
ring at A.
Given:
L AC := 1.60m
a := 3
d := 2mm
L AB := 2.00m
b := 4
M := 150kg
L AD := 2.00m
c := 5
g = 9.81
2
s
a
c
h :=
Solution:
m
v :=
⎛ π ⎞ d2
⎝ 4⎠
b
c
A := ⎜
W := M⋅ g
Compatibility:
( )
In triangle AA'B, since the displacement δA is very small, cos θ A' = v
( )
δ AC⋅ cos θ A' = δ AD
δ AC⋅ ( v) = δ AD
(FAC)⋅ LAC ⋅ ( v) = (FAD)⋅ LAD
A⋅ E
[1]
A⋅ E
Equations of equilibrium:
+ ΣF x=0;
+
ΣF y=0;
FAB⋅ ( h) − FAD⋅ ( h) = 0
FAB = FAD
FAB⋅ ( v) + FAC + FAD⋅ ( v) − W = 0
Given
From [1]:
2⋅ FAD⋅ ( v) = W − FAC
[2]
FAC⋅ L AC⋅ ( v) = FAD⋅ LAD
[3]
Initial guess: FAD := 1N
Solving [2] and [3]:
FAC := 2N
⎛⎜ FAD ⎞
:= Find ( FAD , FAC)
⎜ FAC
⎝
⎠
⎛⎜ FAD ⎞ ⎛ 465.1 ⎞
=⎜
N
⎜ FAC
726.8
⎝
⎠
⎝
⎠
FAB = FAD
Ans
Problem 4-49
The A-36 steel wires AB and AD each have a diameter of 2 mm and the unloaded lengths of each wire
are LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the required diameter of wire AC so that each
wire is subjected to the same force caused by the 150-kg mass suspended from the ring at A.
Given:
L AC := 1.60m
a := 3
dAB := 2mm
L AB := 2.00m
b := 4
dAD := 2mm
L AD := 2.00m
c := 5
g = 9.81
h :=
a
c
v :=
b
c
⎛π⎞
2
AAB := ⎜ dAB
4
⎝ ⎠
2
s
M := 150kg
Solution:
m
W := M⋅ g
⎛π⎞
2
AAD := ⎜ dAD
4
⎝ ⎠
Equations of equilibrium:
Since each wire is required to carry the same amount of load.
Hence,
FAB = F
FAC = F
FAD = F
Compatibility:
( )
In triangle AA'B, since the displacement δA is very small, cos θ A' = v
( )
δ AC⋅ cos θ A' = δ AD
δ AC⋅ ( v) = δ AD
(FAC)⋅ LAC ⋅ ( v) = (FAD)⋅ LAD
AAC⋅ E
AAD⋅ E
L AC
AAC :=
⋅ ( v) ⋅ AAD
L AD
2
AAC = 2.010619 mm
⎛π⎞
2
AAC = ⎜ dAC
4
⎝ ⎠
dAC :=
4
π
(AAC)
dAC = 1.60 mm
Ans
Problem 4-50
The three suspender bars are made of the same material and have equal cross-sectional areas A.
Determinethe average normal stress in each bar if the rigid beam ACE is subjected to the force P .
Solution:
δC − δE
Compatibility:
d
=
δA − δE
2d
2δ C = δ A + δ E
2
(FCD)⋅ L = (FAB)⋅ L + (FEF)⋅ L
A⋅ E
A⋅ E
A⋅ E
2FCD = FAB + FEF
[1]
Equations of equilibrium:
ΣΜA=0;
+
ΣF y=0;
⎛ d⎞ = 0
⎝ 2⎠
FCD⋅ ( d) + FEF⋅ ( 2⋅ d) − P⋅ ⎜
FCD + 2FEF = 0.5P
[2]
FAB + FCD + FEF − P = 0
[3]
Solving [1], [2] and [3]:
FAB =
7
P
12
FCD =
1
P
3
FEF =
1
P
12
σ AB =
7P
12A
σ AB =
P
3A
σ AB =
P
12A
Ans
Problem 4-51
The assembly consists of an A-36 steel bolt and a C83400 red brass tube. If the nut is drawn up snug
against the tube so that L = 75 mm, then turned an additional amount so that it advances 0.02 mm on
the bolt, determine the force in the bolt and the tube. The bolt has a diameter of 7 mm and the tube has
a cross-sectional area of 100 mm2.
Given:
Solution:
L := 75mm
E br := 101⋅ GPa
∆L := 0.02mm
E st := 200⋅ GPa
dst := 7mm
Abr := 100mm
2
⎛π⎞ 2
Ast := ⎜ ⋅ dst
4
⎝ ⎠
( )
Kbr := E br ⋅ Abr
Since no external load is applie, the force acting on the
tube and the bolt is the same.
Equations of equilibrium:
Compatibility:
( )
Kst := Est ⋅ Ast
∆L = δ st + δ br
∆L =
P⋅ L P⋅ L
+
Kst
Kbr
⎛ Kst + Kbr ⎞
∆L = P⋅ L⋅ ⎜
⎝ Kst⋅ Kbr ⎠
P :=
∆L ⎛ Kst⋅ Kbr ⎞
⎜
L Kst + Kbr
⎝
P = 1.165 kN
⎠
Ans
Problem 4-52
The assembly consists of an A-36 steel bolt and a C83400 red brass tube. The nut is drawn up snug
against the tube so that L = 75 mm. Determine the maximum additional amount of advance of the nut
on the bolt so that none of the material will yield. The bolt has a diameter of 7 mm and the tube has a
cross-sectional area of 100 mm2.
Given:
L := 75mm
E br := 101⋅ GPa
dst := 7mm
E st := 200⋅ GPa
2
Abr := 100mm
σ Y_st := 250MPa
σ Y_br := 70MPa
Solution:
⎛π⎞ 2
Ast := ⎜ ⋅ dst
4
( )
( )
Kst := Est ⋅ Ast
⎝ ⎠
Kbr := E br ⋅ Abr
Allowable Normal Stress:
(
)
Pst := σ Y_st ⋅ Ast
(
Pst = 9.621 kN
)
Pbr := σ Y_br ⋅ Abr
Pbr = 7.000 kN
Since Pst > Pbr , by comparison, the brass will yield first.
(
P := min Pst , Pbr
)
P = 7.00 kN
Compatibility:
∆L = δ st + δ br
∆L =
P⋅ L P⋅ L
+
Kst
Kbr
∆L :=
P⋅ L P⋅ L
+
Kst
Kbr
∆L = 0.120 mm
Ans
Problem 4-53
The 10-mm-diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is
20 mm, and its inner diameter is 10 mm. If the bolt is subjected to a compressive force of P = 20 kN,
determine the average normal stress in the steel and the bronze. Est = 200 GPa, Ebr = 100 GPa.
Given:
Solution:
ds := 10mm
do := 20mm
E st := 200GPa
P := 20kN
di := 10mm
E br := 100GPa
π
2
Ast := ⋅ ⎛⎝ ds ⎞⎠
4
π
2
2
Abr := ⋅ ⎛⎝ do − di ⎞⎠
4
(Pbr)⋅ L = (Pst)⋅ L
Compatibility:
δb = δs
Abr⋅ Ebr
Given
Pbr
Abr⋅ Ebr
Ast⋅ E st
=
Pst
Ast⋅ E st
[1]
Equations of equilibrium:
+
ΣF y=0;
Pbr := 1kN
Initial guess:
Solving [1] and [2]:
σ st :=
σ br :=
Pbr + Pst − P = 0
Pst := 2kN
⎛⎜ Pbr ⎞
:= Find ( Pbr , Pst)
⎜ Pst
⎝ ⎠
Pst
Ast
Pbr
Abr
[2]
⎛⎜ Pbr ⎞ ⎛ 12.000 ⎞
=⎜
kN
⎜ Pst
8.000
⎝
⎠
⎝ ⎠
σ st = 101.86 MPa
Ans
σ br = 50.93 MPa
Ans
Problem 4-54
The 10-mm-diameter steel bolt is surrounded by a bronze sleeve.The outer diameter of this sleeve is 20
mm,and its inner diameter is 10 mm. If the yield stress for the steel is (σY )st = 640 MPa, and for the
bronze (σY )br = 520 MPa, determine the magnitude of the largest elastic load P that can be applied to
the assembly. E st = 200 GPa, E br = 100 GPa.
Given:
do := 20mm
E st := 200GPa
σ Y_st := 640MPa
di := 10mm
E br := 100GPa
σ Y_br := 520MPa
ds := 10mm
Solution:
π
2
Ast := ⋅ ⎛⎝ ds ⎞⎠
4
π
2
2
Abr := ⋅ ⎛⎝ do − di ⎞⎠
4
Equations of equilibrium:
+
ΣF y=0;
Pbr + Pst − P = 0
δb = δs
(Pbr)⋅ L = (Pst)⋅ L
Compatibility:
Abr⋅ Ebr
Pbr
Abr⋅ Ebr
Assume failure of bolt, then
(
[1]
Ast⋅ E st
=
Pst
Ast⋅ E st
)
Pst := σ Y_st Ast
⎛ Abr⋅ Ebr ⎞
From [2]:
Pbr := ⎜
From [1]:
P := Pbr + Pst
⎝ Ast⋅ Est ⎠
⋅ Pst
[2]
Pst = 50.265 kN
Pbr = 75.398 kN
P = 125.66 kN
Assume failure of sleeve, then
Pbr := σ Y_br Abr
⎛ Ast⋅ Est ⎞
From [2]:
Pst := ⎜
From [1]:
P := Pbr + Pst
⎝ Abr⋅ Ebr ⎠
⋅ Pbr
Pbr = 122.522 kN
Pst = 81.681 kN
P = 204.20 kN
(Controls!):
Ans
Problem 4-55
The rigid member is held in the position shown by three A-36 steel tie rods. Each rod has an
unstretched length of 0.75 m and a cross-sectional area of 125 mm2. Determine the forces in the rods
if a turnbuckle on rod EF undergoes one full turn. The lead of the screw is 1.5 mm. Neglect the size of
the turnbuckle and assume that it is rigid. Note: The lead would cause the rod, when unloaded, to
shorten 1.5 mm when the turnbuckle is rotated one revolution.
2
L := 0.75m
b := 0.5m
∆L := 1.5mm
E := 200⋅ GPa
Given:
A := 125mm
Solution:
Equations of equilibrium:
ΣΜE=0;
T CD⋅ ( b) + T AB⋅ ( b) = 0
T CD = T AB
+
ΣF y=0;
[1]
T AB + T CD − T EF = 0
T EF = 2T AB
[1]
[2]
Compatibility:
Rod EF shortens 1.5mm causing AB (and CD) to elongate. Thus,:
+
∆L = δ A + δ C
∆L =
(TAB)⋅ L + (TEF)⋅ L
∆L =
(TAB)⋅ L + (2TAB)⋅ L
E⋅ A
T AB :=
E⋅ A
E⋅ A
⋅ ∆L
3L
E⋅ A
E⋅ A
T AB = 16.667 kN
Ans
From [1]:
T CD := TAB
T CD = 16.667 kN
Ans
From [2]:
T EF := 2TAB
T EF = 33.333 kN
Ans
Problem 4-56
The bar is pinned at A and supported by two aluminum rods, each having a diameter of 25 mm and a
modulus of elasticity Eal = 70 GPa. If the bar is assumed to be rigid and initially vertical, determine the
displacement of the end B when the force of 10 kN is applied.
Given:
L CD := 0.6m
L EF := 0.3m
d := 25mm
a := 0.3m
P := 10kN
E al := 70GPa
A :=
Solution:
π 2
⋅d
4
Compatibility:
δE
3⋅ a
Given
δC
=
a
(FEF)⋅ LEF = (FCD)⋅ LCD
3A⋅ Eal
A⋅ Eal
FEF⋅ LEF = 3FCD⋅ LCD
[1]
Equations of equilibrium:
ΣΜA=0;
−FEF⋅ ( 3⋅ a) − FCD⋅ ( a) + P⋅ ( 2a) = 0
−3FEF − FCD + 2P = 0
FEF := 1kN
Initial guess:
Solving [1] and [2]:
δ E :=
δB
4⋅ a
=
δE
3a
FCD := 2kN
⎛⎜ FEF ⎞
:= Find ( FEF , FCD)
⎜ FCD
⎝
⎠
Displacement:
(FEF)⋅ LEF
δ B :=
A⋅ E al
4
δ
3 E
[2]
⎛⎜ FEF ⎞ ⎛ 6.31579 ⎞
=⎜
kN
⎜ FCD
1.05263
⎝
⎠
⎝
⎠
δ E = 0.055142 mm
δ B = 0.073522 mm
Ans
Problem 4-57
The bar is pinned at A and supported by two aluminum rods, each having a diameter of 25 mm and a
modulus of elasticity Eal = 70 GPa. If the bar is assumed to be rigid and initially vertical, determine the
force in each rod when the 10-kN load is applied.
Given:
Solution:
L CD := 0.6m
L EF := 0.3m
d := 25mm
a := 0.3m
P := 10kN
E al := 70GPa
A :=
Compatibility:
Given
π 2
⋅d
4
δE
3⋅ a
=
δC
a
(FEF)⋅ LEF = (FCD)⋅ LCD
3A⋅ Eal
A⋅ Eal
FEF⋅ LEF = 3FCD⋅ LCD
[1]
Equations of equilibrium:
ΣΜA=0;
−FEF⋅ ( 3⋅ a) − FCD⋅ ( a) + P⋅ ( 2a) = 0
−3FEF − FCD + 2P = 0
Initial guess:
Solving [1] and [2]:
FEF := 1kN
[2]
FCD := 2kN
⎛⎜ FEF ⎞
:= Find ( FEF , FCD)
⎜ FCD
⎝
⎠
⎛⎜ FEF ⎞ ⎛ 6.316 ⎞
=⎜
kN
⎜ FCD
⎝
⎠ ⎝ 1.053 ⎠
Problem 4-58
The assembly consists of two posts made from material 1 having a modulus of elasticity of E 1 and
each a cross-sectional area A1 , and a material 2 having a modulus of elasticity E 2 and cross-sectional
area A2. If a central load P is applied to the rigid cap, determine the force in each material.
Solution:
FAB = F1
FEF = F2
Equations of equilibrium:
ΣΜE=0;
FCD⋅ ( d) − FAB⋅ ( d) = 0
FCD = FAB
FCD = F1
+
ΣF y=0;
FAB + FCD + FEF − P = 0
2F1 + F2 = P
Compatibility:
δA = δB
[1]
δB = δC
(F1)⋅ L = (F2)⋅ L
A1⋅ E 1
A2⋅ E2
⎛ A1⋅ E1 ⎞
F1 = ⎜
⎝ A2⋅ E2 ⎠
Solving [1] and [2]:
⎛
F1 = ⎜
⎝
⎛
F2 = ⎜
⎝
⋅ F2
[2]
A1⋅ E 1
⎞
P
2A1⋅ E1 + A2⋅ E2
Ans
A2⋅ E 2
⎞
P
2A1⋅ E1 + A2⋅ E2
Ans
⎠
⎠
Problem 4-59
The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticity
E 1 ofand each a cross-sectional area A 1, and a central post EF made from material 2 having a modulus
of elasticity E2 and a cross-sectional area A 2. If posts AB and CD are to be replaced by those having a
material 2, determine the required cross-sectional area of these new posts so that both assemblies
deform the same amount when loaded.
Solution:
FAB = F1
FEF = F2
Equations of equilibrium:
ΣΜE=0;
FCD⋅ ( d) − FAB⋅ ( d) = 0
FCD = FAB
FCD = F1
+
ΣF y=0;
FAB + FCD + FEF − P = 0
2F1 + F2 = P
[1]
Compatibility:
δA = δB
δB = δC
δA = δ
(F1)⋅ L = (F2)⋅ L
A1⋅ E 1
⎛ A1⋅ E1 ⎞
F1 = ⎜
⎝ A2⋅ E2 ⎠
A2⋅ E2
⋅ F2
[2]
Solving [1] and [2]:
⎛
F1 = ⎜
⎝
A1⋅ E 1
⎞
P
2A1⋅ E1 + A2⋅ E2
⎛
F2 = ⎜
⎠
⎝
A2⋅ E 2
⎞
P
2A1⋅ E1 + A2⋅ E2
⎠
When material 1 has been replaced by material 2 for two side posts, then
Equations of equilibrium:
[1] becomes
Compatibility: [2] becomes
Solving [1'] and [2']:
Requires,
( )
[1']
⎛ A'1 ⎞
F'1 = ⎜
⋅ F'2
⎝ A2 ⎠
[2']
⎛ A2 ⎞
F'2 = ⎜
P
⎝ 2A'1 + A2 ⎠
δ B = δ' B
(F2)⋅ L = (F'2)⋅ L
A2⋅ E 2
⎛ A'1 ⎞
F'1 = ⎜
P
⎝ 2A'1 + A2 ⎠
2 F'1 + F'2 = P
A2⋅ E 2
A2⋅ E2
⎛
⎛ A2 ⎞
⎞
P
=
P
⎜
⎜
2A'1 + A2
2A1⋅ E 1 + A2⋅ E 2
⎝
⎠
⎝
⎠
⎛ E1 ⎞
A'1 = ⎜
⋅ A1
E2
⎝ ⎠
Ans
Problem 4-60
The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticity
E 1 ofand each a cross-sectional area A 1, and a central post EF made from material 2 having a modulus
of elasticity E2 and a cross-sectional area A 2. If post EF is to be replaced by one having a material 1,
determine the required cross-sectional area of this new post so that both assemblies deform the same
amount when loaded.
Solution:
FAB = F1
FEF = F2
Equations of equilibrium:
ΣΜE=0;
FCD⋅ ( d) − FAB⋅ ( d) = 0
FCD = FAB
FCD = F1
+
ΣF y=0;
FAB + FCD + FEF − P = 0
2F1 + F2 = P
[1]
Compatibility:
δA = δB
δB = δC
δA = δ
(F1)⋅ L = (F2)⋅ L
A1⋅ E 1
⎛ A2⋅ E2 ⎞
F2 = ⎜
⎝ A1⋅ E1 ⎠
A2⋅ E2
⋅ F1
[2]
Solving [1] and [2]:
⎛
F1 = ⎜
⎝
A1⋅ E 1
⎞
P
2A1⋅ E1 + A2⋅ E2
⎠
⎛
F2 = ⎜
⎝
A2⋅ E 2
⎞
P
2A1⋅ E1 + A2⋅ E2
⎠
When material 2 has been replaced by material 1 for central post, then
Equations of equilibrium:
[1] becomes
Compatibility: [2] becomes
Solving [1'] and [2']:
Requires,
( )
[1']
⎛ A'2 ⎞
F'2 = ⎜
⋅ F'1
⎝ A1 ⎠
[2']
⎛ A'2 ⎞
F'2 = ⎜
P
⎝ 2A1 + A'2 ⎠
δ A = δ' A
(F1)⋅ L = (F'1)⋅ L
A1⋅ E 1
⎛ A1 ⎞
F'1 = ⎜
P
⎝ 2A1 + A'2 ⎠
2 F'1 + F'2 = P
A1⋅ E 1
A1⋅ E1
⎛
⎛ A1 ⎞
⎞
P= ⎜
P
⎜
⎝ 2A1⋅ E1 + A2⋅ E2 ⎠
⎝ 2A1 + A'2 ⎠
⎛ E2 ⎞
A'2 = ⎜
⋅ A2
⎝ E1 ⎠
Ans
Problem 4-61
The bracket is held to the wall using three A-36 steel bolts at B, C, and D. Each bolt has a diameter of
12.5 mm and an unstretched length of 50 mm. If a force of 4 kN is placed on the bracket as shown,
determine the force developed in each bolt. For the calculation, assume that the bolts carry no shear;
rather, the vertical force of 4 kN is supported by the toe at A. Also, assume that the wall and bracket
are rigid. A greatly exaggerated deformation of the bolts is shown.
Given: a := 12.5mm
b := 25mm
d := 12.5mm
e := 50mm
L := 50mm
P := 4kN
Solution:
c := 50mm
E := 200GPa
L AD := a + b + c
L AC := a + b
A := π ⋅
L AB := a
4
2
d
Compatibility:
δD
L AD
δB
L AB
=
=
(FD)⋅ L = (FC)⋅ L
(LAD)⋅ A⋅ E (LAC)⋅ A⋅ E
δC
LAC
(FB)⋅ L = (FC)⋅ L
(LAB)⋅ A⋅ E (LAC)⋅ A⋅ E
δC
L AC
Given
⎛ LAD ⎞
FD = ⎜
⋅ FC
[1]
⋅ FC
[2]
FD⋅ L AD + FC⋅ LAC + FB⋅ LAB − P⋅ ( e) = 0
[3]
⎝
LAC
⎠
⎛ LAB ⎞
FB = ⎜
⎝
Equations of equilibrium:
ΣΜA=0;
(
Initial guess:
)
(
FB := 10kN
)
(
FC := 20kN
L AC
⎠
)
FD := 30kN
Solving [1], [2] and [3]:
⎛ FB ⎞
⎜
⎜ FC ⎟ := Find ( FB , FC , FD)
⎜F
⎝ D⎠
⎛ FB ⎞ ⎛ 0.2712 ⎞
⎜
⎜
⎜ FC ⎟ = ⎜ 0.8136 kN
⎜F
1.8983 ⎠
⎝ D⎠ ⎝
Ans
Problem 4-62
The bracket is held to the wall using three A-36 steel bolts at B, C, and D. Each bolt has a diameter of
12.5 mm and an unstretched length of 50 mm. If a force of 4 kN is placed on the bracket as shown,
determine how far, s, the top bracket at bolt D moves away from the wall. For the calculation, assume
that the bolts carry no shear; rather, the vertical force of 4 kN is supported by the toe at A. Also,
assume that the wall and bracket are rigid. A greatly exaggerated deformation of the bolts is shown.
Given:
a := 12.5mm
b := 25mm
d := 12.5mm
e := 50mm
L := 50mm
P := 4kN
c := 50mm
E := 200GPa
L AD := a + b + c
Solution:
L AC := a + b
L AB := a
A :=
π 2
⋅d
4
Compatibility:
δD
L AD
δB
L AB
=
=
δC
LAC
(FD)⋅ L = (FC)⋅ L
(LAD)⋅ A⋅ E (LAC)⋅ A⋅ E
(FB)⋅ L = (FC)⋅ L
(LAB)⋅ A⋅ E (LAC)⋅ A⋅ E
δC
L AC
Given
⎛ LAD ⎞
FD = ⎜
⋅ FC
[1]
⋅ FC
[2]
FD⋅ L AD + FC⋅ LAC + FB⋅ LAB − P⋅ ( e) = 0
[3]
⎝
LAC
⎠
⎛ LAB ⎞
FB = ⎜
⎝
Equations of equilibrium:
ΣΜA=0;
(
Initial guess:
)
(
FB := 10kN
)
(
L AC
⎠
)
FC := 20kN
FD := 30kN
Solving [1], [2] and [3]:
⎛ FB ⎞
⎜
⎜ FC ⎟ := Find ( FB , FC , FD)
⎜F
⎝ D⎠
Displacement:
δ D :=
(FD)⋅ L
A⋅ E
⎛ FB ⎞ ⎛ 0.2712 ⎞
⎜
⎜
⎜ FC ⎟ = ⎜ 0.8136 kN
⎜F
1.8983 ⎠
⎝ D⎠ ⎝
δ D = 0.003867 mm
Ans
Problem 4-63
The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the
posts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has a
stiffness of k = 2 MN/m and an unstretched length of 1.02 m, determine the force in each post after
the load is applied to the bar.
L := 1m
Given:
w := 50
L k := 1.02m
( 3) kN
m
kN
m
k := 2⋅ 10 ⋅
2
A := 600mm
Solution:
b := 2m
E := 9.65GPa
∆Lk := L k − L
∆Lk = 0.02 m
Compatibility:
+
δ A + ∆Lk = δ k
[1]
Equations of equilibrium:
ΣΜC=0;
FB⋅ ( b) − FA⋅ ( b) = 0
FA = FB
+
ΣF y=0;
FA + FB + Fsp − w⋅ ( b) = 0
Given
2FA + Fsp − w⋅ ( b) = 0
[2]
From [1]:
(FA)⋅ L + ∆L = Fsp
[1']
Initial guess:
FA := 1kN
Solving [2] and [1']:
k
E⋅ A
k
Fsp := 2kN
⎛⎜ FA ⎞
:= Find ( FA , Fsp)
⎜ Fsp
⎝ ⎠
⎛⎜ FA ⎞ ⎛ 25.58 ⎞
=⎜
kN
⎜ Fsp
⎝ ⎠ ⎝ 48.84 ⎠
FB := FA
FB = 25.58 kN
Ans
Ans
Problem 4-64
The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the
posts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has a
stiffness of k = 2 MN/m and an unstretched length of 1.02 m, determine the vertical displacement of A
and B after the load is applied to the bar.
L := 1m
Given:
w := 50
L k := 1.02m
( 3) kN
m
kN
m
k := 2⋅ 10 ⋅
2
A := 600mm
Solution:
b := 2m
E := 9.65GPa
∆Lk := L k − L
∆Lk = 0.02 m
Compatibility:
+
δ A + ∆Lk = δ k
[1]
Equations of equilibrium:
ΣΜC=0;
FB⋅ ( b) − FA⋅ ( b) = 0
FA = FB
+
ΣF y=0;
FA + FB + Fsp − w⋅ ( b) = 0
Given
2FA + Fsp − w⋅ ( b) = 0
[2]
From [1]:
(FA)⋅ L + ∆L = Fsp
[1']
Initial guess:
FA := 1kN
Solving [2] and [1']:
E⋅ A
k
k
Fsp := 2kN
⎛⎜ FA ⎞
:= Find ( FA , Fsp)
⎜ Fsp
⎝ ⎠
FB := FA
Displacement:
δ A :=
δ B :=
(FA)⋅ L
A⋅ E
(FB)⋅ L
A⋅ E
⎛⎜ FA ⎞ ⎛ 25.58 ⎞
=⎜
kN
⎜ Fsp
⎝ ⎠ ⎝ 48.84 ⎠
FB = 25.58 kN
δ A = 4.418 mm
Ans
δ B = 4.418 mm
Ans
Problem 4-65
The wheel is subjected to a force of 18 kN from the axle. Determine the force in each of the three
spokes. Assume the rim is rigid and the spokes are made of the same material, and each has the same
cross-sectional area.
Given:
Solution:
L := 0.4m
θ := 120deg
P := 18kN
φ := 180deg − θ
Compatibility:
δ AB⋅ cos ( φ ) = δ AC
(FAB)⋅ L ⋅ cos (φ) = (FAC)⋅ L
E⋅ A
At A :
Equations of equilibrium:
+ ΣF x=0;
[1]
E⋅ A
FAC⋅ sin ( φ ) − FAD⋅ sin ( b) = 0
FAC = FAD
+
ΣF y=0;
FAB + FAC⋅ cos ( φ ) + FAD⋅ cos ( φ ) − P = 0
Given
(
[2]
FAB⋅ cos ( φ ) = FAC
[1']
From [1]:
Initial guess:
)
FAB + 2 FAC⋅ cos ( φ ) − P = 0
FAB := 1kN
Solving [2] and [1']:
FAC := 2kN
⎛⎜ FAB ⎞
:= Find ( FAB , FAC)
⎜ FAC
⎝
⎠
FAD := FAC
⎛⎜ FAB ⎞ ⎛ 12 ⎞
=⎜
kN
⎜ FAC
⎝
⎠ ⎝6⎠
Ans
FAD = 6 kN
Ans
Problem 4-66
The post is made from 6061-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A and
B, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originally
uncompressed, determine the reactions at A and B when the force P = 40 kN is applied to the collar.
Given:
LCA := 0.25m
P := 40kN
d := 50mm
LBC := 0.25m
E := 68.9GPa
k := 200 10
π 2
⋅d
4
E⋅ A
kCA :=
L CA
E⋅ A
kBC :=
L BC
FA − P + Fk + FB = 0
[1]
Solution:
A :=
( 3) kN
m
Equations of equilibrium:
+
ΣFy=0;
Compatibility:
Considera a combined force F (=FB+Fk) acted at free end B .
∆ B = δP − δF
Given
0=
P
kBC
⎛ FB + Fk
−⎜
⎝
kCA
+
Fk
Also, ∆ sp=∆ BC
Initial guess:
∆B = 0
k
FB := 1kN
Solving [1] and [2]:
From [1]:
FB + Fk ⎞
kBC + k
⎠
=
FB + Fk
kBC + k
[2]
[3]
Fk := 2kN
⎛⎜ FB ⎞
:= Find ( FB , Fk)
⎜ Fk
⎝ ⎠
FA := P − Fk − FB
⎛⎜ FB ⎞ ⎛ 16.88 ⎞
=⎜
kN
⎜ Fk
6.24
⎝
⎠
⎝ ⎠
Ans
FA = 16.88 kN
Ans
Problem 4-67
The post is made from 6061-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A and
B, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originally
uncompressed, determine the compression in the spring when the load of P = 50 kN is applied to the
collar.
Given:
L CA := 0.25m
P := 50kN
d := 50mm
L BC := 0.25m
E := 68.9GPa
k := 200 10
Solution:
A :=
π 2
⋅d
4
kCA :=
( 3) kN
m
E⋅ A
LCA
kBC :=
E⋅ A
LBC
Equations of equilibrium:
+
ΣF y=0;
Compatibility:
FA − P + Fk + FB = 0
[1]
Considera a combined force F (=FB+Fk) acted at free end B .
∆ B = δP − δF
Given
0=
P
kBC
⎛ FB + Fk
−⎜
⎝
kCA
k
FB := 1kN
Solving [1] and [2]:
Thus,
FB + Fk ⎞
+
Fk
Also, ∆sp=∆BC
Initial guess:
∆B = 0
kBC + k
=
[2]
⎠
FB + Fk
[3]
kBC + k
Fk := 2kN
⎛⎜ FB ⎞
:= Find ( FB , Fk)
⎜ Fk
⎝ ⎠
∆ sp :=
⎛⎜ FB ⎞ ⎛ 21.101 ⎞
=⎜
kN
⎜ Fk
7.799
⎝
⎠
⎝ ⎠
Fk
k
∆ sp = 0.03899 mm
Ans
Problem 4-68
The rigid bar supports the uniform distributed load of 90 kN/m. Determine the force in each cable if
each cable has a cross-sectional area of 36 mm2, and E = 200 GPa.
a := 1m
Given:
b := 2m
2
A := 36mm
w := 90
kN
m
c :=
Solution:
E := 200GPa
2
a +b
LAC :=
2
v :=
2
2
4a + b
LBC := c
b
c
h :=
a
c
L := 3a
LDC := c
Equations of equilibrium:
ΣΜA=0;
TBC⋅ ( v) ⋅ a + T DC⋅ ( v) ⋅ L − w⋅ L⋅ ( 0.5L) = 0
[1]
Compatibility:
2
2
2
2
2
2
(
( )
)
In triangle AB'C:
LB'C = a + LAC − 2⋅ a⋅ L AC ⋅ cos θ' A
In triangle AD'C:
LD'C = L + LAC − 2⋅ L ⋅ LAC ⋅ cos θ' A
( )
Thus, eliminating cos θ' A
:
LD'C
2
2
−
L
LB'C
)
( )
2 ⎛1
−
(
= L − a + LAC ⋅ ⎜
⎝L
a
2
2
2
2
2
2
2
2
2
2
LD'C − 3L B'C = 6a − 2L AC
3L B'C − LD'C = 2a + 2b
But,
Thus,
2
3L B'C − LD'C = 2c
LB'C = LBC + ∆ BC
LD'C = L DC + ∆ DC
LB'C = c + ∆ BC
LD'C = c + ∆ DC
3 ( c + ∆ BC) − ( c + ∆ DC) = 2c
2
2
2
Neglect squares of ∆'s since small strain occurs:
3⎛⎝ c + 2c∆ BC⎞⎠ − ⎛⎝ c + 2c∆ DC⎞⎠ = 2c
2
2
2
3⎛⎝ c + 2c∆ BC⎞⎠ − ⎛⎝ c + 2c∆ DC⎞⎠ = 2c
2
2
3∆ BC − ∆ DC = 0
2
1⎞
a⎠
3
T BC⋅ ( c)
E⋅ A
−
TDC⋅ ( c)
E⋅ A
= 0
TDC = 3T BC
[2]
Substituting [2] into [1]:
TBC⋅ ( v) ⋅ a + 3TBC⋅ ( v) ⋅ L − w⋅ L ⋅ ( 0.5L ) = 0
⎛ 9a⋅ c ⎞ ⋅ w
⎝ 20⋅ b ⎠
From [2] :
TBC := ⎜
TBC = 45.2804 kN
Ans
TDC := 3TBC
TDC = 135.8411 kN
Ans
Problem 4-69
The rigid bar is originally horizontal and is supported by two cables each having a cross-sectional area
of 36 mm2, and E = 200 GPa. Determine the slight rotation of the bar when the uniform load is applied.
Given:
a := 1m
b := 2m
2
A := 36mm
Solution:
E := 200GPa
w := 90
kN
m
c :=
a +b
2
2
v :=
2
L AC :=
4a + b
L BC := c
2
b
c
h :=
a
c
L := 3a
L DC := c
Equations of equilibrium:
ΣΜA=0;
Compatibility:
T BC⋅ ( v) ⋅ a + TDC⋅ ( v) ⋅ L − w⋅ L ⋅ ( 0.5L ) = 0
[1]
See solution of Prob. 4-68.
T DC = 3TBC
Solving [1] and [2]:
T DC := 27.1682 kN
∆ DC :=
Geometry:
In triangle AD'C:
( )
tan θ A =
2
2
[2]
T DC⋅ ( c)
∆ DC = 8.4374919 mm
E⋅ A
⎛b⎞
⎝ 2a ⎠
b
2a
θ A := atan ⎜
2
(
)
θ A = 45.000 deg
( )
L D'C = L + L AC − 2⋅ L⋅ L AC ⋅ cos θ' A
(LDC + ∆ DC)2 = L2 + LAC2 − 2⋅ L⋅ (LAC)⋅ cos (θ'A)
⎡L2 + L 2 − ( L + ∆ DC) 2⎤
⎢
AC
DC
⎥
θ' A := acos ⎢
⎥
2⋅ L⋅ ( L AC)
⎣
⎦
∆θ := θ' A − θ A
θ' A = 45.180 deg
∆θ = 0.180 deg
Ans
Problem 4-70
The electrical switch closes when the linkage rods CD and AB heat up, causing the rigid arm BDE
both to translate and rotate until contact is made at F. Originally, BDE is vertical, and the temperature
is 20°C. If AB is made of bronze C86100 and CD is made of aluminum 6061-T6, determine the gap s
required so that the switch will close when the temperature becomes 110°C.
Unit used:
°C := deg
Given: L BD := 400mm
L DE := 200mm
T 1 := 20°C
L := 300mm
T 2 := 110°C
( − 6) 1
α cu := 17.0 ⋅ 10
°C
( − 6) 1
α st := 24.0 ⋅ 10
Solution:
∆T := T2 − T1
∆T = 90 °C
Thermal Expansion:
δ AB := α cu⋅ ( ∆T ) ⋅ L
δ AB = 0.4590 mm
δ CD := α st⋅ ( ∆T ) ⋅ L
δ CD = 0.6480 mm
Geomotry:
θ :=
δ CD − δ AB
L BD
(
s := δ AB + θ ⋅ LBD + LDE
s = 0.7425 mm
)
Ans
°C
Problem 4-71
A steel surveyor's tape is to be used to measure the length of a line. The tape has a rectangular cross
section of 1.25 mm by 5 mm and a length of 30 m when T1 = 20°C and the tension or pull on the tape
is 100 N. Determine the true length of the line if the tape shows the reading to be 139 m when used
with a pull of 175 N at T2 = 40°C. The ground on which it is placed is flat. αst = 17(10-6)/°C, Est =
200 GPa.
°C := deg
Unit used:
a := 5mm
b := 1.25mm
T 1 := 20°C
P1 := 100N
L 1 := 30m
T 2 := 40°C
P2 := 175N
L 2 := 139m
Given:
( − 6) 1
α st := 17⋅ 10
E st := 200GPa
°C
A := a⋅ b
Solution:
∆T := T2 − T1
∆T = 20 °C
Thermal Expansion:
δ T := α st⋅ ( ∆T ) ⋅ L2
δ P :=
(P2 − P1)⋅ L2
A⋅ E st
δ T = 47.2600 mm
δ T = 0.0473 m
δ P = 8.3400 mm
δ P = 0.0083 m
L' := L 2 + δ T + δ P
L' = 139.056 m
Ans
Problem 4-72
The assembly has the diameters and material make-up indicated. If it fits securely between its fixed
supports when the temperature is T1 = 20°C, determine the average normal stress in each material
when the temperature reaches T2 = 40°C.
Unit used:
°C := deg
Given: T 1 := 20°C
T 2 := 40°C
L 1 := 1.2m
d1 := 300mm
L 2 := 1.8m
d2 := 200mm
L 3 := 0.9m
d3 := 100mm
( − 6) 1
α 1 := 23⋅ 10
°C
E 1 := 73.1GPa
Solution:
( − 6) 1
α 2 := 17⋅ 10
( − 6) 1
α 3 := 17⋅ 10
°C
E 2 := 103GPa
∆T := T2 − T1
∆T = 20 °C
⎛π⎞ 2
A1 := ⎜ ⋅ d1
4
⎛π⎞ 2
A2 := ⎜ ⋅ d2
4
⎝ ⎠
°C
E 3 := 193GPa
⎛π⎞ 2
A3 := ⎜ ⋅ d3
4
⎝ ⎠
⎝ ⎠
Equations of equilibrium:
+ ΣF x=0;
FA − FD = 0
FA = FD
Let FA=F. Then, FA = FAB = FBC = FCD = FD = F
Thermal Expansion:
δ T := α 1⋅ ( ∆T ) ⋅ L1 + α 2⋅ ( ∆T) ⋅ L 2 + α 3⋅ ( ∆T ) ⋅ L3
δ T = 1.470000 mm
Elongation: negative sign indicates shortening
δF = −
F⋅ L 1
A1⋅ E 1
−
F⋅ L 2
A2⋅ E 2
−
F⋅ L 3
A3⋅ E 3
Compatibility:
F⋅ L 1
0 = δT + δF
Given
0 = δT −
Initial guess:
F := 1kN
Solving [1] :
Average Normal Stress:
A1⋅ E 1
−
F⋅ L 2
A2⋅ E 2
−
F := Find ( F)
F⋅ L 3
[1]
A3⋅ E 3
F = 1063.49 kN
σ al :=
F
A1
σ al = 15.05 MPa
Ans
σ br :=
F
A2
σ br = 33.85 MPa
Ans
σ st :=
F
A3
σ st = 135.41 MPa
Ans
Problem 4-73
A high-strength concrete driveway slab has a length of 6 m when its temperature is 10°C. If there is a
gap of 3 mm on one side before it touches its fixed abutment, determine the temperature required to
close the gap. What is the compressive stress in the concrete if the temperature becomes 60°C?
°C := deg
Unit used:
Given:
L := 6m
T 1 := 10°C
∆ gap := 3mm
T 2 := 60°C
( − 6) 1
α := 11⋅ 10
°C
E := 29GPa
Solution:
Require,
(
)
∆ gap = α ⋅ T' − T1 ⋅ L
⎛ ∆ gap ⎞
T'1 := ⎜
⎝ α⋅L ⎠
For
+ T1
∆T := T2 − T1
T'1 = 55.45 °C
Ans
⎛ E⎞
⎝ L⎠
σ = 1.45 MPa
∆T = 50 °C
∆ gap = δ T − δ F
∆ gap = α ⋅ ( ∆T) ⋅ L −
F⋅ L
A⋅ E
F
⎛ E⎞
= α ⋅ ( ∆T) ⋅ E − ∆ gap⋅ ⎜
A
⎝ L⎠
σ=
F
A
σ := α ⋅ ( ∆T) ⋅ E − ∆ gap⋅ ⎜
Ans
Problem 4-74
A 1.8-m-long steam pipe is made of steel with σY = 280 MPa. It is connected directly to two turbines
A and B as shown.The pipe has an outer diameter of 100 mm and a wall thickness of 6 mm.The
connection was made at T1 = 20°C. If the turbines' points of attachment are assumed rigid, determine
the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 =
135°C.
Unit used: °C := deg
Given:
L := 1.8m
T 1 := 20°C
T 2 := 135°C
σ Y := 280MPa
do := 100mm
t := 6mm
E := 200GPa
Solution:
( − 6) 1
α := 12⋅ 10
°C
π ⎛ 2
2
⋅ ⎝ do − di ⎞⎠
4
di := do − 2t
A :=
∆T := T2 − T1
∆T = 115 °C
Compatibility:
0 = δT + δF
0 = α ⋅ ( ∆T ) ⋅ L −
F⋅ L
E⋅ A
F := α ⋅ ( ∆T ) ⋅ E⋅ A
F = 489.03 kN
Ans
F
A
σ = 276.00 MPa
Check stress:
σ :=
(< σY = 280 MPa)
ok.
Problem 4-75
A 1.8-m-long steam pipe is made of steel with σY = 280 MPa. It is connected directly to two turbines
A and B as shown.The pipe has an outer diameter of 100 mm and a wall thickness of 6 mm. The
connection was made at T1 = 20°C. If the turbines' points of attachment are assumed to have a
stiffness of k = 16 MN/m., determine the force the pipe exerts on the turbines when the steam and
thus the pipe reach a temperature of T2 = 135°C.
Unit used:
°C := deg
Given: L := 1.8m
T 1 := 20°C
σ Y := 280MPa
do := 100mm
E := 200GPa
Solution:
T 2 := 135°C
t := 6mm
( − 6) 1
α := 12⋅ 10
°C
kN
( 3) mm
k := 16⋅ 10
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
di := do − 2t
A :=
∆T := T2 − T1
∆T = 115 °C
Compatibility:
2x = δ T + δ F
2x = α ⋅ ( ∆T) ⋅ L −
x :=
( k⋅ x) ⋅ L
E⋅ A
α ⋅ ( ∆T) ⋅ E ⋅ A⋅ L
k⋅ L + 2E⋅ A
x = 0.029830 mm
F := k⋅ x
F = 477.29 kN
Ans
F
A
σ = 269.37 MPa
Check stress:
σ :=
(< σY = 280 MPa)
ok.
Problem 4-76
The 12-m-long A-36 steel rails on a train track are laid with a small gap δ between them to allow
forthermal expansion. Determine the required gap so that the rails just touch one another when the
temperature is increased from T1 = -30°C to T2 = 30°C. Using this gap, what would be the axial force
in the rails if the temperature were to rise to T3 = 40°C? The cross-sectional area of each rail is 3200
mm2.
Unit used:
°C := deg
2
L := 12m
A := 3200mm
T 1 := −30°C
T 2 := 30°C
Given:
T 3 := 40°C
( − 6) 1
E := 200GPa
α := 12⋅ 10
°C
Solution:
Require,
(
)
∆ gap := α ⋅ T2 − T1 ⋅ L
∆ gap = 8.640 mm
For
∆T := T3 − T1
Ans
∆T = 70 °C
Compatibility:
∆ gap = δ T + δ F
∆ gap = α ⋅ ( ∆T) ⋅ L −
F⋅ L
E⋅ A
F := ⎡⎣α ⋅ ( ∆T) ⋅ L − ∆ gap⎤⎦ ⋅
F = 76.80 kN
E⋅ A
L
Ans
Problem 4-77
The two circular rod segments, one of aluminum and the other of copper, are fixed to the rigid walls
such that there is a gap of 0.2 mm between them when T1 = 15°C. What larger temperature T2 is
required in order to just close the gap? Each rod has a diameter of 30 mm, αal = 24(10-6)/°C, Eal = 70
GPa, αcu = 17(10-6)/°C, E cu = 126 GPa. Determine the average normal stress in each rod if T2 = 95°C.
°C := deg
Unit used:
Given: L cu := 100mm L al := 200mm
T 1 := 15°C
T 2 := 95°C
d := 30mm
∆ gap := 0.2mm
( − 6) 1
°C
− 6) 1
(
α al := 24⋅ 10
E cu := 126GPa
α cu := 17⋅ 10
E al := 70GPa
°C
⎛π⎞ 2
A := ⎜ ⋅ d
⎝ 4⎠
Solution:
Require,
(
)
(
)
∆ gap = α cu⋅ T' − T 1 ⋅ L cu + α al⋅ T' − T1 ⋅ Lal
∆ gap
T' := T 1 +
α cu⋅ L cu + α al⋅ Lal
T' = 45.77 °C
For
∆T := T2 − T1
Compatibility:
Ans
∆T = 80 °C
δ cu = δ 1T + δ 1F
δ al = δ 2T + δ 2F
∆ gap = δ cu + δ al
∆ gap = α cu⋅ ( ∆T ) ⋅ Lcu −
F :=
F⋅ Lcu
Ecu⋅ A
+ α al⋅ ( ∆T) ⋅ L al −
α cu⋅ ( ∆T) ⋅ L cu + α al⋅ ( ∆T) ⋅ L al − ∆ gap
Lcu
E cu⋅ A
+
L al
Eal⋅ A
F = 61.958 kN
Ans
σ = 87.65 MPa
Ans
Average Normal Stress:
σ :=
F
A
F⋅ Lal
Eal⋅ A
Problem 4-78
The two circular rod segments, one of aluminum and the other of copper, are fixed to the rigid walls
such that there is a gap of 0.2 mm between them when T1 = 15°C. Each rod has a diameter of 30 mm,
αal = 24(10-6)/°C, Eal = 70 GPa, αcu = 17(10-6)/°C, E cu = 126 GPa. Determine the average normal
stress in each rod if T2 = 150°C, and also calculate the new length of the aluminum segment.
Unit used:
°C := deg
Given: L cu := 100mm L al := 200mm
T 1 := 15°C
T 2 := 150°C
α cu := 17⋅ 10
E al := 70GPa
For
∆ gap := 0.2mm
( − 6) 1
°C
1
−
6
α al := 24⋅ ( 10 )
E cu := 126GPa
Solution:
d := 30mm
°C
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
∆T := T2 − T1
∆T = 135 °C
Compatibility:
δ cu = δ 1T + δ 1F
δ al = δ 2T + δ 2F
∆ gap = δ cu + δ al
∆ gap = α cu⋅ ( ∆T ) ⋅ Lcu −
F :=
F⋅ Lcu
Ecu⋅ A
+ α al⋅ ( ∆T) ⋅ L al −
F⋅ Lal
Eal⋅ A
α cu⋅ ( ∆T) ⋅ L cu + α al⋅ ( ∆T) ⋅ L al − ∆ gap
Lcu
E cu⋅ A
+
F = 131.176 kN
L al
Eal⋅ A
Average Normal Stress:
σ :=
F
A
σ = 185.58 MPa
Ans
Displacement:
δ al := α al⋅ ( ∆T ) ⋅ Lal −
L'al := L al + δ al
F⋅ L al
E al⋅ A
δ al = 0.117783 mm
L'al = 200.117783 mm Ans
Problem 4-79
Two bars, each made of a different material, are connected and placed between two walls when the
temperature is T1 = 10°C. Determine the force exerted on the (rigid) supports when the temperature
becomes T2 = 20°C. The material properties and cross-sectional area of each bar are given in the
figure.
Unit used:
°C := deg
Given: T 1 := 10°C
T 2 := 20°C
2
L := 300mm
2
Ast := 200mm
Abr := 450mm
α st := 12⋅ 10
α br := 21⋅ 10
( − 6) 1
( − 6) 1
°C
E st := 200GPa
Solution:
°C
E br := 100GPa
∆T := T2 − T1
∆T = 10 °C
Equations of equilibrium:
+
ΣF x=0;
FA − FC = 0
FA = FC
Let FA=F. Then, FA = FAB = FBC = FC = F
Compatibility:
0 = δT + δF
0 = α st⋅ ( ∆T) ⋅ L + α br⋅ ( ∆T ) ⋅ L −
F :=
F⋅ L
F⋅ L
−
Ast⋅ E st Abr⋅ Ebr
(α st + α br)⋅ ∆T
1
Ast⋅ Est
+
1
Abr⋅ E br
F = 6.988 kN
Ans
Problem 4-80
The center rod CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance
heating. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the
force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel,
and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectional
area of and 375 mm2. E st = 200 GPa, Eal = 70 GPa, αst = 12(10-6)/°C, αal = 23(10-6)/°C.
Unit used:
°C := deg
Given: T 1 := 30°C
T 2 := 180°C
2
∆ gap := 0.7mm
2
Ast := 125mm
Aal := 375mm
α st := 12⋅ 10
α al := 23⋅ 10
( − 6) 1
( − 6) 1
°C
E st := 200GPa
E al := 70GPa
L st := 300mm
L al := 240mm
Solution:
∆T := T2 − T1
°C
∆T = 150 °C
Equations of equilibrium:
ΣΜC=0;
FAB⋅ ( b) − FEF⋅ ( b) = 0
FAB = FEF
Let FAB= Fst. Then, FAB = FEF = Fst
+
ΣF y=0;
FAB + FEF − Fal = 0
Fal = 2Fst
[1]
Compatibility:
δ st = δ al − ∆ gap
Fst⋅ Lst
E st⋅ Ast
⎡
Fal⋅ L al⎤
⎣
E al⋅ Aal
= ⎢α al⋅ ( ∆T ) ⋅ Lal −
⎥ − ∆ gap
⎦
[2]
Substituting [1] into [2]:
Fst :=
α al⋅ ( ∆T) Lal − ∆ gap
Lst
Ast⋅ Est
From [1]:
+
2Lal
Fst = 4.226 kN
Ans
Fal = 8.453 kN
Ans
Aal⋅ E al
Fal := 2Fst
Problem 4-81
The center rod CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance
heating. Also, the two end rods AB and EF are heated from T1 = 30°C to T2 = 50°C. At the lower
temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF
caused by the increase in temperature. Rods AB and EF are made of steel, and each has a
cross-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectional area of and 375
mm2. Est = 200 GPa, E al = 70 GPa, αst = 12(10-6)/°C, αal = 23(10-6)/°C.
Unit used:
°C := deg
Given: T 1 := 30°C
T 2 := 180°C
2
T'2 := 50°C
2
Ast := 125mm
Aal := 375mm
α st := 12⋅ 10
α al := 23⋅ 10
( − 6) 1
( − 6) 1
°C
E st := 200GPa
E al := 70GPa
L st := 300mm
L al := 240mm
Solution:
°C
∆ gap := 0.7mm
∆T := T2 − T1
∆T = 150 °C
∆T' := T'2 − T1
∆T' = 20 °C
Equations of equilibrium:
ΣΜC=0;
FAB⋅ ( b) − FEF⋅ ( b) = 0
FAB = FEF
Let FAB= Fst. Then, FAB = FEF = Fst
+
ΣF y=0;
FAB + FEF − Fal = 0
Fal = 2Fst
[1]
Compatibility:
δ st = δ al − ∆ gap
⎛ Fst⋅ Lst ⎞
α st⋅ ( ∆T') ⋅ Lst + ⎜
⎝
E st⋅ Ast
⎠
⎡
Fal⋅ Lal⎤
⎣
Eal⋅ Aal
= ⎢α al⋅ ( ∆T) ⋅ L al −
⎥ − ∆ gap
⎦
[2]
Substituting [1] into [2]:
Fst :=
α al⋅ ( ∆T) Lal − ∆ gap − α st⋅ ( ∆T') ⋅ Lst
Lst
Ast⋅ Est
From [1]:
+
2Lal
Fst = 1.849 kN
Ans
Fal = 3.698 kN
Ans
Aal⋅ E al
Fal := 2Fst
Problem 4-82
The pipe is made of A-36 steel and is connected to the collars at A and B.When the temperature is 15°
C, there is no axial load in the pipe. If hot gas traveling through the pipe causes its temperature to rise
by ∆T = (20 + 30x)°C, where x is in meter, determine the average normal stress in the pipe. The inner
diameter is 50 mm, the wall thickness is 4 mm.
Unit used:
Given:
°C := deg
L := 2.4m
di := 50m
T 1 := 15°C
t := 4mm
∆T = ( 20 + 30x)°C
( − 6) 1
E := 200GPa
Solution:
α := 12⋅ 10
do := di + 2t
Compatibility:
A :=
°C
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
0 = δT + δF
L
⌠
F⋅ L
0 = ⎮ α ⋅ ( ∆T) dx −
⌡0
E⋅ A
unit := 1m⋅ °C
L x :=
L
m
Lx
⎛ α ⋅ E⋅ A ⎞ ⋅ ( unit) ⋅ ⌠
⎮ ( 20 + 30x) dx
F := ⎜
⌡0
⎝ L ⎠
F = 84452.766 kN
Average Normal Stress:
σ :=
F
A
σ = 134.40 MPa
Ans
Problem 4-83
The bronze 86100 pipe has an inner radius of 12.5 mm and a wall thickness of 5 mm. If the gas
flowing through it changes the temperature of the pipe uniformly from TA = 60°C at A to TB = 15°C at
B, determine the axial force it exerts on the walls. The pipe was fitted between the walls when T =
15°C.
Unit used: °C := deg
Given:
L := 2.4m
ri := 12.5mm
T A := 60°C
t := 5mm
T B := 15°C
T o := 15°C
( − 6) 1
E := 103GPa
Solution:
α := 17⋅ 10
°C
A := π ⋅ ⎛⎝ ro − ri ⎞⎠
2
ro := ri + t
⎛
⎝
TB − TA
(
)
T x = ⎜ 60 +
L
∆T = Tx − To
2
⎞
⎠
⋅ x °C
T x = ( 60 − 18.75x)°C
∆T = ( 45 − 18.75x)°C
Compatibility:
0 = δT + δF
L
⌠
F⋅ L
0 = ⎮ α ⋅ ( ∆T) dx −
⌡0
E⋅ A
unit := 1m⋅ °C
L x :=
L
m
Lx
⎛ α ⋅ E⋅ A ⎞ ⋅ ( unit) ⋅ ⌠
⎮ ( 45 − 18.75x) dx
F := ⎜
⌡0
⎝ L ⎠
F = 18.566 kN
Ans
Problem 4-84
The rigid block has a weight of 400 kN and is to be supported by posts A and B, which are made of
A-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original
length before they are loaded, determine the average normal stress developed in each post when post C is
heated so that its temperature is increased by 10°C. Each post has a cross-sectional area of 5000 mm2.
Unit used:
Given:
Solution:
°C := deg
2
∆T := 10°C
A := 5000mm
b := 1m
E st := 200GPa
E br := 101GPa
W := 400kN
α br := 18⋅ 10
( − 6) 1
°C
Set: L := 1m
Equations of equilibrium:
ΣΜC=0;
FA⋅ ( b) − FB⋅ ( b) = 0
FA = FB
Let FA= Fst. Then, FA = FB = Fst
+
ΣF y=0;
FA + FB + Fbr − W = 0
Fbr = W − 2Fst
Given
Compatibility:
Fst⋅ L
E st⋅ A
Initial guess:
[1]
δst = δbr
=
Fbr⋅ L
Ebr⋅ A
− α br⋅ ( ∆T ) ⋅ L
Fst := 1kN
[2]
Fbr := 2kN
Solving [1] and [2]:
⎛⎜ Fst ⎞
:= Find ( Fst , Fbr)
⎜ Fbr
⎝ ⎠
Average Normal Stress:
σ st :=
σ br :=
Fst
A
Fbr
A
⎛⎜ Fst ⎞ ⎛ 123.393 ⎞
=⎜
kN
⎜ Fbr
153.214
⎝
⎠
⎝ ⎠
σ st = 24.68 MPa
Ans
σ br = 30.64 MPa
Ans
Ans
Problem 4-85
The bar has a cross-sectional area A, length L, modulus of elasticity E, and coefficient of thermal
expansion α. The temperature of the bar changes uniformly along its length from TA at A to TB at B so
that at any point x along the bar T = TA + x (TB - TA)/L. Determine the force the bar exerts on the rigid
walls. Initially no axial force is in the bar.
Solution:
Compatibility:
0 = δT + δF
Tx = TA +
[1]
TB − TA
L
(
⋅x
)
∆T = Tx − TA
x
∆T = TB − TA ⋅
L
(
However,
)
d ( ∆T ) =
TB − TA
L
( )
d δ T = α ⋅ ( ∆T ) dx
) L⎤⎥⎦ dx
⎡
(
⎣
( )
d δ T = α ⋅ ⎢ TB − TA ⋅
L
x
⌠
x⎤
⎡
δ T = ⎮ α ⋅ ⎢ T B − T A ⋅ ⎥ dx
⎮
L⎦
⎣
⌡
(
)
0
From [1]:
δT =
α⋅L
⋅ TB − TA
2
0=
α⋅L
F⋅ L
⋅ TB − TA −
2
E⋅ A
(
F=
)
(
)
α ⋅ E⋅ A
⋅ TB − TA
2
(
)
Ans
⋅x
Problem 4-86
The rod is made of A-36 steel and has a diameter of 6 mm. If the springs are compressed 12 mm
when the temperature of the rod is T = 10°C, determine the force in the rod when its temperature is T
= 75°C.
Unit used:
°C := deg
N
mm
L := 1.2m
d := 6mm
k := 200
T 1 := 10°C
T 2 := 75°C
xo := 12mm
Given:
E := 200GPa
Solution:
For
( − 6) 1
α := 12⋅ 10
°C
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
∆T := T2 − T1
(
∆T = 65 °C
)
F = k⋅ x + xo
Compatibility:
2x = δ T + δ F
2x = α ⋅ ( ∆T) ⋅ L −
x = α ⋅ ( ∆T ) ⋅
x :=
F⋅ L
E⋅ A
(
)
L k⋅ x + xo ⋅ L
−
2E ⋅ A
2
α ⋅ ( ∆T) ⋅ E ⋅ A⋅ L − k⋅ xo⋅ L
2E ⋅ A + k⋅ L
(
F := k⋅ x + xo
)
x = 0.20892 mm
F = 2.442 kN
Ans
Problem 4-87
Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8
kN.
P := 8kN
d := 20mm
w := 40mm
h := 20mm
t := 5mm
r := 10mm
Given:
Solution:
For the fillet:
w
=2
h
A := h⋅ t
r
= 0.5
h
K := 1.4
From Fig. 4-24,
σ max = K⋅ σ avg
σ max := K⋅
P
A
σ max = 112 MPa
For the hole:
Ao := ( w − d) ⋅ t
ro
= 0.25
w
ro :=
Ans
d
2
Ko := 2.375
From Fig. 4-25,
σ max = K⋅ σ avg
σ max := Ko⋅
P
Ao
σ max = 190 MPa
Ans
Problem 4-88
If the allowable normal stress for the bar is σallow = 120 MPa, determine the maximum axial force P
that can be applied to the bar.
d := 20mm
t := 5mm
w := 40mm
h := 20mm
Given:
r := 10mm
σ allow := 120MPa
Solution:
A := h⋅ t
Assume failure of the fillet:
w
=2
h
r
= 0.5
h
K := 1.4
From Fig. 4-24,
σ allow = σ max
P :=
σ max = K⋅ σ avg
(σallow)⋅ A
K
Ao := ( w − d) ⋅ t
From Fig. 4-25,
Ko := 2.375
P :=
(σallow)⋅ Ao
Ko
P
A
σ max := Ko⋅
P
Ao
P = 8.57 kN
Assume failure of the hole:
ro
= 0.25
w
σ allow = σ max
σ allow = K⋅
ro :=
σ max = K⋅ σ avg
P = 5.05 kN
(Controls!)
d
2
Ans
Problem 4-89
The steel bar has the dimensions shown. Determine the maximum axial force P that can be applied so
as not to exceed an allowable tensile stress of σallow =150 MPa.
d := 24mm
t := 20mm
w := 60mm
h := 30mm
Given:
r := 15mm
σ allow := 150MPa
Solution:
Assume failure occurs at the fillet:
w
=2
h
r
= 0.5
h
From Fig. 4-24,
σ allow = σ max
P :=
A := h⋅ t
(σallow)⋅ A
K
K := 1.4
σ max = K⋅ σ avg
σ allow = σ max
P :=
(σallow)⋅ Ao
Ko
P
A
P = 64.29 kN
Assume failure occrs at the hole: Ao := ( w − d) ⋅ t
ro
= 0.2
w
From Fig. 4-25,
σ allow = K⋅
ro :=
d
2
Ko := 2.45
σ max = K⋅ σ avg
P = 44.08 kN
(Controls!)
σ max := Ko⋅
Ans
P
Ao
Problem 4-90
Determine the maximum axial force P that can be applied to the bar. The bar is made from steel and
has an allowable stress of σallow = 147 MPa.
d := 15mm
t := 4mm
w := 37.5mm
h := 25mm
Given:
r := 5mm
σ allow := 147MPa
Solution:
Assume failure of the fillet:
w
= 1.5
h
A := h⋅ t
r
= 0.2
h
K := 1.73
From Fig. 4-24,
σ allow = σ max
P :=
σ max = K⋅ σ avg
(σallow)⋅ A
K
Ao := ( w − d) ⋅ t
From Fig. 4-25,
Ko := 2.45
P :=
(σallow)⋅ Ao
Ko
P
A
σ max := Ko⋅
P
Ao
P = 8.497 kN
Assume failure of the hole:
ro
= 0.2
w
σ allow = σ max
σ allow = K⋅
ro :=
σ max = K⋅ σ avg
P = 5.4 kN
(Controls!)
d
2
Ans
Problem 4-91
Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8
kN.
d := 15mm
t := 4mm
w := 37.5mm
h := 25mm
Given:
r := 5mm
P := 8kN
Solution:
For the fillet:
w
= 1.5
h
A := h⋅ t
r
= 0.2
h
K := 1.73
From Fig. 4-24,
σ max = K⋅ σ avg
σ max := K⋅
P
A
σ max = 138.4 MPa
For the hole:
Ao := ( w − d) ⋅ t
ro
= 0.20
w
ro :=
d
2
Ko := 2.45
From Fig. 4-25,
σ max = K⋅ σ avg
σ max := Ko⋅
P
Ao
σ max = 217.78 MPa
(Controls!)
Ans
Problem 4-92
Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8
kN.
P := 8kN
d := 12mm
w := 60mm
h := 30mm
t := 5mm
r := 15mm
Given:
Solution:
At the fillet:
w
=2
h
A := h⋅ t
r
= 0.5
h
K := 1.4
From Fig. 4-24,
σ max = K⋅ σ avg
σ max := K⋅
P
A
σ max = 74.67 MPa
At the hole:
Ao := ( w − d) ⋅ t
ro
= 0.1
w
ro :=
d
2
Ko := 2.65
From Fig. 4-25,
σ max = K⋅ σ avg
σ max := Ko⋅
P
Ao
σ max = 88.33 MPa
(Controls!)
Ans
Problem 4-93
The resulting stress distribution along section AB for the bar is shown. From this distribution,
determine the approximate resultant axial force P applied to the bar. Also, what is the
stress-concentration factor for this geometry?
Given:
h := 80mm
t := 10mm
u := 5MPa
v := 20mm
Solution:
⌠
P = ⎮ σ dA
⌡
P = Volume under curve
Number of sqaures,
P := n⋅ ( u⋅ v) ⋅ t
σ avg :=
P
h⋅ t
From the Figure,
K :=
σ max
σ avg
n := 19
P = 19.00 kN
Ans
σ avg = 23.75 MPa
Ans
σ max := 30MPa
K = 1.26
Ans
Problem 4-94
The resulting stress distribution along section AB for the bar is shown. From this distribution,
determine the approximate resultant axial force P applied to the bar.Also, what is the
stress-concentration factor for this geometry?
Given:
h := 80mm
t := 20mm
u := 18MPa
v := 20mm
Solution:
⌠
P = ⎮ σ dA
⌡
P = Volume under curve
Number of sqaures,
P := n⋅ ( u⋅ v) ⋅ t
σ avg :=
P
h⋅ t
From the Figure,
K :=
σ max
σ avg
n := 10
P = 72.00 kN
Ans
σ avg = 45.00 MPa
Ans
σ max := 72MPa
K = 1.60
Ans
Problem 4-95
The A-36 steel plate has a thickness of 12 mm. If there are shoulder fillets at B and C, and σallow = 150
MPa, determine the maximum axial load P that it can support. Compute its elongation neglecting the
effect of the fillets.
Given: t := 12mm
r := 30mm
w := 120mm
h := 60mm
L 1 := 200mm
L 2 := 800mm
σ allow := 150MPa
E := 200GPa
Solution:
A1 := h⋅ t
A2 := w⋅ t
Maximum Normal Stress at fillet:
w
=2
h
r
= 0.5
h
From Fig. 4-23,
K := 1.4
σ allow = σ max
P :=
σ max = K⋅ σ avg
(σallow)⋅ A1
K
σ allow = K⋅
P = 77.14 kN
Ans
δ = 0.429 mm
Ans
Displacement:
δ = δ1 + δ2 + δ1
δ := 2
P⋅ L 1
E ⋅ A1
+
P⋅ L 2
E⋅ A2
P
A1
Problem 4-96
The 1500-kN weight is slowly set on the top of a post made of 2014-T6 aluminum with an A-36 steel
core. If both materials can be considered elastic perfectly plastic, determine the stress in each material.
Given:
ro := 50mm
E st := 200⋅ GPa
ri := 25mm
E al := 73.1 ⋅ GPa
rs := 25mm
σ Y_st := 250MPa
P := 1500kN
σ Y_al := 414MPa
Ast := π ⋅ ⎛⎝ rs ⎞⎠
Aal := π ⋅ ⎛⎝ ro − ri ⎞⎠
2
Solution:
2
2
Equations of equilibrium:
ΣF y=0;
Pal + Pst − P = 0
Compatibility: δst = δal
(Pal)⋅ L = (Pst)⋅ L
Given
+
Aal⋅ Eal
Pal
Aal⋅ Eal
Ast⋅ E st
=
Pst
[2]
Ast⋅ E st
Pal := 1kN
Initial guess:
[1]
Pst := 2kN
⎛⎜ Pal ⎞
:= Find ( Pal , Pst)
⎜ Pst
⎝ ⎠
Solving [1] and [2]:
⎛⎜ Pal ⎞ ⎛ 784.5218 ⎞
=⎜
kN
⎜ Pst
⎝ ⎠ ⎝ 715.4782 ⎠
Average Normal stress:
σ al :=
Pal
σ al = 133.18 MPa ( < σ Y_al = 414MPa ) o.k.!
Aal
Pst
σ st :=
σ st = 364.39 MPa
Ast
( > σ Y_st = 250MPa )
Thererfore, the steel core yields and so the elastic analysis is invalid.
Plastic Analysis:
The stress in the steel is σ st := σ Y_st
σ st = 250 MPa
(
From [1]:
Pst := σ Y_st Ast
)
Pst = 490.87 kN
Pal := P − Pst
Pal = 1009.13 kN
σ al :=
Pal
Aal
Ans
σ al = 171.31 MPa ( < σ Y_al = 414MPa ) o.k.!
Ans
Problem 4-97
The 10-mm-diameter shank of the steel bolt has a bronze sleeve bonded to it. The outer diameter of
this sleeve is 20 mm. If the yield stress for the steel is (σY)st = 640 MPa, and for the bronze (σY)br =
520 MPa, determine the magnitude of the largest elastic load P that can be applied to the assembly. Est
= 200 GPa, Ebr = 100 GPa.
Given:
do := 20mm
E st := 200GPa
σ Y_st := 640MPa
di := 10mm
E br := 100GPa
σ Y_br := 520MPa
ds := 10mm
Solution:
π
2
Ast := ⋅ ⎛⎝ ds ⎞⎠
4
π
2
2
Abr := ⋅ ⎛⎝ do − di ⎞⎠
4
Equations of equilibrium:
+
ΣF y=0;
Compatibility:
δb = δs
Pbr + Pst − P = 0
(Pbr)⋅ L
Abr⋅ Ebr
Pbr
Abr⋅ Ebr
Assume yielding of bolt, then
(
=
=
[1]
(Pst)⋅ L
Ast⋅ E st
Pst
Ast⋅ E st
)
Pst := σ Y_st Ast
⎛ Abr⋅ Ebr ⎞
From [2]:
Pbr := ⎜
From [1]:
P := Pbr + Pst
⎝ Ast⋅ Est ⎠
⋅ Pst
[2]
Pst = 50.265 kN
Pbr = 75.398 kN
P = 125.66 kN
(Controls!):
Assume yielding of sleeve, then
Pbr := σ Y_br Abr
⎛ Ast⋅ Est ⎞
From [2]:
Pst := ⎜
From [1]:
P := Pbr + Pst
⎝ Abr⋅ Ebr ⎠
⋅ Pbr
Pbr = 122.522 kN
Pst = 81.681 kN
P = 204.20 kN
Ans
Problem 4-98
The weight is suspended from steel and aluminum wires, each having the same initial length of 3 m and
cross-sectional area of 4 mm2. If the materials can be assumed to be elastic perfectly plastic, with
(σY )st = 120 MPa and (σY )al = 70 MPa, determine the force in each wire if the weight is (a) 600 N and
(b) 720 N. Eal = 70 GPa, E st = 200 GPa.
L := 3m
Given:
2
A := 4mm
Solution:
(a)
E st := 200GPa
σ Y_st := 120MPa
E al := 70GPa
σ Y_al := 70MPa
W := 600N
Equations of equilibrium:
ΣF y=0;
Pal + Pst − W = 0
Compatibility: δst = δal
(Pal)⋅ L = (Pst)⋅ L
Given
+
A⋅ Eal
Pal
A⋅ Eal
Initial guess:
Solving [1] and [2]:
[1]
A⋅ Est
=
Pst
[2]
A⋅ Est
Pal := 1N
Pst := 2N
⎛⎜ Pal ⎞
:= Find ( Pal , Pst)
⎜ Pst
⎝ ⎠
⎛⎜ Pal ⎞ ⎛ 155.56 ⎞
=⎜
N
⎜ Pst
444.44
⎝
⎠
⎝ ⎠
Ans
Average Normal stress:
σ al :=
Pal
A
Pst
σ al = 38.89 MPa
( < σ Y_al = 70MPa ) o.k.!
σ st = 111.11 MPa ( < σ Y_st = 120MPa ) o.k.!
A
The average normal stress for both wires do not exceed their respective yield
stress. Thererfore, the elastic analysis is valid for both wires.
σ st :=
Solution:
(b)
W := 720N
Equations of equilibrium:
ΣF y=0;
Pal + Pst − W = 0
Compatibility: δst = δal
(Pal)⋅ L = (Pst)⋅ L
Given
+
A⋅ Eal
Pal
A⋅ Eal
Initial guess:
Solving [1] and [2]:
[1]
A⋅ Est
=
Pst
[2]
A⋅ Est
Pal := 1N
Pst := 2N
⎛⎜ Pal ⎞
:= Find ( Pal , Pst)
⎜ Pst
⎝ ⎠
⎛⎜ Pal ⎞ ⎛ 186.67 ⎞
=⎜
N
⎜ Pst
⎝ ⎠ ⎝ 533.33 ⎠
Average Normal stress:
σ al :=
Pal
A
σ al = 46.67 MPa
( < σ Y_al = 70MPa ) o.k.!
Pst
σ st = 133.33 MPa ( > σ Y_st = 120MPa )
A
Thererfore, the steel wire yields and so the elastic analysis is invalid.
σ st :=
Plastic Analysis: The stress in the steel is σ st := σ Y_st
σ st = 120 MPa
(
From [1]:
Pst := σ Y_st A
)
Pst = 480.00 N
Ans
Pal := W − Pst
Pal = 240.00 N
Ans
σ al :=
Pal
A
σ al = 60 MPa
( < σ Y_al = 70MPa ) o.k.!
Problem 4-99
The bar has a cross-sectional area of 625 mm2. If a force of P = 225 kN is applied at B and then
removed, determine the residual stress in sections AB and BC. σY = 210 MPa.
L := 1m
Given:
2
A := 625mm
L AB := 0.75L
L BC := 0.25L
σ Y := 210MPa
P := 225kN
Solution:
By superposition :
∆ C − δC = 0
+
( P) ⋅ LAB
A⋅ E
−
(FC)⋅ L = 0
⎛ LAB ⎞
⎝ L ⎠
FC = 168.75 kN
FA := P − FC
FA = 56.25 kN
FC := P⋅ ⎜
A⋅ E
Equations of equilibrium:
+
ΣF y=0;
FA + FC − P = 0
Average Normal stress:
FA
σ AB :=
σ AB = 90 MPa
A
[1]
( < σ Y = 210MPa ) o.k.!
FC
( > σ Y = 210MPa )
σ BC = 270 MPa
A
Thererfore, the segment BC yields and so the elastic analysis is invalid.
σ BC :=
σ BC := σ Y
Plastic Analysis: The stress in the BC is
σ BC = 210 MPa
( )
From [1]:
FC := σ Y A
FC = 131.25 kN
FA := P − FC
FA = 93.75 kN
σ AB :=
FA
σ AB = 150 MPa ( < σ Y = 210MPa ) o.k.!
A
A reversal of force of 45kip applied results in a reversed FC=270 MPa and FA=90 MPa,
which produce
(T)
σ' AB := 90 MPa
σ' BC := 270 MPa
(C)
Hence,
∆σ AB := σ AB − σ' AB
(
∆σ BC := − σ BC − σ' BC
)
∆σ AB = 60 MPa (T)
Ans
∆σ BC = 60 MPa (T)
Ans
Problem 4-100
The bar has a cross-sectional area of 300 mm2 and is made of a material that has a stressstrain diagram
that can be approximated by the two line segments shown. Determine the elongation of the bar due to
the applied loading.
Given:
2
L AB := 1.5m
L BC := 0.6m
A := 300mm
PB := 40kN
PC := 25kN
σ 1 := 140MPa
mm
mm
σ 2 := 280MPa ε 2 := 0.021
ε 1 := 0.001
mm
mm
Solution:
PBC := PC
PAB := PC + PB
Average Normal stress and Strain: For segment BC
PBC
σ BC :=
σ BC = 83.33 MPa (< 140MPa )
A
⎛ σ BC ⎞
ε BC := ⎜
⎝ σ1 ⎠
⋅ ε1
ε BC = 0.00060
mm
mm
Average Normal stress and Strain: For segment AB
PAB
σ AB :=
σ AB = 216.67 MPa (> 140MPa )
A
⎛ σ AB − σ 1 ⎞
ε AB := ε 1 + ⎜
⎝ σ2 − σ1 ⎠
(
⋅ ε2 − ε1
)
ε AB = 0.01195
Elongation:
δ AB := ε AB⋅ LAB
δ AB = 17.92857 mm
δ BC := ε BC⋅ LBC
δ BC = 0.35714 mm
δ Total := δ AB + δ BC
δ Total = 18.286 mm
Ans
mm
mm
Problem 4-101
The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the
yield stress for the wires is σY = 530 MPa, and Est = 200 GPa, determine the intensity of the
distributed load w that can be placed on the beam and will just cause wire EB to yield. What is the
displacement of point G for this case? For the calculation, assume that the steel is elastic perfectly
plastic.
Given:
a := 400mm
b := 250mm
d := 4mm
L wire := 800mm
E st := 200GPa
Solution:
c := 150mm
σ Y := 530MPa
L := a + b + c
A :=
π 2
⋅d
4
Plastic Analysis:
Wire CD will yield first followed by wire BE. When
both wores yield,
( )
FCD := ( σ Y) A
FBE := σ Y A
FBE = 6.660 kN
FCD = 6.660 kN
Equations of equilibrium:
ΣΜA=0;
FBE⋅ ( a) + FCD⋅ ( a + b) − w⋅ L⋅ ( 0.5L) = 0
⎛ 2a ⎞ + F
w := FBE⋅ ⎜
2
⎝L ⎠
w = 21.85
kN
m
CD⋅
2( a + b)
L
2
Ans
Displacement:
When wire BE achieves yield stress, the corresponding
yield strain is
σY
mm
ε Y :=
ε Y = 0.002650
E st
mm
(
δ BE := ε Y⋅ L wire
Geometry:
δ BE
a
=
)
δ BE = 2.120 mm
δG
L
L
δ G := ⋅ δ BE
a
(
)
δ G = 4.24 mm
Ans
Problem 4-102
The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the
yield stress for the wires is σY = 530 MPa, and Est = 200 GPa, determine (a) the intensity of the
distributed load w that can be placed on the beam that will cause only one of the wires to start to yield
and (b) the smallest intensity of the distributed load that will cause both wires to yield. For the
calculation, assume that the steel is elastic perfectly plastic.
Given:
a := 400mm
b := 250mm
d := 4mm
L wire := 800mm
E st := 200GPa
σ Y := 530MPa
L := a + b + c
Solution:
c := 150mm
A :=
π 2
⋅d
4
Equations of equilibrium:
ΣΜA=0;
FBE⋅ ( a) + FCD⋅ ( a + b) − w⋅ L⋅ ( 0.5L) = 0
a) By observation, wire CD will yield first. Then,
( )
FCD := σ Y A
FCD = 6.660 kN
Geometry:
δ BE
a
=
δ CD
⎛ a ⎞⋅ δ
⎝ a + b ⎠ CD
δ BE = ⎜
a+b
FBE⋅ Lwire
E st⋅ A
=
a ⎛ FCD⋅ L wire ⎞
⋅⎜
E st⋅ A
a+b
⎝
⎠
⎛ a ⎞⋅ F
⎝ a + b ⎠ CD
FBE := ⎜
From [1]:
⎛ 2a ⎞ + F
w := FBE⋅ ⎜
2
⎝L ⎠
w = 18.65
FBE = 4.0986 kN
CD⋅
2( a + b)
L
kN
m
2
Ans
b) When both wires yield,
( )
FCD := ( σ Y) A
FBE := σ Y A
From [1]:
FBE = 6.660 kN
FCD = 6.660 kN
⎛ 2a ⎞ + F
w := FBE⋅ ⎜
2
⎝L ⎠
w = 21.85
kN
m
CD⋅
2( a + b)
L
2
Ans
[1]
Problem 4-103
The rigid beam is supported by the three posts A,B, and C of equal length. Posts A and C have a
diameter of 75 mm and are made of aluminum, for which E al = 70 GPa and (σY)al = 20 MPa. Post B
has a diameter of 20 mm and is made of brass, for which E br = 100 GPa and (σY)br = 590 MPa.
Determine the smallest magnitude of P so that (a) only rods A and C yield and (b) all the posts yield.
Given:
dA := 75mm
dB := 20mm
E al := 70GPa
σ Y_al := 20MPa
E br := 100GPa
σ Y_br := 590MPa
π
Solution: AA := ⋅ dA2
4
dC := 75mm
π
2
AB := ⋅ dB
4
L := 4m
π
2
AC := ⋅ dC
4
Equations of equilibrium:
ΣΜB=0;
FC⋅ ( L ) − P⋅ ( 0.5L ) − FA⋅ ( L ) + P⋅ ( 0.5L ) = 0
FA = FC
Let FA= Fal. Then, FA = FC = Fal
+
ΣF y=0;
FA + FC + Fbr − 2P = 0
2Fal + Fbr − 2P = 0
[1]
a) Post A and C will yield,
(
)
Fal := σ Y_al AA
σ Y_al
ε al :=
E al
Fal = 88.357 kN
ε al = 0.0002857
δ al = δ br
Compatibility condition:
Fal⋅ Lpost
Eal⋅ AA
=
mm
mm
Fbr⋅ L post
Ebr⋅ AB
⎛ Ebr⋅ AB ⎞
Fbr := ⎜
⎝
E al⋅ AA
σ br :=
⎠
⋅ Fal
Fbr = 8.976 kN
Fbr
AB
(< σ Y ) o.k.!
σ br = 28.57 MPa
From [1]:
P := Fal + 0.5Fbr
P = 92.85 kN
Ans
b) All the posts yield. Then,
( )
Fbr := ( σ Y_br) AB
Fal := σ Y_al AA
From [1]:
Fal = 88.357 kN
Fbr = 185.354 kN
P := Fal + 0.5Fbr
P = 181.0 kN
Ans
Problem 4-104
The rigid beam is supported by the three posts A,B, and C. Posts A and C have a diameter of 60 mm
and are made of aluminum, for which Eal = 70 GPa and (σY)al = 20 MPa. Post B is made of brass, for
which E br = 100 GPa and (σY)br = 590 MPa. If P = 130 kN, determine the largest diameter of post B
so that all the posts yield at the same time.
Given:
dA := 60mm
dC := 60mm
E al := 70GPa
σ Y_al := 20MPa
E br := 100GPa
σ Y_br := 590MPa
π
Solution: AA := ⋅ dA2
4
P := 130kN
L := 4m
π
2
AC := ⋅ dC
4
Equations of equilibrium:
ΣΜB=0;
FC⋅ ( L ) − P⋅ ( 0.5L ) − FA⋅ ( L ) + P⋅ ( 0.5L ) = 0
FA = FC
Let FA= Fal. Then, FA = FC = Fal
+
ΣF y=0;
FA + FC + Fbr − 2P = 0
2Fal + Fbr − 2P = 0
When all the posts yield,
(
)
Fal := σ Y_al AA
Fal = 56.549 kN
From [1]:
Fbr := 2P − 2Fal
Fbr = 146.90 kN
Also,
Fbr = σ Y_br AB
(
π
2
AB = ⋅ dB
4
)
⎛ π 2⎞
Fbr = σ Y_br ⎜ ⋅ dB
4
(
dB :=
)⎝
⎠
4Fbr
π ⋅ σ Y_br
dB = 17.805 mm
Ans
[1]
Problem 4-105
The rigid beam is supported by three A-36 steel wires, each having a length of 1.2 m. The crosssectional area of AB and EF is 10 mm2, and the cross-sectional area of CD is 4 mm2. Determine the
largest distributed load w that can be supported by the beam before any of the wires begin to yield. If
the steel is assumed to be elastic perfectly plastic, determine how far the beam is displaced downward
just before all the wires begin to yield.
Given: L := 1.2m b := 1.5m AAB := 10mm2
2
2
AEF := 10mm
ACD := 4mm
E := 200⋅ GPa
σ Y := 250MPa
Solution:
Compatibility:
Beam remains horizontal after the displacement
since the loading and the system are symmetrical.
δ AB = δ CD
(FAB)⋅ L = (FCD)⋅ L
E ⋅ AAB
⎛ ACD ⎞
FCD = ⎜
E⋅ ACD
⎝ AAB ⎠
⋅ FAB
[1]
Equations of equilibrium:
+
ΣΜC=0;
FEF⋅ ( b) − FAB⋅ ( b) = 0
ΣF y=0;
FAB + FCD + FEF − w⋅ ( 2b) = 0
FAB = FEF
2⋅ FAB = w⋅ ( 2b) − FCD
Plastic Analysis:
Assume wires AB and EF yield.
( )
FAB := σ Y ⋅ AAB
⎛ ACD ⎞
From [1]:
FCD := ⎜
From [2]:
w :=
⎝ AAB ⎠
Plastic Analysis:
[2]
FAB
b
+
FAB = 2.500 kN
⋅ FAB
FCD
FCD = 1.000 kN
w = 2.0000
2⋅ b
kN
m
Assume failure of CD.
( )
FCD := σ Y ⋅ ACD
⎛ AAB ⎞
From [1]:
FAB := ⎜
From [2]:
w :=
⎝ ACD ⎠
FAB
b
+
FCD = 1.000 kN
⋅ FCD
FCD
2⋅ b
FAB = 2.500 kN
w = 2.0000
kN
m
The three wires AB, CD and EF yield simultaneously. Hence,
Displacement:
δ :=
FCD⋅ L
E⋅ ACD
δ = 1.500 mm
Ans
w = 2.00
kN
m
Ans
Problem 4-106
A material has a stressstrain diagram that can be described by the curve σ = cε 1/2. Determine the
deflection δ of the end of a rod made from this material if it has a length L, cross-sectional area A, and
a specific weight γ.
Solution:
Stress-strain relationship :
σ = c⋅ ε
2
2
σ = c ⋅ε
σ=
However,
P
Thus,
2
2
P
A
and
2 dδ
= c ⋅
A
dx
P = γ ⋅ A⋅ x
Since
dδ
dx
ε=
2
dδ
P
=
2 2
dx
c A
dδ ⎛⎜ γ ⎞ 2
=
⋅x
dx ⎜ 2
⎝c ⎠
2
Displacement:
L
⌠
⎮ ⎛⎜ γ 2 ⎞ 2
δ= ⎮
⋅ x dx
⎮ ⎜⎝ c2 ⎠
⌡0
δ=
⎜⌠ 2 ⎞
⋅ ⎮ x dx
2 ⎜⌡
c ⎝ 0
⎠
γ
δ=
2 ⎛ L
γ
2⋅ L
3
2
3⋅ c
Ans
Problem 4-107
Solve Prob. 4-106 if the stressstrain diagram is defined by σ = cε 3/2.
Solution:
Stress-strain relationship :
σ = c⋅ ε
3
2
2 3
However,
σ=
σ = c ⋅ε
P
Thus,
2
2
A
P
A
and
2 ⎛ dδ ⎞
= c ⋅⎜
⎝ dx ⎠
3
ε=
dδ
dx
2
dδ ⎛ P ⎞
= ⎜
dx ⎝ c⋅ A ⎠
2
P = γ ⋅ A⋅ x
Since
Displacement:
L
⌠
2
⎮
2
⎮
3
γ
⎛ ⎞ ⋅ x 3 dx
δ= ⎮ ⎜
⎮ ⎝c⎠
⌡
0
3
3
2
dδ ⎛ γ ⎞
3
= ⎜
⋅x
dx ⎝ c ⎠
2 ⎛ L
⎞
⌠
2
⎜
3 ⎮
⎛ γ ⎞ ⋅ ⎜⎮ x 3 dx⎟
δ= ⎜
⎝ c ⎠ ⎜⎝⌡0
⎠
2
3
5
3 ⎛γ⎞
3
δ = ⋅⎜
⋅L
5 ⎝c⎠
Ans
Problem 4-108
The bar having a diameter of 50 mm is fixed connected at its ends and supports the axial load P. If
the material is elastic perfectly plastic as shown by the stressstrain diagram, determine the smallest load
P needed to cause segment AC to yield. If this load is released, determine the permanent displacement
of point C.
Given: L AC := 0.6m L CB := 0.9m
σ Y := 140MPa
Solution:
d := 50mm
ε Y := 0.001
mm
mm
⎛ π 2⎞
A := ⎜ ⋅ d
⎝4 ⎠
When P is increased, the segment AC will become
plastic first, then CB will become plastic. Thus,
( )
FB := ( σ Y) A
FA := σ Y A
FA = 274.889 kN
FB = 274.889 kN
Equations of equilibrium:
+ ΣF x=0;
FA + FB − P = 0
[1]
From the figure,
P := FA + FB
P = 549.78 kN
E :=
σY
εY
Ans
The deflection of point C is,
( )
δ C := ε Y ⋅ L CB
δ C = 0.900 mm
Consider the reverse of P on the bar.
(F'A)⋅ LAC = (F'B)⋅ LCB
A⋅ E
F'A = 1.5F'B
[2]
F'A + F'B − P = 0
[1']
A⋅ E
Equations of equilibrium:
+ ΣF x=0;
Substituting [2] into [1']:
F'A := 0.6P
F'A = 329.87 kN
F'B := 0.4P
F'B = 219.91 kN
The deflection of point C is,
δ' C :=
Hence,
(F'B)⋅ LCB
A⋅ E
∆δ := δ C − δ' C
δ' C = 0.72000 mm
∆δ = 0.180 mm
Ans
3
E = 140 × 10 MPa
Problem 4-109
Determine the elongation of the bar in Prob. 4-108 when both the load P and the supports are removed.
Given: L AC := 0.6m L CB := 0.9m
σ Y := 140MPa
Solution:
d := 50mm
mm
mm
ε Y := 0.001
⎛ π 2⎞
A := ⎜ ⋅ d
⎝4 ⎠
L := L AC + L CB
When P is increased, the segment AC will become
plastic first, then CB will become plastic. Thus,
( )
FB := ( σ Y) A
FA := σ Y A
FA = 274.889 kN
FB = 274.889 kN
Equations of equilibrium:
+ ΣF x=0;
FA + FB − P = 0
[1]
From the figure,
P := FA + FB
P = 549.78 kN
E :=
Ans
The deflection of point C is,
( )
δ C := ε Y ⋅ L CB
δ C = 0.900 mm
Consider the reverse of P on the bar.
(F'A)⋅ LAC = (F'B)⋅ LCB
A⋅ E
A⋅ E
F'A = 1.5F'B
[2]
F'A + F'B − P = 0
[1']
Equations of equilibrium:
+ ΣF x=0;
Substituting [2] into [1']:
The resultant reactions are:
F'A := 0.6P
F'A = 329.87 kN
F'B := 0.4P
F'B = 219.91 kN
F''A := F'A − FA
(
F''A = 54.978 kN
)
F''B := − F'B − FB
F''B = 54.978 kN
When the supports are removed, the elongation will be
F''A⋅ L
δ''C :=
A⋅ E
δ''C = 0.300 mm
Ans
σY
εY
3
E = 140 × 10 MPa
Problem 4-110
A 6-mm-diameter steel rivet having a temperature of 800°C is secured between two plates such that at
this temperature it is 50 mm long and exerts a clamping force of 1.25 kN between the plates.
Determine the approximate clamping force between the plates when the rivet cools to 5°C. For the
calculation, assume that the heads of the rivet and the plates are rigid. Take αst = 14(10-6)/°C, E st =
200 GPa. Is the result a conservative estimate of the actual answer? Why or why not?
Unit used:
°C := deg
Given: T 1 := 800°C
T 2 := 5°C
L := 50mm
d := 6mm
( − 6) 1
E st := 200GPa
∆T := T1 − T2
∆T = 795 °C
α st := 14⋅ 10
Solution:
P := 1.250kN
°C
⎛ π ⎞ ⋅ d2
⎝ 4⎠
A := ⎜
By superposition:
+
0 = δT + δF
0 = α st⋅ ( ∆T) ⋅ L −
( )
(
( FT ) ⋅ L
A⋅ E st
)
FT := α st ⋅ ∆T⋅ A⋅ Est
F := P + FT
F = 64.189 kN
FT = 62.939 kN
Ans
Yes. Because as the rivet cools, the plates and the rivet head will also deform.
Consequently, the force FT on the rivets will not be as great.
Problem 4-111
Determine the maximum axial force P that can be applied to the steel plate. The allowable stress is
σallow = 150 MPa.
d := 24mm
t := 6mm
w := 120mm
h := 60mm
Given:
r := 6mm
σ allow := 150MPa
Solution:
Assume failure of the fillet:
w
=2
h
A := h⋅ t
r
= 0.1
h
K := 2.4
From Fig. 4-24,
σ allow = σ max
P :=
σ max = K⋅ σ avg
(σallow)⋅ A
K
P = 22.5 kN
Assume failure of the hole:
ro
= 0.1
w
Ao := ( w − d) ⋅ t
From Fig. 4-25,
Ko := 2.65
σ allow = σ max
P :=
(σallow)⋅ Ao
Ko
σ allow = K⋅
(Controls!)
ro :=
σ max = K⋅ σ avg
P = 32.604 kN
P
A
Ans
d
2
σ max := Ko⋅
P
Ao
Problem 4-112
The rigid link is supported by a pin at A and two A-36 steel wires, each having an unstretched length of
300 mm and cross-sectional area of 7.8 mm2. Determine the force developed in the wires when the
link supports the vertical load of 1.75 kN.
Given:
L := 300mm
A := 7.8mm
2
P := 1.75kN
a := 150mm
b := 100mm
c := 125mm
Solution:
Compatibility:
δB
b
=
δC
b+c
(FB)⋅ L =
b⋅ ( A⋅ E)
Given
FB
b
=
(FC)⋅ L
( b + c) ⋅ ( A⋅ E)
FC
b+c
[1]
Equations of equilibrium:
ΣΜA=0;
−FC⋅ ( b + c) − FB⋅ ( b) + P⋅ ( a) = 0
Initial guess:
Solving [1] and [2]:
FC := 1kN
[2]
FB := 2kN
⎛⎜ FC ⎞
:= Find ( FC , FB)
⎜ FB
⎝ ⎠
⎛⎜ FC ⎞ ⎛ 0.974 ⎞
=⎜
kN
⎜ FB
⎝ ⎠ ⎝ 0.433 ⎠
Ans
Problem 4-113
The force P is applied to the bar, which is composed of an elastic perfectly plastic material. Construct
a graph to show how the force in each section AB and BC (ordinate) varies as P (abscissa) is
increased. The bar has cross-sectional areas of 625 mm2 in region AB and 2500 mm2 in region BC,
and σY = 210 MPa.
Given: L AB := 150mm AAB := 625mm2
2
L BC := 50mm
ABC := 2500mm
σ Y := 210MPa
Solution:
Equations of equilibrium:
+ ΣF x=0;
P − FA − FC = 0
[1]
Elastic behavior:
+
0 = ∆ C − δC
0=
( P) ⋅ L AB
⎡ ( FC) ⋅ LBC
−⎢
⎣ E⋅ ABC
E ⋅ AAB
( )
0 = 6P − FC ( 0.5 + 6)
Substituting [2] into [1]:
+
(FC)⋅ LAB⎤
E⋅ AAB
⎥
⎦
FC =
12
P
13
[2]
FA =
1
P
13
[3]
By comparison, segment BC will yield first. Hence,
( )
FC := σ Y ⋅ ABC
From [2]:
From [3]:
13
⋅F
12 C
1
FA :=
⋅P
13
P :=
FC = 525 kN
P = 568.75 kN
FA = 43.75 kN
When segment AB yields,
( )
FC := ( σ Y) ⋅ ABC
FA := σ Y ⋅ AAB
From [1]:
P := FA + FC
FA = 131.25 kN
FC = 525 kN
P = 656.25 kN
Problem 4-114
The 2014-T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at A
and B when T1 = 25°C. If the temperature becomes T2 = -20°C, and an axial force of P = 80 kN is
applied to the rigid collar as shown, determine the reactions at A and B.
Unit used:
°C := deg
Given: L AC := 125mm T 1 := 25°C
d := 12mm
L CB := 200mm T 2 := −20°C P := 80N
( − 6) 1
E := 73.1 ⋅ GPa
Solution:
α := 23⋅ 10
∆T := T2 − T1
∆T = −45 °C
L := L AC + L CB
A :=
°C
( )
π 2
⋅ d
4
By superposition :
+
0 = ∆ B + ∆ T + δB
0=
( P) ⋅ L AC
A⋅ E
+ α ⋅ ( ∆T) ⋅ L +
(FB)⋅ L
A⋅ E
⎛ LAC ⎞
− α ⋅ ( ∆T ) ⋅ ( A⋅ E )
⎝ L ⎠
FB := −P⋅ ⎜
FB = 8.526 kN
Ans
Equations of equilibrium:
+
ΣF x=0;
− FA + P + FB = 0
FA := P + FB
FA = 8.606 kN
Ans
Problem 4-115
The 2014-T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at A
and B when T1 = 40°C. Determine the force P that must be applied to the collar so that, when T = 0°C,
the reaction at B is zero.
Unit used:
°C := deg
Given: L AC := 125mm T 1 := 40°C
d := 12mm
L CB := 200mm T 2 := 0°C
FB := 0kN
E := 73.1 ⋅ GPa
( − 6) 1
Solution:
α := 23⋅ 10
∆T := T2 − T1
∆T = −40 °C
L := L AC + L CB
A :=
°C
( )
π 2
⋅ d
4
By superposition :
+
0 = ∆ B + ∆ T + δB
0=
( P) ⋅ L AC
P :=
A⋅ E
L
+ α ⋅ ( ∆T) ⋅ L +
(FB)⋅ L
A⋅ E
⋅ ⎡−α ⋅ ( ∆T ) ⋅ ( A⋅ E ) − FB⎤⎦
LAC ⎣
P = 19.776 kN
Ans
Problem 4-116
The A-36 steel column, having a cross-sectional area of 11250 mm2, is encased in high-strength
concrete as shown. If an axial force of 300 kN is applied to the column, determine the average
compressive stress in the concrete and in the steel. How far does the column shorten? It has an original
length of 2.4 m.
Given:
Solution:
2
ac := 225mm bc := 400mm
Ast := 11250mm
P := 300kN
E c := 29⋅ GPa
E st := 200⋅ GPa
(
L := 2.4m
)
Ac := ac⋅ bc − Ast
δ st = δ conc
Compatibility:
Given
(Pst)⋅ L = (Pc)⋅ L
Ast⋅ E st
Ac⋅ E c
[1]
Pst + Pc − P = 0
[2]
Equations of equilibrium:
+
ΣF y=0;
Initial guess:
Solving [1] and [2]:
Pc := 1kN
Pst := 2kN
⎛⎜ Pc ⎞
:= Find ( Pc , Pst)
⎜ Pst
⎝ ⎠
⎛⎜ Pc ⎞ ⎛ 151.117 ⎞
=⎜
kN
⎜ Pst
⎝ ⎠ ⎝ 148.883 ⎠
Average Normal Stress:
σ st :=
σ c :=
Displacement:
δ :=
Pst
Ast
Pc
Ac
σ st = 13.23 MPa
Ans
σ c = 1.92 MPa
Ans
Either the concrete or steel can be used for the calculation.
(Pst)⋅ L
Ast⋅ Est
δ = 0.15881 mm
Ans
Problem 4-117
The A-36 steel column is encased in high-strength concrete as shown. If an axial force of 300 kN is
applied to the column, determine the required area of the steel so that the force is shared equally
between the steel and concrete. How far does the column shorten? It has an original length of 2.4 m.
Given:
Solution:
ac := 225mm bc := 400mm
L := 2.4m
P := 300kN
E st := 200⋅ GPa
E c := 29⋅ GPa
Pc := 0.5P
Pst := 0.5P
(
)
Ac = ac⋅ bc − Ast
Compatibility:
δ st = δ conc
(Pst)⋅ L = (Pc)⋅ L
Ast⋅ E st
Ac⋅ E c
Ec
Ast :=
⋅ a ⋅b
E c + Est c c
(
)
2
Ast = 11397.38 mm
Displacement:
δ :=
Ans
Either the concrete or steel can be used for the calculation.
(Pst)⋅ L
Ast⋅ Est
δ = 0.15793 mm
Ans
Problem 4-118
The assembly consists of a 30-mm-diameter aluminum bar ABC with fixed collar at B and a
10-mm-diameter steel rod CD. Determine the displacement of point D when the assembly is loaded as
shown. Neglect the size of the collar at B and the connection at C. Est = 200 GPa, Eal = 70 GPa.
Given:
L AB := 300mm dAB := 30mm
L BC := 500mm dBC := 30mm
E al := 70GPa
L CD := 700mm dCD := 10mm
E st := 200GPa
PB := −8kN
PD := 20kN
Solution:
Internal Force:
As shown on FBD.
Displacement:
⎛π⎞
2
AAB := ⎜ ⋅ dAB
4
δ AB :=
⎛π⎞
2
ABC := ⎜ ⋅ dBC
4
δ BC :=
⎝ ⎠
⎝ ⎠
⎛π⎞
2
ACD := ⎜ ⋅ dCD
4
⎝ ⎠
(PD + PB)⋅ (LAB)
E al⋅ ( AAB)
(PD)⋅ (LBC)
E al⋅ ( ABC)
(PD)⋅ (LCD)
δ CD :=
E st⋅ ( ACD)
δ D := δ AB + δ BC + δ CD
δ D = 1.166 mm
Ans
Problem 4-119
The joint is made from three A-36 steel plates that are bonded together at their seams. Determine the
displacement of end A with respect to end B when the joint is subjected to the axial loads shown. Each
plate has a thickness of 5 mm.
b := 100mm
Given:
t := 5mm
L AB := 600mm AAB := t⋅ b
L BC := 200mm ABC := 3t⋅ b
L CD := 800mm ACD := 2t⋅ b
P := 46kN
E := 200GPa
Solution:
Internal Force:
As shown on FBD.
δ A_D = δ AB + δ BC + δ CD
δ A_D :=
( ) + P⋅ (LBC) + P⋅ (LCD)
E ⋅ ( AAB) E⋅ ( ABC) E ⋅ ( ACD)
P⋅ LAB
δ A_D = 0.491 mm
Ans
Problem 5-1
A shaft is made of a steel alloy having an allowable shear stress of τallow = 84 MPa. If the diameter of
the shaft is 37.5 mm, determine the maximum torque T that can be transmitted. What would be the
maximum torque T' if a 25-mm-diameter hole is bored through the shaft? Sketch the shear-stress
distribution along a radial line in each case.
do := 37.5mm
Given:
di := 25mm
τ allow := 84MPa
Solution:
c :=
do
2
Aplying the torsion formula
a) Allowable shear srtess :
⎛ π ⎞ ⎛ do ⎞
J := ⎜ ⋅ ⎜
⎝ 2⎠ ⎝ 2 ⎠
4
τ allow =
T :=
T⋅ c
J
(τ allow)⋅ J
c
T = 0.87 kN⋅ m
b) Allowable shear srtess :
Ans
Aplying the torsion formula
4
4
⎡
⎛ di ⎞ ⎤⎥
π ⎞ ⎢⎛ do ⎞
⎛
J' := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
τ allow =
T' :=
T'⋅ c
J'
(τ allow)⋅ J
c
T' = 0.87 kN⋅ m
Shera stress at
ρ := 0.5 ⋅ di
τ ρ :=
T'⋅ ρ
J'
τ ρ = 69.78 MPa
Ans
Problem 5-2
The solid shaft of radius r is subjected to a torque T. Determine the radius r' of the inner core of the
shaft that resists one-half of the applied torque (T/2). Solve the problem two ways: (a) by using the
torsion formula, (b) by finding the resultant of the shear-stress distribution.
Problem 5-3
The solid shaft of radius r is subjected to a torque T. Determine the radius r' of the inner core of the
shaft that resists one-quarter of the applied torque (T/4). Solve the problem two ways: (a) by using the
torsion formula, (b) by finding the resultant of the shear-stress distribution.
Problem 5-4
The tube is subjected to a torque of 750 N·m. Determine the amount of this torque that is resisted by
the gray shaded section. Solve the problem two ways: (a) by using the torsion formula, (b) by finding
the resultant of the shear-stress distribution.
ro := 100mm
Given:
ri := 25mm
T := 750N⋅ m
rc := 75mm
Solution:
c := ro
a) Aplying the torsion formula:
⎛ π ⎞ ⋅ ⎛ r 4 − r 4⎞
i ⎠
⎝ 2⎠ ⎝ o
⎛ π ⎞ ⋅ ⎛r 4 − r 4⎞
c ⎠
⎝ 2⎠ ⎝ o
J := ⎜
τ max :=
J' := ⎜
T⋅ c
J
τ max =
τ max = 0.479 MPa
T' :=
T'⋅ c
J'
τ max⋅ J'
c
T' = 0.515 kN⋅ m
Ans
b) Integartion Method :
dT' = ρ ⋅ τ ⋅ dA
dA = 2π ρ ⋅ dρ
⎛ ρ ⎞ ⋅ τ ⋅ ( 2π ρ ⋅ dρ)
⎝ c ⎠ max
dT' = ρ ⋅ ⎜
dT' =
2π
3
⋅ τ maxρ ⋅ dρ
c
⎛⌠ o
⎞
2π
3
⎜
⎮ ρ dρ
T' :=
⋅τ
c max⎜ ⌡r
r
⎝ c
T' = 0.515 kN⋅ m
⎠
Ans
⎛ ρ ⎞⋅ τ
⎝ c ⎠ max
τ= ⎜
Problem 5-5
The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the
absolute maximum shear stress on the shaft.
Given: a := 300mm
T A := −300N⋅ m
b := 400mm
T C := 500N⋅ m
c := 500mm
T D := 200N⋅ m
do := 30mm
T B := −400N⋅ m
Solution:
c' :=
do
2
Internal Torque : As shown in the torque diagram.
Allowable shear srtess :
Aplying the torsion formula
From the torque diagram,
⎛ π ⎞ ⎛ do ⎞
J := ⎜ ⋅ ⎜
⎝ 2⎠ ⎝ 2 ⎠
T max := T B
4
τ max :=
T max⋅ c'
J
τ max = 75.45 MPa
x1 := 0 , 0.01 ⋅ a .. a
( )
T 1 x1 := T A⋅
1
N⋅ m
Ans
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
T 2 x2 := TA + T C ⋅
N⋅ m
( ) (
)
1
T 3 x3 := TA + T C + T D ⋅
N⋅ m
( ) (
)
Torque (Nm)
500
( )
T2 ( x 2 )
T3 ( x 3 )
T1 x 1
0
500
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
0.8
1
Problem 5-6
The solid 32-mm-diameter shaft is used to transmit the torques applied to the gears. If it is supported
by smooth bearings at A and B, which do not resist torque, determine the shear stress developed in the
shaft at points C and D. Indicate the shear stress on volume elements located at these points.
Given:
Solution:
do := 32mm
T 1 := 185N⋅ m
T 2 := −260N⋅ m
T 3 := 75N⋅ m
⎛ π ⎞ ⎛ do ⎞
J := ⎜ ⋅ ⎜
⎝ 2⎠ ⎝ 2 ⎠
do
c :=
2
4
T C := T1
T D := T 1 + T 2
τ C :=
τ D :=
(TC)⋅ c
J
(TD)⋅ c
J
τ C = 28.75 MPa
Ans
τ D = −11.66 MPa
Ans
Problem 5-7
The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm. If it is subjected to the
applied torques as shown, determine the absolute maximum shear stress developed in the shaft. The
smooth bearings at A and B do not resist torque.
Given:
do := 32mm
di := 25mm
T 1 := 185N⋅ m
T 2 := −260N⋅ m
Solution:
c :=
T 3 := 75N⋅ m
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
T max := T 1
τ max :=
(Tmax)⋅ c
J
τ max = 45.82 MPa
Ans
Problem 5-8
The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm. If it is subjected to the
applied torques as shown, plot the shear-stress distribution acting along a radial line lying within region
EA of the shaft. The smooth bearings at A and B do not resist torque.
Given:
do := 32mm
di := 25mm
T 1 := 185N⋅ m
T 2 := −260N⋅ m
Solution:
c :=
T 3 := 75N⋅ m
do
2
4
4
⎡
⎛ di ⎞ ⎤⎥
π ⎞ ⎢⎛ do ⎞
⎛
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
T EA := T1
τ max :=
(TEA)⋅ c
J
Ans
τ ρ = 35.80 MPa
Ans
ρ := 0.5 ⋅ di
Shera stress at
τ ρ :=
τ max = 45.82 MPa
TEA⋅ ρ
J
Problem 5-9
The assembly consists of two sections of galvanized steel pipe connected together using a reducing
coupling at B. The smaller pipe has an outer diameter of 18.75 mm and an inner diameter of 17 mm,
whereas the larger pipe has an outer diameter of 25 mm and an inner diameter of 21.5 mm. If the pipe
is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the
pipe when the couple shown is applied to the handles of the wrench.
Given:
Solution:
d1o := 18.75mm
d1i := 17mm
d2o := 25mm
d2i := 21.5mm
aL := 150mm
aR := 200mm
F := 75N
T := F⋅ aL + F⋅ aR
T = 26.25 N⋅ m
Segment AB :
c1 :=
d1o
2
4
4
⎡
⎛ d1i ⎞ ⎤⎥
π ⎞ ⎢⎛ d1o ⎞
⎛
J1 := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
τ AB :=
Segment BC :
c2 :=
T⋅ c1
J1
τ AB = 62.55 MPa
Ans
d2o
2
4
4
⎡
⎛ d2i ⎞ ⎤⎥
π ⎞ ⎢⎛ d2o ⎞
⎛
J2 := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
τ BC :=
T⋅ c2
J2
τ BC = 18.89 MPa
Ans
Problem 5-10
The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an
inner diameter of 25 mm and a wall thickness of 5 mm, determine the maximum shear stress in the
tube when the cable force of 600 N is applied to the cables. Also, sketch the shear-stress distribution
over the cross section.
Given:
di := 25mm
t := 5mm
P := 600N
a := 75mm
Solution:
do := di + 2t
c :=
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
T := P⋅ ( 2a)
τ max :=
T⋅ c
J
Shera stress at
τ ρ :=
T⋅ ρ
J
T = 90.00 N⋅ m
τ max = 14.45 MPa
Ans
ρ := 0.5 ⋅ di
τ ρ = 10.32 MPa
Ans
Problem 5-11
The shaft consists of three concentric tubes, each made from the same material and having the inner
and outer radii shown. If a torque of T = 800 N·m is applied to the rigid disk fixed to its end, determin
the maximum shear stress in the shaft.
Given:
Solution:
ri1 := 20mm
ro1 := 25mm
ri2 := 26mm
ro2 := 30mm
ri3 := 32mm
ro3 := 38mm
T := 800N⋅ m
L := 2m
cmax := ro3
⎛π⎞
4
4
J1 := ⎜ ⋅ ⎛⎝ ro1 − ri1 ⎞⎠
2
⎝ ⎠
⎛π⎞
4
4
J2 := ⎜ ⋅ ⎛⎝ ro2 − ri2 ⎞⎠
2
⎝ ⎠
⎛π⎞
4
4
J3 := ⎜ ⋅ ⎛⎝ ro3 − ri3 ⎞⎠
2
⎝ ⎠
J := J1 + J2 + J3
τ max :=
T ⋅ cmax
J
τ max = 11.94 MPa
Ans
Problem 5-12
The solid shaft is fixed to the support at C and subjected to the torsional loadings shown. Determine
the shear stress at points A and B and sketch the shear stress on volume elements located at these
points.
Given:
ro := 35mm
ρ A := 35mm
ρ B := 20mm
T D := 800N⋅ m
T B := −300N⋅ m
Solution: J :=
π 4
⋅r
2 o
T BA := TD + T B
τ A :=
T BA⋅ ρ A
J
τ A = 7.42 MPa
T AB := TD
τ B :=
TAB⋅ ρ B
J
τ B = 6.79 MPa
Ans
Problem 5-13
A steel tube having an outer diameter of 62.5 mm is used to transmit 3 kW when turning at 27 rev/min.
Determine the inner diameter d of the tube to the nearest multiples of 5mm if the allowable shear stress
is τallow = 70 MPa.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
do := 62.5mm
ω := 27⋅ rpm
P := 3kW
τ allow := 70MPa
ω = 2.83
T :=
Max. stress :
rad
s
P
T = 1061.03 N⋅ m
ω
c :=
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J= ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
0.25
⎤
⎡⎛ d ⎞ 4
o
2
T
⋅
c
⎢
⎛ ⎞⎛
⎞⎥
di := 2 ⎜
⎢⎝ 2 ⎠ − ⎜⎝ π ⎠ ⋅ ⎜ τ
⎥
⎣
⎝ allow ⎠⎦
T⋅ c
τ allow =
J
di = 56.83 mm
Use
di = 60mm
Ans
Problem 5-14
The solid aluminum shaft has a diameter of 50 mm and an allowable shear stress of τallow = 6 MPa.
Determine the largest torque T1 that can be applied to the shaft if it is also subjected to the other
torsional loadings. It is required that T1 act in the direction shown. Also, determine the maximum she
stress within regions CD and DE.
Given:
do := 50mm
τ allow := 6MPa
T A := 68N⋅ m
T C := 49N⋅ m
T D := 35N⋅ m
do
π
4
⋅ do
c :=
32
2
Assume failure at region BC. T BC = − T1 + TA
Solution: J :=
(
)
Aplying the torsion formula
τ allow =
(TBC)⋅ c
J
τ allow =
T 1 := −
(
)
− T1 + TA ⋅ c
J
τ allow⋅ J
− TA
T 1 = −215.26 N⋅ m Ans
c
Internal Torque : Maximum torque occurs ithin region BC as indicated on the torque diagram.
Maximum shear srtesses at Other Regions :
T CD := TA + T 1 + T C
T CD = −98.26 N⋅ m
τ max.CD :=
T DE := TCD + TD
(TCD)⋅ c
J
τ max.CD = −4.00 MPa
Ans
τ max.DE = −2.58 MPa
Ans
T DE = −63.26 N⋅ m
τ max.DE :=
(TDE)⋅ c
J
Let a := 1m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
1
T'1 x1 := T A⋅
N⋅ m
( )
x4 := 3a , 1.01 ⋅ ( 3a) .. 4a
1
T'2 x2 := TA + T 1 ⋅
N⋅ m
( ) (
)
1
T'3 x3 := TA + T 1 + T C ⋅
N⋅ m
( ) (
)
1
T'4 x3 := TA + T 1 + T C + T D ⋅
N⋅ m
( ) (
)
100
( )
0
T'2 ( x 2 )
T'3 ( x 3 )
T'4 ( x 4 ) 100
Torque (Nm)
T'1 x 1
200
0
1
2
x1 , x2 , x3 , x4
Distance (m)
3
4
Problem 5-15
The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress in
the shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stress
is maximum. Set T1 = 20 N·m.
Given:
do := 50mm
T 1 := −20N⋅ m
T A := 68N⋅ m
T C := 49N⋅ m
π
4
⋅ do
32
Maximum Torque :
Solution: J :=
c :=
T D := 35N⋅ m
do
2
Maximum torque occurs ithin region DE
as indicated on the torque diagram.
T DE := TA + T 1 + T C + T D
T DE = 132.00 N⋅ m
T max := T DE
Aplying the torsion formula
τ max :=
(Tmax)⋅ c
τ max = 5.38 MPa
J
Ans
Let a := 1m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
1
T'1 x1 := T A⋅
N⋅ m
( )
x4 := 3a , 1.01 ⋅ ( 3a) .. 4a
1
T'2 x2 := TA + T 1 ⋅
N⋅ m
( ) (
)
1
T'3 x3 := TA + T 1 + T C ⋅
N⋅ m
( ) (
)
1
T'4 x3 := TA + T 1 + T C + T D ⋅
N⋅ m
( ) (
)
150
( )
T'2 ( x 2 )100
T'3 ( x 3 )
T'4 ( x 4 ) 50
Torque (Nm)
T'1 x 1
0
0
1
2
x1 , x2 , x3 , x4
Distance (m)
3
4
Problem 5-16
The motor delivers a torque of 50 N·m to the shaft AB. This torque is transmitted to shaft CD using
the gears at E and F. Determine the equilibrium torque T' on shaft CD and the maximum shear stress
each shaft. The bearings B, C, and D allow free rotation of the shafts.
dAB := 30mm
Given:
RE := 50mm
RF := 125mm
T AB := 50N⋅ m
( )
TAB
dCD := 35mm
Solution:
Equilibrium :
ΣΜE=0;
T AB − F⋅ RE = 0
ΣΜF=0;
T' − F⋅ RF = 0
F :=
RE
T' := F⋅ RF
T' = 125 N⋅ m
Internal Torque :
Ans
As shown in FBD.
π
4
JAB :=
⋅d
32 AB
cAB := 0.5dAB
π
4
JCD :=
⋅ dCD
32
cCD := 0.5dCD
Maximum shear srtesses:
τ max.AB :=
τ max.CD :=
(TAB)⋅ cAB
JAB
T'⋅ cCD
JCD
τ max.AB = 9.43 MPa
Ans
τ max.CD = 14.85 MPa
Ans
Problem 5-17
If the applied torque on shaft CD is T' = 75 N·m, determine the absolute maximum shear stress in eac
shaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed
from rotating.
dEA := 30mm
Given:
dCD := 35mm
RE := 50mm
RF := 125mm
T AB := 50N⋅ m
T' := 75N⋅ m
Solution:
Equilibrium :
ΣΜF=0;
T' − F⋅ RF = 0
ΣΜE=0;
T AB − F⋅ RE − TA = 0
F :=
( )
T'
RF
T A := T AB − F⋅ RE
T A = 20 N⋅ m
Internal Torque :
Ans
As shown in FBD.
T EA := TAB − TA
π
4
JEA :=
⋅ dEA
32
cEA := 0.5dEA
π
4
JCD :=
⋅d
32 CD
cCD := 0.5dCD
T EA = 30.00 N⋅ m
Maximum shear srtesses:
τ max.EA :=
τ max.CD :=
(TEA)⋅ cEA
JEA
T'⋅ cCD
JCD
τ max.EA = 5.66 MPa
Ans
τ max.CD = 8.91 MPa
Ans
Problem 5-18
The copper pipe has an outer diameter of 62.5 mm and an inner diameter of 57.5 mm. If it is tightly
secured to the wall at C and a uniformly distributed torque is applied to it as shown, determine the
shear stress developed at points A and B. These points lie on the pipe's outer surface. Sketch the shear
stress on volume elements located at A and B.
N⋅ m
m
LOA := 300mm LAB := 225mm LBC := 100mm
Given: d := 62.5 mm
o
di := 57.5 mm
q := 625
Solution:
Internal Torque : As shown on FBD
TA = 187.50 N⋅ m
( )
TB := q⋅ ( LOA + LAB) TB = 328.13 N⋅ m
TA := q⋅ LOA
Max. shear stress :
c :=
do
2
τ=
T⋅ c
J
4
4
⎡
⎛ di ⎞ ⎤
π ⎞ ⎢⎛ do ⎞
⎛
⎥
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
τ A :=
τ B :=
TA ⋅ c
J
TB⋅ c
J
τ A = 13.79 MPa
Ans
τ B = 24.14 MPa
Ans
Problem 5-19
The copper pipe has an outer diameter of 62.5 mm and an inner diameter of 57.5 mm. If it is tightly
secured to the wall at C and it is subjected to the uniformly distributed torque along its entire length,
determine the absolute maximum shear stress in the pipe. Discuss the validity of this result.
Given: do := 62.5mm
L OA := 300mm
di := 57.5mm
q := 625
L AB := 225mm
N⋅ m
m
L BC := 100mm
Solution:
Internal Torque : The maximum torque occurs at the
support C
(
T C := q⋅ L OA + LAB + LBC
Max. shear stress :
do
c :=
2
τ=
)
T C = 390.63 N⋅ m
T⋅ c
J
4
4
⎡
⎛ di ⎞ ⎤⎥
π ⎞ ⎢⎛ do ⎞
⎛
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
τ C :=
TC⋅ c
J
τ C = 28.73 MPa
Ans
According to Saint-Venant's principle, application of the torsion formula should be at points
sufficiently removed from the supports or points of concentrated loading.
Problem 5-20
The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings
shown. Determine the shear stress at points A and B, and sketch the shear stress on volume elements
located at these points.
Given: L 1 := 0.3m
L 2 := 0.4m
T 1 := 400N⋅ m
T 2 := −600N⋅ m
do := 60mm
q := 2
kN⋅ m
m
Solution:
Internal Torque : As shown on FBD.
T A := T 1
T A = 400.00 N⋅ m
T B := T1 + T2 + q⋅ L2
Maximum shear stress :
τ B :=
τ=
π ⎛ do ⎞
J := ⋅ ⎜
2 ⎝2⎠
do
c :=
2
τ A :=
T B = 600.00 N⋅ m
T A⋅ c
J
TB⋅ c
J
4
T⋅ c
J
τ A = 9.43 MPa
Ans
τ B = 14.15 MPa
Ans
Problem 5-21
The 60-mm diameter solid shaft is subjected to the distributed and concentrated torsional loadings
shown. Determine the absolute maximum and minimum shear stresses in the shaft and specify their
locations, measured from the fixed end.
Given: L 1 := 0.3m
L 2 := 0.4m
T 1 := 400N⋅ m
T 2 := −600N⋅ m
do := 60mm
q := 2
kN⋅ m
m
Solution:
Internal Torque :
( )
T C := T1 + T2 + q⋅ 2L2
T C = 1400.00 N⋅ m
The maximum torque occurs at the fixed support C.
T max := T C
T max = 1400.00 N⋅ m
The minimum torque occurs in segment loaded with q.
xo
2L2
Shear stress : τ =
=
TC
( )
q⋅ 2L2
T⋅ c
J
τ abs.min :=
τ abs.max :=
xo :=
c :=
do
2
T min⋅ c
J
T max⋅ c
J
TC
T min := 0
xo = 0.700 m
q
⎛ π ⎞⋅ d 4
⎝ 32 ⎠ o
J := ⎜
τ abs.min = 0.00 MPa
Ans
τ abs.max = 33.01 MPa
Ans
According to Saint-Venant's principle, application of the torsion formula should be at points
sufficiently removed from the supports or points of concentrated loading. Therefore, the
absolute τmax is not valid.
( )
1
T a ( x1) := ( −T C + q⋅ x1) ⋅
N⋅ m
x1 := 0 , 0.01 ⋅ 2L2 .. 2L2
( ) (
)
x2 := 2L 2 , 1.01 ⋅ 2L 2 .. 2L2 + 2L 1
1
T b x2 := −TC + 2q⋅ L 2 + T 2 ⋅
N⋅ m
( ) (
)
Torque (Nm)
0
( )
500
Tb ( x 2 )
Ta x 1
1000
1500
0
0.2
0.4
0.6
0.8
x1 , x2
Distance (m)
1
1.2
Problem 5-22
The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the
required diameter d of the shaft if the allowable shear stress for the material is τallow = 175 MPa.
Given: L1 := 0.3m
L2 := 0.4m
T1 := 400N⋅ m
T2 := −600N⋅ m
τ allow := 175MPa
q := 2
kN⋅ m
m
Solution:
Internal Torque :
( )
TC := T 1 + T 2 + q⋅ 2L 2
TC = 1400.00 N⋅ m
The maximum torque occurs at the fixed support C.
Tmax := TC
Allowable Shear stress :
τ allow =
3
do :=
Tmax = 1400.00 N⋅ m
τ=
16Tmax
π ⋅ do
T⋅ c
J
c=
do
2
⎛ π ⎞⋅ d 4
⎝ 32 ⎠ o
J= ⎜
3
16T max
do = 34.41 mm
π ⋅ τ allow
Ans
According to Saint-Venant's principle, application of the torsion formula should be at points
sufficiently removed from the supports or points of concentrated loading. Therefore, the
above analysis is not valid.
( )
1
Ta ( x1) := ( −TC + q⋅ x1) ⋅
N⋅ m
x1 := 0 , 0.01 ⋅ 2L 2 .. 2L 2
( ) (
)
x2 := 2L2 , 1.01⋅ 2L2 .. 2L 2 + 2L1
1
Tb x2 := −T C + 2q⋅ L2 + T2 ⋅
N⋅ m
( ) (
)
Torque (Nm)
0
( )
500
Tb ( x 2 )
Ta x 1
1000
1500
0
0.2
0.4
0.6
0.8
x1 , x2
Distance (m)
1
1.2
Problem 5-23
The steel shafts are connected together using a fillet weld as shown. Determine the average shear
stress in the weld along section aa if the torque applied to the shafts is T = 60 N·m. Note: The critical
section where the weld fails is along section aa.
Given: do := 50mm
a := 12mm
T := 60N⋅ m
θ a := 45deg
Solution:
The maen radius of weld is:
do + a
rweld :=
2
rweld = 31.00 mm
Shear stress :
V :=
T
rweld
τ avg :=
(
V
Aweld
τ avg = 1.17 MPa
( ))
Aweld := 2π ⋅ rweld⋅ a⋅ sin θ a
Ans
Problem 5-24
The rod has a diameter of 12 mm and a weight of 80 N/m. Determine the maximum torsional stress in
the rod at a section located at A due to the rod's weight.
Given: d := 12mm
L x := 0.9m
N
m
L A := 0.3m
L y := 0.9m
L z := 0.3m
w := 80
Solution:
Equilibrium :
Σ Mx = 0;
(
)
( )
T A − w⋅ L y⋅ 0.5L y − w⋅ L z⋅ Ly = 0
(
)
( )
T A := w⋅ Ly⋅ 0.5 ⋅ L y + w⋅ L z⋅ Ly
T A = 54.00 N⋅ m
Max. shear stress :
τ=
T⋅ c
J
c :=
d
2
τ A :=
J :=
T A⋅ c
J
π ⎛ d⎞
⋅⎜
2 ⎝ 2⎠
4
τ A = 159.15 MPa
Ans
Problem 5-25
Solve Prob. 5-24 for the maximum torsional stress at B.
Given: d := 12mm
L x := 0.9m
N
m
L A := 0.3m
L y := 0.9m
L z := 0.3m
w := 80
Solution:
Equilibrium :
Σ Mx = 0;
(
)
( )
T B − w⋅ Ly⋅ 0.5Ly − w⋅ Lz⋅ L y = 0
(
)
( )
T B := w⋅ L y⋅ 0.5 ⋅ Ly + w⋅ Lz⋅ L y
T B = 54.00 N⋅ m
Max. shear stress :
τ=
T⋅ c
J
π ⎛ d⎞
J := ⋅ ⎜
2 ⎝ 2⎠
d
c :=
2
τ B :=
TB⋅ c
J
4
τ B = 159.15 MPa
Ans
Problem 5-26
Consider the general problem of a circular shaft made from m segments each having a radius of cm. If
there are n torques on the shaft as shown, write a computer program that can be used to determine the
maximum shearing stress at any specified location x along the shaft. Show an application of the
program using the values L1 = 0.6 m, c1 = 50 mm, L2 = 1.2 m, c2 = 25 mm, T1 = 1200 N·m, d1 = 0, T2
= -900 N·m, d2 = 1.5 m.
Problem 5-27
The wooden post, which is half buried in the ground, is subjected to a torsional moment of 50 N·m
that causes the post to rotate at constant angular velocity.This moment is resisted by a linear
distribution of torque developed by soil friction, which varies from zero at the ground to t0 N·m/m at
its base. Determine the equilibrium value for t0 , and then calculate the shear stress at points A and B,
which lie on the outer surface of the post.
Given: do := 100mm
L := 0.75m
L AB := 0.5 ⋅ m
T := 50N⋅ m
Solution:
Equilibrium :
(0.5to)⋅ L − T = 0
Σ Mz = 0;
to := 2
T
L
to = 133.33
N⋅ m
m
Internal Torque : As shown on FBD.
T A := T
⎛ LAB ⎞
T B := T − 0.5to⋅ ⎜
⋅L
⎝ L ⎠ AB
Maximum Shear Stress :
c :=
do
2
τ A :=
τ B :=
J :=
T A⋅ c
J
TB⋅ c
J
τ=
T⋅ c
J
T A = 50.00 N⋅ m
T B = 27.78 N⋅ m
π
4
⋅ do
32
τ A = 0.255 MPa
Ans
τ B = 0.141 MPa
Ans
Problem 5-28
A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held
fixed and a torque T is applied to the shaft, determine the maximum shear stress in the rubber.
Solution:
Shear stress :
τ=
F
A
F=
T
r
A = 2π ⋅ r⋅ h
T
τ=
2
2π ⋅ r ⋅ h
Shear stress is maximm when r is the smallest, i.e. r = ri.
Hence,
τ max =
T
2
2π ⋅ ri ⋅ h
Ans
Problem 5-29
The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a
variable torque described by the function t = (25x ex2) N·m/m, where x is in meters. Determine the
minimum torque T0 needed to overcome friction and cause it to twist. Also, determine the absolute
maximum stress in the shaft.
2
x N⋅ m
Given: do := 80mm
L := 2m
t = 25x⋅ e
m
unit := N⋅ m
Solution:
L
Equilibrium :
Σ Mz = 0;
⌠m
2
⎮
x
T o := unit⋅ ⎮ 25x⋅ e dx
⌡0
T o = 669.98 N⋅ m
Maximum Shear Stress :
c :=
do
2
τ abs.max :=
J :=
τ=
T⋅ c
J
π
4
⋅ do
32
T o⋅ c
J
τ abs.max = 6.664 MPa
Ans
Ans
Problem 5-30
The solid shaft has a linear taper from rA at one end to rB at the other. Derive an equation that gives th
maximum shear stress in the shaft at a location x along the shaft's axis.
Problem 5-31
When drilling a well at constant angular velocity, the bottom end of the drill pipe encounters a torsion
resistance TA. Also, soil along the sides of the pipe creates a distributed frictional torque along its
length, varying uniformly from zero at the surface B to tA at A. Determine the minimum torque TB tha
must be supplied by the drive unit to overcome the resisting torques, and compute the maximum shea
stress in the pipe. The pipe has an outer radius ro and an inner radius ri .
Problem 5-32
The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow= 56
MPa. If the outer diameter of the shaft is 62.5 mm and the engine delivers 165 kW to the shaft when
is turning at 1140 rev/min, determine the minimum required thickness of the shaft's wall.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
do := 62.5mm
ω := 1140⋅ rpm
P := 165kW
τ allow := 56MPa
ω = 119.38
T :=
rad
s
P
T = 1382.14 J
ω
c :=
Max. shear stress :
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J= ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
0.25
⎤
⎡⎛ d ⎞ 4
2
T
⋅
c
⎢ o
⎥
⎛ ⎞⎛
⎞
di := 2 ⎜
⎢⎝ 2 ⎠ − ⎜⎝ π ⎠ ⋅ ⎜ τ
⎥
⎣
⎝ allow ⎠⎦
T⋅ c
τ allow =
J
di = 52.16 mm
t :=
do − di
2
t = 5.17 mm
Ans
Problem 5-33
The drive shaft AB of an automobile is to be designed as a thin-walled tube. The engine delivers 125
kW when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft's wall if
the shaft's outer diameter is 62.5 mm.The material has an allowable shear stress of τallow = 50 MPa.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
do := 62.5mm
ω := 1500⋅ rpm
P := 125kW
τ allow := 50MPa
ω = 157.08
T :=
rad
s
P
T = 795.77 J
ω
c :=
Max. shear stress :
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J= ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
0.25
⎤
⎡⎛ d ⎞ 4
⎢ o
⎛ 2 ⎞ ⎛ T⋅ c ⎞⎥
di := 2 ⎜
⎢⎝ 2 ⎠ − ⎜⎝ π ⎠ ⋅ ⎜ τ
⎥
⎣
⎝ allow ⎠⎦
T⋅ c
τ allow =
J
di = 56.50 mm
t :=
do − di
2
t = 2.998 mm
Ans
Problem 5-34
The gear motor can develop 100 W when it turns at 300 rev/min. If the shaft has a diameter of 12 m,
determine the maximum shear stress that will be developed in the shaft.
Unit used:
Given:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
d := 12mm
ω := 300⋅ rpm
P := 100W
Solution:
ω = 31.42
T :=
rad
s
P
T = 3.183 N⋅ m
ω
Max. shear stress :
d
c :=
2
τ max :=
π ⎛ d⎞
J := ⋅ ⎜
2 ⎝ 2⎠
4
T⋅ c
J
τ max = 9.382 MPa
Ans
Problem 5-35
The gear motor can develop 100 W when it turns at 80 rev/min. If the allowable shear stress for the
shaft is τallow = 28 MPa, determine the smallest diameter of the shaft to the nearest multiples of 5mm
that can be used.
Unit used:
Given:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
P := 100W
ω := 80⋅ rpm
τ allow := 28MPa
Solution:
ω = 8.38
T :=
rad
s
P
T = 11.937 N⋅ m
ω
Max. shear stress :
π ⎛ d⎞
J = ⋅⎜
2 ⎝ 2⎠
d
c=
2
τ allow =
4
T⋅ c
J
4
T⋅ d
π ⎛ d⎞
⋅⎜
=
2 ⎝ 2⎠
2⋅ τ allow
1
3
⎛ 16T ⎞
d := ⎜
⎝ π ⋅ τ allow ⎠
d = 12.95 mm
Use
d = 15mm
Ans
Problem 5-36
The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 42 MPa
If the outer diameter is 75 mm and the engine delivers 145 kW to the shaft when it is turning at 1250
rev/min., determine the minimum required thickness of the shaft's wall.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
do := 75mm
ω := 1250⋅ rpm
P := 145kW
τ allow := 42MPa
ω = 130.90
T :=
rad
s
P
T = 1107.718 N⋅ m
ω
c :=
Max. shear stress :
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J= ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
0.25
⎤
⎡⎛ d ⎞ 4
⎢ o
⎛ 2 ⎞ ⎛ T⋅ c ⎞⎥
di := 2 ⎜
⎢⎝ 2 ⎠ − ⎜⎝ π ⎠ ⋅ ⎜ τ
⎥
⎣
⎝ allow ⎠⎦
T⋅ c
τ allow =
J
di = 68.15 mm
t :=
do − di
2
t = 3.427 mm
Ans
Problem 5-37
The 2.5-kW reducer motor can turn at 330 rev/min. If the shaft has a diameter of 20 mm, determine
the maximum shear stress that will be developed in the shaft.
Unit used:
Given:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
d := 20mm
ω := 330⋅ rpm
P := 2.5kW
Solution:
ω = 34.56
T :=
rad
s
P
T = 72.343 N⋅ m
ω
Max. shear stress :
c :=
d
2
τ max :=
J :=
π ⎛ d⎞
⋅⎜
2 ⎝ 2⎠
4
T⋅ c
J
τ max = 46.055 MPa
Ans
Problem 5-38
The 2.5-kW reducer motor can turn at 330 rev/min. If the allowable shear stress for the shaft is τallow
= 56 MPa, determine the smallest diameter of the shaft to the nearest multiples of 5mm that can be
used.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
d := 20mm
ω := 330⋅ rpm
P := 2.5kW
τ allow := 56MPa
ω = 34.56
T :=
Max. shear stress :
rad
s
P
T = 72.343 N⋅ m
ω
π ⎛ d⎞
J = ⋅⎜
2 ⎝ 2⎠
d
c=
2
τ allow =
4
T⋅ c
J
4
T⋅ d
π ⎛ d⎞
⋅⎜
=
2 ⎝ 2⎠
2⋅ τ allow
1
3
⎛ 16T ⎞
d := ⎜
⎝ π ⋅ τ allow ⎠
d = 18.74 mm
Use
d = 20mm
Ans
Problem 5-39
The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It
is coupled to a motor at C, which delivers 3 kW of power to the shaft while it is turning at 50 rev/s. If
gears A and B remove 1 kW and 2 kW, respectively, determine the maximum shear stress developed i
the shaft within regions AB and BC. The shaft is free to turn in its support bearings D and E.
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
Unit used:
Given:
do := 25mm
ω := 50⋅ ( 60) ⋅ rpm
PA := −1kW
Po := 3kW
PB := −2kW
Solution:
ω = 314.16
T C :=
T A :=
Po
ω
PA
Po
rad
s
T C = 9.549 N⋅ m
⋅ TC
T A = 3.183 N⋅ m
Maximum Shear Stress :
do
c :=
2
τ AB.max :=
τ BC.max :=
π ⎛ do ⎞
J := ⋅ ⎜
2 ⎝2⎠
TA⋅ c
J
T C⋅ c
J
4
τ AB.max = 1.038 MPa
Ans
τ BC.max = 3.113 MPa
Ans
Problem 5-40
A ship has a propeller drive shaft that is turning at 1500 rev/min. while developing 1500 kW. If it is 2
m long and has a diameter of 100 mm, determine the maximum shear stress in the shaft caused by
torsion.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
d := 100mm
ω := 1500⋅ rpm
P := 1500kW
L := 2.4m
ω = 157.08
T :=
rad
s
P
T = 9549.297 N⋅ m
ω
Max. shear stress :
d
c :=
2
τ max :=
π ⎛ d⎞
J := ⋅ ⎜
2 ⎝ 2⎠
4
T⋅ c
J
τ max = 48.634 MPa
Ans
Problem 5-41
The motor A develops a power of 300 W and turns its connected pulley at 90 rev/min. Determine the
required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is τallow =
85 MPa.
2π rad
Unit used: rpm := ⎛⎜ ⎞
⎝ 60 ⎠ s
Given: rA := 60mm
ωA := 90⋅ rpm
rB := 150mm
P := 300W
τ allow := 85MPa
Solution:
⎛ rA ⎞
ωB := ωA⋅ ⎜
ωB = 3.77
⎝ rB ⎠
T A :=
T B :=
P
ωA
P
ωB
rad
s
T A = 31.831 N⋅ m
T B = 79.577 N⋅ m
Allowable Shear Stress :
For shaft A:
τ allow =
τ allow =
3
dA :=
For shaft B:
c=
J
dA
2
τ allow =
3
J=
π
4
⋅ dA
32
16T A
3
π ⋅ dA
16TA
dA = 12.40 mm
π ⋅ τ allow
τ allow =
dB :=
T A⋅ c
T B⋅ c
c=
J
dB
2
J=
Ans
π
4
⋅ dB
32
16T B
3
π ⋅ dB
16TB
π ⋅ τ allow
dB = 16.83 mm
Ans
Problem 5-42
The motor delivers 400 kW to the steel shaft AB, which is tubular and has an outer diameter of 50 mm
and an inner diameter of 46 mm. Determine the smallest angular velocity at which it can rotate if the
allowable shear stress for material is τallow = 175 MPa.
Unit used:
Given:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
do := 50mm
di := 46mm
P := 400kW
τ allow := 175MPa
Solution:
c :=
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
τ allow =
T :=
T⋅ c
J
(
)
J⋅ τ allow
c
T = 1218.13 N⋅ m
ω :=
P
T
ω = 328.37
rad
s
ω = 3135.714 rpm
Ans
Problem 5-43
The motor delivers 40 kW while turning at a constant rate of 1350 rpm at A. Using the belt and pulley
system this loading is delivered to the steel blower shaft BC. Determine to the nearest multiples of 5mm
the smallest diameter of this shaft if the allowable shear stress for steel if τallow = 84 MPa.
Unit used:
Given:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
rA := 100mm
rB := 200mm
P := 40kW
ω := 1350⋅ rpm
τ allow := 84MPa
Solution:
ω = 141.37
T A :=
rad
s
P
T A = 282.942 N⋅ m
ω
T A = 2rA⋅ ( F' − F)
T B = 2rB⋅ ( F − F')
⎛ rB ⎞
T B := ⎜
⋅ TA
Max. shear stress :
c=
⎝ rA ⎠
T B = 565.88 N⋅ m
d
2
τ allow =
J=
π ⎛ d⎞
⋅⎜
2 ⎝ 2⎠
4
T B⋅ c
J
4
T B⋅ d
π ⎛ d⎞
⋅⎜
=
2 ⎝ 2⎠
2⋅ τ allow
1
⎛ 16TB ⎞
d := ⎜
⎝ π ⋅ τ allow ⎠
3
d = 32.49 mm
Use
d = 35mm
Ans
Problem 5-44
The propellers of a ship are connected to a solid A-36 steel shaft that is 60 m long and has an outer
diameter of 340 mm and inner diameter of 260 mm. If the power output is 4.5 MW when the shaft
rotates at 20 rad/s, determine the maximum torsional stress in the shaft and its angle of twist.
Given:
do := 340mm
di := 260mm
L := 60m
P := 4500kW
G := 75GPa
ω := 20
Solution:
T :=
P
ω
T = 225 kN⋅ m
Maximum Shear Stress :
do
π ⎛ 4
4
c :=
J :=
⋅ ⎝ do − di ⎞⎠
2
32
T⋅ c
τ max :=
τ max = 44.31 MPa
J
Angle of Twist :
φ :=
T⋅ L
G⋅ J
φ = 0.2085 rad
φ = 11.946 deg
Ans
Ans
rad
s
Problem 5-45
A shaft is subjected to a torque T. Compare the effectiveness of using the tube shown in the figure
with that of a solid section of radius c. To do this, compute the percent increase in torsional stress and
angle of twist per unit length for the tube versus the solid section.
Given:
ro = c
ri = 0.5c
Solution:
Maximum Shear Stress :
For solid shaft::
τ s.max =
τ s.max =
T⋅ c
J
2T
c = ro
π 4
Js = ⋅ ro
2
c = ro
π
4
4
Jt = ⋅ ⎛⎝ ro − ri ⎞⎠
2
3
π ⋅ ro
For the tube::
τ t.max =
τ t.max =
T⋅ c
J
15π 4
Jt =
⋅r
32 o
32T
3
15π ⋅ ro
% increase in shear stress:
τ% =
τ t.max − τ s.max
τ s.max
32
τ % :=
Angle of Twist :
For solid shaft::
φs =
T⋅ L
G⋅ Js
For the tube::
φt =
T⋅ L
G⋅ Jt
% increase in angle of twist:
φ% =
−2
15
⋅ 100
2
τ % = 6.67
φ t − τ s.max
τ s.max
π
φ % :=
⋅ 100
2
−
⋅ 100
φ% =
Ans
1
1
−
Js Jt
1
⋅ 100
Js
15π
32
15π
32
⋅ 100
φ % = 6.67
Ans
Problem 5-46
The tubular drive shaft for the propeller of a hover-craft is 6 m long. If the motor delivers 4 MW of
power to the shaft when the propellers rotate at 25 rad/s, determine the required inner diameter of the
shaft if the outer diameter is 250 mm.What is the angle of twist of the shaft when it is operating? Tak
τallow = 90 MPa and G = 75 GPa.
Given:
do := 250mm
L := 6m
P := 4000kW
τ allow := 90MPa
G := 75GPa
ω := 25
Solution:
T :=
P
rad
s
T = 160 kN⋅ m
ω
Maximum Shear Stress :
τ allow =
τ allow =
T⋅ c
J
do
2
J=
π ⎛ 4
4
⋅ d − di ⎞⎠
32 ⎝ o
16T ⋅ do
4
π ⋅ ⎛⎝ do − di ⎠
Angle of Twist :
φ :=
c :=
4
J :=
T⋅ L
G⋅ J
4⎞
π ⎛ 4
4
⋅ d − di ⎞⎠
32 ⎝ o
φ = 0.05760 rad
φ = 3.30 deg
di :=
Ans
4
do −
16T ⋅ do
π ⋅ τ allow
di = 201.3 mm
Ans
Problem 5-47
The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth
bearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N·m torques,
determine the angle of twist of gear A relative to gear D. The tubes have an outer diameter of 30 mm
and an inner diameter of 20 mm. The solid section has a diameter of 40 mm.
Given: do := 30mm
di := 20mm
ds := 40mm
L AB := 0.4m
L BC := 0.25m
L CD := 0.4m
T := 85N⋅ m
G := 75GPa
Solution:
T AB := T
T BC := T
T CD := T
Angle of Twist :
π ⎛ 4
4
JAB :=
⋅ ⎝ do − di ⎞⎠
32
φ := 2
T AB⋅ LAB
G⋅ JAB
+
TBC⋅ L BC
G⋅ JBC
φ = 0.01534 rad
φ = 0.879 deg
π
4
JBC :=
⋅ ds
32
Ans
JCD := JAB
Problem 5-48
The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth
bearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N·m torques,
determine the angle of twist of the end B of the solid section relative to end C. The tubes have an oute
diameter of 30 mm and an inner diameter of 20 mm.The solid section has a diameter of 40 mm.
Given: do := 30mm
di := 20mm
ds := 40mm
L AB := 0.4m
L BC := 0.25m
L CD := 0.4m
T := 85N⋅ m
G := 75GPa
Solution:
T AB := T
T BC := T
T CD := T
Angle of Twist :
φ :=
π
4
JBC :=
⋅ ds
32
TBC⋅ L BC
G⋅ JBC
φ = 0.001127 rad
φ = 0.0646 deg
Ans
Problem 5-49
The hydrofoil boat has an A-36 steel propeller shaft that is 30 m long. It is connected to an in-line
diesel engine that delivers a maximum power of 2000 kW and causes the shaft to rotate at 1700 rpm. If
the outer diameter of the shaft is 200 mm and the wall thickness is 10 mm, determine the maximum
shear stress developed in the shaft. Also, what is the “wind up,” or angle of twist in the shaft at full
power?
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
do := 200mm
t := 10mm
P := 2000kW
ω := 1700rpm
G := 75GPa
L := 30m
ω = 178.02
T :=
rad
s
P
T = 11234.467 N⋅ m
ω
Max. shear stress :
di := do − 2t
c :=
do
2
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
τ max :=
T⋅ c
J
τ max = 20.797 MPa
Ans
Angle of Twist :
φ :=
T⋅ L
G⋅ J
φ = 0.0832 rad
φ = 4.766 deg
Ans
Problem 5-50
The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown.
Determine the angle of twist of gear C with respect to gear D. The shaft has a diameter of 40 mm.
Given: L AC := 0.3m
L CD := 0.4m
do := 40mm
G := 75GPa
T A := −300N⋅ m
T C := 500N⋅ m
T D := 200N⋅ m
T E := −400N⋅ m
Solution:
T AC := TA
T CD := TA + T C
T DE := TA + T C + T D
Angle of Twist :
φ C_D :=
J :=
π
4
⋅ do
32
TCD⋅ L CD
G⋅ J
φ C_D = 0.004244 rad
φ C_D = 0.243 deg
Ans
L DE := 0.5m
Problem 5-51
The 20-mm-diameter A-36 steel shaft is subjected to the torques shown. Determine the angle of twist
of the end B.
Given: L AD := 0.2m
L DC := 0.6m
L CB := 0.8m
T B := 80N⋅ m
T C := −20N⋅ m
T D := 30N⋅ m
do := 20mm
G := 75GPa
Solution:
T CB := TB
T DC := TB + TC
T AD := T B + T C + T D
Angle of Twist :
φ :=
TCB⋅ L CB
G⋅ J
J :=
+
π
4
⋅ do
32
T DC⋅ LDC
G⋅ J
φ = 0.100162 rad
φ = 5.739 deg
Ans
+
TAD⋅ LAD
G⋅ J
Problem 5-52
The 8-mm-diameter A-36 bolt is screwed tightly into a block at A. Determine the couple forces F that
should be applied to the wrench so that the maximum shear stress in the bolt becomes 18 MPa. Also,
compute the corresponding displacement of each force F needed to cause this stress. Assume that the
wrench is rigid.
Given:
do := 8mm
L := 80mm
a := 150mm
G := 75GPa
τ allow := 18MPa
Solution:
Allowable Shear Stress :
c :=
do
2
τ allow =
⎛ π ⎞⋅ d 4
⎝ 32 ⎠ o
J := ⎜
T⋅ c
J
T :=
Equilibrium :
T − F ( 2a) = 0
F :=
Angle of Twist :
φ :=
T⋅ L
G⋅ J
φ = 0.00480 rad
Displacement :
s' := a⋅ φ
s' = 0.720 mm
Ans
τ allow⋅ J
c
T
2a
T = 1.8096 N⋅ m
F = 6.03 N
Ans
Problem 5-53
The turbine develops 150 kW of power, which is transmitted to the gears such that C receives 70%
and D receives 30%. If the rotation of the 100-mm-diameter A-36 steel shaft is ω = 800 rev/min.,
determine the absolute maximum shear stress in the shaft and the angle of twist of end E of the shaft
relative to B. The journal bearing at E allows the shaft to turn freely about its axis.
Unit used: rpm := ⎛⎜
2π ⎞ rad
⎝ 60 ⎠ s
Given: do := 100mm
ω := 800⋅ rpm
L BC := 3m
L CD := 4m
L DE := 2m
P := 150kW
G := 75GPa
T C = 0.7T
Solution: ω = 83.78
T :=
T D = 0.3T
rad
s
P
ω
T C := 0.7T
T D := 0.3T
T = 1.790 kN⋅ m
T C = 1.253 kN⋅ m
T D = 0.537 kN⋅ m
T BC := T
T CD := 0.3T
T DE := 0
Maximum Shear Stress :
Maximum torque occurs in region BC.
do
π
4
c :=
J :=
⋅ do
2
32
τ max :=
T⋅ c
J
τ max = 9.119 MPa
Ans
Angle of Twist :
φ E_B :=
T BC⋅ LBC
G⋅ J
+
TCD⋅ L CD
G⋅ J
φ E_B = 0.010213 rad
φ E_B = 0.5852 deg
Ans
+
T DE⋅ LDE
G⋅ J
Problem 5-54
The turbine develops 150 kW of power, which is transmitted to the gears such that both C and D
receive an equal amount. If the rotation of the 100-mm-diameter A-36 steel shaft is ω = 500 rev/min.
determine the absolute maximum shear stress in the shaft and the rotation of end B of the shaft relativ
to E. The journal bearing at C allows the shaft to turn freely about its axis.
Unit used: rpm := ⎛⎜
2π ⎞ rad
⎝ 60 ⎠ s
Given: do := 100mm
ω := 500⋅ rpm
L BC := 3m
L CD := 4m
L DE := 2m
P := 150kW
G := 75GPa
T C = 0.5T
Solution: ω = 52.36
T :=
T D = 0.5T
rad
s
P
ω
T C := 0.5T
T D := 0.5T
T = 2.865 kN⋅ m
T C = 1.432 kN⋅ m
T D = 1.432 kN⋅ m
T BC := T
T CD := 0.5T
T DE := 0
Maximum Shear Stress :
Maximum torque occurs in region BC.
do
π
4
c :=
J :=
⋅d
2
32 o
τ max :=
T⋅ c
J
τ max = 14.59 MPa
Ans
Angle of Twist :
φ B_E :=
T BC⋅ LBC
G⋅ J
+
TCD⋅ L CD
G⋅ J
φ B_E = 0.019454 rad
φ B_E = 1.1146 deg
Ans
+
T DE⋅ LDE
G⋅ J
Problem 5-55
The motor delivers 33 kW to the 304 stainless steel shaft while it rotates at 20 Hz. The shaft is
supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D
fixed to the shaft remove 20 kW and 12 kW, respectively. Determine the diameter of the shaft to the
nearest mm if the allowable shear stress is τallow = 56 MPa and the allowable angle of twist of C with
respect to D is 0.20°.
Given: LAC := 250mm
LCD := 200mm
LDB := 150mm
P := 32kW
PC := −20kW
PD := −12kW
f := 20Hz
G := 75GPa
φ allow := 0.20deg
τ allow := 56MPa
Solution:
ω := 2⋅ π ⋅ f
TAC :=
TCD :=
ω = 125.6637
P
ω
P + PC
ω
(
TAC = 254.648 N⋅ m
TCD = 95.493 N⋅ m
)
Tmax := max TAC , TCD
Max. shear stress :
c=
d
2
J=
rad
s
Tmax = 254.6479 N⋅ m
Assume failure due to shear stress
π ⎛ d⎞
⋅⎜
2 ⎝ 2⎠
4
τ allow =
Tmax⋅ c
J
1
4
Tmax⋅ d
π ⎛ d⎞
⋅⎜
=
2 ⎝ 2⎠
2⋅ τ allow
Angle of Twist :
⎛ 16Tmax ⎞
d := ⎜
⎝ π ⋅ τ allow ⎠
3
d = 28.5041 mm
Assume failure due to angle of twist limitation (occured between C and D)
π ⎛ d⎞
J = ⋅⎜
2 ⎝ 2⎠
4
φ allow =
T CD⋅ LCD
G⋅ J
4
TCD⋅ L CD
π ⎛ d⎞
⋅⎜
=
2 ⎝ 2⎠
G⋅ φ allow
Use
d = 30mm
Ans
φ allow = 0.003491 rad
⎛ 2TCD⋅ LCD ⎞
d := 2 ⎜
⎝ π ⋅ G⋅ φ allow ⎠
0.25
d = 29.3601 mm
Problem 5-56
The motor delivers 33 kW to the 304 stainless steel solid shaft while it rotates at 20 Hz. The shaft has a
diameter of 37.5 mm and is supported on smooth bearings at A and B, which allow free rotation of the
shaft. The gears C and D fixed to the shaft remove 20 kW and 12 kW, respectively. Determine the
absolute maximum stress in the shaft and the angle of twist of gear C with respect to gear D.
Given: L AC := 250mm
L CD := 200mm
L DB := 150mm
P := 32kW
PC := −20kW
PD := −12kW
f := 20Hz
G := 75GPa
d := 37.5mm
Solution:
ω := 2⋅ π ⋅ f
T AC :=
T CD :=
ω = 125.66
P
rad
s
T AC = 254.648 N⋅ m
ω
P + PC
T CD = 95.493 N⋅ m
ω
(
)
T max := max T AC , T CD
Max. shear stress :
d
c :=
2
T max = 254.65 N⋅ m
Maximum shear stress occured between A and C
π ⎛ d⎞
J := ⋅ ⎜
2 ⎝ 2⎠
4
τ max :=
T max⋅ c
J
τ max = 24.59 MPa
Angle of Twist :
φ CD :=
Ans
T CD⋅ LCD
G⋅ J
φ CD = 0.001312 rad
φ CD = 0.075152 deg
Ans
Problem 5-57
The motor produces a torque of T = 20 N·m on gear A. If gear C is suddenly locked so it does not
turn, yet B can freely turn, determine the angle of twist of F with respect to E and F with respect to D
of the L2-steel shaft, which has an inner diameter of 30 mm and an outer diameter of 50 mm. Also,
calculate the absolute maximum shear stress in the shaft. The shaft is supported on journal bearings at
G and H.
Given: do := 50mm
di := 30mm
rA := 30mm
rF := 100mm
G := 75GPa
T := 20N⋅ m
L EF := 0.6m
Solution:
Equilibrium :
T − F⋅ r A = 0
F :=
T
rA
F = 666.67 N
T' − F⋅ rF = 0
T' := F⋅ rF
T' = 66.67 N⋅ m
Angle of Twist :
J :=
π ⎛ 4
4
⋅ d − di ⎞⎠
32 ⎝ o
Since shaft is held fixed at C, the torque is only in region EF of the shaft.
φ F_E :=
T'⋅ L EF
G⋅ J
φ F_E = 9.986 × 10
−4
rad
Ans
rad
Ans
Since the torque in region ED is zero,
φ F_D := φ F_E
φ F_D = 9.986 × 10
−4
Maximum Shear Stress :
c :=
do
2
τ max :=
T'⋅ c
J
τ max = 3.121 MPa
Ans
Problem 5-58
The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by
bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of
twist of end B when the torques are applied to the assembly as shown.
Given: L DH := 250mm
L HE := 750mm
L AG := 200mm
L GF := 250mm
L FB := 300mm
d := 25mm
rE := 150mm
rF := 100mm
T H := 120N⋅ m
T G := 60N⋅ m
G := 75GPa
Solution:
AS shown on FBD
Internal Torque :
At F : T F =TG
P :=
TG
( )
At E :
P = 600 N
rF
T E := −P⋅ rE
T E = −90 N⋅ m
Angle of Twist :
J :=
π ⎛ d⎞
⋅⎜
2 ⎝ 2⎠
φ E :=
4
T H⋅ L DH
G⋅ J
+
(
TE ⋅ LDH + L HE
)
G⋅ J
φ E = −0.020861 rad
( )
( )
φ E ⋅ r E = − φ F⋅ r F
φ F := −
rE
rF
⋅ φE
φ F = 0.031291 rad
Since there is no torque applied between F and B,
φ B := φ F
φ B = 0.031291 rad
φ B = 1.793 deg
Ans
Problem 5-59
The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by
bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of
twist of end A when the torques are applied to the assembly as shown.
Given: L DH := 250mm
L HE := 750mm
L AG := 200mm
L GF := 250mm
L FB := 300mm
d := 25mm
rE := 150mm
rF := 100mm
T H := 120N⋅ m
T G := 60N⋅ m
G := 75GPa
Solution:
AS shown on FBD
Internal Torque :
At F : T F =TG
TG
P :=
( )
At E :
P = 600 N
rF
T E := −P⋅ rE
T E = −90 N⋅ m
Angle of Twist :
π ⎛ d⎞
J := ⋅ ⎜
2 ⎝ 2⎠
φ E :=
4
T H⋅ L DH
G⋅ J
+
(
TE ⋅ LDH + L HE
)
G⋅ J
φ E = −0.020861 rad
( )
( )
φ E ⋅ r E = − φ F⋅ r F
φ F := −
rE
rF
⋅ φE
φ F = 0.031291 rad
Since there is no torque applied between A and G,
φ A_F :=
T G⋅ L GF
G⋅ J
φ A := φ A_F + φ F
φ A_F = 0.005215 rad
φ A = 0.036506 rad
φ A = 2.092 deg
Ans
Problem 5-60
Consider the general problem of a circular shaft made from m segments, each having a radius of cm
and shearing modulus Gm. If there are n torques on the shaft as shown, write a computer program that
can be used to determine the angle of twist of its end A. Show an application of the program using the
values L1 = 0.5 m, c1 = 0.02 m, G1 = 30 GPa, L2 = 1.5 m, c2 = 0.05 m, G2 = 15 GPa, T1 = -450 N·m,
d1 = 0.25 m, T2 = 600 N·m , d2 = 0.8 m.
Problem 5-61
The 30-mm-diameter shafts are made of L2 tool steel and are supported on journal bearings that allow
the shaft to rotate freely. If the motor at A develops a torque of T = 45 N·m on the shaft AB, while the
turbine at E is fixed from turning, determine the amount of rotation of gears B and C.
Given: L AB := 1.5m
L DC := 0.5m
L CE := 0.75m
T A := 45N⋅ m
rB := 50mm
rC := 75mm
do := 30mm
G := 75GPa
Solution:
T AB := TA
Equilibrium :
TAB
T AB − F⋅ rB = 0
F :=
T' − F⋅ rC = 0
T' := F⋅ rC
Angle of Twist :
φ B :=
J :=
π
4
⋅d
32 o
T AB⋅ LAB
F = 900.00 N
T' = 67.50 N⋅ m
T CE := T'
φ C :=
G⋅ J
φ B = 0.011318 rad
φ B = 0.648 deg
rB
T CE⋅ L CE
G⋅ J
φ C = 0.008488 rad
Ans
φ C = 0.486 deg
Ans
Problem 5-62
The 60-mm-diameter solid shaft is made of A-36 steel and is subjected to the distributed and
concentrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft due
to these loadings.
Given: L AC := 0.6m
L CB := 0.8m
T A := 400N⋅ m
T C := −600N⋅ m
G := 75GPa
q := 2000
do := 60mm
N⋅ m
m
Solution:
As shown in the torque diagram.
Internal Torque :
T AC := TA
T CB = T A + TC + q⋅ x
Angle of Twist :
φA =
φ A :=
J :=
T AC⋅ LAC
G⋅ J
TAC⋅ L AC
G⋅ J
π
4
⋅d
32 o
L
CB
1 ⌠
+
⋅⎮
T CB dx
G⋅ J ⌡
0
L
CB
1 ⌠
+
⋅⎮
T A + T C + q⋅ x dx
G⋅ J ⌡
0
φ A = 0.007545 rad
φ A = 0.432 deg
Ans
Problem 5-63
When drilling a well, the deep end of the drill pipe is assumed to encounter a torsional resistance TA.
Furthermore, soil friction along the sides of the pipe creates a linear distribution of torque per unit
length, varying from zero at the surface B to t0 at A. Determine the necessary torque TB that must be
supplied by the drive unit to turn the pipe. Also, what is the relative angle of twist of one end of the
pipe with respect to the other end at the instant the pipe is about to turn? The pipe has an outer radius
ro and an inner radius ri . The shear modulus is G.
Problem 5-64
The assembly is made of A-36 steel and consists of a solid rod 15 mm in diameter connected to the
inside of a tube using a rigid disk at B. Determine the angle of twist at A. The tube has an outer
diameter of 30 mm and wall thickness of 3 mm.
Given: do := 30mm
t := 3mm
dr := 15mm
L AB := 0.3m
L BC := 0.3m
G := 75GPa
T A := 50N⋅ m
T B := 30N⋅ m
Solution: di := do − 2t
T AB := TA
T BC := TA + T B
Angle of Twist :
π
4
JAB :=
⋅ dr
32
φ A :=
TAB⋅ L AB
G⋅ JAB
π ⎛ 4
4
JBC :=
⋅ ⎝ do − di ⎞⎠
32
+
T BC⋅ LBC
G⋅ JBC
φ A = 0.04706 rad
φ A = 2.696 deg
Ans
Problem 5-65
The device serves as a compact torsional spring. It is made of A-36 steel and consists of a solid inner
shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can
also be assumed rigid and is fixed from rotating. If a torque of T = 0.25 N·m. is applied to the shaft,
determine the angle of twist at the end C and the maximum shear stress in the tube and shaft.
