Problem 6.63 The tension in cable BD is 500 lb.
Determine the reactions at A for cases (1) and (2).
G
E
6 in
D
6 in
A
B
C
300 lb
8 in
8 in
(1)
Solution: Case (a) The complete structure as a free body: The
sum of the moments about G:
MG = −16(300) + 12Ax = 0,
E
G
6 in
D
from which Ax = 400 lb . The sum of the forces:
Fx = Ax + Gx = 0,
6 in
from which Gx = −400 lb.
Fy = Ay − 300 + Gy = 0,
A
300 lb
ments about E:
ME = −16Gy = 0,
from which Ax = 400 lb .
Element ABC: The tension at the lower end of the cable is up and to
the right, so that the moment exerted by the cable tension about point
C is negative. The sum of the moments about C:
MC = −8B sin α − 16Ay = 0,
noting that B = 500 lb and α = tan−1 68 = 36.87◦ ,
Ay = −150 lb.
Ay
E
B
α
8 in
Cy
Cx
300 lb
6 in
D
6 in
6 in
A
C
300 lb
8 in
E
G
6 in
D
A
B
8 in
8 in
(a)
8 in
(b)
Gy
(a) 12 in
Gx
Ax
Ax
8 in
(2)
G
from which Gy = 0, and from above Ay = 300 lb.
Case (b) The complete structure as a free body: The free body diagram,
except for the position of the internal pin, is the same as for case (a).
The sum of the moments about G is
MC = −16(300) + 12Ax = 0,
(b)
C
8 in
8 in
from which Ay = 300 − Gy . Element GE: The sum of the mo-
then
B
Gy
Ay
Gx
16 in
300 lb
Ey
Ex
C
300 lb
Problem 6.64
Determine the forces on member
ABCD, presenting your answers as shown in Fig. 6.35.
E
3 ft
400 lb
3 ft
B
A
D
C
4 ft
4 ft
Solution: The complete structure as a free body: The sum of the
moments about A:
MA = −400(3) + 12Dy = 0,
4 ft
E
3 ft
400 lb
from which Dy = 100 lb . The sum of the forces:
Fx = Ax + 400 = 0,
A
B
from which Ax = −400 lb .
Fy = Ay + Dy = 0,
4 ft
from which the cable tension is E = 360.6 lb . The sum of the
forces:
Fx = −Cx + 400 − E cos α = 0,
from which Cx = 200 lb.
Fy = −Cy − E sin α = 0,
from which Cy = −300 lb.
Element ABCD: The tension in the cable acts on element ABCD
with equal and opposite tension to the reaction on element EB, up
and to the right at an angle of 56.31◦ , Cx = 200 lb to the right, and
Cy = −300 lb downward.
4 ft
E
F
Cy
Ay
The angle of cable element EB is
6
α = tan−1
= 56.31◦ ,
4
D
4 ft
from which Ay = −100 lb .
Element EC: The sum of the moments about C:
MC = 6E cos α − 3(400) = 0.
C
B
Cx
Cy
Ax
Cx
Dy
360
lb
400 lb
A
100 lb
56.3°
300 lb
200 lb
100 lb
D
3 ft
Problem 6.65 The mass m = 50 kg. Determine the
forces on member ABCD, presenting your answers as
shown in Fig. 6.35.
1m
1m
D
E
1m
C
m
1m
B
1m
A
F
Solution: The weight of the mass hanging is W = mg =
50(9.81) = 490.5 N The complete structure as a free body: The
sum of the moments about A:
MA = −2W + Fy = 0,
from which Fy = 981 N. The sum of the forces:
Fy = Ay + Fy − W = 0,
1m
D
C
from which Ax = −Fx . Element BF: The sum of the moments
about F :
MF = −Bx − By = 0,
1m
from which By = −Bx . The sum of the forces:
Fy = By + Fy = 0,
1m
W
B
Dy
from which Fx = −981 N, and from above, Ax = 981 N ,
Element DE: The sum of the moments about D:
MD = −Ey − 2W = 0,
Ex
Dx
Dx
Dy
Cy
Cx
Cy
Bx
By
Ey
W
Ex
By
Ax
Fx = −Dx − Ex = 0,
from which Ex = −981 N, and from above Dx = 981 N .
Fy = Ey + Cy = 0,
from which Cy = 981 N.
Fx = Ex + Cx = 0,
from which Cx = 981 N, and
Element ABCD: All reactions on ABCD have been determined
above.
The components at B and C have the magnitudes
√
B = C = 9812 + 9812 = 1387 N , at angles of 45◦ .
Ey
Cx
Bx
Fy
Ay
from which Dy = 490.5 N .
from which Dx = −Ex . Element CE: The sum of the moments
about C:
MC = Ey − Ex = 0,
F
A
from which By = −981 N, and Bx = 981 N.
Fx = Bx + Fx = 0,
E
1m
from which Ay = −490.5 N,
Fx = Ax + Fx = 0,
from which Ey = −981 N. The sum of the forces:
Fy = −Dy − Ey − W = 0,
1m
490.5 N
D
C
981 N
1387 N
45°
B
A
45°
1387 N
981 N
490.5 N
Fx
Problem 6.66
ber BCD.
Determine the forces on mem-
400 lb
6 ft
B
A
4 ft
C
4 ft
D
E
8 ft
Solution: The following is based on free body diagrams of the
elements: The complete structure as a free body: The sum of the
moments about D:
MD = −(6)400 + 8Ey = 0,
B
from which Ey = 300 lb. The sum of the forces:
Fx = Dx = 0.
Fy = Ey + Dy − 400 = 0,
from which Dy = 100 lb. Element AB: The sum of the moments
about A:
MA = −8By − (6)400 = 0,
4 ft
C
4 ft
E
D
from which By = −300 lb. The sum of forces:
Fy = −By − Ay − 400 = 0,
from which Ay = −100 lb.
Fx = −Ax − Bx = 0,
from which (1) Ax + Bx = 0 Element ACE: The sum of the moments
about E:
ME = −8Ax + 4Cx − 8Ay + 4Cy = 0,
from which (2) −2Ax +Cx −2Ay +Cy = 0. The sum of the forces:
Fy = Ay + Ey − Cy = 0,
from which Cy = 200 lb .
Fx = Ax − Cx = 0,
from which (3) Ax = Cx . The three numbered equations are solved:
Ax = −400 lb, Cx = 400 lb , and Bx = −400 lb .
Element BCD:
The reactions are now known:
By = −300 lb , Bx = −400 lb , Cy = 200 lb ,
Dx = 0 , Dy = 100 lb ,
where negative sign means that the force is reversed from the direction
shown on the free body diagram.
400 lb
6 ft
A
8 ft
400 lb
Ay
Ax
Ax
By
Bx
Cy
Cx E
Ay
Bx
Cy
Dy
Cx
Dx
By
Problem 6.67
Determine the forces on member ABC.
E
6 kN
1m
C
D
1m
A
B
2m
Solution:
2m
The frame as a whole: The equations of equilibrium are
E
FX = AX + EX = 0,
EX
FY = AY + EY − 6000 N = 0,
1m
and, with moments about E,
ME = 2AX − (5)6000 = 0.
Solving for the support reactions, we get AX = 15,000 N and EX =
−15,000 N. We cannot yet solve for the forces in the y direction at A
and E.
Member ABC: The equations of equilibrium are
FX = AX − BX = 0,
FY = AY − BY − CY = 0,
and summing moments about A,
MA = −2BY − 4CY = 0.
Member BDE: The equations of equilibrium are
FX = EX + DX + BX = 0,
FY = EY + DY + BY = 0,
and, summing moments about E,
ME = (1)DY + (1)DX + (2)BY + (2)BX = 0.
Member CD: The equations of equilibrium are
FX = −DX = 0,
FY = −DY + CY − 6000 = 0,
and summing moments about D,
MD = −(4)6000 + 3CY = 0.
Solving these equations simultaneously gives values for all of the forces
in the frame. The values are AX = 15,000 N, AY = −8,000 N,
BX = 15,000 N, BY = −16,000 N, CY = 8,000 N, DX = 0,
and DY = 2,000 N.
6 kN
D
EY
C
1m
AX
B
2m
1m
2m
AY
EX
EY
E DY
DX D
DX D
DY
BX B
BY BX
B
AX A
BY
AY
CY
C
CY
C
6 kN
1m
Problem 6.68
ber ABD.
Determine the forces on mem8 in
8 in
8 in
A
60 lb
8 in
60 lb
B
E
8 in
C
Solution:
and
D
The equations of equilibrium for the truss as a whole are
AY
FX = AX + CX = 0,
FY = AY − 60 − 60 = 0,
8 in.
AX
8 in.
8 in.
A
MA = 16CX − 16(60) − 24(60) = 0.
60 lb
8 in.
60 lb
Solving these three equations yields
B
AX = −150 lb,
AY = 120 lb,
8 in.
C
and CX = 150 lb.
Member ABD: The equilibrium equations for this member are:
FX = AX − BX − DX = 0,
FY = AY − BY − DY = 0,
and
MA = −8BY − 8DY − 8BX − 16DX = 0.
Member BE: The equilibrium equations for this member are:
FX = BX + EX = 0,
FY = BY + EY − 60 − 60 = 0,
and
MB = −8(60) − 16(60) + 16EY = 0.
Member CDE: The equilibrium equations for this member are:
FX = CX + DX − EX = 0,
FY = DY − EY = 0,
and
MD = 8EX − 16EY = 0.
Solving these equations, we get BX = −180 lb, BY = 30 lb,
DX = 30 lb, DY = 90 lb, EX = 180 lb, and EY = 90 lb. Note
that we have 12 equations in 9 unknowns. The extra equations provide
a check.
E
D
CX
AY
AX
BX
DX
BY
60 lb
BY
BX B
EX
EY
DY
DY
CX
60 lb
DX
EY
EX
Problem 6.69 The mass m = 12 kg. Determine the
forces on member CDE.
A
200 mm
100 mm
E
B
200 mm
C
D
200 mm
Solution:
and
m
400 mm
The equations of equilibrium for the entire truss are:
AY
FX = AX + CX = 0,
FY = AY − mg = 0,
MA = 0.4CX − 0.7mg = 0.
AX
A
200 mm
From these equations we get
100 mm
E
B
AX = −206.0 N, AY = 117.7 N,
200 mm
and Cx = 206.0 N.
Member ABD: The equations are
FX = AX + BX + DX + T = 0,
FY = AY + BY + DY = 0,
and
MA = 0.2BY + 0.2BX + 0.2DY + 0.4DX + 0.1T = 0.
C
CX
The Pulley: The equations are
FX = −T − PX = 0,
and
FY = −T − PY = 0.
The Weight: The equation is
FY = T − mg = 0.
Solving the equations simultaneously, we get
BX = 117.7 N, BY = 0, DX = −29.4 N, DY = −117.7 N,
EX = −235.4 N, EY = −117.7 N, T = 117.7 N,
PX = −117.7 N, PY = −117.7 N
200 mm
m
400 mm
AY
AX
T
PY
PX
T
Member CDE: The equations are
FX = CX − DX + EX = 0,
FY = −DY + EY = 0,
and
MD = 0.4EY − 0.2EX = 0.
Member BE: The equations are
FX = −BX + PX − EX = 0,
FY = −BY + PY − EY = 0,
and
ME = 0.4BY = 0.
F
D
BX
DX
T
BY
DY
CX
DX
BX
BY
EX
EY
PY PX
EY
T
EX
DY
mg
Problem 6.70 The weight W = 80 lb. Determine the
forces on member ABCD.
11 in
5 in
12 in
3 in
A
B
D
C
8 in
W
F
E
Solution: The complete structure as a free body: The sum of the
moments about A:
MA = −31W + 8Ex = 0,
from which Ex = 310 lb. The sum of the forces:
Fx = Ex + Ax = 0,
3 in.
from which Cx = 310 lb . The sum of the moments about E:
ME = 8F − 16Cy + 8Cx = 0.
For frictionless pulleys, F = W , and thus Cy = 195 lb . The sum
of forces parallel to y:
Fy = Ey − Cy + F = 0,
from which Ey = 115 lb .
Equation (1) above is now solvable: Ay = −35 lb .
Element ABCD: The forces exerted by the pulleys on element
ABCD are, by inspection: Bx = W = 80 lb , By = 80 lb ,
Dx = 80 lb , and Dy = −80 lb , where the negative sign means
that the force is reversed from the direction of the arrows shown on the
free body diagram.
B
A
D
C
8 in.
from which Ax = −310 lb .
Fy = Ey + Ay − W = 0,
from which (1) Ey + Ay = W .
Element CFE: The sum of the forces parallel to x:
Fx = Ex − Cx = 0,
12 in.
5 in.
11 in.
W
F
E
Ay
Ax
Cx
By
Cx
Ey
F
Ex
Dy
Dx
Cy
Bx
Cy
W
Problem 6.71 The man using the exercise machine
is holding the 80-lb weight stationary in the position
shown. What are the reactions at the built-in support E
and the pin support F ? (A and C are pinned connections.)
2 ft 2 in
1 ft 6 in
2 ft
D
B
C
A
9 in
60°
6 ft
80 lb
E
Solution: The complete structure as a free body: The sum of the
moments about E:
M = −26W − 68W sin 60◦ + 50Fy − 81W cos 60◦ + ME = 0
from which (1) 50Fy + ME = 10031. The sum of the forces:
Fx = Fx + W cos 60◦ + Ex = 0,
F
2 ft 2 in
2 ft
1 ft 6 in D
B
9 in
A
C
60°
from which (2) Fx + Ex = −40.
Fy = −W − W sin 60◦ + Ey + Fy = 0,
from which (3) Ey + Fy = 149.28
Element CF: The sum of the moments about F :
M = −72Cx = 0,
from which Cx = 0. The sum of the forces:
Fx = Cx + Fx = 0,
6 ft
80 lb
F
E
from which Fx = 0 . From (2) above, Ex = −40 lb
Element AE: The sum of the moments about E:
M = ME − 72Ax = 0, .
26
in
42
in
from which (4) ME = 72Ax . The sum of the forces:
Fy = Ey + Ay = 0,
from which (5) Ey + Ay = 0.
Fx = Ax + Ex = 0;
from which Ax
=
ME = 2880 in lb = 240 ft lb .
60°
W
W
81 in
ME
40
lb,
and from (4)
From (1) Fy = 143.0 lb ,
Ey
Ex
and from (2) Ey = 6.258 lb . This completes the determination of
Fy
Fx
50 in
the 5 reactions on E and F .
Ay
Cy
Ax
Cx
ME
Fx
Fy
Ex
Ey
Problem 6.72 The frame supports a horizontal load F
at C. The resulting compressive axial force in the twoforce member CD is 2400 N. Determine the magnitude
of the reaction exerted on member ABC at B.
C
F
100 mm
B
100 mm
Solution: First, write eqns to determine the support reactions at
A and E (CD is a two force member)
C
D
F
100 mm
A
B
E
0.3 m
100 mm
200 mm
D
100 mm
E
AX
EY
0.4 m
C
F
AY
+
100 mm
Fx :
Ax + F = 0
(1)
Fy :
Ay + Ey = 0
(2)
MA :
B
100 mm
0.4Ey − 0.3F = 0 (3)
D
We have these eqns in four Unknowns (Ax , Ay , Ey , and F ) Now
write eqns for member ABC
100 mm
A
E
CY
F
BY
200 mm
100 mm
100 mm
0.1 m
BY
BX
0.2 m
AX
BX
DY
AY
0.1 m
CY = 2400 N
0.2 m
0.1
m
Cy = 2400 N
+
Fx :
Ax + Bx + F = 0
(4)
Fy :
Ay + By + Cy = 0
(5)
MA :
0.2By − 0.2Bx + 0.3Cy − 0.3F = 0 (6)
We now have 6 eqns in unknowns: (Ax , Ay , Bx , By Ey , F ) Next,
write the equations for member BDE.
Solving, we get Bx = 0 By = −1200 N
|B| = 1200 N
0.1
m
EY
Cy = −Dy
(two force member)
Fx :
−Bx = 0
(7)
Fy :
−By + Dy + Ey = 0 (8)
MB :
0.1Dy + 0.2Ey = 0 (9)
We now have 9 equations in 8 unknowns. Obviously, if they are compatible,
one is a linear combination of the others. We could also have more than one
redundant equation and still need another equation.
Combining Eqs. (4) and (7) gives Eq. (1). Thus, one of these three equations is
not need.
Problem 6.73 The two-force member CD of the
frame shown in Problem 6.72 will safely support a compressive axial load of 3 kN. Based on this criterion, what
is the largest safe magnitude of the horizontal load F ?
Solution: In the solution to Problem 6.72, we derived the equation’s listed below for the loads shown on the frame.
Ax + F = 0
Ay + Ey = 0
0.4Ey − 0.3F = 0
Ax + Bx + F = 0
Ay + By + Cy = 0
0.2By − 0.2Bx + 0.3Cy − 0.3F = 0
C










F
100 mm
B
Entire Frame
100 mm
Member ABC
D
100 mm
A
Cy = −Dy − 2 force member
E
Set Cy = 3000 N and solve.
200 mm
We get F = 2000 N = 2 kN
Ax = −2 kN
Ay = −1.5 kN
Ey = 1.5 kN
Bx = 0
By = −1.5 kN
100 mm
100 mm
Problem 6.74 The unstretched length of the string is
LO . Show that when the system is in equilibrium the
angle α satisfies the relation sin α = 2(LO − 2F/k)L.
F
1– L
4
1– L
4
k
1– L
2
α
α
Solution: Since the action lines of the force F and the reaction
E are co-parallel and coincident, the moment on the system is zero,
and the system is always in equilibrium, for a non-zero force F . The
object is to find an expression for the angle α for any non-zero force F .
The complete structure as a free body:
The sum of the moments about A
MA = −F L sin α + EL sin α = 0,
F
1− L
4
1− L
4
C
from which E = F . The sum of forces:
Fx = Ax = 0,
1− L
2
from which Ax = 0.
from which Ay = 0, which completes a demonstration that F does
not exert a moment on the system. The spring C: The elongation of the
spring is ∆s = 2 L
sin α − LO , from which the force in the spring is
4
L
T =k
sin α − LO
2
Element BE: The strategy is to determine Cy , which is the spring force
on BE. The moment about E is
L
L
L
ME = − Cy cos α − By cos α − Bx cos α = 0,
4
2
2
α
F
L
α
Ax
Ay
E
C
from which 2y + By = −Bx . The sum of forces:
Fx = Bx = 0,
from which Bx = 0.
Fy = Cy + By + E = 0,
from which k
Solve: sin α =
L
sin α − LO
2
2(LO − 2F
k )
L
By
Cy
Bx
L
4
from which Cy +By = −E = −F . The two simultaneous equations
are solved: Cy = −2F , and By = F .
The solution for angle α: The spring force is
L
Cy = T = k
sin α − LO ,
2
= −2F .
D
α
A
Fy = Ay + E − F = 0,
k
B
α
L
4
E
E
Problem 6.75 The pin support B will safely support a
force of 24-kN magnitude. Based on this criterion, what
is the largest mass m that the frame will safely support?
C
500 mm
100 mm
E
D
B
300 mm
m
A
300 mm
Solution: The weight is given by W = mg = 9.81 g
The complete structure as a free body:
Sum the forces in the x-direction:
Fx = Ax = 0,
from which Ax = 0
Element ABC: The sum of the moments about A:
MA = +0.3Bx + 0.9Cx − 0.4W = 0,
4
W.
7
from which By =
The magnitude of the reaction at B is
2
5 2
4
|B| = W
+
= 1.0104W.
6
7
24
1.0104
For a safe value of |B| = 24 kN, W =
= 23.752 kN is the
maximum load that can be carried. Thus, the largest mass that can be
supported is m = W/g = 23752 N/9.81 m/s2 = 2421 kg.
400 mm
400 mm
C
500 mm
100 mm
300 mm
E
D
B
F
W
A
from which (1) 0.3Bx + 0.9Cx = 0.4W . The sum of the forces:
Fx = −Bx − Cx + W + Ax = 0,
from which (2) Bx + Cx = W . Solve the simultaneous equations (1)
and (2) to obtain Bx = 56 W
Element BE: The sum of the moments about E:
ME = 0.4W − 0.7By = 0,
F
300 mm 400 mm 400 mm
Cy
Cy
Cx
Cx
W
By
W
Bx
By Bx
W
Ay
Ax
Ey
Ex
Ey
Ex
F
Problem 6.76
Determine the reactions at A and C.
C
A
3 ft
72 ft-lb
36 lb
3 ft
B
18 lb
4 ft
Solution: The complete structure as a free body:
The sum of the moments about A:
MA = −4(18) + 3(36) + 12Cy − 72 = 0,
8 ft
A
C
from which Cy = 3 lb. The sum of the forces:
Fy = Ay + Cy − 18 = 0,
3 ft
72 ft-lb
36 lb
from which Ay = 15 lb.
Fx = Ax + Cx + 36 = 0,
18 lb
from which (1) Cx = −Ax − 36
Element AB: The sum of the forces:
Fy = Ay − By − 18 = 0,
4 ft
from which By = −3 lb. The sum of the moments:
MA = 6Bx − 4(18) − 4By − 72 = 0,
from which Bx = 22 lb. The sum of the forces:
Fx = Ax + Bx = 0,
3 ft
B
8 ft
Cy
Ay
Ax
Cx 3 ft
72 ft-lb
36 lb
from which Ax = −22 lb From equation (1) Cx = −14 lb
18 lb
8 ft
4 ft
Ay
Ax
72 ft-lb
By 6 ft
Bx
18 lb
Problem 6.77
Determine the forces on member AD.
200 N
130 mm
D
400 mm
C
A
400 N
B
400 mm
Denote the reactions of the support by Rx and Ry . The
complete structure as a free body:
Fx = Rx − 400 = 0,
400 mm
Solution:
from which Rx = 400 N. The sum of moments:
MA = 800C − 400(800) − 400(400) − 400(200) = 0,
200 N
D
400 mm
from which C = 300 N.
Fy = C + Ry − 400 − 200 = 0,
A
400 mm
Ax = −200 N, and Dx = 200 N.
Element BD: The sum of forces:
Fx = Bx − Dx − 400 = 0
from which Bx = 600 N. This completes the solution of the nine
equations in nine unknowns, of which Ax , Ay , Dx , and Dy are the
values required by the Problem.
400 mm
Dy
200 N
from which By = 600 N. Element BD: The sum of the forces:
Fy = By − Dy − 400 = 0,
from which Ay = 0: Element AD: The sum of the forces:
Fx = Ax + Dx = 0
and
MA = −400(200) + 800Dy − 400Dx = 0
400 N
C
B
from which Ry = 300 N. Element ABC: The sum of the moments:
MA = −4By + 8C = 0,
from which Dy = 200 N.
Element AD: The sum of the forces:
Fy = Ay + Dy − 200 = 0,
130 mm
Ay
Ay
Ax
Ry
Dy
400 N
By
400 N
Ax
Rx
Bx
By
Dx
Dx
Bx
C
Problem 6.78 The frame shown is used to support
high-tension wires. If b = 3 ft, α = 30◦ , and W =
200 lb, what is the axial force in member HJ?
A
B
α
C
D
α
E
G
F
W
H
α
I
J
α
W
W
b
Solution: Joints B and E are sliding joints, so that the reactions
are normal to AC and BF , respectively. Member HJ is supported
by pins at each end, so that the reaction is an axial force. The distance
h = b tan α = 1.732 ft
Member ABC. The sum of the forces:
Fx = Ax + B sin α = 0,
Fy = Ay − W − B cos α = 0.
The sum of the moments about B:
MB = bAy − hAx + bW = 0.
These three equations have the solution: Ax = 173.21 lb, Ay =
−100 lb, and B = −346.4 lb.
Member BDEF: The sum of the forces:
Fx = Dx − B sin α − E sin α = 0,
Fy = Dy − W + B cos α − E cos α = 0.
The sum of the moments about H:
MH = bGy − hGx + bW + 2bE cos α − 2hE sin α = 0.
These three equations have the solution: Gx = 346.4 lb, Gy =
200 lb, and H = 300 lb. This is the axial force in HJ.
b
b
A
B
C
D
α
α
E
G
W
F
H
I
W
J
α
α
W
The sum of the moments about D:
MD = −2bW − bE cos α − hE sin α − bB cos α + hB sin α = 0.
These three equations have the solution: Dx = −259.8 lb, Dy =
350 lb, E = −173.2 lb.
Member EGHI: The sum of the forces:
Fx = Gx + E sin α − H cos α = 0,
Fy = Gy − W + E cos α + H sin α = 0.
b
b
B
b
Ay
h
Ax
B
Dy
Dx E
b G
y
W
Gx
H
W
b
W
b
Problem 6.79 What are the magnitudes of the forces
exerted by the pliers on the bolt at A when 30-lb forces
are applied as shown? (B is a pinned connection.)
6 in
45°
2
in
30 lb
B
A
30 lb
Solution:
Element AB: The sum of the moments about B:
MB = 2F − (6)30 = 0,
from which F = 90 lb.
6 in
45°
30 lb
2 in
B
A
30 lb
6 in
By
30 lb
2 in
F
Bx
Problem 6.80 The weight W = 60 kip. What is the
magnitude of the force the members exert on each other
at D?
A
3 ft
B
C
2 ft
D
3 ft
W
3 ft
3 ft
Solution: Assume that a tong half will carry half the weight, and
denote the vertical reaction to the weight at A by R. The complete
structure as a free body: The sum of the forces:
Fy = R − W = 0,
A
from which R = W
3 ft
C
B
Tong-Half ACD:
2 ft
Element AC: The sum of the moments about A:
(1)
MA = 3Cy + 3Cx = 0.
D
(3)
The sum of the forces:
R
Fy =
+ Cy + Ay = 0, and
2
Fx = Cx + Ax = 0.
W
(4)
Element CD: The sum of the forces:
Fx = Dx − P − Cx = 0, and
(2)
(5)
(6)
(7)
(8)
3 ft
3 ft
3 ft
W
= 0.
Fy = Dy − Cy −
2
The sum of the moments:
3
MD = 2Cx − 3Cy − 3P + W = 0
2
Element AB: The sum of the forces:
R
Fy = −Ay +
− By = 0, and
2
Fx = −Ax − Bx = 0.
Element BD: The sum of the forces:
W
(9)
Fy = By − Dy −
= 0, and
2
(10)
Fx = Bx − Dx + P = 0.
These are ten equations in ten unknowns. These have the solution
R = 60 kip. Check, Ax = −30 kip, Ay = 0, Bx = 30 kip,
By = 30 kip, Cx = 30 kip, Cy = −30 kip, Dx = 110 kip,
Dy = 0, and P = 80 kip. The magnitude of the force the
members exert on each other at D is D = 110 kip.
Ay
R
2
R
2
Ay
Ax
Ax
By
Cy
Bx
Dy
Cx
Dy
Bx
Cx
Dx
By
Dx
Cy
P
P
W
2
W
2
Problem 6.81 Figure a is a diagram of the bones and
biceps muscle of a person’s arm supporting a mass. Tension in the biceps muscle holds the forearm in the horizontal position, as illustrated in the simple mechanical
model in Fig. b. The weight of the forearm is 9 N, and
the mass m = 2 kg.
(a) Determine the tension in the biceps muscle AB.
(b) Determine the magnitude of the force exerted on the
upper arm by the forearm at the elbow joint C.
B
290
mm
(a)
A
50
mm
9N
m
200 mm
150 mm
(b)
Solution: Make a cut through AB and BC just above the elbow
joint C. The angle formed
muscle with respect to the
by the biceps
forearm is α = tan−1 290
= 80.2◦ . The weight of the mass is
50
W = 2(9.81) = 19.62 N.
The section as a free body: The sum of the moments about C is
MC = −50T sin α + 150(9) + 350W = 0,
from which T = 166.76 N is the tension exerted by the biceps muscle
AB. The sum of the forces on the section is
FX = Cx + T cos α = 0,
(a)
from which Cx = −28.33 N.
FY = Cy + T sin α − 9 − W = 0,
B
from which Cy = −135.72. The magnitude of the force exerted by
the forearm on the upper arm at joint C is
F = Cx2 + Cy2 = 138.65 N
290
mm
(b)
A
W 200 mm
C
150 mm
50
9N
mm
T
W
9N
200
mm
α Cy
50
150 mm
mm
Cx
C
Problem 6.82 The clamp presses two blocks of wood
together. Determine the magnitude of the force the members exert on each other at C if the blocks are pressed
together with a force of 200 N.
125 mm
125 mm
125 mm
B
50 mm
E
A
50 mm
C
50 mm
D
Solution: Consider the upper jaw only.
The section ABC as a free body:
The sum of the moments about C is
MC = 100B − 250A = 0,
from which, for A = 200 N, B = 500 N. The sum of the forces:
Fx = Cx − B = 0,
from which Cx = 500 N,
Fy = Cy + A = 0,
from which Cy = −200 N. The magnitude of the reaction at C:
C = Cx2 + Cy2 = 538.52 N
125
mm
125
mm
125
mm
B
50
mm
E
A
50
mm
C
50
mm
D
B
Cy
100 mm
A
Cx
250 mm
Problem 6.83 The pressure force exerted on the piston is 2 kN toward the left. Determine the couple M
necessary to keep the system in equilibrium.
B
300 mm
350 mm
45°
A
C
M
400 mm
Solution: From the diagram, the coordinates of point B are (d, d)
where d = 0.3 cos(45◦ ). Thedistance b can be determined from the
Pythagorean Theorem as b = (0.35)2 − d2 . From the diagram, the
angle θ = 37.3◦ . From these calculations, the coordinates of points
B and C are B (0.212, 0.212), and C (0.491, 0) with all distances
being measured in meters. All forces will be measured in Newtons.
The unit vector from C toward B is uCB = −0.795i + 0.606j.
The equations of force equilibrium at C are
FX = FBC cos θ − 2000 = 0,
and
FY = N − FBC sin θ = 0.
B
d
45°
A
d
θ
b
C
y
FBC
FBCY
c
2000 N
x
FBCX
Solving these equations, we get N = 1524 Newtons(N), and FBC =
2514 N.
N
The force acting at B due to member BC is FBC uBC = −2000i +
1524j N.
The position vector from A to B is rAB = 0.212i + 0.212j m, and
the moment of the force acting at B about A, calculated from the cross
product, is given by MF BC = 747.6k N-m (counter - clockwise).
The moment M about A which is necessary to hold the system in
equilibrium, is equal and opposite to the moment just calculated. Thus,
M = −747.6k N-m (clockwise).
0.35 m
0.3 m
B
y
FBC uCB
M
rAB
A
x
Problem 6.84 In Problem 6.83, determine the forces
on member AB at A and B.
Solution: In the solution of Problem 6.83, we found that the force
acting at point B of member AB was FBC uBC = −2000i +
1524j N, and that the moment acting on member BC about point
A was given by M = −747.6k N-m (clockwise). Member AB must
be in equilibrium, and we ensured moment equilibrium in solving
Problem 6.83.
From the free body diagram, the equations for force equilibrium are
FX = AX + FBC uBCX = AX − 2000 N = 0,
and
FY = AY + FBC uBCY = AY + 1524 N = 0.
Thus, AX = 2000 N, and AY = −1524 N.
FBC uCB
y
B
M
AX
A
AY
x
Problem 6.85 The mechanism is used to weigh mail.
A package placed at A causes the weighted point to rotate
through an angle α. Neglect the weights of the members
except for the counterweight at B, which has a mass of
4 kg. If α = 20◦ , what is the mass of the package at A?
A
100
mm
100
mm
B
30°
α
Solution: Consider the moment about the bearing connecting the
motion of the counter weight to the motion of the weighing platform. The moment arm of the weighing platform about this bearing
is 100 cos(30 − α). The restoring moment of the counter weight is
100 mg sin α. Thus the sum of the moments is
M = 100 mB g sin α − 100 mA g cos(30 − α) = 0.
Define the ratio of the masses of the counter weight to the mass of the
B
package to be RM = m
. The sum of moments equation reduces to
mA
M = RM sin α − cos(30 − α) = 0,
A
100
mm
100
mm
cos(30−α)
from which RM =
= 2.8794, and the mass of the packsin α
age is mA = R4 = 1.3892 = 1.39 kg
M
B
30°
α
Problem 6.86 The scoop C of the front-end loader
is supported by two identical arms, one on each side of
the loader. One of the two arms (ABC) is visible in
the figure. It is supported by a pin support at A and
the hydraulic actuator BD. The sum of the other loads
exerted on the arm, including its own weight, is F =
1.6 kN. Determine the axial force in the actuator BD
and the magnitude of the reaction at A.
A
C
0.8 m
D
B
F
0.2 m
1m
1m
Solution: The section ABC as a free body: The sum of the moments about A:
MA = 0.8BD − 2F = 0,
from which BD = 4 kN.
The sum of the forces:
Fx = Ax + BD = 0,
B
from which Ax = −4 kN.
Fy = Ay − F = 0,
from which Ay = 1.6 kN. The magnitude of the reaction at A is
A = A2x + A2y = 4.308 kN
C
80 cm
D
F
20 cm
1m
1m
Ay
Ax
0.8 m
BD
F
2m
Problem 6.87 The mass of the scoop is 220 kg, and its
weight acts at G. Both the scoop and the hydraulic actuator BC are pinned to the horizontal member at B. The
hydraulic actuator can be treated as a two-force member.
Determine the forces exerted on the scoop at B and D.
1m
D
C
1m
0.6 m
G
B
A
0.6 m
0.15 m
Solution: We need to know the locations of various points in
the Problem . Let us use horizontal and vertical axes and define the
coordinates of point A as (0,0). All distances will be in meters (m)
and all forces will be in Newtons (N). From the figure in the text,
the coordinates in meters of the points in the problem are A (0, 0),
B (0.6, 0), C (−0.15, 0.6), D (0.85, 1), and the x coordinate of
point G is 0.9 m. The unit vector from C toward D is given by
uCD = 0.928i + 0.371j, and the force acting on the scoop at D is
given by D = DX i + DY j = 0.928Di + 0.371Dj. From the free
body diagram of the scoop, the equilibrium equations are
FX = BX + DX = 0,
FY = BY + DY − mg = 0,
and
MB = −0.3 mg + xBD DY − yBD DX = 0.
From the geometry, xBD = 0.25 m, and yBD = 1 m. Solving the
equations of equilibrium, we obtain BX = 719.4 N, BY = 2246 N,
and D = −774.8 N (member CD is in tension).
D
C
1m
0.6 m
G
B
A
0.15 m
0.6 m
DY
y
D
DX
C
mg
B
x
BX
BY
D uCD
DY
DX
Problem 6.88 In Problem 6.87, determine the axial
force in the hydraulic actuator BC.
Solution:
Solving these equations, we get TBC = −1267 N(compression), and
TAC = 1112 N(tension).
Scoop
1m
A
The unit vectors in the directions of the forces acting
at C are uCD = 0.928i + 0.371j, uCA = 0.243i − 0.970j, and
uCB = 0.781i − 0.625j. The force equilibrium equations at C are
FX = TBC uCBX + TAC uCAX + TCD uCDX = 0,
and
FY = TBC uCBY + TAC uCAY + TCD uCDY = 0.
0.3 m
D
TAC
TCD
C
C
TCD
TBC
A
TBC
TAC
TCD
C
TBC
TAC
0.3 m Scoop
100 N
Problem 6.89 Determine the force exerted on the bolt
by the bolt cutters.
A
75
mm
40 mm
C 55 mm
B
D
90 mm
60 mm 65 mm
300 mm
100 N
Solution: The equations of equilibrium for each of the members
will be developed.
Member AB: The equations of equilibrium are:
FX = AX + BX = 0,
FY = AY + BY = 0,
and
MB = 90F − 75AX − 425(100) = 0
100 N
F
90 mm 60 mm 65 mm
300 mm
λ
40 mm
B
55 mm
75 mm
A
BX
BY
90 mm
60 mm 65 mm
AY
AX
300 mm
CY
40 mm
75 mm
90 mm
C
55 mm X
C
D
F
60 mm 65 mm
300 mm
BY
DX
B
BX
100 N
100 N
AY
AX
F
Member CD: The equations are:
FX = −CX − DX = 0,
FY = −CY − DY = 0.
Solving the equations simultaneously (we have extra (but compatible)
equations, we get F = 1051 N, AX = 695 N, AY = 1586 N,
BX = −695 N, BY = −435 N, CX = 695 N, CY = 535 N,
DX = −695 N, and Dy = −535 N
40 mm
C 55 mm
B
D
Member BD: The equations are
FX = −BX + DX = 0,
FY = −BY + DY + 100 = 0,
and
MB = 15DX + 60DY + 425(100) = 0.
Member AC: The equations are
FX = −AX + CX = 0,
FY = −AY + CY + F = 0,
and
MA = −90F + 125CY + 40CX = 0.
A
75
mm
D
DY
60 mm 65 mm
300 mm
100 N
CY
C
DX
D
DY
CX
Problem 6.90 For the bolt cutters in Problem 6.89,
determine the magnitude of the force the members exert
on each other at the pin connection B and the axial force
in the two-force member CD.
Solution: From the solution to 6.107, we know BX = −695 N,
and BY = −435 N. We also know that CX = 695 N, and CY =
535 N, from which the axial load in member
CD can be calculated.
2 + C 2 = 877 N
CX
Y
Problem 6.91 The device is designed to exert a large
force on the horizontal bar at A for a stamping operation.
If the hydraulic cylinder DE exerts an axial force of
800 N and α = 80◦ , what horizontal force is exerted on
the horizontal bar at A?
90°
D
0
25
α
m
B
mm
25
0m
The load in CD is given by TCD =
25
0m
m
A
E
C
400 mm
Solution: Define the x-y coordinate system with origin at C. The
projection of the point D on the coordinate system is
90°
Ry = 250 sin α = 246.2 mm,
The angle formed by member DE with the positive x axis is θ =
Ry
180 − tan−1 400−R
= 145.38◦ . The components of the
x
force produced by DE are Fx = F cos θ = −658.3 N, and
Fy = F sin θ = 454.5 N. The angle of the element AB with the
positive x axis is β = 180 − 90 − α = 10◦ , and the components
of the force for this member are Px = P cos β and Py = P sin β,
where P is to be determined. The angle of the arm BC with the positive x axis is γ = 90 + α = 170◦ . The projection of point B is
Lx = 250 cos γ = −246.2 mm, and Ly = 250 sin γ = 43.4 mm.
Sum the moments about C:
MC = Rx Fy − Ry Fx + Lx Py − Ly Px = 0.
Substitute and solve: P = 2126.36 N, and Px = P cos β = 2094 N
is the horizontal force exerted at A.
α
D
and Rx = 250 cos α = 43.4 mm.
B
250 mm
C
A
250 mm
250 mm
400 mm
Fy
B
Py
D
Px
Fx
Cx
Cy
E
Problem 6.92 This device raises a load W by extending the hydraulic actuator DE. The bars AD and BC
are 4 ft long, and the distances b = 2.5 ft and h = 1.5 ft.
If W = 300 lb, what force must the actuator exert to
hold the load in equilibrium?
b
W
A
B
h
D
C
The angle ADC is α = sin−1
distance CD is d = 4 cos α.
Solution:
h
4
E
= 22.02◦ . The
b
The complete structure as a free body: The sum of the forces:
Fy = −W + Cy + Dy = 0.
Fx = Cx + Dx = 0.
W
F
B
A
The sum of the moments about C:
MC = −bW + dDy = 0.
h
These have the solution:
Cy = 97.7 lb,
D
C
E
Dy = 202.3 lb,
and Cx = −Dx .
Divide the system into three elements: the platform carrying the
weight, the member AB, and the member BC.
W
The Platform: (See Free body diagram) The moments about the
point A:
MA = −bW − dB = 0.
A
B
Ex
The sum of the forces:
Fy = A + B + W = 0.
B
A
Cy
Ey
Cx E
These have the solution:
B = −202.3 lb,
Ex
Dx
and A = −97.7 lb.
Element BC: The sum of the moments about E is
h
d
d
MC = −
Cy +
Cx +
B = 0, from which
2
2
2
(3)
Dy
Cy − Ey + B = 0
Element AD: The sum of the moments about E:
d
h
d
ME =
Dy +
Dx −
A = 0,
2
2
2
(1)
dCx − hCy − dB = 0. The sum of the forces:
Fx = Cx − Ex = 0, from which
(2)
Ex − Cx = 0,
Fy = Cy − Ey + B = 0,
These are four equations in the four unknowns: EX , EY , Dx , CX and
DX
from which
Solving, we obtain Dx = −742 lb.
from which
(4)
dDy + hDx − dA = 0.
Problem 6.93 The linkage is in equilibrium under the
action of the couples MA and MB . If αA = 60◦ and
αB = 70◦ , what is the ratio MA /MB ?
250 mm
MB
αB
MA
αA
150 mm
350 mm
Solution: Make a cut through the linkage connecting the two
cranks, and treat each system as a free body. The equilibrium condition occurs when the reaction forces in the linkage are equal and
opposite.
200 mm
The position vector of the end of the system B crank is
αB
MB
rB = RB (i cos αB + i sin αB ) = 85.51i + 234.92j (mm).
MA
The position vector at the end of the system A crank is
150 mm
rA = RA (i cos αA + j sin αA ) = 75i + 129.9j (mm).
The angle of the linkage from the end of the system B crank with
respect to the horizontal is
yA − yB
β = tan−1
= −17.19◦ .
xA − xB + 350
350 mm
The unit vector parallel to the linkage, originating at the B crank, is
|F|
eBA = i cos β + j sin β = 0.9553i − 0.2955j.
The unit vector originating at A crank is eAB = −eBA . The components of the forces in the linkage are |F|eAB , and |F|eBA .
System B: When the system is in equilibrium,
0
0
1
MB + |F| 85.5
234.9
0 = 0,
0.9553 −0.2955 0 from which MB = 249.7|F|.
System A: When the system is in equilibrium:
0
0
1
MA + |F| 75
129.9 0 = 0,
−0.9553 0.2955 0 from which MA = −146.27|F|.
Complete system: Both systems are in equilibrium for the value |F|.
Take the ratio of the two moments to eliminate |F|.
MA
146.27
=−
= −0.5858
MB
249.7
αA
MB
|F|
MA
Problem 6.94 A load W = 2 kN is supported by the
members ACG and the hydraulic actuator BC. Determine the reactions at A and the compressive axial force
in the actuator BC.
A
0.75 m
B
C
1m
G
0.5 m
W
1.5 m
Solution:
1.5 m
The sum of the moments about A is
MA = 0.75BC − 3(2) = 0,
from which BC = 8 kN is the axial force. The sum of the forces
FX = AX + BC = 0,
from which AX = −8 kN.
FY = AY − 2 = 0,
A
0.75 m
B
C
1.0 m
from which AY = 2 kN.
G
0.5 m
W
1.5 m
1.5 m
AY
0.75 m
AX
BC
W
3m
Problem 6.95 The dimensions are a = 260 mm, b =
300 mm, c = 200 mm, d = 150 mm, e = 300 mm,
and f = 520 mm. The ground exerts a vertical force
F = 7000 N on the shovel. The mass of the shovel is
90 kg and its weight acts at G. The weights of the links
AB and AD are negligible. Determine the horizontal
force P exerted at A by the hydraulic piston and the
reactions on the shovel at C.
Solution: The free-body diagram of the shovel is from which we
obtain the equations
Fx = Cx − T cos β = 0,
(1)
Fy = Cy + T sin β + F − mg = 0, (2)
M(ptC) = f F − emg + (b − c)T sin β
+dT cos β = 0.
b
P
Shovel
A
D
a
d
B
C
(3)
G
c
The angle β = arctan[(a − d)/b].
e
From the free-body diagram of joint A,
B
P
f
F
T
T
we obtain the equation
F = P + T cos β = 0. (4)
β
d
b−c
Substituting the given information into Eqs. (1)–(4) and solving, we
obtain
T = −19, 260 N,
CX
CY
mg
P = 18, 080 N,
Cx = −18, 080 N,
and Cy = 513 N.
e
f
F
Problem 6.96 The truss supports a load F = 10 kN.
Determine the axial forces in the members AB, AC,
and BC.
B
3m
C
A
D
4m
3m
F
Solution:
+
Solving,
Find the support reactions at A and D.
Fx :
Ax = 0
Fy :
Ay + Dy − 10 = 0
MA :
B
(−4)(10) + 7Dy = 0
Ax = 0,
3m
C
A
D
Ay = 4.29 kN
Dy = 5.71 kN
4m
Joint A:
tan θ =
3m
3
4
F
θ = 36.87◦
(Ay = 4.29 kN)
Fx :
FAB cos θ + FAC = 0
Fy :
Ay + FAB sin θ = 0
Solving,
3m
AX
4m
FAB = −7.14 kN (C)
3m
AY
10 kN
FAC = 5.71 kN (T )
Joint C:
Fx :
Fy :
FAB
FCD − FAC = 0
y
FBC − 10 kN = 0
Solving FBC = 10 kN (T )
θ
FAC
x
FCD = +5.71 kN (T )
AY
FBC
FAC
FCD
10 kN
DY
Problem 6.97 Each member of the truss shown in
Problem 6.96 will safely support a tensile force of 40 kN
and a compressive force of 32 kN. Based on this criterion,
what is the largest downward load F that can safely be
applied at C?
B
2
5
1
3m
C
A
D
3
4
4m
3m
F
Solution: Assume a unit load F and find the magnitudes of the
tensile and compressive loads in the truss. Then scale the load F
up (along with the other loads) until either the tensile limit or the
compressive limit is reached.
External Support Loads:
Fx :
Ax = 0
(1)
Fy :
Ay + Dy − F = 0 (2)
MA :
−4F + 7Dy = 0 (3)
AX
F
AY
Joint A:
tan θ =
3
2
y
FAB
θ = 36.87◦
Fx :
FAC + FAB cos θ = 0 (4)
Fy :
FAB sin θ + Ay = 0
θ
(5)
x
FAC
Joint C
Fx :
Fy :
FCD − FAC = 0 (6)
FBC − F = 0
AY
(7)
Joint D
tan φ =
3
3
y
FBD
φ = 45◦
Fx :
−FCD − FBD cos φ = 0 (8)
Fy :
FBD sin φ + Dy = 0
φ
(9)
Setting F = 1 and solving, we get the largest tensile load of 0.571 in
AC and CD. The largest compressive load is 0.808 in member BD.
x
FCD
Largest Tensile is in member BC. BC = F = 1
DY
The compressive load will be the limit
Fmax
32
=
1
0.808
y
FBC
Fmax = 40 kN
FCD
FAC
F
x
DY
Problem 6.98 The Pratt bridge truss supports loads at
F , G, and H. Determine the axial forces in members
BC, BG, and F G.
Solution:
The angles of the cross-members are α = 45◦ .
The complete structure as a free body:
B
C
D
4m
E
A
F
G
The sum of the moments about A:
MA = −60(4) − 80(8) − 20(12) + 16E = 0,
60 kN
80 kN
20 kN
4m
4m
4m
4m
from which E = 70 kN. The sum of the forces:
Fx = Ax = 0.
Fy = Ay − 60 − 80 − 20 + E = 0,
from which Ay = 90 kN
H
B
C
D
The method of joints: Joint A:
FY = Ay + AB sin α = 0,
from which AB = −127.3 kN (C),
Fx = AB cos α + AF = 0,
4m
A
E
F
from which AF = 90 kN (T ). Joint F:
Fx = −AF + F G = 0,
G
H
60 kN
80 kN
20 kN
4m
4m
4m
4m
from which F G = 90 kN (T ) .
Fy = BF − 60 = 0,
from which BF = 60 kN (C). Joint B:
Fx = −AB cos α + BC + BG cos α = 0,
and
Fy = −AB sin α − BF − BG sin α = 0,
4m
Ay
−AB sin α − BF − BG sin α = 0.
and − AB cos α + BC + BG cos α = 0,
from which BC = −120 kN (C)
4m
4m
4m
Ax
from which:
Solve: BG = 42.43 kN (T ) ,
4m
Ay
AB
α
AF
Joint A
60 kN 80 kN 20 kN
BF
AF FG
60 kN
Joint F
E
α
α BC
AB BF BG
Joint B
Problem 6.99 Consider the truss in Problem 6.98. Determine the axial forces in members CD, GD, and GH.
BC
CD
BG
α
CG
Joint C
Solution:
CG
GD
α
GH
80 kN
Joint G
Use the results of the solution of Problem 6.98:
BC = −120 kN (C),
BG = 42.43 kN (T ),
and F G = 90 kN (T ).
The angle of the cross-members with the horizontal is α = 45◦ .
Joint C:
Fx = −BC + CD = 0,
from which CD = −120 kN (C)
FY = −CG = 0,
from which CG = 0.
Joint G:
Fy = BG sin α + GD sin α + CG − 80 = 0,
from which GD = 70.71 kN (T ) .
Fy = −BG cos α + GD cos α − F G + GH = 0,
from which GH = 70 kN (T )
Problem 6.100 The truss supports a 400-N load at
G. Determine the axial forces in members AC, CD,
and CF
400 N
A
C
E
G
300 mm
600 mm
H
F
D
B
300 mm
Solution: The complete structure as a free body: The sum of the
moments about A:
MA = −900(400) + 600B = 0,
from which B = 600 N. The sum of forces:
Fx = Ax + B = 0,
from which BD = −632.5 N (C)
Fy = AB + BD sin θ = 0,
from which AB = 200 N (T )
Joint A:
Fy = Ay − AD sin αAD − AB = 0,
from which AD = 233.2 N (T )
Fx = Ax + AC + AD cos αAD = 0,
from which AC = 480 N (T )
E
G
300 mm
H
F
D
B
300
mm
300
mm
300
mm
900 mm
Ay
400 N
Ax
600 mm
The angle from the horizontal of element AD is
300
αAD = 90 − tan−1
= 59.04◦ .
600 − 300 tan θ
Joint B:
Fx = B + BD cos θ = 0,
C
600 mm
from which Ay = 400 N.
The angle from the horizontal of element CF is
300
αCF = 90 − tan−1
= 53.13◦ .
600(1 − tan θ)
300 mm
400 N
A
from which Ax = −600 N.
Fy = Ay − 400 = 0,
The method of joints: The angle from the horizontal of element BD
is
300
θ = tan−1
= 18.43◦ .
900
300 mm
B
AB
B
θ
BD
AC
αAD
AD
AB
Joint A
Joint B
CD
AD
αAD
DF
θ
BD
AY
AX
AC
Joint D
CE
αCF
CF
CD
Joint C
Joint D:
Fx = −AD cos αAD − BD cos θ + DF cos θ = 0,
from which DF = −505.96 N (C)
Fy = AD sin αAD + CD − BD sin θ + DF sin θ = 0,
from which CD = −240 N (C)
Joint C:
Fy = −CD − CF sin αCF = 0,
from which CF = 300 N (T )
Problem 6.101 Consider the truss in Problem 6.100.
Determine the axial forces in members CE, EF ,
and EH.
Solution:
Use the results of the solution of Problem 6.100:
AC = 480 N (T ),
AC
CF = 300 N (T ),
DF = −505.96 N (C),
CD
αCF = 53.1◦ .
The method of joints: The angle from the horizontal of element EH
is
300
αEH = 90 − tan−1
= 45◦
600 − 900 tan θ
CF
Joint C
θ = 18.4◦ ,
EF
CF
αCF
θ
DF
CE
αCF
FH
EF
Joint F
Joint E
from which F H = −316.2 N (C)
Fy = EF + CF sin αCF − DF sin θ + F H sin θ = 0,
from which EF = −300 N (C)
from which CE = 300 N (T )
Joint E:
Fy = −EH sin αEH − EF = 0,
from which EH = 424.3 N (T )
Problem 6.102 The mass m = 120 kg. Determine
the forces on member ABC.
The weight of the hanging mass is given by
m
W = mg = 120 kg 9.81 2 = 1177 N.
s
A
B
C
300 mm
D
The complete structure as a free body: The equilibrium equations are:
FX = AX + EX = 0,
FY = AY − W = 0,
and
MA = 0.3EX − 0.4W = 0.
200 mm
A
AX = −1570 N,
200 mm
B
AY = 1177 N,
Element ABC: The equilibrium equations are
FX = Ax + CX = 0,
FY = AY + CY − BY − W = 0,
and:
MA = −0.2BY + 0.4cY − 0.4W = 0.
Solution gives BY = 2354 N (member BD is in tension),
CX = 1570 N,
m
E
Solving, we get
and EX = 1570 N.
EH
Joint F:
Fy = −CF cos αCF − DF cos θ + F H cos θ = 0,
Joint C:
Fx = −AC + CE + CF cos αCF = 0,
Solution:
EG
αEH
CE
300 mm
D
E
200
mm
W
200
mm
Cy
Ay
Ax
Cx
B
W
B
and CY = 2354 N.
Cx
B
Cy
B
Ex
Problem 6.103
Determine the forces on member
ABC, presenting your answers as shown in Fig. 6.35.
D
400 lb
2 ft
200 ft-lb
A
1 ft
B
C
100 lb
1 ft
E
2 ft
2 ft
Solution: The complete structure as a free body: The sum of the
moments:
MA = 100(1) − 400(6) − 200 + 4E = 0,
D
from which E = 625 lb. The sum of the forces:
Fy = Ay + E − 400 = 0,
2 ft
from which Ay = −225 lb.
Fx = Ax + 100 = 0,
1 ft
400 lb
A
B
2 ft
2 ft
2 ft
Dy
Dx
By
Bx
Ay
Ax
Dy
Bx
By
Cx
Dx
400 lb
Cy200 ft-lb
from which (6) By − Dy = 0
Element ABC: The sum of the moments about A:
MA = −2By − 4Cy − 200 − 6(400) = 0,
from which (7) By + 2Cy = −1300.
Cx
E
Element BD: The sum of the moments about B:
MB = 2Dx − 2Dy = 0,
from which (5) Bx − Dx = 0.
Fy = By − Dy = 0,
Cy
100 lb
thus (3) Dy + Cy + E = 0.
from which (4) Dx − Dy = 0. The sum of the forces:
Fx = Bx − Dx = 0,
200
ft-lb
E
Element ECD: (See the free body diagram.) The sum of the moments
about E:
ME = −4Dx − 2Cx − 100 = 0,
from which (2) Dx + Cx = −100.
Fy = E + Cy + Dy = 0,
C
100 lb
1 ft
from which Ax = −100 lb. These results are used as a check on the
solution below.
from which (1) 4Dx + 2Cx = −100. The sum of the forces:
Fx = Dx + Cx + 100 = 0,
2 ft
100 lb
675 lb
400 lb
150
lb
200
ftÐlb
50 lb
225
lb
50 lb
The sum of the forces:
Fx = Ax − Bx − Cx = 0,
from which (8) Ax − Bx − Cx = 0.
Fy = Ay − By − Cy − 400 = 0,
from which (9) Ay − By − Cy = 400. These nine equations are solved for
the nine reactions The reactions are DX = 50 lb, DY = 50 lb,:
CX = −150 lb, CY = −675 lb, BX = 50 lb,
BY = 50 lb, AX = −100 lb, AY = −225 lb,
and E = 625 lb.
Problem 6.104 Determine the force exerted on the
bolt by the bolt cutters and the magnitude of the force
the members exert on each other at the pin connection A.
90 N
A
Solution:
Element AB: The moment about A is
MA = −10B − 54F = 0,
80
mm
160
mm
540 mm
100
mm
90 N
where F = 90 N. From which B = −486 N. The sum of the forces:
Fy = A + B − F = 0,
from which A = 576 N
Element BC: The moment about C:
MC = −16B − 8FC = 0,
from which the cutting force is FC = 972 N
Problem 6.105 The 600-lb weight of the scoop acts at
a point 1 ft 6 in. to the right of the vertical line CE. The
line ADE is horizontal. The hydraulic actuator AB can
be treated as a two-force member. Determine the axial
force in the hydraulic actuator AB and the forces exerted
on the scoop at C and E.
C
FC
8
cm
B
90 N
A
B
16
cm
54 cm
10
cm
B
C
2 ft
A
B
C
2 ft
A
5 ft
1 ft 6 in
E
2 ft 6 in
D
uBC = 0.981i − 0.196j,
1 ft
and uBD = −0.447i − 0.894j.
The scoop: The equilibrium equations for the scoop are
FX = −TCB uBCX + EX = 0,
FY = −TCB uBCY + EY − 600 = 0,
and
MC = 1.5EX − 1.5(600 lb) = 0.
Solving, we get
1 ft 6"
TCB
C
1.5 ft
1.5 ft
G
EX
EX = 600 lb,
E
EY
EY = 480 lb,
600 lb
and TCB = 611.9 lb.
Joint B: The equilibrium equations for the scoop are
FX = TBA uBAX + TBD uBDX + TCB uBCX = 0,
and
FY = TBA uBAY + TBD uBDY + TCB uBCY = 0.
Solving, we get
TBA = 835 lb,
and TBD = −429 lb.
2 ft 6 in
1 ft
Solution: The free body diagrams are shown at the right. Place the
coordinate origin at A with the x axis horizontal. The coordinates (in
ft) of the points necessary to write the needed unit vectors are A (0, 0),
B (6, 2), C (8.5, 1.5), and D (5, 0). The unit vectors needed for this
problem are
uBA = −0.949i − 0.316j,
E
D
5 ft
y
TBA
x
TCB
TBD
1 ft 6 in
600 lb
Scoop
Problem 7.1 If a = 2, what is the x coordinate of the
centroid of the area?
Strategy: The x coordinate of the centroid is given by
Eq. (7.6). For the element of area dA, use a vertical strip
of width dx. (See Example 7.1).
y
y = x2
x
a
Solution:
y
x dA
x = x =
(1, 1)
dA
a
0
x(y dx)
a
y = x2
y dx
0
Substituting y = x2 , we get
a
4 a
x
x3 dx
3a
4 x = 0a
= 3 =
x
4
2
x dx
3
y
y = x2
0
0
For
x
a
a =2
3
2
x =
dA = y dx
x
a
Problem 7.2 Determine the y coordinate of the centroid of the area shown in Problem 7.1 if a = 3.
Solution:
y
y dA
y = y =
(1, 1)
dA
a
0
1
y(y dx)
2
a
y = x2
(y dx)
0
Substituting y = x2 , we get
a
a
1 4
1 x5
x dx
2
5
2
0
y = a
= 3 a0 =
x
2
x dx
3
0
0
a
a5
2.5
a3
3
x
y
dA = y dx
3 2
y =
a
10
For a = 3,
y=
27
10
Midpoint
y/2
a
dx
x
Problem 7.3 If the x coordinate of the centroid of the
area is x = 2, what is the value of a?
y
y = x3
0
Solution:
x dA
x = A
dA
A
a
x = 0
x(y dx)
a
.
(y dx)
0
Substituting y = x3 ,
a

x4 dx
0
x = a
=
x3 dx
0
If x = 2,
2=
4a
5
x5
5
x4
4
and
a
 = 4a
5
0
a=
10
4
= 2.5
y
y = x3
0
a
x
y
y = x3
dA = y dx
a
x
a
x
Problem 7.4 The x coordinate of the centroid of the
area shown in Problem 7.3 is x = 2. What is the y
coordinate of the centroid?
From Problem 7.4, a = 2.5
2.5
2.5
1
1 6
y(y dx)
x dx
y dA
2
2
0
0
y = =
= 2.5
2.5
dA
y dx
x3 dx
Solution:
0
2.5
2.5
4x3 y = 4 =
= 4.46
x
14 0
4
y
0
y = x3
x7
14
0
y = 4.46
y = x3
a
0
x
y
y/2
a
x
Problem 7.5 Consider the area in Problem 7.3. The
“center of the area” is defined to be the point for which
there is as much area to the right of the point as to the
left of it and as much area above the point as below it.
If a = 4, what are the x coordinate of the center of area
and the x coordinate of the centroid?
Center of Area: Let X be the coordinate of the center
of the area. The area to the left of X is
X
X4
AL =
x3 dx =
.
4
0
Solution:
The area to the right is
a
a4
X4
AR =
x3 dx =
−
.
4
4
X
Equating the two areas,
a4
X4
X4
−
=
,
4
4
4
from which a4 = 2X 4
or
1
X = a(2)− 4 .
For a = 4, X = 0.8408(4) = 3.3636
The centroid: The centroid is
a
x4 dx
4a
0
x= a
=
.
5
x3 dx
0
For a = 4, x =
16
5
= 3.200
Problem 7.6 Determine the x coordinate of the centroid of the area and compare your answer to the value
given in Appendix B.
y
y = cx n
0
Solution:
a
x(y dx)
=
a
y dx
0
x =
a
xn+2
(n+2) 0
n+1 a
x
(n+1) 0
x =a
a
x
y = cx n
dA
0
x =
x
y
x dA
x = a
a
0
a
0
=
xC/xn dx
0
C/xn dx
y = cxn
y
(n + 1)
a
(n + 2)
dA = y dx
(n + 1)
(n + 2)
Checks with result in Appendix
x
a
dx
Problem 7.7 Determine the y coordinate of the centroid of the area and compare your answer to the value
given in Appendix B.
Solution:
y
y dA
y = a
0
y =
1
y
y dx
2
a
y dx
a
0
2
c x
dx
cxn dx
c2
=
c
0
y =
0
a
x
0
2 2n
a
a
y 2 dx
1 0
a
=
2
y dx
0
y =
y = cx n
dA
x2n+1
(2n+1)
xn+1
(n+1)
a
0
c (n + 1) n
a
2 (2n + 1)
y/2
Checks with Appendix
a
Problem 7.8 Suppose that an art student wants to paint
a panel of wood as shown, with the horizontal and vertical lines passing through the centroid of the painted
area, and asks you to determine the coordinates of the
centroid. What are they?
y
y = x + x3
0
Solution:
A =
1
The area:
x2
x4
+
2
4
(x + x3 ) dx =
0
The x-coordinate:
1
x(x + x3 ) dx =
x3
x5
+
3
5
0
Divide by the area: x =
32
45
1
1
=
0
=
0
3
.
4
8
.
15
= 0.711
The y-coordinate: The element of area is dA = (1 − x) dy. Note that
dy = (1 + 3x2 ) dx, hence dA = (1 − x)(1 + 3x2 ) dx. Thus
4
yA =
y dA =
(x + x3 )(1 − x)(1 + 3x2 ) dx,
A
0
from which
1
(x − x2 + 4x3 − 4x4 + 3x5 − 3x6 ) dx
0
=
1
1
4
4
3
3
− + − + − = 0.4381.
2
3
4
5
6
7
Divide by A y = 0.5841
y
y = x + x3
0
1 ft
x
1 ft
x
Problem 7.9 The y coordinate of the centroid of the
area is y = 1.063. Determine the value of the constant
c and the x coordinate of the centroid.
y
y = cx 2
0
2
x
4
Solution:
y
y dA
y = dA
dA = y dx
4
y = 2
y = cx 2
y
y dx
2
4
y dx
2
4
24
C 2 x4 dx
Cx2 dx
0
2
c2/
x5
5
2C/
x3
3
y =
=
1
2
4
C 1024
− 32
2
5
5
64
4 =
2 3 − 83
2
4
x
y = cx2
y
2
y = 5.314C
But y = 1.063
∴ C = 0.200
Now we have C known and y = Cx2
4
x4
C/x3 dx
x dA
4
2
x = = 4
=
x3
dA
C/x2 dx
3
2
x =
(256−16)
4
(64−8)
3
x = 3.214
= 3.214
4
2
4
2
y/2
1
2
3
4
x
Problem 7.10 Determine the coordinates of the centroid of the metal plate’s cross-sectional area.
y
1
y = 4 – – x 2 ft
4
x
Let dA bea vertical strip:
The area dA = y dx = 4 − 14 x2 dx. The curve intersects the x
1 2
axis where 4 − 4 x = 0, or x = ±4.
Therefore
4
1
4 4
4x − x3 dx
x dA
2x2 − x16
4
−4
−4
x = A
= 4
= 4 = 0.
1 2
x3
4x − 12
dA
4− x
dx
−4
4
A
−4
Solution:
To determine y, let y in equation (7.7) be the height of the midpoint of
the vertical strip:
4
1
1
1
4 − x2
4 − x2 dx
y dA
4
4
−4 2
=
y = A
4
1 2
dA
4− x
dx
4
A
−4
4
4
1 4
3
x5
8 − x2 +
x
dx
8x − x3 + 5(32)
32
−4
−4
=
=
4
4
x2
x3
4x − 12
4−
dx
−4
4
−4
=
34.1
= 1.6 ft.
21.3
y
1
y = 4 – – x 2 ft
4
x
y
dA
y
x
x
dx
4
3
y, m
Problem 7.11 An architect wants to build a wall with
the profile shown. To estimate the effects of wind loads,
he must determine the wall’s area and the coordinates of
its centroid. What are they?
y = 2 + 0.02x2
2
1
0
0
2
4
6
8
10
x, m
Solution:
10
10
y dx =
0
4
(2 + 0.02x2 ) dx
3
0
Area = 2x + 0.02
10
x3
3
y, m
Area =
= 26.67 m2
y = 2 + 0.02x2
2
0
1
dA = y dx = (2 + 0.02x2 ) dx
0
0
2
4
6
x, m
Y
X
10
0
x = 10
x dA
=
10
26.67
dA
0
2
x =
(2x + 0.02x3 ) dx
0
10
4
2/ x2/ + 0.02 x4
0
26.67
m
4
x =
100 + (0.02) 104
26.67
x = 5.62 m
10
y =
0
y
y dx
2
10
dA
=
150
26.67
1
=
(2)
10
(2 + 0.02x2 )2 dx
0
(26.67)
0
y =
1
2(26.67)
10
4x + 0.08
y =
y =
(4 + 0.08x2 + 0.0004x4 ) dx
0
x3
3
+ 0.0004
2(26.67)
74.67
53.34
y = 1.40 m
x5
5
10
0
8
10
Problem 7.12 Determine the x coordinate of the centroid of the area.
y
y = – x 2 + 8x – 12
x
Solution: First, we must determine where the curve intersects the
x-axis. These will be the limits of our integration.
y
Set y = 0
y = – x 2 + 8x – 12
0 = −x2 + 8x − 12
or
x2 − 8x + 12 = 0
(x − 6)(x − 2) = 0
Thus, y = 0 at x = 2 and x = 6.
6
6
x dA
x(y dx)
x = 2 6
= 2 6
dA
(y dx)
2
6
2
x =
x
2
(−x3 + 8x2 − 12x) dx
6
2
(−x2 + 8x − 12) dx
4
3
2 6
− x4 + 8 x3 − 12 x2
42.67
2
x = =4
6 =
10.67
x3
x2
− 3 + 8 2 − 12x
2
Note: Once we had the limits of integration, the result was apparent
due to symmetry.
Problem 7.13 Determine the y coordinate of the centroid of the area shown in Problem 7.12.
From Problem 7.12, the limits of integration are x = 2
and x = 6. The area is 10.67 units.
6
6
y
1
dA
y 2 dx
2
2
2
2
y =
=
Area
Area
6
(−x2 + 8x − 12)2 dx
1
2
y =
2
Area
6
(x4 − 16x3 + 88x2 − 192x + 144) dx
1 2
y =
2
Area
5
6
4
x
x
x3
x2
−
16
+
88
−
192
+
144x
4
3
2
1 5
2
y =
2
(10.67)
Solution:
y =
34.13
= 1.6
21.33
y = 1.6
y
y = – x 2 + 8x – 12
x
Problem 7.14 Determine the x coordinate of the centroid of the area.
y
y = x3
y=x
x
Solution:
Work this problem like Example 7.2
1
x dA
x(x − x3 ) dx
0
0
x = 1
= 1
dA
(x − x3 ) dx
x = 1
0
3
x
3
x2
2
5
−
x
5
−
x4
4
y
y = x3
0
1
y=x
1
0
3
1 = 1
2
0
−
−
1
5
1
4
=
2
15
1
4
= 0.533
dA
x = 0.533
x
Problem 7.15 Determine the y coordinate of the centroid of the area shown in Problem 7.14.
Solution:
Solve this problem like example 7.2.
1
1
(x + x3 ) (x − x3 ) dx
y dA
2
y = A
= 0
1
dA
(x − x3 ) dx
A
y =
1
2
6
(x − x ) dx
1 0
=
1
2
(x − x3 ) dx
1
2
y = x3
0
y =
y
0
1
7
1
−
2
4
31
−
y = 0.381
=
4
21
2
=
y=x
7 1
− x7
0
2
4 1
2 x2 − x4
0
x3
3
8
= 0.381
21
½(x+x3)
x
Problem 7.16 Determine the coordinates of the centroid of the area.
y
(1, 1)
y = x 1/2
y = x2
Solution:
Let dA be a vertical strip: The area dA =
x1/2 − x2 dx, so
1
5/2
4 1
x
x dA
x3/2 − x3 dx
− x4
5/2
0
x = A
= 01
= 1 = 0.45.
x3/2
x3
−
dA
x1/2 − x2 dx
3/2
3
A
y
y = x 1/2
(1, 1)
0
0
If we use a horizontal strip to obtain y, we obtain
1
y dA
y 3/2 − y 3 dy
y = A
= 01
= 0.45
dA
y 1/2 − y 2 dy
A
x
0
y = x2
x
y
y = x1/2
(1, 1)
y = x2
dA
x
dx
x
Problem 7.17 Determine the x coordinate of the centroid of the area.
y
y = x 2 – 20
y=x
x
Solution: The intercept of the straight line with the parabola
occurs at the roots of the simultaneous equations: y = x, and
y = x2 − 20. This is equivalent to the solution of the quadratic
x2 − x − 20 = 0, x1 = −4, and x2 = 5. These establish the limits
on the integration. The area: Choose a vertical strip dx wide. The
length of the strip is (x − x2 + 20), which is the distance between the
straight line y = x and the parabola y = x2 − 20. Thus the element
of area is dA = (x − x2 + 20) dx and
+5
+5
x2
x3
A =
(x − x2 + 20) dx =
−
+ 20x
= 121.5.
2
3
−4
−4
The x-coordinate:
xA =
x dA =
A
=
x=
+5
−4
= 0.5
y = x 2 – 20
y=x
2
3
(x − x + 20x) dx
x3
x4
−
+ 10x2
3
4
60.75
121.5
y
+5
= 60.75.
−4
x
Problem 7.18 Determine the y coordinate of the centroid of the area in Problem 7.17.
Solution:
Use the results of the solution to Problem 7.17 in the
following.
The y-coordinate: The centroid of the area element occurs at the midpoint of the strip enclosed by the parabola and the straight line, and
the y-coordinate is:
1
2
y =x−
yA =
1
2
(x − x2 + 20) =
y dA =
A
=
=
1
2
1
2
5
−4
(x + x2 − 20)(x − x2 + 20) dx
+5
−4
1
2
(x + x2 − 20).
(−x4 + 41x2 − 400) dx
x5
41x3
+
− 400x
5
3
−
5
−4
= −923.4.
y = − 923.4
= −7.6
121.5
Problem 7.19 Determine the y coordinate of the centroid of the area.
y
(3, 7)
Solution: The area: The area element is the horizontal strip xlong and dy wide. The length is determined by the straight line, which
has the equation y = mx + b, where
5
y 1 − y2
(7 − 0)
=
= −2.3333,
x1 − x 2
(3 − 6)
m =
and b = y1 − mx1 = 7 − (2.3333)3 = 14.
2
The length of the strip is
1
m
x =
(y − b).
0
The element of area is dA =
A =
1
m
5
2
A
=
1
m
(y − b) dy, from which
1
m
(y − b) dy =
The y-coordinate:
yA =
y dA =
y2
− by
2
y
5
2
1
m
5
2
y(y − b) dy
5
= 46.2857.
2
y = 3.4286
Problem 7.20 Determine the x coordinate of the centroid of the area in Problem 7.19.
Solution: From the solution to Problem 7.19, the area A = 13.5
is bounded by the straight line y = mx + b, where m = −2.3333,
and b = 14. The horizontal strip is x-wide and dy high, where
x=
1
m
(y − b).
The centroid of x is one half this strip, hence
1
1
xA =
(y − b)2 dy =
[(y − b)3 ]52 = 30.583,
2m2
6m2
from which x =
(3, 7)
= 13.5.
5
y3
y2
−b
3
2
1
m
x
6
30.583
13.5
= 2.265
2
0
6
Problem 7.21 An agronomist wants to measure the
rainfall at the centroid of a plowed field between two
roads. What are the coordinates of the point where the
rain gauge should be placed?
y
0.5 mi
0.3 mi
0.3 mi
x
0.5 mi
Solution: The area: The element of area is the vertical strip (yt −
yb ) long and dx wide, where yt = mt x + bt and yb = mb x + bb
are the two straight lines bounding the area, where
0.5
miles
and bt = 0.8 − 1.3 mt = 0.3.
(0.3 − 0)
= 0.2308,
(1.3 − 0)
0.5 miles
and bb = 0.
The element of area is
dA = (yt − yb ) dx = ((mt − mb )x + bt − bb ) dx
= (0.1538x + 0.3) dx,
from which
1.1
A =
(0.1538x + 0.3) dx
0.5
= 0.1538
x2
+ 0.3x
2
1.1
= 0.2538 sq mile.
0.5
The x-coordinate:
1.1
x dA =
(0.1538x + 0.3)x dx
0.5
= 0.1538
x3
x2
+ 0.3
3
2
1.1
= 0.2058.
0.5
x = 0.8109 mi
The y-coordinate: The y-coordinate of the centroid of the elemental
area is
1
1
y = yb + ( )(yt − yb ) = ( )(yt + yb ) = 0.3077x + 0.15.
2
2
Thus, yA =
y dA
=
=
A
1.1
(0.3077x + 0.15)(0.1538x + 0.3) dx
0.5
1.1
(0.0473x2 + 0.1154x + 0.045) dx
0.5
= 0.0471
x3
x2
+ 0.1153
+ 0.045x
3
2
Divide by the area: y =
0.3
miles
x
0.3
miles
Similarly:
mb =
0.1014
0.2538
0.2 mi
y
(0.8 − 0.3)
= 0.3846,
(1.3 − 0)
mt =
A
0.6 mi
= 0.3995 mi
1.1
= 0.1014.
0.5
0.6 miles
0.2
miles
Problem 7.22 The cross section of an earth-fill dam
is shown. Determine the coefficients a and b so that the
y coordinate of the centroid of the cross section is 10 m.
y
y = ax – bx3
x
100 m
Solution:
The area: The elemental area is a vertical strip of length
y and width dx, where y = ax − bx3 . Note that y = 0 at x = 100,
thus b = a × 10−4 . Thus
100
A =
dA = a
(x − (10−4 )x3 ) dx
A
y
y = ax – bx3
0
2
= (0.5a)[x − (0.5 × 10−4 )x4 ]100
0
x
= 0.5a × 104 − 0.25b × 108 ,
100 m
and the area is A = 0.25a×104 . The y-coordinate: The y-coordinate
of the centroid of the elemental area is
y = (0.5)(ax − bx3 ) = (0.5a)(x − (10−4 )x3 ),
from which
yA =
y dA
A
= (0.5)a2
= (0.5)a2
100
0
100
0
= (0.5a2 )
(x − (10−4 )x3 )2 dx
(x2 − 2(10−4 )x4 + (10−8 )x6 ) dx
x3
2x5
x7
− (10−4 )
+ (10−8 )
3
5
7
100
0
= 3.81a2 × 104 .
Divide by the area:
y =
3.810a2 × 104
= 15.2381a.
0.25a × 104
For y = 10, a = 0.6562 , and b = 6.562 × 10−5 m−2
Problem 7.23 The Supermarine Spitfire used by Great
Britain in World War II had a wing with an elliptical
profile. Determine the coordinates of its centroid.
y
y2 = 1
x2 + —
—
a2
b2
x
2b
a
Solution:
y
x2 + —
y2 =
—
y
a2
1
b2
x2
y2
a2
b2
— + — =
1
b
x
b
x
2b
a
By symmetry, y = 0.
a
From the equation of the ellipse,
b 2
y=
a − x2
a
By symmetry, the x centroid of the wing is the same as the x centroid
of the upper half of the wing. Thus, we can avoid dealing with ±
values for y.
y = ab a2 – x2
y
b
dA = y dx
0
a
dx
x dA
x = =
dA
b
a
b
a
a
0
x
a
0
a2 − x2 dx
a2 − x2 dx
Using integral tables
(a2 − x2 )3/2
x a2 − x2 dx = −
3
√
2 − x2
x
a
a2
a2 − x2 dx =
+
sin−1
2
2
Substituting, we get
3/2 a
− a 2 − x2
/3
0 x =
√
a
x a2 −x2
a2
−1 x
+
sin
2
2
a
0
x = x =
−0 + a3 /3
a3 /3
=
2
a2 π/4
0 + a2 π2 − 0 − 0
4a
3π
x
a
x
Problem 7.24 Determine the coordinates of the centroid of the area.
Strategy: Write the equation for the circular boundary
in the form y = (R2 − x2 )1/2 and use a vertical “strip”
of width dx as the element of area dA.
y
Solution: The area: The equation of the circle is x2 +√y2 = R2 .
2
2
Take the elemental area to be a vertical strip of height
√ y = R −x
and width dx, hence the element of area is dA = R2 − x2 dx. The
2
area is A = Acircle
= πR
. The x-coordinate:
4
4
xA =
x=
R
x dA =
A
x
R2
x2
−
0
(R2 − x2 )3/2
dx = −
3
R
x
R
=
0
R3
:
3
y
4R
3π
The y-coordinate: The y-coordinate of the centroid of the element of
area is at the midpoint:
1 y = ( ) R 2 − x2 ,
2
hence yA =
=
y=
A
1
2
1
2
y dA =
0
x3
3
R2 x −
R
x
(R2 − x2 ) dx
R
=
0
R
R3
3
4R
3π
Problem 7.25 Determine the x coordinate of the centroid of the area. By setting h = 0, confirm the answer
to Problem 7.24.
y
Solution:
Use a vertical strip:
R
x dA
x(R2 − x2 )1/2 dx
A
h
= R
.
x = dA
(R2 − x2 )1/2 dx
A
R
x
h
h
The upper integral is
R
x(R2 − x2 )1/2 dx = R
h
R
h
1/2
x 2
x 1− 2
R
1
=R −
(R2 − x2 )3/2
3R
The area is
A =
dA =
A
R
h
R
=R
h
1−
R
=
2
2 1/2
(R − x )
x2
R2
R
x2
=
x 1− 2
2
R
2
dx = R
h
dx
R
h
dA
1
= (R2 − h2 )3/2 .
3
2 1/2
(1 − x )
dx
1/2
x
+ R arcsin
R
− R arcsin
The centroid is x = (R2 − h2 )3/2 /(3A).
R3
πR
3( R
2 )( 2 )
=
R
1/2
4R
3π
h
y = (R2 – x2)½
x
2
1/2
πR
h2
−h 1− 2
2
R
If h = 0, x =
R
y
h
h
R
.
dx
x
dx
Problem 7.26 Determine the y coordinate of the centroid of the area in Problem 7.25.
Let y in Equation (7.7) be the height of the midpoint
of a vertical strip:
R
1 2
y dA
(R − x2 )1/2 (R2 − x2 )1/2 dx
2
y = A
= h
.
dA
dA
Solution:
A
y
h
y = ½(R2 – x2)½
x
A
The upper integral is
R
R
1 2
1
x3
(R − x2 ) dx =
R2 x −
2
3 h
h 2
1
=
2
x
2R3
h3
− R2 h +
3
3
dx
From the solution of Problem 7.25,
R πR
h2
A =
dA =
−h 1− 2
2
2
R
A
.
The centroid is y =
Problem 7.27
troids.
Solution:
1
2A
2R3
3
− R2 h +
1/2
− R arcsin
h3
3
Determine the coordinates of the ceny
Let us solve this by parts.
40 mm
A1
h
A2
b
b = 60 mm
l = 40 mm
h = 40 mm
x
60 mm
40 mm
l
y
b = 60 mm
l = 40 mm
h = 40 mm
A1 =
1
1
bh = (60)(40) = 1200 mm2
2
2
A2 = lh = (40)(40) = 1600 mm
40 mm
2
A1 + A2 = 2800 mm2
From the tables and inspection
x1 =
2
b x2 = b + l/z
3
y1 =
1
h
3
y2 =
x1 = 40 mm
y1 = 13.33 mm
1
h
2
x2 = 80 mm
y2 = 20 mm
For the composite, substituting,
x =
x1 A1 + x2 A2
= 62.9 mm
A1 + A2
y =
y1 A1 + y2 A2
= 17.1 mm
A1 + A2
x
60 mm
40 mm
h
R
.
Problem 7.28
troids.
Determine the coordinates of the ceny
20 mm
60 mm
x
30 mm
70 mm
Solution: Let us solve this problem by using symmetry and by
breaking the composite shape into parts.
l1
20 mm
y
A1
20 mm
h1
l1 = 70 mm
h1 = 70 mm
l2 = 70 mm
h2 = 70 mm
A2
h2
60 mm
60 mm
x
l2
h1 = 20 mm
l2 = 30 mm
h2 = 60 mm
A1 = l1 h1 = 1400 mm2
A2 = l2 h2 = 1800 mm2
By symmetry,
y1 = 70 mm
x2 = 0
y2 = 30 mm
For the composite,
x =
0+0
x1 A1 + x2 A2
=
=0
A1 + A2
320 mm2
y =
y1 A1 + y2 A2
A1 + A2
y =
(70)(1400) + (30)(1800)
152000
=
3200
3200
y = 47.5 mm
x =0
30 mm
70 mm
l1 = 70 mm
x1 = 0
x
Problem 7.29
troids.
Determine the coordinates of the cen-
y
Solution: Divide the shape up into a rectangle, a semicircle, and
a circular cutout as shown. Note that the y coordinates of the centroids
of all three component areas lie on the x axis. Thus, y = 0 for the
combined area.
20 mm
x
40 mm
Rectangle:
2
Area1 = a(2R) = 9600 mm ,
120 mm
and x1 = a/2 = 60 mm.
Semicircle: See example 7.3 or 7.4 for the value of the x coordinate
of the centroid of a semicircle. Also note the x displacement of the
centroid relative to the y axis.
y
Area2 = πR2 /2 = 2513 mm2 ,
20 mm
x2 = a + (4R)/(3π) = 137.0 mm.
x
Cutout:
40 mm
Area3 = πr 2 = 1257 mm2 ,
120 mm
x3 = 120 mm.
Combined Area:
R
x = (x1 Area1 + x2 Area2 − x3 Area3 )/(Area1 + Area2 − Area3 )
1
= 70.9 mm
Determine the coordinates of the cen-
r
3
y
Solution: The strategy is to find the centroid for the half circle
area, and use the result in the composite algorithm. The area: The
element of area is a vertical√strip y high and dx wide. From the
equation of the circle, y = ± R2 − x2 . √
The height of the strip will
be twice the positive value, so that dA = 2 R2 − x2 dx, from which
R
A=
dA = 2
(R2 − x2 )1/2 dx
0
√
x R 2 − x2
R2
=2
+
sin−1
2
2
x
R
R
=
0
10 in
x
πR2
2
20 in
The x-coordinate:
R x dA = 2
x R2 − x2 dx
A
2
a
Problem 7.30
troids.
A
+
2R
0
=2 −
Divide by A: x =
(R2 − x2 )3/2
3
R
=
0
2R3
.
3
y
10 in
4R
3π
The y-coordinate: From symmetry, the y-coordinate is zero.
x
4(20)
The composite: For a complete half circle x1 = 3π = 8.488 in..
For the inner half circle x2 = 4.244 in. The areas are
A1 = 628.32 in.2
2
and A2 = 157.08 in .
20 in
Problem 7.31
troids.
Solution:
Determine the coordinates of the cen-
y
Use Appendix B:
800
mm
y
B
a
600 m
m
x
h
m
m
0
80
600
mm
400
mm
θ
C
X = (a + b) / 3
Y = h/ 3
400 mm
x
D
y
b
We need to know h and a. This is equivalent to knowing the coordinates
of point B.
We can use the law of cosines to find the angle θ and then use θ to find
(xB , yB ).
800
mm
600
mm
B
x
d
c
C
b
D
b = 400 mm
c = 600 mm
d = 800 mm
From the law of cosines
c2 = b2 + d2 − 2 bd cos θ
Substituting, θ = 46.57◦
xB = d cos θ = 550.0 mm
yB = d sin θ = 580.9 mm
∴ h = 580.9 mm
a = 550.0 mm b = 400 mm
x = (a + b)/3 = 316.67 mm
y = h/3 = 193.6 mm
400
mm
Problem 7.32 Determine the coordinates of the centroids.
Solution: The results for a half-circle of radius R:
y
πR2
4R
, x1 =
, y1 = 0.
2
3π
A1 =
Consider three figures: The complete circle, (1) the half circle cut out,
and (2) the composite figure. The centroid of the complete circle is at
the origin; x = 0 and y = 0. The product of its centroid coordinates
and its area is zero. From the composite algorithm, it follows that
12 in
x
20 in
0 = A1 x1 + A2 x2 ,
from which
x2 = −
A1
A2
x1 ,
and y2 = −
A1
A2
y1 .
y
12 in
The areas:
A1 =
πR12
,
2
A2 =
πR22
x
πR12
−
.
2
For R1 = 12 in. and R2 = 20 in., A1 = 226.2 in.2 , and A2 =
1030.4 in.2 , and x1 = 5.093 in.. Thus
x2 = −
226.2
1030.4
20 in
(5.093) = −1.18 in.,
and y2 = 0 , since y1 = 0 .
Problem 7.33
troids.
Determine the coordinates of the cen-
y
Solution: Divide the object into three areas: (1) A rectangle on
the left, 100 mm by 60 mm. (2) A rectangle at the lower right, 80 mm
by 40 mm. (3) A semi circle to the far lower right, radius 20 mm.
60 mm
The areas and centroid coordinates are
(1)
A1 = 6 × 103 mm2 ,
100 mm
x1 = 30 mm,
40 mm
y1 = 50 mm
(2)
x
A2 = 3.2 × 103 mm2 ,
140 mm
x2 = 100 mm,
y2 = 20 mm, and
(3)
y
A3 = 628.32 mm2 ,
60
mm
x3 = 148.49 mm,
y3 = 20 mm.
100 mm
(1) The composite area is
A =
3
1
Ai = 9.828 × 103 .
The centroid coordinates for the composite are
3
x=
1
3
and
y=
1
Ai xi
A
= 60.37 mm ,
Ai yi
A
= 38.31 mm
140
mm
40
mm
x
Problem 7.34
troids.
Determine the coordinates of the cen-
y
18 in
x
6 in
6 in
6 in
Solution: Divide the object into four areas: (1) The rectangle
18 in by 18 in, (2) The triangle of altitude 18 in and base 6 in, and (3)
the semi circle with radius 9 in and (4) The object itself.
The areas and their centroids are determined by inspection:
(1)
(2)
(3)
A1 = 182 = 324 in.2 , x1 = 9 in., y1 = 9 in.
A2 = ( 12 )(18)(6) = 54 in.2 , x2 = 9 in., y2 = 6 in.
A3 = π9
2
21.8 in.
2
= 127.2 in.2 , x3 = 9 in., y3 = 18 +
4(9)
3π
The composite area: A = A1 − A2 + A3 = 397.2 in.2 .
The composite centroid:
x=
A1 x1 −A2 x2 +A3 x3
A
= 9 in.
y=
A1 y1 −A2 y2 +A3 y3
A
= 13.51 in.
y
18 in.
x
6 in. 6 in. 6 in.
=
Problem 7.35
troids.
Determine the coordinates of the cen-
y
20 mm
30 mm
20 mm
10
mm
30 mm
x
90 mm
Solution:
Determine this result by breaking the compound object
y
into parts
y
m
20 m
m
10 m
30 mm
=
A1
A2
+
40 mm
30 mm
30 mm
10
mm
20 mm
A1 = (30)(90) = 2700 mm
x
90 mm
2
For the composite:
y1 = 15 mm
x
=
x1 A1 + x2 A2 + x3 A3 − x4 A4
(A1 + A2 + A3 − A4 )
x
=
155782
= 35.3 mm
4414.2
y
=
y1 A1 + y2 A2 + y3 A3 − y4 A4
A1 + A2 + A3 − A4
y
=
146675
= 33.2 mm
4414.2
(sits on top of A1 )
A2
= (40)(50) = 2000 mm2
x2
= 20 mm
y2
= 30 + 25 = 55 mm
A3 =
1 2
π
πr = (20)2 = 628.3 mm2
2 0
2
x3 = 20 mm
y3 = 80 mm +
A4 :
A4
x
x1 = 45 mm
A3 :
–
30 mm
90 mm
A2 :
A3
+
20 mm
A1 :
20 mm
m
50 m
4r0
= 88.49 mm
3π
A4 = (30)(20) + πri2
A4 = 600 + π(10)2 = 914.2 mm2
x4 = 20 mm
y4 = 50 + 15 = 65 mm
Area (composite)
= A1 + A2 + A3 − A4
= 4414.2 mm2
The value for y is not the same as in the new problem statement. This value
seems correct. (The x value checks).
Problem 7.36
troids.
Determine the coordinates of the cen-
y
5 mm
15 mm
50 mm
5 mm
5 mm
15 mm
x
15 mm
10 15 15 10
mm mm mm mm
Solution: Comparison of the solution to Problem 7.29 and our
areas 1, 2, and 3, we see that in order to use the solution of Problem 7.29,
we must set a = 25 mm, R = 15 mm, and r = 5 mm. If we do this,
we find that for this shape, measuring from the y axis, x = 18.04 mm.
The corresponding areas for regions 1, 2, and 3 is 1025 mm2 . The
centroids of the rectangular areas are at their geometric centers. By
inspection, we how have the following information for the five areas
y
1
15
15 mm
Area 1: Area1 = 1025 mm2 , x1 = 18.04 mm, and y1 = 50 mm.
50 mm
y 4
Area 2: Area2 = 1025 mm2 , x2 = 18.04 mm, and y2 = 0 mm.
Area 3: Area3 = 1025 mm2 , x3 = −18.04 mm, and y3 = 0 mm.
5 mm
5
5 mm
3
2
15
5 mm
2
Area 4: Area4 = 600 mm , x4 = 0 mm, and y4 = 25 mm.
Area 5: Area5 = 450 mm2 , x5 = −7.5 mm, and y5 = 50 mm.
Combining the properties of the five areas, we can calculate the centroid
of the composite area made up of the five regions shown.
AreaTOTAL = Area1 + Area2 + Area3 + Area4 + Area5
= 4125 mm2 .
Then, x = (x1 Area1 + x2 Area2 + x3 Area3 + x4 Area4
+x5 Area5 )/AreaTOTAL = 3.67 mm,
and y = (y1 Area1 + y2 Area2 + y3 Area3 + y4 Area4
+y5 Area5 )/AreaTOTAL = 21.52 mm.
15 mm
15 mm
10 15 15 10
mm mm mm mm
x
Problem 7.37 The dimensions b = 42 mm and h =
22 mm. Determine the y coordinate of the centroid of
the beam’s cross section.
y
200 mm
h
120 mm
x
b
Solution:
Work as a composite shape
y
y
100 mm
100 mm
A2
h
b = 42 mm
h = 42 mm
A
1
200 mm
h
120 mm
120 mm
x
b
x
b = 42 mm
b
h = 22 mm
A1 = 120 b mm2 = 5040 mm2
x1 = 0
by symmetry
y1 = 60 mm
A2 = 200 h = 4400 mm2
x2 = 0
y2 = 120 +
x =
h
= 131 mm
2
A1 x1 + A2 x2
0+0
=
A1 + A2
9440
x =0
y =
A1 y1 + A2 y2
= 93.1 mm
A1 + A2
Problem 7.38 If the cross-sectional area of the beam
shown in Problem 7.37 is 8400 mm2 and the y coordinate
of the centroid of the area is y = 90 mm, what are the
dimensions b and h?
Solution: From the solution to Problem 7.37
A1 = 120 b, A2 = 200 h
and y =
y =
y1 A1 + y2 A2
A1 + A2
(60)(120 b) + 120 +
h
2
(200 h)
120 b + 200 h
where y1 = 60 mm
y = 90 mm
A1 + A2 = 8400 mm2
Also, y2 = 120 + h/2
Solving these equations simultaneously we get
h = 18.2 mm
b = 39.7 mm
y
200 mm
h
120 mm
x
b
200 mm
h
A2
A1
b
120 mm
Problem 7.39 Determine the x coordinate of the centroid of the Boeing 747’s vertical stabilizer.
y
11 m
48°
x
70°
12.5 m
Solution: We can treat the stabilizer as a rectangular area (1) with
two triangular cutouts (2 and 3): The dimensions a and b are
y
•
a = 11 tan 48 = 12.22 m
and b =
11
= 4.00 m.
tan 70•
The areas are
11 m
48°
70°
x
12.5 m
A1 = (11)(12.5 + b) = 181.5,
A2 =
1
(11)a = 67.2 m,
2
A3 =
1
(11)b = 22.0 m.