Given: L BC := 600mm
L BA := 300mm
rti := 18.75mm
rto := 25mm
T C := 0.25kN⋅ m
r := 12.5mm
G := 75GPa
Solution:
Internal Torque :
AS shown on FBD
Inner shaft :
T BC := TC
Outer tube :
T BA := TC
Max. Shear Stress :
Inner shaft :
π 4
JBC := ⋅ r
2
cBC := r
τ BC :=
Outer tube :
cBA := rto
φ B :=
τ BC = 81.49 MPa
Ans
π
4
4
JBA := ⋅ ⎛⎝ rto − rti ⎞⎠
2
T BA⋅ LBA
φ C_B :=
)
JBC
τ BA :=
Angle of Twist :
(
TBC⋅ cBC
G⋅ JBA
TBC⋅ L BC
G⋅ JBC
φ C := φ C_B + φ B
(
)
TBC⋅ cBA
JBA
τ BA = 14.9 MPa
φ B = 0.002384 rad
φ C_B = 0.052152 rad
φ C = 0.054536 rad
φ C = 3.125 deg
Ans
Ans
Problem 5-66
The device serves as a compact torsion spring. It is made of A-36 steel and consists of a solid inner
shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can
also be assumed rigid and is fixed from rotating. If the allowable shear stress for the material is τallow =
84 MPa and the angle of twist at C is limited to φallow = 3°, determine the maximum torque T that can
be applied at the end C.
Given: L BC := 600mm
L BA := 300mm
rti := 18.75mm
rto := 25mm
τ allow := 84MPa
r := 12.5mm
φ allow := 3deg
G := 75GPa
Solution:
Internal Torque :
AS shown on FBD
Inner shaft : T BC=T C
Outer tube : T BA=TC
Assume failure due to shear stress.
Allowable Shear Stress :
Inner shaft :
π 4
JBC := ⋅ r
2
cBC := r
T BC :=
Outer tube :
cBC
BC = 257.71 N⋅ m
(
) T
τ allow⋅ JBA
cBA
BA = 1409.34 N⋅ m
Assume failure due to angle of twist limitation (maximum occurred at C).
φ C = φ C_B + φ B
Thus,
) T
π
4
4
JBA := ⋅ ⎛⎝ rto − rti ⎞⎠
2
cBA := rto
T BA :=
Angle of Twist :
(
τ allow⋅ JBC
φ allow =
T C :=
TC⋅ L BC
G⋅ JBC
φ C_B =
+
T C⋅ LBC
G⋅ JBC
φB =
TC⋅ L BA
G⋅ JBA
T C⋅ LBA
G⋅ JBA
G⋅ φ allow
LBC
JBC
+
LBA
T C = 240.02 N⋅ m
JBA
(controls !)
Ans
Problem 5-67
The shaft has a radius c and is subjected to a torque per unit length of t0 , which is distributed
uniformly over the shaft's entire length L. If it is fixed at its far end A, determine the angle of twist φ o
end B. The shear modulus is G.
Solution:
Internal Torque :
As shown in the torque diagram.
T ( x) = −to⋅ x
J :=
Angle of Twist :
π 4
⋅c
2
L
φB =
1 ⌠
⋅ ⎮ T ( x) dx
G⋅ J ⌡0
− to ⌠ L
φB =
⋅ ⎮ x dx
G⋅ J ⌡0
φB =
φB =
−to⋅ L
2
2G⋅ J
−to⋅ L
2
4
πG⋅ c
φB =
to⋅ L
2
4
πG⋅ c
Ans
Problem 5-68
The A-36 bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed
by the equation t = (k x2) N·m /m, where x is in meters. If a torque of T = 50 N·m is applied to the bol
head, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume the
shank has a constant radius of 4 mm.
Given: ro := 4mm
L := 50mm
T A := 50N⋅ m
t = k⋅ x
G := 75GPa
2 N⋅ m
m
Solution: c := ro
Internal Torque :
As shown in the torque diagram.
t ( x) = k⋅ x
2
Equilibrium :
L
⌠
T A − ⎮ t ( x) dx = 0
⌡0
L
⌠
2
T A − ⎮ k⋅ x dx = 0
⌡0
3
TA −
k⋅ L
= 0
3
k :=
3TA
L
3
⌠
2
⎮
Hence, T ( x) = TA −
⎮ k⋅ x dx
⌡
3
k⋅ x
T ( x) = T A −
3
π 4
Angle of Twist : J := ⋅ c
2
L
1 ⌠
φ=
⋅ ⎮ T ( x) dx
G⋅ J ⌡0
⌠
1 ⎮
⋅
φ :=
G⋅ J ⎮
⌡
L
0
3
⎛
k⋅ x ⎞
⎜TA −
dx
3 ⎠
⎝
φ = 0.06217 rad
φ = 3.562 deg
Ans
k = 1.200 MPa
Ans
Problem 5-69
Solve Prob. 5-68 if the distributed torque is t = (k x2/3) N·m /m.
Given: ro := 4mm
L := 50mm
T A := 50N⋅ m
G := 75GPa
2
t = k⋅ x
3 N⋅ m
m
Solution: c := ro
Internal Torque :
As shown in the torque diagram.
L
Equilibrium :
L
⌠
T A − ⎮ t ( x) dx = 0
⌡0
−4
Let
unit := m
3
⌠
2
⎮
3
T A − ⎮ k⋅ x dx = 0
⌡0
⎛ 5⎞
⎜ 3
3k⋅ ⎝ L ⎠
TA −
= 0
5
−5
⎛ 5TA ⎞ 3
k := ⎜
⋅L
⋅ unit
⎝ 3 ⎠
⌠
2
⎮
⎮
3
Hence, T ( x) = TA − ⎮ k⋅ x dx
⌡
k = 0.01228 MPa Ans
⎛ 5⎞
⎜ 3
3k⋅ ⎝ x ⎠
T ( x) = T A −
5
Angle of Twist :
L
J :=
π 4
⋅c
2
1 ⌠
φ=
⋅ ⎮ T ( x) dx
G⋅ J ⌡0
L
⌠
⎮
⎛ 5⎞
⎜ 3
⎮
1 ⎮
3k⋅ ⎝ x ⎠
φ :=
⋅
TA −
dx
G⋅ J ⎮
5⋅ unit
⌡
0
φ = 0.05181 rad
φ = 2.968 deg
Ans
Problem 5-70
The contour of the surface of the shaft is defined by the equation y = e ax, where a is a constant. If the
shaft is subjected to a torque T at its ends, determine the angle of twist of end A with respect to end B.
The shear modulus is G.
Problem 5-71
The A-36 steel shaft has a diameter of 50 mm and is subjected to the distributed and concentrated
loadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of the
angle of twist of the shaft in radians versus x.
Given: do := 50mm
L := 0.5m
T C := −250N⋅ m
q := 200
G := 75GPa
N⋅ m
m
Solution:
Support Reaction:
RA := −T C − q⋅ L
RA = 150.00 N⋅ m
Internal Torque : As shown in the torque diagram.
T ( x) = RA + q⋅ x
The maximum torque occurs at C.
T max := RA + q⋅ L
T max = 250.00 N⋅ m
c := 0.5 ⋅ do
Maximum Shear Stress :
τ max :=
T max⋅ c
J :=
π 4
⋅c
2
τ max = 10.19 MPa
J
Ans
Angle of Twist :
x
φx =
1 ⌠
⋅ ⎮ T ( x) dx
G⋅ J ⌡0
φx =
1 ⌠
⋅⎮
G⋅ J ⌡
x
(RA + q⋅ x) dx
0
1 ⎛
q⋅ x ⎞
φx =
⋅ ⎜ RA⋅ x +
G⋅ J ⎝
2 ⎠
2
1 ⎛
q⋅ L ⎞
⋅ ⎜ RA⋅ L +
G⋅ J ⎝
2 ⎠
2
At C, x = L
φ C :=
φ C = 0.00217 rad
φ C = 0.125 deg
For L < x < 2L
Since T(x) = 0, then, φ x := φ C
x1 := 0 , 0.01 ⋅ L .. L
( )
φ 1 x1 :=
Ans
⎛
2⎞
q⋅ x1
1
1 ⎜
⋅ RA⋅ x1 +
⋅
G⋅ J ⎝
2 ⎠ N⋅ m
x2 := L , 1.01 ⋅ L .. 2L
1
φ 2 x2 := φ C ⋅
N⋅ m
( ) ( )
Twist (rad)
0.002
( )
φ2 ( x 2 )
φ1 x 1
0.001
0
0
0.2
0.4
0.6
x1 , x2
Distance (m)
0.8
Problem 5-72
A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held
fixed and a torque T is applied to the rigid shaft, determine the angle of twist of the shaft. The shear
modulus of the rubber is G. Hint: As shown in the figure, the deformation of the element at radius r
can be determined from rdθ = drγ. Use this expression along with τ = T/(2π r2h) from Prob. 5-28, to
obtain the result.
Problem 5-73
The A-36 steel shaft has a diameter of 50 mm and is fixed at its ends A and B. If it is subjected to the
couple, determine the maximum shear stress in regions AC and CB of the shaft.
Given: do := 50mm
L AC := 0.4m
G := 75GPa
T C := 300N⋅ m
Solution:
c := 0.5 ⋅ do
J :=
L BC := 0.8m
π 4
⋅c
2
Equilibrium : Given
TA + TB − TC = 0
Compatibility :
(1)
φC/A = φC/B
T A⋅ L AC
G⋅ J
Solving Eqs. (1) and (2):
=
T B⋅ LBC
G⋅ J
(2)
Guess T A := 1N⋅ m
⎛⎜ TA ⎞
:= Find ( TA , T B)
⎜ TB
⎝ ⎠
T B := 1N⋅ m
⎛⎜ TA ⎞ ⎛ 200 ⎞
=⎜
N⋅ m
⎜ TB
⎝ ⎠ ⎝ 100 ⎠
Maximum Shear Stress :
τ AC.max :=
τ BC.max :=
TA⋅ c
J
T B⋅ c
J
τ AC.max = 8.15 MPa
Ans
τ BC.max = 4.07 MPa
Ans
Problem 5-74
The bronze C86100 pipe has an outer diameter of 37.5 mm and a thickness of 3 mm. The coupling on
it at C is being tightened using a wrench. If the torque developed at A is 16 N·m, determine the
magnitude F of the couple forces. The pipe is fixed supported at end B.
Given: L CB := 200mm
L CA := 250mm
L w := 300mm
do := 37.5mm
T A := 16N⋅ m
G := 38GPa
ro := 0.5do
ri := ro − t
Solution:
Compatibility :
t := 3mm
φ C_B = φ C_A
T B⋅ LCB
G⋅ J
=
TB⋅ L CA
G⋅ J
T B⋅ LCB = TA⋅ LCA
⎛ LCA ⎞
T B := ⎜
⎝ LCB ⎠
Equilibrium:
⋅ TA
T B = 20.00 N⋅ m
( )
F⋅ L w − T B − T A = 0
F :=
TB + TA
Lw
F = 120.00 N
Ans
Problem 5-75
The bronze C86100 pipe has an outer diameter of 37.5 mm and a thickness of 3 mm. The coupling on
it at C is being tightened using a wrench. If the applied force is F = 100 N, determine the maximum
shear stress in the pipe.
Given: L CB := 200mm
L CA := 250mm
L w := 300mm
do := 37.5mm
F := 100N
G := 38GPa
Solution:
ro := 0.5do
Compatibility :
ri := ro − t
φ C_B = φ C_A
T B⋅ LCB
G⋅ J
Given
Equilibrium:
Initial guess:
t := 3mm
=
TB⋅ L CA
G⋅ J
T B⋅ LCB = TA⋅ LCA
( )
F⋅ L w − T B − T A = 0
T A := 1N⋅ m
[1]
[2]
T B := 2N⋅ m
⎛⎜ TA ⎞
:= Find ( TA , T B)
⎜ TB
⎝ ⎠
Solving [1] and [2]:
⎛⎜ TA ⎞ ⎛ 13.33 ⎞
=⎜
N⋅ m
⎜ TB
16.67
⎝
⎠
⎝ ⎠
Max. Shear Stress :
c := ro
J :=
π ⎛ 4
4
⋅ r − ri ⎞⎠
2 ⎝o
τ max :=
T B⋅ ( c)
J
τ max = 3.21 MPa
Ans
Problem 5-76
The steel shaft is made from two segments: AC has a diameter of 12 mm, and CB has a diameter of 25
mm. If it is fixed at its ends A and B and subjected to a torque of 750 N·m, determine the maximum
shear stress in the shaft. Gst = 75 GPa.
Given: L AC := 125mm
L CD := 200mm
L DB := 300mm
dAC := 12mm
dCD := 25mm
dDB := 25mm
T D := 750N⋅ m
G := 75GPa
Solution:
π ⎛ dAC ⎞
JAC := ⋅ ⎜
2 ⎝ 2 ⎠
4
π ⎛ dCD ⎞
JCD := ⋅ ⎜
2 ⎝ 2 ⎠
4
4
φ D_A = φ D_B
Compatibility :
T A⋅ L AC
Given
G⋅ JAC
+
T A⋅ L CD
G⋅ JCD
=
T B⋅ LDB
G⋅ JDB
TD − TB − TA = 0
Equilibrium:
T A := 1N⋅ m
Initial guess:
Solving [1] and [2]:
Max. Shear Stress :
dAC
cAC :=
2
cDB :=
π ⎛ dDB ⎞
JDB := ⋅ ⎜
2 ⎝ 2 ⎠
dDB
2
[1]
[2]
T B := 2N⋅ m
⎛⎜ TA ⎞
:= Find ( TA , T B)
⎜ TB
⎝ ⎠
τ AC :=
τ DB :=
(
)
TA⋅ cAC
τ AC = 232.3 MPa
JAC
(
)
TB⋅ cDB
τ DB = 218.77 MPa
JDB
(
⎛⎜ TA ⎞ ⎛ 78.82 ⎞
=⎜
N⋅ m
⎜ TB
⎝ ⎠ ⎝ 671.18 ⎠
)
τ max := max τ AC , τ DB
τ max = 232.30 MPa
Ans
Problem 5-77
The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is
subjected to the couple, determine the maximum shear stress in regions AC and CB.
Given: do := 40mm
G := 75GPa
Solution:
c := 0.5 ⋅ do
L AC := 0.4m
L BC := 0.6m
PC := 2kN
rC := 50mm
J :=
π 4
⋅c
2
Equilibrium : Given
( )
T A + TB − PC⋅ 2rC = 0
Compatibility :
(1)
φC/A = φC/B
T A⋅ L AC
G⋅ J
Solving Eqs. (1) and (2):
=
T B⋅ LBC
G⋅ J
(2)
Guess T A := 1N⋅ m
⎛⎜ TA ⎞
:= Find ( TA , T B)
⎜ TB
⎝ ⎠
T B := 1N⋅ m
⎛⎜ TA ⎞ ⎛ 120 ⎞
=⎜
N⋅ m
⎜ TB
⎝ ⎠ ⎝ 80 ⎠
Maximum Shear Stress :
τ AC.max :=
τ BC.max :=
TA⋅ c
J
T B⋅ c
J
τ AC.max = 9.55 MPa
Ans
τ BC.max = 6.37 MPa
Ans
Problem 5-78
The composite shaft consists of a mid-section that includes the 20-mm-diameter solid shaft and a tube
that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the
angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 800 N·m.
The material is A-36 steel.
Given: L CA := 100mm
L AB := 150mm
L BD := 100mm
ds := 20mm
dto := 60mm
t := 5mm
T := 800N⋅ m
G := 75GPa
Solution:
dti := dto − 2t
π ⎛ ds ⎞
Js := ⋅ ⎜
2 ⎝2⎠
Compatibility :
Given
Equilibrium:
Initial guess:
⎡
⎛ dti ⎞ ⎤⎥
π ⎢⎛ dto ⎞
Jt := ⋅ ⎜
−⎜
2 ⎣⎝ 2 ⎠
⎝2⎠⎦
4
4
T s⋅ LAB
φs = φt
Ts
Js
=
G⋅ Js
Tt
Tt⋅ LAB
G⋅ Jt
[1]
Jt
T − Ts − Tt = 0
[2]
T s := 1N⋅ m
Solving [1] and [2]:
=
T t := 2N⋅ m
⎛⎜ Ts ⎞
:= Find ( T s , T t)
⎜ Tt
⎝ ⎠
Angle of Twist:
φ C_D :=
Ts⋅ L CA
G⋅ Js
+
T t⋅ L AB
G⋅ Jt
φ C_D = 0.005535 rad
φ C_D = 0.317 deg
4
Ans
+
T s⋅ LBD
G⋅ Js
⎛⎜ Ts ⎞ ⎛ 18.63 ⎞
=⎜
N⋅ m
⎜ Tt
⎝ ⎠ ⎝ 781.37 ⎠
Problem 5-79
The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass
core. If it is fixed to a rigid support at A, and a torque of T = 50 N·m is applied to it at C, determine the
angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in
the brass and steel. Take Gst = 80 GPa, Gbr = 40 GPa.
Given: LAB := 1.5m
LBC := 1m
rs := 20mm
rto := 20mm
rti := 10mm
T := 50N⋅ m
Gst := 80GPa
Gbr := 40GPa
π 4
Js := ⋅ rs
2
Solution:
π
4
4
Jt := ⋅ ⎛⎝ rto − rti ⎞⎠
2
π
4
Jbr := ⋅ rti
2
Tst⋅ L BC
φ st = φ br
Compatibility :
T st
Given
Gst⋅ Jt
=
Gst⋅ Jt
T br⋅ L BC
=
Gbr⋅ Jbr
Tbr
[1]
Gbr⋅ Jbr
T − T st − T br = 0
Equilibrium:
Initial guess:
Tst := 1N⋅ m
Solving [1] and [2]:
[2]
Tbr := 2N⋅ m
⎛⎜ Tst ⎞
:= Find ( T st , T br)
⎜ Tbr
⎝
⎠
⎛⎜ Tst ⎞ ⎛ 48.39 ⎞
=⎜
N⋅ m
⎜ Tbr
⎝
⎠ ⎝ 1.61 ⎠
Angle of Twist:
φ C :=
T⋅ LAB
Gst⋅ Js
+
Tbr⋅ LBC
Max. Shear Stress :
cAB := rs
τ AB :=
cBC := rto
τ BC :=
τ br :=
γ br :=
( )
Tbr⋅ rti
Jbr
τ br
Gbr
φ C = 0.006297 rad
Gbr⋅ Jbr
(
)
T ⋅ cAB
Js
T st⋅ cBC
(
Jt
φ C = 0.361 deg
Ans
τ st = 4.11 MPa
Ans
τ AB = 3.98 MPa
)
τ BC = 4.11 MPa
(
τ st := max τ AB , τ BC
γ st :=
τ st
γ br = 25.67 × 10
γ st = 51.34 × 10
Gst
τ br = 1.03 MPa
Ans
−6
rad
)
Ans
−6
rad
Ans
Problem 5-80
The two 1-m-long shafts are made of 2014-T6 aluminum. Each has a diameter of 30 mm and they are
connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and
B. They are also supported by bearings at C and D, which allow free rotation of the shafts along their
axes. If a torque of 900 N·m is applied to the top gear as shown, determine the maximum shear stress
in each shaft.
Given: L := 1000mm
d := 30mm
rA := 80mm
rB := 40mm
T := 900N⋅ m
G := 27GPa
Solution:
π ⎛ d⎞
J := ⋅ ⎜
2 ⎝ 2⎠
Compatibility :
4
( )
T A⋅ L
G⋅ J
Given
( )
φ E ⋅ r A = φ F⋅ r B
( )
⋅ rA =
TB⋅ L
G⋅ J
( )
( )
⋅ rB
( )
T A⋅ rA = TB⋅ rB
( )
T B − F⋅ ( rB) = 0
T − T A − F⋅ r A = 0
Equilibrium:
Initial guess:
T A := 1N⋅ m
Solving [1], [2] and [3]:
[1]
[2]
[3]
T B := 2N⋅ m
F := 1N
⎛ TA ⎞
⎜
⎜ TB ⎟ := Find ( TA , TB , F)
⎜
⎝ F ⎠
⎛⎜ TA ⎞ ⎛ 180 ⎞
=⎜
N⋅ m
⎜ TB
360
⎝
⎠
⎝ ⎠
F = 9000.00 N
Max. Shear Stress :
c :=
d
2
τ AC :=
τ BD :=
TA⋅ ( c)
J
TB⋅ ( c)
J
τ AC = 33.95 MPa
Ans
τ BD = 67.91 MPa
Ans
Problem 5-81
The two shafts are made of A-36 steel. Each has a diameter of 25 mm and they are connected using
the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also
supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a
torque of 500 N·m is applied to the gear at E as shown, determine the reactions at A and B.
Given: L AE := 1.5m
L BF := 0.75m
rE := 100mm
rF := 50mm
T E := 500N⋅ m
G := 75GPa
Solution:
J :=
do := 25mm
π
4
⋅ do
32
Compatibility :
( )
T A⋅ L AE
G⋅ J
Given
Equilibrium:
Initial guess:
( )
φ E ⋅ r E = φ F⋅ r F
( )
⋅ rE =
TB⋅ L BF
G⋅ J
( )
( )
⋅ rF
( )
T A⋅ L AE⋅ rE = TB⋅ L BF⋅ rF
( )
T B − F⋅ ( r F) = 0
T A + F⋅ r E − T E = 0
T A := 1N⋅ m
Solving [1], [2] and [3]:
T B := 2N⋅ m
[1]
[2]
[3]
F := 1N
⎛ TA ⎞
⎜
⎜ TB ⎟ := Find ( TA , TB , F)
⎜
⎝ F ⎠
⎛⎜ TA ⎞ ⎛ 55.56 ⎞
=⎜
N⋅ m
⎜ TB
222.22
⎝
⎠
⎝ ⎠
F = 4444.44 N
A
Problem 5-82
Determine the rotation of the gear at E in Prob. 5-81.
Given: L AE := 1.5m
L BF := 0.75m
rE := 100mm
rF := 50mm
T E := 500N⋅ m
G := 75GPa
Solution:
J :=
do := 25mm
π
4
⋅ do
32
( )
( )
φ E ⋅ r E = φ F⋅ r F
Compatibility :
T A⋅ L AE
G⋅ J
( )
⋅ rE =
TB⋅ L BF
G⋅ J
( )
( )
⋅ rF
( )
T A⋅ L AE⋅ rE = TB⋅ L BF⋅ rF
Given
( )
T B − F⋅ ( r F) = 0
T A + F⋅ r E − T E = 0
Equilibrium:
Initial guess:
T A := 1N⋅ m
Solving [1], [2] and [3]:
T B := 2N⋅ m
[1]
[2]
[3]
F := 1N
⎛ TA ⎞
⎜
⎜ TB ⎟ := Find ( TA , TB , F)
⎜
⎝ F ⎠
⎛⎜ TA ⎞ ⎛ 55.56 ⎞
=⎜
N⋅ m
⎜ TB
222.22
⎝
⎠
⎝ ⎠
F = 4444.44 N
Angle of Twist :
φ E :=
T A⋅ L AE
G⋅ J
φ E = 0.02897 rad
φ E = 1.660 deg
Ans
Problem 5-83
The A-36 steel shaft is made from two segments: AC has a diameter of 10 mm and CB has a diameter
of 20 mm. If the shaft is fixed at its ends A and B and subjected to a uniform distributed torque of 300
N·m/m along segment CB, determine the absolute maximum shear stress in the shaft.
Given: L AC := 0.1m
L CB := 0.4m
dAC := 10mm
dCB := 20mm
G := 75GPa
q := 300
Solution:
π ⎛ dAC ⎞
JAC := ⋅ ⎜
2 ⎝ 2 ⎠
Compatibility :
T A⋅ L AC
G⋅ JAC
=
4
T B⋅ LCB − 0.5q⋅ LCB
2
[1]
G⋅ JCB
q⋅ L CB − T A − TB = 0
Initial guess: T A := 1N⋅ m
Solving [1] and [2]:
Max. Shear Stress :
dAC
cAC :=
2
cCB :=
π ⎛ dCB ⎞
JCB := ⋅ ⎜
2 ⎝ 2 ⎠
φ C_A = φ C_B
Given
Equilibrium:
4
N⋅ m
m
dCB
2
[2]
T B := 2N⋅ m
⎛⎜ TA ⎞
:= Find ( TA , T B)
⎜ TB
⎝ ⎠
τ AC :=
τ CB :=
(
)
TA⋅ cAC
JAC
(
TB⋅ cCB
JCB
)
⎛⎜ TA ⎞ ⎛ 12 ⎞
=⎜
N⋅ m
⎜ TB
⎝ ⎠ ⎝ 108 ⎠
τ AC = 61.12 MPa
Ans
τ CB = 68.75 MPa
Ans
(
τ max := max τ AC , τ CB
)
τ max = 68.75 MPa
Ans
Problem 5-84
The tapered shaft is confined by the fixed supports at A and B. If a torque T is applied at its mid-poin
determine the reactions at the supports.
Problem 5-85
A portion of the A-36 steel shaft is subjected to a linearly distributed torsional loading. If the shaft has
the dimensions shown, determine the reactions at the fixed supports A and C. Segment AB has a
diameter of 30 mm and segment BC has a diameter of 15 mm.
Given: LAB := 1.2m
LBC := 0.96m
dAB := 30mm
dBC := 15mm
kN⋅ m
q := 1.5
m
G := 75GPa
Solution:
π ⎛ dAB ⎞
JAB := ⋅ ⎜
2 ⎝ 2 ⎠
x ⎞
⎛
qx = ⎜ 1 −
⋅q
Lab
⎝
⎠
4
π ⎛ dBC ⎞
JBC := ⋅ ⎜
2 ⎝ 2 ⎠
( )
LAB
Given
Equilibrium:
)
⎛
⎝
x ⎞
⎛ x ⎞ ⋅ q⋅ x
⋅ q⋅ x + 0.5 ⎜
Lab
Lab
⎛
⎝
0.5x ⎞
⋅ q⋅ x
L ab
TR = ⎜1 −
Compatibility :
(
TR = qx ⋅ x + 0.5 q − qx ⋅ x
TR = ⎜1 −
⌠
β := ⎮
⎮
⌡0
4
⎠
⎝
⎠
⎠
⎛ 1 − 0.5x ⎞ ⋅ x dx
⎜ L
AB ⎠
⎝
β = 0.48 m
2
φ B_A = φ B_C
TA⋅ LAB − q⋅ β
G⋅ JAB
=
T C⋅ LBC
G⋅ JBC
0.5q⋅ LAB − TA − T C = 0
Initial guess: TA := 1kN⋅ m
Solving [1] and [2]:
[1]
[2]
TC := 2kN⋅ m
⎛⎜ TA ⎞
:= Find ( T A , TC)
⎜ TC
⎝ ⎠
⎛⎜ TA ⎞ ⎛ 878.26 ⎞
=⎜
N⋅ m
⎜ TC
21.74
⎝
⎠
⎝ ⎠
Ans
Problem 5-86
Determine the rotation of joint B and the absolute maximum shear stress in the shaft in Prob. 5-85.
Given: L AB := 1.2m
L BC := 0.96m
dAB := 30mm
dBC := 15mm
G := 75GPa
q := 1.5
π ⎛ dAB ⎞
JAB := ⋅ ⎜
2 ⎝ 2 ⎠
Solution:
⎛
⎝
qx = ⎜ 1 −
x ⎞
⋅q
L ab
⎠
4
kN⋅ m
m
π ⎛ dBC ⎞
JBC := ⋅ ⎜
2 ⎝ 2 ⎠
( )
4
(
)
T R = qx ⋅ x + 0.5 q − qx ⋅ x
x ⎞
⎛ x ⎞ ⋅ q⋅ x
⋅ q⋅ x + 0.5 ⎜
L ab
L ab
⎛
⎝
TR = ⎜1 −
⎠
⎝
⎠
⎛
⎝
TR = ⎜1 −
L
⌠ AB ⎛
0.5x ⎞
β := ⎮
1−
⋅ x dx
⎜
LAB
⎮
⎝
⎠
⌡0
Compatibility :
Given
β = 0.48 m
φ B_A = φ B_C
T A⋅ L AB − q⋅ β
G⋅ JAB
=
TC⋅ L BC
[1]
G⋅ JBC
0.5q⋅ L AB − T A − TC = 0
Equilibrium:
2
[2]
Initial guess: T A := 1kN⋅ m T C := 2kN⋅ m
⎛⎜ TA ⎞
Solving [1] and [2]:
:= Find TA , T C
⎜ TC
⎝ ⎠
(
)
⎛⎜ TA ⎞ ⎛ 878.26 ⎞
=⎜
N⋅ m
⎜ TC
⎝ ⎠ ⎝ 21.74 ⎠
Angle of Twist:
φ B :=
T C⋅ LBC
G⋅ JBC
φ B = 0.055987 rad
φ B = 3.208 deg
Max. Shear Stress :
dAB
TA⋅ cAB
cAB :=
τ AB :=
2
JAB
dBC
TC⋅ cBC
cBC :=
τ BC :=
2
JBC
(
)
(
)
(
Ans
τ AB = 165.66 MPa
τ BC = 32.8 MPa
τ max := max τ AB , τ BC
) τ max = 165.66 MPa
Ans
0.5x ⎞
⋅ q⋅ x
Lab
⎠
Problem 5-87
The shaft of radius c is subjected to a distributed torque t, measured as torque/length of shaft.
Determine the reactions at the fixed supports A and B.
Problem 5-88
Compare the values of the maximum elastic shear stress and the angle of twist developed in 304
stainless steel shafts having circular and square cross sections. Each shaft has the same cross-sectiona
area of 5600 mm2, length of 900 mm, and is subjected to a torque of 500 N·m.
2
Given: L := 900mm
A := 5600mm
G := 75GPa
T := 500N⋅ m
Solution:
Max. Shear Stress :
For circular shaft: A = π ⋅ r
c := r
J :=
2
π 4
⋅r
2
r :=
A
π
τ C_max :=
T ⋅ ( c)
J
τ C_max = 4.23 MPa
2
For squarr shaft:
A= a
a :=
Ans
A
τ S_max :=
T⋅ ( 4.81 )
3
a
τ S_max = 5.74 MPa
Ans
φ C = 0.001202 rad
φ C = 0.0689 deg
Ans
φ S = 0.001358 rad
φ S = 0.0778 deg
Ans
Angle of Twist:
For circular shaft:
φ C :=
T⋅ L
G⋅ J
For sqaure shaft:
φ S :=
( 7.10 )T ⋅ L
4
G⋅ a
Note: The sqaure shaft has a greater maximum shear stress and angle of twist.
Problem 5-89
The shaft is made of red brass C83400 and has an elliptical cross section. If it is subjected to the
torsional loading shown, determine the maximum shear stress within regions AC and BC, and the ang
of twist φ of end B relative to end A.
Given: L AC := 2m
L CB := 1.5m
a := 50mm
b := 20mm
T A := 50N⋅ m
T B := 30N⋅ m
G := 37GPa
T C := 20N⋅ m
Solution:
Max. Shear Stress :
τ BC :=
τ AC :=
2TB
π ⋅ a⋅ b
2
2T A
π ⋅ a⋅ b
Angle of Twist:
2
τ BC = 0.955 MPa
Ans
τ AC = 1.592 MPa
Ans
φ B_A =
∑
n
φ B_A :=
(a2 + b2)Tn⋅ Ln
3 3
π⋅a ⋅b ⋅G
(a2 + b2)(−TB⋅ LCB − TA⋅ LAC)
3 3
π⋅a ⋅b ⋅G
φ B_A = −0.003618 rad
φ B_A = 0.2073 deg
Ans
Problem 5-90
Solve Prob. 5-89 for the maximum shear stress within regions AC and BC, and the angle of twist φ of
end B relative to C.
Given: L AC := 2m
L CB := 1.5m
a := 50mm
b := 20mm
T A := 50N⋅ m
T B := 30N⋅ m
G := 37GPa
T C := 20N⋅ m
Solution:
Max. Shear Stress :
τ BC :=
τ AC :=
2TB
π ⋅ a⋅ b
2
2T A
π ⋅ a⋅ b
Angle of Twist:
2
τ BC = 0.955 MPa
Ans
τ AC = 1.592 MPa
Ans
φ B_C =
∑
n
φ B_C :=
(a2 + b2)Tn⋅ Ln
3 3
π⋅a ⋅b ⋅G
(a2 + b2)(−TB⋅ LCB)
3 3
π⋅a ⋅b ⋅G
φ B_C = −0.001123 rad
φ B_C = 0.0643 deg
Ans
Problem 5-91
The steel shaft is 300 mm long and is screwed into the wall using a wrench. Determine the largest
couple forces F that can be applied to the shaft without causing the steel to yield. τY = 56 MPa.
Given: L := 300mm
a := 25mm
L w := 400mm
τ Y := 56MPa
Solution:
Max. Shear Stress :
τ max =
4.81T
3
3
T :=
a
a τY
4.81
T = 181.91 N⋅ m
Equilibrium:
( )
F⋅ L w − T = 0
F :=
T
Lw
F = 454.78 N
Ans
Problem 5-92
The steel shaft is 300 mm long and is screwed into the wall using a wrench. Determine the maximum
shear stress in the shaft and the amount of displacement that each couple force undergoes if the couple
forces have a magnitude of F = 150 N. Gst = 75 GPa.
Given: L := 300mm
a := 25mm
L w := 400mm
F := 150N
Gst := 75GPa
Solution:
Equilibrium:
( )
F⋅ L w − T = 0
( )
T := F⋅ L w
T = 60 N⋅ m
Max. Shear Stress :
τ max :=
4.81T
3
a
τ max = 18.47 MPa
Ans
Angle of Twist:
φ :=
( 7.10 )T ⋅ L
4
φ = 0.004362 rad
Gst⋅ a
δ F := φ ⋅
Lw
2
δ F = 0.872 mm
Ans
Problem 5-93
The shaft is made of plastic and has an elliptical cross-section. If it is subjected to the torsional loadin
shown, determine the shear stress at point A and show the shear stress on a volume element located at
this point. Also, determine the angle of twist φ at the end B. Gp = 15 GPa.
Given: L OC := 2m
L CB := 1.5m
a := 50mm
b := 20mm
T B := 50N⋅ m
T C := 40N⋅ m
G := 15GPa
Solution:
T OC := TB + TC
Shear Stress :
τ A :=
2TOC
2
π ⋅ a⋅ b
Angle of Twist:
τ A = 2.865 MPa
φB =
∑
n
φ B :=
Ans
(a2 + b2)Tn⋅ Ln
3 3
π⋅a ⋅b ⋅G
(a2 + b2)(TB⋅ LCB + TOC⋅ LOC)
3 3
π⋅a ⋅b ⋅G
φ B = 0.015693 rad
φ B = 0.8991 deg
Ans
Problem 5-94
The square shaft is used at the end of a drive cable in order to register the rotation of the cable on a
gauge. If it has the dimensions shown and is subjected to a torque of 8 N·m, determine the shear stres
in the shaft at point A. Sketch the shear stress on a volume element located at this point.
Given:
a := 5mm
T := 8N⋅ m
Solution:
Maximum Shear Stress :
τ A.max :=
4.81T
3
a
τ A.max = 307.8 MPa
Ans
Problem 5-95
The brass wire has a triangular cross section, 2 mm on a side. If the yield stress for brass is τY = 205
MPa, determine the maximum torque T to which it can be subjected so that the wire will not yield. If
this torque is applied to a segment 4 m long, determine the greatest angle of twist of one end of the
wire relative to the other end that will not cause permanent damage to the wire. Gbr = 37 GPa.
Given:
a := 2mm
L := 4m
G := 37GPa
τ Y := 205MPa
Solution:
Allowable Shear Stress :
τ allow =
20T
3
a
τ allow := τ Y
3
T :=
τ Y⋅ a
20
T = 0.0820 N⋅ m
Ans
Angle of Twist :
φ :=
46T⋅ L
4
G⋅ a
φ = 25.49 rad
Ans
Problem 5-96
It is intended to manufacture a circular bar to resist torque; however, the bar is made elliptical in the
process of manufacturing, with one dimension smaller than the other by a factor k as shown.
Determine the factor by which the maximum shear stress is increased.
Given:
a = 0.5 ⋅ do
b = 0.5 ⋅ k⋅ do
Solution:
Maximum Shear Stress :
For the circular shaft::
τ c.max =
τ c.max =
T⋅ c
J
16T
π ⋅ do
c = 0.5 ⋅ do
π
4
⋅ do
32
J=
3
For the elliptical shaft:
τ e.max =
τ t.max =
2T
τ e.max =
2
π ⋅ a⋅ b
16T
2
π ⋅ k ⋅ do
2T
(
)(
π ⋅ 0.5 ⋅ do ⋅ 0.5 ⋅ k⋅ do
)2
3
Factor of increase in shear stress:
16T
Fτ =
τ e.max
τ c.max
2
Fτ =
π ⋅ k ⋅ do
16T
π ⋅ do
3
3
Fτ =
1
2
k
Ans
Problem 5-97
The 2014-T6 aluminum strut is fixed between the two walls at A and B. If it has a 50 mm by 50 mm
square cross section, and it is subjected to the torsional loading shown, determine the reactions at the
fixed supports. Also, what is the angle of twist at C?
Given: L AC := 600mm L CD := 600mm
L DB := 600mm T C := 60N⋅ m
T D := 30N⋅ m
Solution:
a := 50mm
G := 27GPa
T := T C + T D
T = 90 N⋅ m
L := L AC + L CD + L DB
φT − φB = 0
Compatibility :
(
) + 7.10(TD⋅ LCD) − 7.10(TB⋅ L) = 0
7.10 T ⋅ L AC
4
4
G⋅ a
T B :=
G⋅ a
4
G⋅ a
T⋅ LAC + TD⋅ LCD
L
T B = 40 N⋅ m
Ans
T A = 50 N⋅ m
Ans
Equilibrium:
TA + TB − T = 0
T A := T − TB
Angle of Twist:
φ C :=
( 7.10 )TA⋅ LAC
4
G⋅ a
φ C = 0.001262 rad
φ C = 0.0723 deg
Ans
Problem 5-98
The 304 stainless steel tube has a thickness of 10 mm. If the allowable shear stress is τallow = 80 MPa,
determine the maximum torque T that it can transmit. Also, what is the angle of twist of one end of th
tube with respect to the other if the tube is 4 m long? Neglect the stress concentrations at the corners.
The mean dimensions are shown.
Given:
a := 70mm
b := 30mm
L := 4m
τ allow := 80⋅ MPa
t := 10mm
G := 75GPa
Solution:
Section properties : S = Σds
Am := a⋅ b
S := 2a + 2⋅ b
2
Am = 2100 mm
S = 200 mm
Average shear stress:
τ avg =
T
2t⋅ Am
τ avg := τ allow
T := 2⋅ t⋅ Am⋅ τ allow
T = 3.36 kN⋅ m
Ans
Angle of Twist :
φ=
⌠
⋅⎮
2
⎮
4Am ⋅ G ⌡
φ :=
⎛ S⎞
2
t
4A ⋅ G ⎝ ⎠
T⋅ L
T⋅ L
1
ds
t
⋅⎜
m
φ = 0.203175 rad
φ = 11.641 deg
Ans
Problem 5-99
The 304 stainless steel tube has a thickness of 10 mm. If the applied torque is T = 50 N·m, determine
the average shear stress in the tube. Neglect the stress concentrations at the corners. The mean
dimensions are shown.
Given:
a := 70mm
b := 30mm
t := 10mm
T := 50N⋅ m
G := 75GPa
Solution:
Section properties :
Am := a⋅ b
2
Am = 2100 mm
Average shear stress:
τ avg :=
T
2t⋅ Am
τ avg = 1.19 MPa
Ans
Problem 5-100
Determine the constant thickness of the rectangular tube if the average shear stress is not to exceed
84 MPa when a torque of T = 2.5 kN·m is applied to the tube. Neglect stress concentrations at the
corners. The mean dimensions of the tube are shown.
Given:
a := 100mm
b := 50mm
T := 2.5kN⋅ m
τ allow := 84MPa
Solution:
Section properties :
Am := a⋅ b
2
Am = 5000 mm
Average shear stress:
T
τ avg =
2t⋅ Am
t :=
T
2⋅ Am⋅ τ allow
t = 2.98 mm
Ans
Problem 5-101
Determine the torque T that can be applied to the rectangular tube if the average shear stress is not to
exceed 84 MPa. Neglect stress concentrations at the corners. The mean dimensions of the tube are
shown and the tube has a thickness of 3 mm.
Given:
a := 100mm
b := 50mm
t := 3mm
τ allow := 84MPa
Solution:
Section properties :
Am := a⋅ b
2
Am = 5000 mm
Average shear stress:
T
τ avg =
2t⋅ Am
T := 2⋅ t⋅ Am⋅ τ allow
T = 2.52 kN⋅ m
Ans
Problem 5-102
A tube having the dimensions shown is subjected to a torque of T = 50 N·m. Neglecting the stress
concentrations at its corners, determine the average shear stress in the tube at points A and B. Show
the shear stress on volume elements located at these points.
Given:
ao := 50mm
ta := 3mm
bo := 50mm
tb := 5mm
T := 50N⋅ m
Solution:
Section properties :
a := ao − ta
b := bo − tb
Am := a⋅ b
Am = 2115 mm
2
Average shear stress:
τ A.avg :=
T
2ta⋅ Am
τ A.avg = 3.94 MPa
Ans
τ B.avg :=
T
2tb⋅ Am
τ B.avg = 2.36 MPa
Ans
Problem 5-103
The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine the
average shear stress at points A and B if it is subjected to the torque of T = 5 N·m. Show the shear
stress on volume elements located at these points.
Given: a := 80mm
b := 110mm
c1 := 40mm
c2 := 30mm
t := 5mm
T := 5N⋅ m
Solution:
Section properties :
c :=
2
2
c1 + c2
1
Am := a⋅ b + ⋅ a⋅ c2
2
c = 50 mm
2
Am = 10000 mm
Average shear stress:
τ A.avg :=
T
2t⋅ Am
τ A.avg = 0.05 MPa
Ans
τ B.avg :=
T
2t⋅ Am
τ B.avg = 0.05 MPa
Ans
Problem 5-104
The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of
t = 5 mm. If the allowable shear stress is τallow = 56 MPa, and the tube is to resist a torque of T = 375
N·m, determine the necessary dimension b. The mean area for the ellipse is A m = π b (0.5b).
T := 375N⋅ m
Given:
t := 5mm
τ allow := 56MPa
Solution:
Section properties :
Am = π ⋅ b⋅ ( 0.5 ⋅ b)
Average shear stress:
T
τ avg =
2t⋅ Am
τ avg =
b :=
T
2t⋅ ⎡⎣π ⋅ b⋅ ( 0.5 ⋅ b)⎤⎦
T
t⋅ π ⋅ τ allow
b = 20.65 mm
Ans
Problem 5-105
The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine the
average shear stress at points A and B if the tube is subjected to the torque of T = 500 N·m. Show the
shear stress on volume elements located at these points. Neglect stress concentrations at the corners.
Given: a := 40mm
b := 100mm
c1 := 20mm
c2 := 30mm
t := 5mm
T := 500N⋅ m
Solution:
Section properties :
c :=
2
c1 + c2
2
c = 36.06 mm
⎞
⎛1
Am := a⋅ b + 2⎜ ⋅ a⋅ c2
2
⎝
⎠
2
Am = 5200 mm
Average shear stress:
τ A.avg :=
T
2t⋅ Am
τ A.avg = 9.62 MPa
Ans
τ B.avg :=
T
2t⋅ Am
τ B.avg = 9.62 MPa
Ans
Problem 5-106
The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of
t = 5 mm. If the allowable shear stress is τallow = 56 MPa, determine the necessary dimension b needed
to resist the torque shown.. The mean area for the ellipse is A m = π b (0.5b).
Given:
T1 := 75N⋅ m
τ allow := 56MPa
T2 := −120N⋅ m
t := 5mm
T3 := 450N⋅ m
Solution:
Section properties :
Am = π ⋅ b⋅ ( 0.5 ⋅ b)
Internal torque :
Tmax := T1 + T2 + T3
Tmax = 405 N⋅ m
Average shear stress:
T
τ avg =
2t⋅ Am
τ avg =
b :=
T
2t⋅ ⎡⎣π ⋅ b⋅ ( 0.5 ⋅ b)⎤⎦
T max
t⋅ π ⋅ τ allow
b = 21.46 mm
Ans
Problem 5-107
The symmetric tube is made from a high-strength steel, having the mean dimensions shown and a
thickness of 5 mm. If it is subjected to a torque of T = 40 N·m, determine the average shear stress
developed at points A and B. Indicate the shear stress on volume elements located at these points.
Given: a := 40mm
b := 60mm
t := 5mm
T := 40N⋅ m
Solution:
Section properties :
Am := a⋅ a + 4( a⋅ b)
2
Am = 11200 mm
Average shear stress:
τ A.avg :=
T
2t⋅ Am
τ A.avg = 0.357 MPa
Ans
τ B.avg :=
T
2t⋅ Am
τ B.avg = 0.357 MPa
Ans
Problem 5-108
Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle. By
what percentage is the torsional strength reduced when the eccentricity e is one-fourth of the
difference in the radii?
a−b
Given: e =
4
Solution:
For the aligned tube::
Section properties :
⎛ a + b⎞
Am = π ⋅ ⎜
⎝ 2 ⎠
t= a−b
2
Average shear stress:
τ A.avg =
T
2t⋅ Am
(
T = τ A.avg⋅ 2t⋅ Am
)
⎛ a + b⎞
⎝ 2 ⎠
2
T = τ A.avg⋅ ( 2) ( a − b) ⋅ π ⋅ ⎜
For the eccentric tube:
Section properties :
t' = a −
e ⎛e
⎞
−⎜ +b
2 ⎝2
⎠
⎛ a + b⎞
A'm = π ⋅ ⎜
⎝ 2 ⎠
2
t' = a − e − b
t' = a −
a−b
−b
4
t' =
Average shear stress:
τ A.avg =
T'
2t'⋅ A'm
(
T' = τ A.avg⋅ 2t'⋅ A'm
T' = τ A.avg⋅ ( 2)
)
3
⎛ a + b⎞
( a − b) ⋅ π ⋅ ⎜
4
⎝ 2 ⎠
2
Factor of increase in shear stress:
FT =
T'
T
τ A.avg⋅ ( 2)
Fτ =
3
4
( a − b) ⋅ π ⋅ ⎛⎜
a + b⎞
⎝ 2 ⎠
⎛ a + b⎞
⎝ 2 ⎠
τ A.avg⋅ ( 2) ( a − b) ⋅ π ⋅ ⎜
% reduction in strength :
(
)
T% = 25.00
Ans
T% := 1 − FT ⋅ 100
2
2
FT :=
3
4
3
( a − b)
4
Problem 5-109
For a given average shear stress, determine the factor by which the torque-carrying capacity is
increased if the half-circular sections are reversed from the dashed-line positions to the section shown.
The tube is 2.5 mm thick.
Given:
a := 30mm
b := 45mm
r := 15mm
t := 2.5mm
Solution:
Section properties :
am := a − t
rm := r − 0.5t
⎛ π 2⎞
A'm := am ⋅ b − 2⋅ ⎜ ⋅ rm
2
A'm = 643.54 mm
⎛ π 2⎞
Am := am ⋅ b + 2⋅ ⎜ ⋅ rm
2
Am = 1831.46 mm
( )
( )
Average shear stress:
T
τ avg =
2t⋅ Am
⎝
2
⎠
⎝
2
⎠
(
)
(
)
T = τ avg⋅ 2t⋅ Am
T' = τ avg⋅ 2t⋅ A'm
Hence, the factor of increase is:
α=
T
T'
α :=
Am
A'm
α = 2.85
Ans
Problem 5-110
For a given maximum shear stress, determine the factor by which the torque carrying capacity is
increased if the half-circular section is reversed from the dashed-line position to the section shown.
The tube is 2.5 mm thick.
Given:
a := 30mm
b := 45mm
r := 15mm
t := 2.5mm
Solution:
Section properties :
am := a − t
bm := b − 0.5t
rm := r − 0.5t
⎛ π 2⎞
A'm := am ⋅ bm − ⎜ ⋅ rm
2
A'm = 906.15 mm
⎛ π 2⎞
Am := am ⋅ bm + ⎜ ⋅ rm
2
Am = 1500.10 mm
( )
( )
Average shear stress:
T
τ avg =
2t⋅ Am
⎝
2
⎠
⎝
2
⎠
(
)
(
)
T = τ avg⋅ 2t⋅ Am
T' = τ avg⋅ 2t⋅ A'm
Hence, the factor of increase is:
α=
T
T'
α :=
Am
A'm
α = 1.66
Ans
Problem 5-111
The steel used for the shaft has an allowable shear stress of τallow = 8 MPa. If the members are
connected with a fillet weld of radius r = 4 mm, determine the maximum torque T that can be applied
Given:
D := 50mm
d := 20mm
r := 4mm
τ allow := 8MPa
Solution:
Stress Concentration Factor :
D
r
= 2.50
= 0.20
d
d
From Fig. 5-36,
K := 1.25
Allowable Shear Stress:
⎛ 0.5T⋅ c ⎞
⎝ J ⎠
τ allow = K⋅ ⎜
T :=
τ allow⋅ J
0.5K⋅ c
T = 20.11 N⋅ m
Ans
c :=
d
2
J :=
π 4
⋅d
32
Problem 5-112
The shaft is used to transmit 660 W while turning at 450 rpm. Determine the maximum shear stress in
the shaft. The segments are connected together using a fillet weld having a radius of 1.875 mm.
Unit used: rpm :=
Given:
2π rad
60 s
D := 25mm
d := 12.5mm
r := 1.875mm
ω := 450rpm
Solution:
T :=
P := 660W
P
T = 14.01 N⋅ m
ω
Stress Concentration Factor :
D
r
= 2.00
= 0.15
d
d
From Fig. 5-36,
K := 1.30
Max. shear stress:
⎛ T⋅ c ⎞
⎝ J ⎠
τ max := K⋅ ⎜
c :=
d
2
J :=
π ⎛ d⎞
⋅⎜
2 ⎝ 2⎠
4
τ max = 47.48 MPa
Ans
Problem 5-113
The shaft is fixed to the wall at A and is subjected to the torques shown. Determine the maximum
shear stress in the shaft. A fillet weld having a radius of 4.5 mm is used to connect the shafts at B.
Given:
D := 60mm
d := 30mm
r := 4.5mm
T C := 250N⋅ m
T D := −300N⋅ m
T E := 800N⋅ m
Solution:
Internal Torque : As shown in the torque diagram.
T CD := TC
T DB := TC + TD
T BE := T C + T D
T EA := TC + TD + T E
Maximum Shear Stress :
For segment CD:
c :=
d
2
τ CD :=
J :=
π 4
⋅d
32
TCD⋅ ( c)
J
τ CD = 47.16 MPa
Ans
(Max. )
For segment EA:
c' :=
D
2
τ EA :=
J' :=
π
4
⋅D
32
TEA⋅ ( c')
J'
τ EA = 17.68 MPa
For the fillet:
Ans
Stress Concentration Factor :
D
r
= 2.00
= 0.15
d
d
From Fig. 5-36,
K := 1.30
Max. shear stress:
⎛ TDB⋅ c ⎞
⎝ J ⎠
τ max := K⋅ ⎜
τ max = 12.26 MPa
Ans
Problem 5-114
The built-up shaft is to be designed to rotate at 720 rpm while transmitting 30 kW of power. Is this
possible? The allowable shear stress is τallow = 12 MPa.
Unit used: rpm := ⎛⎜
2π ⎞ rad
⎝ 60 ⎠ s
Given:
d := 60mm
D := 75mm
P := 30kW
τ allow := 12MPa
Solution:
ω = 75.40
rad
s
T :=
ω := 720rpm
P
ω
T = 397.89 N⋅ m
Maximum Shear Stress :
τ allow = K⋅
T⋅ c
J
c :=
K :=
d
2
J :=
τ allow⋅ J
T⋅ c
π 4
⋅d
32
K = 1.28
Stress Concentration Factor :
D
= 1.25
d
K = 1.28
From Fig. 5-36,
r
= 0.133
d
r := 0.133d
Check:
D−d
= 7.50 mm
2
< r = 7.98mm
No. It si impossible.
r = 7.98 mm
Problem 5-115
The built-up shaft is designed to rotate at 540 rpm. If the radius of the fillet weld connecting the shaft
is r = 7.20 mm, and the allowable shear stress for the material is τallow = 55 MPa, determine the
maximum power the shaft can transmit.
Unit used: rpm := ⎛⎜
2π ⎞ rad
⎝ 60 ⎠ s
Given: d := 60mm
D := 75mm
r := 7.20mm
ω := 540rpm
τ allow := 55MPa
Solution:
Stress Concentration Factor :
D
r
= 1.25
= 0.12
d
d
From Fig. 5-36,
K := 1.30
c :=
Maximum Shear Stress :
τ allow = K⋅
T⋅ c
J
T :=
Maximum Power :
ω = 56.55
rad
s
P := T⋅ ω
P = 101.47 kW
Ans
d
2
J :=
τ allow⋅ J
K⋅ c
π 4
⋅d
32
T = 1.7943 kN⋅ m
Problem 5-116
The steel used for the shaft has an allowable shear stress of τallow = 8 MPa. If the members are
connected together with a fillet weld of radius r = 2.25 mm, determine the maximum torque T that can
be applied.
Given: d := 15mm
D := 30mm
r := 2.25mm
τ allow := 8MPa
Solution:
Stress Concentration Factor :
D
r
= 2.00
= 0.15
d
d
From Fig. 5-36,
K := 1.30
Maximum Shear Stress :
c :=
0.5T⋅ c
J
T :=
τ allow = K⋅
d
2
J :=
π 4
⋅d
32
τ allow⋅ J
0.5K⋅ c
T = 8.156 N⋅ m
Ans
Problem 5-117
A solid shaft is subjected to the torque T, which causes the material to yield. If the material is
elastic-plastic, show that the torque can be expressed in terms of the angle of twist φ of the shaft as
T = 4/3 TY (1 - φY3/ 4φ3), where TY and φY are the torque and angle of twist when the material begins
to yield.
Problem 5-118
A solid shaft having a diameter of 50 mm is made of elastic-plastic material having a yield stress of τY
= 112 MPa and shear modulus of G = 84 GPa. Determine the torque required to develop an elastic core
in the shaft having a diameter of 25 mm. Also, what is the plastic torque?
Given:
d := 50mm
G := 84GPa
dY := 25mm
τ Y := 112MPa
c :=
Solution:
d
2
ρ Y :=
dY
2
Elastic-plastic torque:
Use Eq. 5-26 from the text :
T :=
( ) ⋅ ⎛4c3 − ρ 3⎞
π ⋅ τY
6
⎝
Y ⎠
T = 3.551 kN⋅ m
Ans
TP = 3.665 kN⋅ m
Ans
Plastic torque:
Use Eq. 5-27 from the text :
TP :=
( ) ⋅ (c3)
2π ⋅ τ Y
3
Problem 5-119
Determine the torque needed to twist a short 3-mm-diameter steel wire through several revolutions if it
is made from steel assumed to be elastic-plastic and having a yield stress of τY = 80 MPa. Assume that
the material becomes fully plastic.
Given:
d := 3mm
τ Y := 80MPa
Solution:
c :=
d
2
Plastic torque:
Use Eq. 5-27 from the text :
TP :=
( ) ⋅ (c3)
2π ⋅ τ Y
3
TP = 0.565 N⋅ m
Ans
Problem 5-120
A solid shaft has a diameter of 40 mm and length of 1 m. It is made from an elastic-plastic material
having a yield stress of τY = 100 MPa. Determine the maximum elastic torque TY and the
corresponding angle of twist. What is the angle of twist if the torque is increased to T = 1.2TY ?
G = 80 GPa.
Given:
d := 40mm
G := 80GPa
L := 1m
τ Y := 100MPa
d
π 4
J :=
⋅d
2
32
Maximum Elastic Torque:
c :=
Solution:
TY⋅ c
τY =
T Y :=
J
(τ Y)⋅ J
c
T Y = 1.257 kN⋅ m
Ans
Angle of Twist:
γ Y :=
φ :=
τY
γ Y = 0.00125 rad
G
(γ Y)⋅ L
φ = 0.0625 rad
c
φ = 3.581 deg
Ans
T = 1.2TY
Elastic-plastic torque:
Use Eq. 5-26 from the text :
T=
( ) ⋅ ⎛4c3 − ρ 3⎞
π ⋅ τY
6
1.2T Y =
⎝
( ) ⋅ ⎛4c3 − ρ 3⎞
π ⋅ τY
6
3
ρ Y :=
Y ⎠
3
4c −
⎝
Y ⎠
7.2 ⋅ TY
( )
π ⋅ τY
ρ Y = 14.74 mm
Angle of Twist:
φ' :=
(γ Y)⋅ L
ρY
φ' = 0.084826 rad
φ' = 4.86 deg
Ans
Problem 5-121
The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diamet
segment. Determine the radius of the elastic core produced in the smaller diameter segment. Neglect
the stress concentration at the fillet.
Given:
do := 60mm
ds := 55mm
Solution:
do
co :=
2
π
4
Jo :=
⋅d
32 o
ds
cs :=
2
π
4
Js :=
⋅d
32 s
Set
τ Y := 1MPa
For the larger diameter segment.
Maximum Elastic Torque:
τY =
TY⋅ c
T Y :=
J
co
( ) ⋅ ⎛c 3⎞
2π ⋅ τ Y
T P = 43.56 N⋅ m
⎝s⎠
3
Applying Eq. 5-26 from the text :
T=
( ) ⋅ ⎛4c 3 − ρ 3⎞
π ⋅ τY
TY =
6
⎝ s
6
⎝ s
3
ρ Y :=
Y ⎠
( ) ⋅ ⎛4c 3 − ρ 3⎞
π ⋅ τY
3
4cs −
ρ Y = 12.98 mm
T Y = 42.41 N⋅ m
For the smaller diameter segment.
Elastic-plastic torque:
T P :=
(τ Y)⋅ Jo
Y ⎠
6⋅ T Y
( )
π ⋅ τY
Ans
> TY
Problem 5-122
A bar having a circular cross section of 75 mm diameter is subjected to a torque of 12 kN·m. If the
material is elastic-plastic, with τY = 120 MPa, determine the radius of the elastic core.
d := 75mm
Given:
T := 12kN⋅ m
τ Y := 120MPa
c :=
Solution:
d
2
Elastic-plastic torque:
Use Eq. 5-26 from the text :
T=
( ) ⋅ ⎛4c3 − ρ 3⎞
π ⋅ τY
6
3
ρ Y :=
⎝
Y ⎠
6T
3
4c −
( )
π ⋅ τY
ρ Y = 27.12 mm
Ans
Problem 5-123
A tubular shaft has an inner diameter of 20 mm, an outer diameter of 40 mm, and a length of 1 m. It i
made of an elastic perfectly plastic material having a yield stress of τY = 100 MPa. Determine the
maximum torque it can transmit.What is the angle of twist of one end with respect to the other end if
the shear strain on the inner surface of the tube is about to yield? G = 80 GPa.
Given:
do := 40mm
G := 80GPa
di := 20mm
τ Y := 100MPa
Solution:
ρ Y :=
Plastic Torque:
c
L := 1m
di
2
co := 0.5do
ci := 0.5di
⌠ o
2
T P := 2π ⋅ ⎮
τ Y ⋅ ρ dρ
⎮
⌡c
( )
i
T P = 1.466 kN⋅ m
Ans
Angle of Twist:
γ Y :=
φ :=
τY
G
(γ Y)⋅ L
ρY
γ Y = 0.00125 rad
φ = 0.125 rad
φ = 7.162 deg
Ans
Problem 5-124
The 2-m-long tube is made from an elastic-plastic material as shown. Determine the applied torque T,
which subjects the material of the tube's outer edge to a shearing strain of γmax = 0.008 rad. What
would be the permanent angle of twist of the tube when the torque is removed? Sketch the residual
stress distribution of the tube.
Given: ro := 45mm
γ Y := 0.003
ri := 40mm
L := 2m
τ Y := 240MPa
γ max := 0.008
Solution:
Determine if it is fully plastic :
γ max⋅ L
φ max :=
φ max = 0.35556 rad
ro
However,
γ Y⋅ L
γ Y⋅ L
φ max =
ρ Y :=
ρY
φ max
ρ Y = 16.88 mm
< ri = 40mm
Therefore, the tube is filly plastic.
Also, at r = ri :
γ max
ro
=
γr
ri
γ r :=
ri
⋅γ
ro max
γ r = 0.00711
> γ Y = 0.003
Again, the tube is filly plastic.
Plastic Torque :
r
⌠o
2
Tp := 2π ⋅ ⎮
τ Y ⋅ ρ dρ
⎮
⌡r
( )
Tp = 13.63 kN⋅ m
Ans
i
Angle of Twist:
When the torque is removed: The equal but opposite torque TP is applied.
G :=
τY
γY
φ' :=
J :=
π ⎛ 4
4
⋅ r − ri ⎞⎠
2 ⎝o
T p⋅ L
φ' = 0.14085 rad
G⋅ J
Permanent angle of twist: φ r := φ max − φ'
φ r = 0.21470 rad
Shear Stresses :
At r = ro :
τ' po :=
At r = ri :
τ' pi :=
T p⋅ r o
J
T p⋅ ri
J
φ r = 12.30 deg
τ' po = 253.5 MPa
τ' pi = 225.4 MPa
Ans
Problem 5-125
The tube has a length of 2 m and is made of an elastic-plastic material as shown. Determine the torqu
needed to just cause the material to become fully plastic. What is the permanent angle of twist of the
tube when this torque is removed?
Given: ro := 100mm ri := 60mm
γ Y := 0.007 τ Y := 350MPa
L := 2m
Solution:
At Just Fully Plastic : ρ Y := ri
γ Y⋅ L
φ p :=
φ p = 0.23333 rad
ρY
Plastic Torque :
r
⌠o
2
T p := 2π ⋅ ⎮
τ Y ⋅ ρ dρ
⎮
⌡r
( )
T p = 574.70 kN⋅ m
Ans
i
Angle of Twist:
When the torque is removed: The equal but opposite torque TP is applied.
τY
π
4
4
G :=
J := ⋅ ⎛⎝ ro − ri ⎞⎠
2
γY
T p⋅ L
φ' :=
φ' = 0.16814 rad
G⋅ J
Permanent angle of twist:
φ r := φ p − φ'
φ r = 0.06520 rad
φ r = 3.74 deg
Ans
Problem 5-126
The shaft is made from a strain-hardening material having a τ -γ diagram as shown. Determine the
torque T that must be applied to the shaft in order to create an elastic core in the shaft having a radius
of ρc = 12.5 mm.
Given:
r := 15mm
ρ c := 12.5mm
γ a := 0.005
τ a := 70MPa
γ b := 0.010
τ b := 105MPa
c := r
Solution:
ρ a := ρ c
For linear strain variations against ρ :
γ
ρ
Path 1:
Path 2:
=
⎛ γa ⎞
γa
ρa
γ= ⎜
⋅ρ
⋅γ
τ1 = ⎜
⎝ ρa⎠
⎛ τa ⎞
τ1 = ⎜
⎝ γa⎠
τ2 − τa
γ − γa
=
⎛ τa ⎞
⎝ ρa⎠
⋅ρ
⎛ τb − τa ⎞
τb − τa
τ2 = ⎜
⎝ γb − γa⎠
γb − γa
(
)
⋅ γ ⋅ −γ a + τ a
⎛ τb − τa ⎞ ⎛ ρ
⎞
⋅⎜
− 1 ⋅ γ a + τa
γ
−
γ
ρ
⎝ b a⎠ ⎝ a ⎠
τ2 = ⎜
c
⌠
2
T = 2π ⋅ ⎮ τ ⋅ ρ dρ
⌡0
ρ
⌠ a
⌠
2
T = 2π ⋅ ⎮ τ 1⋅ ρ dρ + 2π ⎮
⎮
⌡
⌡
c
ρa
0
ρ
2
τ 2⋅ ρ dρ
c
c
⎛ τa ⎞ ⌠ a 3
⎛ τb − τa ⎞ ⌠
⌠
⎮ ⎛ρ
2
2
⎞
T := 2π ⋅ ⎜
⋅ ⎮ ρ dρ + 2π ⋅ γ a⋅ ⎜
ρ dρ
⎮ ⎜ ρ − 1 ⋅ ρ dρ + 2π ⋅ τ a⋅ ⎮
⌡
ρ
γ
−
γ
⌡ρ
⎝ a⎠ 0
⎝ b a⎠ ⎮ ⎝ a ⎠
a
⌡ρ
T = 434.27 N⋅ m
Ans
a
Problem 5-127
The 2-m-long tube is made of an elastic perfectly plastic material as shown. Determine the applied
torque T that subjects the material at the tube's outer edge to a shear strain of γmax = 0.006 rad. What
would be the permanent angle of twist of the tube when this torque is removed? Sketch the residual
stress distribution in the tube.
Given: ro := 35mm
γ Y := 0.003
ri := 30mm
L := 2m
τ Y := 210MPa
γ max := 0.006
Solution:
At Fully Plastic : ρ Y := ri
γ max⋅ L
φ p :=
φ p = 0.34286 rad
ro
Also, at r = ri :
γ max
ro
=
γr
γ r :=
ri
ri
ro
⋅ γ max
γ r = 0.00514
> γ Y = 0.003 Confirm that tube is filly plastic.
Plastic Torque :
r
⌠o
2
T p := 2π ⋅ ⎮
τ Y ⋅ ρ dρ
⎮
⌡r
( )
T p = 6.98 kN⋅ m
Ans
i
Angle of Twist:
When the torque is removed: The equal but opposite torque TP is applied.
τY
π
4
4
G :=
J := ⋅ ⎛⎝ ro − ri ⎞⎠
2
γY
T p⋅ L
φ' :=
φ' = 0.18389 rad
G⋅ J
Permanent angle of twist:
φ r := φ p − φ'
φ r = 0.15897 rad
φ r = 9.11 deg
Residual Shear Stresses :
T p⋅ ro
At r = ro : τ' po :=
J
τ ro := −τ Y + τ' po
At r = ri :
τ' pi :=
T p⋅ r i
Ans
τ' po = 225.3 MPa
τ ro = 15.3 MPa
J
τ' pi = 193.1 MPa
τ ri := −τ Y + τ' pi
τ ri = −16.9 MPa
Problem 5-128
The shear stressstrain diagram for a solid 50-mm diameter shaft can be approximated as shown in the
figure. Determine the torque required to cause a maximum shear stress in the shaft of 125 MPa. If the
shaft is 3 m long, what is the corresponding angle of twist?
Given: do := 50mm
L := 3m
γ 1 := 0.0025
τ 1 := 50MPa
γ max := 0.010
τ max := 125MPa
Solution: ro := 0.5 ⋅ do
τ-ρ Function :
At r = ρ1 :
γ max
ro
γ1
=
ρ 1 :=
ρ1
γ1
γ max
⋅ ro
ρ 1 = 6.25 mm
For the region 0 < r < ρ1:
k1 :=
τ1 − 0
ρ1 − 0
τ−0
k1 =
MPa
k1 = 8.00
mm
τ = k1⋅ ρ
ρ−0
For the region ρ1 < r < r0:
k2 :=
k2 =
τ max − τ 1
MPa
k2 = 4.00
mm
ro − ρ 1
τ' − τ 1
(
τ' = τ 1 + k2⋅ ρ' − ρ 1
ρ' − ρ 1
)
Plastic Torque :
ρ
r
⌠ 1
⌠o
2
2
Tp = 2π ⋅ ⎮ ( τ' ) ⋅ ρ' dρ' + 2π ⋅ ⎮ ( τ ) ⋅ ρ dρ
⌡0
⌡
ρ1
ρ
r
⌠ 1
⌠o
2
2
Tp := 2π ⋅ ⎮
k1⋅ ρ ⋅ ρ dρ + 2π ⋅ ⎮ ⎡⎣τ 1 + k2⋅ ρ' − ρ 1 ⎤⎦ ⋅ ρ' dρ'
⎮
⌡
⌡
0
(
Tp = 3.27 kN⋅ m
)
ρ1
(
Ans
Angle of Twist:
φ max :=
γ max⋅ L
ro
φ max = 1.20000 rad
φ max = 68.75 deg
Ans
)
Problem 5-129
The shaft consists of two sections that are rigidly connected. If the material is elastic perfectly plastic
as shown, determine the largest torque T that can be applied to the shaft. Also, draw the shear-stress
distribution over a radial line for each section. Neglect the effect of stress concentration.
Given:
r1 := 20mm
r2 := 25mm
γ Y := 0.002
τ Y := 70MPa
Solution:
Plastic Torque :
For the smaller-diameter segment
c := r1
⌠
T p := 2π ⋅ ⎮
⌡
c
0
(τ Y)⋅ ρ 2 dρ
T p = 1.173 kN⋅ m
For the bigger-diameter segment
Max. Shear Stress :
c := r2
τ max :=
Ans
J :=
π 4
⋅r
2 2
T p⋅ ( c)
J
τ max = 47.79 MPa
Ans
Problem 5-130
The shaft is made of an elastic-perfectly plastic material as shown. Plot the shear-stress distribution
acting along a radial line if it is subjected to a torque of T = 2 kN·m. What is the residual stress
distribution in the shaft when the torque is removed?
Given:
ro := 20mm
T := 2kN⋅ m
γ Y := 0.001875
τ Y := 150MPa
c := ro
Solution:
J :=
π 4
⋅r
2 o
Maximum Elastic Torque:
τY =
TY⋅ c
T Y :=
J
(τ Y)⋅ J
c
T Y = 1.885 kN⋅ m
Plastic Torque:
2π ⋅ τ Y
3
T P :=
⋅ c
3
( )
T P = 2.513 kN⋅ m
< T = 2 kNm
> T = 2 kNm
Therefore, it is elastic-plastic.
Elastic-plastic Torque:
Applying Eq. 5-26 from the text :
T=
( ) ⋅ ⎛4c3 − ρ 3⎞
π ⋅ τY
6
3
ρ Y :=
⎝
Y ⎠
6⋅ T
3
4c −
( )
π ⋅ τY
ρ Y = 18.70 mm
Residual Shear Stresses :
When the torque is removed:
The equal but opposite torque T is applied.
T⋅ ro
At r = ro : τ' po :=
τ' po = 159.15 MPa
J
τ ro := −τ Y + τ' po
At r = ρY :
τ' pi :=
T⋅ ρ Y
J
τ ri := −τ Y + τ' pi
τ ro = 9.15 MPa
τ' pi = 148.78 MPa
τ ri = −1.22 MPa
Problem 5-131
A 40-mm-diameter shaft is made from an elasticplastic material as shown. Determine the radius of its
elastic core if it is subjected to a torque of T = 300 N·m. If the shaft is 250 mm long, determine the
angle of twist.
Given:
d := 40mm
L := 250mm
γ Y := 0.006
τ Y := 21MPa
Solution:
c :=
T := 300N⋅ m
d
2
Elastic-plastic torque:
Use Eq. 5-26 from the text :
T=
( ) ⋅ ⎛4c3 − ρ 3⎞
π ⋅ τY
6
3
ρ Y :=
⎝
Y ⎠
6T
3
4c −
( )
π ⋅ τY
ρ Y = 16.77 mm
Ans
Angle of Twist:
φ :=
(γ Y)⋅ L
ρY
φ = 0.089445 rad
φ = 5.1248 deg
Ans
Problem 5-132
A torque is applied to the shaft having a radius of 100 mm. If the material obeys a shear stress-strain
relation of τ = 20γ 1/3 MPa, determine the torque that must be applied to the shaft so that the maximum
shear strain becomes 0.005 rad.
Given: ro := 100mm
3
γ max := 0.005
Solution:
c := ro
τ-ρ Function :
γ=
τ = 20 γ ⋅ unit
unit := MPa
ρ
⋅γ
c max
3
τ = 20 γ ⋅ unit
3
τ = 20
ρ
⋅γ
⋅ unit
c max
Ultimate Torque :
c
⌠
2
T = 2π ⋅ ⎮ τ ⋅ ρ dρ
⌡0
c
⌠
3
ρ
2
⎮
T := 2π ⋅ unit⋅ ⎮ 20
⋅ γ max⋅ ρ dρ
c
⌡
( )
0
T = 6.446 kN⋅ m
Ans
Problem 5-133
The shaft is made of an elastic-perfectly plastic material as shown. Determine the torque that the shaf
can transmit if the allowable angle of twist is 0.375 rad. Also, determine the permanent angle of twist
once the torque is removed. The shaft is 2-m-long.
Given: ro := 20mm
φ allow := 0.375rad
γ Y := 0.001875
τ Y := 150MPa
c := ro
Solution:
Angle of Twist:
φ allow⋅ c
γ max :=
L
γ max
c
L := 2m
=
γY
γ max = 0.00375 rad
ρ Y :=
ρY
γY
γ max
⋅c
ρ Y = 10.00 mm
Elastic-plastic Torque:
Applying Eq. 5-26 from the text :
T :=
( ) ⋅ ⎛4c3 − ρ 3⎞
π ⋅ τY
6
⎝
T = 2.435 kN⋅ m
Y ⎠
Ans
Permanent Angle of Twist:
When the torque is removed: The equal but opposite torque T is applied.
G :=
φ' :=
T⋅ L
G⋅ J
τY
γY
J :=
π 4
⋅r
2 o
φ' = 0.24219 rad
Permanent angle of twist:
φ r := φ allow − φ'
φ r = 0.13281 rad
φ r = 7.61 deg
Ans
Problem 5-134
Consider a thin-walled tube of mean radius r and thickness t. Show that the maximum shear stress in
the tube due to an applied torque T approaches the average shear stress computed from Eq. 5-18 as r/t
approaches infinity.
Problem 5-135
The 304 stainless steel shaft is 3 m long and has an outer diameter of 60 mm. When it is rotating at 60
rad/s, it transmits 30 kW of power from the engine E to the generator G. Determine the smallest
thickness of the shaft if the allowable shear stress is τallow = 150 MPa and the shaft is restricted not to
twist more than 0.08 rad.
rad
Given: do := 60mm
L := 3m
ω := 60⋅
s
P := 30kW
G := 75GPa
τ allow := 150MPa
Solution:
T :=
P
φ allow := 0.08rad
T = 500 N⋅ m
ω
Allowable Shear Stress : Assume failure due to shear strss.
do
π
4
4
ro :=
c := ro
J = ⋅ ⎛⎝ ro − ri ⎞⎠
2
2
τ allow =
T⋅ c
J
J=
T⋅ c
τ allow
4
T⋅ ro
π ⎛ 4
4
⋅ ⎝ ro − ri ⎞⎠ =
2
τ allow
Angle of Twist :
φ=
T⋅ L
G⋅ J
2T ⋅ ro
4
ri :=
ro −
π ⋅ τ allow
ri = 29.392 mm
Assume failure due to angle of twist limitation.
J' :=
T⋅ L
G⋅ φ allow
T⋅ L
π ⎛ 4
4
⋅ ⎝ ro − r'i ⎞⎠ =
2
G⋅ φ allow
Choose the smallest value of ri :
4
r'i :=
(
ri := min ri , r'i
2T ⋅ L
4
ro −
)
π ⋅ G⋅ φ allow
ri = 28.403 mm
t := ro − ri
t = 1.597 mm
Ans
r'i = 28.403 mm
Problem 5-136
The 304 stainless solid steel shaft is 3 m long and has a diameter of 50 mm. It is required to transmit
40 kW of power from the engine E to the generator G. Determine the smallest angular velocity the
shaft can have if it is restricted not to twist more than 1.5°.
Given: do := 50mm
L := 3m
P := 40kW
G := 75GPa
φ allow := 1.5deg
Solution:
Angle of Twist :
ro :=
do
2
c := ro
φ=
T⋅ L
G⋅ J
T :=
J :=
π 4
⋅r
2 o
G⋅ J
⋅ φ allow
L
T = 401.60 N⋅ m
Angular Velocity:
T=
P
ω
ω :=
P
T
ω = 99.6
rad
s
Ans
Problem 5-137
The drilling pipe on an oil rig is made from steel pipe having an outside diameter of 112 mm and a
thickness of 6 mm. If the pipe is turning at 650 rev/min while being powered by a 12-kW motor,
determine the maximum shear stress in the pipe.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
do := 112mm
t := 6mm
P := 12kW
τ allow := 70MPa
ω = 68.07
T :=
Max. stress :
ω := 650⋅ rpm
rad
s
P
T = 176.29 N⋅ m
ω
do
c :=
2
τ max :=
di := do − 2t
T⋅ c
J
τ max = 1.75 MPa
Ans
4
4
⎡
⎛ di ⎞ ⎥⎤
π ⎞ ⎢⎛ do ⎞
⎛
J := ⎜ ⋅ ⎜
−⎜
⎝ 2 ⎠ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦
Problem 5-138
The tapered shaft is made from 2014-T6 aluminum alloy, and has a radius which can be described by
the function r = 0.02 (1 + x3/2) m, where x is in meters. Determine the angle of twist of its end A if it i
subjected to a torque of 450 N·m.
Given: T := 450N⋅ m
L := 4m
(
G := 27GPa
r = 0.02 1 +
3
)
x ⋅ unit
unit := m
Solution:
Angle of Twist :
J=
(
π 4
⋅r
2
⌠
φ= ⎮
⎮
⌡
L
r = 0.02 1 +
3
)
x ⋅ unit
T
dx
G⋅ J
0
L
⌠m
2T
⎮
φ := unit⋅ ⎮
dx
4
3
⎮
G⋅ π ⋅ ⎡⎣0.02 1 + x ⋅ unit⎤⎦
⌡
(
0
φ = 0.02771 rad
φ = 1.588 deg
Ans
)
Problem 5-139
The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at
1500 rev/min. Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowabl
shear stress is τallow = 56 MPa and the vibrations limit the angle of twist of the shaft to 0.05 rad. The
shaft is 0.6 m long and made of L2 tool steel.
Unit used:
Given:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
L := 600mm
ω := 1500⋅ rpm
P := 660kW
τ allow := 56MPa
G := 75GPa
φ allow := 0.05rad
Solution:
ω = 157.08
T :=
rad
s
P
T = 4.202 kN⋅ m
ω
Allowable shear stress :
Assume failure due to shear stress
π ⎛ d⎞
J = ⋅⎜
2 ⎝ 2⎠
d
c=
2
4
τ allow =
T⋅ c
J
1
⎡⎛ 2 ⎞ ⋅ ⎛ T ⎞⎤
⎜
⎥
⎣⎝ π ⎠ ⎝ τ allow ⎠⎦
3
d := 2 ⎢⎜
Thus,
d = 72.57 mm
Angle of Twist :
Assume failure due to angle of twist limitation
π ⎛ d⎞
J = ⋅⎜
2 ⎝ 2⎠
Thus,
4
φ=
⎡⎛ 2 ⎞ ⎛ T⋅ L ⎞⎤
d := 2 ⎢⎜ ⋅ ⎜
⎥
⎣⎝ π ⎠ ⎝ G⋅ φ allow ⎠⎦
d = 51.15 mm
Shear stress failure controls the design. Hence,
Use
d = 75mm
T⋅ L
G⋅ J
Ans
0.25
Problem 5-140
The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at
1500 rev/min. Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowabl
shear stress is τallow = 75 MPa and the vibrations limit the angle of twist of the shaft to 0.03 rad. The
shaft is 0.6 m long and made of L2 tool steel.
Unit used:
Given:
Solution:
⎛ 2π ⎞ rad
⎝ 60 ⎠ s
rpm := ⎜
L := 600mm
ω := 1500⋅ rpm
P := 660kW
τ allow := 75MPa
G := 75GPa
φ allow := 0.03rad
ω = 157.08
T :=
rad
s
P
T = 4.202 kN⋅ m
ω
Allowable shear stress :
Assume failure due to shear stress
π ⎛ d⎞
J = ⋅⎜
2 ⎝ 2⎠
d
c=
2
4
τ allow =
T⋅ c
J
1
Thus,
⎡⎛ 2 ⎞ ⋅ ⎛ T ⎞⎤
⎜
⎥
⎣⎝ π ⎠ ⎝ τ allow ⎠⎦
3
d := 2 ⎢⎜
d = 65.83 mm
Angle of Twist :
Assume failure due to angle of twist limitation
π ⎛ d⎞
J = ⋅⎜
2 ⎝ 2⎠
Thus,
4
φ=
⎡⎛ 2 ⎞ ⎛ T⋅ L ⎞⎤
d := 2 ⎢⎜ ⋅ ⎜
⎥
⎣⎝ π ⎠ ⎝ G⋅ φ allow ⎠⎦
d = 58.12 mm
Shear stress failure controls the design. Hence,
Use
d = 70mm
T⋅ L
G⋅ J
Ans
0.25
Problem 5-141
The material of which each of three shafts is made has a yield stress of τY and a shear modulus of G.
Determine which shaft geometry will resist the largest torque without yielding. What percentage of th
torque can be carried by the other two shafts? Assume that each shaft is made of the same amount of
material and that it has the same cross-sectional area A.
Given: AΟ = A
Asq = A
A∆ = A
Solution:
For circular shaft::
2
AΟ = πc
A
c=
π 4
⋅c
2
T⋅ c
τ max =
J
J=
π
2
J=
A
2π
τ max =
2⋅ π T
A A
T=
A A
2 π
α Ο :=
⋅ τY
1
2 π
α Ο = 0.2821
For square shaft:
2
Asq = a
τ max =
a=
4.81T
3
a
A
τ max =
4.81T
A⋅ A
T=
A A
⋅τ
4.81 Y
α sq :=
For triangular shaft:
θ := 60deg
a
A∆ = ⋅ a⋅ sin ( θ )
2
a=
τ max =
20T
3
a
4
3
5⋅ 27 T
2A⋅ A
T=
2A A
4
5⋅ 27
α ∆ :=
The circular shaft will carry the largest torque.
For square shaft:
%sq :=
For triangular shaft:
%∆ :=
α sq = 0.2079
2 A
4
τ max =
1
4.81
α sq
αΟ
α∆
αΟ
⋅ τY
2
4
5⋅ 27
α ∆ = 0.1755
Ans
⋅ 100
%sq = 73.7
Ans
⋅ 100
%∆ = 62.2
Ans
Problem 5-142
The A-36 steel circular tube is subjected to a torque of 10 kN·m. Determine the shear stress at the
mean radius ρ = 60 mm and compute the angle of twist of the tube if it is 4 m long and fixed at its far
end. Solve the problem using Eqs. 5-7 and 5-15 and by using Eqs. 5-18 and 5-20.
Given: ρ := 60mm
t := 5mm
T := 10kN⋅ m
L := 4m
G := 75GPa
Solution:
Section properties :
ro := ρ + 0.5t
ri := ρ − 0.5t
Σds := 2π ρ
Σds = 376.99 mm
2
2
Am := π ρ
π
4
4
J := ⋅ ⎛⎝ ro − ri ⎞⎠
2
Shear Stress:
Applying Eq. 5-7,
c := ρ
Am = 11309.73 mm
4
J = 6797621.10 mm
T⋅ c
J
τ :=
τ = 88.27 MPa
Ans
τ avg = 88.42 MPa
Ans
Applying Eq. 5-18,
τ avg :=
T
2⋅ t Am
Angle of Twist :
Applying Eq. 5-20,
φ' :=
T⋅ L
J⋅ G
φ' = 0.07846 rad
φ' = 4.495 deg Ans
Applying Eq. 5-20,
φ=
⌠
⋅⎮
2
⎮
4Am ⋅ G ⌡
φ :=
⎛ Σds ⎞
2
t ⎠
4A ⋅ G ⎝
T⋅ L
T⋅ L
m
⋅⎜
1
ds
t
φ = 0.07860 rad
φ = 4.503 deg Ans
Problem 5-143
The aluminum tube has a thickness of 5 mm and the outer cross-sectional dimensions shown.
Determine the maximum average shear stress in the tube. If the tube has a length of 5 m, determine th
angle of twist. Gal = 28 GPa.
Given: ao := 150mm
bo := 100mm
L := 4m
t := 5mm
T A := 280N⋅ m
T B := 135N⋅ m
L AB := 2m
L BC := 3m
G := 28GPa
Solution:
Section properties : S = Σds
a := ao − t
b := bo − t
Am := a⋅ b
S := 2a + 2⋅ b
Am = 13775 mm
S = 480 mm
2
Maximum Average shear stress:
T AB := TA
T BC := TA − T B
(
T max := max T AB , T BC
τ avg_max :=
)
T max = 280 N⋅ m
Tmax
2t⋅ Am
τ avg_max = 2.03 MPa
Ans
Angle of Twist :
⌠
⋅⎮
φ=
2
⎮
4Am ⋅ G ⌡
T⋅ L
φ :=
1
ds
t
TAB⋅ L AB + T BC⋅ LBC
2
4Am ⋅ G
φ = 0.004495 rad
φ = 0.258 deg
Ans
⎛ S⎞
⎝t⎠
⋅⎜
Problem 6-1
Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical
reactions on the shaft.
Given:
a := 250mm
b := 800mm
F := 24kN
Solution:
Given
Equilibrium :
+
ΣFy=0;
A+B−F= 0
ΣΜA=0;
−F⋅ a − B⋅ b = 0
Guess A := 1kN B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 31.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ −7.50 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
V1 x1 := −F⋅
kN
1
V2 x2 := ( −F + A) ⋅
kN
−F⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣−F⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
( )
( )
(
)
Moment (kN-m)
Shear (kN)
20
( ) 0
V2 ( x 2 )
20
V1 x 1
40
0
0.5
x1 , x2
Distance (m)
1
0
( )
M2(x2) 5
M1 x1
0
0.5
x1 , x2
Distane (m)
1
Problem 6-2
The load binder is used to support a load. If the force applied to the handle is 250 N, determine the
tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm
ABC.
Given:
a := 300mm
b := 75mm
F := 250N
Given
Solution:
Equilibrium :
+
ΣF y=0;
T1 − T2 − F = 0
ΣΜB=0;
− F ⋅ a + T 2⋅ b = 0
Guess T 1 := 1N
T 2 := 1N
⎛⎜ T1 ⎞
:= Find ( T 1 , T 2)
⎜ T2
⎝ ⎠
⎛⎜ T1 ⎞ ⎛ 1.25 ⎞
=⎜
kN
⎜ T2
⎝ ⎠ ⎝ 1.00 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
V1 x1 := −F⋅
kN
1
V2 x2 := −F + T 1 ⋅
kN
− F ⋅ x1
M1 x1 :=
N⋅ m
1
M2 x2 := ⎡⎣−F⋅ x2 + T1⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( ) (
( )
1
( )
V2 ( x 2 ) 0
V1 x 1
1
)
( )
Moment (N-m)
( )
Shear (kN)
Ans
(
)
0
( )
M 2 ( x 2 ) 50
M1 x1
100
0
0.2
x1 , x2
Distance (m)
0
0.2
x1 , x2
Distane (m)
Problem 6-3
Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only vertical
reactions on the shaft. The loading is applied to the pulleys at B and C and E.
Given:
a := 350mm
b := 500mm
c := 375mm
d := 300mm
B := 400N
C := 550N
E := 175N
Solution:
Given
Equilibrium :
+
ΣF y=0;
A+D−B−C−E= 0
ΣΜD=0; A⋅ ( a + b + c) − B⋅ ( b + c) − C⋅ c + E⋅ d = 0
Guess A := 1N D := 1N
⎛A⎞
⎛ A ⎞ ⎛ 411.22 ⎞
:= Find ( A , D)
=⎜
N
⎜
⎜
⎝D⎠
⎝ D ⎠ ⎝ 713.78 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
x4 := a + b + c , 1.01 ⋅ ( a + b + c) .. a + b + c + d
1
V1 x1 := A⋅
N
( )
1
V2 x2 := ( A − B) ⋅
N
1
V3 x3 := ( A − B − C) ⋅
N
( )
( )
1
V4 x4 := ( A − B − C + D) ⋅
N
( )
A⋅ x1
M1 x1 :=
N⋅ m
( )
1
M2 x2 := ⎡⎣A⋅ x2 − B⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − B⋅ x3 − a − C⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡A⋅ x4 − B⋅ x4 − a − C⋅ x4 − a − b + D⋅ x4 − a − b − c ⎤ ⋅
⎣
⎦ N⋅ m
( )
( )
(
)
0.2
0.4
0.6
(
)
(
)
500
( )
0
V2 ( x 2 )
V3 ( x 3 )
V4 ( x 4 ) 500
Shear (N)
V1 x 1
1000
0
0.8
1
1.2
1.4
1
1.2
1.4
x1 , x2 , x3 , x4
Distance (m)
Moment (N-m)
200
( )
M 2 ( x 2 ) 100
M3( x3)
0
M4( x4)
M1 x1
100
0
0.2
0.4
0.6
0.8
x1 , x2 , x3 , x4
Distance (m)
Problem 6-4
Draw the shear and moment diagrams for the beam.
Given:
a := 1m
F := 10kN
Solution:
Equilibrium :
+
ΣF y=0;
Given
A − 4F + B = 0
ΣΜB=0; A⋅ ( 5a) − F⋅ ( 4a + 3a + 2a + a) = 0
Guess A := 1kN B := 1kN
⎛A⎞
⎛ A ⎞ ⎛ 20 ⎞
:= Find ( A , B)
=⎜
kN
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 20 ⎠
x1 := 0 , 0.01 ⋅ a .. a
1
V1 x1 := A⋅
kN
( )
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
x4 := 3a , 1.01 ⋅ ( 3a) .. 4a
x5 := 4a , 1.01 ⋅ ( 4a) .. 5a
1
V2 x2 := ( A − F) ⋅
kN
1
V3 x3 := ( A − 2F) ⋅
kN
1
V4 x4 := ( A − 3F) ⋅
kN
1
V5 x5 := ( A − 4F) ⋅
kN
( )
( )
( )
A⋅ x1
M1 x1 :=
kN⋅ m
( )
1
M2 x2 := ⎡⎣A⋅ x2 − F⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − F⋅ x3 − a − F⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡⎣A⋅ x4 − F⋅ x4 − a − F⋅ x4 − 2a − F⋅ x4 − 3a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
(
)
1
M5 x5 := ⎡A⋅ x5 − F⋅ x5 − a − F⋅ x5 − 2a − F⋅ x5 − 3a − F⋅ x5 − 4a ⎤ ⋅
⎣
⎦ kN⋅ m
( )
(
)
(
)
(
)
(
)
20
( )
V2 ( x 2 )
V3 ( x 3 )
0
V4 ( x 4 )
V5 ( x 5 )
V1 x 1
Shear (kN)
( )
20
0
1
2
3
4
5
4
5
x1 , x2 , x3 , x4 , x5
Distance (m)
( )30
M2( x2)
M3( x3)
20
M4( x4)
M 5 ( x 5 ) 10
Moment (kN-m)
M1 x1
0
0
1
2
3
x1 , x2 , x3 , x4 , x5
Distance (m)
Problem 6-5
A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and
moment diagrams for the pier when it is subjected to the stringer loads shown. Assume the columns a
A and B exert only vertical reactions on the pier.
Given: a := 1m
Solution:
F1 := 60kN
F2 := 35kN
L := 4a + 2⋅ b
Equilibrium :
+
b := 1.5m
ΣF y=0;
Given
A + B − 2F1 − 3F2 = 0
ΣΜB=0; A⋅ ( 2a + 2⋅ b) − F1⋅ ( L − 2a) − F2⋅ ( 3⋅ b + 3a) = 0
Guess A := 1kN B := 1kN
⎛A⎞
⎛ A ⎞ ⎛ 112.5 ⎞
:= Find ( A , B)
=⎜
kN
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 112.5 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. ( 2a + b)
x4 := ( 2a + b) , 1.01 ⋅ ( 2a + b) .. ( 2a + 2⋅ b)
x5 := ( 2a + 2⋅ b) , 1.01 ⋅ ( 2a + 2⋅ b) .. ( 3a + 2⋅ b)
x6 := ( 3a + 2⋅ b) , 1.01 ⋅ ( 3a + 2⋅ b) .. ( 4a + 2⋅ b)
1
V1 x1 := −F1⋅
kN
( )
1
V2 x2 := A − F1 ⋅
kN
( ) (
1
V3 x3 := A − F1 − F2 ⋅
kN
)
( ) (
1
V4 x4 := A − F1 − 2F2 ⋅
kN
( ) (
)
1
V5 x5 := A − F1 − 3F2 ⋅
kN
)
( ) (
)
1
V6 x6 := A − F1 − 3F2 + B ⋅
kN
( ) (
−F1⋅ x1
M1 x1 :=
kN⋅ m
( )
)
1
M2 x2 := ⎡⎣A⋅ x2 − a − F1⋅ x2 ⎤⎦ ⋅
kN⋅ m
( )
(
)
( )
1
M3 x3 := ⎡⎣A⋅ x3 − a − F1⋅ x3 − F2⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
(
)
( )
(
)
1
M4 x4 := ⎡⎣A⋅ x4 − a − F1⋅ x4 − F2⋅ ⎡⎣2 x4 − 2⋅ a − b⎤⎦⎤⎦ ⋅
kN⋅ m
( )
(
)
( )
(
)
1
M5 x5 := ⎡⎣A⋅ x5 − a − F1⋅ x5 − F2⋅ ⎡⎣3 x5 − 2⋅ a − 3⋅ b⎤⎦⎤⎦ ⋅
kN⋅ m
( )
(
)
( )
(
)
1
M6 x6 := ⎡⎣A⋅ x6 − a − F1⋅ x6 − F2⋅ ⎡⎣3⋅ x6 − 2⋅ a − 3⋅ b⎤⎦ + B⋅ ⎡⎣x6 − ( L − a)⎤⎦⎤⎦ ⋅
kN⋅
( )
(
)
( )
(
)
( )
( )
V3 ( x 3 )
V4 ( x 4 ) 0
V5 ( x 5 )
V6 ( x 6 )
Shear (kN)
V1 x 1
50
V2 x 2
50
0
1
2
3
4
5
6
7
5
6
7
x1 , x2 , x3 , x4 , x5 , x6
Distance (m)
20
( )
M2( x2) 0
M3( x3)
20
M4( x4)
M5( x5)
40
M6( x6)
Moment (kN-m)
M1 x1
60
0
1
2
3
4
x1 , x2 , x3 , x4 , x5 , x6
Distance (m)
Problem 6-6
Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical
reactions on the shaft. Also, express the shear and moment in the shaft as a function of x within the
region 125 mm < x < 725 mm.
Given: a := 125mm
b := 600mm
F1 := 0.8kN
F2 := 1.5kN
L := a + b + c
Solution:
Equilibrium :
Given
ΣF y=0;
A − F1 − F2 + B = 0
ΣΜB=0;
A⋅ ( L) − F1⋅ ( b + c) − F2⋅ ( c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 0.8156 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 1.4844 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
V1 x1 := A⋅
kN
1
V2 x2 := A − F1 ⋅
kN
1
V3 x3 := A − F1 − F2 ⋅
kN
A⋅ x1
M1 x1 :=
N⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
( ) (
( )
)
( )
( ) (
(
)
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a − F2⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
0.3
0.4
0.5
0.6
0.7
1
Shear (kN)
+
c := 75mm
( )
V2 ( x 2 )
V3 ( x 3 )
1
V1 x 1 0
2
0
0.1
0.2
x1 , x2 , x3
Distance (m)
Moment (N-m)
150
( )
M2( x2)
M3( x3)
50
M 1 x 1 100
0
0
0.2
0.4
x1 , x2 , x3
Distance (m)
0.6
Problem 6-7
Draw the shear and moment diagrams for the shaft and determine the shear and moment throughout
the shaft as a function of x. The bearings at A and B exert only vertical reactions on the shaft.
Given:
a := 0.9m
b := 0.6m
c := 0.3m
d := 0.15m
F1 := 4kN
F2 := 2.5kN
Solution:
Equilibrium :
ΣF y=0;
A − F1 + B − F2 = 0
A⋅ ( a + b) − F1⋅ ( b) + F2⋅ ( c + d) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 0.85 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 5.65 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
V1 x1 := A⋅
kN
1
V2 x2 := A − F1 ⋅
kN
1
V3 x3 := A − F1 + B ⋅
kN
A⋅ x1
M1 x1 :=
N⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
( ) (
( )
)
( )
( ) (
(
)
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a + B⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
4
Shear (kN)
+
ΣΜB=0;
Given
2
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
2
4
0
0.5
1
x1 , x2 , x3
Distance (m)
1.5
Moment (N-m)
1000
( )
M2( x2)
M3( x3)
500
M1 x1
0
500
1000
0
0.5
1
x1 , x2 , x3
Distance (m)
1.5
Problem 6-8
Draw the shear and moment diagrams for the pipe. The end screw is subjected to a horizontal force of
5 kN. Hint: The reactions at the pin C must be replaced by equivalent loadings at point B on the axis
the pipe.
Given: a := 400mm
h := 80mm
F := 5kN
Given
Solution:
Equilibrium :
+
ΣF y=0;
A+C= 0
ΣΜC=0;
A⋅ a + F⋅ h = 0
Guess A := 1N
C := 1N
⎛A⎞
:= Find ( A , C)
⎜
⎝C⎠
⎛ A ⎞ ⎛ −1.00 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 1.00 ⎠
Ans
x1 := 0 , 0.01 ⋅ a .. a
1
V1 x1 := A⋅
kN
( )
A⋅ x1
M1 x1 :=
N⋅ m
( )
0
( )
V1 x 1
Moment (N-m)
Shear (kN)
1
0
1
2
( )
200
M1 x1
400
600
0
0.2
x1
Distance (m)
0.4
0
0.2
x1
Distane (m)
0
Problem 6-9
Draw the shear and moment diagrams for the beam. Hint: The 100-kN load must be replaced by
equivalent loadings at point C on the axis of the beam.
Given:
a := 1m
b := 1m
c := 1m
d := 0.25m
F1 := 75kN
F2 := 100kN
Solution:
Equilibrium :
ΣF y=0;
A − F1 + B = 0
ΣΜC=0;
A⋅ ( a + b + c) − F1⋅ ( b + c) − F2⋅ ( d) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 58.33 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 16.67 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
V1 x1 := A⋅
kN
1
V2 x2 := A − F1 ⋅
kN
1
V3 x3 := A − F1 ⋅
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( ) (
( )
)
( )
( ) (
(
)
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a − F2⋅ d⎤⎦ ⋅
kN⋅ m
( )
Shear (kN)
+
Given
( )
(
)
( )
V2 ( x 2 )
V3 ( x 3 )
V1 x 1 50
0
0
0.5
1
1.5
x1 , x2 , x3
Distance (m)
2
2.5
3
Moment (kN-m)
60
( )
M2( x2)
M3( x3)
20
M 1 x 1 40
0
0
0.5
1
1.5
x1 , x2 , x3
Distance (m)
2
2.5
3
Problem 6-10
The engine crane is used to support the engine, which has a weight of 6 kN. Draw the shear and
moment diagrams of the boom ABC when it is in the horizontal position shown.
Given:
a := 0.9m
b := 1.5m
c := 1.2m
W := 6kN
2
d :=
Solution:
v :=
2
a +c
c
d
h :=
a
d
Equilibrium :
Given
+
ΣFy=0;
−Ay + B⋅ v − W = 0
ΣΜA=0;
( −B⋅ v) ⋅ a + W⋅ ( a + b) = 0
ΣFx=0;
Ax − B⋅ h = 0
+
Guess Ax := 1kN
Ay := 1kN
⎛ Ax ⎞
⎜
⎜ Ay ⎟ := Find ( Ax , Ay , B)
⎜
⎝B⎠
B := 1kN
⎛ Ax ⎞ ⎛ 12 ⎞
⎜
⎜
⎜ Ay ⎟ = ⎜ 10 kN
⎜
⎝ B ⎠ ⎝ 20 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
V1 x1 := −Ay⋅
kN
1
V2 x2 := −Ay + B⋅ v ⋅
kN
−Ay⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣−Ay⋅ x2 + ( B⋅ v) ⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( ) (
( )
( )
)
( )
(
Moment (kN-m)
Shear (kN)
10
( ) 0
V2 ( x 2 )
V1 x 1
10
0
1
x1 , x2
Distance (m)
2
)
0
( )
M2(x2) 5
M1 x1
10
0
1
x1 , x2
Distane (m)
2
Problem 6-11
Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at A
which slides within the groove and so it cannot support a vertical force, although it can support a
moment and axial load.
Set:
a := 1m
P := 1kN
Solution:
Equilibrium :
ΣF y=0;
A−P+C−P= 0
ΣΜB=0; P⋅ ( 2a) − C⋅ a = 0
Guess A := 1kN
C := 1kN
⎛A⎞
⎛A⎞ ⎛0⎞
:= Find ( A , C)
=⎜
kN
⎜
⎜
⎝C⎠
⎝C⎠ ⎝2⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
x4 := 3a , 1.01 ⋅ ( 3a) .. 4a
1
V1 x1 := A⋅
kN
( )
1
V2 x2 := ( A − P) ⋅
kN
1
V3 x3 := ( A − P) ⋅
kN
( )
( )
1
V4 x4 := ( A − P + C) ⋅
kN
( )
2
( )
V2 ( x 2 )
V3 ( x 3 ) 0
V4 ( x 4 )
V1 x 1 1
Shear (P kN)
+
Given
1
2
0
0.5
1
1.5
2
x1 , x2 , x3 , x4
Distance (m)
2.5
3
3.5
4
MA := C⋅ ( 3a) − P⋅ ( 4a) − P⋅ ( a)
MA = 1.00 kN⋅ m
MA + A⋅ x1
1
M1 x1 :=
M2 x2 := ⎡⎣MA + A⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣( A − P) ⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
M4 x4 := ⎡⎣( A − P) ⋅ x4 − 2a + C⋅ x4 − 3a ⎤⎦ ⋅
kN⋅ m
( )
(
1
2
)
(
)
Moment (P kN-m)
1
( )
M2( x2)
M3( x3) 0
M4( x4)
M1 x1
1
0
x1 , x2 , x3 , x4
Distance (m)
3
4
Problem 6-12
Draw the shear and moment diagrams for the compound beam which is pin connected at B.
Given:
a := 1m
b := 1.5m
c := 1m
d := 1m
F1 := 30kN
F2 := 40kN
Solution:
Given
Equilibrium :
ΣF y=0;
− F1 + A − F2 + C = 0
ΣΜB=0; −F1⋅ ( a + b) + A⋅ b = 0
Guess A := 1kN
C := 1kN
⎛A⎞
⎛ A ⎞ ⎛ 50 ⎞
:= Find ( A , C)
=⎜
kN
⎜
⎜
⎝C⎠
⎝ C ⎠ ⎝ 20 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
x4 := a + b + c , 1.01 ⋅ ( a + b + c) .. a + b + c + d
1
V1 x1 := −F1⋅
kN
( )
1
1
V2 x2 := −F1 + A ⋅
V3 x3 := −F1 + A ⋅
kN
kN
( ) (
)
( ) (
)
1
V4 x4 := −F1 + A − F2 ⋅
kN
( ) (
)
20
( )
V2 ( x 2 ) 0
V3 ( x 3 )
V4 ( x 4 )
V1 x 1
Shear (kN)
+
20
40
0
1
2
x1 , x2 , x3 , x4
Distance (m)
3
4
−F1⋅ x1
M1 x1 :=
kN⋅ m
( )
1
M2 x2 := ⎡⎣−F1⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
M3 x3 := ⎡⎣ −F1 + A ⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
(
)(
)
1
M4 x4 := ⎡⎣ −F1 + A ⋅ x4 − a − b − F2⋅ x4 − a − b − c ⎤⎦ ⋅
kN⋅ m
( )
(
)(
)
(
)
20
Moment (N-m)
10
( )
M2( x2) 0
M3( x3)
10
M4( x4)
M1 x1
20
30
0
1
2
x1 , x2 , x3 , x4
Distance (m)
3
4
Problem 6-13
Draw the shear and moment diagrams for the beam.
Set:
a := 1m
Mo := 1kN⋅ m
Solution:
Equilibrium :
Given
ΣF y=0;
A+B= 0
+
ΣΜB=0; A⋅ ( 3a) + 2Mo − Mo = 0
Guess A := 1kN
B := 1kN
⎛A⎞
⎛ A ⎞ ⎛ −0.33 ⎞
:= Find ( A , B)
=⎜
kN
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 0.33 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
1
V1 x1 := A⋅
kN
1
V2 x2 := ( A) ⋅
kN
1
V3 x3 := ( A) ⋅
kN
( )
( )
M1 x1 :=
( )
Mo + A⋅ x1
kN⋅ m
( )
1
M2 x2 := ⎡⎣2⋅ Mo + A⋅ x2 ⎤⎦ ⋅
kN⋅ m
( )
( )
1
M3 x3 := ⎡⎣2⋅ Mo + A⋅ x3 − Mo⎤⎦ ⋅
kN⋅ m
( )
( )
( )
V2 ( x 2 )
V3 ( x 3 )
2
V1 x 1
0
0.5
0
1
2
x1 , x2 , x3
Distance (m)
3
Moment (Mo kN-m)
Shear (Mo/a kN)
0.5
( )
M2( x2)
1
M3( x3)
M1 x1
0
0
1
2
x1 , x2 , x3
Distance (m)
Problem 6-14
Consider the general problem of a simply supported beam subjected to n concentrated loads. Write a
computer program that can be used to determine the internal shear and moment at any specified
location x along the beam, and plot the shear and moment diagrams for the beam. Show an applicatio
of the program using the values P1 = 2.5 kN, d1 = 1.5 m, P 2 = 4 kN, d2 = 4.5 m, L1 = 3 m, L = 4.5 m.
Problem 6-15
The beam is subjected to the uniformly distributed moment m (Moment/length). Draw the shear and
moment diagrams for the beam.
Set:
L := 1m
Solution:
mo := 1
Given
kN⋅ m
m
Equilibrium :
+
ΣF y=0;
A := 0
ΣΜA=0;
MA := mo⋅ L
x1 := 0 , 0.01 ⋅ L .. L
1
V1 x1 := A⋅
kN
( )
1
M1 x1 := MA + A⋅ x1 − mo⋅ x1 ⋅
kN⋅ m
( ) (
)
1
Moment m*L (kN-m)
Shear (kN)
1.5
1
( )
V1 x 1
0.5
0
( )
M 1 x 1 0.5
0
0
0.5
1
0
0.5
x1
x1
Distance (m)
Distane (m)
1
Problem 6-16
Draw the shear and moment diagrams for the beam.
a := 2.5m
Given:
b := 2.5m
w := 10
kN
m
Solution:
Equilibrium :
+
A := w⋅ a − w⋅ a
ΣF y=0;
A = 0 kN
ΣΜA=0;
MA := ( w⋅ a) ⋅ ( 0.5a) − ( w⋅ b) ⋅ ( a + 0.5b)
MA = −62.50 kN⋅ m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
V1 x1 := A − w⋅ x1 ⋅
kN
( ) (
)
1
V2 x2 := ⎡⎣A − w⋅ a + w⋅ x2 − a ⎤⎦ ⋅
kN
( )
(
)
1
2
M1 x1 := ⎛⎝ −MA + A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
( )
1
2
M2 x2 := ⎡⎣−MA + A⋅ x2 − ( w⋅ a) ⋅ x2 − 0.5a + 0.5w⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
Moment (kN-m)
Shear (kN)
0
( )
V2 ( x 2 )
20
V1 x 1
0
2
x1 , x2
Distance (m)
4
( )50
M2( x2)
M1 x1
0
0
2
x1 , x2
Distane (m)
4
Problem 6-17
The 75-kg man sits in the center of the boat, which has a uniform width and a weight per linear foot of
50 N/m. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a
uniform distributed load upward on the bottom of the boat.
a := 2.5m
Given:
b := 2.5m
W := Mw⋅ g
Solution:
Mw := 75kg
N
w := 50
m
Equilibrium :
+
ΣF y=0;
q :=
W + w⋅ ( 2a)
2a
q = 197.1
N
m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
V1 x1 := ⎡⎣( q − w) ⋅ x1⎤⎦ ⋅
N
( )
1
V2 x2 := ⎡⎣( q − w) ⋅ x2 − W⎤⎦ ⋅
N
( )
1
2
M1 x1 := ⎡⎣0.5 ( q − w) ⋅ x1 ⎤⎦ ⋅
N⋅ m
( )
1
2
M2 x2 := ⎡⎣0.5 ( q − w) ⋅ x2 − W⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
(
)
( )
V2 ( x 2 )
Moment (N-m)
Shear (N)
500
V1 x 1
0
500
0
2
4
400
( )
M 2 ( x 2 )200
M1 x1
0
0
2
x1 , x2
x1 , x2
Distance (m)
Distane (m)
4
Problem 6-18
Draw the shear and moment diagrams for the beam. It is supported by a smooth plate at A which
slides within the groove and so it cannot support a vertical force, although it can support a moment an
axial load.
Set:
L := 1m
Solution:
w := 1
Given
kN
m
A := 0
Equilibrium :
+
ΣF y=0;
A + B − w⋅ L = 0
B := w⋅ L
L
ΣΜA=0; MA + A⋅ L − ( w⋅ L) ⋅ = 0
2
w⋅ L
MA :=
2
2
x1 := 0 , 0.01 ⋅ L .. L
1
V1 x1 := A − w⋅ x1 ⋅
kN
( ) (
)
w 2⎞
1
⎛
M1 x1 := ⎜ MA + A⋅ x1 − ⋅ x1 ⋅
2
kN⋅ m
⎝
⎠
Moment w*L*L (kN-m)
Shear w*L (kN)
( )
0
( )
V1 x 1
1
0
0.5
x1
Distance (m)
1
0.6
0.4
( )
M1 x1
0.2
0
0
0.5
x1
Distane (m)
1
Problem 6-19
Draw the shear and moment diagrams for the beam.
a := 1.5m
Given:
w := 30
Mo := 45kN⋅ m
kN
m
Solution:
Given
Equilibrium :
ΣF y=0;
−w⋅ a + A + B = 0
ΣΜA=0;
−( w⋅ a) ⋅ ( 0.5a) + Mo − B⋅ ( 2a) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
−w⋅ x1
V1 x1 :=
kN
( )
( )
M1 x1 :=
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
1
V2 x2 := ( −w⋅ a + A) ⋅
kN
1
V3 x3 := ( −w⋅ a + A) ⋅
kN
( )
2
−0.5 w⋅ x1
kN⋅ m
⎛ A ⎞ ⎛ 41.25 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 3.75 ⎠
( )
1
M2 x2 := ⎡⎣−( w⋅ a) ⋅ x2 − 0.5a + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
1
M3 x3 := ⎡⎣−( w⋅ a) ⋅ x3 − 0.5a + A⋅ x3 − a + Mo⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
0
Shear (kN)
+
( )
20
V2 ( x 2 )
V3 ( x 3 )
V1 x 1
40
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Moment (kN-m)
0
( )
M2( x2)
20
M3( x3)
M1 x1
40
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Problem 6-20
Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout
the beam as functions of x.
Given:
a := 2.4m
b := 1.2m
P1 := 50kN
w := 30
P2 := 40kN
kN
m
M2 := 60kN⋅ m
Solution:
Equilibrium :
+
ΣF y=0;
A := w⋅ a + P1 + P2
A = 162 kN
ΣΜA=0;
MA := ( w⋅ a) ⋅ ( 0.5a) + P1⋅ a + P2⋅ ( a + b) + M2
MA = 410.40 kN⋅ m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
V1 x1 := A − w⋅ x1 ⋅
kN
( ) (
1
V2 x2 := A − w⋅ a − P1 ⋅
kN
)
( ) (
)
1
2
M1 x1 := ⎛⎝ −MA + A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
( )
1
M2 x2 := ⎡⎣−MA + A⋅ x2 − ( w⋅ a) ⋅ x2 − 0.5 ⋅ a − P1⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
150
( )
100
V2 ( x 2 )
V1 x 1
50
0
)
0
Moment (kN-m)
Shear (kN)
200
(
0
1
2
x1 , x2
Distance (m)
3
( )
M2( x2)
M 1 x 1 200
400
0
1
2
x1 , x2
Distane (m)
3
Problem 6-21
Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam
as functions of x, where 1.2 m < x < 3 m.
a := 1.2m
Given:
Mo := 300N⋅ m
b := 1.8m
w := 2.5
c := 1.2m
kN
m
Solution:
Equilibrium :
ΣF y=0;
−w⋅ b + A + B = 0
ΣΜB=0;
−Mo + A⋅ b − ( w⋅ b) ⋅ ( 0.5b) + Mo = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 2.25 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 2.25 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
0
V1 x1 :=
kN
1
V2 x2 := ⎡⎣A − w⋅ x2 − a ⎤⎦ ⋅
kN
−Mo
M1 x1 :=
N⋅ m
1
2
M2 x2 := ⎡⎣−Mo + A⋅ x2 − a − 0.5w⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
( )
(
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + b + c)
)
( )
(
)
1
V3 x3 := ( A − w⋅ b + B) ⋅
kN
( )
(
)
1
M3 x3 := ⎡⎣−Mo + A⋅ x3 − a − ( w⋅ b) ⋅ x3 − a − 0.5 ⋅ b + B⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
(
)
(
)
(
)
2
Shear (kN)
+
Given
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
2
0
1
2
x1 , x2 , x3
Distance (m)
3
4
800
Moment (N-m)
600
( ) 400
M2( x2)
200
M3( x3)
M1 x1
0
200
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Problem 6-22
Draw the shear and moment diagrams for the compound beam.The three segments are connected by
pins at B and E.
Given: a := 2m
b := 1m
L BE := a + 2⋅ b
F := 3kN
w := 0.8
L AB := a + b
kN
m
L := 3a + 4⋅ b
Solution:
Equilibrium : Consider segment AB:
+
b
⋅F
a+b
ΣΜB=0; A⋅ ( a + b) − F⋅ ( b) = 0
A :=
ΣF y=0;
B := F − A
A+B−F= 0
A = 1.00 kN
B = 2.00 kN
Consider segment BE:
By symmetry,
+
ΣF y=0;
E= B
D= C
2C − 2B − w⋅ ( a + 2⋅ b) = 0
x1 := 0 , 0.01 ⋅ a .. a
C := B +
w
⋅ ( a + 2⋅ b)
2
C = 3.60 kN
D := C
D = 3.60 kN
E := B
E = 2.00 kN
x2 := a , 1.01 ⋅ a .. ( a + b)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + 2⋅
x4 := ( a + 2⋅ b) , 1.01 ⋅ ( a + 2⋅ b) .. ( 2a + 2⋅ b)
x5 := ( 2a + 2⋅ b) , 1.01 ⋅ ( 2a + 2⋅ b) .. ( 2a + 3⋅ b)
x6 := ( 2a + 3⋅ b) , 1.01 ⋅ ( 2a + 3⋅ b) .. ( 2a + 4⋅ b)
x7 := ( 2a + 4⋅ b) , 1.01 ⋅ ( 2a + 4⋅ b) .. ( 3a + 4⋅ b)
1
V1 x1 := A⋅
kN
1
V2 x2 := ( A − F) ⋅
kN
( )
1
V3 x3 := ⎡⎣A − F − w⋅ x3 − L AB ⎤⎦ ⋅
kN
( )
1
V4 x4 := ⎡⎣A − F + C − w⋅ x4 − LAB ⎤⎦ ⋅
kN
( )
(
)
( )
(
( )
)
(
)
V5 x5 := ⎡⎣A − F + C + D − w⋅ x5 − LAB ⎤⎦ ⋅
1
1
V6 x6 := ⎡⎣A − F + C + D − w⋅ LBE ⎤⎦ ⋅
V7 x7 := ⎡⎣A − 2F + C + D − w⋅ L BE ⎤⎦ ⋅
kN
kN
A⋅ x1
1
M1 x1 :=
M2 x2 := ⎡⎣A⋅ x2 − F⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
(
( )
)
( )
( )
( )
(
(
)
)
1
2
M3 x3 := ⎡⎣−B⋅ x3 − LAB − 0.5w⋅ x3 − L AB ⎤⎦ ⋅
kN⋅ m
( )
( )
(
(
)
)
(
(
)
( )
(
)
(
)2 + C⋅ (x5 − LAB − b) + D⋅ (x5 − LAB − b − a)⎤⎦ ⋅
M5 x5 := ⎡⎣−B⋅ x5 − LAB − 0.5w⋅ x5 − L AB
1
M6 x6 := ⎡⎣E ⋅ x6 − LAB − LBE ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
)2 + C⋅ (x4 − LAB − b)⎤⎦ ⋅ kN⋅ m
M4 x4 := ⎡⎣−B⋅ x4 − LAB − 0.5w⋅ x4 − L AB
1
M7 x7 := ⎡E ⋅ x7 − LAB − LBE − F⋅ x7 − LAB − LBE − b ⎤ ⋅
⎣
⎦ kN⋅ m
( )
(
)
(
)
4
( )
V2 ( x 2 )
2
V3 ( x 3 )
V4 ( x 4 )
0
V5 ( x 5 )
V6 ( x 6 )
V7 ( x 7 ) 2
Shear (kN)
V1 x 1
4
0
2
4
6
8
10
8
10
x1 , x2 , x3 , x4 , x5 , x6 , x7
Distance (m)
( )
M2( x2)
1
M3( x3)
M4( x4) 0
M5( x5)
M6( x6) 1
M7( x7)
2
Moment (kN-m)
M1 x1 2
3
0
2
4
6
x1 , x2 , x3 , x4 , x5 , x6 , x7
Distance (m)
Problem 6-23
Draw the shear and moment diagrams for the beam.
a := 1.5m
Given:
Mo := 30kN⋅ m
w := 30
kN
m
Solution:
Equilibrium :
+
Given
ΣF y=0;
−w⋅ a + A − w⋅ a + B = 0
ΣΜB=0;
Mo − ( w⋅ a) ⋅ ( 2.5a) + A⋅ ( 2a) − ( w⋅ a) ⋅ ( 0.5a) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
−w⋅ x1
V1 x1 :=
kN
⎛ A ⎞ ⎛ 57.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 32.50 ⎠
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ 2a .. 3a
1
V2 x2 := [ −( w⋅ a) + A] ⋅
kN
( )
( )
1
V3 x3 := ⎡⎣−( w⋅ a) + A − w⋅ x3 − 2a ⎤⎦ ⋅
kN
( )
Mo − 0.5w⋅ x1
M1 x1 :=
kN⋅ m
( )
2
(
)
1
M2 x2 := ⎡⎣Mo − ( w⋅ a) ⋅ x2 − 0.5a + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
1
2
M3 x3 := ⎡⎣Mo − ( w⋅ a) ⋅ x3 − 0.5a + A⋅ x3 − a − 0.5w⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
(
)
Shear (kN)
20
0
( )
V2 ( x 2 )
V3 ( x 3 ) 20
V1 x 1
40
0
1
2
3
4
x1 , x2 , x3
Distance (m)
Moment (kN-m)
30
20
( )
M2( x2)
M 3 ( x 3 ) 10
M1 x1
0
10
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Problem 6-24
The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed
loading on the beam over its 0.6-m length. Draw the shear and moment diagrams for the beam if it
supports a uniform loading of 30 kN/m.
a := 0.3m
Given:
c := 0.6m
b := 2.4m
kN
m
w := 30
Solution:
Equilibrium :
( )
ΣF y=0;
A − w⋅ b + qB ⋅ c = 0
ΣΜA=0;
( w⋅ b) ⋅ ( a + 0.5b) − qB⋅ c ⋅ ( a + b + 0.5c) = 0
(
Guess A := 1kN
⎛A⎞
⎜ q := Find ( A , qB)
⎝ B⎠
)
qB := 1
kN
m
A = 36.00 kN
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
A
V1 x1 :=
kN
1
V2 x2 := ⎡⎣A − w⋅ x2 − a ⎤⎦ ⋅
kN
( )
( )
(
qB = 60.00
kN
m
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + b + c)
)
1
V3 x3 := ⎡⎣A − w⋅ b + qB⋅ x3 − a − b ⎤⎦ ⋅
kN
( )
A⋅ x1
M1 x1 :=
kN⋅ m
(
)
1
2
M2 x2 := ⎡⎣A⋅ x2 − 0.5w⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
2
M3 x3 := ⎡⎣A⋅ x3 − ( w⋅ b) ⋅ x3 − a − 0.5 ⋅ b + 0.5qB⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
40
20
Shear (kN)
+
Given
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
20
40
0
0.5
1
1.5
2
x1 , x2 , x3
Distance (m)
2.5
3
40
Moment (kN-m)
30
( )
M2( x2)
20
M3( x3)
M1 x1
10
0
0
0.5
1
1.5
2
x1 , x2 , x3
Distance (m)
2.5
3
Problem 6-25
Draw the shear and moment diagrams for the beam. The two segments are joined together at B.
a := 0.9m
Given:
P := 40kN
b := 1.5m
w := 50
c := 2.4m
kN
m
Solution:
Equilibrium :
ΣF y=0;
A − P − w⋅ c + C = 0
ΣΜB=0;
( w⋅ c) ⋅ ( 0.5c) − C⋅ ( c) = 0
Guess A := 1kN
C := 1kN
⎛A⎞
:= Find ( A , C)
⎜
⎝C⎠
⎛ A ⎞ ⎛ 100 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 60 ⎠
MA := P⋅ a − ( C − w⋅ c) ⋅ ( a + b)
MA = 180 kN⋅ m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + b + c)
A
V1 x1 :=
kN
1
V2 x2 := ( A − P) ⋅
kN
1
V3 x3 := ⎡⎣A − P − w⋅ x3 − a − b ⎤⎦ ⋅
kN
( )
( )
M1 x1 :=
( )
−MA + A⋅ x1
kN⋅ m
( )
(
1
M2 x2 := ⎡⎣−MA + A⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
2
M3 x3 := ⎡⎣−MA + A⋅ x3 − P⋅ x3 − a − 0.5w⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
100
Shear (kN)
+
Given
( )
V2 ( x 2 )
V3 ( x 3 ) 0
V1 x 1
50
50
0
1
2
3
x1 , x2 , x3
Distance (m)
4
)
Moment (kN-m)
100
( ) 0
M2( x2)
M 3 ( x 3 ) 100
M1 x1
200
0
1
2
3
x1 , x2 , x3
Distance (m)
4
Problem 6-26
Consider the general problem of a cantilevered beam subjected to n concentrated loads and a constant
distributed loading w. Write a computer program that can be used to determine the internal shear and
moment at any specified location x along the beam, and plot the shear and moment diagrams for the
beam. Show an application of the program using the values P 1 = 4 kN, d1 = 2 m, w = 800 N/m, a1 = 2
m, a2 = 4 m, L = 4 m.
Problem 6-27
Determine the placement distance a of the roller support so that the largest absolute value of the
moment is a minimum. Draw the shear and moment diagrams for this condition.
Problem 6-28
Draw the shear and moment diagrams for the rod. Only vertical reactions occur at its ends A and B.
a := 900mm
Given:
B := 720N
w' := 2.4
Solution:
+
A := 360N
kN
m
⎛ xo ⎞
⋅x = 0
⎝a⎠ o
A − 0.5w'⋅ ⎜
ΣF y=0;
xo :=
A⋅ a
0.5w'
xo = 519.62 mm
ΣΜ
⎛ xo ⎞ ⎛ xo ⎞
Mmax := A⋅ xo − 0.5w'⋅ ⎜
⋅ xo⋅ ⎜
a
3
⎝
⎠
⎝
⎠
Mmax = 124.71 N⋅ m
Shear (N)
⎛ x ⎞ ⋅ x⎤ ⋅ 1
⎥
⎝a⎠ ⎦ N
⎡
⎣
⎛ x ⎞ ⋅ x⋅ ⎛ x ⎞⎤ 1
⎜ ⎥
⎝ a ⎠ ⎝ 3 ⎠⎦ N⋅ m
⎡
⎣
V ( x) := ⎢A − 0.5w'⋅ ⎜
x := 0 , 0.01 ⋅ a .. a
M ( x) := ⎢A⋅ x − 0.5w'⋅ ⎜
0
V( x )
500
0
0.2
0.4
0.6
0.8
0.6
0.8
x
Moment (N-m)
Distance (m)
100
M( x)
50
0
0
0.2
0.4
x
Distance m)
Problem 6-29
Draw the shear and moment diagrams for the beam.
Given:
Set
L := 1m
a :=
L
3
kN
wo := 1
m
Solution:
Equilibrium :
+
Given
(
)
ΣF y=0;
A − 2 0.5wo ⋅ a − wo⋅ a + B = 0
ΣΜB=0;
A⋅ ( 3⋅ a) − 0.5 ⋅ wo⋅ a ⋅ ⎜ 2a +
) ⎛⎝
(
Guess A := 1kN
⎝
)
(
)
⎛ A ⎞ ⎛ 0.33 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 0.33 ⎠
x1 := 0 , 0.01 ⋅ a .. a
wo ⎛ x1 ⎞ ⎤ 1
⎡
V1 x1 := ⎢A −
⋅⎜
⋅ x1⎥ ⋅
2
a
kN
⎣
(
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
( )
a⎞
⎛ 2a ⎞ = 0
− wo⋅ a ⋅ ( 1.5a) − 0.5 ⋅ wo⋅ a ⋅ ⎜
3⎠
⎝3⎠
⎠
x2 := a , 1.01 ⋅ a .. ( 2a)
1
V2 x2 := ⎡⎣A − 0.5 ⋅ wo⋅ a − wo⋅ x2 − a ⎤⎦ ⋅
kN
( )
⎦
( )
x3 := ( 2a) , 1.01 ⋅ ( 2a) .. ( 3a
(
⎡
⎣
(
)
⎛
)⎝
V3 x3 := ⎢A − 0.5 ⋅ wo⋅ a − wo⋅ a − wo⋅ x3 − 2a ⋅ ⎜ 1 − 0.5 ⋅
Shear (kN)
0.4
( )
V2 ( x 2 )
V3 ( x 3 )
0.2
V1 x 1
0
0.2
0.4
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
0.8
x3 − 2a ⎞⎤ 1
⎥⋅
a ⎠⎦ kN
wo ⎛ x1 ⎞
x1⎤ 1
⎡
M1 x1 := ⎢A⋅ x1 −
⋅⎜
⋅ x1⋅ ⎥ ⋅
2 ⎝ a⎠
3 ⎦ N⋅ m
⎣
( )
( )
⎡
⎣
M2 x2 := ⎢A⋅ x2 −
( )
M'3 x3 :=
( )
wo⋅ a
2a ⎞
2⎤ 1
⎛
⋅ ⎜ x2 −
− 0.5wo⋅ x2 − a ⎥ ⋅
2 ⎝
3⎠
⎦ N⋅ m
(
)
wo
⎛ x3 − 2⋅ a ⎞ 1⎤
2 ⎡
⋅ x3 − 2⋅ a ⋅ ⎢1 − ⎜
⋅ ⎥
2
⎣ ⎝ a ⎠ 3⎦
⎡
⎣
(
M3 x3 := ⎢A⋅ x3 −
)
wo⋅ a
⎤ 1
2⋅ a ⎞
⎛
⋅ ⎜ x3 −
− wo⋅ a ⋅ x3 − 1.5 ⋅ a − M'3 x3 ⎥ ⋅
2 ⎝
3 ⎠
⎦ N⋅ m
(
)(
)
( )
120
Moment (N-m)
100
80
( )
M 2 ( x 2 ) 60
M3( x3)
M1 x1
40
20
0
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
0.8
Problem 6-30
Draw the shear and moment diagrams for the beam.
kN
Set:
L := 1m
wo := 1
m
Solution:
Equilibrium :
ΣF y=0;
A + B − 0.5 ⋅ wo⋅ L = 0
ΣΜB=0;
A⋅
2⋅ L ⎛ wo ⎞ ⎛ L ⎞
−⎜
⋅L ⋅⎜
= 0
3
⎝ 2 ⎠ ⎝ 3⎠
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
Let a :=
⎛ A ⎞ ⎛ 0.25 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 0.25 ⎠
L
3
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 3a
⎡ wo x1
⎤ 1
V1 x1 := ⎢−
⋅ ⋅ x1 ⎥ ⋅
2 L
kN
wo x2
⎡
⎤ 1
V2 x2 := ⎢A −
⋅ ⋅ x2 ⎥ ⋅
2 L
⎣
⎦ kN
( )
( )⎦
⎣
( )
( )
⎡ wo x1
⎛ x1 ⎞⎤ 1
M1 x1 := ⎢−
⋅ ⋅ x1 ⋅ ⎜
⎥⋅
3
2 L
kN⋅ m
( )
( )⎝
⎣
⎠⎦
wo x2
⎡
⎛ x2 ⎞⎤ 1
M2 x2 := ⎢A⋅ x2 − a −
⋅ ⋅ x2 ⋅ ⎜
⎥⋅
2 L
⎣
⎝ 3 ⎠⎦ kN⋅ m
( )
(
)
( )
0.2
Shear Wo (kN)
+
Given
( )
V2 ( x 2 )
V1 x 1
0
0.2
0
0.2
0.4
0.6
x1 , x2
Distance (m)
0.8
Moment Wo*L*L (kN-m)
0.04
( )
0.02
M2( x2)
M1 x1
0
0
0.2
0.4
0.6
x1 , x2
Distance (m)
0.8
Problem 6-31
The T-beam is subjected to the loading shown. Draw the shear and moment diagrams.
a := 2m
Given:
P := 10kN
b := 3m
w := 3
c := 3m
kN
m
Solution:
Equilibrium :
ΣF y=0;
−P + A − w⋅ c + B = 0
ΣΜB=0;
−P⋅ ( a + b + c) + A⋅ ( b + c) − ( w⋅ c) ⋅ ( 0.5c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 15.58 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 3.42 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + b + c)
−P
V1 x1 :=
kN
1
V2 x2 := ( −P + A) ⋅
kN
1
V3 x3 := ⎡⎣−P + A − w⋅ x3 − a − b ⎤⎦ ⋅
kN
− P ⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣−P⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
( )
(
( )
(
)
)
1
2
M3 x3 := ⎡⎣−P⋅ x3 + A⋅ x3 − a − 0.5 ⋅ w⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
5
Shear (kN)
+
Given
( )
0
V2 ( x 2 )
V3 ( x 3 ) 5
V1 x 1
10
0
1
2
3
4
x1 , x2 , x3
Distance (m)
5
6
7
8
5
Moment (kN-m)
0
5
( )
M2( x2)
10
M3( x3)
M1 x1
15
20
25
0
2
4
x1 , x2 , x3
Distance (m)
6
8
Problem 6-32
The ski supports the 900-N (~90-kg) weight of the man. If the snow loading on its bottom surface is
trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the
ski.
a := 0.5m
Given:
P := 900N
Solution:
Equilibrium :
ΣF y=0;
w
⋅ ( 2a + 4a) = 0
2
P
N
w :=
w = 600
3a
m
−P +
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
x4 := 3a , 1.01 ⋅ ( 3a) .. 4a
⎡ w ⎛ x1 ⎞ ⎤ 1
V1 x1 := ⎢ ⋅ ⎜
⋅ x1⎥ ⋅
a
2
N
( )
⎣
⎝
⎠
⎦
1
V2 x2 := ⎡⎣0.5 ⋅ w⋅ a + w⋅ x2 − a ⎤⎦ ⋅
N
( )
(
)
1
V3 x3 := ⎡⎣0.5 ⋅ w⋅ a + w⋅ a − P + w⋅ x3 − 2a ⎤⎦ ⋅
N
( )
( )
(
⎡
⎣
)
(
)⎝
⎛
x4 − 3a ⎞⎤ 1
⎥⋅
a ⎠⎦ N
1
1.5
V4 x4 := ⎢0.5 ⋅ w⋅ a + w⋅ a − P + w a + w⋅ x4 − 3a ⋅ ⎜ 1 − 0.5 ⋅
400
( )
V2 ( x 2 )
V3 ( x 3 )
V4 ( x 4 )
V1 x 1
200
Shear (N)
+
0
200
400
0
0.5
x1 , x2 , x3 , x4
Distance (m)
2
x1⎤ 1
⎡ w ⎛ x1 ⎞
M1 x1 := ⎢ ⋅ ⎜
⋅ x1⋅ ⎥ ⋅
3 N⋅ m
2
a
( )
⎣
⎝
⎠
⎦
⎡w⋅ a ⋅ ⎛ x − 2a ⎞ + w ⋅ x − a 2⎤ ⋅ 1
M2 x2 := ⎢
⎜ 2
⎥
3
2 2
2
N⋅ m
( )
⎣
⎝
(
⎠
)⎦
⎡w⋅ a ⋅ ⎛ x − 2a ⎞ + w ⋅ x − a 2 − P⋅ x − 2a ⎤ ⋅ 1
M3 x3 := ⎢
⎜ 3
⎥
3
3
2 3
2
N⋅ m
( )
⎣
( )
⎝
M'4 x4 :=
(
⎠
)
(
)⎦
⎛ x4 − 3⋅ a ⎞ 1⎤
w
2 ⎡
⋅ x4 − 3⋅ a ⋅ ⎢1 − ⎜
⋅ ⎥
2
⎣ ⎝ a ⎠ 3⎦
(
)
⎡w⋅ a ⋅ ⎛ x − 2⋅ a ⎞ + ( 2w⋅ a) ⋅ x − 2⋅ a − P⋅ x − 2a + M' x ⎤ ⋅ 1
M4 x4 := ⎢
⎜ 4
4
4
4 4 ⎥ N⋅ m
3
2
( )
⎣
⎝
(
⎠
)
(
)
( )⎦
Moment (N-m)
150
( )
M2( x2)
100
M3( x3)
M4( x4)
M1 x1
50
0
0
0.5
1
x1 , x2 , x3 , x4
Distance (m)
1.5
2
Problem 6-33
Draw the shear and moment diagrams for the beam.
L := 9m
Given:
kN
wo := 50
m
a := 0.5L
Solution:
Equilibrium :
+
Given
(
ΣF y=0;
)
A − 2 0.5wo ⋅ a + B = 0
ΣΜB=0; A⋅ L − 0.5 ⋅ wo⋅ a ⋅ ⎛⎜ a +
(
)⎝
Guess A := 1kN
a⎞
⎛ 2a ⎞ = 0
− 0.5 ⋅ wo⋅ a ⋅ ⎜
3⎠
⎝3⎠
(
)
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 112.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 112.50 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( 2a)
x1 ⎞⎤ 1
⎡
⎛
V1 x1 := ⎢A − wo⋅ x1⋅ ⎜ 1 − 0.5 ⋅
⎥⋅
a
kN
( )
⎣
⎝
⎠⎦
wo ⎛ x2 − a ⎞
⎡
⎤ 1
V2 x2 := ⎢A − 0.5 ⋅ wo⋅ a −
⋅⎜
⋅ x2 − a ⎥ ⋅
2 ⎝ a ⎠
⎣
⎦ kN
( )
(
wo
⎡
⎛ x1 ⎞ 1⎤⎤ 1
2 ⎡
M1 x1 := ⎢A⋅ x1 −
⋅ x1 ⋅ ⎢1 − ⎜
⋅ ⎥⎥ ⋅
2
⎣
⎣ ⎝ a ⎠ 3⎦⎦ kN⋅ m
( )
( )
M'2 x2 :=
⎡
⎣
( )
wo ⎛ x2 − a ⎞
⎛ x2 − a ⎞
⋅⎜
⋅ x2 − a ⋅ ⎜
2 ⎝ a ⎠
⎝ 3 ⎠
(
M2 x2 := ⎢A⋅ x2 −
wo⋅ a
⎤ 1
a⎞
⎛
⋅ ⎜ x2 −
− M'2 x2 ⎥ ⋅
2 ⎝
3⎠
⎦ kN⋅ m
( )
200
Moment (N-m)
Shear (kN)
100
( )
V2 ( x 2 )
)
V1 x 1
0
100
0
5
( )
100
M2( x2)
M1 x1
0
0
5
x1 , x2
x1 , x2
Distance (m)
Distance (m)
)
Problem 6-34
Draw the shear and moment diagrams for the wood beam, and determine the shear and moment
throughout the beam as functions of x.
a := 1m
Given:
P := 1kN
b := 1.5m
w := 2
c := 1m
kN
m
Solution:
Equilibrium :
+
Given
ΣF y=0;
A − w⋅ b + B − 2P = 0
ΣΜB=0;
−P⋅ ( a + b) + A⋅ b − ( w⋅ b) ⋅ ( 0.5b) + P⋅ c = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
⎛ A ⎞ ⎛ 2.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 2.50 ⎠
x2 := a , 1.01 ⋅ a .. ( a + b)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + b + c)
−P
V1 x1 :=
kN
1
V2 x2 := ⎡⎣−P + A − w⋅ x2 − a ⎤⎦ ⋅
kN
− P ⋅ x1
M1 x1 :=
kN⋅ m
1
2
M2 x2 := ⎡⎣−P⋅ x2 + A⋅ x2 − a − 0.5w⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
( )
1
V3 x3 := ( −P + A − w⋅ b + B) ⋅
kN
)
(
( )
)
(
)
1
M3 x3 := ⎡⎣−P⋅ x3 + A⋅ x3 − a − ( w⋅ b) ⋅ x3 − a − 0.5 ⋅ b + B⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
(
)
2
Shear (kN)
1
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
1
2
0
0.5
1
1.5
2
x1 , x2 , x3
Distance (m)
2.5
3
3.5
0
Moment (kN-m)
0.2
( ) 0.4
M2( x2)
M 3 ( x 3 ) 0.6
M1 x1
0.8
1
0
0.5
1
1.5
2
x1 , x2 , x3
Distance (m)
2.5
3
3.5
Problem 6-35
The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kN/m
caused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and draw
the shear and moment diagrams for the pin.
kN
Given: L := 100mm
w := 0.4
m
a := 0.2L
Solution:
Equilibrium :
(
ΣF y=0;
)
2 0.5wo ⋅ a − w⋅ ( 3a) = 0
kN
wo = 1.20
m
wo := 3w
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( 4a)
⎡wo ⎛ x1 ⎞ ⎤ 1
V1 x1 := ⎢ ⋅ ⎜
⋅ x1⎥ ⋅
2
a
N
( )
⎣
⎝
⎠
x3 := ( 4a) , 1.01 ⋅ ( 4a) .. L
⎡wo
⎤ 1
V2 x2 := ⎢ ⋅ a − w⋅ x2 − a ⎥ ⋅
2
N
( )
⎦
⎣
(
x3 − 4a ⎞⎤ 1
⎡wo
⎛
V3 x3 := ⎢ ⋅ a − w⋅ ( 3a) + wo⋅ x3 − 4a ⋅ ⎜ 1 − 0.5 ⋅
⎥⋅
a
2
N
( )
(
⎣
)⎝
⎠⎦
10
Shear (N)
+
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
10
0
0.02
0.04
0.06
x1 , x2 , x3
Distance (m)
0.08
)⎦
x1⎤ 1
⎡wo ⎛ x1 ⎞
M1 x1 := ⎢ ⋅ ⎜
⋅ x1⋅ ⎥ ⋅
3 N⋅ m
2
a
( )
⎣
⎝
⎠
⎦
⎡wo⋅ a ⎛
2a ⎞
2⎤ 1
M2 x2 := ⎢
⋅ ⎜ x2 −
− 0.5w⋅ x2 − a ⎥ ⋅
3⎠
⎣ 2 ⎝
⎦ N⋅ m
( )
(
( )
M'3 x3 :=
)
wo
⎛ x3 − 4⋅ a ⎞ 1⎤
2 ⎡
⋅ x3 − 4⋅ a ⋅ ⎢1 − ⎜
⋅ ⎥
2
⎣ ⎝ a ⎠ 3⎦
(
)
⎡⎡ wo⋅ a ⎛
⎤
⎤ 1
2⋅ a ⎞
M3 x3 := ⎢⎢
⋅ ⎜ x3 −
− w⋅ ( 3a) ⋅ x3 − 2.5 ⋅ a ⎥ + M'3 x3 ⎥ ⋅
3 ⎠
⎣⎣ 2 ⎝
⎦
⎦ N⋅ m
Moment (N-m)
( )
(
)
( )
( )0.2
M2( x2)
M 3 ( x 3 )0.1
M1 x1
0
0
0.02
0.04
0.06
x1 , x2 , x3
Distance (m)
0.08
Problem 6-36
Draw the shear and moment diagrams for the beam.
Given:
a := 3.6m
b := 1.8m
MA := 2.25kN⋅ m
w := 45
kN
m
Solution:
Given
Equilibrium :
ΣF y=0;
A + B − 0.5w⋅ b = 0
ΣΜB=0;
⎛ b⎞ = 0
MA + A⋅ a + ( 0.5w⋅ b) ⋅ ⎜
3
⎝ ⎠
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ −7.38 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 47.88 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
A
V1 x1 :=
kN
V2 x2 := ⎢A + B − w⋅ x2 − a ⋅ ⎜ 1 − 0.5 ⋅
( )
⎡
⎣
( )
⎛
)⎝
(
1
M1 x1 := MA + A⋅ x1 ⋅
kN⋅ m
( ) (
x2 − a ⎞⎤ 1
⎥⋅
b ⎠⎦ kN
)
⎡
⎛ x2 − a ⎞ 1⎤⎤ 1
w
2 ⎡
M2 x2 := ⎢MA + A⋅ x2 + B⋅ x2 − a − ⋅ x2 − a ⋅ ⎢1 − ⎜
⋅ ⎥⎥ ⋅
2
b
3 kN⋅ m
( )
(
⎣
)
(
) ⎣
⎝
⎠ ⎦⎦
40
Shear (kN)
+
( ) 20
V2 ( x 2 )
V1 x 1
0
20
0
1
2
3
x1 , x2
Distance (m)
4
5
Moment (kN-m)
0
( )
M2( x2)
M 1 x 1 10
20
30
0
1
2
3
x1 , x2
Distane (m)
4
5
Problem 6-37
The compound beam consists of two segments that are pinned together at B. Draw the shear and
moment diagrams if it supports the distributed loading shown.
L := 1m
Set:
a :=
L
3
w := 1
kN
m
Solution:
Consider segment AB.
ΣMB=0;
A⋅ ( 2a) −
w ⎛ 2a ⎞
⎛ 2a ⎞ = 0
⋅⎜
⋅ ( 2⋅ a) ⋅ ⎜
2 ⎝L⎠
⎝3⎠
2w⋅ L
27
Consider whole beam ABC.
A :=
MC = A⋅ L −
MC :=
A = 0.0741 kN
w⋅ L ⎛ L ⎞
⋅⎜
2 ⎝ 3⎠
−5w⋅ L
54
2
MC = −0.09259 kN⋅ m
x1 := 0 , 0.01 ⋅ ( 2a) .. ( 2a)
⎡
⎣
V ( x) := ⎢A −
w ⎛ x⎞ ⎤ 1
⋅ ⎜ ⋅ x⎥ ⋅
2 ⎝ L ⎠ ⎦ kN
⎡
⎣
M ( x) := ⎢A⋅ x −
w ⎛ x ⎞ ⎛ x ⎞⎤ 1
⋅ ⎜ ⋅ x⋅ ⎜ ⎥
2 ⎝ L ⎠ ⎝ 3 ⎠⎦ kN⋅ m
Moment w*L*L (kN-m)
Shear W*L (kN)
0.2
0
V( x )
0.2
0.4
0
0.5
1
0
M( x)
0.05
0.1
0
0.5
x
x
Distance (m)
Distance m)
1
Problem 6-38
Draw the shear and moment diagrams for the beam.
Given: L := 3m
kN
wo := 12
m
kN
w1 := 18
m
Solution:
Equilibrium :
+
ΣF y=0;
L
B := wo + w1 ⋅
2
ΣΜB=0;
L
L ⎛ L⎞
MB := wo⋅ L ⋅ + w1 − wo ⋅ ⋅ ⎜
2
2 3
(
)
(
)
(
B = 45.00 kN
)
⎝ ⎠
MB = 63.00 kN⋅ m
w' := w1 − wo
x1 := 0 , 0.01 ⋅ L .. L
⎡
⎣
V ( x) := ⎢−wo⋅ x −
w' ⎛ x ⎞ ⎤ 1
⋅ ⎜ ⋅ x⎥ ⋅
2 ⎝ L ⎠ ⎦ kN
0
Moment (kN-m)
Shear (kN)
0
V( x )
⎛ x ⎞ − w' ⋅ ⎛ x ⎞ ⋅ x⋅ ⎛ x ⎞⎤ ⋅ 1
⎜
⎜ ⎥
⎝ 2 ⎠ 2 ⎝ L ⎠ ⎝ 3 ⎠⎦ kN⋅ m
⎡
⎣
M ( x) := ⎢−wo⋅ x⋅ ⎜
20
40
0
1
2
x
Distance (m)
3
M( x)
50
0
1
2
x
Distance (m)
3
Problem 6-39
Draw the shear and moment diagrams for the beam and determine the shear and moment as functions
of x.
N
N
Given: a := 3m wo := 200
w1 := 400
m
m
Solution: L := 2a w' := w1 − wo
Equilibrium :
Given
+
ΣF y=0;
(
+
ΣΜB=0; A⋅ ( 2⋅ a) − wo⋅ a ⋅
a
) 2 +B= 0
A − wo + w1 ⋅
) 2 − w'⋅ 2 ⋅ ⎛⎜⎝ 3 ⎞⎠ = 0
(
a
a
a
Guess A := 1N
B := 1N
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 200 ⎞
=⎜
N
⎜
⎝ B ⎠ ⎝ 700 ⎠
Shear and Moment Functions :
For 0 < x < 3m,
V := A
V = 200 N
Ans
M = A⋅ x
M = ( 200x) ⋅ N⋅ m
Ans
For 3m < x < 6m,
V' = A − wo⋅ ( x − a) −
⎛
⎝
V' = ⎜ 500 −
w' ⎛ x − a ⎞
⋅⎜
⋅ ( x − a)
2 ⎝ a ⎠
100 2⎞
⋅x ⋅N
3
⎠
M' = A⋅ x − wo⋅ ( x − a) ⋅
⎛
⎝
M' = ⎜ 600 − 500x +
Ans
x − a w' ⎛ x − a ⎞
⎛ x − a⎞
− ⋅⎜
⋅ ( x − a) ⋅ ⎜
2
2 ⎝ a ⎠
⎝ 3 ⎠
100 3⎞
⋅ x ⋅ N⋅ m
9
⎠
Ans
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. L
A
V1 x1 :=
N
⎡
⎤ 1
w' ⎛ x2 − a ⎞
V2 x2 := ⎢A − wo⋅ x2 − a − ⋅ ⎜
⋅ x2 − a ⎥ ⋅
2
a
N
( )
( )
⎣
(
)
⎝
⎠
(
1
M1 x1 := A⋅ x1 ⋅
N⋅ m
( ) (
( )
⎡
⎣
)
(
)
M2 x2 := ⎢A⋅ x2 − wo⋅ x2 − a ⋅
x2 − a
2
−
⎛ x2 − a ⎞⎤ 1
w' ⎛ x2 − a ⎞
⋅⎜
⋅ x2 − a ⋅ ⎜
⎥⋅
2 ⎝ L ⎠
⎝ 3 ⎠⎦ N⋅ m
(
)
)⎦
500
( )
V2 ( x 2 )
V1 x 1
Moment (N-m)
Shear (N)
600
0
500
0
2
4
6
( )400
M2( x2)
M1 x1
200
0
0
2
4
x1 , x2
x1 , x2
Distance (m)
Distance (m)
6
Problem 6-40
Determine the placement distance a of the roller support so that the largest absolute value of the
moment is a minimum. Draw the shear and moment diagrams for this condition.
Solution:
Equilibrium :
ΣF y=0;
+
A − 2P + B = 0
L
− B⋅ a + P⋅ L = 0
2
3⋅ L
4a − 3L
B=
⋅P
A=
⋅P
2a
2a
ΣΜA=0; P⋅
+
Internal Moment :
For positive moment,
L
Mmax = A⋅
2
For negative moment,
Mmin = −P⋅ ( L − a)
When
Mmax = Mmin
A⋅
L
= P⋅ ( L − a)
2
⎛ 4a − 3L ⋅ P⎞ ⋅ L = P⋅ ( L − a)
⎜
⎝ 2a
⎠ 2
( 4a − 3L ) ⋅ L = 4a( L − a)
a=
Set:
L := 1m
P := 1kN
3
L
2
A :=
a :=
a' := 0.5L
3
L
2
4a − 3L
⋅P
2a
b' := a − a'
Ans
B :=
3⋅ L
⋅P
2a
c' := L − a
x1 := 0 , 0.01 ⋅ a' .. a'
x2 := a' , 1.01 ⋅ a' .. ( a' + b')
x3 := ( a' + b') , 1.01 ⋅ ( a' + b') .. L
1
V1 x1 := A⋅
kN
1
V2 x2 := ( A − P) ⋅
kN
1
V3 x3 := ( A − P + B) ⋅
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − P⋅ x2 − a' ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − P⋅ x3 − a' + B⋅ x3 − a' − b' ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
Shear P (kN)
1
( )
V2 ( x 2 )
V3 ( x 3 )
V1 x 1
0.5
0
0.5
1
0
0.2
0.4
0.6
0.8
x1 , x2 , x3
Moment P*L (N-m)
Distance (m)
( )
M2( x2)
M3( x3)
M1 x1
0.1
0
0.1
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
0.8
Problem 6-41
Draw the shear and moment diagrams for the beam.
Given:
kN
w0 := 8
m
a := 2m
kN
Solution: unit :=
m
w = ( unit) ⋅ 2x
3
2
x
⌠
Wx = ⎮ w dx
⌡
0
⎛⎜⌠x 2 ⎞
Wx = ( unit) ⋅ ⎮ 2x dx
⎜⌡0
⎝
⎠
2 3
Wx = ( unit) ⋅ ⋅ x
3
2 3
Wa = ( unit) ⋅ ⋅ a
3
a
ΣF y=0;
xc =
( )
⌠a
⎮ w⋅ x dx
⌡0
a
⌠
2
A := ( unit) ⋅ ⎮ 2x dx
⌡0
A = 5.33 kN
a
unit ⌠
3
xc :=
⎮ 2⋅ x dx
A ⌡0
A
MA := A⋅ xc
MA = 8.00 kN⋅ m
x := 0 , 0.01 ⋅ a .. a
V ( x) := ⎢A − ( unit) ⋅
⎡
⎣
xc = 1.500 m
2 3⎤ 1
x ⎥⋅
3 ⎦ kN
xc ⎞⎤
⎡
1
2 3 ⎛
M ( x) := ⎢−MA + A⋅ x − ( unit) ⋅ ⋅ x ⋅ x⋅ ⎜ 1 −
⎥⋅
3
a ⎠⎦ kN⋅ m
⎣
⎝
6
Shear (kN)
+
⌠
A − ⎮ w dx = 0
⌡0
4
V( x )
2
0
0
0.5
1
x
Distance (m)
1.5
2
0
Moment (kN-m)
2
M( x)
4
6
8
0
0.5
1
x
Distance m)
1.5
2
Problem 6-42
The truck is to be used to transport the concrete column. If the column has a uniform weight of w
(force/length), determine the equal placement a of the supports from the ends so that the absolute
maximum bending moment in the column is as small as possible. Also, draw the shear and moment
diagrams for the column.
Solution:
Support Reactions: By symmetry, A=B=R
ΣF y=0;
+
2R − wL = 0
R = 0.5w⋅ L
Internal Moment :
2
Mmin = −0.5 w⋅ a
⎛ L ⎞ ⋅ L − R ⎛ L − a⎞
Mmax = ⎜ w⋅
⎜
⎝ 2⎠ 4
⎝2 ⎠
For negative moment,
For positive moment,
(at mid-span)
w⋅ L
⋅ ( 4a − L )
8
Mmax =
For optimal minimum:
Mmax = Mmin
1
w⋅ L
2
⋅ ( 4a − L ) = − ⋅ w⋅ a
2
8
a
Let
α=
L
2
( 4a − L) ⋅ L = −4a
2
α + α − 0.25 = 0
α :=
1 ⎡
⋅ ⎣−1 +
2
1 − 4⋅ ( −0.25 )⎤⎦
2
α = 0.2071
a = 0.2071L
Set:
L := 1m
w := 1
a := αL
kN
m
R := 0.5w⋅ L
x1 := 0 , 0.01 ⋅ a .. a
−w⋅ x1
V1 x1 :=
kN
b := L − 2a
x2 := a , 1.01 ⋅ a .. ( a + b)
( ) (
1
V3 x3 := 2R − w⋅ x3 ⋅
kN
)
⎛ −w ⋅ x 2⎞ ⋅ 1
M1 x1 := ⎜
2 1
kN⋅ m
⎝
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
1
V2 x2 := R − w⋅ x2 ⋅
kN
( )
( )
Ans
( ) (
w 2⎤
1
⎡
M2 x2 := ⎢R⋅ x2 − a − ⋅ x2 ⎥ ⋅
2
kN⋅ m
( )
⎠
⎣
(
)
⎦
w 2⎤
1
⎡
M3 x3 := ⎢R⋅ x3 − a + R⋅ x3 − a − b − ⋅ x3 ⎥ ⋅
2
kN⋅ m
( )
⎣
(
)
(
)
⎦
)
Shear w*L (kN)
0.4
( )
V2 ( x 2 )
V3 ( x 3 )
0.2
V1 x 1
0
0.2
0.4
0
0.2
0.4
0.6
0.8
x1 , x2 , x3
Distance (m)
Moment w*L*L (kN-m)
0.04
0.02
( )
M2( x2)
M3( x3)
M1 x1
0
0.02
0.04
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
0.8
Problem 6-43
A member having the dimensions shown is to be used to resist an internal bending moment of M = 2
kN·m. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b)
about the y axis. Sketch the stress distribution for each case.
Given:
d := 120mm
b := 60mm
Mz := 2kN⋅ m
My := 2kN⋅ m
Solution:
Iz :=
1
3
⋅ b⋅ d
12
Maximum Stress:
σ = M⋅
Iy :=
1
3
⋅ d⋅ b
12
c
I
(a) About the z axis
ymax :=
d
2
ymax
σ max := Mz ⋅
Iz
( )
σ max = 13.89 MPa
Ans
(b) About the y axis
b
zmax :=
2
zmax
σ max := My ⋅
Iy
( )
σ max = 27.78 MPa
Ans
Problem 6-44
The steel rod having a diameter of 20 mm is subjected to an internal moment of M = 300 N·m.
Determine the stress created at points A and B. Also, sketch a threedimensional view of the stress
distribution acting over the cross section.
Given:
d := 20mm
M := 300N⋅ m
θ := 45deg
Solution:
π ⎛ d⎞
I := ⋅ ⎜
4 ⎝ 2⎠
yA :=
d
2
4
σ = M⋅
σ A := M⋅
y
I
yA
I
σ A = 381.97 MPa
⎛ d ⎞ ⋅ sin ( θ )
⎝ 2⎠
yB := ⎜
σ B := M⋅
Ans
yB
I
σ B = 270.09 MPa
Ans
Problem 6-45
The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by th
stresses acting on both the top and bottom boards,A and B, of the beam.
Given: bf := 200mm
tf := 25mm
tw := 25mm
Solution:
dw := 150mm
D := dw + 2tf
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − 2tw ⋅ dw ⎤⎦
12
c
Bending Stress:
σ = M⋅
I
Set
M := 1kN⋅ m
co
co := 0.5D
σ o := M⋅
σ o = 1.097143 MPa
I
(
I :=
ci := 0.5dw
σ i := M⋅
)
ci
I
Resultant Force and Moment:
σ i = 0.822857 MPa
For board A or B.
1
⋅ σ o + σ i ⋅ bf⋅ tf
F = 4.800 kN
2
tf 1
tf
Centroid of force:
σ i⋅ bf⋅ tf ⋅ + ⋅ σ o − σ i ⋅ bf⋅ tf ⋅ = F⋅ yc
2 2
3
F :=
(
)
(
yc :=
(
)
(
)
M' := F⋅ D − 2yc
M' = 0.8457 kN⋅ m
Hence,
%M :=
M'
⋅ 100
M
%M = 84.57
)
tf 1
tf⎤
1⎡
⎢σ i⋅ bf⋅ tf ⋅ + ⋅ σ o − σ i ⋅ bf⋅ tf ⋅ ⎥
F⎣
2 2
3⎦
yc = 11.905 mm
(
)(
Ans
)
(
)(
)
Problem 6-46
Determine the moment M that should be applied to the beam in order to create a compressive stress at
point D of σD = 30 MPa. Also sketch the stress distribution acting over the cross section and compute
the maximum stress developed in the beam.
Given: bf := 200mm
tw := 25mm
Solution:
tf := 25mm
dw := 150mm
σ D := 30MPa
D := dw + 2tf
I :=
1 ⎡
3
3
⋅ b ⋅ D − bf − 2tw ⋅ dw ⎤⎦
12 ⎣ f
(
)
4
I = 91145833.33 mm
c
I
Bending Stress:
σ = M⋅
cD := 0.5dw
σ D = M⋅
cD
I
M :=
σ D⋅ I
cD
M = 36.46 kN⋅ m
cmax := 0.5D
σ max := M⋅
Ans
cmax
I
σ max = 40.00 MPa
Ans
Problem 6-47
The slab of marble, which can be assumed a linear elastic brittle material, has a specific weight of 24
kN/m3 and a thickness of 20 mm. Calculate the maximum bending stress in the slab if it is supported
(a) on its side and (b) on its edges. If the fracture stress is σ f = 1.5 MPa, explain the consequences of
supporting the slab in each position.
Given:
t := 20mm
γ := 24
kN
m
Solution:
L := 1.5m
d := 0.5m
σ f := 1.5MPa
3
w := γ ⋅ d⋅ t
w = 0.24
kN
m
1
2
Mmax := ⋅ w⋅ L
8
Is :=
1
3
⋅ t⋅ d
12
Maximum Stress:
σ = M⋅
Ie :=
1
3
⋅ d⋅ t
12
c
I
(a) Supported on its side
d
c1 :=
2
c1
σ max := Mmax ⋅
Is
(
)
σ max = 0.081 MPa
Ans
(b) Supported on its edges
t
c2 :=
2
c2
σ max := Mmax ⋅
Ie
(
)
σ max = 2.025 MPa
Ans
> σf = 1.5 MPa
The marble slab will break if it is supported as in case (b).
Problem 6-48
The slab of marble, which can be assumed a linear elastic brittle material, has a specific weight of 24
kN/m3. If it is supported on its edges as shown in (b), determine the minimum thickness it should have
without causing it to break.The fracture stress is σ f = 1.5 MPa.
Given:
L := 1.5m
γ := 24
kN
m
Solution:
d := 0.5m
σ f := 1.5MPa
3
w = γ ⋅ d⋅ t
1
2
Mmax = ⋅ w⋅ L
8
1
3
⋅ d⋅ t
12
c=
t
2
Ie =
σ max = Mmax⋅
c
Ie
σ max = Mmax⋅
Maximum Stress:
Thus,
1
2
Mmax = ⋅ ( γ ⋅ d⋅ t) ⋅ L
8
σ max =
t :=
1
2 6
⋅ ( γ ⋅ d⋅ t ) ⋅ L ⋅
2
8
d⋅ t
3⋅ γ ⋅ L
2
4⋅ σ f
t = 27 mm
Ans
6
d⋅ t
2
Problem 6-49
A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow = 170
MPa, determine the largest internal moment the beam can resist if the moment is applied (a) about the z
axis, (b) about the y axis.
Given:
Solution:
bf := 120mm
tf := 5mm
d := 120mm
tw := 5mm
σ allow := 170MPa
D := d + 2tf
Iz :=
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ d ⎤⎦
12 ⎣ f
(
)
⎛ 1 ⋅ t ⋅ b 3⎞ + 1 ⋅ d⋅ t 3
⎝ 12 f f ⎠ 12 w
Iy := 2⎜
Bending Stress:
σ allow = M⋅
c
I
(a) About the z axis
D
cz :=
2
Iz
Mz := σ allow ⋅
cz
(
)
Mz = 14.15 kN⋅ m
Ans
(b) About the y axis
bf
cy :=
2
Iy
My := σ allow ⋅
cy
(
)
My = 4.08 kN⋅ m
Ans
Problem 6-50
Two considerations have been proposed for the design of a beam. Determine which one will support a
moment of with the least amount of M = 150 kN·m bending stress. What is that stress? By what
percentage is it more effective?
Given: bf := 200mm
dw := 300mm
tf.a := 15mm
tw.a := 30mm
tf.b := 30mm
tw.b := 15mm
M := 150kN⋅ m
Solution:
Section Property:
For section (a):
Da := dw + 2tf.a
Ia :=
For section (b):
Db := dw + 2tf.b
1 ⎡
3
3
⋅ ⎣bf⋅ Da − bf − tw.a ⋅ dw ⎤⎦
12
(
1 ⎡
3
3
⋅ b ⋅ D − bf − tw.b ⋅ dw ⎤⎦
12 ⎣ f b
c
σ = M⋅
I
(
Ib :=
Maximum Bending Stress:
For section (a):
cmax := 0.5Da
)
σ max := M⋅
)
cmax
Ia
σ max = 114.35 MPa
For section (b):
c'max := 0.5Db
σ' max := M⋅
c'max
Ib
σ' max = 74.72 MPa
Ans
By comparison, section (b) will have the least amount of bending stress.
%eff :=
σ max − σ' max
σ' max
%eff = 53.03
Ans
⋅ 100
Problem 6-51
The aluminum machine part is subjected to a moment of Determine the bending stress M = 75 N·m.
created at points B and C on the cross section. Sketch the results on a volume element located at each
of these points.
Given:
bf := 80mm
tf := 10mm
tw := 10mm
dw := 40mm
M := 75N⋅ m
Solution: D := dw + tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
yc :=
)
(bf⋅ tf)⋅ 0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)
bf⋅ tf + 2dw⋅ tw
yc = 17.50 mm
If :=
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ yc − 0.5tf
12
Iw :=
1
3
2
⋅ tw⋅ dw + dw⋅ tw ⋅ ⎡⎣yc − 0.5dw + tf ⎤⎦
12
(
)(
(
)
)
(
)
I := If + 2Iw
Bending Stress:
σ = M⋅
c
I
At B:
cB := yc
σ B := M⋅
cB
I
σ B = 3.612 MPa
Ans
At C:
cC := yc − tf
σ C := M⋅
cC
I
σ C = 1.548 MPa
Ans
Problem 6-52
The aluminum machine part is subjected to a moment of M = 75 N·m. Determine the maximum tensil
and compressive bending stresses in the part.
Given:
bf := 80mm
tf := 10mm
tw := 10mm
dw := 40mm
M := 75N⋅ m
Solution: D := dw + tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
yc :=
)
(bf⋅ tf)⋅ 0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)
bf⋅ tf + 2dw⋅ tw
yc = 17.50 mm
If :=
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ yc − 0.5tf
12
Iw :=
1
3
2
⋅ tw⋅ dw + dw⋅ tw ⋅ ⎡⎣yc − 0.5dw + tf ⎤⎦
12
(
)(
(
)
)
(
)
I := If + 2Iw
Bending Stress:
σ = M⋅
For compression:
cc := yc
For tension:
ct := D − yc
c
I
σ c_max := M⋅
cc
I
σ c_max = 3.612 MPa
Ans
σ t_max := M⋅
ct
I
σ t_max = 6.709 MPa
Ans
Problem 6-53
A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on
the cross section is M = 450 N·m, determine the resultant force the bending stress produces on the top
board A and on the side board B.
Given: bf := 240mm
tw := 20mm
tf := 15mm
dw := 200mm
M := 450N⋅ m
Solution:
D := dw + 2tf
1 ⎡
3
3
⋅ D⋅ b − dw⋅ bf − 2tw ⎤⎦
12 ⎣ f
c
Bending Stress:
σ = M⋅
I
Iy :=
(
)
co := 0.5bf
σ o := M⋅
co
Iy
σ o = 0.410251 MPa
ci := 0.5bf − tw
σ i := M⋅
ci
Iy
σ i = 0.341876 MPa
Resultant Force : For board A or B.
FA :=
σ o ⎛ bf ⎞ σ o ⎛ bf ⎞
⋅⎜ ⋅t −
⋅⎜ ⋅t
2 ⎝ 2 f⎠
2 ⎝ 2 f⎠
FA = 0 kN
Ans
FB :=
1
⋅ σ o + σ i ⋅ dw⋅ tw
2
FB = 1.504 kN
Ans
(
)
Problem 6-54
The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment
M = 8 kN·m, determine the bending stress acting at points A and B, and show the results acting on
volume elements located at these points.
b'f := 50mm
tw := 20mm
Given:
tf := 20mm
dw := 220mm
M := 8kN⋅ m
Solution:
I :=
1
3
3⎞
⎛1
⋅ tw⋅ dw + 2⎜ ⋅ b'f⋅ tf
12
12
⎝
⎠
4
I = 17813333.33 mm
Bending Stress:
σ = M⋅
c
I
At A:
cA := 0.5dw
σ A := M⋅
cA
I
σ A = 49.401 MPa
Ans
At B:
cB := 0.5tf
σ B := M⋅
cB
I
σ B = 4.491 MPa
Ans
Problem 6-55
The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment
M = 8 kN·m, determine the maximum bending stress in the beam, and sketch a three-dimensional view
of the stress distribution acting over the entire cross-sectional area.
b'f := 50mm
tw := 20mm
Given:
tf := 20mm
dw := 220mm
M := 8kN⋅ m
Solution:
I :=
1
3
3⎞
⎛1
⋅ tw⋅ dw + 2⎜ ⋅ b'f⋅ tf
12
12
⎝
⎠
4
I = 17813333.33 mm
Bending Stress:
cmax := 0.5dw
σ = M⋅
c
I
σ max := M⋅
At B:
cB := 0.5tf
σ B := M⋅
cmax
I
cB
I
σ max = 49.401 MPa
σ B = 4.491 MPa
Ans
Problem 6-56
The beam is made from three boards nailed together as shown. If the moment acting on the cross
section is M = 1.5 kN·m, determine the maximum bending stress in the beam. Sketch a
three-dimensional view of the stress distribution acting over the cross section.
Given:
bf := 250mm
b'f := 150mm
d := 300mm
tw := 25mm
tf := 38mm
M := 1.5kN⋅ m
Solution: D := d + 2tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
yc :=
)
(bf⋅ tf)⋅ 0.5tf + (d⋅ tw)⋅ (0.5d + tf) + (b'f⋅ tf)⋅ (D − 0.5tf)
bf⋅ tf + d⋅ tw + b'f⋅ tf
yc = 159.71 mm
If :=
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ yc − 0.5tf
12 f f
Iw :=
1
3
2
⋅ t ⋅ d + d⋅ tw ⋅ ⎡⎣yc − 0.5d + tf ⎤⎦
12 w
(
)(
(
)
)
(
)
1
3
2
I'f :=
⋅ b' ⋅ t + b'f⋅ tf ⋅ ⎡⎣yc − D − 0.5tf ⎤⎦
12 f f
(
)
(
)
σ max := M⋅
cmax
I
σ max = 0.684 MPa
I := If + Iw + I'f
Bending Stress:
σ = M⋅
c
I
At B:
cmax := D − yc
At A:
cA := cmax − tf
σ A := M⋅
cA
I
σ A = 0.564 MPa
At C:
cC := yc − tf
σ C := M⋅
cC
I
σ C = 0.385 MPa
At D:
cD := yc
σ D := M⋅
cD
I
σ D = 0.505 MPa
Ans
Problem 6-57
Determine the resultant force the bending stresses produce on the top board A of the beam if M = 1.5
kN·m.
Given:
bf := 250mm
b'f := 150mm
d := 300mm
tw := 25mm
tf := 38mm
M := 1.5kN⋅ m
Solution: D := d + 2tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
yc :=
)
(bf⋅ tf)⋅ 0.5tf + (d⋅ tw)⋅ (0.5d + tf) + (b'f⋅ tf)⋅ (D − 0.5tf)
bf⋅ tf + d⋅ tw + b'f⋅ tf
yc = 159.71 mm
If :=
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ yc − 0.5tf
12
Iw :=
1
3
2
⋅ tw⋅ d + d⋅ tw ⋅ ⎡⎣yc − 0.5d + tf ⎤⎦
12
(
)(
(
)
)
(
)
1
3
2
I'f :=
⋅ b'f⋅ tf + b'f⋅ tf ⋅ ⎡⎣yc − D − 0.5tf ⎤⎦
12
(
)
(
)
I := If + Iw + I'f
Bending Stress:
σ = M⋅
c
I
At C:
cC := yc − tf
σ C := M⋅
cC
I
σ C = 0.385 MPa
At D:
cD := yc
σ D := M⋅
cD
I
σ D = 0.505 MPa
F = 4.23 kN
Ans
The resultant Force: For top board A
(
)(
F := 0.5 σ C + σ D ⋅ bf⋅ tf
)
Problem 6-58
The control level is used on a riding lawn mower. Determine the maximum bending stress in the lever
at section a-a if a force of 100 N is applied to the handle. The lever is supported by a pin at A and a
wire at B. Section a-a is square, 6 mm by 6 mm.
Given:
Solution:
L := 50mm
b := 6mm
d := 6mm
F := 100N
I :=
(
1
3
⋅ b⋅ d
12
)
M := F⋅ L
M = 5.00 N⋅ m
Bending Stress:
c :=
d
2
σ = M⋅
c
I
σ max := M⋅
c
I
σ max = 138.89 MPa
Ans
Problem 6-59
Determine the largest bending stress developed in the member if it is subjected to an internal bending
moment of M = 40 kN·m.
Given:
bf := 100mm
tf := 10mm
rf := 30mm
tw := 10mm
dw := 180mm
M := 40kN⋅ m
Solution: D := dw + tf + 2rf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
)
2
bf⋅ tf) ⋅ 0.5 tf + ( dw⋅ tw) ⋅ ( 0.5dw + tf) + ⎛ π ⋅ rf ⎞ ⋅ ( rf + dw + tf)
(
⎝
⎠
y :=
c
bf⋅ tf + dw⋅ tw + π ⋅ rf
yc = 143.41 mm
2
If :=
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ yc − 0.5tf
12
Iw :=
1
3
2
⋅ tw⋅ dw + dw⋅ tw ⋅ ⎡⎣yc − 0.5dw + tf ⎤⎦
12
(
)(
(
)
)
(
)
π 4
2
2
I'f := ⋅ rf + ⎛⎝ π ⋅ rf ⎞⎠ ⋅ ⎡⎣yc − rf + dw + tf ⎤⎦
4
(
)
I := If + Iw + I'f
c
I
Maximum stress occurs at the bottom fibre.
cmax
cmax := yc
σ max := M⋅
I
Bending Stress:
σ = M⋅
σ max = 128.51 MPa
Ans
Problem 6-60
The tapered casting supports the loading shown. Determine the bending stress at points A and B. The
cross section at section a-a is given in the figure.
Given:
L a := 250mm
P := 750N
L b := 375mm
L c := 125mm
b := 100mm
t := 25mm
d := 75mm
Solution:
Equilibrium :
(
) (
P⋅ ( L c + Lb) + P⋅ ( L b)
F :=
ΣΜC=0;
)
( )
F1⋅ 2L b + L c − P⋅ L c + Lb − P⋅ L b = 0
1
2⋅ Lb + Lc
F1 = 750.00 N
D := d + 2t
Section a-a :
M := F1⋅ La
I :=
M = 187.50 N⋅ m
(
1
3
3
⋅ b⋅ D − b⋅ d
12
Bending Stress:
σ = M⋅
c
I
)
D
cA :=
2
σ A := M⋅
cA
I
σ A = 0.918 MPa
Ans
d
cB :=
2
σ B := M⋅
cB
I
σ B = 0.551 MPa
Ans
Problem 6-61
If the shaft in Prob. 6-1 has a diameter of 100 mm, determine the absolute maximum bending stress in
the shaft.
Given:
a := 250mm
b := 800mm
F := 24kN
do := 100mm
Solution:
Equilibrium :
ΣF y=0;
A+B−F= 0
ΣΜA=0;
−F⋅ a − B⋅ b = 0
Guess A := 1kN B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 31.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ −7.50 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
− F ⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣−F⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
Moment (kN-m)
+
Given
)
0
( )
M2( x2) 5
M1 x1
0
0.5
x1 , x2
Max. Moment :
Distane (m)
unit := kN⋅ m
M1 ( a) = −6.00
Bending Stress:
c'
σ = M⋅
I
4
M := M1 ( a) ⋅ unit
c'
σ max := M⋅
I
I :=
π ⋅ do
64
c' :=
do
2
σ max = 61.12 MPa
Ans
1
Problem 6-62
If the shaft in Prob. 6-3 has a diameter of 40 mm, determine the absolute maximum bending stress in
the shaft.
Given: a := 350mm
b := 500mm
c := 375mm
d := 300mm
do := 40mm
C := 550N
E := 175N
B := 400N
Solution:
Given
Equilibrium :
+
ΣFy=0;
A+D−B−C−E= 0
ΣΜD=0; A⋅ ( a + b + c) − B⋅ ( b + c) − C⋅ c + E ⋅ d = 0
Guess A := 1N D := 1N
⎛A⎞
⎛ A ⎞ ⎛ 411.22 ⎞
:= Find ( A , D)
=⎜
N
⎜
⎜
⎝D⎠
⎝ D ⎠ ⎝ 713.78 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
x4 := a + b + c , 1.01 ⋅ ( a + b + c) .. a + b + c + d
A⋅ x1
M1 x1 :=
N⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − B⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − B⋅ x3 − a − C⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡⎣A⋅ x4 − B⋅ x4 − a − C⋅ x4 − a − b + D⋅ x4 − a − b − c ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
(
)
Moment (N-m)
200
( )
M 2 ( x 2 ) 100
M3(x3)
0
M4(x4)
M1 x1
100
0
0.2
0.4
0.6
0.8
1
1.2
x1 , x2 , x3 , x4
Max. Moment :
unit := N⋅ m
Distance (m)
M3 ( a + b) = 149.54
Bending Stress:
σ = M⋅
c'
I
M := M3 ( a + b) ⋅ unit
σ max := M⋅
c'
I
I :=
π ⋅ do
64
4
c' :=
σ max = 23.8 MPa
do
2
Ans
1.4
Problem 6-63
If the shaft in Prob. 6-6 has a diameter of 50 mm, determine the absolute maximum bending stress in
the shaft.
Given: a := 125mm
b := 600mm
c := 75mm
F1 := 0.8kN
F2 := 1.5kN
do := 50mm
Solution:
L := a + b + c
Equilibrium :
ΣF y=0;
A − F1 − F2 + B = 0
ΣΜB=0;
A⋅ ( L) − F1⋅ ( b + c) − F2⋅ ( c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
A⋅ x1
M1 x1 :=
N⋅ m
( )
⎛ A ⎞ ⎛ 0.8156 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 1.4844 ⎠
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a − F2⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
0
0.2
)
(
)
150
Moment (N-m)
+
Given
( )
100
M2( x2)
M 3 ( x 3 ) 50
M1 x1
0
0.4
0.6
x1 , x2 , x3
Distance (m)
Max. Moment :
unit := N⋅ m
M2 ( a + b) = 111.33
Bending Stress:
M := M2 ( a + b) ⋅ unit
c'
σ = M⋅
I
c'
σ max := M⋅
I
4
I :=
π ⋅ do
64
c' :=
σ max = 9.072 MPa
do
2
Ans
Problem 6-64
If the shaft in Prob. 6-8 has a diameter of 30 mm and thickness of 10 mm, determine the absolute
maximum bending stress in the shaft.
Given: a := 400mm
h := 80mm
do := 30mm
t := 10mm
Given
di := do − 2t
Solution:
F := 5kN
Equilibrium :
ΣF y=0;
A+C= 0
ΣΜC=0;
A⋅ a + F⋅ h = 0
Guess A := 1N
C := 1N
⎛A⎞
:= Find ( A , C)
⎜
⎝C⎠
⎛ A ⎞ ⎛ −1.00 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 1.00 ⎠
A⋅ x1
M1 x1 :=
N⋅ m
( )
x1 := 0 , 0.01 ⋅ a .. a
0
Moment (N-m)
+
( )
200
M1 x1
400
600
0
0.2
x1
Max. Moment :
Distane (m)
unit := N⋅ m
M1 ( a) = −400.00
Bending Stress:
c'
σ = M⋅
I
M := M1 ( a) ⋅ unit
c'
σ max := M⋅
I
I :=
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
64
c' :=
σ max = 152.8 MPa
do
2
Ans
0
Problem 6-65
If the beam ACB in Prob. 6-9 has a square cross section, 150 mm by 150 mm, determine the absolute
maximum bending stress in the beam.
Given:
a := 1m
b := 1m
c := 1m
d := 0.25m
ao := 150mm
F1 := 75kN
F2 := 100kN
Solution:
Equilibrium :
ΣF y=0;
A − F1 + B = 0
ΣΜC=0;
A⋅ ( a + b + c) − F1⋅ ( b + c) − F2⋅ ( d) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
A⋅ x1
M1 x1 :=
kN⋅ m
⎛ A ⎞ ⎛ 58.33 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 16.67 ⎠
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a − F2⋅ d⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
100
Moment (kN-m)
+
Given
( )
M2( x2)
50
M3( x3)
M1 x1
0
0
0.5
1
1.5
2
2.5
x1 , x2 , x3
Distance (m)
Max. Moment :
unit := kN⋅ m
M1 ( a) = 58.33
4
Bending Stress:
M := M1 ( a) ⋅ unit
c'
I
σ max := M⋅
σ = M⋅
ao
I :=
12
c'
I
c' :=
σ max = 103.7 MPa
ao
2
Ans
3
Problem 6-66
If the crane boom ABC in Prob. 6-10 has a rectangular cross section with a base of 60 mm, determine
its required height h to the nearest multiples of 5 mm if the allowable bending stress is σallow = 170 MPa.
Given: a := 0.9m
b := 1.5m
c := 1.2m
2
d :=
v :=
W := 6kN σ allow := 170MPa
2
a +c
c
d
h :=
a
d
Equilibrium :
Given
+
ΣF y=0;
−Ay + B⋅ v − W = 0
ΣΜA=0;
( −B⋅ v) ⋅ a + W⋅ ( a + b) = 0
ΣF x=0;
Ax − B⋅ h = 0
+
Guess Ax := 1kN
Ay := 1kN
⎛ Ax ⎞
⎜
⎜ Ay ⎟ := Find ( Ax , Ay , B)
⎜
⎝B⎠
B := 1kN
⎛ Ax ⎞ ⎛ 12 ⎞
⎜
⎜
⎜ Ay ⎟ = ⎜ 10 kN
⎜
⎝ B ⎠ ⎝ 20 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
−Ay⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣−Ay⋅ x2 + ( B⋅ v) ⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
unit := kN⋅ m
Max. Moment :
M1 ( a) = −9.00
M := M1 ( a) ⋅ unit
Bending Stress:
3
bo⋅ ho
I=
12
c' =
ho
2
Moment (kN-m)
Solution:
bo := 60mm
(
0
( )
M2( x2) 5
M1 x1
10
c'
σ = M⋅
I
ho :=
)
0
1
x1 , x2
Distane (m)
6 M
(
)
bo⋅ σ allow
ho = 72.76 mm
Use ho = 75mm
Ans
2
Problem 6-67
If the crane boom ABC in Prob. 6-10 has a rectangular cross section with a base of 50 mm and a
height of 75 mm, determine the absolute maximum bending stress in the boom.
Given: a := 0.9m
b := 1.5m
bo := 50mm
c := 1.2m
W := 6kN
ho := 75mm
2
d :=
v :=
2
a +c
c
d
h :=
a
d
Equilibrium :
Given
+
ΣF y=0;
−Ay + B⋅ v − W = 0
ΣΜA=0;
( −B⋅ v) ⋅ a + W⋅ ( a + b) = 0
ΣF x=0;
Ax − B⋅ h = 0
+
Guess Ax := 1kN
Ay := 1kN
⎛ Ax ⎞
⎜
⎜ Ay ⎟ := Find ( Ax , Ay , B)
⎜
⎝B⎠
B := 1kN
⎛ Ax ⎞ ⎛ 12 ⎞
⎜
⎜
⎜ Ay ⎟ = ⎜ 10 kN
⎜
⎝ B ⎠ ⎝ 20 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
−Ay⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣−Ay⋅ x2 + ( B⋅ v) ⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
unit := kN⋅ m
Max. Moment :
M1 ( a) = −9.00
M := M1 ( a) ⋅ unit
Bending Stress:
I :=
bo⋅ ho
12
c'
σ = M⋅
I
3
c' :=
ho
2
( )
Moment (kN-m)
Solution:
)
0
( )
M2( x2) 5
M1 x1
10
c'
σ max := M⋅
I
σ max = 192 MPa
(
0
1
x1 , x2
Ans
Distane (m)
2
Problem 6-68
Determine the absolute maximum bending stress in the beam in Prob. 6-24. The cross section is
rectangular with a base of 75 mm and height of 100 mm.
Given: a := 0.3m
bo := 75mm
Solution:
c := 0.6m
ho := 100mm
Given
Equilibrium :
+
b := 2.4m
w := 30
kN
m
( )
ΣFy=0;
A − w⋅ b + qB ⋅ c = 0
ΣΜA=0;
( w⋅ b) ⋅ ( a + 0.5b) − qB⋅ c ⋅ ( a + b + 0.5c) = 0
(
Guess A := 1kN
⎛A⎞
⎜ q := Find ( A , qB)
⎝ B⎠
x1 := 0 , 0.01 ⋅ a .. a
A⋅ x1
M1 x1 :=
kN⋅ m
)
kN
qB := 1
m
A = 36.00 kN
x2 := a , 1.01 ⋅ a .. ( a + b)
kN
qB = 60.00
m
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + b + c)
1
2
M2 x2 := ⎡⎣A⋅ x2 − 0.5w⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
2
M3 x3 := ⎡⎣A⋅ x3 − ( w⋅ b) ⋅ x3 − a − 0.5 ⋅ b + 0.5qB⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
( )
unit := kN⋅ m
Max. Moment :
Moment (kN-m)
M2 ( a + b') = 32.40
Bending Stress:
M := M2 ( a + b') ⋅ unit
3
bo⋅ ho
I :=
12
c'
I
c' :=
ho
2
σ max := M⋅
)
(
)
40
b' := 0.5 ⋅ b
σ = M⋅
(
( )
M2(x2)
20
M3(x3)
M1 x1
0
1
2
x1 , x2 , x3
c'
I
σ max = 259.2 MPa
0
Distance (m)
Ans
3
Problem 6-69
Determine the absolute maximum bending stress in the beam in Prob. 6-25. Each segment has a
rectangular cross section with a base of 100 mm and height of 200 mm.
Given: a := 0.9m
b := 1.5m
bo := 100mm
ho := 200mm
P := 40kN
w := 50
Solution:
kN
m
Given
Equilibrium :
+
c := 2.4m
ΣF y=0;
A − P − w⋅ c + C = 0
ΣΜB=0;
( w⋅ c) ⋅ ( 0.5c) − C⋅ ( c) = 0
Guess A := 1kN
C := 1kN
⎛A⎞
:= Find ( A , C)
⎜
⎝C⎠
⎛ A ⎞ ⎛ 100 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 60 ⎠
MA := P⋅ a − ( C − w⋅ c) ⋅ ( a + b)
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
−MA + A⋅ x1
( )
M1 x1 :=
MA = 180 kN⋅ m
kN⋅ m
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. ( a + b + c)
1
M2 x2 := ⎡⎣−MA + A⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
2
M3 x3 := ⎡⎣−MA + A⋅ x3 − P⋅ x3 − a − 0.5w⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
(
Moment (kN-m)
c' := 0.5 ⋅ c
M3 ( a + b + c') = 36.00
Bending Stress:
M := M2 ( a + b + c') ⋅ unit
I :=
3
( )
M2( x2)
M3( x3)
M1 x1
)
12
c'
I
σ max := M⋅
0
200
ho
co :=
2
σ = M⋅
(
unit := kN⋅ m
Max. Moment :
bo⋅ ho
)
0
2
x1 , x2 , x3
co
I
σ max = 108 MPa
Distance (m)
Ans
4
Problem 6-70
Determine the absolute maximum bending stress in the 20-mm-diameter pin in Prob. 6-35.
Given: L := 100mm
a := 0.2L
w := 0.4
do := 20mm
kN
m
Solution:
Equilibrium :
(
ΣF y=0;
+
)
2 0.5wo ⋅ a − w⋅ ( 3a) = 0
kN
wo = 1.20
m
wo := 3w
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( 4a)
⎡wo⋅ a ⎛
2a ⎞
2⎤ 1
M2 x2 := ⎢
⋅ ⎜ x2 −
− 0.5w⋅ x2 − a ⎥ ⋅
3⎠
⎣ 2 ⎝
⎦ N⋅ m
x1⎤ 1
⎡wo ⎛ x1 ⎞
M1 x1 := ⎢ ⋅ ⎜
⋅ x1⋅ ⎥ ⋅
3 N⋅ m
2
a
( )
⎣
⎝
( )
M'3 x3 :=
⎠
x3 := ( 4a) , 1.01 ⋅ ( 4a) .. L
( )
⎦
(
wo
⎛ x3 − 4⋅ a ⎞ 1⎤
2 ⎡
⋅ x3 − 4⋅ a ⋅ ⎢1 − ⎜
⋅ ⎥
2
⎣ ⎝ a ⎠ 3⎦
(
)
⎡⎡ wo⋅ a ⎛
⎤
⎤ 1
2⋅ a ⎞
M3 x3 := ⎢⎢
⋅ ⎜ x3 −
− w⋅ ( 3a) ⋅ x3 − 2.5 ⋅ a ⎥ + M'3 x3 ⎥ ⋅
3 ⎠
⎣⎣ 2 ⎝
⎦
⎦ N⋅ m
Moment (N-m)
( )
(
)
( )
( )0.2
M2( x2)
M 3 ( x 3 )0.1
M1 x1
0
0
0.02
0.04
0.06
0.08
x1 , x2 , x3
Distance (m)
Max. Moment :
unit := N⋅ m
M2 ( .5L) = 0.260
Bending Stress:
σ = M⋅
c'
I
M := M2 ( .5L ) ⋅ unit
σ max := M⋅
c'
I
I :=
π
4
⋅d
64 o
c' :=
σ max = 0.331 MPa
do
2
Ans
)
Problem 6-71
The member has a cross section with the dimensions shown. Determine the largest internal moment M
that can be applied without exceeding allowable tensile and compressive stresses of (σ t )allow = 150
MPa and (σ c )allow = 100 MPa, respectively.
Given:
bf := 100mm
tf := 10mm
tw := 10mm
dw := 180mm
σ t.allow := 150MPa
rf := 30mm
σ c.allow := 100MPa
Solution: D := dw + tf + 2rf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
)
2
bf⋅ tf) ⋅ 0.5 tf + ( dw⋅ tw) ⋅ ( 0.5dw + tf) + ⎛ π ⋅ rf ⎞ ⋅ ( rf + dw + tf)
(
⎝
⎠
y :=
c
bf⋅ tf + dw⋅ tw + π ⋅ rf
yc = 143.41 mm
2
If :=
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ yc − 0.5tf
12 f f
Iw :=
1
3
2
⋅ t ⋅ d + dw⋅ tw ⋅ ⎡⎣yc − 0.5dw + tf ⎤⎦
12 w w
(
)(
)
(
)
(
)
π 4
2
2
I'f := ⋅ rf + ⎛⎝ π ⋅ rf ⎞⎠ ⋅ ⎡⎣yc − rf + dw + tf ⎤⎦
4
(
I := If + Iw + I'f
)
4
I = 44639608.23 mm
Maximum Bending Stress:
σ = M⋅
c
I
Assume failure due to tensile stress.
ct.max
ct.max := yc
σ t.max = M⋅
I
Mt :=
Assume failure due to compressive stress.
cc.max
cc.max := D − yc σ c.max = M⋅
I
Mc :=
(
)
Mallow := min Mt , Mc
Mallow = 41.88 kN⋅ m
Ans
σ t.allow ⋅ I
ct.max
σ c.allow⋅ I
cc.max
Mt = 46.69 kN⋅ m
Mc = 41.88 kN⋅ m
Problem 6-72
Determine the absolute maximum bending stress in the 30-mm-diameter shaft which is subjected to th
concentrated forces. The sleeve bearings at A and B support only vertical forces.
Given: a := 0.8m
b := 1.2m
F1 := 0.6kN
c := 0.6m
F2 := 0.4kN
do := 30mm
Solution:
L := a + b + c
Equilibrium :
ΣF y=0;
A − F1 − F2 + B = 0
ΣΜB=0;
A⋅ ( b) − F1⋅ ( a + b) + F2⋅ ( c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
−F1⋅ x1
M1 x1 :=
N⋅ m
( )
⎛ A ⎞ ⎛ 0.8 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 0.2 ⎠
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
M2 x2 := ⎡⎣−F1⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
1
M3 x3 := ⎡⎣−F1⋅ x3 + A⋅ x3 − a + B⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
0
( ) 200
M2( x2)
M 3 ( x 3 ) 400
Moment (N-m)
+
Given
M1 x1
600
0
0.5
1
1.5
2
x1 , x2 , x3
Distance (m)
Max. Moment :
unit := N⋅ m
M1 ( a) = −480.000
Bending Stress:
M := M1 ( a) ⋅ unit
c'
I
σ max := M⋅
σ = M⋅
I :=
c'
I
π
4
⋅d
64 o
c' :=
do
2
σ max = 181.1 MPa
Ans
2.5
Problem 6-73
Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces.
The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow
160 MPa.
Given: a := 0.8m
b := 1.2m
c := 0.6m
F1 := 0.6kN
F2 := 0.4kN
σ allow := 160MPa
Solution:
L := a + b + c
Equilibrium :
ΣF y=0;
A − F1 − F2 + B = 0
ΣΜB=0;
A⋅ ( b) − F1⋅ ( a + b) + F2⋅ ( c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
−F1⋅ x1
M1 x1 :=
N⋅ m
( )
⎛ A ⎞ ⎛ 0.8 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 0.2 ⎠
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
M2 x2 := ⎡⎣−F1⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
1
M3 x3 := ⎡⎣−F1⋅ x3 + A⋅ x3 − a + B⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
0
Moment (N-m)
+
Given
( )
200
M2( x2)
M 3 ( x 3 ) 400
M1 x1
600
0
0.5
1
1.5
2
2.5
x1 , x2 , x3
Distance (m)
Max. Moment :
unit := N⋅ m
M1 ( a) = −480.000
Bending Stress:
c'
σ = M⋅
I
M := M1 ( a) ⋅ unit
σ allow = M⋅
I=
3
32
π ⋅ do
π
4
⋅d
64 o
3
do :=
c' =
32 M
π σ allow
do
2
do = 31.26 mm
Ans
Problem 6-74
Determine the absolute maximum bending stress in the 40-mm-diameter shaft which is subjected to the
concentrated forces. The sleeve bearings at A and B support only vertical forces.
Given:
a := 300mm
b := 450mm
c := 375mm
do := 40mm
F1 := 2kN
F2 := 1.5kN
Solution:
Equilibrium :
ΣF y=0;
A − F1 + B − F2 = 0
ΣΜB=0;
A⋅ ( a + b) − F1⋅ ( b) + F2⋅ ( c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 0.45 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 3.05 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
V1 x1 := A⋅
kN
1
V2 x2 := A − F1 ⋅
kN
1
V3 x3 := A − F1 + B ⋅
kN
A⋅ x1
M1 x1 :=
N⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
( ) (
( )
)
( )
( ) (
(
)
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a + B⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
0.8
1
2
Shear (kN)
+
Given
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1 1
1
2
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
Moment (N-m)
500
( ) 0
M2( x2)
500
M3( x3)
M1 x1
1000
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
Max. Moment :
unit := N⋅ m
M3 ( a + b) = −562.50
Bending Stress:
4
M := M2 ( a + b) ⋅ unit
σ = M⋅
c'
I
I :=
σ max := M⋅
π ⋅ do
64
do
co :=
2
co
I
σ max = 89.52 MPa
Ans
0.8
1
Problem 6-75
Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces.
The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow =
150 MPa.
Given:
a := 300mm
b := 450mm
c := 375mm
σ allow := 150MPa
F1 := 2kN
F2 := 1.5kN
Solution:
Given
Equilibrium :
+
ΣF y=0;
A − F1 + B − F2 = 0
ΣΜB=0;
A⋅ ( a + b) − F1⋅ ( b) + F2⋅ ( c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 0.45 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 3.05 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
A⋅ x1
M1 x1 :=
N⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
( )
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a + B⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
unit := N⋅ m
M3 ( a + b) = −562.50
Bending Stress:
M := −M3 ( a + b) ⋅ unit
π ⋅ do
4
do
co =
2
64
(
do :=
(
)
( ) 0
M2( x2)
500
M3( x3)
M1 x1
1000
0
co
σ = M⋅
I
3
)
500
Moment (N-m)
Max. Moment :
I=
( )
0.5
x1 , x2 , x3
Distance (m)
32M
π ⋅ σ allow
do = 33.68 mm
Ans
1
Problem 6-76
The bolster or main supporting girder of a truck body is subjected to the uniform distributed load.
Determine the bending stress at points A and B.
Given:
L 1 := 2.4m
L 2 := 3.6m
b := 150mm
tf := 20mm
d := 300mm
tw := 12mm
w := 25
Solution:
By symmetry :
F1 = R
kN
m
F2 = R
Equilibrium :
+
ΣF y=0;
(
)
−w⋅ L 1 + L 2 + 2R = 0
(
R := 0.5 ⋅ w⋅ L1 + L2
MAB := R⋅ L 1 − 0.5w⋅ L1
2
D := d + 2⋅ tf
Section properties :
I :=
)
1 ⎡ 3
3
⋅ b⋅ D − b − tw ⋅ d ⎤⎦
12 ⎣
Bending Stress:
d
cB :=
2
(
)
σ = M⋅
c
I
σ B := MAB⋅
cB
I
σ B = 89.6 MPa
d
cA := − tf
2
σ A := MAB⋅
Ans
cA
I
σ A = 77.65 MPa
Ans
Problem 6-77
A portion of the femur can be modeled as a tube having an inner diameter of 9.5 mm and an outer
diameter of 32 mm. Determine the maximum elastic static force P that can be applied to its center
without causing failure. Assume the bone to be roller supported at its ends. The σ-ε diagram for the
bone mass is shown and is the same in tension as in compression.
Given:
L 1 := 100mm
L 2 := 100mm
di := 9.5mm
mm
ε e := 0.02
mm
mm
ε r := 0.06
mm
do := 32mm
σ e := 8.75MPa
σ r := 16.1MPa
Solution:
By symmetry :
R = 0.5P
Mmax = R⋅ L 1
Mmax = 0.5P⋅ L1
Section properties :
I :=
π ⎛ 4
4
⋅ d − di ⎞⎠
64 ⎝ o
Bending Stress:
σ = M⋅
M=
Requires:
c
I
c=
do
2
2σ ⋅ I
do
σ max = σ e
P :=
2σ e⋅ I
(0.5⋅ L1)⋅ do
P = 558.6 N
Ans
Problem 6-78
If the beam in Prob. 6-20 has a rectangular cross section with a width of 200 mm and a height of 400
mm, determine the absolute maximum bending stress in the beam.
a := 2.4m
Given:
b := 1.2m
P1 := 50kN
P2 := 40kN
bo := 200mm
w := 30
kN
m
M2 := 60kN⋅ m
do := 400mm
Solution:
Equilibrium :
+
ΣF y=0;
A := w⋅ a + P1 + P2
A = 162 kN
ΣΜA=0;
MA := ( w⋅ a) ⋅ ( 0.5a) + P1⋅ a + P2⋅ ( a + b) + M2
MA = 410.40 kN⋅ m
As indicated in the moment diagram, the maximum moment is MA.
Section properties :
I :=
1
3
⋅ bo⋅ do
12
Bending Stress:
do
co :=
2
σ = M⋅
c
I
σ max := MA⋅
co
I
σ max = 76.95 MPa
Ans
Problem 6-79
If the shaft has a diameter of 37.5 mm, determine the absolute maximum bending stress in the shaft.
Given:
a := 450mm
b := 600mm
c := 300mm
do := 37.5mm
F1 := 1000N
F2 := 750N
Solution:
Given
Equilibrium :
+
ΣF y=0;
A − 2F1 + B − 2F2 = 0
ΣΜB=0;
−2F1⋅ ( a + b) + A⋅ ( b) + 2F2⋅ ( c) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 2.75 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 0.75 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
−2F1⋅ x1
1
M1 x1 :=
M2 x2 := ⎡⎣−2F1⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣−2F1⋅ x3 + A⋅ x3 − a + B⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
Moment (kN-m)
0
( )
M2( x2)
0.5
M3( x3)
M1 x1
1
0
0.2
0.4
0.6
0.8
1
x1 , x2 , x3
Max. Moment :
Distance (m)
unit := kN⋅ m
M1 ( a) = −0.900
Bending Stress:
4
Mmax := M1 ( a) ⋅ unit
σ = M⋅
c'
I
I :=
π ⋅ do
σ max := Mmax⋅
64
do
co :=
2
co
I
σ max = 173.84 MPa Ans
1.2
Problem 6-80
If the beam has a square cross section of 225 mm on each side, determine the absolute maximum
bending stress in the beam.
Given:
a := 2.5m
b := 2.5m
P := 6kN
w := 15
bo := 225mm
do := 225mm
kN
m
Solution:
Equilibrium :
+
A := w⋅ a + P
ΣFy=0;
A = 43.5 kN
MA := ( w⋅ a) ⋅ ( 0.5a) + P⋅ ( a + b)
ΣΜA=0;
MA = 76.88 kN⋅ m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
2
M1 x1 := ⎛⎝ −MA + A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
1
M2 x2 := ⎡⎣−MA + A⋅ x2 − ( w⋅ a) ⋅ x2 − 0.5 ⋅ a ⎤⎦ ⋅
kN⋅ m
( )
(
As indicated in the moment diagram, the
maximum moment is MA.
Section properties :
I :=
1
3
⋅ bo⋅ do
12
Bending Stress:
do
co :=
2
σ = M⋅
)
0
Moment (kN-m)
( )
c
I
0
2
x1 , x2
co
σ max := MA⋅
I
σ max = 40.49 MPa
( )
M 2 ( x 2 ) 50
M1 x1
Distane (m)
Ans
4
Problem 6-81
The beam is subjected to the load P at its center. Determine the placement a of the supports so that th
absolute maximum bending stress in the beam is as large as possible. What is this stress?
Solution:
Equilibrium : By symmetry, A=B=R
+
ΣF y=0;
2R − P = 0
R = 0.5P
Max. Moment :
Mmax = R⋅ ( 0.5L − a)
Mmax = 0.5P⋅ ( 0.5L − a)
For the largest Mmax require,
a := 0
Mmax =
Bending Stress:
σ = M⋅
3
I=
b⋅ d
12
c' =
d
2
Ans
P⋅ L
4
c'
I
σ max = Mmax⋅
σ max =
c'
I
P⋅ L 3
⋅
2
2
b⋅ d
Ans
Problem 6-82
If the beam in Prob. 6-23 has a cross section as shown, determine the absolute maximum bending
stress in the beam.
kN
Given:
Mo := 30kN⋅ m
w := 30
m
a := 1.5m
b := 100mm
tf := 12mm
d := 168mm
tw := 6mm
Solution:
Given
Equilibrium :
+
ΣF y=0;
−w⋅ a + A − w⋅ a + B = 0
ΣΜB=0;
Mo − ( w⋅ a) ⋅ ( 2.5a) + A⋅ ( 2a) − ( w⋅ a) ⋅ ( 0.5a) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ a .. a
⎛ A ⎞ ⎛ 57.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 32.50 ⎠
x2 := a , 1.01 ⋅ a .. 2a
Mo − 0.5w⋅ x1
M1 x1 :=
kN⋅ m
( )
2
x3 := 2a , 1.01 ⋅ 2a .. 3a
1
M2 x2 := ⎡⎣Mo − ( w⋅ a) ⋅ x2 − 0.5a + A⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
1
2
M3 x3 := ⎡⎣Mo − ( w⋅ a) ⋅ x3 − 0.5a + A⋅ x3 − a − 0.5w⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
(
)
(
)
(
( )
20
M2( x2)
M3( x3)
M1 x1
0
0
1
2
3
x1 , x2 , x3
As indicated in the moment
diagram, the maximum
moment is Mo.
Distance (m)
Section properties :
D := d + 2⋅ tf
Bending Stress:
σ = M⋅
D
co :=
2
)
I :=
1 ⎡ 3
3
⋅ ⎣b⋅ D − b − tw ⋅ d ⎤⎦
12
(
)
c
I
σ max := Mo⋅
co
I
σ max = 131.87 MPa
Ans
4
Problem 6-83
The pin is used to connect the three links together. Due to wear, the load is distributed over the top and
bottom of the pin as shown on the free-body diagram. If the diameter of the pin is 10 mm, determine
the maximum bending stress on the cross-sectional area at the center section a-a. For the solution it is
first necessary to determine the load intensities w1 and w2 .
P := 2kN
Given:
a := 25mm
b := 37.5mm
do := 10mm
Solution:
⎛ a b ⎞ + P⋅ ⎛ b ⎞
Maa := −P⋅ ⎜ +
⎜
3 2
2
⎝
⎠
⎝ ⎠
Maa = −16.6667 N⋅ m
Bending Stress:
4
I :=
π ⋅ do
σ = M⋅
64
do
co :=
2
c'
I
σ max := Maa⋅
co
I
σ max = 169.77 MPa Ans
Problem 6-84
A shaft is made of a polymer having an elliptical cross-section. If it resists an internal moment of M =
50 N·m determine the maximum bending stress developed in the material (a) using the flexure formul
where Iz = 1/4 π (0.08 m)(0.04 m)3, (b) using integration. Sketch a three-dimensional view of the
stress distribution acting over the cross-sectional area.
Given: a := 80mm
b := 40mm
Mz := 50N⋅ m
Solution:
a) Using the flexure formula,
cmax := b
3
Iz :=
π ⋅ a⋅ b
4
σ max := Mz⋅
4
Iz = 4021238.60 mm
cmax
Iz
σ max = 0.497 MPa
Ans
b) Using integration,
y
b
2
2
2
+
z
2
= 1
a
⎛⎜ ⌠ 2 ⎞
Izo = ⎮ y dA
⎜ ⌡A
⎝
⎠
z=
a
2
2
⋅ b −y
b
b
⌠
2
Izo = ⎮ y ⋅ ( 2z) dy
⌡− b
⌠
Izo := ⎮
⎮
⌡
b
−b
2 ⎛a
2
2⎞
2y ⋅ ⎜ ⋅ b − y dy
b
⎝
⎠
4
Izo = 4021631.98 mm
Bending Stress:
σ' max := Mz⋅
cmax
Izo
σ' max = 0.497 MPa
Ans
Problem 6-85
Solve Prob. 6-84 if the moment is M = 50 N·m, applied about the y axis instead of the z axis. Here
Iy = 1/4 π (0.04 m)(0.08 m)3.
Given: a := 80mm
b := 40mm
My := 50N⋅ m
Solution:
cmax := a
a) Using the flexure formula,
3
Iy :=
π ⋅ b⋅ a
4
σ max := My⋅
4
Iy = 16084954.39 mm
cmax
Iy
σ max = 0.249 MPa
Ans
b) Using integration,
y
b
2
2
2
+
z
2
= 1
a
⎛⎜⌠ 2 ⎞
Iyo = ⎮ z dA
⎜⌡A
⎝
⎠
y=
b
2
2
⋅ a −z
a
b
⌠
2
Iyo = ⎮ z ⋅ ( 2y) dz
⌡− b
⌠
Iyo := ⎮
⎮
⌡
a
−a
2 ⎛b
2
2⎞
2z ⋅ ⎜ ⋅ a − z dz
a
⎝
⎠
4
Iyo = 16086527.94 mm
Bending Stress:
σ' max := My⋅
cmax
Iyo
σ' max = 0.249 MPa
Ans
Problem 6-86
The simply supported beam is made from four 16-mm-diameter rods, which are bundled as shown.
Determine the maximum bending stress in the beam due to the loading shown.
Given:
L 1 := 0.5m
L 2 := 1.5m
do := 16mm
P := 400N
Solution:
F1 = R
By symmetry :
F2 = R
Equilibrium :
+
ΣF y=0;
−2P + 2⋅ R = 0
(
R := P
)
(
Mmax := R⋅ L 1 + 0.5L 2 − P⋅ 0.5L2
Section properties :
A :=
)
π
2
⋅ do
4
2
⎡
⎛ do ⎞ ⎥⎤
π
4
⎢
I := 4⋅
⋅ d + A⋅ ⎜
⎣ 64 o
⎝2⎠⎦
Bending Stress:
cmax := do
σ = M⋅
c
I
σ max := Mmax⋅
cmax
I
σ max = 49.74 MPa
Ans
Problem 6-87
Solve Prob. 6-86 if the bundle is rotated 45° and set on the supports.
Given:
L 1 := 0.5m
L 2 := 1.5m
do := 16mm
P := 400N
Solution:
By symmetry :
F1 = R
F2 = R
Equilibrium :
+
ΣF y=0;
−2P + 2⋅ R = 0
(
R := P
)
(
Mmax := R⋅ L 1 + 0.5L 2 − P⋅ 0.5L2
A :=
Section properties :
π
2
⋅ do
4
)
2
d' := 0.5 ⋅ do + do
2
⎛ π ⋅ d 4⎞ + 2⋅ ⎛ π ⋅ d 4 + A⋅ d'2⎞
⎜
⎝ 64 o ⎠
⎝ 64 o
⎠
I := 2⋅ ⎜
Bending Stress:
σ = M⋅
do
cmax := d' +
2
c
I
σ max := Mmax⋅
cmax
I
σ max = 60.04 MPa
Ans
Problem 6-88
The steel beam has the cross-sectional area shown. Determine the largest intensity of distributed load
w0 that it can support so that the maximum bending stress in the beam does not exceed σ max = 150
MPa.
Given:
L 1 := 4m
L 2 := 4m
b := 200mm
t := 8mm
d := 250mm
σ max := 150MPa
Solution:
By symmetry :
F1 = R
F2 = R
Equilibrium :
+
ΣF y=0;
(
)
−0.5 wo⋅ L1 + L2 + 2R = 0
(
)
R = 0.25 ⋅ wo⋅ L 1 + L 2
L1
Mmax = R⋅ L 1 − 0.5wo⋅ L1 ⋅
3
(
)
L1
Mmax = 0.25 ⋅ wo⋅ L1 + L2 ⋅ L1 − 0.5wo⋅ L 1 ⋅
3
(
Section properties :
I :=
)
(
)
D := d + 2⋅ t
1 ⎡ 3
3
⋅ ⎣b⋅ D − ( b − t) ⋅ d ⎤⎦
12
Bending Stress:
σ = M⋅
D
cmax :=
2
c
I
⎡
⎣
(
)
(
wo :=
σ max⋅ ( 2I)
D
kN
wo = 13.47
m
L 1⎤ D
) 3 ⎥⎦ ⋅ 2I
σ max = ⎢0.25 ⋅ wo⋅ L1 + L2 ⋅ L1 − 0.5wo⋅ L 1 ⋅
1
⋅
(
)
(
Ans
L1
) 3
0.25 ⋅ L 1 + L 2 ⋅ L 1 − 0.5L1 ⋅
Problem 6-89
The steel beam has the cross-sectional area shown. If w0 = 10 kN/m, determine the maximum bending
stress in the beam.
Given:
L 1 := 4m
L 2 := 4m
b := 200mm
t := 8mm
d := 250mm
kN
wo := 10
m
Solution:
By symmetry :
F1 = R
F2 = R
Equilibrium :
+
ΣF y=0;
(
)
−0.5 wo⋅ L1 + L2 + 2R = 0
(
R := 0.25 ⋅ wo⋅ L1 + L2
)
L1
Mmax := R⋅ L1 − 0.5wo⋅ L 1 ⋅
3
(
Section properties :
I :=
)
D := d + 2⋅ t
1 ⎡ 3
3
⋅ ⎣b⋅ D − ( b − t) ⋅ d ⎤⎦
12
Bending Stress:
D
cmax :=
2
σ = M⋅
c
I
σ max := Mmax⋅
cmax
I
σ max = 111.38 MPa
Ans
Problem 6-90
The beam has a rectangular cross section as shown. Determine the largest load P that can be
supported on its overhanging ends so that the bending stress in the beam does not exceed σ max = 10
MPa.
Given: a := 0.5m
bo := 50mm
σ allow := 10MPa do := 100mm
Solution:
By symmetry :
A= R
B= R
Equilibrium :
+
ΣFy=0;
−2P + 2⋅ R = 0
R= P
Mmax = −P⋅ ( 1.5a) + R⋅ ( 0.5a)
Mmax = −P⋅ a
Section properties :
1 ⎛
3
⋅ ⎝ bo⋅ do ⎞⎠
12
σ = M⋅
Bending Stress:
do
cmax :=
2
c
I
σ = ( P⋅ a) ⋅
do
P :=
2I
σ allow⋅ ( 2I)
P = 1.67 kN
a⋅ do
Ans
R := P
x1 := 0 , 0.01 ⋅ a .. a
−P⋅ x1
M1 x1 :=
kN⋅ m
( )
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
1
M2 x2 := ⎡⎣−P⋅ x2 + R⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
M3 x3 := ⎡⎣−P⋅ x3 + R⋅ x3 − a + R⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
0
Moment (N-m)
I :=
( )
M 2 ( x 2 ) 0.5
M3(x3)
M1 x1
1
0
0.5
1
x1 , x2 , x3
Distance (m)
1.5
Problem 6-91
The beam has the rectangular cross section shown. If P = 1.5 kN, determine the maximum bending
stress in the beam. Sketch the stress distribution acting over the cross section.
Given: a := 0.5m
bo := 50mm
P := 1.5kN
do := 100mm
Solution:
By symmetry :
A= R
B= R
Equilibrium :
+
ΣF y=0;
−2P + 2⋅ R = 0
R := P
Mmax = −P⋅ ( 1.5a) + R⋅ ( 0.5a)
Mmax := −P⋅ a
Mmax = −0.75 kN⋅ m
Section properties :
I :=
1 ⎛
3
⋅ b ⋅d ⎞
12 ⎝ o o ⎠
Bending Stress:
do
cmax :=
2
σ max := Mmax⋅
x1 := 0 , 0.01 ⋅ a .. a
− P ⋅ x1
M1 x1 :=
kN⋅ m
( )
cmax
I
σ max = 9.00 MPa
x2 := a , 1.01 ⋅ a .. 2a
Ans
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
1
M2 x2 := ⎡⎣−P⋅ x2 + R⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
M3 x3 := ⎡⎣−P⋅ x3 + R⋅ x3 − a + R⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
Moment (N-m)
0
( )
M 2 ( x 2 ) 0.5
M3( x3)
M1 x1
1
0
0.5
1
x1 , x2 , x3
Distance (m)
1.5
Problem 6-92
The beam is subjected to the loading shown. If its cross-sectional dimension a = 180 mm, determine
the absolute maximum bending stress in the beam.
Given:
L 1 := 2m
L 2 := 1m
P := 60kN
a := 180mm
w := 40
Solution:
Given
Equilibrium :
ΣF y=0;
A + B − P − w⋅ L 1 = 0
ΣΜA=0;
P⋅ L 1 + L 2 − B⋅ L1 + 0.5w⋅ L 1 = 0
(
)
2
Guess A := 1kN B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ L 1 .. L 1
⎛ A ⎞ ⎛ 10.00 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 130.00 ⎠
x2 := L1 , 1.01 ⋅ L1 .. L1 + L2
1
2
M1 x1 := ⎛⎝ A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
( )
1
M2 x2 := ⎡⎣A⋅ x2 − w⋅ L1 ⋅ x2 − 0.5 ⋅ L1 + B⋅ x2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
Max. Moment :
( )
(
)(
)
( )
Mmax = 60.00 kN⋅ m
Section properties :
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
)
( )
M2( x2)
M 1 x 1 20
40
60
⎛ a⋅ a ⎞ ⋅ a + ⎛ a ⋅ 2a ⎞ ⋅ ⎛ 2a ⎞
⎜
⎜
⎜
3⎠ 6 ⎝ 2 3 ⎠ ⎝ 3 ⎠
⎝
yc :=
⎛ a⋅ a ⎞ + ⎛ a ⋅ 2a ⎞
⎜
⎜
⎝ 3⎠ ⎝ 2 3 ⎠
0
yc = 75.00 mm
3
3
1
2
x1 , x2
1 ⎛ a ⎞ ⎛ 2a ⎞
⎛ a 2a ⎞ ⋅ ⎡y − ⎛ 2a ⎞⎤
Iw :=
⋅⎜ ⋅⎜
+⎜ ⋅
⎢
⎜ ⎥
12 ⎝ 2 ⎠ ⎝ 3 ⎠
⎝ 2 3 ⎠ ⎣ c ⎝ 3 ⎠⎦
If :=
)
0
M1 L1 = −60.000
Mmax := −M1 L 1 ⋅ unit
(
(
unit := kN⋅ m
Moment (kN-m)
+
kN
m
1
⎛ a ⎞ + ⎛a⋅ a ⎞ ⋅ ⎡y − ⎛ a ⎞⎤
⋅ ( a) ⋅ ⎜
⎜
⎢
⎜ ⎥
12
⎝ 3 ⎠ ⎝ 3 ⎠ ⎣ c ⎝ 6 ⎠⎦
2
2
Distane (m)
3
I := If + Iw
Bending Stress:
σ = M⋅
c
I
cmax := a − yc
(
)
σ max := Mmax ⋅
cmax
I
σ max = 105.11 MPa
Ans
Problem 6-93
The beam is subjected to the loading shown. Determine its required cross-sectional dimension a, if the
allowable bending stress for the material is σallo w = 150 MPa.
Given:
L 1 := 2m
L 2 := 1m
P := 60kN
σ allow := 150MPa
w := 40
Solution:
Given
Equilibrium :
ΣF y=0;
A + B − P − w⋅ L 1 = 0
ΣΜA=0;
P⋅ L 1 + L 2 − B⋅ L1 + 0.5w⋅ L 1 = 0
(
)
2
Guess A := 1kN B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ L 1 .. L 1
⎛ A ⎞ ⎛ 10.00 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 130.00 ⎠
x2 := L1 , 1.01 ⋅ L1 .. L1 + L2
1
2
M1 x1 := ⎛⎝ A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
( )
1
M2 x2 := ⎡⎣A⋅ x2 − w⋅ L1 ⋅ x2 − 0.5 ⋅ L1 + B⋅ x2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
Max. Moment :
( )
(
)(
)
( )
Mmax = 60.00 kN⋅ m
Section properties :
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
)
( )
M2( x2)
M 1 x 1 20
40
60
⎛a⋅ a ⎞ ⋅ a + ⎛ a ⋅ 2a ⎞ ⋅ ⎛ 2a ⎞
⎜
⎜
⎜
3⎠ 6 ⎝ 2 3 ⎠ ⎝ 3 ⎠
⎝
yc =
⎛a⋅ a ⎞ + ⎛ a ⋅ 2a ⎞
⎜
⎜
⎝ 3⎠ ⎝ 2 3 ⎠
0
3
3
1
1
⎛ a ⎞ + ⎛ a⋅ a ⎞ ⋅ ⎡y − ⎛ a ⎞⎤
⋅ ( a) ⋅ ⎜
⎜
⎢
⎜ ⎥
12
⎝ 3 ⎠ ⎝ 3 ⎠ ⎣ c ⎝ 6 ⎠⎦
2
2
x1 , x2
5a
yc =
12
1 ⎛ a ⎞ ⎛ 2a ⎞
⎛ a 2a ⎞ ⋅ ⎡y − ⎛ 2a ⎞⎤
Iw =
⋅⎜ ⋅⎜
+⎜ ⋅
⎢
⎜ ⎥
12 ⎝ 2 ⎠ ⎝ 3 ⎠
⎝ 2 3 ⎠ ⎣ c ⎝ 3 ⎠⎦
If =
)
0
M1 L1 = −60.000
Mmax := −M1 L 1 ⋅ unit
(
(
unit := kN⋅ m
Moment (kN-m)
+
kN
m
Distane (m)
4
43a
Iw =
1296
2
4
If =
31a
1296
3
4
I = If + Iw
Bending Stress:
I=
σ = M⋅
(
37a
648
c
I
cmax = a − yc
3
a :=
cmax
)σ
I = Mmax ⋅
allow
7
648Mmax⋅ ⎛⎜ ⎞
12
⎝
⎠
37σ allow
a = 159.88 mm
Ans
7a
cmax =
12
Problem 6-94
The wing spar ABD of a light plane is made from 2014T6 aluminum and has a cross-sectional area of
1000 mm2, a depth of 80 mm, and a moment of inertia about its neutral axis of 1.662 (106) mm4.
Determine the absolute maximum bending stress in the spar if the anticipated loading is to be as shown.
Assume A, B, and C are pins. Connection is made along the central longitudinal axis of the spar.
Given:
a := 1m
b := 2m
A := 1000m
2
( 6)
kN
wo := 15
m
4
I := 1.662 ⋅ 10 mm
do := 80mm
Solution:
L := a + b
Given
Equilibrium :
+
ΣF y=0;
A − B + 0.5wo⋅ ( L) = 0
ΣΜA=0;
B⋅ a − 0.5wo⋅ L ⋅ ⎜
) ⎛⎝ 3 ⎞⎠ = 0
(
L
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ −0.00 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 22.50 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
wo
⎡
⎛ x1 ⎞ 1⎤⎤ 1
2 ⎡
M1 x1 := ⎢A⋅ x1 +
⋅ x1 ⋅ ⎢1 − ⎜
⋅ ⎥⎥ ⋅
2
⎣
⎣ ⎝ L ⎠ 3⎦⎦ kN⋅ m
( )
wo
⎡
⎛ x2 ⎞ 1⎤⎤ 1
2 ⎡
M2 x2 := ⎢⎡⎣A⋅ x2 − B⋅ x2 − a ⎤⎦ +
⋅ x2 ⋅ ⎢1 − ⎜
⋅ ⎥⎥ ⋅
2
⎣
⎣ ⎝ L ⎠ 3⎦⎦ kN⋅ m
( )
(
unit := kN⋅ m
M1 ( a) = 6.667
Mmax := M1 ( a) ⋅ unit
Bending Stress:
σ = M⋅
c
I
do
cmax :=
2
( < σY = 414 MPa)
( )
M2( x2)
M1 x1 5
0
cmax
σ max := Mmax⋅
I
σ max = 160.45 MPa
Moment (kN-m)
Max. Moment :
)
0
1
2
x1 , x2
Ans
Distane (m)
3
Problem 6-95
The boat has a weight of 11.5 kN and a center of gravity at G. If it rests on the trailer at the smooth
contact A and can be considered pinned at B, determine the absolute maximum bending stress
developed in the main strut of the trailer. Consider the strut to be a box-beam having the dimensions
shown and pinned at C.
Given:
a := 0.9m
b := 1.8m
c := 1.2m
d := 0.3m
bo := 45mm
do := 75mm
bi := 38mm
di := 45mm
Solution:
W := 11.5kN
Equilibrium (for boat) :
Given
+
ΣF y=0;
A−W+B= 0
ΣΜB=0; A⋅ ( a + b) − W⋅ ( b − d) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
⎛ A ⎞ ⎛ 6.389 ⎞
:= Find ( A , B)
=⎜
kN
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 5.111 ⎠
Equilibrium (for assembly) :
+
ΣF y=0;
Given
D−A−B+C= 0
ΣΜC=0; −A⋅ ( a + b + c) + D⋅ ( b + c) − B⋅ c = 0
Guess C := 1kN
D := 1kN
⎛C⎞
⎛ C ⎞ ⎛ 1.15 ⎞
:= Find ( C , D)
=⎜
kN
⎜
⎜
⎝D⎠
⎝ D ⎠ ⎝ 10.35 ⎠
x1 := 0 , 0.01 ⋅ a .. a
−A⋅ x1
M1 x1 :=
kN⋅ m
( )
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
M2 x2 := ⎡⎣−A⋅ x2 + D⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
M3 x3 := ⎡−A⋅ x3 + D⋅ x3 − a − B⋅ x3 − a − b ⎤ ⋅
⎣
⎦ kN⋅ m
unit := kN⋅ m
( )
Max. Moment :
Moment (kN-m)
M1 ( a) = −5.750
Mmax := −M1 ( a) ⋅ unit
Section properties :
I :=
1
⋅ ⎛ bo⋅ do − bi⋅ di ⎞
12 ⎝
3
3
⎠
Bending Stress:
σ = M⋅
c
I
do
cmax :=
2
σ max := Mmax⋅
cmax
I
σ max = 166.7 MPa
( )
Ans
(
)
(
)
( )
M2( x2)
M3( x3)
M1 x1 0
5
0
1
2
x1 , x2 , x3
Distance (m)
3
Problem 6-96
The beam supports the load of 25 kN. Determine the absolute maximum bending stress in the beam if
the sides of its triangular cross section are a = 150 mm.
Given:
a := 150mm
L := 0.6m
P := 25kN
Solution:
Mmax := P⋅ L
Section Property :
I :=
1
3
⋅ a⋅ ( a⋅ sin ( 60deg) )
36
σ = M⋅
Maximum Bending Stress:
c
I
2
cmax := ⋅ a⋅ sin ( 60deg)
3
(
)
σ max := Mmax ⋅
cmax
I
σ max = 142.2 MPa
Ans
Problem 6-97
The beam supports the load of 25 kN. Determine the required size a of the sides of its triangular cross
section if the allowable bending stress is σ allow = 126 MPa.
Given:
L := 0.6m
P := 25kN
σ allow := 126MPa
Solution:
Mmax := P⋅ L
Section Property :
I=
1
3
⋅ a⋅ ( a⋅ sin ( 60deg) )
36
Maximum Bending Stress:
σ = M⋅
c
I
2
cmax = ⋅ a⋅ sin ( 60deg)
3
(
)
σ = Mmax ⋅
(
cmax
I
cmax
)σ
I = Mmax ⋅
allow
1
Mmax ⎞
⎛
24
a := ⎜
⋅
2
⎝ sin ( 60deg) σ allow ⎠
a = 156.2 mm
3
Ans
Problem 6-98
The wood beam is subjected to the uniform load of w = 3 kN/m. If the allowable bending stress for the
material is σ allow = 10 MPa, determine the required dimension b of its cross section. Assume the
support at A is a pin and B is a roller.
Given:
L1 := 2m
σ allow := 10MPa
L2 := 1m
w := 3
kN
m
Solution:
Given
Equilibrium :
+
ΣFy=0;
A + B − w⋅ L1 = 0
ΣΜA=0;
−B⋅ L 1 + L 2 + 0.5w⋅ L 1 = 0
(
)
2
Guess A := 1kN B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 4.00 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 2.00 ⎠
x1 := 0 , 0.01 ⋅ L1 .. L1
x2 := L 1 , 1.01 ⋅ L 1 .. L 1 + L 2
1
2
M1 x1 := ⎛⎝ A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − w⋅ L 1 ⋅ x2 − 0.5 ⋅ L 1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)(
)
Moment (kN-m)
As indicated in the moment diagram, the
maximum moment occurs in L1 such that
V1=0:
( )
V1 = A − w⋅ xc
A
xc :=
xc = 1.333 m
w
Max. Moment : unit := kN⋅ m
( )
( )
M2(x2)
M1 x1 2
0
M1 xc = 2.667
Mmax := M1 xc ⋅ unit
0
( )
Bending Stress:
2
x1 , x2
Distane (m)
Mmax = 2.667 kN⋅ m
Section properties :
1
1
3
⋅ b⋅ ( 1.5 ⋅ b)
12
co
1.5 ⋅ b
σ = M⋅
co =
2
I
I=
1
I = Mmax⋅
co
σ allow
⎛ 6⋅ Mmax ⎞
b := ⎜
⎝ 2.25 ⋅ σ allow ⎠
3
b = 89.3 mm
Ans
3
Problem 6-99
The wood beam has a rectangular cross section in the proportion shown. Determine its required
dimension b if the allowable bending stress is σ allow = 10 MPa.
Given:
L 1 := 2m
σ allow := 10MPa
L 2 := 2m
w := 0.5
kN
m
Solution:
Given
Equilibrium :
+
ΣF y=0;
A + B − w⋅ L 1 = 0
ΣΜA=0;
−B⋅ L1 + L2 + 0.5w⋅ L1 = 0
(
)
2
Guess A := 1kN B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
x1 := 0 , 0.01 ⋅ L 1 .. L 1
⎛ A ⎞ ⎛ 0.75 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 0.25 ⎠
x2 := L1 , 1.01 ⋅ L1 .. L1 + L2
1
2
M1 x1 := ⎛⎝ A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
( )
1
M2 x2 := ⎡⎣A⋅ x2 − w⋅ L1 ⋅ x2 − 0.5 ⋅ L1 ⎤⎦ ⋅
kN⋅ m
(
)(
)
As indicated in the moment diagram, the
maximum moment occurs in L1 such that
V1=0:
( )
V1 = A − w⋅ xc
A
xc :=
xc = 1.500 m
w
Max. Moment : unit := kN⋅ m
( )
1
Moment (kN-m)
( )
( )
0.5
M2( x2)
M1 x1
0
M1 xc = 0.5625
Mmax := M1 xc ⋅ unit
0
( )
x1 , x2
Distane (m)
Mmax = 0.5625 kN⋅ m
Section properties :
Bending Stress:
I = Mmax⋅
1
3
⋅ b⋅ ( 1.5 ⋅ b)
12
co
1.5 ⋅ b
σ = M⋅
co =
I
2
I=
co
σ allow
2
3
b :=
6⋅ Mmax
2.25 ⋅ σ allow
b = 53.1 mm
Ans
Problem 6-100
A beam is made of a material that has a modulus of elasticity in compression different from that given
for tension. Determine the location c of the neutral axis, and derive an expression for the maximum
tensile stress in the beam having the dimensions shown if it is subjected to the bending moment M.
Problem 6-101
The beam has a rectangular cross section and is subjected to a bending moment M. If the material
from which it is made has a different modulus of elasticity for tension and compression as shown,
determine the location c of the neutral axis and the maximum compressive stress in the beam.
Problem 6-102
The box beam is subjected to a bending moment of M = 25 kN·m directed as shown. Determine the
maximum bending stress in the beam and the orientation of the neutral axis.
ay := −3
Given:
az := 4
ar := 5
M := 25kN⋅ m
bo := 150mm
do := 150mm
bi := 100mm
di := 100mm
Solution:
Internal Moment Components :
ay
az
My :=
⋅M
Mz :=
⋅M
ar
ar
Section Property :
Iy :=
1 ⎛
3
3
⋅ ⎝ do⋅ bo − di⋅ bi ⎞⎠
12
At B :
Mz⋅ y My⋅ z
+
Iz
Iy
yB := 0.5do
σ B := −
zB := 0.5bo
Mz⋅ yB My⋅ zB
+
Iz
Iy
σ B = −77.5 MPa (C)
At D :
1 ⎛
3
3
⋅ ⎝ bo⋅ do − bi⋅ di ⎞⎠
12
By inspection, maximum bending stress occurs at B and D.
Maximum Bending Stress:
σ= −
Iz :=
yD := −0.5 do
σ D := −
Ans
zD := −0.5 bo
Mz⋅ yD My⋅ zD
+
Iz
Iy
σ D = 77.5 MPa
(T)
Orientation of Neutral Axis :
⎛ ay ⎞
θ := atan ⎜
⎝ az ⎠
Ans
tan ( α ) =
⎛ Iz
α := atan ⎜
⎝
Iy
Iz
Iy
⋅ tan ( θ )
y' := bo⋅ tan ( α )
⎞
⎠
⋅ tan ( θ )
α = −36.87 deg
y' = −112.50 mm
Ans
Problem 6-103
Determine the maximum magnitude of the bending moment M so that the bending stress in the member
does not exceed 100 MPa.
Given:
ay := −3
az := 4
ar := 5
σ allow := 100MPa
bo := 150mm
do := 150mm
bi := 100mm
di := 100mm
Solution:
Internal Moment Components :
ay
az
My =
⋅M
Mz =
⋅M
ar
ar
Section Property :
Iy :=
1 ⎛
3
3
⋅ d ⋅ b − di⋅ bi ⎞⎠
12 ⎝ o o
Maximum Bending Stress:
Iz :=
1 ⎛
3
3
⋅ b ⋅ d − bi⋅ di ⎞⎠
12 ⎝ o o
By inspection, maximum bending stress occurs at B and D.
Apply the flexure formula for biaxial bending at either point B or D. σ = −
At B :
yB := 0.5do
σB = −
zB := 0.5bo
Mz⋅ yB My⋅ zB
+
Iz
Iy
⎛ az
σ allow = −⎜
⎝ ar
⎞ yB
⋅M ⋅
⎠ Iz
⎛ ay
+⎜
⎝ ar
⎞ zB
⋅M ⋅
⎠ Iy
σ allow
M :=
−
az yB ⎛ ay ⎞ zB
⋅
+⎜
⋅
ar Iz
ar Iy
⎝ ⎠
M = 32.24 kN⋅ m
Ans
Mz⋅ y My⋅ z
+
Iz
Iy
Problem 6-104
The beam has a rectangular cross section. If it is subjected to a bending moment of M = 3500 N·m
directed as shown, determine the maximum bending stress in the beam and the orientation of the
neutral axis.
Given:
M := 3.5kN⋅ m
θ' := 30deg
b := 150mm
d := 300mm
Solution:
θ := ( 180deg − θ' )
θ = 150 deg
Internal Moment Components :
My := M⋅ sin ( θ )
Mz := M⋅ cos ( θ )
Section Property :
Iz :=
1
3
⋅ b⋅ d
12
Maximum Bending Stress:
σ= −
Iy :=
At A :
1
3
⋅ d⋅ b
12
yA := 0.5d
σ A := −
At B :
At C :
Mz⋅ yA My⋅ zA
+
Iz
Iy
σ A = 2.903 MPa (T)
Mz⋅ yB My⋅ zB
+
Iz
Iy
σ B = −2.903 MPa (C)
zC := −0.5 b
Mz⋅ yC My⋅ zC
+
Iz
Iy
σ C = −0.208 MPa (C)
yD := −0.5 d
zD := 0.5b
σ D := −
Mz⋅ yD My⋅ zD
+
Iz
Iy
Orientation of Neutral Axis :
⎛ Iz
α := atan ⎜
⎝ Iy
z' := 0.5 ⋅ b −
⋅ tan ( θ )
⎞
σ D = 0.208 MPa (T)
tan ( α ) =
Iz
Iy
⋅ tan ( θ )
⎠
α = −66.59 deg
tan ( α )
z' = 10.05 mm
0.5d
Ans
zB := −0.5 b
yC := 0.5d
σ C := −
At D :
zA := 0.5b
yB := −0.5 d
σ B := −
Mz⋅ y My⋅ z
+
Iz
Iy
Ans
Ans
Problem 6-105
The T-beam is subjected to a bending moment of M = 15 kN·m. directed as shown. Determine the
maximum bending stress in the beam and the orientation of the neutral axis. The location of the
centroid, C, must be determined.
Given:
M := 15kN⋅ m
θ' := 60deg
bf := 300mm
tf := 50mm
tw := 50mm
dw := 200mm
Solution:
θ := ( 180deg − θ' )
θ = 120 deg
Internal Moment Components :
My := M⋅ sin ( θ )
Mz := M⋅ cos ( θ )
Section Property :
1 ⎛
3
3
4
⋅ t ⋅ b − dw⋅ tw ⎞⎠
Iy = 110416666.67 mm
12 ⎝ f f
⎯
bf⋅ tf ⋅ 0.5tf + tw⋅ dw ⋅ 0.5dw + tf
Σ ⋅ yi⋅ Ai
yc =
yc :=
bf⋅ tf + tw⋅ dw
Σ ⋅ ( Ai)
Iy :=
(
Iz :=
)
(
)(
) (
)(
( ) (
)
)
1
3
2 ⎡1
3
2⎤
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc + ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + tf − yc ⎥
12
⎣ 12
⎦
(
)(
)
(
)(
4
Iz = 130208333.33 mm
Maximum Bending Stress:
At A :
yA := yc
σ A := −
At B :
yc = 75.00 mm
σ= −
zA := 0.5bf
Mz⋅ yA My⋅ zA
+
Iz
Iy
yB := yc
Mz⋅ y My⋅ z
+
Iz
Iy
σ A = 21.97 MPa (T)
zB := −0.5 bf
Ans
)
σ B := −
At D :
Mz⋅ yB My⋅ zB
+
Iy
Iz
(
σ B = −13.33 MPa (C)
)
yD := − tf + dw − yc
σ D := −
Mz⋅ yD My⋅ zD
+
Iz
Iy
Orientation of Neutral Axis :
⎛ Iz
α := atan ⎜
⎝ Iy
z' := 0.5bf −
⋅ tan ( θ )
yc
tan ( α )
⎞
⎠
zD := −0.5 tw
σ D = −13.02 MPa (C)
tan ( α ) =
Iz
Iy
⋅ tan ( θ )
α = −63.91 deg
z' = 186.72 mm
Ans
Problem 6-106
If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of
M = 520 N·m and is directed as shown, determine the bending stress at points A and B. The location y
of the centroid C of the strut's cross-sectional area must be determined.Also, specify the orientation o
the neutral axis.
⎛5⎞
θ' := atan ⎜
Given: M := 520N⋅ m
⎝ 12 ⎠
bf := 400mm
tf := 20mm
tw := 20mm
dw := 180mm
Solution: D := dw + tf
θ := ( 180deg − θ' )
Internal Moment Components :
My := M⋅ sin ( θ )
Mz := M⋅ cos ( θ )
Section Property :
1 ⎡
3
3
4
⋅ D⋅ b − dw⋅ bf − 2tw ⎤⎦
Iy = 366826666.67 mm
12 ⎣ f
⎯
bf⋅ tf ⋅ 0.5tf + 2 tw⋅ dw ⋅ 0.5dw + tf
Σ ⋅ yi⋅ Ai
yc =
yc :=
bf⋅ tf + 2 tw⋅ dw
Σ ⋅ ( Ai)
(
Iy :=
(
Iz :=
)
)
(
)(
) (
)(
( ) (
)
)
yc = 57.37 mm
1
3
2
3
2⎤
⎡1
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc + 2⋅ ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + tf − yc ⎥
12
12
⎣
⎦
(
)(
)
(
)(
)
4
Maximum Bending Stress:
At A :
yA := yc − D
σ A := −
At B :
Mz⋅ yA My⋅ zA
+
Iz
Iy
Orientation of Neutral Axis :
⎛ Iz
⎝ Iy
⋅ tan ( θ )
σ A = −1.298 MPa (C) Ans
zB := 0.5bf
Mz⋅ yB My⋅ zB
+
Iz
Iy
α := atan ⎜
Mz⋅ y My⋅ z
+
Iz
Iy
zA := −0.5 bf
yB := yc
σ B := −
σ= −
Iz = 57601403.51 mm
⎞
⎠
σ B = 0.587 MPa (T)
tan ( α ) =
Iz
Iy
⋅ tan ( θ )
α = −3.74 deg
Ans
Problem 6-107
The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of
M = 520 N·m and is directed as shown. Determine the maximum bending stress in the strut. The
location y of the centroid C of the strut's cross-sectional area must be determined.Also, specify the
orientation of the neutral axis.
⎛5⎞
θ' := atan ⎜
Given: M := 520N⋅ m
⎝ 12 ⎠
bf := 400mm
tf := 20mm
tw := 20mm
dw := 180mm
Solution: D := dw + tf
θ := ( 180deg − θ' )
Internal Moment Components :
My := M⋅ sin ( θ )
Mz := M⋅ cos ( θ )
Section Property :
1 ⎡
3
3
4
⋅ ⎣D⋅ bf − dw⋅ bf − 2tw ⎤⎦
Iy = 366826666.67 mm
12
⎯
bf⋅ tf ⋅ 0.5tf + 2 tw⋅ dw ⋅ 0.5dw + tf
Σ ⋅ yi⋅ Ai
yc =
yc :=
bf⋅ tf + 2 tw⋅ dw
Σ ⋅ ( Ai)
(
Iy :=
(
Iz :=
)
)
(
)(
) (
)(
( ) (
)
)
yc = 57.37 mm
1
3
2
3
2⎤
⎡1
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc + 2⋅ ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + tf − yc ⎥
12
⎣ 12
⎦
(
)(
)
(
)(
)
4
Iz = 57601403.51 mm
Mz⋅ y My⋅ z
+
Iz
Iy
By inspection, the maximum bending stress can occur at either point A or B.
Maximum Bending Stress:
At A :
yA := yc − D
σ A := −
At B :
zA := −0.5 bf
Mz⋅ yA My⋅ zA
+
Iz
Iy
yB := yc
σ B := −
σ= −
zB := 0.5bf
Mz⋅ yB My⋅ zB
+
Iz
Iy
Orientation of Neutral Axis :
⎛ Iz
α := atan ⎜
⎝ Iy
⋅ tan ( θ )
σ A = −1.298 MPa (C) Ans
⎞
⎠
σ B = 0.587 MPa (T)
tan ( α ) =
Iz
Iy
⋅ tan ( θ )
α = −3.74 deg
Ans
Problem 6-108
The 30-mm-diameter shaft is subjected to the vertical and horizontal loadings of two pulleys as shown
It is supported on two journal bearings at A and B which offer no resistance to axial loading.
Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft.
Determine the maximum bending stress developed in the shaft.
Given: a := 1m
Fy := 150N
Fz := 400N
do := 30mm
Solution:
Equilibrium : In x-y plane. Given
+
ΣF y=0;
Ay + By − 2Fy = 0
ΣΜA=0;
−2Fy⋅ a − By⋅ ( 2a) = 0
Guess Ay := 1N
⎛⎜ Ay ⎞
:= Find ( Ay , By)
⎜ By
⎝ ⎠
By := 1N
⎛⎜ Ay ⎞ ⎛ 0.45 ⎞
=⎜
kN
⎜ By
−
0.15
⎝
⎠
⎝ ⎠
Equilibrium : In x-z plane, by symmetry: Az = Bz = Rz.
+
ΣF z=0;
2Rz − 2Fz = 0
Rz := Fz
Rz = 400 N
Internal Moment Components:
The shaft is subjected to two bending moment components M y and M z. The moment disgram
for each component is drawn.
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( 2a)
x3 := ( 2a) , 1.01 ⋅ ( 2a) .. 3a
( )
Mz1 x1 :=
−2Fy⋅ x1
1
Mz2 x2 := ⎡⎣−2Fy⋅ x2 + Ay⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
N⋅ m
( )
(
)
1
Mz3 x3 := ⎡⎣−2Fy⋅ x3 + Ay⋅ x3 − a ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
Rz⋅ ( x2 − a)
1
My2 ( x2) :=
My3 ( x3) := ⎡⎣Rz⋅ ( x3 − a) − 2Fz⋅ ( x3 − 2a)⎤⎦ ⋅
N⋅ m
N⋅
( )
My1 x1 := 0
Moment (N-m)
Moment (N-m)
0
( )
M z2( x 2 )
200
M z3( x 3 )
M z1 x 1
400
500
( )
M y3( x 3 )
M y2 x 2
0
0
1
2
x1 , x2 , x3
Distane (m)
3
0
1
2
x2 , x3
Distane (m)
Maximum Bending Stress:
unit := N⋅ m
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at E (x=2a).
Mmax :=
2
2
Mz2 ( 2a) + My2 ( 2a) ⋅ unit
Mmax = 427.20 N⋅ m
π
4
⋅ do
64
cmax
I
Hence,
do
cmax :=
2
I :=
σ max := Mmax⋅
σ max = 161.2 MPa
M1 x1 :=
( )
Mz1 x1
M2 x2 :=
( )
Mz2 x2
( )
Mz3 x3
( )2 + My1 (x1)2
( )2 + My2 (x2)2
( )2 + My3 (x3)2
Moment (N-m)
M3 x3 :=
Ans
( )
M2( x2)
200
M3( x3)
M 1 x 1 400
0
0
0.5
1
1.5
x1 , x2 , x3
Distane (m)
2
2.5
3
Problem 6-109
The shaft is subjected to the vertical and horizontal loadings of two pulleys D and E as shown. It is
supported on two journal bearings at A and B which offer no resistance to axial loading.
Furthermore,the coupling to the motor at C can be assumed not to offer any support to the shaft.
Determine the required diameter d of the shaft if the allowable bending stress for the material is σ allow
= 180 MPa.
Given: a := 1m Fy := 150N Fz := 400N
σ allow := 180MPa
Solution:
Equilibrium : In x-y plane. Given
+
ΣF y=0;
Ay + By − 2Fy = 0
ΣΜA=0;
−2Fy⋅ a − By⋅ ( 2a) = 0
Guess Ay := 1N
⎛⎜ Ay ⎞
:= Find ( Ay , By)
⎜ By
⎝ ⎠
By := 1N
⎛⎜ Ay ⎞ ⎛ 0.45 ⎞
=⎜
kN
⎜ By
−
0.15
⎝
⎠
⎝ ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 3a
( )
Mz1 x1 :=
−2Fy⋅ x1
1
Mz2 x2 := ⎡⎣−2Fy⋅ x2 + Ay⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
N⋅ m
( )
(
)
Equilibrium : In x-z plane, by symmetry: Az = Bz = Rz.
+
ΣF z=0;
2Rz − 2Fz = 0
Rz := Fz
Rz = 400 N
x'1 := a , 1.01 ⋅ a .. ( 2a)
x'2 := ( 2a) , 1.01 ⋅ ( 2a) .. 3a
Rz⋅ x'1 − a
1
My1 x'1 :=
My2 x'2 := ⎡⎣Rz⋅ x'2 − a − 2Fz⋅ x'2 − 2a ⎤⎦ ⋅
N⋅ m
N⋅ m
( )
(
)
( )
(
)
(
)
Internal Moment Components:
The shaft is subjected to two bending moment components M y and M z. The moment disgram
for each component is drawn.
Moment (N-m)
Moment (N-m)
0
( )
200
M z2( x 2 )
M z1 x 1
400
0
1
2
3
500
( )
M y2( x'2 )
M y1 x'1
0
0
1
2
x1 , x2
x'1 , x'2
Distane (m)
Distane (m)
Maximum Bending Stress:
unit := N⋅ m
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at E (x=2a).
Mmax :=
2
2
Mz2 ( 2a) + My1 ( 2a) ⋅ unit
Mmax = 427.20 N⋅ m
π
4
⋅ do
64
cmax
I
Hence,
do
cmax =
2
I=
σ allow = Mmax⋅
M
d
π
4 ⎛ max ⎞ o
⋅ do = ⎜
⋅
64
σ allow 2
⎝
3
do :=
⎠
32Mmax
π ⋅ σ allow
do = 28.91 mm
Ans
Problem 6-110
The board is used as a simply supported floor joist. If a bending moment of M = 1.2 kN·m is applied
3° from the z axis, determine the stress developed in the board at the corner A. Compare this stress
with that developed by the same moment applied along the z axis (θ = 0°). What is the angle a for the
neutral axis when θ = 3°? Comment: Normally, floor boards would be nailed to the top of the beam so
that θ = 0° (nearly) and the high stress due to misalignment would not occur.
Given:
M := 1.2kN⋅ m
θ := 3deg
b := 50mm
d := 150mm
Solution:
Internal Moment Components :
My := −M⋅ sin ( θ ) Mz := M⋅ cos ( θ )
yc := 0.5d
Section Property :
Iz :=
1
3
⋅ b⋅ d
12
Maximum Bending Stress:
σ= −
Iy :=
At A :
1
3
⋅ d⋅ b
12
yA := −yc
σ A := −
zA := −0.5 b
Mz⋅ yA My⋅ zA
+
Iz
Iy
⎛ Iz
⎝ Iy
When θ = 0 :
σ' A := −
⋅ tan ( θ )
⎞
Iz
+
Iz
Iy
(T)
Ans
⋅ tan ( θ )
α = 25.25 deg
⎠
M'y := 0
M'z⋅ yA
σ A = 7.40 MPa
tan ( α ) =
Orientation of Neutral Axis :
α := atan ⎜
Mz⋅ y My⋅ z
+
Iz
Iy
Ans
M'z := M
M'y⋅ zA
Iy
σ' A = 6.40 MPa
(T)
Ans
Problem 6-111
Consider the general case of a prismatic beam subjected to bending-moment components My and Mz,
as shown, when the x, y, z axes pass through the centroid of the cross section. If the material is
linear-elastic, the normal stress in the beam is a linear function of position such that σ = a + by + cz.
Using the equilibrium conditions0 = ∫A σ dA , M y = ∫A zσ dA , M z = ∫A − yσ dA
, determine the
constants a, b, and c, and show that the normal stress can be determined from the equation
σ = [- (Mz Iy + My Iyz) y + (My Iz + Mz Iyz) z] / (Iy Iz - Iyz2), where the moments and products of
inertia are defined in Appendix A.
Given: Linear function: σ x = a + by + z
Solution:
Equilibrium Conditios :
0 = ∫ σ x dA , 0 = ∫ (a + by + cz) dA
A
A
0 = a ∫ dA + b ∫ y dA + c ∫ z dA
A
A
A
M y = ∫ zσ x dA , M y = ∫ z (a + by + cz) dA
A
A
M y = a ∫ z dA + b ∫ yz dA + c ∫ z 2 dA
A
A
A
M z = ∫ - yσ x dA , M z = ∫ - y (a + by + cz) dA
A
A
M z = −a ∫ y dA − b ∫ y 2 dA − c ∫ yz dA
A
A
The integrals are defined in Appendix A. Note that
A
∫ y dA = ∫ z dA = 0
A
A
0 = a ∫ dA + b ⋅ 0 + c ⋅ 0
Thus,
A
M y = a ⋅ 0 + b ⋅ I yz + c ⋅ I y
M z = −a ⋅ 0 − b ⋅ I y − c ⋅ I yz
Solving for a, b and c :
a = 0 (since A ≠ 0)
1
(M y ⋅ I yz + M z ⋅ I y )
D
where D = I y ⋅ I z - I yz
1
(M y ⋅ I z + M z ⋅ I yz )
D
Ans
b=c=
Bending Stress:
σx =
Ans
1
1
(M z ⋅ I y + M y ⋅ I yz ) ⋅ (− y) + (M z ⋅ I yz + M y ⋅ I z ) ⋅ z
D
D
In matrix form,
⎛⎜ Iy Iyz ⎞ ⎛ −y ⎞
1
σ=
M My ⋅
⋅
⎜ Iyz Iz ⎜⎝ z ⎠
D z
⎝
⎠
(
)
QED
2
Ans
Problem 6-112
The 65-mm-diameter steel shaft is subjected to the two loads that act in the directions shown. If the
journal bearings at A and B do not exert an axial force on the shaft, determine the absolute maximum
bending stress developed in the shaft.
Given: a := 1.25m
b := 1m
F := 4kN
L := 2a + b do := 65mm
Solution:
θ := 30deg
Equilibrium : In x-y plane, by symmetry: Ay = By = Ry.
+
2Ry − 2F⋅ cos ( θ ) = 0
Ry := F⋅ cos ( θ )
Ry = 3.464 kN
ΣF y=0;
Equilibrium : In x-z plane, by anti-symmetry: Az = -Bz = Rz.
Az⋅ L − F⋅ sin ( θ ) ⋅ ( b) = 0
⎛ b⎞
Rz := F⋅ sin ( θ ) ⋅ ⎜
Rz = 0.571 kN
⎝L⎠
ΣΜB=0;
Internal Moment Components:
The shaft is subjected to two bending moment components M y and M z. The moment disgram
for each component is drawn.
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
Ry⋅ x1
Mz1 x1 :=
kN⋅ m
1
Mz2 x2 := ⎡⎣Ry⋅ x2 − F⋅ cos ( θ ) ⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
Mz3 x3 := ⎡⎣Ry⋅ x3 − F⋅ cos ( θ ) ⋅ x3 − a − F⋅ cos ( θ ) ⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
( )
(
Rz⋅ x1
My1 x1 :=
kN⋅ m
)
(
)
1
My2 x2 := ⎡⎣Rz⋅ x2 − F⋅ sin ( θ ) ⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
My3 x3 := ⎡⎣Rz⋅ x3 − F⋅ sin ( θ ) ⋅ x3 − a + F⋅ sin ( θ ) ⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
Moment (kN-m)
Moment (kN-m)
1
( )
M z2( x 2 )
M z3( x 3 )
M z1 x 1 5
0
0
2
( )
M y2( x 2 )
0
M y3( x 3 )
M y1 x 1
1
0
2
x1 , x2 , x3
x1 , x2 , x3
Distane (m)
Distane (m)
unit := kN⋅ m
Maximum Bending Stress:
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at x=a.
Mmax :=
2
2
Mz1 ( a) + My1 ( a) ⋅ unit
Mmax = 4.389 kN⋅ m
Hence,
do
cmax :=
2
I :=
π
4
⋅ do
64
σ max := Mmax⋅
cmax
I
σ max = 162.8 MPa
M1 x1 :=
( )
Mz1 x1
M2 x2 :=
( )
Mz2 x2
( )
Mz3 x3
( )2 + My1 (x1)2
( )2 + My2 (x2)2
( )2 + My3 (x3)2
Moment (N-m)
M3 x3 :=
Ans
( )5
M2( x2)
M3( x3)
M1 x1
0
0
1
2
x1 , x2 , x3
Distane (m)
3
Problem 6-113
The steel shaft is subjected to the two loads that act in the directions shown. If the journal bearings at
A and B do not exert an axial force on the shaft, determine the required diameter of the shaft if the
allowable bending stress is σ allow = 180 MPa.
Given: a := 1.25m
b := 1m
L := 2a + b θ := 30deg
F := 4kN
σ allow := 180MPa
Solution:
Equilibrium : In x-y plane, by symmetry: Ay = By = Ry.
+
2Ry − 2F⋅ cos ( θ ) = 0
Ry := F⋅ cos ( θ )
Ry = 3.464 kN
ΣF y=0;
Equilibrium : In x-z plane, by anti-symmetry: Az = -Bz = Rz.
Az⋅ L − F⋅ sin ( θ ) ⋅ ( b) = 0
⎛ b⎞
Rz := F⋅ sin ( θ ) ⋅ ⎜
Rz = 0.571 kN
⎝L⎠
ΣΜB=0;
Internal Moment Components:
The shaft is subjected to two bending moment components M y and M z. The moment disgram
for each component is drawn.
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
Ry⋅ x1
Mz1 x1 :=
kN⋅ m
1
Mz2 x2 := ⎡⎣Ry⋅ x2 − F⋅ cos ( θ ) ⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
Mz3 x3 := ⎡⎣Ry⋅ x3 − F⋅ cos ( θ ) ⋅ x3 − a − F⋅ cos ( θ ) ⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
( )
Rz⋅ x1
My1 x1 :=
kN⋅ m
(
)
(
)
1
My2 x2 := ⎡⎣Rz⋅ x2 − F⋅ sin ( θ ) ⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
My3 x3 := ⎡⎣Rz⋅ x3 − F⋅ sin ( θ ) ⋅ x3 − a + F⋅ sin ( θ ) ⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
Moment (kN-m)
Moment (kN-m)
1
( )
M z2( x 2 )
M z3( x 3 )
M z1 x 1 5
0
0
2
( )
M y2( x 2 )
0
M y3( x 3 )
M y1 x 1
1
0
2
x1 , x2 , x3
x1 , x2 , x3
Distane (m)
Distane (m)
Maximum Bending Stress:
unit := kN⋅ m
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at x=a.
Mmax :=
2
2
Mz1 ( a) + My1 ( a) ⋅ unit
Mmax = 4.389 kN⋅ m
Hence,
do
cmax =
2
I=
π
4
⋅ do
64
σ allow = Mmax⋅
cmax
I
M
d
π
4 ⎛ max ⎞ o
⋅ do = ⎜
⋅
64
σ allow 2
⎝
3
do :=
⎠
32Mmax
π ⋅ σ allow
do = 62.86 mm
Ans
Problem 6-114
Using the techniques outlined in Appendix A, Example A.4 or A.5, the Z section has principal mome
of inertia of Iy = 0.060(10-3) m4 and Iz = 0.471(10-3) m4, computed about the principal axes of inertia y
and z, respectively. If the section is subjected to an internal moment of M = 250 N·m directed
horizontally as shown, determine the stress produced at point A. Solve the problem using Eq. 6-17.
Given:
M := 250N⋅ m
θ := 32.9deg
bf := 300mm
tf := 50mm
tw := 50mm
dw := 150mm
( − 3 ) m4
Iy := 0.060 10
( − 3 ) m4
Iz := 0.471 10
Solution:
Internal Moment Components :
My := M⋅ cos ( θ )
Mz := M⋅ sin ( θ )
Coordinates of Point A :
(
y'A := 0.5bf
z'A := − dw + 0.5tf
⎛⎜ yA ⎞ ⎛ cos ( θ ) −sin ( θ ) ⎞ ⎛⎜ y'A ⎞
:= ⎜
⋅
⎜ zA
⎜ z'A
(
)
(
)
sin
θ
cos
θ
⎝
⎠
⎝ ⎠
⎝ ⎠
Bending Stress:
⎛ Mz⋅ yA
σ A := ⎜ −
⎝
Iz
+
My⋅ zA ⎞
Iy
σ A = −0.293 MPa (C)
⎠
Ans
)
⎛⎜ yA ⎞ ⎛ 221.00 ⎞
=⎜
mm
⎜ zA
−
65.46
⎝
⎠
⎝ ⎠
Problem 6-115
Solve Prob. 6-114 using the equation developed in Prob. 6-111.
Given:
M := 250N⋅ m
θ := 32.9deg
bf := 300mm
tf := 50mm
tw := 50mm
dw := 150mm
( − 3 ) m4
Iy := 0.060 10
( − 3 ) m4
Iz := 0.471 10
Solution:
Internal Moment Components :
My' := M
Mz' := 0
Section Property :
cz.w := 0.5dw + 0.5tf
cy.w := 0.5bf − 0.5tw
Iy' :=
1
3
3
2⎞
⎛1
⋅ bf⋅ tf + 2⎜ ⋅ tw⋅ dw + tw⋅ dw⋅ cz.w
12
12
⎝
⎠
Iy' = 181.25 × 10
Iz' :=
1
3
3
2⎞
⎛1
⋅ tf⋅ bf + 2⎜ ⋅ dw⋅ tw + dw⋅ tw⋅ cy.w
12
⎝ 12
⎠
Iz' = 350.00 × 10
(
)(
)
(
)(
Iy'z' := tw⋅ dw⋅ cz.w ⋅ −cy.w + tw⋅ dw⋅ −cz.w ⋅ cy.w
)
Coordinates of Point A :
(
y'A := 0.5bf
y'A = 150 mm
z'A := − dw + 0.5tf
)
z'A = −175 mm
Bending Stress: Using formula developed in Prob. 6-111.
2
D := Iy'⋅ Iz' − Iy'z'
⎛⎜ Iy' Iy'z' ⎞ ⎛⎜ −y'A ⎞
1
σ A :=
M My' ⋅
⋅
⎜ Iy'z' Iz' ⎜ z'A
D z'
⎝
⎠⎝
⎠
(
)
σ A = −0.293 MPa (C)
Ans
−6
m
−6
m
4
4
−6
Iy'z' = −187.50 × 10
m
4
Problem 6-116
Using the techniques outlined in Appendix A, Example A.4 or A.5, the Z section has principal mome
of inertia of Iy = 0.060(10-3) m4 and Iz = 0.471(10-3) m4, computed about the principal axes of inertia y
and z, respectively. If the section is subjected to an internal moment of M = 250 N·m directed
horizontally as shown, determine the stress produced at point B. Solve the problem using Eq. 6-17.
Given:
M := 250N⋅ m
θ := 32.9deg
bf := 300mm
tf := 50mm
tw := 50mm
dw := 150mm
( − 3 ) m4
Iy := 0.060 10
( − 3 ) m4
Iz := 0.471 10
Solution:
Internal Moment Components :
My := M⋅ cos ( θ )
Mz := M⋅ sin ( θ )
Coordinates of Point B :
y'B := −0.5 bf
z'B := dw + 0.5tf
⎛⎜ yB ⎞ ⎛ cos ( θ ) −sin ( θ ) ⎞ ⎛⎜ y'B ⎞
:= ⎜
⋅
⎜ zB
⎜ z'B
(
)
(
)
sin
θ
cos
θ
⎝
⎠
⎝ ⎠
⎝ ⎠
Bending Stress:
⎛ Mz⋅ yB
σ B := ⎜ −
⎝
Iz
+
My⋅ zB ⎞
Iy
σ B = 0.293 MPa (T)
⎠
Ans
⎛⎜ yB ⎞ ⎛ −221.00 ⎞
=⎜
mm
⎜ zB
65.46
⎝
⎠
⎝ ⎠
Problem 6-117
For the section, Iy' = 31.7110-62 m4, Iz' = 114(10-6) m4, Iy'z' = 15.1(10-6) m4. Using the techniques
outlined in Appendix A, the member's cross-sectional area has principal moments of inertia of Iy =
29.0(10-6) m4 and Iz = 117(10-6) m4, computed about the principal axes of inertia y and z, respectively
If the section is subjected to a moment of M = 2500 N·m directed as shown, determine
the stress produced at point A, using Eq. 6-17.
Given: M := 2500N⋅ m θ' := 10.10deg
h1 := 80mm
h2 := 140mm
b1 := 60mm
b2 := 60mm
( − 6 ) m4
Iy := 29.0 10
Solution:
( − 6 ) m4
Iz := 117 10
θ := −θ'
Internal Moment Components :
My := M⋅ sin ( θ )
Mz := M⋅ cos ( θ )
Coordinates of Point A :
y'A := −h2
z'A := −b2
⎛⎜ yA ⎞ ⎛ cos ( θ ) −sin ( θ ) ⎞ ⎛⎜ y'A ⎞
:= ⎜
⋅
⎜ zA
⎜
⎝ ⎠ ⎝ sin ( θ ) cos ( θ ) ⎠ ⎝ z'A ⎠
Bending Stress:
⎛ Mz⋅ yA
σ A := ⎜ −
⎝
Iz
+
My⋅ zA ⎞
Iy
σ A = 2.599 MPa (T)
⎠
Ans
⎛⎜ yA ⎞ ⎛ −148.35 ⎞
=⎜
mm
⎜ zA
⎝ ⎠ ⎝ −34.52 ⎠
Problem 6-118
Solve Prob. 6-117 using the equation developed in Prob. 6-111.
Given:
M := 2500N⋅ m
θ' := 10.10deg
h1 := 80mm
h2 := 140mm
b1 := 60mm
b2 := 60mm
( − 6 ) m4
−6 4
Iy'z' := 15.1 ( 10 ) m
Iy' := 31.7 10
Solution:
( − 6 ) m4
Iz' := 114 10
θ := −θ'
Internal Moment Components :
My' := 0
Mz' := M
Coordinates of Point A :
y'A := −h2
z'A := −b2
⎛⎜ yA ⎞ ⎛ cos ( θ ) −sin ( θ ) ⎞ ⎛⎜ y'A ⎞
:= ⎜
⋅
⎜ zA
⎜
⎝ ⎠ ⎝ sin ( θ ) cos ( θ ) ⎠ ⎝ z'A ⎠
⎛⎜ yA ⎞ ⎛ −148.35 ⎞
=⎜
mm
⎜ zA
⎝ ⎠ ⎝ −34.52 ⎠
Bending Stress: Using formula developed in Prob. 6-111.
2
D := Iy'⋅ Iz' − Iy'z'
σ A :=
⎛⎜ Iy' Iy'z' ⎞ ⎛⎜ −y'A ⎞
1
Mz' My' ⋅
⋅
⎜ Iy'z' Iz' ⎜ z'A
D
⎝
⎠⎝
⎠
(
)
σ A = 2.608 MPa (T)
Ans
Problem 6-119
The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). Determine the
dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two
metals. What maximum moment will this beam support if the allowable bending stress for the
aluminum is (σ allow) al = 128 MPa and for the brass (σ allow) br = 35 MPa?
Given: b := 150mm
hal := 50mm
E al := 68.9GPa
E br := 101GPa
σ al_allow := 128MPa σ br_allow := 35MPa
Solution:
Section Property :
n :=
Eal
n = 0.682178
E br
Abr = b⋅ h
yc =
Σyi⋅ Ai
ΣA
Given yc := hal
A'al := ( n⋅ b) ⋅ hal
(
)
(
)
(
)
(
)
yc =
A'al⋅ 0.5 ⋅ hal + Abr⋅ hal + 0.5 ⋅ h
A'al + Abr
hal =
A'al⋅ 0.5 ⋅ hal + b⋅ h⋅ hal + 0.5 ⋅ h
A'al + b⋅ h
Guess h := 10mm
h := Find ( h)
h = 41.30 mm
Ans
Abr := b⋅ h
1
3
2
⋅ b⋅ h + Abr⋅ ( 0.5h)
12
1
3
2
I'al :=
⋅ ( n⋅ b) ⋅ hal + A'al⋅ yc − 0.5hal
12
Ibr :=
(
I := Ibr + I'al
)
4
I = 7785108.17 mm
Allowable Bending Stress:
σ=
M⋅ y
I
Assume failure of red brass:
ybr := h
Mbr :=
Assume failure of aluminum:
yal := hal
Mal :=
(σbr_allow)⋅ I
ybr
(σal_allow)⋅ I
n⋅ yal
Mbr = 6.60 kN⋅ m
Mal = 29.22 kN⋅ m
(
)
Mallow := min Mbr , Mal
Mallow = 6.60 kN⋅ m
Ans
Problem 6-120
The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). If the height h = 4
mm, determine the maximum moment that can be applied to the beam if the allowable bending stress
for the aluminum is (σ allow) al = 128 MPa and for the brass (σ allow) br = 35 MPa?
Given: b := 150mm
hal := 50mm
E al := 68.9GPa
hbr := 40mm
E br := 101GPa
σ al_allow := 128MPa σ br_allow := 35MPa
Solution:
Section Property :
n :=
Eal
n = 0.682178
E br
Abr := b⋅ hbr
yc =
Σyi⋅ Ai
ΣA
A'al := ( n⋅ b) ⋅ hal
(
)
(
A'al⋅ 0.5 ⋅ hal + Abr⋅ hal + 0.5 ⋅ hbr
A'al + Abr
yc :=
)
yc = 49.289 mm
1
3
2
⋅ b⋅ hbr + Abr⋅ hal + 0.5hbr − yc
12
1
3
2
I'al :=
⋅ ( n⋅ b) ⋅ hal + A'al⋅ yc − 0.5hal
12
(
Ibr :=
)
(
I := Ibr + I'al
)
4
I = 7457987.63 mm
Allowable Bending Stress:
σ=
M⋅ y
I
Assume failure of red brass:
ybr := hal + hbr − yc
Mbr :=
Assume failure of aluminum:
yal := yc
Mal :=
(σbr_allow)⋅ I
ybr
(σal_allow)⋅ I
n⋅ yal
Mbr = 6.41 kN⋅ m
Mal = 28.39 kN⋅ m
(
)
Mallow := min Mbr , Mal
Mallow = 6.41 kN⋅ m
Ans
Problem 6-121
A wood beam is reinforced with steel straps at its top and bottom as shown. Determine the maximum
bending stress developed in the wood and steel if the beam is subjected to a bending moment of M =
kN·m. Sketch the stress distribution acting over the cross section. Take E w = 11 GPa, E st = 200 GPa.
Given: b := 200mm
hw := 300mm
E w := 11GPa
E st := 200GPa
tst := 20mm
M := 5kN⋅ m
Solution:
Section Property :
n :=
E st
Ew
Aw := b⋅ hw
n = 18.181818
A'st := ( n⋅ b) ⋅ tst
1
3
2
I'st :=
⋅ ( n⋅ b) ⋅ tst + A'st⋅ 0.5hw + 0.5tst
12
(
I :=
1
3
⋅ b⋅ hw + 2I'st
12
Maximum Bending Stress:
)
4
I = 4178484848.48 mm
σ=
M⋅ y
I
For wood beam,
yw := 0.5hw
M⋅ yw
σ w.max :=
I
For steel straps,
yst := 0.5hw + tst
At
y'st := 0.5hw
n⋅ M⋅ yst
σ st.max :=
σ' st :=
I
n⋅ M⋅ y'st
I
σ w.max = 0.179 MPa
Ans
σ st.max = 3.699 MPa
Ans
σ' st = 3.263 MPa
Problem 6-122
The Douglas Fir beam is reinforced with A-36 steel straps at its center and sides. Determine the
maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz =
10 kN·m. Sketch the stress distribution acting over the cross section.
Given: bst := 12mm
d := 150mm
bw := 50mm
E w := 13.1GPa
Mz := 10kN⋅ m
E st := 200GPa
Solution:
Section Property :
Iz :=
n :=
Ew
1 ⎡
3
3
⋅ ⎣ 3bst ⋅ d + n⋅ 2bw ⋅ d ⎤⎦
12
( )
(
Maximum Bending Stress:
ymax := 0.5d
σ st :=
n = 0.0655
E st
)
σ=
Mz⋅ y
Iz
Mz⋅ ymax
Iz
σ st = 62.7 MPa
Ans
( )
σ w = 4.10 MPa
Ans
σ w := n⋅ σ st
Problem 6-123
The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and
in the wood if the beam is subjected to a moment of M = 1.2 kN·m. E st = 200 GPa, Ew = 12GPa.
Given: bst := 399mm
dst := 100mm
tst := 12mm
bw := 375mm
dw := 88mm
M := 1.2kN⋅ m
E w := 12GPa
E st := 200GPa
df := dw
Solution:
Section Property :
As1 := 2tst⋅ df
yc :=
Ew
n :=
n = 0.06
E st
As2 := bst⋅ tst
A'w := n⋅ bw⋅ dw
(A'w + As1)⋅ (0.5dw + tst) + As2⋅ (0.5tst)
yc = 29.04 mm
A'w + As1 + As2
1 ⎛
3
2
⋅ ⎝ n⋅ bw⋅ dw ⎞⎠ + A'w⋅ 0.5dw + tst − yc
12
1 ⎛
3
2
Is1 :=
⋅ 2t ⋅ d ⎞ + As1⋅ 0.5dw + tst − yc
12 ⎝ st f ⎠
1 ⎛
3
2
Is2 :=
⋅ ⎝ bst⋅ tst ⎞⎠ + As2⋅ 0.5tst − yc
12
(
Iw :=
)
(
)
(
)
I := Iw + Is1 + Is2
Maximum Bending Stress:
cmax := dst − yc
σ st :=
M⋅ cmax
σ=
M⋅ c
I
I
σ st = 10.37 MPa
Ans
( )
σ w = 0.62 MPa
Ans
σ w := n⋅ σ st
Problem 6-124
The Douglas Fir beam is reinforced with A-36 steel straps at its sides. Determine the maximum stress
developed in the wood and steel if the beam is subjected to a bending moment of Mz = 4 kN·m. Sketc
the stress distribution acting over the cross section.
Given: bst := 15mm
d := 350mm
bw := 200mm
E w := 13.1GPa
Mz := 4kN⋅ m
E st := 200GPa
Solution:
Section Property :
Iz :=
n :=
Ew
E st
1 ⎡
3
3
⋅ ⎣ 2bst ⋅ d + n⋅ bw⋅ d ⎤⎦
12
( )
Maximum Bending Stress:
ymax := 0.5d
σ st :=
n = 0.0655
σ=
Mz⋅ y
Iz
Mz⋅ ymax
Iz
σ st = 4.546 MPa
Ans
( )
σ w = 0.298 MPa
Ans
σ w := n⋅ σ st
Problem 6-125
The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the cross
section shown. If it is subjected to a moment of M = 6.5 kN·m, determine the maximum stress in the
brass and steel. Also, what is the stress in each material at the seam where they are bonded together?
Given: b := 125mm
E st := 200GPa
hst := 100mm
hbr := 100mm
E br := 101GPa
M := 6.5kN⋅ m
Solution:
Section Property :
n :=
E br
n = 0.505
Est
Ast := b⋅ hst
yc =
Σyi⋅ Ai
ΣA
yc :=
A'br := ( n⋅ b) ⋅ hbr
(
)
(
)
A'br⋅ 0.5 ⋅ hbr + Ast⋅ hbr + 0.5 ⋅ hst
A'br + Ast
yc = 116.445 mm
1
3
2
⋅ b⋅ hst + Ast⋅ hbr + 0.5hst − yc
12
1
3
2
I'br :=
⋅ ( n⋅ b) ⋅ hbr + A'br⋅ yc − 0.5hbr
12
(
Ist :=
)
(
I := Ist + I'br
)
4
I = 57620604.93 mm
Maximum Bending Stress:
σ=
M⋅ y
I
For steel,
yst := hbr + hst − yc
σ st.max :=
σ st.max = 9.426 MPa
Ans
σ br.max = 6.634 MPa
Ans
I
σ' st = 1.855 MPa
Ans
σ' br := n⋅ σ' st
σ' br = 0.937 MPa
Ans
For red brass,
ybr := yc
M⋅ yst
σ br.max :=
I
n⋅ M⋅ ybr
I
Bending Stress at the Seam:
yseam := yc − hbr
σ' st :=
M⋅ yseam
Problem 6-126
The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the cross
section shown. If the allowable bending stress for the steel is (σ allow)st = 180 MPa and for the brass
(σ allow)br = 60 MPa, determine the maximum moment M that can be applied to the beam.
Given: b := 125mm
hst := 100mm
E st := 200GPa
hbr := 100mm
E br := 101GPa
σ st_allow := 180MPa
σ br_allow := 60MPa
Solution:
Section Property :
n :=
E br
n = 0.505
Est
Ast := b⋅ hst
yc =
Σyi⋅ Ai
yc :=
ΣA
A'br := ( n⋅ b) ⋅ hbr
(
)
(
)
A'br⋅ 0.5 ⋅ hbr + Ast⋅ hbr + 0.5 ⋅ hst
A'br + Ast
yc = 116.445 mm
1
3
2
⋅ b⋅ hst + Ast⋅ hbr + 0.5hst − yc
12
1
3
2
I'br :=
⋅ ( n⋅ b) ⋅ hbr + A'br⋅ yc − 0.5hbr
12
(
Ist :=
)
(
I := Ist + I'br
)
4
I = 57620604.93 mm
Allowable Bending Stress:
σ=
M⋅ y
I
Assume failure of red brass:
ybr := yc
Mbr :=
Assume failure of steel:
yst := hbr + hst − yc
Mst :=
(σbr_allow)⋅ I
n⋅ ybr
(σst_allow)⋅ I
yst
Mbr = 58.79 kN⋅ m
Mst = 124.13 kN⋅ m
(
)
Mallow := min Mbr , Mst
Mallow = 58.79 kN⋅ m
Ans
Problem 6-127
The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stress
for the steel is (σ st) allow = 280 MPa and the allowable compressive stress for the concrete is (σ conc )
allow = 21 MPa, determine the maximum moment M that can be applied to the section. Assume the
concrete cannot support a tensile stress. E st = 200 GPa, Econc = 26.5 GPa.
Given: bf := 550mm
df := 100mm
bw := 150mm
dw := 450mm
dr := 25mm
hr := 50mm
E conc := 26.5GPa
E st := 200GPa
σ c_allow := 21MPa
σ st_allow := 280MPa
Est
Solution:
Section Property :
n :=
(
)2
A'st := n⋅ 2π ⋅ 0.5dr
n = 7.54717
E conc
(
)
(
)
(
)
Given A'st⋅ dw − h'w − hr − bf⋅ df⋅ 0.5df + h'w − bw⋅ h'w⋅ 0.5h'w = 0
Guess h'w := 10mm
( )
h'w := Find h'w
h'w = 3.41 mm
(
Ist := A'st⋅ dw − h'w − hr
)2
1 ⎛
3
2
⋅ ⎝ bf⋅ df ⎞⎠ + bf⋅ df ⋅ 0.5df + h'w
12
1 ⎛
3
2
I'w :=
⋅ b ⋅ h' ⎞ + bw⋅ h'w ⋅ 0.5h'w
12 ⎝ w w ⎠
(
If :=
)(
(
)
)(
)
I := Ist + If + I'w
Assume concrete fails:
ymax := df + h'w
σ=
M⋅ y
I
Mconc :=
Assume steel fails:
yst := dw − h'w − hr
(
Mst :=
)
Mallow := min Mconc , Mst
Mallow = 127.98 kN⋅ m
Ans
(σc_allow)⋅ I
ymax
Mconc = 277.83 kN⋅ m
(σst_allow)⋅ I
n⋅ yst
Mst = 127.98 kN⋅ m
Problem 6-128
Determine the maximum uniform distributed load w0 that can be supported by the reinforced concrete
beam if the allowable tensile stress for the steel is (σ st) allow = 200 MPa and the allowable compressive
stress for the concrete is (σ conc ) allow = 20 MPa. Assume the concrete cannot support a tensile stress.
Take E st = 200 GPa, Econc = 25 GPa.
Given: b := 250mm
d := 500mm
dr := 16mm
hr := 50mm
E conc := 25GPa
E st := 200GPa
σ c_allow := 20MPa
L := 2.5m
σ st_allow := 200MPa
Est
Solution:
Section Property :
n :=
(
)2
n=8
E conc
A'st := n⋅ 2π ⋅ 0.5dr
Given A'st⋅ d − h' − hr − b⋅ h'⋅ ( 0.5h') = 0
(
)
Guess h' := 10mm
h' := Find ( h')
(
I := A'st⋅ d − h' − hr
Assume concrete fails:
ymax := h'
h' = 95.51 mm
)2 + ⎡⎢⎣12 ⋅ (b⋅ h'3) + ( b⋅ h') ⋅ ( 0.5h') 2⎤⎥⎦
1
σ=
Mconc :=
M⋅ y
I
(σc_allow)⋅ I
ymax
Assume steel fails:
yst := d − h' − hr
Mst :=
Mconc = 99.85 kN⋅ m
(σst_allow)⋅ I
n⋅ yst
Mst = 33.63 kN⋅ m
Thus, steel fails first.
Maximum moment ocurs over the middle support:
wo :=
2⋅ Mst
L
2
kN
wo = 10.76
m
Ans
wo⋅ L
Mmax =
2
2
Problem 6-129
A bimetallic strip is made from pieces of 2014-T6 aluminum and C83400 red brass, having the cross
section shown. A temperature increase causes its neutral surface to be bent into a circular arc having a
radius of 400 mm. Determine the moment that must be acting on its cross section due to the thermal
stress. Take E al = 74 GPa, E br = 102 GPa.
Given: bbr := 6mm
dbr := 2mm
bal := 6mm
dal := 2mm
E al := 74GPa
ρ := 400mm
E br := 102GPa
Solution: Transform the section to brass
Section Property :
n :=
A'al := n⋅ bal⋅ dal
yc :=
(
Eal
n = 0.72549
E br
Abr := bbr⋅ dbr
)
(
)
A'al⋅ 0.5dal + Abr⋅ 0.5dbr + dal
A'al + Abr
yc = 2.16 mm
1 ⎛
3
2
⋅ ⎝ n⋅ bal⋅ dal ⎞⎠ + A'al⋅ 0.5dal − yc
12
1 ⎛
3
2
Ibr :=
⋅ b ⋅ d ⎞ + Abr⋅ 0.5dbr + dal − yc
12 ⎝ br br ⎠
(
Ial :=
)
(
)
I := Ial + Ibr
Maximum Bending Stress:
M :=
(Ebr)⋅ I
ρ
M = 6.91 N⋅ m
Ans
1
ρ
=
M
E⋅ I
Problem 6-130
The fork is used as part of a nosewheel assembly for an airplane. If the maximum wheel reaction at the
end of the fork is 4.5 kN, determine the maximum bending stress in the curved portion of the fork at
section a-a. There the cross-sectional area is circular, having a diameter of 50 mm.
Given:
rc := 250mm
a := 150mm
θ := 30deg
do := 50mm
F := 4.5kN
Solution:
Internal Moment :
(
)
M − F⋅ a − rc⋅ sin ( θ ) = 0
ΣΜC=0;
(
M := F⋅ a − rc⋅ sin ( θ )
)
M = 112.5 N⋅ m
Section Property :
ro := 0.5do
A := π ⋅ ro
IA_r = Σ ∫
A
R=
dA
r
A
IA_r
2
IA_r = 2⋅ π ⋅ ⎛⎝ rc −
R :=
rc − ro ⎞⎠
2
π ⋅ ro
2⋅ π ⋅ ⎛⎝ rc −
2
2
rc − ro ⎞⎠
2
2
R = 249.373 mm
M⋅ ( R − r)
A⋅ r⋅ rc − R
Bending Stress:
σ=
rA := rc − ro
σ A :=
rB := rc + ro
σ B :=
(
)
(
)
M⋅ R − rA
(
)
σ A = 9.91 MPa
)
σ B = −8.52 MPa
A⋅ rA⋅ rc − R
(
)
M⋅ R − rB
(
A⋅ rB⋅ rc − R
(
σ max := max σ A , σ B
σ max = 9.91 MPa (T)
)
Ans
Problem 6-131
Determine the greatest magnitude of the applied forces P if the allowable bending stress is (σ allow)c =
50 MPa in compression and (σ allow)t = 120 MPa in tension.
Given: bf := 75mm
b'f := 150mm
dw := 150mm
tw := 10mm
σ c.allow := −50MPa
tf := 10mm
ri := 250mm
σ t.allow := 120MPa
Solution:
Internal Moment :
M = P(dw+tf ) kN-m is positive since it tends
to increase the beam's radius of curvature.
Section Property : re := ri + dw + 2tf
2
A := bf⋅ tf + dw⋅ tw + b'f⋅ tf
⎯
⎯ Σ ⋅ ri⋅ Ai
r=
Σ ⋅ ( Ai)
(
)
rc :=
A = 3750 mm
(b'f⋅ tf)⋅ (ri + 0.5tf) + (dw⋅ tw)⋅ (ri + 0.5dw + tf) + (bf⋅ tf)⋅ (re − 0.5tf)
A
rc = 319.000 mm
dA
A r
IA_r = Σ ∫
⎛ r i + tf ⎞
IA_r := b'f⋅ ln ⎜
⎝ ri ⎠
⎛ r e − tf ⎞
+ tw⋅ ln ⎜
⎝ r i + tf ⎠
⎛ re ⎞
+ bf⋅ ln ⎜
⎝ r e − tf ⎠
IA_r = 12.245 mm
R :=
A
IA_r
Normal Stress:
R = 306.243 mm
σ=
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
)
Assume tension failure.
r := ri
(
)
P ⋅ dw + t f =
(
)
σ ⋅ A⋅ r⋅ rc − R
M=
R−r
(
)
σ t.allow ⋅ A⋅ r⋅ rc − R
P :=
R−r
(
)
σ t.allow ⋅ A⋅ r⋅ rc − R
( dw + t f ) ⋅ ( R − r )
P = 159.48 kN
Assume compression failure.
r' := re
(
)
P'⋅ dw + tf =
(
)
σ c.allow⋅ A⋅ r'⋅ rc − R
R − r'
P' :=
(
)
σ c.allow⋅ A⋅ r'⋅ rc − R
(dw + tf)⋅ ( R − r')
P' = 55.2 kN
Pallow := min ( P , P')
Pallow = 55.20 kN
Ans
Problem 6-132
If P = 6 kN, determine the maximum tensile and compressive bending stresses in the beam.
Given: bf := 75mm
b'f := 150mm
dw := 150mm
tw := 10mm
tf := 10mm
ri := 250mm
P := 6kN
Solution:
Internal Moment :
(
M := P⋅ dw + tf
)
M = 0.960 kN⋅ m
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re := ri + dw + 2tf
A := bf⋅ tf + dw⋅ tw + b'f⋅ tf
⎯
⎯ Σ ⋅ ri⋅ Ai
r=
Σ ⋅ ( Ai)
(
)
2
A = 3750 mm
(b'f⋅ tf)⋅ (ri + 0.5tf) + (dw⋅ tw)⋅ (ri + 0.5dw + tf) + (bf⋅ tf)⋅ (re − 0.5tf)
rc :=
A
rc = 319.000 mm
dA
A r
IA_r = Σ ∫
⎛ r i + tf ⎞
⎛ r e − tf ⎞
IA_r := b'f⋅ ln ⎜
+ tw⋅ ln ⎜
⎝ r i + tf ⎠
⎝ ri ⎠
⎛ re ⎞
+ bf⋅ ln ⎜
⎝ r e − tf ⎠
IA_r = 12.245 mm
R :=
A
IA_r
Normal Stress:
R = 306.243 mm
σ=
M⋅ ( R − r)
A⋅ r⋅ rc − R
Maximum tensile stress:
(
)
r := ri
σ t_max :=
Maximum compressive stress:
σ c_max :=
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
)
σ t_max = 4.51 MPa
(T)
Ans
σ c_max = −5.44 MPa
(C)
Ans
r := re
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
)
Problem 6-133
The curved beam is subjected to a bending moment of M = 900 N·m as shown. Determine the stress
at points A and B, and show the stress on a volume element located at each of these points.
Given: bf := 100mm
tf := 20mm
ri := 400mm
tw := 15mm
dw := 150mm
M := −900N⋅ m
θ := 30deg
Solution:
Internal Moment :
M is negative since it tends to decrease the
beam's radius of curvature.
Section Property : re := ri + dw + tf
2
A := bf⋅ tf + dw⋅ tw
⎯
⎯ Σ ⋅ ri⋅ Ai
r=
Σ ⋅ ( Ai)
(
)
A = 4250 mm
(dw⋅ tw)⋅ (ri + 0.5dw) + (bf⋅ tf)⋅ (re − 0.5tf)
rc :=
A
rc = 515.000 mm
dA
A r
IA_r = Σ ∫
⎛ r e − tf ⎞
IA_r := tw⋅ ln ⎜
⎝ ri ⎠
⎛ re ⎞
+ bf⋅ ln ⎜
⎝ r e − tf ⎠
IA_r = 8.349 mm
R :=
A
IA_r
R = 509.067 mm
M⋅ ( R − r)
A⋅ r⋅ rc − R
Normal Stress:
σ=
At A: rA := re
σ A :=
At B: rB := ri
σ B :=
(
)
(
)
M⋅ R − rA
(
)
σ A = 3.82 MPa
(T)
Ans
)
σ B = −9.73 MPa (C)
Ans
A⋅ rA⋅ rc − R
(
)
M⋅ R − rB
(
A⋅ rB⋅ rc − R
Problem 6-134
The curved beam is subjected to a bending moment of M = 900 N·m. Determine the stress at point C.
Given: bf := 100mm
tf := 20mm
ri := 400mm
tw := 15mm
dw := 150mm
M := −900N⋅ m
θ := 30deg
Solution:
Internal Moment :
M is negative since it tends to decrease the
beam's radius of curvature.
Section Property : re := ri + dw + tf
2
A := bf⋅ tf + dw⋅ tw
⎯
⎯ Σ ⋅ ri⋅ Ai
r=
Σ ⋅ ( Ai)
(
)
rc :=
A = 4250 mm
(dw⋅ tw)⋅ (ri + 0.5dw) + (bf⋅ tf)⋅ (re − 0.5tf)
A
rc = 515.000 mm
dA
A r
IA_r = Σ ∫
⎛ r e − tf ⎞
IA_r := tw⋅ ln ⎜
⎝ ri ⎠
⎛ re ⎞
+ bf⋅ ln ⎜
⎝ r e − tf ⎠
IA_r = 8.349 mm
R :=
A
IA_r
Normal Stress:
R = 509.067 mm
σ=
At C: rC := re − tf
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
σ C :=
)
(
)
M⋅ R − rC
(
)
A⋅ rC⋅ rc − R
σ C = 2.66 MPa
(T)
Ans
Problem 6-135
The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple
as shown, determine the maximum tensile and compressive stress acting at section a-a. Sketch the
stress distribution on the section in three dimensions.
Given: b := 50mm
h := 75mm
ri := 162.5mm
P := 250N
θ := 60deg
a := 150mm
Solution:
Internal Moment :
M := P⋅ ( a⋅ sin ( θ ) + h⋅ cos ( θ ) )
M = 41.851 N⋅ m
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re := ri + h
dA
A r
IA_r = Σ ∫
R :=
A
IA_r
2
A := b⋅ h
A = 3750 mm
rc := ri + 0.5 ⋅ h
rc = 200 mm
⎛ re ⎞
IA_r := b⋅ ln ⎜
IA_r = 18.974 mm
ri
⎝ ⎠
R = 197.634 mm
M⋅ ( R − r)
A⋅ r⋅ rc − R
Normal Stress:
σ=
At A: rA := re
σ A :=
At B: rB := ri
σ B :=
(
)
(
)
M⋅ R − rA
(
)
σ A = −0.792 MPa (C)
Ans
)
σ B = 1.020 MPa
Ans
A⋅ rA⋅ rc − R
(
)
M⋅ R − rB
(
A⋅ rB⋅ rc − R
(T)
Problem 6-136
The circular spring clamp produces a compressive force of 3 N on the plates. Determine the maximum
bending stress produced in the spring at A. The spring has a rectangular cross section as shown.
Given: b := 20mm
P := 3N
h := 10mm
ri := 200mm
a := 220mm
Solution:
Internal Moment :
M := P⋅ a
M = 0.660 N⋅ m
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re := ri + h
dA
A r
IA_r = Σ ∫
R :=
A
IA_r
Normal Stress:
2
A := b⋅ h
A = 200 mm
rc := ri + 0.5 ⋅ h
rc = 205 mm
⎛ re ⎞
IA_r := b⋅ ln ⎜
IA_r = 0.976 mm
ri
⎝ ⎠
R = 204.959 mm
σ=
M⋅ ( R − r)
A⋅ r⋅ rc − R
Maximum tensile stress:
(
)
rA := ri
σ t_max :=
Maximum compressive stress:
σ c_max :=
(
)
M⋅ R − rA
(
)
A⋅ rA⋅ rc − R
σ t_max = 2.01 MPa
(T)
Ans
σ c_max = −1.95 MPa
(C)
Ans
r'A := re
(
)
M⋅ R − r'A
(
)
A⋅ r'A⋅ rc − R
Problem 6-137
Determine the maximum compressive force the spring clamp can exert on the plates if the allowable
bending stress for the clamp is σ allow = 4 MPa.
Given: b := 20mm
h := 10mm
σ t.allow := 4MPa
ri := 200mm
a := 220mm
σ c.allow := −4MPa
Solution:
Section Property : re := ri + h
dA
A r
IA_r = Σ ∫
R :=
A
IA_r
2
A := b⋅ h
A = 200 mm
rc := ri + 0.5 ⋅ h
rc = 205 mm
⎛ re ⎞
IA_r := b⋅ ln ⎜
⎝ ri ⎠
IA_r = 0.976 mm
R = 204.959 mm
Internal Moment :
Mmax = P⋅ ( a + R)
M is positive since it tends to increase the
beam's radius of curvature.
Normal Stress:
σ=
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
)
M=
Assume tension failure.
r := ri
P⋅ ( a + R) =
(
(
)
σ ⋅ A⋅ r⋅ rc − R
R−r
)
σ t.allow ⋅ A⋅ r⋅ rc − R
P :=
R−r
(
)
σ t.allow ⋅ A⋅ r⋅ rc − R
( a + R) ⋅ ( R − r)
P = 3.087 N
Assume compression failure.
r' := re
P'⋅ ( a + R) =
(
)
σ c.allow⋅ A⋅ r'⋅ rc − R
R − r'
P' :=
(
)
σ c.allow⋅ A⋅ r'⋅ rc − R
( a + R) ⋅ ( R − r')
P' = 3.189 N
Pallow := min ( P , P')
Pallow = 3.087 N
Ans
Problem 6-138
While in flight, the curved rib on the jet plane is subjected to an anticipated moment of M = 16 N·m a
the section. Determine the maximum bending stress in the rib at this section, and sketch a twodimensional view of the stress distribution.
Given: bf := 30mm
tf := 5mm
tw := 5mm
dw := 20mm
M := 16N⋅ m
ri := 600mm
Solution:
Internal Moment :
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re := ri + dw + 2tf
2
A := 2bf⋅ tf + dw⋅ tw
⎯
⎯ Σ ⋅ ri⋅ Ai
r=
Σ ⋅ ( Ai)
(
)
A = 400 mm
(bf⋅ tf)⋅ (ri + 0.5tf) + (dw⋅ tw)⋅ (ri + 0.5dw + tf) + (bf⋅ tf)⋅ (re − 0.5tf)
rc :=
A
rc = 615.000 mm
dA
A r
IA_r = Σ ∫
⎛ r i + tf ⎞
IA_r := bf⋅ ln ⎜
⎝ ri ⎠
⎛ r e − tf ⎞
+ tw⋅ ln ⎜
⎝ r i + tf ⎠
⎛ re ⎞
+ bf⋅ ln ⎜
⎝ r e − tf ⎠
IA_r = 0.651 mm
R :=
A
IA_r
Normal Stress:
R = 614.793 mm
σ=
M⋅ ( R − r)
A⋅ r⋅ rc − R
Maximum tensile stress:
(
)
r := ri
σ t_max :=
Maximum compressive stress:
σ c_max :=
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
)
σ t_max = 4.77 MPa
(T)
Ans
σ c_max = −4.67 MPa
(C)
Ans
r' := re
M⋅ ( R − r')
A⋅ r'⋅ rc − R
(
)
Problem 6-139
The steel rod has a circular cross section. If it is gripped at its ends and a couple moment of M = 1.5
N·m is developed at each grip, determine the stress acting at points A and B and at the centroid C.
Given:
rci := 50mm
rce := 75mm
ro := 12mm
M := 1.5N⋅ m
Solution:
Internal Moment :
M = 1.5Nm is positive since it tends to
increase the beam's radius of curvature.
Section Property :
(
)
A := π ⋅ ro
2
rc := 0.5 rci + rce
IA_r = Σ ∫
dA
r
IA_r := 2⋅ π ⋅ ⎛⎝ rc −
A
R :=
A
IA_r
rc − ro ⎞⎠
2
2
)
σ A = 1.3594 MPa
(T)
Ans
)
σ B = −0.9947 MPa
(C)
Ans
)
σ C = −0.0531 MPa
(C)
Ans
R = 61.919 mm
Normal Stress:
σ=
rA := rci
σ A :=
rB := rce
σ B :=
rC := rc
σ C :=
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
(
)
)
M⋅ R − rA
(
A⋅ rA⋅ rc − R
(
)
M⋅ R − rB
(
A⋅ rB⋅ rc − R
(
)
M⋅ R − rC
(
A⋅ rC⋅ rc − R
Problem 6-140
The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple
as shown, determine the maximum tensile and compressive stresses acting at section a-a. Sketch the
stress distribution on the section in three dimensions.
Given: b := 50mm
h := 75mm
ri := 100mm
P := 250N
d' := 50mm
d := 150mm
Solution:
Internal Moment :
M := P⋅ d
M = 37.5 N⋅ m
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re := ri + h
dA
A r
IA_r = Σ ∫
R :=
A
IA_r
Normal Stress:
2
A := b⋅ h
A = 3750 mm
rc := ri + 0.5 ⋅ h
rc = 137.5 mm
⎛ re ⎞
IA_r := b⋅ ln ⎜
IA_r = 27.981 mm
⎝ ri ⎠
R = 134.021 mm
σ=
M⋅ ( R − r)
A⋅ r⋅ rc − R
Maximum tensile stress:
(
)
r := ri
σ t_max :=
Maximum compressive stress:
σ c_max :=
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
)
σ t_max = 0.978 MPa
(T)
Ans
σ c_max = −0.673 MPa
(C)
Ans
r' := re
M⋅ ( R − r')
A⋅ r'⋅ rc − R
(
)
Problem 6-141
The member has an elliptical cross section. If it is subjected to a moment of M = 50 N·m, determine
the stress at points A and B. Is the stress at point A', which is located on the member near the wall, the
same as that at A? Explain.
Given: ri := 100mm
L a := 150mm
L b := 75mm
M := 50N⋅ m
Solution:
Internal Moment :
M is positive since it tends to increase the
member's radius of curvature.
Section Property :
dA
A r
IA_r = Σ ∫
R=
A
IA_r
ao := 0.5L a
re := ri + La
bo := 0.5Lb
2
A := π ⋅ ao⋅ bo
A = 8835.73 mm
rc := ri + ao
rc = 175 mm
IA_r =
R :=
2π ⋅ bo
ao
⋅ ⎛⎝ rc −
rc − ao ⎞⎠
2
2
( )
π ⋅ ao⋅ bo⋅ ao
2⋅ π ⋅ bo⋅ ⎛⎝ rc −
rc − ao ⎞⎠
2
2
R = 166.557 mm
M⋅ ( R − r)
A⋅ r⋅ rc − R
Bending Stress:
σ=
rA := ri
σ A :=
rB := re
σ B :=
(
)
(
)
M⋅ R − rA
(
)
σ A = 0.446 MPa
(T)
Ans
)
σ B = −0.224 MPa (C)
Ans
A⋅ rA⋅ rc − R
(
)
M⋅ R − rB
(
A⋅ rB⋅ rc − R
No. The stress at point A' is not the same as that at A,
because of localized stress concentration.
Ans
Problem 6-142
The member has an elliptical cross section. If the allowable bending stress is σ allow = 125 MPa,
determine the maximum moment M that can be applied to the member.
Given: ri := 100mm
L a := 150mm
σ t.allow := 125MPa
L b := 75mm
σ c.allow := −125MPa
Solution:
Internal Moment :
M is positive since it tends to increase the
member's radius of curvature.
Section Property :
dA
A r
IA_r = Σ ∫
R=
A
IA_r
ao := 0.5L a
re := ri + La
bo := 0.5Lb
2
A := π ⋅ ao⋅ bo
A = 8835.73 mm
rc := ri + ao
rc = 175 mm
IA_r =
R :=
2π ⋅ bo
ao
⋅ ⎛⎝ rc −
rc − ao ⎞⎠
2
2
( )
π ⋅ ao⋅ bo⋅ ao
2⋅ π ⋅ bo⋅ ⎛⎝ rc −
rc − ao ⎞⎠
2
2
R = 166.557 mm
Normal Stress:
σ=
M⋅ ( R − r)
A⋅ r⋅ rc − R
(
)
Assume tension failure.
r := ri
M=
M=
(
(
)
σ ⋅ A⋅ r⋅ rc − R
)
σ t.allow ⋅ A⋅ r⋅ rc − R
R−r
M :=
R−r
(
)
σ t.allow ⋅ A⋅ r⋅ rc − R
R−r
M = 14.01 kN⋅ m
Assume compression failure.
σ c.allow⋅ A⋅ r'⋅ rc − R
r' := re
M' =
R − r'
(
)
(
)
Mallow = 14.01 kN⋅ m
Ans
M' :=
σ c.allow⋅ A⋅ r'⋅ rc − R
R − r'
M' = 27.94 kN⋅ m
Mallow := min ( M , M')
Problem 6-143
The bar has a thickness of 6.25 mm and is made of a material having an allowable bending stress of
σ allow = 126 MPa. Determine the maximum moment M that can be applied.
Given: t := 6.25mm
r := 6.25mm
w := 100mm
h := 25mm
σ allow := 126MPa
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 8138.02 mm
Stres s Concentration Factor :
w
=4
h
r
= 0.25
h
K := 1.45
From Fig. 6-48,
Maximum Moment :
c := 0.5 ⋅ h
σ = K⋅
Mmax :=
M⋅ c
I
(σallow)⋅ I
K⋅ c
Mmax = 56.57 N⋅ m
Ans
Problem 6-144
The bar has a thickness of 12.5 mm and is subjected to a moment of 90 N·m. Determine the maximum
bending stress in the bar.
Given: t := 12.5mm
r := 6.25mm
w := 100mm
h := 25mm
M := 90N⋅ m
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 16276.04 mm
Stres s Concentration Factor :
w
=4
h
r
= 0.25
h
From Fig. 6-48,
K := 1.45
Maximum Bending Stress :
c := 0.5 ⋅ h
σ max := K⋅
M⋅ c
I
σ max = 100.2 MPa
Ans
Problem 6-145
The bar is subjected to a moment of M = 40 N·m. Determine the smallest radius r of the fillets so that
an allowable bending stress of σ allow = 124 MPa is not exceeded.
Given:
w := 80mm
h := 20mm
t := 7mm
M := 40N⋅ m
σ allow := 124MPa
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 4666.67 mm
Allowable Bending Stres s : σ = K⋅
c := 0.5 ⋅ h
K :=
M⋅ c
I
σ allow⋅ I
M⋅ c
K = 1.45
Stres s Concentration Factor :
From Fig. 6-48, with
then,
K = 1.45
r
= 0.25
h
r := 0.25 ⋅ h
r = 5.00 mm
Ans
w
=4
h
Problem 6-146
The bar is subjected to a moment of M = 17.5 N · m. If r = 5 mm, determine the maximum bending
stress in the material.
Given: w := 80mm
t := 7mm
h := 20mm
r := 5mm
M := 17.5N⋅ m
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 4666.67 mm
Stres s Concentration Factor :
w
=4
h
r
= 0.25
h
From Fig. 6-48,
K := 1.45
Maximum Bending Stress :
c := 0.5 ⋅ h
σ max := K⋅
M⋅ c
I
σ max = 54.4 MPa
Ans
Problem 6-147
The bar is subjected to a moment of M = 20 N·m. Determine the maximum bending stress in the bar
and sketch, approximately, how the stress varies over the critical section.
Given: w := 30mm
t := 5mm
h := 10mm
r := 1.5mm
M := 20N⋅ m
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 416.67 mm
Stres s Concentration Factor :
w
=3
h
r
= 0.15
h
From Fig. 6-48,
K := 1.6
Maximum Bending Stress :
c := 0.5 ⋅ h
σ max := K⋅
M⋅ c
I
σ max = 384 MPa
Ans
Problem 6-148
The allowable bending stress for the bar is σ allow = 175 MPa. Determine the maximum moment M
that can be applied to the bar.
Given: w := 30mm
h := 10mm
t := 5mm
r := 1.5mm
σ allow := 175MPa
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 416.67 mm
Stres s Concentration Factor :
w
=3
h
r
= 0.15
h
K := 1.6
From Fig. 6-48,
Maximum Moment :
c := 0.5 ⋅ h
σ = K⋅
M :=
M⋅ c
I
(σallow)⋅ I
K⋅ c
M = 9.11 N⋅ m
Ans
Problem 6-149
Determine the maximum bending stress developed in the bar if it is subjected to the couples shown.
The bar has a thickness of 6 mm.
Given: t := 6mm
w := 108mm
h1 := 72mm
h2 := 36mm
r1 := 7.2mm
r2 := 27mm
M1 := 20N⋅ m
Mw := 12.5N⋅ m
M2 := 7.5N⋅ m
Solution:
Section Property:
For the larger section 1:
I1 :=
1
3
⋅ t⋅ h1
12
For the smaller section 2:
4
I1 = 186624 mm
I2 :=
1
3
⋅ t⋅ h2
12
4
I2 = 23328 mm
Stres s Concentration Factor :
For the larger section 1:
r1
w
= 1.5
= 0.1
h1
h1
From Fig. 6-48,
Maximum Moment :
K1 := 1.755
σ = K⋅
For the larger section 1:
c1 := 0.5 ⋅ h1
For the smaller section 2:
r2
w
=3
= 0.75
h2
h2
σ 1 := K1⋅
From Fig. 6-48,
M⋅ c
I
K2 := 1.15
For the smaller section 2:
M1⋅ c1
I1
c2 := 0.5 ⋅ h2
σ 1 = 6.77 MPa
σ 2 := K2⋅
M2⋅ c2
I2
σ 2 = 6.66 MPa
(
)
σ max := max σ 1 , σ 2
σ max = 6.77 MPa
Ans
Problem 6-150
Determine the length L of the center portion of the bar so that the maximum bending stress at A, B,
and C is the same.The bar has a thickness of 10 mm.
Given: w := 60mm
t := 10mm
h := 40mm
r := 7mm
a := 200mm
P := 350N
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 53333.33 mm
Support Reaction :
+
Stres s Concentration Factor :
w
= 1.5
h
MA := R⋅ a
K := 1.5
MC = R⋅ ( a + 0.5L)
σ B.max := σ A.max
At Section C-C:
σ C.max := σ A.max
1
3
⋅ t⋅ w
12
4
I' = 180000 mm
Maximum Bending Stress :
c' := 0.5 ⋅ w
σ C.max =
MC⋅ c'
I'
σ A.max =
R⋅ ( a + 0.5L) ⋅ c'
I'
⎛ 2σ A.max⋅ I'
L := ⎜
⎝
R⋅ c'
L = 950 mm
MA = 35.00 N⋅ m
MB := MA
σ A.max = 19.688 MPa
I' :=
R = 175 N
Internal Moment :
Maximum Bending Stresses at A and B :
MA⋅ c
c := 0.5 ⋅ h
σ A.max := K⋅
I
Require,
2R − P = 0
R := 0.5P
r
= 0.175
h
From Fig. 6-48,
ΣF y=0;
By symmetry, A =B = R
⎞
− 2a
⎠
Ans
Problem 6-151
If the radius of each notch on the plate is r = 10 mm, determine the largest moment M that can be
applied. The allowable bending stress for the material is σ allow = 180 MPa.
Given: w := 165mm
h := 125mm
t := 20mm
r := 10mm
σ allow := 180MPa
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 3255208.33 mm
Stres s Concentration Factor :
b := 0.5 ⋅ ( w − h)
b
=2
r
r
= 0.08
h
K := 2.1
From Fig. 6-50,
Maximum Moment :
c := 0.5 ⋅ h
σ = K⋅
M :=
M⋅ c
I
(σallow)⋅ I
K⋅ c
M = 4.464 kN⋅ m
Ans
Problem 6-152
The stepped bar has a thickness of 15 mm. Determine the maximum moment that can be applied to its
ends if it is made of a material having an allowable bending stress of σ allow = 200 MPa.
Given: w := 45mm
t := 15mm
h1 := 30mm
h2 := 10mm
r1 := 3mm
r2 := 6mm
σ allow := 200MPa
Solution:
Section Property:
For the larger section 1:
I1 :=
1
3
⋅ t⋅ h1
12
For the smaller section 2:
4
I1 = 33750 mm
I2 :=
1
3
⋅ t⋅ h2
12
4
I2 = 1250 mm
Stres s Concentration Factor :
For the larger section 1:
r1
w
= 1.5
= 0.1
h1
h1
From Fig. 6-48,
Maximum Moment :
K1 := 1.75
σ = K⋅
For the larger section 1:
c1 := 0.5 ⋅ h1
For the smaller section 2:
h1
r2
=3
= 0.6
h2
h2
M1 :=
From Fig. 6-48,
M⋅ c
I
K2 := 1.2
For the smaller section 2:
(σallow)⋅ I1
c2 := 0.5 ⋅ h2
K1⋅ c1
M1 = 257.14 N⋅ m
M2 :=
(σallow)⋅ I2
K2⋅ c2
M2 = 41.67 N⋅ m
(
)
Mallow := min M1 , M2
Mallow = 41.67 N⋅ m
Ans
Problem 6-153
The bar has a thickness of 12.5 mm and is made of a material having an allowable bending stress of
σ allow = 140 MPa. Determine the maximum moment M that can be applied.
Given: t := 12.5mm
r := 7.5mm
w := 150mm
h := 50mm
σ allow := 140MPa
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 130208.33 mm
Stres s Concentration Factor :
w
=3
h
r
= 0.15
h
K := 1.6
From Fig. 6-48,
Maximum Moment :
c := 0.5 ⋅ h
σ = K⋅
M :=
M⋅ c
I
(σallow)⋅ I
K⋅ c
M = 455.73 N⋅ m
Ans
Problem 6-154
The bar has a thickness of 12.5 mm and is subjected to a moment of 900 N·m. Determine the
maximum bending stress in the bar.
Given: t := 12.5mm
r := 7.5mm
w := 150mm
h := 50mm
M := 900N⋅ m
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 130208.33 mm
Stres s Concentration Factor :
w
=3
h
r
= 0.15
h
From Fig. 6-48,
K := 1.6
Maximum Bending Stress :
c := 0.5 ⋅ h
σ max := K⋅
M⋅ c
I
σ max = 276.5 MPa
Ans
Problem 6-155
The simply supported notched bar is subjected to two forces P . Determine the largest magnitude of P
that can be applied without causing the material to yield. The material is A-36 steel. Each notch has a
radius of r = 3 mm.
Given: t := 12mm
r := 3mm
w := 42mm
h := 30mm
a := 500mm
σ Y := 250MPa
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 27000.00 mm
Support Reaction :
Stres s Concentration Factor :
+
b := 0.5 ⋅ ( w − h)
b
=2
r
Maximum Moment :
c := 0.5 ⋅ h
2R − 2P = 0
Internal Moment :
r
= 0.1
h
From Fig. 6-50,
ΣF y=0;
MC = R⋅ a
K := 1.92
MC = P⋅ a
σ Y = K⋅
MC =
P⋅ a =
P :=
MC⋅ c
I
(σY)⋅ I
K⋅ c
(σY)⋅ I
K⋅ c
(σY)⋅ I
a⋅ K⋅ c
P = 468.75 N
Ans
By symmstry, R1=R2=R
R= P
At mid-span,
Problem 6-156
The simply supported notched bar is subjected to the two loads, each having a magnitude of P = 500
N. Determine the maximum bending stress developed in the bar, and sketch the bending-stress
distribution acting over the cross section at the center of the bar. Each notch has a radius of r = 3 mm.
Given: t := 12mm
r := 3mm
w := 42mm
h := 30mm
a := 500mm
P := 500N
Solution:
Section Property:
I :=
1
3
⋅ t⋅ h
12
4
I = 27000.00 mm
Support Reaction :
Stres s Concentration Factor :
+
b := 0.5 ⋅ ( w − h)
b
=2
r
MC := R⋅ a
K := 1.92
MC = 250 N⋅ m
Maximum Bending Stress :
c := 0.5 ⋅ h
σ max := K⋅
2R − 2P = 0
R := P
Internal Moment : At mid-span,
r
= 0.1
h
From Fig. 6-50,
ΣF y=0;
By symmstry, R1=R2=R
MC⋅ c
I
σ max = 266.7 MPa
Ans
Problem 6-157
A rectangular A-36 steel bar has a width of 25 mm and height of 75 mm. Determine the moment
applied about the horizontal axis that will cause half the bar to yield.
b := 25mm
Given:
σ Y := 250MPa
d := 75mm
Solution:
de := 0.5d
dp := 0.5d
Elastic-plastic Moment:
dp ⎞
⎛ de ⎞ 2de
⎛ dp ⎞ ⎛
⋅
+ σ Y⋅ ⎜ b⋅
⋅ ⎜ de +
2⎠
⎝ 2⎠ 3
⎝ 2⎠⎝
M := σ Y⋅ ⎜ b⋅
M = 9.52 kN⋅ m
Ans
Problem 6-158
The box beam is made of an elastic perfectly plastic material for which σY = 250 MPa. Determine the
residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then
released.
Given:
bo := 200mm
do := 200mm
bi := 150mm
di := 150mm
σ Y := 250MPa
Solution:
Section Property:
tb := 0.5 do − di
(
I :=
)
(
td := 0.5 bo − bi
1 ⎛
3
3
⋅ b ⋅ d − bi⋅ di ⎞⎠
12 ⎝ o o
)
4
I = 91145833.33 mm
Plastic Moment:
⎛ di ⎞ ⎛ di ⎞
Mp := σ Y⋅ bo⋅ tb ⋅ do − tb + σ Y⋅ ⎜ 2td⋅
⋅⎜
2
2
(
)(
)
⎝
⎠⎝ ⎠
Mp = 289062.50 N⋅ m
Modulus of Rupture:
The modulus of rupture σr can be determined using the flexure formula
with the application of reverse plastic moment Mp.
c := 0.5do
σ r :=
Mp⋅ c
I
σ r = 317.14 MPa
Residul Bending Stress:
σ' t := σ r − σ Y
σ' t = 67.14 MPa
Ans
σ' b := σ r − σ Y
σ' b = 67.14 MPa
Ans
Problem 6-159
The box beam is made of an elastic plastic material for which σY = 250 MPa. Determine the residual
stress in the top and bottom of the beam after the plastic moment Mp is applied and then released.
Given:
bf := 200mm
tf := 15mm
dw := 200mm
tw := 20mm
σ Y := 250MPa
Solution:
Section Property:
D := dw + 2tf
I :=
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
(
)
4
I = 82783333.33 mm
Plastic Moment:
⎛ dw ⎞ ⎛ dw ⎞
Mp := σ Y⋅ bf⋅ tf ⋅ D − tf + σ Y⋅ ⎜ tw⋅
⋅⎜
2
2
(
)(
)
⎝
⎠⎝
⎠
Mp = 211.25 kN⋅ m
Modulus of Rupture:
The modulus of rupture σr can be determined using the flexure formula
with the application of reverse plastic moment Mp.
c := 0.5D
σ r :=
Mp⋅ c
I
σ r = 293.46 MPa
Residul Bending Stress:
σ' t := σ r − σ Y
σ' t = 43.46 MPa
Ans
σ' b := σ r − σ Y
σ' b = 43.46 MPa
Ans
Problem 6-160
Determine the plastic section modulus and the shape factor of the beam's cross section.
Set
a := mm
Given: bf := 2a
tf := a
dw := 2a
tw := a
Solution:
Section Property :
A := bf⋅ tf + dw⋅ tw
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
(
)
yc :=
(bf⋅ tf)⋅ (0.5tf) + (dw⋅ tw)⋅ (0.5dw + tf)
A
yc = 1.25 a
I :=
1
3
2 ⎡1
3
2⎤
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc + ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + tf − yc ⎥
12
12
⎣
⎦
(
)(
)
(
)(
)
4
I = 3.08 a
Maximum Elastic Moment :
c := dw + tf − yc
σY =
c = 1.75 a
MY⋅ c
I
MY
σY
=
⎛ I ⎞ = 1.7619 a3
⎜
⎝ c⎠
I
c
(
3
)
MY = 1.7619 a ⋅ σ Y
Plastic Moment :
∫ σ dA = 0
A
(
)
(
)
σ Y⋅ tw⋅ ( d) − σ Y⋅ tw⋅ dw − d − σ Y⋅ bf⋅ tf = 0
d :=
tw⋅ dw + bf⋅ tf
d = 2a
2tw
darm := 0.5tf + dw − 0.5d
Mp = σ Y⋅ tw⋅ ( d) ⋅ darm
Mp
σY
darm = 1.50 a
= tw⋅ ( d) ⋅ darm
3
tw⋅ ( d) ⋅ darm = 3.00 a
(
3
)
Mp = 3.00 a ⋅ σ Y
3
Mp
MY
k :=
Plastic Section Modulus :
z=
Shape Factor : k =
3.00 a
3
k = 1.70
Ans
1.7619 a
Mp
σY
3
z = 3.00 a
Ans
Problem 6-161
The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and th
plastic moment that can be applied to the cross section. Take a = 50 mmand σY = 230 MPa.
Given: a := 50mm
σ Y := 230MPa
bf := 2a
tf := a
dw := 2a
tw := a
Solution:
Section Property :
A := bf⋅ tf + dw⋅ tw
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
(
)
yc :=
(bf⋅ tf)⋅ (0.5tf) + (dw⋅ tw)⋅ (0.5dw + tf)
A
yc = 62.5 mm
I :=
1
3
2 ⎡1
3
2⎤
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc + ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + tf − yc ⎥
12
⎣ 12
⎦
(
)(
)
(
)(
)
4
I = 19270833.33 mm
Maximum Elastic Moment :
c := dw + tf − yc
σY =
MY⋅ c
I
c = 87.50 mm
MY :=
σ Y⋅ I
MY = 50.65 kN⋅ m
c
Ans
Plastic Moment :
∫ σ dA = 0
A
(
)
(
)
σ Y⋅ tw⋅ ( d) − σ Y⋅ tw⋅ dw − d − σ Y⋅ bf⋅ tf = 0
d :=
tw⋅ dw + bf⋅ tf
2tw
darm := 0.5tf + dw − 0.5d
Mp := σ Y⋅ tw⋅ ( d) ⋅ darm
Mp = 86.25 kN⋅ m
Ans
d = 100 mm
darm = 75.00 mm
Problem 6-162
The rod has a circular cross section. If it is made of an elastic plastic material, determine the shape
factor and the plastic section modulus Z.
Set
r := mm
Solution:
Section Property :
A := πr
2
I :=
( )
π 4
r
4
Maximum Elastic Moment :
c := r
σY =
MY⋅ c
I
MY
=
σY
⎛ I ⎞ = 0.7854 r3
⎜
⎝ c⎠
I
c
(
3
)
MY = 0.7854 r ⋅ σ Y
Plastic Moment :
⎛ 4r ⎞
⎝ 3π ⎠
darm := 2⎜
⎛ A ⎞⋅ d
Mp = σ Y⋅ ⎜
arm
2
⎝ ⎠
Mp
σY
⎛ A ⎞⋅ d
⎝ 2 ⎠ arm
= ⎜
3
⎛ A⎞⋅ d
= 1.3333 r
⎜
arm
⎝2⎠
(
3
)
Mp = 1.3333 r ⋅ σ Y
Mp
MY
k :=
Plastic Section Modulus :
z=
Shape Factor : k =
1.3333 r
0.7854 r
Mp
σY
3
3
k = 1.70
z = 1.333 r
Ans
3
Ans
Problem 6-163
The rod has a circular cross section. If it is made of an elastic plastic material, determine the maximum
elastic moment and plastic moment that can be applied to the cross section.Take r = 75 mm, σY = 250
MPa.
Given: r := 75mm
σ Y := 250MPa
Solution:
Section Property :
A := πr
2
I :=
( )
π 4
r
4
Maximum Elastic Moment :
c := r
σY =
MY⋅ c
I
MY :=
σ Y⋅ I
c
MY = 82.83 kN⋅ m
Ans
Plastic Moment :
⎛ 4r ⎞
⎝ 3π ⎠
darm := 2⎜
⎛ A ⎞⋅ d
Mp := σ Y⋅ ⎜
arm
2
⎝ ⎠
Mp = 140.63 kN⋅ m
Ans
Problem 6-164
Determine the plastic section modulus and the shape factor of the cross section.
Set
a := mm
Given: bf := 3a
tf := a
dw := 3a
tw := a
Solution:
Section Property :
(
)
2
A := bf⋅ tf + dw − tf ⋅ tw
I :=
A = 5a
1
1
3
3
3
⋅ bf⋅ tf +
⋅ tw⋅ ⎛⎝ dw − tf ⎞⎠
12
12
4
I = 2.41667 a
Maximum Elastic Moment :
c := 0.5dw
σY =
c = 1.5 a
MY⋅ c
I
MY
σY
⎛ I ⎞ = 1.61111 a3
⎜
⎝ c⎠
I
c
=
(
3
)
MY = 1.61111 a ⋅ σ Y
Plastic Moment :
(
dw.arm := tf + 0.5 dw − tf
)
dw.arm = 2.00 a
df.arm := 0.5tf
df.arm = 0.50 a
⎛ tf ⎞
⎛ dw − tf ⎞
Mp = σ Y⋅ ⎜ bf⋅
⋅ df.arm + σ Y⋅ tw⋅ ⎜
⋅ dw.arm
2
2
⎝
Mp
σY
⎛
⎝
= ⎜ bf ⋅
⎠
tf ⎞
2⎠
⎝
⎠
⎛ dw − tf ⎞
⋅d
⎝ 2 ⎠ w.arm
⋅ df.arm + tw⋅ ⎜
⎛ tf ⎞
⎛ dw − t f ⎞
3
⎜ bf⋅ ⋅ df.arm + tw⋅ ⎜
⋅ dw.arm = 2.75 a
2
2
⎝
⎠
⎝
⎠
(
3
)
Mp = 2.75 a ⋅ σ Y
3
Mp
MY
k :=
Plastic Section Modulus :
z=
Shape Factor : k =
2.75 a
3
k = 1.71
Ans
1.61111 a
Mp
σY
3
z = 2.75 a
Ans
Problem 6-165
The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and th
plastic moment that can be applied to the cross section. Take a = 50 mm and σY = 250 MPa.
Given: a := 50mm
σ Y := 250MPa
bf := 3a
tf := a
dw := 3a
tw := a
Solution:
Section Property :
(
)
2
A := bf⋅ tf + dw − tf ⋅ tw
I :=
A = 5a
1
1
3
3
3
⋅ bf⋅ tf +
⋅ tw⋅ ⎛⎝ dw − tf ⎞⎠
12
12
4
I = 2.41667 a
Maximum Elastic Moment :
c := 0.5dw
σY =
c = 1.5 a
MY⋅ c
I
MY :=
σ Y⋅ I
c
MY = 50.35 kN⋅ m
Plastic Moment :
(
dw.arm := tf + 0.5 dw − tf
)
df.arm := 0.5tf
dw.arm = 2.00 a
df.arm = 0.50 a
⎛ tf ⎞
⎛ dw − t f ⎞
Mp := σ Y⋅ ⎜ bf⋅
⋅ df.arm + σ Y⋅ tw⋅ ⎜
⋅ dw.arm
2
2
⎝
⎠
Mp = 85.94 kN⋅ m
⎝
Ans
⎠
Ans
Problem 6-166
The beam is made of an elastic perfectly plastic material. Determine the plastic moment Mp that can be
supported by a beam having the cross section shown. σY = 210 MPa.
Given: ro := 50mm
tw := 25mm
ri := 25mm
σ Y := 210MPa
dw := 250mm
Solution:
Plastic Moment :
A1 := π ⋅ ⎛⎝ ro − ri ⎞⎠
darm1 := dw + 2ro
(
darm2 := 0.5dw
2
2
)
A2 := tw 0.5dw
(
)
(
)
Mp := σ Y⋅ A1 ⋅ darm1 + σ Y⋅ A2 ⋅ darm2
Mp = 515 kN⋅ m
Ans
Problem 6-167
Determine the plastic moment Mp that can be supported by a beam having the cross section shown.
σY = 210 MPa.