2
3
The x coordinate of the centroid is
A1 x1 − A2 x2 − A3 x3
x =
A1 − A2 − A3
− A2 a3 − A3 (12.5 + 2b/3)
A1 12.5+b
2
=
= 9.64 m
A1 − A2 − A3
Problem 7.40 Determine the y coordinate of the centroid of the vertical stabilizer in Problem 7.39.
Solution:
Treating the stabilizer as a rectangular area (1) with
triangular cutouts (2 and 3) as shown in the sol of Problem 7.39, the y
coordinate of the centroid is
y =
=
A1 y1 − A2 y2 − A3 y3
A1 − A2 − A3
A1 (11/2) − A2 [2(11/3)] − A3 (11/3)
.
A1 − A2 − A3
From the solution of Problem 7.39,
A1 = 181.5 m,
A2 = 67.2 m,
and A3 = 22.0 m,
giving the result y = 4.60 m.
y
a
2
11 m
48°
70°
x
b
12.5 m
Problem 7.41 The area has elliptical boundaries. If
a = 30 mm, b = 15 mm, and ε = 6 mm, what is the x
coordinate of the centroid of the area?
Solution:
x2
(a + ε)2
y
The equation of the outer ellipse is
y2
=1
(b + ε)2
+
and for the inner ellipse
x2
a2
+
y2
b2
b
=1
x
We will handle the problem by considering two solid ellipses
For any ellipse
x dA
x = =
dA
β
/
α
β
/
α
α
0
x
α2 − x2 dx
α2
−
x2
y
dx
From integral tables
(α2 − x2 )3/2
x α2 − x2 dx = −
3
√
x α 2 − x2
α2
α2 − x2 dx =
+
sin−1
2
2
3/2 α
− α 2 − x2
0 Substituting x =
√
x α
x α2 −x2
α2
+ 2 sin α
3
x
α
a
−0 + α3 /3
α3 /3
=
x = 2 π/4
α2 π
α
0+ 2 2 −0−0
4α
3π
Also Area =
dA =
β
=
α
Area =
β
α
β
α
α
α2
2
π
2
y
ε
b
α2 + x2 dx
0
√
x α 2 − x2
α2
+
sin−1
2
2
b
x
0
x =
a
x
α
α
ε
a
= παβ/4
A1
=
–
(The area of a full ellipse is παβ so this checks.
Now for the composite area.
y
For the outer ellipse, α = a + ε β = b + ε and for the inner ellipse
α=a β=b
β
y = α α 2 – x2
dA = y dx
β
Outer ellipse
x1
4(a + ε)
=
3π
A1 =
π(a + ε)(b + ε)
4
Inner Ellipse
x2 =
A2
4a
3π
πab
=
4
x
0
x
α
For the composite
x =
x1 A1 − x2 A2
A1 − A2
Substituting, we get
x1 = 15.28 mm
A1 = 2375 mm
and x = 19.0 mm
2
x2 = 12.73 mm
A2 = 1414 mm2
A2
Problem 7.42 By determining the x coordinate of the
centroid of the area shown in Problem 7.41 in terms of a,
b, and ε, and evaluating its limit as ε → 0, show that the
x coordinate of the centroid of a quarter-elliptical line is
4a(a + 2b)
.
3π(a + b)
x=
Solution:
From the solution to 7.41, we have
x1 =
4(a + ε)
3π
x2 =
4a
3π
so x1 A1
x2 A2
A1 − A2
A1 − A2
A1 =
(x1 A1 − x2 A2 ) =
π(a + ε)(b + ε)
4
A2 =
1 2
(a b + 2abε + bε2
3
+a2 ε + 2aε2 + ε3 − a2 b)
πab
4
(x1 A1 − x2 A2 ) =
1
((2ab + a2 )ε
3
+(2a + b)ε2 + ε3 )
(a + ε)2 (b + ε)
=
3
Finally x =
a2 b
=
3
π
= (ab + aε + bε + ε2 − ab)
4
π
= (aε + bε + ε2 )
4
x =
x =
x1 A1 − x2 A2
A1 − A2
1
(2ab + a2 ) + (2a + b)ε + ε2 ε/
3
π
4
[(a + b) + ε] ε/
4a(a + 2b)
4(2a + b)ε
4 2
+
+
ε
3π(a + b)
3π
3π
Taking the limit as ε → 0
x=
Problem 7.43
Three sails of a New York pilot
schooner are shown. The coordinates of the points are
in feet. Determine the centroid of sail 1.
Solution: Divide the object into three areas: (1) The triangle with altitude
21 ft and base 20 ft. (2) The triangle with altitude 21 ft and base (20−16) = 4ft,
and (3) the composite sail. The areas and coordinates are:
(1)
2
A1 = 210 ft2 ,
x1 =
2
3
20 = 13.33 ft,
y1 =
1
3
21 = 7 ft.
(2)
1
4a(a + 2b)
3π(a + b)
A2 = 42 ft2 ,
x2 = 16 +
3
2
3
4 = 18.67 ft,
y2 = 7 ft.
(a)
y
(3)
y
y
(14, 29)
(12.5, 23)
(20, 21)
(3, 20)
1
2
3
x
(16, 0)
(3.5, 21)
x
(10, 0)
(b)
x
(23, 0)
The composite area: A = A1 − A2 = 168 ft2 .
The composite centroid:
x=
A1 x1 −A2 x2
A
= 12 ft ,
y=
A1 y1 −A2 y2
A
= 7 ft
Problem 7.44
lem 7.43.
Determine the centroid of sail 2 in Prob-
Solution:
Divide the object into five areas: (1) a triangle on the
left with altitude 20 ft and base 3 ft, (2) a rectangle in the middle 23 ft
by 9.5 ft, (3) a triangle at the top with base of 9.5 ft and altitude of
3 ft. (4) a triangle on the right with altitude of 23 ft and base of 2.5 ft.
(5) the composite sail. The areas and centroids are:
(1)
A1 =
3(20)
2
= 30 ft2 ,
x1 =
2
3
3 = 2 ft,
y1 =
1
3
20 = 6.67 ft.
(2)
y2 =
9.5
2
= 7.75 ft,
23
= 11.5 ft
2
Problem 7.45
lem 7.43.
2
(3)(9.5) = 14.25 ft2 ,
1
3
x3 = 3 +
9.5 = 6.167 ft,
2
3 = 22 ft
3
A4 = 12 (2.5)(23) = 28.75 ft2 ,
y3 = 20 +
(4)
y4 =
2
3
1
3
(2.5) = 11.67 ft,
23 = 7.66 ft
The composite area: A = A1 + A2 − A3 − A4 = 205.5 ft2 .
The composite centroid:
(5)
x=
A1 x1 +A2 x2 −A3 x3 −A4 x4
A
= 6.472 ft ,
y=
A1 y1 +A2 y2 −A3 y3 −A4 y4
A
= 10.603 ft
Determine the centroid of sail 3 in Prob-
Solution: Divide the object into six areas: (1) The triangle Oef,
with base 3.5 ft and altitude 21 ft. (2) The rectangle Oabc, 14 ft by
29 ft. (3) The triangle beg, with base 10.5 ft and altitude 8 ft. (4) The
triangle bcd, with base 9 ft and altitude 29 ft. (5) The rectangle agef
3.5 ft by 8 ft. (6) The composite, Oebd. The areas and centroids are:
(1)
1
x4 = 10 +
A2 = (23)(9.5) = 218.5 ft2 ,
x2 = 3 +
A3 =
(3)
a
f
g
b
e
2
A1 = 36.75 ft ,
x1 = 1.167 ft,
y1 = 14 ft.
(2)
A2 = 406 ft2 ,
o
x2 = 7 ft,
(5)
y2 = 14.5 ft.
(3)
(6)
The composite area:
A = −A1 + A2 − A3 + A4 − A5 = 429.75 ft2 .
2
A4 = 130.5 ft ,
y4 = 9.67 ft.
A5 = 28 ft2 ,
y5 = 25 ft.
y3 = 26.33 ft
x4 = 17 ft,
d
x5 = 1.75 ft,
A3 = 42 ft2 ,
x3 = 7 ft,
(4)
c
The composite centroid:
x=
−A1 x1 +A2 x2 −A3 x3 +A4 x4 −A5 x5
A
= 10.877 ft
y=
−A1 y1 +A2 y2 −A3 y3 +A4 y4 −A5 y5
A
= 11.23 ft
Problem 7.46 The value of the distributed load w at
x = 6 m is 240 N/m.
(a) The equation for the loading curve is w = 40x N/m.
Use Eq. (7.10) to determine the magnitude of the
total force exerted on the beam by the distributed
load.
(b) If you use the area analogy to represent the distributed load by an equivalent force, what is the
magnitude of the force and where does it act?
(c) Determine the reactions at A and B.
Solution:
(a) ω(x) = 40x N/m
6
F =
ω(x) dx =
0
6
0
6
x2 40x dx = 40 2 0
F = 720 N
(b)
Using the area analogy
6
6
ω(x)x dx
40x2 dx
0
0
x= 6
=
720
ω(x) dx
x=
0
3
40 x3
6
720
0
= 4.00 m
x = 4.00 m (720 N force)
(c)
Fx :
Ax = 0
Fy :
Ay + By − 720 = 0
MA :
(−4)(720) + (6)By = 0
Solving,
By = 480 N
Ay = 240 N
Ax = 0
y
240 N/m
A
B
x
6m
y
720 N
4m
AX
2m
x
AY
BY
y
240 N/m
A
B
6m
x
Problem 7.47 In a preliminary design study for a
pedestrian bridge, an engineer models the combined
weight of the bridge and maximum expected load due
to traffic by the distributed load shown.
y
= 50 kN/m
(a) Use Eq. (7.10) to determine the magnitude of the
total force exerted on the bridge by the distributed
load.
(b) If you use the area analogy to represent the distributed load by an equivalent force, what is the
magnitude of the force and where does it act?
(c) Determine the reactions at A and B.
Solution:
(a) F =
10
ω(x) dx =
0
10
0
A
10 m
10
50 dx = 50x kN
0
F = 500 kN
(b)
Area analogy
10
10
xω(x) dx
50x dx
x = 0 10
= 0
500 kN
ω(x) dx
0
10
25x2 0
x=
500 kN
=
2500 kN · m
500 kN
x = 5 m (500 kN force)
(c)
Fx :
Ax = 0
Fy :
Ay + By − 500 = 0
MA :
(−5)(500) + 10By = 0
By = 250 kN
Ax = 0
Ay = 250 kN
y
= 50 kN/m
x
B
A
10 m
y
500 kN
5m
5m
Ax
x
Ay
By
B
x
Problem 7.48
support A.
Determine the reactions at the built-in
y
200 N/m
x
A
3m
Total distributed force acting on the beam ω(x) =
− 3) for 3 ≤ x ≤ 6. The total force acting on the beam
due to the distributed load is
6
6
200
200 x2
F =
(x − 3) dx =
− 3x
3
3
2
3
3
200 36
9
200 9
F =
− 18 − + 9 =
·
3
2
2
3
2
Solution:
200
(x
3
F = 300 N.
Using the area analogy and the fact that the load is triangular, x = 5 m.
300 N
mA
5m
Ax
Ay
Fx :
Ax = 0
Fy :
Ay − 300 N = 0
MA :
MA − (5)(300) = 0
Ax = 0
Ay = 300 N
MA = 1500 N-m
y
200 N/m
x
A
3m
3m
3m
Problem 7.49
Determine the reactions at A and B.
y
x
A
B
L/2
Solution:
L/2
Let us break the load into two parts and use the area
analogy.
y
x
A
L1
A
B
L2
L
2
L
2
B
L/2
L/2
x
Now we can find the support reactions
3L
4
For Load L1
2ω0
x for (0 ≤ x ≤ L/2)
L
ω(x) =
L
3
L
≤x≤L
2
L1 =
L1 =
L/2
0
L
L/2
L
ω0 dx = ω0 x
L/2
=
ω0 L
2
And from the area analogy, L2 acts half way between L/2 and L.
3L
x2 =
.
4
x1 = L/3
By
Ay
using the area analogy, load L1 acts 2/3 of the distance from the origin
to L/2. Thus
L2 =
L
2
L/2
2ω0
2/ω0 x2 x dx =
L
L 2/ 0
ω0 L2
Lω0
=
L 4
4
Load 2
4
Ax
Load 1
2
L
For Load L2
ω(x) = ω0 for
L
Fx :
Ax = 0
Fy :
Ay + By −
MA :
By
L
2
Lω0
Lω0
−
=0
4
2
−
Lω0
4
Solving the third eqn.
By =
Lω0
3Lω0
11
+
=
Lω0
6
4
12
From the second eqn,
Ay + By =
Hence Ay =
3
Lω0
4
3
lω0
4
0
− By = − Lω
6
Ax = 0 Ay = −Lω0 /6
By = 11 Lω0 /12
L
3
−
Lω0
2
3L
4
=0
Problem 7.50
support A.
Determine the reactions at the built-in
y
Solution: The free-body diagram of the beam is: The downward
force exerted by the distributed load is
5
x2
w dx =
3 1−
dx
25
L
0
5
x3
=3 x−
= 10 kN.
75 0
= 3(1 – x 2/25) kN/m
The clockwise moment about the left end of the beam due to the distributed load is
5
x3
xw dx =
3 x−
dx
25
L
0
5
x2
x4
= 18.75 kN-m.
=3
−
2
100 0
x
A
5m
= 3(1 – x 2/25) kN/m
From the equilibrium equations
Fx = Ax = 0,
Fy = Ay − 10 = 0,
m(leftend) = Ma + 5Ay − 18.75 = 0,
x
A
5m
= 3(1 – x 2/25) kN/m
we obtain
Ma
Ax = 0,
Ay = 10 kN,
Ax
5m
x
Ay
and Ma = −31.25 kN-m.
Problem 7.51 An engineer measures the forces exerted by the soil on a 10-m section of a building foundation and finds that they are described by the distributed
load w = −10x − x2 + 0.2x3 kN[/]m.
(a) Determine the magnitude of the total force exerted
on the foundation by the distributed load.
(b) Determine the magnitude of the moment about A
due to the distributed load.
y
2m
10 m
A
x
Solution:
(a)
The total force is
12
F
=−
(10x + x2 − 0.2x3 ) dx
0
= −5x2 −
|F |
x3
3
+
0.2 4
x
4
y
10
A
0
x
= 333.3 kN
(b)
The moment about the origin is
10
M
=−
(10x + x2 − 0.2x3 )x dx
0
= −
|M |
10 3
1
0.2 5
x − x4 +
x
3
4
5
10
,
0
= 1833.33 kN.
The distance from the origin to the equivalent force is
d=
10 m
2m
|M |
= 5.5 m,
F
from which
|MA | = (d + 2)F = 2500 kN m.
w
Problem 7.52 The distributed load is w = 6x +
0.4x2 N/m. Determine the reactions at A and B.
A
B
2m
4m
Solution:
6
F =
The total distributed load is
6
w(x) dx =
(6x + 0.4x3 ) dx
0
F = 6
0
x2
2
+ 0.4
6
x4
4
= 3.36 +
0
0.4(36)2
4
F = 237.6 N
A
Using the area analogy, the point of application of the equivalent concentrated force F is
6
6
xw(x) dx
(6x2 + 0.4x4 ) dx
x = 0 6
= 0
237.6
w(x) dx
x =
0
3
5
6 x3 + 0.4 x5
6
237.6
0
= 4.436 m
Now to determine the support reactions
y
x
F
Ax
x
4m
Ay
By
Fx :
Ax = 0
Fy :
Ay + By − F = 0
MA :
4By − xF = 0
Solving, we get
Ax = 0
Ay = −25.9 N
By = 263.5 N
B
4m
2m
Problem 7.53 The aerodynamic lift of the wing is
described by the distributed load
w = −300 1 − 0.04x2 N/m.
y
The mass of the wing is 27 kg, and its center of mass is
located 2 m from the wing root R.
x
R
(a) Determine the magnitudes of the force and the moment about R exerted by the lift of the wing.
(b) Determine the reactions on the wing at R.
2m
5m
Solution:
y
(a)
The force due to the lift is
5
F
= −w =
300(1 − 0.04x2 )1/2 dx,
w
0
300 5
=
(25 − x2 )1/2 dx
5 0
√
x 25 − x2
25
= 60
+
sin−1
2
2
F
F
|F |
R
x
5
2m
5
5m
= 375π N,
0
= 1178.1 N.
w
The moment about the root due to the lift is
5
M
= 300
(1 − 0.04x2 )1/2 x dx,
0
(25 − x2 )3/2
= −60
3
M
|M |
5
0
MR
mg
FR
60(25)3/2
=
= 2500
3
2m
3m
= 2500 Nm.
(b) The sum of the moments about the root:
M = M R + 2500 − 27g(2) = 0,
from which M R = −1970 N-m. The sum of forces
Fy = FR + 1178.1 − 27g = 0,
from which FR = −1178.1 + 27g = −913.2 N
Problem 7.54 The force F = 2000 lb. Determine the
reactions at A and B.
Solution:
The free-body diagram of the beam is: The downward
force exerted by the distributed load is
3
3
x3
w dx =
400x2 dx = 400
= 3600 lb.
3 0
L
0
The clockwise moment about the left end of the beam due to the distributed load is
3
3
x4
xw dx =
400x3 dx = 400
= 8100 ft-lb.
4 0
L
0
From the equilibrium equations
Fx = Ax = 0,
Fy = Ay + B − 3600 − 2000 = 0,
m(leftend) = 3Ay − 6(2000) + 8B − 8100 = 0,
we obtain Ax = 0, Ay = 4940 lb, B = 660 lb.
y
w = 400x2 lb/ft
F
A
B
x
3 ft
3 ft
w = 400x2 lb/ft
2 ft
2000 lb
AX
x
3 ft
AY
3 ft
2 ft
B
Determine the reactions at A and B.
Problem 7.55
y
4 kN/m
20 kN-m
A
6 kN
Solution: Break the load into two parts and find the equivalent
concentrated load for each part. Then find the reactions at A and B
x
6
12
6 kN
18
w1 (x) = 4 kN/m (6 m ≤ × ≤ 12 m)
F1 =
12
6
kN/m (12 m ≤ × ≤ 18 m)
w1 (x) dx =
12
F1 = 4x = 24 kN
12
4 dx
6
6
By symmetry, F1 is applied at x = 9 m
18
18
2
F2 =
w2 (x) dx =
− x + 12
3
12
12
18
x2
F2 = −
+ 12x
= 108 − 96 kN
3
12
dx
F2 = 12 kN
By the area analogy, this load is applied at x = 14 m ( 13 of the way
from 12 to 18).
y
24 kN
3m
6m
5m
12 kN
20 kN-m
x
AX
6 kN
6m
AY
BY
Fx :
Ax = 0
Fy :
Ay + By − 24 − 12 − 6 = 0
MA :
Solving
x
B
w2 (x)
4 kN/m
2
− x + 12
3
6m
4 kN/m
A
w1 (x)
w2 (x) =
6m
y
20 k N-m
y
6m
x
B
(6)(6) + 20 − 3(24) + 6By − 8(12) = 0
Ax = 0, Ay = 23.3 kN, By = 18.7 kN
6m
6m
6m
Problem 7.56 Determine the reactions on member
AB at A and B.
600 N/m
C
A
B
1m
1m
Solution: Divide the beams at the pin, B. Find the equivalent
concentrated load on AB and find two equivalent concentrated loads
on BC. Then find the support reactions
600 N/m
Load 1 w1 (x) = 300x N/m
1
1
F1 =
300x dx = 150x2 = 150 N
0
C
A
B
1m
0
1m
Similarly,
w2 (x)
F2 = 150 N
2
2
F3 =
300 dx = 300x = 300 N
1
300 N/m
w1 (x)
1m
A
1
300 N/m
x
B
w3 (x) B
using the area analogy,
F1
F3
MA
5
m
3
AX
3
= 300 N applied at x = m
2
MA :
For BC
Fx :
Fy :
MB :
Ay + By − 150 = 0
2
3
(150) + MA = 0
−Bx = 0
−By − 300 − 150 + Cy = 0
(1)Cy − (0.5)(300) −
Solving, we get, acting on AB
Ax = 0
Ay = 350 N
MA = 300 N-m
By = −200 N
Bx = 0
also Cy = 250 N
C
1m
BY
300 N
BX
0.5 m
BX
Ax + Bx = 0
(1)By −
2m
3
AY
we now need to write the equilibrium equations.
For AB
Fx :
Fy :
1m
150 N
2
= 150 N applied at x = m
3
F2 = 150 N applied at x =
300 N/m
x
2
3
(150) = 0
BY
1m
150 N
1 m
3
CY
Problem 7.57 Determine the reactions on member
ABCD at A and D.
2 kN/m
2 kN/m
D
E
1m
1m
C
1m
B
1m
A
F
1m
Solution: First, replace the distributed forces with equivalent concentrated forces, then solve for the loads. Note that BF and CE are
two force members.
Distributed Load on ABCD, F1
By area analogy, concentrated load is applied at y = ±m. The load
is 12 (2)(3) kN
E
D
1m
1m
C
F1 = 3 kN
By the area analogy,
1m
F2 = 4 kN applied at x = 1 m
Assume FCE and FBF are tensions
For ABCD:
Fx :
Ax + FBF cos 45◦ + FCE cos 45◦ + Dx + 3 kN = 0
Fy :
Ay + Dy − FBF sin 45◦ + FCE sin 45◦ = 0
MA :
−1(FBF cos 45◦ ) − 2(FCE cos 45◦ ) − 2(3) − 3Dx = 0
For DE:
Fx :
Fy :
ME :
2 kN/m
2 kN/m
B
1m
A
F
1m
2 kN / m
y
−Dx − FCE cos 45◦ = 0
−Dy − FCE sin 45◦ − 4 = 0
(1)Dy = 0
3m
x
y
2 kN/ m
2m
6.57 Contd.
Solving, we get
Ax = 7 kN
Ay = −6 kN
Dx = 4 kN
Dy = 0
also FBF = −14.14 kN(c)
FCE = −5.66 kN(c)
DY
DX
FCE
45
3 kN
°
C
B
2m
FBF
45°
AX
AY
1
4 kN
1m
DX
E
DY
45°
FCE
2
Problem 7.58
of the frame.
Determine the forces on member ABC
A
1m
3 kN/m
B
1m
C
2m
2m
1m
Solution: The free body diagram of the member on which the
distributed load acts is
From the equilibrium equations
Fx = Bx = 0,
Fy = By + E − 12 = 0,
m(leftend) = 3E − (2)(12) = 0,
we find that Bx = 0, By = 4 kN, and E = 8 kN. From the lower
fbd, writing the equilibrium equation
m(leftend) = −2Cy − (4)(8) = 0,
we obtain Cy = −16 kN. Then from the middle free body diagram,
we write the equilibrium equations
Fx = Ax + Cx = 0,
Fy = Ay − 4 − 16 = 0,
m(rightend) = −2Ax − 2Ay + (1)(4) = 0
A
1m
3 kN/m
B
1m
C
2m
1m 1m
1m
(4 m)(3 kN/m) = 12 kN
BX
2m
BY
1m
E
AX
4 kN
AY
obtaining Ax = −18 kN, Ay = 20 kN, Cx = 18 kN.
CX
8 kN
CY
CX
DX
DY
2m
CY
2m
Problem 7.59 Determine the coordinates of the centroid of the truncated conical volume.
Strategy: Use the method described in Example 7.8.
y
z
R
x
h–
2
Solution:
Refer to Example 7.8.
h–
2
y
y
dV
z
z
R
x
x
x
dx
(a) An element dV in the form of a disk.
y
h
R
–x
h
h–
2
R
x
x
dx
(b) The radius of the element is (R/h)x.
xdV
x = V
dV
= V
x =
h
x4 /4 h/2
[x3 /3]h
h/2
x =
45
h
56
h
h/2
h
h/2
=
π
R2 3
x dx
h2
π
R2 2
x dx
h2
h4 /4 − h4 /64
[h3 /3 − h3 /24]
h–
2
h–
2
Problem 7.60 A grain storage tank has the form of a
surface of revolution with the profile shown. The height
of the tank is 7 m and its diameter at ground level is 10 m.
Determine the volume of the tank and the height above
ground level of the centroid of its volume.
y
y = ax1/2
7m
10 m
x
Solution:
y
O
y
y = ax1/2
1/2
y = ax
7m
y
dx
dV = π y2dx
10 m
x
x
dV = πy 2 dx
7
7
χπy 2 dx
χπa2 x dx
x = 0 7
= 0 7
πy 2 dx
πa2 x dx
x =
0
7
x3 /3 0
[x2 /2]70
0
= 4.67 m
The height of the centroid above the ground is 7 m − x
h = 2.33 m
The volume is
7
V =
πa2 x dx = πa2
0
49
2
To determine a,
y = 5, m when x = 7 m.
√
y = ax1/2 , 5 = a 7
√
a = 5/ 7a2 = 25/7
V =π
25
7
V = 275 m3
49
2
= 275 m3
m3
Problem 7.61 The object shown, designed to serve
as a pedestal for a speaker, has a profile obtained by
revolving the curve y = 0.167x2 about the x axis. What
is the x coordinate of the centroid of the object?
y
z
x
0.75 m
0.75 m
Solution:
y
y = 0.167 x2
dV = π y2dx
x
dv
xdV
x = V
V
2 2
xπ(0.167x ) dx
= 0.75
1.50
dV
π(0.167x2 )2 dx
x =
z
1.50
0.75
1.5
π(0.167)2
· 0.75
1.5
π(0.167)2
0.75
x = 1.27 m
x
0.75 m
0.75 m
x5 dx
=
4
x dx
x6 /6
1.5
0.75
[x5 /5]1.5
0.75
Problem 7.62
the pyramid.
Determine the volume and centroid of
y
400 mm
z
600 mm
x
400 mm
Solution:
The volume: The element of volume is a square of
thickness dx. The length of a side is a linear function of the height of
the pyramid. Thus L = Ax + B. For x = 0, L = 0, and therefore
B = 0. For x = 600, L = 400, therefore A = 23 . The area is
L2 = 49 x2 . The volume element is dV = 49 x2 dx, from which
600
4
V =
dV =
x2 dx
9
V
0
=
4
27
[x3 ]600
= 3.2 × 107 mm3 = 0.032 m3
0
The x-coordinate:
600
4
x dV =
x3 dx
9
V
0
=
1
9
[x4 ]600
= 1.44 × 1010 .
0
Divide by V : x = 0.45 m.
By symmetry, the y- and z-coordinates of the centroid are zero.
y
y
x
600
mm
z
400
mm
400
mm
Problem 7.63 Determine the centroid of the hemispherical volume.
y
R
z
x
Solution:
z 2 = R2 .
The equation of the surface of a sphere is x2 + y 2 +
The volume: The element of volume is a disk of radius ρ and thickness
dx. The radius of the disk at any point within the hemisphere is ρ2 =
y 2 + z 2 . From the equation of the surface of the sphere, ρ2 = (R2 −
x2 ). The area is πρ2 , and the element of volume is dV = π(R2 −
x2 ) dx, from which
V =
Vsphere
2π 3
=
R .
2
3
The x-coordinate is:
R
x dV = π
(R2 − x2 )x dx
V
0
=π
R 2 x2
x4
−
2
4
R
0
π
= R4 .
4
Divide by the volume:
x =
πR4
4
3
2πR3
=
3
R.
8
By symmetry, the y- and z-coordinates of the centroid are zero.
y
x
R
Problem 7.64 The volume consists of a segment of a
sphere of radius R. Determine its centroid.
y
x
R
R
2
z
Solution: The volume: The element of volume is a disk of radius
ρ and thickness dx. The area of the disk is πρ2 , and the element of
volume is πρ2 dx. From the equation of the surface of a sphere (see
solution to Problem 7.63) ρ2 = R2 − x2 , from which the element of
volume is dV = π(R2 − x2 ) dx. Thus
R
dV = π
(R2 − x2 ) dx
V =
V
R/2
= π R2 x −
x3
3
R
5π
24
=
R/2
R3 .
The x-coordinate:
R
x dV = π
(R2 − x2 )x dx
V
R/2
=π
R 2 x2
x4
−
2
4
R
=
R/2
9π 4
R .
64
Divide by the volume:
x =
9πR4
64
24
5πR3
=
27
R = 0.675R.
40
By symmetry the y- and z-coordinates are zero.
y
R
–
2
x
R
Problem 7.65 A volume of revolution is obtained by
2
2
revolving the curve xa2 + yb2 = 1 about the x axis. Determine its centroid.
y
y2
x 2 + ––
––
=1
2
a
b2
z
x
Solution: The volume: The element of volume is a disk of radius
y and thickness dx. The area of the disk is πy 2 . From the equation
for the surface of the ellipse,
x2
1− 2
a
πy 2 = πb2
and dV = πy 2 dx = πb2
from which
V =
dV = πb2
V
a
0
= πb2 x −
x3
3a2
The x-coordinate:
x dV = πb2
V
= πb2
a
a
=
0
1−
1−
x2
a2
x2
a2
πb2 a2
4
x
dx
2πb2 a
.
3
x2
x dx
a2
0
a
x2
x4
πb2 a2
− 2
=
.
2
4a 0
4
1−
3
2πb2 a
=
3
8
x2 + y2 = 1
a2 b2
dx,
Divide by volume:
x =
y
a.
By symmetry, the y- and z-coordinates of the centroid are zero.
Problem 7.66 The volume of revolution has a cylindrical hole of radius R. Determine its centroid.
y
z
h
R
R+a
x
Solution: The volume: The element of volume is a disk of radius
y and thickness
dx. The area of the disk is π(y 2 − R2 ). The radius
a
is y = h
x + R. The volume element is
dV = π
a
x+R
h
2
y
dx − πR2 dx.
R
Denote
m=
a
, dV = π(m2 x2 + 2mRx) dx,
h
from which
V =
dV = πm
V
h
(mx2 + 2Rx) dx
0
= πm m
x3
+ Rx2
3
The x-coordinate:
x dV = πm
V
h
h
= πmh2
0
mh
+R .
3
(mx3 + 2Rx2 ) dx
0
= πm m
x4
2Rx3
+
4
3
= πmh3
m
h
2R
+
4
3
h
0
.
Divide by the volume:
x =h
mh
+
4
2R
3
mh
+R
3
a
4
= h a
3
+
2R
3
+R
.
By symmetry, the y- and z-coordinates of the centroid are zero.
h
R+a
Problem 7.67 Determine the y coordinate of the centroid of the line (see Example 7.9).
y
(1, 1)
y = x2
L
x
Solution:
line is
The length of the line: The elementary length of the
y
dL =
1+
dy
dx
2
(1, 1)
dx.
dy
Noting that dx
= 2x, the element of length is dL = (1+4x2 )1/2 dx,
from which
1
L=
dL =
(1 + 4x2 )1/2 dx
L
0
1
1
(1 + 4x2 )1/2 x + loge (2x + (1 + 4x2 )1/2 )
2
2
√
√
5
1
=
+ loge (2 + 5) = 1.4789.
2
4
=
y = x2
L
x
1
0
The y-coordinate: The coordinate of the centroid of the length element
is y = y = x2 , from which
1
x(1 + 4x2 )3/2
x(1 + 4x2 )1/2
y =
(1 + 4x2 )1/2 x2 dx =
−
16
32
0
−
1
loge (2x + (1 + 4x2 )1/2 )
64
1
0
√
1
1
1
=
(5)3/2 −
(5)1/2 −
loge (2 + 5) = 0.6063.
16
32
64
Divide by the length: y =
0.6063
1.4789
= 0.410
Problem 7.68 Determine the x coordinate of the centroid of the line.
Solution:
y
dL =
is
The length: Noting that
dy
dx
1+
2
dx =
√
L
L
0
Divide by the length: x =
0
x
5
2 5/2
x
5
21.961
6.7869
5
5
= 6.7869.
1
= 21.961.
1
= 3.2357
2
y = – (x – 1)3/2
3
y
0
Problem 7.69 Determine the x coordinate of the centroid of the line.
2
(x)3/2
3
1
The x-coordinate:
5
x dL =
x3/2 dx =
= (x−1)1/2 , the element of length
x dx
from which
5
L =
dL =
(x)1/2 dx =
2
y = – (x – 1)3/2
3
dy
dx
x
5
y
2
y = – x 3/2
3
0
Solution:
length is
dy
dx
2
dx =
√
= x1/2 the element of
2
y = – x 3/2
3
1 + x dx
from which
2
L =
dL =
(1 + x)1/2 dx =
L
dy
dx
y
1+
dL =
The length: Noting that
2
0
2
(1 + x)3/2
3
2
= 2.7974
0
The x-coordinate:
2
2
(1 + x)5/2
(1 + x)3/2
x dL =
x(1 + x)1/2 dx = 2
−
5
3
L
0
0
35/2
33/2
1
1
=2
−
−
+
= 3.0379.
5
3
5
3
Divide by the length: x = 1.086
0
2
x
x
Problem 7.70
arc.
Determine the centroid of the circular
y
α
x
R
Solution:
y=
The length: From the equation for the circle,
R2 − x2 and
dy
= −(R2 − x2 )−1/2 x.
dx
The element of length
dy 2
dL = 1 +
dx = R(R2 − x2 )−1/2 dx,
dx
from which
L =
dL =
L
= R sin−1
R
R cos α
R(R2 − x2 )−1/2 dx
x R
R R cos α
π
− sin−1 (cos α)
2
π
π
=R
− sin−1 sin
−α
2
2
=R
= Rα
Check: L = Rα from the definition of α. check.
The x-coordinate:
R
x dL = R
L
R cos α
x(R2 − x2 )−1/2 dx
R
= R −(R2 − x2 )1/2
R cos α
Divide by the length: x =
R
α
= R2 sin α
sin α.
The y-coordinate:
The y-coordinate of the centroid of each element is
√
y = y = R2 − x2 . Hence
R
y dL = R
(R2 − x2 )1/2 (R2 − x2 )−1/2 dx
L
=R
R cos α
Rc
dx
R cos α
= R2 (1 − cos α).
Divide by the length:
y =
R
α
(1 − cos α)
y
α
x
R
Problem 7.71
Determine the centroids of the volumes.
y
60 mm
40 mm
x
40 mm
40 mm
60 mm
z
Solution: Divide the object into two volumes: (1) The left-most
volume with dimensions 40 by 80 by 60 mm. (2) The right-most
volume with dimensions 60 by 60 by 40 mm. The volumes and centroids are:
(1)
y
V1 = 1.92 × 105 mm3 ,
60 mm
x1 = 20 mm,
y1 = 40 mm,
40 mm
z1 = 30 mm.
(2)
V2 = 1.44 × 105 mm3 ,
x2 = 70 mm,
x
40 mm
y2 = 20 mm,
z2 = 30 mm.
(3)
The composite volume: V = V1 + V2 = 3.36 × 105 mm3 .
The composite centroid:
x=
V1 x1 + V2 x2
= 41.43 mm.
V
y=
V1 y1 + V2 y2
= 31.43 mm.
V
z=
V1 z1 + V2 z2
= 30 mm
V
z
40 mm
60 mm
Problem 7.72
Determine the centroids of the volumes.
y
15 in
20 in
15 in
x
15 in
35 in
z
Solution:
The Rectangle:
AreaR = 30 × 35 = 1050 in2 ,
y
15 in
xR = 35/2 = 17.5 in,
yR = 30/2 = 15 in
The Triangle:
20 in
15 in
AreaT = (20)(15)/2 = 150 in2 ,
xT = 15 +
2
(20) = 28.33 in,
3
yT = 30 − 15/3 = 25 in
x
15 in
35 in
z
The Solid:
x = (xR AreaR − xT AreaT )/(AreaR − AreaT ) = 15.7 in,
y
y = y = (yR AreaR − yT AreaT )/(AreaR − AreaT ) = 13.3 in,
15 in
20 in
and from symmetry, z = 0.
15 in
30 in
15 in
35 in
x
Problem 7.73
Determine the centroids of the volumes.
y
160
mm
80 mm
60 mm
40 mm
40 mm
50 mm
80 mm
z
Solution: For the block L = 240 mm, H = 160 mm, and
D = 50 mm. For each hole, r = 20 mm. Centroids of parts are at
the geometric centers.
Component
Block
Hole 1
Hole 2
Hole 3
Hole 4
V
LHD
πr 2
πr 2
πr 2
πr 2
Component Properties
V (mm3 )
x (mm)
2 × 106
120
62832
80
62832
80
62832
180
62832
180
x =
x1 V1 − x2 V2 − x3 V3 − x4 V4 − x5 V5
V1 − V2 − V3 − V4 − V5
x =
2.073 × 108 mm4
= 118.56 mm
1.748 × 106 mm3
y (mm)
80
40
120
100
40
z (mm)
25
25
25
25
25
y = SIMILAR Eqn = 80.72 mm
z = 25 mm
y
2
3
160
mm
80 mm
60 mm
1
40 mm
4
40 mm
50 mm
80 mm
100 mm
60 mm
z
Holes are 40 mm in diameter.
x
100 mm
60 mm
Holes are 40 mm in diameter.
x
Problem 7.74
Determine the centroids of the volumes.
y
120 mm
100
mm
40 mm
z
30 mm
20 mm
Solution: Divide the object into four volumes: (1) The left-most
cylinder, with diameter 80 mm and length 120 mm, (2) the cylinder to
the right, with diameter 60 mm and length 100 mm, (3) the cylinder
bored through the center, with diameter 40 mm and length 220 mm,
and (4) the composite cylinder. The volumes and centroids are:
(1)
V1 = 6.032 × 105 mm3 ,
x1 = 60 mm,
y1 = z1 = 0,
V2 = 2.83 × 105 mm3 ,
(2)
x2 = 170 mm,
y2 = z2 = 0,
(3)
V3 = 2.76 × 105 mm3 ,
x3 = 110 mm,
y3 = z3 = 0.
(4)
The composite volume: V = V1 + V2 − V3 = 6.095 ×
105 mm3 . The composite centroid:
x=
V1 x1 + V2 x2 − V3 x3
= 88.4 mm,
V
y=z=0
y
z
x
y
60
40 mm
mm
x
80
mm
x
120 mm
100 mm
Problem 7.75
Determine the centroids of the volumes.
y
z
60 mm
90 mm
360 mm
x
460 mm
Solution: This is a composite shape. Let us consider a solid
cylinder and then subtract the cone. Use information from the appendix
Cylinder
Cone
Volume
πR2 L
1
πr 2 h
3
Volume (mm3 )
1.1706 × 107
1.3572 × 106
R = 90 mm
L = 460 mm
x
L/2
L-h/4
y
x (mm)
230
370
z
r = 60 mm
60 mm
90 mm
h = 360 mm
x =
XCyL VCyL − XCONE VCONE
VCyL − VCONE
x = 211.6 mm
y = z = 0 mm
360 mm
460 mm
x
Problem 7.76
Determine the centroids of the volumes.
y
20 mm
25 mm
75 mm
x
120 mm
25 mm
100 mm
z
Solution: Break the composite object into simple shapes, find
the volumes and centroids of each, and combine to find the required
centroid.
Object
1
2
3
4
Volume (V)
LW H
hW D
x
0
0
y
H/2
(H + h/2)
z
L/2
D/2
πR2 D/2
πr 2 D
0
0
H + h + 4R
3π
(H + h)
D/2
D/2
y
20 mm
25 mm
75 mm
where R = W/2. For the composite,
x=
x1 V1 + x2 V2 + x3 V3 − x4 V4
V1 + V2 + V3 − V4
x
with similar eqns for y and z
120 mm
25 mm
The dimensions, from the figure, are
100 mm
L = 120 mm
z
W = 100 mm
y
H = 25 mm
x
r = 20 mm
h = 75 mm
(H)
25 mm
D = 25 mm
+
R = 50 mm
12
V mm3
300000
187500
98175
31416
x (mm)
0
0
0
0
y (mm)
12.5
62.5
121.2
100
z (mm)
60
12.5
12.5
12.5
0m
(L)
m
100
mm (W)
y
z
1
+
50
mm
Object
+1
+2
+3
−4
–
3
x
7.76 Contd.
Substituting into the formulas for the composite, we get
y
x =0
y = 43.7 mm
z = 38.2 mm
mm
25
mm
100
(D)
mm
75
x
(h)
2
H
y
z
r = 20 mm
x
4
z
Problem 7.77
Determine the centroids of the volumes.
y
1.75 in
1 in
5 in
z
4 in
1 in
x
Solution: Divide the object into six volumes: (1) A cylinder 5 in.
long of radius 1.75 in., (2) a cylinder 5 in. long of radius 1 in., (3) a
block 4 in. long, 1 in. thick, and 2(1.75) = 3.5 in. wide. (4) Semicylinder 1 in long with a radius of 1.75 in., (5) a semi-cylinder 1 in
long with a radius of 1.75 in. (6) The composite object. The volumes
and centroids are:
Volume
V1
V2
V3
V4
V5
Vol, cu in
48.1
15.7
14
4.81
4.81
x, in.
0
0
2
0.743
0
y, in.
2.5
2.5
0.5
0.5
4.743
z, in.
0
0
0
0
0
The composite volume is V = V1 − V2 + V3 − V4 + V5 = 46.4 in3 .
The composite centroid:
x =
V1 x1 − V2 x2 + V3 x3 − V4 x4 + V5 x5
= 1.02 in.,
V
y =
V1 y1 − V2 y2 + V3 y3 − V4 y4 + V5 y5
= 1.9 in.,
V
z =0
1 in
x
1.75 in
z
z
x 5 in
4 in
y
1 in
x
Problem 7.78
Determine the centroids of the volumes.
y
30 mm
60 mm
z
x
180
mm
Solution: Consider the composite volume as being made up of
three volumes, a cylinder, a large cone, and a smaller cone which is
removed
Object
Cylinder
Cone 1
Cone 2
Cylinder
Cone 1
Cone 2
V
πr 2 L/2
1
πR2 L
3
1
πr 2 L
3
2
(mm3 )
5.089 × 105
1.357 × 106
1.696 × 105
180
mm
y
x
L/4
3L/4
30 mm
3(L/2)/4
(mm)
90
270
135
60 mm
z
x
180
mm
L = 360 mm
180
mm
r = 30 mm
R = 60 mm
y
For the composite shape
x = 229.5 mm
2R
xCyl VCyL + x1 V1 − x2 V2
VCyL + V1 − V2
2r
L/2
L/2
y
60 mm
Cylinder
1
y
x
360 mm
cone
+
120 mm
x =
2
y
cone
–
60 mm
3
180 mm
x
x
Problem 7.79 The dimensions of the Gemini spacecraft (in meters) are a = 0.70, b = 0.88, c = 0.74,
d = 0.98, e = 1.82, f = 2.20, g = 2.24, and h = 2.98.
Determine the centroid of its volume.
y
g
e
b
a
Solution: The spacecraft volume consists of three truncated cones
and a cylinder. Consider the truncated cone of length L with radii at
the ends R1 and R2 , where R2 > R1 . Choose the origin of the x-y
coordinate system at smaller end. The radius of the cone is a linear
function of the length; from geometry, the length of the cone before
truncations was
R2 L
(R2 −R1 )
(1)
H=
(2)
2
πR2
H
.
3
η=
(4)
2
πR1
η
.
3
(5)
(6)
(7)
(8)
The length of the truncated portion is
R1 L
(R2 −R1 )
(3)
with volume
with volume
The volume of the truncated cone is the difference of the
two volumes,
V =
πL
3
3
3
R2
−R1
R2 −R1
. The centroid of the removed part of the
cone is
xη = 34 η, and the centroid of the complete cone is
xh = 34 H, measured from the pointed end. From the composite theorem, the centroid of the truncated cone is
V x −V x
x = h hV η η −η+x, where x is the x-coordinate of the left
hand edge of the truncated cone in the specific coordinate system.
These eight equations are the algorithm for the determination of
the volumes and centroids of the truncated cones forming the
spacecraft.
c
f
d
h
x
Beginning from the left, the volumes are (1) a truncated cone, (2) a cylinder,
(3) a truncated cone, and (4) a truncated cone. The algorithm and the data for
these volumes were entered into TK Solver Plus and the volumes and centroids
determined. The volumes and x-coordinates of the centroids are:
Volume
V1
V2
V3
V4
Composite
Vol, cu m
0.4922
0.5582
3.7910
11.8907
16.732
x, m
0.4884
1.25
2.752
4.8716
3.999
The last row is the composite volume and x-coordinate of the centroid of the
composite volume.
The total length of the spacecraft is 5.68 m, so the centroid of the volume
lies at about 69% of the length as measured from the left end of the spacecraft.
Discussion: The algorithm for determining the centroid of a system of truncated
cones may be readily understood if it is implemented for a cone of known
dimensions divided into sections, and the results compared with the known
answer. Alternate algorithms (e.g. a Pappus-Guldinus algorithm) are useful for
checking but arguably do not simplify the computations End discussion.
Problem 7.80 Two views of a machine element are
shown. Determine the centroid of its volume.
y
y
24 mm
8 mm
18 mm
60 mm
8 mm
z
x
20
mm
16
mm
50 mm
Solution: We divide the volume into six parts as shown. Parts 3
and 6 are the “holes”, which each have a radius of 8 mm. The volumes
are
y
V1 = (60)(48)(50) = 144,000 mm3 ,
V2 =
2
1
Π(24)2 (50) = 45, 239 mm3 ,
2
2
3
3
V3 = Π(8) (50) = 10, 053 mm ,
V4 = (16)(36)(20) = 11, 520 mm3 ,
V5 =
1
Π(18)2 (20) = 10, 179 mm3 ,
2
1
6
5
4
z
V6 = Π(8)2 (20) = 4021 mm3 .
4(18)
= 47.6 mm,
3Π
The coordinates of the centroids are
z5 = 24 + 16 +
x1 = 25 mm,
x6 = 10 mm,
y1 = 30 mm,
y6 = 18 mm,
z1 = 0,
z6 = 24 + 16 = 40 mm.
x2 = 25 mm,
The x coordinate of the centroid is
y2
4(24)
= 60 +
= 70.2 mm,
3Π
z2 = 0,
x3 = 25 mm,
y3 = 60 mm,
z3 = 0,
x4 = 10 mm,
y4 = 18 mm,
z4 = 24 + 8 = 32 mm,
x5 = 10 mm,
y5 = 18 mm,
x =
x1 V1 + x2 V2 − x3 V3 + x4 V4 + x5 V5 − x6 V6
= 23.65 mm.
V1 + V2 − V3 + V4 + V5 − V6
Calculating the y and z coordinates in the same way, we obtain y = 36.63 mm
and z = 3.52 mm
Problem 7.81
Determine the centroids of the lines.
y
4m
Solution:
Part 1
Part 2
Break the line into two parts
xi
4
11
yi
2
2
x
L
√i
√80 = 8.94 m
52 = 7.21 m
6m
8m
y
x1 L1 + x2 L2
x =
= 7.12 m
L1 + L2
By inspection, since y1 = y2 = 2 m we get
y = 2 m.
4m
x
8m
Problem 7.82
Determine the centroids of the lines.
Solution:
(1)
(2)
(3)
y
6m
The object is divided into two lines and a composite.
L1 = 6 m, x1 = 3 m, y1 = 0.
L2 = 3π m, x2 = 6 + π6 m (Note: See Example 7.13) y2 = 3.
The composite length: L = 6 + 3π m. The composite centroid:
x=
L1 x1 + L2 x2
= 6 m,
L
y=
3π
= 1.83 m
2+π
3m
y
3m
x
6m
x
6m
Problem 7.83
Determine the centroids of the lines.
Solution: Break the composite line into two parts (the quarter circle and
the straight line segment). (see Appendix B)
y
xi
Part 1
Part 2
yi
2R/π
3m
Li
2R/π
0
πR/2
2m
(R = 2 m)
2R
= 1.273 πR/2 = 3.14 m
π
2m
x =
x1 L1 + x2 L2
= 1.94 m
L1 + L2
y =
y1 L1 + y2 L2
= 0.778 m
L1 + L2
2m
x
2m
2m
2m
2m
x
2m
2m
Problem 7.84 The semicircular part of the line lies in
the x − z plane. Determine the centroid of the line.
y
Solution: The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The composite
length is:
L =
3
i=1
100 mm
160 mm
Li .
x
120 mm
The composite coordinates are:
3
x =
i=1
and y =
,
L
3
i=1
Segment
L1
L2
L3
Composite
z
Li xi
y
Li yi
L
100 mm
Length, mm
120π
100
188.7
665.7
x, mm
240
π
0
80
65.9
y, mm
0
50
50
21.7
z, mm
120
0
0
68.0
2
3
160 mm
z
1
120 mm
x
Problem 7.85 The mass of the homogeneous flat plate
is 450 kg. What are the reactions at A and B?
Strategy: The center of mass of the plate is coincident
with the centroid of its area. Determine the horizontal
coordinate of the centroid and assume that the plate’s
weight acts there.
2m
1m
A
B
5m
Solution: To find the location of the center of mass, find the centroid by breaking the plate into a triangle and a rectangle.
X1
2m
1m
1m
A
5m
B
X2
5m
1m
x
5m
Area1 = 5/2 m2
x1 = 5/3 m
Area2 = 5 m2
x2 =
x =
5
m
2
x1 A1 + x2 A2
= 2.22 m
A1 + A2
m = 450 kg
x = 2.22 m
mg
X
AX
5m
AY
Fx :
Ax = 0
Fy :
Ay + By − mg = 0
MA :
−x mg + 5By = 0
Solving:
Ax = 0,
Ay = 2.45 kN
By = 1.96 kN
BY
Problem 7.86 The mass of the homogeneous flat plate
is 50 kg. Determine the reactions at the supports A
and B.
100 mm
400 mm
200 mm
A
B
600 mm
800 mm
Solution: Divide the object into three areas and the composite.
Since the distance to the action line of the weight is the only item of
importance, and since there is no horizontal component of the weight,
it is unnecessary to determine any centroid coordinate other than the
x-coordinate. The areas and the x-coordinate of the centroid are tabulated. The last row is the composite area and x-coordinate of the
centroid.
Area
Rectangle
Circle
Triangle
Composite
A, sq mm
3.2 × 105
3.14 × 104
1.2 × 105
4.09 × 105
400
mm
200
mm
100 mm
A
x
400
600
1000
561
600 mm
B
600 mm
800 mm
600 mm
The composite area is A = Arect − Acirc + Atriang . The composite
x-coordinate of the centroid is
x =
Arect xrect − Acirc xcirc + Atriang xtriang
.
A
The sum of the moments about A:
Fy = Ay + B − 500 = 0,
from which Ay = 300 N.
X
B
AY
1400 mm
MA = −500(561) + 1400B = 0,
from which B = 200 N. The sum of the forces:
500 N
AX
Fx = Ax = 0
Problem 7.87 The suspended sign is a homogeneous
flat plate that has a mass of 130 kg. Determine the axial
forces in members AD and CE. (Notice that the y axis
is positive downward.)
A
2m
4m
C
1m
E
B
D
y = 1 + 0.0625x2
y
Solution: The strategy is to determine the distance to the action
line of the weight (x-coordinate of the centroid) from which to apply
the equilibrium conditions to the method of sections.
The area: The element of area is the vertical strip of length y and
width dx. The element of area dA = y dx = (1 + ax2 ) dx, where
a = 0.0625. Thus
4
4
ax3
A =
dA =
(1 + ax2 ) dx = x +
= 5.3333 sq ft.
3 0
A
0
The x-coordinate:
4
x dA =
x(1 + ax2 ) dx =
A
0
Divide A: x =
12
5.3333
C
4m
1m
E
B
D
y
x2
2
+
4
ax4
4
y = 1 + 0.0625x2
= 12.
0
= 2.25 ft.
The equilibrium conditions: The angle of the member CE is
1
α = tan−1 ( ) = 14.04◦ .
4
The weight of the sign is W = 130(9.81) = 1275.3 N. The sum of
the moments about D is
MD = −2.25W + 4CE sin α = 0,
from which CE = 2957.7 N (T ) .
Method of sections: Make a cut through members AC, AD and BD
and consider the section to the right. The angle of member AD is
1
β = tan−1 ( ) = 26.57◦ .
2
The section as a free body: The sum of the vertical forces:
FY = AD sin β − W = 0
from which AD = 2851.7 N (T )
2m
A
x
Problem 7.88 The bar has a mass of 80 kg. What are
the reactions at A and B?
A
2m
2m
B
Solution: Break the bar into two parts and find the masses and
centers of masses of the two parts. The length of the bar is
A
L = L1 + L2 = 2 m + 2πR/4(R = 2 m)
z
L =2+πm
2m
Part
Lengthi (m)
1
2
2
π
Massi (kg)
2
2+π
π
2+π
m1 = 31.12 kg
x1 = 1 m
m2 = 48.88 kg
x2 = 3.27 m
xi (m)
80
1
80
2+
B
2R
π
y
X1
m1g
m2g
AX
Fx :
Ax = 0
Fy :
Ay + By − m1 g − m2 g = 0
MA :
2m
X2
AY
−x1 m1 g − x2 m2 g + 4By = 0
x
4m
Solving
BY
Ax = 0, Ay = 316 N, B = 469 N
Problem 7.89 The semicircular part of the homogeneous slender bar lies in the x − z plane. Determine the
center of mass of the bar.
y
Solution: The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The composite
length is:
L =
3
i=1
3
i=1
12 in
z
i=1
Segment
L1
L2
L3
Composite
x
Li xi
,
L
3
and y =
16 in
Li .
The composite coordinates are:
x =
10 in
y
Li yi
10 in
L
16 in
Length, in.
12π
10
18.868
66.567
x, in.
24
π
0
8
6.594
y, in.
0
5
5
2.168
z, in.
12
0
0
6.796
z
x
12 in
Problem 7.90 When the truck is unloaded, the total
reactions at the front and rear wheels are A = 54 kN
and B = 36 kN. The density of the load of gravel is
ρ = 1600 kg/m3 . The dimension of the load in the z
direction is 3 m, and its surface profile, given by the
function shown, does not depend on z. What are the
total reactions at the front and rear wheels of the loaded
truck?
y
y = 1.5 – 0.45x + 0.062x2
x
A
B
2.8 m
3.6 m
5.2 m
Solution: First, find the location of the center of mass of the unloaded truck (and its mass). Then find the center of mass and mass of
the load. Combine to find the wheel loads on the loaded truck.
Unloaded Truck
Fx :
no forces
Fy :
54000 + 36000 − mT g = 0(N )
MA :
−xT mT g + 5.2(36) = 0
y
y = 1.5 – 0.45x + 0.062 x 2
x
Solving xT = 2.08 m, mT = 9174 kg
A
B
Next, find xL and mL (for the load)
x dm
Num
m
xL = L
=
mL
dm
mL
where mL =
mL =
3.6
mL
2.8 m
dm, Num =
3ρy dx = 3ρ
0
3.6
0
mL = 3ρ 1.5x − 0.45
x2
2
3.6 m
5.2 m
XT
x dm
mTg
mL
5.2 m
(1.5 − 0.45x + 0.062x2 ) dx
+ 0.062
x3
3
B
36 kN
54 kN
3.6
mL g
y
0
mL = 16551 kg
Num = 3ρ
3.6
0
Num = 3ρ 1.5
y = 1.5 − 0.45x − 0.062x2
dm = p(3)y dx
(1.5x − 0.45x2 + 0.062x3 ) dx
x2
2
− 0.45
x3
3
+ 0.062
XL
x4
4
3.6
0
Num = 25560 kg · m
xL
0
Num
=
= 1.544 m
mL
dL
measured from the front of the load
XT
The horizontal distance from A to the center of mass of the load is
dL = xL + 2.8 m = 4.344 m
Now we can find the wheel loads on the loaded truck
Fx :
no forces
Fy :
Ay + By − mT g − mL g = 0
MA :
5.2By − xT mT g − dL mL g = 0
Solving Ay = 80.7 kN, By = 171.6 kN
3.6 m
AY
mL g
mT g
5.2 m
BY
x
Problem 7.91 The 10-ft horizontal cylinder with 1-ft
radius is supported at A and B. Its weight density is
γ = 100(1 − 0.002x2 ) lb/ft3 . What are the reactions at
A and B?
y
10 ft
A
1 ft
z
x
B
Solution: The weight: Denote a = 0.002. The element of volume is a disk of radius R = 1 ft, thickness dx, and weight
y
1 ft
2
dW = γπR dx,
from which
W =
dW = 100πR2
x
L
0
= 100πR2
ax3
x−
3
(1 − ax2 ) dx
L
= 2932.15 lb
0
The x-coordinate of the mass center:
L
x dW = 100πR2
(1 − ax2 )x dx
W
0
L
25πR2 (1 − ax2 )2 0 = 14137.17.
=−
a
Divide by W : x = 4.8214 ft.
The equilibrium conditions: The sum of the moments about A is
MA = −W x + 10B = 0,
from which
B=
Wx
(2932.15)(4.8214)
=
L
10
= 1413.7 lb .
The sum of the vertical forces:
FY = A + B − W = 0,
from which A = 1518.4 lb . The horizontal components of the reactions are zero:
FX = 0
A
B
10 ft
Problem 7.92 A horizontal cone with 800-mm length
and 200-mm radius has a built-in support at A. Its mass
density is ρ = 6000(1 + 0.4x2 ) kg/m3 , where x is in
meters. What are the reactions at A?
y
200 mm
A
x
800 mm
Solution: The strategy is to determine the distance to the line of
action of the weight, from which to apply the equilibrium conditions.
The mass: The element of volume is a disk of radius y and thickness
dx. y varies linearly with x: y = 0.25x. Denote a = 0.4. The mass
of the disk is
y
200
mm
x
A
dm = ρπy 2 dx = 6000π(1 + ax2 )(0.25x)2 dx
= 375π(1 + ax2 )x2 dx,
800 mm
from which
m = 375π
0.8
(1 + ax2 )x2 dx = 375π
0
x3
x5
+a
3
5
0.8
0
= 231.95 kg
The x-coordinate of the mass center:
0.8
0.8
x4
x6
x dm = 375π
(1 + ax2 )x3 dx = 375π
+a
4
6 0
m
0
= 141.23.
Divide by the mass: x = 0.6089 m
The equilibrium conditions: The sum of the moments about A:
M = MA − mgx = 0,
from which
MA = mgx = 231.94(9.81)(0.6089)
= 1385.4 N-m .
The sum of the vertical forces:
FY = AY − mg = 0
from which AY = 2275.4 N . The horizontal component of the
reaction is zero,
FX = 0.
Problem 7.93 The circular cylinder is made of aluminum (Al) with mass density 2700 kg/m3 and iron (Fe)
with mass density 7860 kg/m3 .
y
(a) Determine the centroid of the volume of the cylinder.
(b) Determine the center of mass of the cylinder.
Al
z
Fe
200 mm
600 mm
x
600 mm
Solution:
(a)
y
The volume of the cylinder is
AI
Fe
V = π(0.1)2 (1.2) = 0.0377 m3 .
The volume of the parts:
VAl = VF e
V
=
= 0.0188 m3 .
2
The centroid of the first part is xAl = 0.3 m, yAl = zAl = 0.
The centroid of the iron part is
xF e = 0.6 + 0.3 = 0.9 m,
yF e = zF e = 0.
The composite centroid
x=
VAl (0.3) + VF e (0.9)
1.2
=
= 0.6 m,
V
2
y = z = 0.
(b)
The mass center: The mass of the aluminum part is mAl =
VAl (2700) = 50.89 kg. The mass of the iron part is mF e =
VF e (7860) = 148.16 kg. The composite mass is m = mAl +
mF e = 199.05 kg. The composite center of mass is
xm =
(50.89)(0.3) + (148.16)(0.9)
= 0.7466 m
199.05
ym = zm = 0
200 mm
x
600 mm
600 mm
z
y
Problem 7.94 The cylindrical tube is made of aluminum with mass density 2700 kg/m3 . The cylindrical
plug is made of steel with mass density 7800 kg/m3 .
Determine the coordinates of the center of mass of the
composite object.
y
z
x
y
y
A
20 mm
x
z
35 mm
100
mm
100
mm
A
Section A-A
Solution:
The volume of the aluminum tube is
y
VAl = π(0.0352 − 0.022 )(0.2) = 5.18 × 10−4 m3 .
The mass of the aluminum tube is mAl = (2700)VAl = 1.4 kg. The
centroid of the aluminum tube is xAL = 0.1 m, yAl = zAl = 0.
The volume of the steel plug is VF e = π(0.02)2 (0.1) = 1.26 ×
10−4 m3 . The mass of the steel plug is mF e = (7800)VF e =
0.9802 kg. The centroid of the steel plug is xF e = 0.15 m, yF e =
zF e = 0.
The composite mass is m = 2.38 kg. The composite centroid is
mAl (0.1) + mF e (0.15)
x =
= 0.121 m
m
y =z=0
Problem 7.95 A machine consists of three parts. The
masses and the locations of the centers of mass of the
parts are:
Part
1
2
3
Mass (kg)
2.0
4.5
2.5
x (mm)
100
150
180
y (mm)
50
70
30
z (mm)
−20
0
0
Determine the coordinates of the center of mass of the
machine.
Solution:
The composite mass is 9 kg.
x =
2(100) + 4.5(150) + 2.5(180)
= 147 mm
9
y =
2(50) + 4.5(70) + 2.5(30)
= 54.4 mm
9
z =
2(−20)
= −4.4 mm
9
z
x
y
y
A
x
100 100
mm mm
Section A-A
20 mm
z
A
35 mm
Problem 7.96 A machine consists of three parts. The Solution: The composite mass is m = 2.0 + 4.5 + 2.5 = 9 kg. The
masses and the locations of the centers of mass of two location of the third part is
of the parts are:
120(9) − 2(100) − 4.5(150)
x3 =
Part
1
2
Mass (kg)
2.0
4.5
x (mm)
100
150
y (mm)
50
70
z (mm)
−20
0
The mass of part 3 is 2.5 kg. The design engineer wants
to position part 3 so that the center of mass of location
of the machine is x = 120 mm, y = 80 mm, and z = 0.
Determine the necessary position of the center of mass
of part 3.
Problem 7.97 Two views of a machine element are
shown. Part 1 is aluminum alloy with mass density
2800 kg/m3 , and part 2 is steel with mass density
7800 kg/m3 . Determine the x coordinate of its center
of mass.
= 82 mm
2.5
y3 =
80(9) − 2(50) − 4.5(70)
= 122 mm
2.5
z3 =
2(20)
= 16 mm
2.5
y
y
1
24 mm
2
8 mm
18 mm
60 mm
8 mm
z
x
20
mm
16
mm
50 mm
Solution:
V1
V2
The volumes of the parts are
1
= (60)(48) + π(24)2 − π(8)2 (50)
2
y
y
1
24 mm
= 179, 186 mm3 = 17.92 × 10−5 m3 ,
1
= (16)(36) + π(18)2 − π(8)2 (20) = 17, 678 mm3
2
2
= 1.77 × 10−5 m3 ,
so their masses are
x
m1 = S1 V1 = (2800)(17.92 × 10−5 ) = 0.502 kg,
m2 = S2 V2 = (7800)(1.77 × 10−5 ) = 0.138 kg.
The x coordinates of the centers of mass of the parts are x1 = 25 mm,
x2 = 10 mm, so
x =
x1 m1 + x2 m2
= 21.8 mm
m1 + m2
8 mm
18 mm
20
mm
50 mm
60 mm
8 mm 8 mm
z
16
mm
Problem 7.98 Determine the y and z coordinates of
the center of mass of the machine element in Problem 7.97.
The z coordinate of the center of mass of part 1 is
z1 = 0. To determine the y coordinate, we divide it into three parts
a, b, and c. (Part c is the “hole” with radius 8 mm). The volumes are
Solution:
y
Va = (60)(48)(50) = 144,000 mm3 ,
Vb =
y
24 mm
b
c
1
π(24)2 (50) = 45, 239 mm3 ,
2
a
60 mm
Vc = π(8)2 (50) = 10, 053 mm3
50
mm
z
so
y1
ya Va + yb Vb − yc Vc
=
Va + Vb − Vc
4(24)
(30)Va + 60 + 3π Vb − (60)Vc
=
Va + Vb − Vc
y
Vd = (16)(36)(20) = 11,520 mm3 ,
Ve =
1
π(18)2 (20) = 10,179 mm3 ,
2
Vf = π(8)2 (20) = 4021 mm3
so
z2 =
=
zd Vd + ze Ve − zf Vf
Vd + Ve − Vf
(24 + 8)Vd + 24 + 16 +
4(18)
3π
Ve − (24 + 16)Vf
Vd + Ve + Vf
= 39.2 mm.
From the solution of Problem 7.97, the masses are m1 = 0.502 kg,
m2 = 0.138 kg, Therefore,
y =
y1 m1 + y2 m2
= 34.05 mm,
m1 + m2
z =
z1 m1 + z2 m2
= 8.45 mm.
m1 + m2
y
16 mm
e
f
= 38.5 mm.
The y coordinate of the center of mass of part 2 is y2 = 18 mm. To
determine the z coordinate, we divide it into three parts d, e, and f .
The volumes are
x
d
z
x
18 mm 24 mm
20 mm
Problem 7.99 With its engine removed, the mass of
the car is 1100 kg and its center of mass is at C. The
mass of the engine is 220 kg.
(a) Suppose that you want to place the center of mass
E of the engine so that the center of mass of the car
is midway between the front wheels A and the rear
wheels B. What is the distance b?
(b) If the car is parked on a 15◦ slope facing up the
slope, what total normal force is exerted by the road
on the rear wheels B?
E
C
0.6 m
0.45 m
A
B
1.14 m
b
2.60 m
Solution:
(a)
The composite mass is m = mC + mE = 1320 kg. The
x-coordinate of the composite center of mass is given:
2.6
x=
= 1.3 m,
2
from which the x-coordinate of the center of mass of the engine is
xE = b =
(1.3 m − 1.14 mC )
= 2.1 m.
mE
The y-coordinate of the composite center of mass is
y=
(b)
0.45 mC + 0.6 mE
= 0.475 m.
m
Assume that the engine has been placed in the new position, as
given in Part (a). The sum of the moments about B is
MA = 2.6A + ymg sin(15◦ )
−(2.6 − x)mg cos(15◦ ) = 0,
from which A = 5641.7 N. This is the normal force exerted by
the road on A. The normal force exerted on B is obtained from;
FN = A − mg cos(15◦ ) + B = 0,
from which B = 6866 N
C
E
0.6 m
0.45 m
A
1.14 m
b
2.60 m
B
Problem 7.100 The airplane is parked with its landing
gear resting on scales. The weights measured at A, B,
and C are 30 kN, 140 kN, and 146 kN, respectively. After
a crate is loaded onto the plane, the weights measured at
A, B, and C are 31 kN, 142 kN, and 147 kN, respectively.
Determine the mass and the x and y coordinates of the
center of mass of the crate.
B
6m
A
x
C
6m
10 m
y
Solution: The weight of the airplane is WA = 30+140+146 =
316 kN. The center of mass of the airplane:
My−axis = 30(10) − xA WA = 0,
x
A
from which xA = 0.949 m.
Mx−axis = (140 − 146)(6) + yA WA = 0,
from which yA = 0.114 m. The weight of the loaded plane:
10 m
B
W = 31 + 142 + 147 = 320 kN.
The center of mass of the loaded plane:
My−axis = (31)10 − xW = 0,
from which x = 0.969 m.
Mx−axis = (142 − 147)(6) + yW = 0,
from which y = 0.0938 m. The weight of the crate is Wc = W −
WA = 4 kN. The center of mass of the crate:
xc =
W x − WA xA
= 2.5 m,
Wc
yc =
W y − WA yA
= −1.5 m.
Wc
The mass of the crate:
mc =
Wc × 103
= 407.75 kg
9.81
6m
C
6m
y
Determine the Iy and ky .
Problem 7.101
y
h
x
b
Let dA = x dy = b dy. A =
Solution:
Iy =
x2 dA = h
A
b
0
x3
x2 dx = h
3
h
b dy = hb.
0
b
0
b3 h
=
ky =
3
y
Iy
b
= √
A
3
h
x
b
Problem 7.102 Determine Ix and kx by letting dA be
(a) a horizontal strip of height dy; (b) a vertical strip of
width dx.
Solution:
(See figure in Problem 7.101.)
(a) A
h
=
Ix
b dy = hb.
0
y 2 dA = b
=
A
kx
=
(b) A
0
=h
kx
=
bh3
,
3
h dx = hb.
0
Ix
y 2 dy =
Ix
h
= √
A
3
b
=
h
b
y 2 dx = h
0
h
0
y2
b
h
dy =
bh3
,
3
Ix
h
= √
A
3
Problem 7.103
Determine Ixy .
Solution:
(See figure in Problem 7.101.)
b h
dA =
dx dy = hb.
A =
A
0
Ixy =
0
xy dA =
A
b
0
h
0
xy dx dy =
y2
2
h
0
x2
2
b
=
0
h2 b2
4
Problem 7.104 Determine Ix , kx , Iy , and ky for the
beam’s rectangular cross section.
y
60 mm
x
40 mm
Solution:
Ix =
y
y 2 dA
A
30
Let dA = 40 dy = 60 mm
30
y3
=
40y dy = 40
3
−30
27000
27000
= 40
+
3
3
Ix
2
dA = 40 dy
30
−30
60 mm
= 40[18000] mm4
–30
Ix = 7.2 × 105 mm4
A = 2400 mm2
kx =
Ix
=
A
40 mm
7.2 × 105
mm
2.4 × 103
kx = 17.3 mm
y
–20
(mm)
120
30
x
–30
dA = 60 dx
Iy =
Iy =
x2 dA
A
20
x2 (60) dx =
−20
Iy = 60
x3
3
20
ky =
ky =
√
60x2 dx
−20
20
= 60
−20
8000
8000
+
3
3
Iy = 3.2 × 105 mm4
Iy
=
A
x
3.2 × 105
2.4 × 103
133.3 = 11.5 mm
mm4
Problem 7.105 Determine Ixy and JO for the beam’s
rectangular cross section.
Solution:
Ixy =
y
xy dA
Ixy =
A
20
xy dx dy
−30
−20
30
x2
y
2
Ixy =
30
−30
+20
30
dy =
dA
0 dy = 0
60 mm
−30
−20
x
Ixy = 0
JO =
r2 dA =
A
−30
30
=
−30
30
=
JO =
30
−30
30
−30
x3
+ y2 x
3
20
(x2 + y 2 ) dx dy
−20
20
40 mm
dy
−20
8000
8000
+ 20y 2 +
+ 20y 2 dy
3
3
16000
+ 40y 2 dy
3
16000
y3
y + 40
3
3
JO =
(16000)
(60) + 80
3
b
dx
0
−h
x+h
b
h
− x+h
b
0
Iy =
x2 dA =
A
ky =
b
dx = −
x2 dx
mm4
y
hx2
+ hx
2b
−h
x+h
b
b
=
0
h
hb
2
x
dy
b
0
h
− x3 + hx2
b
0
dy
0
b
=
303
3
0
b
=
−30
JO = 10.4 × 105 mm4
Problem 7.106 Determine Iy and ky .
Solution: y = − hb x + h, dA = dy dx, therefore
A =
30
JO =
dx = −
hx4
hx3
+
4b
3
b
=
0
hb3
12
y
Iy
b
= √
A
6
h
x
b
Determine JO and kO .
Problem 7.107
kx =
Solution: (See figure in Problem 7.106.)
h
b
y =−
Ix =
=
=
x + h, dA = dx dy
y 2 dA =
A
1
3
b
dx
b
0
bh3
h
− x+h
b
3
−h
x+h
b
dx = −
bh
(b2 + h2 )
12
(b2 + h2 )
2
2
= kx + ky =
6
JO = Ix + Iy =
y 2 dy
0
12
Ix
h
= √
A
6
b
12h
h
− x+h
b
4 b
0
kO
Determine Ixy .
Problem 7.108
Solution:
(See figure for Problem 7.106.)
Ixy =
b
xy dA =
A
1
2
=
0
b
0
−h
x+h
b
x dx
2
h
x − x+h
b
dx
h 2 x4
h 2 x3
x2
−2
+ h2
2
b 4
b 3
2
2
2
2
2
h b
1
2
1
h b
=
− +
=
2
4
3
2
24
1
2
=
y dy
0
b
0
Determine Iy .
Problem 7.109
y
y = cxn
x
O
b
Solution:
Iy =
Iy =
x2 dA
b
Iy
x2 (cxn ) dx
y = cx n
0
Iy =
y
A
b
cxn+3
(n + 3)
cx(n+2) dx =
0
b
0
cb(n+3)
=
(n + 3)
x
b
dA = ydx
= cx ndx
Determine Ix .
Problem 7.110
Solution:
Ix =
y 2 dA =
A
Ix =
b
0
Ix =
Ix =
b
0
3 0
cx
y3 b
0
cxn
y
y 2 dy dx
0
n
3n c3 x
3
c3
x3n+1
3
(3n + 1)
dx
dx =
b
=
0
Ix = c3 b3n+1 /(9n + 3)
y = cx n
c3
3
b
dA = dydx
x3n dx
0
c3 b3n+1
3(3n + 1)
x
b
Determine JO .
Problem 7.111
Solution:
JO =
2
r dA =
A
JO =
JO =
b
0
b
x2 y +
0
b
y3
3
cxn
0
2
(x + y ) dy dx
0
cxn
dx
0
c3 x3n
3
cxn+2 +
y
2
y = cx n
dx
dA
b
cxn+3
c3 x3n+1 +
(n + 3)
3 (3n + 1) 0
JO =
cb(n+3)
JO =
(n + 3)
+
x
c3 b(3n+1)
b
(9n + 3)
dA = dy dx
Determine Ixy .
Problem 7.112
Solution:
Ixy =
A
Ixy =
b
0
cx
xy 2
b
2
0
dx
xc2 x2n
dx =
2
0
b
c2 x2n+2 =
2 (2n + 2) 0
Ixy = c b
xy dy dx
0
0
2 (2n+2)
y
cxn
n
b
Ixy =
Ixy
xy dA =
b
0
y = cx n
c2 (2n+1)
x
dx
2
dA
x
/(4n + 4)
b
Determine Iy and ky .
Problem 7.113
y
y = x3
Solution:
Iy =
Iy =
x2 dA =
A
0
1
0
1
[x2 y]x
dx =
x3
x
x3
1
0
y=x
x2 dy dx
(x3 − x5 ) dx
1
x4
x6 1
1
−
= −
4
6 0
4
6
Iy =
x
Iy = 0.0833
Area =
1
0
x
x3
dy dx =
1
0
x
y
dx
x3
1
1
x2
x4 Area =
(x − x3 ) dx =
−
= 0.25
2
4 0
0
ky =
Iy
=
Area
ky = 0.577
0.0833
0.25
y=x
y
(1, 1)
y = x3
dA = dy dx
x
Determine Ix and kx .
Problem 7.114
Solution:
Ix =
Ix =
A
1
y
x
x
1
y 3 x3
x9
dx
=
−
dx
3 x3
3
3
0
1
x4
x10 1 1
1
−
=
−
4
10 0
3 4
10
1
1
3
y=x
y 2 dy dx
x3
0
0
Ix =
y 2 dA =
(1, 1)
y = x3
dA = dx dy
Ix = 0.0500
1
Area =
x
x3
0
dy dx =
0
−
= 0.25
2
4 0
Ix
0.050
=
=
= 0.447
A
0.25
Determine JO and kO .
Problem 7.115
Solution:
JO =
JO =
JO =
JO =
JO =
JO
0
x
1
x4
x2
Area =
ky
x
1
1 x
y dx =
(x − x3 ) dx
3
r2 dA =
A
1
0
1
x2 y +
0
1
x3 +
0
y3
3
x3
3
x
x
x3
y
(x2 + y 2 ) dy dx
dx
(1, 1)
x3
x9
− x5 −
3
9
1
0
4 3
x
x − x5 −
3
3
4/
3
x4
4/
−
x6
x10
−
6
30
dx
y = x3
dx
1
x
0
Area = 0.25
JO
0.133
kO =
=
= 0.730
Area
0.25
1
1
1
= − −
3
6
30
JO = 0.133
Area =
0
=
1
1
0
x
x3
y=x
dy dx =
(x − x3 ) dx =
1
0
kO = 0.730
x
[y] dx
3
x
x2
2
−
x4
4
1
0
Determine Ixy .
Problem 7.116
Solution:
Ixy =
Ixy =
1
x
x3
0
1
x
0
y
xy dy dx
y2
2
x
x3
dx =
0
1
x4
x8 −
=
8
16 0
Ixy =
1
x3
x7
−
2
2
y=x
1
(1, 1)
dx
1
1
−
8
16
Ixy = 0.0625
x
1
Problem 7.117 Determine the moment of inertia Iy
of the metal plate’s cross-sectional area.
y
1
y = 4 – – x 2 ft
4
x
Solution:
Iy =
Iy =
Iy =
x2 dA =
A
4x2 −
−4
x3
3
43
3
43
−
−
−
Iy = 68.3 ft4
20
y
x2 dy dx
dx
1
y = 4 – – x2
4
1 4
x dx
4
5 4
1x
4 5
45
(4− 1
x2 )
4
0
[x2 y]0
4
3
(4− 1
x2 )
4
−4
Iy = 4
4
−4
4
Iy = 4
Iy = 8
−
−4
4(−43 )
(−4)5
+
3
20
2(45 )
= 170.67 − 102.4
20
4
dA = dy dx
x
–4
4
Problem 7.118 Determine the moment of inertia Ix
and the radius of gyration kx of the cross-sectional area
of the metal plate.
Solution:
Ix =
y 2 dA =
A
Ix =
(4− 1
x2 )
4
y
y 2 dy dx
1
y = 4 – – x2
4
0
(4− 14 x2 )
y 3 dx
3 0
−4
1
=
3
−4
4
1
=
3
4
4
3
1
−4 − x2
4
−4
dA = dy dx
dx
x
4
–4
3
1 6
64 − 12x2 + x4 −
x
dx
4
64
−4
4
1
12x3
3 x5
1 x7 =
64x −
+
−
3
3
4 5
64 7 −4
4
1 2
1 x3
x dx = 4x −
4
4 3 −4
−4
16
16
Area = 16 −
+ 16 −
= 21.33 ft2
3
3
Area =
Ix = 78.0 ft4
Area =
dA =
A
Area =
4
−4
4
−4
(4− 1
x2 )
4
4
dy dx
kx =
0
4− 1 x2
4
[y]
dx
4−
Ix
=
Area
78.02
ft
21.33
kx = 1.91 ft
0
Problem 7.119 (a) Determine Iy and ky by letting dA
be a vertical strip of width dx.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO = 12 πR4 . Explain how you
can use this information to confirm your answer to (a).
y
The equation of the circle is x2 +y 2 = R2 , from which
y = ± R2 √
− x2 . The strip dx wide and y long has the elemental
area dA = 2 R2 − x2 dx. The area of the semicircle is
R πR2
A=
Iy =
x2 dA = 2
x2 R2 −x2 dx
2
A
0
R
x(R2 −x2 )3/2 R2 x(R2 −x2 )1/2 R4
x
=2 −
+
+
sin−1
4
8
8
R
x
Solution:
√
0
=
4
πR4
8
Iy R
ky =
=
A
2
R
(b) If the integration were done for a circular area with the center at the origin,
the limits of integration for the variable x would be from −R to R, doubling
the result. Hence, doubling the answer above,
Iy =
πR4
.
4
By symmetry, Ix = Iy , and the polar moment would be
y
JO = 2Iy =
x
R
πR4
,
2
which is indeed the case. Also, since kx = ky by symmetry for the full circular
area,
Ix
JO
Iy
Iy
kO =
+
= 2
=
A
A
A
A
as required by the definition. Thus the result checks.
Problem 7.120 (a) Determine Ix and kx for the area
in Problem 7.119 by letting dA be a horizontal strip of
height dy.
(b) The polar moment of inertia of a circular area with
its center at the origin is JO = 12 πR4 . Explain how you
can use this information to confirm your answer to (a).
Use the results of the solution to Problem 7.119, A =
The equation for the circle is x2 + y 2 = R2 , from which
x = ± R2 − y 2 . The horizontal strip is from 0 to R, hence the
element of area is
dA = R2 −y 2 dy.
+R Ix =
y 2 dA =
y 2 R2 −y 2 dy
Solution:
πR2
.
2
A
y(R2 −y 2 )3/2 R2 y(R2 −y 2 )1/2 R4
= −
+
+
sin−1
4
8
8
R4
R4
y
R
R
−R
πR4
π
π
+
=
8 2
8 2
8
kx =
Ix =
πR4
.
4
By symmetry Iy = Ix ,
and JO = 2Ix =
−R
=
(b) If the area were circular, the strip would be twice as long, and the moment
of inertia would be doubled:
πR4
,
2
which is indeed the result. Since kx = ky by symmetry for the full circular
area, the
Ix
JO
Iy
Ix
kO =
+
= 2
=
A
A
A
A
as required by the definition. This checks the answer.
Ix R
= .
A
2
Problem 7.121 Determine the moments of inertia Ix
and Iy .
Strategy: Use the procedure described in Example 8.2
to determine JO , then use the symmetry of the area to
determine Ix and Iy .
y
Ro
x
Ri
Solution:
JO =
Ro
Ri
JO = 2π
Let dA = 2πr dr
Ro
r2 dA = 2π
r 2 r dr
y
Ri
RO
R
Ri4
r4 o
Ro4
=
2π
−
4 R
4
4
i
From symmetry Ix = Iy
Also
JO = Ix + Iy
∴ Ix = Iy =
π 4
(R − Ri4 )
4 0
x
Ri
Problem 7.122 If a = 5 m and b = 1 m, what are the
values of Iy and ky for the elliptical area of the airplane’s
wing?
y
x2
y2
— + — =1
a2
b2
x
2b
a
Solution:
Iy =
A
Iy = 2
0
Iy = 2
[x
a
a
x
2b
a
2
1−
x2
a2
dx
1/2
dx
a
y = b 1– x2
a2
x2
dx
a2
a
x2
0
0
a2 − x2 dx
√
a4
sin−1
8
x
a
a
0
(from the integral tables)

0
√ 0

a3 a2 − a2
2b  a(a2 − a2 )3/2
Iy =
/+
/
−
a 
4
8


a4
+
8
sin−1

a  
−
a
0√
+
0
0(a2 )3/2
0
a2 · 0 a2 a4
/+
sin−1
8
8
Iy =
2/b a4 π
a 8 2/
Iy =
2a3 bπ
8
4
The area of the ellipse (half ellipse) is
A =2
0
2 1/2
b 1− xa
0
/
2b
a
b 1−
a
0
dy dx
x2
a
dx
(a2 − x2 )1/2 dx
√
a
x a 2 − x2
a2
x
+
sin−1
2
2
a
0
$ √
%
2
2b
a 0
a
a
=
+
sin−1
a
2
2
a
√
0 a
a2
0
−
+
sin−1
2
2
a
=
2b
a
2/b a2/ π
πab
=
a/ 2/ 2
2
Evaluating, we get
A = 7.85 m2
Finally
ky =
Iy
=
A
ky = 2.5 m
Iy = 49.09 m4
a
=
A =


0 
/


a

Evaluating, we get
a
A =2
2b
x(a2 − x2 )3/2
a 2 x a 2 − x2
=
−
+
a
4
8
+
x
2
b(1− x2 )1/2
a
y]0
0
Rewriting
x2 + y2 = 1
a2 b2
−y
x2 dy dx
x2 b 1 −
0
Iy = 2b
y
x2 dy dx
2b
2
0
Iy = 2
Iy
y
y
0
a
a
−0
a
Iy =
x2 dA =
49.09
7.85
Problem 7.123 What are the values of Ix and kx
for the elliptical area of the airplane’s wing in Problem 7.122?
Solution:
Ix =
a
y dA = 2
A
Ix = 2
a
y3 a
b
3
0
√
0
b
y= a
√
a2 −x2
y
y dy dx
0
a2 −x2
dx
2b
x
0
y = b a2 – x2
a
a
0
√
a(0)
3a3 0
3 π
+
+ a4
4
8
8 2
√
0(a2 )
3a2 · 0 a2
−
−
+0
4
8
Ix =
x2
y2
— + — =1
a2
b2
2
b3
(a2 − x2 )3/2 dx
3
0 3a
√
2b3 x(a2 − x2 )3/2
3a2 x a2 − x2
=
+
3a3
4
8
a
3
x + a4 sin−1
8
a Ix = 2
Ix
2
2b3
3a3
a
Ix =
2/b3
·
3/a3/
Ix =
3/ab3 π
ab3 π
=
3/.8
8
3/
8
π
2/
Evaluating (a = 5, b = 1)
Ix =
5π
= 1.96 m4
8
From Problem 7.122, the area of the wing is A = 7.85 m2
Ix
1.96
kx =
=
kx = 0.500 m
A
7.85
Determine Iy and ky .
Problem 7.124
a4/
y
y = x 2 – 20
Solution: The straight line and curve intersect where x = x2 −
20. Solving this equation for x, we obtain
√
1 ± 1 + 80
x =
= −4, 5.
2
y=x
x
If we use a vertical strip: the area
dA = [x − (x2 − 20)] dx.
Therefore
Iy =
x2 dA =
A
5
−4
x2 (x − x2 + 20) dx
x4
x5
20x3
−
+
4
5
3
=
The area is
dA =
A =
A
5
= 522.
−4
y
5
−4
(x − x2 + 20) dx
5
So
ky
x2
x3
=
−
+ 20x
= 122.
2
3
−4
Iy
522
=
=
= 2.07.
A
122
y = x2 – 20
y=x
x
dA
dx
Problem 7.125
Problem 7.124.
Determine Ix and kx for the area in
Solution: Let us determine the moment of inertia about the x axis
of a vertical strip holding x and dx fixed:
x
x
y3
(Ix )(strip) =
y 2 dAs =
y 2 (dx dy) = dx
3 x2 −20
As
x2 −20
y
y=x
y = x2 – 20
dx
=
(−x6 + 60x4 + x3 − 1200x2 + 8000).
3
dAs
x
Integrating this value from x = −4 to x = 5 (see the solution to
Problem 7.124), we obtain Ix for the entire area:
5
1
Ix =
(−x6 + 60x4 + x3 − 1200x2 + 8000) dx
−4 3
5
x7
x4
400x3
8000x
= −
+ 4x5 +
−
+
= 10,900.
21
12
3
3
−4
x
dx
From the solution to Problem 7.124, A = 122 so
Ix
10,900
kx =
=
= 9.48.
A
122
Problem 7.126 A vertical plate of area A is beneath
the surface of a stationary body of water. The pressure of
the water subjects each element dA of the surface of the
plate to a force (p0 + γy) dA, where p0 is the pressure
at the surface of the water and γ is the weight density
of the water. Show that the magnitude of the moment
about the x axis due to the pressure on the front face of
the plate is
x
A
y
M(x axis) = p0 yA + γIx ,
where y is the y coordinate of the centroid of A and Ix
is the moment of inertia of A about the x axis.
The moment about the x-axis is dM = y(p0 +γy) dA
integrating over the surface of the plate:
M =
(p0 + γy)y dA.
Solution:
x
A
Noting that p0 and γ are constants over the area,
M = p0
y dA + γ y 2 dA.
A
By definition,
y dA
A
y =
and Ix =
A
y 2 dA,
A
then M = p0 yA + γIX , which demonstrates the result.
A
y
Problem 7.127 Determine Ix and kx for the composite
area by dividing it into rectangles 1 and 2 as shown, and
compare your results to those of Example 8.4.
y
1m
1
4m
2
1m
x
3m
Solution:
Ix =
Ix =
A1
1
4
A2
3
y3
3
0
1
1
0
3
dx +
1m
0
4
y3
3
0
y
y 2 dA2
y 2 dy dx +
1
1
y 2 dy dx
1
dx
0
A1
3m
4m
3
1
64
1
=
−
dx +
dx
3
3
0
0 3
63 1
1 3
63
3
=
x + x =
+
3
3 0
3
3
0
Ix
0
Ix =
Ix
y 2 dA1 +
1
A2
1m
x
3m
Ix = 21 + 1 = 22 ft4
Area = 3 + 3 = 6 ft2
Ix
22
kx =
=
= 1.91 ft
Area
6
Determine Iy and ky for the composite
Problem 7.128
area.
Solution:
Iy =
Iy =
Iy =
Iy =
=
A1
x2 dA1 +
1
0
[x
0
1
2
A2
x2 dy dx +
1
1
0
4
y]41
3
dx +
[x
0
(4x2 − x2 ) dx +
1
3x3
3
0
+
3
x3
3
3
0
0
Iy = 1 + 9 = 10.00 m4
10
ky =
= 1.29 m
6
y
x2 dA2
x2 dy dx
0
2
1
1m
y]10
3
0
A1
dx
Area = 6m2
3m
4m
x2 dx
A2
1m
x
3m
Determine Ix and kx .
Problem 7.129
y
6m
x
2m
3m
12 m
Solution:
Ix =
y 2 dA =
A
Ix =
3
9
Ix =
9
−3
y
4
y 2 dy dx
Area = (6)(12) m2
Area = 72 m2
−2
4m
dx
−2
(4)3
(−2)3
−
3
3
−3
−3
4
y3
9
−3
Ix =
9
72
dx = 24
3
9
dx =
−3
9
−3
64
8
+
3
3
x
2m
dx
3m
9
dx = 24x
9m
−3
Ix = 24[9 − (−3)] = (24)(12)
Ix = 288 m4
√
Ix
288
ky =
=
= 4=2m
Area
72
Determine Iy and ky .
Problem 7.130
Solution:
Iy =
x2 dA =
A
Iy =
−3
Iy =
9
9
9
−3
4
[x2 y]
x2 dy dx
−2
dx =
9
−3
−2
6x2 dx = 6
−3
3
y
4
x3
3
9
−3
x2 (4 − (−2)) dx
= [2x3 ]9−3
3
Iy = 2[9 − (−3) ]
Iy = 1512 m4
ky =
Iy
=
Area
ky = 4.58 m
√
1512
= 21
72
Area = (6)(12) m2
Area = 72 m2
4m
x
2m
3m
9m
Determine Ix and kx .
Problem 7.131
Solution:
Ix =
Ix =
A1
−15
−45
70
y3
3
−45
45
Ix =
−15
Ix
15
703 =
x
3
703
Ix =
3
703
100
y 2 dy dx
70
70
mm
100
y3
3
dx
70
dx
x
0
45
1003
−45
3
dx +
3
−
703
30 mm
dx
3
90 mm
dx
y
45
x
1003
703
−
3
3
−45
45
x
A2
−15
1003
703
−
3
3
(30) +
3
70
+
−45
+
−45
70
3
15
45
dx +
0
3
45
45
y 2 dA3
A3
100
mm
703
−45
+
y 2 dy dx
70
y3
3
15
y 2 dy dx +
70
0
−15
+
A2
−45
15
Ix =
y 2 dA2 +
0
45
+
y 2 dA1 +
y
Area = (100)(90)
–(30)(70)
= 9000 – 2100
Area = 6900 mm2
703
90 +
(30)
3
100 mm
70 mm
A1
30 mm
Ix = 2.66 × 107 mm4
kx =
Ix
=
Area
A3
30 mm
2.66 × 107
= 62.1 mm
6900
30 mm
x
90 mm
kx = 62.1 mm
Determine Iy and ky .
Problem 7.132
Solution:
Iy =
Iy =
x2 dA1 +
A1
−15
−45
15
Iy =
−15
−45
15
Iy =
Iy
−15
−45
x2 dy dx +
45
−45
70
A3
[x2 y]70
0 dx +
100
45
−45
0
70x dx +
100 mm
A1
45
−45
(15)3
(45)3
−
3
3
A3
70
30
mm
2
(100 − 70)x dx +
(−15)3
(−453 )
−
3
3
+ 70
Area = 9000 mm2
– 2100 mm2
Area = 6900 mm2
100
(x2 y)
dx
−15
45
45
x3 x3 x3 = 70 + 30 + 70 3 −45
3 −45
3 15
Iy = 70
A2
x2 dy dx
70
70
2 (x y) dx
2
y
x2 dA3
x2 dy dx
0
45
+
A2
x2 dA2 +
0
45
+
70
+ 30
45
2
70x dx
15
453
(−45)3
−
3
3
Iy = 140
453
153
−
3
3
+ 60
30
mm
453
3
Iy = 5.92 × 106 mm4
Iy
5.92 × 106
ky =
=
= 29.3 mm
Area
6900
ky = 29.3 mm
30
mm
70 mm
x
Determine JO and kO .