Given: ro := 50mm
tw := 25mm
ri := 25mm
σ Y := 210MPa
dw := 250mm
Solution:
A1 = π ⋅ ⎛⎝ ro − ri ⎞⎠
2
2
(
)
A2 = tw dw − d'
A3 = tw d'
∫ σ dA = 0
σ Y⋅ A1 + σ Y⋅ A2 − σ Y⋅ A3 = 0
A
A1 + A2 − A3 = 0
(
)
π ⋅ ⎛⎝ ro − ri ⎞⎠ + tw dw − d' − tw d' = 0
2
2
π ⋅ ⎛ r o − r i ⎞ + t w dw
⎝
⎠
d' :=
2tw
2
( )
2
d' = 242.81 mm
Plastic Moment :
(
)
A1 := π ⋅ ⎛⎝ ro − ri ⎞⎠
darm1 := ro + dw − d'
A2 := tw dw − d'
(
darm2 := 0.5 dw − d'
A3 := tw d'
darm3 := 0.5d'
2
(
2
)
)
(
(
)
)
(
)
Mp := σ Y⋅ A1 ⋅ darm1 + σ Y⋅ A2 ⋅ darm2 + σ Y⋅ A3 ⋅ darm3
Mp = 225.6 kN⋅ m
Ans
Problem 6-168
Determine the plastic section modulus and the shape factor for the member having the tubular cross
section.
Set
d := mm
Solution:
Section Property :
π⎡
2
2
⎣( 2d) − d ⎤⎦
4
π ⎡
4
4
I :=
⎣( 2d) − d ⎤⎦
64
A :=
A = 2.35619 d
2
4
I = 0.73631 d
Maximum Elastic Moment :
c := d
σY =
MY⋅ c
I
MY
=
σY
⎛ I ⎞ = 0.73631 d3
⎜
⎝ c⎠
I
c
(
3
)
MY = 0.73631 d ⋅ σ Y
Plastic Moment :
yc =
(
⎯
Σ ⋅ yi⋅ Ai
Σ ⋅ ( Ai)
)
⎛ π ⎞ ( 2d) 2⋅ ⎛ 4d ⎞ − 0.5⎛ π ⎞ d2⋅ 1 ⎛ 4d ⎞
⎜
⎜
⎜
⎝ 4⎠
⎝ 4 ⎠ 2 ⎝ 3π ⎠
⎝ 3π ⎠
0.5 ⎜
yc :=
0.5A
yc = 0.49515 d
darm := 2yc
⎛ A⎞⋅ d
Mp = σ Y⋅ ⎜
arm
2
⎝ ⎠
Mp
σY
⎛ A⎞⋅ d
⎝ 2 ⎠ arm
= ⎜
3
⎛ A ⎞⋅ d
= 1.16667 d
⎜
arm
⎝2⎠
(
3
)
Mp = 1.16667 d ⋅ σ Y
3
Mp
MY
k :=
Plastic Section Modulus :
z=
Shape Factor : k =
1.16667 d
3
k = 1.58
Ans
0.73631 d
Mp
σY
3
z = 1.16667 d
Ans
Problem 6-169
Determine the plastic section modulus and the shape factor for the member.
Solution:
Set
b := mm
h := mm
Section Property :
1
1
3
A = ( b⋅ h)
I=
b⋅ h
2
36
Maximum Elastic Moment :
MY⋅ c
2
c := ⋅ h
σY =
3
I
(
Plastic Moment :
From the geometry,
b' =
)
MY
σY
⎛ I ⎞ = 1 ⋅ b⋅ h2
⎜
⎝ c ⎠ 24
2⎞
⎛1
MY = ⎜ ⋅ b⋅ h ⋅ σ Y
24
⎝
⎠
I
c
=
d
⋅b
h
1
A∆ = ⋅ b'⋅ d
2
1
Atrp = ⋅ ( b + b') ⋅ ( h − d)
2
1 ⎛d ⎞
A∆ = ⋅ ⎜ ⋅ b ⋅ d
2 h
1 ⎛
d ⎞
Atrp = ⋅ ⎜ b + ⋅ b ⋅ ( h − d)
2
h
⎝
∫ σ dA = 0
A
A∆ = Atrp
⎠
⎝
( )
(
⎠
)
σ Y⋅ A∆ − σ Y⋅ Atrp = 0
1 ⎛
d ⎞
1 ⎛d ⎞
⋅ ⎜ ⋅ b ⋅ d = ⋅ ⎜ b + ⋅ b ⋅ ( h − d)
2 ⎝
h ⎠
2 ⎝h ⎠
d=
h
b' =
2
b
Note: The centroid of a trapezoidal area was used
in the calculation.
h − d 2⋅ b' + b
hc =
⋅
3
b' + b
1
darm = ⋅ d + ⎡⎣( h − d) − hc⎤⎦
3
1
b' + 2⋅ b
darm = ⋅ d + ( h − d)
3
3( b' + b)
1 h
h ⎞ b + 2 2⋅ b
⎛
darm = ⋅
+ ⎜h −
⋅
3 2 ⎝
2 ⎠ 3( b + 2⋅ b)
⎛ A ⎞⋅ d
Mp = σ Y⋅ ⎜
arm
2
⎝ ⎠
Mp
σY
darm =
⎛ A ⎞⋅ d
⎝ 2 ⎠ arm
4( 2 −
2) ⋅ h
6
2− 2
A
2
⋅ darm =
⋅ b⋅ h
2
6
= ⎜
⎛2 − 2
2⎞
⋅ b⋅ h ⋅ σ Y
⎝ 6
⎠
Mp = ⎜
2−
Mp
Shape Factor : k =
MY
6
1
k :=
24
Plastic Section Modulus :
z=
Mp
σY
2
2
⋅ b⋅ h
2
k = 2.34
Ans
2− 2
2
⋅ b⋅ h
6
Ans
⋅ b⋅ h
z=
2
Problem 6-170
The member is made of elastic perfectly plastic material for which σY = 230 MPa. Determine the
maximum elastic moment and the plastic moment that can be applied to the cross section.Take b = 50
mm and h = 80 mm.
Given:
b := 50mm
h := 80mm
σ Y := 230MPa
Solution:
Section Property :
1
( b⋅ h)
2
1
3
I :=
b⋅ h
36
2
A :=
(
A = 2000 mm
)
4
I = 711111.11 mm
Maximum Elastic Moment :
c :=
2
⋅h
3
σY =
MY⋅ c
I
Plastic Moment :
From the geometry, b' =
MY :=
d
⋅b
h
σ Y⋅ I
MY = 3.07 kN⋅ m
c
1
A∆ = ⋅ b'⋅ d
2
1
Atrap = ⋅ ( b + b') ⋅ ( h − d)
2
1 ⎛d ⎞
A∆ = ⋅ ⎜ ⋅ b ⋅ d
2 h
1 ⎛
d ⎞
Atrap = ⋅ ⎜ b + ⋅ b ⋅ ( h − d)
2
h
⎝
( )
∫ σ dA = 0
(
⎠
)
σ Y⋅ A∆ − σ Y⋅ Atrap = 0
A
1 ⎛
d ⎞
1 ⎛d ⎞
⋅ ⎜ ⋅ b ⋅ d = ⋅ ⎜ b + ⋅ b ⋅ ( h − d)
2 ⎝
h ⎠
2 ⎝h ⎠
Note: The centroid of a trapezoidal area was used
in the calculation.
h − d 2⋅ b' + b
hc =
⋅
3
b' + b
1
⋅ d + ⎡⎣( h − d) − hc⎤⎦
3
1
b' + 2⋅ b
darm = ⋅ d + ( h − d)
3
3( b' + b)
darm =
darm =
darm :=
1 h
h ⎞ b + 2 2⋅ b
⎛
⋅
+ ⎜h −
⋅
3 2 ⎝
2 ⎠ 3( b + 2⋅ b)
4( 2 −
2) ⋅ h
6
⎛ A ⎞⋅ d
Mp := σ Y⋅ ⎜
arm
2
⎝ ⎠
Mp = 7.19 kN⋅ m
Ans
Ans
darm = 31.24 mm
⎝
⎠
A∆ = Atrap
d=
h
2
b' =
b
2
Problem 6-171
The wide-flange member is made from an elasticplastic material. Determine the shape factor and the
plastic section modulus Z.
b := mm
h := mm
Given: bf := b
tf := t
D := h
tw := t
Set
t := mm
Solution: dw = D − 2⋅ tf
Section Property :
(
dw := h − 2t
)
A = bf ⋅ D − bf − t w ⋅ dw
I=
A := b⋅ h − ( b − t) ⋅ ( h − 2t)
1
1
3
3
⋅ bf ⋅ D −
⋅ bf − t w ⋅ dw
12
12
(
)
I :=
Maximum Elastic Moment :
c := 0.5D
σY =
1
1
3
3
⋅ b⋅ h −
⋅ ( b − t) ⋅ ( h − 2t)
12
12
c := 0.5h
MY⋅ c
I
MY
=
σY
1
I
1
2
3
= ⋅ b⋅ h −
⋅ ( b − t) ⋅ ( h − 2t)
6⋅ h
c
6
1 ⎡ 3
3
MY =
⋅ ⎣b⋅ h − ( b − t) ⋅ ( h − 2⋅ t) ⎤⎦ ⋅ σ Y
6⋅ h
I
c
Plastic Moment :
h − 2t
2
dw.arm = 0.5dw
dw.arm =
df.arm = D − tf
df.arm = h − t
⎛ dw ⎞
Mp = σ Y⋅ bf⋅ tf ⋅ df.arm + σ Y⋅ tw⋅ ⎜
⋅ dw.arm
2
(
Mp
σY
)
⎝
⎠
⎛ h − 2t ⎞ ⋅ h − 2t
⎝ 2 ⎠ 2
= ( b⋅ t) ⋅ ( h − t) + t⋅ ⎜
t
2⎤
⎡
Mp = ⎢( b⋅ t) ⋅ ( h − t) + ⋅ ( h − 2t) ⎥ ⋅ σ Y
4
⎣
Mp
Shape Factor : k =
MY
⎦
( b⋅ t) ⋅ ( h − t) +
k=
1
6⋅ h
t
4
⋅ ( h − 2t)
2
⋅ ⎡⎣b⋅ h − ( b − t) ⋅ ( h − 2⋅ t) ⎤⎦
3
3
3⋅ h ⎡⎢ 4b⋅ t⋅ ( h − t) + t⋅ ( h − 2t) ⎤⎥
3⎥
2 ⎢ 3
⎣ b⋅ h − ( b − t) ⋅ ( h − 2⋅ t) ⎦
2
k=
Plastic Section Modulus :
z=
Mp
σY
z = ( b⋅ t ) ⋅ ( h − t ) +
Ans
t
2
⋅ ( h − 2t)
4
Ans
Problem 6-172
The beam is made of an elastic-plastic material for which σY = 200 MPa. If the largest moment in the
beam occurs within the center section a-a, determine the magnitude of each force P that causes this
moment to be (a) the largest elastic moment and (b) the largest plastic moment.
Given: a := 2m
b := 100mm
h := 200mm
σ Y := 200MPa
Solution:
Section Property :
2
A := b⋅ h
I :=
A = 20000 mm
1
3
⋅ b⋅ h
12
4
I = 66666666.67 mm
a) Maximum Elastic Moment :
c := 0.5 ⋅ h
MY = P⋅ a
σY =
MY⋅ c
I
σY =
P⋅ a⋅ c
I
P :=
σ Y⋅ I
a⋅ c
P = 66.67 kN
Ans
b) Plastic Moment :
darm :=
h
2
⎛ A ⎞⋅ d
Mp := σ Y⋅ ⎜
arm
2
⎝ ⎠
P' :=
Mp
a
Mp = 200.00 kN⋅ m
P' = 100.00 kN
Ans
Problem 6-173
The beam is made of phenolic, a structural plastic, that has the stress-strain curve shown. If a portion
of the curve can be represented by the equation σ = (5(106)ε )1/2 MPa, determine the magnitude of w
the distributed load that can be applied to the beam without causing the maximum strain in its fibers a
the critical section to exceed εmax = 0.005 mm/mm.
Given:
b := 150mm
2
h := 150mm
( 6) ε
σ = 5 10
L := 2m
ε max := 0.005
Solution:
mm
mm
Stress-strain Relationship : unit := MPa
When εmax = 0.005,
( 6)
σ max := unit⋅ 5 10 ε max
σ max = 158.11 MPa
Resultant Internal Forces :
The resultant internal forces T and C can be evaluated from the volume of
stress block which is a paraboloid, T = C.
2
⎛ b⋅ h ⎞
⋅ σ max⋅ ⎜
3
⎝ 2 ⎠
T = 1185.85 kN
⎡ 3 ⎛ h ⎞⎤
darm := 2⋅ ⎢ ⋅ ⎜ ⎥
⎣ 5 ⎝ 2 ⎠⎦
darm = 90 mm
T :=
Maximum Internal Moment :
Mmax := T⋅ darm
Mmax = 106.73 kN⋅ m
By observation, the maximum moment occurs over the middle support.
Mmax = w⋅ L
w :=
Mmax
L
w = 53.36 m
kN
m
Ans
Problem 6-174
The box beam is made from an elastic plastic material for which σY = 175 MPa. Determine the
intensity of the distributed load w0 that will cause the moment to be (a) the largest elastic moment and
(b) the largest plastic moment.
Given:
bo := 200mm
do := 400mm
bi := 150mm
di := 300mm
L := 3m
σ Y := 175MPa
Solution:
Support Reaction :
+
By symmstry, R1=R2=R
(
ΣF y=0;
)
2R − 2 0.5 ⋅ wo⋅ L = 0
R = 0.5 ⋅ wo⋅ L
Maximum Moment :
L
M = R⋅ L −
0.5 ⋅ wo⋅ L
3
(
M⋅ c
I
1 ⎛
3
3
I :=
⋅ b ⋅ d − bi⋅ di ⎞⎠
12 ⎝ o o
c := 0.5do
wo :=
σ Y⋅ I
b)
MY = 638.02 kN⋅ m
c
3⋅ MY
L
)
kN
wo = 212.67
m
2
Plastic Analysis :
(
tw := 0.5 bo − bi
Ans
) tf := 0.5(do − di)
Af := bo⋅ tf
darm1 := do − tf
Aw := di⋅ tw
darm2 := 0.5di
(
2
σ=
a) Elastic Analysis :
MY :=
wo⋅ L
M=
3
)
(
)
Mp := σ Y⋅ Af ⋅ darm1 + σ Y⋅ Aw ⋅ darm2
Mp = 809.4 kN⋅ m
w'o :=
3⋅ Mp
L
2
kN
w'o = 269.79
m
Ans
Problem 6-175
The beam is made of a polyester that has the stress-strain curve shown. If the curve can be
represented by the equation σ = [140 tan-1(15ε )] MPa, where tan-1(15ε ) is in radians, determine the
magnitude of the force P that can be applied to the beam without causing the maximum strain in its
fibers at the critical section to exceed εmax = 0.003 mm/mm.
Given:
b := 50mm
h := 100mm
σ = 140 atan ( 15ε )
L := 2.4m
mm
mm
ε max := 0.003
Solution:
By symmstry, R1=R2=R
Support Reaction :
+
ΣF y=0;
2R − 2P = 0
R= P
Maximum Moment :
M = R⋅ L
M = P⋅ L
unit := MPa
Stress-strain Relationship :
The bending stress can bs expressed
in terms of y using
ε max
ε=
⋅y
0.5h
⎛
ε max
⎝
0.5h
σ = 140 atan ⎜ 15
When εmax = 0.003,
⎞
⋅ y ⋅ unit
⎠
ymax := 0.5h
⎛
ε max
⎝
0.5h
σ max := 140⋅ atan ⎜ 15
⎞
⋅ ymax ⋅ unit
⎠
σ max = 6.30 MPa
Resultant Internal Moment : M = ∫ y σ dA
A
⌠
⎮
M := 2 ⎮
⌡
0.5h
0
⎛
⎛
ε max
⎝
⎝
0.5h
y⋅ ⎜ 140 atan ⎜ 15
⎞
⎞
⎠
⎠
⋅ y ⋅ unit ⋅ b dy
M = 524.79 N⋅ m
P :=
M
L
P = 218.66 N
Ans
Problem 6-176
The stress-strain diagram for a titanium alloy can be approximated by the two straight lines. If a strut
made of this material is subjected to bending, determine the moment resisted by the strut if the
maximum stress reaches a value of (a) σA and (b) σB.
Given: b := 50mm
d := 75mm
σ A := 980MPa
ε A := 0.01
σ B := 1260MPa
mm
mm
mm
mm
ε B := 0.04
Solution:
Maximum Elastic Moment :
Since the stress is linearly related to strain up
to point A, the flexure formula can be applied.
σ=
M⋅ c
I
c := 0.5d
MY :=
I := b⋅ d
σ A⋅ I
c
3
MY = 551.25 kN⋅ m Ans
UItimate Moment :
εA
yA :=
⋅ ( 0.5d)
εB
σ
not zero.
h := 0.5d − yA
C1 = T 1
T 1 :=
C2 = T 2
T 2 :=
σA + σB
2
σA
2
(
⋅ ( b⋅ h)
)
⋅ b⋅ yA
Note: The centroid of a trapezoidal area was used
in the calculation of moment.
(
)
2
darm2 := ( 2yA)
3
hc :=
darm1 := 2 yA + hc
( )
h 2σ B + σ A
⋅
3 σB + σA
( )
M := T 1 ⋅ darm1 + T2 ⋅ darm2
M = 78.54 kN⋅ m
Ans
Problem 6-177
A beam is made from polypropylene plastic and has a stress-strain diagram that can be approximated
by the curve shown. If the beam is subjected to a maximum tensile and compressive strain of ε = 0.02
mm/mm, determine the maximum moment M.
Given:
b := 30mm
σ = 10
4
h := 100mm
ε MPa
ε max := 0.02
mm
mm
Solution:
Stress-strain Relationship :
unit := MPa
The bending stress can bs expressed
in terms of y using
ε max
ε=
⋅y
0.5h
4
ε max
σ = 10
0.5h
⋅ y⋅ unit
Resultant Internal Moment : M = ∫ y σ dA
A
⌠
⎮
M := 2 ⎮
⎮
⌡
0.5h
0
⎛⎜ 4 ε
⎞
max
y⋅ ⎜ 10
⋅ y⋅ unit ⋅ b dy
0.5h
⎝
⎠
M = 0.251 kN⋅ m
Ans
Problem 6-178
The bar is made of an aluminum alloy having a stress-strain diagram that can be approximated by the
straight line segments shown. Assuming that this diagram is the same for both tension and
compression, determine the moment the bar will support if the maximum strain at the top and bottom
fibers of the beam is εmax = 0.03.
Given: b := 75mm
d := 100mm
σ A := 420MPa
ε A := 0.006
mm
mm
σ C := 630MPa
ε max := 0.03
mm
mm
σ B := 560MPa
mm
mm
mm
ε C := 0.05
mm
ε B := 0.025
Solution:
Maximum Stress :
σ max − σ B
ε max − ε B
=
σC − σB
σ max :=
εC − εB
Maximum Moment :
εA
yA :=
⋅ ( 0.5d)
ε max
C1 = T 1
T 1 :=
C2 = T 2
T 2 :=
yB :=
εB
ε max
σ B + σ max
2
σA + σB
σA
(
(
)
)
h1 2σ max + σ B
⋅
3 σ max + σ B
⋅ b⋅ yA
2
Note: The centroid of a trapezoidal area was used
in the calculation of moment.
hc2 :=
h2 2σ B + σ A
⋅
3 σB + σA
(
)
darm2 := 2( yA + hc2)
2
darm3 := ( 2yA)
3
M := ( T 1) ⋅ darm1 + ( T2) ⋅ darm2 + ( T 3) ⋅ darm3
darm1 := 2 yB + hc1
M = 96.48 kN⋅ m
Ans
)
h1 := 0.5d − yB
)
(
hc1 :=
(
⋅ ε max − ε B + σ B
⋅ b⋅ h1
⋅ b⋅ h2
T 3 :=
εC − εB
⋅ ( 0.5d)
2
C3 = T 3
σC − σB
σ max = 574 MPa
h2 := yB − yA
Problem 6-179
The bar is made of an aluminum alloy having a stress-strain diagram that can be approximated by the
straight line segments shown. Assuming that this diagram is the same for both tension and
compression, determine the moment the bar will support if the maximum strain at the top and bottom
fibers of the beam is εmax = 0.05.
Given: b := 75mm
d := 100mm
σ A := 420MPa
σ B := 560MPa
mm
ε A := 0.006
mm
ε B := 0.025
σ C := 630MPa
ε C := 0.05
mm
mm
mm
mm
ε max := ε C
Solution:
Stress-strain Relationship :
σ1
ε
σ2 − σA
ε − εA
σ3 − σB
ε − εB
Strain :
σA
=
σ1 =
εA
σB − σA
=
σ2 =
εB − εA
σC − σB
=
σ3 =
εC − εB
ε max
ε=
0.5d
σA
εA
σB − σA
εB − εA
σC − σB
εC − εB
yA :=
ε
⋅y
⋅ε
εA
max
(
)
(
)
⋅ ε − ε A + σA
⋅ ε − ε B + σB
⋅ ( 0.5d)
yB :=
ε
εB
σ1 =
σ A ⎛ ε max ⎞
⋅⎜
⋅y
ε A ⎝ 0.5d ⎠
σ2 =
σ B − σ A ⎛ ε max
⎞
⋅⎜
⋅ y − ε A + σA
ε B − ε A ⎝ 0.5d
⎠
for yA < y < yB
σ3 =
σ C − σ B ⎛ ε max
⎞
⋅⎜
⋅ y − ε B + σB
ε C − ε B ⎝ 0.5d
⎠
for yB < y < 0.5d
Resultant Moment :
⋅ ( 0.5d)
max
for 0 < y < yA
M = ∫ y σ dA
A
y
⌠ A ⎡σ ⎛ ε
A
max ⎞⎤
M1 := 2 ⎮
y⋅ ⎢
⋅⎜
⋅ y ⎥ ⋅ b dy
⎮
ε A ⎝ 0.5d ⎠
⎣
⎦
⌡0
M1 = 0.76 kN⋅ m
y
⌠ B ⎡σ − σ ⎛ ε
⎤
⎞
⎮
B
A
max
M2 := 2 ⎮
y⋅ ⎢
⋅⎜
⋅ y − ε A + σ A⎥ ⋅ b dy
⎠
⎣ ε B − ε A ⎝ 0.5d
⎦
⎮
⌡y
M2 = 22.28 kN⋅ m
A
y
⌠ B ⎡σ − σ ⎛ ε
⎤
⎞
⎮
C
B
max
M3 := 2 ⎮
y⋅ ⎢
⋅⎜
⋅ y − ε B + σ B⎥ ⋅ b dy
⎠
⎣ ε C − ε B ⎝ 0.5d
⎦
⎮
⌡y
A
M3 = 23.8 kN⋅ m
M := M1 + M2 + M3
M = 46.84 kN⋅ m
Ans
Note: The solution can also be obtained from stress blocks as in Prob, 6-178
Problem 6-180
The beam is made of a material that can be assumed perfectly plastic in tension and elastic perfectly
plastic in compression. Determine the maximum bending moment M that can be supported by the bea
so that the compressive material at the outer edge starts to yield.
Solution:
1
σ ⋅A
2 Y 1
A1 = a⋅ d
C=
A2 = a( h − d)
T = σ Y⋅ A2
∫ σ dA = 0
C−T= 0
A
1
σ ⋅ A − σ Y⋅ A2 = 0
2 Y 1
1
A − A2 = 0
2 1
1
a⋅ d − a( h − d) = 0
2
2⋅ h
d=
3
Plastic Moment :
darm =
2
1
d + ( h − d)
3
2
(
darm =
)
Mp = σ Y⋅ A2 ⋅ darm
11⋅ h
Mp = σ Y⋅ a⋅ ( h − d) ⋅
18
2
Mp =
11⋅ a⋅ h
⋅ σY
54
Ans
2 2⋅ h 1 ⎛
2⋅ h ⎞
+ ⎜h −
3 3
3 ⎠
2⎝
darm =
11⋅ h
18
Problem 6-181
The plexiglass bar has a stress-strain curve that can be approximated by the straight-line segments
shown. Determine the largest moment M that can be applied to the bar before it fails.
Given: b := 20mm
h := 20mm
σ t1 := 40MPa
ε t1 := 0.02
σ t2 := 60MPa
mm
mm
ε t2 := 0.04
σ c1 := −80MPa
ε c1 := −0.04
mm
mm
σ c2 := −100MPa
mm
mm
ε c2 := −0.06
mm
mm
Solution:
Assume failure due to tension and ε c < ε c1
1
A1 = b( h − d)
C = σ c⋅ A1
2
1
1
A2 = ⋅ b⋅ d
T 1 = σ t1⋅ A2
2
2
1
1
A3 = ⋅ b⋅ d
T2 =
σ + σ t2 ⋅ A3
2
2 t1
(
∫ σ dA = 0
)
C − T1 − T2 = 0
A
1
1
⎛1 ⎞ 1
⎛1 ⎞
σ c⋅ [ b( h − d) ] − σ t1⋅ ⎜ ⋅ b⋅ d − σ t1 + σ t2 ⋅ ⎜ ⋅ b⋅ d = 0
2
2
⎝2 ⎠ 2
⎝2 ⎠
(
⎛ h − 1⎞ = σ + 0.5σ
t.1
t.2
⎝d ⎠
)
σ c⋅ ⎜
Try σ c := 74.833MPa
σ c < σ c1
Check :
From the strain diagram, ε c :=
From the σ-ε diagram,
Hence,
σ :=
h−d
⋅ ε t2
d
εc
ε c1
1
σ ⋅ [ b( h − d) ]
2 c
1
⎛1 ⎞
T 1 := σ t1⋅ ⎜ ⋅ b⋅ d
2
⎝2 ⎠
1
⎛1 ⎞
T 2 :=
σ t1 + σ t2 ⋅ ⎜ ⋅ b⋅ d
2
⎝2 ⎠
C :=
(
)
d :=
then
⋅ σ c1
h⋅ σ c
σ t1 + 0.5σ t2 + σ c
ε c = 0.037417
mm
mm
σ = 74.833 MPa
C = 7.2336 kN
T 1 = 2.0667 kN
T 2 = 5.1668 kN
d = 10.334 mm
O.K! ε c < ε c1
O.K! Close to assumed value.
Ultimate Moment :
darm1 :=
2
( h − d)
3
darm1 = 6.4442 mm
darm2 :=
2 ⎛ d⎞
⎜
3 ⎝ 2⎠
darm2 = 3.4446 mm
Note: The centroid of a trapezoidal area was used
in the calculation.
0.5d 2σ t1 + σ t2
hc :=
⋅
hc = 2.4112 mm
3
σ t1 + σ t2
darm3 := d − hc
darm3 = 7.9225 mm
Mult := C⋅ darm1 + T 1⋅ darm2 + T2⋅ darm3
Mult = 94.67 N⋅ m
Ans
Problem 6-182
The beam is made from three boards nailed together as shown. If the moment acting on the cross
section is M = 650 N·m, determine the resultant force the bending stress produces on the top board.
Given:
bf := 290mm
tf := 15mm
tw := 20mm
dw := 125mm
M := 650N⋅ m
Solution: D := dw + tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
yc :=
)
(bf⋅ tf)⋅ 0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)
bf⋅ tf + 2dw⋅ tw
yc = 44.933 mm
If :=
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ yc − 0.5tf
12
Iw :=
1
3
2
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ ⎡⎣yc − 0.5dw + tf ⎤⎦
12
(
)(
( )
(
)
(
)
4
I := If + Iw
Bending Stress:
)
I = 17990374.89 mm
σ = M⋅
c
I
At B:
cB := yc − tf
σ B := M⋅
cB
I
σ B = 1.0815 MPa
At A:
cA := yc
σ A := M⋅
cA
I
σ A = 1.6235 MPa
F = 5.883 kN
Ans
Resultant Force: For the top board.
(
)(
F := 0.5 σ A + σ B ⋅ bf⋅ tf
)
Problem 6-183
The beam is made from three boards nailed together as shown. Determine the maximum tensile and
compressive stresses in the beam.
Given:
bf := 290mm
tf := 15mm
tw := 20mm
dw := 125mm
M := 650N⋅ m
Solution: D := dw + tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
yc :=
)
(bf⋅ tf)⋅ 0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)
bf⋅ tf + 2dw⋅ tw
yc = 44.933 mm
If :=
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ yc − 0.5tf
12
Iw :=
1
3
2
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ ⎡⎣yc − 0.5dw + tf ⎤⎦
12
(
( )
I := If + Iw
Maximum Bending Stress:
)(
(
)
)
(
)
4
I = 17990374.89 mm
σ = M⋅
c
I
For compression:
cc := yc
σ c_max := M⋅
cc
I
σ c_max = 1.623 MPa
Ans
For tension:
ct := D − yc
σ t_max := M⋅
ct
I
σ t_max = 3.435 MPa
Ans
Problem 6-184
Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as
functions of x, where 0 ≤ x < 1.8 m.
Given:
a := 2.4m
b := 1.2m
P := 40kN
M := 75kN⋅ m
w := 30
kN
m
Solution:
Equilibrium :
+
A := w⋅ a + P
ΣF y=0;
ΣΜA=0;
A = 112 kN
30
5.
4k
N.
m
MA := ( w⋅ a) ⋅ ( 0.5a) + P⋅ ( a + b) + M
1
1
2k
N
30x
MA = 305.40 kN⋅ m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
1
V1 x1 := A − w⋅ x1 ⋅
kN
( ) (
1
V2 x2 := ( A − w⋅ a) ⋅
kN
)
( )
1
2
M1 x1 := ⎛⎝ −MA + A⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
kN⋅ m
( )
Ans
Ans
1
M2 x2 := ⎡⎣−MA + A⋅ x2 − ( w⋅ a) ⋅ x2 − 0.5 ⋅ a + M⎤⎦ ⋅
kN⋅ m
( )
(
)
100
( )
V2 ( x 2 )
V1 x 1
0
Moment (kN-m)
Shear (kN)
150
50
0
0
1
2
x1 , x2
Distance (m)
3
( )
M2( x2)
M 1 x 1 200
400
0
1
2
x1 , x2
Distane (m)
3
Problem 6-185
Draw the shear and moment diagrams for the beam. Hint: The 100-kN load must be replaced by
equivalent loadings at point C on the axis of the beam.
Given:
a := 1.2m
b := 1.2m
c := 1.2m
d := 0.3m
F1 := 75kN
F2 := 100kN
Solution:
Equilibrium :
+
Given
ΣF y=0;
A − F1 + B = 0
ΣΜC=0;
A⋅ ( a + b + c) − F1⋅ ( b + c) − F2⋅ ( d) = 0
Guess A := 1kN
B := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 58.33 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 16.67 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
1
V1 x1 := A⋅
kN
1
V2 x2 := A − F1 ⋅
kN
1
V3 x3 := A − F1 ⋅
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − F1⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( ) (
( )
( )
)
( )
( ) (
(
)
)
1
M3 x3 := ⎡⎣A⋅ x3 − F1⋅ x3 − a − F2⋅ d⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
Moment (kN-m)
Shear (kN)
100
( )
V2 ( x 2 )
V3 ( x 3 )
0
V1 x 1
50
0
1
2
3
( )
M2( x2)
50
M3( x3)
M1 x1
0
0
1
2
x1 , x2 , x3
x1 , x2 , x3
Distance (m)
Distance (m)
3
Problem 6-186
Determine the plastic section modulus and the shape factor for the wide-flange beam.
Given: bf := 180mm
tf := 20mm
dw := 180mm
tw := 30mm
Solution: D := dw + 2⋅ tf
Section Property :
(
)
A = 12600 mm
⎡ 1 ⋅ b ⋅ D3 − 1 ⋅ b − t ⋅ d 3⎤
(
) ⎥
12 f w w ⎦
⎣ 12 f
I = 86820000 mm
2
A := bf⋅ D − bf − tw ⋅ dw
4
I := ⎢
Set
σ Y := MPa
Maximum Elastic Moment :
MY⋅ c
c := 0.5D
σY =
I
⎛I⎞
MY := ⎜ ⋅ σ Y
c
⎝ ⎠
MY
σY
3
= 789272.73 mm
Plastic Moment :
dw.arm := 0.5dw
dw.arm = 90 mm
df.arm := D − tf
df.arm = 200 mm
⎡
⎛ dw ⎞
⎤
Mp := ⎢ bf⋅ tf ⋅ df.arm + tw⋅ ⎜
⋅ dw.arm⎥ ⋅ σ Y
2
(
⎣
Mp
σY
)
⎝
⎠
⎦
3
= 963000.00 mm
Plastic Section Modulus :
Shape Factor : k :=
Mp
MY
z :=
Mp
σY
k = 1.22
−6
z = 963 × 10
Ans
m
3
Ans
Problem 6-187
Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt,
gear, and flywheel. The bearings at A and B exert only vertical reactions on the shaft.
Given:
a := 200mm
b := 400mm
c := 300mm
d := 200mm
C := 450N
D := −300N
E := 150N
Solution:
Given
Equilibrium :
ΣFy=0;
A+B−C−D−E= 0
ΣΜB=0; A⋅ ( a + b + c) − C⋅ ( b + c) − D⋅ c + E⋅ d = 0
Guess A := 1N B := 1N
⎛A⎞
⎛ A ⎞ ⎛ 216.67 ⎞
:= Find ( A , B)
=⎜
N
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 83.33 ⎠
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
x4 := a + b + c , 1.01 ⋅ ( a + b + c) .. a + b + c + d
1
V1 x1 := A⋅
N
1
V2 x2 := ( A − C) ⋅
N
( )
1
V3 x3 := ( A − C − D) ⋅
N
( )
( )
1
V4 x4 := ( A − C − D + B) ⋅
N
( )
A⋅ x1
M1 x1 :=
N⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − C⋅ x2 − a ⎤⎦ ⋅
N⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − C⋅ x3 − a − D⋅ x3 − a − b ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡⎣A⋅ x4 − C⋅ x4 − a − D⋅ x4 − a − b + B⋅ x4 − a − b − c ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
(
)
400
( )
V2 ( x 2 )
0
V3 ( x 3 )
V4 ( x 4 )
200
V1 x 1
200
Shear (N)
+
400
0
0.2
0.4
0.6
x1 , x2 , x3 , x4
Distance (m)
0.8
1
Moment (N-m)
50
( )
M2(x2)
M3(x3) 0
M4(x4)
M1 x1
50
0
0.2
0.4
0.6
x1 , x2 , x3 , x4
Distance (m)
0.8
1
Problem 6-188
The beam is constructed from four pieces of wood, glued together as shown. If the internal bending
moment is M = 120 kN·m, determine the maximum bending stress in the beam. Sketch a
three-dimensional view of the stress distribution acting over the cross section.
Given:
bo := 300mm
do := 300mm
bi := 250mm
di := 250mm
M := 120kN⋅ m
Solution:
Section Property :
I :=
1 ⎛
3
3
⋅ ⎝ bo⋅ do − bi⋅ di ⎞⎠
12
Maximum Bending Stress:
σ=
M⋅ c
I
cmax := 0.5do
M⋅ cmax
σ max :=
I
σ max = 51.51 MPa
Ans
Problem 6-189
The beam is constructed from four pieces of wood, glued together as shown. If the internal bending
moment is M = 120 kN·m, determine the resultant force the bending moment exerts on the top and
bottom boards of the beam.
Given:
bo := 300mm
do := 300mm
bi := 250mm
di := 250mm
M := 120kN⋅ m
Solution:
Section Property :
I :=
(
t := 0.5 do − di
1 ⎛
3
3
⋅ ⎝ bo⋅ do − bi⋅ di ⎞⎠
12
σ=
Maximum Bending Stress:
co := 0.5do
σ o :=
ci := 0.5di
σ i :=
M⋅ co
I
M⋅ ci
I
Resultant Force :
F :=
)
1
⋅ σ o + σ i ⋅ t⋅ bo
2
(
F = 354.10 kN
)( )
Ans
M⋅ c
I
σ o = 51.505 MPa
σ i = 42.921 MPa
Problem 6-190
For the section, Iy =31.7(10-6) m4, Iz = 114(10-6) m4, Iyz = 15.1(10-6) m4. Using the techniques outline
in Appendix A, the member's cross-sectional area has principal moments of inertia of Iy' = 29(10-6) m
and Iz' = 117(10-6) m4, computed about the principal axes of inertia y' and z', respectively. If the sectio
is subjected to a moment of M = 2 kN·m directed as shown, determine the stress produced at point A,
(a) using Eq. 6-11 and (b) using the equation developed in Prob. 6-111.
Given:
M := 2000N⋅ m
θ := 10.10deg
b1 := 80mm
b2 := 140mm
h1 := 60mm
h2 := 60mm
( − 6 ) m4
−6 4
Iyz := 15.1 ( 10 ) m
−6 4
Iy' := 29.0 ( 10 ) m
Iy := 31.7 10
Solution:
( − 6 ) m4
Iz := 114 10
( − 6 ) m4
Iz' := 117 10
θ' := −θ
yA := b2
Coordinates of Point A :
zA := h2
⎛⎜ y'A ⎞ ⎛ cos ( θ' ) −sin ( θ' ) ⎞ ⎛⎜ yA ⎞
:= ⎜
⋅
⎜ z'A
⎜ zA
(
)
(
)
sin
θ'
cos
θ'
⎝
⎠
⎝ ⎠
⎝ ⎠
⎛⎜ y'A ⎞ ⎛ 148.35 ⎞
=⎜
mm
⎜ z'A
34.52
⎝
⎠
⎝ ⎠
a) Using Eq. 6-11
Internal Moment Components :
My' := M⋅ sin ( θ )
Mz' := M⋅ cos ( θ )
Bending Stress:
⎛ Mz'⋅ y'A My'⋅ z'A ⎞
σ A := ⎜ −
+
Iz'
Iy'
⎝
⎠
σ A = −2.079 MPa (C)
b) Using the equation developed in Prob. 6-111
Internal Moment Components :
My := 0
Mz := M
Bending Stress: Using formula developed in Prob. 6-111.
2
D := Iy⋅ Iz − Iyz
⎛⎜ Iy Iyz ⎞ ⎛⎜ −yA ⎞
1
σ A :=
M My ⋅
⋅
⎜ Iyz Iz ⎜ zA
D z
⎝
⎠⎝
⎠
(
)
σ A = −2.086 MPa (C)
Ans
Ans
Problem 6-191
The strut has a square cross section a by a and is subjected to the bending moment M applied at an
angle θ as shown. Determine the maximum bending stress in terms of a, M, and θ. What angle θ wil
give the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.
Solution:
Internal Moment Components :
My = −M⋅ sin ( θ )
Mz = −M⋅ cos ( θ )
Section Property :
4
4
a
Iy =
12
a
Iz =
12
Maximum Bending Stress:
By inspection, maximum bending stress occurs at A (and B).
At A :
yA = 0.5a
σ= −
Mz⋅ y My⋅ z
+
Iz
Iy
σ= −
zA = −0.5 a
−12M⋅ cos ( θ ) ⋅ ( 0.5a)
4
+
−12M⋅ sin ( θ ) ⋅ ( −0.5 a)
4
a
σ=
6⋅ M
3
a
⋅ ( cos ( θ ) + sin ( θ ) )
Ans
a
d
dθ
d
dθ
σ=
6⋅ M
3
⋅ ( −sin ( θ ) + cos ( θ ) )
a
σ= 0
−sin ( θ ) + cos ( θ ) = 0
tan ( θ ) = 1
Orientation of Neutral Axis :
tan ( α ) =
Iz
Iy
θ := 45deg
Ans
θ' := −θ
⋅ tan ( θ' )
Iz
Iy
α := atan ( 1⋅ tan ( θ' ) )
α = −45.00 deg
Ans
= 1
Problem 7-1
If the beam is subjected to a shear of V = 15 kN, determine the web's shear stress at A and B. Indicate
the shear-stress components on a volume element located at these points. Set w = 125 mm. Show that
the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182(10-3) m4.
Given: bf := 200mm b'f := 125mm
tf := 30mm
tw := 25mm
dw := 250mm
V := 15kN
Solution:
Section Property : D := dw + 2tf
2
A := bf⋅ tf + dw⋅ tw + b'f⋅ tf
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
)
yc :=
A = 16000 mm
(b'f⋅ tf)⋅ (0.5tf) + (dw⋅ tw)⋅ (0.5dw + tf) + (bf⋅ tf)⋅ (D − 0.5tf)
A
yc = 174.69 mm
1
3
2
I'f :=
⋅ b'f⋅ tf + b'f⋅ tf ⋅ 0.5tf − yc
12
(
)(
)
1
3
2
⋅ t ⋅ d + tw⋅ dw ⋅ 0.5dw + tf − yc
12 w w
1
3
2
If :=
⋅ bf⋅ tf + bf⋅ tf ⋅ D − 0.5tf − yc
12
(
Iw :=
(
)(
)(
)
)
−6
4
I := If + Iw + I'f
I = 218.18 × 10
⎯
Q = Σ ⋅ yi⋅ A1
QA := D − 0.5tf − yc ⋅ bf⋅ tf
Shear Stress: τ =
τ A :=
τ B :=
V⋅ QA
I ⋅ tw
V⋅ QB
I ⋅ tw
m
(
)( )
QB := ( yc − 0.5tf) ⋅ ( b'f⋅ tf)
V⋅ Q
I⋅ t
τ A = 1.99 MPa
Ans
τ B = 1.65 MPa
Ans
3
QA = 721875.00 mm
3
QB = 598828.12 mm
Problem 7-2
If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in
the beam. Set w = 200 mm.
Given: bf := 200mm
tf := 30mm
tw := 25mm
dw := 250mm
V := 30kN
Solution:
Section Property : D := dw + 2tf
I :=
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ dw ⎤⎦
12 ⎣ f
(
I = 268.65 × 10
−6
⎯
Q = Σ ⋅ yi⋅ A1
)
m
4
⎛ D tf ⎞ ⎛ dw ⎞ ⎛ dw ⎞
Qmax := bf⋅ tf ⋅ ⎜ −
+⎜
⋅t ⋅⎜
2 w
2
2
4
(
)⎝
⎠
⎝
⎠⎝
⎠
3
Qmax = 1035312.50 mm
V⋅ Q
I⋅ t
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Shear Stress: τ =
τ max :=
V⋅ Qmax
I ⋅ tw
τ max = 4.62 MPa
Ans
Problem 7-3
If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the
web of the beam. Set w = 200 mm.
bf := 200mm
tf := 30mm
Given:
dw := 250mm
tw := 25mm
V := 30kN
Solution:
D := dw + 2tf
Section Property :
I :=
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
(
I = 268.65 × 10
−6
)
m
4
Af = ( 0.5D − y) ⋅ bf
yf = 0.5 ( 0.5D − y) + y
yf = 0.5 ( 0.5D + y)
Qf = Af⋅ yf
Qf = 0.5 ( 0.5D − y) ⋅ bf⋅ ( 0.5D + y)
(
2
2
Qf = 0.5bf 0.25D − y
Shear Stress: τ =
)
V⋅ Q
I⋅ t
⎛ V ⎞⋅ Q
f
⎝ I ⋅ bf ⎠
τf = ⎜
⎛ V ⎞ ⋅ ⎡0.5b ( 0.25D2 − y2)⎤
⎣ f
⎦
⎝ I ⋅ bf ⎠
τf = ⎜
Resultant Shear Force: V =
∫ τ dA = 0
A
For the flange,
⎤
⎡0.5⋅ b ⋅ ( 0.25⋅ D2 − y2)⎤ ⋅ b dy⎥
f
⎣
⎦ f ⎥
0.5d w
⎣
⎦
V ⎡⎢⌠
Vf :=
⋅ ⎮
I⋅ bf ⎢⌡
0.5D
Vf = 1.46 kN
Vw := V − 2Vf
Vw = 27.09 kN
Ans
Problem 7-4
If the wide-flange beam is subjected to a shear of V = 125 kN, determine the maximum shear stress in
the beam.
Given:
bf := 200mm
dw := 250mm
tf := 25mm
tw := 25mm
V := 125kN
Solution:
Section Property :
I :=
D := dw + 2tf
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
(
)
⎛ D tf ⎞ ⎛ dw ⎞ ⎛ dw ⎞
Qmax := bf⋅ tf ⋅ ⎜ −
+⎜
⋅t ⋅⎜
2
4
2
2 w
(
)⎝
Maximum Shear Stress:
⎠
⎝
τ=
V⋅ Q
I⋅ t
⎠⎝
⎠
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τ max :=
V⋅ Qmax
I ⋅ tw
τ max = 19.87 MPa
Ans
Problem 7-5
If the wide-flange beam is subjected to a shear of V = 125 kN, determine the shear force resisted by
the web of the beam.
Given:
bf := 200mm
dw := 250mm
tf := 25mm
tw := 25mm
V := 125kN
Solution:
D := dw + 2tf
Section Property :
I :=
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
(
)
( )
A2 = ( 0.5dw − y) ⋅ tw
y2c = 0.5 ( 0.5dw − y) + y
y2c = 0.5 ( 0.5dw + y)
Qw = A1⋅ ( y1c) + A2⋅ ( y2c)
Qw = 0.5bf⋅ tf⋅ ( D − tf) + 0.5 ( 0.5dw − y) ⋅ tw⋅ ( 0.5dw. + y)
2
2
Qw = 0.5bf⋅ tf⋅ ( D − tf) + 0.5tw⎛⎝ 0.25dw − y ⎞⎠
A1 = bf⋅ tf
y1c = 0.5 D − tf
Shear Stress: τ =
V⋅ Q
I⋅ t
⎛ V ⎞⋅ Q
w
⎝ I ⋅ tw ⎠
τw = ⎜
⎛ V ⎞ ⋅ ⎡0.5⋅ b ⋅ t ⋅ D − t + 0.5⋅ t ⋅ ⎛ 0.25⋅ d 2 − y2⎞⎤
f f(
f)
w⎝
w
⎣
⎠⎦
⎝ I ⋅ tw ⎠
τw = ⎜
Resultant Shear Force:
0.5d
w
⌠
Vw := ⎮
⎮
⌡− 0.5d
For the web.
V = ∫ τ dA = 0
A
⎛ V ⎞ ⋅ ⎡0.5⋅ b ⋅ t ⋅ D − t + 0.5⋅ t ⋅ ⎛ 0.25⋅ d 2 − y2⎞⎤ ⋅ t dy
f f(
f)
w⎝
w
⎜ I⋅ t ⎣
⎠⎦ w
⎝ w⎠
w
Vw = 115.04 kN
Ans
Problem 7-6
The beam has a rectangular cross section and is made of wood having an allowable shear stress of
τallow = 11.2 MPa. If it is subjected to a shear of V = 20 kN, determine the smallest dimension a of its
bottom and 1.5a of its sides.
V := 20kN
Given:
τ allow := 11.2MPa
Solution:
Section Property :
I=
1
3
⋅ a⋅ ( 1.5a)
12
t= a
⎛ 1.5a ⋅ a⎞ ⋅ ⎛ 1.5a ⎞
Qmax = ⎜
⎜
⎝ 2 ⎠⎝ 4 ⎠
Allowablwe Shear Stress:
I⋅ t =
τ=
V⋅ Q
I⋅ t
V⋅ Qmax
τ allow
1
3
⎛ V ⎞ ⋅ ⎛ 1.5a ⋅ a⎞ ⋅ ⎛ 1.5a ⎞
⋅ a⋅ ( 1.5a) ⋅ a = ⎜
⎜
⎜
12
τ allow ⎝ 2
⎠⎝ 4 ⎠
⎝
a :=
⎠
V
τ allow
a = 42.26 mm
Ans
Problem 7-7
The beam has a rectangular cross section and is made of wood. If it is subjected to a shear of V = 20
kN, and a = 250 mm, determine the maximum shear stress and plot the shearstress variation over the
cross section. Sketch the result in three dimensions.
Given:
a := 250mm
V := 20kN
Solution:
Section Property :
I :=
b := a
d := 1.5a
1
3
⋅ b⋅ d
12
⎛ d ⎞ ⎛ d⎞
Qmax := ⎜ ⋅ b ⋅ ⎜
2
4
⎝
⎠⎝ ⎠
Maximum Shear Stress:
τ=
V⋅ Q
I⋅ b
Maximum shear stress occurs at the point where
the neutral axis passes through the section.
τ max :=
V⋅ Qmax
I⋅ b
τ max = 0.320 MPa
Ans
Problem 7-8
Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.
Given: bf := 120mm
tf := 12mm
tw := 80mm
dw := 60mm
V := 20kN
Solution:
Section Property : D := dw + 2tf
I :=
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ dw ⎤⎦
12 ⎣ f
(
I = 5.21 × 10
−6
⎯
Q = Σ ⋅ yi⋅ A1
m
)
4
⎛ D tf ⎞ ⎛ dw ⎞ ⎛ dw ⎞
Qmax := bf⋅ tf ⋅ ⎜ −
+⎜
⋅t ⋅⎜
2 w
2
2
4
(
)⎝
⎠
⎝
⎠⎝
⎠
3
Qmax = 87840.00 mm
V⋅ Q
I⋅ t
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Shear Stress: τ =
τ max :=
V⋅ Qmax
I ⋅ tw
τ max = 4.22 MPa
Ans
Problem 7-9
Determine the maximum shear force V that the strut can support if the allowable shear stress for the
material is τallow = 40 MPa.
Given: bf := 120mm
tf := 12mm
tw := 80mm
dw := 60mm
τ allow := 40MPa
Solution:
Section Property : D := dw + 2tf
I :=
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ dw ⎤⎦
12 ⎣ f
(
I = 5.21 × 10
−6
⎯
Q = Σ ⋅ yi⋅ A1
m
)
4
⎛ D tf ⎞ ⎛ dw ⎞ ⎛ dw ⎞
Qmax := bf⋅ tf ⋅ ⎜ −
+⎜
⋅t ⋅⎜
2 w
2
2
4
(
)⎝
⎠
⎝
⎠⎝
⎠
3
Qmax = 87840.00 mm
V⋅ Q
I⋅ t
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Shear Stress: τ =
τ allow =
V⋅ Qmax
I ⋅ tw
V :=
I⋅ tw⋅ τ allow
Qmax
V = 189.69 kN
Ans
Problem 7-10
Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to a
shear force of V = 15 kN.
Given: bf := 120mm
tf := 12mm
tw := 80mm
dw := 60mm
V := 15kN
Solution:
Section Property : D := dw + 2tf
I :=
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ dw ⎤⎦
12 ⎣ f
(
I = 5.21 × 10
−6
m
)
4
⎯
Q = Σ ⋅ yi⋅ A1
⎛ D tf ⎞
QA := bf⋅ tf ⋅ ⎜ −
2
2
(
)⎝
3
QA = 51840.00 mm
⎠
⎛ D tf ⎞ ⎛ dw ⎞ ⎛ dw ⎞
Qmax := bf⋅ tf ⋅ ⎜ −
+⎜
⋅t ⋅⎜
2 w
2
4
2
(
)⎝
⎠
⎝
⎠⎝
⎠
3
Qmax = 87840.00 mm
V⋅ Q
I⋅ t
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Shear Stress: τ =
τ max :=
τ w_A :=
τ f_A :=
V⋅ Qmax
I ⋅ tw
V⋅ QA
I ⋅ tw
V⋅ QA
I ⋅ bf
τ max = 3.16 MPa
Ans
τ w_A = 1.87 MPa Ans
τ f_A = 1.24 MPa
Ans
Problem 7-11
If the pipe is subjected to a shear of V = 75 kN, determine the maximum shear stress in the pipe.
Given:
ro := 60mm
ri := 50mm
V := 75kN
Solution:
Section Property :
I :=
t := ro − ri
π ⎛ 4
4
⋅ r − ri ⎞⎠
4 ⎝o
4ro ⎛ π ⋅ ro ⎞ 4ri ⎛ π ⋅ ri ⎞
Qmax :=
⋅⎜
−
⋅⎜
3π ⎝ 2 ⎠ 3π ⎝ 2 ⎠
2
Maximum Shear Stress:
2
τ=
V⋅ Q
I⋅ b
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τ max :=
V⋅ Qmax
I⋅ ( 2t)
τ max = 43.17 MPa
Ans
Problem 7-12
The strut is subjected to a vertical shear of V = 130 kN. Plot the intensity of the shear-stress
distribution acting over the cross-sectional area, and compute the resultant shear force developed in the
vertical segment AB.
Given: bf := 350mm
tf := 50mm
tw := 50mm
dw := 350mm
V := 130kN
Solution:
(
)
Section Property : a := 0.5 bf − tw
I :=
1 ⎛
3
3
3
⋅ ⎝ bf⋅ tf + tw⋅ dw − tw⋅ tf ⎞⎠
12
I = 181.77 × 10
⎯
Q = Σ ⋅ yi⋅ A1
−6
m
4
⎛ a tf ⎞
QC := a⋅ tw ⋅ ⎜ +
2
2
QC = 750000 mm
⎛a
QD := a⋅ tw ⋅ ⎜ +
2
2
QD = 859375 mm
(
)⎝
(
3
⎠
⎛ tf ⎞ ⎛ tf ⎞
+ ⎜ ⋅ bf ⋅ ⎜
⎠ ⎝ 2 ⎠ ⎝ 4⎠
tf ⎞
)⎝
3
V⋅ Q
Shear Stress: τ = I⋅ t
τ w_C :=
V⋅ QC
τ f_C :=
I ⋅ tw
τ w_C = 10.73 MPa
I⋅ bf
τ f_C = 1.53 MPa
)
Aw = 0.5dw − y ⋅ tw
τ D :=
V⋅ QD
I ⋅ bf
τ D = 1.76 MPa
∫ τ dA = 0
yw = 0.5 ( 0.5dw − y) + y
yw = 0.5 ( 0.5dw + y)
Resultant Shear Force: V =
(
V⋅ QC
A
Qw = Aw⋅ yw
(
)
(
Qw = 0.5 0.5dw − y ⋅ tw⋅ 0.5dw + y
)
Q = 0.5tw⎛⎝ 0.25dw − y ⎞⎠
2
⎛ V ⎞⋅ Q
w
⎝ I ⋅ tw ⎠
τw = ⎜
⎡
V ⎢⌠
VAB :=
⋅ ⎮
I⋅ tw ⎢⎮
⌡
⎣
0.5d w
0.5tf
2
⎛ V ⎞ ⋅ ⎡0.5t ⎛0.25d 2 − y2⎞⎤
w
⎣ w⎝
⎠⎦
⎝ I ⋅ tw ⎠
τw = ⎜
⎤
⎡0.5⋅ t ⋅ ⎛ 0.25⋅ d 2 − y2⎞⎤ ⋅ t dy⎥
w
⎣ w⎝
⎠⎦ w ⎥
⎦
VAB = 50.29 kN
Ans
Problem 7-13
The steel rod has a radius of 30 mm. If it is subjected to a shear of V = 25 kN, determine the
maximum shear stress.
Given:
r := 30mm
V := 25kN
Solution:
Section Property :
I :=
π 4
⋅r
4
4r ⎛ π ⋅ r ⎞
⋅⎜
3π ⎝ 2 ⎠
2
Qmax :=
Maximum Shear Stress:
τ=
V⋅ Q
I⋅ b
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τ max :=
V⋅ Qmax
I⋅ ( 2r)
τ max = 11.79 MPa
Ans
Problem 7-14
If the T-beam is subjected to a vertical shear of V = 60 kN, determine the maximum shear stress in the
beam. Also, compute the shear-stress jump at the flange-web junction AB. Sketch the variation of the
shear-stress intensity over the entire cross section.
Given:
bf := 300mm
dw := 150mm
tf := 75mm
tw := 100mm
V := 60kN
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw
bf⋅ tf + dw⋅ tw
yc = 82.50 mm
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc
12
1
3
2
I2 :=
⋅ t ⋅ d + dw⋅ tw ⋅ 0.5dw + tf − yc
12 w w
(
I1 :=
)(
)
(
)(
)
I := I1 + I2
(
)
Qmax := tw⋅ D − yc ⋅
(
)(
Shear Stress:
τ=
D − yc
2
QAB := bf⋅ tf ⋅ yc − 0.5tf
τ max :=
V⋅ Q
I⋅ b
V⋅ Qmax
τ f_AB :=
τ w_AB :=
I ⋅ tw
V⋅ QAB
I ⋅ bf
V⋅ QAB
I ⋅ tw
)
τ max = 3.993 MPa
Ans
τ f_AB = 1.327 MPa
Ans
τ w_AB = 3.982 MPa
Ans
Problem 7-15
If the T-beam is subjected to a vertical shear of V = 60 kN, determine the vertical shear force resisted
by the flange.
Given:
bf := 300mm
dw := 150mm
tf := 75mm
tw := 100mm
V := 60kN
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw
bf⋅ tf + dw⋅ tw
yc = 82.50 mm
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc
12
1
3
2
I2 :=
⋅ t ⋅ d + dw⋅ tw ⋅ 0.5dw + tf − yc
12 w w
(
I1 :=
)(
(
)
)(
)
I := I1 + I2
(
)
(
)
yfc = 0.5 ( yc + y)
A'f = yc − y ⋅ bf
yfc = 0.5 yc − y + y
Q = A'f⋅ yfc
(
) (
)
Q = 0.5 yc − y ⋅ bf yc + y
Q = 0.5bf⎛⎝ yc − y ⎞⎠
2
Shear Stress: τ =
2
V⋅ Q
I⋅ b
⎛ V ⎞⋅ Q
⎝ I⋅ bf ⎠
τf = ⎜
⎛ V ⎞ ⋅ ⎡0.5b ⎛ y 2 − y2⎞⎤
⎣ f⎝ c
⎠⎦
⎝ I⋅ bf ⎠
τf = ⎜
Resultant Shear Force:
yc
⌠
Vf := ⎮
⎮
⌡y
For the flange.
yo := yc − tf
⎛ V ⎞ ⋅ ⎡0.5b ⎛ y 2 − y2⎞⎤ ⋅ b dy
⎜ I⋅ b ⎣ f⎝ c
⎠⎦ f
⎝ f⎠
o
Vf = 19.08 kN
Ans
V = ∫ τ dA = 0
A
Problem 7-16
The T-beam is subjected to the loading shown. Determine the maximum transverse shear stress in the
beam at the critical section.
L 1 := 2m
Given:
L 2 := 2m
L 3 := 3m
bf := 100mm
dw := 100mm
tf := 20mm
tw := 20mm
P := 20kN
w := 8
Solution:
kN
m
L := L 1 + L 2 + L 3
Support Reaction :
Given
Equilibrium :
+ ΣF y=0;
A − P − w⋅ L3 + B = 0
(
ΣΜB=0;
) (
)(
)
A L − P⋅ L2 + L3 − w⋅ L 3 ⋅ 0.5L3 = 0
Guess A := 1kN
B := 1kN
⎛A⎞
⎛ A ⎞ ⎛ 19.43 ⎞
:= Find ( A , B)
=⎜
kN
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 24.57 ⎠
Section Property :
yc :=
(
D := dw + tf
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw
bf⋅ tf + dw⋅ tw
yc = 40.00 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ tw⋅ dw + dw⋅ tw ⋅ 0.5dw + tf − yc
12
(
I1 :=
)(
(
)
)(
I := I1 + I2
4
(
)
Qmax := tw⋅ D − yc ⋅
D − yc
V⋅ Q
I⋅ b
Vmax⋅ Qmax
( I ) ⋅ tw
τ max = 14.74 MPa
I = 5333333.33 mm
3
Qmax = 64000 mm
2
Maximum Shear Stress: τ =
τ max :=
)
Ans
Vmax := B
Shear Force Diagram:
(
)
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
A
V1 x1 :=
kN
1
V2 x2 := ( A − P) ⋅
kN
( )
(
)
( )
(
Shear (kN)
20
( )
V2 ( x 2 ) 0
V3 ( x 3 )
V1 x 1
20
2
)
1
V3 x3 := ⎡⎣A − P − w⋅ x3 − L 1 − L 2 ⎤⎦ ⋅
kN
( )
0
(
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. ( L)
4
x1 , x2 , x3
Distance (m)
6
)
Problem 7-17
Determine the largest end forces P that the member can support if the allowable shear stress is τallow =
70 MPa. The supports at A and B only exert vertical reactions on the beam.
L 1 := 1m
Given:
τ allow := 70MPa
L 2 := 2m
w := 3
L 3 := 1m
kN
m
do := 100mm
bo := 160mm
di := 60mm
bi := 80mm
Solution:
Section Property :
yc :=
(
) (
)(
0.5do bo⋅ do − 0.5di bi⋅ di
bo⋅ do − bi⋅ di
)
yc = 58.57 mm
1
3
2
⋅ b ⋅ d + bo⋅ do ⋅ 0.5do − yc
12 o o
1
3
2
I2 :=
⋅ bi⋅ di + bi⋅ di ⋅ 0.5di − yc
12
(
I1 :=
)(
(
)
)(
)
(
Qmax = A'f⋅ yfc
A'f = yc⋅ bo − bi
Maximum Shear Stress: τ =
V⋅ Q
I⋅ b
)
I := I1 − I2
y'c = 0.5yc
Vmax = P
P
⎤ ⋅ ⎡0.5y 2 b − b ⎤
⎡
c ( o
i)⎦
I⋅ ( bo − bi)⎥ ⎣
⎣
⎦
τ allow = ⎢
⎛ P ⎞ ⋅ ⎛0.5y 2⎞
⎝I⎠ ⎝ c ⎠
τ allow = ⎜
P :=
⎛ I ⎞⋅ τ
( allow)
⎜
2
0.5y
c ⎠
⎝
P = 373.42 kN
Shear Force Diagram:
Equilibrium :
+
Ans
L := L 1 + L 2 + L 3
Given
ΣF y=0;
A − w⋅ L2 + B − 2P = 0
ΣΜB=0;
−P⋅ L 1 + L 2 + A⋅ L 2 − w⋅ L 2 ⋅ 0.5L2 + P⋅ L3 = 0
(
)
(
)(
2
(
Qmax = 0.5yc × bo − bi
)
Guess A := 1kN
B := 1kN
⎛A⎞
⎛ A ⎞ ⎛ 376.42 ⎞
:= Find ( A , B)
=⎜
kN
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 376.42 ⎠
)
x1 := 0 , 0.01 ⋅ L 1 .. L 1
−P
V1 x1 :=
kN
Shear (kN)
( )
(
)
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
(
1
V2 x2 := ⎡⎣−P + A − w⋅ x2 − L 1 ⎤⎦ ⋅
kN
( )
(
)
(
1
V3 x3 := −P + A − w⋅ L2 + B ⋅
kN
)
( ) (
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
0
1
2
x1 , x2 , x3
Distance (m)
)
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. ( L)
3
4
)
Problem 7-18
If the force P = 4 kN, determine the maximum shear stress in the beam at the critical section. The
supports at A and B only exert vertical reactions on the beam.
L 1 := 1m
Given:
P := 4kN
L 2 := 2m
w := 3
L 3 := 1m
kN
m
do := 100mm
bo := 160mm
di := 60mm
bi := 80mm
Solution:
Section Property :
yc :=
(
) (
)(
0.5do bo⋅ do − 0.5di bi⋅ di
bo⋅ do − bi⋅ di
)
yc = 58.57 mm
1
3
2
⋅ bo⋅ do + bo⋅ do ⋅ 0.5do − yc
12
1
3
2
I2 :=
⋅ b ⋅ d + bi⋅ di ⋅ 0.5di − yc
12 i i
(
I1 :=
)(
(
)
)(
)
(
Qmax = A'f⋅ yfc
A'f = yc⋅ bo − bi
Maximum Shear Stress: τ =
V⋅ Q
I⋅ b
)
I := I1 − I2
y'c = 0.5yc
Vmax = P
P
⎤ ⋅ ⎡0.5y 2 b − b ⎤
⎡
⎥ ⎣ c ( o i)⎦
⎣I⋅ ( bo − bi)⎦
τ max = ⎢
⎛ P ⎞ ⋅ ⎛0.5y 2⎞
⎝I⎠ ⎝ c ⎠
τ max := ⎜
τ max = 0.750 MPa
Shear Force Diagram:
Equilibrium :
+
Ans
L := L 1 + L 2 + L 3
Given
ΣF y=0;
A − w⋅ L2 + B − 2P = 0
ΣΜB=0;
−P⋅ L 1 + L 2 + A⋅ L 2 − w⋅ L 2 ⋅ 0.5L2 + P⋅ L3 = 0
(
)
(
)(
)
Guess A := 1kN
B := 1kN
A
⎛ ⎞
⎛ A ⎞ ⎛ 7.00 ⎞
:= Find ( A , B)
=⎜
kN
⎜
⎜
⎝B⎠
⎝ B ⎠ ⎝ 7.00 ⎠
2
(
Qmax = 0.5yc × bo − bi
)
(
x1 := 0 , 0.01 ⋅ L 1 .. L 1
−P
V1 x1 :=
kN
( )
)
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
1
V2 x2 := ⎡⎣−P + A − w⋅ x2 − L 1 ⎤⎦ ⋅
kN
( )
(
)
(
)
(
1
V3 x3 := −P + A − w⋅ L2 + B ⋅
kN
( ) (
)
Shear (kN)
10
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
10
0
1
2
x1 , x2 , x3
Distance (m)
)
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. ( L)
3
4
Problem 7-19
Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is
the maximum shear stress greater than the average shear stress acting over the cross section?
Problem 7-20
Develop an expression for the average vertical component of shear stress acting on the horizontal plane
through the shaft, located a distance y from the neutral axis.
Problem 7-21
Railroad ties must be designed to resist large shear loadings. If the tie is subjected to the 150-kN rail
loadings and the gravel bed exerts a distributed reaction as shown, determine the intensity w for
equilibrium, and find the maximum shear stress in the tie.
Given:
L 1 := 0.45m
L 2 := 0.90m
L 3 := 0.45m
P := 150kN
w := 3
d := 150mm
kN
m
b := 200mm
Solution:
Equilibrium :
+
ΣF y=0;
0.5w⋅ L1 + w⋅ L 2 + 0.5w⋅ L3 − 2P = 0
w :=
2P
0.5L1 + L2 + 0.5L3
w = 222.22
Section Property :
kN
m
I :=
1
3
⋅ b⋅ d
12
Qmax := ( 0.5 ⋅ b⋅ d) ⋅ 0.25 d
Shear Force Diagram:
L := L 1 + L 2 + L 3
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
w⋅ x1 ⎞ 1
⎛
V1 x1 := ⎜ 0.5 ⋅ x1
⋅
L1
⎝
⎠ kN
)
(
( )
⎡
(
(
)
⎛
V3 x3 := ⎢0.5 ⋅ w⋅ L 1 − 2⋅ P + w⋅ L2 + w⋅ x3 − L1 − L2 ⋅ ⎜ 1 −
⎣
Shear (kN)
Maximum Shear Stress:
V⋅ Q
τ=
I⋅ b
( )
Vmax := V2 L 1 ⋅ kN
(
)
( )
V2 ( x 2 )
V3 ( x 3 )
⎝
)
x3 − L1 − L2 ⎞⎤ 1
⎥⋅
kN
2L3
⎠⎦
100
V1 x 1
0
100
⎛ Vmax ⎞
τ max := ⎜
⋅Q
⎝ I⋅ b ⎠ max
τ max = 5 MPa
)
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. ( L)
1
V2 x2 := ⎡⎣0.5w⋅ L1 − P + w⋅ x2 − L1 ⎤⎦ ⋅
kN
( )
( )
(
0
0.5
1
x1 , x2 , x3
Ans
Distance (m)
1.5
Problem 7-22
The beam is subjected to a uniform load w. Determine the placement a of the supports so that the shear
stress in the beam is as small as possible. What is this stress?
Set
w :=
kN
m
a := m
Given:
L 1 := a
L 3 := a
L := 5m
L 2 := L − 2a
Solution:
Equilibrium : By equilibrium, A = B = R
+
ΣF y=0;
2R − w⋅ L = 0
R := 0.5w⋅ L
Shear Force Diagram:
(
)
(
)
(
)
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. ( L)
1
V1 x1 := w⋅ x1 ⋅
kN
1
V2 x2 := w⋅ x2 − R ⋅
kN
1
V3 x3 := w⋅ x3 − 2⋅ R ⋅
kN
( ) (
)
( ) (
)
Shear (kN)
Require,
V1 ( a) = −V2 ( a)
w⋅ a = −( w⋅ a − R)
2w⋅ a = 0.5w⋅ L
Section Property :
( )
V2 ( x 2 )
0
V3 ( x 3 )
0
Maximum Shear Stress:
⎛ Vmax ⎞
⋅Q
⎝ I⋅ b ⎠ max
τ max = ⎜
1
τ=
⎛ d⎞ d
Qmax = ⎜ b⋅ ⋅
⎝ 2⎠ 4
V⋅ Q
I⋅ b
⎛ 0.25w⋅ L ⎞ ⋅ ⎡⎛ b⋅ d ⎞ ⋅ d⎤
⎢⎜
⎥
1
2 ⎠ 4⎦
3
⎣
⎝
⎜ ⋅ b⋅ d ⋅ b
⎝ 12
⎠
τ max = ⎜
τ max =
2
3
x1 , x2 , x3
Vmax := 0.25w⋅ L
1
3
I=
⋅ b⋅ d
12
)
V1 x 1
a = 0.25L
Vmax = w⋅ a
( ) (
3w⋅ L
8⋅ b⋅ d
Ans
Distance (m)
4
5
Problem 7-23
The timber beam is to be notched at its ends as shown. If it is to support the loading shown, determine
the smallest depth d of the beam at the notch if the allowable shear stress is τallow = 3 MPa. The beam
has a width of 200 mm.
Given:
L1 := 1.2m
P1 := 12.5kN
L2 := 1.8m
P2 := 25.0kN
L3 := 1.8m
P3 := 12.5kN
L4 := 1.2m
b := 200mm
τ allow := 3MPa
Solution:
Equilibrium : By symmetry, R1=R , R2=R
+
ΣFy=0;
P1 + P2 + P3 − 2R = 0
(
R := 0.5 P1 + P2 + P3
Section Property : I =
)
R = 25.00 kN
1
3
⋅ b⋅ d
12
Qmax = ( 0.5 ⋅ b⋅ d) ⋅ 0.25 d
Maximum Shear Stress:
τ=
V⋅ Q
I⋅ b
Vmax := R
(
2
)
R⋅ 0.125b ⋅ d
τ allow =
⎛ 1 ⋅ b⋅ d3⎞ ⋅ b
⎜
⎝ 12
⎠
d :=
12⋅ ( 0.125 ) ⋅ R
(τ allow)⋅ b
d = 62.5 mm
Ans
Qmax = 0.125 ⋅ b⋅ d
2
Problem 7-24
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the shear stress developed in the glued joints at section a-a. The supports at
C and D exert only vertical reactions on the beam.
Given:
bf := 150mm
dw := 200mm
tf := 40mm
tw := 50mm
P := 25kN
Solution:
Equilibrium : By symmetry, RC=R , RD=R
+
ΣF y=0;
3P − 2R = 0
R := 1.5P
R = 37.50 kN
D := dw + 2tf
Section Property :
I :=
1
1
3
3
⋅ bf⋅ D −
⋅t ⋅d
12
12 w w
(
)(
QA := bf⋅ tf ⋅ 0.5D − 0.5tf
)
QB := QA
Shear Stress:
τ=
V⋅ Q
I⋅ b
Vaa := R − P
τ A :=
Vaa⋅ QA
I ⋅ tw
τ A = 0.747 MPa
Ans
τ B :=
Vaa⋅ QB
I ⋅ tw
τ B = 0.747 MPa
Ans
Problem 7-25
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the maximum shear stress developed in the glued joints. The supports at C
and D exert only vertical reactions on the beam.
Given:
bf := 150mm
dw := 200mm
tf := 40mm
tw := 50mm
P := 25kN
Solution:
Equilibrium : By symmetry, RC=R , RD=R
+
ΣF y=0;
3P − 2R = 0
R := 1.5P
R = 37.50 kN
Section Property :
I :=
D := dw + 2tf
1
1
3
3
⋅ bf⋅ D −
⋅t ⋅d
12
12 w w
(
)(
QA := bf⋅ tf ⋅ 0.5D − 0.5tf
)
QB := QA
Shear Stress:
τ=
V⋅ Q
I⋅ b
Vmax := R
τ A :=
Vmax⋅ QA
I ⋅ tw
τ A = 2.24 MPa
Ans
τ B :=
Vmax⋅ QB
I ⋅ tw
τ B = 2.24 MPa
Ans
Problem 7-26
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the maximum vertical shear force resisted by the top flange of the beam. The
supports at C and D exert only vertical reactions on the beam.
Given: bf := 150mm
tf := 40mm
dw := 200mm
tw := 50mm
P := 25kN
Solution:
Equilibrium : By symmetry, RC=R , RD=R
+
ΣF y=0;
3P − 2R = 0
R := 1.5P
D := dw + 2tf
Section Property :
I :=
1
1
3
3
⋅ bf⋅ D −
⋅ tw⋅ dw
12
12
yc := 0.5D
(
)
(
)
yfc = 0.5 ( yc + y)
A'f = yc − y ⋅ bf
yfc = 0.5 yc − y + y
Q = A'f⋅ yfc
(
) (
)
Q = 0.5 yc − y ⋅ bf yc + y
Q = 0.5bf⎛⎝ yc − y ⎞⎠
2
2
Shear Stress in flange:
τ=
V⋅ Q
I⋅ b
V= R
⎛ R ⎞⋅ Q
⎝ I ⋅ bf ⎠
τ= ⎜
⎛ R ⎞ ⋅ ⎡0.5⎛ y 2 − y2⎞⎤
⎠⎦
⎝I⎠ ⎣ ⎝ c
τ= ⎜
Resultant Shear Force:
For the flange.
y
⌠ c ⎛ R⎞
2
2
Vf := ⎮ ⎜ ⋅ ⎡⎣0.5 ⎛⎝ yc − y ⎞⎠⎤⎦ ⋅ bf dy
⎮ ⎝I⎠
⌡y
o
Vf = 2.36 kN
Ans
yo := yc − tf
⌠
Vf = ⎮ τ dA
⌡
A
Problem 7-27
Determine the shear stress at points B and C located on the web of the fiberglass beam.
Given:
bf := 100mm
tf := 18mm
dw := 150mm
tw := 12mm
L1 := 2m
L2 := 0.6m
L3 := 2m
kN
wo := 2.5
m
L := L1 + L2 + L3
Solution:
Equilibrium :
+
kN
w1 := 3
m
Given
ΣFy=0;
A − wo⋅ L1 − 0.5 ⋅ w1⋅ L 3 + D = 0
ΣΜD=0;
A⋅ L − wo⋅ L1⋅ L − 0.5L 1 − 0.5 ⋅ w1⋅ L 3⋅ ⎜
(
⎛ 2L3 ⎞
= 0
⎝ 3 ⎠
)
Guess A := 1kN D := 1kN
⎛A⎞
:= Find ( A , D)
⎜
⎝D⎠
D := dw + 2tf
Section Property :
I :=
⎛ A ⎞ ⎛ 4.783 ⎞
=⎜
kN
⎜
⎝ D ⎠ ⎝ 3.217 ⎠
1
0
0
mm
1
1
3
3
⋅ bf ⋅ D −
⋅ tw⋅ dw
12
12
(
)(
QB := bf⋅ tf ⋅ 0.5D − 0.5tf
1
8mm
84mm
)
75mm
75mm
1
2mm
QC := QB
Shear Stress:
1
8mm
τ=
V⋅ Q
I⋅ b
(
)
VBC := A − wo⋅ 0.5L 1
τ B :=
VBC⋅ QB
I ⋅ tw
τ B = 0.572 MPa
Ans
τ C :=
VBC⋅ QC
I ⋅ tw
τ C = 0.572 MPa
Ans
Problem 7-28
Determine the maximum shear stress acting in the fiberglass beam at the critical section.
Given:
bf := 100mm
tf := 18mm
dw := 150mm
tw := 12mm
L1 := 2m
L2 := 0.6m
L3 := 2m
kN
wo := 2.5
m
L := L1 + L2 + L3
Solution:
Equilibrium :
+
kN
w1 := 3
m
Given
ΣFy=0;
A − wo⋅ L1 − 0.5 ⋅ w1⋅ L 3 + D = 0
ΣΜD=0;
A⋅ L − wo⋅ L1⋅ L − 0.5L 1 − 0.5 ⋅ w1⋅ L 3⋅ ⎜
(
⎛ 2L3 ⎞
= 0
⎝ 3 ⎠
)
Guess A := 1kN D := 1kN
⎛A⎞
:= Find ( A , D)
⎜
⎝D⎠
D := dw + 2tf
Section Property :
I :=
⎛ A ⎞ ⎛ 4.783 ⎞
=⎜
kN
⎜
⎝ D ⎠ ⎝ 3.217 ⎠
1
1
3
3
⋅ bf ⋅ D −
⋅ tw⋅ dw
12
12
(
)(
) (
Shear Stress:
τ=
)(
)
Qmax := bf⋅ tf ⋅ 0.5D − 0.5tf + 0.5dw⋅ tw ⋅ 0.25dw
τ max :=
V⋅ Q
I⋅ b
Vmax := A
Vmax⋅ Qmax
I ⋅ tw
τ max = 1.467 MPa
Shear Force Diagram:
L := L1 + L2 + L3
x1 := 0 , 0.01 ⋅ L1 .. L1
x2 := L 1 , 1.01 ⋅ L 1 .. L1 + L2
1
V1 x1 := A − wo⋅ x1 ⋅
kN
( ) (
( )
⎡
(
)
(
)
(
)
x3 := L1 + L2 , 1.01 ⋅ L1 + L2 .. ( L )
1
V2 x2 := A − wo⋅ L1 ⋅
kN
)
( ) (
(
)
)
⎛
V3 x3 := ⎢A − wo⋅ L1 − w1⋅ x3 − L 1 − L 2 ⋅ ⎜ 1 −
⎝
x3 − L 1 − L 2 ⎞⎤ 1
⎥⋅
kN
2L 3
⎠⎦
5
Shear (kN)
⎣
Ans
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
5
0
1
2
3
x1 , x2 , x3
Distance (m)
4
Problem 7-29
The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected to
the loading shown, determine the shear stress developed in the glued joints at the critical section. The
supports at C and D exert only vertical reactions on the beam.
Given:
Solution:
bf := 200mm
dw := 200mm
tf := 50mm
tw := 50mm
kN
wo := 3
m
L := 2.5m
Equilibrium : By symmetry, RC=R , RD=R
+
ΣF y=0;
wo⋅ L − 2R = 0
R := 0.5 wo⋅ L
(
)
R = 3.75 kN
Section Property :
I :=
D := dw + 2tf
1
1
3
3
⋅ bf⋅ D −
⋅t ⋅d
12
12 w w
(
)(
QA := bf⋅ tf ⋅ 0.5D − 0.5tf
)
QB := QA
Shear Stress:
τ=
V⋅ Q
I⋅ b
Vmax := R
τ A :=
Vmax⋅ QA
I ⋅ tw
τ A = 0.225 MPa
Ans
τ B :=
Vmax⋅ QB
I ⋅ tw
τ B = 0.225 MPa
Ans
Problem 7-30
The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected to
the loading shown, determine the vertical shear force resisted by the top flange of the beam at the
critical section. The supports at C and D exert only vertical reactions on the beam.
Given:
bf := 200mm
dw := 200mm
tf := 50mm
tw := 50mm
L := 2.5m
kN
wo := 3
m
Solution:
Equilibrium : By symmetry, RC=R , RD=R
+
ΣF y=0;
wo⋅ L − 2R = 0
R := 0.5 wo⋅ L
(
)
R = 3.75 kN
D := dw + 2tf
Section Property :
I :=
1
1
3
3
⋅ bf⋅ D −
⋅t ⋅d
12
12 w w
yc := 0.5D
(
)
(
)
yfc = 0.5 ( yc + y)
A'f = yc − y ⋅ bf
yfc = 0.5 yc − y + y
Q = A'f⋅ yfc
(
) (
)
Q = 0.5 yc − y ⋅ bf yc + y
Q = 0.5bf⎛⎝ yc − y ⎞⎠
2
2
Shear Stress in flange:
τ=
V⋅ Q
I⋅ b
Vmax = R
⎛ R ⎞⋅ Q
⎝ I ⋅ bf ⎠
τ= ⎜
⎛ R ⎞ ⋅ ⎡0.5⎛ y 2 − y2⎞⎤
⎠⎦
⎝I⎠ ⎣ ⎝ c
τ= ⎜
Resultant Shear Force:
yc
⌠
Vf := ⎮
⎮
⌡y
For the flange.
⎛ R ⎞ ⋅ ⎡0.5⎛y 2 − y2⎞⎤ ⋅ b dy
⎜ ⎣ ⎝ c
⎠⎦ f
⎝I⎠
o
Vf = 0.30 kN
Ans
yo := yc − tf
⌠
Vf = ⎮ τ dA
⌡
A
Problem 7-31
Determine the variation of the shear stress over the cross section of a hollow rivet. What is the
maximum shear stress in the rivet? Also, show that if r 1 → r 0then τ max = 2( V/A ).
Problem 7-32
The beam has a square cross section and is subjected to the shear force V. Sketch the shear-stress
distribution over the cross section and specify the maximum shear stress. Also, from the neutral axis,
locate where a crack along the member will first start to appear due to shear.
Problem 7-33
Write a computer program that can be used to determine the maximum shear stress in the beam that
has the cross section shown, and is subjected to a specified constant distributed load w and
concentrated force P. Show an application of the program using the values L = 4 m, a = 2 m, P = 1.5
kN, d1 = 0, d2 = 2 m, w = 400 N/m, t1 = 15 mm, t2 = 20 mm, b = 50 mm, and h = 150 mm.
Problem 7-34
The beam has a rectangular cross section and is subjected to a load P that is just large enough to
develop a fully plastic moment Mp = PL at the fixed support. If the material is elastic-plastic, then at a
distance x < L the moment M = P x creates a region of plastic yielding with an associated elastic core
having a height 2y'. This situation has been described by Eq. 6-30 and the moment M is distributed
over the cross section as shown in Fig. 6-54e. Prove that the maximum shear stress developed in the
beam is given by τ max = 3/2 (P/A'),where A' = 2y'b, the cross-sectional area of the elastic core.
Problem 7-35
The beam in Fig. 6-54f is subjected to a fully plastic moment Mp. Prove that the longitudinal and
transverse shear stresses in the beam are zero. Hint: Consider an element of the beam as shown in Fig.
7-4d.
Problem 7-36
The beam is constructed from two boards fastened together at the top and bottom with two rows of
nails spaced every 150 mm. If each nail can support a 2.5-kN shear force, determine the maximum
shear force V that can be applied to the beam.
Given:
b := 150mm
d1 := 50mm
d2 := 50mm
sn := 150mm
Fallow := 2.5kN
Solution:
D := d1 + d2
Section Property :
I :=
1
3
⋅ b⋅ D
12
(
)(
Q := b⋅ d1 ⋅ 0.5d1
)
V⋅ Q
I
There are two rows of nails. Hence, the allowable shear flow is
Shear Flow :
q=
2Fallow
sn
Vmax⋅ Q
=
I
Vmax :=
2⋅ I Fallow
Q⋅ sn
Vmax = 2.222 kN
Ans
qallow :=
2Fallow
sn
Problem 7-37
The beam is constructed from two boards fastened together at the top and bottom with two rows of
nails spaced every 150 mm. If an internal shear force of V = 3 kN is applied to the boards, determine
the shear force resisted by each nail.
Given:
b := 150mm
d1 := 50mm
d2 := 50mm
sn := 150mm
V := 3kN
Solution:
Section Property :
I :=
D := d1 + d2
1
3
⋅ b⋅ D
12
(
)(
Q := b⋅ d1 ⋅ 0.5d1
Shear Flow :
q :=
)
V⋅ Q
I
q = 45.00
kN
m
There are two rows of nails. Hence, the shear force resisted by each nail is
F = 3.37 kN
Ans
F :=
q
⋅s
2 n
Problem 7-38
A beam is constructed from five boards bolted together as shown. Determine the maximum shear
force developed in each bolt if the bolts are spaced s = 250 mm apart and the applied shear is V = 35
kN.
Given: d1 := 250mm
d2 := 350mm
Solution:
t := 25mm
a := 100mm
s := 250mm
V := 35kN
(
)
a' := d2 − d1 − a
a' = 200 mm
h := a' + 0.5d1
h = 325 mm
Section Property :
yc :=
( )(
) ( )(
2( d1⋅ t) + 3( d2⋅ t)
)
2 d1⋅ t ⋅ 0.5d1 + a' + 3 d2⋅ t ⋅ 0.5d2
yc = 223.39 mm
1
3
2
⋅ ( 2t) ⋅ d1 + 2t⋅ d1 ⋅ 0.5d1 + a' − yc
12
1
3
2
I2 :=
⋅ ( 3⋅ t) ⋅ d2 + 3⋅ t⋅ d2 ⋅ 0.5d2 − yc
12
(
I1 :=
)(
(
)
)(
)
4
I := I1 + I2
I = 523597110.22 mm
(
)
Q := d1⋅ ( 2t) ⋅ h − yc
3
Q = 1270161.29 mm
Shear Flow :
q :=
V⋅ Q
I
q = 84.90
kN
m
There are four planes on the bolt. Hencs, the shear
force resisted by each shear plane of the bolt is
⎛q ⎞
F := ⎜ ⋅ s
⎝4 ⎠
F = 5.31 kN
Ans
Problem 7-39
A beam is constructed from five boards bolted together as shown. Determine the maximum spacing s
of the bolts if they can each resist a shear of 20 kN and the applied shear is V = 45 kN.
Given:
Solution:
d1 := 250mm
d2 := 350mm
t := 25mm
a := 100mm
V := 45kN
Fallow := 20kN
(
)
a' := d2 − d1 − a
a' = 200 mm
h := a' + 0.5d1
h = 325 mm
Section Property :
yc :=
( )(
) ( )(
2( d1⋅ t) + 3( d2⋅ t)
)
2 d1⋅ t ⋅ 0.5d1 + a' + 3 d2⋅ t ⋅ 0.5d2
yc = 223.39 mm
1
3
2
⋅ ( 2t) ⋅ d1 + 2t⋅ d1 ⋅ 0.5d1 + a' − yc
12
1
3
2
I2 :=
⋅ ( 3⋅ t) ⋅ d2 + 3⋅ t⋅ d2 ⋅ 0.5d2 − yc
12
(
I1 :=
)(
(
)(
)
4
I := I1 + I2
I = 523597110.22 mm
(
)
q :=
V⋅ Q
I
Q := d1⋅ ( 2t) ⋅ h − yc
Shear Flow :
)
3
Q = 1270161.29 mm
q = 109.16
kN
m
Since there are four planes on the bolt, the allowable shear flow is
s :=
4Fallow
q
s = 732.9 mm
Ans
qallow =
4Fallow
s
Problem 7-40
The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in the
nails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2
mm.
Given: bf := 250mm
dw := 150mm
t := 30mm
a := 100mm
s := 100mm
V := 800N
do := 2mm
h' := 0.5t
Solution:
h' = 15 mm
Section Property :
yc :=
(
)
bf⋅ t⋅ ( 0.5t) + dw⋅ ( 2t) ⋅ 0.5dw
(
bf⋅ t + 2 dw⋅ t
)
yc = 47.73 mm
1
3
2
⋅ bf⋅ t + bf⋅ t ⋅ 0.5t − yc
12
1
3
2
I2 :=
⋅ ( 2⋅ t) ⋅ dw + 2⋅ t⋅ dw ⋅ 0.5dw − yc
12
( )(
I1 :=
)
(
)(
4
I := I1 + I2
(
)
I = 32164772.73 mm
)
Q = 245454.55 mm
q = 6.105
kN
m
3
Q := bf⋅ t⋅ yc − h'
Shear Flow :
q :=
V⋅ Q
I
F := q⋅ s
F = 0.6105 kN
Since each side of the beam resists this shear force, then
π
2
Ao := ⋅ do
4
τ avg :=
F
2Ao
τ avg = 97.16 MPa
Ans
Problem 7-41
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of
150 mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determine
the maximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.
bf := 75mm
dw := 75mm
dp := 150mm
dg := 50mm
V := 250kN
Fallow := 75kN
Section Property :
D := 2dw + 2t + dg
Given:
t := 12mm
Solution:
IT :=
1
1
1
3
3
3
⋅ bf ⋅ D −
⋅ t ⋅ dg −
⋅ bf − t ⋅ ( D − 2t)
12
12
12
IP :=
1
3
⋅ ( 2t) ⋅ dp
12
(
)
I := IT + IP
( )
(
)(
)
Q := bf⋅ t ⋅ ( 0.5D − 0.5t) + t⋅ dw ⋅ 0.5dw + 0.5dg
V⋅ Q
I
Since there are two shear planes on the bolt, the allowable shear flow is
Shear Flow :
q=
2Fallow
sn
sn :=
=
V⋅ Q
I
2⋅ I Fallow
V⋅ Q
sn = 138.0 mm
Ans
q=
2F
sn
Problem 7-42
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of
150 mm and a thickness of 12 mm. If the bolts are spaced at s = 200 mm, determine the maximum
shear force V that can be applied to the cross section. Each bolt can resist a shear force of 75 kN.
Given:
bf := 75mm
dw := 75mm
t := 12mm
dp := 150mm
dg := 50mm
sn := 200mm
Fallow := 75kN
Solution:
D := 2dw + 2t + dg
Section Property :
IT :=
1
1
1
3
3
3
⋅ bf ⋅ D −
⋅ t ⋅ dg −
⋅ bf − t ⋅ ( D − 2t)
12
12
12
IP :=
1
3
⋅ ( 2t) ⋅ dp
12
(
)
I := IT + IP
( )
(
)(
)
Q := bf⋅ t ⋅ ( 0.5D − 0.5t) + t⋅ dw ⋅ 0.5dw + 0.5dg
V⋅ Q
I
Since there are two shear planes on the bolt, the allowable shear flow is
Shear Flow :
q=
2Fallow
sn
V :=
=
V⋅ Q
I
2⋅ I Fallow
sn⋅ Q
V = 172.5 kN
Ans
q=
2F
sn
Problem 7-43
The double-web girder is constructed from two plywood sheets that are secured to wood members at
its top and bottom. If each fastener can support 3 kN in single shear, determine the required spacing s
of the fasteners needed to support the loading P = 15 kN. Assume A is pinned and B is a roller.
Given:
bi := 150mm
di := 250mm
t := 12mm
do := 350mm
P := 15kN
Fallow := 3kN
Solution:
Equilibrium : By symmetry, A=R , B=R
+
ΣF y=0;
2R − P = 0
R := 0.5P
bo := bi + 2t
Section Property :
I :=
(
d' := 0.5 do − di
)
1
1
3
3
⋅ bo⋅ do −
⋅b ⋅d
12
12 i i
(
)(
)
Q := bi⋅ d' ⋅ 0.5do − 0.5d'
V⋅ Q
Vmax := R
I
Since there are two shear planes on the bolt, the allowable shear flow is
Shear Flow :
q=
2Fallow
sn
=
Vmax⋅ Q
I
2⋅ I Fallow
sn :=
Vmax⋅ Q
sn = 303.2 mm
Ans
q=
2F
sn
Problem 7-44
The double-web girder is constructed from two plywood sheets that are secured to wood members at
its top and bottom. The allowable bending stress for the wood is σallow = 56 MPa and the allowable
shear stress is τallow = 21 MPa. If the fasteners are spaced s = 150 mm and each fastener can support
3 kN in single shear, determine the maximum load P that can be applied to the beam.
Given:
bi := 150mm
di := 250mm
t := 12mm
do := 350mm
sn := 150mm
Fallow := 3kN
σ allow := 56MPa
τ allow := 21MPa
Solution:
Equilibrium : By symmetry, A=R , B=R
+
ΣF y=0;
2R − P = 0
R = 0.5P
Section Property :
bo := bi + 2t
(
d' := 0.5 do − di
I :=
)
1
1
3
3
⋅ bo⋅ do −
⋅ bi ⋅ di
12
12
(
)(
)
Q := bi⋅ d' ⋅ 0.5do − 0.5d'
V⋅ Q
Vmax = R
I
Since there are two shear planes on the bolt, the allowable shear flow is
Shear Flow :
q=
2Fallow
Vmax⋅ Q
I
sn
=
2Fallow
sn
P :=
=
0.5P⋅ Q
I
4⋅ I Fallow
sn⋅ Q
P = 30.32 kN
Ans
q=
2F
sn
Problem 7-45
The beam is made from three polystyrene strips that are glued together as shown. If the glue has a
shear strength of 80 kPa, determine the maximum load P that can be applied without causing the glue
to lose its bond.
Given: bf := 30mm
tf := 40mm
tw := 20mm
dw := 60mm
τ allow := 0.080MPa
Solution:
Equilibrium : By equilibrium, A = B = R
1
ΣFy=0; 2R − P − 2⎛⎜ ⋅ P⎞ = 0
+
⎝4 ⎠
R = 0.75P
Maximum Shear :
3
Vmax = P
4
Vmax = R
Section Property : D := dw + 2tf
I :=
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
(
I = 6.68 × 10
−6
m
4
⎛ D tf ⎞
Q := bf⋅ tf ⋅ ⎜ −
2
2
⎯
Q = Σ ⋅ yi⋅ A1
Shear Stress: τ =
τ allow =
)
(
)⎝
⎠
3
Q = 60000 mm
V⋅ Q
I⋅ t
Vmax⋅ Q
I ⋅ tw
Vmax =
τ allow⋅ I⋅ tw
Q
τ allow⋅ I⋅ tw
3
P=
Q
4
P :=
4τ allow⋅ I⋅ tw
3Q
P = 0.238 kN
Ans
Problem 7-46
The beam is made from four boards nailed together as shown. If the nails can each support a shear
force of 500 N., determine their required spacings s' and s if the beam is subjected to a shear of V =
3.5 kN.
Given:
bf := 250mm
dw := 250mm
tf := 25mm
tw := 40mm
tb := 25mm
db := 75mm
V := 3.5kN
Fallow := 0.5kN
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
(
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw + 0.5db 2tb⋅ db
bf⋅ tf + dw⋅ tw + 2 tb⋅ db
(
)
)
yc = 85.94 mm
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc
12
1
3
2
I2 :=
⋅ t ⋅ d + dw⋅ tw ⋅ 0.5dw + tf − yc
12 w w
1
3
2
I3 :=
⋅ 2tb ⋅ db + 2tb⋅ db ⋅ 0.5db − yc
12
(
I1 :=
)(
(
( )
)
)(
(
)
)(
)
I := I1 + I2 + I3
( )(
)
QD := ( dw⋅ tw) ⋅ ( D − yc − 0.5dw)
QC := tb⋅ db ⋅ yc − 0.5db
V⋅ Q
I
The allowable shear flow at points C and D are :
Shear Flow :
q=
qC =
V⋅ QC
Fallow
I
s'n
I Fallow
sn :=
V⋅ QC
s'n :=
sn = 216.6 mm
s'n = 30.7 mm
Fallow
sn
=
=
V⋅ QD
F
sn
I
I Fallow
V⋅ QD
Ans
qD =
F
s'n
Problem 7-47
The beam is fabricated from two equivalent channels and two plates. Each plate has a height of 150
mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determine the
maximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.
bf := 300mm
dw := 88mm
dp := 150mm
dg := 50mm
V := 250kN
Fallow := 75kN
Section Property :
D := 2dw + 2t + dg
Given:
t := 12mm
Solution:
IU :=
1
1
1
3
3
3
⋅ bf⋅ D −
⋅ ( 2t) ⋅ dg −
⋅ b − 2t ⋅ ( D − 2t)
12
12
12 f
IP :=
1
3
⋅ ( 2t) ⋅ dp
12
(
)
I := IU + IP
( )
(
)(
Q := bf⋅ t ⋅ ( 0.5D − 0.5t) + 2t⋅ dw ⋅ 0.5dw + 0.5dg
)
V⋅ Q
I
Since there are two rows of bolts, the allowable shear flow is
Shear Flow :
q=
2Fallow
sn
sn :=
=
V⋅ Q
I
2⋅ I Fallow
V⋅ Q
sn = 137.6 mm
Ans
q=
2F
sn
Problem 7-48
A built-up timber beam is made from n boards, each having a rectangular cross section. Write a
computer program that can be used to determine the maximum shear stress in the beam when it is
subjected to any shear V. Show an application of the program using a cross section that is in
the form of a “T” and a box.
Problem 7-49
The timber T-beam is subjected to a load consisting of n concentrated forces Pn ., If the allowable
shear Vnail for each of the nails is known, write a computer program that will specify the nail spacing
between each load. Show an application of the program using the values L = 4.5 m, a1 = 1.2 m, P1 = 3
kN, a2 = 2.4 m, P2 = 7.5 kN, b1 = 37.5 mm, h1 = 250 mm, b2 = 200 mm, h2 = 25 mm, and Vnail = 1
kN.
Problem 7-50
The strut is constructed from three pieces of plastic that are glued together as shown. If the allowable
shear stress for the plastic is τallow = 5.6 MPa and each glue joint can withstand 50 kN/m, determine
the largest allowable distributed loading w that can be applied to the strut.
Given:
L 1 := 1m
τ allow := 5.6MPa
L 2 := 2m
qallow := 50
L 3 := 1m
kN
m
bf := 74mm
tf := 25mm
dw := 75mm
tw := 12mm
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf 2tw⋅ dw
bf⋅ tf + 2tw⋅ dw
yc = 37.16 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ 0.5dw + tf − yc
12
(
I1 :=
)(
( )
)
(
)(
)
I := I1 + I2
( )(
)
Qmax := 2tw ⋅ D − yc ⋅
(
)(
QA := bf⋅ tf ⋅ yc − 0.5tf
Allowable Shear Stress:
D − yc
2
)
V⋅ Q
I⋅ t
τ=
Vmax = w⋅ L1
D − yc⎤
⎡ w⋅ L1 ⎤ ⎡
⎥
⎥ ⋅ ⎢( 2tw) ⋅ ( D − yc) ⋅
2 ⎦
⎣ I⋅ (2tw)⎦ ⎣
τ allow = ⎢
τ allow =
w⋅ L1
2I
(
)2
⋅ D − yc
2I
w :=
(
)
L1⋅ D − yc
(
2
)
⋅ τ allow
w = 9.13
kN
m
Shear Flow : Assume the beam fails at the glue joint and the allowable shear flow is 2⋅ qallow
w⋅ L1 ⋅ QA
V⋅ Q
2qallow =
2qallow =
I
I
(
w :=
2⋅ I qallow
L1⋅ QA
)
w = 7.06
kN
m
(Controls !)
Ans
Shear Force Diagram:
L := L 1 + L 2 + L 3
Equilibrium : By symmetry, R1 = R
2R − w⋅ L = 0
+ ΣF y=0;
R2 = R
R := 0.5w⋅ L
x1 := 0 , 0.01 ⋅ L 1 .. L 1
−w⋅ x1
kN
)
(
)
(
)
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. ( L)
1
V2 x2 := −w⋅ x2 + R ⋅
kN
1
V3 x3 := −w⋅ x3 + 2R ⋅
kN
( ) (
)
( ) (
)
10
Shear (kN)
( )
V1 x1 :=
(
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
10
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Problem 7-51
The strut is constructed from three pieces of plastic that are glued together as shown. If the distributed
load w = 3 kN/m, determine the shear flow that must be resisted by each glue joint.
Given:
L 1 := 1m
L 2 := 2m
w := 3
L 3 := 1m
kN
m
bf := 74mm
tf := 25mm
dw := 75mm
tw := 12mm
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf 2tw⋅ dw
bf⋅ tf + 2tw⋅ dw
yc = 37.16 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ 0.5dw + tf − yc
12
(
I1 :=
( )
)(
(
)
)(
)
I := I1 + I2
(
)(
QA := bf⋅ tf ⋅ yc − 0.5tf
)
Shear Flow : Since there are two glue joints, hence
Vmax := w⋅ L 1
q :=
Vmax⋅ QA
2I
q = 21.24
kN
m
Ans
2q =
V⋅ Q
I
Shear Force Diagram:
L := L 1 + L 2 + L 3
Equilibrium : By symmetry, R1 = R
2R − w⋅ L = 0
+ ΣF y=0;
R2 = R
R := 0.5w⋅ L
x1 := 0 , 0.01 ⋅ L 1 .. L 1
−w⋅ x1
kN
)
(
)
(
)
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. ( L)
1
V2 x2 := −w⋅ x2 + R ⋅
kN
1
V3 x3 := −w⋅ x3 + 2R ⋅
kN
( ) (
)
( ) (
)
10
Shear (kN)
( )
V1 x1 :=
(
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
10
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Problem 7-52
The beam is subjected to the loading shown, where P = 7 kN. Determine the average shear stress
developed in the nails within region AB of the beam. The nails are located on each side of the beam and
are spaced 100 mm apart. Each nail has a diameter of 5 mm.
Given:
bf := 250mm
dw := 150mm
t := 30mm
a := 2m
s := 100mm
do := 5mm
P' := 3kN
P := 7kN
Solution:
Section Property :
I :=
1 ⎡
3
3
⋅ ⎣ bf + 2⋅ t ⋅ dw − bf⋅ dw − 2⋅ t ⎤⎦
12
(
)
(
)
4
I = 72000000 mm
(
Q := bf⋅ t⋅ 0.5dw − 0.5t
)
3
Q = 450000 mm
Maximum Shear :
Shear Flow :
Vmax⋅ Q
q :=
I
Vmax := P' + P
q = 62.500
Vmax = 10 kN
kN
m
There are two rows of nails. Hence, the sher force
resisted by each nail is
q
⋅s
2
π
2
Ao := ⋅ do
4
F :=
F = 3.125 kN
τ avg :=
F
Ao
τ avg = 159.2 MPa
Ans
Problem 7-53
The beam is constructed from four boards which are nailed together. If the nails are on both sides of
the beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to the
end of the beam.
Given:
bf := 250mm
dw := 150mm
t := 30mm
a := 2m
s := 100mm
P' := 3kN
Fallow := 3kN
Solution:
Section Property :
I :=
1 ⎡
3
3
⋅ b + 2⋅ t ⋅ dw − bf⋅ dw − 2⋅ t ⎤⎦
12 ⎣ f
(
)
(
)
4
I = 72000000 mm
(
Q := bf⋅ t⋅ 0.5dw − 0.5t
)
3
Q = 450000 mm
Vmax = P' + P
There are two rows of nails. Hence, the allowable sher flow is
Maximum Shear :
qallow :=
2Fallow
s
qallow = 60.00
kN
m
Shear Flow :
qallow =
Vmax⋅ Q
I
qallow =
( P' + P) ⋅ Q
I
⎛ qallow⋅ I ⎞
− P'
⎝ Q ⎠
P := ⎜
P = 6.60 kN
Ans
Problem 7-54
The member consists of two plastic channel strips 12 mm thick, bonded together at A and B. If the
glue can support an allowable shear stress of τallow = 4.2 MPa, determine the maximum intensity w0 of
the triangular distributed loading that can be applied to the member based on the strength of the glue.
Given:
bo := 150mm
t := 12mm
do := 150mm
L := 4m
τ allow := 4.2MPa
Solution:
Equilibrium : By symmetry, A=R , B=R
+
ΣF y=0;
2R − 0.5wo⋅ L = 0
R = 0.25wo⋅ L
bi := bo − 2t
Section Property :
I :=
di := do − 2t
1
1
3
3
⋅ bo⋅ do −
⋅b ⋅d
12
12 i i
( )(
)
(
) (
Q := bo⋅ t ⋅ 0.5do − 0.5t + ⎡⎣2⋅ t⋅ 0.5 ⋅ di ⎤⎦ ⋅ 0.5di
)
V⋅ Q
Vmax = R
I
Since there are two planes of glue, the allowable shear flow is
Shear Flow :
q=
( 2t) ⋅ τ allow =
( 2⋅ t)τ allow =
wo :=
Vmax⋅ Q
I
(0.25wo⋅ L)⋅ Q
I
8t⋅ Iτ allow
L⋅ Q
kN
wo = 9.73
m
Ans
2t⋅ τ allow
Problem 7-55
The member consists of two plastic channel strips 12 mm thick, glued together at A and B. If the
distributed load has a maximum intensity of w0 = 50 kN/m, determine the maximum shear stress
resisted by the glue.
Given: bo := 150mm do := 150mm
L := 4m
kN
wo := 50
m
t := 12mm
Solution:
Equilibrium : By symmetry, A=R , B=R
+
ΣF y=0;
2R − 0.5wo⋅ L = 0
R := 0.25wo⋅ L
Section Property :
I :=
bi := bo − 2t
di := do − 2t
1
1
3
3
⋅ bo⋅ do −
⋅b ⋅d
12
12 i i
( )(
)
(
) (
Q := bo⋅ t ⋅ 0.5do − 0.5t + ⎡⎣2⋅ t⋅ 0.5 ⋅ di ⎤⎦ ⋅ 0.5di
Allowable Shear Stress:
τ=
V⋅ Q
I⋅ b
τ max :=
)
Vmax := R
Vmax⋅ Q
I⋅ ( 2t)
τ max = 21.58 MPa
Ans
Problem 7-56
A shear force of V = 18 kN is applied to the symmetric box girder. Determine the shear flow at A and
B.
Given:
bf := 125mm
dw := 300mm
t := 10mm
dm := 200mm
dg := 30mm
V := 18kN
Solution:
Section Property :
1
3
2
⋅ b ⋅ t + bf⋅ t ⋅ 0.5dw − 0.5t
12 f
1
3
2
I2 :=
⋅ bf⋅ t + bf⋅ t ⋅ 0.5t + 0.5dm
12
1
3
I3 :=
⋅ t⋅ dw
12
I1 :=
( )(
)
( )(
)
4
I := 2I1 + 2I2 + 2I3
I = 125166666.67 mm
( )(
)
QA = 181250 mm
( )(
)
QB = 131250 mm
QA := bf⋅ t ⋅ 0.5dw − 0.5t
QB := bf⋅ t ⋅ 0.5t + 0.5dm
3
3
Shear Flow :
qA :=
1 V⋅ QA
2 I
qA = 13.03
qB :=
1 V⋅ QB
2 I
qB = 9.44
kN
m
kN
m
Ans
Ans
Problem 7-57
A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at C.
Given:
bf := 125mm
dw := 300mm
t := 10mm
dm := 200mm
dg := 30mm
V := 18kN
Solution:
Section Property :
1
3
2
⋅ b ⋅ t + bf⋅ t ⋅ 0.5dw − 0.5t
12 f
1
3
2
I2 :=
⋅ bf⋅ t + bf⋅ t ⋅ 0.5t + 0.5dm
12
1
3
I3 :=
⋅ t⋅ dw
12
I1 :=
( )(
)
( )(
)
I := 2I1 + 2I2 + 2I3
4
I = 125166666.67 mm
( )(
) ( )(
) (
)(
)
QC := bf⋅ t ⋅ 0.5dw − 0.5t + bf⋅ t ⋅ 0.5t + 0.5dm + 2 0.5dw⋅ t ⋅ 0.25dw
3
QC = 537500 mm
Shear Flow :
qC :=
1 V⋅ QC
2 I
qC = 38.65
kN
m
Ans
Problem 7-58
The channel is subjected to a shear of V = 75 kN. Determine the shear flow developed at point A.
Given:
bf := 400mm
dw := 200mm
tf := 30mm
tw := 30mm
V := 75kN
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf 2tw⋅ dw
bf⋅ tf + 2tw⋅ dw
yc = 72.50 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ 0.5dw + tf − yc
12
(
I1 :=
( )
)(
(
)
)(
)
I := I1 + I2
4
I = 120250000 mm
(
)(
QA := bf⋅ tf ⋅ yc − 0.5tf
)
3
QA = 690000 mm
Shear Flow :
qA :=
1 V⋅ QA
2 I
qA = 215.2
kN
m
Ans
Problem 7-59
The channel is subjected to a shear of V = 75 kN. Determine the maximum shear flow in the channel.
Given:
bf := 400mm
dw := 200mm
tf := 30mm
tw := 30mm
V := 75kN
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf 2tw⋅ dw
bf⋅ tf + 2tw⋅ dw
yc = 72.50 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ 0.5dw + tf − yc
12
(
I1 :=
( )
)(
(
)
)(
)
I := I1 + I2
4
I = 120250000 mm
1
Qmax := tw⋅ D − yc ⋅ ⋅ D − yc
2
(
) (
)
3
Qmax = 372093.75 mm
Shear Flow :
qmax :=
V⋅ Qmax
I
qmax = 232.1
kN
m
Ans
Problem 7-60
The beam supports a vertical shear of V = 35 kN. Determine the resultant force developed in segment
AB of the beam.
Given:
bf := 125mm
dw := 250mm
tf := 12mm
tw := 12mm
V := 35kN
Solution:
D := dw + 2tf
Section Property :
I :=
1
1
3
3
⋅ 2tf ⋅ bf +
⋅ dw⋅ tw
12
12
( )
Q = A'f⋅ yfc
(
)
(
)
Q = tf⋅ 0.5bf − y ⋅ ⎡⎣0.5 0.5bf − y + y⎤⎦
Q = 0.5tf⎛⎝ 0.25bf − y ⎞⎠
Shear Flow:
2
2
q=
V⋅ Q
I
q=
V⋅ tf
2I
Resultant Shear Force:
0.5b f
⌠
VAB := ⎮
⎮
⌡y
V⋅ tf
2I
⋅ ⎛⎝ 0.25bf − y ⎞⎠
2
yo := 0.5tw
For AB:
⋅ ⎛⎝ 0.25bf − y ⎞⎠ dy
2
o
VAB = 7.43 kN
2
Ans
2
⌠
VAB = ⎮ q dy
⌡
A
Problem 7-61
The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V
= 150 N, determine the shear flow at points A and B.
Given: bf := 60mm
tf := 10mm
b'f := 80mm
dw := 40mm
tw := 10mm
V := 150N
Solution:
Section Property : D := dw + 2tf
A := bf⋅ tf + 2dw⋅ tw + b'f⋅ tf
2
A = 2200 mm
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
) y := (b'f⋅ tf)⋅ (0.5tf) + 2(dw⋅ tw)⋅ (0.5dw + tf) + (bf⋅ tf)⋅ (D − 0.5tf)
c
A
yc = 27.73 mm
1
3
2
I'f :=
⋅ b' ⋅ t + b'f⋅ tf ⋅ 0.5tf − yc
12 f f
(
)(
)
1
3
2
⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + tf − yc
12
1
3
2
If :=
⋅ b ⋅ t + bf⋅ tf ⋅ D − 0.5tf − yc
12 f f
(
Iw :=
(
)(
)
)(
)
I := If + 2Iw + I'f
I = 981.9697 × 10
(
−9
)(
m
4
QA := 0.5b'f⋅ tf ⋅ yc − 0.5tf
(
)(
)
3
QA = 9090.91 mm
)
QB := bf⋅ tf ⋅ D − 0.5tf − yc
3
QB = 16363.64 mm
Shear Flow :
qA :=
qB :=
V⋅ QA
I
qA = 1.39
kN
m
1 V⋅ QB
2 I
qB = 1.25
kN
m
Ans
Ans
Problem 7-62
The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V
= 150 N, determine the maximum shear flow in the strut.
Given: bf := 60mm
tf := 10mm
b'f := 80mm
dw := 40mm
tw := 10mm
V := 150N
Solution:
Section Property : D := dw + 2tf
A := bf⋅ tf + 2dw⋅ tw + b'f⋅ tf
2
A = 2200 mm
⎯
⎯ Σ ⋅ yi⋅ Ai
y=
Σ ⋅ ( Ai)
(
) y := (b'f⋅ tf)⋅ (0.5tf) + 2(dw⋅ tw)⋅ (0.5dw + tf) + (bf⋅ tf)⋅ (D − 0.5tf)
c
A
yc = 27.73 mm
1
3
2
I'f :=
⋅ b' ⋅ t + b'f⋅ tf ⋅ 0.5tf − yc
12 f f
(
)(
)
1
3
2
⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + tf − yc
12
1
3
2
If :=
⋅ b ⋅ t + bf⋅ tf ⋅ D − 0.5tf − yc
12 f f
(
Iw :=
)(
(
)
)(
)
I := If + 2Iw + I'f
I = 981.9697 × 10
−9
m
4
1
Qmax := bf⋅ tf ⋅ D − 0.5tf − yc + 2⋅ tw⋅ D − yc − tf ⋅ ⋅ D − yc − tf
2
(
)(
)
(
) (
3
Qmax = 21324.38 mm
Shear Flow :
qmax :=
1 V⋅ Qmax
2
I
qmax = 1.63
kN
m
Ans
)
Problem 7-63
The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks.
Given:
L := 125mm
t := 6mm
θ := 45deg
V := 10kN
Solution:
Section Property :
h := L ⋅ cos ( θ )
t
b :=
sin ( θ )
1
3
I :=
⋅ ( 2b) ⋅ h
12
Q = A'⋅ y'c
⎛ 0.5h − y ⎞ ⋅ [ 0.5( 0.5h − y) + y]
⎝ sin ( θ ) ⎠
Q = t⋅ ⎜
Q=
t
2
2
(
0.25h − y )
( )
2 sin θ
Shear Flow:
q=
q=
At y = 0,
V⋅ Q
I
(
( )
2
(
2
V⋅ t
2I⋅ sin θ
⋅ 0.25h − y
2
q = qmax
qmax :=
V⋅ t
2I⋅ sin ( θ )
qmax = 84.85
kN
m
⋅ 0.25h
)
Ans
)
Problem 7-64
The beam is subjected to a shear force of V = 25 kN. Determine the shear flow at points A and B.
Given: bf := 274mm
tf := 12mm
b'f := 250mm
dw := 200mm
t'f := 12mm
tw := 12mm
d'w := 50mm
Solution:
V := 25kN
D := dw + tf
Section Property :
D' := D − d'w
yc :=
(
) (
)(
) (
)(
0.5tf bf⋅ tf + 0.5dw + tf 2tw⋅ dw + D' − 0.5t'f b'f⋅ t'f
bf⋅ tf + 2tw⋅ dw + b'f⋅ t'f
yc = 92.47 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ 0.5dw + tf − yc
12
1
3
2
I3 :=
⋅ b' ⋅ t' + b'f⋅ t'f ⋅ D' − 0.5t'f − yc
12 f f
(
I1 :=
)(
( )
(
(
)
)(
)(
)
)
I := I1 + I2 + I3
( )(
)
QB := ( b'f⋅ t'f) ⋅ ( D' − 0.5t'f − yc)
QA := bf⋅ tf ⋅ yc − 0.5tf
Shear Flow :
qA :=
qB :=
V⋅ QA
2I
V⋅ QB
2I
qA = 65.09
kN
m
Ans
qB = 43.63
kN
m
Ans
)
Problem 7-65
The beam is constructed from four plates and is subjected to a shear force of V = 25 kN. Determine
the maximum shear flow in the cross section.
Given: bf := 274mm
tf := 12mm
b'f := 250mm
dw := 200mm
t'f := 12mm
tw := 12mm
d'w := 50mm
Solution:
V := 25kN
D := dw + tf
Section Property :
D' := D − d'w
yc :=
(
) (
)(
) (
)(
0.5tf bf⋅ tf + 0.5dw + tf 2tw⋅ dw + D' − 0.5t'f b'f⋅ t'f
bf⋅ tf + 2tw⋅ dw + b'f⋅ t'f
yc = 92.47 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ 2tw ⋅ dw + 2tw⋅ dw ⋅ 0.5dw + tf − yc
12
1
3
2
I3 :=
⋅ b' ⋅ t' + b'f⋅ t'f ⋅ D' − 0.5t'f − yc
12 f f
(
I1 :=
( )
)(
)
(
(
)(
)
)(
)
)
⎛ yc − tf ⎞
⋅ (y − t )
⎝ 2 ⎠ c f
I := I1 + I2 + I3
(
)(
Qmax := bf⋅ tf ⋅ yc − 0.5tf + 2tw⋅ ⎜
Maximum Shear Flow :
qmax :=
V⋅ Qmax
2I
qmax = 82.88
kN
m
Ans
)
Problem 7-66
A shear force of V = 18 kN is applied to the box girder. Determine the position d of the stiffener plates
BE and FG so that the shear flow at A is twice as great as the shear flow at B. Use the centerline
dimensions for the calculation. All plates are 10 mm thick.
Given:
t := 10mm
bf := 135mm − t
V := 18kN
dw := t + 290mm
Solution:
Section Property :
( )(
QA := bf⋅ t ⋅ 0.5dw − 0.5t
)
3
QA = 181250 mm
( )
QB = bf⋅ t ⋅ ( d)
Shear Flow :
qA =
1 V⋅ QA
2 I
qB =
Require, qA = 2qB
⎛ 1 V⋅ QB ⎞
1 V⋅ QA
= 2⋅ ⎜
2
I
⎝2 I ⎠
QA = 2QB
( )
QA = 2 bf⋅ t ⋅ ( d)
d :=
QA
2bf⋅ t
d = 72.50 mm
Ans
1 V⋅ QB
2 I
Problem 7-67
The pipe is subjected to a shear force of V = 40 kN. Determine the shear flow in the pipe at points A
and B.
ri := 150mm
Given:
t := 5mm
V := 40kN
Solution:
Section Property :
I :=
ro := ri + t
π ⎛ 4
4
⋅ r − ri ⎞⎠
4 ⎝o
Since a' -> 0, then
QA := 0
4ro ⎛ π ⋅ ro ⎞
2
2
4ri ⎛ π ⋅ ri ⎞
⎜
⎜
QB :=
−
3π ⎝ 2 ⎠ 3π ⎝ 2 ⎠
Shear Flow :
qA :=
qB :=
V⋅ QA
I
V⋅ QB
2I
qA = 0.00
kN
m
Ans
kN
m
Ans
qB = 83.48
Problem 7-68
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown where b2 > b1. The member segments have the same thickness t.
Problem 7-69
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.The member segments have the same thickness t.
Problem 7-70
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-71
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-72
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-73
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-74
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Given:
dθ := 150mm
dv := 150mm
θ := 30deg
Set
Solution:
t := 1mm
Section Property :
hθ := dθ ⋅ sin ( θ )
t
bθ :=
sin ( θ )
⎛ dv + hθ ⎞
1
3
Iθ :=
⋅ bθ ⋅ hθ + dθ ⋅ t ⋅ ⎜
12
⎝ 2 ⎠
(
Iv :=
)
2
1
3
⋅ t⋅ dv
12
I := 2Iθ + Iv
y'c = 0.5dv + hθ − 0.5x⋅ sin ( θ )
Q = A'⋅ y'c
(
Q = ( x⋅ t) ⋅ 0.5dv + hθ − 0.5x⋅ sin ( θ )
)
Shear Flow Resultant:
q=
V⋅ Q
I
V= P
q
x⋅ t
=
⋅ 0.5dv + hθ − 0.5x⋅ sin ( θ )
P
I
(
d
)
⌠ θ x⋅ t
⋅ 0.5dv + hθ − 0.5x⋅ sin ( θ ) dx
= ⎮
P
I
⎮
⌡
F1
(
)
0
Shear Center: Summing moment about point A
P⋅ e = F1⋅ dv⋅ cos ( θ )
e=
F1
P
⋅ dv⋅ cos ( θ )
d
⌠ θ
⎛ x⋅ t ⎞ ⋅ 0.5d + h − 0.5x⋅ sin ( θ ) dx
e := ⎮ dv⋅ cos ( θ ) ⋅ ⎜
v
θ
⎮
⎝ I ⎠
⌡
(
0
e = 43.30 mm
Ans
)
Problem 7-75
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its
side.
a := 100mm
Given:
b := 100mm
Solution:
Set
t := mm
P := kN
Section Property :
h := 2a
1
3
2
⋅ ( 2t) ⋅ h + 2( b⋅ t) ⋅ a
12
1
Q1 = ( y⋅ t) ⋅ ( y)
2
1
Q2 = ( a⋅ t) ⋅ ( a) + ( x⋅ t) ⋅ ( a)
2
3
I :=
I = 3.3333 a t
t 2
Q1 = y
2
a⋅ t
Q1 =
( a + 2x)
2
Shear Flow Resultant:
2
V⋅ Q1
P ⋅ t⋅ y
q1 =
q1 =
I
2I
q2 =
V⋅ Q2
q2 =
I
P⋅ ( a⋅ t) ⋅ ( a + 2x)
2I
a
a
⌠
Fw = ⎮ q1 dy
⌡
0
b
⌠
Ff = ⎮ q2 dx
⌡
0
⌠
2
⎮ P ⋅ t⋅ y
dy
Fw := ⎮
2I
⌡
Fw = 0.05 P
⌠
Ff := ⎮
⎮
⌡
Ff = 0.3 P
0
b
P⋅ ( a⋅ t) ⋅ ( a + 2x)
dx
2I
0
Shear Center: Summing moment about point A
P⋅ e = 2Fw⋅ b + Ff⋅ h
e :=
1
⋅ 2Fw⋅ b + Ff⋅ h
P
(
e = 70 mm
Ans
)
Problem 7-76
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its
side. Each element has aconstant thickness t.
a := mm
Given:
b := a
Solution:
Set
t := mm
P := kN
Section Property :
h := 2a
1
3
2
⋅ ( 2t) ⋅ h + 2( b⋅ t) ⋅ a
12
1
Q1 = ( y⋅ t) ⋅ ( y)
2
1
Q2 = ( a⋅ t) ⋅ ( a) + ( x⋅ t) ⋅ ( a)
2
3
I :=
I = 3.3333 a t
t 2
Q1 = y
2
a⋅ t
Q1 =
( a + 2x)
2
Shear Flow Resultant:
2
V⋅ Q1
P ⋅ t⋅ y
q1 =
q1 =
I
2I
q2 =
V⋅ Q2
q2 =
I
P⋅ ( a⋅ t) ⋅ ( a + 2x)
2I
a
a
⌠
Fw = ⎮ q1 dy
⌡
0
⌠
2
⎮ P ⋅ t⋅ y
dy
Fw := ⎮
2I
⌡
0
Fw = 0.05 P
b
⌠
Ff := ⎮
⎮
⌡
⌠
Ff = ⎮ q2 dx
⌡
0
b
P⋅ ( a⋅ t) ⋅ ( a + 2x)
dx
2I
0
Ff = 0.3 P
Shear Center: Summing moment about point A
P⋅ e = 2Fw⋅ b + Ff⋅ h
e :=
1
⋅ 2Fw⋅ b + Ff⋅ h
P
(
e = 0.7 a
Ans
)
Problem 7-77
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
a := mm
Given:
θ := 60deg
Solution:
Set
t := mm
P := kN
Section Property :
b := a⋅ sin ( θ )
t' :=
b = 0.86603 a
t
t' = 2 t
cos ( θ )
h :=
a
2
1
1
3
3
3
⋅ ( t) ⋅ a +
( t') ⋅ a
I = 0.25 a t
12
12
1
t' 2
Q1 = ( y'⋅ t') ⋅ ( y')
Q1 = y'
2
2
1
⎤
⎡ 1
Q2 = ( h⋅ t') ⋅ ( h) + ( h − y) ⋅ t⎢y + ( h − y)⎥
2
⎣ 2
⎦
1⎡ 2
2
2
Q2 = ⎣t' h + t h − y ⎤⎦
2
I :=
(
)
Shear Flow Resultant:
2
V⋅ Q1
P⋅ t'⋅ y'
q1 =
q1 =
I
2I
q2 =
V⋅ Q2
I
(2
)
P⋅ ⎡⎣t' h + t h − y ⎤⎦
2I
2
q2 =
⌠
⎮
F' := ⎮
⌡
h
⌠
F' = ⎮ q1 dy
⌡
0
h
0
h
⌠
⎮
F := ⎮
⌡
h
⌠
F = ⎮ q2 dy
⌡
0
2
2
P⋅ t'⋅ y'
dy'
2I
(2
2
Shear Center: Summing moment about point A
e :=
1
⋅ ( 2F⋅ b)
P
e = 1.1547 a
Ans
)
P⋅ ⎡⎣t' h + t h − y ⎤⎦
dy
2I
0
P⋅ e = 2( F⋅ b + F'⋅ 0)
F' = 0.1667 P
2
F = 0.6667 P
Problem 7-78
If the angle has a thickness of 3 mm, a height h = 100 mm, and it is subjected to a shear of V = 50 N,
determine the shear flow at point A and the maximum shear flow in the angle.
Given:
h := 100mm
t := 3mm
θ := 45deg
V := 50N
Solution:
Section Property :
t
t' :=
t' = 4.2426 mm
sin ( θ )
1
3
I :=
⋅ ( 2t') ⋅ h
12
Q = A'⋅ y'c
⎤
⎡1
Q = t'⋅ ( 0.5h − y) ⋅ ⎢ ( 0.5h − y) + y⎥
2
⎣
⎦
Q=
(
t'
2
2
0.25h − y
2
Shear Flow:
)
q=
V⋅ Q
I
q=
V⋅ t'
2
2
⋅ 0.25h − y
2I
(
At A,
yA := 0.5 ⋅ h
At y = 0,
q = qmax
qmax :=
qA := 0
(
V⋅ t'
2
⋅ 0.25h
2I
qmax = 375
N
m
)
Ans
)
Ans
Problem 7-79
The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks. The thickness is 6 mm and the legs (AB) are 125 mm.
Given:
L := 125mm
t := 6mm
θ := 45deg
V := 10kN
Solution:
Section Property :
h := L ⋅ cos ( θ )
I :=
t
t' :=
t' = 8.4853 mm
sin ( θ )
1
3
⋅ ( 2t') ⋅ h
12
Q = A'⋅ y'c
⎤
⎡1
Q = t'⋅ ( 0.5h − y) ⋅ ⎢ ( 0.5h − y) + y⎥
2
⎣
⎦
Q=
(
t'
2
2
0.25h − y
2
Shear Flow:
At y = 0,
)
q=
V⋅ Q
I
q=
V⋅ t'
2
2
⋅ 0.25h − y
2I
(
)
q = qmax
qmax :=
(
V⋅ t'
2
⋅ 0.25h
2I
qmax = 84.85
kN
m
)
Ans
Ans
Problem 7-80
Determine the placement e for the force P so that the beam bends downward without twisting. Take
h = 200 mm.
Given: h1 := 100mm
bf := 300mm
h2 := 200mm
Solution:
Set
t := mm
P := kN
Section Property :
I :=
1 ⎛
3
3
3
⋅ ⎝ bf⋅ t + t⋅ h1 + t⋅ h2 ⎞⎠
12
⎤
⎡ 1
Qw2 = t⋅ 0.5h2 − y ⋅ ⎢y + 0.5h2 − y ⎥
2
(
)⎣
(
)⎦
t
2
2
Qw2 = ⋅ ⎛⎝ 0.25h2 − y ⎞⎠
2
Shear Flow Resultant:
qw2 =
V⋅ Qw2
qw2 =
I
P⋅ t⋅ ⎛ 0.25h2 − y ⎞
2
⎝
2I
0.5h
2
⌠
⎮
Fw2 =
qw2 dx
⎮
⌡− 0.5h
2
0.5h
2
⌠
⎮
Fw2 := ⎮
⎮
⌡− 0.5h
P⋅ t⋅ ⎛ 0.25h2 − y ⎞
2
⎝
2I
2
⎠ dy
2
Fw2 = 0.8889 P
Shear Center: Summing moment about point A
(
P⋅ e = Fw2⋅ bf + t
e :=
)
1
⋅ ⎡F ⋅ b + t ⎤⎦
P ⎣ w2 f
(
e = 267.5 mm
)
Ans
2
⎠
Problem 7-81
A force P is applied to the web of the beam as shown. If e = 250 mm, determine the height h of the
right flange so that the beam will deflect downward without twisting. The member segments have the
same thickness t.
Given: h1 := 100mm
bf := 300mm
e := 250mm
Solution:
Set
t := mm
P := kN
Shear Center: Summing moment about point A
(
P⋅ e = Fw2⋅ bf + t
)
Assume bf+t equal to bf :
Section Property :
I=
Fw2 :=
e
⋅P
bf + t
Fw2 :=
e
⋅P
bf
Assume bf t3 negligible.
1 ⎛
3
3
⋅ t⋅ h + t⋅ h2 ⎞⎠
12 ⎝ 1
⎤
⎡ 1
Qw2 = t⋅ 0.5h2 − y ⋅ ⎢y + 0.5h2 − y ⎥
2
(
)⎣
(
)⎦
t
2
2
Qw2 = ⋅ ⎛⎝ 0.25h2 − y ⎞⎠
2
Shear Flow Resultant:
qw2 =
V⋅ Qw2
I
qw2 =
P⋅ t⋅ ⎛ 0.25h2 − y ⎞
2
⎝
2
⎠
2I
0.5h
0.5h 2
⌠
Fw2 = ⎮
qw2 dx
⎮
⌡− 0.5h
2
2
⌠
⎮
e
⋅P = ⎮
bf
⎮
⌡− 0.5h
6P⋅ t⋅ ⎛ 0.25h2 − y ⎞
2
⎝
3
t⋅ h1 + t⋅ h2
2
3
⎠ dy
2
Given
e⋅ ⎛ h1 + h2 ⎞
⎝
⎠ =⌠
⎮
⎮
6bf
⌡
3
3
0.5h 2
− 0.5h 2
Guess h2 := 10mm
⎛0.25h 2 − y2⎞ dy
2
⎝
⎠
( )
h2 := Find h2
h2 = 171.0 mm
Ans
Problem 7-82
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Problem 7-83
Determine the location e of the shear center, point O, for the tube having a slit along its length.
Problem 7-84
The beam is fabricated from four boards nailed together as shown. Determine the shear force each nail
along the sides C and the top D must resist if the nails are uniformly spaced s = 75 mm. The beam is
subjected to a shear of V = 22.5 kN.
Given:
bf := 250mm
dw := 300mm
tf := 25mm
tw := 25mm
tb := 25mm
db := 100mm
V := 22.5kN
sn := 75mm
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
(
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw + 0.5db 2tb⋅ db
bf⋅ tf + dw⋅ tw + 2 tb⋅ db
(
)
)
yc = 87.50 mm
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc
12
1
3
2
I2 :=
⋅ t ⋅ d + dw⋅ tw ⋅ 0.5dw + tf − yc
12 w w
1
3
2
I3 :=
⋅ 2tb ⋅ db + 2tb⋅ db ⋅ 0.5db − yc
12
(
I1 :=
)(
(
( )
)
)(
(
)
)(
)
I := I1 + I2 + I3
( )(
)
QD := ( dw⋅ tw) ⋅ ( D − yc − 0.5dw)
QC := tb⋅ db ⋅ yc − 0.5db
V⋅ Q
I
The allowable shear flow at points C and D are :
q=
Shear Flow :
FC
sn
=
FC :=
V⋅ QC
FD
I
sn
V⋅ QC⋅ sn
I
FC = 0.987 kN
=
FD :=
V⋅ QD
qC =
FC
sn
I
V⋅ QD⋅ sn
I
FD = 6.906 kN
Ans
qD =
FD
sn
Problem 7-85
The beam is constructed from four boards glued together at their seams. If the glue can withstand 15
kN/m, what is the maximum vertical shear V that the beam can support?
Given:
bf := 100mm
dw := 249mm
tf := 12mm
tw := 12mm
qallow := 15
di := 75mm
kN
m
Solution:
Section Property :
⎛ di + tf ⎞
2
2
3
3
I :=
⋅ tw⋅ dw +
⋅ bf⋅ tf + 2bf⋅ tf⋅ ⎜
12
12
⎝ 2 ⎠
(
)
Q := bf⋅ tf ⋅
di + t f
2
Shear Flow : Since there are two glue joints, hence
Vmax :=
2
2I⋅ qallow
Q
Vmax = 20.37 kN
Ans
2q =
V⋅ Q
I
Problem 7-86
Solve Prob. 7-85 if the beam is rotated 90° from the position shown.
Given:
bf := 100mm
dw := 249mm
tf := 12mm
tw := 12mm
qallow := 15
di := 75mm
kN
m
Solution:
Section Property :
⎛ bf + tw ⎞
2
2
3
3
I :=
⋅ tf⋅ bf +
⋅ dw⋅ tw + 2dw⋅ tw⋅ ⎜
12
12
⎝ 2 ⎠
(
)
Q := dw⋅ tw ⋅
bf + t w
2
Shear Flow : Since there are two glue joints, hence
Vmax :=
2
2I⋅ qallow
Q
Vmax = 3.731 kN
Ans
2q =
V⋅ Q
I
Problem 7-87
The member is subjected to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C.
The thickness of each thin-walled segment is 15 mm.
Given:
bf := 200mm
dw := 300mm
tf := 15mm
tw := 15mm
tb := 15mm
db := 115mm
V := 2kN
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
(
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw + 0.5db 2tb⋅ db
bf⋅ tf + dw⋅ tw + 2 tb⋅ db
(
)
)
yc = 87.98 mm
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc
12
1
3
2
I2 :=
⋅ t ⋅ d + dw⋅ tw ⋅ 0.5dw + tf − yc
12 w w
1
3
2
I3 :=
⋅ 2tb ⋅ db + 2tb⋅ db ⋅ 0.5db − yc
12
(
I1 :=
)(
(
)
)(
( )
(
)
)(
)
I := I1 + I2 + I3
4
I = 86939045.38 mm
QA := 0
(
)(
)
QC := QB + tf⋅ ( 0.5bf − 0.5tw) ⋅ ( yc − 0.5tf)
3
QB := tb⋅ db ⋅ yc − 0.5db
Shear Flow :
q=
qC :=
V⋅ QB
I
V⋅ QC
I
3
QC = 164242.29 mm
V⋅ Q
I
qA := 0
qB :=
QB = 52577.05 mm
Ans
qB = 1.210
kN
m
Ans
qC = 3.778
kN
m
Ans
Problem 7-88
The member is subjected to a shear force of V = 2 kN. Determine the maximum shear flow in the
member. All segments of the cross section are 15 mm thick.
Given:
bf := 200mm
dw := 300mm
tf := 15mm
tw := 15mm
tb := 15mm
db := 115mm
V := 2kN
Solution:
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
(
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw + 0.5db 2tb⋅ db
bf⋅ tf + dw⋅ tw + 2 tb⋅ db
(
)
)
yc = 87.98 mm
1
3
2
⋅ bf⋅ tf + bf⋅ tf ⋅ 0.5tf − yc
12
1
3
2
I2 :=
⋅ t ⋅ d + dw⋅ tw ⋅ 0.5dw + tf − yc
12 w w
1
3
2
I3 :=
⋅ 2tb ⋅ db + 2tb⋅ db ⋅ 0.5db − yc
12
(
I1 :=
)(
(
)
)(
( )
(
)
)(
)
I := I1 + I2 + I3
4
I = 86939045.38 mm
1
Qmax := tw⋅ D − yc ⋅ ⋅ D − yc
2
(
) (
)
3
Qmax = 386537.47 mm
V⋅ Q
I
Maximum shear flow occurs at the point where the neutral axis passes through the section.
Shear Flow :
qmax :=
q=
V⋅ Qmax
I
qmax = 8.892
kN
m
Ans
Problem 7-89
The beam is made from three thin plates welded together as shown. If it is subjected to a shear of
V = 48 kN, determine the shear flow at points A and B. Also, calculate the maximum shear stress in
the beam.
Given: bf := 215mm
tf := 15mm
tw := 15mm
dw := 315mm
V := 48kN
h := 200mm
Solution:
(
)
⎡⎣( bf − tw) ⋅ tf⎤⎦ ( h + 0.5tf) + ( dw⋅ tw) ( 0.5dw)
yc :=
(bf − tw)⋅ tf + dw⋅ tw
Section Property : a := 0.5 bf − tw
yc = 176.92 mm
If :=
1
3
2
⋅ b − t ⋅ t + bf − tw ⋅ tf⋅ h + 0.5 ⋅ tf − yc
12 f w f
Iw :=
1
3
2
⋅ t ⋅ d + tw⋅ dw ⋅ yc − 0.5dw
12 w w
(
)
(
(
I := If + Iw
) (
)(
)
)
I = 43.71347 × 10
−6
m
( )(
)
QB := ( a⋅ tf) ⋅ ( h + 0.5 ⋅ tf − yc)
1
Qmax := tw⋅ ( yc) ⋅ ⋅ ( yc)
2
3
QA := a⋅ tw ⋅ dw − yc − 0.5a
q=
Shear Flow :
qA :=
qB :=
4
QA = 132123.79 mm
3
QB = 45873.79 mm
3
Qmax = 234748.45 mm
V⋅ Q
I
V⋅ QA
I
V⋅ QB
I
Maximum Shear Stress:
qA = 145.1
kN
m
Ans
qB = 50.37
kN
m
Ans
τ=
V⋅ Q
I⋅ b
Maximum shear stress occurs at the point where the neutral axis passes through the section.
τ max :=
V⋅ Qmax
( )
I ⋅ tw
τ max = 17.18 MPa
Ans
Problem 7-90
A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjected
to a shear force of V = 1.25 kN, determine the shear stress at points A and C. Indicate the results on
volume elements located at these points.
Given:
bf := 100mm
dw := 50mm
t := 6mm
V := 1.25kN
b'f := 25mm
Solution:
D := dw + 2t
Section Property :
yc :=
(
) (
)(
)
( )
0.5t 2b'f⋅ t + 0.5dw + t 2dw⋅ t + ( D − 0.5t) bf⋅ t
2b'f⋅ t + 2dw⋅ t + bf⋅ t
yc = 36.60 mm
1
3
2
⋅ b'f⋅ t + b'f⋅ t ⋅ 0.5t − yc
12
1
3
2
I2 :=
⋅ t⋅ dw + dw⋅ t ⋅ 0.5dw + t − yc
12
1
3
2
I3 :=
⋅ bf⋅ t + bf⋅ t ⋅ D − 0.5t − yc
12
( )(
I1 :=
(
)
)(
( )(
)
)
I := 2I1 + 2I2 + I3
( )(
)
mm
QC := ( b'f⋅ t) ⋅ ( yc − 0.5t) + ( dw⋅ t) ⋅ 0.5dw + t − yc − ( 0.5bf⋅ t) ⋅ ( D − 0.5t − yc)
QA := b'f⋅ t ⋅ yc − 0.5t
3
QC = 0.00 mm
Shear Stress :
τ A :=
τ C :=
V⋅ QA
I⋅ t
V⋅ QC
I⋅ t
τ=
(since A' = 0)
V⋅ Q
I⋅ t
τ A = 1.335 MPa
Ans
τ C = 0 MPa
Ans
mm
Problem 7-91
A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjected
to a shear force of V = 1.25 kN, determine the shear stress at point B.
Given:
bf := 100mm
dw := 50mm
t := 6mm
V := 1.25kN
b'f := 25mm
Solution:
D := dw + 2t
Section Property :
yc :=
(
) (
)(
)
( )
0.5t 2b'f⋅ t + 0.5dw + t 2dw⋅ t + ( D − 0.5t) bf⋅ t
2b'f⋅ t + 2dw⋅ t + bf⋅ t
yc = 36.60 mm
1
3
2
⋅ b'f⋅ t + b'f⋅ t ⋅ 0.5t − yc
12
1
3
2
I2 :=
⋅ t⋅ dw + dw⋅ t ⋅ 0.5dw + t − yc
12
1
3
2
I3 :=
⋅ bf⋅ t + bf⋅ t ⋅ D − 0.5t − yc
12
( )(
I1 :=
(
)
)(
)
( )(
)
I := 2I1 + 2I2 + I3
( )(
)
QB := bf⋅ t ⋅ D − 0.5t − yc
Shear Stress :
τ B :=
V⋅ QB
I⋅ ( 2t)
τ=
V⋅ Q
I⋅ b
τ B = 1.781 MPa
Ans
Problem 7-92
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Problem 7-93
Sketch the intensity of the shear-stress distribution acting over the beam's cross-sectional area, and
determine the resultant shear force acting on the segment AB. The shear acting at the section is V =
175 kN. Show that INA = 340.82(106) mm4.
Given: b1 := 200mm
d1 := 200mm
V := 175kN
b2 := 50mm
d2 := 150mm
Solution:
Section Property :
yc :=
(
D := d1 + d2
) (
)(
)
0.5d1 b1⋅ d1 + 0.5d2 + d1 b2⋅ d2
b1⋅ d1 + b2⋅ d2
yc = 127.63 mm
y'c := D − yc
1
3
2
⋅ b1⋅ d1 + b1⋅ d1 ⋅ 0.5d1 − yc
12
1
3
2
I2 :=
⋅ b ⋅ d + b2⋅ d2 ⋅ D − 0.5d2 − yc
12 2 2
I1 :=
(
)(
(
)(
)
)
6
I := I1 + I2
4
I = 340.82 × 10 mm
(
)
(
)
Q1 = 0.5 ( yc − y1) ⋅ b1( yc + y1)
y2c = 0.5 ( y'c − y2) + y2
Q2 = 0.5 ( y'c − y2) ⋅ b2( y'c + y2)
A'1 = yc − y1 ⋅ b1
y1c = 0.5 yc − y1 + y1
Q1 = A'1⋅ y'1c
(
)
A'2 = y'c − y2 ⋅ b2
Q2 = A'2⋅ y'2c
τ=
Shear Stress :
(Q.E.D)
V⋅ Q
I⋅ b
(
y1c = 0.5 yc + y1
)
Q1 = 0.5b1⎛⎝ yc − y1 ⎞⎠
2
(
y2c = 0.5 y'c + y2
2
)
Q2 = 0.5b2⎛⎝ y'c − y2 ⎞⎠
2
2
⎛ V ⎞ ⋅ ⎡0.5⎛y 2 − y 2⎞⎤
1 ⎠⎦
⎝I⎠ ⎣ ⎝ c
τ CB = ⎜
⎛ V ⎞ ⋅ ⎛ 0.5y 2⎞
⎝I⎠ ⎝ c ⎠
τ 1B := ⎜
At B: y1 = 0
τ 1B = 4.18 MPa
At C: y1 := yc − d1
⎛ V ⎞ ⋅ ⎡0.5⎛y 2 − y 2⎞⎤
1 ⎠⎦
⎝I⎠ ⎣ ⎝ c
τ 1C := ⎜
τ 1C = 2.84 MPa
⎛ V ⎞ ⋅ ⎡0.5⎛ y' 2 − y 2⎞⎤
2 ⎠⎦
⎣ ⎝ c
⎝ INA ⎠
V
2
2
At C: y2 := y'c − d2 τ 2C := ⎛⎜ ⎞ ⋅ ⎡0.5 ⎛ y'c − y2 ⎞⎤
⎣
⎝
⎠⎦
⎝I⎠
τ AB = ⎜
τ 2C = 11.35 MPa
Resultant Shear Force:
For segment AB.
y'
⌠ c ⎛ V⎞
2
2
VAB := ⎮ ⎜
⋅ ⎡⎣0.5 ⎛⎝ y'c − y ⎞⎠⎤⎦ ⋅ b2 dy
⎮ ⎝I⎠
⌡y
o
VAB = 49.78 kN
Ans
yo := y'c − d2
⌠
VAB = ⎮ τ AB dA
⌡
A
Problem 8-1
A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p =
300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa.
Given:
r := 1.5m
p := 0.3MPa
Solution:
Normal Stress :
σ allow =
t :=
σ allow := 12MPa
p⋅ r
2⋅ t
p⋅ r
2⋅ σ allow
t = 18.75 mm
Ans
Problem 8-2
A pressurized spherical tank is to be made of 125mm-thick steel. If it is subjected to an internal
pressure of p = 1.4 MPa, determine its outer radius if the maximum normal stress is not to exceed 105
MPa.
Given:
t := 125mm
p := 1.4MPa
Solution:
Normal Stress :
σ=
ri :=
σ allow := 105MPa
p⋅ r
2⋅ t
2⋅ tσ allow
p
ri = 18.750 m
ro := ri + t
ro = 18.875 m
Ans
Problem 8-3
The thin-walled cylinder can be supported in one of two ways as shown. Determine the state of stress
in the wall of the cylinder for both cases if the piston P causes the internal pressure to be 0.5 MPa.
The wall has a thickness of 6 mm and the inner diameter of the cylinder is 200 mm.
Given:
t := 6mm
p := 0.5MPa
ri := 200mm
Solution:
Case (a) :
Hoop Stress :
σ 1 :=
p⋅ ri
t
Normal Stress :
Case (b) :
Hoop Stress :
σ 1 :=
Normal Stress :
σ 2 :=
p⋅ ri
t
p⋅ ri
2⋅ t
σ 1 = 16.67 MPa
Ans
σ 2 := 0
Ans
σ 1 = 16.67 MPa
Ans
σ 2 = 8.33 MPa
Ans
Problem 8-4
The tank of the air compressor is subjected to an internal pressure of 0.63 MPa. If the internal diameter
of the tank is 550 mm, and the wall thickness is 6 mm, determine the stress components acting at point
A. Draw a volume element of the material at this point, and show the results on the element.
Given:
t := 6mm
p := 0.63MPa
di := 550mm
Solution:
ri := 0.5di
α :=
ri
α = 45.83
t
Since α > 10. then thin-wall analysis can be used.
Hoop Stress :
σ 1 :=
p⋅ ri
t
σ 1 = 28.88 MPa
Ans
σ 2 = 14.44 MPa
Ans
Longitudinal Stress :
σ 2 :=
p⋅ ri
2⋅ t
Problem 8-5
The open-ended pipe has a wall thickness of 2 mm and an internal diameter of 40 mm. Calculate the
pressure that ice exerted on the interior wall of the pipe to cause it to burst in the manner shown. The
maximum stress that the material can support at freezing temperatures is σmax = 360 MPa. Show the
stress acting on a small element of material just before the pipe fails.
Given:
t := 2mm
σ allow := 360MPa
di := 40mm
Solution:
ri := 0.5di
α :=
ri
α = 10.00
t
Since α > 10. then thin-wall analysis can be used.
Hoop Stress :
σ 1 := σ allow
σ1 =
p⋅ r i
t
p :=
σ allow⋅ t
ri
p = 36.0 MPa
Longitudinal Stress :
Since the pipe is open at both neds, then
σ 2 := 0
Ans
Ans
Problem 8-6
The open-ended polyvinyl chloride pipe has an inner diameter of 100 mm and thickness of 5 mm. If it
carries flowing water at 0.42 MPa pressure, determine the state of stress in the walls of the pipe.
Given:
t := 5mm
p := 0.42MPa
di := 100mm
Solution:
ri := 0.5di
Hoop Stress :
Normal Stress :
σ 1 :=
p⋅ ri
t
σ 1 = 4.2 MPa
Ans
σ 2 := 0
Ans
There is no stress componenet in the longitudinal direction since pipe has open ends.
Problem 8-7
If the flow of water within the pipe in Prob. 8-6 is stopped due to the closing of a valve, determine the
state of stress in the walls of the pipe. Neglect the weight of the water. Assume the supports only exert
vertical forces on the pipe.
Given:
t := 5mm
p := 0.42MPa
di := 100mm
Solution:
ri := 0.5di
Hoop Stress :
σ 1 :=
Normal Stress :
σ 2 :=
p⋅ ri
t
p⋅ ri
2⋅ t
σ 1 = 4.2 MPa
Ans
σ 2 = 2.1 MPa
Ans
Problem 8-8
The A-36-steel band is 50 mm wide and is secured around the smooth rigid cylinder. If the bolts are
tightened so that the tension in them is 2 kN, determine the normal stress in the band, the pressure
exerted on the cylinder, and the distance half the band stretches.
Given:
t := 3mm
b := 50mm
r := 200mm
F := 2kN
E := 200GPa
Solution:
rb := r + 0.5t
L b := π ⋅ rb
Tensile Stress in the Band :
σ 1 :=
F
b⋅ t
σ=
Hoop Stress :
p :=
σ 1 = 13.33 MPa
tσ 1
rb
Ans
p⋅ r
t
p = 0.199 MPa
Ans
Stectch :
δ = ε 1⋅ Lb
δ :=
σ 1⋅ L b
E
ε1 =
σ1
E
δ = 0.0422 mm
Ans
Problem 8-9
The 304 stainless steel band initially fits snugly around the smooth rigid cylinder. If the band is then
subjected to a nonlinear temperature drop of ∆T = 12 sin2θ °C, where θ is in radians, determine the
circumferential stress in the band.
Unit used:
°C := deg
Given: t := 0.4mm
b := 25mm
r' := 250mm
E := 193GPa
∆ T = 12⋅ sin ( θ )
( − 6) 1
2
α := 17⋅ 10
°C
Solution:
Compatibility: Since the band is fitted to a rigid cylinder
(which does not deform under load), then
δF − δT = 0
P⋅ ( 2π ⋅ r) ⌠
−⎮
A⋅ E
⌡
2π
( )
α ⋅ ∆ T ⋅ r dθ = 0
0
2π
⌠
2π ⋅ r ⎛ P ⎞
⋅⎜
− 12 ⎮
⌡0
E ⎝ A⎠
However,
2
α ⋅ sin ( θ ) ⋅ r dθ = 0
P
= σc
A
2π
⌠
2π ⋅ r
⋅ σ c = 12⋅ α ⋅ r ⎮
⌡0
E
⎛ 2π
6⋅ α E ⎜ ⌠
σ c :=
⋅ ⎮
π ⎜ ⌡0
⎝
σ c = 19.69 MPa
2
sin ( θ ) dθ
⎞
2
sin ( θ ) dθ ⋅ °C
⎠
Ans
Problem 8-10
The barrel is filled to the top with water. Determine the distance s that the top hoop should be placed
from the bottom hoop so that the tensile force in each hoop is the same. Also, what is the force in each
hoop? The barrel has an inner diameter of 1.2 m. Neglect its wall thickness. Assume that only the
hoops resist the water pressure. Note: Water develops pressure in the barrel according to Pascal's law,
p = (0.01z) MPa, where z is the depth from the surface of the water in meter.
Given:
d := 1.2m
p = 0.01z⋅ MPa
h := 2.4m
h' := 0.6m
r := 0.5d
Solution:
h
⌠
P = ⎮ p⋅ ( 2r) dz
⌡0
⎛⎜ ⌠
⎞ MPa
P := 2r⋅ ⎮ 0.01z dz ⋅
⎜ ⌡0
m
h
⎝
⎠
P = 34.56 kN
Equilibrium for the Steel Hoop :
ΣF y=0;
P − 4F = 0
ΣΜBase=0;
P⋅
F := 0.25P
h
− 2F⋅ h' − 2F⋅ ( h' + s) = 0
3
P⋅ h
s :=
− h' − h'
6F
s = 400 mm
Ans
F = 8.64 kN
Problem 8-11
A wood pipe having an inner diameter of 0.9 m is bound together using steel hoops having a
cross-sectional area of 125 mm2. If the allowable stress for the hoops is σallow = 84 MPa, determine
their maximum spacing s along the section of pipe so that the pipe can resist an internal gauge pressure
of 28 kPa. Assume each hoop supports the pressure loading acting along the length s of the pipe.
Given:
( − 3)⋅ MPa
d := 0.9m
p := 28⋅ 10
2
As := 125mm
Solution:
σ allow := 84MPa
r := 0.5d
P = p⋅ ( 2r⋅ s)
( )
F = σ allow⋅ As
Equilibrium for the Steel Hoop :
From the FBD,
+ ΣF y=0;
P − 2F = 0
( )
p⋅ ( 2r⋅ s) − 2σ allow⋅ As = 0
s :=
( )
σ allow⋅ As
p⋅ r
s = 833.33 mm
Ans
Problem 8-12
A boiler is constructed of 8-mm thick steel plates that are fastened together at their ends using a butt
joint consisting of two 8-mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm
apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress
in the boiler's plate apart from the seam, (b) the circumferential stress in the outer cover plate along the
rivet line a-a, and (c) the shear stress in the rivets.
Given:
to := 8mm
ri := 750mm
p := 1.35MPa
tc := 8mm
db := 10mm
s := 50mm
Solution:
a) Hoop Stress :
p⋅ ri
σ 1 :=
to
b)
σ 1 = 126.6 MPa
Ans
Hoop Stress in cover plate along line a-a :
Consider a width of s (mm), Fo (in boiler plate) = F'o (in cover plates)
( )
(
)( )
σ 1⋅ s⋅ to = σ' 1⋅ s − db ⋅ 2⋅ tc
σ' 1 :=
( )
(s − db)⋅ (2⋅ tc)
σ 1⋅ s⋅ to
σ' 1 = 79.1 MPa
c)
Ans
Shear Stress in Rivet :
From the FBD,
+
ΣF y=0;
τ avg :=
rb := 0.5db
( )
Fb := σ 1⋅ ( s⋅ to)
Fb − σ 1⋅ s⋅ to = 0
1 ⎛⎜ Fb ⎞
⋅
2
2 ⎜
⎝ π ⋅ rb ⎠
τ avg = 322.3 MPa
Ans
Problem 8-13
The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with a
pressure p. Determine the change in the internal radius of the ring after this pressure is applied. The
modulus of elasticity for the ring is E.
Solution:
Equilibrium for the Ring :
+ ΣF y=0;
2P − 2p⋅ ri⋅ w = 0
P = p⋅ r i ⋅ w
Hoop Stress and Strain for the Ring :
P
A
σ1 =
σ1 =
p⋅ ri⋅ w
σ1 =
( r o − r i) ⋅ w
p⋅ r i
ro − ri
Using Hooke's Law,
ε1 =
σ1
ε1 =
E
1
2⋅ π ⋅ r
ε1 =
Then, from Eq.(1)
δri
r
=
δri =
p⋅ r i
( r o − r i) ⋅ E
p⋅ r i
2
( r o − r i) ⋅ E
(1)
( r o − r i) ⋅ E
However,
2⋅ π ⋅ r i − 2⋅ π ⋅ r i
ε1 =
p⋅ r i
Ans
ri − ri
1
r
ε1 =
δri
r
Problem 8-14
A closed-ended pressure vessel is fabricated by cross-winding glass filaments over a mandrel, so that
the wall thickness t of the vessel is composed entirely of filament and an epoxy binder as shown in the
figure. Consider a segment of the vessel of width w and wrapped at an angle θ. If the vessel is subjected
to an internal pressure p, show that the force in the segment is F θ = σ0 wt , where σ0 is the stress in the
filaments. Also, show that the stresses in the hoop and longitudinal directions are σh = σ0 sin2 θ and σl
= σ0 cos2 θ , respectively. At what angle θ (optimum winding angle) would the filaments have to be
wound so that the hoop and longitudinal stresses are equivalent?
Problem 8-15
The steel bracket is used to connect the ends of two cables. If the allowable normal stress for the steel
is σallow = 168 MPa, determine the largest tensile force P that can be applied to the cables. The bracket
has a thickness of 12 mm and a width of 18 mm.
Given:
b := 18mm
a := 50mm
t := 12mm
σ allow := 168MPa
Solution:
Internal Force and Moment :
ao := a + 0.5 ⋅ b
N= P
M = P⋅ ao
Section Property :
A := t⋅ b
I :=
1
3
⋅ t⋅ b
12
N M⋅ c
+
I
A
The maximum normal stress occurs at the bottom of the steel bracket.
Alowable Normal Stress:
σ=
cmax := 0.5 ⋅ b
σ allow =
P :=
σ allow
ao⋅ cmax
1
+
I
A
P = 1.756 kN
Ans
P⋅ ao⋅ cmax
P
+
A
I
Problem 8-16
The steel bracket is used to connect the ends of two cables. If the applied force P = 2.5 kN, determine
the maximum normal stress in the bracket. The bracket has a thickness of 12 mm and a width of 18
mm.
Given:
b := 18mm
a := 50mm
t := 12mm
P := 2.5kN
Solution:
Internal Force and Moment :
ao := a + 0.5 ⋅ b
N := P
M := P⋅ ao
Section Property :
A := t⋅ b
I :=
1
3
⋅ t⋅ b
12
N M⋅ c
+
I
A
The maximum normal stress occurs at the bottom of the steel bracket.
Alowable Normal Stress:
σ=
cmax := 0.5 ⋅ b
σ max :=
N M⋅ cmax
+
A
I
σ max = 239.2 MPa
Ans
Problem 8-17
The joint is subjected to a force of 1.25 kN as shown. Sketch the normal-stress distribution acting over
section a-a if the member has a rectangular cross section of width 12 mm and thickness 18 mm.
b := 18mm
Given:
av := 32mm
t := 12mm
ah := 50mm
P := 1.25kN
v := 3
h := 4
r := 5
Solution:
Internal Force and Moment :
⎛ h⎞ − N = 0
⎝r⎠
N := P⋅ ⎜
⎛ v⎞ = 0
⎝r⎠
V := P⋅ ⎜
+
ΣF x=0;
P⋅ ⎜
+
ΣF y=0;
V − P⋅ ⎜
+ ΣΜA=0;
M + P⋅ ⎜
⎛ h⎞
⎝r⎠
⎛ v⎞
⎝r⎠
⎛ h ⎞ ⋅ a − P⋅ ⎛ v ⎞ ⋅ a = 0
⎜
⎝r⎠ v
⎝r⎠ h
⎛ v⎞
⎛ h⎞
M := P⋅ ⎜ ⋅ ah − P⋅ ⎜ ⋅ av
⎝r⎠
⎝r⎠
Section Property :
A := b⋅ t
I :=
1
3
⋅ b⋅ t
12
Normal Stress:
σ=
N M⋅ c
+
I
A
ctop := 0.5 ⋅ t
σ top :=
N M⋅ ctop
+
A
I
σ top = 17.36 MPa (T)
Ans
cbot := −0.5 ⋅ t
σ bot :=
N M⋅ cbot
+
A
I
σ bot = −8.10 MPa (C)
Ans
Location of zero stress:
σ top
σ bot
yo :=
=
yo
t − yo
t⋅ σ top
σ bot + σ top
yo = 8.18 mm
Problem 8-18
The joint is subjected to a force of 1.25 kN as shown. Determine the state of stress at points A and B,
and sketch the results on differential elements located at these points. The member has a rectangular
cross-sectional area of width 12 mm and thickness 18 mm.
b := 18mm
Given:
av := 32mm
t := 12mm
ah := 50mm
P := 1.25kN
v := 3
h := 4
r := 5
Solution:
Internal Force and Moment :
⎛ h⎞ − N = 0
⎝r⎠
N := P⋅ ⎜
⎛ v⎞ = 0
⎝r⎠
V := P⋅ ⎜
+
ΣF x=0;
P⋅ ⎜
+
ΣF y=0;
V − P⋅ ⎜
+ ΣΜA=0;
M + P⋅ ⎜
⎛ h⎞
⎝r⎠
⎛ v⎞
⎝r⎠
⎛ h ⎞ ⋅ a − P⋅ ⎛ v ⎞ ⋅ a = 0
⎜
⎝r⎠ v
⎝r⎠ h
⎛ v⎞
⎛ h⎞
M := P⋅ ⎜ ⋅ ah − P⋅ ⎜ ⋅ av
⎝r⎠
⎝r⎠
Section Property :
A := b⋅ t
I :=
1
3
⋅ b⋅ t
12
QA := ( 0.5 ⋅ t⋅ b) ⋅ ( 0.25t)
QB := 0
(since A' = 0)
Normal Stress:
σ=
N M⋅ c
+
I
A
cA := 0
σ A :=
N M⋅ cA
+
A
I
σ A = 4.63 MPa
(T)
Ans
cB := −0.5 ⋅ t
σ B :=
N M⋅ cB
+
A
I
σ B = −8.10 MPa
(C)
Ans
Shear Stress :
τ=
τ A :=
τ B :=
V⋅ QA
I⋅ b
V⋅ QB
I⋅ b
V⋅ Q
I⋅ b
τ A = 5.21 MPa
Ans
τ B = 0 MPa
Ans
Problem 8-19
The coping saw has an adjustable blade that is tightened with a tension of 40 N. Determine the state of
stress in the frame at points A and B.
Given:
t := 3mm
a := 100mm
h := 8mm
P := 40N
Solution:
Internal Force and Moment :
At A:
NA := −P
MA := P⋅ a
At B:
NB := 0
MB := P⋅ ( 0.5a)
Section Property :
I :=
1
3
⋅ t⋅ h
12
State of Stress: σ =
N M⋅ c
+
A
I
A := t⋅ h
At A:
cA := 0.5 ⋅ h
σ A :=
NA MA⋅ cA
+
A
I
σ A = 123.3 MPa (T)
At B:
cB := 0.5 ⋅ h
σ B :=
Ans
NB MB⋅ cB
+
A
I
σ B = 62.5 MPa
(T)
Ans
Problem 8-20
Determine the maximum and minimum normal stress in the bracket at section a when the load is
applied at x = 0.
Given:
a := 30mm
b := 20mm
xe := 0.5a
P := 4kN
Solution:
Internal Force and Moment :
N := −P
M := P⋅ xe
M = 0.060 kN⋅ m
Section Property :
A := a⋅ b
I :=
1
3
⋅ b⋅ a
12
2
A = 600 mm
4
I = 45000 mm
Normal Stress:
σ=
N M⋅ c
+
A
I
cmax := 0.5a
σ t :=
N M⋅ cmax
+
A
I
N M⋅ cmin
cmin := −0.5 ⋅ a σ c :=
+
A
I
(
)
σ min := min ( σ t , σ c )
σ max := max σ t , σ c
σ t = 13.33 MPa
(T)
σ c = −26.67 MPa
(C)
σ max = 26.67 MPa Ans
σ min = 13.33 MPa Ans
Problem 8-21
Determine the maximum and minimum normal stress in the bracket at section a when the load is
applied at x = 50 mm.
Given:
a := 30mm
b := 20mm
P := −4kN
xe := 0.5a − 50mm
Solution:
Internal Force and Moment :
N := P
M := P⋅ xe
M = 0.140 kN⋅ m
Section Property :
A := a⋅ b
I :=
1
3
⋅ b⋅ a
12
2
A = 600 mm
4
I = 45000 mm
Normal Stress:
σ=
N M⋅ c
+
A
I
cmax := 0.5a
σ t :=
N M⋅ cmax
+
A
I
N M⋅ cmin
cmin := −0.5 ⋅ a σ c :=
+
A
I
(
)
σ min := min ( σ t , σ c )
σ max := max σ t , σ c
σ t = 40.00 MPa
(T)
σ c = −53.33 MPa
(C)
σ max = 53.33 MPa
Ans
σ min = 40.00 MPa
Ans
Problem 8-22
The vertical force P acts on the bottom of the plate having a negligible weight. Determine the maximum
distance d to the edge of the plate at which it can be applied so that it produces no compressive
stresses on the plate at section a-a. The plate has a thickness of 10 mm and P acts along the centerline
of this thickness.
t := 10mm
Given:
b := 150mm
Solution:
Internal Force and Moment :
N= P
xe = d − 0.5b
M = P⋅ xe
Section Property :
2
A := t⋅ b
A = 1500 mm
1
3
⋅ t⋅ b
12
I = 2812500 mm
Normal Stress:
Require σ min := 0
I :=
σ=
4
N Mc
±
A
I
cmin := −0.5 ⋅ b
(
)
0=
P⋅ xe ⋅ cmin
P
+
A
I
0=
P
P⋅ ( d − 0.5b) ⋅ ( −0.5 ⋅ b)
+
A
I
d :=
2I
b
+
A⋅ b 2
d = 100 mm
Ans
Problem 8-23
The vertical force P = 600 N acts on the bottom of the plate having a negligible weight. The plate has a
thickness of 10 mm and P acts along the centerline of this thickness such that d = 100 mm. Plot the
distribution of normal stress acting along section a-a.
Given:
t := 10mm
b := 150mm
d := 100mm
P := 0.6kN
Solution:
Internal Force and Moment :
N := P
xe := d − 0.5b
M := P⋅ xe
M = 0.015 kN⋅ m
Section Property :
A := t⋅ b
I :=
1
3
⋅ t⋅ b
12
2
A = 1500 mm
4
I = 2812500 mm
N M⋅ c
+
A
I
Normal Stress:
σ=
cA := −0.5 b
σ A :=
N M⋅ cA
+
A
I
σ A = 0 MPa
(C)
Ans
cB := 0.5 ⋅ b
σ B :=
N M⋅ cB
+
A
I
σ B = 0.800 MPa
(T)
Ans
Problem 8-24
The gondola and passengers have a weight of 7.5 kN and center of gravity at G. The suspender arm
AE has a square cross-sectional area of 38 mm by 38 mm, and is pin connected at its ends A and E.
Determine the largest tensile stress developed in regions AB and DC of the arm.
Given:
b := 38mm
d := 38mm
L 1 := 1.2m
L 2 := 1.65m
ah := 375mm
W := 7.5kN
Solution:
Section Property :
A := b⋅ d
I :=
1
3
⋅ b⋅ d
12
Segment AB :
NAB := W
MAB := 0
Maximum Normal Stress: σ =
σ AB :=
NAB
A
N M⋅ c
+
I
A
σ AB = 5.19 MPa (T)
Ans
Segment DC :
NDC := W
MDC := W⋅ ah
Maximum Normal Stress: σ =
cmax := 0.5d
σ DC :=
N M⋅ c
+
I
A
NDC MDC⋅ cmax
+
A
I
σ DC = 312.73 MPa (T)
Ans
Problem 8-25
The stepped support is subjected to the bearing load of 50 kN. Determine the maximum and minimum
compressive stress in the material.
Given:
a := 100mm
b := 100mm
P := 50kN
xe := 0.5a − 30mm
Solution:
Internal Force and Moment :
N := −P
M := P⋅ xe
M = 1.000 kN⋅ m
Section Property :
For the bottom portion of the stepped support.
A := a⋅ b
I :=
1
3
⋅ b⋅ a
12
2
A = 10000 mm
4
I = 8333333.33 mm
Normal Stress:
σ=
N M⋅ c
+
A
I
cmax := 0.5a
σ t :=
N M⋅ cmax
+
A
I
N M⋅ cmin
cmin := −0.5 ⋅ a σ c :=
+
A
I
(
)
σ c_min := min ( 0 , σ c )
σ c_max := max 0 , σ c
σ t = 1 MPa
(T)
σ c = −11.00 MPa
(C)
σ c_max = 11 MPa
Ans
σ c_min = 0 MPa
Ans
Problem 8-26
The bar has a diameter of 40 mm. If it is subjected to a force of 800 N as shown, determine the stress
components that act at point A and show the results on a volume element located at this point.
Given: do := 40mm
a := 200mm
P := 0.8kN
θ := 30deg
Solution:
Internal Force and Moment : At section AB.
N := P⋅ sin ( θ )
N = 0.400 kN
V := P⋅ cos ( θ )
V = 0.693 kN
M := V⋅ a
M = 0.1386 kN⋅ m
Section Property : ro := 0.5do
A := π ⋅ ro
2
2
π 4
⋅r
4 o
4ro A
⎛ ⎞
QA :=
⋅⎜
3π ⎝ 2 ⎠
I :=
A = 1256.64 mm
4
I = 125663.71 mm
3
QA = 5333.33 mm
N M⋅ c
+
A
I
Normal Stress:
σ=
cA := 0
σ A :=
N M⋅ cA
+
A
I
σ A = 0.318 MPa
Shear Stress :
τ A :=
V⋅ QA
I⋅ do
τ=
(T)
Ans
V⋅ Q
I⋅ b
τ A = 0.735 MPa
Ans
Problem 8-27
Solve Prob. 8-26 for point B.
Given: do := 40mm
a := 200mm
P := 0.8kN
θ := 30deg
Solution:
Internal Force and Moment : At section AB.
N := P⋅ sin ( θ )
N = 0.400 kN
V := P⋅ cos ( θ )
V = 0.693 kN
M := V⋅ a
M = 0.1386 kN⋅ m
Section Property : ro := 0.5do
A := π ⋅ ro
I :=
2
2
A = 1256.64 mm
π 4
⋅r
4 o
QB := 0
4
I = 125663.71 mm
(since A'=0 )
Normal Stress:
cB := −ro
σ=
N M⋅ c
+
A
I
σ B :=
N M⋅ cB
+
A
I
σ B = −21.73 MPa (C)
Shear Stress :
τ A :=
V⋅ QB
I⋅ do
τ=
V⋅ Q
I⋅ b
τ A = 0 MPa
Ans
Ans
Problem 8-28
Since concrete can support little or no tension, this problem can be avoided by using wires or rods to
prestress the concrete once it is formed. Consider the simply supported beam shown, which has a
rectangular cross section of 450 mm by 300 mm. If concrete has a specific weight of 24 kN/m3,
determine the required tension in rod AB, which runs through the beam so that no tensile stress is
developed in the concrete at its center section a-a. Neglect the size of the rod and any deflection of the
beam.
Given: b := 300mm
d := 450mm
d' := 400mm
γ := 24
L := 2.4m
Solution:
kN
m
3
a := d − d'
w := γ ⋅ b⋅ d
By symmetry, RA= R ; RA= R
Support Reactions :
2R − w⋅ L = 0
+
R := 0.5w⋅ L
Internal Force and Moment :
+
ΣF x=0;
+ ΣΜO=0;
T−N= 0
N= T
M + T ⋅ ( 0.5d − a) − R⋅ ( 0.5L) + ( 0.5w⋅ L) ⋅ ( 0.25L ) = 0
M = R⋅ ( 0.25 ⋅ L ) − T⋅ ( 0.5 ⋅ d − a)
Section Property :
A := b⋅ d
I :=
1
3
⋅ b⋅ d
12
Normal Stress:
σa =
N M⋅ c
+
A
I
Requires σa= 0
0=
−T M⋅ ca
+
I
A
ca := 0.5d
0=
−T [ R⋅ ( 0.25 ⋅ L ) − T⋅ ( 0.5 ⋅ d − a) ] ⋅ ca
+
I
A
T :=
R⋅ ( 0.25 ⋅ L)
( 0.5 ⋅ d − a) +
T = 9.331 kN
I
A⋅ ca
Ans
Problem 8-29
Solve Prob. 8-28 if the rod has a diameter of 12 mm. Use the transformed area method discussed in
Sec. 6.6. E st = 200 GPa, E c = 25 GPa.
Given: b := 300mm
d := 450mm
d' := 400mm
do := 12mm
E st := 200GPa
E c := 25GPa
L := 2.4m
Solution:
γ := 24
kN
m
a := d − d'
3
w := γ ⋅ b⋅ d
By symmetry, RA= R ; RA= R
Support Reactions :
2R − w⋅ L = 0
+
R := 0.5w⋅ L
Internal Force and Moment :
+
ΣF x=0;
+ ΣΜO=0;
T−N= 0
N= T
(
)
M + T ⋅ d − yc − a − R⋅ ( 0.5L ) + ( 0.5w⋅ L ) ⋅ ( 0.25L) = 0
(
)
M = R⋅ ( 0.25 ⋅ L ) − T⋅ d − yc − a
Section Property :
E st
⎛ π 2⎞
n :=
A'conc := ( n − 1) ⋅ ⎜ ⋅ do
Ec
⎝4
⎠
A := b⋅ d + A'conc
yc :=
b⋅ d⋅ ( 0.5d) + A'conc ⋅ d'
A
1
3
2
2
I :=
⋅ b⋅ d + b⋅ d⋅ 0.5d − yc + A'conc ⋅ d' − yc
12
(
)
(
)
Normal Stress:
σa =
N M⋅ c
+
A
I
Requires σa= 0
0=
−T M⋅ ca
+
I
A
ca := d − yc
0=
−T ⎡⎣R⋅ ( 0.25 ⋅ L) − T ⋅ d − yc − a ⎤⎦ ⋅ ca
+
I
A
T :=
(
R⋅ ( 0.25 ⋅ L)
I
(d − yc − a) + A⋅ c
a
T = 9.343 kN
Ans
)
Problem 8-30
The block is subjected to the two axial loads shown. Determine the normal stress developed at points A
and B. Neglect the weight of the block.
Given: b := 50mm
d := 75mm
P1 := 250N
P2 := 500N
Solution:
Internal Force and Moment :
+
ΣF x=0;
+ ΣΜz=0;
− N − P1 − P2 = 0
Mz + P1⋅ ( 0.5d) − P2⋅ ( 0.5d) = 0
(
Mz := 0.5 ⋅ d⋅ P2 − P1
+ ΣΜy=0;
N := −P1 − P2
)
My + P1⋅ ( 0.5b) − P2⋅ ( 0.5b) = 0
(
)
My := 0.5 ⋅ b⋅ P2 − P1
Section Property :
A := b⋅ d
Iz :=
1
3
⋅ b⋅ d
12
Normal Stress:
σ=
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
At A:
At B:
Iy :=
1
3
⋅ d⋅ b
12
yA := 0.5d
zA := 0.5b
N Mz⋅ yA My⋅ zA
σ A :=
−
+
A
Iz
Iy
zB := −0.5 b
N Mz⋅ yB My⋅ zB
σ B :=
−
+
A
Iz
Iy
σ A = −0.200 MPa (C)
Ans
σ B = −0.600 MPa (C)
Ans
yB := 0.5d
Problem 8-31
The block is subjected to the two axial loads shown. Sketch the normal stress distribution acting over
the cross section at section a-a. Neglect the weight of the block.
Given: b := 50mm
d := 75mm
P1 := 250N
P2 := 500N
Solution:
Internal Force and Moment :
+
ΣF x=0;
+ ΣΜz=0;
− N − P1 − P2 = 0
Mz + P1⋅ ( 0.5d) − P2⋅ ( 0.5d) = 0
(
Mz := 0.5 ⋅ d⋅ P2 − P1
+ ΣΜy=0;
N := −P1 − P2
)
My + P1⋅ ( 0.5b) − P2⋅ ( 0.5b) = 0
(
)
My := 0.5 ⋅ b⋅ P2 − P1
Section Property :
A := b⋅ d
Iz :=
1
3
⋅ b⋅ d
12
Normal Stress:
σ=
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
At A:
At B:
At C:
At D:
Iy :=
1
3
⋅ d⋅ b
12
yA := 0.5d
zA := 0.5b
N Mz⋅ yA My⋅ zA
σ A :=
−
+
A
Iz
Iy
zB := −0.5 b
N Mz⋅ yB My⋅ zB
σ B :=
−
+
A
Iz
Iy
σ A = −0.200 MPa (C)
Ans
σ B = −0.600 MPa (C)
Ans
σ C = −0.200 MPa (C)
Ans
σ D = 0.200 MPa
Ans
yB := 0.5d
yC := −0.5 d
zC := −0.5 b
N Mz⋅ yC My⋅ zC
σ C :=
−
+
A
Iz
Iy
yD := −0.5 d
zD := 0.5b
N Mz⋅ yD My⋅ zD
σ D :=
−
+
A
Iz
Iy
(T)
Problem 8-32
A bar having a square cross section of 30 mm by 30 mm is 2 m long and is held upward. If it has a
mass of 5 kg/m, determine the largest angle θ, measured from the vertical, at which it can be
supported before it is subjected to a tensile stress near the grip.
Given:
b := 30mm
L := 2m
t := 30mm
mo := 5
Solution: W := mo⋅ g⋅ L
kg
m
W = 0.0981 kN
Internal Force and Moment :
ΣF y=0;
+
+
−N − W⋅ cos ( θ ) = 0
ΣΜO=0; M − W⋅ sin ( θ ) ⋅ ( 0.5 ⋅ L) = 0
N = −W⋅ cos ( θ )
M = 0.5W⋅ L ⋅ sin ( θ )
Section Property :
A := b⋅ t
Normal Stress:
σ=
I :=
1
3
⋅ b⋅ t
12
Require σ max := 0
N Mc
±
A
I
cmax := 0.5 ⋅ b
0=
N M⋅ cmax
+
A
I
−W⋅ cos ( θ ) 0.5W⋅ L⋅ sin ( θ ) ⋅ ( 0.5 ⋅ b)
+
b⋅ t
1
3
⋅ b⋅ t
12
2
t
tan ( θ ) =
3⋅ L ⋅ b
0=
⎛ t2 ⎞
θ := atan ⎜
⎝ 3⋅ L ⋅ b ⎠
θ = 0.00500 rad
θ = 0.286 deg Ans
Problem 8-33
Solve Prob. 8-32 if the bar has a circular cross section of 30-mm diameter.
do := 30mm
Given:
L := 2m
mo := 5
Solution: W := mo⋅ g⋅ L
kg
m
W = 0.0981 kN
Internal Force and Moment :
ΣF y=0;
+
+
−N − W⋅ cos ( θ ) = 0
N = −W⋅ cos ( θ )
ΣΜO=0; M − W⋅ sin ( θ ) ⋅ ( 0.5 ⋅ L) = 0
Section Property :
A := π ⋅ ro
2
π 4
⋅r
4 o
Normal Stress:
I :=
σ=
ro := 0.5do
2
A = 706.86 mm
4
I = 39760.78 mm
Require σ max := 0
N Mc
±
A
I
cmax := ro
M = 0.5W⋅ L ⋅ sin ( θ )
0=
N M⋅ cmax
+
A
I
0=
−W⋅ cos ( θ ) 0.5W⋅ L⋅ sin ( θ ) ⋅ ro
+
A
I
tan ( θ ) =
( )
2I
A⋅ L⋅ ro
⎛ 2I ⎞
⎝ A⋅ L⋅ ro ⎠
θ := atan ⎜
θ = 0.00375 rad
θ = 0.215 deg Ans
Problem 8-34
The wide-flange beam is subjected to the loading shown. Determine the stress components at points A
and B and show the results on a volume element at each of these points. Use the shear formula to
compute the shear stress.
Given:
Solution:
b := 100mm
d := 150mm
t := 12mm
P1 := 2.5kN
d'B := 50mm
P2 := 12.5kN
P3 := 15kN
L1 := 0.5m
L2 := 1m
L3 := 1.5m
L := 3⋅ L1 + L2 + L3
Given
Support Reactions :
+
ΣFy=0;
R1 + R2 − P1 − P2 − P3 = 0
(
+ ΣΜR2=0;
)
(
)
−R1⋅ L + P1⋅ L − L1 + P2⋅ L2 + L3 + P3⋅ L 3 = 0
Guess R1 := 1N
R2 := 1N
⎛⎜ R1 ⎞
⎛⎜ R1 ⎞ ⎛ 15.63 ⎞
:= Find R1 , R2
=⎜
kN
⎜ R2
⎜ R2
⎝ ⎠
⎝ ⎠ ⎝ 14.38 ⎠
(
)
At Section A-B:
( )
M := R1⋅ 2L1
V := R1 − P1
D := d + 2t
Section Property :
A := 2⋅ b⋅ t + d⋅ t
I :=
1
1
3
3
⋅ b⋅ D −
⋅ ( b − t) ⋅ d
12
12
(
)
QB := b⋅ t⋅ ( 0.5D − 0.5t) + d'B⋅ t⋅ 0.5D − t − 0.5d'B
QA := 0
(since A' = 0)
Normal Stress:
σ=
M⋅ y
I
yA := 0.5D
σ A :=
yB := 0.5d − d'B
σ B :=
Shear Stress : τ =
V⋅ Q
I⋅ t
τ A :=
τ B :=
−M⋅ yA
I
M⋅ yB
I
V⋅ QA
I⋅ b
V⋅ QB
I⋅ t
σ A = −70.98 MPa (C) Ans
σ B = 20.4 MPa (T)
Ans
τ A = 0.00 MPa
Ans
τ B = 7.265 MPa
Ans
Problem 8-35
The cantilevered beam is used to support the load of 8 kN. Determine the state of stress at points A
and B, and sketch the results on differential elements located at each of these points.
bo := 100mm d := 100mm
t := 10mm
L := 3m
Given:
dA := 25mm
P := 8kN
dB := 45mm
Solution:
Internal Force and Moment : At Section A-B:
V := P
V = 8 kN
M := P⋅ L
M = 24 kN⋅ m
bi := bo − 2t
Section Property :
2
A := 2⋅ d⋅ t + bi⋅ t
I :=
A = 2800 mm
1
1
3
3
⋅ bi ⋅ t +
⋅ ( 2t) ⋅ d
12
12
4
I = 1673333.33 mm
( )(
)
QB := ( dB⋅ t) ⋅ ( 0.5d − 0.5dB)
3
QA := dA⋅ t ⋅ 0.5d − 0.5dA
Normal Stress:
σ=
M⋅ y
I
yA := 0.5d − dA
σ A :=
yB := 0.5d − dB
σ B :=
Shear Stress : τ =
V⋅ Q
I⋅ t
τ A :=
τ B :=
QA = 9375 mm
3
QB = 12375 mm
M⋅ yA
I
M⋅ yB
I
V⋅ QA
I⋅ t
V⋅ QB
I⋅ t
σ A = 358.6 MPa (T) Ans
σ B = 71.7 MPa (T)
Ans
τ A = 4.48 MPa
Ans
τ B = 5.92 MPa
Ans
Problem 8-36
The cylinder of negligible weight rests on a smooth floor. Determine the eccentric distance ey at which
the load can be placed so that the normal stress at point A is zero.
Solution:
Internal Force and Moment :
V = −P
( )
M = P⋅ ey
Section Property :
A = π⋅r
I=
2
π 4
⋅r
4
Normal Stress:
σ=
Require σ max := 0
N Mc
±
A
I
cA = r
0=
0=
N M⋅ cr
+
A
I
−P
π⋅r
r
ey =
4
2
+
( )
4P⋅ ey ⋅ r
π⋅r
Ans
4
Problem 8-37
The beam supports the loading shown. Determine the state of stress at points E and F at section a-a,
and represent the results on a differential volume element located at each of these points.
Given:
bf := 150mm
dw := 200mm
tf := 10mm
tw := 15mm
a := 1m
b := 2m
yB := 3.3m
w := 10
yD := 0.3m
L := 2⋅ a + b
Solution:
Support Reactions :
+
kN
m
ΣF y=0;
+ ΣΜC=0;
Given
⎛ L ⎞⋅ D
Dx = ⎜
y
⎝ yB − yD ⎠
(1)
Cy + Dy − w⋅ ( 2a) = 0
(2)
( )
−w⋅ ( 2a) ⋅ a + Dy⋅ ( L) + Dx⋅ yD = 0
Solving Eqs.(1), (2) and (3). Guess Cy := 1N
⎛ Cy ⎞
⎛ Cy ⎞
(3)
Dx := 1N
Dy := 1N
⎛ 15.4545 ⎞
⎜
⎜
⎜ Dx ⎟ = ⎜ 6.0606 kN
⎜D
4.5455 ⎠
⎝ y⎠ ⎝
⎜
⎜ Dx ⎟ := Find ( Cy , Dx , Dy)
⎜D
⎝ y⎠
Internal Force and Moment : At Section a-a:
N := −Dx
N = −6.0606 kN
V := w⋅ a − Dy
V = 5.4545 kN
M := −Dy⋅ ( a + b) − Dx⋅ yD + w⋅ a⋅ ( 0.5a)
M = −10.4545 kN⋅ m
Section Property :
do := dw + 2tf
2
A := 2⋅ bf⋅ tf + dw⋅ tw
I :=
A = 6000 mm
1
1
3
3
⋅ bf⋅ do −
⋅ bf − tw ⋅ dw
12
12
(
(
)
)
4
I = 43100000 mm
(
)
QE := bf⋅ tf⋅ 0.5do − 0.5tf + 0.5dw⋅ tw⋅ 0.25dw
QF := 0
(since A' = 0)
Normal Stress:
cE := 0
3
QE = 232500 mm
N Mc
±
A
I
N M⋅ cE
σ E :=
+
A
I
σ=
σ E = −1.01 MPa
(C)
Ans
cF := 0.5 ⋅ do
σ F :=
Shear Stress : τ =
V⋅ Q
I⋅ t
τ E :=
τ F :=
N M⋅ cF
+
A
I
V⋅ QE
I ⋅ tw
V⋅ QF
I⋅ bf
σ F = −27.69 MPa
τ E = 1.96 MPa
Ans
τ F = 0 MPa
Ans
(C)
Ans
Problem 8-38
The metal link is subjected to the axial force of P = 7 kN. Its original cross section is to be altered by
cutting a circular groove into one side. Determine the distance a the groove can penetrate into the cross
section so that the tensile stress does not exceed σallow = 175 MPa. Offer a better way to remove this
depth of material from the cross section and calculate the tensile stress for this case. Neglect the
effects of stress concentration.
Given: h := 80mm
t := 25mm
P := 7kN
σ allow := 175MPa
Solution:
Internal Force and Moment : At narrow section.
+
ΣF x=0;
P−N= 0
+ ΣΜbase=0;
M + N⋅
h−a
h
− P⋅ = 0
2
2
Section Property :
A = ( h − a) ⋅ t
N := P
I=
M = 0.5P⋅ a
t
3
⋅ ( h − a)
12
Require σ max := σ allow
cmax = 0.5 ( d − a)
N M⋅ cmax
σ max =
+
A
I
Normal Stress:
σ allow =
Given
P
0.5P⋅ a⋅ [ 0.5 ( h − a) ]
+
( h − a) ⋅ t
t
3
⋅ ( h − a)
12
(σallow⋅ t)⋅ ( h − a) 2 − P⋅ ( h − a) − 3P⋅ a = 0
Solving Eq.(1),.
(1)
Guess a := 1mm
a := Find ( a)
a = 61.94 mm
Better way:
Ans
To remove material equally from both sides such that M=0.
A := ( h − a) ⋅ t
σ' max :=
N
+0
A
σ' max = 15.50 MPa
Ans
Problem 8-39
Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN. Indicate
the result as a differential volume element.
Given:
bf := 150mm
tf := 20mm
dw := 200mm
tw := 15mm
a := 2m
b := 0.75m
c := 1m
hG := 375mm
rG := 250mm
P := 4kN
Solution:
L := a + b + c
Given
Support Reactions :
+
ΣFx=0;
+ ΣΜD=0;
Cx − P = 0
Cx := P
P⋅ hg + rG ⋅ a + Cy⋅ ( L) = 0
(
)
Cy :=
hG + r G
L
P
Internal Force and Moment : At Section A-B:
N := Cx
N = 4 kN
V := Cy
V = 0.6667 kN
M := Cy⋅ c
M = 0.6667 kN⋅ m
do := dw + 2tf
A := 2⋅ bf⋅ tf + dw⋅ tw
Section Property :
I :=
1
1
3
3
⋅ bf⋅ do −
⋅ bf − t w ⋅ dw
12
12
(
(
2
A = 9000 mm
)
)
4
I = 82800000 mm
(
)
(T)
Ans
QA := bf⋅ tf⋅ 0.5do − 0.5tf + 0.5dw⋅ tw⋅ 0.25dw
Normal Stress:
cA := 0
3
QA = 405000 mm
N Mc
±
A
I
N M⋅ cA
σ A :=
+
A
I
σ=
σ A = 0.444 MPa
Shear Stress : τ =
V⋅ Q
I⋅ t
τ A :=
V⋅ QA
I ⋅ tw
τ A = 0.217 MPa
Ans
Problem 8-40
Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN. Indicate
the result as a differential volume element.
Given:
bf := 150mm
dw := 200mm
tf := 20mm
tw := 15mm
a := 2m
b := 0.75m
c := 1m
hG := 375mm
rG := 250mm
P := 4kN
Solution:
L := a + b + c
Support Reactions :
+
ΣF x=0;
+ ΣΜD=0;
Given
Cx − P = 0
Cx := P
P⋅ hg + rG ⋅ a + Cy⋅ ( L ) = 0
(
)
Cy :=
hG + rG
P
L
Internal Force and Moment : At Section A-B:
N := Cx
N = 4 kN
V := Cy
V = 0.6667 kN
M := Cy⋅ c
M = 0.6667 kN⋅ m
Section Property :
do := dw + 2tf
A := 2⋅ bf⋅ tf + dw⋅ tw
I :=
2
A = 9000 mm
1
1
3
3
⋅ bf⋅ do −
⋅ bf − tw ⋅ dw
12
12
(
)
4
I = 82800000 mm
QB := 0 (since A' = 0)
Normal Stress:
cB := −0.5 do
N Mc
±
A
I
N M⋅ cB
σ B :=
+
A
I
σ=
σ B = −0.522 MPa
Shear Stress : τ =
(C) Ans
V⋅ Q
I⋅ t
τ B :=
V⋅ QB
I ⋅ tw
τ B = 0 MPa
Ans
Problem 8-41
The bearing pin supports the load of 3.5 kN. Determine the stress components in the support member
at point A. The support is 12 mm thick.
Given: b := 12mm
d := 18mm
W := 3.5kN
L 1 := 50mm
L 2 := 75mm
a := 32mm
θ := 30deg
Solution:
Internal Force and Moment :
ΣF x=0;
N − W⋅ cos ( θ ) = 0
N := W⋅ cos ( θ )
ΣF y=0;
V − W⋅ sin ( θ ) = 0
V := W⋅ sin ( θ )
+ ΣΜ=0;
(
)
M − W⋅ a − L1⋅ sin ( θ ) = 0
(
M := W⋅ a − L 1⋅ sin ( θ )
Section Property :
1
3
⋅ b⋅ d
12
QA := 0 (since A' = 0)
A := b⋅ d
I :=
Normal Stress:
σ=
At A:
N M⋅ y
+
I
A
yA := −0.5 d
σ A :=
N M⋅ yA
+
A
I
σ A = −23.78 MPa
Shear Stress :
τ=
(C)
V⋅ Q
I⋅ t
τ A :=
V⋅ QA
I⋅ b
τ A = 0 MPa
Ans
Ans
)
Problem 8-42
The bearing pin supports the load of 3.5 kN. Determine the stress components in the support member
at point B. The support is 12 mm thick.
Given: b := 12mm
d := 18mm
W := 3.5kN
L 1 := 50mm
L 2 := 75mm
a := 32mm
θ := 30deg
Solution:
Internal Force and Moment :
ΣF x=0;
N − W⋅ cos ( θ ) = 0
N := W⋅ cos ( θ )
ΣF y=0;
V − W⋅ sin ( θ ) = 0
V := W⋅ sin ( θ )
(
)
M − W⋅ a − L1⋅ sin ( θ ) = 0
+ ΣΜ=0;
(
M := W⋅ a − L 1⋅ sin ( θ )
Section Property :
A := b⋅ d
QB := 0
1
3
⋅ b⋅ d
12
(since A' = 0)
Normal Stress:
At B:
I :=
σ=
N M⋅ y
+
I
A
yB := 0.5d
σ B :=
N M⋅ yB
+
A
I
σ B = 51.84 MPa
Shear Stress :
τ=
(T)
V⋅ Q
I⋅ t
τ B :=
V⋅ QB
I⋅ b
τ B = 0 MPa
Ans
Ans
)
Problem 8-43
The uniform sign has a weight of 7.5 kN and is supported by the pipe AB, which has an inner radius of
68 mm and an outer radius of 75 mm. If the face of the sign is subjected to a uniform wind pressure
of p = 8 kN/m2, determine the state of stress at points C and D. Show the results on a differential
volume element located at each of these points. Neglect the thickness of the sign, and assume that it is
supported along the outside edge of the pipe.
Given: b := 3.6m
h := 1.8m
ro := 75mm
W := 7.5kN
ri := 68mm
d := 0.9m
p := 8
kN
m
2
P := p⋅ b⋅ h
Solution:
Internal Force and Moment :
ΣF x=0;
W + Nx = 0
ΣFy=0;
Vy − P = 0
Nx := −W
Vy := P
ΣFz=0;
Vz := 0
ΣΜx=0;
Mx − P⋅ ( 0.5b) = 0 Mx := 0.5 ⋅ P⋅ b
My − W⋅ ( 0.5b) = 0
My := 0.5 ⋅ W⋅ b
Mz + P⋅ ( 0.5h + d) = 0
Mz := −P⋅ ( 0.5h + d)
ΣΜy=0;
ΣΜz=0;
Nx = −7.5 kN
Mx = 93.312 kN⋅ m
2
π ⎛ 4
4
⋅ ⎝ ro − ri ⎞⎠
4
QC_y :=
2
Iz := Iy
4ro
⎛ π 2⎞ 4ri ⋅ ⎛ π ⋅ r 2⎞
⋅ ⎜ ⋅ ro −
⎜
3π ⎝ 2
⎠ 3π ⎝ 2 i ⎠
QD_z := QC_y
J :=
Vz = 0 kN
Mz = −93.312 kN⋅ m
A := π ⋅ ⎛⎝ ro − ri ⎞⎠
Section Property :
Iy :=
Vy = 51.84 kN
My = 13.5 kN⋅ m
QD_y := 0
π ⎛ 4
4
⋅ r − ri ⎞⎠
2 ⎝o
Normal Stress:
At C:
At D:
σ=
QC_z := 0
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
yC := 0
zC := ri
N Mz⋅ yC My⋅ zC
σ C :=
−
+
A
Iz
Iy
zD := 0
N Mz⋅ yD My⋅ zD
σ D :=
−
+
A
Iz
Iy
σ C = 113.9 MPa
(T)
Ans
σ D = 868.5 MPa
(T)
Ans
yD := ro
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for τ v and the torsional formula for τ t respectively.
V⋅ Q
T⋅ ρ
τv =
τt =
I⋅ t
J
At C:
(
)
ρ := ri
Vy⋅ QC_y
Iz⋅ t
τ t :=
t := 2⋅ ro − ri
τ v_y :=
τ C_xy := τ v_y − τ t
At D:
(
τ C_xy = −360.8 MPa
Ans
τ C_xz := 0
Ans
τ C_yz := 0
Ans
)
ρ' := ro
Vz⋅ QD_z
I y⋅ t
τ' t :=
t := 2⋅ ro − ri
τ v_z :=
Mx⋅ ρ
J
τ D_xz := τ v_z + τ' t
Mx⋅ ρ'
J
τ D_xz = 434.3 MPa
Ans
τ D_xy := 0
Ans
τ D_yz := 0
Ans
Problem 8-44
Solve Prob. 8-43 for points E and F.
Given: b := 3.6m
h := 1.8m
ro := 75mm
ri := 68mm
W := 7.5kN
d := 0.9m
p := 8
kN
m
2
P := p⋅ b⋅ h
Solution:
Internal Force and Moment :
ΣF x=0;
W + Nx = 0
ΣF y=0;
Vy − P = 0
Nx := −W
Vy := P
ΣF z=0;
Vz := 0
ΣΜx=0;
Mx − P⋅ ( 0.5b) = 0 Mx := 0.5 ⋅ P⋅ b
My − W⋅ ( 0.5b) = 0
My := 0.5 ⋅ W⋅ b
Mz + P⋅ ( 0.5h + d) = 0
Mz := −P⋅ ( 0.5h + d)
ΣΜy=0;
ΣΜz=0;
Nx = −7.5 kN
Mx = 93.312 kN⋅ m
2
π ⎛ 4
4
⋅ r − ri ⎞⎠
4 ⎝o
QF_y :=
2
Iz := Iy
4ro
⎛ π 2⎞ 4ri ⋅ ⎛ π ⋅ r 2⎞
⋅ ⎜ ⋅ ro −
⎜
3π ⎝ 2
⎠ 3π ⎝ 2 i ⎠
QE_z := QF_y
J :=
Vz = 0 kN
Mz = −93.312 kN⋅ m
A := π ⋅ ⎛⎝ ro − ri ⎞⎠
Section Property :
Iy :=
Vy = 51.84 kN
My = 13.5 kN⋅ m
QE_y := 0
π ⎛ 4
4
⋅ ⎝ ro − ri ⎞⎠
2
Normal Stress:
At F:
At E:
σ=
QF_z := 0
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
yF := 0
zF := −ro
N Mz⋅ yF My⋅ zF
σ F :=
−
+
A
Iz
Iy
zE := 0
N Mz⋅ yE My⋅ zE
σ E :=
−
+
A
Iz
Iy
σ F = −125.7 MPa
(C)
Ans
σ E = −868.5 MPa (C)
Ans
yE := −ro
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for τ v and the torsional formula for τ t respectively.
V⋅ Q
T⋅ ρ
τv =
τt =
I⋅ t
J
At F:
(
)
ρ := ro
Vy⋅ QF_y
Iz⋅ t
τ t :=
t := 2⋅ ro − ri
τ v_y :=
τ F_xy := τ v_y + τ t
At E:
(
τ F_xy = 467.2 MPa
Ans
τ F_xz := 0
Ans
τ F_yz := 0
Ans
)
ρ' := ro
Vz⋅ QE_z
Iy⋅ t
τ' t :=
t := 2⋅ ro − ri
τ v_z :=
Mx⋅ ρ
J
τ E_xz := τ v_z − τ' t
Mx⋅ ρ'
J
τ E_xz = −434.3 MPa
Ans
τ E_xy := 0
Ans
τ E_yz := 0
Ans
Problem 8-45
The bar has a diameter of 40 mm. If it is subjected to the two force components at its end as shown,
determine the state of stress at point A and show the results on a differential volume element located at
this point.
Given: a := 100mm
b := 150mm
do := 40mm
Py := 0.3kN
Pz := −0.5 kN
Solution:
Internal Force and Moment :
ΣF x=0;
ΣF y=0;
Nx := 0
Vy := −Py
Vy + Py = 0
Vz + Pz = 0
ΣF z=0;
Vz := −Pz
Mx := 0
ΣΜx=0;
ΣΜy=0;
My − Pz⋅ ( −b) = 0
Mz + Py⋅ ( −b) = 0
ΣΜz=0;
Nx = 0 kN
Mx = 0 kN⋅ m
A := π ⋅ ro
Iy :=
Vy = −0.3 kN
My = 0.075 kN⋅ m
ro := 0.5do
Section Property :
2
J :=
π
4
⋅d
64 o
At A:
Vz = 0.5 kN
Mz = 0.045 kN⋅ m
π 4
⋅r
2 o
Iz := Iy
QA_z := 0
Normal Stress:
My := −Pz⋅ b
Mz := Py⋅ b
QA_y :=
σ=
4ro
⎛ A⎞
3π ⎝ 2 ⎠
⋅⎜
Nx Mz⋅ y My⋅ z
−
+
A
Iy
Iz
yA := 0
σ A :=
zA := ro
Nx Mz⋅ yA My⋅ zA
−
+
A
Iy
Iz
σ A = 11.9 MPa (T)
Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for τ v and the torsional formula for τ t respectively.
V⋅ Q
T⋅ ρ
τv =
τt =
I⋅ t
J
At A:
ty := do
ρ := ro
τ v_y :=
Vy⋅ QA_y
Iz⋅ ty
τ t :=
Mx⋅ ρ
J
τ v_y = −0.318 MPa
τ t = 0 MPa
τ A_xy := τ v_y − τ t
τ A_xy = −0.318 MPa
Ans
τ A_xz := 0
Ans
τ A_yz := 0
Ans
Problem 8-46
Solve Prob. 8-45 for point B.
Given: a := 100mm
b := 150mm
do := 40mm
Py := 0.3kN
Pz := −0.5 kN
Solution:
Internal Force and Moment :
ΣF x=0;
ΣF y=0;
Nx := 0
Vy := −Py
Vy + Py = 0
Vz + Pz = 0
ΣF z=0;
Vz := −Pz
Mx := 0
ΣΜx=0;
ΣΜy=0;
My − Pz⋅ ( −b) = 0
Mz + Py⋅ ( −b) = 0
ΣΜz=0;
Nx = 0 kN
Mx = 0 kN⋅ m
A := π ⋅ ro
Iy :=
Vy = −0.3 kN
My = 0.075 kN⋅ m
ro := 0.5do
Section Property :
2
J :=
π
4
⋅ do
64
At B:
Vz = 0.5 kN
Mz = 0.045 kN⋅ m
π 4
⋅r
2 o
Iz := Iy
QB_y := 0
Normal Stress:
My := −Pz⋅ b
Mz := Py⋅ b
QB_z :=
σ=
4ro
⎛ A⎞
3π ⎝ 2 ⎠
⋅⎜
Nx Mz⋅ y My⋅ z
−
+
A
Iy
Iz
yB := ro
σ B :=
zB := 0
Nx Mz⋅ yB My⋅ zB
−
+
A
Iy
Iz
σ B = −7.16 MPa (C)
Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for τ v and the torsional formula for τ t respectively.
V⋅ Q
T⋅ ρ
τv =
τt =
I⋅ t
J
At B:
tz := do
τ v_z :=
ρ := ro
Vz⋅ QB_z
I y⋅ t z
τ t :=
Mx⋅ ρ
J
τ v_z = 0.531 MPa
τ t = 0 MPa
τ B_xz := τ v_z + τ t
τ B_xz = 0.531 MPa
Ans
τ B_xy := 0
Ans
τ B_yz := 0
Ans
Problem 8-47
The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mg
and center of mass at G. If the pipe has an outer diameter of 70 mm and a wall thickness of 10
mm,determine the state of stress acting at point C. Show the results on a differential volume element
located at this point. Neglect the weight of the pipe.
Given:
do := 70mm
t := 10mm
hA := 75mm
L := 3m
Mo := 3000kg
θ := 45deg
W := Mo⋅ g
W = 29.42 kN
Solution:
Equilibrium :
P−W= 0
P := W
+ ΣF y=0;
Support Reactions : By symmetry, Ay = By = R
P + 2R = 0
R := −0.5 P
R = −14.71 kN
Also, Ax = -Bx and tan ( θ ) =
Ax :=
R
Ay
Ax
Ax = −14.71 kN
tan ( θ )
Internal Force and Moment : At Section C-D:
N := Ax
N = −14.71 kN
V := R + 0.5W
V = 0 kN
( )
M := V⋅ ( 0.5 ⋅ L) + Ax⋅ hA
M = −1.103 kN⋅ m
di := do − 2t
Section Property :
A :=
π ⎛ 2
2
⋅ ⎝ do − di ⎞⎠
4
A = 1884.96 mm
I :=
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
64
I = 871791.96 mm
Normal Stress:
2
N Mc
±
A
I
N M⋅ cC
σ C :=
+
A
I
4
σ=
cC := 0.5do
Shear Stress : τ =
σ C = −52.10 MPa
V⋅ Q
I⋅ t
τ C := 0 (since V = 0)
Ans
(C)
Ans
Problem 8-48
The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mg
and center of mass at G. If the pipe has an outer diameter of 70 mm and a wall thickness of 10
mm,determine the state of stress acting at point D. Show the results on a differential volume element
located at this point. Neglect the weight of the pipe.
Given:
do := 70mm
t := 10mm
hA := 75mm
L := 3m
Mo := 3000kg
θ := 45deg
W := Mo⋅ g
W = 29.42 kN
Solution:
Equilibrium :
P−W= 0
P := W
+ ΣF y=0;
Support Reactions : By symmetry, Ay = By = R
P + 2R = 0
R := −0.5 P
R = −14.71 kN
Also, Ax = -Bx and tan ( θ ) =
Ax :=
R
Ay
Ax
Ax = −14.71 kN
tan ( θ )
Internal Force and Moment : At Section C-D:
N := Ax
N = −14.71 kN
V := R + 0.5W
V = 0 kN
( )
M := V⋅ ( 0.5 ⋅ L) + Ax⋅ hA
M = −1.103 kN⋅ m
di := do − 2t
Section Property :
A :=
π ⎛ 2
2
⋅ ⎝ do − di ⎞⎠
4
A = 1884.96 mm
I :=
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
64
I = 871791.96 mm
Normal Stress:
2
N Mc
±
A
I
N M⋅ cD
σ D :=
+
A
I
4
σ=
cD := 0
Shear Stress : τ =
σ D = −7.80 MPa (C)
V⋅ Q
I⋅ t
τ D := 0 (since V = 0)
Ans
Ans
Problem 8-49
The sign is subjected to the uniform wind loading. Determine the stress components at points A and B
on the 100-mm-diameter supporting post. Show the results on a volume element located at each of
these points.
do := 100mm
bo := 2m
Given:
a := 3m
po := 1500Pa
ho := 1m
Solution:
Px := −po⋅ bo⋅ ho
Px = −3.00 kN
Internal Force and Moment :
ΣFx=0; Vx + Px = 0
ΣFy=0;
Vx := −Px
Vy := 0
ΣFz=0;
Nz := 0
ΣΜx=0;
Mx := 0
My := −Px⋅ a + 0.5ho
(
(
ΣΜy=0;
)
My + Px⋅ a + 0.5ho = 0
Mz − Px⋅ 0.5bo
ΣΜz=0;
Vx = 3 kN
Mx = 0 kN⋅ m
Section Property :
)
(
Vy = 0 kN
My = 10.5 kN⋅ m
Nz = 0 kN
Mz = −3 kN⋅ m
ro := 0.5 ⋅ do
A = 7853.98 mm
I :=
I = 4908738.52 mm
π ⎛ 4⎞
⋅ r
4 ⎝ o ⎠
π
4
J := ⋅ ⎛⎝ ro ⎞⎠
2
)
)
Mz := Px⋅ 0.5bo
A := π⋅ ⎛⎝ ro ⎞⎠
2
(
2
4
4
J = 9817477.04 mm
(since A' = 0)
QA := 0
4ro A
⎛ ⎞
QB :=
⋅⎜
3π ⎝ 2 ⎠
Normal Stress:
σ=
N Mc
±
A
I
At A: xA := ro
ρ A := ro
Nz My⋅ xA
+
A
I
σ A = 107.0 MPa (T)
At B:
σ A :=
σ B :=
Ans
xB := 0
ρ B := ro
Nz My⋅ xB
+
A
I
σ B = 0.0 MPa
Ans
Shear Stress :
τ=
T⋅ ρ
V⋅ Q
+
J
I⋅ t
τ A :=
Mz⋅ ρ A
(since QA = 0)
J
τ A = 15.28 MPa
Ans
τ B :=
Mz⋅ ρ B Vx⋅ QB
+
J
I⋅ do
τ B = 14.77 MPa Ans
Problem 8-50
The sign is subjected to the uniform wind loading. Determine the stress components at points C and D
on the 100-mm-diameter supporting post. Show the results on a volume element located at each of
these points.
do := 100mm
bo := 2m
Given:
a := 3m
po := 1500Pa
ho := 1m
Solution:
Px := −po⋅ bo⋅ ho
Px = −3.00 kN
Internal Force and Moment :
ΣFx=0; Vx + Px = 0
ΣFy=0;
Vx := −Px
Vy := 0
ΣFz=0;
Nz := 0
ΣΜx=0;
Mx := 0
My := −Px⋅ a + 0.5ho
(
(
ΣΜy=0;
)
My + Px⋅ a + 0.5ho = 0
Mz − Px⋅ 0.5bo
ΣΜz=0;
Vx = 3 kN
Mx = 0 kN⋅ m
Section Property :
A := π ⋅ ⎛⎝ ro ⎞⎠
2
)
(
(
Mz := Px⋅ 0.5bo
Vy = 0 kN
My = 10.5 kN⋅ m
)
)
Nz = 0 kN
Mz = −3 kN⋅ m
ro := 0.5 ⋅ do
2
A = 7853.98 mm
π ⎛ 4⎞
4
⋅ r
I = 4908738.52 mm
4 ⎝o ⎠
π
4
4
J := ⋅ ⎛⎝ ro ⎞⎠
J = 9817477.04 mm
2
(since A' = 0)
QC := 0
I :=
QD :=
4ro
⎛ A⎞
3π ⎝ 2 ⎠
⋅⎜
Normal Stress:
σ=
N Mc
±
A
I
At C: xC := −ro
ρ C := ro
σ C :=
At D:
Nz My⋅ xC
+
A
I
σ C = −107.0 MPa (C)
xD := 0
ρ D := ro
Nz My⋅ xD
+
A
I
Ans
σ D = 0 MPa
σ D :=
Ans
Shear Stress :
τ=
T⋅ ρ
V⋅ Q
+
J
I⋅ t
τ C :=
Mz⋅ ρ C
(since QC = 0)
J
τ C = 15.28 MPa
Ans
−Mz⋅ ρ D
+
Vx⋅ QD
I ⋅ do
τ D = 15.79 MPa
Ans
τ D :=
J
Problem 8-51
The 18-mm-diameter shaft is subjected to the loading shown. Determine the stress components at
point A. Sketch the results on a volume element located at this point. The journal bearing at C can exert
only force components Cy and Cz on the shaft, and the thrust bearing at D can exert force components
Dx, Dy, and Dz on the shaft.
Given: do := 18mm
L := 500mm
ax := 50mm
ay := 200mm
P := 600N
a := 250mm
ro := 0.5do
Solution:
Support Reactions at C :
T x := 0
My := 0
Cy := 0
Cx := 0
Mz := 0
Cz := P
Internal Force and Moment at A :
N := 0
Vy := 0
My := −Cz⋅ a
T x := 0
Section Property :
A := π ⋅ ro
2
QA := 0
Normal Stress:
At A:
Shear Stress :
At A:
σ=
Vz := Cz
Mz := 0
Iy :=
π 4
⋅r
4 o
J :=
π 4
⋅r
2 o
Iz := Iy
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
yA := 0
zA := ro
N Mz⋅ yA My⋅ zA
σ A :=
−
+
A
Iz
Iy
τ=
σ A = −262.0 MPa (C)
V⋅ Q
I⋅ b
b := 0
τ A :=
Vz⋅ QA
Iy⋅ b
τ A = 0.00
Ans
Ans
Problem 8-52
Solve Prob. 8-51 for the stress components at point B.
Given: do := 18mm
L := 500mm
ax := 50mm
ay := 200mm
P := 600N
a := 250mm
ro := 0.5do
Solution:
Support Reactions at C :
T x := 0
My := 0
Cy := 0
Cx := 0
Mz := 0
Cz := P
Internal Force and Moment at B :
N := 0
Vy := 0
My := −Cz⋅ a
T x := 0
Section Property :
A := π ⋅ ro
QB :=
Iy :=
2
Vz := Cz
Mz := 0
π 4
⋅r
4 o
4ro
⎛ π 2⎞
⋅ ⎜ ⋅ ro
3π ⎝ 2
⎠
Normal Stress:
At B:
Shear Stress :
At B:
σ=
Iz := Iy
J :=
π 4
⋅r
2 o
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
yB := ro
zB := 0
N Mz⋅ yB My⋅ zB
σ A :=
−
+
A
Iz
Iy
τ=
σ A = 0.0 MPa
Ans
τ B = 3.14 MPa
Ans
V⋅ Q
I⋅ b
b := 2⋅ ro
τ B :=
Vz⋅ QB
I y⋅ b
Problem 8-53
The solid rod is subjected to the loading shown. Determine the state of stress developed in the material
at point A, and show the results on a differential volume element at this point.
Given:
ro := 30mm
a := 150mm
Px := −10kN
Py := 1kN
Pz := 15kN
Tx := 0.2kN⋅ m
Solution:
Internal Force and Moment : At Section A
ΣFx=0;
Nx + Px = 0
Nx := −Px
Vy := 0
ΣFy=0;
ΣFz=0;
ΣΜx=0;
Mx + T x = 0
ΣΜy=0;
ΣΜz=0;
Vz := 0
Mx := −Tx
( )
Mz − Px⋅ −ro = 0
My := 0
Mz := −Px⋅ ro
ρ := ro
Section Property :
A := π ⋅ ⎛⎝ ro ⎞⎠
A = 2827.43 mm
I :=
I = 636172.51 mm
2
Vz = 0 kN
Mx = −0.2 kN⋅ m
My = 0 kN⋅ m
Mz = 0.3 kN⋅ m
2
π ⎛ 4⎞
⋅ r
4 ⎝o ⎠
π
4
J := ⋅ ⎛⎝ ro ⎞⎠
2
4ro A
⎛ ⎞
QA_z :=
⋅⎜
3π ⎝ 2 ⎠
Normal Stress: σ =
yA := −ro
Nx = 10 kN
Vy = 0 kN
4
4
J = 1272345.02 mm
QA_y := 0 (since A' = 0)
Nx Mz y M yz
−
+
A
Iz
Iy
zA := 0
σ A :=
Nx Mz⋅ yA My⋅ zA
−
+
A
I
I
σ A = 17.7 MPa (T)
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for τ v and the torsional formula for τ t respectively.
V⋅ Q
τv =
τ v_z := 0 (since Vz= 0)
I⋅ t
τt =
T⋅ ρ
J
τ xz := τ v_z − τ t
τ t :=
Mx⋅ ρ
J
τ t = −4.72 MPa
τ xz = 4.716 MPa
Ans
τ xy := 0
Ans
τ yz := 0
Ans
Ans
Problem 8-54
The solid rod is subjected to the loading shown. Determine the state of stress at point B, and show the
results on a differential volume element at this point.
Given:
ro := 30mm
a := 150mm
Px := −10kN
Py := 10kN
Pz := 15kN
Tx := 0.2kN⋅ m
Solution:
Internal Force and Moment : At Section B
ΣFx=0;
ΣFy=0;
Nx + Px = 0
Vy + Py = 0
Nx := −Px
Vy := −Py
ΣFz=0;
ΣΜx=0;
Vz := 0
Mx − Py⋅ ro + T x = 0
ΣΜy=0;
Mx := Py⋅ ro − Tx
My := 0
( )
ΣΜz=0; Mz − Px⋅ −ro + Py⋅ ( −a) = 0
Mz := −Px⋅ ro + Py⋅ a
A := π ⋅ ⎛⎝ ro ⎞⎠
A = 2827.43 mm
I :=
I = 636172.51 mm
2
2
π ⎛ 4⎞
⋅ r
4 ⎝o ⎠
π
4
J := ⋅ ⎛⎝ ro ⎞⎠
2
4ro A
⎛ ⎞
QB_z :=
⋅⎜
3π ⎝ 2 ⎠
4
4
J = 1272345.02 mm
QB_y := 0 (since A' = 0)
Normal Stress: σ = N x − M z y +
yB := ro
Vz = 0 kN
Mx = 0.1 kN⋅ m
My = 0 kN⋅ m
Mz = 1.8 kN⋅ m
ρ := ro
Section Property :
Nx = 10 kN
Vy = −10 kN
A
Iz
zB := 0
σ B :=
M yz
Iy
Nx Mz⋅ yB My⋅ zB
−
+
A
I
I
σ B = −81.3 MPa (C)
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for τ v and the torsional formula for τ t respectively.
V⋅ Q
τv =
τ v_z := 0 (since Vz= 0)
I⋅ t
τt =
T⋅ ρ
J
τ xz := τ v_z + τ t
τ t :=
Mx⋅ ρ
J
τ t = 2.36 MPa
τ xz = 2.358 MPa
Ans
τ xy := 0
Ans
τ yz := 0
Ans
Ans
Problem 8-55
The solid rod is subjected to the loading shown. Determine the state of stress at point C, and show the
results on a differential volume element at this point.
Given:
ro := 30mm
a := 150mm
Px := −10kN
Py := 10kN
Pz := 15kN
Tx := 0.2kN⋅ m
Solution:
Internal Force and Moment : At Section A
ΣFx=0;
ΣFy=0;
ΣFz=0;
Nx + Px = 0
Vy + Py = 0
Nx := −Px
Vy := −Py
ΣΜy=0;
Vz + Pz = 0
Vz := −Pz
Mx − Py⋅ ro + Pz⋅ ro + Tx = 0
Mx := Py⋅ ro − Pz⋅ ro − T x
My − Pz⋅ ( −a) = 0
My := −Pz⋅ a
ΣΜz=0;
Mz − Px⋅ −ro + Py⋅ ( −3a) = 0
ΣΜx=0;
( )
Mz := −Px⋅ ro + 3Py⋅ a
ρ := ro
Section Property :
A = 2827.43 mm
I :=
I = 636172.51 mm
2
π ⎛ 4⎞
⋅ r
4 ⎝o ⎠
π
4
J := ⋅ ⎛⎝ ro ⎞⎠
2
4ro A
⎛ ⎞
QC_y :=
⋅⎜
3π ⎝ 2 ⎠
4
4
J = 1272345.02 mm
QC_z := 0 (since A' = 0)
Normal Stress: σ = N x − M z y +
yC := 0
Vz = −15 kN
Mx = −0.35 kN⋅ m
My = −2.25 kN⋅ m
Mz = 4.8 kN⋅ m
A := π ⋅ ⎛⎝ ro ⎞⎠
2
Nx = 10 kN
Vy = −10 kN
A
Iz
zC := ro
σ C :=
M yz
Iy
Nx Mz⋅ yC My⋅ zC
−
+
A
I
I
σ C = −102.6 MPa (C)
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for τ v and the torsional formula for τ t respectively.
Vy⋅ QC_y
V⋅ Q
τv =
τ v_y :=
τ v_y = −4.72 MPa
I⋅ t
I⋅ 2ro
τt =
T⋅ ρ
J
Mx⋅ ρ
τ t :=
J
τ xy := τ v_y − τ t
( )
τ t = −8.25 MPa
τ xy = 3.54 MPa
Ans
τ xz := 0
Ans
τ yz := 0
Ans
Ans
Problem 8-56
The 25-mm-diameter rod is subjected to the loads shown. Determine the state of stress at point A, and
show the results on a differential element located at this point.
Given: ax := −200mm az := 75mm
Px := −375N Py := −400N
do := 25mm
Pz := 500N
ro := 0.5do
Solution:
Internal Force and Moment :
ΣF x=0;
ΣF y=0;
Nx + Px = 0
Vy + Py = 0
Nx := −Px
Vy := −Py
ΣF z=0;
Vz + Pz = 0
Vz := −Pz
ΣΜx=0;
T x − Py⋅ az = 0
ΣΜy=0;
ΣΜz=0;
T x := Py⋅ ( az)
( )
My − Pz⋅ ( ax) + Px⋅ ( az) = 0 My := Pz⋅ ( ax) − Px⋅ ( az)
Mz + Py⋅ ( ax) = 0
Mz := −Py⋅ ( ax)
A := π ⋅ ro
Section Property :
Iy :=
π ⎛ 4⎞
⋅ r
4 ⎝o ⎠
Iz := Iy
QA_y := 0
J :=
QA_z :=
π ⎛ 4⎞
⋅ r
2 ⎝o ⎠
Normal Stress:
At A:
Shear Stress :
At A:
2
σ=
4ro
⎛ π 2⎞
⋅ ⎜ ⋅ ro
3π ⎝ 2
⎠
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
yA := ro
zA := 0
Nx Mz⋅ yA My⋅ zA
σ A :=
−
+
A
Iz
Iy
τv =
V⋅ Q
I⋅ b
bz := 2⋅ ro
τ v_z :=
Vz⋅ QA_z
Iy⋅ bz
τ A_xz := τ v_z + τ t
τt =
σ A = 52.9 MPa
T⋅ ρ
J
ρ := ro
τ t :=
T x⋅ ρ
J
τ A_xz = 11.14 MPa
Ans
τ A_xy := 0
Ans
(T)
Ans
Problem 8-57
The 25-mm-diameter rod is subjected to the loads shown. Determine the state of stress at point B, and
show the results on a differential element located at this point.
Given: ax := −200mm az := 75mm
Px := −375N Py := −400N
do := 25mm
Pz := 500N
ro := 0.5do
Solution:
Internal Force and Moment :
ΣF x=0;
ΣF y=0;
Nx + Px = 0
Vy + Py = 0
Nx := −Px
Vy := −Py
ΣF z=0;
Vz + Pz = 0
Vz := −Pz
ΣΜx=0;
T x − Py⋅ az = 0
ΣΜy=0;
ΣΜz=0;
T x := Py⋅ ( az)
( )
My − Pz⋅ ( ax) + Px⋅ ( az) = 0 My := Pz⋅ ( ax) − Px⋅ ( az)
Mz + Py⋅ ( ax) = 0
Mz := −Py⋅ ( ax)
A := π ⋅ ro
Section Property :
Iy :=
π ⎛ 4⎞
⋅ r
4 ⎝o ⎠
Iz := Iy
QB_z := 0
J :=
QB_y :=
π ⎛ 4⎞
⋅ r
2 ⎝o ⎠
Normal Stress:
At B:
Shear Stress :
At B:
2
σ=
4ro
⎛ π 2⎞
⋅ ⎜ ⋅ ro
3π ⎝ 2
⎠
N Mz⋅ y My⋅ z
−
+
Iz
A
Iy
yB := 0
zB := ro
Nx Mz⋅ yB My⋅ zB
σ B :=
−
+
A
Iz
Iy
τv =
V⋅ Q
I⋅ b
by := 2⋅ ro
τ v_y :=
Vy⋅ QB_y
Iz⋅ by
τ B_xy := τ v_y − τ t
τt =
σ B = −46.1 MPa
T⋅ ρ
J
ρ := ro
τ t :=
T x⋅ ρ
J
τ B_xy = 10.86 MPa
Ans
τ B_xz := 0
Ans
(C)
Ans
Problem 8-58
The crane boom is subjected to the load of 2.5 kN. Determine the state of stress at points A and B.
Show the results on a differential volume element located at each of these points.
Given:
b := 75mm
d := 76mm
t := 12mm
ah := 1.5m
v := 4
av := 2.4m
h := 3
P := 2.5kN
r := 5
Solution:
Internal Force and Moment :
⎛ h⎞ = 0
⎝r⎠
V := P⋅ ⎜
⎛ v⎞ = 0
⎝r⎠
N := −P⋅ ⎜
+
ΣF x=0;
− V + P⋅ ⎜
+
ΣF y=0;
− N − P⋅ ⎜
+ ΣΜO=0;
M − P⋅ ⎜
⎛ h⎞
⎝r⎠
⎛ v⎞
⎝r⎠
⎛ h ⎞ ⋅ a − P⋅ ⎛ v ⎞ ⋅ a = 0
⎜
⎝r⎠ v
⎝r⎠ h
⎛ v⎞
⎛ h⎞
M := P⋅ ⎜ ⋅ ah + P⋅ ⎜ ⋅ av
⎝r⎠
⎝r⎠
D := d + 2t
Section Property :
A := b⋅ D − t⋅ d
I :=
1
1
3
3
⋅ b⋅ D −
⋅ ( b − t) ⋅ d
12
12
QA := 0
QB := 0
Normal Stress:
σ=
cA := 0.5D
σ A :=
N M⋅ cA
+
A
I
σ A = 83.34 MPa
(T)
Ans
cB := −0.5 D
σ B :=
N M⋅ cB
+
A
I
σ B = −83.95 MPa (C)
Ans
Shear Stress :
τ=
τ A :=
τ B :=
V⋅ QA
I⋅ b
V⋅ QB
I⋅ b
(since A' = 0)
N M⋅ c
+
I
A
V⋅ Q
I⋅ b
τ A = 0 MPa
Ans
τ B = 0 MPa
Ans
Problem 8-59
The masonry pier is subjected to the 800-kN load. Determine the equation of the line y = f (x) along
which the load can be placed without causing a tensile stress in the pier. Neglect the weight of the pier.
Given: a := 1.5m
b := 2.25m
xA := −a
Pz := −800kN
yA := −b
Solution:
Section Property :
A := ( 2a) ⋅ ( 2⋅ b)
A = 13.5 m
1
3
⋅ ( 2a) ( 2⋅ b)
12
1
3
Iy :=
⋅ ( 2b) ( 2⋅ a)
12
Force and Moment :
Ix :=
2
Ix = 22.78125 m
Iy = 10.125 m
Mx = Pz⋅ y
My = −Pz⋅ x
Normal Stress: Require σ A := 0
Pz Mx⋅ yA My⋅ xA
σA =
+
−
A
Iy
Ix
0=
Pz
A
+
(Pz⋅ y)⋅ yA + (Pz⋅ x)⋅ xA
Ix
Iy
0=
⎛ xA ⎞
1 ⎛ yA ⎞
+⎜
⋅y + ⎜
⋅x
A
Iy
Ix
0=
1
⎛
−
4a⋅ b ⎜
y=
b b
− ⋅x
3 a
⎝
⎠
⎝
⎠
⎞⋅ y − ⎛ 3 ⎞⋅ x
⎜
2
2
⎝ 4a⋅ b ⎠
⎝ 4b⋅ a ⎠
3
y = 0.75 − 1.5 ⋅ x
Ans
4
4
Problem 8-60
The masonry pier is subjected to the 800-kN load. If x = 0.25 m and y = 0.5 m, determine the normal
stress at each corner A, B, C, D (not shown) and plot the stress distribution over the cross section.
Neglect the weight of the pier.
Unit Used:
3
kPa := 10 Pa
Given: a := 1.5m
b := 2.25m
Pz := −800kN
xA := −a
xB := a
xC := a
xD := −a
yA := −b
yB := −b
yC := b
yD := b
x := 0.25m
y := 0.5m
Solution:
Section Property :
A := ( 2a) ⋅ ( 2⋅ b)
A = 13.5 m
1
3
⋅ ( 2a) ( 2⋅ b)
12
1
3
Iy :=
⋅ ( 2b) ( 2⋅ a)
12
Force and Moment :
Ix :=
Ix = 22.78125 m
Iy = 10.125 m
Mx := Pz⋅ y
Mx = −400 kN⋅ m
My := −Pz⋅ x
My = 200 kN⋅ m
Normal Stress:
σ A :=
σ B :=
σ C :=
σ D :=
Pz
A
Pz
A
Pz
A
Pz
A
σ=
Pz
A
2
+
4
4
Mx⋅ y My⋅ x
−
Iy
Ix
+
Mx⋅ yA My⋅ xA
−
Iy
Ix
σ A = 9.877 kPa
+
Mx⋅ yB My⋅ xB
−
Iy
Ix
σ B = −49.38 kPa (C) Ans
+
Mx⋅ yC My⋅ xC
−
Iy
Ix
σ C = −128.4 kPa (C) Ans
+
Mx⋅ yD My⋅ xD
−
Iy
Ix
σ D = −69.1 kPa
(T) Ans
(C) Ans
Problem 8-61
The symmetrically loaded spreader bar is used to lift the 10-kN (~1-tonne) tank. Determine the state of
stress at points A and B, and indicate the results on a differential volume elements.
Given:
b := 25mm
d := 50mm
a := 0.45m
L := 1.2m
W := 10kN
θ := 30deg
Solution:
Support Reactions :
−W + 2F⋅ cos ( θ ) = 0
ΣFy=0;
+
F :=
W
2⋅ cos ( θ )
Internal Force and Moment :
+
ΣFx=0;
F⋅ sin ( θ ) − N = 0
N := F⋅ sin ( θ )
+
ΣFy=0;
V − F⋅ cos ( θ ) = 0
V := F⋅ cos ( θ )
+ ΣΜB=0;
M − F⋅ cos ( θ ) ⋅ a = 0
M := F⋅ cos ( θ ) ⋅ a
Section Property :
A := b⋅ d
I :=
1
3
⋅ b⋅ d
12
QB := ( 0.5 ⋅ d⋅ b) ⋅ ( 0.25d)
QA := 0
(since A' = 0)
N M⋅ c
+
I
A
Normal Stress:
σ=
cA := 0.5d
σ A :=
N M⋅ cA
+
A
I
σ A = 218.31 MPa (T)
Ans
cB := 0
σ B :=
N M⋅ cB
+
A
I
σ B = 2.31 MPa
Ans
Shear Stress :
τ=
τ A :=
τ B :=
V⋅ QA
I⋅ b
V⋅ QB
I⋅ b
V⋅ Q
I⋅ b
τ A = 0 MPa
Ans
τ B = 6.00 MPa
Ans
(T)
Problem 8-62
A post having the dimensions shown is subjected to the bearing load P. Specify the region to which
this load can be applied without causing tensile stress to be developed at points A, B, C, and D.
Problem 8-63
The man has a mass of 100 kg and center of mass at G. If he holds himself in the position shown,
determine the maximum tensile and compressive stress developed in the curved bar at section a-a. He
is supported uniformly by two bars, each having a diameter of 25 mm. Assume the floor is smooth.
Given: Ri := 150mm
e := 0.3m
do := 25mm
mo := 100kg
a := 0.35m
b := 1m
Solution:
Equilibrium: For the man.
(mo⋅ g)⋅ b − 2P⋅ ( a + b) = 0
b
P := ( mo⋅ g) ⋅
2( a + b)
+ ΣΜtoe=0;
P = 0.3632 kN
Section Property :
ro := 0.5do
rc := Ri + ro
2
A = 490.87 mm
dA
A r
IA_r := 2π ⋅ ⎛⎝ rc −
A := π ⋅ ro
2
IA_r = Σ ∫
rc − ro ⎞⎠
2
2
IA_r = 3.0252 mm
R :=
A
R = 162.259 mm
IA_r
Internal Force and Moment :
As shown on BFBD. The internal moment must be
computed about the neutral axis. M is negative since it
tends to decrease the bar's radius of curvature.
N := −P
N = −0.3632 kN
M := −P⋅ ( R + e)
M = −0.16790 kN⋅ m
Maximum Normal Stress:
For tensile stress,
r2 := rc + ro
σt =
)
)
M⋅ ( R − r2)
N
σ t :=
+
A A⋅ r2⋅ ( rc − R)
For compressive stress,
r1 := rc − ro
(
M⋅ R − r2
N
+
A A⋅ r1⋅ rc − R
σc =
(
(
σ t = 102.7 MPa
(T)
Ans
)
M⋅ R − r1
N
+
A A⋅ r1⋅ rc − R
(
)
M⋅ ( R − r1)
N
σ c :=
+
A A⋅ r1⋅ ( rc − R)
σ c = −116.9 MPa
(C)
Ans
Problem 8-64
The block is subjected to the three axial loads shown. Determine the normal stress developed at points
A and B. Neglect the weight of the block.
Given: b := 100mm
b' := 50mm
d := 75mm
d' := 125mm
P1 := 500N
P2 := 1250N
Solution:
B := b + 2b'
P3 := 250N
D := d + 2d'
Internal Force and Moment :
+
ΣF z=0;
− N − P1 − P2 − P3 = 0
(
N := − P1 + P2 + P3
+ ΣΜx=0;
)
Mx − P1⋅ ( 0.5d) − P2⋅ ( 0.5d) + P3⋅ ( 0.5D) = 0
Mx := ⎡⎣P1⋅ ( 0.5 ⋅ d) + P2⋅ ( 0.5 ⋅ d)⎤⎦ − P3⋅ ( 0.5 ⋅ D)
+ ΣΜy=0;
My + P1⋅ ( 0.5B) − P2⋅ ( 0.5B) − P3⋅ ( 0.5 ⋅ b) = 0
My := −P1⋅ ( 0.5 ⋅ B) + P2⋅ ( 0.5 ⋅ B) + P3⋅ ( 0.5 ⋅ b)
Section Property :
A := B⋅ D − 4b'⋅ d'
Normal Stress:
At A:
σ=
1
2
3
3
⋅ b⋅ D +
⋅ b'⋅ d
12
12
1
2
3
3
Iy :=
⋅ d⋅ B +
⋅ d'⋅ b
12
12
Ix :=
N Mx⋅ y My⋅ x
−
−
Ix
A
Iy
xA := 0.5B
σ A :=
At B:
N Mx⋅ yA My⋅ xA
−
−
A
Ix
Iy
xB := 0.5b
σ B :=
yA := −0.5 d
σ A = −0.1703 MPa (C)
Ans
σ B = −0.0977 MPa (C)
Ans
yB := −0.5 D
N Mx⋅ yB My⋅ xB
−
−
A
Ix
Iy
Problem 8-65
If P = 15 kN, plot the distribution of stress acting over the cross section a-a of the offset link.
Given: ho := 50mm
P := 15kN
to := 10mm
a := 30mm
Solution:
Section Property :
2
A := ho⋅ to
A = 500 mm
1
3
⋅ t o ho
12
I :=
4
I = 104166.67 mm
Moment :
(
M := P⋅ a + 0.5ho
)
M = 0.825 kN⋅ m
Normal Stress:
σ=
N Mc
±
A
I
y
σA
=
ho − y
σB
(
)
(
)
σ A :=
M⋅ 0.5ho
P
+
A
I
σ A = 228 MPa
(T)
Ans
σ B :=
M⋅ 0.5ho
P
−
A
I
σ B = −168 MPa
(C)
Ans
y := ho⋅
σA
σA + σB
y = 28.79 mm
Problem 8-66
Determine the magnitude of the load P that will cause a maximum normal stress σmax = 200 MPa of in
the link at section a-a.
Given: ho := 50mm
a := 30mm
to := 10mm
σ allow := 200MPa
Solution:
Section Property :
2
A := ho⋅ to
I :=
A = 500 mm
1
3
⋅t h
12 o o
4
I = 104166.67 mm
Moment :
(
)
M = P⋅ a + 0.5ho
Normal Stress:
σ=
N Mc
±
A
I
The maximum normal stress occurs at A.
σA =
(
M⋅ 0.5ho
P
+
A
I
σ allow =
)
(
)(
P⋅ a + 0.5ho ⋅ 0.5ho
P
+
A
I
)
I
P := σ allow⋅ A ⋅
I + A⋅ a + 0.5ho ⋅ 0.5ho
(
P = 13.16 kN
)
(
Ans
)(
)
Problem 8-67
Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a
radius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall
of the cylinder.
Given:
t := 2mm
P := 2kN
ri := 45mm
Solution:
A := πri
2
p :=
P
A
p = 0.3144 MPa
α :=
ri
α = 22.50
t
Since α > 10. then thin-wall analysis can be used.
Hoop Stress :
σ 1 :=
p⋅ ri
σ 1 = 7.07 MPa
t
Ans
Longitudinal Stress :
σ 2 := 0
Ans
The pressure p is supported by the surface of the
pistons in the longitudinal direction.
Problem 8-68
Determine the maximum force P that can be exerted on each of the two pistons so that the
circumferential stress component in the cylinder does not exceed 3 MPa. Each piston has a radius of
45 mm and the cylinder has a wall thickness of 2 mm.
t := 2mm
Given:
σ allow := 3MPa
ri := 45mm
Solution:
A := πri
2
p=
α :=
P
A
ri
α = 22.50
t
Since α > 10. then thin-wall analysis can be used.
Hoop Stress :
σ1 =
p⋅ r i
t
σ allow =
P :=
P⋅ r i
A⋅ t
σ allow
ri
⋅ A⋅ t
P = 0.848 kN
Ans
Problem 8-69
The screw of the clamp exerts a compressive force of 2.5 kN on the wood blocks. Determine the
maximum normal stress developed along section a-a. The cross section there is rectangular, 18 mm by
12 mm.
Given: b := 12mm
d := 18mm
a := 100mm
P := 2.5kN
Solution:
Internal Force and Moment :
N := P
M := P⋅ a
Section Property :
A := b⋅ d
I :=
1
3
⋅ b⋅ d
12
Normal Stress:
σ=
N M⋅ c
+
I
A
cmax := 0.5d
σ max :=
N M⋅ cmax
+
A
I
σ max = 397.4 MPa (T)
Ans
Problem 8-70
The wall hanger has a thickness of 6 mm and is used to support the vertical reactions of the beam that
is loaded as shown. If the load is transferred uniformly to each strap of the hanger, determine the state
of stress at points C and D of the strap at B. Assume the vertical reaction F at this end acts in the
center and on the edge of the bracket as shown.
Given: t := 6mm
L 1 := 1.2m
d := 50mm
P := 50kN
L 2 := 1.8m
w := 30
kN
m
L := L 1 + L 2
Solution:
Support Reactions : Given
+
ΣF y=0;
FA − P − w⋅ L2 + FB = 0
ΣΜB=0;
FA⋅ L − P⋅ L − 0.5L1 − w⋅ L2⋅ 0.5L2 = 0
(
)
(
)
Guess FA := 1kN
FB := 1kN
⎛⎜ FA ⎞
:= Find FA , FB
⎜ FB
⎝ ⎠
(
)
⎛⎜ FA ⎞ ⎛ 56.20 ⎞
=⎜
kN
⎜ FB
47.80
⎝
⎠
⎝
⎠
Section Property :
A := 2t⋅ d
I :=
2
3
⋅ t⋅ d
12
At Section CD:
P := FB
Stresses:
σ=
At C:
yC := 0
M := P⋅ ( 0.5d)
M⋅ y
P
+
I
A
σ C :=
τ=
V := 0
V⋅ Q
I⋅ b
M⋅ yC
P
+
A
I
σ C = 79.67 MPa
(T)
τ C := 0
At D:
yD := −0.5 d
σ D :=
Ans
Ans
M⋅ yD
P
+
A
I
σ D = −159.33 MPa
τ D := 0
(C)
Ans
Ans
Problem 8-71
The support is subjected to the compressive load P. Determine the absolute maximum and minimum
normal stress acting in the material.
Problem 8-72
The support has a circular cross section with a radius that increases linearly with depth. If it is
subjected to the compressive load P, determine the maximum and minimum normal stress acting in the
material.
Problem 8-73
The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter
of 1.5 m and a wall thickness of 18 mm. If the largest normal stress is not to exceed 150 MPa,
determine the maximum pressure the tank can sustain. Also, compute the number of bolts required to
attach the cap to the tank if each bolt has a diameter of 20 mm. The allowable stress for the bolts is
(σallow)b = 180 MPa.
Given:
t := 18mm
di := 1.5m
σ allow := 150MPa
db := 20mm
σ b.allow := 180MPa
Solution: ri := 0.5di
ri
Hoop Stress :
α :=
α = 41.67
t
Since α > 10. then thin-wall analysis can be used.
σ1 =
p⋅ r i
t
σ allow =
p :=
p⋅ r i
t
σ allow⋅ t
ri
p = 3.6 MPa
Force Equilibrium for the Cap :
+
ΣF y=0;
Ans
A := πri
2
p⋅ A − F b = 0
Fb := p⋅ A
Fb = 6361.73 kN
Allowable Normal Stress for Bolts :
σ b.allow =
n :=
Fb
n⋅ Ab
Fb
(σb.allow)⋅ Ab
n = 112.5
Use
n := 113
Ans
π 2
Ab := db
4
Problem 8-74
The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter
of 1.5 m and a wall thickness of 18 mm. If the pressure in the tank is p = 1.20 MPa, determine the
force in the 16 bolts that are used to attach the cap to the tank. Also, specify the state of stress in the
wall of the tank.
Given:
t := 18mm
di := 1.5m
p := 1.20MPa
n := 16
Solution: ri := 0.5di
ri
Hoop Stress :
α :=
α = 41.67
t
Since α > 10. then thin-wall analysis can be used.
σ 1 :=
p⋅ ri
t
σ 1 = 50 MPa
Ans
Longitudinal Stress :
σ 2 :=
p⋅ ri
2⋅ t
σ 2 = 25 MPa
Ans
Force Equilibrium for the Cap :
+
ΣF y=0;
A := πri
p⋅ A − 16⋅ Fb = 0
Fb :=
p⋅ A
16
Fb = 132.5 kN
Ans
2
Problem 8-75
The crowbar is used to pull out the nail at A. If a force of 40 N is required, determine the stress
components in the bar at points D and E. Show the results on a differential volume element located at
each of these points. The bar has a circular cross section with a diameter of 12 mm. No slipping
occurs at B.
Given: do := 12mm
dA := 60mm
a := 125mm
dP := 300mm
F := 40N
hP := 300mm
rp :=
2
dP + hP
Support Reactions :
Solution:
+ ΣΜB=0;
hA := 75mm
2
( )
F⋅ hA − P⋅ rp = 0
P :=
F ⋅ hA
rp
Internal Force and Moment :
+
ΣF x=0;
+
ΣF y=0;
V−P= 0
V := P
+ ΣΜO=0;
M − P⋅ a = 0
M := P⋅ a
N := 0
Section Property :
A := π ⋅ ro
QE :=
2
4ro
ro := 0.5do
I :=
π 4
⋅r
4 o
⋅ ⎛⎝ 0.5π ⋅ ro ⎞⎠
2
3π
QD := 0 (since A' = 0)
N M⋅ c
+
I
A
Normal Stress:
σ=
cD := ro
σ D :=
N M⋅ cD
+
A
I
σ D = 5.21 MPa (T) Ans
cE := 0
σ E :=
N M⋅ cE
+
A
I
σ E = 0.00 MPa
Shear Stress :
τ=
bE := 2⋅ ro
Ans
V⋅ Q
I⋅ b
τ E :=
V⋅ QE
I⋅ bE
τ E = 0.0834 MPa
Ans
τ D := 0
Ans
Problem 8-76
The screw of the clamp exerts a compressive force of 2.5 kN on the wood blocks. Sketch the stress
distribution along section a-a of the clamp. The cross section there is rectangular, 18 mm by 12 mm.
Given: b := 12mm
d := 18mm
a := 100mm
P := 2.5kN
Solution:
Internal Force and Moment :
N := P
M := P⋅ a
Section Property :
A := b⋅ d
I :=
1
3
⋅ b⋅ d
12
Normal Stress:
σ=
N M⋅ c
+
I
A
cmax := 0.5d
σ max :=
N M⋅ cmax
+
A
I
σ max = 397.4 MPa (T)
cmin := 0.5d
σ min :=
Ans
N M⋅ cmin
−
A
I
σ min = −374.2 MPa (C) Ans
σ min
y
=
d−y
σ max
y :=
d σ min
σ min + σ max
y = 8.73 mm
Problem 8-77
The clamp is made from members AB and AC, which are pin connected at A. If the compressive force
at C and B is 180 N, determine the state of stress at point F, and indicate the results on a differential
volume element. The screw DE is subjected only to a tensile force along its axis.
Given:
h := 15mm
t := 15mm
a := 30mm
b := 40mm
P := 180N
Solution:
Support Reactions :
+ ΣΜO=0;
P⋅ ( b + a) − FDE⋅ ( a) = 0
b+a
⋅P
a
FDE :=
FDE = 0.420 kN
Internal Force and Moment :
+ ΣF y=0;
+ ΣF x=0;
+ ΣΜO=0;
N' := 0
V + FDE − P = 0
V := P − FDE
M + P⋅ ( b + 0.5a) − FDE⋅ ( 0.5a) = 0
M := FDE⋅ ( 0.5a) − P⋅ ( b + 0.5a)
M = −3.60 N⋅ m
Section Property:
2
A := h⋅ t
A = 225 mm
1
3
⋅ t⋅ h
12
QF := 0 (since A' = 0)
I :=
4
I = 4218.75 mm
N' M⋅ c
+
A
I
Normal Stress:
σ=
cF := 0.5 ⋅ h
σ F :=
N' M⋅ cF
+
A
I
σ F = −6.40 MPa (C)
Shear Stress :
τ F :=
V⋅ QF
I⋅ t
τ=
V⋅ Q
I⋅ b
τ F = 0 MPa
Ans
Ans
Problem 8-78
The eye is subjected to the force of 250 N. Determine the maximum tensile and compressive stresses
at section a-a. The cross section is circular and has a diameter of 6 mm. Use the curved-beam formula
to compute the bending stress.
Ri := 30mm
do := 6mm
Solution: ro := 0.5do
Section Property :
rc := Ri + ro
Given:
P := 0.250kN
2
A = 28.27 mm
dA
A r
IA_r := 2π ⋅ ⎛⎝ rc −
A := π ⋅ ro
2
IA_r = Σ ∫
rc − ro ⎞⎠
2
2
IA_r = 0.8586 mm
R :=
A
R = 32.932 mm
IA_r
Internal Force and Moment :
As shown on BFBD. The internal moment must be
computed about the neutral axis. M is positive since it
tends to increase the beam's radius of curvature.
N := P
M := P⋅ R
Maximum Normal Stress:
For tensile stress,
r1 := rc − ro
σt =
(
)
M⋅ ( R − r1)
N
σ t :=
+
A A⋅ r1⋅ ( rc − R)
σ t = 425.3 MPa
For compressive stress,
r2 := rc + ro
)
M⋅ R − r1
N
+
A A⋅ r1⋅ rc − R
σc =
(
(T)
Ans
(
)
M⋅ R − r2
N
+
A A⋅ r1⋅ rc − R
(
)
M⋅ ( R − r2)
N
σ c :=
+
A A⋅ r2⋅ ( rc − R)
σ c = −354.4 MPa
(C)
Ans
Problem 8-79
Solve Prob. 8-78 if the cross section is square, having dimensions of 6 mm by 6 mm.
Ri := 30mm
do := 6mm
Solution: ro := 0.5do
Section Property :
rc := Ri + ro
Given:
A := do
2
P := 0.250kN
2
A = 36.00 mm
⎛ rc + ro ⎞
IA_r := do ⋅ ln ⎜
⎝ rc − ro ⎠
( )
dA
IA_r = Σ ∫
A r
IA_r = 1.0939 mm
R :=
A
R = 32.91 mm
IA_r
Internal Force and Moment :
As shown on BFBD. The internal moment must be
computed about the neutral axis. M is positive since it
tends to increase the beam's radius of curvature.
N := P
M := P⋅ R
Maximum Normal Stress:
For tensile stress,
r1 := rc − ro
σt =
(
)
M⋅ ( R − r1)
N
σ t :=
+
A A⋅ r1⋅ ( rc − R)
σ t = 250.2 MPa
For compressive stress,
r2 := rc + ro
)
M⋅ R − r1
N
+
A A⋅ r1⋅ rc − R
σc =
(
(T)
Ans
(
)
M⋅ R − r2
N
+
A A⋅ r1⋅ rc − R
(
)
M⋅ ( R − r2)
N
σ c :=
+
A A⋅ r2⋅ ( rc − R)
σ c = −208.4 MPa
(C)
Ans
Problem 9-1
Prove that the sum of the normal stresses σx + σy = σx' + σy' is constant. See Figs. 9-2a and 9-2b.
Solution:
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text.
σ x' =
σ y' =
σx + σy
2
σx + σy
2
+
−
σx − σy
2
σx − σy
2
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
(1)
⋅ cos ( 2θ ) − τ xy⋅ sin ( 2θ )
(2)
(1) + (2) :
LHS = σ x' + σ y'
RHS =
σx + σy
2
+
σx + σy
2
Hence,
σ x' + σ y' = σ x + σ y
(Q.E.D.)
RHS = σ x + σ y
Problem 9-2
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σ x := 3MPa
Solution:
Set
σ y := 5MPa
∆ A := m
Force Equilibrium:
2
τ xy := −8MPa φ := 40deg
θ := 180deg + φ
For the sectioned element,
∆ Ax := ∆ A⋅ cos ( φ )
∆ Ay := ∆ A⋅ sin ( φ )
Fxx := σ x⋅ ∆ Ax
Fxy := τ xy⋅ ∆ Ax
Fyy := σ y⋅ ∆ Ay
Fyx := τ xy⋅ ∆ Ay
Given
+
+ ΣF x'=0;
∆ Fx' + Fxy⋅ sin ( θ ) + Fxx⋅ cos ( θ ) + Fyx⋅ cos ( θ ) + Fyy⋅ sin ( θ ) = 0
ΣF y'=0;
∆ Fy' + Fxy⋅ cos ( θ ) − Fxx⋅ sin ( θ ) − Fyx⋅ sin ( θ ) + Fyy⋅ cos ( θ ) = 0
Guess
∆ Fx' := 1kN
∆ Fy' := 1kN
⎛⎜ ∆ Fx' ⎞
:= Find ( ∆ Fx' , ∆ Fy')
⎜ ∆ Fy'
⎝
⎠
Normal and Shear Stress:
σ x' :=
∆ Fx'
τ x'y' :=
∆A
∆ Fy'
∆A
σ=
⎛⎜ ∆ Fx' ⎞ ⎛ −4052.11 ⎞
=⎜
kN
⎜ ∆ Fy'
−404.38 ⎠
⎝
⎝
⎠
⎛ F⎞
⎜
A → 0 ⎝ A⎠
lim
σ x' = −4.052 MPa
Ans
τ x'y' = −0.404 MPa
Ans
The negative signs indicate that the sense of σx' and τx'y' are opposite to that shown in FBD.
Problem 9-3
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σ x := −0.200 MPa σ y := 0.350MPa
φ := 50deg
Solution:
Set
τ xy := 0MPa
∆ A := m
Force Equilibrium:
2
θ := 180deg + φ
For the sectioned element,
∆ Ax := ∆ A⋅ cos ( φ )
∆ Ay := ∆ A⋅ sin ( φ )
Fxx := σ x⋅ ∆ Ax
Fxy := τ xy⋅ ∆ Ax
Fxy = 0.00
Fyy := σ y⋅ ∆ Ay
Fyx := τ xy⋅ ∆ Ay
Fyx = 0.00
Given
+
+ ΣF x'=0;
∆ Fx' + Fxy⋅ sin ( θ ) + Fxx⋅ cos ( θ ) + Fyx⋅ cos ( θ ) + Fyy⋅ sin ( θ ) = 0
ΣF y'=0;
∆ Fy' + Fxy⋅ cos ( θ ) − Fxx⋅ sin ( θ ) − Fyx⋅ sin ( φ ) + Fyy⋅ cos ( θ ) = 0
Guess
∆ Fx' := 1kN
∆ Fy' := 1kN
⎛⎜ ∆ Fx' ⎞
:= Find ( ∆ Fx' , ∆ Fy')
⎜ ∆ Fy'
⎝
⎠
Normal and Shear Stress:
σ x' :=
∆ Fx'
τ x'y' :=
∆A
∆ Fy'
∆A
σ=
⎛⎜ ∆ Fx' ⎞ ⎛ 122.75 ⎞
=⎜
kN
⎜ ∆ Fy'
270.82 ⎠
⎝
⎝
⎠
⎛ F⎞
⎜
A → 0 ⎝ A⎠
lim
σ x' = 0.123 MPa
Ans
τ x'y' = 0.271 MPa
Ans
Problem 9-4
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σ x := −0.650 MPa σ y := 0.400MPa
φ := 60deg
Solution:
Set
τ xy := 0MPa
∆ A := m
θ := −( 90deg + φ )
2
For the sectioned element,
Force Equilibrium:
∆ Ax := ∆ A⋅ sin ( φ )
∆ Ay := ∆ A⋅ cos ( φ )
Fxx := σ x⋅ ∆ Ax
Fxy := τ xy⋅ ∆ Ax
Fxy = 0.00
Fyy := σ y⋅ ∆ Ay
Fyx := τ xy⋅ ∆ Ay
Fyx = 0.00
Given
+
ΣF x'=0;
∆ Fx' + Fxy⋅ sin ( θ ) + Fxx⋅ cos ( θ ) + Fyx⋅ cos ( θ ) + Fyy⋅ sin ( θ ) = 0
+ ΣF y'=0;
∆ Fy' + Fxy⋅ cos ( θ ) − Fxx⋅ sin ( θ ) − Fyx⋅ sin ( θ ) + Fyy⋅ cos ( θ ) = 0
Guess
∆ Fx' := 1kN
∆ Fy' := 1kN
⎛⎜ ∆ Fx' ⎞
:= Find ( ∆ Fx' , ∆ Fy')
⎜ ∆ Fy'
⎝
⎠
Normal and Shear Stress:
σ x' :=
∆ Fx'
τ x'y' :=
∆A
∆ Fy'
∆A
σ=
⎛⎜ ∆ Fx' ⎞ ⎛ −387.50 ⎞
=⎜
kN
⎜ ∆ Fy'
⎝
⎠ ⎝ 454.66 ⎠
⎛ F⎞
⎜
A → 0 ⎝ A⎠
lim
σ x' = −0.387 MPa
Ans
τ x'y' = 0.455 MPa
Ans
The negative signs indicate that the sense of σx' is opposite to that shown in FBD.
Problem 9-5
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σ x := −60MPa
σ y := −50MPa
φ := 30deg
Solution:
Set
τ xy := 28MPa
∆ A := m
Force Equilibrium:
2
θ := 180deg + φ
For the sectioned element,
∆ Ax := ∆ A⋅ cos ( φ )
∆ Ay := ∆ A⋅ sin ( φ )
Fxx := σ x⋅ ∆ Ax
Fxy := τ xy⋅ ∆ Ax
Fxy = 24248.71 kN
Fyy := σ y⋅ ∆ Ay
Fyx := τ xy⋅ ∆ Ay
Fyx = 14000.00 kN
Given
+
ΣF x'=0;
∆ Fx' + Fxy⋅ sin ( θ ) + Fxx⋅ cos ( θ ) + Fyx⋅ cos ( θ ) + Fyy⋅ sin ( θ ) = 0
+
ΣF y'=0;
∆ Fy' + Fxy⋅ cos ( θ ) − Fxx⋅ sin ( θ ) − Fyx⋅ sin ( θ ) + Fyy⋅ cos ( θ ) = 0
Guess
∆ Fx' := 1kN
∆ Fy' := 1kN
⎛⎜ ∆ Fx' ⎞
:= Find ( ∆ Fx' , ∆ Fy')
⎜ ∆ Fy'
⎝
⎠
Normal and Shear Stress:
σ x' :=
∆ Fx'
τ x'y' :=
∆A
∆ Fy'
∆A
σ=
⎛⎜ ∆ Fx' ⎞ ⎛ −33251.29 ⎞
=⎜
kN
⎜ ∆ Fy'
18330.13 ⎠
⎝
⎝
⎠
⎛ F⎞
⎜
A → 0 ⎝ A⎠
lim
σ x' = −33.251 MPa
Ans
τ x'y' = 18.330 MPa
Ans
The negative signs indicate that the sense of σx' is opposite to that shown in FBD.
Problem 9-6
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σ x := 90MPa
σ y := 50MPa
τ xy := −35MPa
Solution:
Set
φ := 60deg
∆ A := m
θ := −( 90deg + φ )
2
For the sectioned element,
Force Equilibrium:
∆ Ax := ∆ A⋅ sin ( φ )
∆ Ay := ∆ A⋅ cos ( φ )
Fxx := σ x⋅ ∆ Ax
Fxy := τ xy⋅ ∆ Ax
Fyy := σ y⋅ ∆ Ay
Fyx := τ xy⋅ ∆ Ay
Given
+
ΣF x'=0;
∆ Fx' + Fxy⋅ sin ( θ ) + Fxx⋅ cos ( θ ) + Fyx⋅ cos ( θ ) + Fyy⋅ sin ( θ ) = 0
+ ΣF y'=0;
∆ Fy' + Fxy⋅ cos ( θ ) − Fxx⋅ sin ( θ ) − Fyx⋅ sin ( θ ) + Fyy⋅ cos ( θ ) = 0
Guess
∆ Fx' := 1kN
∆ Fy' := 1kN
⎛⎜ ∆ Fx' ⎞
:= Find ( ∆ Fx' , ∆ Fy')
⎜ ∆ Fy'
⎝
⎠
Normal and Shear Stress:
σ x' :=
∆ Fx'
τ x'y' :=
∆A
∆ Fy'
∆A
σ=
⎛⎜ ∆ Fx' ⎞ ⎛ 49689.11 ⎞
=⎜
kN
⎜ ∆ Fy'
⎝
⎠ ⎝ −34820.51 ⎠
⎛ F⎞
⎜
A → 0 ⎝ A⎠
lim
σ x' = 49.69 MPa
Ans
τ x'y' = −34.82 MPa
Ans
The negative signs indicate that the sense of τx'y' isopposite to that shown in FBD.
Problem 9-7
Solve Prob. 9-2 using the stress-transformation equations developed in Sec. 9.2.
Given: σ x := 5MPa
Solution:
σ y := 3MPa
τ xy := 8MPa
φ := 40deg
θ := 90deg + φ
Normal Stress:
σ x' :=
σx + σy
2
+
σx − σy
2
σ x' = −4.05 MPa
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
The negative signs indicate that the sense of σx' is a compressive stress.
Shear Stress:
τ x'y' := −
σx − σy
2
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
τ x'y' = −0.404 MPa
Ans
The negative signs indicate that the sense of τx'y' is in the -y' direction.
Problem 9-8
Solve Prob. 9-4 using the stress-transformation equations developed in Sec. 9.2.
Given: σ x := −0.650 MPa σ y := 0.400MPa
φ := 60deg
τ xy := 0MPa
Solution: θ := 90deg − φ
Normal Stress:
σ x' :=
σx + σy
2
+
σx − σy
σ x' = −0.387 MPa
2
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
The negative signs indicate that the sense of σx' is a compressive stress.
Shear Stress:
τ x'y' := −
σx − σy
2
τ x'y' = 0.455 MPa
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
Ans
Problem 9-9
Solve Prob. 9-6 using the stress-transformation equations developed in Sec. 9.2. Show the result on a
sketch.
Given: σ x := 90MPa
σ y := 50MPa
τ xy := −35MPa
Solution:
φ := 60deg
θ := −( 90deg + φ )
Normal Stress:
σ x' :=
σx + σy
2
+
σx − σy
2
σ x' = 49.69 MPa
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
Shear Stress:
τ x'y' := −
σx − σy
2
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
τ x'y' = −34.82 MPa
Ans
The negative signs indicate that the sense of τx'y' is in the -y' direction.
Problem 9-10
Determine the equivalent state of stress on an element if the element is oriented 30° counterclockwise
from the element shown. Use the stress-transformation equations.
Unit Used: kPa := 1000Pa
Given: σ x := 0kPa
σ y := −300kPa
θ := 30deg
τ xy := 950kPa
Solution:
Normal Stress:
σ x' :=
σx + σy
2
+
σx − σy
2
σ x' = 747.7 kPa
σ y' :=
σx + σy
2
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
−
σx − σy
σ y' = −1047.7 kPa
2
⋅ cos ( 2θ ) − τ xy⋅ sin ( 2θ )
Ans
Shear Stress:
τ x'y' := −
σx − σy
2
τ x'y' = 345.1 kPa
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
Ans
Problem 9-11
Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from the
element shown.
Given: σ x := 0.300MPa
σ y := 0MPa
θ := −60deg
τ xy := 0.120MPa
Solution:
Normal Stress:
σ x' :=
σx + σy
2
+
σx − σy
2
σ x' = −0.0289 MPa
σ y' :=
σx + σy
2
−
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
σx − σy
2
σ y' = 0.329 MPa
⋅ cos ( 2θ ) − τ xy⋅ sin ( 2θ )
Ans
Shear Stress:
τ x'y' := −
σx − σy
2
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
τ x'y' = 0.0699 MPa
Ans
Problem 9-12
Solve Prob. 9-6 using the stress-transformation equations.
Given: σ x := 90MPa
σ y := 50MPa
φ := 60deg
Solution:
τ xy := −35MPa
θ := −( 90deg + φ )
Normal Stress:
σ x' :=
σx + σy
2
+
σx − σy
2
σ x' = 49.69 MPa
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
Shear Stress:
τ x'y' := −
σx − σy
2
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
τ x'y' = −34.82 MPa
Ans
Problem 9-13
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σ x := 45MPa
σ y := −60MPa
τ xy := 30MPa
Solution:
(a) Principal Stress:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
σ 1 = 52.97 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 2 = −67.97 MPa
Ans
Orientation of Principal Stress:
( )
tan 2θ p =
2τ xy
θ p :=
σx − σy
⎛ 2τ xy ⎞
1
atan ⎜
2
⎝ σx − σy ⎠
θ' p := θ p − 90deg
θ p = 14.87 deg
θ' p = −75.13 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x' :=
σx + σy
2
+
σx − σy
2
( )
( )
⋅ cos 2θ p + τ xy⋅ sin 2θ p
σ x' = 52.97 MPa
Therefore,
θ p1 := θ p
θ p1 = 14.87 deg
Ans
θ p2 := θ' p
θ p2 = −75.13 deg
Ans
(b)
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
2
⎝
⎠
σ avg :=
σx + σy
2
τ max = 60.47 MPa
Ans
σ avg = −7.50 MPa
Ans
Orientation of Maximum In-plane Shear Stress:
( )
tan 2θ s = −
σx − σy
2τ xy
θ s :=
⎛ σx − σy ⎞
1
atan ⎜ −
2
2τ xy
⎝
θ s = −30.13 deg
⎠
θ' s := θ s + 90deg
θ' s = 59.87 deg
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-14
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σ x := 180MPa
σ y := 0MPa
τ xy := −150MPa
Solution:
(a) Principal Stress:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
σ 1 = 264.93 MPa
Ans
σ 2 = −84.93 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
Orientation of Principal Stress:
( )
tan 2θ p =
2τ xy
θ p :=
σx − σy
⎛ 2τ xy ⎞
1
atan ⎜
2
⎝ σx − σy ⎠
θ' p := θ p + 90deg
θ p = −29.52 deg
θ' p = 60.48 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x' :=
σx + σy
2
+
σx − σy
2
( )
( )
⋅ cos 2θ p + τ xy⋅ sin 2θ p
σ x' = 264.93 MPa
Therefore,
θ p1 := θ p
θ p1 = −29.52 deg
Ans
θ p2 := θ' p
θ p2 = 60.48 deg
Ans
(b)
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
2
⎝
⎠
σ avg :=
σx + σy
2
τ max = 174.93 MPa
Ans
σ avg = 90.00 MPa
Ans
Orientation of Maximum In-plane Shear Stress:
( )
tan 2θ s = −
σx − σy
2τ xy
θ s :=
⎛ σx − σy ⎞
1
atan ⎜ −
2
2τ xy
⎝
θ s = 15.48 deg
⎠
θ' s := θ s − 90deg
θ' s = −74.52 deg
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-15
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σ x := −30MPa
σ y := 0MPa
τ xy := −12MPa
Solution:
(a) Principal Stress:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
σ 1 = 4.21 MPa
Ans
σ 2 = −34.21 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
Orientation of Principal Stress:
( )
tan 2θ p =
2τ xy
θ p :=
σx − σy
⎛ 2τ xy ⎞
1
atan ⎜
2
⎝ σx − σy ⎠
θ' p := θ p − 90deg
θ p = 19.33 deg
θ' p = −70.67 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x' :=
σx + σy
2
+
σx − σy
2
( )
( )
⋅ cos 2θ p + τ xy⋅ sin 2θ p
σ x' = −34.21 MPa
Therefore,
θ p1 := θ' p
θ p1 = −70.67 deg
Ans
θ p2 := θ p
θ p2 = 19.33 deg
Ans
(b)
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
2
⎝
⎠
σ avg :=
σx + σy
2
τ max = 19.21 MPa
Ans
σ avg = −15.00 MPa
Ans
Orientation of Maximum In-plane Shear Stress:
( )
tan 2θ s = −
σx − σy
2τ xy
θ s :=
⎛ σx − σy ⎞
1
atan ⎜ −
2
2τ xy
⎝
θ s = −25.67 deg
⎠
θ' s := θ s + 90deg
θ' s = 64.33 deg
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-16
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σ x := −200MPa
σ y := 250MPa
τ xy := 175MPa
Solution:
(a) Principal Stress:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
σ 1 = 310.04 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 2 = −260.04 MPa
Ans
Orientation of Principal Stress:
( )
tan 2θ p =
2τ xy
θ p :=
σx − σy
⎛ 2τ xy ⎞
1
atan ⎜
2
⎝ σx − σy ⎠
θ' p := θ p + 90deg
θ p = −18.94 deg
θ' p = 71.06 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x' :=
σx + σy
2
+
σx − σy
2
( )
( )
⋅ cos 2θ p + τ xy⋅ sin 2θ p
σ x' = −260.04 MPa
Therefore,
θ p1 := θ' p
θ p1 = 71.06 deg
Ans
θ p2 := θ p
θ p2 = −18.94 deg
Ans
(b)
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
2
⎝
⎠
σ avg :=
σx + σy
2
τ max = 285.04 MPa
Ans
σ avg = 25.00 MPa
Ans
Orientation of Maximum In-plane Shear Stress:
( )
tan 2θ s = −
σx − σy
2τ xy
θ s :=
⎛ σx − σy ⎞
1
atan ⎜ −
2
2τ xy
⎝
θ s = 26.06 deg
⎠
θ' s := θ s − 90deg
θ' s = −63.94 deg
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-17
A point on a thin plate is subjected to the two successive states of stress shown. Determine the
resultant state of stress represented on the element oriented as shown on the right.
Given:
(a)
(a)
σ x'_a := −200MPa
τ x'y'_a := 0MPa
σ y'_a := −350MPa
θ a := −30deg
σ x'_b := 0
τ x'y'_b := 58MPa
σ y'_b := 0
θ b := 25deg
Solution:
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 of the text.
For element (a):
⎛ σ x'_a + σ y'_a σ x'_a − σ y'_a
⎞
σ x_a := ⎜
+
⋅ cos 2θ a + τ x'y'_a⋅ sin 2θ a
2
2
( )
( )
⎝
⎠
⎛ σ x'_a + σ y'_a σ x'_a − σ y'_a
⎞
σ y_a := ⎜
−
⋅ cos ( 2θ a) − τ x'y'_a⋅ sin ( 2θ a)
2
2
⎝
⎠
⎛ σ x'_a − σ y'_a
⎞
τ xy_a := ⎜ −
⋅ sin ( 2θ a) + τ x'y'_a⋅ cos ( 2θ a)
2
⎝
⎠
For element (b):
⎛ σ x'_b + σ y'_b σ x'_b − σ y'_b
⎞
σ x_b := ⎜
+
⋅ cos 2θ b + τ x'y'_b⋅ sin 2θ b
2
2
( )
( )
⎝
⎠
σ
+
σ
σ
−
σ
⎛ x'_b y'_b
⎞
x'_b
y'_b
σ y_b := ⎜
−
⋅ cos ( 2θ b) − τ x'y'_b⋅ sin ( 2θ b)
2
2
⎝
⎠
σ x_a = −237.50 MPa
σ y_a = −312.50 MPa
τ xy_a = 64.95 MPa
σ x_b = 44.43 MPa
σ y_b = −44.43 MPa
⎛ σ x'_b − σ y'_b
τ xy_b := ⎜ −
⎝
2
( )
⎞
( )⎠
⋅ sin 2θ b + τ x'y'_b⋅ cos 2θ b
Combining the stress componenets of two elements yields
σ x := σ x_a + σ x_b
σ x = −193.1 MPa
Ans
σ y := σ y_a + σ y_b
σ y = −356.9 MPa
Ans
τ xy := τ xy_a + τ xy_b
τ xy = 102.2 MPa
Ans
τ xy_b = 37.28 MPa
Problem 9-18
The steel bar has a thickness of 12 mm and is subjected to the edge loading shown. Determine the
principal stresses developed in the bar.
d := 50mm
Given:
q := 4
kN
m
t := 12mm
L := 500mm
Solution:
Normal and Shear Stress:
In accordance with the established sign convention.
σ x := 0MPa
τ xy :=
q
t
σ y := 0MPa
τ xy = 0.333 MPa
In-plane Principal Stress:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
Apply Eq. 9-5.
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
σ 1 = 0.333 MPa
Ans
σ 2 = −0.333 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
Problem 9-19
The steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the
maximum in-plane shear stress and the average normal stress developed in the steel.
ax := 300mm
kN
qx := 30
m
Given:
ay := 100mm
kN
qy := 40
m
t := 10mm
Solution:
Normal and Shear Stress:
In accordance with the established sign convention.
qx
σ x :=
σ x = 3.00 MPa
t
σ y :=
qy
t
σ y = 4.00 MPa
τ xy := 0
Maximum In-plane Shear Stress:
Apply Eq. 9-7.
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
⎝ 2 ⎠
Average Normal Stress:
σ avg :=
σx + σy
2
τ max = 0.500 MPa
Ans
σ avg = 3.50 MPa
Ans
Apply Eq. 9-8.
Problem 9-20
The stress acting on two planes at a point is indicated. Determine the shear stress on plane a-a and the
principal stresses at the point.
Given: σ a := 80MPa
σ b := 60MPa
θ := 45deg
β := 60deg
Solution:
σ x := σ b⋅ sin ( β )
τ xy := σ b⋅ cos ( β )
Given
σa =
σx + σy
τa = −
Guess
2
+
σx − σy
2
σx − σy
2
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
σ y := 1MPa
⎛⎜ σ y ⎞
:= Find ( σ y , τ a)
⎜ τa
⎝ ⎠
τ a := 1MPa
⎛⎜ σ y ⎞ ⎛ 48.04 ⎞
=⎜
MPa
⎜ τa
⎝ ⎠ ⎝ −1.96 ⎠
Ans
Principal Stress:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 80.06 MPa
Ans
σ 2 = 19.94 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Problem 9-21
The stress acting on two planes at a point is indicated. Determine the normal stress σb and the principal
stresses at the point.
Given: σ a := 4MPa
τ x'y' := −2MPa
φ bb := 45deg
β := 60deg
Solution:
Stress Transformation Equations :
Applying Eqs. 9-3 and 9-1 with θ := −φ bb − 90deg
σ y := σ a⋅ sin ( β )
Given
σ x' =
σx + σy
2
τ x'y' = −
Guess
τ xy := σ a⋅ cos ( β )
+
σx − σy
2
σx − σy
2
σ x' = σ b
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
σ x := 1MPa
⎛⎜ σ x ⎞
:= Find ( σ x , σ x')
⎜ σ x'
⎝ ⎠
σ x' := 1MPa
⎛⎜ σ x ⎞ ⎛ 7.464 ⎞
=⎜
MPa
⎜ σ x'
⎝ ⎠ ⎝ 7.464 ⎠
σ b := σ x'
σ b = 7.464 MPa
In-plane Principal Stress:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
Ans
Applying Eq. 9-5,
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 8.29 MPa
Ans
σ 2 = 2.64 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Problem 9-22
The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the bolt
is 40 kN, determine the principal stresses at points A and B and show the results on elements located at
each of these points. The cross-sectional area at A and B is shown in the adjacent figure.
Given: h := 50mm
t := 30mm
a := 100mm
P := 40kN
b := 200mm
Solution: L := 3a + b
Support Reactions :
+ ΣΜO=0;
E'⋅ L − P⋅ ( a + b) = 0
P⋅ ( a + b)
E' = 24 kN
L
Internal Force and Moment : At Section A-B:
E' :=
+
ΣF x=0;
+ ΣΜO=0;
E+V= 0
V := −E'
M − E ⋅ ( a) = 0
M := E'⋅ a
Section Property :
2
A := h⋅ t
I :=
A = 1500 mm
1
3
⋅ t⋅ h
12
4
I = 312500 mm
QB := ( 0.5 ⋅ h⋅ t) ( 0.25 ⋅ h)
QA := 0 (since A' = 0)
M⋅ c
Normal Stress: σ =
I
M⋅ cA
cA := −0.5 h σ A :=
I
cB := 0
σ B :=
Shear Stress :
τ=
M⋅ cB
I
3
QB = 9375 mm
σ A = −192.00 MPa
σ B = 0 MPa
V⋅ Q
I⋅ t
τ A :=
τ B :=
V⋅ QA
I⋅ t
V⋅ QB
I⋅ t
τ A = 0.00 MPa
τ B = −24.00 MPa
In-plane Principal Stress:
At A:
σ xA := σ A
σ yA := 0
τ xy := τ A
Since no shear stress acts upon the element,
Ans
σ A1 := σ yA
σ A1 = 0 MPa
σ A2 := σ xA
σ A2 = −192 MPa
Ans
At B:
σ xB := σ B
σ 1 .2 =
σ yB := 0
σx +σy
2
τ xy := τ B
⎛σ −σy ⎞
2
⎟⎟ + τ xy
± ⎜⎜ x
2
⎝
⎠
2
σ B1 := −τ xy
σ B1 = 24 MPa
Ans
σ B2 := τ xy
σ B2 = −24 MPa
Ans
Applying Eq. 9-4 for point B,
Orientation of Principal Plane:
( )
tan 2θ p = −
2τ B
θ p :=
σ xB − σ yB
1
( 90deg)
2
θ' p := θ p − 90deg
θ p = 45 deg
θ' p = −45 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x'_B :=
σ xB + σ yB
2
+
σ xB − σ yB
2
( )
( )
⋅ cos 2θ p + τ B⋅ sin 2θ p
σ x'_B = −24.00 MPa
Therefore,
θ p1 := θ' p
θ p1 = −45.00 deg
Ans
θ p2 := θ p
θ p2 = 45.00 deg
Ans
Problem 9-23
Solve Prob. 9-22 for points C and D.
Given: h := 50mm
a := 100mm
t := 30mm
P := 40kN
b := 200mm
hD := 10mm
Solution: L := 3a + b
Support Reactions :
+ ΣΜE=0;
P⋅ ( 2a) − R⋅ L = 0
2P⋅ a
R = 16 kN
L
Internal Force and Moment : At Section C-D:
R :=
+
ΣF x=0;
+ ΣΜO=0;
R−V= 0
V := R
M − R⋅ ( b) = 0
M := R⋅ ( b)
Section Property :
2
A := h⋅ t
I :=
A = 1500 mm
1
3
⋅ t⋅ h
12
(
4
I = 312500 mm
)(
)
QD := hD⋅ t 0.5 ⋅ h − 0.5 ⋅ hD
QC := 0
3
QD = 6000 mm
(since A' = 0)
Normal Stress:
σ=
M⋅ c
I
cC := 0.5h
σ C :=
cD := −0.5 ⋅ h + hD
σ D :=
Shear Stress :
τ=
M⋅ cC
I
M⋅ cD
I
σ C = 256.00 MPa
σ D = −153.6 MPa
V⋅ Q
I⋅ t
τ C :=
τ D :=
V⋅ QC
I⋅ t
V⋅ QD
I⋅ t
τ C = 0.00 MPa
τ D = 10.24 MPa
In-plane Principal Stress:
At C:
σ xC := 0
σ yC := σ C
τ xy := τ C
Since no shear stress acts upon the element,
Ans
σ C1 := σ yC
σ C1 = 256 MPa
σ C2 := σ xC
At D:
σ xD := 0
σ C2 = 0 MPa
σ yD := σ D
τ xy := τ D
Ans
σ D1 :=
σ D2 :=
σ xD + σ yD
2
σ xD + σ yD
2
2
⎛ σ xD − σ yD ⎞
2
+ ⎜
+ τD
2
⎝
⎠
( )
2τ D
σ xD − σ yD
Ans
σ D2 = −154.28 MPa
Ans
2
⎛ σ xD − σ yD ⎞
2
− ⎜
+ τD
2
⎝
⎠
Applying Eq. 9-4 for point D,
Orientation of Principal Plane:
tan 2θ p =
σ D1 = 0.680 MPa
θ p :=
⎛ 2τ D ⎞
1
atan ⎜
2
⎝ σ xD − σ yD ⎠
θ p = 3.797 deg
θ' p = −86.203 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x'_D :=
σ xD + σ yD
2
+
σ xD − σ yD
2
( )
( )
⋅ cos 2θ p + τ D⋅ sin 2θ p
σ x'_D = 0.680 MPa
Therefore,
θ' p := θ p − 90deg
θ p1 := θ p
θ p1 = 3.80 deg
Ans
θ p2 := θ' p
θ p2 = −86.20 deg
Ans
Problem 9-24
The grains of wood in the board make an angle of 20° with the horizontal as shown. Determine the
normal and shear stress that act perpendicular to the grains if the board is subjected to an axial load of
250 N.
kPa := 1000Pa
Unit Used:
Given: P := 250N
φ := 20deg
h := 60mm
t := 25mm
Solution:
σ x :=
P
h⋅ t
σ x = 0.1667 MPa
σ y := 0
τ xy := 0
θ := 90deg + φ
σ x' :=
σx + σy
τ x'y' := −
2
+
σx − σy
2
σx − σy
2
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
σ x' = 19.50 kPa
Ans
τ x'y' = 53.57 kPa
Ans
Problem 9-25
The wooden block will fail if the shear stress acting along the grain is 3.85 MPa. If the normal stress
σx = 2.8 MPa, determine the necessary compressive stress σy that will cause failure.
Given: σ x := 2.8MPa
τ xy := 0MPa
θ grain := 58deg
τ x'y' := 3.85MPa
Solution: θ := θ grain + 90deg
Shear Stress:
τ x'y' = −
τ x'y' = −
σ y :=
σx − σy
2
σx − σy
2
2τ x'y'
sin ( 2θ )
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
⋅ sin ( 2θ )
+ σx
σ y = −5.767 MPa
Ans
Problem 9-26
The T-beam is subjected to the distributed loading that is applied along its centerline. Determine the
principal stresses at points A and B and show the results on elements located at each of these points.
Given: bf := 150mm
tf := 20mm
dw := 150mm
tw := 20mm
a := 2m
b := 1m
hB := 50mm
w := 12
Solution:
kN
m
Internal Force and Moment : At Section A-B:
ΣF y=0;
+
+ ΣΜA=0;
V − w⋅ a = 0
V := w⋅ a
M − w⋅ a⋅ ( 0.5a + b) = 0
M := w⋅ a⋅ ( 0.5a + b)
D := dw + tf
Section Property :
yc :=
(
) (
)(
)
0.5tf bf⋅ tf + 0.5dw + tf dw⋅ tw
bf⋅ tf + dw⋅ tw
yc = 52.50 mm
1
3
2
⋅ b ⋅ t + bf⋅ tf ⋅ 0.5tf − yc
12 f f
1
3
2
I2 :=
⋅ tw⋅ dw + dw⋅ tw ⋅ 0.5dw + tf − yc
12
(
I1 :=
)(
(
)
)(
)
4
I := I1 + I2
I = 16562500.00 mm
QA := 0 (since A' = 0)
(
)(
)
3
QB := hB⋅ tw D − yc − 0.5 ⋅ hB
Normal Stress:
σ=
M⋅ c
I
cA := yc
σ A :=
(
)
σ B :=
Shear Stress : τ =
V⋅ Q
I⋅ t
τ A :=
cB := − D − yc − hB
τ B :=
QB = 92500 mm
M⋅ cA
I
M⋅ cB
I
V⋅ QA
I ⋅ bf
V⋅ QB
I ⋅ tw
σ A = 152.15 MPa
σ B = −195.62 MPa
τ A = 0.00 MPa
τ B = 6.702 MPa
In-plane Principal Stress:
At A:
σ xA := σ A
σ yA := 0
τ xy := τ A
Since no shear stress acts upon the element,
Ans
σ A1 := σ xA
σ A1 = 152.15 MPa
σ A2 := σ yA
At B:
σ xB := σ B
σ B1 :=
σ B2 :=
σ A2 = 0 MPa
σ yB := 0
σ xB + σ yB
2
σ xB + σ yB
2
( )
2τ B
σ xB − σ yB
τ xy := τ B
2
⎛ σ xB − σ yB ⎞
2
+ ⎜
+ τB
2
⎝
⎠
σ B1 = 0.229 MPa
Ans
σ B2 = −195.852 MPa
Ans
2
⎛ σ xB − σ yB ⎞
2
− ⎜
+ τB
2
⎝
⎠
Applying Eq. 9-4 for point B,
Orientation of Principal Plane:
tan 2θ p = −
Ans
θ p :=
⎛ −2τ B ⎞
1
atan ⎜
2
σ xB − σ yB
⎝
⎠
θ p = 1.96 deg
θ' p := θ p − 90deg
θ' p = −88.04 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x'_B :=
σ xB + σ yB
2
+
σ xB − σ yB
2
( )
( )
⋅ cos 2θ p + τ B⋅ sin 2θ p
σ x'_B = −194.94 MPa
Therefore,
θ p1 := θ' p
θ p1 = −88.04 deg
Ans
θ p2 := θ p
θ p2 = 1.96 deg
Ans
Problem 9-27
The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal
stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the
results on properly oriented elements located at these points.
Given: do := 15mm
a := 50mm
P := 0.6kN
Solution:
Internal Force and Moment : At Section A-B:
+
ΣF x=0;
+ ΣΜO=0;
N−P= 0
N := P
M − P⋅ ( a) = 0
M := P⋅ a
Section Property :
A :=
I :=
π ⋅ do
4
2
2
A = 176.71 mm
4
π ⋅ do
64
4
I = 2485.05 mm
Normal Stress:
σ 1.2 =
N Mc
±
A
I
cA := 0.5do
σ A :=
N M⋅ cA
−
A
I
σ A = −87.15 MPa
cB := 0.5do
σ B :=
N M⋅ cB
+
A
I
σ B = 93.94 MPa
In-plane Principal Stress:
At A:
σ xA := σ A
σ yA := 0
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ A1 := σ yA
σ A1 = 0 MPa
σ A2 := σ xA
At B:
σ xB := σ B
σ A2 = −87.15 MPa
σ yB := 0
Ans
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ B1 := σ xB
σ B1 = 93.94 MPa
σ B2 := σ yB
σ B2 = 0 MPa
Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7
2
⎛ σ xA − σ yA ⎞
2
τ A.max := ⎜
+ τ xy
2
⎝
⎠
τ A.max = 43.6 MPa
Ans
τ B.max = 47.0 MPa
Ans
2
⎛ σ xB − σ yB ⎞
2
τ B.max := ⎜
+ τ xy
2
⎝
⎠
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
σ xA − σ yA
θ s_A := 45deg
tan( 2θ S_A ) = ∞
tan 2θ s_A = −
2τ xy
Ans
θ' s_A := θ s_A − 90deg
θ' s_A = −45 deg
(
(
)
)
tan 2θ s_B = −
σ xB − σ yB
2τ xy
σ avg_B :=
σ xB + σ yB
2
Ans
tan( 2θ S_B ) = ∞
θ s_B := −45deg
Ans
θ' s_B := θ s_B + 90deg
Ans
θ' s_B = 45 deg
Ans
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Average Normal Stress:
σ xA + σ yA
σ avg_A :=
2
Ans
Applying Eq. 9-8
σ avg_A = −43.57 MPa
σ avg_B = 46.97 MPa
Problem 9-28
The simply supported beam is subjected to the traction stress τ0 on its top surface. Determine the
principal stresses at points A and B.
Problem 9-29
The beam has a rectangular cross section and is subjected to the loadings shown. Determine the
principal stresses and the maximum in-plane shear stress that are developed at point A and point B.
These points are just to the left of the 10-kN load. Show the results on properly oriented elements
located at these points.
Given: b := 150mm
d := 375mm
F := 10kN
P := 5kN
L := 1.2m
Solution:
Support Reactions : By symmetry, R1=R ; R2= R
+
+
ΣF y=0;
+
ΣF x=0;
+
ΣF y=0;
2R − F = 0
R := 0.5F
ΣF x=0;
H1 − P = 0
H1 := P
Internal Force and Moment : At Section A-B:
+ ΣΜO=0;
Section Property :
A := b⋅ d
QA := 0
H1 + N = 0
R+V= 0
N := −H1
M − R⋅ ( 0.5L ) = 0
M := 0.5R⋅ L
V := −R
1
3
⋅ b⋅ d
12
(since A' = 0)
QB := 0
I :=
cA := −0.5 d
N M⋅ c
+
I
A
N M⋅ cA
σ A :=
+
A
I
σ A = −0.942 MPa
cB := 0.5d
σ B :=
N M⋅ cB
+
A
I
σ B = 0.764 MPa
Normal Stress:
Shear Stress :
σ=
τ A := 0
Since QA = QB = 0,
τ B := 0
In-plane Principal Stress:
At A:
σ xA := σ A
σ yA := 0
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ A1 := σ yA
σ A1 = 0 MPa
σ A2 := σ xA
At B:
σ xB := σ B
σ A2 = −0.942 MPa
σ yB := 0
Ans
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ B1 := σ xB
σ B1 = 0.764 MPa
σ B2 := σ yB
σ B2 = 0 MPa
Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σ xA − σ yA ⎞
2
τ max_A := ⎜
+ τ xy
2
⎝
⎠
τ max_A = 0.471 MPa
Ans
τ max_B = 0.382 MPa
Ans
2
⎛ σ xB − σ yB ⎞
2
τ max_B := ⎜
+ τ xy
2
⎝
⎠
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
(
)
tan 2θ s_A = −
(
)
tan 2θ s_B = −
σ xA − σ yA
2τ xy
σ xB − σ yB
2τ xy
tan( 2θ S_A ) = ∞
θ s_A := 45deg
Ans
θ' s_A := θ s_A − 90deg
Ans
θ' s_A = −45 deg
Ans
tan( 2θ S_B ) = ∞
θ s_B := −45deg
Ans
θ' s_B := θ s_B + 90deg
Ans
θ' s_B = 45 deg
Ans
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Average Normal Stress:
σ avg_A :=
σ avg_B :=
Applying Eq. 9-8,
σ xA + σ yA
2
σ xB + σ yB
2
σ avg_A = −0.471 MPa
Ans
σ avg_B = 0.382 MPa
Ans
Problem 9-30
The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at
point A and at point B. These points are located at the top and bottom of the web, respectively.
Although it is not very accurate, use the shear formula to compute the shear stress.
Given: bf := 200mm
tf := 10mm
tw := 10mm
dw := 200mm
P := 25kN
θ := 30deg
a := 3m
w := 8
kN
m
Solution:
Internal Force and Moment : At Section A-B:
+ ΣF x=0;
+
ΣF y=0;
+ ΣΜO=0;
P⋅ cos ( θ ) − N = 0
V − P⋅ sin ( θ ) − w⋅ a = 0
V := P⋅ sin ( θ ) + w⋅ a
M − P⋅ sin ( θ ) ⋅ ( a) − ( w⋅ a) ⋅ ( 0.5a) = 0
M := P⋅ a⋅ sin ( θ ) + 0.5w⋅ a
2
Section Property :
(
D := dw + 2tf
)
A := bf⋅ D − bf − tw ⋅ dw
(
)
)
cB := −0.5 dw
σ B :=
Shear Stress :
τ=
2
A = 6000 mm
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ dw ⎤⎦
12 ⎣ f
⎛ D tf ⎞
QA := bf⋅ tf ⋅ ⎜ −
⎝ 2 2⎠
N M⋅ c
Normal Stress: σ =
+
A
I
N M⋅ cA
cA := 0.5dw
σ A :=
+
A
I
I :=
(
N := P⋅ cos ( θ )
N M⋅ cB
+
A
I
I = 50.80 × 10
τ B :=
V⋅ QA
I ⋅ tw
V⋅ QB
I ⋅ tw
m
4
3
QA = 210000 mm
σ A = 148.293 MPa
σ B = −141.077 MPa
V⋅ Q
I⋅ t
τ A :=
−6
τ A = 15.09 MPa
τ B = 15.09 MPa
QB := QA
In-plane Principal Stress:
At A:
σ xA := σ A
σ A1 :=
σ A2 :=
At B:
σ yA := 0
σ xA + σ yA
2
σ xA + σ yA
2
σ xB := σ B
σ B1 :=
σ B2 :=
2
⎛ σ xA − σ yA ⎞
2
+ ⎜
+ τA
2
⎝
⎠
2
σ xB + σ yB
2
σ A1 = 149.8 MPa
Ans
σ A2 = −1.52 MPa
Ans
σ B1 = 1.60 MPa
Ans
σ B2 = −142.7 MPa
Ans
2
⎛ σ xA − σ yA ⎞
2
− ⎜
+ τA
2
⎝
⎠
σ yB := 0
σ xB + σ yB
τ xy := τ A
τ xy := τ B
2
⎛ σ xB − σ yB ⎞
2
+ ⎜
+ τB
2
⎝
⎠
2
⎛ σ xB − σ yB ⎞
2
− ⎜
+ τB
2
⎝
⎠
Problem 9-31
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses and
the maximum in-plane shear stress that is developed anywhere on the surface of the shaft.
Problem 9-32
A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown.
Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is
subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30
mm.
kPa := 1000Pa
Unit used:
Given: do := 30mm
t := 1mm
θ := 30deg
P := 10N
Solution:
Section Property :
A :=
di := do − 2t
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
2
A = 91.11 mm
Normal Stress:
σ x :=
P
A
σ x = 109.76 kPa
σ y := 0
τ xy := 0
Shear stress along the seam:
τ x'y' := −
σx − σy
2
τ x'y' = −47.53 kPa
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
Ans
Problem 9-33
Solve Prob. 9-32 for the normal stress acting perpendicular to the seam.
kPa := 1000Pa
Unit used:
Given: do := 30mm
t := 1mm
θ := 30deg
P := 10N
Solution:
di := do − 2t
Section Property :
A :=
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
2
A = 91.11 mm
Normal Stress:
σ x :=
P
A
σ x = 109.76 kPa
σ y := 0
τ xy := 0
Normal stress perpendicular to the seam:
σ x' :=
σx + σy
2
σ x' = 82.32 kPa
+
σx − σy
2
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
Problem 9-34
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses and
the maximum in-plane shear stress that is developed at point A. The bearings only support vertical
reactions.
Problem 9-35
The drill pipe has an outer diameter of 75 mm, a wall thickness of 6 mm, and a weight of 0.8 kN/m. If
it is subjected to a torque and axial load as shown, determine (a) the principal stresses and (b) the
maximum in-plane shear stress at a point on its surface at section a.
Given: do := 75mm
P := 7.5kN
t := 6mm
L := 6m
Mx := 1.2kN⋅ m
w := 0.8
Solution:
kN
m
Internal Force and Moment : At section a:
ΣF x=0;
N + P + w⋅ L = 0
N := −P − w⋅ L
ΣΜ x=0; T − Mx = 0
Section Property :
A :=
T := Mx
di := do − 2t
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
Normal Stress:
σ :=
J :=
N
A
π ⎛ 4
4
⋅ d − di ⎞⎠
32 ⎝ o
σ = −9.457 MPa
Shear Stress :
T⋅ c
J
a) In-plane Principal Stresses:
c := 0.5do
τ :=
σ x := 0
σ y := σ
τ = 28.850 MPa
τ xy := τ
for any point on the shaft's surface. Applying Eq. 9-5,
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 24.51 MPa
Ans
σ 2 = −33.96 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
b) Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
⎝ 2 ⎠
τ max = 29.24 MPa
Ans
Problem 9-36
The internal loadings at a section of the beam are shown. Determine the principal stresses at point A.
Also compute the maximum in-plane shear stress at this point.
Given: bf := 200mm
tf := 50mm
tw := 50mm
dw := 200mm
Px := −500kN
My := −30kN⋅ m
Mz := 40kN⋅ m
Py := −800kN
Solution:
D := dw + 2tf
Section Property :
(
)
2
A := bf⋅ D − bf − tw ⋅ dw
A = 30000 mm
Iz :=
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
)
Iz = 350.00 × 10
Iy :=
1 ⎛
3
3
⋅ ⎝ 2tf⋅ bf + dw⋅ tw ⎞⎠
12
Iy = 68.75 × 10
(
−6
−6
m
m
4
4
QA := 0 (since A' = 0)
yA := 0.5D
Normal Stress:
σ A :=
Shear Stress :
Px
A
−
zA := 0.5bf
Mz⋅ yA My⋅ zA
+
Iy
Iz
Since QA = 0,
σ A = −77.446 MPa
τ A := 0
In-plane Principal Stress:
σ x := σ A
σ y := 0
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ 1 := σ y
σ 1 = 0 MPa
σ 2 := σ x
σ 2 = −77.45 MPa
Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
⎝ 2 ⎠
τ max = 38.72 MPa
Ans
Problem 9-37
Solve Prob. 9-36 for point B.
Given: bf := 200mm
tf := 50mm
tw := 50mm
dw := 200mm
Px := −500kN
My := −30kN⋅ m
Mz := 40kN⋅ m
Py := −800kN
Solution:
Section Property :
(
D := dw + 2tf
)
2
A := bf⋅ D − bf − tw ⋅ dw
A = 30000 mm
Iz :=
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
)
Iz = 350.00 × 10
Iy :=
1 ⎛
3
3
⋅ ⎝ 2tf⋅ bf + dw⋅ tw ⎞⎠
12
Iy = 68.75 × 10
(
QB := 0
−6
−6
m
m
4
4
(since A' = 0)
yB := −0.5 D
Normal Stress:
σ B :=
Shear Stress :
Px
A
−
zB := −0.5 bf
Mz⋅ yB My⋅ zB
+
Iy
Iz
Since QB = 0,
σ B = 44.113 MPa
τ B := 0
In-plane Principal Stress:
σ x := σ B
σ y := 0
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ 1 := σ x
σ 1 = 44.113 MPa
σ 2 := σ y
σ 2 = 0.00 MPa
Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
⎝ 2 ⎠
τ max = 22.06 MPa
Ans
Problem 9-38
Solve Prob. 9-36 for point C, located in the center on the bottom of the web.
Given: bf := 200mm
tf := 50mm
tw := 50mm
dw := 200mm
Px := −500kN
My := −30kN⋅ m
Mz := 40kN⋅ m
Py := −800kN
Solution:
D := dw + 2tf
Section Property :
(
)
2
A := bf⋅ D − bf − tw ⋅ dw
A = 30000 mm
1 ⎡
3
3
⋅ ⎣bf⋅ D − bf − tw ⋅ dw ⎤⎦
12
1 ⎛
3
3
Iy :=
⋅ 2t ⋅ b + dw⋅ tw ⎞⎠
12 ⎝ f f
(
Iz :=
)
⎛ D tf ⎞
QC := bf⋅ tf ⋅ ⎜ −
2
2
(
)⎝
Shear Stress :
Px
A
V⋅ Q
τ=
I⋅ t
τ C :=
−6
Iy = 68.75 × 10
m
m
4
4
3
yC := −0.5 dw
σ C :=
−6
QC = 1250000 mm
⎠
Normal Stress:
Iz = 350.00 × 10
−
zC := 0
Mz⋅ yC My⋅ zC
+
Iy
Iz
Py⋅ QC
σ C = −5.238 MPa
τ C = −57.14 MPa
Iz⋅ tw
In-plane Principal Stress:
σ x := σ C
σ 1 :=
σ 2 :=
σ y := 0
σx + σy
2
σx + σy
2
τ xy := τ C
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 54.58 MPa
Ans
σ 2 = −59.82 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
⎝ 2 ⎠
τ max = 57.20 MPa
Ans
Problem 9-39
The wide-flange beam is subjected to the 50-kN force. Determine the principal stresses in the beam at
point A located on the web at the bottom of the upper flange. Although it is not very accurate, use the
shear formula to calculate the shear stress.
Given: bf := 200mm
tf := 12mm
tw := 10mm
dw := 250mm
P := 50kN
a := 3m
Solution:
Internal Force and Moment : At Section A-B:
+
ΣF y=0;
+ ΣΜO=0;
V−P= 0
V := P
M − P⋅ ( a) = 0
M := P⋅ a
D := dw + 2tf
Section Property :
(
)
A := bf⋅ D − bf − tw ⋅ dw
I :=
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ dw ⎤⎦
12 ⎣ f
(
)
⎛ D tf ⎞
QA := bf⋅ tf ⋅ ⎜ −
2
2
(
)⎝
2
A = 7300 mm
I = 95.45 × 10
−6
QA = 314400 mm
⎠
m
4
3
M⋅ c
I
Normal Stress:
σ=
cA := 0.5dw
σ A :=
Shear Stress :
τ=
M⋅ cA
σ A = 196.435 MPa
I
V⋅ Q
I⋅ t
τ A :=
V⋅ QA
τ A = 16.47 MPa
I ⋅ tw
In-plane Principal Stress:
σ x := σ A
σ 1 :=
σ 2 :=
σ y := 0
σx + σy
2
σx + σy
2
τ xy := τ A
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 197.81 MPa
Ans
σ 2 = −1.37 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Problem 9-40
Solve Prob. 9-39 for point B located on the web at the top of the bottom flange.
Given: bf := 200mm
tf := 12mm
tw := 10mm
dw := 250mm
P := 50kN
a := 3m
Solution:
Internal Force and Moment : At Section A-B:
+
ΣF y=0;
+ ΣΜO=0;
V−P= 0
V := P
M − P⋅ ( a) = 0
M := P⋅ a
Section Property :
(
D := dw + 2tf
)
A := bf⋅ D − bf − tw ⋅ dw
I :=
1 ⎡
3
3
⋅ b ⋅ D − bf − tw ⋅ dw ⎤⎦
12 ⎣ f
(
)
⎛ D tf ⎞
QB := bf⋅ tf ⋅ ⎜ −
2
2
(
)⎝
Normal Stress:
cB := −0.5 dw
Shear Stress :
−6
m
4
3
M⋅ c
I
M⋅ cB
σ B :=
τ=
I = 95.45 × 10
QB = 314400 mm
⎠
σ=
2
A = 7300 mm
σ B = −196.435 MPa
I
V⋅ Q
I⋅ t
τ B :=
V⋅ QB
τ B = 16.47 MPa
I ⋅ tw
In-plane Principal Stress:
σ x := σ B
σ 1 :=
σ 2 :=
σ y := 0
σx + σy
2
σx + σy
2
τ xy := τ B
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 1.37 MPa
Ans
σ 2 = −197.81 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Problem 9-41
The bolt is fixed to its support at C. If a force of 90 N is applied to the wrench to tighten it, determine
the principal stresses developed in the bolt shank at point A. Represent the results on an element located
at this point. The shank has a diameter of 6 mm.
Given: do := 6mm
a := 150mm
L := 50mm
P := 90N
Solution:
Internal Force and Moment : At section AB:
Mx := P⋅ L
T y := P⋅ a
Section Property :
I :=
π
4
⋅d
64 o
J :=
π
4
⋅d
32 o
Normal Stress:
σ A :=
Shear Stress :
τ A :=
cA := 0.5do
Mx⋅ cA
σ A = 212.21 MPa
I
T y⋅ cA
τ A = 318.31 MPa
J
In-plane Principal Stresses: Applying Eq. 9-5,
σ x := σ A
σ 1 :=
σ 2 :=
σ y := 0
σx + σy
2
σx + σy
2
τ xy := τ A
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 441.63 MPa
Ans
σ 2 = −229.42 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Orientation of Principal Stress:
( )
tan 2θ p =
⎛ 2τ xy ⎞
1
atan ⎜
2
⎝ σx − σy ⎠
θ p = 35.783 deg
2τ xy
θ p :=
σx − σy
θ' p := θ p − 90deg
θ' p = −54.217 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x' :=
σx + σy
2
+
σx − σy
2
( )
( )
⋅ cos 2θ p + τ xy⋅ sin 2θ p
σ x' = 441.63 MPa
Therefore,
θ p1 := θ p
θ p1 = 35.78 deg
Ans
θ p2 := θ' p
θ p2 = −54.22 deg
Ans
Problem 9-42
Solve Prob. 9-41 for point B.
Given: do := 6mm
a := 150mm
L := 50mm
P := 90N
Solution:
Internal Force and Moment : At section AB:
Mx := P⋅ L
T y := P⋅ a
Vz := P
Section Property :
I :=
π
4
⋅d
64 o
ro := 0.5do
J :=
π
4
⋅d
32 o
4ro ⎛ π ⋅ ro ⎞
QB :=
⋅⎜
3π ⎝ 2 ⎠
2
Mx⋅ cBσ
I
Normal Stress:
cBσ := 0
σ B :=
Shear Stress :
bB := do
cBτ := ro
τ B :=
Vz⋅ QB Ty⋅ cBτ
−
I ⋅ bB
J
σ B = 0 MPa
τ B = −314.07 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
σ x := σ B
σ 1 :=
σ 2 :=
σ y := 0
σx + σy
2
σx + σy
2
τ xy := τ B
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 314.07 MPa
Ans
σ 2 = −314.07 MPa
Ans
tan(2θP ) = ∞
θ p1 := 45deg
Ans
θ p2 := θ p1 − 90deg
θ p2 = −45 deg
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Orientation of Principal Stress:
( )
tan 2θ p =
2τ xy
σx − σy
Problem 9-43
The beam has a rectangular cross section and is subjected to the loadings shown. Determine the
principal stresses that are developed at point A and point B, which are located just to the left of the
20-kN load. Show the results on elements located at these points.
Given: b := 100mm
d := 200mm
F := 20kN
P := 10kN
L := 4m
Solution:
Support Reactions : By symmetry, R1=R ; R2= R
+
+
ΣF y=0;
+
ΣF x=0;
+
ΣF y=0;
2R − F = 0
R := 0.5F
ΣF x=0;
H1 − P = 0
H1 := P
Internal Force and Moment : At Section A-B:
+ ΣΜO=0;
H1 + N = 0
R+V= 0
N := −H1
M − R⋅ ( 0.5L ) = 0
M := 0.5R⋅ L
V := −R
Section Property :
1
3
⋅ b⋅ d
12
QA := 0 (since A' = 0)
QB := b⋅ ( 0.5d) ⋅ ( 0.25d)
A := b⋅ d
I :=
Normal Stress:
σ=
cA := −0.5 d
σ A :=
N M⋅ cA
+
A
I
σ A = −30.5 MPa
cB := 0
σ B :=
N M⋅ cB
+
A
I
σ B = −0.5 MPa
N M⋅ c
+
A
I
Shear Stress :
Since QA = 0,
τ B :=
V⋅ QB
I⋅ b
τ A := 0
τ B = −0.75 MPa
In-plane Principal Stress:
At A:
σ xA := σ A
Applying Eq. 9-5
σ yA := 0
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ A1 := σ yA
σ A1 = 0 MPa
σ A2 := σ xA
At B:
σ xB := σ B
σ A2 = −30.50 MPa
σ yB := 0
τ xy := τ B
Ans
σ B1 :=
σ B2 :=
σ xB + σ yB
2
σ xB + σ yB
2
2
⎛ σ xB − σ yB ⎞
2
+ ⎜
+ τB
2
⎝
⎠
( )
Ans
σ B2 = −1.041 MPa
Ans
2
⎛ σ xB − σ yB ⎞
2
− ⎜
+ τB
2
⎝
⎠
Applying Eq. 9-4 for point B,
Orientation of Principal Plane:
tan 2θ p =
σ B1 = 0.541 MPa
2τ B
σ xB − σ yB
θ p :=
⎛ 2τ B ⎞
1
atan ⎜
2
⎝ σ xB − σ yB ⎠
θ' p := θ p − 90deg
θ' p = −54.217 deg
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σ x'_B :=
σ xB + σ yB
2
+
σ xB − σ yB
2
( )
( )
⋅ cos 2θ p + τ B⋅ sin 2θ p
σ x'_B = −1.04 MPa
Therefore,
θ p = 35.783 deg
θ p1 := θ' p
θ p1 = −54.22 deg
Ans
θ p2 := θ p
θ p2 = 35.78 deg
Ans
Problem 9-44
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at ω = 15
rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.
If the shaft has an outer diameter of 250 mm, determine the principal stresses at any point located on
the surface of the shaft.
Given: do := 250mm
L := 0.75m
F := 1230kN
P := 900kW
ω := 15
Solution:
Internal Force and Moment : As shown on FBD
T o :=
P
rad
s
T o = 60.00 kN⋅ m
ω
N := −F
Section Property :
π ⋅ do
A :=
J :=
4
2
2
A = 49087.39 mm
4
π ⋅ do
4
J = 383495196.97 mm
32
Normal Stress:
σ a :=
N
A
σ a = −25.06 MPa
cmax := 0.5 ⋅ do
Shear Stress :
τ o :=
To⋅ cmax
J
τ o = 19.56 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
σ x := σ a
σ 1 :=
σ 2 :=
σ y := 0
σx + σy
2
σx + σy
2
τ xy := τ o
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 10.70 MPa
Ans
σ 2 = −35.75 MPa
Ans
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Problem 9-45
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at ω = 15
rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.
If the shaft has a diameter of 250 mm, determine the maximum in-plane shear stress at any point
located on the surface of the shaft.
Given: do := 250mm
L := 0.75m
F := 1230kN
P := 900kW
ω := 15
Solution:
Internal Force and Moment : As shown on FBD
T o :=
P
rad
s
T o = 60.00 kN⋅ m
ω
N := −F
Section Property :
π ⋅ do
A :=
J :=
4
2
2
A = 49087.39 mm
4
π ⋅ do
4
J = 383495196.97 mm
32
Normal Stress:
σ a :=
N
A
σ a = −25.06 MPa
cmax := 0.5 ⋅ do
Shear Stress :
τ o :=
To⋅ cmax
J
τ o = 19.56 MPa
Maximum In-plane Shear Stress: Applying Eq. 9-7
σ x := σ a
σ y := 0
τ xy := τ o
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
⎝ 2 ⎠
τ max = 23.2 MPa
Ans
Problem 9-46
The steel pipe has an inner diameter of 68mm and an outer diameter of 75 mm. If it is fixed at C and
subjected to the horizontal 100-N force acting on the handle of the pipe wrench at its end, determine
the principal stresses in the pipe at point A which is located on the surface of the pipe.
Given: do := 75mm
P := 100N
di := 68mm
L := 250mm
a := 300mm
Solution:
Internal Force and Moment : At section AB:
T x := P⋅ a
My := P⋅ L
Vz := P
Section Property :
I :=
ro := 0.5do
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
64
ri := 0.5di
J :=
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
32
4ro ⎛ π ⋅ ro ⎞ 4ri ⎛ π ⋅ ri ⎞
QAz :=
⋅⎜
−
⋅⎜
3π ⎝ 2 ⎠ 3π ⎝ 2 ⎠
2
2
Normal Stress:
cAσ := 0
Shear Stress :
bA := do − di
τ A :=
σ A :=
My⋅ cAσ
I
σ A = 0 MPa
cAτ := ro
Vz⋅ QAz Tx⋅ cAτ
−
I ⋅ bA
J
τ A = −0.863 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
σ x := σ A
σ 1 :=
σ 2 :=
σ z := 0
σx + σz
2
σx + σz
2
τ xz := τ A
2
⎛ σx − σz ⎞
2
+ ⎜
+ τ xz
⎝ 2 ⎠
σ 1 = 0.863 MPa
Ans
σ 2 = −0.863 MPa
Ans
2
⎛ σx − σz ⎞
2
− ⎜
+ τ xz
⎝ 2 ⎠
Problem 9-47
Solve Prob. 9-46 for point B, which is located on the surface of the pipe.
Given: do := 75mm
P := 100N
di := 68mm
L := 250mm
a := 300mm
Solution:
Internal Force and Moment : At section AB:
T x := P⋅ a
My := P⋅ L
Vz := P
Section Property :
I :=
ro := 0.5do
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
64
QBz := 0
(Since A'=0)
Normal Stress:
cBσ := ro
ri := 0.5di
J :=
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
32
σ B :=
My⋅ cBσ
I
σ B = 1.862 MPa
Shear Stress :
cBτ := ro
τ B := −
Tx⋅ cBτ
J
τ B = −1.117 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
σ x := σ B
σ 1 :=
σ 2 :=
σ z := 0
σx + σz
2
σx + σz
2
τ xz := τ B
2
⎛ σx − σz ⎞
2
+ ⎜
+ τ xz
⎝ 2 ⎠
σ 1 = 2.385 MPa
Ans
σ 2 = −0.523 MPa
Ans
2
⎛ σx − σz ⎞
2
− ⎜
+ τ xz
2
⎝
⎠
Problem 9-48
The cantilevered beam is subjected to the load at its end. Determine the principal stresses in the beam at
points A and B.
Given: b := 120mm
h := 150mm
P := 15kN
yA := 45mm
zA := 60mm
zB := −20mm
yθ :=
yB := 75mm
−4
5
L := 1.2m
3
zθ :=
5
Solution:
Internal Force and Moment : At Section A-B:
+ ΣF z0;
+
Vz + P⋅ zθ = 0
Vy + P⋅ yθ = 0
ΣF y=0;
Vz := −P⋅ zθ
Vy := −P⋅ yθ
Mz := P⋅ yθ ⋅ L
My := −P⋅ zθ ⋅ L
Section Property :
1
3
⋅ b⋅ h
12
1
3
Iy :=
⋅ h⋅ b
12
−6
Iz :=
Iz = 33.75 × 10
−6
Iy = 21.60 × 10
m
4
m
4
(
)
(
)
QB.z := h⋅ ( 0.5b − zB ) ⋅ ⎡⎣ zB + 0.5 ( 0.5b − zB )⎤⎦
QA.y := b⋅ 0.5h − yA ⋅ ⎡⎣yA + 0.5 0.5h − yA ⎤⎦
QA.z := 0
QB.y := 0
(since A' = 0)
(since A' = 0)
Normal Stress:
Mz⋅ yA My⋅ zA
σ A := −
+
Iz
Iy
σ B := −
Mz⋅ yB My⋅ zB
+
Iz
Iy
Shear Stress :
τ A :=
τ B :=
τ=
Vy⋅ QA.y
Iz⋅ b
Vz⋅ QB.z
Iy⋅ h
σ xA := σ A
σ B = 42.0 MPa
V⋅ Q
I⋅ t
τ A = 0.640 MPa
τ B = −0.667 MPa
In-plane Principal Stress:
At A:
σ A = −10.8 MPa
Applying Eq. 9-5
σ yA := 0
τ xy := 0
3
QA.y = 216000 mm
3
QB.z = 240000 mm
σ A1 :=
σ A2 :=
At B:
σ xA + σ yA
2
σ xA + σ yA
2
σ xB := σ B
σ B1 :=
σ B2 :=
σ xB + σ zB
2
σ xB + σ zB
2
2
⎛ σ xA − σ yA ⎞
2
+ ⎜
+ τA
2
⎝
⎠
σ A1 = 0.0378 MPa
Ans
σ A2 = −10.84 MPa
Ans
σ B1 = 42.01 MPa
Ans
σ B2 = −0.0106 MPa
Ans
2
⎛ σ xA − σ yA ⎞
2
− ⎜
+ τA
2
⎝
⎠
σ zB := 0
τ xz := τ B
2
⎛ σ xB − σ zB ⎞
2
+ ⎜
+ τB
2
⎝
⎠
2
⎛ σ xB − σ zB ⎞
2
− ⎜
+ τB
2
⎝
⎠
Problem 9-49
The box beam is subjected to the loading shown. Determine the principal stresses in the beam at points
A and B.
Given: bo := 200mm bi := 150mm
do := 200mm di := 150mm
L 1 := 0.9m
L 2 := 1.5m
P1 := 4kN
P2 := 6kN
Solution: L := L 1 + 2L2
Support Reactions :
+
ΣF y=0;
Given
R1 + R2 − P1 − P2 = 0
(
+ ΣΜR2=0;
)
P1⋅ L − R1⋅ L − L 1 + P2⋅ L2 = 0
Guess R1 := 1kN R2 := 1kN
⎛⎜ R1 ⎞
⎛⎜ R1 ⎞ ⎛ 8.2 ⎞
:= Find R1 , R2
=⎜
kN
⎜ R2
⎜ R2
1.8
⎝
⎠
⎝ ⎠
⎝ ⎠
(
)
Internal Force and Moment : At section A-B:
N := 0
V := P1 − R1
(
)
(
M := P1⋅ L 1 + 0.5L 2 − R1⋅ 0.5L2
)
Section Property :
1 ⎛
3
3
⋅ b ⋅ d − bi⋅ di ⎞⎠
12 ⎝ o o
QA := 0
QB := 0 (Since A'=0)
I :=
For point A:
τ A := 0
cA := 0.5 ⋅ do
σ A :=
σ 1 := σ A
σ 1 = 0.494 MPa
Ans
σ 2 := 0
σ 2 = 0.000 MPa
Ans
M⋅ cA
I
σ A = 0.494 MPa
For point B:
τ B := 0
cB := −0.5 ⋅ di
σ B :=
σ 1 := 0
σ 1 = 0.000 MPa
Ans
σ 2 := σ B
σ 2 = −0.370 MPa
Ans
M⋅ cB
I
σ B = −0.37 MPa
Problem 9-50
A bar has a circular cross section with a diameter of 25 mm. It is subjected to a torque and a bending
moment. At the point of maximum bending stress the principal stresses are 140 MPa and -70 MPa.
Determine the torque and the bending moment.
Given: do := 6mm
σ 1 := 140MPa
σ 2 := −70MPa
Solution:
In-plane Principal Stresses: Applying Eq. 9-5,
Given σ y := 0
σ1 =
σ2 =
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
(1)
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
(2)
Solving Eqs. (1) and (2):
Guess σ x := 2MPa
τ xy := 1MPa
⎛⎜ σ x ⎞
:= Find ( σ x , τ xy)
⎜ τ xy
⎝
⎠
⎛⎜ σ x ⎞ ⎛ 70.00 ⎞
=⎜
MPa
⎜ τ xy
⎝
⎠ ⎝ 98.99 ⎠
Section Property :
ro := 0.5do
π
4
⋅ do
64
J :=
π
4
⋅ do
32
Normal Stress:
σ=
M⋅ c
I
c := ro
M :=
Shear Stress:
τ=
c := ro
T :=
I :=
σ x⋅ I
c
M = 1.484 N⋅ m
Ans
T = 4.199 N⋅ m
Ans
T⋅ c
J
τ xy⋅ J
c
Problem 9-51
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point A.
Also calculate the maximum in-plane shear stress at this point.
kPa := 1000Pa
Unit Used:
Given: b := 100mm
h := 200mm
Px := 0.5kN
yA := 100mm
zA := 0mm
Py := 0.8kN
My := −0.04 kN⋅ m
Mz := 0.03kN⋅ m
Solution:
Section Property :
2
A := b⋅ h
A = 20000 mm
1
3
−6 4
⋅ b⋅ h
Iz = 66.67 × 10 m
12
1
3
−6 4
Iy :=
⋅ h⋅ b
Iy = 16.67 × 10 m
12
QA.y := 0 (since A' = 0)
Iz :=
Normal Stress:
Px Mz⋅ yA My⋅ zA
σ A :=
−
+
A
Iy
Iz
Shear Stress :
Since QA = 0,
σ A = −20.0 kPa
τ A := 0
In-plane Principal Stress:
σ x := σ A
σ y := 0
τ xy := 0
Since no shear stress acts upon the element,
Ans
σ 1 := σ y
σ 1 = 0 kPa
σ 2 := σ x
σ 2 = −20 kPa
Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σx − σy ⎞
2
τ max := ⎜
+ τ xy
2
⎝
⎠
τ max = 10 kPa
Ans
Problem 9-52
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point B.
Also calculate the maximum in-plane shear stress at this point.
Unit Used:
kPa := 1000Pa
Given: b := 100mm
h := 200mm
Px := 0.5kN
yB := 0mm
zB := −50mm Py := 0.8kN
My := −0.04 kN⋅ m
Mz := 0.03kN⋅ m
Solution:
Section Property :
2
A := b⋅ h
A = 20000 mm
1
3
⋅ b⋅ h
12
1
3
Iy :=
⋅ h⋅ b
12
−6
Iz :=
Iz = 66.67 × 10
m
−6
Iy = 16.67 × 10
4
m
4
QB.y := b⋅ ( 0.5h) ⋅ ( 0.25h)
Normal Stress:
Px Mz⋅ yB My⋅ zB
σ B :=
−
+
A
Iy
Iz
Shear Stress :
Py⋅ QB.y
τ B :=
Iz⋅ b
σ B = 145.0 kPa
τ B = 60.00 kPa
In-plane Principal Stress:
σ xB := σ B
σ B1 :=
σ B2 :=
σ yB := 0
σ xB + σ yB
2
σ xB + σ yB
2
τ xy := τ B
2
⎛ σ xB − σ yB ⎞
2
+ ⎜
+ τB
2
⎝
⎠
σ B1 = 166.6 kPa
Ans
σ B2 = −21.61 kPa
Ans
2
⎛ σ xB − σ yB ⎞
2
− ⎜
+ τB
2
⎝
⎠
Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σ xB − σ yB ⎞
2
τ max := ⎜
+ τ xy
2
⎝
⎠
τ max = 94.11 kPa Ans
Problem 9-53
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point C.
Also calculate the maximum in-plane shear stress at this point.
Unit Used:
kPa := 1000Pa
Given: b := 100mm
h := 200mm
Px := 0.5kN
yC := −50mm
zC := 0mm
Py := 0.8kN
My := −0.04 kN⋅ m
Mz := 0.03kN⋅ m
Solution:
Section Property :
2
A := b⋅ h
A = 20000 mm
1
3
⋅ b⋅ h
12
1
3
Iy :=
⋅ h⋅ b
12
−6
Iz :=
Iz = 66.67 × 10
m
−6
Iy = 16.67 × 10
(
)
4
m
4
(
)
QC.y := b⋅ 0.5h − yC ⋅ ⎡⎣ yC + 0.5 0.5h − yC ⎤⎦
Normal Stress:
Px Mz⋅ yC My⋅ zC
σ C :=
−
+
A
Iy
Iz
Shear Stress :
Py⋅ QC.y
τ C :=
Iz⋅ b
3
QC.y = 375000 mm
σ C = 47.5 kPa
τ C = 45.00 kPa
In-plane Principal Stress:
σ xC := σ C
σ C1 :=
σ C2 :=
σ yC := 0
σ xC + σ yC
2
σ xC + σ yC
2
τ xy := τ C
2
⎛ σ xC − σ yC ⎞
2
+ ⎜
+ τC
2
⎝
⎠
σ C1 = 74.63 kPa
Ans
σ C2 = −27.13 kPa
Ans
2
⎛ σ xC − σ yC ⎞
2
− ⎜
+ τC
2
⎝
⎠
Maximum In-plane Shear Stress: Applying Eq. 9-7,
2
⎛ σ xC − σ yC ⎞
2
τ max := ⎜
+ τ xy
2
⎝
⎠
τ max = 50.88 kPa Ans
Problem 9-54
The beam has a rectangular cross section and is subjected to the loads shown. Write a computer
program that can be used to determine the principal stresses at points A, B, C, and D. Show an
application of the program using the values h = 300 mm, b = 200 mm, Nx = 2 kN, Vy = 1.5 kN, Vz = 0,
My = 0, and Mz = -225 kN·m.
Problem 9-55
The member has a rectangular cross section and is subjected to the loading shown. Write a computer
program that can be used to determine the principal stresses at points A, B, and C. Show an application
of the program using the values b = 150 mm, h = 200 mm, P = 1.5 kN, x = 75 mm, z = -50 mm, Vx =
300 N, and V z = 600 N.
Problem 9-56
Solve Prob. 9-4 using Mohr's circle.
Given: σ x := −0.650 MPa
σ y := 0.400MPa
φ' := 60deg
Solution:
τ xy := 0MPa
θ := 90deg − φ'
θ = 30.00 deg
Center :
σ c :=
σx + σy
σ c = −0.125 MPa
2
Radius :
R := σ x − σ c
R = 0.525 MPa
Coordinates:
(
A σx , 0
)
(
)
B σy , 0
(
C σc , 0
)
Stresses:
σ x' := σ c − R⋅ cos ( 2θ )
σ x' = −0.387 MPa
Ans
τ x'y' := R⋅ sin ( 2θ )
τ x'y' = 0.455 MPa
Ans
Problem 9-57
Solve Prob. 9-2 using Mohr's circle.
Given: σ x := 5MPa
σ y := 3MPa
τ xy := 8MPa
φ' := 40deg
Solution:
Center :
σ c :=
σx + σy
2
σ c = 4 MPa
Radius :
R :=
(σx − σc)2 + τ xy2
Angles: θ := 90deg + φ'
⎛ τ xy ⎞
R = 8.062 MPa
θ = 130 deg
φ := atan ⎜
φ = 82.875 deg
α := 180deg − ( 2θ − φ )
α = 2.875 deg
⎝ σx − σc ⎠
Stresses:
σ x' := σ c − R⋅ cos ( α )
σ x' = −4.052 MPa
Ans
τ x'y' := −R⋅ sin ( α )
τ x'y' = −0.404 MPa
Ans
Problem 9-58
Solve Prob. 9-3 using Mohr's circle.
Given: σ x := 0.350MPa
φ' := 50deg
σ y := −0.200 MPa
τ xy := 0MPa
Solution: θ := 90deg + φ'
θ = 140 deg
Center :
σ c :=
σx + σy
σ c = 0.075 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 0.275 MPa
Coordinates:
(
A σx , 0
)
(
C σc , 0
)
Angles:
α := 360deg − 2θ
α = 80 deg
Stresses: (represented by coordinates of point P)
σ x' := σ c + R⋅ cos ( α )
σ x' = 0.123 MPa
Ans
τ x'y' := R⋅ sin ( α )
τ x'y' = 0.271 MPa
Ans
Problem 9-59
Solve Prob. 9-10 using Mohr's circle.