Problem 7.133
Solution:
JO =
JO =
r2 dA =
A
A
2
x dA +
A
y
(x2 + y 2 ) dA
A
y 2 dA = Ix + Iy
From the solutions to 8.31 and 8.32
7
Ix = 2.66 × 10 mm
7
Area = 9000 mm2
– 2100 mm2
Area = 6900 mm2
4
and Iy = 0.592 × 10 mm
4
100
mm
70 mm
Hence,
JO = 3.25 × 107 mm4
kO =
JO
=
A
30
mm
3.25 × 107
6900
kO = 68.6 mm
Problem 7.134 If you design the beam cross section
so that Ix = 6.4×105 mm4 , what are the resulting values
of Iy and JO ?
y
h
x
Solution:
The area moment of inertia for a triangle about the
base is
h
1
12
Ix =
bh3 ,
1
12
from which Ix = 2
30
mm
(60)h3 = 10h3 mm4 ,
30
mm
Ix = 10h3 = 6.4 × 105 mm4 ,
y
from which h = 40 mm.
Iy = 2
1
12
(2h)(303 ) =
1
3
h(303 )
h
x
from which Iy =
1
3
3
5
(40)(30 ) = 3.6 × 10 mm
4
and JO = Ix + Iy = 3.6 × 105 + 6.4 × 105 = 1 × 106 mm4
h
30 mm 30 mm
30
mm
30
mm
x
Problem 7.135
Determine Iy and ky .
y
160
mm
40 mm
Solution:
Divide the area into three parts:
200
mm
40
mm
Part (1): The top rectangle.
40 mm
A1 = 160(40) = 6.4 × 103 mm2 ,
dx1 =
160
= 80 mm,
2
1
12
Iyy1 =
x
120
mm
(40)(1603 ) = 1.3653 × 107 mm4 .
y
From which
160
mm
Iy1 = d2x1 A1 + Iyy1 = 5.4613 × 107 mm4 .
Part (2): The middle rectangle:
40 mm
A2 = (200 − 80)(40) = 4.8 × 103 mm2 ,
200
mm
dx2 = 20 mm,
1
12
Iyy2 =
40
mm
(120)(403 ) = 6.4 × 105 mm4 .
40 mm
x
From which,
120
mm
Iy2 = d2x2 A2 + Iyy2 = 2.56 × 106 mm4 .
Part (3) The bottom rectangle:
A3 = 120(40) = 4.8 × 103 mm2 ,
dx3 =
120
= 60 mm,
2
1
12
Iyy3 =
40(1203 ) = 5.76 × 106 mm4
Problem 7.136
Solution:
Use the solution to Problem 7.135. Divide the area into
Part (1): The top rectangle.
A1 = 6.4 × 103 mm2 ,
dy1 = 200 − 20 = 180 mm,
1
12
(160)(403 ) = 8.533 × 105 mm4 .
From which
Ix1 = d2y1 A1 + Ixx1 = 2.082 × 108 mm4
Part (2): The middle rectangle:
A2 = 4.8 × 103 mm2 ,
dy2 =
Ixx2 =
120
+ 40 = 100 mm,
2
1
12
Iy3 = d2X3 A3 + Iyy3 = 2.304 × 107 mm4
The composite:
Iy = Iy1 + Iy2 + Iy3 = 8.0213 × 107 mm4
Iy
ky =
= 70.8 mm.
(A1 + A2 + A3 )
Determine Ix and kx .
three parts:
Ixx1 =
From which
(40)(1203 ) = 5.76 × 106 mm4
from which
Ix2 = d2y2 A2 + Ixx2 = 5.376 × 107 mm4
Part (3) The bottom rectangle:
A3 = 4.8 × 103 mm2 ,
dy3 = 20 mm,
Ixx3 =
1
12
120(403 ) = 6.4 × 105 mm4
and Ix3 = d2y3 A3 + Ixx3 = 2.56 × 106 mm4 .
The composite:
Ix = Ix1 + Ix2 + Ix3 = 2.645 × 108 mm4
Ix
kx =
= 128.6 mm
(A1 + A2 + A3 )
Problem 7.137
Determine Ixy .
Solution: (See figure in Problem 7.135). Use the solutions in
Problems 7.135 and 8.36. Divide the area into three parts:
Part (1): A1 = 160(40) = 6.4 ×
dx1
103
mm2 ,
Ixy2 = dx2 dy2 A2 = 9.6 × 106 mm4 .
Part (3): A3 = 120(40) = 4.8 × 103 mm2 ,
160
=
= 80 mm,
2
dx3 =
dy1 = 200 − 20 = 180 mm,
120
= 60 mm,
2
dy3 = 20 mm,
Ixxyy1 = 0,
from which
from which
7
4
Ixy1 = dx1 dy1 A1 + Ixxyy1 = 9.216 × 10 mm .
Part (2) A2 = (200 − 80)(40) = 4.8 × 103 mm2 ,
dx2 = 20 mm,
dy2 =
from which
Ixy3 = dx3 dy3 A3 = 5.76 × 106 .
The composite:
Ixy = Ixy1 + Ixy2 + Ixy3 = 1.0752 × 108 mm4
120
+ 40 = 100 mm,
2
Problem 7.138
Determine Ix and kx .
y
160
mm
Solution: The strategy is to use the relationship Ix = d2 A+Ixc ,
where Ixc is the area moment of inertia about the centroid. From this
Ixc = −d2 A + Ix . Use the solutions to Problems 7.135, 7.136, and
7.137. Divide the area into three parts and locate the centroid relative
to the coordinate system in the Problems 7.135, 7.136, and 7.137.
40 mm
x
200
mm
40
mm
40 mm
Part (1) A1 = 6.4 × 103 mm2 ,
120
mm
dy1 = 200 − 20 = 180 mm.
Part (2) A2 = (200 − 80)(40) = 4.8 × 103 mm2 ,
y
dx1 =
160
= 80 mm,
2
dy2 =
120
+ 40 = 100 mm,
2
Part (3) A3 = 120(40) = 4.8 × 103 mm2 ,
dx3 =
120
= 60 mm,
2
160
mm
dx2 = 20 mm,
40 mm
200
mm
x
40 mm
40 mm
dy3 = 20 mm.
120 mm
The total area is
A = A1 + A2 + A3 = 1.6 × 104 mm2 .
from which
The centroid coordinates are
Ixc = −y2 A + Ix = −1.866 × 108 + 2.645 × 108
x =
A1 dx1 + A2 dx2 + A3 dx3
= 56 mm,
A
y =
A1 dy1 + A2 dy2 + A3 dy3
= 108 mm
A
Problem 7.139
Determine Iy and ky .
Solution: The strategy is to use the relationship Iy = d2 A+Iyc ,
where Iyc is the area moment of inertia about the centroid. From this
Iyc = −d2 A + Iy . Use the solution to Problem 7.138. The centroid
coordinates are x = 56 mm, y = 108 mm, from which
Iyc = −x2 A + Iy = −5.0176 × 107 + 8.0213 × 107
kyc
= 3.0 × 107 mm4 ,
Iyc
=
= 43.33 mm
A
kxc
= 7.788 × 107 mm4
Ixc
=
= 69.77 mm
A
Determine Ixy .
Problem 7.140
Solution:
Use the solution to Problem 7.137. The centroid coor-
dinates are
x = 56 mm,
y = 108 mm,
from which Ixyc = −xyA + Ixy = −9.6768 × 107 + 1.0752 × 108
= 1.0752 × 107 mm4
Problem 7.141 Determine Ix and kx .
Solution: Divide the area into two parts:
y
Part (1): a triangle and Part (2): a rectangle. The area moment of
inertia for a triangle about the base is
Ix =
1
12
3 ft
4 ft
bh3 .
3 ft
The area moment of inertia about the base for a rectangle is
Ix =
1
3
x
bh3 .
Part (1) Ix1 =
Part (2) Ix2 =
1
12
1
3
4(33 ) = 9 ft2 .
3(33 )
= 27.
y
4 ft
3 ft
The composite: Ix = Ix1 + Ix2 = 36 ft4 . The area:
3 ft
1
4(3) + 3(3) = 15 ft2 .
2
Ix
=
= 1.549 ft.
A
A =
kx
Problem 7.142
x
Determine JO and kO .
Solution: (See Figure in Problem 7.141.) Use the solution to
Problem 7.141.
from which
Part (1): The area moment of inertia about the centroidal axis parallel
to the base for a triangle is
where A2 = 9 ft2 .
Iyc =
1
36
bh3 =
1
36
3(43 ) = 5.3333 ft4 ,
from which
Iy1 =
8
3
2
A1 + Iyc = 48 ft4 .
where A1 = 6 ft2 .
Part (2): The area moment of inertia about a centroid parallel to the
base for a rectangle is
Iyc =
1
12
bh3 =
1
12
3(33 ) = 6.75 ft4 ,
Iy2 = (5.5)2 A2 + Iyc = 279 ft4 ,
The composite: Iy = Iy1 + Iy2 = 327 ft4 , from which, using a result from
Problem 7.141,
JO = Ix + Iy = 327 + 36 = 363 ft4
JO
and kO =
= 4.92 ft
A
Problem 7.143
Determine Ixy .
Solution: (See Figure in Problem 7.141.) Use the results of the
solutions to Problems 7.141 and 7.142. The area cross product of
the moment of inertia about centroidal axes parallel to the bases for a
1 2 2
triangle is Ix y = 72
b h , and for a rectangle it is zero. Therefore:
Ixy1 =
1
72
(42 )(32 ) +
8
3
3
3
A1 = 18 ft4
and Ixy2 = (1.5)(5.5)A2 = 74.25 ft4 ,
Ixy = Ix y 1 + Ixy2 = 92.25 ft4
Problem 7.144
Determine Ix and kx .
y
4 ft
3 ft
x
3 ft
Solution: Use the results of Problems 7.141, 7.142, and 7.143.
The strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate
y =
A1 (1) + A2 (1.5)
= 1.3 ft.
A
From which
Ixc = −y2 A + Ix = 10.65 ft4
Ixc
and kxc =
= 0.843 ft
A
Problem 7.145
Determine JO and kO .
Solution: Use the results of Problems 7.141, 7.142, and 7.143.
The strategy is to use the parallel axis theorem and solve for the area
moment of inertia about the centroidal axis. The centroidal coordinate:
A1 83 + A2 (5.5)
x =
= 4.3667 ft,
A
from which
IY C = −x2 A + IY = 40.98 ft4 .
Using a result from Problem 7.144,
JO = IXC + IY C = 10.65 + 40.98 = 51.63 ft4
JO
and kO =
= 1.855 ft
A
y
4 ft
3 ft
3 ft
x
Determine IXY .
Problem 7.146
Solution: Use the results of Problems 7.141–7.145. The strategy
is to use the parallel axis theorem and solve for the area moment of
inertia about the centroidal axis. Using the centroidal coordinates
determined in Problems 7.144 and 7.145,
Ixyc = −xyA + Ixy = −85.15 + 92.25 = 7.1 ft4
Determine Ix and kx .
Problem 7.147
y
120
mm
20
mm
80
mm
x
40 mm
80 mm
Solution: Let Part 1 be the entire rectangular solid without the
hole and let part 2 be the hole.
Ix1 =
1 3
bh
3
y
m
20 m
where b = 80 mm
h = 120 mm
Ix1
y′
1
= (80)(120)3 = 4.608 × 107 mm4
3
For Part 2,
Ix 2
40 mm
1
1
= πR4 = π(20)4 mm4
4
4
Ix 2 = 1.257 × 105 mm4
Area = (80)(120)
– π (20)2
= 8343 mm2
120 mm
40 mm
Part 2
x′
Ix2 = Ix 2 + d2y A
dy = 80 mm
where A = πR2 = 1257 mm2
d = 80 mm
Ix2 = 1.257 × 105 + π(20)2 (80)2
6
Part 1
6
x
4
Ix2 = 0.126 × 10 + 8.042 × 10 mm
= 8.168 × 106 mm4 = 0.817 × 107 mm4
7
4
Ix = Ix1 − Ix2 = 3.79 × 10 mm
Area = hb − πR2 = (80)(120) − πR2
Area = 8343 mm2
Ix
kx =
= 67.4 mm
Area
80 mm
Determine JO and kO .
Problem 7.148
Solution: For the rectangle,
JO1 = Ix1 + Iy1 =
1 3
1
bh + hb3
3
3
7
7
JO1 = 4.608 × 10 + 2.048 × 10 mm
y
40 mm
4
y′
JO1 = 6.656 × 107 mm4
R = 20 mm
A1 = bh = 9600 mm2
For the circular cutout about x y x′
1
1
= πR4 + πR4
4
4
JO2
= Ix 2 + Iy
2
120 mm
A2
JO2
= 1.257 × 105 + 1.257 × 105 mm4
JO2
= 2.513 × 105 mm2
(h) 80 mm
A1
Using the parallel axis theorem to determine JO2 (about x, y)
JO2 = J0 2 + (d2x + d2y )A2
x
A2 = πR2 = 1257 mm2
JO2 = 1.030 × 107 mm4
80 mm
JO = JO1 − JO2
(b)
7
7
JO = 6.656 × 10 − 1.030 × 10 mm
4
JO = 5.63 × 107 mm4
JO
JO
kO =
=
Area
A1 − A2
kO = 82.1 mm
Determine Ixy .
Problem 7.149
Solution:
y
A1 = (80)(120) = 9600 mm2
80 mm
A2 = πR2 = π(20)2 = 1257 mm2
For the rectangle (A1 )
Ixy1 =
R = 20 mm
y′
1 2 2
1
b h = (80)2 (120)2
4
4
x′
Ixy1 = 2.304 × 107 mm2
A2
A1
For the cutout
Ix y 2 = 0
120 mm
A2
dy = 80 mm
A1
and by the parallel axis theorem
Ixy2 = Ix y 2 + A2 (dx )(dy )
Ixy2 = 0 + (1257)(40)(80)
6
4
Ixy2 = 4.021 × 10 mm
Ixy = Ixy1 − Ixy2
Ixy = 2.304 × 107 − 0.402 × 107 mm4
Ixy = 1.90 × 107 mm4
x
dx = 40 mm
Determine Ix and kx .
Problem 7.150
y
20
mm
120
mm
x
80
mm
40 mm
80 mm
Solution: We must first find the location of the centroid of the
total area. Let us use the coordinates XY to do this. Let A1 be
the rectangle and A2 be the circular cutout. Note that by symmetry
Xc = 40 mm
Rectangle1
Circle2
Area
9600 mm2
1257 mm2
Xc
40 mm
40 mm
y
Y
80 mm
Yc
60 mm
80 mm
R = 20 mm
A1 = 9600 mm2
120 mm
A2 = 1257 mm2
For the composite,
A1 Xc1 − A2 Xc2
Xc =
= 40 mm
A1 − A2
Yc =
x
80 mm
A1 Yc1 − A2 Yc2
= 57.0 mm
A1 − A2
Now let us determine Ix and kx about the centroid of the composite body.
X
40 mm
40 mm
Rectangle about its centroid (40, 60) mm
Ix1 =
1 3
1
bh =
(80)(120)3
12
12
7
3
Now to C → dy2 = 80 − 57 = 23 mm
Ixc2 = Ix2 + (dy2 )2 A2
Ix1 = 1.152 × 10 mm ,
Ixc2 = 7.91 × 105 mm4
Now to C
For the composite about the centroid
Ixc1 = Ix1 + (60 − Yc )2 A1
7
4
Ixc1 = 1.161 × 10 mm
Circular cut out about its centroid
A2 = πR2 = (20)2 π = 1257 mm2
Ix2 =
1
πR4 = π(20)4 /4
4
Ix2 = 1.26 × 105 mm4
Ix = Ixc1 − Ixc2
Ix = 1.08 × 107 mm4
The composite Area = 9600 − 1257 mm2
kx
= 8343 mm2
Ix
=
= 36.0 mm
A
Problem 7.151
Determine Iy and ky .
Solution: From the solution to Problem 7.150, the centroid of the
composite area is located at (40, 57.0) mm.
y
The area of the rectangle, A1 , is 9600 mm2 .
The area of the cutout, A2 , is 1257 mm2 .
The area of the composite is 8343 mm2 .
(1)
Rectangle about its centroid (40, 60) mm.
Iy1 =
1 3
1
hb =
(120)(80)3
12
12
+
Iy1 = 5.12 × 106 mm4
dx1 = 0
(2)
x
Circular cutout about its centroid (40, 80)
80 mm
Iy2 = πR4 /4 = 1.26 × 105 mm4
dx2 = 0
40 mm
Since dx1 and dx2 are zero. (no translation of axes in the xdirection), we get
80 mm
Iy = Iy1 − Iy2
6
4
Iy = 4.99 × 10 mm
Finally,
Iy
=
A1 − A2
ky
=
ky
= 24.5 mm
Problem 7.152
4.99 × 106
8343
Determine JO and kO .
Solution: From the solutions to Problems 7.151 and 7.152,
y
Ix = 1.07 × 107 mm4
Iy = 4.99 × 106 mm4
and A = 8343 mm2
JO = Ix + Iy = 1.57 × 107 mm4
JO
kO =
= 43.4 mm
A
20
mm
x
120 mm
80 mm
Problem 7.153
Determine Iy and ky .
y
12 in
x
20 in
Solution: Treat the area as a circular area with a half-circular
cutout: From Appendix B,
(Iy )1 =
and (Iy )2
1
π(20)4 in4
4
1
1
1
π(20)4 − π(12)4 = 1.18 × 105 in4 .
4
8
The area is A = π(20)2 − 12 π(12)2 = 1030 in2
Iy
1.18 × 105
so, ky =
=
A
1.03 × 103
= 10.7 in
Problem 7.154
Determine JO and kO .
Solution: Treating the area as a circular area with a half-circular
cutout as shown in the solution of Problem 7.153, from Appendix B,
(JO )1 = (Ix )1 + (Iy )1 =
and (JO )2 = (Ix )2 + (Iy )2 =
Therefore JO =
y
y
12
in.
1
= π(12)4 in4 ,
8
so Iy =
y
1
π(20)4 in4
2
1
π(12)4 in4 .
4
1
1
π(20)4 − π(12)4
2
4
= 2.35 × 105 in4 .
From the solution of Problem 7.153,
JO
2
A = 1030 in Ro =
A
2.35 × 105
=
= 15.1 in.
1.03 × 103
20 in.
2
x
x
20 in.
2 in.
x
Determine Iy and ky if h = 3 m.
Problem 7.155
Solution:
y
Break the composite into two parts, a rectangle and a
semi-circle.
1.2 m
For the semi-circle
Ix c =
Iy c =
9
π
−
8
8π
1
πR4
8
R4
d=
4R
3π
h
y′
x
y
d
x′
1.2 m
AC
d = 4R
3π
To get moments about the x and y axes, the (dxc , dyc ) for the semicircle are
dxc = 0,
dyc = 3 m +
4R
3π
AR
3m = h
and Ac = πR2 /2 = 2.26 m2
1
πR4
8
Iy c =
x
and Iyc = Iy c + d2xc A (dx = 0)
To get moments of area about the x, y axes, dxR = 0, dyR = 1.5 m
Iyc = Iy c = π(1.2)4 /8
0
IyR = Iy R + (dxR )2 (/ bh)
Iyc = 0.814 m4
For the Rectangle
Ix R =
1 3
bh
12
Iy R =
1 3
hb
12
IyR = Iy R =
1
(3)(2, 4)3 m4
12
IyR = 3.456 m2
AR = bh = 7.2 m2
AR = bh
Iy = Iyc + IyR
Iy = 4.27 m2
y′ y
To find ky , we need the total area, A = AR + Ac
A = 7.20 + 2.26 m2
2.4 m
h
3m
b
A = 9.46 m2
Iy
ky =
= 0.672 m
A
x′
x
Determine Ix and kx if h = 3 m.
Problem 7.156
Solution: Break the composite into two parts, the semi-circle and
the rectangle. From the solution to Problem 7.155,
Ix c =
π
9
−
8
8π
dyc =
3+
4R
3π
y
R4
Ac
m
Ac = 2.26 m2
Ixc = Ix c + Ac d2yc
Substituting in numbers, we get
R = 1.2 m
h=3m
b = 2.4 m
AR
3m
h
Ix c = 0.0717 m4
dyc = 3.509 m
and Ixc = Ix c + Ac d2y
Ixc = 27.928 m2
For the Rectangle h = 3 m, b = 2.4 m
x
2.4 m
yc′
Area: AR = bh = 7.20 m2
Ix R =
1 3
bh , dyR = 1.5 m
12
IxR = Ix R +
d2yR AR
xc′
R
Substituting, we get
Ix R = 5.40 m4
IxR = 21.6 m4
For the composite,
Ix = IxR + Ixc
Ix = 49.5 m4
Also kx =
Ix
= 2.29 m
AR + Ac
kx = 2.29 m
4R
3π
Problem 7.157
Determine the centroid of the area.
y
60 cm
x
60 cm
80 cm
Solution: The strategy is to develop useful general results for the
triangle and the rectangle.
y
The rectangle: The area of the rectangle of height h and width w is
w
A =
h dx = hw = 4800 cm2 .
0
60 cm
The x-coordinate:
w
w
x2
hx dx = h
=
2 0
0
Divide by the area: x =
w
2
1
2
80 cm
= 40 cm
Divide by the area: y =
The y-coordinate:
w
1
h2 dx =
2
0
1
h2 w.
2
Divide by the area: y = 12 h = 30 cm
The triangle: The area of the triangle of altitude a and base b is (assuming that the two sides a and b meet at the origin)
b
b
b
a
ax2
A =
y(x) dx =
− x + a dx = −
− ax
b
2b
0
0
0
ab
ab
= −
+ ab =
= 1800 cm2
2
2
Check: This is the familiar result. check.
The x-coordinate:
b
b
a
ax3
ax2
ab2
− x + a x dx = −
+
=
.
b
3b
2 0
6
0
Divide by the area: x =
x
hw2 .
b
3
= 20 cm
The y-coordinate:
b
2
1
a
y dA =
− x+a
dx
2
b
A
0
3 b
b
a
ba2
=−
− x+a
=
.
6a
b
6
0
x =
a
20
3
60 cm
cm. The composite:
xR AR + xT AT
40(4800) + 100(1800)
=
AR + AT
4800 + 1800
= 56.36 cm
y =
(30)(4800) + (20)(1800)
4800 + 1800
= 27.27 cm
Problem 7.158
Determine the centroid of the area.
y
40 mm
20 mm
40 mm
Solution:
Divide the object into five areas:
80 mm
(1)
(2)
(3)
(4)
(5)
The rectangle 80 mm by 80 mm,
The rectangle 120 mm by 80 mm,
the semicircle of radius 40 mm,
The circle of 20 mm radius, and
the composite object. The areas and centroids:
(1)
A1 = 6400 mm2 ,
x1 = 40 mm, y1 = 40 mm,
(2)
A2 = 9600 mm2 ,
x2 = 120 mm, y2 = 60 mm,
(3)
A3 = 2513.3 mm2 ,
x3 = 120 mm, y3 = 136.98 mm,
(4)
A4 = 1256.6 mm2 ,
x4 = 120 mm, y4 = 120 mm.
40 mm
(5)
The composite area: A = A1 + A2 + A3 − A4 =
17256.6 mm2 . The composite centroid:
80 mm
x
120 mm
160 mm
y
x=
A1 x1 +A2 x2 +A3 x3 −A4 x4
A
= 90.3 mm .
y=
A1 y1 +A2 y2 +A3 y3 −A4 y4
A
= 59.4 mm
20 mm
40 mm
x
120 mm
160 mm
Problem 7.159 The cantilever beam is subjected to a
triangular distributed load. What are the reactions at A?
y
200 N/m
x
A
10 m
line with intercept
Solution: The load distribution is a straight
w = 200 N/m at x = 0, and slope − 200
= −20 N/m2 . The sum
10
of the moments is
10
M = MA −
(−20x + 200)x dx = 0,
y
200
N/m
0
from which
MA = −
20 3
x + 100x2
3
x
10
10 m
= 3333.3 Nm.
0
The sum of the forces:
10
Fy = Ay −
(−20x + 200) dx = 0,
0
from which
and
10
Ay = −10x2 + 200x 0 = 1000 N,
Fx = Ax = 0
200 N/m
AX
MA
AY
10 m
Problem 7.160
of the frame?
What is the axial load in member BD
C 100 N/m
5m
B
D
5m
A
E
10 m
Solution: The distributed load is two straight lines: Over the interval 0 ≤ y ≤ 5 the intercept is w = 0 at y = 0 and the slope is
+ 100
= 20.
5
Over the interval 5 ≤ y ≤ 10, the load is a constant w = 100 N/m.
The moment about the origin E due to the load is
5
10
ME =
(20y)y dy +
100y dy,
0
from which
ME =
20 3
y
3
100 N/m
C
5m
B
D
5m
5
E
A
10 m
5
0
+
100 2
y
2
10
= 4583.33 N-m.
Cy
Cx
5
Cx
Check: The area of the triangle is
1
F1 = ( )(5)(100) = 250 N.
2
The area of the rectangle: F2 = 500 N. The centroid distance for the
triangle is
d1
2
= ( )5 = 3.333 m.
3
The centroid distance of the rectangle is d2 = 7.5 m. The moment
about E is
ME = d1 F1 + d2 F2 = 4583.33 Nm check.
The Complete Structure: The sum of the moments about E is
M = −10AR + ME = 0,
where AR is the reaction at A, from which AR = 458.33 N.
The element ABC: Element BD is a two force member, hence By = 0.
The sum of the moments about C:
MC = −5Bx − 10Ay = 0,
where Ay is equal and opposite to the reaction of the support, from
which
Bx = −2Ay = 2AR = 916.67 N.
Since the reaction in element BD is equal and opposite, Bx =
−916.67 N, which is a tension in BD.
By
Bx
Ay
Cy
Bx
By
Dy
Dx
Dx Dy
Ey
Ex
Problem 7.161 An engineer estimates that the maximum wind load on the 40-m tower in Fig. a is described
by the distributed load in Fig. b. The tower is supported
by three cables A, B, and C from the top of the tower to
equally spaced points 15 m from the bottom of the tower
(Fig. c). If the wind blows from the west and cables B
and C are slack, what is the tension in cable A? (Model
the base of the tower as a ball and socket support.)
200 N/m
B
N
A
40 m
15 m
C
400 N/m
(a)
Solution: The load distribution is a straight line with the intercept
w = 400 N/m, and slope −5. The moment about the base of the tower
due to the wind load is
40
MW =
(−5y + 400)y dy,
0
5
MW = − y 3 + 200y 2
3
40
θ = 90◦ − tan−1
15
40
200 N/m
B
40 m
A
15 m
= 213.33 kN-m,
0
clockwise about the base, looking North. The angle formed by the
cable with the horizontal at the top of the tower is
= 69.44◦ .
The sum of the moments about the base of the tower is
M = −MW + 40TA cos θ = 0,
(a)
400 N/m
(b)
200 N/m
θ
40 m
TA
from which
TA =
1
40 cos θ
MW = 15.19 kN
Fx
Fy
Problem 7.162 If the wind in Problem 7.161 blows
from the east and cable A is slack, what are the tensions
in cables B and C?
Solution: From Problem 7.161, the moment about the base of the
tower is MW = 213.33 kN-m, counterclockwise if the wind is from
the east and the observer is looking North. The angle in the horizontal
plane between the cables and the east is 60◦ . The sum of the moments
about the base is
M = MW − 40TB cos θ cos 60◦ − 40TC cos θ cos 60◦ = 0.
From symmetry, the tensions in the two cables are equal, from which
TB = TC =
1
80 cos θ cos 60◦
MW = 15.19 kN
(c)
(b)
400 N/m
N
C
(c)
Problem 7.163 Determine the y coordinate of the center of mass of the homogeneous steel plate.
y
20 mm
Solution: Divide the object into five areas: (1) The lower rectangle 20 by 80 mm, (2) an upper rectangle, 20 by 40 mm, (3) the
semicircle of radius 20 mm, (4) the circle of radius 10 mm, and (5)
the composite part. The areas and the centroids are tabulated. The last
row is the composite and the centroid of the composite. The composite
area is
A =
3
1
10 mm
20 mm
20 mm
Ai − A4 .
x
80 mm
The centroid:
3
x =
1
,
A
3
and y =
Ai xi − A4 x4
1
10 mm
y
20 mm
Ai yi − A4 y4
.
A
20 mm
The following relationships were used for the centroids: For a rectangle: the centroid is at half the side and half the base. For a semicircle,
the centroid is on the centerline and at 4R
from the base. For a circle,
3π
the centroid is at the center.
A, sq mm
1600
800
628.3
314.2
2714
Area
A1
A2
A3
A4
Composite
Problem 7.164
x, mm
40
60
60
60
48.2
20 mm
x
y, mm
10
30
48.5
40
21.3
80 mm
Determine Iy and ky .
Solution: Divide the section into two parts: Part (1) is the upper
rectangle 40 mm by 200 mm, Part (2) is the lower rectangle, 160 mm
by 40 mm.
Part (1) A1 = 0.040(0.200) = 0.008 m2 ,
y
40
mm
y1 = 0.180 m
x1 = 0,
Iy1 =
1
12
160
mm
0.04(0.2)3 = 2.6667 × 10−5 m4 .
Part (2): A2 = (0.04)(0.16) = 0.0064 m2 ,
x
80
mm
y2 = 0.08 m,
40
mm
x2 = 0,
Iy2 =
1
12
(0.16)(0.04)3 = 8.5 × 10−7 m4 .
The composite:
y
40 mm
2
A = A1 + A2 = 0.0144 m ,
160 mm
Iy = Iy1 + Iy2 ,
Iy = 2.752 × 10−5 m4 = 2.752 × 107 mm4 ,
Iy
and ky =
= 0.0437 m = 43.7 mm
A
80 40 80
mm mm mm
80
mm
Problem 7.165
Problem 7.164.
Solution:
Determine Ix and kx for the area in
Use the results in the solution to Problem 7.164. Part (1)
A1 = 0.040(0.200) = 0.008 m2 ,
y1 = 0.180 m,
1
12
Ix1 =
0.2(0.043 ) + (0.18)2 A1 = 2.603 × 10−4 m4 .
Part (2):
A2 = (0.04)(0.16) = 0.0064 m2 ,
y2 = 0.08 m,
Ix2 =
1
12
(0.04)(0.16)3 + (0.08)2 A2 = 5.461 × 10−5 m4 .
The composite: A = A1 + A2 = 0.0144 m2 , The area moment of
inertia about the x axis is
Ix = Ix1 + Ix2 = 3.15 × 10−4 m4 = 3.15 × 108 mm4 ,
Ix
and kx =
= 0.1479 m = 147.9 mm
A
Problem 7.166
y
Determine Ix and kx .
40
mm
x
160
mm
80
mm
Solution:
Use the results of the solutions to Problems 7.164–
7.165. The centroid is located relative to the base at
xc =
x1 A1 + x2 A2
= 0,
A
yc =
y1 A1 + y2 A2
= 0.1356 m.
A
The moment of inertia about the x-axis is
2
Ixc = −yC
A + IX = 5.028 × 107 mm4
Ixc
and kxc =
= 59.1 mm
A
Problem 7.167
Problem 7.166.
Determine JO and kO for the area in
Solution: Use the results of the solutions to Problems 7.164–
7.165. The area moments of inertia about the centroid are
Ixc = 5.028 × 10−5 m4
and Iyc = Iy = 2.752 × 10−5 m4 ,
from which
JO = Ixc + Iyc = 7.78 × 10−5 m4 = 7.78 × 107 mm4
JO
and kO =
= 0.0735 m
A
= 73.5 mm
40
mm
80
mm
y
40 mm
x
160 mm
80 40 80
mm mm mm
Problem 8.1 The coefficients of static and kinetic friction between the 0.4-kg book and the table are µs = 0.30
and µk = 0.28. A person exerts a horizontal force on
the book as shown.
(a) If the magnitude of the force is 1 N and the book
remains stationary, what is the magnitude of the friction force exerted on the book by the table?
(b) What is the largest force the person can exert without
causing the book to slip?
(c) If the person pushes the book across the table at a
constant speed, what is the magnitude of the friction
force?
Solution:
(a)
(b)
Fx :
F −f =0
Fy :
N −W =0
F =1N
Solving f = 1 N
fmax = µs N, µs = 0.3
N = 3.92 N, fmax = 1.18 N
Fmax = 1.18 N
(c)
For constant speed,
f = µk N = (0.28)(3.92)
f = 1.10 N
Thus, since F − f = 0
F = 1.10 N
W = mg = (0.4) 9.81 = 3.92 N
F
N
f
Problem 8.2 The 10.5-kg Sojourner rover, placed on
the surface of Mars by the Pathfinder Lander on July
4, 1997, was designed to negotiate a 45◦ slope without
tipping over.
(a) What minimum static coefficient of friction between
the wheels of the rover and the surface is necessary
for it to rest on a 45◦ slope? The acceleration due
to gravity at the surface of Mars is 3.69 m/s2 .
(b) Engineers testing the Sojourner on Earth want to
confirm that it will negotiate a 45◦ slope without
tipping over. What minimum static coefficient of
friction between the wheels of the rover and the surface is necessary for it to rest on a 45◦ slope on
Earth?
(a) Assume that slip is impending when the rover is on a 45◦
slope. The friction force f = µs N , and the free-body diagram is: The equilibrium equations are:
Fx = −µs N + mg sin 45◦ = 0,
Fy = N − mg cos 45◦ = 0.
Solution:
Summing the two equations, we obtain N − µs N = 0 so µs = 1 is the
minimum static coefficient of friction. (b) the solution in (a) is independent of
the value of g so µs = 1 is the minimum on earth also.
y
45°
µsN
mg
N
x
Problem 8.3 The coefficient of static friction between
the tires of the 8000-kg truck and the road is µs = 0.6.
(a) If the truck is stationary on the incline and α =
15◦ , what is the magnitude of the total friction force
exerted on the tires by the road?
(b) What is the largest value of α for which the truck
will not slip?
Solution:
y
α
W = mg
x
α
α
f
(a)
Fy :
Fx :
N
f − mg sin α = 0
N − mg cos α = 0
g = 9.81 m/s2 , α = 15◦ , m = 8000 kg
Solving, f = 20.3 kN also N = 75.8 kN
(b) Set f = µs N and solve for α in the basic eqns.
Fx : µs N − mg sin α = 0
Fy : N − mg cos α = 0
m = 8000 kg, g = 9.81 m/s2
Solving, α = 30.96◦
Problem 8.4 The coefficient of static friction between
the 5-kg box and the inclined surface is µs = 0.3. The
force F is horizontal and the box is stationary.
(a) If F = 40 N, what friction force is exerted on the
box by the inclined surface?
(b) What is the largest value of F for which the box will
not slip?
F
30°
Solution:
30°
F
y
mg
30°
x
30°
F
30°
f
N
(a) F = 40 N
Fx : f + F cos 30◦ − mg sin 30◦ = 0
Fy : N − F sin 30◦ − mg cos 30◦ = 0
(b)
f = −10.1 N (down the plane)
In the equilibrium eqns, set f = −µs N and treat F as unknown.
F is negative to resist F .
Solving, F = 52.0 N
Problem 8.5 In Problem 9.4, what is the smallest value
of the force F for which the box will not slip?
Solution: For this problem, facts up the plane
30°
mg
F
y
x
F
30°
30°
f
N
Fx :
f + F cos 30◦ − mg sin 30◦ = 0
Fy :
= N − mg cos 30◦ − F sin 30◦ = 0
Solving, F = 11.6 N
Problem 8.6 The device shown is designed to position
pieces of luggage on a ramp. It exerts a force parallel
to the ramp. The mass of the suitcase S is 9 kg. The
coefficients of friction between the suitcase and ramp are
µs = 0.20 and µk = 0.18.
(a) Will the suitcase remain stationary on the ramp
when the device exerts no force on it?
(b) What force must the device exert to start the suitcase
moving up the ramp?
(c) What force must the device exert to move the suitcase up the ramp at a constant speed?
S
20°
Solution:
y
(a)
20°
mg
x
f
F
(b)
(c)
Fx :
F − f − mg sin(20◦ ) = 0
Fy :
N − mg cos(20◦ ) = 0
Set f = −µs N (up the plane). If F ≤ 0, it will remain stationary.
Solving, we get F = 13.6 N (required to hold it stationary. No, it will not
remain stationary if F = 0.
Set f = µs N (down the plane) Solving, we get F = 46.8 N
Set f = µk N (down the plane) Solving, F = 45.1 N
20°
20°
N
µs = 0.20
µk = 0.18
m = 9 kg
Problem 8.7 The mass of the stationary crate is 40 kg.
The length of the spring is 180 mm, its unstretched length
is 200 mm, and the spring constant is k = 2500 N/m.
The coefficient of static friction between the crate and the
inclined surface is µs = 0.6. Determine the magnitude
of the friction force exerted on the crate.
20°
Solution: The magnitude of the force exerted on the crate by the
compressed spring is. (2500 N/m)(0.2 m − 0.18 m) = 50 N. The free
body diagram of the crate is shown. From the equilibrium equations
Fx = 50 − f + (40)(9.81) sin 20◦ = 0,
Fy = N − (40)(9.81) cos 20◦ = 0,
20°
we obtain N = 369 N, f = 184 N.
y
50 N
20°
mg
f
N
x
Problem 8.8 The coefficient of kinetic friction between the 40-kg crate and the floor is µk = 0.3. If the
angle α = 20◦ , what tension must the person exert on
the rope to move the crate at constant speed?
α
Solution:
mg
Fx :
T cos 20◦ − f = 0
Fy :
N + T sin 20◦ − mg = 0
Solving, T = 112.94 N
T
also f = 106.13 N
N = 353.77 N
α
α
f
N
m = 40 kg
α = 20◦
f = µk N
Problem 8.9 In Problem 9.8, for what angle α is the
tension necessary to move the crate at constant speed a
minimum? What is the necessary tension?
Solution: From the solution to Problem 9.8, we have
Differentiating with respect to α, we get
dT
T (sin α − µk cos α)
=
dα
(cos α + µk sin α)
mg
T
α
= 0, we get tan α = µk
Setting dT
dα
Solving, α = 16.7◦ . Substituting back into the equilibrium equations, we can
now solve for N and T .
T = 112.76 N,
N = 360 N
f = µkN
f = 108 N
N
Fx :
T cos α − µk N = 0
Fy :
+ T sin α + N − mg = 0
Solving the second eqn. for N and substituting into the first, we get
T cos α − µk mg + T µk sin α = 0.
α
Problem 8.10 Box A weighs 100 lb, and box B weighs
30 lb. The coefficients of friction between box A and
the ramp are µs = 0.30 and µk = 0.28. What is the
magnitude of the friction force exerted on box A by the
ramp?
A
B
30°
Solution:
The sum of the forces parallel to the inclined surface is
F = −A sin α + B + f = 0,
from which f = A sin α − B = 100 sin 30◦ − 30 = 20 lb
A
B
30°
B
α
A
f
N
Problem 8.11 In Problem 9.10, box A weighs 100 lb,
and the coefficients of friction between box A and the
ramp are µs = 0.30 and µk = 0.28. For what range of
the weights of the box B will the system remain stationary?
Solution: The upper and lower limits on the range are determined
by the weight required to move the box up the ramp, and the weight that
will allow the box to slip down the ramp. Assume impending slip. The
friction force opposes the impending motion. For impending motion
up the ramp the sum of forces parallel to the ramp are
F = A sin α − BMAX + µS A cos α = 0,
from which
BMAX
A
α
N
µ sN
BMAX = A(sin α + µs cos α)
= 100(sin 30◦ + 0.3 cos 30◦ ) = 75.98 lb
BMIN
from which
B = A(sin α − µs cos α)
= 100(sin 30◦ − 0.3 cos 30◦ ) = 24.02 lb
α
A
For impending motion down the ramp:
F = A sin α − BMIN − µs A cos α = 0,
µ sN
N
Problem 8.12 The mass of the box on the left is 30 kg,
and the mass of the box on the right is 40 kg. The coefficient of static friction between each box and the inclined
surface is µs = 0.2. Determine the minimum angle α
for with the boxes will remain stationary.
α
Solution: If the boxes slip when α is decreased, they will slip
toward the right. Assume that slip toward the right impends, the free
body diagrams are as shown.
The equilibrium equations are
Fx = T − 0.2 NA − (30)(9.81) sin α = 0,
(1)
Fy = NA − (30)(9.81) cos α = 0,
(2)
◦
Fx = −T − 0.2 NB + (40)(9.81) sin 30 = 0, (3)
Fy = NB − (40)(9.81) cos 30◦ = 0,
(4)
Summing Equations (1) and (3), we obtain −0.2 NA − 0.2 NB −
(30)(9.81) sin α + (40)(9.81) sin 30◦ = 0. Solving Equation (2)
for NA and Equation (4) for NB and substituting the results into
Equation (5) gives 15sin α + 3 cos α = 10 − 4 cos 30◦ . (6) Using
the identity cos α = 1 − sin2 α and solving Equation (6) for sin α,
we obtain sin α = 0.242, so α = 14.0◦
30°
α
30°
y
y
T
T
x
α
30°
(40)(9.81)
(30)(9.81)
0.2 NB
0.2 NA
NA
NB
x
Problem 8.13 In Problem 9.12, determine the maximum angle α for with the boxes will remain stationary.
If the boxes slip when α is increased, they will slip
toward the left. When slip toward the left impends, The free body
diagrams are as shown.
The equilibrium equations are
Fx = T + 0.2 NA − (30)(9.81) sin α = 0,
(1)
Fy = NA − (30)(9.81) cos α = 0,
(2)
Fx = −T + 0.2 NB + (40)(9.81) sin 30◦ = 0, (3)
Fy = NB − (40)(9.81) cos 30◦ = 0,
(4)
Solution:
Summing Equations (1) and (3), we obtain
0.2NA + 0.2 NB − (30)(9.81) sin α + (40)(9.81) sin 30◦ = 0. (5)
Solving Equation (2) for NA and Equation (4) for NB and substituting the results into Equation (5) gives 15 sin α − 3 cos α =
10 + 4 cos 30◦ . (6) Using the identity cos α =
1 − sin2 α
and solving Equation (6) for sin α, we obtain sin α = 0.956, so
α = 73.0◦
y
y
α
(30)(9.81)
T
T
30°
x
(40)(9.81)
0.2 NA
NA
NB
0.2 NB
x
Problem 8.14 The box is stationary on the inclined
surface. The coefficient of static friction between the
box and the surface is µs .
(a) If the mass of the box is 10 kg, α = 20◦ , β = 30◦ ,
and µs = 0.24, what force T is necessary to start
the box sliding up the surface?
(b) Show that the force T necessary to start the box
sliding up the surface is a minimum when tan β =
µs .
T
β
α
Solution:
T
T
β
mg
y
α
β
α
α
f
x
N
α = 20◦
µs = 0.24
m = 10 kg
g = 9.81 m/s2
(a)
Fx :
− T cos β + f + mg sin α = 0
Fy :
N + T sin β − mg cos α = 0
β = 30◦ , f = µs N
Substituting the known values and solving, we get
T = 56.5 N,
N = 64.0 N,
f = 15.3 N
Solving the 2nd equilibrium eqn for N and substituting for
f (f = µs N ) in the first eqn, we get
−T cos β + µs mg cos α − µs T sin β + mg sin α = 0
Differentiating with respect to β, we get
dT
T (sin β − µs cos β)
=
dβ
(cos β + µs sin β)
Setting
tan β = µs
dT
dβ
= 0, we get
Problem 8.15 To explain observations of ship launchings at the port of Rochefort in 1779, Coulomb analyzed
the system shown in Problem 9.14 to determine the minimum force T necessary to hold the box stationary on
the inclined surface. Show that the result is
(sin α − µs cos α)mg
T =
.
cos β − µs sin β
T
β
Solution:
mg
y
T
β
α
α
α is fixed, β is variable. Solve the second eqn for N and substitute into the first.
We get
0 = T (µs sin β − cos β) = mg(sin α − µs cos α)
α
or T =
x
mg(sin α − µs cos α)
(cos β − µs sin β)
To get the conditions for the minimum, set
N
f = µsN
dT
dβ
dT
T (sin β + µs cos β)
=
=0
dβ
(cos β − µs sin β)
Fx :
− T cos β + mg sin α − µs N = 0
For the min.
Fy :
N + T sin β − mg cos α = 0
tan β = −µs .
Note β is negative!
Problem 8.16 Two sheets of plywood A and B lie on
the bed of a truck. They have the same weight W , and
the coefficient of static friction between the two sheets
of wood and between sheet B and the truck bed is µs .
(a) If you apply a horizontal force to sheet A and apply
no force to sheet B, can you slide sheet A off the
truck without causing sheet B to move? What force
is necessary to cause sheet A to start moving?
(b) If you prevent sheet A from moving by applying a
horizontal force on it, what horizontal force on sheet
B is necessary to start it moving?
A
B
Solution:
(a)
The friction force exerted by sheet A on B at impending motion
is fAB = µs W . The friction force exerted by sheet B on the bed
of the truck is fBT = µs (2W ), since the normal force is due to
the weight of both sheets. Since fBT > fAB , the top sheet will
begin moving before the bottom sheet. Yes
The force required to start sheet A to move is
fAB
(a)
The force on B is the friction between A and B and the friction
between B and the truck bed. Thus the force required to start B
in motion is
fBT
FB = fAB + fBT = 3µs W.
(b)
fAB
W
F = fAB = µs W.
(b)
F
W
fAB
2W
W
FB
fBT
2W
=0
Problem 8.17 Suppose that the truck in Problem 9.16
is loaded with N sheets of plywood of the same weight
W , labeled (from the top) sheets 1, 2, . . . , N . The coefficient of static friction between the sheets of wood and
between the bottom sheet and the truck bed is µs . If you
apply a horizontal force to the sheets above it to prevent them from moving, can you pull out the ith sheet,
1 ≤ i ≤ N , without causing any of the sheets below it
to move? What force must you apply to cause it to start
moving?
The force holding the sheets below the ith sheet from
moving is the friction force between the bed of the truck and the bottom
sheet. The weight is N W , hence the force opposing motion of the
sheets below the ith sheet is FN = µs N W . The force causing motion
to start is the friction between the ith sheet and those below it, which
is FB = µs (iW ). The resultant force is FR = µs W (N − i). The
bottom sheets will begin to move when the resultant is zero, which
can only occur for the last sheet, i = N . Thus the ith sheet can be
extracted without the sheets below it moving. Yes The force required
to move the ith sheet is the force required to overcome the friction
force due to sheet above it and the sheet below it. The weight of the
sheets above is (i − 1)W and the weight on the sheet below it is iW .