Given: σ x := 0MPa
σ y := −0.300 MPa
θ := 30deg
τ xy := 0.950MPa
Solution:
Center :
σ c :=
σx + σy
2
σ c = −0.15 MPa
Radius :
R :=
(σx − σc)2 + τ xy2
R = 0.962 MPa
Angles:
⎛ τ xy ⎞
φ := atan ⎜
φ = 81.027 deg
α := 2θ − φ
α = −21.027 deg
⎝ σx − σc ⎠
Stresses:
σ x' := σ c + R⋅ cos ( α )
σ x' = 0.748 MPa
Ans
σ y' := σ c − R⋅ cos ( α )
σ y' = −1.048 MPa
Ans
τ x'y' := −R⋅ sin ( α )
τ x'y' = 0.345 MPa
Ans
Problem 9-60
Solve Prob. 9-6 using Mohr's circle.
Given: σ x := 90MPa
σ y := 50MPa
τ xy := −35MPa
φ' := 60deg
Solution:
Center :
σ c :=
σx + σy
2
σ c = 70 MPa
Radius :
R :=
Angles:
(σx − σc)2 + τ xy2
R = 40.311 MPa
θ := 90deg − φ'
⎛ τ xy ⎞
φ := atan ⎜
φ = −60.255 deg
α := 180deg − ( 2θ − φ )
α = 59.745 deg
⎝ σx − σc ⎠
Stresses:
σ x' := σ c − R⋅ cos ( α )
σ x' = 49.689 MPa
Ans
τ x'y' := −R⋅ sin ( α )
τ x'y' = −34.821 MPa
Ans
Problem 9-61
Solve Prob. 9-11 using Mohr's circle.
Given: σ x := 0.300MPa
θ := −60deg
σ y := 0MPa
τ xy := 0.120MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 0.15 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 0.1921 MPa
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
Angles:
⎛ τ xy ⎞
φ := atan ⎜
φ = 38.660 deg
α := 180deg + 2θ − φ
α = 21.340 deg
⎝ σx − σc ⎠
Stresses: (represented by coordinates of points P and Q)
σ x' := σ c − R⋅ cos ( α )
σ x' = −0.0289 MPa
Ans
σ y' := σ c + R⋅ cos ( α )
σ y' = 0.329 MPa
Ans
τ x'y' := R⋅ sin ( α )
τ x'y' = 0.0699 MPa
Ans
Problem 9-62
Solve Prob. 9-13 using Mohr's circle.
Given: σ x := 45MPa
σ y := −60MPa
τ xy := 30MPa
Solution:
Center :
σ c :=
σx + σy
σ c = −7.5 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 60.467 MPa
Coordinates:
(
)
(
A σ x , τ xy
C σc , 0
)
Stresses:
σ 1 := σ c + R
σ 1 = 52.97 MPa
Ans
σ 2 := σ c − R
σ 2 = −67.97 MPa
Ans
τ max := R
τ max = 60.47 MPa
Ans
σ avg := σ c
σ avg = −7.5 MPa
Ans
Angles:
θ p1 :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p1 = 14.872 deg
⎠
(Counter-clockwise) Ans
2θ s1 = 90deg − 2θ p1
θ s1 := 45deg − θ p1
θ s1 = 30.128 deg
(Clockwise)
Ans
Problem 9-63
Solve Prob. 9-14 using Mohr's circle.
Given: σ x := 180MPa
σ y := 0MPa
τ xy := −150MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 90 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 174.929 MPa
Coordinates:
(
)
A σ x , τ xy
(
B σ y , −τ xy
)
(
C σc , 0
)
Stresses:
σ 1 := σ c + R
σ 1 = 264.93 MPa
Ans
σ 2 := σ c − R
σ 2 = −84.93 MPa
Ans
τ max := R
τ max = 174.93 MPa Ans
σ avg := σ c
σ avg = 90 MPa
Ans
Angles:
θ p :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p = 29.518 deg
⎠
(Clockwise)
Ans
2θ s = 90deg − 2θ p
θ s := 45deg − θ p
θ s = 15.482 deg
(Counter-clockwise)
Ans
Problem 9-64
Solve Prob. 9-16 using Mohr's circle.
Given: σ x := −200MPa
σ y := 250MPa
τ xy := 175MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 25 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 285.044 MPa
Coordinates:
(
)
A σ x , τ xy
(
B σ y , −τ xy
)
(
C σc , 0
)
Stresses:
σ 1 := σ c + R
σ 1 = 310.04 MPa
Ans
σ 2 := σ c − R
σ 2 = −260.04 MPa
Ans
τ max := R
τ max = 285.04 MPa Ans
σ avg := σ c
σ avg = 25 MPa
Ans
Angles:
θ p :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p = 18.937 deg
⎠
(Clockwise)
Ans
2θ s = 90deg − 2θ p
θ s := 45deg − θ p
θ s = 26.063 deg
(Counter-clockwise)
Ans
Problem 9-65
Solve Prob. 9-15 using Mohr's circle.
Given: σ x := −30MPa
σ y := 0MPa
τ xy := −12MPa
Solution:
Center :
σ c :=
σx + σy
σ c = −15 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 19.209 MPa
Coordinates:
(
)
(
A σ x , τ xy
C σc , 0
)
Stresses:
σ 1 := σ c + R
σ 1 = 4.21 MPa
Ans
σ 2 := σ c − R
σ 2 = −34.21 MPa
Ans
τ max := R
τ max = 19.21 MPa
Ans
σ avg := σ c
σ avg = −15 MPa
Ans
Angles:
θ p2 :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p2 = 19.330 deg
⎠
(Countr-clockwise)
Ans
(Countr-clockwise)
Ans
2θ s2 = 2θ p2 + 90deg
θ s2 := θ p2 + 45deg
θ s2 = 64.330 deg
Problem 9-66
Determine the equivalent state of stress if an element is oriented 20° clockwise from the element
shown. Show the result on the element.
Given: σ x := 3MPa
σ y := −2MPa
θ := −20deg
τ xy := −4MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 0.5 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 4.717 MPa
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
Angles:
⎛
φ := atan ⎜
τ xy
⎞
⎝ σx − σc ⎠
α := φ + 2θ
φ = 57.995 deg
α = 17.995 deg
Stresses: (represented by coordinates of points P and Q)
σ x' := σ c + R⋅ cos ( α )
σ x' = 4.986 MPa
Ans
σ y' := σ c − R⋅ cos ( α )
σ y' = −3.986 MPa
Ans
τ x'y' := −R⋅ sin ( α )
τ x'y' = −1.457 MPa
Ans
Problem 9-67
Determine the equivalent state of stress if an element is oriented 60° counterclockwise from the
element shown.
Given: σ x := 0.750MPa
σ y := −0.800 MPa
θ := 60deg
τ xy := 0.450MPa
Solution:
Center :
σ c :=
σx + σy
σ c = −0.025 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 0.8962 MPa
Coordinates:
(
)
A σ x , τ xy
(
B σ y , −τ xy
)
(
C σc , 0
)
Angles:
⎛ τ xy ⎞
φ := atan ⎜
φ = 30.141 deg
α := 2θ − φ
α = 89.859 deg
⎝ σx − σc ⎠
Stresses:
σ x' := σ c + R⋅ cos ( α )
σ x' = −0.0228 MPa
Ans
σ y' := σ c − R⋅ cos ( α )
σ y' = −0.0272 MPa
Ans
τ x'y' := −R⋅ sin ( α )
τ x'y' = −0.896 MPa
Ans
Problem 9-68
Determine the equivalent state of stress if an element is oriented 30° clockwise from the element
shown.
Given: σ x := 350MPa
σ y := 230MPa
θ := 30deg
τ xy := −480MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 290 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 483.7355 MPa
Coordinates:
(
)
A σ x , τ xy
(
B σ y , −τ xy
)
(
C σc , 0
)
Angles:
⎛ τ xy ⎞
φ := atan ⎜
φ = −82.875 deg
α := −2θ − φ
α = 22.875 deg
⎝ σx − σc ⎠
Stresses:
σ x' := σ c + R⋅ cos ( α )
σ x' = 735.6922 MPa
σ y' := σ c − R⋅ cos ( α )
σ y' = −155.6922 MPa Ans
τ x'y' := −R⋅ sin ( α )
τ x'y' = −188.038 MPa Ans
Ans
Problem 9-69
Determine the equivalent state of stress if an element is oriented 30° clockwise from the element
shown. Show the result on the element.
Given: σ x := 7MPa
σ y := 8MPa
θ := −30deg
τ xy := 15MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 7.5 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 15.0083 MPa
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
Angles:
⎛
φ := atan ⎜
τ xy
⎞
⎝ σx − σc ⎠
α := 180deg − ( 2 θ + φ )
φ = 88.091 deg
α = 31.909 deg
Stresses: (represented by coordinates of points P and Q)
σ x' := σ c − R⋅ cos ( α )
σ x' = −5.240 MPa
Ans
σ y' := σ c + R⋅ cos ( α )
σ y' = 20.240 MPa
Ans
τ x'y' := R⋅ sin ( α )
τ x'y' = 7.933 MPa
Ans
Problem 9-70
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σ x := 350MPa
σ y := −200MPa
τ xy := 500MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 75 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 570.636 MPa
Coordinates:
(
)
(
A σ x , τ xy
C σc , 0
)
a) Principal Stresses:
σ 1 := σ c + R
σ 1 = 645.64 MPa
Ans
σ 2 := σ c − R
σ 2 = −495.64 MPa
Ans
Angles:
θ p1 :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p1 = 30.595 deg
⎠
(Counter-clockwise) Ans
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τ max := R
τ max = 570.64 MPa Ans
σ avg := σ c
σ avg = 75 MPa
Ans
2θ s = 90deg − 2⋅ θ p1
θ s := 45deg − θ p1
θ s = 14.405 deg
(Clockwise)
Ans
Problem 9-71
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σ x := 10MPa
σ y := 80MPa
τ xy := −60MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 45 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 69.462 MPa
Coordinates:
(
)
(
A σ x , τ xy
C σc , 0
)
a) Principal Stresses:
σ 1 := σ c + R
σ 1 = 114.46 MPa
Ans
σ 2 := σ c − R
σ 2 = −24.46 MPa
Ans
Angles:
θ p2 :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
⎠
θ p1 := 90deg − θ p2
θ p1 = 60.128 deg
(Clockwise)
Ans
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τ max := −R
τ max = −69.46 MPa Ans
σ avg := σ c
σ avg = 45 MPa
Ans
2θ s2 = 90deg − 2⋅ θ p2
θ s2 := 45deg − θ p2
θ s2 = 15.128 deg
(Clockwise)
Ans
Problem 9-72
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σ x := 0MPa
σ y := 50MPa
τ xy := −30MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 25 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 39.051 MPa
Coordinates:
(
)
(
A σ x , τ xy
B σ y , −τ xy
)
(
C σc , 0
)
a) Principal Stresses:
σ 1 := σ c + R
σ 1 = 64.05 MPa
Ans
σ 2 := σ c − R
σ 2 = −14.05 MPa
Ans
Angles:
θ p :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p = 25.097 deg
⎠
(Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τ max := R
τ max = 39.05 MPa
Ans
σ avg := σ c
σ avg = 25 MPa
Ans
2θ s2 = 90deg − 2⋅ θ p
θ s2 := 45deg − θ p
θ s2 = 19.903 deg
(Counter-clockwise)
Problem 9-73
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σ x := −12MPa
σ y := −8MPa
τ xy := 4MPa
Solution:
Center :
σ c :=
σx + σy
σ c = −10 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 4.472 MPa
Coordinates:
(
)
(
A σ x , τ xy
B σ y , −τ xy
)
(
C σc , 0
)
a) Principal Stresses:
σ 1 := σ c + R
σ 1 = −5.53 MPa
Ans
σ 2 := σ c − R
σ 2 = −14.47 MPa
Ans
Angles:
θ p :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p = 31.717 deg
⎠
(Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τ max := R
τ max = 4.47 MPa
Ans
σ avg := σ c
σ avg = −10 MPa
Ans
2θ s2 = 90deg − 2⋅ θ p
θ s2 := 45deg − θ p
θ s2 = 13.283 deg
(Counter-clockwise)
Problem 9-74
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σ x := 45MPa
σ y := 30MPa
τ xy := −50MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 37.5 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 50.559 MPa
Coordinates:
(
)
(
A σ x , τ xy
B σ y , −τ xy
)
(
C σc , 0
)
a) Principal Stresses:
σ 1 := σ c + R
σ 1 = 88.06 MPa
Ans
σ 2 := σ c − R
σ 2 = −13.06 MPa
Ans
Angles:
θ p :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p = 40.735 deg
⎠
(Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τ max := R
τ max = 50.56 MPa
Ans
σ avg := σ c
σ avg = 37.5 MPa
Ans
2θ s2 = 90deg − 2⋅ θ p
θ s2 := 45deg − θ p
θ s2 = 4.265 deg
(Counter-clockwise)
Problem 9-75
The square steel plate has a thickness of 12 mm and is subjected to the edge loading shown. Determine
the principal stresses developed in the steel.
kN
Given: σ x := 0MPa
σ y := 0MPa
qxy := 3.2
m
L := 100mm
t := 12mm
Solution:
τ xy :=
qxy
τ xy = 0.267 MPa
t
Center :
σ c :=
σx + σy
σ c = 0 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 0.267 MPa
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 0.267 MPa
Ans
σ 2 := σ c − R
σ 2 = −0.267 MPa
Ans
Problem 9-76
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σ x := 105MPa
σ y := 0MPa
τ xy := −35MPa
Solution:
Center :
σ c :=
σx + σy
σ c = 52.5 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
R = 63.097 MPa
Coordinates:
(
A σ x , τ xy
)
(
C σc , 0
)
a) In-planePrincipal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 115.60 MPa
Ans
σ 2 := σ c − R
σ 2 = −10.60 MPa
Ans
Angles:
θ p :=
⎛ τ xy ⎞
1
atan ⎜
2
σx − σc
⎝
θ p = 16.845 deg
⎠
(Clockwise)
b) Maximum In-plane Shear Stresses:
Represented by the coordinates of point E on the circle.
τ max := −R
τ max = −63.10 MPa Ans
2θ s = 90deg − 2⋅ θ p
θ s := 45deg − θ p
θ s = 28.155 deg
(Counter-clockwise) Ans
Problem 9-77
Draw Mohr's circle that describes each of the following states of stress.
Given: σ x := −30MPa
σ y := 30MPa
τ xy := 0MPa
Solution:
Cases (a) and (b) are the same.
Center :
σ c :=
σx + σy
σ c = 0 MPa
2
Radius :
R :=
(σx − σc)2 + τ xy2
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
R = 30 MPa
Problem 9-78
Draw Mohr's circle that describes each of the following states of stress.
a) Given: σ x := 0.8MPa
τ xy := 0MPa
σ y := −0.6 MPa
Solution:
Center : σ c :=
σx + σy
2
σ c = 0.1 MPa
Radius : R :=
(σx − σc)2 + τ xy2
R = 0.7 MPa
Coordinates:
(
)
A σ x , τ xy
b) Given: σ x := 0MPa
(
C σc , 0
)
τ xy := 0MPa
σ y := −2MPa
Solution:
Center : σ c :=
σx + σy
2
σ c = −1 MPa
Radius : R :=
(σx − σc)2 + τ xy2
R = 1 MPa
Coordinates:
(
)
A σ x , τ xy
c) Given: σ x := 0MPa
(
C σc , 0
)
τ xy := 20MPa
σ y := 0MPa
Solution:
Center : σ c :=
σx + σy
2
σ c = 0 MPa
Radius : R :=
(σx − σc)2 + τ xy2
R = 20 MPa
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
Problem 9-79
A point on a thin plate is subjected to two successive states of stress as shown. Determine the resulting
state of stress with reference to an element oriented as shown on the bottom.
Unit used: kPa := 1000Pa
Given: σ x_a := 50kPa
σ y_a := 0
τ xy_a := 0
σ y_b := −18kPa
σ x_b := 0
τ xy_b := −45kPa
θ a := 30deg
β b := 50deg
Solution:
For element a:
Center : σ c_a :=
σ x_a + σ y_a
σ c_a = 25 kPa
2
(σx_a − σc_a)2 + τ xy_a2 Ra = 25 kPa
Coordinates:
A ( σ x_a , τ xy_a) B ( σ x_a , −τ xy_a)
C ( σ c_a , 0)
σ' x_a := σ c_a + Ra⋅ cos ( 2θ a)
σ' y_a := σ c_a − Ra⋅ cos ( 2θ a)
τ' xy_a := Ra⋅ sin ( 2θ a)
Radius : Ra :=
For element b:
(
Center : σ c_b :=
Radius : Rb :=
Coordinates:
Angles :
)
θ b := − 90deg − β b
σ x_b + σ y_b
σ c_b = −9 kPa
2
(σx_b − σc_b)2 + τ xy_b2 Rb = 45.891 kPa
A ( σ x_b , τ xy_b) B ( σ x_a , −τ xy_b) C ( σ c_b , 0)
⎛
φ := atan ⎜
τ xy_b
⎞
⎝ σ x_b − σ c_b ⎠
φ = 78.69 deg
α b := 2θ b + φ
α b = −1.310 deg
( )
σ' y_b := σ c_b − Rb⋅ cos ( α b)
τ' xy_b := Rb⋅ sin ( α b )
σ' x_b := σ c_b + Rb⋅ cos α b
Resultants:
σ' x := σ' x_a + σ' x_b
σ' x = 74.38 kPa
Ans
σ' y := σ' y_a + σ' y_b
σ' y = −42.38 kPa
Ans
τ' xy := τ' xy_a + τ' xy_b
τ' xy = 22.70 kPa
Ans
Problem 9-80
Mohr's circle for the state of stress in Fig. 9-15a is shown in Fig. 9-15b. Show that finding the
coordinates of point P(σx' , τx'y' ) on the circle gives the same value as the stress-transformation Eqs.
9-1 and 9-2.
Problem 9-81
The cantilevered rectangular bar is subjected to the force of 25 kN. Determine the principal stresses at
point A.
Given: b := 80mm
d := 160mm
P := 25kN
cA := 40mm
bA := −40mm
L := 0.4m
rv :=
Solution:
3
5
4
5
rh :=
Internal Force and Moment : At Section A-B:
+
ΣF x=0;
− P⋅ r h + N = 0
N := P⋅ rh
+
ΣF y=0;
− P⋅ r v + V = 0
V := P⋅ rv
M − P⋅ r v ⋅ L = 0
M := P⋅ rv ⋅ L
(
+ ΣΜO=0;
)
(
)
dA := 0.5d − cA
Section Property :
A := b⋅ d
I :=
(
1
3
⋅ b⋅ d
12
)
QA := b⋅ dA⋅ 0.5dA + cA
Normal Stress:
σ A :=
N M⋅ cA
+
A
I
Shear Stress :
V⋅ QA
τ A :=
I⋅ b
σ A = 10.352 MPa
τ A = 1.318 MPa
Construction of Mohr's Circle :
σ x := σ A
σ y := 0MPa
Center : σ c :=
Radius : R :=
τ xy := −τ A
σx + σy
σ c = 5.176 MPa
2
(σx − σc)2 + τ xy2
R = 5.341 MPa
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 10.52 MPa
Ans
σ 2 := σ c − R
σ 2 = −0.165 MPa
Ans
Problem 9-82
Solve Prob. 9-81 for the principal stresses at point B.
Given: b := 80mm
d := 160mm
P := 25kN
cB := −25mm
bB := 25mm
L := 0.4m
rv :=
Solution:
3
5
4
5
rh :=
Internal Force and Moment : At Section A-B:
+
ΣF x=0;
− P⋅ r h + N = 0
N := P⋅ rh
+
ΣF y=0;
− P⋅ r v + V = 0
V := P⋅ rv
M − P⋅ r v ⋅ L = 0
M := P⋅ rv ⋅ L
(
+ ΣΜO=0;
)
(
)
dB := 0.5d − cB
1
3
I :=
⋅ b⋅ d
12
Section Property :
A := b⋅ d
(
QB := b⋅ dB⋅ 0.5dB + cB
)
Normal Stress:
σ B :=
N M⋅ cB
+
A
I
Shear Stress :
V⋅ QB
τ B :=
I⋅ b
σ B = −3.931 MPa
τ B = 1.586 MPa
Construction of Mohr's Circle :
σ x := σ B
σ y := 0MPa
Center : σ c :=
Radius : R :=
τ xy := −τ B
σx + σy
σ c = −1.965 MPa
2
(σx − σc)2 + τ xy2
R = 2.526 MPa
Coordinates:
(
)
A σ x , τ xy
(
C σc , 0
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 0.560 MPa
Ans
σ 2 := σ c − R
σ 2 = −4.491 MPa
Ans
Problem 9-83
The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller at
B. If a man having a weight of 1500 N (~150 kg) stands in the center of the tread, determine the
principal stresses developed in the supporting truck on the cross section at point C. The stairs move at
constant velocity.
Given: b := 12mm
d := 50mm
W := 1.5kN
hC := 150mm
cC := 0mm
θ := 30deg
vA := 450mm
hA := 375mm
hB := 150mm
Solution:
Support Reactions :
+ ΣΜA=0; W⋅ hA − By⋅ hB = 0
By :=
W⋅ hA
hB
Bx := −By⋅ tan ( θ )
Internal Force and Moment : At Section C:
+
ΣF x=0;
+
ΣF y=0;
Bx + V = 0
By + N = 0
V := −Bx
N := −By
+ ΣΜO=0; Bx⋅ hC − M = 0
dC := 0.5d
Section Property :
A := b⋅ d
M := Bx⋅ hC
I :=
1
3
⋅ b⋅ d
12
(
)
QC := b⋅ dC⋅ 0.5dC
Normal Stress:
N M⋅ cC
+
σ C = −6.250 MPa
A
I
Shear Stress :
V⋅ QC
τ C :=
τ C = 5.413 MPa
I⋅ b
σ C :=
Construction of Mohr's Circle :
σ x := 0MPa
Center : σ c :=
σ y := σ C
τ xy := τ C
σx + σy
σ c = −3.125 MPa
2
Radius : R :=
(σx − σc)2 + τ xy2
Coordinates:
A σ x , τ xy
(
)
(
C σc , 0
R = 6.25 MPa
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 3.125 MPa
Ans
σ 2 := σ c − R
σ 2 = −9.375 MPa
Ans
Problem 9-84
The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not
rotate while subjected to a force of 400 N, determine the principal stresses in the material on the cross
section at point C.
Given: b := 7.5mm
d := 20mm
cC := 5mm
P := 0.4kN
L := 100mm
Solution:
Internal Force and Moment : At Section C:
+
ΣF y=0;
V−P= 0
V := P
+ ΣΜO=0; M + P⋅ L = 0
dC := 0.5d − cC
Section Property :
A := b⋅ d
M := −P⋅ L
I :=
(
1
3
⋅ b⋅ d
12
)
QC := b⋅ dC⋅ cC + 0.5dC
Normal Stress:
M⋅ cC
σ C := −
I
Shear Stress :
V⋅ QC
τ C :=
I⋅ b
σ C = 40.000 MPa
τ C = 3.000 MPa
Construction of Mohr's Circle :
σ x := σ C
Center : σ c :=
σ y := 0MPa
σx + σy
τ xy := τ C
σ c = 20 MPa
2
Radius : R :=
(σx − σc)2 + τ xy2
Coordinates:
A σ x , τ xy
(
)
(
C σc , 0
R = 20.224 MPa
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 40.224 MPa
Ans
σ 2 := σ c − R
σ 2 = −0.224 MPa
Ans
Problem 9-85
The frame supports the distributed loading of 200 N/m. Determine the normal and shear stresses at
point D that act perpendicular and parallel, respectively, to the grains. The grains at this point make an
angle of 30° with the horizontal as shown.
Unit Used: kPa := 1000Pa
Given: b := 100mm
a := 1m
h := 200mm
w := 0.2
L := 2.5m
kN
m
hD := 75mm
θ := 60deg
Solution:
Support Reactions: Due to symmetry, B=C=R
+
ΣFy=0; 2R − w⋅ L = 0
R := 0.5w⋅ L
Internal Force and Moment : At Section D:
+
ΣFy=0;
−V + R − w⋅ a = 0
V := R − w⋅ a
+ ΣΜO=0;
M − R⋅ a + w a⋅ ( 0.5a) = 0
M := R⋅ a − w⋅ a⋅ ( 0.5a)
cD := 0.5h − hD
1
3
A := b⋅ h
I :=
⋅ b⋅ h
12
QD := b⋅ hD⋅ cD + 0.5hD
Section Property :
(
Normal Stress:
)
M⋅ ( −cD)
σ := −
Shear Stress :
τ D :=
D
σ D = 56.25 kPa
I
V⋅ QD
τ D = 3.516 kPa
I⋅ b
Construction of Mohr's Circle :
σ x := σ D
σ y := 0MPa
Center : σ c :=
Radius : R :=
Coordinates:
Angles:
σx + σy
τ xy := τ D
σ c = 28.125 kPa
2
(σx − σc)2 + τ xy2 R = 28.344 kPa
A ( σ x , −τ xy) C ( σ c , 0)
⎛
τ xy
⎞
⎝
σx − σc
⎠
φ := atan ⎜
α := 180deg − ( 2 θ + φ )
φ = 7.125 deg
α = 52.875 deg
Stresses on the Rotated Element: (represented by coordinates of point P)
σ x' := σ c − R⋅ cos ( α )
σ x' = 11.02 kPa
Ans
τ x'y' := −R⋅ sin ( α )
τ x'y' = −22.60 kPa
Ans
Problem 9-86
The frame supports the distributed loading of 200 N/m. Determine the normal and shear stresses at
point E that act perpendicular and parallel, respectively, to the grains. The grains at this point make an
angle of 60° with the horizontal as shown.
Unit Used: kPa := 1000Pa
Given: b := 50mm
w := 0.2
h := 100mm
L := 2.5m
kN
m
θ := 60deg
Solution:
Support Reactions: Due to symmetry, B=C=R
+
ΣFy=0; 2R − w⋅ L = 0
Internal Force:
+
ΣFy=0;
R := 0.5w⋅ L
At Section E:
−N − R = 0
N := −R
Section Property :
2
A := b⋅ h
A = 5000 mm
N
A
Normal Stress:
σ E :=
Shear Stress :
τ D := 0
σ E = −50 kPa
Construction of Mohr's Circle :
σ x := σ E
σ y := 0MPa
Center : σ c :=
Radius : R :=
Coordinates:
Angles:
σx + σy
τ xy := τ D
σ c = −25 kPa
2
(σx − σc)2 + τ xy2 R = 25 kPa
A ( σ x , τ xy)
C ( σ c , 0)
⎛
φ := atan ⎜
τ xy
⎞
⎝ σx − σc ⎠
α := 180deg − ( 2 θ + φ )
φ = 0.000 deg
α = 60.000 deg
Stresses on the Rotated Element: (represented by coordinates of point P)
σ x' := σ c + R⋅ cos ( α )
σ x' = −12.50 kPa
Ans
τ x'y' := R⋅ sin ( α )
τ x'y' = 21.65 kPa
Ans
Problem 9-87
The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal
stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the
results on elements located at these points.
Given: do := 15mm
P := 0.6kN
a := 50mm
L := 0.1m
Solution:
Internal Force:
At Section AB:
N := P
N = 0.600 kN
M := P⋅ a
M = 0.030 kN⋅ m
Section Property :
π
2
2
A := ⋅ do
A = 176.71 mm
4
π
4
2
2
I :=
⋅d
I = 0 m mm
64 o
(since A' = 0)
QA := 0
(since A' = 0)
QB := 0
Normal Stress:
N M⋅ cA
cA := 0.5do
σ A :=
−
A
I
σ A = −87.15 MPa
N M⋅ cB
+
A
I
σ B = 93.94 MPa
cB := 0.5do
σ B :=
Shear Stress :
τ A := 0
(since QA = 0)
τ B := 0
(since QB = 0)
For A:
Construction of Mohr's Circle :
σ x := σ A
Center : σ c :=
Radius : R :=
Coordinates:
σ y := 0MPa
σx + σy
2
τ xy := τ A
σ c = −43.573 MPa
(σx − σc)2 + τ xy2 R = 43.573 MPa
A ( σ x , τ xy)
C ( σ c , 0)
In-plane Principal Stresses:
The coordinates of points B and A represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 0 MPa
Ans
σ 2 := σ c − R
σ 2 = −87.15 MPa
Ans
Maximum In-plane Shear Stresses:
Represented by the coordinates of point E on the circle.
τ max := R
τ max = 43.57 MPa
2θ s = 90deg
θ s := 45deg
Ans
(Counter-clockwise) Ans
For B:
Construction of Mohr's Circle :
σ x := σ B
Center : σ c :=
Radius : R :=
Coordinates:
σ y := 0MPa
σx + σy
2
τ xy := τ B
σ c = 46.968 MPa
(σx − σc)2 + τ xy2 R = 46.968 MPa
A ( σ x , τ xy)
C ( σ c , 0)
In-plane Principal Stresses:
The coordinates of points A and B represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 93.94 MPa
Ans
σ 2 := σ c − R
σ 2 = 0 MPa
Ans
Maximum In-plane Shear Stresses:
Represented by the coordinates of point E on the circle.
τ max := R
τ max = 46.97 MPa
2θ s = 90deg
θ s := 45deg
Ans
(Clockwise) Ans
Problem 9-88
Draw the three Mohr's circles that describe each of the following states of stress.
a)
σ max := 6MPa
σ int := 0
σ min := 0
b) σ max := 50MPa
σ min := −40MPa
c) Unit used:
σ int := 0
kPa := 1000Pa
σ max := 600kPa
σ int := 200kPa
σ min := 100kPa
d) σ max := 0
σ min := −9MPa
σ int := −7MPa
σ max := −30MPa
σ int := −30MPa
e)
σ min := −30MPa
Problem 9-89
Draw the three Mohr's circles that describe each of the following states of stress.
a)
σ 1 := 15MPa
σ 2 := 0
σ 3 := −15MPa
τ max := 15MPa
b) σ 1 := 65MPa
σ 3 := −65MPa
τ max := 65MPa
σ 2 := −65MPa
Problem 9-90
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: σ x := −100MPa
τ xy := 0MPa
σ y := 90MPa
σ z := −80MPa
τ yz := 40MPa
τ xz := 0MPa
Solution:
Construction of Mohr's Circle in y-z Plane :
Center : σ c :=
σy + σz
σ c = 5 MPa
2
Radius : R :=
(σy − σc)2 + τ yz2
Coordinates:
A σ y , τ yz
(
)
(
R = 93.941 MPa
C σc , 0
)
In-plane Principal Stresses:
The coordinates of points A and B represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 98.941 MPa
σ 2 := σ c − R
σ 2 = −88.941 MPa
Construction of Three Circles :
From the results obtained above,
σ max := σ 1
σ int := σ 2
σ max = 98.94 MPa
Ans
σ int = −88.94 MPa
Ans
σ min = −100.00 MPa
Ans
Absolute Maximum Shear Stress :
From the three Mohr's circles,
τ abs.max :=
σ max − σ min
2
τ abs.max = 99.47 MPa
Ans
σ min := σ x
Problem 9-91
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: σ x := 0MPa
τ xy := 0MPa
σ y := 7MPa
σ z := 0MPa
τ yz := 50MPa
τ xz := 5MPa
Solution:
Construction of Mohr's Circle in x-z Plane :
Center : σ c :=
σx + σz
σ c = 0 MPa
2
Radius : R :=
(σx − σc)2 + τ xz2
Coordinates:
A σ y , τ xz
(
)
R = 5 MPa
(
C σc , 0
)
In-plane Principal Stresses:
The coordinates of points A and B represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 5 MPa
σ 2 := σ c − R
σ 2 = −5 MPa
Construction of Three Circles :
From the results obtained above,
σ max := σ y
σ int := σ 1
σ max = 7 MPa
Ans
σ int = 5 MPa
Ans
σ min = −5 MPa
Ans
Absolute Maximum Shear Stress :
From the three Mohr's circles,
τ max :=
σ max − σ min
τ max = 6 MPa
2
Ans
σ min := σ 2
Problem 9-92
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: σ x := 150MPa
τ xy := 0MPa
σ y := 0MPa
σ z := 90MPa
τ yz := −80MPa
τ xz := 0MPa
Solution:
Construction of Mohr's Circle in y-z Plane :
Center : σ c :=
σy + σz
σ c = 45 MPa
2
Radius : R :=
(σz − σc)2 + τ yz2
Coordinates:
A σ y , τ yz
(
)
(
R = 91.788 MPa
)
B σ z , −τ yz
(
C σc , 0
)
In-plane Principal Stresses:
The coordinates of points A and B represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 136.788 MPa
σ 2 := σ c − R
σ 2 = −46.788 MPa
Construction of Three Circles :
From the results obtained above,
σ max := σ x
σ int := σ 1
σ max = 150.00 MPa
Ans
σ int = 136.79 MPa
Ans
σ min = −46.79 MPa
Ans
Absolute Maximum Shear Stress :
From the three Mohr's circles,
τ max :=
σ max − σ min
2
τ max = 98.39 MPa
Ans
σ min := σ 2
Problem 9-93
The principal stresses acting at a point in a body are shown. Draw the three Mohr's circles that
describe this state of stress, and find the maximum in-plane shear stresses and associated average
normal stresses for the x-y, y-z, and x-z planes. For each case, show the results on the element oriented
in the appropriate direction.
Given: σ x := 40MPa
τ xy := 0MPa
σ y := −40MPa
σ z := −40MPa
τ yz := 0MPa
τ xz := 0MPa
Solution:
Construction of Three Circles :
σ max := σ x
σ int := σ y
σ min := σ z
σ max = 40.00 MPa
σ int = −40.00 MPa
σ min = −40.00 MPa
For x-y Plane :
σx + σy
σ avg :=
2
τ max :=
σx − σy
2
For y-z Plane :
σy + σz
σ' avg :=
2
τ' max :=
σy − σz
2
For x-y Plane :
σx + σz
σ''avg :=
2
τ''max :=
σx − σz
2
σ avg = 0 MPa
Ans
τ max = 40 MPa
Ans
σ' avg = −40 MPa
Ans
τ' max = 0 MPa
Ans
σ''avg = 0 MPa
Ans
τ''max = 40 MPa
Ans
Problem 9-94
Consider the general case of plane stress as shown. Write a computer program that will show a plot of
the three Mohr's circles for the element, and will also calculate the maximum in-plane shear stress and
the absolute maximum shear stress.
Problem 9-95
The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the
principal stresses acting at points A and B and the absolute maximum shear stress.
Given: ro := 25mm
L := 450mm
P := 0.8kN
T o := 0.045kN⋅ m
Mo := 0.3kN⋅ m
ρ := ro
Solution:
Internal Force and Moment : At Section AB:
Vy := P
T x := To
Section Property :
A := π ⋅ ro
2
Iz :=
Mz := Mo − P⋅ L
π 4
⋅r
4 o
J :=
QA := 0 (Since A'=0)
π 4
⋅r
2 o
QB :=
4⋅ r o
3⋅ π
⋅ ( 0.5A)
Normal Stress: cA := ro
cB := 0
Mz⋅ cA
σ A := −
σ A = 4.889 MPa
Iz
Mz⋅ cB
σ B := −
σ B = 0 MPa
Iz
Shear Stress : bB := 2⋅ ro
T x⋅ ρ
τ A :=
J
τ B :=
Vy⋅ QB T x⋅ ρ
−
Iz⋅ bB
J
τ A = 1.833 MPa
τ B = −1.290 MPa
For Point A:
Construction of Mohr's Circle:
σ x := σ A
σ z := 0MPa
Center : σ c :=
σx + σz
τ xz := −τ A
σ c = 2.445 MPa
2
Radius : R :=
(σz − σc)2 + τ xz2
Coordinates:
A σ z , τ xz
(
)
(
C σc , 0
R = 3.056 MPa
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 5.500 MPa
σ 2 := σ c − R
σ 2 = −0.611 MPa
Three Mohr's Circles:σ y := 0
From the results obtained above,
σ max := σ 1
σ int := σ y
σ max = 5.50 MPa
Ans
σ int = 0.00 MPa
Ans
σ min = −0.611 MPa
Ans
σ min := σ 2
Absolute Maximum Shear Stress :
From the three Mohr's circles,
σ max − σ min
τ max :=
τ max = 3.06 MPa
2
For Point B:
Construction of Mohr's Circle:
σ x := σ B
σ z := 0MPa
Center : σ c :=
σx + σz
τ xz := τ B
σ c = 0 MPa
2
Radius : R :=
(σz − σc)2 + τ xz2
Coordinates:
A σ z , τ xz
(
Ans
)
(
C σc , 0
R = 1.290 MPa
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 1.290 MPa
σ 2 := σ c − R
σ 2 = −1.290 MPa
Three Mohr's Circles:σ y := 0
From the results obtained above,
σ max := σ 1
σ int := σ y
σ max = 1.29 MPa
Ans
σ int = 0.00 MPa
Ans
σ min = −1.29 MPa
Ans
σ min := σ 2
Absolute Maximum Shear Stress :
From the three Mohr's circles,
σ max − σ min
τ max :=
τ max = 1.29 MPa
2
Ans
Problem 9-96
The bolt is fixed to its support at C. If a force of 90 kN is applied to the wrench to tighten it, determine
the principal stresses and the absolute maximum shear stress developed in the bolt shank at point A.
Represent the results on an element located at this point. The shank has a diameter of 6 mm.
Given: do := 6mm
a := 50mm
P := 90N
Solution:
L := 150mm
ρ := 0.5do
Internal Force and Moment : At Section AB:
Vy := P
Mz := −P⋅ a
Tx := P⋅ L
Section Property :
π
π
2
4
⋅ do Iz :=
⋅ do
64
4
QA := 0 (Since A'=0)
A :=
Normal Stress:
σ A := −
J :=
π
4
⋅ do
32
cA := 0.5do
Mz⋅ cA
Iz
Shear Stress :
T x⋅ ρ
τ A :=
J
σ A = 212.21 MPa
τ A = 318.31 MPa
Construction of Mohr's Circle in x-z Plane :
σ x := σ A
Center : σ c :=
σ z := 0MPa
σx + σz
τ xz := −τ A
σ c = 106.1 MPa
2
Radius : R :=
(σz − σc)2 + τ xz2
Coordinates:
A σ z , τ xz
(
)
(
R = 335.53 MPa
)
C σc , 0
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 441.631 MPa
σ 2 := σ c − R
σ 2 = −229.425 MPa
Angles:
θ p2 :=
⎛ τ xz ⎞
1
atan ⎜
2
⎝ σz − σc ⎠
2θ p1 = 180deg − 2⋅ θ p2
θ p2 = 35.78 deg
θ p1 := 90deg − θ p2
Construction of Three Circles : σ y := 0
From the results obtained above,
θ p1 = 54.22 deg
(Clockwise)
σ max := σ 1
σ int := σ y
σ max = 441.63 MPa
Ans
σ int = 0.00 MPa
Ans
σ min := σ 2
σ min = −229.42 MPa Ans
Absolute Maximum Shear Stress :
From the three Mohr's circles,
τ max :=
σ max − σ min
2
τ max = 335.53 MPa
Ans
And the orientation is,
θ s :=
⎛ σz − σc ⎞
1
atan ⎜
2
τ xz
⎝
⎠
θ s = 9.22 deg
Problem 9-97
Solve Prob. 9-96 for point B.
Given: do := 6mm
a := 50mm
P := 90N
L := 150mm
ρ := 0.5do
Solution:
Internal Force and Moment : At Section AB:
Vy := P
Mz := −P⋅ a
T x := P⋅ L
Section Property :
π
π
2
4
⋅ do Iz :=
⋅ do
64
4
4ρ ⎛ A ⎞
QB :=
⋅⎜
3π ⎝ 2 ⎠
A :=
τ B :=
π
4
⋅ do
32
cB := 0
Normal Stress:
Mz⋅ cB
σ B := −
Iz
Shear Stress :
J :=
σ B = 0 MPa
bB := do
Vy⋅ QB T x⋅ ρ
−
Iz⋅ bB
J
τ B = −314.07 MPa
Construction of Mohr's Circle in x-z Plane :
σ x := σ B
Center : σ c :=
σ z := 0MPa
σx + σz
τ xz := τ B
σ c = 0 MPa
2
Radius : R :=
(σz − σc)2 + τ xz2
Coordinates:
A σ z , τ xz
(
)
(
C σc , 0
R = 314.07 MPa
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 314.07 MPa
σ 2 := σ c − R
σ 2 = −314.07 MPa
Angles:
θ p1 :=
1
( 90deg)
2
θ p1 = 45.00 deg
(Clockwise)
Construction of Three Circles : σ y := 0
From the results obtained above,
σ max := σ 1
σ int := σ y
σ max = 314.07 MPa
Ans
σ int = 0.00 MPa
Ans
σ min := σ 2
σ min = −314.07 MPa Ans
Absolute Maximum Shear Stress :
From the three Mohr's circles,
τ max :=
σ max − σ min
2
τ max = 314.07 MPa
Ans
Problem 9-98
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: σ x := −90MPa
σ y := 70MPa
σ z := −120MPa
τ xy := 30MPa
τ yz := 0MPa
τ xz := 0MPa
Solution:
Construction of Mohr's Circle in x-y Plane :
Center : σ c :=
σx + σy
σ c = −10 MPa
2
Radius : R :=
(σx − σc)2 + τ xy2
Coordinates:
A σ y , −τ xy
(
)
R = 85.44 MPa
(
B σ x , τ xy
)
(
C σc , 0
)
In-plane Principal Stresses:
The coordinates of points representing σ1 and σ2, respectively, are
σ 1 := σ c + R
σ 1 = 75.440 MPa
σ 2 := σ c − R
σ 2 = −95.440 MPa
Construction of Three Circles :
From the results obtained above,
σ max := σ 1
σ int := σ 2
σ max = 75.44 MPa
Ans
σ int = −95.44 MPa
Ans
σ min = −120.00 MPa
Ans
Absolute Maximum Shear Stress :
From the three Mohr's circles,
τ max :=
σ max − σ min
2
τ max = 97.72 MPa
Ans
σ min := σ z
Problem 9-99
The cylindrical pressure vessel has an inner radius of 1.25 m and a wall thickness of 15 mm. It is made
from steel plates that are welded along a 45° seam with the horizontal. Determine the normal and shear
stress components along this seam if the vessel is subjected to an internal pressure of 3 MPa.
Given:
t := 15mm
ri := 1250mm
p := 3MPa
θ := 45deg
Solution:
α :=
ri
α = 83.33
t
Since α > 10. then thin-wall analysis can be used.
p⋅ ri
σ 1 :=
σ 1 = 250 MPa
t
Hoop Stress :
Longitudinal Stress :
p⋅ ri
σ 2 :=
σ 2 = 125 MPa
2⋅ t
Construction of Mohr's Circle:
σ x := σ 2
Center : σ c :=
σ y := σ 1
τ xy := 0
σx + σy
σ c = 187.5 MPa
2
Radius : R :=
(σy − σc)2 + τ xy2
Coordinates:
A σ x , τ xz
(
)
(
R = 62.5 MPa
C σc , 0
)
Stresses: (represented by coordinates of point P on the circle)
σ x' := σ c
σ x' = 187.5 MPa
Ans
τ x'y' := R
τ x'y' = 62.5 MPa
Ans
Problem 9-100
Determine the equivalent state of stress if an element is oriented 40° clockwise from the element
shown. Use Mohr's circle.
Given: σ x := 6MPa
σ y := −10MPa
θ := 40deg
τ xy := 0MPa
Solution:
Center :
σ c :=
σx + σy
σ c = −2 MPa
2
Radius :
R := σ x − σ c
R = 8 MPa
Coordinates:
(
A σx , 0
)
(
)
B σy , 0
(
C σc , 0
)
Stresses:
σ x' := σ c + R⋅ cos ( 2θ )
σ x' = −0.611 MPa
Ans
τ x'y' := R⋅ sin ( 2θ )
τ x'y' = 7.878 MPa
Ans
σ y' := σ c − R⋅ cos ( 2θ )
σ y' = −3.389 MPa
Ans
Problem 9-101
The internal loadings at a cross section through the 150-mm-diameter drive shaft of a turbine consist
of an axial force of 12.5 kN, a bending moment of 1.2 kN·m, and a torsional moment of 2.25 kN·m.
Determine the principal stresses at point A. Also compute the maximum in-plane shear stress at this
point.
Given: do := 150mm
N := −12.5 kN
Mz := 1.2kN⋅ m
Solution: ρ := 0.5do
T x := 2.25kN⋅ m
Section Property :
π
π
π
2
4
4
⋅ do Iz :=
⋅ do
J :=
⋅d
64
4
32 o
Normal Stress: cA := ρ
N Mz⋅ cA
σ A :=
−
σ A = −4.33 MPa
A
Iz
A :=
Shear Stress :
T x⋅ ρ
τ A :=
J
τ A = 3.4 MPa
Construction of Mohr's Circle in x-y Plane :
σ x := σ A
Center : σ c :=
σ y := 0MPa
σx + σy
τ xy := τ A
σ c = −2.16 MPa
2
Radius : R :=
(σx − σc)2 + τ xy2
Coordinates:
A σ x , τ xy
(
)
(
C σc , 0
R = 4.03 MPa
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 1.862 MPa
σ 2 := σ c − R
σ 2 = −6.191 MPa
Maximum In-plane Shear Stress :
Represented by the coordinates of point E on the circle.
τ max := R
τ max = 4.03 MPa
Ans
Problem 9-102
The internal loadings at a cross section through the 150-mm-diameter drive shaft of a turbine consist
of an axial force of 12.5 kN, a bending moment of 1.2 kN·m, and a torsional moment of 2.25 kN·m.
Determine the principal stresses at point B. Also compute the maximum in-plane shear stress at this
point.
Given: do := 150mm
N := −12.5 kN
Mz := 1.2kN⋅ m
Solution: ρ := 0.5do
T x := 2.25kN⋅ m
Section Property :
A :=
π
π
2
4
⋅ do Iz :=
⋅d
64 o
4
Normal Stress:
σ B :=
N
+
A
cB := 0
Mz⋅ cB
Iz
Shear Stress :
T x⋅ ρ
τ B :=
J
J :=
π
4
⋅d
32 o
σ B = −0.707 MPa
τ B = 3.395 MPa
Construction of Mohr's Circle in x-y Plane :
σ x := σ B
σ y := 0MPa
Center : σ c :=
σx + σy
τ xy := τ B
σ c = −0.354 MPa
2
Radius : R :=
(σx − σc )2 + τ xy2
Coordinates:
A σ x , τ xy
(
)
(
C σc , 0
R = 3.414 MPa
)
In-plane Principal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ 1 := σ c + R
σ 1 = 3.060 MPa
σ 2 := σ c − R
σ 2 = −3.767 MPa
Maximum In-plane Shear Stress :
Represented by the coordinates of point E on the circle.
τ max := R
τ max = 3.41 MPa
Ans
Problem 9-103
Determine the equivalent state of stress on an element if it is oriented 30° clockwise from the element
shown. Use the stress-transformation equations.
Given: σ x := 0MPa
σ y := −0.300 MPa
θ := −30deg
τ xy := 0.950MPa
Solution:
Normal Stress:
σ x' :=
σx + σy
2
+
σx − σy
2
σ x' = −0.898 MPa
σ y' :=
σx + σy
2
−
⋅ cos ( 2θ ) + τ xy⋅ sin ( 2θ )
Ans
σx − σy
σ y' = 0.598 MPa
2
⋅ cos ( 2θ ) − τ xy⋅ sin ( 2θ )
Ans
Shear Stress:
τ x'y' := −
σx − σy
2
τ x'y' = 0.605 MPa
⋅ sin ( 2θ ) + τ xy⋅ cos ( 2θ )
Ans
Problem 9-104
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB.
Given: σ x := −50MPa
σ y := −100MPa
φ' := 30deg
τ xy := 28MPa
Solution:
Construction of Mohr's Circle:
Center : σ c :=
σx + σy
θ := 90deg + φ'
σ c = −75 MPa
2
Radius : R :=
(σy − σc)2 + τ xy2
Coordinates:
A σ x , τ xz
Angles:
(
)
⎛ τ xy ⎞
(
R = 37.54 MPa
C σc , 0
)
φ := atan ⎜
φ = 48.240 deg
α := 360deg − 2θ − φ
α = 71.760 deg
⎝ σx − σc ⎠
Stresses: (represented by coordinates of point P on the circle)
σ x' := σ c + R⋅ cos ( α )
σ x' = −63.25 MPa
Ans
τ x'y' := R⋅ sin ( α )
τ x'y' = 35.65 MPa
Ans
Problem 10-1
Prove that the sum of the normal strains in perpendicular directions is constant.
Solution:
Stress Transformation Equations: Applying Eqs. 10-5 and 10-7 of the text.
ε x' =
ε y' =
εx + εy
2
εx + εy
2
+
−
εx − εy
2
εx − εy
2
⋅ cos ( 2θ ) +
⋅ cos ( 2θ ) −
γ xy
2
γ xy
2
⋅ sin ( 2θ )
(1)
⋅ sin ( 2θ )
(2)
(1) + (2) :
LHS = ε x' + ε y'
RHS =
εx + εy
2
+
εx + εy
2
Hence,
ε x' + ε y' = ε x + ε y
(Q.E.D.)
RHS = ε x + ε y
Problem 10-2
The state of strain at the point on the leaf of the caster assembly has components of εx = -400(10-6),
εy = 860(10-6), and γxy = 375(10-6). Use the strain-transformation equations to determine the equivalent
in-plane strains on an element oriented at an angle of θ = 30° counterclockwise from the original
position. Sketch the deformed element due to these strains within the x-y plane.
−6
Given: ε x := −400⋅ 10− 6
γ xy := 375⋅ 10
(
)
( − 6)
ε y := 860⋅ 10
(
)
θ := 30deg
Solution:
Stress Transformation Equations:
εx + εy εx − εy
γ xy
ε x' :=
+
⋅ cos ( 2θ ) +
⋅ sin ( 2θ )
2
2
2
−6
ε x' = 77.38 × 10
ε y' :=
εx + εy
2
−
εx − εy
2
⋅ cos ( 2θ ) −
γ xy
2
Ans
⋅ sin ( 2θ )
ε y' = 382.62 × 10
−6
Ans
γ xy
⎛ εx − εy
⎞
⋅ sin ( 2θ ) +
⋅ cos ( 2θ )
2
2
⎝
⎠
γ x'y' := 2⎜ −
−3
γ x'y' = 1.279 × 10
Ans
Problem 10-3
The state of strain at the point on the pin leaf has components of εx = 200(10-6), εy = 180(10-6), and γxy
= -300(10-6). Use the strain-transformation equations to determine the equivalent in-plane strains on an
element oriented at an angle of θ = 60° counterclockwise from the original position. Sketch the
deformed element due to these strains within the x-y plane.
( )
−6
ε y := 180⋅ ( 10 )
Given: ε x := 200⋅ 10− 6
( − 6)
γ xy := −300⋅ 10
θ := 60deg
Solution:
Stress Transformation Equations:
εx + εy εx − εy
γ xy
ε x' :=
+
⋅ cos ( 2θ ) +
⋅ sin ( 2θ )
2
2
2
−6
ε x' = 55.10 × 10
ε y' :=
εx + εy
2
−
εx − εy
2
⋅ cos ( 2θ ) −
γ xy
2
⋅ sin ( 2θ )
ε y' = 324.90 × 10
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
⋅ sin ( 2θ ) +
γ xy
2
Ans
⋅ cos ( 2θ )
−6
Ans
⎞
⎠
−6
γ x'y' = 132.679 × 10
Ans
Problem 10-4
Solve Prob. 10-3 for an element oriented θ = 30° clockwise.
( )
−6
ε y := 180⋅ ( 10 )
Given: ε x := 200⋅ 10− 6
( − 6)
γ xy := −300⋅ 10
θ := −30deg
Solution:
Stress Transformation Equations:
εx + εy εx − εy
γ xy
ε x' :=
+
⋅ cos ( 2θ ) +
⋅ sin ( 2θ )
2
2
2
ε x' = 324.90 × 10
ε y' :=
εx + εy
2
−
εx − εy
2
⋅ cos ( 2θ ) −
γ xy
2
−6
Ans
⋅ sin ( 2θ )
−6
ε y' = 55.10 × 10
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
⋅ sin ( 2θ ) +
γ xy
2
⋅ cos ( 2θ )
Ans
⎞
⎠
γ x'y' = −132.679 × 10
−6
Ans
Problem 10-5
Due to the load P , the state of strain at the point on the bracket has components of εx = 500(10-6),
εy = 350(10-6), and γxy = -430(10-6). Use the strain-transformation equations to determine the equivalent
in-plane strains on an element oriented at an angle of θ = 30° clockwise from the original position.
Sketch the deformed element due to these strains within the x-y plane.
( )
−6
ε y := 350⋅ ( 10 )
Given: ε x := 500⋅ 10− 6
( − 6)
γ xy := −430⋅ 10
θ := −30deg
Solution:
Stress Transformation Equations:
εx + εy εx − εy
γ xy
ε x' :=
+
⋅ cos ( 2θ ) +
⋅ sin ( 2θ )
2
2
2
ε x' = 648.70 × 10
ε y' :=
εx + εy
2
−
εx − εy
2
⋅ cos ( 2θ ) −
γ xy
2
⎛ εx − εy
⎝
2
⋅ sin ( 2θ ) +
γ xy
2
Ans
⋅ sin ( 2θ )
ε y' = 201.30 × 10
γ x'y' := 2⎜ −
−6
⋅ cos ( 2θ )
−6
Ans
⎞
⎠
−6
γ x'y' = −85.096 × 10
Ans
Problem 10-6
The state of strain at the point on a wrench has components εx = 120(10-6), εy = -180(10-6), γxy =
150(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and (b)
the maximum in-plane shear strain and average normal strain. In each case specify the orientation of
the element and show how the strains deform the element within the x-y plane.
(
Given: ε x := 120⋅ 10− 6
)
( − 6)
( − 6)
ε y := −180⋅ 10
γ xy := 150⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
− ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
2
−6
ε 1 = 137.71 × 10
Ans
2
−6
ε 2 = −197.71 × 10
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
⎠
θ p = 13.283 deg
θ' p := θ p − 90deg
θ' p = −76.717 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
ε x' = 137.71 × 10
Therefore,
−6
θ p1 := θ p
θ p1 = 13.28 deg
Ans
θ p2 := θ' p
θ p2 = −76.72 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
γ max = 335.41 × 10
ε avg :=
εx + εy
2
−6
2
Ans
−6
ε avg = −30 × 10
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
( )
Ans
θ' s := θ s + 90deg
θ s = −31.717 deg
θ' .s = 58.283 deg
Use Eq. 10-6 to determine the sign of γmax.
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = 335.41 × 10
Therefore,
−6
θ s1 := θ s
θ s1 = −31.72 deg
Ans
θ s2 := θ' s
θ s2 = 58.28 deg
Ans
Problem 10-7
The state of strain at the point on the gear tooth has components εx = 850(10-6), εy = 480(10-6), γxy =
650(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and (b)
the maximum in-plane shear strain and average normal strain. In each case specify the orientation of
the element and show how the strains deform the element within the x-y plane.
(
Given: ε x := 850⋅ 10− 6
)
( − 6)
( − 6)
ε y := 480⋅ 10
γ xy := 650⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
εx + εy
−
2
⎛ εx − εy ⎞
⎜
⎝
2
2
⎛ γ xy ⎞
2
−3
ε 1 = 1.039 × 10
Ans
2
−6
+⎜
⎠
ε 2 = 291.03 × 10
⎝ 2 ⎠
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
θ' p := θ p + 90deg
⎠
θ p = 30.175 deg
θ' p = 120.175 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
−3
ε x' = 1.039 × 10
Therefore,
θ p1 := θ p
θ p1 = 30.18 deg
Ans
θ p2 := θ' p
θ p2 = 120.18 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
γ max = 747.93 × 10
ε avg :=
εx + εy
2
−6
2
Ans
−6
ε avg = 665 × 10
Ans
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
θ s = −14.825 deg
( )
θ' s := θ s + 90deg
θ' s = 75.175 deg
Use Eq. 10-6 to determine the sign of γmax.
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = 747.93 × 10
Therefore,
−6
θ s1 := θ s
θ s1 = −14.82 deg
Ans
θ s2 := θ' s
θ s2 = 75.18 deg
Ans
Problem 10-8
The state of strain at the point on the gear tooth has the components εx = 520(10-6), εy = -760(10-6),
γxy = -750(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains
and (b) the maximum in-plane shear strain and average normal strain. In each case specify the
orientation of the element and show how the strains deform the element within the x-y plane.
(
Given: ε x := 520⋅ 10− 6
)
( − 6) γ xy := −750⋅ (10− 6)
ε y := −760⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
εx + εy
2
εx + εy
2
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
−
⎛ εx − εy ⎞
⎜
⎝
2
2
⎛ γ xy ⎞
2
ε 1 = 621.772 × 10
Ans
2
+⎜
⎠
−6
ε 2 = −861.77 × 10
⎝ 2 ⎠
−6
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
⎠
θ p = −15.184 deg
θ' p := θ p + 90deg
θ' p = 74.816 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
ε x' = 621.772 × 10
Therefore,
−6
θ p1 := θ p
θ p1 = −15.18 deg
Ans
θ p2 := θ' p
θ p2 = 74.82 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
γ max = 1.48 × 10
ε avg :=
εx + εy
2
−3
2
Ans
ε avg = −120 × 10
−6
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
θ s = 29.816 deg
( )
Use Eq. 10-6 to determine the sign of γmax.
Ans
θ' s := θ s − 90deg
θ' s = −60.184 deg
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = −1.48 × 10
Therefore,
−3
θ s1 := θ s
θ s1 = 29.82 deg
Ans
θ s2 := θ' s
θ s2 = −60.18 deg
Ans
Problem 10-9
The state of strain at the point on the spanner wrench has components εx = 260(10-6), εy = 320(10-6),
γxy = 180(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains
and (b) the maximum in-plane shear strain and average normal strain. In each case specify the
orientation of the element and show how the strains deform the element within the x-y plane.
(
Given: ε x := 260⋅ 10− 6
)
( − 6)
( − 6)
ε y := 320⋅ 10
γ xy := 180⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
− ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
2
−6
ε 1 = 384.87 × 10
Ans
2
−6
ε 2 = 195.13 × 10
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
θ' p := θ p + 90deg
⎠
θ p = −35.783 deg
θ' p = 54.217 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
−6
ε x' = 195.132 × 10
Therefore,
θ p1 := θ' p
θ p1 = 54.22 deg
Ans
θ p2 := θ p
θ p2 = −35.78 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
ε avg :=
εx + εy
2
2
γ max = 189.74 × 10
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
θ s = 9.217 deg
( )
−6
ε avg = 290 × 10
−6
Ans
Ans
θ' s := θ s − 90deg
θ' s = −80.783 deg
Use Eq. 10-6 to determine the sign of γmax.
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = 189.74 × 10
Therefore,
−6
θ s1 := θ s
θ s1 = 9.22 deg
Ans
θ s2 := θ' s
θ s2 = −80.78 deg
Ans
Problem 10-10
The state of strain at the point on the arm has components εx = 250(10-6), εy = -450(10-6), γxy
= -825(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and
(b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation
of the element and show how the strains deform the element within the x-y plane.
(
Given: ε x := 250⋅ 10− 6
)
( − 6)
( − 6)
ε y := −450⋅ 10
γ xy := −825⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
εx + εy
−
2
⎛ εx − εy ⎞
⎜
⎝
2
2
⎠
⎛ γ xy ⎞
2
−6
ε 1 = 440.98 × 10
Ans
2
−6
+⎜
ε 2 = −640.98 × 10
⎝ 2 ⎠
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
⎠
θ p = −24.843 deg
θ' p := θ p + 90deg
θ' p = 65.157 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
−6
ε x' = 440.977 × 10
Therefore,
θ p1 := θ p
θ p1 = −24.84 deg
Ans
θ p2 := θ' p
θ p2 = 65.16 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
−3
γ max = 1.082 × 10
ε avg :=
εx + εy
2
2
Ans
−6
ε avg = −100 × 10
Ans
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
θ s = 20.157 deg
( )
θ' s := θ s − 90deg
θ' s = −69.843 deg
Use Eq. 10-6 to determine the sign of γmax.
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = −1.082 × 10
Therefore,
−3
θ s1 := θ s
θ s1 = 20.16 deg
Ans
θ s2 := θ' s
θ s2 = −69.84 deg
Ans
Problem 10-11
The state of strain at the point on the fan blade has components εx = 250(10-6), εy = -450(10-6), γxy
= -825(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and
(b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation
of the element and show how the strains deform the element within the x-y plane.
(
Given: ε x := 250⋅ 10− 6
)
( − 6)
( − 6)
ε y := −450⋅ 10
γ xy := −825⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
εx + εy
−
2
⎛ εx − εy ⎞
⎜
⎝
2
2
⎛ γ xy ⎞
2
−6
ε 1 = 440.98 × 10
−6
+⎜
⎠
Ans
2
ε 2 = −640.98 × 10
⎝ 2 ⎠
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
θ' p := θ p + 90deg
⎠
θ p = −24.843 deg
θ' p = 65.157 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
−6
ε x' = 440.977 × 10
Therefore,
θ p1 := θ p
θ p1 = −24.84 deg
Ans
θ p2 := θ' p
θ p2 = 65.16 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
ε avg :=
εx + εy
2
2
−3
γ max = 1.082 × 10
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
θ s = 20.157 deg
( )
−6
ε avg = −100 × 10
Ans
Ans
θ' s := θ s − 90deg
θ' s = −69.843 deg
Use Eq. 10-6 to determine the sign of γmax.
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = −1.082 × 10
Therefore,
−3
θ s1 := θ s
θ s1 = 20.16 deg
Ans
θ s2 := θ' s
θ s2 = −69.84 deg
Ans
Problem 10-12
A strain gauge is mounted on the 25-mm-diameter A-36 steel shaft in the manner shown. When the
shaft isrotating with an angular velocity of ω = 1760 rev/min, using a slip ring the reading on the strain
gauge is ε = 800(10-6). Determine the power output of the motor. Assume the shaft is only subjected to
a torque.
(2π ) rad
Unit used: rpm := 60sec
(
Given: ε x' := 800⋅ 10− 6
ε y := 0
)
ε x := 0
G := 75GPa
ω := 1760rpm
θ := 60deg
do := 25mm
Solution:
Section Property :
π
4
⋅ do
32
Stress Transformation Equations:
εx + εy εx − εy
γ xy
ε x' =
+
⋅ cos ( 2θ ) +
⋅ sin ( 2θ )
2
2
2
ρ := 0.5do
γ xy :=
J :=
εx + εy εx − εy
⎛
⎞
⎜ ε x' −
−
⋅ cos ( 2θ )
2
2
sin ( 2θ ) ⎝
⎠
2
Shear Stress :
τ=
T :=
T⋅ ρ
J
τ := G⋅ γ xy
G⋅ J⋅ γ xy
ρ
P := T⋅ ω
T = 0.4251 kN⋅ m
P = 78.35 kW
Ans
−3
γ xy = 1.848 × 10
Problem 10-13
The state of strain at the point on the support has components εx = 350(10-6), εy = 400(10-6), γxy
= -675(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and
(b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation
of the element and show how the strains deform the element within the x-y plane.
(
Given: ε x := 350⋅ 10− 6
)
( − 6)
( − 6)
ε y := 400⋅ 10
γ xy := −675⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
εx + εy
−
2
⎛ εx − εy ⎞
⎜
⎝
2
2
⎠
⎛ γ xy ⎞
2
−6
ε 1 = 713.42 × 10
Ans
2
−6
+⎜
ε 2 = 36.58 × 10
⎝ 2 ⎠
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
θ' p := θ p − 90deg
⎠
θ p = 42.882 deg
θ' p = −47.118 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
ε x' = 36.575 × 10
Therefore,
−6
θ p1 := θ' p
θ p1 = −47.12 deg
Ans
θ p2 := θ p
θ p2 = 42.88 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
ε avg :=
εx + εy
2
2
−6
γ max = 676.849 × 10
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
θ s = −2.118 deg
( )
−6
ε avg = 375 × 10
Ans
Ans
θ' s := θ s + 90deg
θ' s = 87.882 deg
Use Eq. 10-6 to determine the sign of γmax.
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = −676.849 × 10
Therefore,
−6
θ s1 := θ s
θ s1 = −2.12 deg
Ans
θ s2 := θ' s
θ s2 = 87.88 deg
Ans
Problem 10-14
Consider the general case of plane strain where εx , εy , γxy and are known. Write a computer program
that can be used to determine the normal and shear strain, εx' and γx'y' , on the plane of an element
oriented θ from the horizontal. Also, compute the principal strains and the element's orientation, and the
maximum in-plane shear strain, the average normal strain, and the element's orientation.
Problem 10-15
Solve Prob. 10-2 using Mohr's circle.
(
Given: ε x := −400⋅ 10− 6
)
( − 6)
γ xy := 375⋅ 10
( − 6)
ε y := 860⋅ 10
θ := 30deg
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 230 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 657.31 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
Angles:
⎛ −0.5γ xy ⎞
φ := atan ⎜
φ = −16.574 deg
α := 2θ − φ
α = 76.574 deg
⎝ εy − εc ⎠
Strain on the inclined Element: (represented by coordinates of points P and Q)
−6
ε x' := ε c − R⋅ cos ( α )
ε x' = 77.38 × 10
ε y' := ε c + R⋅ cos ( α )
ε y' = 382.62 × 10
γ x'y' := 2R⋅ sin ( α )
γ x'y' = 1.279 × 10
−6
−3
Ans
Ans
Ans
Problem 10-16
Solve Prob. 10-4 using Mohr's circle.
( )
−6
ε y := 180⋅ ( 10 )
( − 6)
Given: ε x := 200⋅ 10− 6
γ xy := −300⋅ 10
θ := −30deg
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 190 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 150.33 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
Angles:
⎛ −0.5γ xy ⎞
φ := atan ⎜
φ = −86.186 deg
α := 2θ − φ
α = 26.186 deg
⎝ εy − εc ⎠
Strain on the inclined Element: (represented by coordinates of points P and Q)
−6
ε x' := ε c + R⋅ cos ( α )
ε x' = 324.9 × 10
ε y' := ε c − R⋅ cos ( α )
ε y' = 55.1 × 10
γ x'y' := −2R⋅ sin ( α )
γ x'y' = −132.7 × 10
Ans
−6
Ans
−6
Ans
Problem 10-17
Solve Prob. 10-3 using Mohr's circle.
( )
−6
ε y := 180⋅ ( 10 )
Given: ε x := 200⋅ 10− 6
( − 6)
γ xy := −300⋅ 10
θ := 60deg
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 190 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 150.33 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
Angles:
⎛ −0.5γ xy ⎞
φ := atan ⎜
φ = −86.186 deg
α := 2θ − φ
α = 206.186 deg
⎝ εy − εc ⎠
Strain on the inclined Element: (represented by coordinates of points P and Q)
−6
ε x' := ε c + R⋅ cos ( α )
ε x' = 55.1 × 10
ε y' := ε c − R⋅ cos ( α )
ε y' = 324.9 × 10
γ x'y' := −2R⋅ sin ( α )
γ x'y' = 132.7 × 10
Ans
−6
−6
Ans
Ans
Problem 10-18
Solve Prob. 10-5 using Mohr's circle.
( )
−6
ε y := 350⋅ ( 10 )
( − 6)
Given: ε x := 500⋅ 10− 6
γ xy := −430⋅ 10
θ := −30deg
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 425 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 227.71 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
Angles:
⎛ −0.5γ xy ⎞
φ := atan ⎜
φ = −70.769 deg
α := 2θ − φ
α = 10.769 deg
⎝ εy − εc ⎠
Strain on the inclined Element: (represented by coordinates of points P and Q)
−6
ε x' := ε c + R⋅ cos ( α )
ε x' = 648.7 × 10
ε y' := ε c − R⋅ cos ( α )
ε y' = 201.3 × 10
γ x'y' := −2R⋅ sin ( α )
γ x'y' = −85.1 × 10
−6
−6
Ans
Ans
Ans
Problem 10-19
Solve Prob. 10-6 using Mohr's circle.
(
Given: ε x := 120⋅ 10− 6
)
( − 6)
( − 6)
ε y := −180⋅ 10
γ xy := 150⋅ 10
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = −30 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 167.71 × 10
−6
Coordinates:
(
)
(
A ε x , 0.5 ⋅ γ xy
C εc , 0
)
In-plane Principal Strains: (represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 137.71 × 10
ε 2 := ε c − R
ε 2 = −197.71 × 10
Ans
−6
Ans
Orientation of Principal Strain:
(
)
tan 2θ p1 =
0.5γ xy
εx − εc
⎛ 0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p2 := θ p1 − 90deg
θ p1 :=
θ p1 = 13.28 deg
Ans
θ p2 = −76.72 deg
Ans
Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
−6
γ max = −335.41 × 10
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = −31.717 deg
(Clockwise)
Ans
Problem 10-20
Solve Prob. 10-8 using Mohr's circle.
(
Given: ε x := 520⋅ 10− 6
)
( − 6)
( − 6)
ε y := −760⋅ 10
γ xy := −750⋅ 10
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = −120 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 741.77 × 10
−6
Coordinates:
(
)
(
A ε x , 0.5 ⋅ γ xy
C εc , 0
)
a) In-plane Principal Strains:
(represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 621.77 × 10
ε 2 := ε c − R
ε 2 = −861.77 × 10
Ans
−6
Ans
Orientation of Principal Strain:
(
)
tan 2θ p1 =
0.5γ xy
εx − εc
⎛ 0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p2 := θ p1 − 90deg
θ p1 :=
θ p1 = −15.18 deg
Ans
θ p2 = −105.18 deg
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
−3
γ max = −1.484 × 10
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = 29.816 deg
(Clockwise)
Ans
Problem 10-21
Solve Prob. 10-7 using Mohr's circle.
(
Given: ε x := 850⋅ 10− 6
)
( − 6)
( − 6)
ε y := 480⋅ 10
γ xy := 650⋅ 10
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 665 × 10
2
Radius :
2
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
R = 373.97 × 10
−6
Coordinates:
(
)
(
A ε x , 0.5 ⋅ γ xy
)
C εc , 0
a) In-plane Principal Strains:
(represented by coordinates of points B and D)
−3
ε 1 := ε c + R
ε 1 = 1.039 × 10
ε 2 := ε c − R
ε 2 = 291.03 × 10
−6
Ans
Ans
Orientation of Principal Strain:
(
)
tan 2θ p1 =
0.5γ xy
εx − εc
⎛ 0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p2 := θ p1 − 90deg
θ p1 :=
θ p1 = 30.18 deg
Ans
θ p2 = −59.82 deg
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
γ max = −747.93 × 10
−6
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = −14.825 deg
(Clockwise)
Ans
Problem 10-22
Solve Prob. 10-9 using Mohr's circle.
( − 6) ε y := 320⋅ (10− 6)
−6
γ xy := 180⋅ ( 10 )
ε x := 260⋅ 10
Given:
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 290 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 94.87 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
(represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 384.87 × 10
ε 2 := ε c − R
ε 2 = 195.13 × 10
Ans
−6
Ans
Orientation of Principal Strain:
(
)
tan 2θ p2 =
−0.5 γ xy
εx − εc
⎛ −0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p1 := 90deg − θ p2
θ p2 :=
θ p2 = 35.78 deg
Ans
θ p1 = 54.22 deg
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
−6
γ max = −189.74 × 10
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = 9.217 deg
(Counter-clockwise) Ans
Problem 10-23
The strain at point A on the bracket has components εx = 300(10-6), εy = 550(10-6), γxy = -650(10-6), εz
= 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x-y plane, and (c) the
absolute maximum shear strain.
( − 6)
ε x := 300⋅ 10
Given:
( − 6)
ε y := 550⋅ 10
( − 6)
γ xy := −650⋅ 10
ε z := 0
Solution:
Construction of Mohr's Circle for x-y plane :
Center :
εx + εy
−6
ε c :=
ε c = 425 × 10
2
Radius :
2
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
R = 348.21 × 10
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
−6
ε 1 := ε c + R
ε 1 = 773.21 × 10
ε 2 := ε c − R
ε 2 = 76.79 × 10
Ans
−6
Ans
b) Maximum In-plane Shear Strain:
γ max := 2R
γ max = 696.42 × 10
−6
c) Absolute Maximum Shear Strain:
Construction of Three Circles :
From the results obtained above,
ε max := ε 1
ε int := ε 2
ε min := 0
Absolute Maximum Shear Strain :
From the three Mohr's circles,
γ abs.max := ε max − ε min
γ abs.max = 773.21 × 10
−6
Ans
Ans
−6
Problem 10-24
The strain at a point has components of εx = -480(10-6), εy = 650(10-6), γxy = 780(10-6), εz = 0.
Determine (a) the principal strains, (b) the maximum shear strain in the x-y plane, and (c) the absolute
maximum shear strain.
( − 6) ε y := 650⋅ (10− 6)
ε x := −480⋅ 10
Given:
( − 6)
γ xy := 780⋅ 10
ε z := 0
Solution:
Construction of Mohr's Circle for x-y plane :
Center :
εx + εy
−6
ε c :=
ε c = 85 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 686.53 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains: (represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 771.53 × 10
ε 2 := ε c − R
ε 2 = −601.53 × 10
−6
Ans
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := 2R
−3
γ max = 1.373 × 10
c) Absolute Maximum Shear Strain:
Construction of Three Circles :
From the results obtained above,
ε max := ε 1
ε int := 0
ε min := ε 2
Absolute Maximum Shear Strain :
From the three Mohr's circles,
γ abs.max := ε max − ε min
γ abs.max = 1.373 × 10
−3
Ans
Ans
Problem 10-25
The strain at a point on a pressure-vessel wall has components of εx = 350(10-6), εy = -460(10-6), γxy
= -560(10-6), εz = 0. Determine (a) the principal strains at the point, (b) the maximum shear strain in
the x-y plane, and (c) the absolute maximum shear strain.
( − 6)
ε x := 350⋅ 10
Given:
( − 6)
ε y := −460⋅ 10
( − 6)
γ xy := −560⋅ 10
ε z := 0
Solution:
Construction of Mohr's Circle for x-y plane :
Center :
εx + εy
−6
ε c :=
ε c = −55 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 492.37 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains: (represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 437.37 × 10
ε 2 := ε c − R
ε 2 = −547.37 × 10
Ans
−6
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
−6
γ max = −984.7 × 10
c) Absolute Maximum Shear Strain:
Construction of Three Circles :
From the results obtained above,
ε max := ε 1
ε int := 0
ε min := ε 2
Absolute Maximum Shear Strain :
From the three Mohr's circles,
γ abs.max := ε max − ε min
γ abs.max = 984.7 × 10
−6
Ans
Ans
Problem 10-26
The strain at point A on the leg of the angle has components of εx = -140(10-6), εy = 180(10-6), γxy
= -125(10-6), εz = 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x-y
plane, and (c) the absolute maximum shear strain.
( − 6)
−6
γ xy := −125⋅ ( 10 )
( − 6)
ε x := −140⋅ 10
Given:
ε y := 180⋅ 10
ε z := 0
Solution:
Construction of Mohr's Circle for x-y plane :
Center :
εx + εy
−6
ε c :=
ε c = 20 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 171.77 × 10
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
−6
ε 1 := ε c + R
ε 1 = 191.77 × 10
ε 2 := ε c − R
ε 2 = −151.77 × 10
Ans
−6
Ans
b) Maximum In-plane Shear Strain:
γ max := −2R
−6
γ max = −343.5 × 10
c) Absolute Maximum Shear Strain:
Construction of Three Circles :
From the results obtained above,
ε max := ε 1
ε int := 0
ε min := ε 2
Absolute Maximum Shear Strain :
From the three Mohr's circles,
γ abs.max := ε max − ε min
γ abs.max = 343.5 × 10
−6
Ans
Ans
−6
Problem 10-27
The steel bar is subjected to the tensile load of 2.5 kN. If it is 12 mm thick determine the absolute
maximum shear strain. E = 200 GPa, ν = 0.3.
Given: d := 50mm
t := 12mm
L := 375mm
N := 2.5kN
E := 200GPa
ν := 0.3
Solution:
Stress:
σ x :=
Strain:
ε x :=
N
d⋅ t
σ y := 0
σx
σ z := 0
−5
E
ε x = 2.0833 × 10
ε y := −ν ⋅ ε x
ε y = −6.25 × 10
( )
ε z := −ν ⋅ ( ε x)
−6
−6
ε z = −6.25 × 10
γ xy := 0
Construction of Mohr's Circle in x-y Plane :
εx + εy
−6
Center : ε c :=
ε c = 7.2917 × 10
2
Radius : R :=
(ε x − ε c)2 + (0.5γ xy)2
Coordinates:
A εx , 0
(
)
(
C εc , 0
)
In-plane Principal Strains:
−5
ε 1 := ε c + R
ε 1 = 2.0833 × 10
ε 2 := ε c − R
ε 2 = −6.25 × 10
−6
Similarly, from Mohr's Circle in x-z Plane :
ε 3 := ε c − R
−6
ε 3 = −6.25 × 10
Absolute Maximum In-plane Shear Strain :
γ abs.max := ε 1 − ε 2
γ abs.max = 2.7083 × 10
−5
Ans
R = 1.3542 × 10
−5
Problem 10-28
The 45° strain rosette is mounted on the surface of an aluminum plate. The following readings are
obtained for each gauge: εa = 475(10-6), εb = 250(10-6), and εc = -360(10-6). Determine the in-plane
principal strains.
(
Given: ε a := 475⋅ 10− 6
)
( − 6)
θ a := 0deg
( − 6)
ε b := 250⋅ 10
ε c := −360⋅ 10
θ b := −45deg
θ c := −90deg
Solution:
Strain Rosettes (450): Applying Eq. 10-16,
Given
( )2 + ε y⋅ sin(θa)2 + γ xy⋅ sin(θa)⋅ cos (θa)
2
2
ε b = ε x⋅ cos ( θ b) + ε y⋅ sin ( θ b) + γ xy⋅ sin ( θ b) ⋅ cos ( θ b)
2
2
ε c = ε x⋅ cos ( θ c) + ε y⋅ sin ( θ c) + γ xy⋅ sin ( θ c) ⋅ cos ( θ c)
ε a = ε x⋅ cos θ a
−6
(1)
(2)
(3)
−6
Solving Eqs.(1), (2) and (3),
Guess ε x := 10
ε y := 10
⎛ εx ⎞
⎜
⎜ ε y ⎟ := Find ( ε x , ε y , γ xy)
⎜
⎝ γ xy ⎠
⎛ ε x ⎞ ⎛⎜ 475 × 10− 6 ⎞
⎜
⎜ ε y ⎟ = ⎜ −360 × 10− 6 ⎟
⎟
⎜
⎜
−
6
⎜
γ
⎝ xy ⎠ ⎝ −385 × 10 ⎠
γ xy := 10
−6
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 57.5 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 459.74 × 10
−6
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
In-plane Principal Strains: (represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 517.24 × 10
ε 2 := ε c − R
ε 2 = −402.24 × 10
−6
Ans
Ans
Problem 10-29
The 60° strain rosette is mounted on the surface of the bracket. The following readings are obtained
for each gauge: εa = -780(10-6), εb = 400(10-6), and εc = 500(10-6).Determine (a) the principal strains
and (b) the maximum in-plane shear strain and associated average normal strain. In each case show the
deformed element due to these strains.
(
Given: ε a := −780⋅ 10− 6
)
θ a := 0deg
( − 6)
−6
ε c := 500⋅ ( 10 )
ε b := 400⋅ 10
θ b := 60deg
θ c := 120deg
Solution:
Strain Rosettes (600): Applying Eq. 10-16,
Given
( )2 + ε y⋅ sin(θa)2 + γ xy⋅ sin(θa)⋅ cos (θa) (1)
2
2
ε b = ε x⋅ cos ( θ b) + ε y⋅ sin ( θ b) + γ xy⋅ sin ( θ b) ⋅ cos ( θ b) (2)
2
2
ε c = ε x⋅ cos ( θ c) + ε y⋅ sin ( θ c) + γ xy⋅ sin ( θ c) ⋅ cos ( θ c) (3)
ε a = ε x⋅ cos θ a
Solving Eqs.(1), (2) and (3),
−6
Guess ε x := 10
−6
ε y := 10
γ xy := 10
−6
⎛ ε x ⎞ ⎛⎜ −780 × 10− 6 ⎞
⎜
⎜ ε y ⎟ = ⎜ 860 × 10− 6 ⎟
⎟
⎜
⎜
−
6
⎜
γ
⎝ xy ⎠ ⎝ −115.47 × 10 ⎠
⎛ εx ⎞
⎜
⎜ ε y ⎟ := Find ( ε x , ε y , γ xy)
⎜
⎝ γ xy ⎠
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 40 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 822.03 × 10
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
(represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 862.03 × 10
ε 2 := ε c − R
ε 2 = −782.03 × 10
−6
Ans
Ans
−6
Orientation of Principal Strain:
(
)
tan 2θ p2 =
0.5γ xy
εx − εc
⎛ 0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p1 := 90deg − θ p2
θ p2 :=
θ p2 = 2.01 deg
Ans
θ p1 = 87.99 deg
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
−3
γ max = −1.644 × 10
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s =
θ s := atan ⎜
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = 42.986 deg (Clockwise)
Ans
Problem 10-30
The 45° strain rosette is mounted near the tooth of the wrench. The following readings are obtained for
each gauge: εa = 800(10-6), εb = 520(10-6), and εc = -450(10-6). Determine (a) the in-plane principal
strains and (b) the maximum in-plane shear strain and associated average normal strain. In each case
show the deformed element due to these strains.
( )
−6
ε b := 520⋅ ( 10 )
−6
ε c := −450⋅ ( 10 )
Given: ε a := 800⋅ 10− 6
θ a := −135deg
θ b := −90deg
θ c := −45deg
Solution:
Strain Rosettes (450): Applying Eq. 10-16,
Given
( )2 + ε y⋅ sin(θa)2 + γ xy⋅ sin(θa)⋅ cos (θa)
2
2
ε b = ε x⋅ cos ( θ b) + ε y⋅ sin ( θ b) + γ xy⋅ sin ( θ b) ⋅ cos ( θ b)
2
2
ε c = ε x⋅ cos ( θ c) + ε y⋅ sin ( θ c) + γ xy⋅ sin ( θ c) ⋅ cos ( θ c)
ε a = ε x⋅ cos θ a
(1)
(2)
(3)
Solving Eqs.(1), (2) and (3),
−6
Guess ε x := 10
−6
ε y := 10
γ xy := 10
−6
⎛ ε x ⎞ ⎛⎜ −170 × 10− 6 ⎞
⎜
⎜ ε y ⎟ = ⎜ 520 × 10− 6 ⎟
⎟
⎜
⎜
−
3
⎜
γ
⎝ xy ⎠ ⎝ 1.25 × 10 ⎠
⎛ εx ⎞
⎜
⎜ ε y ⎟ := Find ( ε x , ε y , γ xy)
⎜
⎝ γ xy ⎠
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 175 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 713.9 × 10
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
(represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 888.90 × 10
ε 2 := ε c − R
ε 2 = −538.90 × 10
−6
Ans
Ans
−6
Orientation of Principal Strain:
(
)
tan 2θ p2 =
−0.5 γ xy
εx − εc
⎛ −0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p1 := 90deg − θ p2
θ p2 :=
θ p2 = 30.55 deg
Ans
θ p1 = 59.45 deg
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := 2R
−3
γ max = 1.428 × 10
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = 14.449 deg
(Counter-clockwise) Ans
Problem 10-31
The 60° strain rosette is mounted on a beam. The following readings are obtained from each gauge: εa
= 150(10-6), εb = -330(10-6), and εc = 400(10-6). Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain. In each case show the deformed element due
to these strains.
(
Given: ε a := 150⋅ 10− 6
)
θ a := −30deg
( − 6)
−6
ε c := 400⋅ ( 10 )
ε b := −330⋅ 10
θ b := 30deg
θ c := 90deg
Solution:
Strain Rosettes (600): Applying Eq. 10-16,
Given
( )2 + ε y⋅ sin(θa)2 + γ xy⋅ sin(θa)⋅ cos (θa)
2
2
ε b = ε x⋅ cos ( θ b) + ε y⋅ sin ( θ b) + γ xy⋅ sin ( θ b) ⋅ cos ( θ b)
2
2
ε c = ε x⋅ cos ( θ c) + ε y⋅ sin ( θ c) + γ xy⋅ sin ( θ c) ⋅ cos ( θ c)
ε a = ε x⋅ cos θ a
(1)
(2)
(3)
Solving Eqs.(1), (2) and (3),
−6
Guess ε x := 10
−6
ε y := 10
γ xy := 10
−6
⎛ ε x ⎞ ⎛⎜ −253.33 × 10− 6 ⎞
⎜
⎜ ε y ⎟ = ⎜ 400 × 10− 6 ⎟
⎟
⎜
⎜
−
6
⎜
γ
⎝ xy ⎠ ⎝ −554.26 × 10 ⎠
⎛ εx ⎞
⎜
⎜ ε y ⎟ := Find ( ε x , ε y , γ xy)
⎜
⎝ γ xy ⎠
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 73.33 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 428.38 × 10
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
(represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 501.72 × 10
ε 2 := ε c − R
ε 2 = −355.05 × 10
−6
Ans
Ans
−6
Orientation of Principal Strain:
(
)
tan 2θ p2 =
0.5γ xy
εx − εc
⎛ 0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p1 := 90deg − θ p2
θ p2 :=
θ p2 = 20.15 deg
Ans
θ p1 = 69.85 deg
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
−6
γ max = −856.764 × 10
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s =
θ s := atan ⎜
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = 24.845 deg (Clockwise)
Ans
Problem 10-32
The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from each
gauge: εa = 800(10-6), εb = 520(10-6), and εc = -450(10-6). Determine the in-plane principal strains and
their orientation.
(
Given: ε a := 800⋅ 10− 6
)
( − 6)
θ a := −45deg
( − 6)
ε b := 520⋅ 10
ε c := −450⋅ 10
θ b := 0deg
θ c := 45deg
Solution:
Strain Rosettes (450): Applying Eq. 10-16,
Given
( )2 + ε y⋅ sin(θa)2 + γ xy⋅ sin(θa)⋅ cos (θa)
2
2
ε b = ε x⋅ cos ( θ b) + ε y⋅ sin ( θ b) + γ xy⋅ sin ( θ b) ⋅ cos ( θ b)
2
2
ε c = ε x⋅ cos ( θ c) + ε y⋅ sin ( θ c) + γ xy⋅ sin ( θ c) ⋅ cos ( θ c)
ε a = ε x⋅ cos θ a
(1)
(2)
(3)
Solving Eqs.(1), (2) and (3),
−6
Guess ε x := 10
−6
ε y := 10
γ xy := 10
−6
⎛ ε x ⎞ ⎛⎜ 520 × 10− 6 ⎞
⎜
⎜ ε y ⎟ = ⎜ −170 × 10− 6 ⎟
⎟
⎜
⎜
−
3
⎜
γ
⎝ xy ⎠ ⎝ −1.25 × 10 ⎠
⎛ εx ⎞
⎜
⎜ ε y ⎟ := Find ( ε x , ε y , γ xy)
⎜
⎝ γ xy ⎠
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 175 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 713.9 × 10
−6
Coordinates:
(
)
(
A ε x , 0.5 ⋅ γ xy
C εc , 0
)
a) In-plane Principal Strains:
−6
ε 1 := ε c + R
ε 1 = 888.90 × 10
ε 2 := ε c − R
ε 2 = −538.90 × 10
−6
Ans
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
0.5γ xy
εx − εc
θ p :=
⎛ 0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p = −30.55 deg
Ans
Problem 10-33
Consider the general orientation of three strain gauges at a point as shown. Write a computer program
that can be used to determine the principal in-plane strains and the maximum in-plane shear strain at the
point. Show an application of the program using the values θa = 40°, εa = 160(10-6), θb = 125°, εb =
100(10-6), θc = 220°, εc = 80(10-6).
Problem 10-34
For the case of plane stress, show that Hooke's law can be written as
σx =
E
(ε x + νε y )
(1 − ν 2 )
σy =
E
(ε y + νε x )
(1 − ν 2 )
Problem 10-35
Use Hooke's law, Eq. 10-18, to develop the strain-transformation equations, Eqs. 105 and 106, from
the stress-transformation equations, Eqs. 9-1 and 9-2.
Problem 10-36
Abar of copper alloy is loaded in a tension machine and it is determined that εx = 940(10-6) and σx =
100 MPa, σy = 0, σz = 0. Determine the modulus of elasticity, Ecu , and the dilatation, ecu , of the
copper. νcu = 0.35.
Given: σ x := 100MPa
σ y := 0
( − 6)
ε x := 940⋅ 10
σ z := 0
ν cu := 0.35
Solution:
εx =
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
(
E cu :=
ε cu :=
)
(
)
σ x − ν cu⋅ σ y + σ z
E cu = 106.38 GPa
εx
1 − 2ν cu
Ecu
(
)
⋅ σx + σy + σz
ε cu = 2.820 × 10
Ans
−4
Ans
Problem 10-37
The principal plane stresses and associated strains in a plane at a point are σ1 = 250 MPa, σ2 = 112
MPa, ε1 = 1.02(10-3), ε2 = 0.180(10-3). Determine the modulus of elasticity and Poisson's ratio.
Given: σ 1 := 250MPa
σ 2 := 112MPa
−3
ε 1 := 1.02 ⋅ 10
Solution:
−3
ε 2 := 0.180 ⋅ 10
σ 3 := 0MPa
Given
1
⎡σ − ν ⋅ σ 2 + σ 3 ⎤⎦
E⎣ 1
1
ε 2 = ⎡⎣σ 2 − ν ⋅ σ 1 + σ 3 ⎤⎦
E
ε1 =
(
)
(1)
(
)
(2)
Solving (1) and (2):
Guess E := 1GPa
⎛E⎞
⎜ := Find ( E , ν )
⎝ν ⎠
E = 212.77 GPa
Ans
ν = 0.295
Ans
ν := 10
−6
Problem 10-38
Determine the bulk modulus for hard rubber if E r = 5 GPa, νr= 0.43.
Given: E r := 5GPa
ν r := 0.43
Solution:
Applying Eq. 10-25
Bulk Modulus:
κ :=
(
Er
3⋅ 1 − 2ν r
κ = 11.90 GPa
)
Ans
Problem 10-39
The principal strains at a point on the aluminum fuselage of a jet aircraft are ε1 = 780(10-6) and
ε2 = 400(10-6). Determine the associated principal stresses at the point in the same plane. E al = 70
GPa, νal= 0.33. Hint: See Prob. 10-34.
( )
−6
ε 2 := 400⋅ ( 10 )
Given: ε 1 := 780⋅ 10− 6
E := 70GPa
ν := 0.33
Solution:
Plane stress: σ 3 := 0MPa
Use the formula developed in Prob. 10-34,
σ 1 :=
σ 2 :=
E
1−ν
2
( ε 1 + ν ⋅ ε 2)
σ 1 = 71.64 MPa
Ans
2
( ε 2 + ν ⋅ ε 1)
σ 2 = 51.64 MPa
Ans
E
1−ν
Problem 10-40
The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter
of 20 mm, determine the absolute maximum shear strain in the rod at a point on its surface.
Given:
Px := 700N
do := 20mm
σ y := 0
σ z := 0
E := 73.1GPa
ν := 0.35
Solution:
σ x :=
4Px
2
σ x = 2.228 MPa
π ⋅ do
Normal Strains: Apply the general Hooke's Law,
ε x :=
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
)
ε x = 30.481 × 10
ε y :=
1
⎡σ − ν ⋅ σ x + σ z ⎤⎦
E⎣ y
)
ε y = −10.668 × 10
ε z :=
1
⎡σ − ν ⋅ σ x + σ y ⎤⎦
E⎣ z
ε z = −10.668 × 10
(
(
Principal Strains:
(
−6
−6
)
−6
From the results obtained above,
−6
ε max := ε x
ε max = 30.481 × 10
ε min := ε z
ε min = −10.668 × 10
−6
Absolute Maximum Shear Strain :
From the three Mohr's circles,
γ abs.max := ε max − ε min
γ abs.max = 41.149 × 10
−6
Ans
Problem 10-41
The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameter
of 20 mm, determine the principal strains at a point on the surface of the rod.
Given:
Px := 700N
do := 20mm
σ y := 0
σ z := 0
E := 73.1GPa
ν := 0.35
Solution:
σ x :=
4Px
2
σ x = 2.228 MPa
π ⋅ do
Normal Strains: Apply the general Hooke's Law,
ε x :=
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
)
ε x = 30.481 × 10
ε y :=
1
⎡σ − ν ⋅ σ x + σ z ⎤⎦
E⎣ y
)
ε y = −10.668 × 10
ε z :=
1
⎡σ − ν ⋅ σ x + σ y ⎤⎦
E⎣ z
ε z = −10.668 × 10
(
(
Principal Strains:
(
)
−6
−6
−6
From the results obtained above,
−6
ε max := ε x
ε max = 30.481 × 10
ε int := ε y
ε int = −10.668 × 10
ε min := ε z
ε min = −10.668 × 10
−6
−6
Ans
Ans
Ans
Problem 10-42
A rod has a radius of 10 mm. If it is subjected to an axial load of 15 N such that the axial strain in the
rod is εx = 2.75(10-6), determine the modulus of elasticity E and the change in its diameter. ν = 0.23.
Given: Px := 15N
σ y := 0
ro := 10mm
σ z := 0
ν := 0.23
ε x := 2.75 ⋅ 10
( − 6)
Solution:
Normal Stresses :
Px
σ x :=
2
π ⋅ ro
σ x = 0.0477 MPa
Appling the generalized Hooke's Law,
Normal Strains:
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
1
E :=
⎡σ x − ν ⋅ σ y + σ z ⎤⎦
εx ⎣
εx =
(
)
(
)
E = 17.36 GPa
−9
ε y := −ν ⋅ ε x
ε y = −632.50 × 10
ε z := −ν ⋅ ε x
ε z = −632.50 × 10
−9
Thus,
(
)
∆ d := ε y⋅ 2⋅ ro
−6
∆ d = −12.65 × 10
mm
Ans
Ans
Problem 10-43
The principal strains at a point on the aluminum surface of a tank are ε1 = 630(10-6) and ε2 = 350(10-6).
If this is a case of plane stress, determine the associated principal stresses at the point in the same
plane. Eal = 70 GPa, νal= 0.33. Hint: See Prob. 10-34.
( )
−6
ε 2 := 350⋅ ( 10 )
Given: ε 1 := 630⋅ 10− 6
E al := 70GPa
ν al := 0.33
Solution:
Plane stress: σ 3 := 0MPa
Use the formula developed in Prob. 10-34,
σ 1 :=
σ 2 :=
Eal
2
(ε 1 + νal⋅ ε 2)
σ 1 = 58.56 MPa
Ans
2
(ε 2 + νal⋅ ε 1)
σ 2 = 43.83 MPa
Ans
1 − ν al
Eal
1 − ν al
Problem 10-44
A uniform edge load of 100 kN/m and 70 kN/m is applied to the polystyrene specimen. If the specimen
is originally square and has dimensions of a = 50 mm, b = 50 mm, and a thickness of t = 6 mm,
determine its new dimensions a', b', and t' after the load is applied. Ep = 4 GPa, νp= 0.25.
Given: a := 50mm
qa := 100
b := 50mm
kN
m
qb := 70
E := 4GPa
t := 6mm
kN
m
ν := 0.25
Solution:
Plane stress: σ z := 0MPa
qa
σ x :=
σ x = 16.67 MPa
t
σ y :=
qb
t
σ y = 11.67 MPa
Normal Strains: Apply the general Hooke's Law,
ε x :=
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
)
ε x = 3.438 × 10
ε y :=
1
⎡σ − ν ⋅ σ x + σ z ⎤⎦
E⎣ y
)
ε y = 1.875 × 10
ε z :=
1
⎡σ − ν ⋅ σ x + σ y ⎤⎦
E⎣ z
ε z = −1.771 × 10
(
(
(
)
−3
−3
−3
The new dimensions for the new specimen are,
(
)
b' := b⋅ ( 1 + ε x)
t' := t⋅ ( 1 + ε z)
a' := a⋅ 1 + ε y
a' = 50.09 mm
Ans
b' = 50.17 mm
Ans
t' = 5.99 mm
Ans
Problem 10-45
The principal stresses at a point are shown. If the material is graphite for which E g = 5.6 GPa, νg=
0.23, determine the principal strains.
Given:
σ x := 70MPa
σ y := −105MPa
σ z := −182MPa
E := 5.6GPa
ν := 0.23
Solution:
Normal Strains: Apply the general Hooke's Law,
ε x :=
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
)
ε x = 24.287 × 10
ε y :=
1
⎡σ − ν ⋅ σ x + σ z ⎤⎦
E⎣ y
)
ε y = −14.150 × 10
ε z :=
1
⎡σ − ν ⋅ σ x + σ y ⎤⎦
E⎣ z
ε z = −31.063 × 10
(
(
Principal Strains:
(
)
−3
−3
−3
From the results obtained above,
ε max := ε x
ε max = 0.0243
Ans
ε int := ε y
ε int = −0.0142
Ans
ε min := ε z
ε min = −0.0311
Ans
Problem 10-46
The shaft has a radius of 15 mm and is made of L2 tool steel. Determine the strains in the x' and y'
directions if a torque T = 2 kN·m is applied to the shaft.
Given: ro := 15mm
T := 2kN⋅ m
G := 75GPa
θ := 45deg
Solution:
4
Section Property :
Shear Stress:
J :=
τ :=
π ⋅ ro
2
T⋅ ro
J
Shear Stress-strain Relationship:
τ
Applying Hooke's Law, γ xy :=
G
γ xy = 5.03 × 10
Strain Rosettes: For pure shear,
Applying Eq. 10-15,
θ x' := θ
ε x := 0
−3
ε y := 0
θ y' := θ + 90deg
( )2 + ε y⋅ sin(θx')2 + γ xy⋅ sin(θx')⋅ cos (θx')
ε x' = 2.52 × 10
( )2 + ε y⋅ sin(θy')2 + γ xy⋅ sin(θy')⋅ cos (θy')
ε y' = −2.52 × 10
ε x' := ε x⋅ cos θ x'
ε y' := ε x⋅ cos θ y'
−3
−3
Ans
Ans
Problem 10-47
The cross section of the rectangular beam is subjected to the bending moment M. Determine an
expression for the increase in length of lines AB and CD. The material has a modulus of elasticity E
and Poisson's ratio is ν.
Problem 10-48
The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge
having a length of 20mm is attached to it, and it is observed to increase in length by 0.012 mm when
the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum
in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the
vessel. The material is steel, for which E st = 200 GPa and νst = 0.3.
Given:
t := 10mm
di := 2m
E := 200GPa ν := 0.3
L := 20mm
∆L := 0.012mm
Solution:
Appling the generalized Hooke's Law with
Normal Strains:
∆L
L
ε max = 600 × 10
ε int := ε max
ε int = 600 × 10
ε max :=
−6
−6
1
− ν ⋅ σ int + σ min ⎤⎦
⎡σ
E ⎣ max
(
)
σ max = E⋅ ε max + ν ⋅ ( σ int + σ min)
ε max =
50 p = E ⋅ ε max + ν ⋅ ( 50 p + 0)
p :=
E⋅ ε max
p = 3.43 MPa
50( 1 − ν )
Normal Stresses :
ri := 0.5di
α :=
Ans
ri
α = 100
t
Since α > 10. then thin-wall analysis can be used.
σ min := 0
This is a plane stress problem where
since there is no load acting on the outer surface of the wall.
p⋅ r i
σ max :=
σ max = 171.43 MPa
2⋅ t
σ int := σ max
σ int = 171.43 MPa
Maximum In-plane Shear Stress (Sperical Surface):
Mohr's circle is simply a dot. As the result, the state of stress is the same consisting of two
normal stresses with zero shear stress regardless of the orientation of the element.
τ max := 0
Ans
Absolute Maximum Shear Stress :
τ abs.max :=
σ max − σ min
2
τ abs.max = 85.71 MPa
Ans
Problem 10-49
A rod has a radius of 10 mm. If it is subjected to an axial load of 15 N such that the axial strain in the
rod is εx = 2.75(10-6), determine the modulus of elasticity E and the change in its diameter. ν = 0.23.
Given: Px := 15N
σ y := 0
ro := 10mm
σ z := 0
ν := 0.23
ε x := 2.75 ⋅ 10
( − 6)
Solution:
Normal Stresses :
Px
σ x :=
2
π ⋅ ro
σ x = 0.0477 MPa
Appling the generalized Hooke's Law,
Normal Strains:
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
1
E :=
⎡σ x − ν ⋅ σ y + σ z ⎤⎦
εx ⎣
εx =
(
)
(
)
E = 17.36 GPa
−9
ε y := −ν ⋅ ε x
ε y = −632.50 × 10
ε z := −ν ⋅ ε x
ε z = −632.50 × 10
−9
Thus,
(
)
∆ d := ε y⋅ 2⋅ ro
−6
∆ d = −12.65 × 10
mm
Ans
Ans
Problem 10-50
A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 60° to the axis
of the pipe, gives a reading at point A of εA = -250(10-6). Determine the vertical force P if the pipe has
an outer diameter of 25 mm and an inner diameter of 15 mm.The pipe is made of C86100 bronze.
Given: do := 25mm
a := 200mm
di := 15mm
L := 150mm
θ := 60deg
ε A := −250⋅ 10
( − 6)
ρ o := 0.5do
Solution:
G := 38GPa
ρ i := 0.5di
ε x := 0
Strain Rosettes: For pure shear,
ε y := 0
Applying Eq. 10-15,
ε A = ε x + ε y + γ xy⋅ sin ( θ ) ⋅ cos ( θ )
γ xy :=
εA − εx − εy
sin ( θ ) ⋅ cos ( θ )
Shear Stress-strain Relationship:
τ A := G⋅ γ xy
Applying Hooke's Law,
Internal Force and Moment : At Section A:
Mz = −P⋅ a
Vy = P
Iz :=
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
π ⎛ 4
4
J :=
⋅ ⎝ do − di ⎞⎠
32
A :=
Section Property :
T x = P⋅ L
π ⎛ 4
4
⋅ ⎝ do − di ⎞⎠
64
4ρ o ⎛⎜ π ⋅ ρ o ⎞ 4ρ i ⎛⎜ π ⋅ ρ i ⎞
QA :=
⋅⎜
−
⋅⎜
3π ⎝ 2 ⎠
3π ⎝ 2 ⎠
2
2
cA := 0
Normal Stress:
Mz⋅ cA
σA = −
Iz
σ A := 0MPa
Shear Stress in x-y plane :
τA =
Vy⋅ QA T x⋅ ρ o
−
Iz⋅ bA
J
G⋅ γ xy =
P :=
bA := do − di
P⋅ QA
Iz⋅ bA
−
( P⋅ L ) ⋅ ρ o
J
G⋅ γ xy
QA
Iz⋅ bA
−
L⋅ ρ o
J
P = 0.438 kN
Ans
Problem 10-51
A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 60° to the axis
of the pipe, gives a reading at point A of εA = -250(10-6). Determine the principal strains in the pipe at
point A. The pipe has an outer diameter of 25 mm and an inner diameter of 15 mm and is made of
C86100 bronze.
Given: do := 25mm
di := 15mm
a := 200mm
L := 150mm
θ := 60deg
ε A := −250⋅ 10
Solution:
G := 38GPa
( − 6)
ρ o := 0.5do
ρ i := 0.5di
ε x := 0
Strain Rosettes: For pure shear,
ε y := 0
Applying Eq. 10-15,
ε A = ε x + ε y + γ xy⋅ sin ( θ ) ⋅ cos ( θ )
γ xy :=
εA − εx − εy
−6
γ xy = −577.35 × 10
sin ( θ ) ⋅ cos ( θ )
Construction of Mohr's Circle in x-y Plane :
Center : ε c :=
εx + εy
εc = 0
2
Radius : R :=
(ε x − ε c)2 + (0.5γ xy)2
Coordinates:
A ε x , 0.5γ xy
(
)
(
C εc , 0
R = 288.675 × 10
)
In-plane Principal Strains:
The coordinates of points B and D represent ε1
and ε2, respectively.
−6
ε 1 := ε c + R
ε 1 = 288.675 × 10
ε 2 := ε c − R
ε 2 = −288.675 × 10
−6
Principal Stresses:
Since σx=σy=σz=0, then from the generalized Hooke's Law,
ε z := 0
From the results obtained above, we have
−6
ε max := ε 1
ε max = 288.7 × 10
ε int := ε z
ε int = 0
ε min := ε 2
ε min = −288.7 × 10
Ans
Ans
−6
Ans
−6
Problem 10-52
A material is subjected to principal stresses σx and σy . Determine the orientation θ of a strain gauge
placed at the point so that its reading of normal strain responds only to σy and not σx .The material
constants are E and ν.
Problem 10-53
The principal stresses at a point are shown in the figure. If the material is aluminum for which E al =
70 GPa and νal = 0.33, determine the principal strains.
Given:
σ x := 70MPa
σ y := −105MPa
σ z := −182MPa
E := 70GPa
ν := 0.33
Solution:
Apply the general Hooke's Law,
ε x :=
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
)
ε x = 2.353 × 10
ε y :=
1
⎡σ − ν ⋅ σ x + σ z ⎤⎦
E⎣ y
)
ε y = −9.720 × 10
ε z :=
1
⎡σ − ν ⋅ σ x + σ y ⎤⎦
E⎣ z
ε z = −2.435 × 10
(
(
(
)
−3
−4
−3
Ans
Ans
Ans
Ans
Problem 10-54
A thin-walled cylindrical pressure vessel has an inner radius r, thickness t, and length L. If it is
subjected to an internal pressure p, show that the increase in its inner radius is dr = rε1
= pr2(1 - ½ ν) /Et and the increase in its length is ∆L = pLr ( ½ - ν) /Et. Using these results, show that
the change in internal volume becomes dV = π r2(1 + ε1)2 (1 + ε2)L - π r2L. Since ε1and ε2 are small
quantities, show further that the change in volume per unit volume, called volumetric strain, can be
written as dV /V = pr ( 2.5 - 2ν) /Et.
Problem 10-55
The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce the
bending stress that would occur if flat ends were used. The bending stresses at the seam where the
caps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder,
respectively. This requires the radial expansion to be the same for both the hemispheres and cylinder.
Show that this ratio is tc / th = (2 - ν) / (1 - ν). Assume that the vessel is made of the same material and
both the cylinder and hemispheres have the same inner radius. If the cylinder is to have a thickness of
12 mm, what is the required thickness of the hemispheres? Take ν = 0.3.
tc := 12mm
Given:
ν := 0.3
Solution:
For cylindrical vessel:
ο1 =
p⋅ r
tc
ε1 =
1
⎡σ − ν ⋅ σ 2 + σ 3 ⎤⎦
E⎣ 1
ε1 =
1 ⎡ p⋅ r
⎛ p⋅ r ⎞⎤
− ν⋅ ⎜
⎢
E tc
2tc ⎥
ο2 =
p⋅ r
2tc
(
⎣
ο3 = 0
)
⎝
ε1 =
⎠⎦
p⋅ r ⎛
ν⎞
⎜1 −
E ⋅ tc ⎝
2⎠
2
p⋅ r ⎛
ν⎞
dr =
⎜1 −
2⎠
E ⋅ tc ⎝
dr = ε 1⋅ r
(1)
For hemispherical end cap:
ο1 =
p⋅ r
2th
ε1 =
1
⎡σ − ν ⋅ σ 2 + σ 3 ⎤⎦
E⎣ 1
ε1 =
1 ⎡ p⋅ r
⎛ p⋅ r ⎞⎤
− ν⋅ ⎜
⎢
E 2th
2th ⎥
ο2 =
p⋅ r
2th
(
⎣
ο3 = 0
)
⎝
⎠⎦
ε1 =
p⋅ r
(1 − ν)
2E⋅ th
2
p⋅ r
(1 − ν)
dr =
2E⋅ th
dr = ε 1⋅ r
(2)
Equate Eqs. (1) and (2):
2
2
ν⎞
p⋅ r ⎛
p⋅ r
(1 − ν )
=
⎜1 −
2⎠
E ⋅ tc ⎝
2E ⋅ th
ν⎞
1⎛
1
(1 − ν )
=
⎜1 −
2⎠
tc ⎝
2th
Hence,
⎛ 1 − ν ⎞⋅ t
c
⎝2 − ν⎠
th := ⎜
tc
th
=
2−ν
1−ν
th = 4.94 mm
QED
Ans
Problem 10-56
TheA-36 steel pipe is subjected to the axial loading of 60 kN. Determine the change in volume of the
material after the load is applied.
Given:
do := 40mm
di := 30mm
L := 0.5m
P := 60kN
E := 200GPa
ν := 0.32
Solution:
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
The pipe is subjected to uniaxial load. Therefore,
Section Property :
Normal Stress:
σ x :=
P
A
A :=
σ x = 109.13 MPa
σ y := 0
σ z := 0
Dilation:
Apply Eq. 10-23,
δV
1 − 2ν
=
σx + σy + σz
V
E
(
δV :=
)
1 − 2ν
σ x + σ y + σ z ⋅ ( A⋅ L)
E
(
)
3
δV = 54.00 mm
Ans
Problem 10-57
The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.3 mm from
the top of the cavity. If the top of the cavity is covered and the temperature is increased by 110°C,
determine the stress components σx , σy , and σz in the aluminum. Hint: Use Eqs. 10-18 with an
additional strain term of α∆T (Eq. 4-4).
°C := deg
Unit used:
Given: H := 150mm
∆ H := 0.3mm
E := 68.9GPa
ν := 0.35
∆ T := 110°C
α := 24 × 10
−6 1
°C
Solution:
Normal Strains: Since the aluminum is confined at the
sides by a rigid container and allowed to expand in the
z-direction,
∆H
ε x := 0
ε y := 0
ε z :=
H
Applying the generalized Hooke's Law with the additional thernal strain, ε T := α ⋅ ∆ T
( − 3) (Scale down to avoid floating-point error during calculation)
1
ε x = ⎡⎣σ x − ν ⋅ ( σ y + σ z)⎤⎦ + ε T
(1)
E'
E' := E ⋅ 10
Given
εy =
1
⎡σ − ν ⋅ σ x + σ z ⎤⎦ + ε T
E' ⎣ y
εz =
1
⎡σ − ν ⋅ σ y + σ x ⎤⎦ + ε T
E' ⎣ z
(
(
)
)
(2)
(3)
Solving (1), (2) and (3):
Guess
σ x := 1MPa
σ y := 2MPa
σ z := 3MPa
⎛ σx ⎞
⎜
⎜ σ y ⎟ := Find ( σ x , σ y , σ z)
⎜
⎝ σz ⎠
⎛ σx ⎞
⎛ σx ⎞
⎜
⎜
3
⎜ σ y ⎟ := 10 ⋅ ⎜ σ y ⎟
⎜
⎜
⎝ σz ⎠
⎝ σz ⎠
⎛ σx ⎞
⎛ −487.2 ⎞
⎜
⎜
⎜ σ y ⎟ = ⎜ −487.2 MPa
⎜
⎝ −385.2 ⎠
⎝ σz ⎠
Ans
(Scale back up)
Problem 10-58
The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.3 mm from
the top of the cavity. If the top of the cavity is not covered and the temperature is increased by 110°C,
determine the strain components εx , εy , and εz in the aluminum. Hint: Use Eqs. 10-18 with an
additional strain term of α∆T (Eq. 4-4).
°C := deg
Unit used:
Given: H := 150mm
∆ H := 0.3mm
E := 68.9GPa
ν := 0.35
∆ T := 110°C
α := 24 × 10
−6 1
°C
Solution:
Normal Strains: Since the aluminum is confined at the
sides by a rigid container, then
ε x := 0
ε y := 0
Ans
and since it is not restrained in z-direction,
σ z := 0
Applying the generalized Hooke's Law with the additional thernal strain, ε T := α ⋅ ∆ T
Given
εx =
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦ + ε T
E⎣ x
(1)
εy =
1
⎡σ − ν ⋅ σ x + σ z ⎤⎦ + ε T
E⎣ y
(2)
(
(
)
)
Solving (1) and (2):
Guess
σ x := 1MPa
σ y := 2MPa
⎛⎜ σ x ⎞
:= Find ( σ x , σ y)
⎜ σy
⎝ ⎠
⎛⎜ σ x ⎞ ⎛ −279.8 ⎞
=⎜
MPa
⎜ σy
−279.8 ⎠
⎝
⎝ ⎠
⎡ 1 σ − ν⋅ σ + σ + ε ⎤
⎡
( y x)⎤⎦ T⎥⎦
⎣E ⎣ z
ε z := ⎢
ε z = 5.48 × 10
−3
Ans
Problem 10-59
The thin-walled cylindrical pressure vessel of inner radius r and thickness t is subjected to an internal
pressure p. If the material constants are E and ν, determine the strains in the circumferential and
longitudinal directions. Using these results, compute the increase in both the diameter and the length of
a steel pressure vessel filled with air and having an internal gauge pressure of 15 MPa. The vessel is
3 m long, and has an inner radius of 0.5 m and a wall thick of 10 mm. Est = 200 GPa, νst = 0.3.
Given:
t := 10mm
r := 0.5m
E := 200GPa ν := 0.3
L := 3m
p := 15MPa
Solution:
r
α = 50
t
Since α > 10. then thin-wall analysis can be used.
α :=
Normal Stresses :
p⋅ r
t
σ 1 :=
Normal Strains:
σ 2 :=
p⋅ r
2⋅ t
σ 3 := 0
Appling the generalized Hooke's Law,
ε cir =
1
⎡σ − ν ⋅ σ 2 + σ 3 ⎤⎦
E⎣ 1
ε cir =
1 ⎡ p⋅ r
⎛ p⋅ r ⎞⎤
⎢ − ν⋅ ⎜ + 0 ⎥
E⎣ t
⎝ 2⋅ t
⎠⎦
ε cir :=
p⋅ r
(2 − ν)
2E ⋅ t
ε long =
1
⎡σ − ν ⋅ σ 1 + σ 3 ⎤⎦
E⎣ 2
ε long =
1 ⎡ p⋅ r
⎛ p⋅ r ⎞⎤
⎢ − ν⋅ ⎜ + 0 ⎥
E ⎣ 2t
⎝ t
⎠⎦
ε long :=
p⋅ r
( 1 − 2ν )
2E ⋅ t
(
)
ε cir = 3.1875 × 10
(
−3
Ans
)
−6
ε long = 750 × 10
Deformations :
∆d := ε cir⋅ ( 2r)
∆d = 3.19 mm
Ans
∆L := ε long⋅ L
∆L = 2.25 mm
Ans
Ans
Problem 10-60
Estimate the increase in volume of the tank in Prob. 10-59. Suggestion: Use the results of Prob. 10-54
as a check.
Given: t := 10mm
r := 0.5m
L := 3m
E := 200GPa ν := 0.3
p := 15MPa
Solution:
2
V := π ⋅ r ⋅ L
Section Property :
r
α = 50
t
Since α > 10. then thin-wall analysis can be used.
α :=
Normal Stresses :
σ 1 :=
p⋅ r
t
σ 2 :=
p⋅ r
2⋅ t
σ 3 := 0
Appling the generalized Hooke's Law,
Normal Strains:
1
⎡σ − ν ⋅ σ 2 + σ 3 ⎤⎦
E⎣ 1
1
ε long := ⎡⎣σ 2 − ν ⋅ σ 1 + σ 3 ⎤⎦
E
(
ε cir :=
)
(
ε cir = 3.1875 × 10
)
−6
ε long = 750 × 10
Deformations :
∆r := ε cir⋅ ( r)
∆r = 1.59 mm
∆L := ε long⋅ L
∆L = 2.25 mm
2
∆V := π ( r + ∆r) ⋅ ( L + ∆L) − πr ⋅ L
∆V = 0.0168 m
3
2
Ans
Or, appling the result of Prob. 10-54,
∆V
p⋅ r
(2.5 − 2ν )
=
V
E⋅ t
∆V :=
(
p⋅ r
( 2.5 − 2ν ) ⋅ π ⋅ r2⋅ L
E⋅ t
∆V = 0.0168 m
3
Ans
)
−3
Problem 10-61
A soft material is placed within the confines of a rigid cylinder which rests on a rigid support.
Assuming that εx = 0 and εy = 0, determine the factor by which the modulus of elasticity will be
increased when a load is applied if ν = 0.3 for the material.
Given:
ν := 0.3
Solution:
Normal Strains: Since the material is confined in a rigid cylinder,
ε x := 0
ε y := 0
Appling the generalized Hooke's Law,
1
⎡σ − ν ⋅ σ y + σ z ⎤⎦
E⎣ x
1
ε y = ⎡⎣σ y − ν ⋅ σ x + σ z ⎤⎦
E
εx =
(
)
σx = ν⋅ σy + σz
(
)
(1)
(
)
σy = ν⋅ σx + σz
(
)
(2)
⎛ ν ⎞⋅ σ
z
⎝1 − ν⎠
σx = ⎜
Solving Eqs. (1) and (2):
⎛ ν ⎞⋅ σ
z
⎝1 − ν⎠
σy = ⎜
Thus,
εz =
1
⎡σ − ν ⋅ σ x + σ y ⎤⎦
E⎣ z
εz =
1⎡
⎛ ν ⎞⋅ σ + σ ⎤
⎢σ z − ν ⋅ ⎜
z
z⎥
E⎣
⎝1 − ν⎠
⎦
εz =
εz =
(
)
(
σz ⎛
2ν
E⎝
1 − ν⎠
⎜1 −
2 ⎞
σ z ( 1 + ν ) ⋅ ( 1 − 2ν )
⎡
⎢
E⎣
)
1−ν
⎤
⎥
⎦
Hence, when the material is not being confined and undergoes the same normal
strain of εz, then the required modulus of elasticity is
σz
1−ν
E' =
E' =
E
(1 + ν )⋅ (1 − 2ν )
εz
The increased factor is
κ=
κ :=
E'
E
1−ν
( 1 + ν ) ⋅ ( 1 − 2ν )
κ = 1.35
Ans
Problem 10-62
A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to an
internal pressure p. Show that the increase in the volume within the vessel is ∆V = (2pπ r4/Et)(1-ν).
Use a small-strain analysis.
Problem 10-63
A material is subjected to plane stress. Express the distortion-energy theory of failure in terms of σx ,
σy , and τxy .
Problem 10-64
A material is subjected to plane stress. Express the maximum-shear-stress theory of failure in terms of
σx , σy , and τxy . Assume that the principal stresses are of different algebraic signs.
Problem 10-65
The components of plane stress at a critical point on an A-36 structural steel shell are shown.
Determine if failure (yielding) has occurred on the basis of the maximum-shear-stress theory.
Given: σ x := −75MPa
σ y := 125MPa
τ xy := −80MPa
Solution:
In-plane Principal Stress:
σ 1 :=
σx + σy
2
Applying Eq. 9-5,
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 153.06 MPa
σ 2 :=
σx + σy
2
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 2 = −103.06 MPa
Maximum Shear Stress Theory:
σ1 and σ2 have opposite signs. Therefore,
fσ := σ 1 − σ 2
fσ = 256.12 MPa
> σY (= 250 MPa)
Based on the result obtained above, the material yields according
to the maximum shear stress theory.
Ans
Ans
Problem 10-66
The components of plane stress at a critical point on an A-36 structural steel shell are shown.
Determine if failure (yielding) has occurred on the basis of the maximum-distortion-energy theory.
Given: σ x := −75MPa
σ y := 125MPa
τ xy := −80MPa
Solution:
Applying Eq. 9-5,
In-plane Principal Stress:
σ 1 :=
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 153.06 MPa
σ 2 :=
σx + σy
2
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 2 = −103.06 MPa
Maximum Distortion Energy Theory:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
2
fσ := σ 1 − σ 1⋅ σ 2 + σ 2
2
2
fσ = 49825 MPa
< σY2 (= 62500 MPa2)
Based on the result obtained above, the material does not yield according to
the maximum distortion energy theory.
Ans
Ans
Problem 10-67
The yield stress for a zirconium-magnesium alloy is σY = 107 MPa. If a machine part is made of this
material and a critical point in the material is subjected to in-plane principal stresses σ1 and σ2 = -0.5σ1,
determine the magnitude of σ1 that will cause yielding according to the maximum-shear-stress theory.
Given: σ Y := 107MPa
σ 2 = −0.5 σ 1
Solution:
σ1 − σ2 = σY
(
)
σ 1 − −0.5 σ 1 = σ Y
1.5σ 1 = σ Y
σ 1 :=
σY
1.5
σ 1 = 71.33 MPa
Ans
Problem 10-68
Solve Prob. 10-67 using the maximum-distortion-energy theory.
Given: σ Y := 107MPa
σ 2 = −0.5 σ 1
Solution:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
(
2
) (
σ 1 − σ 1⋅ −0.5 σ 1 + −0.5 σ 1
2
2
1.75σ 1 = σ Y
2
σ 1 :=
σY
1.75
σ 1 = 80.88 MPa
Ans
)2 = σY2
Problem 10-69
If a shaft is made of a material of which σY = 350 MPa, determine the maximum torsional shear stress
required to cause yielding using the maximum-distortion-energy theory.
Given: σ Y := 350MPa
Solution:
σ1 = τ
σ 2 = −τ
2
2
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
2
τ − τ ⋅ ( −τ ) + ( −τ ) = σ Y
2
2
3τ = σ Y
2
τ :=
σY
3
τ = 202.1 MPa
Ans
Problem 10-70
Solve Prob. 10-69 using the maximum-shear-stress theory. Both principal stresses have opposite signs.
Given: σ Y := 350MPa
Solution:
σ1 = τ
σ 2 = −τ
σ1 − σ2 = σY
τ − ( −τ ) = σ Y
2τ = σ Y
τ :=
σY
2
τ = 175.0 MPa
Ans
Problem 10-71
The yield stress for a plastic material is σY =110 MPa. If this material is subjected to plane stress and
elastic failure occurs when one principal stress is 120 MPa, what is the smallest magnitude of the other
principal stress? Use the maximum-distortion-energy theory.
Given: σ Y := 110MPa
σ 1 := 120MPa
Solution:
Given
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
Solving Eq. (1),
(1)
Guess σ 2 := 1MPa
( )
σ 2 := Find σ 2
σ 2 = 23.94 MPa
Ans
Problem 10-72
Solve Prob. 10-71 using the maximum-shear-stress theory. Both principal stresses have the same sign.
Given: σ Y := 110MPa
σ 1 := 120MPa
Solution:
Since
σ1 > σY (= 110 MPa) ,
the material will fail for any σ 2.
Ans
Problem 10-73
The plate is made of Tobin bronze, which yields at σY = 175 MPa. Using the maximum-shear-stress
theory, determine the maximum tensile stress σx that can be applied to the plate if a tensile stress σy =
0.75σx is also applied.
Given: σ Y := 175MPa
σ y = 0.75σ x
Solution: σ 1 = σ x
σ 2 = 0.75σ x
σ1 and σ1 have the same signs, so
σ2 = σY
0.75σ x = σ Y
0.75σ x = σ Y
Or,
σ x :=
σY
0.75
σ x = 233.3 MPa
σ x := σ Y
σ x = 175.0 MPa
σ1 = σY
σx = σY
(Controls !)
Ans
Problem 10-74
The plate is made of Tobin bronze, which yields at σY = 175 MPa. Using the maximum-distortionenergy theory, determine the maximum tensile stress σx that can be applied to the plate if a tensile
stress σy = 0.75σx is also applied.
Given: σ Y := 175MPa
σ y = 0.75σ x
σ1 = σx
Solution:
σ 2 = 0.75σ x
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
2
(
) (
σ x − σ x⋅ 0.75σ x + 0.75σ x
2
2
0.8125σ x = σ Y
2
σ 1 :=
σY
0.8175
σ 1 = 193.55 MPa
Ans
)2 = σY2
Problem 10-75
An aluminum alloy 6061-T6 is to be used for a solid drive shaft such that it transmits 33 kW at 2400
rev/min. Using a factor of safety of 2 with respect to yielding, determine the smallest-diameter shaft
that can be selected based on the maximum-shear-stress theory.
2π rad
60 s
Unit Used: rpm :=
Given: P := 33kW
ω := 2400rpm
σ Y := 255MPa
Fsafety := 2
Solution:
Torsion :
P
T :=
T = 0.1313 kN⋅ m
ω
Section Property :
J=
π 4
⋅ρ
2
Shear Stress :
τ=
T⋅ ρ
J
τ=
2T
π⋅ρ
3
Principal Stresses :
σ1 = τ
σ 2 = −τ
Maximum shear stress theory: Both principal stresses have opposite sign, hence
σY
σ1 − σ2 =
Fsafety
σY
τ − ( −τ ) =
Fsafety
4T
π⋅ρ
3
=
3
ρ :=
σY
Fsafety
4T ⎛ Fsafety ⎞
⋅⎜
π ⎝ σY ⎠
do := 2ρ
do = 21.89 mm
Ans
Problem 10-76
Solve Prob. 10-75 using the maximum-distortion-energy theory.
2π rad
60 s
Unit Used: rpm :=
Given: P := 33kW
ω := 2400rpm
σ Y := 255MPa
Fsafety := 2
Solution:
Torsion :
P
T :=
T = 0.1313 kN⋅ m
ω
π 4
⋅ρ
2
J=
Section Property :
Shear Stress :
τ=
T⋅ ρ
J
2T
τ=
π⋅ρ
3
Principal Stresses :
σ1 = τ
σ 2 = −τ
Maximum distortion energy theory:
2
⎛ σY ⎞
σ 1 − σ 1⋅ σ 2 + σ 2 = ⎜
⎝ Fsafety ⎠
2
2
2 ⎛ σY ⎞
τ − τ ⋅ ( −τ ) + ( −τ ) = ⎜
⎝ Fsafety ⎠
2
2
3⋅ τ =
σY
Fsafety
⎛ 2T ⎞ =
3⋅ ⎜
3
⎝ π⋅ρ ⎠
3
ρ :=
σY
Fsafety
2⋅ 3 T ⎛ Fsafety ⎞
⋅⎜
π
⎝ σY ⎠
do := 2ρ
do = 20.87 mm
Ans
Problem 10-77
An aluminum alloy is to be used for a drive shaft such that it transmits 20 kW at 1500 rev/min. Using a
factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can be
selected based on the maximum-distortion-energy theory. σY = 25 MPa.
2π rad
60 s
Unit Used: rpm :=
Given: P := 30kW
ω := 1500rpm
σ Y := 25MPa
Fsafety := 2.5
Solution:
Torsion :
P
T :=
T = 0.1910 kN⋅ m
ω
π 4
⋅ρ
2
J=
Section Property :
Shear Stress :
τ=
T⋅ ρ
J
2T
τ=
π⋅ρ
3
Principal Stresses :
σ1 = τ
σ 2 = −τ
Maximum distortion energy theory:
2
⎛ σY ⎞
σ 1 − σ 1⋅ σ 2 + σ 2 = ⎜
⎝ Fsafety ⎠
2
2
2 ⎛ σY ⎞
τ − τ ⋅ ( −τ ) + ( −τ ) = ⎜
⎝ Fsafety ⎠
2
2
3⋅ τ =
σY
Fsafety
⎛ 2T ⎞ =
3⋅ ⎜
3
⎝ π⋅ρ ⎠
3
ρ :=
σY
Fsafety
2⋅ 3 T ⎛ Fsafety ⎞
⋅⎜
π
⎝ σY ⎠
do := 2ρ
do = 55.23 mm
Ans
Problem 10-78
A bar with a square cross-sectional area is made of a material having a yield stress of σY = 840 MPa. If
the bar is subjected to a bending moment of 10 kN·m., determine the required size of the bar according
to the maximum-distortion-energy theory. Use a factor of safety of 1.5 with respect to yielding.
Given: M := 10kN⋅ m
Solution:
σ Y := 840MPa
Fsafety := 1.5
4
Section Property :
a
I=
12
Normal Stresses : σ y := 0
c=
a
2
σx =
Mc
I
σx =
6M
3
a
In-plane Principal Stresses :
Since no shaer stress acts on the element,
σ1 = σx
σ2 = σy
Maximum distortion energy theory:
2
⎛ σY ⎞
σ 1 − σ 1⋅ σ 2 + σ 2 = ⎜
⎝ Fsafety ⎠
2
σY ⎞
2
2 ⎛
σ x − σ x⋅ ( 0) + 0 = ⎜
⎝ Fsafety ⎠
2
2
σY
σx =
6M
3
Fsafety
=
a
3
a :=
σY
Fsafety
⎛ Fsafety ⎞
6M⋅ ⎜
⎝ σY ⎠
a = 47.50 mm
Ans
Problem 10-79
Solve Prob. 10-78 using the maximum-shear-stress theory.
Given: M := 10kN⋅ m
σ Y := 840MPa
Solution:
Fsafety := 1.5
4
Section Property :
a
12
I=
Normal Stresses : σ y := 0
c=
a
2
σx =
Mc
I
σx =
6M
3
a
In-plane Principal Stresses :
Since no shaer stress acts on the element,
σ1 = σx
σ2 = σy
Maximum shear stress theory:
σ2 = 0
σ1 =
σ2 <
σY
6M
Fsafety
3
a
=
σY
Fsafety
(O.K.!)
σY
Fsafety
3
a :=
⎛ Fsafety ⎞
6M⋅ ⎜
⎝ σY ⎠
a = 47.50 mm
Ans
Problem 10-80
The principal plane stresses acting on a differential element are shown. If the material is machine steel
having a yield stress of σY = 700 MPa, determine the factor of safety with respect to yielding using the
maximum-distortion-energy theory.
Given: σ x := −480MPa
θ := 30deg
σ y := −475MPa
τ xy := 0
σ Y := 700MPa
Solution:
Principal Stresses:
σ 1 := σ y
σ 2 := σ x
Maximum Distortion Eenergy Theory:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
σ allow :=
2
2
σ 1 − σ 1⋅ σ 2 + σ 2
σ allow = 477.5 MPa
Fsafety :=
σY
σ allow
Fsafety = 1.47
Ans
Problem 10-81
The principal plane stresses acting on a differential element are shown. If the material is machine steel
having a yield stress of σY = 700 MPa, determine the factor of safety with respect to yielding if the
maximum-shear-stress theory is considered.
Given: σ x := 80MPa
σ y := −50MPa
θ := 0deg
τ xy := 0
σ Y := 700MPa
Solution:
Principal Stresses:
σ max := σ x
σ min := σ y
Maximum Shear Stress Theory:
τ abs.max :=
τ max :=
σ max − σ min
2
σY
Fsafety :=
τ abs.max = 65 MPa
τ max = 350 MPa
2
τ max
τ abs.max
Fsafety = 5.38
Ans
Problem 10-82
The state of stress acting at a critical point on a machine element is shown in the figure. Determine the
smallest yield stress for a steel that might be selected for the part, based on the maximum-shear-stress
theory.
Given: σ x := 56MPa
σ y := −70MPa
τ xy := 28MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 1 = 61.94 MPa
σ 2 = −75.94 MPa
Maximum shear stress theory: Both principal stresses have opposite sign, hence
σ1 − σ2 = σY
σ Y := σ 1 − σ 2
σ Y = 137.9 MPa
Ans
Problem 10-83
The yield stress for a uranium alloy is σY = 160 MPa. If a machine part is made of this material
and a critical point in the material is subjected to plane stress, such that the principal stresses are σ1
and σ2 = 0.25σ1 , determine the magnitude of σ1 that will cause yielding according to the maximumdistortion-energy theory.
Given: σ 2 = 0.25σ 1
σ Y := 160MPa
Solution:
Maximum Distortion Eenergy Theory:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
2
(
) (
2
2
σ 1 − σ 1⋅ 0.25σ 1 + 0.25σ 1
0.8125 ⋅ σ 1 = σ Y
Principal Stress:
σY
σ 1 :=
0.8125
σ 1 = 177.5 MPa
Ans
)2 = σY2
Problem 10-84
Solve Prob. 10-83 using the maximum-shear-stress theory.
Given:
σ 2 = 0.25σ 1
σ Y := 160MPa
Solution:
Principal Stresses: This is a plane stress case.
σ max = σ 1
σ int = 0.25σ 1
σ min = 0
Maximum Shear Stress Theory:
σY
τ allow :=
τ allow = 80 MPa
2
τ abs.max =
σ max − σ min
2
τ abs.max = τ allow
σ1
2
= 80 MPa
σ 1 := 160MPa
Ans
τ abs.max =
σ1
2
Problem 10-85
An aluminum alloy is to be used for a solid drive shaft such that it transmits 25 kW at 1200 rev/min.
Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can
be selected based on the maximum-shear-stress theory. σY = 70 MPa.
2π rad
60 s
Unit Used: rpm :=
Given: P := 25kW
ω := 1200rpm
σ Y := 70MPa
Fsafety := 2.5
Solution:
Torsion :
P
T :=
T = 0.1989 kN⋅ m
ω
Section Property :
J=
π 4
⋅ρ
2
Shear Stress :
τ=
T⋅ ρ
J
τ=
2T
π⋅ρ
3
Principal Stresses :
σ1 = τ
σ 2 = −τ
Maximum shear stress theory: Both principal stresses have opposite sign, hence
σY
σ1 − σ2 =
Fsafety
σY
τ − ( −τ ) =
Fsafety
4T
π⋅ρ
3
=
3
ρ :=
σY
Fsafety
4T ⎛ Fsafety ⎞
⋅⎜
π ⎝ σY ⎠
do := 2ρ
do = 41.67 mm
Ans
Problem 10-86
The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown
in the figure. Determine the smallest yield stress for a steel that can be selected for the member, based
on the maximum-shear-stress theory.
Given: σ x := 560MPa
σ y := 0MPa
τ xy := 175MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 610.19 MPa
σ 2 = −50.19 MPa
Maximum shear stress theory: Both principal stresses have opposite sign, hence
σ1 − σ2 = σY
σ Y := σ 1 − σ 2
σ Y = 660.4 MPa
Ans
Problem 10-87
Solve Prob. 10-86 using the maximum-distortion-energy theory.
Given: σ x := 560MPa
σ y := 0MPa
τ xy := 175MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 1 = 610.19 MPa
σ 2 = −50.19 MPa
Maximum distortion energy theory:
2
2
2
σ Y :=
σ 1 − σ 1⋅ σ 2 + σ 2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
2
σ Y = 636.8 MPa
Ans
2
Problem 10-88
If a machine part is made of titanium (Ti-6A1-4V) and a critical point in the material is subjected
to plane stress, such that the principal stresses are σ1 and σ2 = 0.5σ1 , determine the magnitude of σ1
in MPa that will cause yielding according to (a) the maximum-shear-stress theory, and (b) the
maximum-distortion-energy theory.
Given: σ 2 = 0.5σ 1
σ Y := 924MPa
Solution:
a) Maximum shear stress theory:
Both principal stresses have the same signs, so
σ1 = σY
(Controls !)
σ2 = σY
0.5σ 1 = σ Y
σ 1 = 2σ Y
Hence σ 1 := σ Y
σ 1 = 924.0 MPa
Ans
b) Maximum Distortion Eenergy Theory:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
(
2
) (
σ 1 − σ 1⋅ 0.5σ 1 + 0.5σ 1
2
2
0.75 ⋅ σ 1 = σ Y
Principal Stress:
σY
σ 1 :=
0.75
σ 1 = 1066.9 MPa
Ans
)2 = σY2
Problem 10-89
Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circular
cross section, would cause the same energy of distortion as the combination of an applied bending
moment M and torque T.
Problem 10-90
An aluminum alloy 6061-T6 is to be used for a drive shaft such that it transmits 40 kW at 1800
rev/min. Using a factor of safety of F.S. = 2, with respect to yielding, determine the smallest-diameter
shaft that can be selected based on the maximum-distortion-energy theory.
Unit Used: rpm :=
2π rad
60 s
Given: P := 40kW
ω := 1800rpm
σ Y := 255MPa
Fsafety := 2
Solution:
Torsion :
P
T :=
T = 0.2122 kN⋅ m
ω
J=
Section Property :
π 4
⋅ρ
2
Shear Stress :
τ=
T⋅ ρ
J
τ=
2T
π⋅ρ
3
Principal Stresses :
σ1 = τ
σ 2 = −τ
Maximum distortion energy theory:
2
⎛ σY ⎞
σ 1 − σ 1⋅ σ 2 + σ 2 = ⎜
⎝ Fsafety ⎠
2
2
2 ⎛ σY ⎞
τ − τ ⋅ ( −τ ) + ( −τ ) = ⎜
⎝ Fsafety ⎠
σ
2
2
3τ =
2⋅ 3 T
π⋅ρ
3
3
ρ :=
Y
Fsafety
=
σY
Fsafety
2⋅ 3 T ⎛ Fsafety ⎞
⋅⎜
π
⎝ σY ⎠
do := 2ρ
do = 24.49 mm
Ans
Problem 10-91
Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a
circular cross section, would cause the same energy of distortion as the combination of an applied
bending moment M and torque T.
Problem 10-92
The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a
torque of 3.45 kN·m, a bending moment of 2.25 kN·m, and an axial thrust of 12.5 kN. If the yield
points for tension and shear are σY = 700 MPa and τY = 350 MPa, respectively, determine the required
diameter of the shaft using the maximum-shear-stress theory.
Given: Nx := −12.5 kN
M := −2.25 kN⋅ m
σ Y := 700MPa
τ Y := 350MPa
Solution:
π 2
⋅ρ
2
Section Property :
A=
Normal Stress :
σ y := 0
σx =
Nx M⋅ ρ
+
A
I
Shear Stress :
T := 3.45kN⋅ m
σx =
τ xy =
T⋅ ρ
J
I=
2Nx
π⋅ρ
2
+
τ xy =
π 4
⋅ρ
4
π 4
⋅ρ
2
J=
4M
π⋅ρ
3
2T
π⋅ρ
3
Principal Stresses :
σ1 =
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σx + σy
σ2 =
2
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Maximum shear stress theory: Assume σ 1 and σ 2 have opposite sign, hence
2
2
2
σY
⎛ σx − σy ⎞
⎛ σx − σy ⎞
2
2
⎜
2 ⎜
+ τ xy = σ Y
+ τ xy =
4
⎝ 2 ⎠
⎝ 2 ⎠
2
2
2
1 ⎛⎜ 2Nx
4M ⎞
⎛ 2T ⎞ = σ Y
+
+⎜
2
3
3
4⎜
4
π⋅ρ ⎠
⎝ π⋅ρ
⎝ π⋅ρ ⎠
Given
(2ρ ⋅ Nx + 4M)2 + ( 4T) 2 = ⎛⎝π ⋅ ρ 3⋅ σY⎞⎠
Solving Eq. (1):
Guess ρ := 10mm
ρ := Find ( ρ )
ρ = 19.68 mm
Check signs:
⎛ 2Nx
4M ⎞
+
⎜ π⋅ρ2 π⋅ρ3
⎝
⎠
σ x := ⎜
2
2
τ xy :=
(1)
2T
π⋅ρ
3
2
⎛ σx ⎞
2
σ 1 :=
+ ⎜
+ τ xy
2
⎝2⎠
⎛ σx ⎞
2
σ 2 :=
− ⎜
+ τ xy
2
⎝2⎠
σ 1 = 151.65 MPa
σ 2 = −548.35 MPa
σx
σx
σ 1 and σ 2 are of opposite sign. (O.K.!)
Therefore,
do := 2ρ
do = 39.35 mm
Ans
σ1 − σ2 = σY
Problem 10-93
The element is subjected to the stresses shown. If σY = 350 ksi, determine the factor of safety for this
loading based on (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory.
Given: σ x := 84MPa
σ y := −56MPa
τ xy := 49MPa
σ Y := 350MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 99.45 MPa
σ 2 = −71.45 MPa
a) Maximum shear stress theory: Both principal stresses have opposite sign, hence
σ allow := σ 1 − σ 2
σ allow = 170.9 MPa
σY
Fsafety :=
Factor of safety is,
Fsafety = 2.05
σ allow
Ans
b) Maximum distortion energy theory:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ' allow
σ' allow :=
2
2
σ 1 − σ 1⋅ σ 2 + σ 2
Factor of safety is,
Fsafety :=
σ' allow = 148.7 MPa
σY
σ' allow
Fsafety = 2.35
Ans
Problem 10-94
The state of stress acting at a critical point on a wrench is shown in the figure. Determine the smallest
yield stress for steel that might be selected for the part, based on the maximum-distortion-energy
theory.
Given: σ x := 175MPa
σ y := 0MPa
τ xy := 70MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 1 = 199.55 MPa
σ 2 = −24.55 MPa
Maximum distortion energy theory:
2
2
2
σ1 − σ1 ⋅ σ2 + σ2 = σY
σ Y :=
2
σ1 − σ1 ⋅ σ2 + σ2
σ Y = 212.9 MPa
2
Ans
Problem 10-95
The state of stress acting at a critical point on a wrench is shown in the figure. Determine the smallest
yield stress for steel that might be selected for the part, based on the maximum-shear-stress theory.
Given: σ x := 175MPa
σ y := 0MPa
τ xy := 70MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 1 = 199.55 MPa
σ 2 = −24.55 MPa
Maximum shear stress theory: Both principal stresses have opposite sign, hence
σ Y := σ 1 − σ 2
σ Y = 224.1 MPa
Ans
Problem 10-96
The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N·m and an
axial compressive force of 2 kN. Determine if it fails according to the maximum-normal-stress theory.
The ultimate stress of the concrete is σult = 28 MPa.
Given: do := 50mm
P := 2kN
σ ult := 28MPa
T := 0.5kN⋅ m
Solution:
Section Property :
A :=
π
2
⋅ do
4
Normal Stress:
J :=
σ :=
P
A
σ = 1.019 MPa
τ :=
T⋅ ρ
J
τ = 20.37 MPa
Shear Stress :
ρ := 0.5do
π
4
⋅ do
32
In-plane Principal Stresses:
σ x := 0
σ y := −σ
τ xy := τ
for any point on the shaft's surface. Applying Eq. 9-5,
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
⎝ 2 ⎠
σ 1 = 19.87 MPa
< σult (=28 MPa)
σ 2 = −20.89 MPa
< σult (=28 MPa)
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
⎝ 2 ⎠
Failure Criteria:
σ 1 < σ ult
OK
σ 2 < σ ult
OK
Based on the result obtained above, the material does not fail according to the
maximum normal-stress theory.
Ans
Problem 10-97
If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by the
maximum-normal-stress theory the maximum allowable principal stress is
σ allow = (16 / πd 3 ) ( M + M 2 + T 2 ) .
Problem 10-98
The principal stresses acting at a point on a thin-walled cylindrical pressure vessel are σ1 = pr/t, σ2 =
pr/2t, and σ3 = 0. If the yield stress is σy , determine the maximum value of p based on (a) the
maximum-shear-stress theory and (b) the maximum-distortion-energy theory.
p⋅ r
p⋅ r
Given: σ 1 =
σ2 =
σ 3 := 0
t
2t
Solution:
a) Maximum Shear Stress Theory:
σ 2 = 0.5σ 1
Both principal stresses have the same signs, so
σ1 = σY
(Controls !)
σ2 = σY
0.5σ 1 = σ Y
Hence
σ 1 = 2σ Y
p⋅ r
= σY
t
t
⋅σ
r Y
p=
Ans
b) Maximum Distortion Eenergy Theory:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
(
2
) (
σ 1 − σ 1⋅ 0.5σ 1 + 0.5σ 1
2
)2 = σY2
2
0.75 ⋅ σ 1 = σ Y
Principal Stress:
σY
σ1 =
0.75
Hence
σY
p⋅ r
=
t
p=
0.75
2t
3⋅ r
⋅ σY
Ans
Problem 10-99
A thin-walled spherical pressure vessel has an inner radius r, thickness t, and is subjected to an internal
pressure p. If the material constants are E and v, determine the strain in the circumferential direction in
terms of the stated parameters.
Solution:
Normal Stresses :
This is a plane stress problem where σ min := 0
since there is no load acting on the outer surface of the wall.
σ1 =
p⋅ r
2⋅ t
σ2 = σ1
Appling the generalized Hooke's Law,
Normal Strains:
1
⎡σ − ν ⋅ σ 2 + σ 3 ⎤⎦
E⎣ 1
1
ε 2 = ⎡⎣σ 2 − ν ⋅ σ 1 + σ 3 ⎤⎦
E
ε1 =
Hence,
ε cir =
ε cir =
(
)
(
)
σ1
E
1
σ − ν ⋅ σ2
E 1
1
ε2 =
σ − ν ⋅ σ1
E 2
ε1 =
(1 − ν)
p⋅ r
(1 − ν )
2E⋅ t
Ans
(
)
(
)
ε1 =
ε2 =
σ1
E
σ1
E
(1 − ν)
(1 − ν)
Problem 10-100
The strain at point A on the shell has components εx = 250(10-6), εy = 400(10-6), γxy = 275(10-6), εz =
0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x-y plane, and (c) the
absolute maximum shear strain.
(
Given: ε x := 250⋅ 10− 6
)
( − 6)
ε y := 400⋅ 10
( − 6)
γ xy := 275⋅ 10
ε z := 0
Solution:
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 325 × 10
2
Radius :
2
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
R = 156.62 × 10
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
−6
ε 1 := ε c + R
ε 1 = 481.62 × 10
ε 2 := ε c − R
ε 2 = 168.38 × 10
Ans
−6
Ans
b) Maximum In-plane Shear Strain:
γ max := 2R
γ max = 313.25 × 10
−6
c) Absolute Maximum Shear Strain :
From the results obtained above,
−6
ε max := ε 1
ε max = 481.625 × 10
ε min := ε z
ε min = 0
γ abs.max := ε max − ε min
γ abs.max = 481.62 × 10
−6
Ans
Ans
−6
Problem 10-101
A differential element is subjected to plane strain that has the following components: εx = 950(10-6), εy
= 420(10-6), γxy = -325(10-6). Use the strain-transformation equations and determine (a) the principal
strains and (b) the maximum in-plane shear strain and the associated average strain. In each case
specify the orientation of the element and show how the strains deform the element.
(
Given: ε x := 950⋅ 10− 6
)
( − 6)
( − 6)
ε y := 420⋅ 10
γ xy := −325⋅ 10
Solution:
a) In-plane Principal Strains: Applying Eq. 10-9,
ε 1 :=
ε 2 :=
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
+ ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
− ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
εx + εy
2
2
−6
ε 1 = 995.86 × 10
Ans
2
−6
ε 2 = 374.14 × 10
Ans
Orientation of Principal Strain:
( )
tan 2θ p =
γ xy
θ p :=
εx − εy
⎛ γ xy ⎞
1
atan ⎜
2
εx − εy
⎝
θ' p := θ p + 90deg
⎠
θ p = −15.758 deg
θ' p = 74.242 deg
Use Eq. 10-5 to determine the direction of ε1 and ε2.
ε x' :=
εx + εy
2
+
εx − εy
2
( )
⋅ cos 2θ p +
γ xy
2
( )
⋅ sin 2θ p
−6
ε x' = 995.856 × 10
Therefore,
θ p1 := θ p
θ p1 = −15.76 deg
Ans
θ p2 := θ' p
θ p2 = 74.24 deg
Ans
b) Maximum In-plane Shear Strain: Applying Eq. 10-11,
2
⎛ ε x − ε y ⎞ ⎛ γ xy ⎞
γ max := 2 ⎜
+⎜
⎝ 2 ⎠ ⎝ 2 ⎠
ε avg :=
εx + εy
2
2
−6
γ max = 621.711 × 10
Orientation of Principal Strain:
εx − εy
⎛ εx − εy ⎞
1
tan 2θ s = −
θ s := atan ⎜ −
2
γ xy
⎝ γ xy ⎠
θ s = 29.242 deg
( )
−6
ε avg = 685 × 10
Ans
Ans
θ' s := θ s + 90deg
θ' s = 119.242 deg
Use Eq. 10-6 to determine the sign of γmax.
⎛ εx − εy
γ x'y' := 2⎜ −
⎝
2
( )
⋅ sin 2θ s +
γ xy
2
⎞
( )⎠
⋅ cos 2θ s
γ x'y' = −621.711 × 10
Therefore,
−6
θ s1 := θ s
θ s1 = 29.24 deg
Ans
θ s2 := θ' s
θ s2 = 119.24 deg
Ans
Problem 10-102
The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure
(yielding) has occurred on the basis of the maximum-distortion-energy theory. The yield stress for the
steel is σY = 650 MPa.
Given: σ x := −55MPa
σ y := 340MPa
τ xy := 65MPa
σ Y := 650MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 1 = 350.42 MPa
σ 2 = −65.42 MPa
Maximum Distortion Energy Theory:
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = σ Y
2
fσ := σ 1 − σ 1⋅ σ 2 + σ 2
2
2
fσ = 150000 MPa
< σY2 (= 422500 MPa2) Ans
Based on the result obtained above, the material does not yield according to
the maximum distortion energy theory.
Ans
Problem 10-103
Solve Prob. 10-102 using the maximum-shear-stress theory.
Given: σ x := −55MPa
σ y := 340MPa
τ xy := 65MPa
σ Y := 650MPa
Solution:
Principal Stresses:
σ 1 :=
σ 2 :=
σx + σy
2
σx + σy
2
2
⎛ σx − σy ⎞
2
+ ⎜
+ τ xy
2
⎝
⎠
2
⎛ σx − σy ⎞
2
− ⎜
+ τ xy
2
⎝
⎠
σ 1 = 350.42 MPa
σ 2 = −65.42 MPa
Maximum Shear Stress Theory: Both principal stresses have opposite sign, hence
σ1 − σ2 = σY
fσ := σ 1 − σ 2
fσ = 415.84 MPa
< σY (= 650 MPa)
Ans
Based on the result obtained above, the material does not yield according to
the maximum shear stress theory.
Ans
Problem 10-104
The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: εa =
600(10-6), εb = -700(10-6), and εc = 350(10-6). Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain. In each case show the deformed element due
to these strains.
(
Given: ε a := 600⋅ 10− 6
)
θ a := 150deg
( − 6)
−6
ε c := 350⋅ ( 10 )
ε b := −700⋅ 10
θ b := −150deg
θ c := −90deg
Solution:
Strain Rosettes (600): Applying Eq. 10-16,
Given
( )2 + ε y⋅ sin(θa)2 + γ xy⋅ sin(θa)⋅ cos (θa)
2
2
ε b = ε x⋅ cos ( θ b) + ε y⋅ sin ( θ b) + γ xy⋅ sin ( θ b) ⋅ cos ( θ b)
2
2
ε c = ε x⋅ cos ( θ c) + ε y⋅ sin ( θ c) + γ xy⋅ sin ( θ c) ⋅ cos ( θ c)
ε a = ε x⋅ cos θ a
(1)
(2)
(3)
Solving Eqs.(1), (2) and (3),
−6
Guess ε x := 10
−6
ε y := 10
γ xy := 10
−6
⎛ ε x ⎞ ⎛⎜ −183.333 × 10− 6 ⎞
⎜
⎜ ε y ⎟ = ⎜ 350 × 10− 6 ⎟
⎟
⎜
⎜
−
3
⎜
γ
⎝ xy ⎠ ⎝ −1.501 × 10
⎠
⎛ εx ⎞
⎜
⎜ ε y ⎟ := Find ( ε x , ε y , γ xy)
⎜
⎝ γ xy ⎠
Construction of Mohr's Circle :
Center :
εx + εy
−6
ε c :=
ε c = 83.33 × 10
2
Radius :
⎛ γ xy ⎞
R := ( ε y − ε c) + ⎜
⎝ 2 ⎠
2
2
R = 796.52 × 10
Coordinates:
(
)
A ε x , 0.5 ⋅ γ xy
(
C εc , 0
)
a) In-plane Principal Strains:
(represented by coordinates of points B and D)
−6
ε 1 := ε c + R
ε 1 = 879.85 × 10
ε 2 := ε c − R
ε 2 = −713.19 × 10
−6
Ans
Ans
−6
Orientation of Principal Strain:
(
)
tan 2θ p2 =
0.5γ xy
εx − εc
⎛ 0.5γ xy ⎞
1
atan ⎜
2
⎝ εx − εc ⎠
θ p1 := 90deg − θ p2
θ p2 :=
θ p2 = 35.22 deg
Ans
θ p1 = 54.78 deg
Ans
b) Maximum In-plane Shear Strain: (represented by coordinates of point E)
γ max := −2R
−3
γ max = −1.593 × 10
Ans
Orientation of Maximum In-plane Shear Strain:
εx − εc
⎛ εx − εc ⎞
1
tan 2θ s =
θ s := atan ⎜
2
0.5γ xy
⎝ 0.5γ xy ⎠
( )
θ s = 9.78 deg
(Clockwise)
Ans
Problem 10-105
The aluminum beam has the rectangular cross section shown. If it is subjected to a bending moment of
M = 7.5 kN·m., determine the increase in the 50-mm dimension at the top of the beam and the
decrease in this dimension at the bottom. Eal = 70 GPa, νal= 0.3.
Given: M := 7.5kN⋅ m
b := 50mm
E := 70GPa
Solution:
ν := 0.3
3
Section Property :
b⋅ h
I :=
12
Normal Stresses :
c :=
h
2
σ z :=
M⋅ c
I
Lateral Strain and deformation :
ν⋅ σz
−6
ε x :=
ε x = 685.71 × 10
E
At the top:
∆ b := ε x⋅ b
∆ b = 0.03429 mm
At the bottom:
∆ b := −ε x⋅ b
∆ b = −0.03429 mm
The negative sign indicates shortening.
h := 75mm
Problem 11-01
The simply supported beam is made of timber that has an allowable bending stress of σallow = 6.5 MPa
and an allowable shear stress of τallow = 500 kPa. Determine its dimensions if it is to be rectangular
and have a height-to-width ratio of 1.25.
Given: σ allow := 6.5MPa
L a := 2m
L b := 4m
kN
wo := 8
m
τ allow := 0.5MPa
h = 1.25 ⋅ b
Solution: L := 2La + L b
Support Reactions : By symmetry, RL=RR=R
+
ΣF y=0;
2R − wo⋅ L = 0
R := 0.5wo⋅ L
R = 32 kN
Maximum Moment and Shear:
Vmax := wo⋅ La
Vmax = 16 kN
(
)
Mmax := wo⋅ La⋅ 0.5 ⋅ La
Mmax = 16 kN⋅ m
3
b⋅ h
I=
12
I
Sx =
0.5h
Section Property :
2
Sx =
b⋅ h
6
b⋅ ( 1.25 ⋅ b)
6
Sx =
2
Sx =
25⋅ b
96
3
Bending Stress:
Sreq'd =
3
Mmax
25⋅ b
=
96
σ allow
Mmax
σ allow
3
b :=
96Mmax
25σ allow
h := 1.25 ⋅ b
3
Shear Check :
τ max :=
b⋅ h
I :=
12
Vmax⋅ Qmax
I⋅ b
Qmax := ( 0.5b⋅ h) ⋅ 0.25 h
τ max = 0.429 MPa
< τ allow =0.5 MPa
(O.K.!)
b = 211.4 mm
Ans
h = 264.3 mm
Ans
(
x1 := 0 , 0.01 ⋅ L a .. La
−wo⋅ x1
V1 x1 :=
kN
x2 := La , 1.01 ⋅ L a .. L a + Lb
( )
( )
M1 x1 :=
2
kN⋅ m
(
)
( ) (
(
)
)
1
2
M2 x2 := ⎡⎣−0.5 ⋅ wo⋅ x2 + R⋅ x2 − La ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
2
M3 x3 := ⎡⎣−0.5 ⋅ wo⋅ x3 + R⋅ x3 − La + R⋅ x3 − L a − Lb ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
Shear (kN)
20
10
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
10
20
0
2
4
6
8
6
8
x1 , x2 , x3
Distance (m)
Moment (kN-m)
0
( )
M2( x2)
10
M3( x3)
M1 x1
20
0
2
4
x1 , x2 , x3
Distance (m)
)
x3 := L a + Lb , 1.01 ⋅ La + L b .. L
1
V3 x3 := −wo⋅ x3 + 2R ⋅
kN
1
V2 x2 := −wo⋅ x2 + R ⋅
kN
( ) (
−0.5 ⋅ wo⋅ x1
)
Problem 11-02
The joists of a floor in a warehouse are to be selected using square timber beams made of oak. If each
beam is to be designed to carry 1.5 kN/m over a simply supported span of 7.5 m, determine the
dimension a of its square cross section to the nearest multiples of 5mm. The allowable bending stress
is σallow = 32 MPa and the allowable shear stress is τallow = 0.875 MPa.
σ allow := 32MPa
Given:
L := 7.5m
τ allow := 0.875MPa
w := 1.5
Solution:
kN
m
Support Reactions : By symmetry, RL=RR=R
ΣF y=0;
+
2R − w⋅ L = 0
R := 0.5w⋅ L
Maximum Moment and Shear:
Vmax := R
Vmax = 5.63 kN
Mmax := R⋅ ( 0.5L) − w⋅ ( 0.5L) ⋅ ( 0.25L )
Mmax = 10.55 kN⋅ m
4
a
I=
12
Section Property :
Bending Stress:
σ max =
M⋅ cmax
I
Qmax = ( 0.5a⋅ a) ⋅ 0.25 a
cmax = 0.5a
12Mmax⋅ ( 0.5a)
σ allow =
4
a
3
a :=
6Mmax
σ allow
a = 125.52 mm
(Use 130mm)
Ans
4
Shear Stress :
τ max :=
a
I :=
12
Qmax := ( 0.5a⋅ a) ⋅ 0.25 a
Vmax⋅ Qmax
I⋅ a
τ max = 0.536 MPa
< τ allow =0.875 MPa
V( x )
0
V ( x) := ( R − w⋅ x) ⋅
2
4
x
Distance (m)
6
1
kN
M ( x) := [ R⋅ x − w⋅ x⋅ ( 0.5x) ]
Moment (kN-m)
Shear (kN)
x := 0 , 0.01 ⋅ L .. L
(O.K.!)
1
kN⋅ m
10
M( x)
5
0
2
4
x
Distance m)
6
Problem 11-03
The timber beam is to be loaded as shown. If the ends support only vertical forces, determine the
greatest magnitude of P that can be applied. σallow = 25 MPa, τallow = 700 kPa.
Given: σ allow := 25MPa
a := 4m
τ allow := 0.7MPa
bf := 150mm
df := 30mm
tw := 40mm
dw := 120mm
L := 2a
Solution:
h := df + dw
Section Property :
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
(
yc :=
I :=
)
(bf⋅ df)⋅ (0.5df) + (tw⋅ dw)⋅ (0.5dw + df)
(bf⋅ df) + (tw⋅ dw)
yc = 53.71 mm
1
3
2 ⎡1
3
2⎤
⋅ bf⋅ df + bf⋅ df ⋅ 0.5df − yc + ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + df − yc ⎥
12
⎣12
⎦
(
)(
)
(
)(
)
4
I = 19162016.13 mm
(
)
(
)
3
Qmax := h − yc ⋅ tw⋅ ⎡⎣0.5 ⋅ h − yc ⎤⎦
Qmax = 185436.52 mm
Support Reactions : By symmetry, A=B=R
+
ΣF y=0;
2R − P = 0
Maximum Load:
Mmax = R⋅ a
σ allow =
R = 0.5P
Assume failure due to bending moment.
Mmax = 0.5P⋅ a
Mmax⋅ cmax
I
cmax := h − yc
σ allow =
(
)
0.5P⋅ a⋅ h − yc
I
P :=
(
)
a⋅ ( h − yc)
2I σ allow
P = 2.49 kN
Check Shear :
τ max :=
R := 0.5P
Vmax⋅ Qmax
I ⋅ tw
Vmax := R
Mmax := R⋅ a
τ max = 0.301 MPa
< τ allow =0.7 MPa
(O.K.!)
Ans
Problem 11-04
Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the machine
loading shown. The allowable bending stress is σallow = 168 MPa and the allowable shear stress is τallow
= 98 MPa.
Given: σ allow := 168MPa a := 0.6m
τ allow := 98MPa
Solution:
P := 25kN
L := 5a
Support Reactions : By symmetry, RL=RR=R
+
ΣF y=0;
2R − 4P = 0
R := 2P
Maximum Moment and Shear:
Vmax := R
Mmax := R⋅ ( 2a) − P⋅ a
Vmax = 50 kN
Mmax = 45 kN⋅ m
Bending Stress:
Sreq'd :=
Mmax
3
Select W 310x24 :
Shear Stress :
( 3) mm3
Sx := 281⋅ 10
d := 305mm
tw := 5.59mm
Provide a shear stress check.
τ max :=
Hence,
3
Sreq'd = 267.86 × 10 mm
σ allow
Vmax
tw⋅ d
Use W 310x24
τ max = 29.33 MPa
< τ allow =98 MPa
Ans
(O.K.!)
x1 := 0 , 0.01 ⋅ a .. a
1
V1 x1 := R⋅
kN
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
x4 := 3a , 1.01 ⋅ ( 3a) .. 4a
x5 := 4a , 1.01 ⋅ ( 4a) .. 5a
1
V2 x2 := ( R − P) ⋅
kN
1
V4 x4 := ( R − 3P) ⋅
kN
1
V3 x3 := ( R − 2P) ⋅
kN
1
V5 x5 := ( R − 4P) ⋅
kN
R⋅ x1
1
M1 x1 :=
M2 x2 := ⎡⎣R⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
kN⋅ m
1
M3 x3 := ⎡⎣R⋅ x3 − P⋅ x3 − a − P⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
(
(
)
)
(
)
1
M4 x4 := ⎡⎣R⋅ x4 − P⋅ x4 − a − P⋅ x4 − 2a − P⋅ x4 − 3a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
(
)
1
M5 x5 := ⎡⎣R⋅ x5 − P⋅ x5 − a − P⋅ x5 − 2a − P⋅ x5 − 3a − P⋅ x5 − 4a ⎤⎦ ⋅
kN⋅ m
( )
( )
V2 ( x 2 )
V3 ( x 3 )
V4 ( x 4 )
V5 ( x 5 )
Shear (kN)
V1 x 1
( )
(
)
(
)
(
)
(
)
50
0
0.5
1
1.5
2
2.5
3
2
2.5
3
50
x1 , x2 , x3 , x4 , x5
Distance (m)
( )40
M2( x2)
M3( x3)
M 4 ( x 4 ) 20
M5( x5)
Moment (kN-m)
M1 x1
0
0.5
1
1.5
x1 , x2 , x3 , x4 , x5
Distance (m)
Problem 11-05
The simply supported beam is made of timber that has an allowable bending stress of σallow = 7 MPa
and an allowable shear stress of τallow = 0.5 MPa. Determine its dimensions if it is to be rectangular and
have a height-to-width ratio of 1.25.
Given: σ allow := 7MPa
τ allow := 0.5MPa
L := 2m
kN
wo := 75
m
h = 1.25 ⋅ b
Solution:
Support Reactions : By symmetry, RL=RR=R
+
ΣF y=0;
2R − 0.5wo⋅ ( 2L) = 0
( )
R := 0.5 wo ⋅ L
Maximum Moment and Shear:
Vmax := R
Vmax = 75 kN
L
Mmax := R⋅ L − 0.5wo⋅ L ⋅
3
(
)
Mmax = 100 kN⋅ m
3
Section Property :
b⋅ h
I=
12
I
Sx =
0.5h
2
Sx =
b⋅ h
6
Sx =
b⋅ ( 1.25 ⋅ b)
6
2
Sx =
25⋅ b
96
3
Bending Stress:
Sreq'd =
3
3
Mmax
25⋅ b
=
96
σ allow
Mmax
σ allow
b :=
96Mmax
25σ allow
Provide a shear stress check.
Shear Stress :
h := 1.25 ⋅ b
τ max :=
1.5Vmax
b⋅ h
τ max = 0.62 MPa
> τ allow =0.5 MPa
Shear controls :
τ allow =
b = 380.0 mm
1.5Vmax
b⋅ h
τ allow =
b :=
1.5Vmax
b⋅ ( 1.25b )
1.5Vmax
1.25τ allow
b = 424.3 mm
Ans
(NG.!)
x1 := 0 , 0.01 ⋅ L .. L
x2 := L , 1.01 ⋅ ( L ) .. ( 2L )
⎡ wo ⎛ x1 ⎞ ⎤ 1
V1 x1 := ⎢R −
⋅⎜
⋅ x ⎥⋅
2 ⎝ L ⎠ 1⎦ kN
⎣
( )
⎡
⎣
( )
⎛
)⎝
(
V2 x2 := ⎢R − 0.5 ⋅ wo⋅ L − wo⋅ x2 − L ⋅ ⎜ 1 − 0.5 ⋅
x2 − L ⎞⎤ 1
⎥⋅
L ⎠⎦ kN
wo ⎛ x1 ⎞
x1⎤
⎡
1
M1 x1 := ⎢R⋅ x1 −
⋅⎜
⋅ x1⋅ ⎥ ⋅
2 ⎝L⎠
3 ⎦ kN⋅ m
⎣
( )
⎡
⎣
( )
Shear (kN)
M2 x2 := ⎢R⋅ x2 −
wo⋅ L
⎛ x2 − L ⎞ 1⎤⎤ 1
2⋅ L ⎞ wo
2 ⎡
⎛
⋅ ⎜ x2 −
−
⋅ x2 − L ⋅ ⎢1 − ⎜
⋅ ⎥⎥ ⋅
2 ⎝
3 ⎠
2
⎣ ⎝ L ⎠ 3⎦⎦ kN⋅ m
(
)
( )
0
V2 ( x 2 )
V1 x 1
0
1
2
3
4
3
4
x1 , x2
Moment (kN-m)
Distance (m)
100
( )
M 2 ( x 2 ) 50
M1 x1
0
0
1
2
x1 , x2
Distane (m)
Problem 11-06
The wooden beam has a rectangular cross section and is used to support a load of 6 kN. If the
allowable bending stress is σallow = 14 MPa and the allowable shear stress is τallow = 5 MPa, determine
the height h of the cross section to the nearest multiples of 5mm if it is to be rectangular and have a
width of b = 75 mm. Assume the supports at A and B only exert vertical reactions on the beam.
Given: σ allow := 14MPa
L 1 := 1.2m
τ allow := 5MPa
L 2 := 1.8m
P := 6kN
b := 75mm
Solution:
L := L 1 + L 2
Support Reactions :
+
ΣF y=0;
A+B−P= 0
(1)
ΣΜB=0;
A⋅ L − P⋅ L2 = 0
(2)
A :=
Solving Eqs. (1) and (2):
P⋅ L 2
L
B :=
P⋅ L 1
L
Maximum Moment and Shear:
Vmax := max ( A , B)
Mmax := A⋅ L 1
Vmax = 3.60 kN
Mmax = 4.32 kN⋅ m
3
b⋅ h
12
Qmax = ( 0.5h⋅ b) ⋅ 0.25 h
Section Property :
I=
Bending Stress:
cmax = 0.5h
12Mmax⋅ ( 0.5h)
σ allow =
3
b⋅ h
σ max =
M⋅ cmax
I
h :=
6Mmax
b⋅ σ allow
h = 157.12 mm
3
Shear Stress :
τ max :=
I :=
b⋅ h
12
Vmax⋅ Qmax
I⋅ b
Qmax := ( 0.5h⋅ b) ⋅ 0.25 h
τ max = 0.458 MPa
< τ allow =5 MPa
(O.K.!)
(Use 160mm)
Ans
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L
1
V1 x1 := A⋅
kN
1
V2 x2 := ( A − P) ⋅
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − P⋅ x2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
( )
( )
(
)
Moment (kN-m)
Shear (kN)
5
( )
0
V2 ( x 2 )
V1 x 1
5
0
1
2
3
5
( )
M2( x2)
M1 x1
0
0
1
2
x1 , x2
x1 , x2
Distance (m)
Distane (m)
3
Problem 11-07
Solve Prob. 11-6 if the cross section has an unknown width but is to be square, i.e., h = b.
Given: σ allow := 14MPa
L 1 := 1.2m
τ allow := 5MPa
L 2 := 1.8m
P := 6kN
b= h
Solution:
L := L 1 + L2
Support Reactions :
+
ΣF y=0;
A+B−P= 0
(1)
ΣΜB=0;
A⋅ L − P⋅ L2 = 0
(2)
P⋅ L 2
A :=
Solving Eqs. (1) and (2):
B :=
L
P⋅ L 1
L
Maximum Moment and Shear:
Vmax := max ( A , B)
Mmax := A⋅ L 1
Vmax = 3.60 kN
Mmax = 4.32 kN⋅ m
3
Section Property :
Bending Stress:
σ max =
M⋅ cmax
I
b⋅ h
I=
12
Qmax = ( 0.5h⋅ b) ⋅ 0.25 h
cmax = 0.5h
b= h
12Mmax⋅ ( 0.5h)
σ allow =
3
h⋅ h
( )
3
h :=
6Mmax
σ allow
h = 122.79 mm
3
Shear Stress :
τ max :=
b := h
Vmax⋅ Qmax
I⋅ b
b⋅ h
I :=
12
Qmax := ( 0.5h⋅ b) ⋅ 0.25 h
τ max = 0.358 MPa
< τ allow =5 MPa
(O.K.!)
(Use 125mm)
Ans
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L
1
V1 x1 := A⋅
kN
1
V2 x2 := ( A − P) ⋅
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − P⋅ x2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
( )
( )
(
)
Moment (kN-m)
Shear (kN)
5
( )
0
V2 ( x 2 )
V1 x 1
5
0
1
2
3
5
( )
M2( x2)
M1 x1
0
0
1
2
x1 , x2
x1 , x2
Distance (m)
Distane (m)
3
Problem 11-08
The simply supported beam is composed of two W310 X 33 sections built up as shown. Determine the
maximum uniform loading w the beam will support if the allowable bending stress is σallow = 160 MPa
and the allowable shear stress is τallow = 100 MPa.
Given:
σ allow := 160MPa
L := 8m
τ allow := 100MPa
Use W 310x33
Solution:
Support Reactions : By symmetry, RL=RR=R
+
ΣF y=0;
2R − w⋅ L = 0
R = 0.5w⋅ L
Maximum Moment and Shear:
Vmax = R
Mmax = R⋅ ( 0.5L ) − w⋅ ( 0.5L ) ⋅ ( 0.25L)
Mmax = ( 0.5w⋅ L) ⋅ ( 0.5L) − w⋅ ( 0.5L) ⋅ ( 0.25L )
Mmax = 0.125 ⋅ w⋅ L
2
One W 310x33
Section Property :
( 6)
tw := 6.60mm
4
2
Ix := 65.0 ⋅ 10 mm
A := 4180mm
Two W 310x33 : Ic := 2⎡Ix + A ( 0.5d) ⎤
⎣
⎦
2
Sc :=
d := 313mm
Ic
d
Assume moment controls.
2
Mmax⋅ cmax
0.125 ⋅ w⋅ L ⋅ ( d)
σ max =
σ allow =
2
Ic
2⎡Ix + A ( 0.5d) ⎤
Maximum Loading:
cmax := d
w :=
16σ allow
2
L ⋅d
⋅ ⎡⎣Ix + A ( 0.5d) ⎤⎦
2
Vmax
Aw
x := 0 , 0.01 ⋅ L .. L
τ max = 20.71 MPa
V ( x) := ( R − w⋅ x) ⋅
1
kN
R := 0.5w⋅ L
Vmax := R
(O.K.!)
M ( x) := [ R⋅ x − w⋅ x⋅ ( 0.5x) ]
Moment (kN-m)
Shear (kN)
Ans
1
kN⋅ m
200
0
100
kN
m
< τ allow =100 MPa
100
V( x )
⎦
w = 21.39
(Neglect area of flanges.) Aw := 2d⋅ tw
Check Shear :
τ max :=
⎣
0
5
x
Distance (m)
M ( x ) 100
0
0
5
x
Distance m)
Problem 11-09
The simply supported beam is composed of two W310 X 33 sections built up as shown. Determine if
the beam will safely support a loading of w = 30 kN/m. The allowable bending stress is σallow = 160
MPa and the allowable shear stress is τallow = 100 MPa.
Given:
σ allow := 160MPa
L := 8m
τ allow := 100MPa
w := 30
Solution:
Use W 310x33
kN
m
Support Reactions : By symmetry, RL=RR=R
+
ΣF y=0;
2R − w⋅ L = 0
R := 0.5w⋅ L
Maximum Moment and Shear:
Vmax := R
Mmax = R⋅ ( 0.5L ) − w⋅ ( 0.5L ) ⋅ ( 0.25L)
Mmax = ( 0.5w⋅ L) ⋅ ( 0.5L) − w⋅ ( 0.5L) ⋅ ( 0.25L )
Mmax := 0.125 ⋅ w⋅ L
One W 310x33
Section Property :
( 6)
tw := 6.60mm
4
2
Ix := 65.0 ⋅ 10 mm
A := 4180mm
Two W 310x33 : Ic := 2⎡Ix + A ( 0.5d) ⎤
⎣
⎦
2
Bending Stress:
cmax := d
σ max :=
Sc :=
d := 313mm
Ic
d
Mmax⋅ cmax
Ic
σ max = 224.4 MPa
> σ allow =160 MPa (Not O.K.!)
The beam fails due to bending stress criteria..
Check Shear :
τ max :=
Ans
(Neglect area of flanges.) Aw := 2d⋅ tw
Vmax
Aw
τ max = 29.04 MPa
V ( x) := ( R − w⋅ x) ⋅
< τ allow =100 MPa
1
kN
V( x ) 0
0
5
x
Distance (m)
(O.K.!)
M ( x) := [ R⋅ x − w⋅ x⋅ ( 0.5x) ]
Moment (kN-m)
x := 0 , 0.01 ⋅ L .. L
Shear (kN)
2
1
kN⋅ m
200
M( x)
0
0
5
x
Distance m)
Problem 11-10
Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading
shown, where w = 100 kN/m and P = 25 kN. The allowable bending stress is σallow = 160 MPa, and
the allowable shear stress is τallow = 100 MPa.
Given:
σ allow := 160MPa
L1 := 2.4m
τ allow := 100MPa
L2 := 1.8m
P := 25kN
Solution:
L := L1 + L2
Support Reactions :
+
w := 100
kN
m
Given
ΣFy=0;
A + B − P − w⋅ L1 = 0
ΣΜB=0;
A⋅ L 1 + P⋅ L2 − 0.5w⋅ L 1 = 0
(1)
2
(2)
Solving Eqs. (1) and (2):
Guess A := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 101.25 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 163.75 ⎠
B := 1kN
Maximum Moment and Shear:
Vmax := P − B
When V=0,
xo
L1
=
A
A+B
2
Mmax := A⋅ xo − 0.5w⋅ xo
Bending Stress:
Sreq'd :=
3
Sreq'd = 317510.1 mm
σ allow
Select W 310x33 :
( 3) mm3 d := 313mm tw := 6.60mm
Sx := 415⋅ 10
Provide a shear stress check.
τ max :=
Hence,
Mmax = 50.8 kN⋅ m
Assume bending controls the design.
Mmax
Shear Stress :
Vmax = −138.75 kN
A⋅ L1
xo :=
xo = 0.917 m
A+B
Vmax
tw⋅ d
Use W 310x33
τ max = 67.17 MPa
< τ allow =100 MPa
Ans
(O.K.!)
x1 := 0 , 0.01 ⋅ L1 .. L1
x2 := L 1 , 1.01 ⋅ L 1 .. L
1
V1 x1 := A − w⋅ x1 ⋅
kN
1
V2 x2 := A − w⋅ L 1 + B ⋅
kN
( ) (
)
( ) (
)
1
2
M1 x1 := ⎛ A⋅ x1 − 0.5w⋅ x1 ⎞ ⋅
⎝
⎠ kN⋅ m
( )
1
M2 x2 := ⎡A⋅ x2 − w⋅ L 1 ⋅ x2 − 0.5 ⋅ L 1 + B⋅ x2 − L 1 ⎤ ⋅
⎣
⎦ kN⋅ m
( )
(
)(
)
(
( )
V2 ( x 2 )
V1 x 1
50
Moment (kN-m)
Shear (kN)
100
)
0
100
0
2
x1 , x2
Distance (m)
4
( )
M2(x2) 0
M1 x1
50
0
2
x1 , x2
Distane (m)
4
Problem 11-11
Select the lightest-weight steel wide-flange beam having the shortest height from Appendix B that will
safely support the loading shown, where w = 0 and P = 50 kN. The allowable bending stress is σallow =
168 MPa, and the allowable shear stress is τallow = 100 MPa.
Given:
σ allow := 168MPa
L1 := 2.4m
τ allow := 100MPa
L2 := 1.8m
P := 50kN
Solution:
L := L1 + L2
Support Reactions :
+
w := 0
kN
m
Given
ΣFy=0;
A+B−P= 0
(1)
ΣΜB=0;
A⋅ L 1 + P⋅ L2 = 0
(2)
Solving Eqs. (1) and (2):
Guess A := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ −37.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 87.50 ⎠
B := 1kN
Maximum Moment and Shear:
Vmax := P
Mmax := P⋅ L2
Vmax = 50 kN
Mmax = 90 kN⋅ m
Bending Stress: Assume bending controls the design.
Mmax
3
Sreq'd :=
Sreq'd = 535714.3 mm
σ allow
Three choices of wide flange section having the weight of 39 kg/m can be
made. W 310x39 , W 360x39 , and W 410x39. However, the shortest is
the W 310x39.
Select W 310x39 :
Shear Stress :
Provide a shear stress check.
τ max :=
Hence,
( 3) mm3 d := 310mm tw := 5.84mm
Sx := 547⋅ 10
Vmax
tw⋅ d
Use W 310x39
τ max = 27.62 MPa
< τ allow =100 MPa
Ans
(O.K.!)
x1 := 0 , 0.01 ⋅ L1 .. L1
1
V1 x1 := ( A) ⋅
kN
x2 := L 1 , 1.01 ⋅ L 1 .. L
1
V2 x2 := ( A + B) ⋅
kN
( )
( )
1
M1 x1 := A⋅ x1 ⋅
kN⋅ m
( ) (
)
1
M2 x2 := ⎡A⋅ x2 + B⋅ x2 − L1 ⎤ ⋅
⎣
⎦ kN⋅ m
( )
(
)
0
Moment (kN-m)
Shear (kN)
50
( )
V2 ( x 2 ) 0
V1 x 1
50
0
2
x1 , x2
Distance (m)
4
( )
50
M2(x2)
M1 x1
100
0
2
x1 , x2
Distane (m)
4
Problem 11-12
Determine the minimum width of the beam to the nearest multiples of 5mm that will safely support the
loading of P = 40 kN. The allowable bending stress is σallow = 168 MPa, and the allowable shear stress
is τallow = 105 MPa.
Given:
Solution:
σ allow := 168MPa
L1 := 2m
τ allow := 105MPa
L2 := 2m
P := 40kN
h := 150mm
L := L1 + L2
Support Reactions :
+
Given
ΣFy=0;
A+B−P= 0
(1)
ΣΜB=0;
A⋅ L 1 − P⋅ L = 0
(2)
Solving Eqs. (1) and (2):
Guess A := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 80 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ −40 ⎠
B := 1kN
Maximum Moment and Shear:
Vmax := P
Mmax := P⋅ L1
Vmax = 40 kN
Mmax = 80 kN⋅ m
3
b⋅ h
I=
12
I
Sx =
0.5h
Section Property :
Sx =
b⋅ h
6
2
Qmax = ( 0.5h⋅ b) ⋅ 0.25 h
Assume bending controls the design.
Bending Stress:
Sreq'd =
2
Mmax
b⋅ h
=
6
σ allow
Mmax
σ allow
3
b = 126.98 mm
Qmax := ( 0.5h⋅ b) ⋅ 0.25 h
Vmax⋅ Qmax
I⋅ b
τ max = 3.150 MPa
I :=
6Mmax
(h2)σallow
b⋅ h
12
Check Shear :
τ max :=
b :=
< τ allow =105 MPa
(Use 130mm)
(O.K.!)
Ans
x1 := 0 , 0.01 ⋅ L1 .. L1
1
V1 x1 := ( −P) ⋅
kN
x2 := L 1 , 1.01 ⋅ L 1 .. L
1
V2 x2 := ( −P + A) ⋅
kN
( )
( )
1
M1 x1 := −P⋅ x1 ⋅
kN⋅ m
( ) (
)
1
M2 x2 := ⎡−P⋅ x2 + A⋅ x2 − L 1 ⎤ ⋅
⎣
⎦ kN⋅ m
( )
(
)
0
Moment (kN-m)
Shear (kN)
50
( )
V2 ( x 2 ) 0
V1 x 1
50
0
2
x1 , x2
Distance (m)
4
( )
50
M2(x2)
M1 x1
100
0
2
x1 , x2
Distane (m)
4
Problem 11-13
Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading
shown. The allowable bending stress is σallow = 168 MPa, and the allowable shear stress is τallow = 100
MPa.
Given: σ allow := 168MPa L 1 := 2m
Solution:
τ allow := 100MPa
L 2 := 3m
P := 75kN
kN
wo := 75
m
L := L 1 + L 2
Support Reactions :
+
Given
ΣF y=0;
A + B − P − 0.5wo⋅ L2 = 0
ΣΜA=0;
−P⋅ L1 + 0.5wo⋅ L2⋅ ⎜
(1)
⎛ L2 ⎞
− B⋅ L 2 = 0
⎝3⎠
Solving Eqs. (1) and (2):
Guess A := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 200.00 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ −12.50 ⎠
(2)
B := 2kN
Maximum Moment and Shear:
Vmax := −P + A
Mmax := −P⋅ L 1
Bending Stress:
Sreq'd :=
Vmax = 125 kN
Mmax = −150 kN⋅ m
Assume bending controls the design.
Mmax
3
Sreq'd = 892857.1 mm
σ allow
Select W 410x53 :
Shear Stress :
Provide a shear stress check.
τ max :=
Hence,
( 3) mm3 d := 403mm tw := 7.49mm
Sx := 923⋅ 10
Vmax
tw⋅ d
Use W 410x53
τ max = 41.41 MPa
< τ allow =100 MPa
Ans
(O.K.!)
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L
−P
V1 x1 :=
kN
( )
⎡
( )
(
)
⎛
V2 x2 := ⎢−P + A − wo⋅ x2 − L1 ⋅ ⎜ 1 − 0.5 ⋅
⎣
1
M1 x1 := −P⋅ x1 ⋅
kN⋅ m
( ) (
)
⎝
x2 − L 1 ⎞⎤ 1
⎥⋅
L2
kN
wo
⎡
⎛ x2 − L1 ⎞ 1⎤⎤ 1
2 ⎡
M2 x2 := ⎢−P⋅ x2 + A⋅ x2 − L1 −
⋅ x2 − L1 ⋅ ⎢1 − ⎜
⋅ ⎥⎥ ⋅
2
L2
3 kN⋅ m
( )
(
⎣
)
(
)
⎣
⎝
⎠ ⎦⎦
Shear (kN)
100
( )
V2 ( x 2 )
V1 x 1
0
100
0
1
2
3
4
5
x1 , x2
Distance (m)
Moment (kN-m)
0
50
( )
M 2 ( x 2 ) 100
M1 x1
150
0
1
2
3
x1 , x2
Distane (m)
4
5
⎠⎦
Problem 11-14
Select the lightest-weight steel structural wide-flange beam with the shortest depth from Appendix B
that will safely support the loading shown. The allowable bending stress is σallow = 168 MPa, and the
allowable shear stress is τallow = 100 MPa.
Given:
σ allow := 168MPa
L := 2m
τ allow := 100MPa
kN
wo := 120
m
Solution:
Support Reactions :
+
Given
ΣF y=0;
R − 0.5wo⋅ L = 0
ΣΜA=0;
⎛ L⎞ = 0
M + 0.5wo⋅ L ⋅ ⎜
⎝ 3⎠
R := 0.5wo⋅ L
M :=
−wo⋅ L
2
6
Maximum Moment and Shear:
Vmax := R
Mmax := M
Vmax = 120 kN
Mmax = −80 kN⋅ m
Bending Stress:
Assume bending controls the design.
Sreq'd :=
Mmax
3
Sreq'd = 476190.5 mm
σ allow
Select W 310x39 :
Shear Stress :
Provide a shear stress check.
τ max :=
Hence,
( 3) mm3 d := 310mm tw := 5.84mm
Sx := 547⋅ 10
Vmax
tw⋅ d
Use W 310x39
τ max = 66.28 MPa
< τ allow =100 MPa
Ans
(O.K.!)
Problem 11-15
Select the shortest and lightest-weight steel wide-flange beam from Appendix B that will safely support
the loading shown. The allowable bending stress is σallow = 160 MPa, and the allowable shear stress is
τallow = 84 MPa.
Given:
Solution:
σ allow := 160MPa
a := 1.2m
τ allow := 84MPa
P1 := 20kN
P2 := 50kN
P3 := 30kN
L := 4a
Support Reactions :
+
Given
ΣF y=0;
A + B − P1 − P2 − P3 = 0
(1)
ΣΜB=0;
A⋅ L − P1⋅ ( 3a) − P2⋅ ( 2⋅ a) − P3⋅ ( a) = 0
(2)
Solving Eqs. (1) and (2):
Guess A := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 47.50 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 52.50 ⎠
B := 2kN
Maximum Moment and Shear:
Vmax := max ( A , B)
Mmax := A⋅ ( 2a) − P1⋅ ( a)
Bending Stress:
Sreq'd :=
Assume bending controls the design.
Mmax
3
Sreq'd = 562500 mm
σ allow
Select W 250x58 :
Shear Stress :
( 3) mm3 d := 252mm tw := 8.00mm
Sx := 693⋅ 10
Provide a shear stress check.
τ max :=
Hence,
Vmax = 52.5 kN
Mmax = 90 kN⋅ m
Vmax
tw⋅ d
Use W 250x58
τ max = 26.04 MPa
< τ allow =84 MPa
Ans
(O.K.!)
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
x4 := 3a , 1.01 ⋅ ( 3a) .. 4a
1
V1 x1 := A⋅
kN
1
V2 x2 := A − P1 ⋅
kN
( )
( ) (
1
V3 x3 := A − P1 − P2 ⋅
kN
)
( ) (
)
1
V4 x4 := ⎡⎣ A − P1 − P2 − P3⎤⎦ ⋅
kN
( )
A⋅ x1
M1 x1 :=
kN⋅ m
(
)
1
M2 x2 := ⎡⎣A⋅ x2 − P1⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − P1⋅ x3 − a − P2⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡⎣⎡⎣A⋅ x4 − P1⋅ x4 − a − P2⋅ x4 − 2a ⎤⎦ − P3⋅ x4 − 3a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
(
50
( )
V2 ( x 2 )
V3 ( x 3 ) 0
V4 ( x 4 )
V1 x 1
Shear (kN)
( )
50
0
1
2
3
4
x1 , x2 , x3 , x4
Distance (m)
Moment (N-m)
100
( )
M2( x2)
M 3 ( x 3 ) 50
M4( x4)
M1 x1
0
0
1
2
x1 , x2 , x3 , x4
Distance (m)
3
4
)
Problem 11-16
The beam is made of a ceramic material having an allowable bending stress of σallow = 5 MPa and an
allowable shear stress of τallow = 2.8 MPa. Determine the width b of the beam if the height h = 2b.
Given:
σ allow := 5MPa
L 1 := 50mm
τ allow := 2.8MPa
L o := 150mm
P1 := 75N
w := 1.2
P2 := 50N
Solution:
Given
Support Reactions :
+
h = 2b
L := 2L1 + Lo
kN
m
ΣF y=0;
A + B − P1 − P2 − w⋅ L o = 0
ΣΜB=0;
A⋅ L o − P1⋅ L 1 + L o − w⋅ L o⋅ 0.5L o + P2⋅ L1 = 0 (2)
( )
(
(1)
)
(
Solving Eqs. (1) and (2):
Guess A := 1N
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 173.33 ⎞
=⎜
N
⎜
⎝ B ⎠ ⎝ 131.67 ⎠
)
B := 1N
Maximum Moment and Shear:
Vmax := −P1 + A
Mmax := P1⋅ L1
Vmax = 98.33 N
Mmax = 3.75 N⋅ m
3
b⋅ h
12
I
Sx =
0.5h
I=
Section Property :
2
b⋅ h
Sx =
6
4b
Sx =
6
3
Assume bending controls the design.
Bending Stress:
Sreq'd =
h = 2b
Mmax
σ allow
3
Mmax
4b
=
6
σ allow
3
b :=
6Mmax
4σ allow
b = 10.40 mm
Check Shear :
τ max :=
h := 2⋅ b
1.5Vmax
h⋅ b
τ max = 0.682 MPa
< τ allow =2.8 MPa
(O.K.!)
Ans
(
)
x1 := 0 , 0.01 ⋅ L 1 .. L 1 x2 := L1 , 1.01 ⋅ L1 .. L 1 + L o
− P1
1
V1 x1 :=
V2 x2 := ⎡⎣−P1 + A − w⋅ x2 − L1 ⎤⎦ ⋅
N
N
( )
( )
( )
M1 x1 :=
−P1⋅ x1
(
)
(
)
( ) (
(
)
x3 := L 1 + L o , 1.01 ⋅ L 1 + L o .. L
1
V3 x3 := −P1 + A − w⋅ L o + B ⋅
N
)
1
2
M2 x2 := ⎡⎣−P1⋅ x2 + A⋅ x2 − L 1 − 0.5w⋅ x2 − L1 ⎤⎦ ⋅
N⋅ m
( )
N⋅ m
(
)
(
)
1
M3 x3 := ⎡⎣−P1⋅ x3 + A⋅ x3 − L 1 − w⋅ Lo ⋅ x3 − L 1 − 0.5 ⋅ Lo + B⋅ x3 − L1 − Lo ⎤⎦ ⋅
N⋅ m
Shear (N)
( )
(
) (
)(
)
100
( )
V2 ( x 2 )
V3 ( x 3 )
V1 x 1
0
100
0
0.05
0.1
0.15
0.2
x1 , x2 , x3
Distance (m)
Moment (N-m)
2
( )
M2( x2)
M3( x3) 2
M1 x1 0
4
0
0.05
0.1
0.15
x1 , x2 , x3
Distance (m)
0.2
(
)
Problem 11-17
The steel cantilevered T-beam is made from two plates welded together as shown. Determine the
maximum loads P that can be safely supported on the beam if the allowable bending stress is σallow =
170 MPa and the allowable shear stress is τallow = 95 MPa.
Given: σ allow := 170MPa
a := 2m
τ allow := 95MPa
bf := 150mm
df := 15mm
tw := 15mm
dw := 150mm
Solution:
L := 2a
h := df + dw
Section Property :
yc =
I :=
⎯
Σ ⋅ yi⋅ Ai
(
Σ ⋅ ( Ai)
)
(bf⋅ df)⋅ (0.5df) + (tw⋅ dw)⋅ (0.5dw + df)
(bf⋅ df) + (tw⋅ dw)
yc :=
yc = 48.75 mm
1
3
2 ⎡1
3
2⎤
⋅ bf⋅ df + bf⋅ df ⋅ 0.5df − yc + ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + df − yc ⎥
12
12
⎣
⎦
(
)(
)
(
)(
)
4
(
)
(
I = 11917968.75 mm
)
3
Qmax := h − yc ⋅ tw⋅ ⎡⎣0.5 ⋅ h − yc ⎤⎦
Qmax = 101355.47 mm
Support Reactions :
+
ΣF y=0;
R−P−P= 0
R = 2P
ΣΜA=0;
MA + P⋅ a + P⋅ ( 2a) = 0
MA = 3P⋅ a
Maximum Load:
Mmax = MA
σ allow =
cmax := h − yc
Mmax⋅ cmax
I
Check Shear :
τ max :=
Assume failure due to bending moment.
R := 2P
Vmax⋅ Qmax
I ⋅ tw
σ allow =
Vmax := R
(
)
3P⋅ a⋅ h − yc
I
MA := 3P⋅ a
P :=
(
(
)
3a⋅ h − yc
P = 2.90 kN
τ max = 3.294 MPa
< τ allow =95 MPa
)
I σ allow
(O.K.!)
Ans
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. L
1
V1 x1 := ( R) ⋅
kN
1
V2 x2 := ( R − P) ⋅
kN
( )
( )
1
M1 x1 := −MA + R⋅ x1 ⋅
kN⋅ m
( ) (
)
1
M2 x2 := ⎡⎣−MA + R⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
0
Moment (kN-m)
Shear (kN)
6
( )
V2 ( x 2 )
V1 x 1 4
2
0
0
2
4
( )
10
M2( x2)
M1 x1
20
0
2
x1 , x2
x1 , x2
Distance (m)
Distane (m)
4
Problem 11-18
Draw the shear and moment diagrams for the W310 X 21 beam and check if the beam will safely
support the loading. The allowable bending stress is σallow = 160 MPa and the allowable shear stress is
τallow = 84 MPa.
Given:
σ allow := 160MPa
L 1 := 1m
τ allow := 84MPa
L 2 := 4m
Mo := 75kN⋅ m
Solution: L := L 1 + L 2
Support Reactions :
+
w := 25
kN
m
Given
ΣF y=0;
A + B − w⋅ L 2 = 0
ΣΜA=0;
−Mo + 0.5w⋅ L 2 − B⋅ L2 = 0
(1)
2
(2)
Solving Eqs. (1) and (2):
Guess A := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 68.75 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 31.25 ⎠
B := 2kN
Maximum Moment and Shear:
Vmax := max ( A , B)
Mmax := Mo
Assume bending controls the design.
Bending Stress:
Use W 310x21 :
Sreq'd :=
Shear Stress :
Vmax = 68.75 kN
Mmax = 75 kN⋅ m
( 3) mm3 d := 303mm tw := 5.08mm
Sx := 244⋅ 10
Mmax
σ allow
3
Sreq'd = 468750 mm
( 3)
3
> Sx =244⋅ 10 mm
(No Good.!)
Provide a shear stress check.
τ max :=
Vmax
tw⋅ d
τ max = 44.66 MPa
< τ allow =84 MPa
(O.K.!)
Hence, the wide flange section W 310x21 fails due to the bending stress
and will not safely support the loading.
Ans
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L
0
V1 x1 :=
kN
( )
1
V2 x2 := ⎡⎣A − w⋅ x2 − L 1 ⎤⎦ ⋅
kN
( )
(
)
1
M1 x1 := −Mo ⋅
kN⋅ m
( ) (
)
w
1
2⎤
⎡
M2 x2 := ⎢−Mo + A⋅ x2 − L1 − ⋅ x2 − L 1 ⎥ ⋅
2
kN⋅ m
Shear (kN)
( )
(
⎣
( )
V2 ( x 2 )
V1 x 1
)
(
)⎦
50
0
0
1
2
3
4
5
4
5
x1 , x2
Moment (kN-m)
Distance (m)
0
( )
M 2 ( x 2 ) 50
M1 x1
100
0
1
2
3
x1 , x2
Distane (m)
Problem 11-19
Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading
shown. The allowable bending stress is σallow = 160 MPa and the allowable shear stress is τallow = 84
MPa.
Given: σ allow := 160MPa L 1 := 1m
τ allow := 84MPa
L 2 := 4m
Mo := 75kN⋅ m
Solution: L := L 1 + L2
w := 25
Support Reactions :
+
kN
m
Given
ΣF y=0;
A + B − w⋅ L 2 = 0
ΣΜA=0;
−Mo + 0.5w⋅ L 2 − B⋅ L2 = 0
(1)
2
(2)
Solving Eqs. (1) and (2):
Guess A := 1kN
⎛A⎞
:= Find ( A , B)
⎜
⎝B⎠
⎛ A ⎞ ⎛ 68.75 ⎞
=⎜
kN
⎜
⎝ B ⎠ ⎝ 31.25 ⎠
B := 2kN
Maximum Moment and Shear:
Vmax := max ( A , B)
Mmax := Mo
Bending Stress:
Sreq'd :=
Vmax = 68.75 kN
Mmax = 75 kN⋅ m
Assume bending controls the design.
Mmax
Sreq'd = 468750 mm
σ allow
Select W 360x33 :
Shear Stress :
d := 349mm
tw := 5.84mm
Provide a shear stress check.
τ max :=
Hence,
( 3) mm3
Sx := 475⋅ 10
3
Vmax
tw⋅ d
Use W 360x33
τ max = 33.73 MPa
< τ allow =84 MPa
Ans
(O.K.!)
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L
0
V1 x1 :=
kN
( )
1
V2 x2 := ⎡⎣A − w⋅ x2 − L 1 ⎤⎦ ⋅
kN
( )
(
)
1
M1 x1 := −Mo ⋅
kN⋅ m
( ) (
)
w
1
2⎤
⎡
M2 x2 := ⎢−Mo + A⋅ x2 − L1 − ⋅ x2 − L 1 ⎥ ⋅
2
kN⋅ m
Shear (kN)
( )
(
⎣
( )
V2 ( x 2 )
V1 x 1
)
(
)⎦
50
0
0
1
2
3
4
5
4
5
x1 , x2
Moment (kN-m)
Distance (m)
0
( )
M 2 ( x 2 ) 50
M1 x1
100
0
1
2
3
x1 , x2
Distane (m)
Problem 11-20
The compound beam is made from two sections, which are pinned together at B. Use Appendix B and
select the light wide-flange beam that would be safe for each section if the allowable bending stress is
σallow = 168 MPa and the allowable shear stress is τallow = 100 MPa. The beam supports a pipe loading
of 6 kN and 9 kN as shown.
Given:
σ allow := 168MPa
L 1 := 3.6m
τ allow := 100MPa
L 2 := 2.4m
P1 := 6kN
L 3 := 3m
P2 := 9kN
Solution:
L := L 1 + L 2 + L 3
Support Reactions :
Given
For segment BC :
+
ΣF y=0;
B + C − P2 = 0
(1)
ΣΜB=0;
−C⋅ L2 + L3 + P2⋅ L 2 = 0
)
(2)
Guess C := 1kN
B := 2kN
(
Solving Eqs. (1) and (2):
⎛C⎞
⎜ := Find ( C , B)
⎝B⎠
⎛C⎞ ⎛4⎞
=⎜
kN
⎜
⎝B⎠ ⎝5⎠
For segment AB :
+
A := P1 + B
(
)
MA := −P1⋅ 0.5L 1 − B⋅ L1
A = 11 kN
MA = −28.80 kN⋅ m
Maximum Moment and Shear:
For segment AB :
Vmax := max ( A , B)
Mmax := MA
Vmax = 11 kN
Mmax = −28.8 kN⋅ m
For segment BC :
V'max := max ( B , C)
M'max := C⋅ L 3
V'max = 5 kN
M'max = 12 kN⋅ m
Assume bending controls the design.
Mmax
For segment AB :
3
Sreq'd :=
Sreq'd = 171428.6 mm
σ allow
Select W 250x18 :
3
3
Sx := 179⋅ 10 mm d := 251mm tw := 4.83mm
Bending Stress:
( )
For segment BC :
Select W 150x14 :
S'req'd :=
M'max
3
S'req'd = 71428.6 mm
σ allow
( 3)
3
S'x := 91.2 ⋅ 10 mm d' := 150mm
t'w := 4.32mm
Shear Stress :
Provide a shear stress check.
For segment AB :
Vmax
tw⋅ d
τ max :=
τ max = 9.07 MPa
< τ allow =100 MPa
Hence, Use W 250x18
For segment BC :
V'max
t'w⋅ d'
τ' max :=
Ans
τ' max = 7.72 MPa
< τ allow =100 MPa
Hence, Use W 150x14
Set
a := 0.5L1
b := 0.5L 1
x1 := 0 , 0.01 ⋅ a .. a
(O.K.!)
c := L2
(O.K.!)
Ans
d := L 3
x2 := a , 1.01 ⋅ a .. a + b
x3 := a + b , 1.01 ⋅ ( a + b) .. a + b + c
x4 := a + b + c , 1.01 ⋅ ( a + b + c) .. a + b + c + d
1
V1 x1 := A⋅
kN
1
V2 x2 := A − P1 ⋅
kN
( )
( ) (
( )
MA + A⋅ x1
kN⋅ m
( ) (
)
1
V4 x4 := A − P1 − P2 ⋅
kN
( ) (
M1 x1 :=
1
V3 x3 := A − P1 ⋅
kN
)
)
1
M2 x2 := ⎡⎣MA + A⋅ x2 − P1⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
M3 x3 := ⎡⎣B⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
M4 x4 := ⎡⎣B⋅ x4 − a − b − P2⋅ x4 − a − b − c ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
( )
V2 ( x 2 )
V3 ( x 3 )
0
V4 ( x 4 )
Shear (kN)
V1 x 1 10
10
0
2
4
x1 , x2 , x3 , x4
Distance (m)
6
8
Moment (kN-m)
20
( )
M2( x2)
0
M3( x3)
M4( x4)
M1 x1
20
0
2
4
x1 , x2 , x3 , x4
Distance (m)
6
8
Problem 11-21
The steel beam has an allowable bending stress σallow = 140 MPa and an allowable shear stress of
τallow = 90 MPa. Determine the maximum load that can safely be supported.
Given: σ allow := 140MPa
a := 2m
τ allow := 90MPa
bf := 120mm
tw := 20mm
df := 20mm
dw := 150mm
L := 2a
Solution:
h := df + dw
Section Property :
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
(
yc :=
I :=
)
(bf⋅ df)⋅ (0.5df) + (tw⋅ dw)⋅ (0.5dw + df)
(bf⋅ df) + (tw⋅ dw)
yc = 57.22 mm
1
3
2 ⎡1
3
2⎤
⋅ bf⋅ df + bf⋅ df ⋅ 0.5df − yc + ⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + df − yc ⎥
12
12
⎣
⎦
(
)(
)
(
)(
)
4
I = 15338333.33 mm
(
)
(
)
3
Qmax := h − yc ⋅ tw⋅ ⎡⎣0.5 ⋅ h − yc ⎤⎦
Qmax = 127188.27 mm
Support Reactions : By symmetry, RR= - P
+
ΣFy=0;
Rc − P − P = 0
Maximum Load:
Mmax = P⋅ a
σ allow =
Rc = 2P
Assume failure due to bending moment.
cmax := h − yc
Mmax⋅ cmax
I
σ allow =
(
)
P⋅ a⋅ h − yc
I
P :=
(
)
a⋅ ( h − yc)
I σ allow
P = 9.52 kN
Check Shear :
τ max :=
Vmax := P
Vmax⋅ Qmax
I ⋅ tw
Rc := 2P
τ max = 3.947 MPa
< τ allow =90 MPa
(O.K.!)
Ans
x1 := 0 , 0.01 ⋅ a .. a
1
V1 x1 := ( −P) ⋅
kN
x2 := a , 1.01 ⋅ a .. L
1
V2 x2 := −P + Rc ⋅
kN
( )
( ) (
)
1
M1 x1 := −P⋅ x1 ⋅
kN⋅ m
( ) (
)
1
M2 x2 := ⎡−P⋅ x2 + Rc⋅ x2 − a ⎤ ⋅
⎣
⎦ kN⋅ m
( )
(
)
0
Moment (kN-m)
Shear (kN)
10
( )
0
V2 ( x 2 )
V1 x 1
10
0
2
x1 , x2
Distance (m)
4
( )
M2(x2)
M 1 x 1 10
20
0
2
x1 , x2
Distane (m)
4
Problem 11-22
The timber beam has a rectangular cross section. If the width of the beam is 150 mm, determine its
height h so that it simultaneously reaches its allowable bending stress of σallow = 10 MPa and an
allowable shear stress of τallow = 0.35 MPa. Also, what is the maximum load P that the beam can then
support?
Given:
σ allow := 10MPa
L := 1.5m
τ allow := 0.35MPa
b := 150mm
Solution:
Support Reactions :
+
ΣFy=0;
By symmetry, A=B=R
2R − P = 0
R = 0.5P
Section Property :
3
I=
b⋅ h
12
Sx =
I
0.5h
Sx =
b⋅ h
6
2
2
Qmax = ( 0.5h⋅ b) ⋅ 0.25 h
Qmax = 0.125b⋅ h
Maximum Moment and Shear:
Vmax = R
Mmax = R⋅ L
Vmax = 0.5P
Mmax = 0.5P⋅ L
If shear conrols :
τ allow =
Vmax
I⋅ b
=
Qmax
τ allow
Vmax⋅ Qmax
I⋅ b
(b⋅ h3)⋅ b = 0.5P
2
τ allow
12⋅ ( 0.125b⋅ h )
If bending conrols : Sreq'd =
Mmax
σ allow
Solving Eqs. (1) and (2):
h :=
From Eq. (1):
P :=
R := 0.5P
R = 7.35 kN
4τ allow⋅ L
σ allow
4 ⎯
⋅ ( b h) τ allow
3
b⋅ h =
3⋅ P
4τ allow
(1)
2
b⋅ h
0.5P⋅ L
=
6
σ allow
h = 210 mm
Ans
P = 14.70 kN
Ans
(2)
x1 := 0 , 0.01 ⋅ L .. L
1
V1 x1 := ( R) ⋅
kN
x2 := L , 1.01 ⋅ L .. 2L
1
V2 x2 := ( R − P) ⋅
kN
( )
( )
1
M1 x1 := R⋅ x1 ⋅
kN⋅ m
( ) (
)
1
M2 x2 := ⎡R⋅ x2 − P⋅ x2 − L ⎤ ⋅
⎣
⎦ kN⋅ m
( )
(
)
( )
0
V2 ( x 2 )
V1 x 1
0
1
2
x1 , x2
Distance (m)
3
Moment (kN-m)
Shear (kN)
20
( )10
M2(x2)
M1 x1
0
0
1
2
x1 , x2
Distane (m)
3
Problem 11-23
The beam is to be used to support the machine, which has a weight of 80 kN and a center of gravity at
G. If the maximum bending stress is not to exceed σallow = 160 MPa, determine the required width b of
the flanges. The supports at B and C are smooth.
Given: W := 80kN
σ allow := 160MPa
L 1 := 1.8m
L 2 := 0.9m
L 3 := 1.5m
L 4 := 1.8m
dw := 175mm
df := 12mm
tw := 12mm
Solution:
L := L 1 + L 2 + L 3 + L 4
Support Reactions :
Given
For machine BC :
+
ΣF y=0;
B+C−W= 0
(
(1)
)
(2)
ΣΜC=0
B⋅ L2 + L3 − W⋅ L 3 = 0
;
Solving
Eqs. (1) and (2): Guess B := 1kN C := 2kN
⎛B⎞
⎜ := Find ( B , C)
⎝C⎠
⎛ B ⎞ ⎛ 50 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 30 ⎠
For beam AD :
+
ΣF y=0;
A+D−W= 0
(
(3)
)
ΣΜD=0
A⋅ L − W⋅ L3 + L4 = 0
;
W⋅ L 3 + L 4
Solving Eq. (4) : A :=
L
From Eq. (3) :
D := W − A
(
1 ⎡ 3
3
⋅ ⎣b⋅ h − b − tw ⋅ dw ⎤⎦
12
Bending Stress:
Mmax = A⋅ L1
)
A = 44 kN
D = 36 kN
h := 2df + dw
Section Property :
I=
(4)
(
)
cmax := 0.5h
I=
1 ⎡ ⎛ 3
3
3
⋅ ⎣b⋅ ⎝ h − dw ⎞⎠ + tw⋅ dw ⎤⎦
12
σ allow =
A⋅ L1⋅ ( 0.5h)
1 ⎡ ⎛ 3
3
3
⋅ ⎣b⋅ ⎝ h − dw ⎞⎠ + tw⋅ dw ⎤⎦ =
12
σ allow
Mmax⋅ cmax
I
I=
Mmax⋅ cmax
σ allow
12A⋅ L1⋅ ( 0.5h)
b :=
σ allow
3
3
− tw⋅ dw
3
h − dw
b = 208.9 mm
Ans
A
V1 x1 :=
kN
(
)
x3 := ( L 1 + L 2 + L 3) , 1.01 ⋅ ( L 1 + L 2 + L 3) .. L
1
1
V2 ( x2) := ( A − B) ⋅
V3 ( x3) := ( A − B − C) ⋅
kN
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − B⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
x1 := 0 , 0.01 ⋅ L 1 .. L 1
( )
( )
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2 + L 3
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − B⋅ x3 − L 1 − C⋅ x3 − L 1 − L 2 − L 3 ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
Shear (kN)
50
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
50
0
1
2
3
4
5
6
4
5
6
x1 , x2 , x3
Distance (m)
Moment (kN-m)
100
( )
M2( x2)
50
M3( x3)
M1 x1
0
0
1
2
3
x1 , x2 , x3
Distance (m)
Problem 11-24
The beam has a flange width b = 200 mm. If the maximum bending stress is not to exceed σallow = 160
MPa, determine the greatest weight of the machine that the beam can support. The center of gravity
for the machine is at G, and the supports at B and C are smooth.
Given: L 1 := 1.8m
L 2 := 0.9m
L 3 := 1.5m
L 4 := 1.8m
dw := 175mm
df := 12mm
tw := 12mm
b := 200mm
σ allow := 160MPa
L := L 1 + L 2 + L 3 + L 4
Solution:
Support Reactions :
For beam AD :
+ ΣF =0;
A+D−W= 0
y
(
)
(1)
ΣΜD=0
A⋅ L − W⋅ L3 + L4 = 0
;
W⋅ L3 + L4
Solving Eq. (2) : A =
L
From Eq. (1) :
D= W−A
(
)
h := 2df + dw
Section Property :
1 ⎡ 3
3
⋅ b⋅ h − b − tw ⋅ dw ⎤⎦
12 ⎣
Bending Stress:
(
I :=
(2)
Mmax = A⋅ L 1
)
I = 47379775.00 mm
cmax := 0.5h
σ allow =
Mmax⋅ cmax
I
⎛ I⋅ σ allow ⎞
I⋅ σ allow
⎡ W⋅ ( L3 + L4)⎤
⎢
⎥ ⋅ L1 =
L
cmax
⎣
⎦
4
W := ⎜
Mmax =
I⋅ σ allow
cmax
L
⋅
⎝ cmax ⎠ L1⋅ ( L3 + L4)
W = 76.96 kN
Ans
Evaluate Support Reactions :
For beam AD :
For machine BC :
From Eqs. (1) and (2) :
A :=
(
)
W⋅ L 3 + L 4
L
D := W − A
A = 42.33 kN
D = 34.63 kN
+
ΣF y=0;
Given
B+C−W= 0
(
(3)
)
(4)
ΣΜC=0 B⋅ L2 + L3 − W⋅ L 3 = 0
;
Solving Eqs. (3) and (4): Guess B := 1kN C := 2kN
⎛B⎞
⎜ := Find ( B , C)
⎝C⎠
⎛ B ⎞ ⎛ 48.1 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 28.86 ⎠
A
V1 x1 :=
kN
(
)
x3 := ( L 1 + L 2 + L 3) , 1.01 ⋅ ( L 1 + L 2 + L 3) .. L
1
1
V2 ( x2) := ( A − B) ⋅
V3 ( x3) := ( A − B − C) ⋅
kN
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − B⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
x1 := 0 , 0.01 ⋅ L 1 .. L 1
( )
( )
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2 + L 3
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − B⋅ x3 − L 1 − C⋅ x3 − L 1 − L 2 − L 3 ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
Shear (kN)
50
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
50
0
1
2
3
4
5
6
4
5
6
x1 , x2 , x3
Distance (m)
Moment (kN-m)
100
( )
M2( x2)
50
M3( x3)
M1 x1
0
0
1
2
3
x1 , x2 , x3
Distance (m)
Problem 11-25
The box beam has an allowable bending stress of σallow = 10 MPa and an allowable shear stress of
τallow = 775 kPa. Determine the maximum intensity w of the distributed loading that it can safely
support. Also, determine the maximum safe nail spacing for each third of the length of the beam. Each
nail can resist a shear force of 200 N.
Given: Vallow := 0.2kN
L := 6m
bf := 210mm
tw := 30mm
df := 30mm
dw := 190mm
σ allow := 10MPa
τ allow := 0.775MPa
Solution:
Section Property :
h := 2df + dw
1
1
3
3
⋅ bf ⋅ h −
⋅ bf − 2tw ⋅ dw
12
12
(
)
QA := ( 0.5 ⋅ h − 0.5df) ⋅ ( bf − 2tw) ⋅ df
1
2 1
2
Qmax := ⋅ bf⋅ h − ⋅ ( bf − 2tw) ⋅ dw
8
8
I :=
4
I = 187700000 mm
3
QA = 495000 mm
3
Qmax = 963750 mm
Support Reactions : By symmetry, RL=RR=R
+
ΣFy=0;
2R − wo⋅ L = 0
R = 0.5wo⋅ L
Maximum Moment and Shear:
Vmax = R
Vmax = 0.5 ⋅ wo⋅ L
Mmax = R⋅ ( 0.5L) − wo⋅ ( 0.5L ) ⋅ ( 0.25 ⋅ L )
Mmax = 0.125 ⋅ wo⋅ L
2
Maximum Load:
Assume failure due to bending:
Mmax⋅ cmax
σ allow =
I
cmax := 0.5 ⋅ h
2
σ allow =
0.125 ⋅ wo⋅ L ⋅ ( 0.5 ⋅ h)
I
Assume failure due to shear:
Vmax⋅ Qmax
τ allow =
I⋅ 2tw
0.5 ⋅ wo⋅ L ⋅ Qmax
τ allow =
I⋅ 2tw
wo :=
( )
( )
w'o :=
(
)
16( I) σ allow
h⋅ L
2
(
)
4( I) τ allow ⋅ tw
L ⋅ Qmax
(
)
w := min wo , w'o
kN
wo = 3.337
m
kN
w'o = 3.019
m
w = 3.019
kN
m
Shear Flow: Since there are two rows of nails, the allowable shear flow is q = 2V/s
Region AB (0 < x < 2m) and CD (4m < x < 6m):
Ans
Vmax := 0.5 ⋅ w⋅ L
q1 :=
Vmax ⋅ QA
I
kN
q1 = 23.88
m
s1 :=
2Vallow
q1
s1 = 16.7 mm
Ans
Region BC (2m < x < 4m) :
L
V'max := Vmax − w⋅
3
q2 :=
V'max ⋅ QA
I
kN
q2 = 7.96
m
s2 :=
2Vallow
q2
s2 = 50.2 mm
Ans
R := 0.5w⋅ L
x := 0 , 0.01 ⋅ L .. L
V ( x) := ( R − w⋅ x) ⋅
1
kN
M ( x) := [ R⋅ x − w⋅ x⋅ ( 0.5x) ]
20
Moment (kN-m)
Shear (kN)
10
V( x )
0
10
1
kN⋅ m
0
2
4
x
Distance (m)
6
M ( x ) 10
0
0
2
4
x
Distance m)
6
Problem 11-26
The beam is constructed from three boards as shown. If each nail can support a shear force of 250 N,
determine the maximum spacing of the nails, s, s', and s”, for regions AB, BC, and CD, respectively.
Given: Vallow := 250N
a := 1.5m
bf := 200mm
df := 25mm
tw := 25mm
dw := 150mm
P1 := 4kN
P2 := 6kN
L := 3a
Solution:
h := df + dw
Section Property :
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
(
)
(bf⋅ df)⋅ (0.5df) + 2(tw⋅ dw)⋅ (0.5dw + df)
(bf⋅ df) + 2(tw⋅ dw)
yc :=
yc = 65.00 mm
1
3
2
3
2⎤
⎡1
⋅ bf⋅ df + bf⋅ df ⋅ 0.5df − yc + 2⎢ ⋅ tw⋅ dw + tw⋅ dw ⋅ 0.5dw + df − yc ⎥
12
12
⎣
⎦
(
I :=
(
)(
)
(
)
)(
)
4
Q := yc − 0.5df ⋅ bf⋅ df
I = 37291666.67 mm
3
Q = 262500.00 mm
Support Reactions :
+
ΣF y=0;
Given
B + D − P1 − P2 = 0
ΣΜD=0 −P1⋅ ( 3a) + B⋅ ( 2a) − P2⋅ a = 0
;
Solving Eqs. (1) and (2): Guess B := 1kN
⎛B⎞
:= Find ( B , D)
⎜
⎝D⎠
(1)
(2)
D := 2kN
⎛B⎞ ⎛9⎞
=⎜
kN
⎜
⎝D⎠ ⎝1⎠
VAB := −P1
VAB ⋅ Q
kN
q :=
q = 28.16
I
m
Region AB :
VBC := −P1 + B
VBC ⋅ Q
kN
q' :=
q' = 35.20
I
m
s :=
Vallow
0.5q
s = 17.76 mm
Ans
s' :=
Vallow
0.5q'
s' = 14.21 mm
Ans
s'' :=
Vallow
0.5q''
s'' = 71.03 mm
Ans
Region BC :
Region CD :
q'' :=
VCD := −P1 + B − P2
VCD ⋅ Q
I
q'' = 7.04
kN
m
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( 2a)
− P1
V1 x1 :=
kN
1
V2 x2 := −P1 + B ⋅
kN
( )
( ) (
)
x3 := ( 2a) , 1.01 ⋅ ( 2a) .. L
1
V3 x3 := −P1 + B − P2 ⋅
kN
( ) (
)
Shear (kN)
10
5
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
5
10
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Problem 11-27
The beam is constructed from two boards as shown. If each nail can support a shear force of 1 kN,
determine the maximum spacing of the nails, s, s', and s”, to the nearest multiples of 5 mm for regions
AB,BC, and CD, respectively.
Given: Vallow := 1kN
a := 1.5m
bf := 200mm
df := 25mm
tw := 25mm
dw := 150mm
P1 := 2.5kN
P2 := 7.5kN
L := 3a
Solution:
h := df + dw
Section Property :
⎯
Σ ⋅ yi⋅ Ai
yc =
Σ ⋅ ( Ai)
(
)
(bf⋅ df)⋅ (0.5df) + (tw⋅ dw)⋅ (0.5dw + df)
(bf⋅ df) + (tw⋅ dw)
yc :=
yc = 50.00 mm
1
1
3
2
3
2
⋅ bf⋅ df + bf⋅ df ⋅ 0.5df − yc +
⋅ t ⋅ d + tw⋅ dw ⋅ 0.5dw + df − yc
12
12 w w
(
I :=
(
)(
)
(
)
)(
)
4
Q := yc − 0.5df ⋅ bf⋅ df
I = 23697916.67 mm
3
Q = 187500.00 mm
Support Reactions :
+
ΣF y=0;
Given
B + D − P1 − P2 = 0
ΣΜD=0 −P1⋅ ( 3a) + B⋅ ( 2a) − P2⋅ a = 0
;
Solving Eqs. (1) and (2): Guess B := 1kN
⎛B⎞
⎛ B ⎞ ⎛ 7.5 ⎞
:= Find ( B , D)
=⎜
kN
⎜
⎜
⎝D⎠
⎝ D ⎠ ⎝ 2.5 ⎠
VAB := −P1
VAB ⋅ Q
kN
q :=
q = 19.78
I
m
(1)
(2)
D := 2kN
Region AB :
VBC := −P1 + B
VBC ⋅ Q
kN
q' :=
q' = 39.56
I
m
s :=
Vallow
q
s = 50.56 mm
Use s = 55mm Ans
Region BC :
Region CD :
q'' :=
s' :=
Vallow
q'
VCD := −P1 + B − P2
VCD ⋅ Q
I
q'' = 19.78
kN
m
s'' :=
Vallow
q''
s' = 25.28 mm
Use s' = 30mm Ans
s'' = 50.56 mm
Use s'' = 55mm Ans
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( 2a)
− P1
V1 x1 :=
kN
1
V2 x2 := −P1 + B ⋅
kN
( )
( ) (
)
x3 := ( 2a) , 1.01 ⋅ ( 2a) .. L
1
V3 x3 := −P1 + B − P2 ⋅
kN
( ) (
)
Shear (kN)
10
5
( )
V2 ( x 2 )
0
V3 ( x 3 )
V1 x 1
5
10
0
1
2
x1 , x2 , x3
Distance (m)
3
4
Problem 11-28
Draw the shear and moment diagrams for the shaft, and determine its required diameter to the nearest
multiples of 5mm. if σallow = 50 kN and τallow = 20 kN. The bearings at A and D exert only vertical
reactions on the shaft. The loading is applied to the pulleys at B, C, and E. Take P = 550 N.
Given:
σ allow := 50MPa
L 1 := 350mm
τ allow := 20MPa
L 2 := 500mm
PB := 400N
L 3 := 375mm
PC := 550N
L 4 := 300mm
PE := 175N
Solution:
L o := L1 + L2 + L3
Support Reactions :
+
ΣF y=0;
L := L o + L 4
Given
A + D − PB − PC − PE = 0
(
(1)
)
ΣΜD=0
A⋅ Lo − PB⋅ L 2 + L 3 − PC⋅ L3 + PE⋅ L4 = 0 (2)
;
Solving Eqs. (1) and (2): Guess A := 1N
D := 1N
⎛A⎞
:= Find ( A , D)
⎜
⎝D⎠
⎛ A ⎞ ⎛ 411.22 ⎞
=⎜
N
⎜
⎝ D ⎠ ⎝ 713.78 ⎠
Maximum Moment and Shear:
Vmax := A − PB − PC
Mmax := A⋅ L1 + L2 − PB⋅ L 2
(
)
Section Property :
π ⋅ do
I=
4
Vmax = −538.78 N
Mmax = 149.54 N⋅ m
I
Sx =
0.5 ⋅ do
64
Sx =
4⋅ 0.5 ⋅ do 1 ⎛ π ⋅ do ⎞
Qmax =
⋅ ⋅⎜
2 ⎝ 4 ⎠
3π
(
3
Mmax
π ⋅ do
σ allow
32
4
τ max :=
32
2
3
do
Qmax =
12
=
3
Mmax
do :=
σ allow
32Mmax
(
I :=
π ⋅ do
64
Vmax ⋅ Qmax
I ⋅ do
)
π σ allow
do = 31.23 mm
Check Shear :
3
Assume bending controls the design.
Bending Stress:
Sreq'd =
)
π ⋅ do
(Use 35mm)
3
do
Qmax :=
12
τ max = 0.938 MPa
< τ allow =20 MPa
(O.K.!)
Ans
(
x1 := 0 , 0.01 ⋅ L 1 .. L 1
)
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
(
)
(
) (
)
x4 := ( L 1 + L 2 + L 3) , 1.01 ⋅ ( L 1 + L 2 + L 3) .. L
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. L 1 + L 2 + L 3
1
V1 x1 := A⋅
N
1
V2 x2 := A − PB ⋅
kN
( )
( ) (
1
V3 x3 := A − PB − PC ⋅
N
)
( ) (
)
1
V4 x4 := A − PB − PC + D ⋅
N
( ) (
A⋅ x1
M1 x1 :=
N⋅ m
)
1
M2 x2 := ⎡⎣A⋅ x2 − PB⋅ x2 − L 1 ⎤⎦ ⋅
N⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − PB⋅ x3 − L 1 − PC⋅ x3 − L 1 − L 2 ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡⎣A⋅ x4 − PB⋅ x4 − L 1 − PC⋅ x4 − L 1 − L 2 + D⋅ L4 − x4 − L ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
(
)
500
( )
V2 ( x 2 )
V3 ( x 3 )
V4 ( x 4 )
Shear (N)
V1 x 1
0
500
0
0.2
0.4
0.6
0.8
1
1.2
1.4
x1 , x2 , x3 , x4
Distance (m)
Moment (N-m)
200
( )
M 2 ( x 2 ) 100
M3( x3)
0
M4( x4)
M1 x1
100
0
0.2
0.4
0.6
0.8
x1 , x2 , x3 , x4
Distance (m)
1
1.2
1.4
Problem 11-29
Draw the shear and moment diagrams for the shaft, and determine its required diameter to the nearest
multiples of 5mm if σallow = 50 kN and τallow = 20 kN. The bearings at A and D exert only vertical
reactions on the shaft. The loading is applied to the pulleys at B, C, and E. Take P = 400 N.
Given:
σ allow := 50MPa
L1 := 350mm
τ allow := 20MPa
L2 := 500mm
PB := 400N
L3 := 375mm
PC := 400N
L4 := 300mm
PE := 175N
Solution:
Lo := L 1 + L 2 + L 3
Support Reactions :
+
ΣFy=0;
L := Lo + L4
Given
A + D − PB − PC − PE = 0
(
(1)
)
ΣΜD=0
A⋅ L o − PB⋅ L2 + L3 − PC⋅ L 3 + PE ⋅ L 4 = 0 (2)
;
Solving Eqs. (1) and (2): Guess A := 1N
D := 1N
⎛A⎞
:= Find ( A , D)
⎜
⎝D⎠
⎛ A ⎞ ⎛ 365.31 ⎞
=⎜
N
⎜
⎝ D ⎠ ⎝ 609.69 ⎠
Maximum Moment and Shear:
Vmax := A − PB − PC
Mmax := A⋅ L1
Vmax = −434.69 N
Mmax = 127.86 N⋅ m
4
Section Property :
π ⋅ do
I=
64
(
4⋅ 0.5 ⋅ do
Qmax =
3π
Sx =
) ⋅ 1 ⋅ ⎛⎜ π ⋅ do2 ⎞
2 ⎝
Mmax
π ⋅ do
σ allow
32
3
⎠
4
τ max :=
32
do
Qmax =
12
3
=
3
Mmax
do :=
σ allow
32Mmax
(
)
π σ allow
do = 29.64 mm
Check Shear :
π ⋅ do
Assume bending controls the design.
Bending Stress:
Sreq'd =
3
I
Sx =
0.5 ⋅ do
I :=
π ⋅ do
4
64
Vmax ⋅ Qmax
I⋅ do
x1 := 0 , 0.01 ⋅ L1 .. L1
(Use 30mm)
3
do
Qmax :=
12
τ max = 0.840 MPa
< τ allow =20 MPa
(
x2 := L 1 , 1.01 ⋅ L 1 .. L1 + L2
(O.K.!)
)
(
)
(
) (
)
x4 := ( L1 + L2 + L3) , 1.01 ⋅ ( L1 + L2 + L3) .. L
x3 := L1 + L2 , 1.01 ⋅ L1 + L2 .. L1 + L2 + L3
Ans
1
V1 x1 := A⋅
N
1
V2 x2 := A − PB ⋅
kN
( )
( ) (
1
V3 x3 := A − PB − PC ⋅
N
)
( ) (
)
1
V4 x4 := A − PB − PC + D ⋅
N
( ) (
A⋅ x1
M1 x1 :=
N⋅ m
)
1
M2 x2 := ⎡⎣A⋅ x2 − PB⋅ x2 − L1 ⎤⎦ ⋅
N⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − PB⋅ x3 − L1 − PC⋅ x3 − L1 − L2 ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡⎣A⋅ x4 − PB⋅ x4 − L1 − PC⋅ x4 − L1 − L2 + D⋅ L 4 − x4 − L ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
(
)
(
)
500
( )
V2 ( x 2 )
V3 ( x 3 )
V4 ( x 4 )
Shear (N)
V1 x 1
0
500
0
0.2
0.4
0.6
0.8
1
1.2
1.4
x1 , x2 , x3 , x4
Distance (m)
Moment (N-m)
200
( )
M 2 ( x 2 ) 100
M3(x3)
0
M4(x4)
M1 x1
100
0
0.2
0.4
0.6
0.8
x1 , x2 , x3 , x4
Distance (m)
1
1.2
1.4
Problem 11-30
The overhang beam is constructed using two 50-mm by 100-mm pieces of wood braced as shown. If
the allowable bending stress is σallow = 4.2 MPa, determine the largest load P that can be applied. Also,
determine the associated maximum spacing of nails, s, along the beam section AC if each nail can resist
a shear force of 4 kN. Assume the beam is pin-connected at A, B, and D. Neglect the axial force
developed in the beam along DA.
Given: Vallow := 4kN
L := 0.9m
b := 100mm
ho := 50mm
σ allow := 4.2MPa
Solution:
h := 2ho
Maximum Moment and Shear:
Vmax = P
Mmax = P⋅ L
3
I :=
Section Property :
b⋅ h
12
I
0.5h
Sx :=
2
Qmax = ( 0.5h⋅ b) ⋅ 0.25 h
Qmax = 0.125b h
Maximum Load :
Sreq'd =
Mmax
σ allow
(
P :=
Nail Spacing :
q :=
Vmax := P
Vmax ⋅ Qmax
I
)
(
Mmax = Sx⋅ σ allow
(
)
Sx⋅ σ allow
P = 777.8 N
L
Qmax := 0.125b h
q = 11.7
kN
m
)
P⋅ L = Sx⋅ σ allow
Ans
2
smax :=
Vallow
q
smax = 342.9 mm
Ans
Problem 11-31
The tapered beam supports a concentrated force P at its center. If it is made from a plate that has a
constant width b, determine the absolute maximum bending stress in the beam.
Problem 11-32
Determine the variation of the radius r of the cantilevered beam that supports the uniform distributed
load so that it has a constant maximum bending stress σmax throughout its length.
Problem 11-33
Determine the variation in the depth d of a cantilevered beam that supports a concentrated force P at
its end so that it has a constant maximum bending stress σallow throughout its length. The beam has a
constant width b0 .
Problem 11-34
The beam is made into the shape of a frustum and has a diameter of 12 mm at A and a diameter of 300
mm at B. If it supports a force of 750 N at A, determine the absolute maximum bending stress in the
beam and specify its location x.
Given: P := 750N
L := 900mm
do := 150mm
d1 := 300mm
Solution:
r − δr
x
=
π⋅r
I=
4
δr
M
σ=
S
L
π⋅r
S=
4
I
S=
r
Bendiug Stress :
2
δ r ⋅ ( L + x)
r=
L
4
d1 − do
δ r :=
Section Property :
3
3
S=
π ⋅ δ r ⋅ ( L + x)
4⋅ L
3
3
M = P⋅ x
4⋅ L
σ = ( P ⋅ x) ⋅
3
3
π ⋅ δ r ⋅ ( L + x)
(1)
3
In order to have the absolute maximum bending stress,
d
σ= 0
dx
3
x
⎡d
⎤=0
⋅⎢
3 dx
3⎥
π ⋅ δ r ⎣ ( L + x) ⎦
4P⋅ L
Differentiate Eq. (1):
⎞
⎡d
3⎤
x − x⋅ ⎢ ( L + x) ⎥ = 0
⎝ dx ⎠
⎣dx
⎦
3 ⎛d
( L + x) ⋅ ⎜
( L + x) − x⋅ ⎡⎣3⋅ ( L + x) ⎤⎦ = 0
3
2
2
( L + x) ⋅ ( L − 2x) = 0
x :=
Substituting into Eq. (1):
σ max := ( P⋅ x) ⋅
4⋅ L
3
L
2
x = 450 mm
Ans
σ max = 0.3018 MPa
Ans
3
π ⋅ δ r ⋅ ( L + x)
3
Problem 11-35
The beam has a width w and a depth that varies as shown. If it supports a concentrated force P at its
end, determine the absolute maximum bending stress in the beam and specify its location x.
Problem 11-36
The tapered beam supports a uniform distributed load w. If it is made from a plate and has a constant
width b, determine the absolute maximum bending stress in the beam.
Problem 11-37
The tapered simply supported beam supports the concentrated force P at its center. Determine the
absolute maximum bending stress in the beam.
Problem 11-38
t0ta0t0nt0The bearings at A and D exert only y and z components of force on the shaft. If τallow = 60
MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use
the maximum-shear-stress theory of failure.
Given: a := 300mm
r := 50mm
τ allow := 60MPa
PB := 5kN
Solution:
PC := 5kN
L := 3a
Support Reactions :
ΣFz=0;
In x-z plane : +
Az + DZ − PB = 0
(1)
ΣΜD=0; Az⋅ ( 3a) − PB⋅ ( 2a) = 0 (2)
Solving Eqs. (1) and (2):
2
Az := PB
3
Az = 3.3333 kN
Dz := PB − Az
Dz = 1.6667 kN
(Ay + Dy) − PC = 0
(3)
ΣΜD=0; −PC⋅ a + Ay⋅ ( 3a) = 0
(4)
In x-y plane : ΣFy=0;
Solving Eqs. (3) and (4):
1
Ay := PC
3
Ay = 1.6667 kN
Dy := PC − Ay
Dy = 3.3333 kN
( )
TBC := PB ⋅ r
Torsion occurs in segment BC :
Critical Section :
My := Dz⋅ a
TBC = 0.250 kN⋅ m
Located just to the left of gear C and just to the right of gear B, where
Mz := Dy⋅ a
M :=
2
2
My + Mz
T := TBC
M = 1.118 kN⋅ m
T = 0.250 kN⋅ m
Maximum Shear Stress Theory :
3
c :=
2
π ⋅ τ allow
2
⋅ M +T
2
c = 22.99 mm
do := 2c
do = 45.99 mm
Use
do = 46mm
Ans
x1 := 0 , 0.01 ⋅ a .. a
Az⋅ x1
My1 x1 :=
kN⋅ m
x2 := a , 1.01 ⋅ a .. 2a
( )
( )
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
1
My2 x2 := ⎡⎣Az⋅ x2 − PB⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
Moment (kN-m)
1
My3 x3 := ⎡⎣Az⋅ x3 − PB⋅ x3 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
0.2
0.4
(
)
( )1
M y2( x 2 )
0.5
M y3( x 3 )
M y1 x 1
0
0
0.6
0.8
x1 , x2 , x3
Distance (m)
( )
Mz1 x1 :=
( )
−Ay⋅ x1
( )
Mz2 x2 :=
kN⋅ m
( )
−Ay⋅ x2
1
Mz3 x3 := ⎡⎣−Ay⋅ x3 + PC⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
( )
0.6
0.8
Moment (kN-m)
0
( )
M z2( x 2 ) 0.5
M z3( x 3 )
M z1 x 1
1
0
0.2
0.4
x1 , x2 , x3
Distance (m)
− r ⋅ PB
Mx2 x2 :=
kN⋅ m
( )
( )
Moment (kN-m)
Mx1 x1 := 0
( )
M x2( x 2 )
M x3( x 3 )
M x1 x 1
( )
Mx3 x3 := 0
0
0.5
0
0.2
0.4
x1 , x2 , x3
Distance (m)
0.6
0.8
(
)
Problem 11-39
Solve Prob. 11-38 using the maximum-distortion-energy theory of failure with σallow = 180 MPa.
Given: a := 300mm
r := 50mm
PB := 5kN
PC := 5kN
Solution:
σ allow := 180MPa
L := 3a
Support Reactions :
In x-z plane : +
ΣF z=0;
Az + DZ − PB = 0
(1)
ΣΜD=0; Az⋅ ( 3a) − PB⋅ ( 2a) = 0 (2)
Solving Eqs. (1) and (2):
2
Az := PB
3
Az = 3.3333 kN
Dz := PB − Az
Dz = 1.6667 kN
(Ay + Dy) − PC = 0
(3)
ΣΜD=0; −PC⋅ a + Ay⋅ ( 3a) = 0
(4)
In x-y plane : ΣF y=0;
Solving Eqs. (3) and (4):
1
Ay := PC
3
Ay = 1.6667 kN
Dy := PC − Ay
Dy = 3.3333 kN
Torsion occurs in segment BC :
Critical Section :
( )
T BC := PB ⋅ r
T BC = 0.250 kN⋅ m
Located just to the left of gear C and just to the right of gear B, where
My := Dz⋅ a
Mz := Dy⋅ a
M :=
2
2
My + Mz
T := T BC
Maximum Distortion Energy Theory : Applying Eq. 9-5:
(
) ⎡⎣0.5(σx' + σy')⎤⎦ 2 + τ x'y'2
2
2
σ 2 = 0.5 ( σ x' + σ y') − ⎡⎣0.5 ⋅ ( σ x' + σ y')⎤⎦ + τ x'y'
σ 1 = 0.5 σ x' + σ y' +
where
σ y' := 0
σ x' =
M⋅ c
M⋅ c
4M⋅ c
=
=
4
I
π 4
π⋅c
⋅c
4
τ x'y' =
T⋅ c
T⋅ c
2T ⋅ c
=
=
4
J
π 4
π⋅c
⋅c
2
M = 1.118 kN⋅ m
T = 0.250 kN⋅ m
Let
a' = 0.5σ x'
and b' =
2
2
Then σ 1 = ( a' + b')
(0.5σx')2 + τ x'y'2
2
σ 2 = ( a − 'b')
2
2
2
(2
σ 1⋅ σ 2 = ( a' + b') ⋅ ( a' − b') = a' − b'
2
2
2
2
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = ( a' + b') − a' − b'
) + ( a' − b') 2 = a'2 + 3b'2
Hence σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
2
(0.5σx') + 3 ⎡⎣ (0.5σx')2 + τ x'y'2⎤⎦ = σallow2
2
2
2
2
σ x' + 3τ x'y' = σ allow
2
2
2
⎛ 4M⋅ c ⎞ + 3 ⎛ 2T⋅ c ⎞ = σ
allow
⎜ 4
⎜ 4
⎝ π⋅c ⎠
⎝ π⋅c ⎠
6
c :=
2
16M + 12T
2
2
2
c = 20.05 mm
do := 2c
do = 40.09 mm
π ⋅ σ allow
Use
do = 41mm
Ans
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. 2a
Az⋅ x1
My1 x1 :=
kN⋅ m
1
My2 x2 := ⎡⎣Az⋅ x2 − PB⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
x3 := 2a , 1.01 ⋅ ( 2a) .. 3a
( )
(
)
Moment (kN-m)
1
My3 x3 := ⎡⎣Az⋅ x3 − PB⋅ x3 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
0.2
0.4
(
)
( )1
M y2( x 2 )
0.5
M y3( x 3 )
M y1 x 1
0
0
0.6
0.8
x1 , x2 , x3
Distance (m)
( )
Mz1 x1 :=
( )
−Ay⋅ x1
( )
Mz2 x2 :=
kN⋅ m
( )
−Ay⋅ x2
1
Mz3 x3 := ⎡⎣−Ay⋅ x3 + PC⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
( )
0.6
0.8
Moment (kN-m)
0
( )
M z2( x 2 ) 0.5
M z3( x 3 )
M z1 x 1
1
0
0.2
0.4
x1 , x2 , x3
Distance (m)
− r ⋅ PB
Mx2 x2 :=
kN⋅ m
( )
( )
Moment (kN-m)
Mx1 x1 := 0
( )
M x2( x 2 )
M x3( x 3 )
M x1 x 1
( )
Mx3 x3 := 0
0
0.5
0
0.2
0.4
x1 , x2 , x3
Distance (m)
0.6
0.8
(
)
Problem 11-40
The bearings at A and D exert only y and z components of force on the shaft. If τallow = 60 MPa,
determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the
maximum-shear-stress theory of failure.
Given: L1 := 200mm L2 := 400mm L3 := 350mm τ allow := 60MPa
rB := 50mm
Solution:
rC := 75mm
PB := 3kN
PC := 2kN
L := L1 + L2 + L3
Support Reactions :
ΣFz=0;
In x-z plane : +
Az + DZ − PC = 0
(1)
( )
ΣΜD=0; Az⋅ ( L ) − PC⋅ L 3 = 0 (2)
Solving Eqs. (1) and (2):
L3
Az :=
P
Az = 0.7368 kN
L C
Dz := PB − Az
Dz = 2.2632 kN
(3)
(Ay + Dy) − PB = 0
ΣΜD=0; −PB⋅ ( L 2 + L 3) + Ay⋅ ( L) =(4)
0
In x-y plane : ΣFy=0;
Solving Eqs. (3) and (4):
L2 + L3
Ay :=
PB Ay = 2.3684 kN
L
Dy := PB − Ay
Dy = 0.6316 kN
( )
TBC := PB ⋅ rB
Torsion occurs in segment BC :
TBC = 0.150 kN⋅ m
Critical Section : Located just to right of gear B, where
My := Az⋅ L 1
Mz := Ay⋅ L 1
M :=
2
2
My + Mz
T := TBC
M = 0.496 kN⋅ m
T = 0.150 kN⋅ m
Maximum Shear Stress Theory :
3
c :=
2
π ⋅ τ allow
2
⋅ M +T
2
c = 17.65 mm
do := 2c
do = 35.30 mm
Use
do = 36mm
Ans
x1 := 0 , 0.01 ⋅ L1 .. L1
Az⋅ x1
My1 x1 :=
kN⋅ m
Moment (kN-m)
( )
(
)
(
)
( )
( )
(
0.6
0.8
(
)
x2 := L 1 , 1.01 ⋅ L 1 .. L1 + L2
x3 := L1 + L2 , 1.01 ⋅ L1 + L2 .. L
Az⋅ x2
1
My2 x2 :=
My3 x3 := ⎡⎣Az⋅ x3 − PC⋅ x3 − L 1 − L 2 ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
( )
( )
)
( )0.4
M y2( x 2 )
M y3( x 3 )0.2
M y1 x 1
0
0
0.2
0.4
x1 , x2 , x3
Distance (m)
( )
Ay⋅ x1
Mz1 x1 :=
kN⋅ m
( )
1
Mz2 x2 := ⎡⎣Ay⋅ x2 − PB⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
Mz3 x3 := ⎡⎣Ay⋅ x3 − PB⋅ x3 − L 1 ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
( )
( )0.4
M z2( x 2 )
M z3( x 3 )0.2
M z1 x 1
0
0
0.2
0.4
0.6
0.8
x1 , x2 , x3
Distance (m)
rB⋅ PB
Mx2 x2 :=
kN⋅ m
( )
( )
Moment (kN-m)
Mx1 x1 := 0
( )
Mx3 x3 := 0
( )
M x2( x 2 )0.1
M x3( x 3 )
M x1 x 1
0
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
0.8
(
)
Problem 11-41
The bearings at A and D exert only y and z components of force on the shaft. If τallow = 60 MPa,
determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the
maximum-distortion-energy theory of failure. σallow = 130 MPa.
Given: L1 := 200mm L2 := 400mm
rB := 50mm
Solution:
rC := 75mm
L3 := 350mm
σ allow := 130MPa
PB := 3kN
PC := 2kN
L := L1 + L2 + L3
Support Reactions :
In x-z plane : +
ΣFz=0;
Az + DZ − PC = 0
(1)
( )
ΣΜD=0; Az⋅ ( L ) − PC⋅ L 3 = 0 (2)
Solving Eqs. (1) and (2):
L3
Az :=
P
Az = 0.7368 kN
L C
Dz := PB − Az
Dz = 2.2632 kN
(3)
(Ay + Dy) − PB = 0
ΣΜD=0; −PB⋅ ( L 2 + L 3) + Ay⋅ ( L) =(4)
0
In x-y plane : ΣFy=0;
Solving Eqs. (3) and (4):
L2 + L3
Ay :=
PB Ay = 2.3684 kN
L
Dy := PB − Ay
Torsion occurs in segment BC :
Dy = 0.6316 kN
( )
TBC := PB ⋅ rB
TBC = 0.150 kN⋅ m
Critical Section : Located just to right of gear B, where
My := Az⋅ L 1
Mz := Ay⋅ L 1
M :=
2
2
My + Mz
T := TBC
Maximum Distortion Energy Theory : Applying Eq. 9-5:
(
) ⎡⎣0.5(σx' + σy')⎤⎦ 2 + τ x'y'2
2
2
σ 2 = 0.5 ( σ x' + σ y') − ⎡⎣0.5 ⋅ ( σ x' + σ y')⎤⎦ + τ x'y'
σ 1 = 0.5 σ x' + σ y' +
where
σ y' := 0
σ x' =
M⋅ c
M⋅ c
4M⋅ c
=
=
4
I
π 4
π⋅c
⋅c
4
τ x'y' =
T⋅ c
T⋅ c
2T⋅ c
=
=
4
J
π 4
π⋅c
⋅c
2
M = 0.496 kN⋅ m
T = 0.150 kN⋅ m
Let
a' = 0.5σ x'
and
2
Then σ 1 = ( a' + b')
(0.5σx')2 + τ x'y'2
b' =
2
2
σ 2 = ( a − 'b')
2
2
2
(2
σ 1⋅ σ 2 = ( a' + b') ⋅ ( a' − b') = a' − b'
2
2
2
2
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = ( a' + b') − a' − b'
) + ( a' − b') 2 = a'2 + 3b'2
Hence σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
2
(0.5σx') + 3 ⎡⎣ (0.5σx')2 + τ x'y'2⎤⎦ = σallow2
2
2
2
2
σ x' + 3τ x'y' = σ allow
2
2
2
⎛ 4M⋅ c ⎞ + 3 ⎛ 2T⋅ c ⎞ = σ
allow
⎜ 4
⎜ 4
⎝ π⋅c ⎠
⎝ π⋅c ⎠
6
c :=
2
16M + 12T
2
2
2
c = 17.13 mm
do := 2c
do = 34.25 mm
π ⋅ σ allow
Use
do = 35mm
Ans
x1 := 0 , 0.01 ⋅ L1 .. L1
Az⋅ x1
My1 x1 :=
kN⋅ m
Moment (kN-m)
( )
(
)
(
)
( )
( )
(
0.6
0.8
(
)
x2 := L 1 , 1.01 ⋅ L 1 .. L1 + L2
x3 := L1 + L2 , 1.01 ⋅ L1 + L2 .. L
Az⋅ x2
1
My2 x2 :=
My3 x3 := ⎡⎣Az⋅ x3 − PC⋅ x3 − L 1 − L 2 ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
( )
( )
)
( )0.4
M y2( x 2 )
M y3( x 3 )0.2
M y1 x 1
0
0
0.2
0.4
x1 , x2 , x3
Distance (m)
( )
Ay⋅ x1
Mz1 x1 :=
kN⋅ m
( )
1
Mz2 x2 := ⎡⎣Ay⋅ x2 − PB⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
Mz3 x3 := ⎡⎣Ay⋅ x3 − PB⋅ x3 − L 1 ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
( )
( )0.4
M z2( x 2 )
M z3( x 3 )0.2
M z1 x 1
0
0
0.2
0.4
0.6
0.8
x1 , x2 , x3
Distance (m)
rB⋅ PB
Mx2 x2 :=
kN⋅ m
( )
( )
Moment (kN-m)
Mx1 x1 := 0
( )
Mx3 x3 := 0
( )
M x2( x 2 )0.1
M x3( x 3 )
M x1 x 1
0
0
0.2
0.4
0.6
x1 , x2 , x3
Distance (m)
0.8
(
)
Problem 11-42
The pulleys attached to the shaft are loaded as shown. If the bearings at A and B exert only horizontal
and vertical forces on the shaft, determine the required diameter of the shaft to the nearest mm. using
the maximum-shear-stress theory of failure. τallow = 84 MPa.
Given: P1 := 1250N
P2 := 750N
r := 150mm
L 1 := 0.3m
L 2 := 1.5m
L 3 := 0.6m
τ allow := 84MPa
Solution:
L := L 1 + L 2 + L 3
Support Reactions :
Ps := P1 + P2
In y-z plane :
+
ΣF z=0;
Az + BZ − P1 − P2 = 0
(
(1)
)(
)
ΣΜB=0; Az⋅ L − P1 + P2 ⋅ L 2 + L 3 = 0 (2)
Solving Eqs. (1) and (2):
1
Az :=
P + P2 ⋅ L 2 + L 3
L 1
Az = 1750 N
Bz := P1 + P2 − Az
Bz = 250 N
(
In x-y plane : ΣF x=0;
)(
)
Ax + Bx − P1 − P2 = 0
(
(3)
)( )
ΣΜB=0; Ax⋅ L − P1 + P2 ⋅ L3 = 0
(4)
Solving Eqs. (3) and (4):
1
Ax :=
P + P2 ⋅ L 3
L 1
Ax = 500 N
Bx := P1 + P2 − Ax
Bx = 1500 N
(
)( )
(
)
Torsion occurs in segment DC : T := P1 − P2 ⋅ r
Critical Section :
T = 75 N⋅ m
Located just to the left of point C.
Mx := Bx⋅ L3
Mz := Bz⋅ L3
M :=
2
2
Mx + Mz
M = 912.41 N⋅ m
Maximum Shear Stress Theory :
3
c :=
2
π ⋅ τ allow
2
2
c = 19.07 mm
do := 2c
do = 38.15 mm
⋅ M +T
Use
do = 39mm
Ans
y1 := 0 , 0.01 ⋅ L 1 .. L 1
1
Mz1 y1 := Az⋅ y1 ⋅
N⋅ m
( ) (
(
y2 := L1 , 1.01 ⋅ L1 .. L1 + L2
)
)
(
)
y3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. L
1
Mz2 y2 := ⎡⎣Az⋅ y2 − Ps⋅ y2 − L 1 ⎤⎦ ⋅
N⋅ m
( )
(
)
1
Mz3 y3 := ⎡⎣Az⋅ y3 − Ps⋅ y3 − L 1 ⎤⎦ ⋅
N⋅ m
( )
(
)
( )500
M z2( y 2 )
M z3( y 3 )
Mz (N-m)
M z1 y 1
0
0
1
2
y1 , y2 , y3
Distance (m)
Ax⋅ y1
Mx1 y1 :=
N⋅ m
Ax⋅ y2
Mx2 y2 :=
N⋅ m
1
Mx3 y3 := ⎡⎣Ax⋅ y3 − Ps⋅ y3 − L 1 − L 2 ⎤⎦ ⋅
N⋅ m
( )
Mx (N-m)
( )
( )
(
( )1000
M x2( y 2 )
500
M x3( y 3 )
M x1 y 1
0
0
1
2
y1 , y2 , y3
Distance (m)
T
My2 y2 :=
N⋅ m
( )
( )
My1 y1 := 0
( )
My3 y3 := 0
My (N-m)
100
( )
M y2( y 2 ) 50
M y3( y 3 )
M y1 y 1
0
0
1
y1 , y2 , y3
Distance (m)
2
)
Problem 11-43
The pulleys attached to the shaft are loaded as shown. If the bearings at A and B exert only horizontal
and vertical forces on the shaft, determine the required diameter of the shaft to the nearest mm. Use
the maximum-ditortion-energy theory of failure. σallow = 140 MPa
Given: P1 := 1250N
P2 := 750N
r := 150mm
L1 := 0.3m
L2 := 1.5m
L3 := 0.6m
σ allow := 140MPa
Solution:
L := L1 + L2 + L3
Support Reactions :
Ps := P1 + P2
In y-z plane :
+
ΣFz=0;
Az + BZ − P1 − P2 = 0
(
(1)
)(
)
ΣΜB=0; Az⋅ L − P1 + P2 ⋅ L2 + L3 = 0 (2)
Solving Eqs. (1) and (2):
1
Az :=
P + P2 ⋅ L 2 + L 3
L 1
(
)(
)
Az = 1750 N
Bz := P1 + P2 − Az
In x-y plane : ΣFx=0;
Bz = 250 N
Ax + Bx − P1 − P2 = 0
(
(3)
)( )
ΣΜB=0; Ax⋅ L − P1 + P2 ⋅ L 3 = 0
(4)
Solving Eqs. (3) and (4):
1
Ax :=
P + P2 ⋅ L 3
L 1
Ax = 500 N
Bx := P1 + P2 − Ax
Bx = 1500 N
(
)( )
(
)
Torsion occurs in segment DC : T := P1 − P2 ⋅ r
Critical Section :
T = 75 N⋅ m
Located just to the left of point C.
Mx := Bx⋅ L 3
Mz := Bz⋅ L 3
M :=
2
2
Mx + Mz
M = 912.41 N⋅ m
Maximum Distortion Energy Theory :
Both states of stress will yield the same result.
σ 1 = 0.5σ +
(0.5σ )2 + τ 2
σ 2 = 0.5σ −
(0.5σ )2 + τ 2
Let
a' = 0.5σ
2
and
Then σ 1 = ( a' + b')
b' =
(0.5σ )2 + τ 2
2
2
σ 2 = ( a − 'b')
2
2
2
2
(2
σ 1⋅ σ 2 = ( a' + b') ⋅ ( a' − b') = a' − b'
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = ( a' + b') − a' − b'
) + ( a' − b') 2 = a'2 + 3b'2
2
2
2
Hence σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
2
(0.5σ ) + 3 ⎡⎣ ( 0.5σ) 2 + τ 2⎤⎦ = σ allow2
2
2
2
2
σ + 3τ = σ allow
σ=
τ=
(5)
M⋅ c
M⋅ c
4M⋅ c
=
=
4
I
π 4
π⋅c
⋅c
4
T⋅ c
T⋅ c
2T ⋅ c
=
=
4
J
π 4
π⋅c
⋅c
2
2
2
4M⋅ c ⎞
2
⎛ 2T⋅ c ⎞ = σ
From Eq. (5), ⎛⎜
+ 3⎜
allow
4
4
⎝ π⋅c ⎠
⎝ π⋅c ⎠
6
c :=
2
16M + 12T
2
2
2
c = 20.26 mm
do := 2c
do = 40.52 mm
π ⋅ σ allow
Use
do = 41mm
Ans
y1 := 0 , 0.01 ⋅ L1 .. L1
1
Mz1 y1 := Az⋅ y1 ⋅
N⋅ m
( ) (
(
y2 := L 1 , 1.01 ⋅ L 1 .. L 1 + L 2
)
)
(
)
y3 := L1 + L2 , 1.01 ⋅ L1 + L2 .. L
1
Mz2 y2 := ⎡⎣Az⋅ y2 − Ps⋅ y2 − L1 ⎤⎦ ⋅
N⋅ m
( )
(
)
1
Mz3 y3 := ⎡⎣Az⋅ y3 − Ps⋅ y3 − L1 ⎤⎦ ⋅
N⋅ m
( )
(
)
( )500
M z2( y 2 )
M z3( y 3 )
Mz (N-m)
M z1 y 1
0
0
1
2
y1 , y2 , y3
Distance (m)
Ax⋅ y1
Mx1 y1 :=
N⋅ m
Ax⋅ y2
Mx2 y2 :=
N⋅ m
1
Mx3 y3 := ⎡⎣Ax⋅ y3 − Ps⋅ y3 − L1 − L2 ⎤⎦ ⋅
N⋅ m
( )
Mx (N-m)
( )
( )
(
( )1000
M x2( y 2 )
500
M x3( y 3 )
M x1 y 1
0
0
1
2
y1 , y2 , y3
Distance (m)
T
My2 y2 :=
N⋅ m
( )
( )
My1 y1 := 0
( )
My3 y3 := 0
My (N-m)
100
( )
M y2( y 2 ) 50
M y3( y 3 )
M y1 y 1
0
0
1
y1 , y2 , y3
Distance (m)
2
)
Problem 11-44
The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowable
normal stress for the shaft is σallow = 80 MPa, determine to the nearest millimeter the smallest diameter
of the shaft that will support the loading. Use the maximum-distortion-energy theory of failure.
Given: La := 250mm Lb := 500mm σ allow := 80MPa
rD := 150mm
rC := 100mm
PD := 0.20kN
PC := 0.35kN
θ := 30deg
∆PD := 0.10kN ∆PC := 0.15kN
Solution:
L := La + L b + L a
Support Reactions :
ΣFz=0;
In y-z plane : +
Az + BZ − PC⋅ sin ( θ ) − PD⋅ sin ( θ ) = 0
(
(1)
)
ΣΜA=0; Bz⋅ ( L) − PC⋅ sin ( θ ) ⋅ L − L a − PD⋅ sin ( θ ) ⋅ La = 0
Solving Eqs. (1) and (2):
La ⎞
⎛ L − La
Bz := ⎜ PC⋅
+ PD⋅
sin ( θ )
Bz = 0.15625 kN
L
L⎠
⎝
(
)
Az := PC + PD ⋅ sin ( θ ) − Bz
Az = 0.11875 kN
Ax + Bx − PC⋅ cos ( θ ) + PD⋅ cos ( θ ) = 0
In x-y plane : ΣFx=0;
(
)
ΣΜA=0; PC⋅ cos ( θ ) ⋅ L − L a − PD⋅ cos ( θ ) ⋅ L a − Bx⋅ L = 0
Solving Eqs. (3) and (4):
La ⎞
⎛ L − La
Bx := ⎜ PC⋅
− PD⋅
cos ( θ )
L
L⎠
⎝
(
)
Ax := PC − PD ⋅ cos ( θ ) − Bx
(
Mx := Bz⋅ La
(3)
(4)
Bx = 0.18403 kN
Ax = −0.05413 kN
)
Torsion occurs in segment CD : TCD := ∆PC ⋅ rC
Critical Section :
(2)
TCD = 0.015 kN⋅ m
Located just to the left of gear C, where.
Mz := Bx⋅ La
M :=
2
T := TCD
σ 1 = 0.5 σ x' + σ y' +
(
)
⎡⎣0.5( σ x' + σ y')⎤⎦ + τ x'y'
(
)
⎡⎣0.5⋅ ( σ x' + σ y')⎤⎦ + τ x'y'
σ 2 = 0.5 σ x' + σ y' −
2
M = 0.060354 kN⋅ m
T = 0.015 kN⋅ m
Maximum Distortion Energy Theory : Applying Eq. 9-5:
2
2
Mx + Mz
2
2
where
σ y' := 0
σ x' =
M⋅ c
M⋅ c
4M⋅ c
=
=
4
I
π 4
π⋅c
⋅c
4
T⋅ c
T⋅ c
2T⋅ c
=
=
4
J
π 4
π⋅c
⋅c
2
τ x'y' =
Let
a' = 0.5σ x'
and
2
Then σ 1 = ( a' + b')
(0.5σx')2 + τ x'y'2
b' =
2
2
σ 2 = ( a − 'b')
2
2
2
(2
σ 1⋅ σ 2 = ( a' + b') ⋅ ( a' − b') = a' − b'
2
2
2
2
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = ( a' + b') − a' − b'
) + ( a' − b') 2 = a'2 + 3b'2
Hence σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
(
)
0.5σ x'
2
2
⎡ 0.5σ 2 + τ 2⎤ = σ
( x') x'y' ⎦ allow2
+ 3⎣
2
2
2
σ x' + 3τ x'y' = σ allow
2
2
2
⎛ 4M⋅ c ⎞ + 3 ⎛ 2T⋅ c ⎞ = σ
allow
⎜ 4
⎜ 4
⎝ π⋅c ⎠
⎝ π⋅c ⎠
6
c :=
2
16M + 12T
2
2
2
c = 9.94 mm
do := 2c
do = 19.88 mm
π ⋅ σ allow
Use
do = 20mm
Ans
(
y1 := 0 , 0.01 ⋅ La .. L a
Az⋅ y1
Mx1 y1 :=
kN⋅ m
( )
( )
)
(
y2 := L a , 1.01 ⋅ La .. La + L b
)
(
)
y3 := La + L b , 1.01 ⋅ L a + Lb .. L
1
Mx2 y2 := ⎡⎣Az⋅ y2 − PD⋅ sin ( θ ) ⋅ y2 − La ⎤⎦
kN⋅ m
( )
(
)
1
Mx3 y3 := ⎡⎣Az⋅ y3 − PD⋅ sin ( θ ) ⋅ y3 − La − PC⋅ sin ( θ ) ⋅ y3 − L a − Lb ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
(
)
(
)
( )
M x2( y 2 )
0.02
M x3( y 3 )
M x1 y 1 0.04
0
0
0.2
0.4
0.6
0.8
y1 , y2 , y3
Distance (m)
( )
Ax⋅ y1
Mz1 y1 :=
kN⋅ m
( )
1
Mz2 y2 := ⎡⎣Ax⋅ y2 + PD⋅ cos ( θ ) ⋅ y2 − L a ⎤⎦
kN⋅ m
( )
(
)
1
Mz3 y3 := ⎡⎣Ax⋅ y3 + PD⋅ cos ( θ ) ⋅ y3 − L a − PC⋅ cos ( θ ) ⋅ y3 − L a − Lb ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
( )
M z2( y 2 )
M z3( y 3 )
M z1 y 1
(
)
(
0.05
0
0
0.2
0.4
0.6
0.8
y1 , y2 , y3
Distance (m)
T
My2 y2 :=
kN⋅ m
( )
( )
Moment (kN-m)
My1 y1 := 0
( )
M y2( y 2 )
M y3( y 3 )
( )
My3 y3 := 0
M y1 y 1 0.02
0
0
0.2
0.4
0.6
y1 , y2 , y3
Distance (m)
0.8
)
Problem 11-45
The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowable
shear stress for the shaft is τallow = 35 MPa, determine to the nearest millimeter the smallest diameter of
the shaft that will support the loading. Use the maximum-shear-stress theory of failure.
Given: La := 250mm
Lb := 500mm
τ allow := 35MPa
rD := 150mm
rC := 100mm
θ := 30deg
PD := 0.20kN
PC := 0.35kN
∆PD := 0.10kN ∆PC := 0.15kN
Solution:
L := La + L b + L a
Support Reactions :
Az + BZ − PC⋅ sin ( θ ) − PD⋅ sin ( θ ) = 0
ΣFz=0;
In y-z plane : +
(
(1)
)
ΣΜA=0; Bz⋅ ( L) − PC⋅ sin ( θ ) ⋅ L − L a − PD⋅ sin ( θ ) ⋅ La = 0
Solving Eqs. (1) and (2):
La ⎞
⎛ L − La
Bz := ⎜ PC⋅
+ PD⋅
sin ( θ )
Bz = 0.15625 kN
L
L
⎝
(
⎠
)
Az := PC + PD ⋅ sin ( θ ) − Bz
Az = 0.11875 kN
Ax + Bx − PC⋅ cos ( θ ) + PD⋅ cos ( θ ) = 0
In x-y plane : ΣFx=0;
(
(3)
)
ΣΜA=0; PC⋅ cos ( θ ) ⋅ L − L a − PD⋅ cos ( θ ) ⋅ L a − Bx⋅ L = 0
Solving Eqs. (3) and (4):
La ⎞
⎛ L − La
Bx := ⎜ PC⋅
− PD⋅
cos ( θ )
L
L
⎝
(
⎠
)
Ax := PC − PD ⋅ cos ( θ ) − Bx
(
(4)
Bx = 0.18403 kN
Ax = −0.05413 kN
)
Torsion occurs in segment CD : TCD := ∆PC ⋅ rC
Critical Section :
(2)
TCD = 0.015 kN⋅ m
Located just to the left of gear C, where.
Mx := Bz⋅ La
Mz := Bx⋅ La
M :=
2
2
Mx + Mz
T := TCD
M = 0.060354 kN⋅ m
T = 0.015 kN⋅ m
Maximum Shear Stress Theory :
3
c :=
2
π ⋅ τ allow
2
2
c = 10.42 mm
do := 2c
do = 20.84 mm
⋅ M +T
Use
do = 21mm
Ans
(
y1 := 0 , 0.01 ⋅ La .. L a
Az⋅ y1
Mx1 y1 :=
kN⋅ m
( )
( )
)
(
y2 := L a , 1.01 ⋅ La .. La + L b
)
(
)
y3 := La + L b , 1.01 ⋅ L a + Lb .. L
1
Mx2 y2 := ⎡⎣Az⋅ y2 − PD⋅ sin ( θ ) ⋅ y2 − La ⎤⎦
kN⋅ m
( )
(
)
1
Mx3 y3 := ⎡⎣Az⋅ y3 − PD⋅ sin ( θ ) ⋅ y3 − La − PC⋅ sin ( θ ) ⋅ y3 − L a − Lb ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
(
)
(
)
( )
M x2( y 2 )
0.02
M x3( y 3 )
M x1 y 1 0.04
0
0
0.2
0.4
0.6
0.8
y1 , y2 , y3
Distance (m)
( )
Ax⋅ y1
Mz1 y1 :=
kN⋅ m
( )
1
Mz2 y2 := ⎡⎣Ax⋅ y2 + PD⋅ cos ( θ ) ⋅ y2 − L a ⎤⎦
kN⋅ m
( )
(
)
1
Mz3 y3 := ⎡⎣Ax⋅ y3 + PD⋅ cos ( θ ) ⋅ y3 − L a − PC⋅ cos ( θ ) ⋅ y3 − L a − Lb ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
( )
M z2( y 2 )
M z3( y 3 )
M z1 y 1
(
)
(
0.05
0
0
0.2
0.4
0.6
0.8
y1 , y2 , y3
Distance (m)
T
My2 y2 :=
kN⋅ m
( )
( )
Moment (kN-m)
My1 y1 := 0
( )
M y2( y 2 )
M y3( y 3 )
( )
My3 y3 := 0
M y1 y 1 0.02
0
0
0.2
0.4
0.6
y1 , y2 , y3
Distance (m)
0.8
)
Problem 11-46
The shaft is supported by bearings at A and B that exert force components only in the x and z
directions on the shaft. If the allowable normal stress for the shaft is σallow = 105 MPa, determine to
the nearest mm the smallest diameter of the shaft that will support the gear loading. Use the
maximum-distortion-energy theory of failure.
Given: L a := 200mm
L b := 100mm
PC := 1kN
r := 100mm
PD := 0.25kN
PE := 1.25kN
σ allow := 105MPa
Solution:
L := 3La + L b
Support Reactions :
In y-z plane : +
ΣF z=0;
Az + BZ − PD = 0
(
)
(1)
(
)
ΣΜB=0; Az⋅ 2La + L b − PD⋅ La + L b = 0
Solving Eqs. (1) and (2):
La + Lb
Az :=
P
2L a + Lb D
Bz := PD − Az
(2)
Az = 150 N
Bz = 100 N
(Ax + Bx) + PC − PE = 0
ΣΜB=0; PC⋅ L + Ax⋅ ( 2La + L b) − PE⋅ ( L b) = 0
In x-y plane : ΣF x=0;
(3)
(4)
Solving Eqs. (3) and (4):
1
Ax :=
P ⋅ L − PC⋅ L
2La + L b E b
Bx := PE − PC − Ax
(
)
Ax = −1150 N
Bx = 1400 N
( )
T DE := ( PE ) ⋅ r
Torsion occurs in segment DC : T DC := PC ⋅ r
in segment DE :
Critical Section :
Mx := PC⋅ L a
T DC = 100 N⋅ m
T DE = 125 N⋅ m
Located at support A (within segment DC).
Mz := 0
M :=
2
2
Mx + Mz
T := T DC
M = 200 N⋅ m
T = 100 N⋅ m
Maximum Distortion Energy Theory : Applying Eq. 9-5:
(
) ⎡⎣0.5(σx' + σy')⎤⎦ 2 + τ x'y'2
2
2
σ 2 = 0.5 ( σ x' + σ y') − ⎡0.5 ⋅ ( σ x' + σ y')⎤ + τ x'y'
⎣
⎦
σ 1 = 0.5 σ x' + σ y' +
where
σ y' := 0
σ x' =
M⋅ c
M⋅ c
=
I
π
4
T⋅ c
J
=
T⋅ c
π
2
Let
4
2T ⋅ c
=
4
π⋅c
⋅c
a' = 0.5σ x'
and b' =
2
2
Then σ 1 = ( a' + b')
4
π⋅c
⋅c
4
τ x'y' =
4M⋅ c
=
(0.5σx')2 + τ x'y'2
2
σ 2 = ( a − 'b')
2
2
2
(2
σ 1⋅ σ 2 = ( a' + b') ⋅ ( a' − b') = a' − b'
2
2
2
2
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = ( a' + b') − a' − b'
) + ( a' − b') 2 = a'2 + 3b'2
Hence σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
(
)
0.5σ x'
2
+ 3⎡
⎣ (
)
0.5σ x'
2
2
+ τ x'y' ⎤ = σ allow
2
2
⎦
2
2
2
σ x' + 3τ x'y' = σ allow
2
2
2
⎛ 4M⋅ c ⎞ + 3 ⎛ 2T⋅ c ⎞ = σ
allow
⎜ 4
⎜ 4
⎝ π⋅c ⎠
⎝ π⋅c ⎠
6
c :=
2
16M + 12T
2
2
2
c = 13.83 mm
do := 2c
do = 27.65 mm
π ⋅ σ allow
Use
do = 28mm
Ans
y1 := 0 , 0.01 ⋅ L a .. La
y3 := 2L a , 1.01⋅ 2La .. 3L a
( )
y4 := 3L a , 1.01⋅ 3La .. L
( )
0⋅ y1
Mz1 y1 :=
N⋅ m
( )
( )
y2 := La , 1.01 ⋅ L a .. 2L a
1
Mz2 y2 := ⎡Az⋅ y2 − L a ⎤ ⋅
⎣
⎦ N⋅ m
( )
(
)
1
Mz3 y3 := ⎡Az⋅ y3 − L a − PD⋅ y3 − 2La ⎤ ⋅
⎣
⎦ N⋅ m
( )
(
)
(
)
1
Mz4 y4 := ⎡Az⋅ y4 − L a − PD⋅ y4 − 2La ⎤ ⋅
⎣
⎦ N⋅ m
Moment (N-m)
( )
(
)
(
)
( )
M z2( y 2 )
20
M z3( y 3 )
M z4( y 4 )
M z1 y 1
0
0
0.2
0.4
0.6
y1 , y2 , y3 , y4
Distance (m)
( )
Mx1 y1 :=
( ) M
PC⋅ y1
1
x2 y2 := ⎡⎣PC⋅ y2 + Ax⋅ y2 − L a ⎤⎦ ⋅ N⋅ m
N⋅ m
( )
( )
(
)
1
Mx3 y3 := ⎡PC⋅ y3 + Ax⋅ y3 − L a ⎤ ⋅
⎣
⎦ N⋅ m
( )
( )
(
)
1
Mx4 y4 := ⎡PC⋅ y4 + Ax⋅ y4 − L a − PE ⋅ y4 − 3L a ⎤ ⋅
⎣
⎦ N⋅ m
Moment (N-m)
( )
( )
(
)
(
200
( )
M x2( y 2 )
M x3( y 3 )100
M x4( y 4 )
M x1 y 1
0
0
0.2
0.4
y1 , y2 , y3 , y4
Distance (m)
0.6
)
− r ⋅ PC
My1 y1 :=
N⋅ m
− r ⋅ PC
My2 y2 :=
N⋅ m
( )
( )
1
My3 y3 := −r⋅ PC − r⋅ PD ⋅
N⋅ m
( ) (
)
1
My4 y4 := −r⋅ PC − r⋅ PD + r⋅ PE ⋅
N⋅ m
( ) (
)
Moment (N-m)
0
( )
M y2( y 2 ) 50
M y3( y 3 )
M y4( y 4 ) 100
M y1 y 1
150
0
0.2
0.4
y1 , y2 , y3 , y4
Distance (m)
0.6
Problem 11-47
The shaft is supported by bearings at A and B that exert force components only in the x and z
directions on the shaft. If the allowable normal stress for the shaft is σallow = 105 MPa, determine to
the nearest mm the smallest diameter of the shaft that will support the gear loading. Use the
maximum-shear-stress theory of failure with τallow = 42 MPa.
Given: L a := 200mm
L b := 100mm
PC := 1kN
r := 100mm
PD := 0.25kN
PE := 1.25kN
τ allow := 42MPa
L := 3La + L b
Solution:
Support Reactions :
In y-z plane :
+
ΣF z=0;
Az + BZ − PD = 0
(
(1)
)
(
)
ΣΜB=0; Az⋅ 2La + L b − PD⋅ La + L b = 0
Solving Eqs. (1) and (2):
La + Lb
Az :=
P
2L a + Lb D
Bz := PD − Az
(2)
Az = 150 N
Bz = 100 N
(Ax + Bx) + PC − PE = 0
ΣΜB=0; PC⋅ L + Ax⋅ ( 2La + L b) − PE⋅ ( L b) = 0
In x-y plane : ΣF x=0;
(3)
(4)
Solving Eqs. (3) and (4):
1
Ax :=
P ⋅ L − PC⋅ L
2La + L b E b
Bx := PE − PC − Ax
(
)
Ax = −1150 N
Bx = 1400 N
( )
T DE := ( PE ) ⋅ r
Torsion occurs in segment DC : T DC := PC ⋅ r
in segment DE :
Critical Section :
T DC = 100 N⋅ m
T DE = 125 N⋅ m
Located at support A (within segment DC).
Mx := PC⋅ L a
Mz := 0
M :=
2
2
Mx + Mz
T := T DC
M = 200.00 N⋅ m
T = 100.00 N⋅ m
Maximum Shear Stress Theory :
3
c :=
2
π ⋅ τ allow
2
⋅ M +T
2
c = 15.0 mm
do := 2c
do = 30.0 mm
Use
do = 31mm
Ans
y1 := 0 , 0.01 ⋅ L a .. La
y3 := 2L a , 1.01⋅ 2La .. 3L a
( )
y4 := 3L a , 1.01⋅ 3La .. L
( )
0⋅ y1
Mz1 y1 :=
N⋅ m
( )
( )
y2 := La , 1.01 ⋅ L a .. 2L a
1
Mz2 y2 := ⎡⎣Az⋅ y2 − L a ⎤⎦ ⋅
N⋅ m
( )
(
)
1
Mz3 y3 := ⎡⎣Az⋅ y3 − L a − PD⋅ y3 − 2La ⎤⎦ ⋅
N⋅ m
( )
(
)
(
)
1
Mz4 y4 := ⎡⎣Az⋅ y4 − L a − PD⋅ y4 − 2La ⎤⎦ ⋅
N⋅ m
Moment (N-m)
( )
(
)
(
)
( )
M z2( y 2 )
20
M z3( y 3 )
M z4( y 4 )
M z1 y 1
0
0
0.2
0.4
0.6
y1 , y2 , y3 , y4
Distance (m)
( )
Mx1 y1 :=
( ) M
PC⋅ y1
1
x2 y2 := ⎡⎣PC⋅ y2 + Ax⋅ y2 − L a ⎤⎦ ⋅ N⋅ m
N⋅ m
( )
( )
(
)
1
Mx3 y3 := ⎡⎣PC⋅ y3 + Ax⋅ y3 − L a ⎤⎦ ⋅
N⋅ m
( )
( )
(
)
1
Mx4 y4 := ⎡⎣PC⋅ y4 + Ax⋅ y4 − L a − PE ⋅ y4 − 3L a ⎤⎦ ⋅
N⋅ m
Moment (N-m)
( )
( )
(
)
(
200
( )
M x2( y 2 )
M x3( y 3 )100
M x4( y 4 )
M x1 y 1
0
0
0.2
0.4
y1 , y2 , y3 , y4
Distance (m)
0.6
)
− r ⋅ PC
My1 y1 :=
N⋅ m
− r ⋅ PC
My2 y2 :=
N⋅ m
( )
( )
1
My3 y3 := −r⋅ PC − r⋅ PD ⋅
N⋅ m
( ) (
)
1
My4 y4 := −r⋅ PC − r⋅ PD + r⋅ PE ⋅
N⋅ m
( ) (
)
Moment (N-m)
0
( )
M y2( y 2 ) 50
M y3( y 3 )
M y4( y 4 ) 100
M y1 y 1
150
0
0.2
0.4
y1 , y2 , y3 , y4
Distance (m)
0.6
Problem 11-48
The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert
only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then
determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use
the maximum-shear-stress theory of failure with τallow = 60 MPa.
Given: L1 := 150mm
L2 := 250mm
L3 := 100mm
τ allow := 60MPa
rD := 100mm rC := 75mm
.
Solution: L := L1 + L2 + L3
rC' := 50mm
PD := 1.5kN
Equilibrium Torque at C :
ΣΜx=0;
−PD⋅ rD + FC⋅ rC = 0
FC :=
Thus,
rD
rC
⋅ PD
FC = 2.00 kN
TC := FC⋅ rC'
TC = 0.100 kN⋅ m
Ans
Support Reactions :
ΣFz=0;
In x-z plane : +
Az + Bz − PD = 0
(
(1)
)
ΣΜB=0; Az⋅ L − L 1 − PD⋅ L = 0 (2)
Solving Eqs. (1) and (2):
L
Az :=
P
L − L1 D
Bz := PD − Az
Az = 2.1429 kN
Bz = −0.6429 kN
(Ay + By) − FC = 0
ΣΜB=0; Ay⋅ ( L − L1) − FC⋅ L 3 = 0
In x-y plane : ΣFy=0;
(3)
(4)
Solving Eqs. (3) and (4):
L3
Ay :=
F
Ay = 0.5714 kN
L − L1 C
By := FC − Ay
By = 1.4286 kN
Torsion occurs in segment D-C : TDC := PD⋅ rD
TDC = 0.150 kN⋅ m
Critical Section : Located just to right of gear A, where
My := PD⋅ L 1
Mz := 0
M :=
2
2
My + Mz
T := TDC
M = 0.225 kN⋅ m
T = 0.150 kN⋅ m
Maximum Shear Stress Theory :
3
c :=
2
π ⋅ τ allow
2
⋅ M +T
2
c = 14.21 mm
do := 2c
do = 28.42 mm
Use
do = 29mm
Ans
(
x1 := 0 , 0.01 ⋅ L1 .. L1
PD⋅ x1
My1 x1 :=
kN⋅ m
x2 := L 1 , 1.01 ⋅ L 1 .. L1 + L2
( )
( )
)
(
)
(
)
x3 := L1 + L2 , 1.01 ⋅ L1 + L2 .. L
1
My2 x2 := ⎡⎣PD⋅ x2 − Az⋅ x2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
My3 x3 := ⎡⎣PD⋅ x3 − Az⋅ x3 − L1 ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
( )
(
)
( )
0.2
M y2( x 2 )
M y3( x 3 )
M y1 x 1
0
0
0.1
0.2
0.3
0.4
x1 , x2 , x3
Distance (m)
( )
Mz1 x1 := 0
1
Mz2 x2 := ⎡⎣Ay⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
Mz3 x3 := ⎡⎣Ay⋅ x3 − L 1 − FC⋅ x3 − L 1 − L 2 ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
(
)
(
0.3
0.4
( )
M z2( x 2 )
0.1
M z3( x 3 )
M z1 x 1 0.2
0
0
0.1
0.2
x1 , x2 , x3
Distance (m)
T
Mx1 x1 :=
kN⋅ m
Moment (kN-m)
( )
T
Mx2 x2 :=
kN⋅ m
( )
( )
Mx3 x3 := 0
( )0.2
M x2( x 2 )
M x3( x 3 )
M x1 x 1
0
0
0.1
0.2
0.3
x1 , x2 , x3
Distance (m)
0.4
)
Problem 11-49
The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert
only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then
determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use
the maximum-distortion-energy theory of failure with σallow = 80 MPa.
Given: L 1 := 150mm
L 2 := 250mm
L 3 := 100mm
σ allow := 80MPa
rD := 100mm rC := 75mm
.
Solution: L := L 1 + L 2 + L 3
rC' := 50mm
PD := 1.5kN
Equilibrium Torque at C :
ΣΜx=0;
−PD⋅ rD + FC⋅ rC = 0
FC :=
Thus,
rD
rC
⋅ PD
T C := FC⋅ rC'
FC = 2.00 kN
T C = 0.100 kN⋅ m
Ans
Support Reactions :
In x-z plane : +
ΣF z=0;
Az + Bz − PD = 0
(
(1)
)
ΣΜB=0; Az⋅ L − L1 − PD⋅ L = 0 (2)
Solving Eqs. (1) and (2):
L
Az :=
P
L − L1 D
Bz := PD − Az
Az = 2.1429 kN
Bz = −0.6429 kN
(Ay + By) − FC = 0
ΣΜB=0; Ay⋅ ( L − L 1) − FC⋅ L3 = 0
In x-y plane : ΣF y=0;
(3)
(4)
Solving Eqs. (3) and (4):
L3
Ay :=
F
Ay = 0.5714 kN
L − L1 C
By := FC − Ay
By = 1.4286 kN
Torsion occurs in segment D-C : T DC := PD⋅ rD
T DC = 0.150 kN⋅ m
Critical Section : Located just to right of gear A, where
My := PD⋅ L1
Mz := 0
M :=
2
2
My + Mz
T := T DC
Maximum Distortion Energy Theory : Applying Eq. 9-5:
(
) ⎡⎣0.5(σx' + σy')⎤⎦ 2 + τ x'y'2
2
2
σ 2 = 0.5 ( σ x' + σ y') − ⎡⎣0.5 ⋅ ( σ x' + σ y')⎤⎦ + τ x'y'
σ 1 = 0.5 σ x' + σ y' +
M = 0.225 kN⋅ m
T = 0.150 kN⋅ m
where
σ y' := 0
σ x' =
M⋅ c
M⋅ c
4M⋅ c
=
=
4
I
π 4
π⋅c
⋅c
4
τ x'y' =
Let
T⋅ c
T⋅ c
2T ⋅ c
=
=
4
J
π 4
π⋅c
⋅c
2
a' = 0.5σ x'
and b' =
2
2
Then σ 1 = ( a' + b')
(0.5σx')2 + τ x'y'2
2
σ 2 = ( a − 'b')
2
2
2
(2
σ 1⋅ σ 2 = ( a' + b') ⋅ ( a' − b') = a' − b'
2
2
2
2
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = ( a' + b') − a' − b'
) + ( a' − b') 2 = a'2 + 3b'2
Hence σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
2
(0.5σx') + 3 ⎡⎣ (0.5σx')2 + τ x'y'2⎤⎦ = σallow2
2
2
2
2
σ x' + 3τ x'y' = σ allow
2
2
2
⎛ 4M⋅ c ⎞ + 3 ⎛ 2T⋅ c ⎞ = σ
allow
⎜ 4
⎜ 4
⎝ π⋅c ⎠
⎝ π⋅c ⎠
6
c :=
2
16M + 12T
2
2
c = 16.05 mm
2
π ⋅ σ allow
do := 2c
x1 := 0 , 0.01 ⋅ L 1 .. L 1
( )
My1 x1 :=
( )
PD⋅ x1
kN⋅ m
do = 32.10 mm
(
)
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
Use
(
)
do = 33mm
Ans
(
)
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. L
1
My2 x2 := ⎡⎣PD⋅ x2 − Az⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
My3 x3 := ⎡⎣PD⋅ x3 − Az⋅ x3 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
Moment (kN-m)
( )
0.2
M y2( x 2 )
M y3( x 3 )
M y1 x 1
0
0
0.1
0.2
0.3
0.4
x1 , x2 , x3
Distance (m)
( )
Mz1 x1 := 0
1
Mz2 x2 := ⎡⎣Ay⋅ x2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
Mz3 x3 := ⎡⎣Ay⋅ x3 − L1 − FC⋅ x3 − L1 − L2 ⎤⎦ ⋅
kN⋅ m
Moment (kN-m)
( )
(
)
(
0.3
0.4
( )
M z2( x 2 )
0.1
M z3( x 3 )
M z1 x 1 0.2
0
0
0.1
0.2
x1 , x2 , x3
Distance (m)
T
Mx1 x1 :=
kN⋅ m
Moment (kN-m)
( )
T
Mx2 x2 :=
kN⋅ m
( )
( )
Mx3 x3 := 0
( )0.2
M x2( x 2 )
M x3( x 3 )
M x1 x 1
0
0
0.1
0.2
0.3
x1 , x2 , x3
Distance (m)
0.4
)
Problem 11-50
Draw the shear and moment diagrams for the shaft, and then determine its required diameter to the
nearest millimeter if σallow = 140 MPa and τallow = 80 MPa. The bearings at A and B exert only vertical
reactions on the shaft.
Given: L 1 := 125mm
L 2 := 600mm
L 3 := 75mm
P1 := 0.8kN
P2 := 1.5kN
τ allow := 80MPa
σ allow := 140MPa
Solution: L := L 1 + L 2 + L 3
Support Reactions :
+
ΣF z=0;
A + B − P1 − P2 = 0
(
(1)
)
( )
ΣΜB=0; A⋅ L − P1⋅ L − L 1 − P2⋅ L 3 = 0
(2)
Solving Eqs. (1) and (2):
A :=
1
⋅ ⎡P ⋅ L − L 1 + P2⋅ L 3 ⎤⎦
L ⎣ 1
(
)
( )
A = 0.815625 kN
B := P1 + P2 − A
B = 1.484375 kN
Maximum Moment and Shear:
Vmax := B
Mmax := B⋅ L3
Section Property :
Vmax = 1.484375 kN
Mmax = 0.111328 kN⋅ m
I=
π ⋅ do
4
64
I
S=
0.5do
3
S=
π ⋅ do
32
Bending Stress:
Sreq'd =
3
Mmax
π ⋅ do
σ allow
32
=
Mmax
σ allow
3
32Mmax
do :=
do = 20.1 mm
π σ allow
Use
2
Shear Check :
τ max :=
Ao :=
Vmax⋅ Qmax
I ⋅ do
π ⋅ do
4
4
I :=
π ⋅ do
64
⎛ 2⋅ do ⎞ Ao
Qmax := ⎜
⋅
⎝ 3π ⎠ 2
τ max = 5.714 MPa
< τ allow =80 MPa
(O.K.!)
do := 21mm
Ans
x1 := 0 , 0.01 ⋅ L 1 .. L 1
(
)
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
(
A
V1 x1 :=
kN
1
V2 x2 := A − P1 ⋅
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 − P1⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
( ) (
( )
( )
)
)
(
1
V3 x3 := A − P1 − P2 ⋅
kN
(
( ) (
)
)
1
M3 x3 := ⎡⎣A⋅ x3 − P1⋅ x3 − L 1 − P2⋅ x3 − L 1 − L 2 ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
Shear (kN)
1
( )
V2 ( x 2 )
V3 ( x 3 ) 1
V1 x 1
0
2
0
0.2
0.4
0.6
x1 , x2 , x3
Moment (kN-m)
Distance (m)
( )
M2( x2)
M3( x3)
0.05
M 1 x 1 0.1
0
0
0.2
0.4
x1 , x2 , x3
Distance (m)
)
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. L
0.6
)
Problem 11-51
The cantilevered beam has a circular cross section. If it supports a force P at its end, determine its
radius y as a function of x so that it is subjected to a constant maximum bending stress σallow
throughout its length.
Problem 11-52
The simply supported beam is made of timber that has an allowable bending stress of σallow = 8 MPa
and an allowable shear stress of τallow = 750 kPa. Determine its dimensions if it is to be rectangular and
have a height-to-width ratio of h/b = 1.25.
Given: σ allow := 8MPa
a := 3m
kN
wo := 0.3
m
τ allow := 0.75MPa
h = 1.25 ⋅ b
Solution: L := 2a
Support Reactions :
+
ΣF y=0; A + B − 0.5 ⋅ wo⋅ a = 0
(1)
(0.5⋅ wo⋅ a)⋅ ⎛⎜⎝ 3 ⎞⎠ − B⋅ L = 0
2a
ΣΜA=0;
A :=
Solving Eqs. (1) and (2):
(2)
wo⋅ a
3
B :=
Maximum Moment and Shear:
Mmax occurs at x', where V(x')=0.
wo x'
wo x'
V = A − x'⋅
⋅
0 = A − x'⋅
⋅
2 a
2 a
⎛ 2a ⎞ ⋅ wo⋅ a
⎜w
⎝ o⎠ 3
x' =
Vmax := max ( A , B)
x' :=
2
⋅a
3
Vmax = 0.300 kN
x' ⎞ ⎛ x' ⎞
⎛
Mmax := A⋅ x' − ⎜ 0.5 ⋅ x'⋅ wo⋅
⋅⎜
a
3
⎝
wo⋅ a
6
Mmax = 0.4899 kN⋅ m
⎠⎝ ⎠
3
b⋅ h
12
I
Sx =
0.5h
I=
Section Property :
2
b⋅ ( 1.25 ⋅ b)
Sx =
6
b⋅ h
Sx =
6
2
25⋅ b
Sx =
96
3
Bending Stress:
Sreq'd =
3
Mmax
25⋅ b
=
96
σ allow
Mmax
σ allow
3
b :=
96Mmax
25σ allow
h := 1.25 ⋅ b
3
Shear Check :
τ max :=
I :=
b⋅ h
12
Vmax⋅ Qmax
I⋅ b
Qmax := ( 0.5b⋅ h) ⋅ 0.25 h
τ max = 0.094 MPa
< τ allow =0.75 MPa
(O.K.!)
b = 61.7 mm
Ans
h = 77.2 mm
Ans
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. L
wo ⎛ x1 ⎞ ⎤ 1
⎡
V1 x1 := ⎢A −
⋅⎜
⋅ x ⎥⋅
2 ⎝ a ⎠ 1⎦ kN
⎣
( )
1
V2 x2 := A − 0.5 ⋅ wo⋅ a ⋅
kN
( ) (
)
wo ⎛ x1 ⎞
x1⎤
⎡
1
M1 x1 := ⎢A⋅ x1 −
⋅⎜
⋅ x1⋅ ⎥ ⋅
2 ⎝ a⎠
3 ⎦ kN⋅ m
⎣
( )
⎡
⎣
( )
M2 x2 := ⎢A⋅ x2 −
wo⋅ a
2⋅ a ⎞⎤
1
⎛
⋅ ⎜ x2 −
⎥⋅
2 ⎝
3 ⎠⎦ kN⋅ m
Shear (kN)
0.5
( )
V2 ( x 2 )
V1 x 1
0
0.5
0
1
2
3
4
5
6
x1 , x2
Moment (kN-m)
Distance (m)
0.5
( )
M2( x2)
M1 x1
0
0
1
2
3
x1 , x2
Distane (m)
4
5
6
Problem 11-53
The beam is made in the shape of a frustum that has a diameter of 0.3 m at A and a diameter of 0.6 m
at B. If it supports a couple moment of 12 kN·m at its end, determine the absolute maximum bending
stress in the beam and specify its location x.
Given: M := 12kN⋅ m
L := 1.8m
do := 0.3m
d1 := 0.6m
Solution:
Section Property :
r − δr
x
=
π⋅r
I=
4
δr
r=
L
4
d1 − do
δ r :=
2
δ r ⋅ ( L + x)
L
π⋅r
S=
4
I
S=
r
3
3
S=
π ⋅ δ r ⋅ ( L + x)
4⋅ L
3
3
Bendiug Stress :
M
σ=
S
4⋅ L
σ = ( M) ⋅
3
3
π ⋅ δ r ⋅ ( L + x)
3
(1)
Since σ is a decreasing function, the maximum bending stress occurs at
Substituting x=0 into Eq. (1):
σ max := ( M) ⋅
4⋅ L
3
3
π ⋅ δ r ⋅ ( L + x)
3
σ max = 4.527 MPa
Ans
x := 0
Ans
Problem 11-54
Select the lightest-weight steel wide-flange overhanging beam from Appendix B that will safely support
the loading. Assume the support at A is a pin and the support at B is a roller. The allowable bending
stress is σallow = 168 MPa and the allowable shear stress is τallow = 100 MPa.
Given:
σ allow := 168MPa
P := 10kN
τ allow := 100MPa
L 1 := 2.4m
L 2 := 0.6m
L 3 := 1.2m
L := L 1 + L 2 + L 3
Solution:
Support Reactions :
+
ΣF y=0;
A + B − 2P = 0
(1)
(
)
ΣΜB=0; A⋅ L1 + P⋅ L 2 + P⋅ L 2 + L 3 = 0
Solving Eqs. (1) and (2):
2L 2 + L 3
A := −
P
L1
B := 2P − A
(2)
A = −10 kN
B = 30 kN
Maximum Moment and Shear:
Vmax := A + B
Mmax := A⋅ L 1
Vmax = 20 kN
Mmax = −24 kN⋅ m
Bending Stress:
Sreq'd :=
Mmax
3
Sreq'd = 142857.14 mm
σ allow
Select W 250x18 :
Shear Stress :
d := 251mm
tw := 4.83mm
Provide a shear stress check.
τ max :=
Hence,
( 3) mm3
Sx := 179⋅ 10
Vmax
tw⋅ d
Use W 250x18
τ max = 16.5 MPa
< τ allow =100 MPa
Ans
(O.K.!)
x1 := 0 , 0.01 ⋅ L 1 .. L 1
(
)
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
(
)
(
)
A
V1 x1 :=
kN
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. L
1
1
V2 x2 := ( A + B) ⋅
V3 x3 := ( A + B − P) ⋅
kN
kN
A⋅ x1
M1 x1 :=
kN⋅ m
1
M2 x2 := ⎡⎣A⋅ x2 + B⋅ x2 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣⎡⎣A⋅ x3 + B⋅ x3 − L1 ⎤⎦ − P⋅ x3 − L1 − L2 ⎤⎦ ⋅
kN⋅ m
Shear (kN)
( )
(
)
(
)
( )
V2 ( x 2 )
V3 ( x 3 ) 0
V1 x 1 20
0
1
2
3
4
3
4
x1 , x2 , x3
Moment (kN-m)
Distance (m)
( ) 0
M2( x2)
M3( x3)
20
M1 x1
0
1
2
x1 , x2 , x3
Distance (m)
Problem 11-55
The bearings at A and B exert only x and z components of force on the steel shaft. Determine the
shaft's diameter to the nearest millimeter so that it can resist the loadings of the gears without
exceeding an allowable shear stress of τallow = 80 MPa. Use the maximum-shear-stress theory of
failure.
Given: PC := 7.5kN
PD := 5kN
rC := 50mm
L 1 := 150mm
L 2 := 350mm
τ allow := 80MPa
Solution:
L 3 := 250mm
rD := 75mm
L := L 1 + L 2 + L 3
Support Reactions :
In y-z plane : +
ΣF z=0;
Az + BZ − PC = 0
(1)
ΣΜB=0; Az⋅ L − PC⋅ L 3 = 0
(2)
Solving Eqs. (1) and (2):
L3
Az :=
⋅P
Az = 2.5 kN
L C
Bz := PC − Az
In x-y plane :
ΣF x=0;
Bz = 5 kN
Ax + Bx − PD = 0
(
(3)
)
ΣΜB=0; Ax⋅ L − PD⋅ L − L 1 = 0
Solving Eqs. (3) and (4):
L − L1
Ax :=
⋅ PD
L
Ax = 4 kN
Bx := PD − Ax
Bx = 1 kN
Torsion occurs in segment DC : T := PC⋅ rC
Critical Section :
(4)
T = 0.375 kN⋅ m
Located just to the left of gear C.
Mx := Bx⋅ L3
Mz := Bz⋅ L3
M :=
2
2
Mx + Mz
M = 1.27475 kN⋅ m
Maximum Shear Stress Theory :
3
c :=
2
π ⋅ τ allow
2
2
c = 21.95 mm
do := 2c
do = 43.90 mm
⋅ M +T
Use
do = 44mm
Ans
y1 := 0 , 0.01 ⋅ L 1 .. L 1
y2 := L1 , 1.01 ⋅ L1 .. L1 + L2
Az⋅ y1
Mz1 y1 :=
kN⋅ m
Az⋅ y2
Mz2 y2 :=
kN⋅ m
( )
(
)
(
)
y3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. L
1
Mz3 y3 := ⎡⎣Az⋅ y3 − PC⋅ y3 − L 1 − L 2 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
( )
1
M z2( y 2 )
M z3( y 3 )
Mz (kN-m)
M z1 y 1
0
0
0.5
y1 , y2 , y3
Distance (m)
Ax⋅ y1
Mx1 y1 :=
kN⋅ m
1
Mx2 y2 := ⎡⎣Ax⋅ y2 − PD⋅ y2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
Mx3 y3 := ⎡⎣Ax⋅ y3 − PD⋅ y3 − L1 ⎤⎦ ⋅
kN⋅ m
Mx (kN-m)
( )
(
( )
M x2( y 2 )0.5
M x3( y 3 )
M x1 y 1
0
0
0.5
y1 , y2 , y3
Distance (m)
T
My2 y2 :=
kN⋅ m
( )
( )
My1 y1 := 0
( )
My3 y3 := 0
My (kN-m)
0.5
( )
M y2( y 2 )
M y3( y 3 )
M y1 y 1
0
0
0.5
y1 , y2 , y3
Distance (m)
)
Problem 11-56
The bearings at A and B exert only x and z components of force on the steel shaft. Determine the
shaft's diameter to the nearest millimeter so that it can resist the loadings of the gears without
exceeding an allowable shear stress of τallow = 80 MPa. Use the maximum-distortion-energy theory of
failure with σallow = 200 MPa.
Given: PC := 7.5kN
L1 := 150mm
PD := 5kN
rC := 50mm
L2 := 350mm
L3 := 250mm
σ allow := 200MPa
Solution:
rD := 75mm
L := L1 + L2 + L3
Support Reactions :
ΣFz=0;
In y-z plane : +
Az + BZ − PC = 0
(1)
ΣΜB=0; Az⋅ L − PC⋅ L3 = 0
(2)
Solving Eqs. (1) and (2):
L3
Az :=
⋅P
Az = 2.5 kN
L C
Bz := PC − Az
In x-y plane :
ΣFx=0;
Bz = 5 kN
Ax + Bx − PD = 0
(
(3)
)
ΣΜB=0; Ax⋅ L − PD⋅ L − L1 = 0
Solving Eqs. (3) and (4):
L − L1
Ax :=
⋅ PD
L
Ax = 4 kN
Bx := PD − Ax
Bx = 1 kN
Torsion occurs in segment DC : T := PC⋅ rC
Critical Section :
(4)
T = 0.375 kN⋅ m
Located just to the left of gear C.
Mx := Bx⋅ L 3
Mz := Bz⋅ L 3
M :=
2
Maximum Distortion Energy Theory : Applying Eq. 9-5:
(
) ⎡⎣0.5(σx' + σy')⎤⎦ 2 + τ x'y'2
2
2
σ 2 = 0.5 ( σ x' + σ y') − ⎡⎣0.5 ⋅ ( σ x' + σ y')⎤⎦ + τ x'y'
σ 1 = 0.5 σ x' + σ y' +
where
σ y' := 0
σ x' =
M⋅ c
M⋅ c
4M⋅ c
=
=
4
I
π 4
π⋅c
⋅c
4
τ x'y' =
T⋅ c
T⋅ c
2T⋅ c
=
=
4
J
π 4
π⋅c
⋅c
2
2
Mx + Mz
M = 1.27475 kN⋅ m
Let
a' = 0.5σ x'
and
2
Then σ 1 = ( a' + b')
(0.5σx')2 + τ x'y'2
b' =
2
2
σ 2 = ( a − 'b')
2
2
2
(2
σ 1⋅ σ 2 = ( a' + b') ⋅ ( a' − b') = a' − b'
2
2
2
2
2
2
2
σ 1 − σ 1⋅ σ 2 + σ 2 = ( a' + b') − a' − b'
) + ( a' − b') 2 = a'2 + 3b'2
Hence σ 1 − σ 1⋅ σ 2 + σ 2 = σ allow
(
)
0.5σ x'
2
2
⎡ 0.5σ 2 + τ 2⎤ = σ
( x') x'y' ⎦ allow2
+ 3⎣
2
2
2
σ x' + 3τ x'y' = σ allow
2
2
2
⎛ 4M⋅ c ⎞ + 3 ⎛ 2T⋅ c ⎞ = σ
allow
⎜ 4
⎜ 4
⎝ π⋅c ⎠
⎝ π⋅c ⎠
6
c :=
2
16M + 12T
2
2
2
c = 20.31 mm
do := 2c
do = 40.61 mm
π ⋅ σ allow
Use
do = 41mm
Ans
(
y1 := 0 , 0.01 ⋅ L1 .. L1
Az⋅ y1
Mz1 y1 :=
kN⋅ m
)
(
)
y2 := L 1 , 1.01 ⋅ L 1 .. L 1 + L 2
y3 := L1 + L2 , 1.01 ⋅ L1 + L2 .. L
Az⋅ y2
1
Mz2 y2 :=
Mz3 y3 := ⎡⎣Az⋅ y3 − PC⋅ y3 − L1 − L2 ⎤⎦ ⋅
kN⋅ m
kN⋅ m
( )
( )
( )
(
)
( )
1
M z2( y 2 )
M z3( y 3 )
Mz (kN-m)
M z1 y 1
0
0
0.5
y1 , y2 , y3
Distance (m)
Ax⋅ y1
Mx1 y1 :=
kN⋅ m
1
Mx2 y2 := ⎡⎣Ax⋅ y2 − PD⋅ y2 − L 1 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
Mx3 y3 := ⎡⎣Ax⋅ y3 − PD⋅ y3 − L 1 ⎤⎦ ⋅
kN⋅ m
Mx (kN-m)
( )
(
( )
M x2( y 2 )0.5
M x3( y 3 )
M x1 y 1
0
0
0.5
y1 , y2 , y3
Distance (m)
T
My2 y2 :=
kN⋅ m
( )
( )
My1 y1 := 0
( )
My3 y3 := 0
My (kN-m)
0.5
( )
M y2( y 2 )
M y3( y 3 )
M y1 y 1
0
0
0.5
y1 , y2 , y3
Distance (m)
)
Problem 11-57
Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading
shown. The allowable bending stress is σallow = 160 MPa and the allowable shear stress is τallow = 84
MPa.
Given: σ allow := 160MPa L 1 := 3m
τ allow := 84MPa
L 2 := 1.5m
P1 := 40kN
P2 := 50kN
Solution:
(
)
L := 2 L 1 + L 2
Support Reactions : By symmetry, RA=RB=R
+
ΣF y=0;
2R − 2P1 − P2 = 0
R := P1 + 0.5P2
Maximum Moment and Shear:
Vmax := R
Mmax := R⋅ L 1 + L 2 − P1⋅ L2
(
Vmax = 65 kN
Mmax = 232.5 kN⋅ m
)
Bending Stress:
Sreq'd :=
Mmax
3
Sreq'd = 1453125.00 mm
σ allow
Select W 460x74 :
Shear Stress :
d := 457mm
tw := 9.02mm
Provide a shear stress check.
τ max :=
Hence,
( 3) mm3
Sx := 1460⋅ 10
Vmax
tw⋅ d
Use W 460x74
τ max = 15.77 MPa
< τ allow =98 MPa
Ans
(O.K.!)
(
x1 := 0 , 0.01 ⋅ L 1 .. L 1
)
x2 := L1 , 1.01 ⋅ L1 .. L 1 + L 2
(
)
(
) (
x4 := ( L 1 + 2L2) , 1.01 ⋅ ( L1 + 2L 2) .. L
x3 := L 1 + L 2 , 1.01 ⋅ L 1 + L 2 .. L 1 + 2L2
1
V1 x1 := R⋅
kN
1
V2 x2 := R − P1 ⋅
kN
( )
( ) (
)
1
V3 x3 := R − P1 − P2 ⋅
kN
)
( ) (
)
1
V4 x4 := R − P1 − P2 − P1 ⋅
kN
( ) (
R⋅ x1
M1 x1 :=
kN⋅ m
)
1
M2 x2 := ⎡⎣R⋅ x2 − P1⋅ x2 − L1 ⎤⎦ ⋅
kN⋅ m
( )
( )
( )
(
)
1
M3 x3 := ⎡⎣R⋅ x3 − P1⋅ x3 − L1 − P2⋅ x3 − L1 − L2 ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
1
M4 x4 := ⎡⎣⎡⎣R⋅ x4 − P1⋅ x4 − L 1 − P2⋅ x4 − L 1 − L 2 ⎤⎦ − P1⋅ L1 − x4 − L ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
(
)
(
)
100
( )
V2 ( x 2 )
V3 ( x 3 )
V4 ( x 4 )
Shear (kN)
V1 x 1
0
100
0
2
4
6
8
6
8
x1 , x2 , x3 , x4
Distance (m)
Moment (kN-m)
300
( )
M 2 ( x 2 )200
M3( x3)
M 4 ( x 4 )100
M1 x1
0
0
2
4
x1 , x2 , x3 , x4
Distance (m)
Problem 12-01
An L2 steel strap having a thickness of 3 mm and a width of 50 mm is bent into a circular arc of radius
15 m. Determine the maximum bending stress in the strap.
Given: b := 50mm
t := 3mm
ρ := 15m
E := 200GPa
Solution:
3
t
b⋅ t
I :=
2
12
Moment - Curvature Relationship :
Section Property :
1
ρ
=
M
E⋅ I
c :=
M :=
E⋅ I
ρ
Bendiug Stress :
σ=
M⋅ c
I
σ=
E⋅ I ⎛ c ⎞
⋅⎜
ρ ⎝I⎠
σ :=
E⋅ c
ρ
σ = 20 MPa
Ans
Problem 12-02
A picture is taken of a man performing a pole vault, and the minimum radius of curvature of the pole is
estimated by measurement to be 4.5 m. If the pole is 40 mm in diameter and it is made of a glassreinforced plastic for which Eg = 131 GPa, determine the maximum bending stress in the pole.
Given: do := 40mm
ρ := 4.5m
E := 131GPa
Solution:
4
do
π ⋅ do
c :=
I :=
2
64
Moment - Curvature Relationship :
Section Property :
1
ρ
=
M
E⋅ I
M :=
E⋅ I
ρ
Bendiug Stress :
σ=
M⋅ c
I
σ=
E⋅ I ⎛ c ⎞
⋅⎜
ρ ⎝I⎠
σ :=
E⋅ c
ρ
σ = 582.2 MPa
Ans
Problem 12-03
Determine the equation of the elastic curve for the beam using the x coordinate that is valid for
0 ≤ x < L / 2Specify the slope at A and the beam's maximum deflection. EI is constant.
Problem 12-04
Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the
beam's maximum deflection. EI is constant.
Problem 12-05
Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant.
Problem 12-06
Determine the equations of the elastic curve for the beam using the x1 and x3 coordinates. Specify the
beam's maximum deflection. EI is constant.
Problem 12-07
Determine the equations of the elastic curve for the shaft using the x1 and x2 coordinates. Specify the
slope at A and the displacement at the center of the shaft. EI is constant.
Problem 12-08
Determine the equations of the elastic curve for the shaft using the x1 and x3 coordinates. Specify the
slope at A and the deflection at the center of the shaft. EI is constant.
Problem 12-09
The beam is made of two rods and is subjected to the concentrated load P. Determine the maximum
deflection of the beam if the moments of inertia of the rods are IAB and IBC, and the modulus of
elasticity is E.
Problem 12-10
The beam is made of two rods and is subjected to the concentrated load P. Determine the slope at C.
The moments of inertia of the rods are IAB and IBC, and the modulus of elasticity is E.
Problem 12-11
The bar is supported by a roller constraint at B, which allows vertical displacement but resists axial
load and moment. If the bar is subjected to the loading shown, determine the slope at A and the
deflection at C. EI is constant.
Problem 12-12
Determine the deflection at B of the bar in Prob. 12-11.
Problem 12-13
The fence board weaves between the three smooth fixed posts. If the posts remain along the same line,
determine the maximum bending stress in the board. The board has a width of 150 mm. and a
thickness of 12 mm. E = 12 GPa. Assume the displacement of each end of the board relative to its
center is 75 mm.
Given: b := 150mm
t := 12mm
E := 12GPa
δ := 75mm
L := 2.4m
Solution:
Support Reactions : By symmetry, RA=RC=R
2R − P = 0
R = 0.5P
+ ΣFy=0;
Moment Function :
M ( x) = R⋅ x
M ( x) = 0.5 ⋅ P⋅ x
Mmax = R⋅ ( 0.5L)
Mmax = 0.25P⋅ L
3
Section Property :
t
2
Slope and Elastic Curve :
c :=
I :=
2
E ⋅ I⋅
d ⋅v
2
b⋅ t
12
2
d v
E ⋅ I⋅
= M ( x)
2
dx
=
dx
P⋅ x
2
2
dv
P⋅ x
E ⋅ I⋅
=
+ C1
4
dx
(1)
3
P⋅ x
E⋅ I⋅ v =
+ C1⋅ x + C2
12
Due to symmetry, dv/dx=0 at x=L/2. Also, v=0 at x=0.
Boundary Conditions :
2
0=
From Eq. (2):
0 = 0 + 0 + C2
3
2
P ⎛ L⎞
⋅⎜
+ C1
4 ⎝2⎠
From Eq. (1):
The Elastic Curve :
(2)
P⋅ L
C1 = −
16
C2 := 0
Substitute the values of C1 and C 2 into Eq. (2),
2
(
)
P⋅ L
P⋅ x
P⋅ x
2
2
E ⋅ I⋅ v =
−
⋅x
v=
⋅ 4x − 3L
12
16
48E⋅ I
Require at x=L/2, v=-δ. From Eq. (3),
P ⎛ L⎞ ⎡ ⎛ L⎞
2⎤
⋅⎜
⋅ ⎢4 ⎜
− 3L ⎥
48E⋅ I ⎝ 2 ⎠ ⎣ ⎝ 2 ⎠
⎦
2
−δ =
Bending Stress:
t
cmax :=
2
(3)
3
−δ =
Mmax := 0.25P⋅ L
Mmax⋅ cmax
σ max :=
I
− P⋅ L
48⋅ E⋅ I
P :=
48⋅ E⋅ I
3
⋅δ
L
σ max = 11.25 MPa
Ans
Problem 12-14
Determine the equation of the elastic curve for the beam using the x coordinate. Specify the slope at A
and the maximum deflection. EI is constant.
Problem 12-15
Determine the deflection at the center of the beam and the slope at B. EI is constant.
Problem 12-16
A torque wrench is used to tighten the nut on a bolt. If the dial indicates that a torque of 90 N·m is
applied when the bolt is fully tightened, determine the force P acting at the handle and the distance s the
needle moves along the scale. Assume only the portion AB of the beam distorts. The cross section is
square having dimensions of 12 mm by 12 mm. E = 200 GPa.
Given: b := 12mm
h := 12mm
L := 0.45m
E := 200GPa δ := 75mm
R := 0.3m
T z := 90N⋅ m
Solution:
Equations of Equilibrium :
+ ΣF =0; A − P = 0
y
y
ΣΜB=0; T z − P⋅ L = 0
(2)
Solving Eqs. (1) and (2):
P :=
Moment Function :
(1)
Tz
Ay := P
L
Ay = 200 N
P = 200 N
Ans
M ( x) = Ay⋅ x − Tz
3
b⋅ h
I :=
12
Slope and Elastic Curve :
Section Property :
2
E ⋅ I⋅
d ⋅v
2
= M ( x)
dx
2
E ⋅ I⋅
d v
2
dx
= Ay⋅ x − T z
2
Ay⋅ x
dv
E ⋅ I⋅
=
− T z⋅ x + C1
2
dx
(1)
Tz⋅ x
Ay⋅ x
E⋅ I⋅ v =
−
+ C1⋅ x + C2
6
2
(2)
3
2
Due to symmetry, dv/dx=0 at x=0, and v=0 at x=0.
Boundary Conditions :
From Eq. (1):
0 = 0 − 0 + C1
From Eq. (2):
0 = 0 − 0 + 0 + C2
The Elastic Curve :
Substitute the values of C1 and C 2 into Eq. (2),
3
2
Tz⋅ x
Ay⋅ x
E⋅ I⋅ v =
−
6
2
At x=R, v=-s. From Eq. (3),
2
s :=
−R
⋅ Ay R − 3Tz
6E⋅ I
(
C1 := 0
C2 := 0
)
2
x
v=
⋅ Ay x − 3T z
6E ⋅ I
(
s = 9.11 mm
)
Ans
(3)
Problem 12-17
The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft and at B
by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Draw the bending-moment
diagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft's
centerline. Determine the equations of the elastic curve using the coordinates x1 and x2 . EI is constant.
Given: h := 60mm
L 1 := 150mm
P := 5kN
L 2 := 400mm
L := L 1 + L 2
Solution:
Support Reactions:
+ ΣF =0; A + B = 0
y
(1)
ΣΜB=0; P⋅ h + A⋅ L2 = 0 (2)
Solving Eqs. (1) and (2):
A :=
− P⋅ h
L2
B := −A
( )
M2 ( x2) := B⋅ x2
Moment Function :
M1 x1 := P⋅ h
Section Property :
EI := kN⋅ m
A = −0.75 kN
B = 0.75 kN
2
Slope and Elastic Curve :
2
EI⋅
d ⋅ v1
dx1
2
2
( )
EI⋅
= M1 x1
dx2
2
EI⋅
d v1
dx1
EI⋅
2
dv1
dx1
EI⋅ v1 =
d ⋅ v2
( )
= M2 x2
2
2
= P⋅ h
EI⋅
d v2
dx2
= ( P⋅ h) ⋅ x1 + C1
P⋅ h⋅ x1
(1)
EI⋅
2
dv2
=
dx2
2
2
= B⋅ x2
B 2
⋅ x + C3
2 2
(3)
3
+ C1⋅ x1 + C2
(2)
EI⋅ v2 =
B⋅ x2
6
+ C3⋅ x2 + C4
(4)
Boundary Conditions :
v1=0 at x1=0.15m, From Eq. (2):
P⋅ h⋅ ( 0.15m)
0=
2
v2=0 at x2=0,
0 = 0 + 0 + C4
From Eq. (4):
2
+ C1⋅ ( 0.15m) + C2
(5)
C4 := 0
Ans
3
v2=0 at x2=0.4m,
From Eq. (4):
B⋅ ( 0.4m)
0=
+ C3⋅ ( 0.4m) + C4
6
B
2
2
C3 := − ⋅ ( 0.4m)
C3 = −0.02 kN⋅ m
6
Ans
Continuity Condition:
dv1/dx1= - dv2/dx2 at A (x1=0.15m and x2=0.4m)
From Eqs. (1) and (3),
P⋅ h⋅ ( 0.15m) + C1 =
B
2
⋅ ( 0.4m) + C3
2
B
2
C1 := ⋅ ( 0.4m) + C3 − P⋅ h⋅ ( 0.15m)
2
From Eq. (5):
The Elastic Curve :
C1 = −0.005 kN⋅ m
2
Ans
P⋅ h⋅ ( 0.15m)
3
Ans
C2 := −
− C1⋅ ( 0.15m)
C2 = −0.00263 kN⋅ m
2
Substitute the values of C1 and C 2 into Eq. (2), and C3 and C 4 into Eq. (4),
v1 =
1 ⎛ P⋅ h
2
⎞
⎜ ⋅ x1 + C1⋅ x1 + C2
EI ⎝ 2
⎠
Ans
v2 =
1 ⎛B 3
⎞
⎜ ⋅ x2 + C3⋅ x2 + C4
EI ⎝ 6
⎠
Ans
BMD :
x'2 := L1 , 1.01 ⋅ L1 .. L
( )
(
1
) kN⋅ m
M'2 x'2 := ⎡⎣P⋅ h + A⋅ x'2 − L1 ⎤⎦ ⋅
Moment (kN-m)
x'1 := 0 , 0.01 ⋅ L 1 .. L 1
P⋅ h
M'1 x'1 :=
kN⋅ m
( )
2
0.4
( )
M'2 ( x'2 )0.2
M'1 x'1
0
0
0.2
0.4
x'1 , x'2
Distane (m)
Problem 12-18
Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope and
deflection at C. EI is constant.
Problem 12-19
Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope at
A. EI is constant.
Problem 12-20
Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope and
deflection at B. EI is constant.
Problem 12-21
Determine the equations of the elastic curve using the coordinates x1 and x3 , and specify the slope and
deflection at B. EI is constant.
Problem 12-22
Determine the maximum slope and maximum deflection of the simply-supported beam which is
subjected to the couple moment M0 . EI is constant.
Problem 12-23
The two wooden meter sticks are separated at their centers by a smooth rigid cylinder having a
diameter of 50 mm. Determine the force F that must be applied at each end in order to just make their
ends touch. Each stick has a width of 20 mm and a thickness of 5 mm. Ew = 11 GPa.
Given: do := 50mm L := 0.5m
b := 20mm
t := 5mm
Solution:
E := 11GPa
3
b⋅ t
12
Section Property:
I :=
Moment Function :
M ( x) = −F⋅ x
Slope and Elastic Curve :
2
E⋅ I⋅
d ⋅v
= M ( x)
2
dx
2
E⋅ I⋅
d v
2
= − F⋅ x
dx
2
E⋅ I⋅
dv
F⋅ x
= −
+ C1
dx
2
(1)
3
E⋅ I⋅ v = −
F⋅ x
+ C1⋅ x + C2
6
(2)
Boundary Conditions :
2
dv/dx=0 at x=L. From Eq. (1):
F⋅ L
0= −
+ C1
2
F⋅ L
C1 :=
2
v=0 at x=L.
0= −
F⋅ L
+ C1⋅ L + C2
6
C2 =
F⋅ L
− C1⋅ L
6
2
3
From Eq. (2):
3
Require :
v := −0.5 do at x=0.
From Eq. (2):
−0.5 do⋅ E⋅ I = −0 + 0 −
F :=
F⋅ L
3
1.5do⋅ E ⋅ I
L
3
F = 1.375 N
Ans
3
F⋅ L
C2 := −
3
3
Problem 12-24
The pipe can be assumed roller supported at its ends and by a rigid saddle C at its center. The saddle
rests on a cable that is connected to the supports. Determine the force that should be developed in the
cable if the saddle keeps the pipe from sagging or deflecting at its center. The pipe and fluid within it
have a combined weight of 2 kN/m. EI is constant.
Given:
e := 0.3m
L := 3.75m
w :=
Solution:
Moment Function :
M ( x) = P ⋅ x −
Slope and Elastic Curve :
2kN
m
1
2
⋅ w⋅ x
2
2
E ⋅ I⋅
d ⋅v
= M ( x)
2
dx
2
E ⋅ I⋅
d v
2
= P⋅ x −
dx
w 2
⋅x
2
2
E ⋅ I⋅
w 3
dv
P⋅ x
=
− ⋅ x + C1
2
6
dx
(1)
3
E ⋅ I⋅ v =
w 4
P⋅ x
−
⋅ x + C1⋅ x + C2
6
24
v=0 at x=0 and at x=L.
Boundary Conditions :
From Eq. (2):
(2)
0 = 0 − 0 + 0 + C2
C2 := 0
3
w 4
P⋅ L
0=
−
⋅ L + C1⋅ L
6
24
Also, dv/dx=0 at x=L.
(3)
2
w 3
P⋅ L
0=
− ⋅ L + C1
2
6
Solving Eqs. (3) and (4) for P,
From Eq. (1):
2
(4)
2
P⋅ L
w 3
w 3
P⋅ L
−
⋅ L + C1 =
− ⋅ L + C1
6
24
2
6
2
P⋅ L
w 3
= ⋅L
3
8
P :=
3w⋅ L
8
P = 2.813 kN
Equations of Equilibrium :
+
ΣFy=0;
2P + F − w⋅ ( 2L ) = 0
F := 2w⋅ L − 2P
At C : +
ΣFy=0;
2T cable ⋅
F = 9375 N
2
e
2
e +L
2
−F= 0
Tcable :=
2
e +L
⋅F
2e
Tcable = 58.78 kN
Ans
Problem 12-25
Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope at C
and displacement at B. EI is constant.