The maximum total friction force opposing motion is
Solution:
(i−1)W
µs(i−1)W
Fi
µsiW
iW
Fi = µs (iW + (i − 1)W ) = µs W (2i − 1)
Problem 8.18 The masses of the two boxes are m1 =
45 kg and m2 = 20 kg. The coefficients of friction
between the left box and the inclined surface are µs =
0.12 and µk = 0.10. Determine the tension the man
must exert on the rope to pull the boxes upward at a
constant rate.
m1
30°
30°
Solution:
m1g
m2
y
30° y
Tman
30°
m1
N1
f = µ kN1
m1
T2
x
30°
T2
m2
m2
m2g
Equilibrium Eqns:
Mass 2:
F:
T2 − m2 g = 0
Mass 1:
Fx :
Fy :
T2 − TMAN + µk N1 + m1 g sin 30◦ = 0
N1 − m1 g cos 30◦ = 0
m1 = 45 kg,
m2 = 20 kg,
g = 9.81 m/s2 ,
and µk = 0.1.
Substituting these values into the eqns. and solving,
TMAN = 455 N
Problem 8.19 In Problem 9.18, for what range of tensions exerted on the rope by the man will the boxes remain stationary?
Solution:
(1)
(2)
We must look at two cases.
Impending slip up the plane
Impending slip down the plane.
In either case, |f | = µs N1 , only the sign changes
30°
m1
30°
m2
y
m1g
Tman
30°
30°
f2
T2
N1
f1
x
T2 = m2 g
f1 is used in case 1
f2 is used in case 2
Equilibrium Eqns:
Fx :
T2 − Tman ± f + m1 g sin(30◦ ) = 0
Fy :
N1 − m1 g cos(30◦ ) = 0
f = µs N1 = 0.12 N1 .
The (+) is used for case 1 and the (−) is used for case 2
Solving case 1, we get
Tman = 462.8 N
Solving case 2, we get
Tman = 371.0 N
Thus, the boxes are stationary for
371 N ≤ Tman ≤ 463 N
Problem 8.20 The coefficient of static friction between the two boxes is µs = 0.2, and between the lower
box and the inclined surface it is µs = 0.32. What is the
largest angle α for which the lower box will not slip?
W
W
α
Solution:
We need free body diagrams of both boxes
W
W
α
y
y
α
m 2g
NU
T
x
x
fU
fU
1
α
fL
α NL
m1g
fU = 0.2 NU
fL = 0.32 NL
m1 = m2 = m
Lower Box
Fx :
fL + fU − mg sin α = 0
Fy :
NL − NU − mg cos α = 0
Upper Box
m1 = m2 = m
Fx :
T − fU − mg sin α = 0
Fy :
NU − mg cos α = 0
Substituting in known values and solving, we get
α = 40.0◦
NU
Problem 8.21 The coefficient of static friction between the two boxes and between the lower box and
the inclined surface is µs . What is the largest force F
that will not cause the boxes to slip?
W
F
2W
α
Solution: At impending motion, the sum of the forces parallel to
the inclined surface for the upper box is
FP = F − µs W cos α + W sin α − T = 0,
where T is the tension in the string, and the friction force opposes
impending motion in the direction of F . For the lower box,
F = 2W sin α + µs (3W cos α) + µs W cos α − T = 0,
W
F
2W
α
where the friction force opposes impending motion in the direction of
T . Combining:
F = µs (5W cos α) + W sin α = W (5µs cos α + sin α)
W
F
Wcos α
µsWcosα
T
µsWcosα
Wcos α
T
2W
3µsWcosα
3Wcosα
Problem 8.22 Consider the system shown in Problem 9.21. The coefficient of static friction between the
two boxes and between the lower box and the inclined
surface is µs . If F = 0, the lower box will slip down the
inclined surface. What is the smallest force F for which
the boxes will not slip?
Solution: The solution is obtained by the same procedure as in
Problem 10.21, with the exception that the friction forces now oppose
impending motion in the direction of T for the upper box, and impending motion down the surface, for the lower box. The sums of forces
parallel to the inclined surface for the two boxes are:
FP = F + µs W cos α + W sin α − T = 0
and
F = 2W sin α − µs (3W cos α) − µs W cos α − T = 0.
Combining:
F = −µs W cos α + W sin α − 4µs W cos α = W (sin α − 5µs cos α)
F
µsWcosα
T
W
Wcos α
Wcos α
µsWcosα
T
2W
3Wcosα
3µsWcosα
Problem 8.23 A sander consists of a rotating disk with
sandpaper bonded to the outer surface. The normal force
exerted on the workpiece A by the sander is 30 lb. The
workpiece A weighs 50 lb. The coefficients of friction
between the sander and the workpiece A are µs = 0.65
and µk = 0.60. The coefficients of friction between the
workpiece A and the table are µs = 0.35 and µk = 0.30.
Will the workpiece remain stationary while it is being
sanded?
Solution:
(1)
(2)
Two possible situations can cause impending motion:
before the contact surface has begun slipping with respect to the
workpiece (that is, as the sanding wheel is brought into contact
with the work piece), so that the static coefficient of friction
applies between the sanding wheel and the workpiece.
After the sanding wheel has begun slipping relative to the workpiece, when the kinetic coefficient of friction applies between
the sander and the workpiece. In the first situation, the force
inducing motion of the workpiece is
F = µs 30 = 19.5 lb.
The force resisting motion is FA = µs (30 + 50) = 28 lb. Thus
the workpiece will not slip. The second situation is less severe,
since the force inducing motion is F = 30µk = 18 lb, and the
force opposing motion is the same. The workpiece will remain
stationary.
A
30 lb
µ (30)
50 lb
f
(30 + 50) lb
A
Problem 8.24 Suppose that you want the bar of length
L to act as a simple brake that will allow the workpiece
A to slide to the left but will not allow it to slide to the
right no matter how large a horizontal force is applied
to it. The weight of the bar is W , and the coefficient
of static friction between it and the workpiece A is µs .
You can neglect friction between the workpiece and the
surface it rests on.
(a) What is the largest angle α for which the bar will
prevent the workpiece from moving to the right?
(b) If α has the value determined in (a), what horizontal
force is required to start the workpiece A toward the
left at a constant rate?
L
α
A
Solution:
(a)
To resist motion to the right no matter how large the horizontal
force requires a very large friction force. For impending motion
to the right, the sum of the moments about the bar hinge is
M =+
L
W
L sin α − FN L sin α + f L cos α = 0,
2
α
where FN is the normal force and f is the friction force resisting
the impending motion.
Noting that f = µs FN , the sum of moments yields
FN =
W sin α
.
2(sin α − µs cos α)
If the denominator vanishes, the normal force and hence the friction force become as large as required. Thus sin α − µs cos α =
0, from which µs = tan α, or α = tan−1 (µs )
(b) For impending motion to the left, the sum of moments about the
bar hinge is
L
M = +W
sin α − FN L sin α − f L cos α = 0,
2
where the friction force opposes the impending motion. Noting
FN = µf , then
s
f =
2
A
W sin α
sin α
µs
+ cos α
=
W sin α
µs W
=
4 cos α
4
is the force required to start motion to the left.
α
α
W
f
FN
W
f
FN
(a)
(b)
Problem 8.25 The coefficient of static friction between the 20-lb bar and the floor is µs = 0.3. Neglect
friction between the bar and the wall.
(a) If α = 20◦ , what is the magnitude of the friction
force exerted on the bar by the floor?
(b) What is the maximum value of α for which the bar
will not slip?
L
α
Solution:
The sum of the moments about the upper end of the
bar is
WL
sin α + FN L sin α − f L cos α = 0,
2
−W
+ FN sin α
2
from which f =
.
cos α
M =−
L
α
The sum of the forces in the vertical direction
FY = −W + FN = 0,
from which FN = W.
Substitute:
f =
W sin α
= 10 tan 20◦ = 3.64 lb
2 cos α
(b) At impending slip, f = µs FN = µs W , from which, substituting
above,
1
µs =
tan α, or α = tan−1 (2µs ) = tan−1 (0.6) = 31◦
2
R
α
W
f
FN
Problem 8.26 The masses of the ladder and person
are 18 kg and 90 kg, respectively. The center of mass
of the 4-m ladder is at its midpoint. If α = 30◦ , what
is the minimum coefficient of static friction between the
ladder and the floor necessary for the person to climb
to the top of the ladder? Neglect friction between the
ladder and the wall.
α
x
Solution: The weight of the ladder is W = 18 g = 176.58 N.
The weight of the person is P = 90 g = 882.9 N. Let h be the
distance along the ladder of the person’s center of mass, and L be
the length of the ladder. The horizontal distance is. The sum of the
moments about the top of the ladder:
M = P (L sin α − x) − FN L sin α + W
L
sin α + f L cos α = 0.
2
From the sum of the forces,
FY = FN − W − P = 0, from
which the normal force at the foot of the ladder is FN = W + P .
Substitute, solve for the friction force, and reduce algebraically:
h
1
f =
P + W tan α.
L
2
α
h
At the top of the ladder, L
= 1, hence
W
f = P+
tan α = (883 + 88.3)(0.5774) = 560.72 N
2
x
At impending slip, f = µs FN = µs (P + W ), from which
µs =
f
560.72
=
= 0.5292
P +W
1059.48
R
P
α
W
f
FN
x
Problem 8.27 In Problem 9.26, the coefficient of static
friction between the ladder and the floor is µs = 0.6. The
masses of the ladder and the person are 18 kg and 100 kg,
respectively. The center of mass of the 4-m ladder is at
its midpoint. What is the maximum value of α for which
the person can climb to the top of the ladder? Neglect
friction between the ladder and the wall.
Solution: The solution in Problem 9.26 for the friction force is
f =
h
1
P+ W
L
2
tan α and f = µs (W + P ).
At impending slip
µs (W + P )
.
tan α = h
P + 12 W
L
At the top of the ladder
h
= 1, and tan α = 0.6495, or α = tan−1 (0.6495) = 33◦
L
Problem 8.28 In Problem 9.26, the coefficient of static
friction between the ladder and the floor is µs = 0.6, and
α = 35◦ . The center of mass of the 4-m ladder is at its
midpoint, and its mass is 18 kg.
(a) If a football player with a mass of 140 kg attempts
to climb the ladder, what maximum value of x will
he reach? Neglect friction between the ladder and
the wall.
(b) What minimum friction coefficient would be required for him to reach the top of the ladder?
The weight of the football player is P = 140 g =
1373.4 N, and the weight of the ladder is W = 176.58 N. From the
solution to Problem 9.26,
h
1
f =
P + W tan α
L
2
Solution:
and f = µs (W + P ).
Substitute and solve for
µs (W + P ) − W
tan α
h
2
=
= 0.90277.
L
P tan α
From 9.26, we see that x = h sin α, from which
h
L sin α = 2.07 m
x =
L
(b) Solve the above for the static friction coefficient:
h
P + 12 W tan α
L
µs =
.
(W + P )
h
At the top of the ladder, L
= 1, and
1
P + 2 W tan α
µs =
,
(P + W )
µs = 0.66
Problem 8.29 The disk weighs 50 lb. Neglect the
weight of the bar. The coefficients of friction between
the disk and the floor are µs = 0.6 and µk = 0.4.
(a) What is the largest couple M that can be applied to
the stationary disk without causing it to start rotating?
(b) What couple M is necessary to rotate the disk at a
constant rate?
Solution: The normal force at the point of contact is found from
the sum of moments about the pin support.
M = −8(100) − 20(50) + 20FN = 0,
8 in
12 in
100 lb
5 in
M
100 lb
8 in
12 in
5 in
from which FN = 90 lb. The friction force is f = µs FN . The
moment exerted by the friction force is
MF = µs RFN = 0.6(5)(90) = 270 in lb
M
This is the moment to be overcome at impending slip.
(b) The moment required to rotate the disk at a constant rate is
MK = µk RFN = 0.4(5)(90) = 180 in lb
8 in
12 in
5 in
100 lb
M
f
Problem 8.30 The cylinder has weight W . The coefficient of static friction between the cylinder and the floor
and between the cylinder and the wall is µs . What is the
largest couple M that can be applied to the stationary
cylinder without causing it to rotate?
R
M
Solution: Assume impending slip. The force opposing rotation
is the sum of the friction force at the wall and at the floor. Denote the
normal force at the wall by FN W and the normal force on the floor by
FN F . From the sum of forces:
Fy = µs FN W + FN F − W = 0,
and
Fx = FN W − µs FN F = 0.
R
M
Solve these two simultaneous equations to obtain:
FN F =
W
,
1 + µ2s
and FN W =
µs W
.
1 + µ2s
µsFNW
The sum of moments about the center of the cylinder is
MC = Mapp − µs RFN W − µs RFN F = 0.
Substitute and solve:
1 + µs
Mapp = µs RW
.
1 + µ2s
At impending slip, this is the maximum moment that can be applied
to the cylinder.
FN
M
FNW
W
µsFNF
FNF
Problem 8.31 The cylinder has weight W . The coefficient of static friction between the cylinder and the floor
and between the cylinder and the wall is µs . What is the
largest couple M that can be applied to the stationary
cylinder without causing it to rotate?
R
α
M
Solution: Assume impending slip. Denote the normal force at
the wall by FN W and the normal force at the floor by FN F . The
projection of the friction force at the wall on an x-y coordinate system
is
f W = µs |FN W |(i cos(90 − α) + j sin(90 − α))
= µs |FN W |(i sin α + j cos α).
R
M
α
The projection of the normal force at the wall on an x-y coordinate
system is
FN W = |FN W |(i cos α − j sin α).
The sum of the forces:
Fy = |FN W |(µs cos α − sin α) + |FN F | − |W| = 0,
and
Fx = |FN W |(cos α + µs sin α) − µs |FN F | = 0.
Solve these simultaneous equations to obtain:
|FN F | =
and |FN W | =
|W|(cos α + µs sin α)
,
(1 + µ2s ) cos α
FNW
α
W
µs |W|
.
(1 + µ2s ) cos α
The sum of the moments about the center of the cylinder is
MC = Mapp − µs R|FN F | − µs R|FN W | = 0.
Substitute and reduce algebraically:
Mapp =
fW
µs RW (cos α + µs sin α + µs )
.
(1 + µ2s ) cos α
This is the maximum applied moment at impending slip. Check: This
reduces to the solution of Problem 9.30 when α = 0, as it should.
check.
µs |FNF |
FNF
Problem 8.32 Suppose that α = 30◦ in Problem 9.31
and that a couple M = 0.5RW is required to turn the
cylinder at a constant rate. What is the coefficient of
kinetic friction?
Solution:
Substitute the angle and the moment into the solution
of Problem 9.31 to obtain:
√
√
µk 23 + µ2k + µk
µk (1 + 3µk )
√
0.5 =
=
.
(1 + µ2k )
(1 + µ2 ) 3
k
2
This reduces to the quadratic equation µ2k + 2bµk − c = 0, where
b =c=
1
√
.
(2 3 − 1)
√
The solution is µk = −b ± b2 + c. Substitute numerical values:
µk = 0.3495, or µk = −1.1612. The negative value has no meaning here.
Problem 8.33 The disk of weight W and radius R is
held in equilibrium on the circular surface by a couple
M . The coefficient of static friction between the disk
and the surface is µs . Show that the largest value M can
have without causing the disk to slip is
M
µs RW
M =
.
1 + µ2s
This is an inclined plane problem. Let α be the angle
at the point of contact. From the sum of forces: the normal force is
FN = W cos α, and the friction force is f = µs FN = W sin α,
from which µs = tan α. The sum of moments about the center of the
disk yields M = f R = µs RW cos α. Noting that
Solution:
cos α = √
1
1 + tan2 α
M
,
µs RW
,
then M = 1 + µ2s
which is the moment at impending slip.
M
f
W
FN
α
Problem 8.34 The coefficient of static friction between the jaws of the pliers and the gripped object is
µs . What is the largest value of the angle α for which
the gripped object will not slip? (Neglect the object’s
weight.)
α
Solution: Choose an x-y coordinate system such that the x-axis
bisects the angle α. Define
β =
α
.
2
The projection of the normal forces on the x-y system is, for the top:
FN T = |FN T |(−i sin β − j cos β).
For the bottom:
α
FN B = |FN B |(−i sin β + j cos β).
The projection of the friction forces on the x-y system is, for the top:
fT = µs |FN T |(i cos β − j sin β).
For the bottom:
fB = µs |FN B |(i cos β + j sin β).
The forces tending to expel the gripped object are the components
of the normal forces in the negative x direction, and the components
tending to retain the gripped object are the friction forces in the positive
x-direction. These must balance:
Fx = −|FN B | sin β − |FN T | sin β + µs |FN T | cos β
+ µs |FN B | cos β = 0,
from which
|FN B |(− sin β + µs cos β) + |FN T |(− sin β + µs cos β) = 0.
From symmetry, |FN B | = |FN t |, since the weight of the object
is neglected. For non-trivial values of the normal forces, − sin β +
µs cos β = 0, from which µs = tan β, or β = tan−1 (µs ). Noting
β =
α
,
2
α = 2 tan−1 (µs )
β
FNT
fT
fB
β
FNB
Problem 8.35 The stationary disk, of 300-mm radius,
is attached to a pin support at D. The disk is held in place
by the brake ABC in contact with the disk at C. The
hydraulic actuator BE exerts a horizontal 400-N force
on the brake at B. The coefficients of friction between
the disk and the brake are µs = 0.6 and µk = 0.5. What
couple must be applied to the stationary disk to cause it
to slip in the counterclockwise direction?
C 300
mm
200 mm
E
D
B
200 mm
A
200
mm
Solution: Assume impending slip. For counterclockwise motion
the friction force f = µs FN opposes the impending slip, so that it acts
on the brake in a downward direction, producing a negative moment
(clockwise) about A. The sum of the moments about A:
MA = −0.2(400) + (0.4 − 0.2µs )FN ,
from which FN = 285.7 N. The sum of the moments about the center
of the disk:
MD = M − 0.3(µs )FN = 0,
C
D
200 mm
E
B
200 mm
300
mm
A
200
mm
from which M = 51.43 N m.
f
M
FN
FN
200 mm
400 N
f
200 mm
200
mm
Problem 8.36 What couple must be applied to the
stationary disk in Problem 9.35 to cause it to slip in a
clockwise direction?
Solution: Assume impending slip. For clockwise motion the friction force f = µs FN opposes the impending slip, so that it acts on
the brake in an upward direction, producing a positive moment (counterclockwise) about A. The sum of the moments about A:
MA = −0.2(400) + (0.4 + 0.2µs )FN = 0,
from which FN = 153.85 N. The sum of the moments about the
center of the disk:
MD = M − 0.3µs FN = 0,
from which M = 27.69 N m
300
mm
Problem 8.37 The mass of block B is 8 kg. The
coefficient of static friction between the surfaces of the
clamp and the block is µs = 0.2. When the clamp is
aligned as shown, what minimum force must the spring
exert to prevent the block from slipping out?
45°
160 mm
200 mm
B
100
mm
Solution: The free-body diagram of the block when slip is impending is shown. From the equilibrium equation
45°
µs FT + µs (FT + W cos α) − W cos α = 0,
160 mm
we obtain
FT =
=
w(1 − µs ) cos α
2µs
200 mm
(8)(9.81)(1 − 0.2) cos 45◦
2(0.2)
= 111 N.
B
100
mm
The free-body diagram of the upper arm of the clamp is shown. Summing moments about the upper end,
0.16Fs + 0.1µs FT − 0.36FT = 0,
the force exerted by the spring is
Fs
W
0.36FT − 0.1µs FT
=
0.16
=
µsFT
FT
FT +Wcosα
100 mm
[0.36 − 0.1(0.2)]111
0.16
= 236 N.
µsFT
Problem 8.38 By altering its dimensions, redesign
the clamp in Problem 9.37 so that the minimum force
the spring must exert to prevent the block from slipping
out is 180 N. Draw a sketch of your new design.
Solution:
This problem does not have a unique solution.
µs (FT + Wcos α )
FS
FT
160
200 mm
mm
Problem 8.39 The horizontal bar is attached to a collar
that slides on the smooth vertical bar. The collar at P
slides on the smooth horizontal bar. The total mass of
the horizontal bar and the two collars is 12 kg. The
system is held in place by the pin in the circular slot. The
pin contacts only the lower surface of the slot, and the
coefficient of static friction between the pin and the slot
is 0.8. If the system is in equilibrium and y = 260 mm,
what is the magnitude of the friction force exerted on the
pin by the slot?
P
y
300 mm
Solution: The free body diagram of the horizontal bar and right
collar is as shown, where m1 is the mass of the horizontal bar and right
collar, N1 is the normal force exerted by the vertical bar, and N2 is
the force exerted by the left collar. From the equilibrium equations
Fx = −N1 = 0,
Fx = N2 − m1 g = 0,
P
we see that N2 = m1 g. The free body diagram of the left collar is
as shown, where m2 is the mass of the left collar and N , f are the
normal and friction forces exerted by the curved slot.
y
300 mm
y = 260 mm = (300 mm) sin θ,
so the angle θ = 60.1◦ .
From the equilibrium equations,
Fx = −f + m2 g cos θ + N2 cos θ = 0,
Fy = N − m2 g sin θ − N2 sin θ = 0,
y
N2
N1
m1g
we obtain
f = (m2 g + N2 ) cos θ = (m2 + m1 ) cos θ = (12)(9.81) cos 60.1◦
y
N2
f
= 58.7 N.
θ
N
x
m2g
Problem 8.40 In Problem 9.39, what is the minimum
height y at which the system can be in equilibrium?
Solution: From the solution of Problem 9.39, the friction and
normal forces exerted on the pin by the circular slot are
f = (m2 g + N2 ) cos θ,
N = (m2 g + N2 ) sin θ,
so
f
N
= cot θ. When slip impends,
f = µs N = 0.8 N,
so 0.8 = cot θ and θ = 51.3◦ . The height y = 300 sin θ = 234 mm.
x
Problem 8.41 The rectangular 100-lb plate is supported by the pins A and B. If friction can be neglected at
A and the coefficient of static friction between the pin at
B and the slot is µs = 0.4, what is the largest angle α
for which the plate will not slip?
α
2 ft 3 in
B
A
2 ft
2 ft
Choose a coordinate system with the x- axis parallel to
the rail. The sum of the moments about A is
MA = −2W cos α − 2.25W sin α + 4B = 0,
Solution:
from which
α
B
A
W
(2 cos α + 2.25 sin α).
B =
4
The component of weight causing the plate to slide is F = W sin α.
This must be balanced by the friction force: 0 = −W sin α + µs B,
from which
W sin α
W
=
(2 cos α + 2.25 sin α).
µs
4
Reduce algebraically to obtain
µs
α = tan−1
= 14.47◦
2 − 1.125µs
2 ft 3 in
2 ft
2 ft
2 ft
2 ft
µ sB
2.25 ft
B
A
W
Problem 8.42 If you can neglect friction at B in Problem 9.41 and the coefficient of static friction between the
pin at A and the slot is µs = 0.4, what is the largest angle
α for which the plate will not slide?
Solution: The normal force acts normally to the slots, and the
friction force acts parallel to the slot. Choose a coordinate system
with the x-axis parallel to the slots. The normal component of the
reaction at A is found from the sum of the moments about B:
MB = −2.25W sin α + 2W cos α − 4A = 0,
from which
AN =
W
(−2.25 sin α + 2 cos α).
4
The force tending to make the plate slide is F = −W sin α. This is
balanced by the friction force at A,
0 = −W sin α + µs AN ,
from which
W sin α
W
=
(−2.25 sin α + 2 cos α).
µs
4
Reduce algebraically to obtain
µs
= 9.27◦
α = tan−1
2 − 1.125µs
Check: The normal reactions at A and B are unequal: as the slots are inclined
from the horizontal, the parallel component of the gravity force reduces the
normal force at A, and increases the normal force at B. check.
Check: The sum of the
reactions at A and B are AN + BN = W cos α.
check. The magnitude (AN + BN )2 + (µs AN )2 = W , hence the system
is in equilibrium at impending slip. check.
2 ft
2 ft
µs AN
2.25 ft
BN
AN
W
Problem 8.43 The airplane’s weight is W = 2400 lb.
Its brakes keep the rear wheels locked, and the coefficient
of static friction between the wheels and the runway is
µs = 0.6. The front (nose) wheel can turn freely and
so exerts only a normal force on the runway. Determine
the largest horizontal thrust force T the plane’s propeller
can generate without causing the rear wheels to slip.
Solution: The free body diagram when slip of the rear wheels
impends is shown. From the equilibrium equations
fx = −T + µs B = 0,
fy = A + B − W = 0,
MptA = 4T − 5W + 7B = 0,
we obtain
A = 1120 lb,
B = 1280 lb,
and T = 766 lb.
T
4 ft
W
A
5 ft
B
2 ft
y
T
W
4 ft
µsB
A
B
5 ft
2 ft
x
T
4 ft
W
A
5 ft
B
2 ft
Problem 8.44 The refrigerator weighs 350 lb. The
distances h = 60 in. and b = 14 in. The coefficient of
static friction at A and B is µs = 0.24.
(a) What force F is necessary for impending slip?
(b) Will the refrigerator tip over before it slips?
F
h
A
B
b
Solution: The normal forces on the right and left supports are
found: The sum of moments about support A:
MA = −bW − hF + 2bB = 0,
b
F
from which
B =
bW + hF
.
2b
h
The sum of forces:
FY = A + B − W = 0,
b
b
from which
A =W −B =
B
A
bW − hF
.
2b
(a)
The friction forces must balance the applied force at impending
slip:
bW − hF
0 = F − µs A − µ s B = F − µ s
2b
bW + hF
− µs
,
2b
from which F = µs W = 0.24(350) = 84 lb.
If the normal force at A approaches zero before motion occurs,
the refrigerator will start to tip:
b
14
Ftip =
W =
350 = 81.67 lb.
h
60
(b)
F
h
W
µs A
A
b
µs B
B
b
Since Ftip < Fmove , the refrigerator will tip.
Yes.
Problem 8.45 If you want the refrigerator in Problem 9.44 to slip before it tips over, what is the maximum
height h at which you can push it?
Solution:
From the solution to Problem 9.44, the tipping force must be
equal to or greater than the moving force, Fmove = 84 lb. Thus when the
normal force at A approaches zero, the tipping force must equal or exceed 84 lb,
from which
W
350
h≤b
= 14
≤ 58.33 in
Fmove
84
Problem 8.46 To obtain a preliminary evaluation of
the stability of a turning car, imagine subjecting the stationary car to an increasing lateral force F at the height
of its center of mass, and determine whether the car will
slip (skid) laterally before it tips over. Show that this will
be the case if b/h > 2µs . (Notice the importance of the
height of the center of mass relative to the width of the
car. This reflects on recent discussions of the stability of
sport utility vehicles and vans that have relatively high
centers of mass.)
F
h
b_
2
b_
2
b_
2
b_
2
Solution:
y
F
mg
h
F
h
A
fL
NL
B
b
2
b
2
fR
NR
EQUILIBRIUM Eqns:
Fx :
F − fL − fR = 0
Fy :
NL + NR − mg = 0
MA :
−hF + bNR −
b
mg = 0
2
Assume skid and tip simultaneously.
f L = µs N L ,
fR = µs NR (skid)
and NL = 0 (tip), ∴ fL = 0.
fR = µs mg.
The equilibrium eqns become
F = fR = µs NR = µs mg
and the moment eqn. uses
−h(µs mg) + b(mg) −
or
b
h
For
b
h
b
h
b
h
b
mg = 0
2
= 2µs
b
h
> 2µs , slip before tip
< 2µs , tip before slip
big N low cm, relative to track width
small N high cm, relative to track width
x
Problem 8.47 The man exerts a force P on the car at an
angle α = 20◦ . The 1760-kg car has front wheel drive.
The driver spins the front wheels, and the coefficient of
kinetic friction is µk = 0.02. Snow behind the rear tires
exerts a horizontal resisting force S. Getting the car to
move requires overcoming a resisting force S = 420 N.
What force P must the man exert?
Solution:
Fx :
S − µk NF − P cos α = 0
Fy :
NR + NF − mg − P sin α = 0
MA :
−(1.62)mg + 2.55 NF
+(0.90)P cos α − (3.40)P sin α = 0
α = 20◦ ,
m = 1760 kg,
g = 9.81 m/s2
S = 420 N,
µk = 0.02
3 eqns in 3 unknowns (NR , NF , and P )
Solving the equations, we get P = 213 N
P
α
NR = 6.34 kN
NF = 11.00 kN
0.90 m
S
1.62 m
2.55 m
3.40 m
P
α
0.90 m
S
1.62 m
2.55 m
3.40 m
y
mg
F
α
1.62 m
S
0.90 m
B
A
2.55 m
µ k NF
NR
NF
x
Problem 8.48 In Problem 9.47, what value of the angle α minimizes the magnitude of the force P the man
must exert to overcome the resisting force S = 420 N
exerted on the rear tires by the snow? What force must
he exert?
Solution: From the solution to Problem 9.47, we have
Use this eqn to find
S − µk NF − P cos α = 0
(1)
NR + NF − mg − P sin α = 0
(2)
−3.40
dP
sin α − 3.40 P cos α = 0
dα
or
dP
2.55
cos α
0.90 cos α − 3.40 sin α −
dα
µk
µk = 0.02,
S = 420 N,
m = 1760 kg,
From Eqn (1),
1
(S − P cos α) (a)
µk
NR = −NF + mg + P sin α
1
(S − P cos α) + mg + P sin α (b)
µk
Substitute (a) and (b) into (3)
We get
−1.62 mg + 2.55
k
tan α = From Eqn (2),
or NR = −
2.55
sin α − 0.90 sin α − 3.40 cos α = 0
µk
2.55
−P
− 0.90 sin α − 3.40 cos α
dP
µk
=0=
dα
cos α
0.90 cos α − 3.40 sin α − 2.55
µ
+P
and g = 9.81 m/s2 .
NF =
and set it to zero.
2.55
dP
dP
−
cos α + P sin α + 0.90
cos α − 0.90 P sin α
µk
dα
dα
−1.62 mg + 2.55 NF + 0.90P cos α − 3.40P sin α = 0 (3)
where
dP
dα
1
µk
(S − P cos α)
+0.90 P cos α − 3.40 P sin α = 0
3.40
2.55
µk
− 0.90
Solving, α = 1.54◦
Substituting this back into eqns (1), (2), and (3), and solving, we get
P = 202 N
Problem 8.49 The coefficient of static friction between the 3000-lb car’s tires and the road is µs = 0.5.
Determine the steepest grade (the largest value of the
angle α) the car can drive up at constant speed if the car
has (a) rear-wheel drive; (b) front-wheel drive; (c) fourwheel drive.
n
19 i
n
35 i
n
72 i
α
Solution: The friction force acts parallel to the incline, and the
normal force is normal to the incline. Choose a coordinate system
with the x-axis parallel to the incline. The component of the weight
that acts parallel to the incline is W sin α, and the component acting
normally to the incline is W cos α.
(a) For rear wheel drive: The moment about the point of contact of
the front wheels:
MF W = 35W cos α + 19W sin α − 107R = 0,
from which the normal reaction of the two rear wheels is
R=
W
(35 cos α + 19 sin α).
107
(c)
For four wheel drive: Use the reactions of the front and rear wheels obtained in Parts (a) and (b). The sum of the forces parallel to the incline
is
FX = −W sin α + µs R + µs F = 0,
from which
W sin α
W
=
(35 cos α + 19 sin α + 72 cos α − 19 sin α).
µs
107
Reduce and solve: α = tan−1 (µs ) = 26.57◦
Check: This result is the same as if the Mercedes with four wheel drive
were a box on an incline, as it should be.
The force causing impending slip is W sin α, which is balanced
by the friction force: 0 = W sin α − µs R, from which
W sin α
W
=
(35 cos α + 19 sin α).
µs
107
Reduce and solve:
α = tan−1
(b)
35
− 19
19 in
35 in
= 10.18◦
107
µs
72 in
is the maximum angle at impending slip.
For front wheel drive: The moments about the point of contact
of the rear wheels is
α
MRW = −72W cos α + 19W sin α + 107F = 0,
α
from which the normal reaction of the two front wheels is
F =
µsF
W
(72 cos α − 19 sin α).
107
The friction force balances the component of gravity parallel to
the incline: 0 = −W sin α + µs F , from which
W sin α
W
=
(72 cos α − 19 sin α).
µs
107
Reduce and solve:
α = tan−1
72
107
+ 19
µ
s
= 17.17◦
W
µsR
R
F
35 in
72 in
Problem 8.50 The stationary cabinet has weight W .
Determine the force F that must be exerted to cause it to
move if (a) the coefficient of static friction at A and B
is µs ; (b) if the coefficient of static friction at A is µsA
and the coefficient of static friction at B is µsB .
F
G
h
H
A
B
b
—
2
(a) The sum of the moments about B is
b
= −hF +
W − bA = 0,
2
b
—
2
Solution:
MB
from which
W
A =
−
2
F
h
h
F.
b
H
A
The sum of forces:
Fy = −W + A + B = 0,
W
h
B =W −A=
+
F.
2
b
Fx = F − µs A − µs B = 0,
from which
W
h
W
h
F = µs
+
F+
−
F = µs W
2
b
2
b
(b) Use the normal reactions found in Part (a). From the sum of forces
parallel to the floor,
W
h
W
h
F = µsA A + µsB B = µsA
−
F + µsB
+
.
2
b
2
b
Reduce and solve:
W
F = 2
1+
b
—
2
B
F
from which
b
—
2
(µsA + µsB )
h
b
(µsA − µsB )
h
W
µSAA A
B
b
2
b
2
µSBB
Problem 8.51 A force F = 200 N is necessary to raise
the block A at a constant rate. The mass of the wedge
B is negligible. Between all of the contacting surfaces,
µs = 0.28 and µk = 0.26. What is the mass of block
A?
A
F
B
10°
Solution: The friction at all surfaces is kinetic. Draw a free body
diagram of each block and write the equilibrium equations.
f1 = µk N1 (1)
f2 = µk N2 (2)
A
f3 = µk N3 (3)
F
F = 200 N
10°
µk = 0.26
Block A:
Fx :
Fy :
Wedge B:
Fx :
Fy :
B
N1 − f2 cos 10◦ − N2 sin 10◦ = 0
y
(4)
mAg
f1
−f1 − mA g − f2 sin 10◦ + N2 cos 10◦ = 0 (5)
f1 = µ KN1 (1)
A
N1
f2 = µ KN2 (2)
f3 − F + f2 cos 10◦ + N2 sin 10◦ = 0 (6)
N3 + f2 sin 10◦ − N2 cos 10◦ = 0
(7)
10°
f3 = µ KN3 (3)
x
f2
N2
N2
f2
Unknowns:
f1 , N1 , f2 , N2 , f3 , N3 , mA
We have 7 eqns. in 7 unknowns.
Solving, we get
mA = 25.0 kg
Also,
f1 = 33.2 N,
N1 = 127.5 N
f2 = 77.2 N,
N2 = 296.7 N
f3 = 72.5 N,
N3 = 278.8 N
B
F = 200 N
µK = 0.26
f3
N3
10°
F
Problem 8.52 In Problem 8.51, suppose that the mass
of block A is 30 kg and the mass of the wedge B is 5 kg.
What force F is necessary to start the wedge B moving
to the left?
Solution:
The solution to this problem is very similar to that of
Problem 8.51.
mA = 30 kg
f1 = µs N1 (1)
A
f2 = µs N2 (2)
f3 = µs N3 (3)
F
B
10°
(impending slip)
mB = 5 kg
y
Block A:
Fx :
Fy :
N1 − f2 cos 10◦ − N2 sin 10◦ = 0
Wedge B:
Fx :
Fy :
f3 − F + f2 cos 10◦ + N2 sin 10◦ = 0
mAg
(4)
−f1 − mA g − f2 sin 10◦ + N2 cos 10◦ = 0 (5)
(6)
N1
A
N3 − mB g + f2 sin 10◦ − N2 cos 10◦ = 0 (7)
f1
Unknowns:
10°
f1 , N1 , f 2 , N 2 , f 3 , N 3 , F
N2
F = 272 N
N2
Also
f1 = 45.7 N
x
f2
We have 7 eqns. in 7 unknowns.
Solving, we get
f2
N1 = 163.2 N
10°
f2 = 101.7 N N2 = 363.2 N
f3 = 108.9 N N3 = 389.0 N
mBg
B
F
f3
N3
Problem 8.53 The wedge shown is being used to split
the log. The wedge weighs 20 lb and the angle α equals
30◦ . The coefficient of kinetic friction between the faces
of the wedge and the log is 0.28. If the normal force
exerted by each face of the wedge must equal 150 lb to
split the log, what vertical force F is necessary to drive
the wedge into the log at a constant rate?
Solution:
µk = 0.28
f = 0.28 N (1)
N = 150 lb (2)
Fx : (N − N ) cos 15◦ + (f − f ) sin 15◦ = 0
(no information here)
Fy : 2f cos 15◦ + 2 N sin 15◦ − F = 0 (3)
Unknowns: N, f, F
We have 3 eqns. in 3 unknowns
Solving
F = 159 lb, f = 42 lb
F
α
F
y
F
15°
15°
f
f
30°
15°
15°
N
N
x
F
F
α
Problem 8.54 The coefficient of static friction between the faces of the wedge and the log in Problem 8.53
is 0.30. Will the wedge remain in place in the log when
the vertical force F is removed?
Solution: For this problem, remove F and solve for the minimum
µs necessary for equilibrium.
The required µs
F
α
F
f = µs N
f
µs =
(1)
N
Fy :
−2f cos 15◦ + 2 N sin 15◦ = 0 (2)
f
= µs = tan 15◦ = 0.268 < 0.30.
N
15°
Yes—the wedge will stay in place
15°
f
f
15°
15°
N
N
Problem 8.55 The masses of A and B are 42 kg and
50 kg, respectively. Between all contacting surfaces,
µs = 0.05. What force F is required to start A moving
to the right?
B
45°
F
A
20°
Solution: If F is decreased until slip of A to the left impends, the
free body diagrams are as shown. The equilibrium equations are left
block:
Fx = F + N sin 20◦ + 0.05 N cos 20◦ − P cos 45◦
+ 0.05P cos 45◦ = 0,
Fy = N cos 20◦ − 0.05 N sin 20◦ − P cos 45◦
Solving, we obtain
N = 955 N,
P = 632 N,
Q = 425 N,
and F = 53.0 N.
− 0.05P cos 45◦ − (42)(9.81) = 0.
Right block:
Fx = P cos 45◦ − 0.05P cos 45◦ − Q = 0,
Fy = P cos 45◦ + 0.05P cos 45◦ + 0.05Q − (50)(9.81) = 0.
P
0.05 Q
0.05 P
(42)(9.81) 0.05 P
F
(50)(9.81)
P
20°
N
0.05 N
45°
Q
Problem 8.56 The stationary blocks A, B, and C each
have a mass of 200 kg. Between all contacting surfaces,
µs = 0.6. What force F is necessary to start B moving
downward?
F
B
A
C
80°
80°
Solution: The wedge angle is 10◦ for each side. The block A
cannot move, hence the friction contact surfaces are the wedge surfaces
plus the bottom surface of block C. Assuming that downward slip of
B impends, the free body diagrams of blocks B and C are as shown.
The equilibrium equations are Block B:
Fx = N sin 80◦ − µs N cos 80◦ − P sin 80◦ + µs P cos 80◦
= 0,
◦
◦
Fy = N cos 80 + µs N sin 80 + P cos 80 + µs P sin 80
Block C:
Fx = P sin 80◦ − µs P cos 80◦ − µs Q = 0,
(3)
◦
◦
Fy = Q − P cos 80 − µs P sin 80 − (200)(9.81) = 0. (4)
Solving them with µs = 0.6, we obtain
Q = 4100 N,
and F = 2300 N
80°
C
80°
◦
− F − (200)(9.81) = 0.
P = 2790 N,
B
A
(1)
◦
N = 2790 N,
F
F
(2)
µsN
(20.0)(9.81)
µsP
P
N
80°
(20.0)(9.81)
B
80°
P
µsP
80°
µsQ
C
Q
Problem 8.57 Small wedges called shims can be used
to hold an object in place. The coefficient of kinetic
friction between the contacting surfaces is 0.4. What
force F is needed to push the shim downward until the
horizontal force exerted on the object A is 200 N?
F
Shims
5°
A
5°
Solution:
F
fL = µk NL (1)
fR = µk NR (2)
NL = 200 N (3)
Fx :
Fy :
Shims
NL − NR cos 5◦ + fR sin 5◦ = 0
◦
(4)
5°
◦
−F + fL + fR cos 5 + NR sin 5 = 0 (5)
Unknowns: fL , NL , FR , NR , F
(5 eqns. in 5 unknowns)
Solving,
A
5°
F = 181 N
y
F
5°
fL
NL
fR
NR
x
Problem 8.58 The coefficient of static friction between the contacting surfaces in Problem 8.57 is 0.44. If
the shims are in place and exert a 200-N horizontal force
on the object A, what upward force must be exerted on
the left shim to loosen it?
Solution:
FL = µs NL (1)
FR = µs NR (2)
µs = 0.44
NL = 200 N
Fx :
NL − NR cos 5◦ − fR sin 5◦ = 0
Fy :
F − fL − fR cos 5◦ + NR sin 5◦ = 0 (4)
(3)
Unknowns F, fL , fR , NR
Solving, F = 156 N
F
Shims
5°
A
5°
5°
y
F
NL
NR
fL
fR
x
Problem 8.59 The crate A weighs 600 lb. Between all
contacting surfaces, µs = 0.32 and µk = 0.30. Neglect
the weights of the wedges. What force F is required to
move A to the right at a constant rate?
F
5°
A
5°
Solution: The active sliding contact surfaces are between the wall
and the left wedge, between the wedges, between the floor and the
bottom of the right wedge, and between the crate and the floor. Leftmost
wedge: Denote the normal force exerted by the wall by Q, and the
normal force between the wedges by N . The equilibrium conditions
for the left wedge moving at a constant rate are:
Fy = −F + µk N cos α + N sin α + µk Q = 0.
Fx = Q − N cos α + µk N sin α = 0.
For the right wedge: Denote the normal force exerted by the crate by
A, and the normal force exerted by the floor by P .
Fy = −N sin α − µk N cos α + P = 0.
Fx = N cos α − µk N sin α − µk P − A = 0.
For the crate: Denote the weight of the crate by W .
Fx = A − µk W = 0.
These five equations are solved for the five unknowns by iteration:
Q = 204.4 lb,
N = 210.7 lb
P = 81.34 lb,
A = 180 lb,
and F = 142.66 lb
F
5°
A
5°
F
µkQ
µkN
Q
N
N
A
W
A
µkN
P
µkP
µkW
W
Problem 8.60 Suppose that between all contacting
surfaces in Problem 8.59, µs = 0.32 and µk = 0.30.
Neglect the weights of the 5◦ wedges. If a force F =
800 N is required to move A to the right at a constant
rate, what is the mass of A?
Solution: The free body diagrams of the left wedge and the combined right wedge and crate are as shown. The equilibrium equations are
Wedge:
Fx = N − P cos 5◦ + 0.3P sin 5◦ = 0,
Fy = 0.3 N + P sin 5◦ + 0.3P cos 5◦ − F = 0,
P = 1180 N,
N = 1150 N,
Q = 3820 N,
and m = 343 kg.
F
Wedge and box:
Fx = P cos 5◦ − 0.3P sin 5◦ − 0.3Q = 0,
Fy = Q − P sin 5◦ − 0.3P cos 5◦ − 9.81 m = 0.
0.3 N
N
0.3 P
P
P
Solving them, we obtain
m (9.81)
0.3 P
5°
0.3 Q
Q
Problem 8.61 The box A has a mass of 80 kg, and the
wedge B has a mass of 40 kg. Between all contacting
surfaces, µs = 0.15 and µk = 0.12. What force F is
required to raise A at a constant rate?
A
10°
F
B
10°
Solution:
From the free-body diagrams shown, the equilibrium
equations are
Box A:
A
Q − N sin 10◦ − µk N cos 10◦ = 0,
N cos 10◦ − µk N sin 10◦ − µk Q − W = 0.
Wedge B:
◦
◦
◦
10°
B
F
10°
◦
P sin 10 + µk P cos 10 + N sin 10 + µk N cos 10 − F = 0
P cos 10◦ − µk P sin 10◦ − N cos 10◦ + µk N sin 10◦ − Ww = 0.
Solving with
W = (80)(9.81) N,
Ww = (40)(9.81) N,
and µk = 0.12,
we obtain
N = 845 N,
Q = 247 N,
P = 1252 N,
and F = 612 N.
Q
N
A
W
µkQ
B WW
P
µkN
N
µkN
µkP
F
Problem 8.62 Suppose that in Problem 8.61, A weighs
800 lb and B weighs 400 lb. The coefficients of friction
between all of the contacting surfaces are µs = 0.15 and
µk = 0.12. Will B remain in place if the force F is
removed?
Solution:
The equilibrium conditions are: For the box A: Denote
the normal force exerted by the wall by Q, and the normal force exerted
by the wedge by N . The friction forces oppose motion.
Fy = −W + N cos α + µs N sin α + µs Q = 0,
Fx = +µs N cos α − N sin α + Q = 0.
(A comparison with the equilibrium conditions for Problem 8.61 will show
that the friction forces are reversed, since for slippage the box A will move
downward, and the wedge B to the right.) The strategy is to solve these equations
for the required µs to keep the wedge B in place when F = 0. The solution
Q = 0, N = 787.8 lb, P = 1181.8 lb and µs = 0.1763. Since the value of
µs required to hold the wedge in place is greater than the value given, the wedge
will slip out.
For the wedge B. Denote the normal force on the lower surface by P .
Fx = −µs N cos α − µs P cos α + P sin α + N sin α = 0.
Fy = −N cos α + P cos α − µs N sin α + µs P sin α − Ww = 0.
µSN
µSQ
WW
W
Q
µSP
N
P
µSN
Problem 8.63 Between A and B, µs = 0.20, and
between B and C, µs = 0.18. Between C and the wall,
µs = 0.30. The weights WB = 20 lb and WC = 80 lb.
What force F is required to start C moving upward?
C
F
B
15°
A
Solution: The active contact surfaces are between the wall and C,
between the wedge B and C, and between the wedge B and A. For the
weight C: Denote the normal force exerted by the wall by Q, and the
normal force between B and C by N . Denote the several coefficients
of static friction by subscripts. The equilibrium conditions are:
Fy = −WC + N − µCW Q = 0,
Fx = −Q + µBC N = 0.
C
F
B
µ sQ
These four equations in four unknowns are solved:
and F = 66.9 lb
µsN
Q
WC
F
N = 84.6 lb,
P = 114.4 lb,
15°
A
For the wedge B: Denote the normal force between A and B by P .