Problem 12-26
Determine the equations of the elastic curve using the coordinates x1 and x3 , and specify the slope at B
and deflection at C. EI is constant.
Problem 12-27
Determine the elastic curve for the simply supported beam using the x coordinate
determine the slope at A and the maximum deflection of the beam. EI is constant.
0 ≤ x ≤ L /. 2Also,
Problem 12-28
Determine the elastic curve for the cantilevered beam using the x coordinate. Also determine the
maximum slope and maximum deflection. EI is constant.
Problem 12-29
The beam is made of a material having a specific weight γ. Determine the displacement and slope at its
end A due to its weight. The modulus of elasticity for the material is E.
Problem 12-30
The beam is made of a material having a specific weight γ. Determine the displacement and slope at its
end A due to its weight. The modulus of elasticity for the material is E.
Problem 12-31
The leaf spring assembly is designed so that it is subjected to the same maximum stress throughout its
length. If the plates of each leaf have a thickness t and can slide freely between each other, show that
the spring must be in the form of a circular arc in order that the entire spring becomes flat when a
large enough load P is applied. What is the maximum normal stress in the spring? Consider the spring
to be made by cutting the n strips from the diamond-shaped plate of thickness t and width b. The
modulus of elasticity for the material is E. Hint: Show that the radius of curvature of the spring is
constant.
Problem 12-32
The beam has a constant width b and is tapered as shown. If it supports a load P at its end, determine
the deflection at B.The load P is applied a short distance s from the tapered end B, where s << L. EI is
constant.
Problem 12-33
A thin flexible 6-m-long rod having a weight of 10 N/m rests on the smooth surface. If a force of 15 N
is applied at its end to lift it, determine the suspended length x and the maximum moment developed in
the rod.
10N
Given: P := 15N
w :=
m
L max := 6m
Solution:
Since the horizontal section has no curvature,
the moment in the rod is zero. Hence, R acts
at the end of the suspended portion and this
portion acts like a simply supported beam.
Thus,
Equations of Equilibrium :
+ ΣF =0; P + R − w⋅ x = 0
(1)
y
ΣΜ0=0; P⋅ x −
w 2
⋅x = 0
2
(2)
x = 3m
Ans
From Eqs. (2) :
x :=
2P
w
Maximum moment occurs at center.
⎛ x⎞ − w ⋅⎛ x⎞
Mmax := P⋅ ⎜
⎜
2
2 2
⎝ ⎠
Mmax = 11.25 N⋅ m
2
⎝ ⎠
Ans
Problem 12-34
The shaft supports the two pulley loads shown. Determine the equation of the elastic curve. The
bearings at A and B exert only vertical reactions on the shaft. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
a := m
Given:
P := kN
EI := kN⋅ m
2
EIo := 1
Solution:
Support Reactions :
+
ΣF y=0;
A + B − P − 2P = 0
(1)
ΣΜB=0;
A⋅ ( 2a) − P⋅ a + 2P⋅ a = 0
(2)
A :=
Solving Eqs. (1) and (2):
−P
2
B :=
7P
2
Moment Function :
M ( x) = A⋅ Ψ ( x − 0) − P⋅ Ψ ( x − a) + B⋅ Ψ ( x − 2a)
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
= M ( x)
dx
2
EI⋅
d v
2
= A⋅ Ψ ( x − 0) − P⋅ Ψ ( x − a) + B⋅ Ψ ( x − 2a)
dx
dv
A
2 P
2 B
2
EI⋅
= ⋅ Ψ ( x − 0) − ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 2a) + C1
dx
2
2
2
A
3 P
3 B
3
EI⋅ v = ⋅ Ψ ( x − 0) − ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 2a) + C1⋅ x + C2
6
6
6
A 3 P
B
3
3
EI⋅ v = ⋅ x − ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 2a) + C1⋅ x + C2
6
6
6
Boundary Conditions :
From Eq. (1):
From Eq. (1):
v=0 at x=0
0 = 0 − 0 + 0 + 0 + C2
Also v=0 at x=2a.
0=
(1)
C2 := 0
A
3
3 P
⋅ ( 2a) − ⋅ Ψ ( 2a − a) + 0 + C1⋅ ( 2a)
6
6
A
P 3
3
C1 := −
⋅ ( 2a) +
⋅a
12a
12a
5
2
C1 =
P⋅ a
12
Equation of Elastic Curve :
v=
1 ⎛A 3 P
3 B
3
⎞
⎜ ⋅ x − ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 2a) + C1⋅ x m
EI ⎝ 6
6
6
⎠
v ( x) :=
1 ⎛ P 3 P
5
3 7P
3
2 ⎞
P⋅ a ⋅ x m
⎜ − ⋅ x − ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 2a) +
EIo ⎝ 12
6
12
12
⎠
Ans
Problem 12-35
Determine the equation of the elastic curve. Specify the slopes at A and B. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given:
a := m
w :=
kN
m
L := 2a
Solution:
Support Reactions :
+ ΣF =0;
A + B − w⋅ a = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) − B⋅ ( 2⋅ a) = 0
B :=
(2)
1
⋅ ( w a)
4
B = 0.25 w⋅ a
Solving Eqs. (1) and (2):
A := w⋅ a − B
A = 0.75 w⋅ a
Moment Function :
2
2
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − 0)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x
EI := kN⋅ m
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
EIo := 1
= M ( x)
2
dx
2
EI⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x
dx
dv
1
3 A 2
3 1
(3)
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + C1
dx
6
6
2
1
1
4 A 3
4
(4)
EI⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ x + C1⋅ x + C2
24
24
6
Boundary Conditions : v=0 at x=0 and x=2a. From Eq. (4):
EI⋅
0 = −0 + 0 + 0 + 0 + C2
0= −
1
24
w⋅ ( 2a) + ⎛⎜
4
C2 := 0
1⎞
⎝ 24 ⎠
4
⋅ w⋅ a +
A
6
3
⋅ ( 2a) + C1⋅ ( 2a)
1
1
3
3 2A 2
C1 := w⋅ a −
w⋅ a −
⋅a
3
48
3
3
3
C1 = − w⋅ a
16
Equations of Elastic Curve and Slope :
v ( x) :=
1 ⎛ 1
1
3
4 3w⋅ a 3
4
3
⋅x −
w⋅ a ⋅ x⎞
⎜− w⋅ x + w⋅ Ψ ( x − a) +
EIo ⎝ 24
24
24
16
⎠
θ ( x) :=
1 ⎛ 1
3 A 2
3 1
⎞
⎜ − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + C1
EIo ⎝ 6
6
2
⎠
Slope at A:
Substitute x=0 into Eq.(3).
θ ( 0) = −
Slope at B:
Substitute x=L into Eq.(3).
θ ( L) =
3 ⎛ w⋅ a ⎞
⎜
Ans
3
16 ⎝ EIo ⎠
7 ⎛ w⋅ a ⎞
Ans
3
⎜
48 ⎝ EIo ⎠
Ans
Problem 12-36
The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 1.5m
P := 20kN
w := 6
b := 3m
kN
m
L := 2a + b
Solution:
Support Reactions :
+ ΣF =0;
A + B − P − w⋅ a = 0
(1)
y
ΣΜA=0;
−( w a) ⋅ ( 0.5a) + P⋅ ( a + b) − B⋅ b = 0
B :=
Solving Eqs. (1) and (2):
1
b
(2)
⋅ [ −( w a) ⋅ ( 0.5a) + P⋅ ( a + b) ]
B = 27.75 kN
A := P + w⋅ a − B
A = 1.25 kN
Moment Function :
2
2
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − a − b)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − a − b)
Slope and Elastic Curve :
2
E ⋅ I⋅
d ⋅v
= M ( x)
2
dx
2
E ⋅ I⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − a − b)
dx
dv
1
3 A
2 B
2
3 1
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ Ψ ( x − a) + ⋅ Ψ ( x − a − b) + C1
dx
6
6
2
2
1
1
4 A
3 B
3
4
E⋅ I⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ Ψ ( x − a) + ⋅ Ψ ( x − a − b) + C1⋅ x + C2
24
24
6
6
Boundary Conditions : v=0 at x=a and x=a+b. From Eq. (4):
1
4
(5)
0 = − w⋅ a + 0 + 0 + 0 + C1⋅ a + C2
24
E ⋅ I⋅
0= −
1
24
4
w⋅ ( a + b) +
(6)-(5):
1
24
4
w⋅ b +
A 3
⋅ b + 0 + C1⋅ ( a + b) + C2
6
C1 = 25.125 kN⋅ m
⎦
1
4
From Eq. (5): C2 :=
⋅ w⋅ a − C1⋅ a
24
w
Equation of Elastic Curve : ao :=
24
kN
ao = 0.25
m
v=
1 ⎛
a1 = 0.2083 kN
(4)
(6)
1 1
1
1
4
4 A 3
4
C1 := ⎡⎢ w⋅ ( a + b) −
w⋅ b − ⋅ b −
⋅ w⋅ a ⎥⎤
b 24
24
6
24
⎣
(3)
2
C2 = −36.422 kN⋅ m
A
a1 :=
6
B
a2 :=
6
a3 := C1
a2 = 4.625 kN a3 = 25.13 kN⋅ m
2
3
a4 := C2
a4 = −36.42 kN⋅ m
4
3
3
4
−ao x + ao⋅ Ψ ( x − a) + a1⋅ Ψ ( x − a) + a2⋅ Ψ ( x − a − b) + a3⋅ x + a4⎞
⎝
⎠
E⋅ I
3
Ans
Ans
Problem 12-37
The shaft supports the two pulley loads shown. Determine the equation of the elastic curve. The
bearings at A and B exert only vertical reactions on the shaft. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given:
a := 0.5m
P1 := 200N
P2 := 300N
Solution:
Support Reactions :
+
ΣF y=0;
A + B − P1 − P2 = 0
(1)
ΣΜB=0;
A⋅ ( 2a) − P1⋅ a + P2⋅ a = 0
(2)
Solving Eqs. (1) and (2):
A :=
P1 − P2
2
A = −50 N
B :=
P1 + 3P2
2
B = 550 N
Moment Function :
M ( x) = A⋅ Ψ ( x − 0) − P1⋅ Ψ ( x − a) + B⋅ Ψ ( x − 2a)
Slope and Elastic Curve :
2
E ⋅ I⋅
d ⋅v
2
= M ( x)
dx
2
E ⋅ I⋅
d v
2
dx
E ⋅ I⋅
= A⋅ Ψ ( x − 0) − P1⋅ Ψ ( x − a) + B⋅ Ψ ( x − 2a)
dv
A
2 P1
2 B
2
= ⋅ Ψ ( x − 0) −
⋅ Ψ ( x − a) + ⋅ Ψ ( x − 2a) + C1
dx
2
2
2
E⋅ I⋅ v =
A
3 P1
3 B
3
⋅ Ψ ( x − 0) −
⋅ Ψ ( x − a) + ⋅ Ψ ( x − 2a) + C1⋅ x + C2
6
6
6
Boundary Conditions :
From Eq. (1):
From Eq. (1):
v=0 at x=0
0 = 0 − 0 + 0 + 0 + C2
Also v=0 at x=2a.
0=
A
Equation of Elastic Curve : ao :=
6
v=
C2 := 0
A
3 P1
3
⋅ Ψ ( 2a − 0) −
⋅ Ψ ( 2a − a) + 0 + C1⋅ ( 2a)
6
6
P1
A
3
3
C1 := −
⋅ ( 2a) +
⋅a
12a
12a
ao = −8.33 N
(1)
a1 = −33.33 N
P1
a1 := −
6
a2 = 91.67 N
C1 = 12.50 N⋅ m
B
a2 :=
6
a3 := C1
a3 = 12.50 N⋅ m
1 ⎛
3
3
3
ao⋅ Ψ ( x − 0) + a1⋅ Ψ ( x − a) + a2⋅ Ψ ( x − 2a) + a3⋅ x⎞⎠
⎝
E⋅ I
2
2
Ans
Ans
Problem 12-38
The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 4m
P := 50kN
kN
w := 3
b := 3m
m
L := a + 2⋅ b
Solution:
Support Reactions :
+ ΣF =0;
A + B − P − w⋅ a = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) + P⋅ ( a + b) − B⋅ L = 0
B :=
Solving Eqs. (1) and (2):
1
L
(2)
⋅ [ ( w a) ⋅ ( 0.5a) + P⋅ ( a + b) ]
A := P + w⋅ a − B
B = 37.4 kN
Moment Function :
2
A = 24.6 kN
2
M ( x) = A⋅ Ψ ( x − 0) − 0.5w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) − P⋅ Ψ ( x − a − b)
2
2
M ( x) = A⋅ x − 0.5w⋅ x + 0.5w⋅ Ψ ( x − a) − P⋅ Ψ ( x − a − b)
EI := kN⋅ m
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
EIo := 1
= M ( x)
2
dx
2
EI⋅
d v
2
2
2
= A⋅ x − 0.5w⋅ x + 0.5w⋅ Ψ ( x − a) − P⋅ Ψ ( x − a − b)
dx
EI⋅
dv
=
dx
A 2 w 3 w
3 P
2
⋅ x − ⋅ x + ⋅ Ψ ( x − a) − ⋅ Ψ ( x − a − b) + C1
2
6
6
2
(3)
A 3 w 4 w
4 P
3
⋅x −
⋅x +
⋅ Ψ ( x − a) − ⋅ Ψ ( x − a − b) + C1⋅ x + C2
6
24
24
6
Boundary Conditions :
EI⋅ v =
v=0 at x=0. From Eq. (4):
v=0 at x=L. From Eq. (4):
0 = 0 − 0 + 0 + 0 + 0 + C2
C2 := 0
A 3 w 4 w
4 P
3
0 = ⋅L −
⋅L +
⋅ ( L − a) − ⋅ ( L − a − b) + C1⋅ L
6
24
24
6
C1 := −
Equation of Elastic Curve :
ao = 4.10 kN
v=
1⎛
(4)
A⋅ L
2
+
w
3
⋅L −
w
4
⋅ ( L − a) +
24
6
2
C1 = −278.70 kN⋅ m
24L
A
ao :=
6
P
a2 :=
6
kN
a1 = 0.125
m
w
a1 :=
24
a2 = 8.33 kN
P
6L
a3 := C1
a3 = 0.00 N⋅ m
4
3
3
4
ao⋅ x − a1⋅ x + a1⋅ Ψ ( x − a) − a2⋅ Ψ ( x − a − b) + a3⋅ x + a4⎞
⎝
⎠
EI
2
⋅ ( L − a − b)
3
a3 := C2
Ans
Ans
Problem 12-39
The beam is subjected to the load shown. Determine the displacement at x = 7 m and the slope at A.
EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 4m
P := 50kN
kN
m
w := 3
b := 3m
L := a + 2⋅ b
Solution:
Support Reactions :
+ ΣF =0;
A + B − P − w⋅ a = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) + P⋅ ( a + b) − B⋅ L = 0
B :=
Solving Eqs. (1) and (2):
(2)
1
⋅ [ ( w a) ⋅ ( 0.5a) + P⋅ ( a + b) ]
L
B = 37.4 kN
A := P + w⋅ a − B
A = 24.6 kN
Moment Function :
2
2
M ( x) = A⋅ Ψ ( x − 0) − 0.5w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) − P⋅ Ψ ( x − a − b)
2
2
M ( x) = A⋅ x − 0.5w⋅ x + 0.5w⋅ Ψ ( x − a) − P⋅ Ψ ( x − a − b)
EI := kN⋅ m
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
2
EIo := 1
= M ( x)
dx
2
EI⋅
d v
2
2
2
= A⋅ x − 0.5w⋅ x + 0.5w⋅ Ψ ( x − a) − P⋅ Ψ ( x − a − b)
dx
EI⋅
dv
A 2 w 3 w
3 P
2
= ⋅ x − ⋅ x + ⋅ Ψ ( x − a) − ⋅ Ψ ( x − a − b) + C1
dx
2
6
6
2
EI⋅ v =
A 3 w 4 w
4 P
3
⋅x −
⋅x +
⋅ Ψ ( x − a) − ⋅ Ψ ( x − a − b) + C1⋅ x + C2
6
24
24
6
(3)
(4)
Boundary Conditions :
v=0 at x=0. From Eq. (4):
v=0 at x=L. From Eq. (4):
0 = 0 − 0 + 0 + 0 + 0 + C2
C2 := 0
A 3 w 4 w
4 P
3
0 = ⋅L −
⋅L +
⋅ ( L − a) − ⋅ ( L − a − b) + C1⋅ L
6
24
24
6
2
A⋅ L
w 3
w
P
4
3
C1 := −
+
⋅L −
⋅ ( L − a) +
⋅ ( L − a − b)
24
24L
6L
6
2
C1 = −278.70 kN⋅ m
Equations of Elastic Curve and Slope :
1 ⎛A 3 w 4 w
4 P
3
⎞
v ( x) :=
⎜ ⋅ x − ⋅ x + ⋅ Ψ ( x − a) − ⋅ Ψ ( x − a − b) + C1⋅ x + C2
EI ⎝ 6
24
24
6
⎠
m
Displacement at x=7m.: Substitute x=7m into Eq.(4).
v ( 7m) = −834.60
EIo
θ ( x) :=
Ans
1 ⎛A 2 w 3 w
3 P
2
⎞
⎜ ⋅ x − ⋅ x + ⋅ Ψ ( x − a) − ⋅ Ψ ( x − a − b) + C1
EI ⎝ 2
6
6
2
⎠
Slope at A:
Substitute x=0 into Eq.(3).
θ ( 0) = −278.70
1
EIo
Ans
Problem 12-40
The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given:
a := 2.4m
P1 := 10kN
MB := 6kN⋅ m
P2 := 20kN
Solution:
Support Reactions :
+ ΣF =0;
A + B − P1 − P2 = 0
y
ΣΜB=0; A⋅ ( 3a) − P1⋅ ( 2a) − P2⋅ a + MB = 0
Solving Eqs. (1) and (2):
A :=
2P1 + P2
3
A = 12.5 kN
−
MB
3a
(1)
(2)
B := P1 + P2 − A
B = 17.5 kN
Moment Function :
M ( x) = A⋅ Ψ ( x − 0) − P1⋅ Ψ ( x − a) − P2⋅ Ψ ( x − 2a)
Slope and Elastic Curve :
2
E ⋅ I⋅
d ⋅v
2
= M ( x)
dx
2
E ⋅ I⋅
d v
2
dx
E ⋅ I⋅
= A⋅ Ψ ( x − 0) − P1⋅ Ψ ( x − a) − P2⋅ Ψ ( x − 2a)
dv
A
2 P1
2 P2
2
= ⋅ Ψ ( x − 0) −
⋅ Ψ ( x − a) −
⋅ Ψ ( x − 2a) + C1
dx
2
2
2
E ⋅ I⋅ v =
A
3 P1
3 P2
3
⋅ Ψ ( x − 0) −
⋅ Ψ ( x − a) −
⋅ Ψ ( x − 2a) + C1⋅ x + C2
6
6
6
Boundary Conditions :
From Eq. (1):
From Eq. (1):
v=0 at x=0
0 = 0 − 0 + 0 + C2
Also v=0 at x=3a.
0=
C2 := 0
A
3 P1
3 P2
3
⋅ Ψ ( 3a − 0) −
⋅ Ψ ( 3a − a) −
⋅ Ψ ( 3a − 2a) + C1⋅ ( 3a)
6
6
6
P1
P2
A
3
3
3
C1 := −
⋅ ( 3a) +
⋅ ( 2a) +
⋅a
18a
18a
18a
A
Equation of Elastic Curve : ao :=
6
ao = 2.08 kN
v=
(1)
a1 = −1.67 kN
P1
a1 := −
6
P2
a2 := −
6
C1 = −76.00 kN⋅ m
a3 := C1
a2 = −3.33 kN a3 = −76.00 kN⋅ m
1 ⎛
3
3
3
ao⋅ Ψ ( x − 0) + a1⋅ Ψ ( x − a) + a2⋅ Ψ ( x − 2a) + a3⋅ x⎞⎠
⎝
E⋅ I
2
Ans
Ans
2
Problem 12-41
The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 1.5m
b := 3m
P := 20kN
w := 6
L := 2a + b
Solution:
kN
m
Support Reactions :
+ ΣF =0;
A + B − P − w⋅ a = 0
(1)
y
ΣΜA=0;
−( w a) ⋅ ( 0.5a) + P⋅ ( a + b) − B⋅ b = 0
Solving Eqs. (1) and (2):
B :=
(2)
1
⋅ [ −( w a) ⋅ ( 0.5a) + P⋅ ( a + b) ]
b
B = 27.75 kN
A := P + w⋅ a − B
A = 1.25 kN
Moment Function :
2
2
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − a − b)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − a − b)
Slope and Elastic Curve :
2
E ⋅ I⋅
d ⋅v
2
= M ( x)
dx
2
E ⋅ I⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − a − b)
dx
dv
1
3 A
2 B
2
3 1
E ⋅ I⋅
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ Ψ ( x − a) + ⋅ Ψ ( x − a − b) + C1
dx
6
6
2
2
1
1
A
B
4
3
3
4
E⋅ I⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ Ψ ( x − a) + ⋅ Ψ ( x − a − b) + C1⋅ x + C2
24
24
6
6
Boundary Conditions : v=0 at x=a and x=a+b. From Eq. (4):
1
4
(5)
0 = − w⋅ a + 0 + 0 + 0 + C1⋅ a + C2
24
1
1
4
4 A 3
(6)
0 = − w⋅ ( a + b) +
w⋅ b + ⋅ b + 0 + C1⋅ ( a + b) + C2
24
24
6
(6)-(5):
1⎡ 1
1
1
4
4 A 3
4⎤
C1 := ⎢ w⋅ ( a + b) −
w⋅ b − ⋅ b −
⋅ w⋅ a ⎥
b 24
24
6
24
⎣
1
4
From Eq. (5): C2 :=
⋅ w⋅ a − C1⋅ a
24
kN
ao = 0.25
m
v=
a1 = 0.2083 kN
(4)
C1 = 25.125 kN⋅ m
⎦
w
Equation of Elastic Curve : ao :=
24
(3)
2
C2 = −36.422 kN⋅ m
A
a1 :=
6
B
a2 :=
6
a3 := C1
a2 = 4.625 kN a3 = 25.13 kN⋅ m
2
3
a4 := C2
a4 = −36.42 kN⋅ m
1 ⎛
4
3
3
4
−ao x + ao⋅ Ψ ( x − a) + a1⋅ Ψ ( x − a) + a2⋅ Ψ ( x − a − b) + a3⋅ x + a4⎞⎠
⎝
E⋅ I
3
Ans
Ans
Problem 12-42
The beam is subjected to the load shown. Determine the equation of the slope and elastic curve. EI is
constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 5m
b := 3m
Mo := 15kN⋅ m
Solution: L := a + b
w := 3
kN
m
Support Reactions :
+ ΣF =0;
A + B − w⋅ a = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) − B⋅ a + Mo = 0
Solving Eqs. (1) and (2):
B :=
(2)
1
⋅ ⎡( w a) ⋅ ( 0.5a) + Mo⎤⎦
a ⎣
B = 10.5 kN
A := w⋅ a − B
A = 4.5 kN
Moment Function :
2
2
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − 0) + B⋅ Ψ ( x − a)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x + B⋅ Ψ ( x − a)
Slope and Elastic Curve :
2
E ⋅ I⋅
d ⋅v
2
= M ( x)
dx
2
E ⋅ I⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x + B⋅ Ψ ( x − a)
dx
dv
1
3 A 2 B
2
3 1
E ⋅ I⋅
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1
dx
6
6
2
2
1
1
A
4
3
4
3 B
E⋅ I⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1⋅ x + C2
24
24
6
6
Boundary Conditions : v=0 at x=0 and x=a. From Eq. (4):
0 = −0 + 0 + 0 + 0 + 0 + C2
0= −
1
A 3
4
w⋅ a + 0 + ⋅ a + 0 + C1⋅ ( a)
24
6
Equation of Elastic Curve and slope :
kN
ao = 0.125
m
a1 = 0.75 kN
C2 := 0
1
3 A 2
C1 :=
w⋅ a − ⋅ a
24
6
w
ao :=
24
a2 = 1.75 kN
A
a1 :=
6
B
a2 :=
6
a3 = −3.125 kN⋅ m
(3)
(4)
C1 = −3.125 kN⋅ m
a3 := C1
2
Ans
v=
1 ⎛
4
3
4
3
−ao x + ao⋅ Ψ ( x − a) + a1⋅ x + a2⋅ Ψ ( x − a) + a3⋅ x⎞⎠
⎝
E⋅ I
Ans
θ=
1 ⎛
3
2
3
2
−4⋅ ao x + 4⋅ ao⋅ Ψ ( x − a) + 3a1⋅ x + 3a2⋅ Ψ ( x − a) + a3⎞⎠
⎝
E⋅ I
Ans
2
Problem 12-43
Determine the equation of the elastic curve. Specify the slope at A and the displacement at C. EI is
constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := m
w :=
kN
m
L := 2a
Solution:
Support Reactions :
+ ΣF =0;
A + B − w⋅ a = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) − B⋅ ( 2⋅ a) = 0
B :=
(2)
1
⋅ ( w a)
4
B = 0.25 w⋅ a
Solving Eqs. (1) and (2):
A := w⋅ a − B
A = 0.75 w⋅ a
Moment Function :
2
2
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − 0)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x
EI := kN⋅ m
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
EIo := 1
= M ( x)
2
dx
2
EI⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x
dx
dv
1
3 A 2
3 1
(3)
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + C1
dx
6
6
2
1
1
4 A 3
4
(4)
EI⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ x + C1⋅ x + C2
24
24
6
Boundary Conditions : v=0 at x=0 and x=2a. From Eq. (4):
EI⋅
0 = −0 + 0 + 0 + 0 + C2
0= −
w⋅ ( 2a) + ⎛⎜
1
4
C2 := 0
1⎞
⎝ 24 ⎠
24
4
⋅ w⋅ a +
A
3
⋅ ( 2a) + C1⋅ ( 2a)
6
1
1
3
3 2A 2
C1 := w⋅ a −
w⋅ a −
⋅a
3
48
3
3
3
C1 = − w⋅ a
16
Equations of Elastic Curve and Slope :
v ( x) :=
1 ⎛ 1
1
3
4 3w⋅ a 3
4
3
⋅x −
w⋅ a ⋅ x⎞
⎜− w⋅ x + w⋅ Ψ ( x − a) +
EIo ⎝ 24
24
24
16
⎠
Displacement at C:
θ ( x) :=
Substitute x=a into Eq.(4).
v ( a) = −
5 ⎛ w⋅ a ⎞
⎜
4
48 ⎝ EIo ⎠
Ans
1 ⎛ 1
3 A 2
3 1
⎞
⎜ − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + C1 m
EIo ⎝ 6
6
2
⎠
Slope at A:
Substitute x=0 into Eq.(3).
⎛ w⋅ a3 ⎞
θ ( 0) = −
m⎜
16 ⎝ EIo ⎠
3
Ans
Problem 12-44
Determine the equation of the elastic curve. Specify the slope at A and B. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given:
a := m
Solution:
w :=
L := 2a
kN
m
Support Reactions :
+ ΣF =0;
A + B − w⋅ a = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) − B⋅ ( 2⋅ a) = 0
Solving Eqs. (1) and (2):
(2)
1
⋅ ( w a)
4
A := w⋅ a − B
B = 0.25 w⋅ a
A = 0.75 w⋅ a
B :=
Moment Function :
2
2
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − 0)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
EI := kN⋅ m
2
EIo := 1
= M ( x)
dx
2
EI⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x
dx
dv
1
3 A 2
3 1
(3)
EI⋅
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + C1
dx
6
6
2
1
1
4 A 3
4
(4)
EI⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ x + C1⋅ x + C2
24
24
6
Boundary Conditions : v=0 at x=0 and x=2a. From Eq. (4):
0 = −0 + 0 + 0 + 0 + C2
C2 := 0
1
4 ⎛ 1 ⎞
4 A
3
w⋅ ( 2a) + ⎜
⋅ w⋅ a + ⋅ ( 2a) + C1⋅ ( 2a)
24
6
⎝ 24 ⎠
1
1
3
3 2A 2
C1 := w⋅ a −
w⋅ a −
⋅a
3
48
3
Equations of Elastic Curve and Slope :
0= −
3
3
C1 = − w⋅ a
16
v ( x) :=
1 ⎛ 1
1
3
4 3w⋅ a 3
4
3 ⎞
⋅x −
w⋅ a ⋅ x
⎜− w⋅ x + w⋅ Ψ ( x − a) +
EIo ⎝ 24
24
24
16
⎠
θ ( x) :=
1 ⎛ 1
3 A 2
3 1
⎞
⎜ − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + C1
EIo ⎝ 6
6
2
⎠
Slope at A: Substitute x=0 into Eq.(3).
3 ⎛ w⋅ a ⎞
θ ( 0) = − ⎜
16 EIo
Ans
Slope at B: Substitute x=L into Eq.(3).
7 ⎛ w⋅ a ⎞
θ ( L) =
⎜
48 EIo
Ans
3
⎝
⎠
3
⎝
⎠
Ans
Problem 12-45
The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 1.5m
P := 20kN
L := 4a
Solution:
Support Reactions :
+
ΣF y=0;
A + B − 2P = 0
ΣΜB=0;
A⋅ ( 2a) − P⋅ ( 3a) + P⋅ a = 0 (2)
Solving Eqs. (1) and (2):
A := P
(1)
B := P
Moment Function :
M ( x) = −P⋅ Ψ ( x − 0) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − 3a)
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
EI := kN⋅ m
2
= M ( x)
dx
2
EI⋅
d v
= −P⋅ Ψ ( x − 0) + A⋅ Ψ ( x − a) + B⋅ Ψ ( x − 3a)
2
dx
dv
P
2 A
2 B
2
EI⋅
= − ⋅ Ψ ( x − 0) + ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 3a) + C1
dx
2
2
2
P
A
B
3
3
3
EI⋅ v = − ⋅ Ψ ( x − 0) + ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 3a) + C1⋅ x + C2
6
6
6
P 3 A
3 B
3
EI⋅ v = − ⋅ x + ⋅ Ψ ( x − a) + ⋅ Ψ ( x − 3a) + C1⋅ x + C2
6
6
6
Boundary Conditions :
From Eq. (1):
(1)
(2)
Due to symmetry, dv/dx=0 at x=2a
P
2
2 A
0 = − ⋅ ( 2a) + ⋅ Ψ ( 2a − a) + 0 + C1
2
2
A⎞ 2
⎛
C1 := ⎜ 2⋅ P −
⋅a
2
P 3
0 = − ⋅ a + 0 + 0 + C1⋅ ( a) + C2
6
C1 = 67.50 kN⋅ m
P 3
C2 := ⋅ a − C1⋅ a
6
⎝
⎠
2
Also v=0 at x=a.
From Eq. (2):
C2 = −90 kN⋅ m
P
Equation of Elastic Curve : ao :=
6
ao = 3333.33 N
v=
a1 = 3.3333 kN
A
a1 :=
6
a2 := C1
a3 := C2
2
a3 = −90 kN⋅ m
a2 = 67.5 kN⋅ m
1 ⎛
3
3
3
−ao⋅ x + a1⋅ Ψ ( x − a) + a1⋅ Ψ ( x − 3a) + a2⋅ x + a3⎞⎠
⎝
E⋅ I
3
3
Ans
Ans
Problem 12-46
The beam is subjected to the load shown. Determine the equation of the slope and elastic curve. EI is
constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 5m
b := 3m
P := 8kN
Solution:
w := 2
L := a + b
kN
m
Support Reactions :
+ ΣF =0;
A + B − w⋅ a − P = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) − B⋅ a + P⋅ L = 0
Solving Eqs. (1) and (2):
B :=
(2)
1
⋅ [ ( w a) ⋅ ( 0.5a) + P⋅ L]
a
B = 17.8 kN
A := w⋅ a + P − B
A = 0.2 kN
2
2
Moment Function :
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − 0) + B⋅ Ψ ( x − a)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x + B⋅ Ψ ( x − a)
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
= M ( x)
dx
2
EI⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x + B⋅ Ψ ( x − a)
dx
dv
1
3 A 2 B
2
3 1
EI⋅
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1
dx
6
6
2
2
1
1
4 A 3 B
3
4
EI⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1⋅ x + C2
24
24
6
6
Boundary Conditions : v=0 at x=0 and x=a. From Eq. (4):
0 = −0 + 0 + 0 + 0 + 0 + C2
0= −
1
A 3
4
w⋅ a + 0 + ⋅ a + 0 + C1⋅ ( a)
24
6
Equation of Elastic Curve and slope :
kN
ao = 0.0833
m
a1 = 0.0333 kN
w
ao :=
24
C2 := 0
1
3 A 2
C1 :=
w⋅ a − ⋅ a
24
6
A
a1 :=
6
B
a2 :=
6
a2 = 2.9667 kN a3 = 9.583 kN⋅ m
(3)
(4)
C1 = 9.583 kN⋅ m
a3 := C1
2
Ans
v=
1⎛
4
3
4
3
−ao x + ao⋅ Ψ ( x − a) + a1⋅ x + a2⋅ Ψ ( x − a) + a3⋅ x⎞⎠
⎝
EI
Ans
θ=
1⎛
3
2
3
2
−4⋅ ao x + 4⋅ ao⋅ Ψ ( x − a) + 3a1⋅ x + 3a2⋅ Ψ ( x − a) + a3⎞⎠
⎝
EI
Ans
2
Problem 12-47
The beam is subjected to the load shown. Determine the slope at A and the displacement at C. EI is
constant.
Use Macaulay Function: Ψ ( z) := Φ ( z) ⋅ ( z)
Given: a := 5m
b := 3m
P := 8kN
Solution:
w := 2
L := a + b
kN
m
Support Reactions :
+ ΣF =0;
A + B − w⋅ a − P = 0
(1)
y
ΣΜA=0;
( w a) ⋅ ( 0.5a) − B⋅ a + P⋅ L = 0
Solving Eqs. (1) and (2):
(2)
1
⋅ [ ( w a) ⋅ ( 0.5a) + P⋅ L]
a
B = 17.8 kN
B :=
Moment Function :
2
A := w⋅ a + P − B
A = 0.2 kN
2
M ( x) = −0.5 w⋅ Ψ ( x − 0) − ( −0.5 )w⋅ Ψ ( x − a) + A⋅ Ψ ( x − 0) + B⋅ Ψ ( x − a)
2
2
M ( x) = −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x + B⋅ Ψ ( x − a)
Slope and Elastic Curve :
2
EI⋅
d ⋅v
2
EI := kN⋅ m
2
EIo := 1
= M ( x)
dx
2
EI⋅
d v
2
2
2
= −0.5 w⋅ x + 0.5w⋅ Ψ ( x − a) + A⋅ x + B⋅ Ψ ( x − a)
dx
dv
1
3 A 2 B
2
3 1
EI⋅
= − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1
dx
6
6
2
2
1
1
4 A 3 B
3
4
EI⋅ v = − w⋅ x +
w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1⋅ x + C2
24
24
6
6
Boundary Conditions : v=0 at x=0 and x=a. From Eq. (4):
0 = −0 + 0 + 0 + 0 + 0 + C2
1
A 3
4
w⋅ a + 0 + ⋅ a + 0 + C1⋅ ( a)
24
6
Equation of Elastic Curve and slope :
0= −
C2 := 0
1
3 A 2
C1 :=
w⋅ a − ⋅ a
24
6
1⎛ 1
1
4 A 3 B
3
4
⎞
⎜ − w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1⋅ x
EI ⎝ 24
24
6
6
⎠
m
Displacement at C: Substitute x=L into Eq.(4).
v ( L) = −160.75
EIo
(3)
(4)
C1 = 9.583 kN⋅ m
v ( x) :=
1⎛ 1
3 A 2 B
2
3 1
⎞
⎜− w⋅ x + w⋅ Ψ ( x − a) + ⋅ x + ⋅ Ψ ( x − a) + C1
EI ⎝ 6
6
2
2
⎠
1
Slope at A: Substitute x=0 into Eq.(3).
θ ( 0) = 9.58
EIo
Ans
θ ( x) :=
Ans
2
Problem 12-48
The beam is subjected to the load shown. Determine the equation of the elastic curve.
Use Macaulay Function:
Given: L 1 := 2m
Ψ ( x) := Φ ( x) ⋅ x
kN
wo := 150
m
L 2 := 3m
L := L 1 + L 2
Solution:
Support Reactions :
+
ΣF y=0;
A + B − wo⋅ L1 −
ΣΜA=0; −
wo
⋅ L2 = 0
2
(1)
wo
wo
⎛ L2 ⎞
2
⋅ L1 +
⋅ L 2⋅ ⎜
− B⋅ L 2 = 0
2
2
⎝3⎠
2
Solving Eqs. (1) and (2):
(2)
2
− 3L 1 + L 2
B :=
⋅ wo
6L 2
(
B = −25 kN
A = 550 kN
Moment Function :
wo
wo
2
3
M ( x) = −
⋅ Ψ ( x − 0) +
⋅ Ψ x − L1 + A⋅ Ψ x − L1
2
6L2
(
)
(
2
Slope and Elastic Curve :
2
EI⋅
EI⋅
d ⋅v
2
= M ( x)
)
A := wo⋅ L 1 + 0.5L 2 − B
EI := kN⋅ m
2
)
EIo := 1
dx
wo
wo
2
3
= −
⋅ Ψ ( x − 0) +
⋅ Ψ x − L 1 + A⋅ Ψ x − L 1
2
2
6L 2
d v
(
)
(
)
dx
wo
wo
dv
3
4 A
2
EI⋅
= −
⋅ Ψ ( x − 0) +
⋅ Ψ x − L1 + ⋅ Ψ x − L1 + C1
dx
6
24⋅ L2
2
(
EI⋅ v = −
)
(
)
(3)
wo
wo
4
5 A
3
⋅ Ψ ( x − 0) +
⋅ Ψ x − L1 + ⋅ Ψ x − L1 + C1⋅ x + C2
24
120⋅ L2
6
(
)
(
)
(4)
v=0 at x=L1 Given
wo
4
0= −
⋅ L1 + 0 + 0 + C1⋅ L1 + C2
24
Boundary Conditions :
From Eq. (4):
Also v=0 at x=L.
From Eq. (4):
0= −
Solving Eqs. (5) and (6):
(5)
wo
wo
4
5 A
3
⋅L +
⋅ L − L 1 + ⋅ L − L1 + C1⋅ L + C2
24
120⋅ L2
6
(
Guess
)
C1 := 1kN⋅ m
⎛⎜ C1 ⎞
:= Find ( C1 , C2)
⎜ C2
⎝ ⎠
2
(
)
C2 := 1kN⋅ m
C1 = 410 kN⋅ m
2
3
C2 = −720 kN⋅ m
3
(6)
wo
Equation of Elastic Curve : ao := −
24
1
ao = −6.25
kN
m
a1 = 0.42
1
m
2
wo
a1 :=
120⋅ L
v=
a3 := C1
a2 = 91.67 kN
kN
a3 = 410.00 kN⋅ m
2
A
a2 :=
6
2
a4 = −720.00 kN⋅ m
Ans
3
1 ⎛
4
5
3
ao⋅ Ψ ( x − 0) + a1⋅ Ψ x − L 1 + a2⋅ Ψ x − L1 + a3⋅ x + a4⎞⎠
⎝
E⋅ I
(
)
(
)
a4 := C2
Ans
Ans
Problem 12-49
Determine the displacement at C and the slope at A of the beam.
Use Macaulay Function:
Given: L 1 := 2m
Ψ ( x) := Φ ( x) ⋅ x
kN
wo := 150
m
L 2 := 3m
L := L 1 + L 2
Solution:
Support Reactions :
+
ΣF y=0;
A + B − wo⋅ L1 −
ΣΜA=0; −
wo
⋅ L2 = 0
2
(1)
wo
wo
⎛ L2 ⎞
2
⋅ L1 +
⋅ L 2⋅ ⎜
− B⋅ L 2 = 0
2
2
⎝3⎠
2
Solving Eqs. (1) and (2):
(2)
2
− 3L 1 + L 2
B :=
⋅ wo
6L 2
(
B = −25 kN
A = 550 kN
Moment Function :
wo
wo
2
3
M ( x) = −
⋅ Ψ ( x − 0) +
⋅ Ψ x − L1 + A⋅ Ψ x − L1
2
6L2
(
)
(
2
Slope and Elastic Curve :
2
EI⋅
EI⋅
d ⋅v
2
= M ( x)
)
A := wo⋅ L 1 + 0.5L 2 − B
EI := kN⋅ m
2
)
EIo := 1
dx
wo
wo
2
3
= −
⋅ Ψ ( x − 0) +
⋅ Ψ x − L 1 + A⋅ Ψ x − L 1
2
2
6L 2
d v
(
)
(
)
dx
wo
wo
dv
3
4 A
2
EI⋅
= −
⋅ Ψ ( x − 0) +
⋅ Ψ x − L1 + ⋅ Ψ x − L1 + C1
dx
6
24⋅ L2
2
(
EI⋅ v = −
)
(
)
(3)
wo
wo
4
5 A
3
⋅ Ψ ( x − 0) +
⋅ Ψ x − L1 + ⋅ Ψ x − L1 + C1⋅ x + C2
24
120⋅ L2
6
(
)
(
)
(4)
v=0 at x=L1 Given
wo
4
0= −
⋅ L1 + 0 + 0 + C1⋅ L1 + C2
24
Boundary Conditions :
From Eq. (4):
Also v=0 at x=L.
From Eq. (4):
0= −
Solving Eqs. (5) and (6):
(5)
wo
wo
4
5 A
3
⋅L +
⋅ L − L 1 + ⋅ L − L1 + C1⋅ L + C2
24
120⋅ L2
6
(
Guess
)
C1 := 1kN⋅ m
⎛⎜ C1 ⎞
:= Find ( C1 , C2)
⎜ C2
⎝ ⎠
2
(
)
C2 := 1kN⋅ m
C1 = 410 kN⋅ m
2
3
C2 = −720 kN⋅ m
3
(6)
Equation of Elastic Curve and Slope :
wo
⎞
1 ⎛ wo
4
5 A
3
v ( x) :=
−
⋅
Ψ
(
x
−
0
)
+
⋅ Ψ x − L 1 + ⋅ Ψ x − L 1 + C1⋅ x + C2
⎜
EI 24
120⋅ L 2
6
⎝
⎠
m
Ans
Displacement at C: Substitute x=0 into Eq.(7).
v ( 0m) = −720
EIo
(
θ ( x) :=
)
(
)
wo
⎞
1 ⎛ wo
3
4 A
2
−
⋅
Ψ
(
x
−
0
)
+
⋅ Ψ x − L 1 + ⋅ Ψ x − L 1 + C1
⎜
EI
24⋅ L 2
2
6
Slope at A:
(
⎝
Substitute x=L1 into Eq.(3).
)
(
)
1
θ L 1 = 210
EIo
( )
⎠
Ans
(7)
Problem 12-50
Determine the equation of the elastic curve. Specify the slope at A. EI is constant.
Ψ ( x) := Φ ( x) ⋅ x
kN
EIo := 1
w :=
m
Use Macaulay Function:
Given: L := m
Solution:
Support Reactions :
+
ΣF y=0;
A + B − w⋅ L = 0
(1)
ΣΜA=0; −( 0.5L ) ⋅ w L − B⋅ L = 0
Solving Eqs. (1) and (2):
(2)
B := −0.5 w L
A := w⋅ L − B
A = 1.5 w⋅ L
Moment Function :
2
2
M ( x) = −( 0.5w) ⋅ Ψ ( x − 0) − ( −0.5 w) ⋅ Ψ ( x − L ) + A⋅ Ψ ( x − L)
2
Slope and Elastic Curve :
2
= M ( x)
EI := kN⋅ m
2
dx
2
EI⋅
EI⋅
d ⋅v
d v
2
2
2
= −( 0.5w) ⋅ Ψ ( x − 0) − ( −0.5 w) ⋅ Ψ ( x − L) + A⋅ Ψ ( x − L )
dx
dv
w
3 w
3 A
2
EI⋅
= − ⋅ Ψ ( x − 0) + ⋅ Ψ ( x − L ) + ⋅ Ψ ( x − L) + C1
dx
6
6
2
EI⋅ v = −
(3)
w
w
4
4 A
3
⋅ Ψ ( x − 0) +
⋅ Ψ ( x − L) + ⋅ Ψ ( x − L ) + C1⋅ x + C2
24
24
6
Boundary Conditions :
v=0 at x=L
v=0 at x=2L
(4)
Given
w 4
⋅ L + 0 + 0 + C1⋅ L + C2
24
w
w 4 A 3
4
From Eq. (4): 0 = − ⋅ ( 2L) +
⋅ L + ⋅ L + C1⋅ ( 2L) + C2
24
24
6
From Eq. (4): 0 = −
Solving Eqs. (5) and (6):
Guess C1 := 1kN⋅ m
⎛⎜ C1 ⎞
:= Find ( C1 , C2)
⎜ C2
⎝ ⎠
1
3
C1 = w⋅ L
3
2
C2 := 1kN⋅ m
(5)
(6)
3
7
4
C2 = − w⋅ L
24
Equation of Elastic Curve and Slope :
v ( x) :=
1⎛ w 4 w
7
4 3w⋅ L
3 1
3
4⎞
⋅ Ψ ( x − L ) + w⋅ L ⋅ x −
w⋅ L
⎜− ⋅ x + ⋅ Ψ ( x − L) +
EI ⎝ 24
24
12
3
24
⎠
θ ( x) :=
1⎛ w
3 w
3 9w⋅ L
2
⎞
⋅ Ψ ( x − L) + C1
⎜ − ⋅ Ψ ( x − 0) + ⋅ Ψ ( x − L ) +
EI ⎝ 6
6
12
⎠
Slope at A:
Substitute x=L into Eq.(3).
θ ( L) =
1 w⋅ L
6 EI
3
Ans
Ans
Problem 12-51
Determine the equation of the elastic curve. Specify the deflection at C. EI is constant.
Use Macaulay Function: Ψ ( x) := Φ ( x) ⋅ x
kN
Given: L := m w :=
m
Solution:
Support Reactions :
+
ΣF y=0;
A + B − w⋅ L = 0
(1)
ΣΜA=0; −( 0.5L ) ⋅ w L − B⋅ L = 0
(2)
B := −0.5 w L
Solving Eqs. (1) and (2):
A := w⋅ L − B
A = 1.5 w⋅ L
Moment Function :
2
2
M ( x) = −( 0.5w) ⋅ Ψ ( x − 0) − ( −0.5 w) ⋅ Ψ ( x − L ) + A⋅ Ψ ( x − L)
2
EI⋅
Slope and Elastic Curve :
2
= M ( x)
EI := kN⋅ m
dx
2
EI⋅
d ⋅v
d v
2
2
EIo := 1
2
= −( 0.5w) ⋅ Ψ ( x − 0) − ( −0.5 w) ⋅ Ψ ( x − L) + A⋅ Ψ ( x − L )
2
dx
dv
w
3 w
3 A
2
EI⋅
= − ⋅ Ψ ( x − 0) + ⋅ Ψ ( x − L ) + ⋅ Ψ ( x − L) + C1
dx
6
6
2
EI⋅ v = −
w
w
4
4 A
3
⋅ Ψ ( x − 0) +
⋅ Ψ ( x − L) + ⋅ Ψ ( x − L ) + C1⋅ x + C2
24
24
6
Boundary Conditions :
v=0 at x=L
v=0 at x=2L
(3)
(4)
Given
w 4
⋅ L + 0 + 0 + C1⋅ L + C2
24
w
w 4 A 3
4
From Eq. (4): 0 = − ⋅ ( 2L) +
⋅ L + ⋅ L + C1⋅ ( 2L) + C2
24
24
6
From Eq. (4): 0 = −
Solving Eqs. (5) and (6):
Guess C1 := 1kN⋅ m
⎛⎜ C1 ⎞
:= Find ( C1 , C2)
⎜ C2
⎝ ⎠
1
3
C1 = w⋅ L
3
2
C2 := 1kN⋅ m
(5)
(6)
3
7
4
C2 = − w⋅ L
24
Equation of Elastic Curve :
v ( x) :=
1⎛ w 4 w
7
4 3w⋅ L
3 1
3
4⎞
⋅ Ψ ( x − L ) + w⋅ L ⋅ x −
w⋅ L
⎜− ⋅ x + ⋅ Ψ ( x − L) +
EI ⎝ 24
24
12
3
24
⎠
v ( 0) = −
7 w⋅ L
24 EI
4
Ans
Ans
Problem 12-52
Determine the equation of the elastic curve. Specify the slope at B. EI is constant.
Ψ ( x) := Φ ( x) ⋅ x
kN
w :=
m
Use Macaulay Function:
Given: L := m
Solution:
Support Reactions :
+
ΣF y=0;
A + B − w⋅ L = 0
(1)
ΣΜA=0; −( 0.5L ) ⋅ w L − B⋅ L = 0
Solving Eqs. (1) and (2):
(2)
B := −0.5 w L
A := w⋅ L − B
A = 1.5 w⋅ L
Moment Function :
2
2
M ( x) = −( 0.5w) ⋅ Ψ ( x − 0) − ( −0.5 w) ⋅ Ψ ( x − L ) + A⋅ Ψ ( x − L)
2
Slope and Elastic Curve :
2
= M ( x)
EI := kN⋅ m
2
EIo := 1
dx
2
EI⋅
EI⋅
d ⋅v
d v
2
2
2
= −( 0.5w) ⋅ Ψ ( x − 0) − ( −0.5 w) ⋅ Ψ ( x − L) + A⋅ Ψ ( x − L )
dx
dv
w
3 w
3 A
2
EI⋅
= − ⋅ Ψ ( x − 0) + ⋅ Ψ ( x − L ) + ⋅ Ψ ( x − L) + C1
dx
6
6
2
EI⋅ v = −
(3)
w
w
4
4 A
3
⋅ Ψ ( x − 0) +
⋅ Ψ ( x − L) + ⋅ Ψ ( x − L ) + C1⋅ x + C2
24
24
6
Boundary Conditions :
v=0 at x=L
v=0 at x=2L
(4)
Given
w 4
⋅ L + 0 + 0 + C1⋅ L + C2
24
w
w 4 A 3
4
From Eq. (4): 0 = − ⋅ ( 2L) +
⋅ L + ⋅ L + C1⋅ ( 2L) + C2
24
24
6
From Eq. (4): 0 = −
Solving Eqs. (5) and (6):
Guess C1 := 1kN⋅ m
⎛⎜ C1 ⎞
:= Find ( C1 , C2)
⎜ C2
⎝ ⎠
1
3
C1 = w⋅ L
3
2
C2 := 1kN⋅ m
(5)
(6)
3
7
4
C2 = − w⋅ L
24
Equation of Elastic Curve and Slope :
v ( x) :=
1⎛ w 4 w
7
4 3w⋅ L
3 1
3
4⎞
⋅ Ψ ( x − L ) + w⋅ L ⋅ x −
w⋅ L
⎜− ⋅ x + ⋅ Ψ ( x − L) +
EI ⎝ 24
24
12
3
24
⎠
θ ( x) :=
1⎛ w
3 w
3 9w⋅ L
2
⎞
⋅ Ψ ( x − L) + C1
⎜ − ⋅ Ψ ( x − 0) + ⋅ Ψ ( x − L ) +
EI ⎝ 6
6
12
⎠
Slope at A:
Substitute x=2L into Eq.(3). θ ( 2L ) = −
1 w⋅ L
12 EI
3
Ans
Ans
Problem 12-53
The shaft is made of steel and has a diameter of 15 mm. Determine its maximum deflection. The
bearings at A and B exert only vertical reactions on the shaft. E st = 200 GPa.
Use Macaulay Function: Ψ ( x) := Φ ( x) ⋅ x
Given:
a := 200mm
P1 := 250N
b := 300mm
P2 := 80N
do := 15mm
E := 200GPa
Solution: L := 2a + b
Section Property:
πdo
I :=
64
4
Support Reactions :
+ ΣF =0;
A + B − P1 − P2 = 0
y
ΣΜB=0; A⋅ L − P1⋅ ( a + b) − P2⋅ b = 0
Solving Eqs. (1) and (2):
A :=
P1⋅ ( a + b) + P2⋅ a
L
B := P1 + P2 − A
Moment Function :
(1)
(2)
A = 201.429 N
B = 128.571 N
M ( x) = A⋅ Ψ ( x − 0) − P1⋅ Ψ ( x − a) − P2⋅ Ψ ( x − a − b)
2
Slope and Elastic Curve :
d v
2
dx
E⋅ I⋅
d ⋅v
2
= M ( x)
dx
2
E⋅ I⋅
E⋅ I⋅
= A⋅ Ψ ( x − 0) − P1⋅ Ψ ( x − a) − P2⋅ Ψ ( x − a − b)
dv
A
2 P1
2 P2
2
= ⋅ Ψ ( x − 0) −
⋅ Ψ ( x − a) −
⋅ Ψ ( x − a − b) + C1
dx
2
2
2
E⋅ I⋅ v =
(1)
A
3 P1
3 P2
3
⋅ Ψ ( x − 0) −
⋅ Ψ ( x − a) −
⋅ Ψ ( x − a − b) + C1⋅ x + C2
6
6
6
(2)
Boundary Conditions :
v=0 at x=0 From Eq. (2): 0 = 0 − 0 + 0 + C2
v=0 at x=L
C2 := 0
P2
A 3 P1
3
3
From Eq. (2): 0 = ⋅ L −
⋅ ( L − a) −
⋅ a + C1⋅ ( L )
6
6
6
P2
A 2 P1
3
3
C1 := − ⋅ L +
⋅ ( L − a) +
⋅a
6
6L
6L
Equation of Elastic Curve :
⎞
1 ⎛ A 3 P1
3 P2
3
v ( x) :=
⎜ ⋅ x − ⋅ Ψ ( x − a) − ⋅ Ψ ( x − a − b) + C1⋅ x
E⋅ I ⎝ 6
6
6
⎠
Maximum Deflection:
Assume v max occurs at a < x < a+b.
Given
C1 = −8.857 N⋅ m
(3)
2
dv/dx=0 at x', From Eq. (1): 0 =
Solving Eq.(3):
A 2 P1
2
⋅ x' −
⋅ ( x' − a) − 0 + C1
2
2
Guess x' := 300⋅ mm
(4)
x' := Find ( x')
x' = 330.05 mm
For vmax,
substitute x' into Eq.(3).
v ( x') = −3.64 mm
Ans
Note: The negative sign indicates downward displacement.
Problem 12-54
Determine the slope and deflection at C. EI is constant.
Given:
L 1 := 8m
L 2 := 4m
P := 75kN
Solution:
L := L 1 + L 2
Support Reactions :
+
ΣF y=0;
A+B−P= 0
(1)
ΣΜB=0;
A⋅ L1 + P⋅ L 2 = 0
(2)
Solving Eqs. (1) and (2):
A := −P⋅
L2
B := P⋅
L1
A = −37.5 kN
L1 + L2
L1
B = 112.5 kN
M/EI Diagram:
1
M'1 x1 := A⋅ x1 ⋅
EI
Moment-Area Theorems :
( ) (
)
( )
(
0
Slopes :
⎛ L1 ⎞
1
tB.A := M'1 L1 ⋅ L1⋅ ⎜
2
⎝3⎠
( )
θ A :=
tB.A
θ A = 400
L1
1
EIo
( )
M'2 ( x 2 ) 200
M'1 x 1
0
( )
θ C := θ C.A − θ A
θ C = 1400
Distance (m)
1
EIo
Ans
Deflections :
tC.A :=
⎛ L1
⎛ 2L2 ⎞
⎞ 1
1
M'1 L1 ⋅ L1⋅ ⎜
+ L2 + M'1 L1 ⋅ L2⋅ ⎜
2
⎝3
⎠ 2
⎝ 3 ⎠
( )
L
∆ C := tC.A −
⋅ t
L1 B.A
5
x1 , x2
1
1
M'1 L1 ⋅ L1 + M'1 L 1 ⋅ L 2
2
2
( )
θ C.A :=
1
) EI
M'2 x2 := ⎡⎣A⋅ x2 + B⋅ x2 − L1 ⎤⎦ ⋅
M/EI (1/m)
EI := kN⋅ m
2
EIo := 1
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L
Set
( )
∆ C = 4800
m
EIo
Ans
10
Problem 12-55
Determine the slope and deflection at B. EI is constant.
Problem 12-56
If the bearings exert only vertical reactions on the shaft, determine the slope at the bearings and the
maximum deflection of the shaft. EI is constant.
Problem 12-57
Determine the slope at B and the deflection at C. EI is constant.
Problem 12-58
Determine the slope at C and the deflection at B. EI is constant.
Problem 12-59
The 60-kg gymnast stands on the center of the simply supported balance beam. If the beam is made of
wood and has the cross section shown, determine the maximum deflection.The supports at A and B
are assumed to be rigid. Ew = 12 GPa.
Given: mo := 60⋅ kg
L := 2.7m
bt := 125mm
bb := 75mm
h := 150mm
E := 12GPa
Solution: W := mo⋅ g
Support Reactions : By symmetry, A=B=R
+
ΣF y=0;
2R − W = 0
R := 0.5W
R = 294.20 N
79
4.
34
∆b := bt − bb
Section Property :
⎯
Σ ⋅ yi⋅ Ai
(
yc =
I :=
Σ ⋅ ( Ai)
2
.
7m
)
yc :=
(bb⋅ h)⋅ ( 0.5h) + 0.5∆b⋅ h⋅ ⎛⎜⎝ 3 ⎞⎠
h
yc = 68.75 mm
bb⋅ h + 0.5∆b⋅ h
1
1
3
2
3
⎞
⎛h
⋅ bb⋅ h + bb⋅ h ⋅ 0.5h − yc +
⋅ ∆b⋅ h + ( 0.5 ⋅ ∆b⋅ h) ⋅ ⎜ − yc
3
12
36
⎝
⎠
(
)(
5
.
4m
)
2
4
I = 27539062.50 mm
M/EI Diagram:
x1 := 0 , 0.01 ⋅ L .. L
( )
⎡
(
⎣
) E⋅ I⎤⎥⎦
M'1 x1 := ⎢ R⋅ x1 ⋅
1
x2 := L , 1.01 ⋅ L .. ( 2L )
( )
⎡
⎣
) E⋅ I⎤⎥⎦
(
M'2 x2 := ⎢⎡⎣R⋅ x2 − W⋅ x2 − L ⎤⎦ ⋅
1
Deflections :
1
⎛ 2L ⎞
tA.C := M'1 ( L ) ⋅ L ⋅ ⎜
2
⎝3⎠
∆ C := tA.C
( )
M'2 ( x 2 )
M'1 x 1 0.002
0
∆ C = 5.84 mm
Due to symmetry, the slope at
midspan (point C) is zero. Hence,
∆ max := ∆ C
∆ max = 5.84 mm
M/EI
Moment-Area Theorems :
Ans
0
2
4
x1 , x2
Distance (m)
m
Problem 12-60
The shaft is supported by a journal bearing at A, which exerts only vertical reactions on the shaft, and
by a thrust bearing at B, which exerts both horizontal and vertical reactions on the shaft. Determine the
slope of the shaft at the bearings. EI is constant.
Given:
L := 0.3m
a := 0.1m
Solution:
Support Reactions :
L := 2L
Due to anti-symmetry,
ΣΜB=0;
P := 400N
B = −A
A⋅ ( 2L ) − P⋅ ( 2a) = 0
⎛ a ⎞P
⎝L⎠
A := ⎜
B := −A
A = 66.67 N
B = −66.67 N
M/EI Diagram:
EI := kN⋅ m
2
EIo := 1
x2 := L , 1.01 ⋅ L .. ( 2L )
x1 := 0 , 0.01 ⋅ L .. L
1
M'1 x1 := A⋅ x1 ⋅
EI
( ) (
)
( )
M'2 x2 := ⎡⎣A⋅ x2 − P⋅ ( 2a)⎤⎦ ⋅
M/EI (1/m)
Set
Moment-Area Theorems :
( )
0
M'2 ( x 2 )
M'1 x 1
0
Slopes :
−1
⎛ 2L ⎞ + 1 M' ( L) ⋅ L⋅ ⎛ L + L⎞
tB.A :=
M'1 ( L ) ⋅ L ⋅ ⎜
⎜
2
⎝3⎠ 2 1
⎝3
⎠
θ A :=
1
EI
tB.A
L
θ A = 0.008
1
EIo
Ans
Due to anti-symmetry (or in a similar manner),
θ B := θ A
θ B = 0.008
1
EIo
Ans
0.5
x1 , x2
Distance (m)
1
Problem 12-61
The beam is subjected to the loading shown. Determine the slope at A and the displacement at C.
Assume the support at A is a pin and B is a roller. EI is constant.
Problem 12-62
The rod is constructed from two shafts for which the moment of inetia of AB is I and BC is 2I.
Determine the maximum slope and deflection of the rod due to the loading. The modulus of elasticity is
E.
Problem 12-63
Determine the deflection and slope at C. EI is constant.
Problem 12-64
If the bearings at A and B exert only vertical reactions on the shaft, determine the slope at A. EI is
constant.
Problem 12-65
If the bearings at A and B exert only vertical reactions on the shaft, determine the slope at C. EI is
constant.
Problem 12-66
Determine the deflection at C and the slope of the beam at A, B, and C. EI is constant.
L 1 := 6m
Given:
L 2 := 3m
Mo := 8kN⋅ m
L := L 1 + L 2
Solution:
Support Reactions :
+ ΣF =0;
A+B= 0
(1)
y
ΣΜB=0;
A⋅ L1 + Mo = 0
(2)
Mo
A := −
L1
Solving Eqs. (1) and (2):
B := −A
A = −1.333 kN
B = 1.333 kN
M/EI Diagram:
EI := kN⋅ m
2
EIo := 1
x1 := 0 , 0.01 ⋅ L 1 .. L 1
x2 := L1 , 1.01 ⋅ L1 .. L
Set
1
M'1 x1 := A⋅ x1 ⋅
EI
( ) (
)
( )
(
0
M/EI (1/m)
Moment-Area Theorems :
Deflections :
tB.A :=
⎛ L1 ⎞
1
M'1 L1 ⋅ L1⋅ ⎜
2
⎝3⎠
( )
( )
5
M'2 ( x 2 )
M'1 x 1
⎛ L1
⎛ L2 ⎞
⎞
1
tC.A := M'1 L1 ⋅ L1⋅ ⎜
+ L2 + M'1 L 1 ⋅ L 2⋅ ⎜
2
⎝3
⎠
⎝2⎠
( )
( )
m
EIo
Ans
Slopes :
θ A :=
θ B.A :=
tB.A
θA = 8
L1
1
EIo
1
M' L ⋅ L
2 1 1 1
Ans
( )
θ B := θ B.A − θ A
θ B = 16
1
EIo
1
M' L ⋅ L + M'1 L1 ⋅ L2
2 1 1 1
1
θ C := θ C.A − θ A
θ C = 40
EIo
θ C.A :=
( )
10
0
5
x1 , x2
Distance (m)
L
∆ C := tC.A −
⋅ t
L1 B.A
∆ C = 84
1
) EI
M'2 x2 := ⎡⎣A⋅ x2 + B⋅ x2 − L1 ⎤⎦ ⋅
Ans
( )
Ans
Problem 12-67
The bar is supported by the roller constraint at C, which allows vertical displacement but resists axial
load and moment. If the bar is subjected to the loading shown, determine the slope and displacement at
A. EI is constant.
Problem 12-68
The acrobat has a weight of 750 N (~75 kg), and suspends himself uniformly from the center of the
high bar. Determine the maximum bending stress in the pipe (bar) and its maximum deflection. The
pipe is made of L2 steel and has an outer diameter of 25 mm and a wall thickness of 3 mm.
Given: a := 0.9m
b := 0.45m
do := 25mm
W := 750N
t := 3mm
E := 200GPa
Solution: L := 2a + b
Support Reactions : By symmetry, A=B=R
+
ΣF y=0;
2R − W = 0
R := 0.5W
R = 375.00 N
ro := 0.5do
Section Property :
I :=
ri := ro − t
π ⎛ 4
4
⋅ r − ri ⎞⎠
4 ⎝o
Maximum Bendiug Stress :
Mmax := R⋅ a
σ max :=
Mmax⋅ ro
I
σ max = 330.17 MPa
< σY = 703 MPa
Ans
(O.K. !)
M/EI Diagram:
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
1
M'1 x1 := R⋅ x1 ⋅
E⋅ I
M'2 x2 := ⎢R⋅ x2 −
( ) (
( )
)
⎡
⎣
( )
⎛
⎝
M'3 x3 := ⎢R⋅ x3 − W⋅ ⎜ x3 −
L ⎞⎤ 1
⎥⋅
2 ⎠⎦ E ⋅ I
⎡
⎣
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
W
⎤ 1
⋅ x2 − a ⎥ ⋅
2
⎦ E⋅ I
(
)
Moment-Area Theorems :
M/EI (1/m)
0.2
( )
M'2 ( x 2 )
0.1
M'3 ( x 3 )
M'1 x 1
0
Deflections :
tA.C :=
0
1
⎛ 2a ⎞ + M' ( a) ⋅ ⎛ b ⎞ ⋅ ⎛ b + a⎞
M'1 ( a) ⋅ a⋅ ⎜
⎜ ⎜
1
2
⎝3⎠
⎝ 2⎠ ⎝ 4 ⎠
∆ C := tA.C
∆ C = 65.74 mm
Due to symmetry, the slope at midspan (point C) is zero. Hence,
∆ max := ∆ C
∆ max = 65.74 mm Ans
1
x1 , x2 , x3
Distance (m)
2
Problem 12-69
Determine the value of a so that the displacement at C is equal to zero. EI is constant.
Problem 12-70
The beam is made of a ceramic material. In order to obtain its modulus of elasticity, it is subjected to
the elastic loading shown. If the moment of inertia is I and the beam has a measured maximum
deflection ∆, determine E. The supports at A and D exert only vertical reactions on the beam.
Problem 12-71
Determine the maximum deflection of the shaft. EI is constant. The bearings exert only vertical
reactions on the shaft.
Problem 12-72
The beam is subjected to the load P as shown. Determine the magnitude of force F that must be
applied at the end of the overhang C so that the deflection at C is zero. EI is constant.
Problem 12-73
At what distance a should the bearing supports at A and B be placed so that the deflection at the center
of the shaft is equal to the deflection at its ends? The bearings exert only vertical reactions on the shaft.
EI is constant.
Problem 12-74
Determine the slope of the 50-mm-diameter A-36 steel shaft at the bearings at A and B. The bearings
exert only vertical reactions on the shaft.
Given: a := 0.5m
b := 0.8m
P1 := 600N
c := 1.2m
P2 := 1200N
do := 50mm
E := 200GPa
Solution: L := a + b + c
Support Reactions :
+ ΣF =0;
A + B − P1 − P2 = 0
y
ΣΜB=0; A⋅ L − P1⋅ ( b + c) − P2⋅ c = 0
I :=
B := P1 + P2 − A
L
A = 1.056 kN
4
Section Property :
(2)
P1⋅ ( b + c) + P2⋅ c
A :=
Solving Eqs. (1) and (2):
(1)
π ⋅ do
B = 0.744 kN
64
M/EI Diagram:
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
1
M'1 x1 := A⋅ x1 ⋅
E⋅ I
M'2 x2 := ⎡⎣A⋅ x2 − P1⋅ x2 − a ⎤⎦ ⋅
( ) (
)
( )
( )
(
)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
(
(
1
) E⋅ I
1
) E⋅ I
M/EI (1/m)
M'3 x3 := ⎡⎣A⋅ x3 − P1⋅ x3 − a − P2⋅ x3 − a − b ⎤⎦ ⋅
Moment-Area Theorems :
( )
M'2 ( x 2 ) 0.01
M'3 ( x 3 )
M'1 x 1
Slopes :
tB.A :=
1
2c
M'3 ( L − c) ⋅ c⋅
...
2
3
1
b
+ M'2 ( a + b) − M'1 ( a) ⋅ b⋅ ⎛⎜ + c⎞ ...
2
3
(
)
⎝
0
0
1
2
x1 , x2 , x3
Distance (m)
⎠
2a ⎞
⎛b ⎞ 1
⎛
+ M'1 ( a) ⋅ b⋅ ⎜ + c + M'1 ( a) ⋅ a⋅ ⎜ L −
3⎠
⎝2 ⎠ 2
⎝
tB.A
θ A :=
θ A = 0.01046 rad
L
tA.B :=
1
2c ⎞
⎛
...
M'3 ( L − c) ⋅ c⋅ ⎜ L −
2
3⎠
⎝
2⋅ b Ans
1
+ M'2 ( a + b) − M'1 ( a) ⋅ b⋅ ⎛⎜
+ a⎞ ...
3
2
θ B :=
+ M'1 ( a) ⋅ ( b) ⋅ ⎜
θ B = 0.00968 rad
(
)
⎝
⎠
⎛ b + a⎞ + 1 M' ( a) ⋅ a⋅ ⎛ 2a ⎞
⎜
⎝2 ⎠ 2 1
⎝3⎠
tA.B
L
Ans
Problem 12-75
Determine the maximum deflection of the 50-mm diameter A-36 steel shaft. It is supported by bearings
at its ends A and B which only exert vertical reactions on the shaft.
Given: a := 0.5m
b := 0.8m
P1 := 600N
c := 1.2m
P2 := 1200N
do := 50mm
E := 200GPa
Solution: L := a + b + c
Support Reactions :
+ ΣF =0;
A + B − P1 − P2 = 0
y
ΣΜB=0; A⋅ L − P1⋅ ( b + c) − P2⋅ c = 0
I :=
B := P1 + P2 − A
L
A = 1.056 kN
4
Section Property :
(2)
P1⋅ ( b + c) + P2⋅ c
A :=
Solving Eqs. (1) and (2):
(1)
π ⋅ do
B = 0.744 kN
64
M/EI Diagram:
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
1
M'1 x1 := A⋅ x1 ⋅
E⋅ I
M'2 x2 := ⎡⎣A⋅ x2 − P1⋅ x2 − a ⎤⎦ ⋅
( ) (
( )
)
( )
(
)
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
(
(
1
) E⋅ I
M/EI (1/m)
M'3 x3 := ⎡⎣A⋅ x3 − P1⋅ x3 − a − P2⋅ x3 − a − b ⎤⎦ ⋅
Moment-Area Theorems :
Deflections :
tB.A :=
( )
M'2 ( x 2 ) 0.01
M'3 ( x 3 )
M'1 x 1
1
2c
M'3 ( L − c) ⋅ c⋅
...
2
3
1
b
+ M'2 ( a + b) − M'1 ( a) ⋅ b⋅ ⎛⎜ + c⎞ ...
2
3
(
1
) E⋅ I
0
0
)
x1 , x2 , x3
⎝
⎠
b
1
2a ⎞
⎞
⎛
⎛
+ M'1 ( a) ⋅ b⋅ ⎜ + c + M'1 ( a) ⋅ a⋅ ⎜ L −
3⎠
⎝2 ⎠ 2
⎝
θ A :=
Distance (m)
tB.A
Ans
θ A = 0.01046 rad
L
Maximum Deflection: Assume v max occurs at point E ( a < x' < a+b ).
Slopes :
1
1
x'
M'1 ( a) ⋅ a + M'1 ( a) ⋅ x' + M'2 ( a + b) − M'1 ( a) ⋅ ⋅ x'
2
2
b
1 A⋅ a
A⋅ a
1
x'
θ E.A =
⋅a +
⋅ x' +
A⋅ ( a + b) − P1⋅ b − A⋅ a⎤⎦ ⋅ ⋅ x'
⎡
⎣
2 E⋅ I
E⋅ I
2E⋅ I
b
θ E.A =
(
1
)
2
θ E.A =
1
⎡A⋅ a + 2A⋅ a⋅ x' + A − P ⋅ x' ⎤
1
⎦
2E⋅ I ⎣
(
2
) 2
θ E = −θ A + θ E.A
θ E = −θ A +
1
⎡A⋅ a2 + 2A⋅ a⋅ x' + ( A − P ) ⋅ x'2⎤
1
⎦
2E ⋅ I ⎣
At point E, where θE=0.
Given
2
(
) 2 (3)
0 = −θ A⋅ ( 2E⋅ I) + A⋅ a + 2A⋅ a⋅ x' + A − P1 ⋅ x'
Solving Eq.(3):
Guess x' := 0.7 ⋅ m
x' := Find ( x')
x' = 733.27 mm
< 0.8m (O.K. !)
Deflection:
1
2a
x'
1
2⋅ x'
x'
tA.E := ⎡⎢ M'1 ( a) ⋅ a⋅ ⎛⎜ ⎞ + M'1 ( a) ⋅ x'⋅ ⎛⎜ + a⎞ + M'2 ( a + b) − M'1 ( a) ⋅ ⋅ x'⋅ ⎛⎜
+ a⎞⎤⎥
b
⎣2
⎝3⎠
⎝2 ⎠ 2
⎝ 3
⎠⎦
(
∆ max := tA.E
∆ max = 8.161 mm
Ans
)
Problem 12-76
Determine the slope of the 20-mm-diameter A-36 steel shaft at the bearings at A and B. The bearings
exert only vertical forces on the shaft.
Given: a := 0.2m
b := 0.3m
P1 := 800N
c := 0.5m
do := 20mm
E := 200GPa
P2 := 350N
Solution: L := a + b + c
Support Reactions :
+ ΣF =0;
A + B − P1 − P2 = 0
y
ΣΜB=0; −P1⋅ L + A⋅ ( b + c) − P2⋅ c = 0
A :=
Solving Eqs. (1) and (2):
Section Property :
I :=
π ⋅ do
M/EI Diagram:
(1)
(2)
P1⋅ L + P2⋅ c
B := P1 + P2 − A
b+c
A = 1218.75 N
4
B = −68.75 N
64
x1 := 0 , 0.01 ⋅ a .. a
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
1
1
M'1 x1 := −P1⋅ x1 ⋅
M'2 x2 := ⎡⎣−P1⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
E⋅ I
E⋅ I
1
M'3 x3 := ⎡⎣−P1⋅ x3 + A⋅ x3 − a − P2⋅ x3 − a − b ⎤⎦ ⋅
E⋅ I
)
( )
( )
(
(
)
(
Moment-Area Theorems :
Slopes :
( )
M'2 ( x 2 ) 0.05
M'3 ( x 3 )
M'1 x 1
1
2c
⎛b ⎞
tB.A := M'3 ( a + b) ⋅ c⋅
+ M'3 ( a + b) ⋅ b⋅ ⎜ + c ...
2
3
⎝2 ⎠
1
2
⋅
b
+ M'2 ( a) − M'3 ( a + b) ⋅ b⋅ ⎛⎜
+ c⎞
2
3
(
θ A :=
tA.B :=
tB.A
b+c
⎝
⎠
θ A = 0.01811 rad
Ans
1
⎛c ⎞
⎛ b ⎞ ...
M'3 ( a + b) ⋅ c⋅ ⎜ + b + M'3 ( a + b) ⋅ b⋅ ⎜
2
⎝3 ⎠
⎝ 2⎠
1
b⎞
⎛
+ M'2 ( a) − M'3 ( a + b) ⋅ b⋅ ⎜
2
3
(
θ B :=
)
tA.B
b+c
)
⎝ ⎠
θ B = 0.00592 rad
)
)
M/EI (1/m)
( ) (
x2 := a , 1.01 ⋅ a .. ( a + b)
Ans
0.1
0
0.5
x1 , x2 , x3
Distance (m)
1
Problem 12-77
Determine the displacement of the 20-mm-diameter A-36 steel shaft at D. The bearings at A and B
exert only vertical reactions on the shaft.
Given: a := 0.2m
b := 0.3m
P1 := 800N
c := 0.5m
do := 20mm
E := 200GPa
P2 := 350N
Solution: L := a + b + c
Support Reactions :
+ ΣF =0;
A + B − P1 − P2 = 0
y
ΣΜB=0; −P1⋅ L + A⋅ ( b + c) − P2⋅ c = 0
A :=
Solving Eqs. (1) and (2):
Section Property :
I :=
π ⋅ do
M/EI Diagram:
(1)
(2)
P1⋅ L + P2⋅ c
B := P1 + P2 − A
b+c
A = 1218.75 N
4
B = −68.75 N
64
x1 := 0 , 0.01 ⋅ a .. a
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
1
1
M'1 x1 := −P1⋅ x1 ⋅
M'2 x2 := ⎡⎣−P1⋅ x2 + A⋅ x2 − a ⎤⎦ ⋅
E⋅ I
E⋅ I
1
M'3 x3 := ⎡⎣−P1⋅ x3 + A⋅ x3 − a − P2⋅ x3 − a − b ⎤⎦ ⋅
E⋅ I
( ) (
x2 := a , 1.01 ⋅ a .. ( a + b)
)
(
(
)
(
)
)
M/EI (1/m)
( )
( )
( )
M'2 ( x 2 ) 0.05
M'3 ( x 3 )
M'1 x 1
0.1
Moment-Area Theorems :
0
Deflection :
tD.B :=
1
2c ⎞
⎛
⎛b ⎞
M'3 ( a + b) ⋅ c⋅ ⎜ L −
+ M'3 ( a + b) ⋅ b⋅ ⎜ + a ...
2
3⎠
⎝
⎝2 ⎠
b
1
2a
1
+ M'2 ( a) − M'3 ( a + b) ⋅ b⋅ ⎛⎜ + a⎞ + M'1 ( a) ⋅ a⋅
3
2
3
2
(
tA.B :=
)
⎝
⎠
1
⎛c ⎞
⎛ b ⎞ ...
M'3 ( a + b) ⋅ c⋅ ⎜ + b + M'3 ( a + b) ⋅ b⋅ ⎜
2
⎝3 ⎠
⎝ 2⎠
1
b⎞
⎛
+ M'2 ( a) − M'3 ( a + b) ⋅ b⋅ ⎜
2
3
(
)
L
∆ D := tD.B −
t
b + c A.B
∆ D = 4.98 mm
Ans
⎝ ⎠
0.5
x1 , x2 , x3
Distance (m)
1
Problem 12-78
The beam is subjected to the loading shown. Determine the slope at B and deflection at C. EI is
constant.
Problem 12-79
Determine the slope at B and the displacement at C. The bearings at A and B exert only vertical
reactions on the shaft. EI is constant.
Problem 12-80
Determine the displacement at D and the slope at C. The bearings at A and B exert only vertical
reactions on the shaft. EI is constant.
Problem 12-81
The two force components act on the tire of the automobile as shown. The tire is fixed to the axle,
which is supported by bearings at A and B. Determine the maximum deflection of the axle. Assume
that the bearings resist only vertical loads. The thrust on the axle is resisted at C. The axle has a
diameter of 32 mm and is made of A-36 steel. Neglect the effect of axial load on deflection.
Given: a := 50mm
do := 32mm
Fv := 4.5kN
Fh := 1kN
b := 650mm
h := 450mm
E := 200GPa
Solution: L := a + b
Support Reactions :
+ ΣF =0; F − A + B = 0
y
v
ΣΜA=0; Fv⋅ a + Fh⋅ h − B⋅ b = 0
B :=
Solving Eqs. (1) and (2):
I :=
Section Property :
(1)
(2)
F v⋅ a + F h⋅ h
A := Fv + B
b
π
4
⋅d
64 o
M/EI Diagram:
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. L
1
M'1 x1 := Fh⋅ h + Fv⋅ x1 ⋅
E⋅ I
( ) (
)
( )
(
1
) E⋅ I
M'2 x2 := ⎡⎣Fh⋅ h + Fv⋅ x2 − A⋅ x2 − a ⎤⎦ ⋅
Moment-Area Theorems :
θB =
tA.B
θ B :=
M/EI (1/m)
Deflections :
1
⎛ b⎞
tA.B := M'1 ( a) ⋅ b⋅ ⎜
2
⎝ 3⎠
M'1 ( a) ⋅ b
( )
0.05
M'2 ( x 2 )
M'1 x 1
6
b
The maximum displacement occurs at point C,
where θC = 0. Let x' = L - x
θ C.B =
0
0
1 ⎡
⎛ x' ⎞⎤
⋅ ⎢M'1 ( a) ⋅ ⎜ ⎥ ⋅ x'
2 ⎣
⎝ b ⎠⎦
M'1 ( a) ⋅ b
6
x' := b⋅
∆ C = tB.C
∆ C :=
1
3
2
+
1
x'
⋅ M'1 ( a) ⋅
b
2
x' = 375.28 mm < b = 650 mm (O.K. !)
⎛ 2x' ⎞
⎝3⎠
tB.C = θ B.C⋅ ⎜
θ B.C = θ C.B
1 ⎡
⎛ x' ⎞⎤ ⎛ 2x' ⎞
⋅ ⎢M'1 ( a) ⋅ ⎜ ⎥ ⋅ x'⋅ ⎜
2 ⎣
⎝ b ⎠⎦ ⎝ 3 ⎠
∆ max := ∆ C
0.4
x1 , x2
Distance (m)
θ C = θ B + θ C.B
0= −
0.2
∆ max = 1.78 mm
Ans
0.6
Problem 12-82
Determine the displacement at D and the slope at C. The bearings at A and B exert only vertical
reactions on the shaft. EI is constant.
Given: a := 0.3m
P := 20kN
b := 0.9m
Solution: L := 2a + b
Support Reactions : By symmetry, A=B=R
ΣF y=0;
+
2R − 2P = 0
R := P
Maximum Bendiug Stress :
Mmax := R⋅ a
M/EI Diagram:
Set
EI := kN⋅ m
2
x1 := 0 , 0.01 ⋅ a .. a
EIo := 1
x2 := a , 1.01 ⋅ a .. ( a + b)
1
M'1 x1 := R⋅ x1 ⋅
EI
( ) (
)
( )
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
( )
(
(
)
(
) EI
M'3 x3 := ⎡⎣R⋅ x3 − P⋅ x3 − a − P⋅ x3 − a − b ⎤⎦ ⋅
1
) EI
M'2 x2 := ⎡⎣R⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
1
M/EI (1/m)
10
( )
M'2 ( x 2 )
5
M'3 ( x 3 )
M'1 x 1
Moment-Area Theorems :
0
Deflections :
tA.C :=
0
1
⎛ 2a ⎞ + M' ( a) ⋅ ⎛ b ⎞ ⋅ ⎛ b + a⎞
M'1 ( a) ⋅ a⋅ ⎜
⎜ ⎜
1
2
⎝3⎠
⎝ 2⎠ ⎝ 4 ⎠
∆ C := tA.C
∆ C = 1.5975
m
EIo
Due to symmetry, the slope at midspan (point C) is zero. Hence,
∆ max := ∆ C
∆ max = 1.5975
m
EIo
Ans
0.5
1
x1 , x2 , x3
Distance (m)
1.5
Problem 12-83
Beams made of fiber-reinforced plastic may one day replace many of those made of A-36 steel since
they are one-fourth the weight of steel and are corrosion resistant. Using the table in Appendix B, with
σallow = 160 MPa and τallow = 84 MPa, select the lightest-weight steel wide-flange beam that will safely
support the 25-kN load, then compute its maximum deflection. What would be the maximum deflection
of this beam if it were made of a fiber-reinforced plastic with Ep = 126 GPa and had the same moment
of inertia as the steel beam?
Given: σ allow := 160MPa
τ allow := 84MPa
P := 25kN
L := 3m
E := 126GPa
Solution:
Support Reactions : By symmetry, A=B=R
+
ΣF y=0;
2R − P = 0
R := 0.5P
R = 12.50 kN
Maximum Moment and Shear:
Vmax := R
Vmax = 12.5 kN
Mmax := R⋅ L
Mmax = 37.5 kN⋅ m
Bending Stress:
Mmax
Sreq'd :=
σ allow
Hence,
( 3) mm3
Sx := 244⋅ 10
Select W 310x21 :
Shear Stress :
3
Sreq'd = 234375.00 mm
d := 303mm
Provide a shear stress check.
Vmax
τ max :=
τ max = 8.12 MPa
tw⋅ d
Use W 310x21
tw := 5.08mm
< τ allow =84 MPa
I := Sx⋅ ( 0.5d)
(O.K.!)
Ans
M/EI Diagram:
x1 := 0 , 0.01 ⋅ L .. L
( )
⎡
(
⎣
) E⋅ I⎤⎥⎦
M'1 x1 := ⎢ R⋅ x1 ⋅
x2 := L , 1.01 ⋅ L .. ( 2L )
1
( )
⎡
⎣
) E⋅ I⎥⎦⎤
(
M'2 x2 := ⎢⎡⎣R⋅ x2 − P⋅ x2 − L ⎤⎦ ⋅
1
Deflections :
1
⎛ 2L ⎞
tA.C := M'1 ( L ) ⋅ L ⋅ ⎜
2
⎝3⎠
∆ C := tA.C
∆ C = 24.15 mm
Due to symmetry, the slope at
midspan (point C) is zero. Hence,
∆ max := ∆ C
∆ max = 24.15 mm Ans
M/EI
Moment-Area Theorems :
( )
0.005
M'2 ( x 2 )
M'1 x 1
0
0
2
4
x1 , x2
Distance (m)
6
Problem 12-84
The simply supported shaft has a moment of inertia of 2I for region BC and a moment of inertia I for
regions AB and CD. Determine the maximum deflection of the shaft due to the load P . The modulus of
elasticity is E.
Problem 12-85
The A-36 steel shaft is used to support a rotor that exerts a uniform load of 5 kN/m within the region
CD of the shaft. Determine the slope of the shaft at the bearings A and B. The bearings exert only
vertical reactions on the shaft.
Problem 12-86
The beam is subjected to the loading shown. Determine the slope at B and deflection at C. EI is
constant.
Problem 12-87
Determine the slope of the shaft at A and the displacement at D. The bearings at A and B exert only
vertical reactions on the shaft. EI is constant.
Problem 12-88
Determine the slope at B and the displacement at C. The member is an A-36 steel structural tee for
which I = 30(106) mm4.
kN
Given: P := 25kN
L := 1m
w := 25
m
6
4
E := 200GPa I := 30 10 mm
( )
Solution:
Support Reactions : By symmetry, A=B=R
ΣF y=0; 2R' − P = 0
+
2R'' − w⋅ ( 2L ) = 0
R' := 0.5P
R'' := w⋅ L
R' = 12.50 kN
R'' = 25.00 kN
M/EI Diagram:
x1 := 0 , 0.01 ⋅ L .. L
x2 := L , 1.01 ⋅ L .. ( 2L )
For point load P:
1
M'1 x1 := R'⋅ x1 ⋅
E⋅ I
For uniform load w :
( ) (
)
( )
(
1
2
M''1 x1 := ⎛⎝ R''⋅ x1 − 0.5w⋅ x1 ⎞⎠ ⋅
E⋅ I
1
2
M''2 x2 := ⎛⎝ R''⋅ x2 − 0.5 ⋅ w⋅ x2 ⎞⎠ ⋅
E⋅ I
( )
( )
M'2 ( x 2 )
M'1 x 1 0.002
M/EI
M/EI
( )
0
0
1
1
) E⋅ I
M'2 x2 := ⎡⎣R'⋅ x2 − P⋅ x2 − L ⎤⎦ ⋅
( )
M''2 ( x 2 )
M''1 x 1 0.002
0
2
0
1
2
x1 , x2
x1 , x2
Distance (m)
Distance (m)
Moment-Area Theorems :
Slope :
Due to symmetry, the slope at midspan (point C) is zero. Hence, θB = θB/C
θ B.C :=
1
2
⋅ M'1 ( L ) ⋅ L + ⋅ M''1 ( L ) ⋅ L
2
3
θ B := θ B.C
Deflection :
⎡1
⎛ 2L ⎞ + 2 ⋅ M'' ( L) ⋅ L⋅ ⎛ 5L ⎞⎤
tA.C := ⎢ ⋅ M'1 ( L ) ⋅ L ⋅ ⎜
⎜ ⎥
⎣2
⎝3⎠ 3 1
⎝ 8 ⎠⎦
∆ C := tA.C
∆ max := ∆ C
∆ max = 1.56 mm
Ans
θ B = 0.00243 rad
Ans
Problem 12-89
The W200 X 71 cantilevered beam is made of A-36 steel and is subjected to the loading shown.
Determine the displacement at its end A.
Given: P := 6kN
Mo := 3kN⋅ m
a := 2.4m
E := 200GPa
Solution:
L := 2a
Use W 200x71 :
( 6)
4
I := 76.6 ⋅ 10 mm
Elsastic Curve : The elastic curves for the concentrated load and
couple moment are drawn separately as shown.
Method of Superposition :
Using the table in Appendix C, the required slope
and displacement are,
2
3
Mo⋅ a
∆ C2 :=
2E⋅ I
Mo⋅ a
θ C2 :=
E⋅ I
P⋅ L
∆ A1 :=
3E ⋅ I
( )
∆ A2 := ∆ C2 + θ C2 ⋅ a
Hence, the displacement at A is
∆ A := ∆ A1 + ∆ A2
∆ A = 16.13 mm
Ans
Problem 12-90
The W200 X 71 cantilevered beam is made of A-36 steel and is subjected to the loading shown.
Determine the displacement at C and the slope at A.
Given: P := 6kN
Mo := 3kN⋅ m
a := 2.4m
E := 200GPa
Solution:
L := 2a
Use W 200x71 :
( 6)
4
I := 76.6 ⋅ 10 mm
Elsastic Curve : The elastic curves for the concentrated load and
couple moment are drawn separately as shown.
Method of Superposition :
Using the table in Appendix C, the required slope
and displacement are,
2
2
Mo⋅ a
∆ C2 :=
2E⋅ I
P⋅ a
∆ C1 :=
⋅ ( 3L − a)
6E ⋅ I
2
P⋅ L
θ A1 :=
2E ⋅ I
θ C2 :=
θ A2 := θ C2
Hence, the slope at A is
θ A := θ A1 + θ A2
θ A = 0.00498 rad
Ans
Hence, the displacement at C is
∆ C := ∆ C1 + ∆ C2
∆ C = 5.08 mm
Ans
Mo⋅ a
E⋅ I
Problem 12-91
The W360 X 64 simply supported beam is made of A-36 steel and is subjected to the loading shown.
Determine the deflection at its center C.
a := 3m
Given:
Mo := 60kN⋅ m
kN
w := 30
m
E := 200GPa
Solution:
L := 2a
Use W 360x64:
( 6)
4
I := 179⋅ 10 mm
Elsastic Curve : The elastic curves for the uniform load and couple
moment are drawn separately as shown.
Method of Superposition :
Using the table in Appendix C, the required displacements are,
4
∆ C1 :=
5w⋅ L
768E ⋅ I
∆ C2 :=
Hence, the displacement at C is
∆ C := ∆ C1 + ∆ C2
∆ C = 10.84 mm
Ans
(
Mo
2
⎛a⎞ 2
⋅ ⎜ ⋅ a − 3⋅ L⋅ a + 2⋅ L
6⋅ E⋅ I ⎝ L ⎠
)
Problem 12-92
The W360 X 64 simply supported beam is made of A-36 steel and is subjected to the loading shown.
Determine the slope at A and B.
a := 3m
Given:
Mo := 60kN⋅ m
kN
w := 30
m
E := 200GPa
Solution:
L := 2a
Use W 360x64:
( 6)
4
I := 179⋅ 10 mm
Elsastic Curve : The elastic curves for the uniform load and couple
moment are drawn separately as shown.
Method of Superposition :
Using the table in Appendix C, the required slopes are,
3
θ A1 :=
7w⋅ L
384E⋅ I
θ A2 :=
Mo⋅ L
6⋅ E⋅ I
Hence, the slope at A is
θ A := θ A1 + θ A2
θ A = 0.00498 rad
θ A = 0.285 deg
Ans
3
θ B1 :=
3w⋅ L
128E ⋅ I
θ B2 :=
Hence, the slope at B is
θ B := θ B1 + θ B2
θ B = 0.00759 rad
θ B = 0.435 deg
Ans
Mo⋅ L
3⋅ E ⋅ I
Problem 12-93
Determine the moment M0 in terms of the load P and dimension a so that the deflection at the center of
the beam is zero. EI is constant.
Problem 12-94
The beam supports the loading shown. Code restrictions, due to a plaster ceiling, require the maximum
deflection not to exceed 1/360 of the span length. Select the lightest-weight A-36 steel wide-flange
beam from Appendix B that will satisfy this requirement and safely support the load. The allowable
bending stress is σallow = 168 MPa and the allowable shear stress is τallow = 100 MPa. Assume A is a
roller and B is a pin.
Given: σ allow := 168MPa
a := 3.6m
τ allow := 100MPa
w := 60
E := 200GPa
kN
m
Solution: L := 2a
+
ΣF y=0;
A + B − w⋅ a = 0
(1)
2
ΣΜB=0; A⋅ L ⋅ −0.5 w⋅ a = 0 (2)
2
Solving Eqs. (1) and (2):
wa
2L
B := w⋅ a − A
A = 54 kN
B = 162 kN
A :=
Maximum Moment and Shear:
Vmax := max ( A , B)
Vmax = 162 kN
Mmax occurs at x where V(x) = 0:
V ( x) = A − w⋅ Φ ( x − a)
0 = A − w⋅ Φ ( x − a)
x :=
Mmax := A⋅ x − 0.5w⋅ ( x − a)
Strength criterion:
Mmax
Sreq'd :=
σ allow
Select W 410x74 :
Shear Stress :
2
A
+a
w
Mmax = 218.7 kN⋅ m
3
Sreq'd = 1301785.71 mm
( 3)
3
Sx := 1330 10 mm
d := 413mm
Provide a shear stress check.
Vmax
τ max :=
τ max = 40.65 MPa < τ allow =100 MPa
tw⋅ d
Deflection criterion:
L
∆ allow :=
360
I := Sx⋅ ( 0.5d)
(O.K.!)
∆ allow = 20 mm
Using the table in Appendix C,
4
w⋅ L
∆ max := 0.006563
E⋅ I
∆ max = 19.27 mm
Hence,
tw := 9.65mm
Use W 410x74
Ans
< ∆ allow =20mm
(O.K.!)
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. L
1
V1 x1 := ( A)
kN
1
V2 x2 := ⎡⎣A − w⋅ x2 − a ⎤⎦ ⋅
kN
1
M1 x1 := A⋅ x1
kN⋅ m
1
2
M2 x2 := ⎡⎣A⋅ x2 − 0.5 ⋅ w⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
V2 ( x 2 )
V1 x 1
)
)
( )
0
200
(
(
M (kN-m)
Shear (kN)
( ) (
( )
0
5
)
200
( )
M2( x2)
M1 x1
0
0
5
x1 , x2
x1 , x2
Distance (m)
Distance (m)
Problem 12-95
The simply supported beam carries a uniform load of 29 kN/m. Code restrictions, due to a plaster
ceiling, require the maximum deflection not to exceed 1/360 of the span length. Select the lightestweight A-36 steel wide-flange beam from Appendix B that will satisfy this requirement and safely
support the load. The allowable bending stress is σallow = 168 MPa and the allowable shear stress is
τallow = 100 MPa. Assume A is a pin and B a roller support.
Given: σ allow := 168MPa
a := 1.2m
τ allow := 100MPa
P := 40kN
E := 200GPa
w := 29
Solution: L := 2a + b
b := 2.4m
kN
m
Support Reactions : By symmetry, A=B=R
+
ΣF y=0;
2R − 2P − w⋅ L = 0
R := 0.5 ( 2P + w⋅ L )
R = 109.60 kN
Maximum Moment and Shear:
Vmax := R
Vmax = 109.6 kN
Mmax occurs at midspan.
Mmax := R⋅ ( 0.5L) − P⋅ ( 0.5L − a) − 0.5w⋅ ( 0.5L)
Mmax = 131.52 kN⋅ m
Strength criterion:
Mmax
Sreq'd :=
σ allow
Select W 360x51 :
Shear Stress :
2
3
Sreq'd = 782857.14 mm
( 3)
3
Sx := 794 10 mm
d := 355mm
Provide a shear stress check.
Vmax
τ max :=
τ max = 42.64 MPa < τ allow =100 MPa
tw⋅ d
Deflection criterion:
L
∆ allow :=
360
∆ allow = 13.33 mm
Using the table in Appendix C,
4
5w⋅ L
⎛ P⋅ a⋅ b ⎞ ⋅ ⎡L2 − a2 − ( 0.5L) 2⎤
∆ max :=
+ 2⋅ ⎜
⎣
⎦
384E ⋅ I
⎝ 6E⋅ I⋅ L ⎠
∆ max = 11.61 mm
Hence,
tw := 7.24mm
< ∆ allow =20mm
Use W 360x51
Ans
(O.K.!)
I := Sx⋅ ( 0.5d)
(O.K.!)
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( a + b)
1
V1 x1 := R − w⋅ x1
kN
( ) (
x3 := ( a + b) , 1.01 ⋅ ( a + b) .. L
1
V2 x2 := R − w⋅ x2 − P ⋅
kN
)
( ) (
1
2
M1 x1 := ⎛⎝ R⋅ x1 − 0.5 ⋅ w x1 ⎞⎠
kN⋅ m
)
1
V3 x3 := R − w⋅ x3 − 2P ⋅
kN
( ) (
)
1
2
M2 x2 := ⎡⎣R⋅ x2 − 0.5 ⋅ w x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
( )
(
)
1
2
M3 x3 := ⎡⎣R⋅ x3 − 0.5 ⋅ w x3 − P⋅ x3 − a − P⋅ x3 − a − b ⎤⎦ ⋅
kN⋅ m
( )
V2 ( x 2 )
V3 ( x 3 )
(
)
(
)
100
V1 x 1
M (kN-m)
Shear (kN)
( )
0
100
0
2
4
( )
100
M2( x2)
M3( x3)
M1 x1
0
0
2
x1 , x2 , x3
x1 , x2 , x3
Distance (m)
Distance (m)
4
Problem 12-96
The W250 X 45 cantilevered beam is made of A-36 steel and is subjected to unsymmetrical bending
caused by the applied moment. Determine the deflection of the centroid at its end A due to the loading.
Hint: Resolve the moment into components and use superposition.
Given:
L := 4.5m
E := 200GPa
Solution:
Use W 250x45:
Mo := 2.5kN⋅ m
θ := 30deg
( 6) 4
6
4
Iy := 7.03 ⋅ ( 10 ) mm
Ix := 71.1 ⋅ 10 mm
Displacements :
Using the table in Appendix C, the required displacements are,
Mo⋅ sin ( θ ) ⋅ L
∆ x_max :=
2E⋅ Iy
Mo⋅ cos ( θ ) ⋅ L
∆ y_max :=
2E⋅ Ix
2
Hence, the displacement at A is
∆ A := ∆ x_max + ∆ y_max
∆ A = 10.54 mm
Ans
2
Problem 12-97
The assembly consists of a cantilevered beam CB and a simply supported beam AB. If each beam is
made of A-36 steel and has a moment of inertia about its principal axis of Ix = 46(106) mm4, determine
the displacement at the center D of beam BA.
Given: P := 75kN
( 6)
4
I := 46⋅ 10 mm
E := 200GPa
L DA := 2.4m
L BC := 4.8m
L BA := 4.8m
Solution:
Consider beam AB :
Support Reactions : By symmetry, A=B=R
+
ΣF y=0;
2R − P = 0
R := 0.5P
Method of Superposition :
Using the table in Appendix C, the required slope
and displacement are,
3
R⋅ L BC
Consider beam BC : ∆ B :=
3E ⋅ I
3
Consider beam AB :
∆ D1 :=
P⋅ LBA
48E⋅ I
Hence, the displacement at D is
∆ D := ∆ D1 + ∆ D2
∆ D = 93.91 mm
Ans
∆ D2 :=
L DA
L BA
⋅∆B
Problem 12-98
The rod is pinned at its end A and attached to a torsional spring having a stiffness k, which measures
the torque per radian of rotation of the spring. If a force P is always applied perpendicular to the end of
the rod, determine the displacement of the force. EI is constant.
Problem 12-99
The relay switch consists of a thin metal strip or armature AB that is made of red brass C83400 and is
attracted to the solenoid S by a magnetic field. Determine the smallest force F required to attract the
armature at C in order that contact is made at the free end B. Also, what should the distance a be for
this to occur? The armature is fixed at A and has a moment of inertia of I = 0.18(10-12) m4.
Given: ∆ B := 2mm
L := 50mm
( − 12) m4
E := 101GPa
I := 0.18 10
Solution: L AC := L
L CB := L
Elsastic Curve : The elastic curves for the concentrated load and
couple moment are drawn separately as shown.
Method of Superposition :
Using the table in Appendix C, the required slope
and displacement are,
2
θC =
F⋅ L AC
2E⋅ I
3
∆C =
F⋅ L AC
(1)
3E⋅ I
Hence, the displacement at B is
∆ B = ∆ C + θ C⋅ L CB
3
∆B =
F⋅ L AC
3E⋅ I
2
+
F⋅ L AC
2E⋅ I
⋅ L CB
6∆ B⋅ E ⋅ I
F :=
3
2
F = 0.349 N
Ans
∆ C = 0.800 mm
Ans
2⋅ LAC + 3⋅ L CB⋅ LAC
3
From Eq.(1),
∆ C :=
F⋅ LAC
3E ⋅ I
Problem 12-100
Determine the vertical deflection and slope at the end A of the bracket. Assume that the bracket is fixed
supported at its base, and neglect the axial deformation of segment AB. EI is constant.
Given: h := 75mm
P := 400N
L := 100mm
w := 4
Solution:
2
Set EI := 1kN m
kN
m
Elsastic Curve :
The elastic curves for the concentrated load,
uniform distributed load and couple moment
are drawn separately as shown.
MB := 0.5w⋅ L
2
Method of Superposition :
Using the table in Appendix C, the required slope
and displacement are,
3
w⋅ L
θ A1 :=
6EI
MB⋅ h
θ B2 :=
θ A2 := θ B2
EI
2
θ B3 :=
P⋅ h
2EI
θ A3 := θ B3
4
∆ A1 :=
w⋅ L
8EI
∆ A2 := θ B2⋅ ( L)
∆ A3 := θ B3⋅ ( L)
Hence, the slope at A is
θ A := θ A1 + θ A2 + θ B3
θ A = 0.00329 rad
θA =
0.00329
EI
Ans
∆A =
0.313
EI
Ans
The displacement at A is
∆ A := ∆ A1 + ∆ A2 + ∆ A3
∆ A = 0.313 mm
Problem 12-101
The W610 X 155 A-36 steel beam is used to support the uniform distributed load and a concentrated
force which is applied at its end. If the force acts at an angle with the vertical as shown, determine the
horizontal and vertical displacement at point A.
Given: L := 3m
P := 25kN
E := 200GPa
3
cx :=
5
Solution:
4
cy :=
5
w := 30
Use W 610x155:
Ix := 1290⋅ 10
kN
m
( 6) mm4
6
4
Iy := 108⋅ ( 10 ) mm
Elsastic Curve :
The elastic curves for the concentrated load and
uniform load are drawn separately as shown.
Py := P⋅ cy
Py = 20 kN
Px := P⋅ cx
Px = 15 kN
Method of Superposition :
Using the table in Appendix C, the required displacement are,
4
∆ A1 :=
∆ A2 :=
w⋅ L
8E⋅ Ix
P y⋅ L
3
3E ⋅ Ix
The vertical displacement at A is
∆ A_y := ∆ A1 + ∆ A2
∆ A_y = 1.875 mm
Ans
The horizontal displacement at A is
∆ A_x :=
Px⋅ L
3
3E⋅ Iy
∆ A_x = 6.250 mm
Ans
Problem 12-102
The framework consists of two A-36 steel cantilevered beams CD and BA and a simply supported
beam CB. If each beam is made of steel and has a moment of inertia about its principal axis of Ix =
46(106) mm4, determine the deflection at the center G of beam CB.
Given: P := 75kN
( 6)
4
I := 46⋅ 10 mm
E := 200GPa
L DC := 4.8m
L BC := 4.8m
L AB := 4.8m
Solution:
Consider beam BC :
Support Reactions : By symmetry, B=C=R
+
ΣF y=0;
2R − P = 0
R := 0.5P
Method of Superposition :
Using the table in Appendix C, the required slope
and displacement are,
3
R⋅ L DC
Consider beam DC :
∆ C :=
Due to symmetry,
∆ B := ∆ C
Consider beam BC :
∆' G :=
3E ⋅ I
P⋅ L BC
48E ⋅ I
Hence, the displacement at G is
∆ G := ∆ C + ∆' G
∆ G = 169.04 mm
Ans
3
Problem 12-103
Determine the reactions at the supports A and B, then draw the moment diagram. EI is constant.
Problem 12-104
Determine the reactions at the supports A and B, then draw the shear and moment diagrams. EI is
constant. Neglect the effect of axial load.
Problem 12-105
Determine the reactions at the supports A, B, and C; then draw the shear and moment diagrams. EI is
constant.
Problem 12-106
Determine the reactions at the supports, then draw the shear and moment diagram. EI is constant.
Problem 12-107
Determine the moment reactions at the supports A and B. EI is constant.
Problem 12-108
Determine the value of a for which the maximum positive moment has the same magnitude as the
maximum negative moment. EI is constant.
Problem 12-109
Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
Problem 12-110
The beam has a constant E 1I1 and is supported by the fixed wall at B and the rod AC. If the rod has a
cross-sectional area A 2 and the material has a modulus of elasticity E2 , determine the force in the rod.
Problem 12-111
Determine the moment reactions at the supports A and B, and then draw the shear and moment
diagrams. Solve by expressing the internal moment in the beam in terms of Ay and MA. EI is constant.
Problem 12-112
Determine the moment reactions at the supports A and B. EI is constant.
Problem 12-113
Determine the moment reactions at the supports A and B, then draw the shear and moment diagrams.
EI is constant.
Problem 12-114
Determine the moment reactions at the supports A and B, then draw the shear and moment diagrams.
EI is constant.
Problem 12-115
Determine the reactions at the supports. EI is constant.
Problem 12-116
Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
Given: P := 25kN
a := 1.2m
b := 1.8m
Solution: L := a + b
Due to symmetry, the slope at support
B is zero. Hence, θB = 0 and tanθB = 0.
Moment-Area Theorems :
Set
EI := 1kN m
2
A := 1kN
x1 := 0 , 0.01 ⋅ a .. a
For point load P:
( )
M'1 x1 := ( 0) ⋅
x2 := a , 1.01 ⋅ a .. L
1
EI
( )
For reaction A :
1
M''1 x1 := A⋅ x1 ⋅
EI
( ) (
(
1
) EI
M'2 x2 := ⎡⎣−P⋅ x2 − a ⎤⎦ ⋅
1
M''2 x2 := A⋅ x2 ⋅
EI
)
( ) (
)
( )
M'2 ( x 2 )
M'1 x 1
M/EI
M/EI
0
50
0
1
2
( )
M''2 ( x 2 )
M''1 x 1
0
3
x1 , x2
tA.B_2 :=
1
⎛ 2L ⎞
⋅ M''2 ( L) ⋅ L⋅ ⎜
2
⎝3⎠
1
2
Distance (m)
Deflection :
1
⎛ 2⋅ b ⎞
⋅ M'2 ( L) ⋅ b⋅ ⎜ a +
2
3 ⎠
⎝
0
x1 , x2
Distance (m)
tA.B_1 :=
5
⎛ Ay ⎞
⎝A⎠
tA.B = tA.B_1 + tA.B_2⋅ ⎜
Compatibility : tA.B := 0
⎛ Ay ⎞
⎝A⎠
0 = tA.B_1 + tA.B_2⋅ ⎜
⎛ tA.B_1 ⎞
Ay := −A⋅ ⎜
⎝ tA.B_2 ⎠
Ay = 10.80 kN
Ans
Cy := Ay
Cy = 10.80 kN
Ans
By := 2P − Ay − Cy
By = 28.40 kN
Ans
Support Reactions : Due to symmetry,
+
ΣF y=0;
Ay + By + Cy − 2P = 0
3
x1 := 0 , 0.01 ⋅ a .. a
x3 := L , 1.01 ⋅ L .. ( L + b)
x2 := a , 1.01 ⋅ a .. L
x4 := ( L + b) , 1.01 ⋅ ( L + b) .. ( 2L )
1
V1 x1 := Ay
kN
1
V3 x3 := Ay − P + By ⋅
kN
1
V2 x2 := Ay − P ⋅
kN
1
V4 x3 := Ay − P + By − P ⋅
kN
( ) ( )
( ) (
( ) (
)
)
( ) (
)
1
M1 x1 := Ay⋅ x1
kN⋅ m
( ) (
)
1
M2 x2 := ⎡⎣Ay⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
( )
(
)
1
M3 x3 := ⎡⎣Ay⋅ x3 − P⋅ x3 − a + By⋅ x3 − L ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
1
M4 x4 := ⎡⎣Ay⋅ x4 − P⋅ x4 − a + By⋅ x4 − L − P⋅ x4 − L − b ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
(
)
20
( )
V2 ( x 2 )
V3 ( x 3 ) 0
V4 ( x 4 )
Shear (kN)
V1 x 1
20
0
1
2
3
4
5
6
x1 , x2 , x3 , x4
Distance (m)
20
( )
M2( x2)
M3( x3) 0
M4( x4)
M (kN-m)
M1 x1
20
0
1
2
3
x1 , x2 , x3 , x4
Distance (m)
4
5
6
Problem 12-117
Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
Support B is a thrust bearing.
Problem 12-118
Determine the reactions at the supports. EI is constant.
Problem 12-119
Determine the value of a for which the maximum positive moment has the same magnitude as the
maximum negative moment. EI is constant.
Problem 12-120
Determine the moment reactions at the supports A and B, then draw the shear and moment diagrams.
EI is constant.
Problem 12-121
Determine the reactions at the supports A and B. EI is constant.
Problem 12-122
Determine the reactions at the bearing supports A, B, and C of the shaft, then draw the shear and
moment diagrams. EI is constant. Each bearing exerts only vertical reactions on the shaft.
Given: a := 1m
P := 0.4kN
b := 2m
L := 4a
Solution:
EI := kN m
Set
2
Method of Superposition :
Using the table in Appendix C, the required displacements are,
(
P⋅ a ⎛ b ⎞ 2
2
2
⋅⎜ ⋅ L − a − b
6⋅ EI ⎝ L ⎠
∆ B1 :=
)
3
B⋅ L
∆ B2 = −
48EI
The compatibity condition at B requires
2∆ B1 + ∆ B2 = 0
∆ B2 = −2∆ B1
3
B⋅ L
= 2∆ B1
48E⋅ I
⎛ 48⋅ EI ⎞ ⋅ 2∆
B := ⎜
B1
3
⎝ L ⎠
(
Support Reactions :
+
)
B = 0.550 kN
Ans
Given
ΣFy=0;
A + B + C − 2P = 0
(1)
ΣΜC=0;
A⋅ L − P⋅ ( 3a) + B⋅ ( b) − P⋅ ( a) = 0
(2)
Solving Eqs. (1) and (2):
Guess A := 1kN
C := 1kN
⎛A⎞
:= Find ( A , C)
⎜
⎝C⎠
⎛ A ⎞ ⎛ 0.125 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 0.125 ⎠
Ans
x1 := 0 , 0.01 ⋅ a .. a
1
V1 x1 := ( A)
kN
x2 := a , 1.01 ⋅ a .. ( 2a)
x3 := ( 2a) , 1.01 ⋅ ( 2a) .. ( 3a) x4 := ( 3a) , 1.01 ⋅ ( 3a) .. L
1
1
V2 x2 := ( A − P) ⋅
V3 x3 := ( A − P + B) ⋅
kN
kN
1
V4 x4 := ( A − P + B − P) ⋅
kN
( )
( )
( )
( )
1
M1 x1 := A⋅ x1
kN⋅ m
( ) (
1
M2 x2 := ⎡⎣A⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
)
( )
(
)
1
M3 x3 := ⎡⎣A⋅ x3 − P⋅ x3 − a + B⋅ x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
1
M4 x4 := ⎡⎣A⋅ x4 − P⋅ x4 − a + B⋅ x4 − 2a − P⋅ x4 − 3a ⎤⎦ ⋅
kN⋅ m
( )
(
( )
V2 ( x 2 )
V3 ( x 3 ) 0
V4 ( x 4 )
(
)
(
)
( )
M2(x2)
0
M3(x3)
M4(x4)
0.1
M 1 x 1 0.1
M (kN-m)
V1 x 1
Shear (kN)
)
0
2
4
0
2
x1 , x2 , x3 , x4
x1 , x2 , x3 , x4
Distance (m)
Distance (m)
4
Problem 12-123
The A-36 steel beam and rod are used to support the load of 40 kN. If it is required that the allowable
normal stress for the steel is σallow = 125 MPa, and the maximum deflection not exceed 1.25 mm,
determine the smallest diameter rod that should be used. The beam is rectangular, having a height of
125 mm and a thickness of 75 mm.
Given: σ allow := 125MPa
L b := 1.2m
L r := 1.5m
δ allow := 1.25mm
b := 75mm
h := 125mm
E := 200GPa
P := 40kN
Solution:
Compatibility at B :
3
δr = δb
Fr ⋅ L r
b⋅ h
Ib :=
12
E⋅ Ar
3
P − Fr ) ⋅ L b
(
=
3E ⋅ Ib
Strength criterion: Assume rod reaches its maximum stress.
σ allow =
Fr
σ allow⋅ Lr
Ar
E
3
P − Fr ) ⋅ L b
(
=
3E⋅ Ib
Fr := P −
3σ allow⋅ L r⋅ Ib
3
Fr = 36.03 kN
Check Bending Stress : c := 0.5 ⋅ h
(
)
Mmax := P − Fr ⋅ Lb
σ max :=
Lb
Mmax⋅ c
Ib
σ max = 24.41 MPa
< σ allow =125 MPa
Check Deflection :
δ b :=
(P − Fr)⋅ Lb3
Fr
Hence, from σ allow =
Ar
σ allow =
do :=
< δ allow =1.25mm
δ b = 0.94 mm
3E ⋅ Ib
Ar =
π ⋅ do
4Fr
2
π ⋅ do
4Fr
π ⋅ σ allow
do = 19.16 mm
Ans
4
2
(O.K.!)
(O.K.!)
Problem 12-124
Determine the reactions at the supports A, B, and C, then draw the shear and moment diagrams. EI is
constant.
kN
Given: a := 2m
P := 60kN
w := 50
m
b := 4m
L := 2a + b
Solution:
EI := 1kN m
Set
2
Method of Superposition :
Using the table in Appendix C, the required displacements are,
4
5w⋅ L
∆ B1 :=
768EI
P⋅ a ⎛ b ⎞ 2
2
2
∆ B2 :=
⋅⎜ ⋅ L − a − b
6⋅ EI ⎝ L ⎠
(
)
3
B⋅ L
48EI
The compatibity condition at C requires
∆ B3 = −
∆ B1 + ∆ B2 + ∆ B3 = 0
(
)
∆ B3 = − ∆ B1 + ∆ B2
3
B⋅ L
= ∆ B1 + ∆ B2
48E ⋅ I
⎛ 48⋅ EI ⎞ ⋅ ∆ + ∆
B := ⎜
B1
B2
3
⎝ L ⎠
(
Support Reactions :
+
)
B = 166.25 kN
Ans
Given
ΣF y=0;
A + B + C − P − w⋅ b = 0
ΣΜC=0;
2
A⋅ L − P⋅ ( a + b) + B⋅ b − 0.5 ⋅ w⋅ b = 0 (2)
(1)
Solving Eqs. (1) and (2):
Guess A := 1kN
C := 1kN
⎛A⎞
:= Find ( A , C)
⎜
⎝C⎠
⎛ A ⎞ ⎛ 11.88 ⎞
=⎜
kN
⎜
⎝ C ⎠ ⎝ 81.88 ⎠
Ans
x1 := 0 , 0.01 ⋅ a .. a
x2 := a , 1.01 ⋅ a .. ( 2a)
x3 := ( 2a) , 1.01 ⋅ ( 2a) .. L
1
V1 x1 := ( A)
kN
1
V2 x2 := ( A − P) ⋅
kN
1
V3 x3 := ⎡⎣A − P + B − w⋅ x3 − 2a ⎤⎦ ⋅
kN
( )
( )
1
M1 x1 := A⋅ x1
kN⋅ m
( ) (
( )
(
1
M2 x2 := ⎡⎣A⋅ x2 − P⋅ x2 − a ⎤⎦ ⋅
kN⋅ m
)
( )
(
)
1
2
M3 x3 := ⎡⎣A⋅ x3 − P⋅ x3 − a + B⋅ x3 − 2a − 0.5 ⋅ w x3 − 2a ⎤⎦ ⋅
kN⋅ m
( )
(
)
(
)
(
)
( )
V2 ( x 2 )
V3 ( x 3 )
100
V1 x 1
M (kN-m)
Shear (kN)
200
0
200
0
5
( )
M2( x2)
M3( x3)
M1 x1
0
100
0
5
x1 , x2 , x3
x1 , x2 , x3
Distance (m)
Distance (m)
)
Problem 12-125
Determine the reactions at support C. EI is constant for both beams.
Problem 12-126
Determine the reactions at A and B. Assume the support at A only exerts a moment on the beam. EI is
constant.
Problem 12-127
Determine the reactions at the supports A and B. EI is constant.
Problem 12-128
Each of the two members is made from 6061-T6 aluminum and has a square cross section 25 mm X
25 mm. They are pin connected at their ends and a jack is placed between them and opened until the
force it exerts on each member is 2.5 kN. Determine the greatest force P that can be applied to the
center of the top member without causing either of the two members to yield. For the analysis neglect
the axial force in each member. Assume the jack is rigid.
Given: b := 25mm
h := 25mm
Ro := 2.5kN
E := 68.9GPa
L := 4m
σ allow := 255MPa
Solution:
δE = δF
Compatibility at B :
⎡⎣P − ( R + Ro)⎤⎦ ⋅ L
(R + Ro)⋅ L
=
3
48E ⋅ I
3
48E⋅ I
(
) (
P − R + Ro = R + Ro
)
R = 0.5P − Ro
Criterion:
Maximum moment occursat center of each member.
Top member:
1
L
Mmax = ⎡⎣P − R + Ro ⎤⎦ ⋅
2
2
(
)
Mmax =
P⋅ L
8
Mmax =
P⋅ L
8
Bottom member:
1
L
Mmax =
R + Ro ⋅
2
2
(
)
Both members will yieldat the same time.
3
I :=
b⋅ h
12
c := 0.5 ⋅ h
σ allow =
Mmax⋅ c
I
σ allow =
( P⋅ L ) ⋅ c
8I
P :=
( 8I) ⋅ σ allow
L⋅ c
P = 1.33 kN
Ans
Problem 12-129
Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.
Problem 12-130
The beam is supported by a pin at A, a spring having a stiffness k at B, and a roller at C. Determine the
force the spring exerts on the beam. EI is constant.
Problem 12-131
The beam AB has a moment of inertia I = 200(106) mm4 and rests on the smooth supports at its ends.
A 18-mm-diameter rod CD is welded to the center of the beam and to the fixed support at D. If the
temperature of the rod is decreased by 80°C, determine the force developed in the rod. The beam and
rod are both made of A-36 steel.
Unit used: °C := deg
( 6) mm4
Given: rc := 18mm
Ib := 200⋅ 10
L c := 1.25m
L b := 3m
∆ T := 80°C
α := 12⋅ 10
( − 6) 1
°C
E := 200GPa
Solution:
π 2
Ac := ⋅ rc
4
Section Property:
Method of Superposition :
Using the table in Appendix C, the required displacements are,
vC =
Fc⋅ Lb
3
48E⋅ Ib
Using the axial force formula,
Fc⋅ Lc
δF = −
Ac E
The thermal contraction is,
( )
δ T := α ⋅ ∆ T ⋅ Lc
The compatibity condition at C requires
vC = δ F + δ T
Fc⋅ Lb
vC − δ F = δ T
3
48E⋅ Ib
Fc :=
−
Lb
−Fc⋅ Lc
Ac E
δT
3
48E⋅ Ib
+
Fc = 31.07 kN
= δT
Lc
Ac E
Ans
Problem 12-132
Determine the deflection at the end B of the clamped A-36 steel strip. The spring has a stiffness of
k = 2 N/mm. The strip is 5 mm wide and 10 mm high. Also, draw the shear and moment diagrams for
the strip.
Given: L := 200mm P := 50N
N
h := 10mm
k := 2
mm
t := 5mm
E := 200GPa
Solution:
3
t⋅ h
I :=
12
Method of Superposition :
Section Property :
Using the table in Appendix C, the required displacements are,
3
P⋅ L
∆ B1 :=
3⋅ E⋅ I
3
F⋅ L
∆ B2 = −
F = k⋅ ∆ B
3( E ⋅ I )
The compatibity condition at B requires
∆ B = ∆ B1 + ∆ B2
3
k⋅ ∆ B⋅ L
P⋅ L
∆B =
−
3⋅ ( E⋅ I)
3( E ⋅ I)
∆ B :=
P⋅ L
3
3
3⋅ ( E ⋅ I) + k⋅ L
∆ B = 1.50 mm
3
Ans
Support Reactions :
B := k⋅ ∆ B − P
A := −B
B = −46.992 N
A = 46.992 N
x1 := 0 , 0.01 ⋅ L .. L
A
V1 x1 :=
N
( )
MA := B⋅ L
MA = −9.40 N⋅ m
1
M1 x1 := MA + A⋅ x1
N⋅ m
( ) (
)
0
M (N-m)
Shear (N)
50
( )
V1 x 1
0
0
0.1
( )
M1 x1
10
0
0.1
x1
x1
Distance (m)
Distance (m)
Problem 12-133
The beam is made from a soft elastic material having a constant EI. If it is originally a distance ∆ from
the surface of its end support, determine the distance a at which it rests on this support when it is
subjected to the uniform load w0 , which is great enough to cause this to happen.
Problem 12-134
The box frame is subjected to a uniform distributed loading w along each of its sides. Determine the
moment developed in each corner. Neglect the deflection due to axial load. EI is constant.
Problem 12-135
Determine the equation of the elastic curve for the beam. Specify the slope and displacement at A. EI is
constant.
Problem 12-136
The wooden beam is subjected to the loading shown. Assume the support at A is a pin and B is a roller.
Determine the slope at A and the displacement at C. Use the moment-area theorems. EI is constant.
Problem 12-137
Determine the maximum deflection between the supports A and B. EI is constant. Use the method of
integration.
Problem 12-138
If the bearings at A and B exert only vertical reactions on the shaft, determine the slope at B and the
deflection at C. EI is constant. Use the moment-area theorems.
Problem 12-139
The W200 X 36 simply supported beam is subjected to the loading shown. Using the method of
superposition, determine the deflection at its center C. The beam is made of A-36 steel.
Given: a := 2.4m
Mo := 7.5kN⋅ m
kN
w := 100
m
E := 200GPa
Solution:
L := 2a
Use W 200x36 :
( 6)
4
I := 34.4 ⋅ 10 mm
Elsastic Curve : The elastic curves for the uniform load and couple
moment are drawn separately as shown.
Method of Superposition :
Using the table in Appendix C, the required displacement are,
4
∆ C1 :=
5w⋅ L
768E ⋅ I
∆ C2 :=
Mo a
⎛ ⎞ ⋅ a2 − 3a⋅ L + 2L2
⎜
6E ⋅ I ⎝ L ⎠
(
Hence, the displacement at C is
∆ C := ∆ C1 + ∆ C2
∆ C = 51.80 mm
Ans
)
Problem 12-140
The shaft is supported by a journal bearing at A, which exerts only vertical reactions on the shaft, and
by a thrust bearing at B, which exerts both horizontal and vertical reactions on the shaft. Draw the
bending-moment diagram for the shaft and then, from this diagram, sketch the deflection or elastic
curve for the shaft's centerline. Determine the equations of the elastic curve using the coordinates x1
and x2 . EI is constant.
Given: L := 0.3m
a := 0.1m
P := 400N
Solution:
Support Reactions :
L o := 2L
B = −A
Due to anti-symmetry,
ΣΜB=0;
A⋅ ( 2L ) − P⋅ ( 2a) = 0
⎛ a ⎞P
⎝L⎠
A := ⎜
B := −A
A = 133.33 N
B = −133.33 N
( )
( )
M1 x1 = A⋅ x1
Slope and Elastic Curve :
Moment Function :
2
E ⋅ I⋅
d ⋅ v1
dx1
M2 x2 = B⋅ x2
2
( )
E ⋅ I⋅
= M x1
2
dx2
2
E ⋅ I⋅
d v1
dx1
E ⋅ I⋅
2
dv1
dx1
E ⋅ I ⋅ v1 =
d ⋅ v2
( )
= M x2
2
2
= A⋅ x1
E ⋅ I⋅
dx2
2
=
A⋅ x1
2
A⋅ x1
d v2
+ C1
E ⋅ I⋅
(1)
2
dv2
=
dx2
3
B⋅ x2
0 = 0 + 0 + C2
(3)
B⋅ x2
6
+ C3⋅ x2 + C4
From Eq. (4):
0 = 0 + 0 + C4
(4)
C4 := 0
dv1
dx1
= −
2
dv2
dx2
at x1=L and at x2=L.
2
From Eqs. (1) and (3),
A⋅ L
B⋅ L
+ C1 = −
− C3
2
2
Continuity Condition 2 :
v1 = v2
From Eqs. (2) and (4),
A⋅ L
B⋅ L
+ C1⋅ L =
+ C3⋅ L
6
6
3
Solving Eqs. (5) and (6),
+ C3
2
C2 := 0
Continuity Condition 1 :
2
3
+ C1⋅ x1 + C2 (2)
E ⋅ I ⋅ v2 =
6
Boundary Conditions : v1=0 at x1=0 and v2=0 at x2=0.
From Eq. (2):
= B⋅ x2
C1 = −C3
(5)
at x1=L and at x2=L.
3
2
C1 =
( B − A) ⋅ L
+ C3 (6)
6
( B − A) ⋅ L
C1 :=
12
C3 := −C1
2
C1 = −2.00 N⋅ m
C3 = 2.00 N⋅ m
A
Hence, Co :=
6
Co = 22.22 N
2
Ans
2
Ans
Ans
v1 =
1⎛
3
C ⋅ x + C1⋅ x1⎞⎠
EI ⎝ o 1
Ans
v2 =
1⎛
3
−Co⋅ x2 + C3⋅ x2⎞⎠
⎝
EI
Ans
M/EI diagram :
Set
EI := 1kN⋅ m
2
x1 := 0 , 0.01 ⋅ L .. L
x2 := L , 1.01 ⋅ L .. ( 2L )
1
M'1 x1 := A⋅ x1 ⋅
EI
M'2 x2 := ⎡⎣A⋅ x2 − P⋅ ( 2a)⎤⎦ ⋅
)
( )
M/EI (1/m)
( ) (
1
EI
( )
0
M'2 ( x 2 )
M'1 x 1
0
0.2
0.4
x1 , x2
Distance (m)
Problem 12-141
The rim on the flywheel has a thickness t, width b, and specific weight γ. If the flywheel is rotating at
a constant rate of ω, determine the maximum moment developed in the rim. Assume that the spokes do
not deform. Hint: Due to symmetry of the loading, the slope of the rim at each spoke is zero. Consider
the radius to be sufficiently large so that the segment AB can be considered as a straight beam fixed at
both ends and loaded with a uniform centrifugal force per unit length. Show that this force is w =
btγω2r/g.
Problem 12-142
Determine the moment reactions at the supports A and B. Use the method of integration. EI is
constant.
Problem 12-143
Using the method of superposition, determine the magnitude of M0 in terms of the distributed load w
and dimension a so that the deflection at the center of the beam is zero. EI is constant.
Problem 13-1
Determine the critical buckling load for the column. The material can be assumed rigid.
Solution:
Equilibrium: The disturbing force F can be deternined by summing
moments about point A.
ΣΜA=0;
⎛ L⎞ = 0
⎝ 2⎠
P⋅ ( L ⋅ θ ) − F⋅ ⎜
F = 2P⋅ θ
Spring Formula :
The restoring spring force F s can be deternined using
the spring formula Fs = kx.
Fs = k⋅
L
k⋅ L⋅ θ
⋅θ =
2
2
Critical Buckling Load :
For the mechanism to be on the verge of buckling, the
disturbing force F must be equal to the spring force Fs.
Thus,
2Pcr⋅ θ =
k⋅ L⋅ θ
2
Pcr =
k⋅ L
4
Ans
Problem 13-2
The column consists of a rigid member that is pinned at its bottom and attached to a spring at its top.
If the spring is unstretched when the column is in the vertical position, determine the critical load that
can be placed on the column.
Solution:
Equilibrium:
ΣΜA=0;
P⋅ L⋅ sin ( θ ) − k⋅ L ⋅ sin ( θ ) ⋅ ( L⋅ cos ( θ ) ) = 0
P = k⋅ L ⋅ cos ( θ )
Since θ is small, cos ( θ )=1.
Pcr = k⋅ L
Ans
Problem 13-3
An A-36 steel column has a length of 4 m and is pinned at both ends. If the cross sectional area has the
dimensions shown, determine the critical load.
Given:
L := 4m
E := 200⋅ GPa
b := 25mm
a := 10mm
σ Y := 250MPa
Solution:
Section Property:
2
bo := 2⋅ b + a
A := 2⋅ bo⋅ a − a
3
Ix :=
a⋅ bo
12
3
2⋅ b⋅ a
+
12
Iy := Ix
Buckling Load :
Applying Euler's formula,
For pinned ends,
(
)
I := min Ix , Iy
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 22.7 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 20.66 MPa
< σY = 250 MPa
Therefore, Euler's formula is valid.
(O.K.!)
Problem 13-4
Solve Prob. 13-3 if the column is fixed at its bottom and pinned at its top.
Given:
L := 4m
E := 200⋅ GPa
b := 25mm
a := 10mm
σ Y := 250MPa
Solution:
Section Property:
2
bo := 2⋅ b + a
A := 2⋅ bo⋅ a − a
3
Ix :=
a⋅ bo
12
3
2⋅ b⋅ a
+
12
Iy := Ix
Buckling Load :
Applying Euler's formula,
(
)
I := min Ix , Iy
For fixed-pinned ends, K := 0.7
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 46.4 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 42.15 MPa
< σY = 250 MPa
Therefore, Euler's formula is valid.
(O.K.!)
Problem 13-5
A square bar is made from PVC plastic that has a modulus of elasticity of E = 9 GPa and a yield strain
of εY = 0.001 mm/mm. Determine its smallest cross-sectional dimensions a so it does not fail from
elastic buckling. It is pinned at its ends and has a length of 1250 mm.
Given:
L := 1.250m
ε Y := 0.001
Solution:
E := 9⋅ GPa
mm
mm
σ Y := E ⋅ ε Y
Stress-strain Relationship:
σ Y = 9.00 MPa
Section Property:
2
A= a
4
I=
a
12
Buckling Load :
2
Applying Euler's formula, Pcr =
π ⋅ E⋅ I
( K⋅ L )
2
2
σ Y⋅ A =
π ⋅ E⋅ I
( K⋅ L)
2
For pinned ends, K := 1.0
Thus,
2
4
σ Y⋅ a ⋅ ( 1.0L )
a
=
2
12
π ⋅E
a :=
12σ Y⋅ L
2
2
2
π ⋅E
a = 43.59 mm
Ans
I=
σ Y⋅ A⋅ ( K⋅ L)
2
π ⋅E
2
Problem 13-6
The rod is made from an A-36 steel rod. Determine the smallest diameter of the rod, to the nearest
mm, that will support the load of P = 25 kN without buckling. The ends are roller-supported.
L := 500mm
Given:
E := 200⋅ GPa
Pcr := 25kN
Solution:
Section Property:
I=
π ⋅ do
4
64
2
A=
π ⋅ do
4
Buckling Load :
2
Applying Euler's formula,
π ⋅ E⋅ I
Pcr =
( K⋅ L )
2
Pcr⋅ ( K⋅ L )
I=
2
2
π ⋅E
For roller ends,
K := 1.0
Thus,
π ⋅ do
4
=
64
2
64Pcr⋅ L
σ cr =
Pcr
A
2
3
π ⋅E
do = 15.94 mm
Critical Stress :
2
π ⋅E
4
do :=
Pcr⋅ ( 1.0 ⋅ L )
Use
do := 16mm
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
4Pcr
π ⋅ do
2
σ cr = 124.34 MPa
< σY = 250 MPa
(O.K.!)
Problem 13-7
The rod is made from a 25-mm-diameter steel rod. Determine the critical buckling load if the ends are
roller supported. E st = 200 GPa, σY = 350 MPa.
Given:
L := 500mm
E := 200⋅ GPa
do := 25mm
σ Y := 350MPa
Solution:
Section Property:
4
I :=
π ⋅ do
64
A :=
π ⋅ do
2
4
Buckling Load :
Applying Euler's formula,
For roller ends,
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 151.4 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 308.43 MPa
< σY = 350 MPa
(O.K.!)
Problem 13-8
An A-36 steel column has a length of 5 m and is fixed at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load.
Given: E := 200⋅ GPa
σ Y := 250MPa
L := 5m
bo := 100mm
ho := 50mm
hi := 30mm
bi := 80mm
Solution:
Section Property:
A := bo⋅ ho − bi⋅ hi
3
3
3
3
bo⋅ ho
bi⋅ hi
Ix :=
−
12
12
ho⋅ bo
hi⋅ bi
Iy :=
−
12
12
Buckling Load :
Applying Euler's formula,
For fixed ends,
(
)
I := min Ix , Iy
K := 0.5
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 272.1 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 104.67 MPa
< σY = 250 MPa
Therefore, Euler's formula is valid.
(O.K.!)
Problem 13-9
An A-36 steel column has a length of 4.5 m and is pinned at both ends. If the cross-sectional area has
the dimensions shown, determine the critical load.
Given: E := 200⋅ GPa
σ Y := 250MPa
L := 4.5m
bf := 200mm
tw := 12mm
dw := 150mm
df := 12mm
Solution:
Section Property: D := dw + 2df
(
)
A := bf⋅ D − dw⋅ bf − tw
(
3
)
3
bf⋅ D
bf − t w ⋅ dw
Ix :=
−
12
12
Iy :=
2df⋅ bf
3
3
+
12
dw⋅ tw
12
Buckling Load :
Applying Euler's formula,
For pinned ends,
(
)
I := min Ix , Iy
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 1561.7 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 236.63 MPa
< σY = 250 MPa
Therefore, Euler's formula is valid.
(O.K.!)
Problem 13-10
The W250 X 67 is made of A-36 steel and is used as a column that has a length of 4.55 m. If its ends
are assumed pin supported, and it is subjected to an axial load of 500 kN, determine the factor of safety
with respect to buckling.
Given: E := 200⋅ GPa
L := 4.55m
σ Y := 250MPa
P := 500kN
Solution:
Use W 250x67 :
( 6) mm4 A := 8560mm2
6
4
Iy := 22.2 ⋅ ( 10 ) mm
Ix := 104⋅ 10
Buckling Load :
(
For pinned ends,
)
I := min Ix , Iy
Applying Euler's formula,
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
2
( K⋅ L)
Pcr = 2116.7 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 247.28 MPa
< σY = 250 MPa
Factor of safety:
Fsafety :=
Pcr
P
Fsafety = 4.23
(O.K.!)
Ans
Problem 13-11
The W250 X 67 is made of A-36 steel and is used as a column that has a length of 4.55 m. If the ends
of the column are fixed supported, can the column support the critical load without yielding?
Given: E := 200⋅ GPa
L := 4.55m
σ Y := 250MPa
P := 500kN
Solution:
Use W 250x67 :
( 6) mm4 A := 8560mm2
6
4
Iy := 22.2 ⋅ ( 10 ) mm
Ix := 104⋅ 10
Buckling Load :
Applying Euler's formula,
For fixed ends,
(
)
I := min Ix , Iy
K := 0.5
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 8466.8 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 989.11 MPa
> σY = 250 MPa
(No !)
Ans
The column will yield before the axial force achieves the critical load Pcr ,
and the Euler's formula is not valid.
Problem 13-12
Determine the maximum force P that can be applied to the handle so that the A-36 steel control rod AB
does not buckle. The rod has a diameter of 30 mm. It is pin connected at its ends.
Given:
L := 0.9m
E := 200⋅ GPa
a := 0.6m
do := 30mm
σ Y := 250MPa
b := 0.9m
Solution:
Section Property:
4
I :=
π ⋅ do
A :=
64
π ⋅ do
2
4
Buckling Load :
For rod AB,
For pinned ends,
K := 1.0
2
Pcr :=
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 96.89 kN
FAB := Pcr
Equations of Equilibrium :
(
)
ΣΜh=0; − FAB ⋅ a + P⋅ b = 0
⎛ a ⎞F
⎝ b ⎠ AB
P := ⎜
P = 64.6 kN
Ans
Critical Stress check :
σ cr :=
Pcr
A
σ cr = 137.08 MPa
< σY = 250 MPa
Therefore, Euler's formula is valid.
(O.K.!)
Problem 13-13
The two steel channels are to be laced together to form a 9-m-long bridge column assumed to be pin
connected at its ends. Each channel has a cross-sectional area of A = 1950 mm2 and moments of
inertia Ix = 21.60(106) mm4, Iy = 0.15(106) mm4. The centroid C of its area is located in the figure.
Determine the proper distance d between the centroids of the channels so that buckling occurs about
the x-x and y' -y' axes due to the same load. What is the value of this critical load? Neglect the effect of
the lacing. E st = 200 GPa, σY = 350 MPa.
Given: E := 200⋅ GPa
L := 9m
2
σ Y := 350MPa
A' := 1950mm
( 6) 4
6
4
I'y := 0.15 ⋅ ( 10 ) mm
I'x := 21.60 ⋅ 10 mm
c'1 := 6.5mm
c'2 := 30mm
Solution:
Section Property:
Ix := 2I'x
A := 2A'
Iy = 2I'y + 2A'⋅ ( 0.5d)
2
In order for the column to buckle about x-x and y-y axes
at the same time, Iy must be equal to Ix.
Iy = Ix
2
2I'y + 2A'⋅ ( 0.5d) = 2I'x
I'x − I'y
d := 2
d = 209.76 mm
A'
> 2c'2 = 60 mm
Iy := 2I'y + 2A'⋅ ( 0.5d)
6
Ans
(O.K. !)
2
4
Iy = 43.20 × 10 mm
Buckling Load :
Applying Euler's formula,
For pinned ends,
(
)
I := min Ix , Iy
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 1052.8 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 269.94 MPa
> σY = 350 MPa
Therefore, Euler's formula is valid.
(O.K. !)
Problem 13-14
The W200 X 100 is used as a structural A-36 steel column that can be assumed fixed at its base and
pinned at its top. Determine the largest axial force P that can be applied without causing it to buckle.
Given: E := 200⋅ GPa
L := 7.5m
σ Y := 250MPa
Solution:
Use W 200x100 :
( 6) mm4 A := 12700mm2
6
4
Iy := 36.6 ⋅ ( 10 ) mm
Ix := 133⋅ 10
Buckling Load :
Applying Euler's formula,
(
)
I := min Ix , Iy
K := 0.7
For fixed-pinned ends,
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 2621.2 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 206.39 MPa
< σY = 250 MPa
(O.K.!)
Problem 13-15
Solve Prob. 13-14 if the column is assumed fixed at its bottom and free at its top.
Given: E := 200⋅ GPa
L := 7.5m
σ Y := 250MPa
Solution:
( 6) mm4 A := 12700mm2
6
4
Iy := 36.6 ⋅ ( 10 ) mm
Ix := 133⋅ 10
Use W 200x100 :
Buckling Load :
Applying Euler's formula,
(
)
I := min Ix , Iy
K := 2.0
For fixed-free ends,
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 321.1 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 25.28 MPa
< σY = 250 MPa
(O.K.!)
Problem 13-16
A steel column has a length of 9 m and is fixed at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load. E st = 200 GPa, σY = 250 MPa.
Given: E := 200⋅ GPa
σ Y := 250MPa
L := 9m
bf := 200mm
tw := 10mm
dw := 150mm
df := 10mm
Solution:
Section Property: D := dw + 2df
(
)
A := bf⋅ D − dw⋅ bf − tw
(
3
)
3
bf⋅ D
bf − t w ⋅ dw
Ix :=
−
12
12
Iy :=
2df⋅ bf
3
3
+
12
dw⋅ tw
12
Buckling Load :
Applying Euler's formula,
For fixed ends,
(
)
I := min Ix , Iy
K := 0.5
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 1300.9 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 236.53 MPa
< σY = 250 MPa
Therefore, Euler's formula is valid.
(O.K.!)
Problem 13-17
Solve Prob. 13-16 if the column is pinned at its top and bottom.
Given: E := 200⋅ GPa
σ Y := 250MPa
L := 9m
bf := 200mm
tw := 10mm
dw := 150mm
df := 10mm
Solution:
Section Property: D := dw + 2df
(
)
A := bf⋅ D − dw⋅ bf − tw
(
3
)
3
bf⋅ D
bf − t w ⋅ dw
Ix :=
−
12
12
Iy :=
2df⋅ bf
3
3
+
12
dw⋅ tw
12
Buckling Load :
Applying Euler's formula,
For pinned ends,
(
)
I := min Ix , Iy
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 325.2 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 59.13 MPa
< σY = 250 MPa
Therefore, Euler's formula is valid.
(O.K.!)
Problem 13-18
The 3.6-m A-36 steel pipe column has an outer diameter of 75 mm and a thickness of 6 mm.
Determine the critical load if the ends are assumed to be pin connected.
Given: E := 200⋅ GPa
L := 3.6m
σ Y := 250MPa
do := 75mm
t := 6mm
Solution:
Section Property: di := do − 2t
A :=
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
I :=
π ⎛ 4
4
⋅ d − di ⎞⎠
64 ⎝ o
Buckling Load :
Applying Euler's formula,
For pinned ends,
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 118.8 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 91.33 MPa
< σY = 250 MPa
(O.K.!)
Problem 13-19
The 3.6-m A-36 steel column has an outer diameter of 75 mm and a thickness of 6 mm. Determine the
critical load if the bottom is fixed and the top is pinned.
Given: E := 200⋅ GPa
L := 3.6m
σ Y := 250MPa
do := 75mm
t := 6mm
Solution:
Section Property: di := do − 2t
A :=
π ⎛ 2
2
⋅ d − di ⎞⎠
4 ⎝ o
I :=
π ⎛ 4
4
⋅ d − di ⎞⎠
64 ⎝ o
Buckling Load :
Applying Euler's formula,
K := 0.7
For fixed-pinned ends,
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 242.4 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 186.38 MPa
< σY = 250 MPa
(O.K.!)
Problem 13-20
The 3-m wooden rectangular column has the dimensions shown. Determine the critical load if the ends
are assumed to be pin connected. E w = 12 GPa, σY = 35 MPa.
Given: E := 12⋅ GPa
L := 3m
σ Y := 35MPa
b := 50mm
h := 100mm
Solution:
Section Property:
3
3
b⋅ h
Ix :=
12
A := b⋅ h
Buckling Load :
h⋅ b
Iy :=
12
(
)
I := min Ix , Iy
Applying Euler's formula,
For pinned ends,
K := 1.0
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 13.7 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 2.74 MPa
< σY = 35 MPa
(O.K.!)
Problem 13-21
The 3-m column has the dimensions shown. Determine the critical load if the bottom is fixed and the
top is pinned. Ew = 12 GPa, σY = 35 MPa.
Given: E := 12⋅ GPa
L := 3m
σ Y := 35MPa
b := 50mm
h := 100mm
Solution:
Section Property:
3
3
b⋅ h
Ix :=
12
A := b⋅ h
Buckling Load :
h⋅ b
Iy :=
12
(
)
I := min Ix , Iy
Applying Euler's formula,
For fixed-pinned ends, K := 0.7
2
Pcr :=
Critical Stress :
π ⋅ E⋅ I
( K⋅ L)
2
Pcr = 28 kN
Ans
Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 5.60 MPa
< σY = 35 MPa
(O.K.!)
Problem 13-22
The members of the truss are assumed to be pin connected. If member BD is an A-36 steel rod of
radius 50 mm, determine the maximum load P that can be supported by the truss without causing the
member to buckle.
Given: a := 4m
E := 200⋅ GPa
h := 3m
σ Y := 250MPa
ro := 50mm
Solution:
Section Property:
4
I :=
π ⋅ ro
4
A := π ⋅ ro
2
Buckling Load : L BD := a
Applying Euler's formula,
For pinned ends, K := 1.0
2
Pcr :=
π ⋅ E⋅ I
(K⋅ LBD)2
Pcr = 605.6 kN
Equations of Equilibrium :
ΣΜC=0;
R⋅ a − FBD⋅ h = 0
Support Reactions :
+
ΣF y=0;
FBD := Pcr
⎛ h⎞F
⎝ a ⎠ BD
R := ⎜
By symmetry, A=B=R
2R − 2P = 0
P := R
P = 454.19 kN
Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 77.11 MPa
< σY = 250 MPa
(O.K.!)
Problem 13-23
Solve Prob. 13-22 in the case of member AB, which has a radius of 50 mm.
Given: a := 4m
E := 200⋅ GPa
h := 3m
σ Y := 250MPa
ro := 50mm
Solution:
Section Property:
4
I :=
π ⋅ ro
4
A := π ⋅ ro
2
2
Buckling Load : L AB :=
Applying Euler's formula,
2
a +h
For pinned ends, K := 1.0
2
Pcr :=
π ⋅ E⋅ I
(K⋅ LAB)2
Pcr = 387.6 kN
At Joint A:
+
ΣF y=0;
Equations of Equilibrium :
⎛ h ⎞=0
⎝ LAB ⎠
R − FAB⋅ ⎜
Support Reactions :
+
ΣF y=0;
FAB := Pcr
⎛ h ⎞
⎝ LAB ⎠
R := FAB⋅ ⎜
By symmetry, A=B=R
2R − 2P = 0
P := R
P = 232.55 kN
Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
σ cr :=
Pcr
A
σ cr = 49.35 MPa
< σY = 250 MPa
(O.K.!)
Problem 13-24
The truss is made from A-36 steel bars, each of which has a circular cross section with a diameter of
40 mm. Determine the maximum force P that can be applied without causing any of the members to
buckle. The members are pin connected at their ends.
Given: a := 1.2m
E := 200⋅ GPa
h := 0.9m
σ Y := 250MPa
Solution:
do := 40mm
4
Section Property: I :=
π ⋅ do
A :=
π
2
⋅ do
4
64
Buckling Load :
Member AB and BC are in compression.
L AB := 2a
2
L BC :=
2
a +h
L AC := LBC
Applying Euler's formula,
Assume failure of rod AB:
Assume failure of rod BC:
For pinned ends, K := 1.0
For pinned ends, K := 1.0
2
Pcr :=
2
π ⋅ E⋅ I
P'cr :=
(K⋅ LAB)2
Pcr = 43.1 kN
+
ΣF y=0;
+ ΣF x=0;
(K⋅ LBC)2
P'cr = 110.2 kN
FBC := P'cr
FAB := Pcr
At Joint A:
π ⋅ E⋅ I
Equations of Equilibrium :
L AC
h
FAC⋅
−P= 0
FAC = P⋅
L AC
h
a
a
FAC⋅
− FAB = 0
FAB = FAC⋅
L AC
LAC
FAB = P⋅
h
a
Ans
P := Pcr⋅
P = 32.30 kN
a
h
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
< σY = 250 MPa
σ cr :=
σ cr = 34.27 MPa
A
Check Rod BC :
Support Reactions :
2a
ΣΜD=0; −P⋅ ( 2a) + Bh⋅ h = 0
Bh = P⋅
h
At Joint B: Equations of Equilibrium :
+ ΣF x=0; FAB + FBC⋅ a − Bh = 0
LAC
a
a
2a
P⋅ + P'cr⋅
− P⋅
= 0
L AC
h
h
a
h
Pcr = P⋅
P := P'cr⋅
h
LAC
P = 66.15 kN
(Not Control !)
(O.K.!)
Problem 13-25
The truss is made from A-36 steel bars, each of which has a circular cross section. If the applied load
P = 50 kN, determine the diameter of member AB to the nearest multiples of 5mm that will prevent this
member from buckling. The members are pin supported at their ends.
Given: a := 1.2m
E := 200⋅ GPa
h := 0.9m
P := 50kN
σ Y := 250MPa
Solution:
At Joint A:
+
Equations of Equilibrium :
ΣF y=0;
+ ΣF x=0;
FAC⋅
h
−P= 0
L AC
a
FAC⋅
− FAB = 0
L AC
FAB = FAC⋅
Buckling Load :
a
FAC = P⋅
FAB := P⋅
LAC
L AC
h
a
h
L AB := 2a
2
Applying Euler's formula,
For pinned ends, K := 1.0
π ⋅ E⋅ I
Pcr =
I=
(K⋅ LAB)2
[ 1.0 ⋅ ( 2a) ]
2
π ⋅E
2
⋅ Pcr
2
Pcr := FAB
π ⋅ do
Section Property: I =
4
A=
4
π
2
⋅ do
4
2
=
64
⎛ a⎞
2
h⎠
π ⋅E ⎝
⋅ ⎜ P⋅
4
64
π ⋅ do
do :=
I=
4a
⎛ a⎞
2
h⎠
π ⋅E ⎝
4a
⋅ ⎜ P⋅
2
⎛ a⎞
3
h⎠
π ⋅E ⎝
256a
⋅ ⎜ P⋅
do = 44.62 mm
Use:
do := 45mm
Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
π
2
A := ⋅ do
σ cr :=
σ cr = 41.92 MPa
4
A
< σY = 250 MPa
(O.K.!)
Problem 13-26
An L-2 tool steel link in a forging machine is pin connected to the forks at its ends as shown.
Determine the maximum load P it can carry without buckling. Use a factor of safety with respect to
buckling of F.S. = 1.75. Note from the figure on the left that the ends are pinned for buckling, whereas
from the figure on the right the ends are fixed.
Given: E := 200⋅ GPa
L := 0.6m
Fsafety := 1.75
σ Y := 703MPa
b := 36mm
t := 12mm
Solution:
Section Property:
3
t⋅ b
Ix :=
12
4
Ix = 46656.00 mm
3
b⋅ t
Iy :=
12
Iy = 5184.00 mm
A := b⋅ t
A = 432.00 mm
4
2
Critical Buckling Load :
With respect to the x-x axis: for pinned ends,
Applying Euler's formula,
K := 1.0
2
Pcr :=
π ⋅ E ⋅ Ix
( K⋅ L)
Pcr = 255.8 kN
2
With respect to the y-y axis: for fixed ends,
Applying Euler's formula,
K := 0.5
2
P'cr :=
π ⋅ E ⋅ Iy
P'cr = 113.7 kN
2
(Controls !)
( K⋅ L)
Critical Stress : Euler's formula is only valid if σcr < σY.
σ cr :=
P'cr
A
σ cr = 263.19 MPa
< σY = 703 MPa
Factor of safety:
Fsafety =
P :=
P'cr
P
P'cr
Fsafety
P = 64.97 kN
Ans
(O.K.!)
Problem 13-27
The linkage is made using two A-36 steel rods, each having a circular cross section. Determine the
diameter of each rod to the nearest multiples of 5mm that will support a load of P = 30 kN. Assume
that the rods are pin connected at their ends. Use a factor of safety with respect to buckling of 1.8.
Given: h := 3.6m
θ AB := 45deg
E := 200⋅ GPa
P := 30kN
θ BC := 30deg
σ Y := 250MPa
Solution:
At Joint B:
Equations of Equilibrium : Given
( )
( )
+ ΣF =0; FAB⋅ sin ( θ AB) − FBC⋅ sin ( θ BC) = 0
+
ΣFy=0; FAB⋅ cos θ AB + FBC⋅ cos θ BC − P = 0 (1)
(2)
x
Guess FAB := 1kN FBC := 1kN
⎛⎜ FAB ⎞
⎛⎜ FAB ⎞ ⎛ 15.53 ⎞
:= Find FAB , FBC
=⎜
kN
⎜ FBC
⎜ FBC
21.96
⎝
⎠
⎝
⎠
⎝
⎠
Solving Eqs. (1) and (2):
(
)
4
π ⋅ do
π
2
⋅ do
64
4
2
π ⋅ E⋅ I
Buckling Load : Applying Euler's formula, Pcr =
2
( K⋅ L)
For pinned ends, K := 1.0
Section Property: I =
I=
[ 1.0 ⋅ ( L) ]
2
2
π ⋅E
⋅ Pcr
For Member AB:
Pcr := FAB
A=
π ⋅ do
LAB :=
4
L
=
64
2
π ⋅E
(
h
4
2
⋅ Pcr
)
cos θ AB
L := LAB
do =
4
dAB :=
64L
2
3
π ⋅E
64L
3
2
π ⋅E
( )
⋅ Pcr
( )
⋅ Pcr
dAB = 45.15 mm
Use:
dAB := 50mm Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
π
2
< σY = 250 MPa
A := ⋅ dAB σ cr :=
σ cr = 7.91 MPa
A
4
For Member BC:
P'cr := FBC
LBC :=
(
h
cos θ BC
L' := LBC
)
4
dBC :=
2
64L'
3
π ⋅E
( )
⋅ P'cr
(O.K.!)
dBC = 44.49 mm
Use:
dBC := 45mm Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
P'cr
π
2
< σY = 250 MPa
A' := ⋅ dBC σ' cr :=
σ' cr = 13.81 MPa
A'
4
(O.K.!)
Problem 13-28
The linkage is made using two A-36 steel rods, each having a circular cross section. If each rod has a
diameter of 20 mm, determine the largest load it can support without causing any rod to buckle.
Assume that the rods are pin-connected at their ends.
Given: h := 3.6m
do := 20mm
Set
Solution:
At Joint B:
θ AB := 45deg
E := 200⋅ GPa
θ BC := 30deg
σ Y := 250MPa
P' := 1
Equations of Equilibrium : Given
( )
( )
+ ΣF =0; F'AB⋅ sin ( θ AB) − F'BC⋅ sin ( θ BC) = 0
+
ΣF y=0; F'AB⋅ cos θ AB + F'BC⋅ cos θ BC − P' = 0 (1)
(2)
x
Guess F'AB := 1
F'BC := 1
⎛⎜ F'AB ⎞
⎛⎜ F'AB ⎞ ⎛ 0.5176 ⎞
⎛⎜ FAB ⎞ ⎛⎜ F'AB⋅ P ⎞
:= Find F'AB , F'BC
=⎜
=
kN
⎜ F'BC
⎜ F'BC
⎜ FBC
⎜ F'BC⋅ P
0.7321
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠ ⎝
⎠
Solving Eqs. (1) and (2):
(
)
4
π ⋅ do
π
2
⋅d
4 o
64
2
π ⋅ E⋅ I
Buckling Load : Applying Euler's formula, Pcr =
2
( K⋅ L )
2
π ⋅ E⋅ I
For pinned ends, K := 1.0
Pcr =
2
L
h
For Member AB: L AB :=
cos θ AB
Section Property: I :=
A :=
(
)
2
L := L AB
Pcr = FAB
F'AB⋅ P =
π ⋅ E⋅ I
2
2
P :=
π ⋅ E⋅ I
2
F'AB⋅ LAB
L AB
P = 1.16 kN
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
Pcr := F'AB⋅ P
σ cr :=
σ cr = 1.90 MPa < σY = 250 MPa
A
Check Member BC:
2
P'cr :=
π ⋅ E⋅ I
L BC
2
L BC :=
(
h
cos θ BC
P'cr = 0.8972 kN
)
Ans
(O.K.!)
FBC := F'BC⋅ P
FBC = 0.846 kN
> FBC = 0.846 kN
(O.K.!)
Problem 13-29
The A-36 steel pipe has an outer diameter of 50 mm and a thickness of 12 mm. If it is held in place by
a guywire, determine the largest vertical force P that can be applied without causing the pipe to buckle.
Assume that the ends of the pipe are pin connected.
Given: L := 4.2m
do := 50mm
Set
Solution:
At Joint B:
θ BC := 30deg
E := 200⋅ GPa
t := 12mm
σ Y := 250MPa
P' := 1
Equations of Equilibrium : Given
(1)
( )
+ ΣF =0; F'BC⋅ cos ( θ BC) − F'AB = 0 (2)
+
ΣF y=0;
F'BC⋅ sin θ BC − P' = 0
x
Guess F'AB := 1
F'BC := 1
⎛⎜ F'AB ⎞
⎛⎜ F'AB ⎞ ⎛ 1.7321 ⎞
⎛⎜ FAB ⎞ ⎛⎜ F'AB⋅ P ⎞
:= Find F'AB , F'BC
=⎜
=
kN
⎜ F'BC
⎜ F'BC
⎜ FBC
⎜ F'BC⋅ P
2.0000
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠ ⎝
⎠
Solving Eqs. (1) and (2):
(
)
Section Property: di := do − 2t
4
4
π
2
2
π ⋅ ⎛ do − di ⎞
A := ⋅ ⎛⎝ do − di ⎞⎠
⎝
⎠
I :=
4
64
2
π ⋅ E⋅ I
Buckling Load : Applying Euler's formula, Pcr =
2
( K⋅ L )
2
π ⋅ E⋅ I
For pinned ends, K := 1.0
Pcr =
2
L
2
2
π ⋅ E⋅ I
π ⋅ E⋅ I
Pcr = FAB
F'AB⋅ P =
P :=
2
2
L
F'AB⋅ L
P = 18.37 kN
Ans
Check Critical Stress : Euler's formula is only valid if σcr < σY.
Pcr
Pcr := F'AB⋅ P
σ cr :=
σ cr = 22.21 MPa < σY = 250 MPa
A
(O.K.!)
Problem 13-30
The A-36 steel pipe has an outer diameter of 55 mm. If it is held in place by a guywire, determine its
required inner diameter to the nearest multiples of 5mm, so that it can support a maximum vertical load
of P = 20 kN without causing the pipe to buckle. Assume the ends of the pipe are pin connected.
Given: L := 4.2m
do := 55mm
θ BC := 30deg
E := 200⋅ GPa
P := 20kN
σ Y := 250MPa
Solution:
At Joint B:
Equations of Equilibrium : Given
(1)
( )
+ ΣF =0; FBC⋅ cos ( θ BC) − FAB = 0 (2)
+
ΣF y=0;
FBC⋅ sin θ BC − P = 0
x
Guess FAB := 1kN
FBC := 1kN
⎛⎜ FAB ⎞
⎛⎜ FAB ⎞ ⎛ 34.64 ⎞
:= Find FAB , FBC
=⎜
kN
⎜ FBC
⎜ FBC
40.00
⎝
⎠
⎝
⎠
⎝
⎠
Solving Eqs. (1) and (2):
(
)
Section Property: di = do − 2t
π ⋅ ⎛ do − di ⎞
π ⎛ 2
2⎞
⎝
⎠
A = ⋅ ⎝ do − di ⎠
I=
64
4
2
π ⋅ E⋅ I
Buckling Load : Applying Euler's formula, Pcr =
2
( K⋅ L )
4
4
2
For pinned ends, K := 1.0
I :=
FAB⋅ L
2
2
π ⋅E
Pcr = FAB
π ⋅ ⎛ do − di ⎞
⎝
⎠ = FAB⋅ L
2
64
π ⋅E
4
4
di