Fy = −N + P cos α − µAB P sin α − WB = 0.
Fx = F − µBC N − µAB P cos α − P sin α = 0.
Q = 15.2 lb,
N
N
µsN
N
WB
µsP
P
Problem 8.64 The masses of A, B, and C are 8 kg,
12 kg, and 80 kg, respectively. Between all contacting
surfaces, µs = 0.4. What force F is required to start C
moving upward?
F
Solution: The active contact surfaces are between A and B, between A and the wall, between B and the floor, and between B and
C. Assume that the roller supports between C and the wall exert no
friction forces. For the wedge A: Denote the normal force exerted by
the wall as Q and the normal force between A and B as N . The weight
is WA = 8 g = 78.48 N. The equilibrium conditions:
Fy = −F + µs Q + µs N cos α + N sin α − WA = 0
Fy = Q − N cos α + µs N sin α = 0.
C
A
10°
B
12°
For wedge B: Denote the normal force exerted on B by the floor by
P , and the normal exerted by the weight C as S. The weight of B is
WB = 12 g = 117.72 N. The equilibrium conditions:
Fy = −N sin α − S cos β + P − µs N cos α + µs S sin β − WB
= 0.
C
F
Fx = N cos α − µs N sin α − µs P − µs S cos β − S sin β = 0.
For the weight C: The weight is WC = 80 g = 784.8 N. The
equilibrium conditions:
Fy = −WC + S cos β − µs S sin β = 0.
A
B
10°
12°
These five equations in five unknowns are solved:
Q = 1157.6 N,
N = 1293.5 N,
F
S = 857.4 N,
µsQ
P = 1677.5,
Q
and F = 1160 N
Problem 8.65 The vertical threaded shaft fits into a
mating groove in the tube C. The pitch of the threaded
shaft is p = 0.1 in., and the mean radius of the thread
is r = 0.5 in. The coefficients of friction between the
thread and the mating groove are µs = 0.15 and µk =
0.10. The weight W = 200 lb. Neglect the weight of
the threaded shaft.
(a) Will the stationary threaded shaft support the weight
if no couple is applied to the shaft?
(b) What couple must be applied to the threaded shaft
to raise the weight at a constant rate?
α
WA
R
µsN
N
N µ
sN µ P
s
W
C
(a) The angle of static friction is θs = tan−1 (0.15) =
8.53◦ . The pitch angle is
p 0.1
α = tan−1
= tan−1
= 1.82◦ .
2πr
2π(0.5)
Solution:
From Eq. (10.14) the moment necessary for the shaft to be on the
verge of rotating is M = rF tan(θs − α). For a zero moment,
θs = α, which is not satisfied. Therefore the shaft will support the
weight when no moment is applied. (b) The angle of kinetic friction
is θk = tan−1 (0.10) = 5.71◦ . From Eq. (9.9) the moment required
to raise the weight at a constant rate is
M = rW tan(θk + α) = 0.5(200) tan(7.533) = 13.2 in lb.
µsS
W
C
S
WB
P
WC
β
S
µsS
Problem 8.66 Suppose that in Problem 8.65, the pitch
of the threaded shaft is p = 2 mm and the mean radius
of the thread is r = 20 mm. The coefficients of friction
between the thread and the mating groove are µs = 0.22,
and µk = 0.20. The weight W = 500 N. Neglect
the weight of the threaded shaft. What couple must be
applied to the threaded shaft to lower the weight at a
constant rate?
Solution:
θk = tan
−1
The angle of kinetic friction is
(0.2) = 11.31◦ .
The angle of pitch is
p 2
α = tan−1
= tan−1
= 0.9118◦ .
2πr
2π(20)
The moment required to lower the weight at a constant rate is
M = 0.02(500) tan(11.31 − 0.9118) = 1.835 N-m.
Problem 8.67 The position of the horizontal beam can
be adjusted by turning the machine screw A. Neglect the
weight of the beam. The pitch of the screw is p = 1 mm,
and the mean radius of the thread is r = 4 mm. The
coefficients of friction between the thread and the mating
groove are µs = 0.20 and µk = 0.18. If the system is
initially stationary, determine the couple that must be
applied to the screw to cause the beam to start moving
(a) upward; (b) downward.
Solution:
400 N
A
100 mm
300 mm
The sum of the moments about the pin support is
M = −0.4F + (0.3)400 = 0,
400 N
from which the force exerted by the screw is F = 300 N. The pitch
angle is
1
α = tan−1
= 2.28◦ .
2π(4)
The static friction angle is θs = tan−1 (0.2) = 11.31◦ . (a) The
moment required to start motion upward is
300
mm
100
mm
M = 0.004(300) tan(11.31◦ + 2.28◦ ) = 0.29 N-m
(b) The moment required to start motion downward is
400 N
M = 0.004(300) tan(11.31◦ − 2.28◦ ) = 0.19 N-m
F
100
mm
300
mm
Problem 8.68 Suppose that in Problem 8.67, the pitch
of the machine screw is p = 1 mm and the mean radius
of the thread is r = 4 mm. What minimum value of the
coefficient of static friction between the thread and the
mating groove is necessary for the beam to remain in the
position shown with no couple applied to the screw?
Solution:
From the solution to Problem 8.67 the force applied to
the screw is F = 300 N. The pitch angle is
1
α = tan−1
= 2.28◦ .
2π(4)
The moment required to start motion downward is M = 0.004(300)
tan(θs − α). For M = 0, tan(θs − α) = 0, from which
θs = α = 2.279◦ ,
and µs = tan(2.279◦ ) = 0.0398
Problem 8.69 The mass of block A is 60 kg. Neglect
the weight of the 5◦ wedge. The coefficient of kinetic
friction between the contacting surfaces of the block A,
the wedge, the table, and the wall is µk = 0.4. The pitch
of the threaded shaft is 5 mm, the mean radius of the
thread is 15 mm, and the coefficient of kinetic friction
between the thread and the mating groove is 0.2. What
couple must be exerted on the threaded shaft to raise the
block A at a constant rate?
A
5°
Denote the wedge angle by β = 5◦ and the normal
force on the top by N and on the lower surface by P . The free body
diagrams of the wedge and block are as shown. The equilibrium equations for wedge:
Fx = F − µk P − N sin 5◦ − µk N cos 5◦ = 0,
Fy = P − N cos 5◦ + µk N sin 5◦ = 0.
Solution:
A
5°
For the Block:
Fx = N sin 5◦ + µk N cos 5◦ − Q = 0,
Fy = N cos 5◦ − µk N sin 5◦ − µk Q − W = 0.
Solving them, we obtain F = 668 N. From Equation (9.9), the couple
necessary to rotate the threaded shaft when it is subjected to the axial
force F is M = rF tan(θk + α) r is the radius 15 mm = 0.015 m.
θk is the angle of kinetic friction θk = arctan(0.2) = 11.31◦ .
From Equation (9.7), the slope is given in terms of the pitch by
P
5
α = arctan
= arctan
= 3.04◦ .
2πr
2π(15)
The couple is
M = (0.015 m)(668 N) tan(11.31◦ + 3.04◦ ) = 2.56 N-m.
µkQ
W
µkN
N
µkP
P
Q
F
N
µkN
Problem 8.70 The vise exerts 80-lb forces on A. The
threaded shafts are subjected only to axial loads by the
jaws of the vise. The pitch of their threads is p = 1/8 in.,
the mean radius of the threads is r = 1 in., and the
coefficient of static friction between the threads and the
mating grooves is 0.2. Suppose that you want to loosen
the vise by turning one of the shafts. Determine the
couple you must apply (a) to shaft B; (b) to shaft C.
A
4 in
B
4 in
C
Solution:
Isolate the left jaw. The sum of the moments about C:
MC = −4B + 8(80) = 0,
A
from which B = 160 lb (T ). The sum of the forces:
Fx = −80 + B − C = 0,
4 in
B
from which C = 80 lb (C). The pitch angle is
1
α = tan−1
= 1.14◦ .
16π
C
The static friction angle is θs = tan−1 (0.2) = 11.31◦ . The moments required to loosen the vise are
1
MB =
(160) tan(11.31◦ − 1.14◦ ) = 2.39 ft lb,
12
80 lb
and MC = rC tan(θs − α) = 1.2 ft-lb.
C
Problem 8.71 Suppose that you want to tighten the
vise in Problem 8.70 by turning one of the shafts. Determine the couple you must apply (a) to shaft B; (c) to
shaft C.
Solution: Use the solution to Problem 8.70. (a) The moment on
shaft B required to tighten the vise is MB = rB tan(θs + α). Note
1
that r = 12
, B = 160 lb,
1
α = tan−1
= 1.14◦
16π
and θs = tan−1 (0.2) = 11.31◦ ,
then MB = 2.94 ft lb (b) For shaft C, MC = rC tan(θs + α),
where C = 80 lb, MC = 1.47 ft-lb.
B
4 in
4 in
4 in
Problem 8.72 The threaded shaft has a ball and socket
support at B. The 400-lb load A can be raised or lowered
by rotating the threaded shaft, causing the threaded collar at C to move relative to the shaft. Neglect the weights
of the members. The pitch of the shaft is p = 14 in., the
mean radius of the thread is r = 1 in., and the coefficient of static friction between the thread and the mating
groove is 0.24. If the system is stationary in the position shown, what couple is necessary to start the shaft
rotating to raise the load?
9 in
C
A
12 in
B
9 in
Solution: Denote the lower right pin support
by D. The length
√
of the connecting member CD is LCD = 92 + 122 = 15 in. The
angle between the threaded shaft and member CD is
9
β = 2 tan−1
= 73.74◦ .
12
The sum of the moments about D is
MD = LCD F cos(90 − β) − 18W = 0,
9 in
12 in
9 in
from which F = 500 lb. The pitch angle is
p α = tan−1
= 2.28◦ .
2πr
The angle of static friction is θs = tan−1 (0.24) = 13.5◦ . The
moment needed to start the threaded collar in motion is
1
M = rF tan(θs + α) =
(500) tan(13.5◦ + 2.28◦ )
12
= 11.77 ft-lb
9 in
9 in
W
C
F
Dy
β
LCD
Dx
18 in
Problem 8.73 In Problem 8.72, if the system is stationary in the position shown, what couple is necessary
to start the shaft rotating to lower the load?
Solution: Use the results of the solution to Problem 8.72. The
moment is M = rF tan(θs − α), where
1
r =
ft,
12
F = 500 lb,
θs = 13.5◦ ,
and α = 2.28◦ ,
from which M = 8.26 ft lb
18 in
18 in
Problem 8.74 The car jack is operated by turning the
threaded shaft at A. The threaded shaft fits into a mating
groove in the collar at B, causing the collar to move
relative to the shaft as the shaft turns. As a result, points
B and D move closer together or farther apart, causing
point C (where the jack is in contact with the car) to
move up or down. The pitch of the threaded shaft is
p = 5 mm, the mean radius of the thread is r = 10 mm,
and the coefficient of kinetic friction between the thread
and the mating groove is 0.15. What couple is necessary
to turn the shaft at a constant rate and raise the jack when
it is in the position shown if F = 6.5 kN?
F
C
150 mm
D
150 mm
300 mm
Solution: Isolate members BC and BD. Assume that half the
car load is carried by these members. The equilibrium conditions for
member BC are:
Fx = Cx − Bx = 0,
F
MB = 0.3
− 0.15Cx = 0.
2
These equations are solved for
F
6.5
=
= 3.25 kN
2
2
C
150 mm B
B
150 mm
D
300 mm
300 mm
F
α
C
C
θs = tan−1 (0.15) = 8.53◦ .
2C sin α = F
C
The moment required to rotate the shaft at a constant rate is
D = 2Ccos α
F = 2F
tan α
D
M = (0.01)(6.5) tan(8.53◦ + 4.55◦ ) = 0.0151 kN
m = 15.1 N m
300 mm
F
to obtain Bx = 6.5 kN, which is the force on the collar to be balanced
by the rotating threaded shaft. The pitch angle is
5
α = tan−1
= 4.55◦ .
2π10
The angle of kinetic friction is
=
C
F
2
CX
BY
BX
BX
DY
BY
DX
Problem 8.75 In Problem 8.74, what couple is necessary to turn the threaded shaft at a constant rate and
lower the jack when it is in the position shown if the
force F = 6.5 kN?
Solution: Use the results of the solution of Problem 8.74. The moment required to lower the jack at a constant rate is M = rB tan(θk −
α), where r = 0.01 m, B = 6500 N, θk = 8.53◦ , α = 4.55◦ , from
which M = 4.52 N-m
A
B
A
Problem 8.76 A turnbuckle, used to adjust the length
or tension of a bar or cable, is threaded at both ends.
Rotating it draws threaded segments of a bar or cable
together or moves them apart. Suppose that the pitch
of the threads is p = 3 mm their mean radius is r =
25 mm, and the coefficient of static friction between the
threads and the mating grooves is 0.24. If T = 800 N,
what couple must be exerted on the turnbuckle to start
tightening it?
T
T
T
T
Solution:
θs = a tan(µs ) = 13.49◦
p α = a tan
= 1.09◦
2πr
M = rT tan(θs + α)
since M tends to create motion opposite to the direction of T .
M = (0.025)(800 N) tan(14.59◦ )
M = 5.21 N-m
for each screw.
There are two screws in the turnbuckle.
∴ M = 10.42 N-m
Problem 8.77 The horizontal shaft is supported by
two journal bearings. The coefficient of kinetic friction
between the shaft and the bearings is µk = 0.2. The
radius of the shaft is 20 mm, and its mass is 5 kg. Determine the couple M necessary to rotate the shaft at a
constant rate.
Strategy: You can obtain the moment necessary to rotate the shaft at a constant rate by replacing θs by θk in
Eq. (9.12).
Solution: The weight of the shaft is W = mg = 5(9.81) =
49 N, divided between two bearings. The angle of kinetic friction is
θk = tan−1 (0.2) = 11.31◦ . The moment per bearing is
W
M =
(0.02) sin θk = 0.096 N m.
2
The total moment is Mt = 0.192 N m
M
M
Problem 8.78 The horizontal shaft is supported by
two journal bearings. The coefficient of static friction
between the shaft and the bearings is µs = 0.3. The
radius of the shaft is 20 mm, and its mass is 5 kg. Determine the largest mass m that can be suspended as
shown without causing the stationary shaft to slip in the
bearings.
m
Solution: The weight of the shaft is W = mg = 5(9.81) =
49 N. This weight is divided between two bearings. The angle of
static friction is θs = tan−1 (0.3) = 16.7◦ . The load per bearing is
F =
W + Wm
,
2
and the moment required to start rotation is
Mm = (W + Wm )r sin θs
where Wm is the suspended weight,
Wm =
Mm
.
r
From which
Wm r = (W + Wm )r sin θs ,
from which
W
sinθs
m =
= 2.02 kg.
g
1 − sin θs
Problem 8.79 Suppose that in Problem 8.78 the mass
m = 8 kg and the coefficient of kinetic friction between
the shaft and bearings is µk = 0.26. What couple must
be applied to the shaft to raise the mass at a constant
rate?
Solution:
From the solution to Problem 8.78 the moment re-
quired is
Mapplied = (W + Wm )r sin θk ,
where Wm = mg is the weight of the suspended mass. The moment
required to raise the suspended mass is Mm = Wm r. The total
moment is the sum of the moment required to turn the shaft and the
moment required to raise the mass:
Mtotal = (W + mg)r sin θk + Wm r
= (49 + 78.5)(0.02) sin(14.6◦ ) + 1.57 = 2.21 N m
m
Problem 8.80 The pulley is mounted on a horizontal shaft supported by journal bearings. The coefficient
of kinetic friction between the shaft and the bearings is
µk = 0.3. The radius of the shaft is 20 mm, and the
radius of the pulley is 150 mm. The mass m = 10 kg.
Neglect the masses of the pulley and shaft. What force
T must be applied to the cable to move the mass upward
at a constant rate?
m
T
The angle of kinetic friction is θk = tan−1 (µk ) =
16.7◦ . The moment required to turn the shaft is M = (mg +
T )r sin θk . The applied moment is M = (T − mg)R where R
is the radius of the pulley. Equating and reducing:
r
1+ R
sin θk
1.0383
T = mg
=
(98.1)
= 105.92 N
r
1− R
sin θk
0.9617
Solution:
m
T
Problem 8.81 In Problem 8.80, what force T must be
applied to the cable to lower the mass at a constant rate?
Solution: Form the solution to Problem 8.80, θk =
tan−1 (µk ) = 16.7◦ , and M = (mg + T )r sin θk . The applied
moment is M = (mg − T )R. Substitute and reduce:
r
1− R
sin θk
0.9617
T = mg
=
(98.1)
= 90.86 N
r
1+ R
sin θk
1.0383
Problem 8.82 The pulley of 8-in. radius is mounted
on a shaft of 1-in. radius. The shaft is supported by
two journal bearings. The coefficient of static friction
between the bearings and the shaft is µs = 0.15. Neglect
the weights of the pulley and shaft. The 50-Ib block A
rests on the floor. If sand is slowly added to the bucket
B, what do the bucket and sand weigh when the shaft
slips in the bearings?
(See Problem 8.80). The angle of static friction is θs =
tan−1 (µs ) = 8.53◦ . The moment required to start rotation for both bearings is M = r(B + W ) sin θs . The applied moment is M = (B − W )R,
where R is the radius of the pulley. Substitute and reduce:
r
1+ R
sin θs
1.0185
B =W
=
(50)
= 51.9 lb
r
1− R
sin θs
0.9815
Solution:
8 in
1 in
8 in
B
A
B
A
Problem 8.83 The pulley of 50-mm radius is mounted
on a shaft of 10-mm radius. The shaft is supported by
two journal bearings. The mass of the block A is 8 kg.
Neglect the weights of the pulley and shaft. If a force
T = 84 N is necessary to raise the block A at a constant
rate, what is the coefficient of kinetic friction between
the shaft and the bearings?
50 mm
20°
10 mm
T
A
The weight is W = mg = 78.5 N. The force on the
pulley is
F = (W + T sin α)2 + (T cos α)2 ,
Solution:
where α = 20◦ .
F = 107.22 + 78.92 = 133.13 N.
The moment required to raise the mass at constant rate for both bearings
is M = rF sin θk = 1.33 sin θk . The applied moment is M =
(T − W )R = 0.276 N m. Substitute and reduce:
sin θk =
(T − W )R
0.276
=
= 0.2073,
rF
1.33
from which
θk = 11.96◦
and µk = tan(11.96◦ ) = 0.2119
50 mm
10 mm
20°
T
A
Problem 8.84 The mass of the suspended object is
4 kg. The pulley has a 100-mm radius and is rigidly
attached to a horizontal shaft supported by journal bearings. The radius of the horizontal shaft is 10 mm and the
coefficient of kinetic friction between the shaft and the
bearings is 0.26. What tension must the person exert on
the rope to raise the load at a constant rate?
25°
100 mm
Solution:
R = 0.1 m
25°
µk = 0.26
Shaft radius 0.01 m
10 mm
µk (shaft) = 0.26
tan θk = µk
θk = 14.57◦
Ms = rF sin θk
R = 0.1 m
m = 4 kg
To Find F , we must find the forces acting on the shaft.
Fx :
Ox − T cos 25◦ = 0
(1)
Fy :
Oy − T sin 25◦ − mg = 0 (2)
F = Ox2 + Oy2
(3)
Mo :
RT − Rmg − Ms = 0
(4)
Ms = rF sin θk
Unknowns: Ox , Oy , T, Ms , F
Solving, we get
T = 40.9 N
Also,
F = 67.6 N,
Ms = 0.170 N-m
Ox = 37.1 N,
Oy = 56.5 N
(5)
25°
T
µ K = 0.26
Shaft radius 0.01 m
µK (shaft) = 0.26
MS
OX
OY
mg
Problem 8.85 The circular flat-ended shaft is pressed
into the thrust bearing by an axial load of 100 N. Neglect
the weight of the shaft. The coefficients of friction between the end of the shaft and the bearing are µs = 0.20
and µk = 0.15. What is the largest couple M that can
be applied to the stationary shaft without causing it to
rotate in the bearing?
100 N
30 mm
M
Solution: The bearing meets the conditions for Eq. (9.14). For
impending rotation, the moment is
2
2
M = µs F r =
(0.2)(100)(0.03) = 0.4 N m
3
3
100 N
30 mm
M
Problem 8.86 In Problem 8.85, what couple M is
required to rotate the shaft at a constant rate?
Solution: The bearing meets the conditions for Eq. (10.17). The
moment required to sustain a constant rate of rotation is
2
2
M =
µk F r =
(0.15)(100)(0.03) = 0.3 N m
3
3
Problem 8.87 Suppose that the end of the shaft in
Problem 8.85 is supported by a thrust bearing of the
type shown in Fig. 9.22, where ro = 30 mm, ri =
10 mm, α = 30◦ , and µk = 0.15. What couple M is
required to rotate the shaft at a constant rate?
Solution: The bearing meets the conditions for Eq (9.13). The
moment required to sustain a constant rate of rotation is
3
ro − r13
2µk F
M =
= 0.3753 N m
3 cos α
ro2 − r12
Problem 8.88 The disk D is rigidly attached to the
vertical shaft. The shaft has flat ends supported by thrust
bearings. The disk and the shaft together have a mass
of 220 kg and the diameter of the shaft is 50 mm. The
vertical force exerted on the end of the shaft by the upper
thrust bearing is 440 N. The coefficient of kinetic friction
between the ends of the shaft and the bearings is 0.25.
What couple M is required to rotate the shaft at a constant
rate?
Solution:
M
M
D
D
There are two thrust bearings, one at the top and one at
the bottom
FU = 440 N
m = 220 kg
Fy :
FL − FU − mg = 0
M
M
D
D
FL = 2598.2 N.
The couple necessary to turn D at a constant rate is the sum of the
couples for the two bearings.
MU =
2
µk F U r
3
ML =
2
µk F L r
3
FU
r = 0.025 m
µk = 0.25
Solving,
MU = 1.833 N-m
D
ML = 10.826
MTOTAL = 12.7 N-m
mg
FL
Problem 8.89 Suppose that the ends of the shaft in
Problem 8.88 are supported by thrust bearings of the
type shown in Fig. 9.22, where ro = 25 mm, ri =
6 mm, α = 45◦ , and µk = 0.25. What couple M is
required to rotate the shaft at a constant rate?
Solution:
There are two thrust bearings, one at the top and one at
the bottom.
FU = 440 N
m = 220 kg
M
Fy :
FL − FU − mg = 0
M
D
D
FL = 2598.2 N.
The couple necessary to turn D at a constant rate is the sum of the
couples for the two bearings.
For the bearings used
m =
2µk F (ro3 − ri3 )
3 cos α (ro2 − ri2 )
FU
α = 45◦ , ro = 0.025 m
µk = 0.25, ri = 0.006 m
Thus,
MU =
2µk FU (ro3 − ri3 )
= 2.7 N-m
3 cos α (ro2 − ri2 )
ML =
2µk FL (ro3 − ri3 )
= 16.0 N-m
3 cos α (ro2 − ri2 )
D
mg
MTOTAL = MU + ML = 18.7 N-m
FL
γi
α
γo
Problem 8.90 The shaft is supported by thrust bearings that subject it to an axial load of 800 N. The coefficients of kinetic friction between the shaft and the left
and right bearings are 0.20 and 0.26, respectively. What
couple is required to rotate the shaft at a constant rate?
15 mm
38 mm
38
mm
Solution:
The left bearing: The parameters are
ro = 38 mm,
38 mm
ri = 0,
15 mm
α = 45◦ ,
µk = 0.2,
and F = 800 N.
The moment required to sustain a constant rate of rotation is
ro3 − ri3
2µk F
Mleft =
= 5.73 N m.
3 cos α ro2 − ri2
38
mm
The right bearing: This is a flat-end bearing. The parameters are
µk = 0.26, r = 15 mm, and F = 800 N. The moment required to
sustain a constant rate of rotation is
Mright =
2µk F r
= 2.08 N m.
3
The sum of the moments: M = 5.73 + 2.08 = 7.81 N m
Problem 8.91 A motor is used to rotate a paddle for
mixing chemicals. The shaft of the motor is coupled to
the paddle using a friction clutch of the type shown in
Fig. 9.25. The radius of the disks of the clutch is 120 mm,
and the coefficient of static friction between the disks is
0.6. If the motor transmits a maximum torque of 15 N-m
to the paddle, what minimum normal force between the
plates of the clutch is necessary to prevent slipping?
Solution:
M =
Clutch
Paddle
The moment necessary to prevent slipping is
2µs F r
2(0.6)(0.12)F
=
= 15 N m.
3
3
Solve: F = 312.5 N
Clutch
Paddle
Problem 8.92 The thrust bearing is supported by contact of the collar C with a fixed plate. The area of contact
is an annulus with an inside diameter D1 = 40 mm and
an outside diameter D2 = 120 mm. The coefficient
of kinetic friction between the collar and the plate is
µk = 0.3. The force F = 400 N. What couple M is
required to rotate the shaft at a constant rate?
F
F
M
M
C
C
D1
D2
Solution:
This is a thrust bearing with parameters
µk = 0.3,
F
F
α = 0,
ro = 60 mm,
ri = 20 mm,
and F = 400 N.
The moment required to sustain rotation at a constant rate is
2µk F ro3 − ri3
M =
= 5.2 N m
3
ro2 − ri2
Problem 8.93 Suppose that you want to lift a 50-lb
crate off the ground by using a rope looped over a tree
limb as shown. The coefficient of static friction between
the rope and the limb is 0.4, and the rope is wound 120◦
around the limb. What force must you exert to lift the
crate?
Strategy: The tension necessary to cause impending
slip of the rope on the limb is given by Eq. (9.17), with
T1 = 50 lb, µs = 0.4, and β = (π/180)(120) rad.
Solution:
radians is
β = (120◦ )
This meets the conditions for Eq. (9.17). The angle in
π = 2.094 radians.
180
The force required is T = 50eµs β = 115.6 lb
C
M
M
C
D1
D2
Problem 8.94 In Problem 8.93, once you have lifted
the crate off the ground, what is the minimum force you
must exert on the rope to keep it suspended?
If T is decreased until slip of the rope toward the left
impends, Equation (9.17) is 50 = T eµs β where µs = 0.4 and β =
(π/180)(120) rad.
Solving for T , we obtain T = 21.6 lb.
Solution:
Problem 8.95 Winches are used on sailboats to help
support the forces exerted by the sails on the ropes
(sheets) holding them in position. The winch shown is a
post that will rotate in the clockwise direction (seen from
above), but will not rotate in the counterclockwise direction. The sail exerts a tension TS = 800 N on the sheet,
which is wrapped two complete turns around the winch.
The coefficient of static friction between the sheet and
the winch is µs = 0.2. What tension TC must the crew
member exert on the sheet to prevent it from slipping on
the winch?
TC
TS
Solution:
Ts = Tc eµs β
Ts = 800 N
µs = 0.2
β = 4π
Solving,
Tc = 64.8 N
TC
TS
TC
TS
Problem 8.96 The coefficient of kinetic friction between the sheet and the winch in Problem 8.95 is µk =
0.16. If the crew member wants to let the sheet slip at a
constant rate, releasing the sail, what initial tension TC
must he exert on the sheet as it begins slipping?
Solution:
Ts = Tc eµk β
Ts = 800 N, µk = 0.16
β = 4π
Solving
Tc = 107.1 N
C
TS
TC
TS
Problem 8.97 The mass of the block A is 18 kg. The
rope is wrapped one and one-fourth turns around the
fixed wooden post. The coefficients of friction between
the rope and post are µs = 0.15 and µk = 0.12. What
force would the person have to exert to raise the block at
a constant rate?
Solution:
β = 2.5π
µk = 0.12
m = 18 kg.
T = (mg)eµk β
A
Solving,
T = 453 N
T
mg
A
Problem 8.98 The weight of the block A is W . The
disk is supported by a smooth bearing. The coefficient of
kinetic friction between the disk and the belt is µk . What
couple M is necessary to turn the disk at a constant rate?
Solution: The angle is β = π radians. The tension in the left belt when
the belt is slipping on the disk is Tleft = W eµk β . The tension in the right belt
is Tright = W . The moment applied to the disk is
M = R(Tleft − Tright ) = R(W eµk β − W ) = RW (eµk π − 1).
This is the moment that is required to rotate the disk at a constant rate.
r
M
r
M
A
A
Problem 8.99 The couple required to turn the wheel of
the exercise bicycle is adjusted by changing the weight
W . The coefficient of kinetic friction between the wheel
and the belt is µk . Assume the wheel turns clockwise.
(a) Show that the couple M required to turn the wheel
is M = W R(1 − e−3.4µk ).
(b) If W = 40 lb and µk = 0.2, what force will the
scale S indicate when the bicycle is in use?
S
15°
R
30°
Solution: Let β be the angle in radians of the belt contact with
wheel. The tension in the top belt when the belt slips is Tupper =
W e−µk β . The tension in the lower belt is Tlower = W . The moment
applied to the wheel is
S
M = R(Tlower − Tupper ) = RW (1 − e−µk β ).
This is the moment required to turn the wheel at a constant rate. The
angle β in radians is
π β = π + (30 − 15)
= 3.40 radians,
180
15°
R
30°
from which M = RW (1 − e−3.4µk ). (b) The upper belt tension is
W
Tupper = 40e−3.4(0.2) = 20.26 lb.
This is also the reading of the scale S.
Problem 8.100 The box B weighs 50 lb. The coefficient of friction between the cable and the fixed round
supports are µs = 0.4 and µk = 0.3.
(a) What is the minimum force F required to support
the box?
(b) What force F is required to move the box upward
at a constant rate?
B
F
Solution: The angle of contact between the cable and each round
support is β = π2 radians.
(a) Denote the tension in the horizontal part of the cable by H. The
tension in H is H = W e−µs β . The force F is
F = He−µs β = W e−2µs β ,
(b)
from which F = 14.23 lb is the force necessary to hold the box
stationary.
As the box is being raised,
H
and F
= W eµk β ,
= Heµk β = W e2µk β ,
from which F = 128.32 lb
B
F
W
Problem 8.101 The 20-kg box A is held in equilibrium
on the inclined surface by the force T acting on the rope
wrapped over the fixed cylinder. The coefficient of static
friction between the box and the inclined surface is 0.1.
The coefficient of static friction between the rope and the
cylinder is 0.05. Determine the largest value of T that
will not cause the box to slip up the inclined surface.
45°
A
20°
T
Solution: Assuming that slip of the box up the surface impends.
The free body diagrams of the box and rope around the cylinder are as
shown.
From the equilibrium equations
Fx = TA cos 45◦ − N sin 20◦ − 0.1 N cos 20◦ = 0,
Fy = TA sin 45◦ + N cos 20◦ − 0.1 N sin 20◦ − (20)(9.81) = 0
45°
A
we obtain TA = 90.2 N. Equation (9.17) is T = TA eµs B where
µs = 0.05 and β = (π/180)(135) rad. Solving for T we obtain
T = 101 N.
T
20°
y
TA
45°
20°
x
(20)(9.81)
0.1 N
N
135°
TA
T
Problem 8.102
In Problem 8.101, determine the
smallest value of T necessary to hold the box in equilibrium on the inclined surface.
Solution: In this case, we assume that slip of the box down the
surface impends. This requires reversing the direction of the friction
force in the free body diagram of Problem 8.101. The friction now
acts up the surface and the friction on the drum is reversed. See the
free body diagrams. From the equilibrium equations,
FX = TA cos(45◦ ) − N sin(20◦ ) + 0.1 N cos(20◦ ) = 0,
and
FY = TA sin(45◦ ) + N cos(20◦ ) + 0.1 N sin(20◦ )
y
TA
20°
45°
(20)(9.81)
− (20)(9.81) = 0.
x
0.1 N
N
Solving, we obtain TA = 56.3 N. We can now use this to find the
force T that must be applied to the rope to keep the box from slipping
down the plane. Eq. (9.17) is TA = T eµs β , where µs = 0.05 and
β = (π/180)(135) rad. Solving for T , we obtain T = 50.1 N.
135°
TA
T
Problem 8.103 The mass of the block A is 14 kg. The
coefficient of kinetic friction between the rope and the
cylinder is 0.2. If the cylinder is rotated at a constant
rate, first in the counterclockwise direction and then in
the clockwise direction, the difference in the height of
block A is 0.3 m. What is the spring constant k?
k
A
Solution:
k
T1 = T2 eµk β
Case 1: Clockwise Rotation
µk = 0.2
m = 14 kg
β = π/2
mg = Ts1 e(0.2)(π/2)
Ts1 = 100.31 N
A
Case 2: Counterclockwise Rotation
Case 1: Clockwise Rotation
Ts2 = mge(0.2)(π/2)
Ts2 = 188.03 N
k = 0.2
m = 14 kg
= /2
TS1
we know
Ts1 = kδ1
mg = TS1 e(0.2)(π/ 2)
Ts2 = kδ2
TS1 = 100.31 N
Ts2 − Ts1 = k(δ2 − δ1 )
mg
and δ2 − δ1 = 0.3 m
k = (Ts2 − Ts1 )/(δ2 − δ1 )
k = 292 N/m
Case 2: Counterclockwise Rotation
TS2
TS2 = mg e(0.2)(π/ 2)
TS2 = 188.03 N
mg
Problem 8.104 The weight of the box is W = 30 lb,
and the force F is perpendicular to the inclined surface.
The coefficient of static friction between the box and the
inclined surface is µs = 0.2.
(a) If F = 30 lb, what is the magnitude of the friction
force exerted on the stationary box?
(b) If F = 10 lb, show that the box cannot remain at
rest on the inclined surface.
F
W
20°
Solution: The maximum friction force is defined to be f = µs N ,
where N is the normal force.
(a) The box is stationary, hence the friction force is equal to the force
acting to move the box down the plane:
FP = f − WP = 0,
(b)
from which f = WP = W sin α = 10.26 lb
The component of force parallel to the surface is WP =
W sin α = 10.26 lb acting to move the box down the plane.
The friction force is f = µs (10 + 30 cos α) = 7.638 lb, acting
to hold the box in place. Since WP > f , the box will move.
F
W
α
F
20°
W
f
N
Problem 8.105 In Problem 8.104, what is the smallest force F necessary to hold the box stationary on the
inclined surface?
Solution: At impending slip, the sum of the forces parallel to the
surface is
FP = f − WP = 0,
from which f = WP . The friction force is f = µs (F + W cos α),
and WP = W sin α. Equate and solve:
sin α
sin 20◦
F =W
− cos α = 30
− cos 20◦ = 23.1 lb
µs
0.2
Problem 8.106 The mass of the van is 2250 kg, and
the coefficient of static friction between its tires and the
road is 0.6. If its front wheels are locked and its rear
wheels can turn freely, what is the largest value of α for
which it can remain in equilibrium?
1m
1.2 m
α
3m
Choose a coordinate system with the x-axis parallel to
the incline. The weight of the van is W = mg = 22072.5 N. The
moment about the point of contact of the rear wheels is
MR = (3 − 1.2)W cos α + 1W sin α − 3 N = 0,
Solution:
from which the normal force at the front wheels is
W (1.8 cos α + sin α)
N =
.
3
1m
1.2
α
m
3m
The sum of the forces parallel to the inclined surface is
Fx = +µs N − W sin α = 0.
Combine and reduce:
µ
1.8
s
µs
cos α +
− 1 sin α = 0,
3
3
from which
α = tan−1
1.8µs
3 − µs
1m
W
R
= tan−1 (0.45) = 24.2◦
µSN
N
1.8 m
1.2 m
Problem 8.107 In Problem 8.106, what is the largest
value of α for which the van can remain in equilibrium
if it points up the slope?
Solution: The sum of the moments about the point of contact of
the rear wheels is
MR = −1.8W cos α + 1W sin α + 3 N = 0.
The normal force is
The sum of forces parallel to the incline is
Fx = +µs N − W sin α = 0.
Combine and reduce:
µ
1.8µs
s
cos α −
+ 1 sin α = 0,
3
3
from which
α = tan−1
1.8µs
µs + 3
µsN
W
W (1.8 cos α − sin α)
N =
.
3
= 16.7◦
1m
N
1.2 m
R
1.8 m
A
Problem 8.108 Each of the uniform 1-m bars has a
mass of 4 kg. The coefficient of static friction between
the bar and the surface at B is 0.2. If the system is in
equilibrium, what is the magnitude of the friction force
exerted on the bar at B?
45°
O
B
30°
Solution:
The free body diagrams of the bars are as shown. The
equilibrium equations are
Left bar:
Fx = Cx + Ax = 0,
Fy = Cy + Ay − mg = 0,
M(leftend) = (1) cos 45◦ Ay − (1) cos 45◦ Ax − (0.5) cos 45◦ mg
A
Right bar:
B
45°
O
= 0,
30°
Ay
Fx = −Ax + f cos 30◦ − N sin 30◦ = 0,
45°
Fy = −Ay − mg + f sin 30◦ + N cos 30◦ = 0,
Cy
M(rightend) = (1) cos 45◦ Ax + (1) cos 45◦ Ay + (0.5) cos 45◦ mg
Ay
Ax
45°
Ax
mg
mg
f
= 0,
Cx
30°
N
Solving, we obtain N = 43.8 N and f = 2.63 N.
Problem 8.109 In Problem 8.108, what is the mini- Solution: From the solution of Problem 8.108, the normal and friction
are N = 43.8 N and f = 2.63 N. Slip impends when f = µs N so,
mum coefficient of static friction between the bar and forces 2.63
µs = 43.8 = 0.06.
the surface at B necessary for the system to be in equilibrium?
Problem 8.110 The clamp presses two pieces of wood
together. The pitch of the threads is p = 2 mm, the mean
radius of the thread is r = 8 mm, and the coefficient
of kinetic friction between the thread and the mating
groove is 0.24. What couple must be exerted on the
thread shaft to press the pieces of wood together with a
force of 200 N?
125 mm
125 mm
125 mm
B
50 mm
E
A
50 mm
C
50 mm
D
Solution: The free-body diagram of the upper arm of the clamp
is shown.
From the equilibrium equation
M(ptc) = −(0.25)(200) − 0.1BE = 0,
we find that BE = −500 N. The compressive load in BE is 500 N.
The slope of the thread is
P
α = arctan
2πr
= arctan
From Eq. (9.9) with θs = θk , the required couple is
M = rF tan(θk + α)
= (0.008)(500) tan(13.496◦ + 2.279◦ )
= 1.13 N-m.
BE
0.002
2π(0.008)
= 2.279◦ .
Cy
200 N
Cx
The angle of friction is
θk = arctan(0.24) = 13.496◦ .
0.1 m
0.25 m
Problem 8.111 In Problem 8.110, the coefficient of
static friction between the thread and the mating groove
is 0.28. After the threaded shaft is rotated sufficiently to
press the pieces of wood together with a force of 200 N,
what couple must be exerted on the shaft to loosen it?
Solution:
First, find the forces in the parts of the clamp. Then
analyze the threaded shaft. BE is a two force member
Fx :
BE + Cx = 0
(1)
Fy :
200 + Cy = 0
(2)
MA :
−0.05BE + 0.05Cx + 0.25Cy = 0 (3)
tan α = P/2πr =
2
2π(8)
α = 2.28◦
F = |BE| = 500 N
Solving
M = 0.950 N-m
Solving, we get
BE = −500 N
(compression)
Cy = −200 N
We don’t have to solve for additional forces because we used the fact
that member BE was a two force member.
From Problem 8.110, P = 2 mm, r = 8 mm. We have µs = 0.28.
We want to loosen the clamp (Turn the clamp such that the motion is
in the direction of the axial force.
To do this,
125 mm
B
50 mm
E
A
50 mm
C
50 mm
D
M = rF tan(θs − α)
y
where
tan θs = µs = 0.28
125 mm
125 mm
Cx = 500 N
0.125 m
B
0.05 m
θs = 15.64◦
A
0.125 m
200 N
BE
x
0.05 m
C
CX
CY
Problem 8.112 The axles of the tram are supported
by journal bearing. The radius of the wheels is 75 mm,
the radius of the axles is 15 mm, and the coefficient of
kinetic friction between the axles and the bearings is
µk = 0.14. The mass of the tram and its load is 160 kg.
If the weight of the tram and its load is evenly divided
between the axles, what force P is necessary to push the
tram at a constant speed?
Solution: Assume that there are two bearings per axle. The weight of the
tram is W = mg = 1569.6 N. This load is divided between four bearings:
F =
W
= 392.4 N.
4
The angle of kinetic friction is θk = tan−1 (µk ) = 7.97◦ . The moment required to turn each bearing at a constant rate is M = F r sin θk = 0.8161 N m,
and the force per wheel is
Pw =
M
0.8161
=
= 10.88 N.
R
0.075
The total force required to push the tram is P = 4Pw = 43.5 N
P
P
Problem 8.113 The two pulleys have a radius of 6 in.
and are mounted on shafts of 1-in. radius supported by
journal bearings. Neglect the weights of the pulleys and
shafts. The coefficient of kinetic friction between the
shafts and the bearings is µk = 0.2. If a force T = 200 lb
is required to raise the man at a constant rate, what is his
weight?
T
Solution: Denote the tension in the horizontal portion of the cable
by H. The angle of kinetic friction is θk = tan−1 (0.2) = 11.31◦
Consider the right Pulley: The force on the right pulley is
F = T 2 + H2.
The magnitude
√ of the moment required to turn the shaft in the bearing is
Mright = r T 2 + H 2 sin θk . The applied moment is Mapplied =
√
(T − H)R, from which (T − H)R = r T 2 + H 2 sin θk . Square
both sides and reduce to obtain the quadratic:
1
H 2 − 2T H
+ T 2 = 0,
r 2
1− R
sin2 θk
or H 2 − 2(200.214)H + 40000 = 0.
√
This has the solutions: H = 200.214 ± 200.2142 − 40000 =
209.46, 190.95. The lesser root corresponds to the horizontal tension,
H = 190.97 = 191 lb.
Consider
the left Pulley: The force on the pulley is Fleft =
√
W 2 + H 2 . The applied moment is Mapplied = −(H − W )R,
√
from which (H − W )R = r W 2 + H 2 sin θk . Square both sides
and reduce to the quadratic:
H
2
W −2
W + H 2 = 0,
r 2
1− R
sin2 θk
or W 2 − 2(191.1)W + 36431.9 = 0.
√
This has the solutions: W1,2 = 191.166± 191.1662 − 36431.9 =
199.9 lb, 182.25 lb. By an analogous argument to that used in
Problem ??.??, the lesser root corresponds to the weight of the man,
Wraised = 182.3 lb
Problem 9.1 The prismatic bar has a circular cross
section with 50-mm radius and is subjected to 4-kN axial
loads. Determine the average normal stress at the plane
P.
Free Body Diagrams:
Solution:
The cross-sectional area of the bar is:
A = πr 2 = π(0.05 m)2 = 0.007854 m2
The average normal stress is the load distributed across the face of the
cross section.
σAV = P/A = (4000 N)/(0.007854 m2 )
ANS:
σAV = 509 kPa
Problem 9.2 In Problem 9.1, what is the average shear
stress at the plane P ?
Solution:
Free Body Diagram:
We see from the FBD that the shearing force is the only force with any
vertical component. Summing vertical forces:
ΣFy = 0 = −τ A = −τ [π(0.025 m)2 ]
ANS:
τ =0
Problem 9.3 The prismatic bar has a cross-sectional
area A = 30 in2 and is subjected to axial loads. Determine the average normal stress (a) at plane P1 ; (b) at
plane P2 .
Free Body Diagrams:
Solution:
At Plane P1 , the average normal stress is:
(σAV )1 = P/A = (2000 lb)/(30 in2 )
ANS: (σAV )1 = 66.7 psi
At Plane P2 , the average normal stress is:
(σAV )2 = P/A = (2000 lb)/(30 in2 )
ANS:
(σAV )2 = 66.7 psi
Problem 9.4 The prismatic bar has a solid circular
cross section with 2-in. radius. Determine the average
normal stress (a) at plane P1 ; (b) at plane P2 .
Free Body Diagram:
Solution:
The cross-sectional area of the prismatic bar is:
A = πr 2 = π(2 in)2 = 12.566 in2
Note that the loads are NOT the same at P1 and P2 . The bar is in
tension at P1 and in compression at P2 .
(a) The average normal stress at P1 is:
(σAV )1 = P/A = (4000 lb)/(12.566 in2 )
ANS: (σAV )1 = 318.3 psi
(b) The average normal stress at P2 is:
(σAV )2 = P/A = (−8000 lb)/(12.566 in2 )
ANS:
(σAV )2 = −636.6 psi
Problem 9.5 A prismatic bar with cross-sectional area
A is subjected to axial loads. Determine the average
normal and shear stresses at the plane P if A = 0.02 m2 ,
P = 4 kN, and θ = 25◦ .
Free Body Diagram:
Solution:
The cross-sectional area of this section is larger than 0.02 m2 because
the bar is cut at an angle. The actual area of the angled cut is:
A = (0.02 m2 )/(cos 25◦ ) = 0.02206 m2
Summing horizontal forces on the bar:
ΣFx = 0 = −4000 N+[σAV (0.02206 m2 )](cos 25◦ )−τAV (0.02206 m2 )(sin 25◦ )
[1]
τAV = −429, 048 N + 2.145σAV
Summing vertical forces on the bar:
Σy = 0 = σAV (0.02206 m2 )(sin 25◦ )+τAV (0.02206 m2 )(cos 25◦ )
[2]
τAV = −0.4663σAV
Solving Equations [1] and [2] together:
ANS: σAV = 164.3 kPa
ANS: τAV = −76.6 kPa Note: The (−) sign indicates the
wrong direction assumed for the FBD.
Problem 9.6 Suppose that the prismatic bar shown in
Problem 9.5 has cross-sectional area A = 0.024 m2 . If
the angle θ = 35◦ and the average normal stress on the
plane P is σAV = 200 kPa, what are τAV and the axial
force P ?
Free Body Diagram:
Solution:
The cross-section area at plane P is:
A = (0.024 m2 )/(cos 35◦ ) = 0.0293 m2
Summing vertical forces on the FBD:
ΣFy = 0 = τAVG A(sin 55◦ ) + σAVG A(sin 35◦ )
τAVG = −0.7σAVG = −(0.7)(200 kPa)
τAVG = −140 kPa (−) indicates opposite direction.
Summing horizontal forces on the FBD:
ANS:
ΣFx = 0 = −P + σAVG A(cos 35◦ ) − τAVG A(cos 55◦ )
0 = −P +(200 kPa)(0.0293 m2 )(cos 35◦ )−(−0.7)(200 kPa)(0.0293 m2 )(sin 35◦ )
ANS:
P = 7.15 kN
Problem 9.7 For the prismatic bar in Problem 9.5, derive equations for the average normal and shear stresses
at the plane P as functions of θ.
Free Body Diagram:
σAV A
Solution:
Summing forces in the horizontal direction:
ΣFx = 0 = [A/(cos θ)]σAV (cos θ)−[A/(cos θ)]τAV (sin θ)−P
[1]
τAV = σAV (cos θ)/(sin θ) − [P (cos θ)]/A(sin θ)
Summing forces in the vertical direction:
ΣFy = 0 = σAV A(sin θ) + τAV A(cos θ)
[2]
τAV = −σAV (sin θ)/(cos θ)
Solving Equations [1] and [2] together:
−σAV (sin θ)/(cos θ) = σAV (cos θ)/(sin θ)−[P (cos θ)]/A(sin θ)
−σAV [(sin θ)/(cos θ)]−[(cos θ)/(sin θ)] = −[P (cos θ)]/A(sin θ)
−[σAV (sin2 θ) + (cos2 θ)] = −[P (cos2 θ)]/A
ANS: σAV = P (cos2 θ)/A
From Equation [2]:
τAV = −σAV (sin θ)/(cos θ) = P [(cos2 θ)/A][(sin θ)/(cos θ)]
ANS:
τAV = (P/A)(sin θ)(cos θ)
θ
Problem 9.8 The prismatic bar has a solid circular Free Body Diagram:
cross section with 30-mm radius. It is suspended from
one end and is loaded only by its own weight. The mass
density of the homogeneous material is 2800 kg/m3 . Determine the average normal stress at the plane P , where
x is the distance from the bottom of the bar in meters.
Strategy: Draw a free-body diagram of the part of the
bar below the plane P .
Solution:
The weight of the cylinder for any distance x from its bottom is:
W = ρgπr2 x = (2800 kg/m3 )(9.81 m/sec2 )π(0.03 m)2 x
W = 77.66x N
The stress produced in supporting this weight is:
σ = W/A = (77.66x N)/π(0.03 m)2
ANS:
σ = 27, 467x Pa = 27.5x kPa
Problem 9.9 The beam has cross-sectional area A =
0.0625 m2 . What are average normal stress and the magnitude of the average shear stress at the plane P ?
Strategy: Draw the free-body diagram of the entire
beam and determine the reactions at the pin and roller
supports. Then determine the average normal and shear
stresses by drawing the free-body diagram of the part of
the beam to the left of plane P .
Free Body Diagram:
Solution:
To find Ax , sum horizontal forces.
ΣFx = 0 = 4 kN − Ax
Ax = 4 kN ←
To find Ay , sum moments about point B:
ΣMB = 0 = (2 kN)(2 m) − (6 kN − m) + Ay (6 m)
Ay = 333.33 N ↓
Σfy = 0 = −333.3 − 2000 + By = 0
By = 2333.33 N ↑
Now cut the beam through the plane P . Using the left-hand portion of
the beam:
ΣFx = 0 = σA − Ax = σAV G (0.0625 m2 ) − 4, 000 N
ANS: σAV G = 64 kPa
Summing vertical forces on the left-hand portion of the beam:
ΣFy = 0 = −Ay + τAV G A = −333.3 lb + τAV G (0.0625 m2 )
ANS:
τAV G = 5.33 kPa
Problem 9.10 Determine the average normal stress
and the magnitude of the average shear stress at the plane
P of the beam in Problem 9.9 by drawing the free-body
diagram of the part of the beam to the right of plane P
and compare your answers to those of Problem 9.9.
Free Body Diagram:
Solution:
Sum moments about the left-hand end of the beam to find By :
ΣMA = 0 = −(2, 000 N)(4 m) − 6, 000 N − m + By (6 m)
By = 2, 333 N ↑
Summing vertical forces on the FBD of the right-hand portion of the
beam:
ΣFy = 0 = 2, 333 N − 2, 000 N − τAV G (0.0625 m2 )
ANS: τAV G = 5.33 kPa
Summing horizontal forces on the FBD of the right-hand portion of
the beam:
ΣFx = 0 = 4, 000 N − σAV G (0.0625 m2 )
ANS:
σAV G = 64 kPa
Problem 9.11 The beams have cross-sectional area
A = 60 in2 . What are the average normal stress and
the magnitude of the average shear stress at the plane P
in cases (a) and (b)?
Solution:
18 kip
Ax
20 kip
Ay
(a)
By
Summing moments about point B on the beam:
ΣMB = 0 = −(18, 000 lb)(4 ft) + Ay (12 ft)
Ay = 6, 000 lb ↑
F.2 = 1/2(12)(3000)
8'
Ax
4'
Ay
20 kip
By
Summing horizontal forces on the beam:
ΣFx = 0 = −20, 000 lb + Ax
Ax = 20, 000 lb →
Summing vertical forces on the left-hand portion of the beam:
ΣFy = 0 = Ay − τavg (area) = 6, 000 lb − τavg (60 in2 )
ANS: τavg = 100 psi
Summing horizontal forces on the left-hand portion of the beam:
ΣFx = 0 = Ax − σavg (area) = 20, 000 lb − σavg (60 in2 )
ANS:
(b)
σavg = +333.3 psi(C)
Summing moments about point B on the beam:
ΣMB = 0 = −1/2(3, 000 lb/ft)(12 ft)(4 ft) + Ay (12 ft)
Ay = 6, 000 lb ↑
Summing horizontal forces on the beam:
ΣFx = 0 = −20, 000 lb + Ax
Ax = 20, 000 lb →
Summing vertical forces on the left-hand portion of the beam:
ΣFy = 0 = Ay −1/2(1, 500 lb/ft)(6 ft)−τavg (area) = 6, 000 lb−4, 500 lb+τavg (60 in2 )
τavg = 650 psi
Summing horizontal forces on the left-hand portion of the beam:
ANS:
ΣFx = 0 = Ax − σavg (area) = 20, 000 lb − σavg (60 in2 )
ANS:
σavg = −333.33 psi(C)
Problem 9.12 Figure (a) is a diagram of the bones and
biceps muscle of a person’s arm supporting a mass. Figure (b) is a biomedical model of the arm in which the
biceps muscle AB is represented by a bar with pin supports. The suspended mass is m = 2 kg and the weight
of the forearm is 9 N. If the cross-sectional area of the tendon connecting the biceps to the forearm at A is 28 mm2 ,
what is the average normal stress in the tendon.
Free Body Diagram:
Solution:
Summing moments about point C:
ΣMc = 0 = −(19.62 N)(0.35 m)−(9 N)(0.15 m)+σavg (28×10−6 m2 )(0.05 m(sin 80.2))
ANS:
σavg = 5.96 MPa
Problem 9.13 The force F exerted on the bar is 20i −
20j − 10k (lb). The plane P is parallel to the y-z plane
and is 5 in. from the origin O. The bar’s cross sectional
area at P is 0.65 in2 . What is the average normal stress
in the bar at P ?
Free Body Diagram:
Solution:
The average normal stress at point P is exerted only by the xcomponent of the applied force. Summing forces in the x-direction:
ΣFx = 0 = 20 lb − σavg (area) = 20 lb − σavg (0.65 in2 )
ANS:
σavg = 30.8 psi
Problem 9.14 In Problem 9.13, what is the magnitude
of the average shear stress in the bar at P ?
Free Body Diagram:
Solution:
The average shear stress at point P is exerted only by the y- and zcomponents of the applied force. Summing forcers in the y-direction:
ΣFy = 0 = −20 lb + Py
Py = 20 lb
Summing forces in the z-direction:
ΣFz = 0 = −10 lb + Pz
Pz = 10 lb
The y- and z-components of the reaction at P are added vectorially.
RP = Py2 + Pz2 = (20 lb)2 + (10 lb)2
RP = 22.36 lb
The average shear stress at point P is:
τavg = RP /A = (22.36 lb)/(0.65 in2 )
ANS:
τavg = 34.4 psi
Problem 9.15 The plane P is parallel to the y-z plane
of the coordinate system. The cross-sectional area of
the tennis racquet at P is 400 mm2 . Including the force
exerted on the racquet by the ball and inertial effects,
the total force on the racquet above the plane P is 35i −
16j − 85k N. What is the average normal stress on the
racquet at P ?
Free Body Diagram:
Solution:
The average normal stress at point P is exerted only by the xcomponent of the applied force. Summing forces in the x-direction:
ΣFy = 0 = 35 N − σavg (area) = 35 N − σavg (400 × 10−6 m2 )
ANS:
σavg = 87, 500 Pa = 87.5 kPa
Problem 9.16 The fixture shown connects a 50-mm Free Body Diagram:
diameter bridge cable to a flange that is attached to the
bridge. A 60-mm diameter circular pin connects the
fixture to the flange. If the average normal stress in the
cable is σAV = 120 MPa, what average shear stress
τAV must the pin support?
Solution:
The average normal stress of 120 MPa in the cable means that the axial
load in the cable is:
P = σavg A = (120 × 106 N/m2 )(π)(0.025 m)2
P = 236 kN
We see that the shear load in the pin is divided between two areas, one
on either side of the flange. The average shear stress which must be
supported by the pin is:
τavg = P/(2A) = (236, 000 N)/(2)[π(0.03 m2 )]
ANS:
τavg = 41.7 MPa
Problem 9.17 Consider the fixture shown in Problem 9.16. The cable will safely support an average normal stress of 700 MPa and the circular pin will safely
support an average shear stress of 220 MPa. Based on
these criteria, what is the largest tensile load the cable
will safely support?
Solution:
Calculating the largest load which can be supported by the cable:
(FMAX )CABLE = (σAVG )ALLOW A = (700×106 N/m2 )[π(0.025 m)2 ]
(FMAX )CABLE = 1.37 MN
Calculating the largest load which can be supported by the pin:
(FMAX )PIN = 2[(τAVG )MAX A] = 2[(220×106 N/m2 )(π)(0.03 m)2 ]
(FMAX )PIN = 1.24 MN
We see that the cable can support a larger load than the pin. Any load
larger than 1.24 MN will cause the pin to fail in shear. The largest load
which can be supported by the cable and pin is:
ANS: FMAX = 1.24 MN
Problem 9.18 The truss is made of prismatic bars with Free Body Diagram:
cross-sectional area A = 0.25 ft2 . Determine the average normal stress in member BE acting on a plane
perpendicular to the axis of the member.
Solution:
We can determine the axial load in member BE directly by summing
vertical forces on the FBD.
ΣFy = 0 = −8, 000 lb − 10, 000 lb + PBE (sin 45◦ )
PBE = 25.5 kip
The average normal stress in member BE of the truss is:
σavg = (PBE )/A = (25, 500 lb)/(0.25 ft2 )
ANS:
σavg = 102 ksf = 708 psi
Problem 9.19 For the truss in Problem 9.18, determine Free Body Diagram:
the average normal stress in member BD acting on a
plane perpendicular to the axis of the member.
Solution:
To solve directly for the axial load in member BD of the truss, sum
moments on the FBD about point E.
ΣME = 0 = −(8, 000 lb)(20 ft)−(10, 000 lb)(10 ft)+PBD (10 ft)
PBD = 26, 000 lb (C) = 26 kip (C)
The average normal stress in member BD of the truss is:
σavg = PBD /A = (26, 000 in)/(0.25 ft2 )
σavg = −104, 000 psf(C) = −722.2 psi NOTE: The
negative sign indicates compressive stress.
ANS:
Problem 9.20 Three views of joint A of the truss in
Problem 9.18 are shown. The joint is supported by a
cylindrical pin 2 in. in diameter. What is the magnitude
of the average shear stress in the pin?
Free Body Diagram:
Solution:
Sum vertical forces at joint A to determine the axial load in member
AC.
ΣFy = 0 = −8, 000 lb + PAC (sin 45◦ )
PAC = 11, 313.7 lb (T) = 11.3 kip (T)
Sum horizontal forces at joint A to find the axial load in member AB.
ΣFx = 0 = PAB −PAC (cos 45◦ ) = PAB −(11, 300 lb)(cos 45◦ )
PAB = 8, 000 lb (C) = 8 kip (C)
When the 8,000 lb load and PAB are added vectorially, we see the load
which must be supported by the pin (right-hand free body diagram).
The average shear stress in the pin is:
τavg = (11, 313.7 lb)/[(π)(1 in)2 ]
ANS:
τavg = 3597 psi
Problem 9.21 The top view of pin A of the pliers is
shown. The cross-sectional area of the pin is 4.5 mm2 .
What is the average shear stress in the pin when 150-N
forces are applied to the pliers as shown?
Free Body Diagram:
Solution:
The easier way in which to find the load in member AB is to draw the
FBD of the lower handle and sum moments about point D.
ΣMD = 0 = (150 N)(0.13 m) − PAB (sin 23.2◦ )(0.03 m)
PAB = 1650 N (C)
The shear load which must be supported by the pin at joint A is the
load in member AB. The average shear stress is:
τavg = (PAB )/(A) = (1650 N)/(4.5 × 10−6 m2 )
ANS:
τavg = 367 × 106 Pa = 367 MPa
Problem 9.22 In Problem 9.21 the vertical plane P is Free Body Diagrams:
30 mm to the left of C. The cross-sectional area of member AC of the pliers at plane P is 50 mm2 . Determine
the average normal stress and the magnitude of the average shear stress at P when 150-N forces are applied to
the pliers as shown.
Solution:
The easier way in which to find the load in member AB is to draw the
FBD of the lower handle and sum moments about point D.
ΣMD = 0 = (150 N)(0.13 m) − PAB (sin 23.2◦ )(0.03 m)
PAB = 1650 N (C)
To find the average normal stress at plane P , sum horizontal forces on
the FBD of the upper handle.
ΣFx = 0 = −PAB (cos 23.2◦ )+σavg A = −(1650 N)(cos 23.2◦ )+σavg (50×10−6 m2 )
ANS: σavg = 30.3 × 106 Pa = 30.3 MPa
To find the average shear stress at plane P , sum vertical forces on the
FBD of the upper handle.
ΣFy = 0 = −150 N+(1650 N)(sin 23.2◦ )−τavg (50×10−6 m2 )
ANS:
τavg = 10 × 106 Pa = 10 MPa
Problem 9.23 The suspended crate weighs 2000 lb and
the angle α = 30◦ . The top view of the pin support A
of the crane’s boom is shown. The cross-sectional area
of the pin is 23 in2 . What is the average shear stress in
the pin?
Free Body Diagram:
Solution:
The Pythagorean Theorem will help to determine the distance, x.
(9 ft)2 = (4.5 ft)2 + (6 ft + x)2
x = 1.79 ft
From the geometric construction
tan−1 (1.79 ft/4.5 ft), or:
we
see
that
β
=
β = 21.7◦
Summing moments about point A:
ΣMA = 0 = −(2000 lb)(15 ft)(cos 30◦ )+PBC (cos 21.7◦ )[(9 ft)(cos 30◦ )]−PBC (sin 21.7◦ )[(9 ft)(sin 30◦ )]
PBC = 4658 lb (C)
To find the reaction at point A, sum horizontal and vertical forces on
the boom.
ΣFx = 0 = PBC (sin 21.7◦ ) − Ax = (4657 lb)(sin 21.7◦ ) − Ax
Ax = 1722 lb ←
ΣFy = 0 = −2, 000 lb+PBC (cos 21.7◦ )−Ay = −2, 000 lb+(4657 lb)(cos 21.7◦ )−Ay
Ay = 2327 lb ↓
Total shearing force at joint A is:
FA = A2x + A2y = (1722 lb)2 + (2327 lb)2
FA = 2895 lb
We see that the shear load is divided between the cross-sections on
either side of the pin where it connects member BC to the boom. The
average shearing load is:
τavg = FA /2A = (2895 lb)/(2)(23 in2 )
ANS:
τavg = 62.9 psi
Problem 9.24 In Problem 9.23, the plane P is 3 ft
from end D of the crane’s boom and is perpendicular to
the boom. The cross-sectional area of the boom at P
is 15 in2 . Determine the average normal stress and the
magnitude of the average shear stress in the boom at P .
Free Body Diagram:
Solution:
Note the directions for the x- and y-axes on the free body diagram.
The average normal stress can be found by summing forces in the
x-direction.
ΣFx = 0 = −(2, 000 lb)(sin 30◦ )+σavg A = −(2, 000 lb)(sin 30◦ )+σavg (15 in2 )
σavg = −66.7 psi NOTE: The negative sign indicates that
the normal stress is compressive.
The average shear stress can be determined by summing forces in the
y-direction.
ANS:
ΣFy = 0 = −(2, 000 lb)(cos 30◦ )+τavg A = −(2, 000 lb)(cos 30◦ )+τavg (15 in2 )
ANS:
τavg = 115.5 psi
Problem 9.25 Three rectangular boards are glued together and subjected to axial loads as shown. What is
the average shear stress on each glued surface?
Free Body Diagram of the Upper Board
Solution:
The glued area of the upper board is:
A = lw = (0.09 m)(0.135 m)
A = 0.01215 m2
The average shear stress across the glued joint is:
τavg = P/A = (1, 000 N)/(0.01215 m2 )
ANS:
τavg = 82, 300 N = 82.3 kPa
Problem 9.26 Two boards with 4 in. × 4 in. square
cross sections are mitered and glued together as shown.
If the axial forces P = 600 lb, what average shear stress
must the glue support?
Free Body Diagram:
Solution:
Total glued area which is subjected to shear stress is:
A = 2[(4 in)(4 in)]
A = 32 in2
The average shear stress across the glued surfaces is:
τavg = P/A = (600 lb)/(32 in2 )
ANS:
τavg = 18.75 psi
Problem 9.27 A 1/8-in. diameter punch is used to cut
blanks out of a 1/16-in. thick plate of aluminum. If an
average shear stress of 20,000 psi must be induced in the
plate to create a blank, what force F must be applied?
Free Body Diagram:
Solution:
The average shear stress in the material to be punched will be exerted
around the entire cylindrical area which is left after the hole has been
punched. The cylindrical area for this process is:
A = 2πrh = 2π(1/16 in)(1/16 in)
A = 0.0245 in2
The force required to produce the needed average shear stress is:
P = τavg A = (20, 000 lb/in2 )(0.0245 in2 )
ANS: P = 490.9 lb
or as per Example 9.3
F = τAV πtD = 20, 000(π)(
1 1
)( ) = 490.9 lbs
16 8
Problem 9.28 Two pipes are connected by bolted
flanges. The bolts are 20 mm in diameter. One pipe has
a built-in support and the other is subjected to a torque
T = 6 kN − m about its axis.
Estimate the resulting average shear stress in each bolt.
Why is the result an estimate?
Free Body Diagram:
Solution:
The free body diagram illustrates the manner in which the bolts resist
an applied counterclockwise torque. The moment exerted by each of
the bolts is:
M = τavg Ar → τavg = M/6 Ar
τavg = (6, 000 N − m)/(6)[(π)(0.01 m)2 (0.15 m)]
ANS: τavg = 21.2 × 106 Pa = 21.2 MPa
The result assumes that the load is evenly distributed among the six
bolts. The load supported by each bolt can be affected by (among
other factors) the degree to which each bolt is (or is NOT) tightened,
the accuracy of their radii from the axis of the flange, the uniformity of
material quality among the bolts, and the precision of their diameters.
Problem 9.29 The bolts in Problem 9.28 will each
safely support an average shear stress of 130 MPa. Based
on this criteria, estimate the largest safe torque that can
be applied.
Free Body Diagram:
Solution:
The free body diagram illustrates the manner in which the bolts resist
an applied counterclockwise torque. The moment exerted by each of
the bolts is:
M = 6(τavg Ar) = (6)[(130×106 N/m2 )[π(0.01 m)2 ](0.15 m)]
ANS:
M = 36, 800 N − m = 36.8 kN − m
Problem 9.30 “Shears”, such as the familiar scissors,
have two blades which subject a material to shear stress.
For the shearing process shown, draw a suitable freebody diagram and determine the average shear stress the
blades exert on the sheet of material of thickness t and
width b.
Strategy: Obtain a free-body diagram by passing a
vertical plane through the material between the two
blades.
Free Body Diagram:
Solution:
The average shear stress is assumed to be evenly distributed across the
entire cross-section of the material.
The area over which the shear load is distributed is:
A = bt
The magnitude of the average shear stress is:
τavg = F/A
ANS:
τavg = F/(bt)
Problem 9.31 A 2-in. diameter cylindrical steel bar
is attached to a 3-in. thick fixed plate by a cylindrical
rubber grommet. If the axial load P = 60 lb, what is the
average shear stress on the cylindrical surface of contact
between the bar and the grommet?
Solution:
The area of contact between the bar and the grommet is:
A = 2πrh = 2π(1 in)(3 in) = 18.8 in2
The average shear stress between the bar and the grommet is:
τavg = P/A = (60 lb)/(18.8 in2 )
ANS:
τavg = 3.19 psi
Problem 9.32 The outer diameter of the cylindrical
rubber grommet in Problem 9.31 is 3.5 in. What is the
average shear stress on the cylindrical surface of contact
between the grommet and the fixed plate?
Solution:
The cylindrical area of contact between the grommet and the fixed
plate is:
A = πdh = π(3.5 in)(3 in) = 32.99 in2
Average shear stress between the grommet and the fixed plate is:
τAVG =
ANS:
P
60 lb
=
A
32.99 in2
τAVG = 1.82 psi
Problem 9.33 The steel bar described in Problem 9.31
is subjected to a torque T = 100 in − lb about its axis.
What is the average shear stress on the cylindrical surface
of contact between the bar and the grommet?
Solution:
The area of contact between the bar and the grommet is:
A = 2πrh = 2π(1 in)(3 in) = 18.8 in2
The force resulting from the shear stress between the bar and the grommet is:
F = τavg A = τavg (2π)(1 in)(3 in)
F = 6πτavg lb
The force exerted by the average shear stress is located at a radius from
the axis of the bar of:
r = 1 in.
The average shear stress which is produced by the applied moment is:
M = F r = (6π)(τavg )(1 in) = 100 in − lb
ANS:
τavg = 5.31 psi
Problem 9.34 A traction distribution t acts on a plane
surface A. The value of t at a given point on A is t =
45i + 40j − 30k kPa.
The unit vector i is perpendicular to A and points away
from the material. What is the normal stress σ at the
given point?
Solution:
The i-component of the applied stress is the only portion which contributes to NORMAL stress. The j- and k-components act in the plane
of the material and so contribute to shear stress.
The normal stress at any point on the plane surface is:
ANS: σavg = 45 kPa
Problem 9.35 In Problem 9.34, what is the magnitude
of the shear stress τ at the given point?
Solution:
The components of the applied stress which produce shear are the jand k-components. The resultant of these two orthogonal components
is:
R = t2j + t2k = (40 kPa)2 + (−30 kPa)2
ANS:
R = 50 kPa
Problem 9.36 A traction distribution t acts on a plane
surface A. The value of t at a given point on A is t =
3000i − 2000j + 6000k psi.
The unit vector e = (6/7)i + (3/7)j + (2/7)k is perpendicular to A and points away from the material. What is
the normal stress σ at the given point?
Solution:
To find that portion of the applied stress which is normal to the plane
surface it is necessary to find the scalar product between the applied
stress and the unit vector which is normal to the plane.
Finding the scalar product between the vectors:
σavg = t·e = (3000i−2000j+6000k)·(6/7i+3/7j+2/7k) = [(3000 psi)(6/7)]+[(−2000 psi)(3/7)]+[(6000 psi)(2/7)]
ANS:
σavg = 3430 psi
Problem 9.37 A line of length dL at a particular point
of a material in a reference state has length dL = 1.2 dL
in a deformed state. What is the extensional strain corresponding to that particular point and the direction of
the line dL?
Solution:
Extensional strain is the ratio of the change in length to the length In
the reference state.
ε = (dL − dL)/dL = (1.2dL − dL)/dL
ANS:
ε = 0.2
Problem 9.38 The extensional strain corresponding to
a point of a material and the direction of a line of length
dL in the reference state is ε = 0.15. What is the length
dL of the line in the deformed state?
Solution:
The definition of extensional strain can be used in this solution.
Using Equation 9.3 from the text:
dL = (1 + ε)dL = (1 + 0.15)dL
ANS:
dL = 1.15dL
Problem 9.39 A straight line within a reference state
of an object is 50 mm long.
In a deformed state, the line is 54 mm long. If the extensional strain ε in the direction tangent to the line is
uniform throughout the line’s length, what is ε?
Solution:
Using Equation 9.3 from the text:
dL = (1 + ε)dL
ε = (dL − dL)/dL = (54 mm − 50 mm)/(50 mm)
ANS:
ε = 0.08
Problem 9.40 The length of the curved line within the
material in the reference state is L = 0.2 m. The material then undergoes a deformation such that the value
of the extensional strain ε in the direction tangent to the
curved line is ε = 0.03 at each point of the line.
What is the length L of the line in the deformed state?
Solution:
Using Equation 9.3 from the text:
L = (1 + ε)L
L = (1 + 0.03)(0.2 m)
ANS:
L = 0.206 m
Problem 9.41 The length of the curved line L in the
reference state shown in Problem 9.40 is 0.5 m. Its length
in the deformed state is L = 0.488 m. If the value of
the extensional strain ε in the direction tangent to the line
is the same at each point of the line, what is ε?
Solution:
Using the definition of strain:
ε=
ANS:
L − L
0.488 m − 0.5 m
=
L
0.5 m
ε = −0.024
Problem 9.42 The coordinate s measures distance
along the curved line in the reference state. The length of
the line is L = 0.2 m. The material then undergoes a deformation such that the value of the extensional strain ε in
the direction tangent to the curved line is ε = 0.03+2s2 .
What is the length L of the line in the deformed state?
Solution:
Since the extensional strain rate is not constant, we need to integrate.
Using Equation 9.4 from the text:
0.2
L = 00.2 (1 + ε)ds = 00.2 1 + 0.03 + 2s2 ds = 00.2 1.03 + 2s2 ds = 1.03s + 23 s3 0
L = 0.211m
Problem 9.43 In Problem 9.42, suppose that the material undergoes a deformation that induces an extensional strain tangent to the line given by the equation
ε = 0.01[1 + (s/L)3 ]. What is the length L of the line
in the deformed state?
Solution:
Since the extensional strain rate is not constant, we need to integrate.
Using Equation 9.4 from the text:
3 s 3 s
L = 0L (1 + ε) ds = 0L 1 + 0.01 1 + L
ds = 0L 1.01 + 0.01 L
ds
3
4 L
4
(L)
s
= 1.01L + 0.01 4L3
L = 1.01s + 0.01 4L
3 L = 1.0125L
0
Problem 9.44 In a shock-wave experiment, the left
side of a 100-mm thick plate of steel is subjected to a
constant velocity of 1.5 km/s to the right at time t = 0.
As a result, a shock wave travels across the plate with
a constant velocity U > 1.5 km/s. To the right of the
shock wave, the material of the plate is stationary and
undeformed. To the left of the shock wave, the material is moving with a uniform velocity of 1.5 km/s and is
subject to a homogeneous (uniform) extensional strain ε.
If optical instrumentation indicates that the shockwave
arrives at the right side of the plate at t = 16 × 10−6 s,
what is ε?
Solution:
The distance traveled by the shock wave during the elapsed time is:
D = RT = (1500 m/sec)(16 × 10−6 sec) = 0.024 m
The strain generated by the shock wave is the distance traveled by the
wave divided by the thickness of the material:
ε = D/t = (0.024 m)/(0.100 m)
ANS:
ε = 0.24
Problem 9.45 In Problem 9.44, suppose that the time
at which the shock wave arrives at the right side of the
plate is unknown, but an embedded strain gage indicates
that the homogeneous extensional strain to the left of the
shock wave is ε = −0.3.
What is the velocity U of the shock wave?
Solution:
The change in dimension of the left-hand portion of the plate is:
∆L = (0.1 m)(−0.3) = −0.03
Calculating the time required for the shock wave to travel through the
plate:
−0.03 = (−1500 m/s) t → t = 2 × 10−5 sec
The speed of the shock wave is:
v=
ANS:
D
0.1 m
=
= 5000 m/s
t
2 × 10−5 sec
v = 5 km/sec
Problem 9.46 When it is unloaded, the nonprismatic
bar is 12 in. long. The loads cause axial strain given by
the equation ε = 0.04/(x + 12), where x is the distance
from the left end of the bar in inches. What is the change
in length of the bar?
Solution:
The change in length is the deformed length minus the reference length.
δ=
12
0
(1 + ε)dx − L =
12
0
(1 +
0.04
x+12
dx − L = [x + 0.04 ln (x + 12)]|12
0 − 12in
δ = [12 + 0.04 ln (12 + 12) − (0 + 0.04 ln (0 + 12))] − 12in
δ = 0.028in
Problem 9.47 The force F causes point B to move Drawing:
downward 0.002 m. If you assume the resulting extensional strain ε parallel to the axis of the bar AB is uniform
along the bar’s length, what is ε?
Solution:
The original length of the bar is:
L = (0.6 m)2 + (0.8 m)2 = 1.0 m
The deformed length of the bar is:
L = (0.6 m)2 + (0.798 m)2 = 0.9984 m
The strain is:
−L
m−1.0 m
ε = LL
= 0.99841.0
m
ε = −0.00160
Problem 9.48 When the truss is subjected to the vertical force F , joint A moves a distance v = 0.3 m vertically and a distance u = 0.1 m horizontally. If the
extensional strain εAB in the direction parallel to member AB is uniform throughout the length of the member,
what is εAB ?
Diagram:
Solution:
The reference length of member AB is:
2m
= 2.309 m
sin 60◦
L=
The deformed length of member AB is:
L = (2.2996 m)2 + (1.2545 m)2 = 2.6195
The strain for member AB is:
−L
εAB = L L
=
εAB = 0.1344
2.6195 m−2.309 m
2.309 m
Problem 9.49 In Problem 9.48, if the extensional
strain εAC in the directional parallel to member AC is
uniform throughout the length of the member, what is
εAC ?
Diagram:
Solution:
The reference length of member AC is:
L = (2 m)2 + (2 m)2 = 2.828 m
The deformed length of member AC is:
L = (2.299 m)2 + (1.899 m)2 = 2.982 m
The strain for member AC is:
−L
ε = LL
=
ε = 0.0548
2.982 m−2.828 m
2.828 m
Problem 9.50 Suppose that a downward force is applied at point A of the truss, causing point A to move
0.360 in. downward and 0.220 in. to the left. If the
resulting extensional strain εAB in the direction parallel
to the axis of bar AB is uniform, what is εAB ?
Drawing:
Solution:
The reference length of member AB is:
L = (16 in)2 + (16 in)2 = 22.627 in
The deformed length of member AB is:
L = (16.36 in)2 + (15.78 in)2 = 22.73 in
The strain in member AB is:
−L
ε = LL
=
ε = 0.00455
22.73 in−22.627 in
22.627 in
Problem 9.51 In Problem 9.50, if the resulting extensional strain εAC in the direction parallel to the axis of
bar AC is uniform, what is εAC ?
Drawing:
Solution:
The reference length for member AC is:
L = (16 in)2 + (24 in)2 = 28.844 in
The deformed length for member AC is:
L = (24.22 in)2 + (16.36 in)2 = 29.228 in
The strain in member AC is:
−L
ε = LL
=
ε = 0.0133
29.228 in−28.844 in
28.844 in
Problem 9.52 A steel tube (a) has an outer radius
r = 20 mm. The tube is then pressurized, increasing
its outer radius to r = 20.04 mm. What is the resulting
extensional strain of the bar’s outer circumference in the
direction tangent to the circumference?
Solution:
The reference circumference of the tube is:
C = 2πr = 2π(20 mm) = 125.664 mm
The deformed circumference of the tube is:
C = 2πr = 2π(20.04 mm) = 125.915 mm
The strain in the bar’s outer circumference is:
mm−125.664 mm
ε = C C−C = 125.915125.664
mm
ε = 0.00199 ≈ 0.002
Problem 9.53 The angle between two infinitesimal
lines dL1 and dL2 which are perpendicular in a reference state is 120◦ in a deformed state. What is the shear
strain at this point corresponding to the directions dL1
and dL2 ?
Solution:
The angle between the two reference lines has increased by 30◦ . The
shear strain between the two lines is this angle measured in radians.
π
τ = −γ
2
τ =
ANS:
π
2
π
− π = − = −0.524
2
3
6
γ = −0.524
Problem 9.54 When the airplane’s wing is unloaded
(the reference state), the perpendicular lines L1 and L2
on the upper surface of the right wing are each 600 mm
long. In the loaded state shown, L1 is 600.2 mm long
and L2 is 595 mm long. If you assume that they are
uniform, what are the longitudinal strains in the L1 and
L2 directions?
Solution:
The longitudinal strain in the direction of L1 is:
L −L
ε1 = 1L 1 =
1
ε1 = 0.000333
600.2 mm−600 mm
600 mm
The longitudinal strain in the direction of L2 is:
L −L
mm
ε2 = 2L 2 = 595 mm−600
600 mm
2
ε2 = −0.008333
Problem 9.55 In Problem 9.54, the angle between the
lines L1 and L2 at the point where they intersect is 90.2◦ .
What is the shear strain referred to the directions L1 and
L2 at that point?
Solution:
The change in the angle between L1 and L2 increases by 0.2◦ . This
angle, measured in radians, is:
π
τ = − γ = 90 − 90.2 = −0.2◦ = −0.00349 rad
2
ANS:
γ = −0.00349
Problem 9.56 Two infinitesimal lines dL1 and dL2
are shown in a reference state and in a deformed state.
(The lines dL1 , dL2 , dL1 and dL2 are contained in
the x-y plane.) What is the shear strain at this point
corresponding to the directions dL1 and dL2 ?
Solution:
The angle between L1 and L2 , originally 90◦ , has been reduced by
40◦ due to deformation. The shear strain in the direction of L1 and
L2 is:
ANS: γ = 40◦ = 0.698
Problem 9.57 Two infinitesimal lines dL1 and dL2
within a material are parallel to the x and y axes in a
reference state (Figure a). After a motion and deformation of the material, dL1 points in the direction of the
unit vector e1 = and dL2 points in the direction of the
unit vector e2 = −0.408i + 0.816j − 0.408k (Figure b).
What is the shear strain referred to the directions dL1
and dL2 ?
Solution:
We need to find the angle between the two unit vectors. The scalar
product is of use:
e1 · e2 = |e1 ||e2 |(cos θ)
The magnitude of each of the unit vectors is obviously 1. The cosine
of the angle between he two unit vectors is:
cos θ = e1 · e2 = [(0.667) (−0.408)] + [(0.667) (0.816)] + [(0.333) (−0.408)]
cos θ = 0.1363
θ = 82.17◦
The original angle of 90◦ between the two unit vectors has been reduced by an amount of:
θ = 90◦ − γ
82.17◦ = 90◦ − γ
γ = 7.83◦ or
ANS:
γ = 0.137 in radians
Problem 9.58 In Problem 9.57, suppose that after the
motion and deformation of the material dL1 points in
the direction of the unit vector e1 = 0.667i + 0.667j +
0.333k and dL2 = −0.514i + 0.686j + 0.514k. What is
the shear strain referred to the directions dL1 and dL2 ?
Solution:
We need to find the angle between the two unit vectors.
The scalar product is of use: e1 · e2 = |e1 ||e2 |(cos θ)
The magnitude of each of the unit vectors is obviously 1.
The cosine of the angle between he two unit vectors is:
cos θ = e1 · e2 = [(0.667) (−0.514) + (0.667) (0.686) + (0.333) (0.514)]
cos θ = 0.2859
θ = 73.4◦
The original angle of 90◦ between the two unit vectors has been reduced by an amount of:
θ = 90◦ − γ
73.4◦ = 90◦ − γ
γ = 16.6◦
ANS:
γ = 0.290 in radians
Problem 9.59 An infinitesimal rectangle at a point in a
reference state of a material is shown. In a deformed state
the extensional strains in the dL1 and dL2 directions are
ε1 = 0.04 and ε2 = −0.02 and the shear strain referred
to the dL1 and dL2 is γ = 0.02. What is the extensional
strain in the dL direction? (See Example 9.6)
Solution:
Using the derivation from Example 2.6, the equation we will use is:
ε=
π
(1 + ε1 )2 cos2 θ + (1 + ε2 )2 sin2 θ − 2 (1 + ε1 ) (1 + ε2 ) (cos θ) (sin θ) cos
+ γ −1
2
Substituting the given values for ε1 , ε2 , and γ:
ε = (1 + 0.04)2 cos2 (40◦ ) + (1 − 0.02)2 sin2 (40◦ ) − 2 (1 + 0.04) (1 − 0.02) cos (40◦ ) sin (40◦ ) cos π2 + 0.02
ε = 0.0255 − 1
Problem 9.60 For the infinitesimal rectangle at a point
in a reference state of a material shown in Problem X.XX,
suppose that in a deformed state the extensional strains
in the dL1 , dL2 , and dL directions are ε1 = 0.030,
ε2 = 0.020, and ε = 0.038. What is the shear strain
referred to the dL1 and dL2 directions?
Solution:
We use the equation derived in Example 9.6:
π
(1 + ε1 )2 cos2 θ + (1 + ε2 )2 sin2 θ − 2 (1 + ε1 ) (1 + ε2 ) (cos θ) (sin θ) cos
+ γ −1
2
ε=
Solving this equation for γ:
cos
cos
π
2
π
2
+γ =
+γ =
−(1+ε)2 +(1+ε1 )2 (cos2 θ )+(1+ε2 )2 (sin2 θ )
2(1+ε1 )(1+ε2 )(cos θ)(sin θ)
−(1+2ε+ε2 )+(1+ε1 )2 (cos2 θ )+(1+ε2 )2 (sin2 θ )
2(1+ε1 )(1+ε2 )(cos θ)(sin θ)
Substituting the given values for ε, ε1 , and ε2 :
π
− 1 + 2 (0.038) + 0.0382 + (1 + 0.03)2 cos2 40◦ + (1 + 0.02)2 sin2 40◦
cos
+γ =
= −0.0242
2
2 (1 + 0.03) (1 + 0.02) (cos 40◦ ) (sin 40◦ )
π
+ γ = 91.39◦
2
γ = 1.39◦ = 0.0242
Problem 9.61 The bar is made of material 1-in. thick.
Its width varies linearly from 2 in. at its left end to 4 in.
at its right end. If the axial load P = 200 lb, what is the
average normal stress (a) at plane P1 ; (b) at plane P2 .
Free Body Diagram:
Solution:
The width of the material at P1 and P2 is:
W1 = 2 in + (1/3)(2 in) = 2.667 in
W2 = 2 in + (2/3)(2 in) = 3.333 in
The cross-sectional areas at P1 and P2 are:
A1 = (2.667 in)(1 in) = 2.667 in2
A2 = (3.333 in)(1 in) = 3.333 in2
The average normal stress at P1 and P2 are:
σ1 = (200 lb)/(2.667 in2 )
σ2 = (200 lb)/(3.333 in2 )
ANS:
ANS:
σ1 = 75 psi
σ2 = 60 psi
Problem 9.62 If the average normal stress at plane P1
of the bar described in Problem 9.61 is σAV = 300 psi,
what is the axial load P and the average normal stress at
plane P2 ?
Solution:
Free Body Diagrams:
The width of the material at P1 and P2 is:
W1 = 2 in+(2 in)(x/12 in) = 2 in+(2 in)(4 in/12 in) = 2.667 in
W2 = 2 in+(2 in)(x/12 in) = 2 in+(2 in)(8 in/12 in) = 3.333 in
The cross-sectional areas at P1 and P2 are:
A1 = W1 t = (2.667 in)(1 in) = 2.667 in2
A2 = W2 t = (3.333 in)(1 in) = 3.333 in2
Summing horizontal forces on the left-hand FBD:
ΣFx = 0 = −P + σAV G A1 = −P + (300 lb/in2 )(2.667 in2 )
ANS:
P = 800 lb
Summing horizontal forces on the right-hand FBD:
ΣFx = 0 = −P + σAV G A2 = −800 lb + (σAV G )2 (3.333 in2 )
ANS:
(σAV G )2 = 240 psi
Problem 9.63 The beam has cross-sectional area A =
0.1 m2 . What are the average normal stress and the magnitude of the average shear stress at the plane P ?
Free Body Diagram:
160 = 40(4)
2m
Ax
2m
Ay
20
By
Solution:
Summing horizontal forces:
ΣFx = 0 = 20 kN − Ax
Ax = 20 kN
Summing moments about the right-hand end of the beam:
ΣM = 0 = (40 kN/m)(4 m)(2 m) − Ay (4 m)
Ay = 80 kN ↑
Cut the beam through plane P and sum forces on the free body diagram:
ΣFy = 0 = −(80 kN)+Ay +τ (0.1 m2 ) = −(80 kN)+(80 kN)+τ (0.1 m2 )
ANS: τ = 0
Summing horizontal forces on the free body diagram:
ΣFx = 0 = −Ax + σA = −20, 000 N + σ(0.1 m2 )
ANS:
σ = 200, 000 N/m2 = 200 kN/m2 = 200 kPa
Problem 9.64 In Problem 9.63, what are the average
normal stress and the magnitude of the average shear
stress at the plane P if the plane is 1 m from the left end
of the beam?
Free Body Diagram:
160 = 40(4)
2m
Ax
2m
Ay
20
By
Solution:
Summing moments about the right-hand end of the beam in order to
find Ay :
ΣMB = 0 = (40, 000 N/m)(4 m)(2 m) − Ay (4 m)
Ay = 80, 000 N
Summing horizontal forces on the FBD:
ΣFx = 0 = −20, 000 N + σAV G (0.1 m2 )
ANS: σAV G = 200 kPa
Summing vertical forces on the FBD:
ΣFy = 0 = 40, 000 N − τAV G (0.1 m2 ) = 0
ANS:
τAV G = 400, 000 N/m2 = 400 kPa
Problem 9.65 The prismatic bar AB has cross- Free Body Diagram:
sectional area A = 0.01 m2 .
If the force F = 6 kN, what is the average normal stress
at the plane P ?
Solution:
Determine the axial load in the prismatic bar by summing vertical
forces at point B.
ΣFy = 0 = −6, 000 N + Paxial (cos 36.9◦ )
Paxial = 7502.95 N
The average normal stress at the plane, P , is:
σavg = (Paxial )/A = (7500 N)/(0.01 m2 )
σavg = −750, 000 Pa = −750 kPa NOTE: The negative
sign indicates a compressive stress.
ANS:
Problem 9.66 The prismatic bar AB in Problem 9.65 Free Body Diagram:
will safely support an average compressive normal stress
of 1.2 MPa on the plane P .
Based on this criterion, what is the largest downward
force F that can safely be applied?
Solution:
For an average compressive normal stress of 1.2 MPa at plane P , the
axial load in the prismatic bar is:
F = σavg A = (1.2 × 106 n/m2 )(0.01 m2 )
F = 12, 000 N = 12 kN
Now the load, F , can be found by summing vertical forces at point B.
ΣFy = 0 = (12 kN)(sin 53.1◦ ) − F
ANS:
F = 9.6 kN ↓
Problem 9.67 The jaws of the bolt cutter are connected
by two links AB.
The cross-sectional area of each link is 750 mm2 . What
average normal stress is induced in each link by the 90-N
forces exerted on the handles?
Free Body Diagram:
FBD Handle
FBD Jaw
Solution:
Inspection of the FBD of jaw A reveals that Cx must be zero. This
information is necessary in the analysis of the handle.
Summing moments about point D on the handle:
ΣMD = 0 = −(90 N)(0.54 m) + Cy (0.1 m)
Cy = 486 N
Summing moments about point P on the jaw:
ΣMP = 0 = Cy (0.24 m)−Ey (0.08 m) = (486 N)(0.24 m)−Ey(0.08 m)
Ey = 1458 N
Here we must realize that there are TWO links connecting the jaws of
the bolt cutter, so HALF of the force Ey is exerted by each of the links.
The average normal stress in link AB is:
σavg = (0.5)Ey /A = (0.5)(1458 N)/(750 × 10−6 m2 )
ANS:
σavg = 972, 000 Pa = 972 kPa
Problem 9.68 The pins connecting the links AB to
the jaws of the bolt cutter in Problem 9.67 are 20-mm
in diameter. What average shear stress is induced in the
bolts by the 90-N forces exerted on the handles?
Free Body Diagrams:
Solution:
From the solution to Problem 9.67 we have seen that the total downward
force on the upper jaw is 1458 N. We see that the shear load produced
in the pin by this downward force is divided between the two portions
of the pin through jaw A which extend from the jaw into the connecting
links.
The average shear stress in each of the pins is:
τavg = (1/2)(1458 N)/[π(0.01 m)2 ]
ANS:
τavg = 2, 320, 000 Pa = 2.32 MPa
Problem 9.69 Suppose that you subject a 2-m prismatic bar to compressive axial forces that cause a uniform extensional strain ε = −0.003 in the axial direction. What is the bar’s length in the deformed state?
Solution:
Using Equation 9.3 from the text:
L = (1 + ε)L
L = (1 − 0.003)(2m)
L = 1.994m
Problem 9.70 A prismatic bar is subjected to loads that
cause uniform axial strains ε = 0.002 in its left half and
ε = −0.004 in its right half.
What is the resulting change in length of the 28-in. bar?
Free Body Diagram:
Solution:
The change in length over the left half of the prismatic bar is:
δL = (0.002)(14 in) = 0.028 in
The change in length of the right half of the prismatic bar is:
δR = (−0.004)(14 in) = −0.056 in
Total change in length for the prismatic bar is:
δ = δL + δR = 0.028 in − 0.056 in
ANS:
δ = −0.028 in
Problem 9.71 The prismatic bar in Problem 9.70 is
subjected to loads that cause a uniform axial strain εL =
0.006 in its left half and a uniform axial strain εR = in
its right half. As a result, the length of the 28-in. bar
increases by 0.032 in. What is εR ?
Solution:
Total change in length of the deformed bar is:
∆L = 0.032 = εL (14) + εR (14)
0.032 = 0.006(14) + εR (14)
Solving for εR :
ANS: εR = −0.00371
Problem 10.1 A prismatic bar with cross-sectional
area A = 0.1 m2 is loaded at the ends in two ways:
(a) by 100-Pa uniform normal tractions; (b) by 10-N axial
forces acting at the centroid of the bar’s cross section.
What are the normal and shear stress distributions at the
plane P in the two cases?
Diagram:
Solution:
(a) Because the applied loads are in the axial direction and the plane
is normal to the axis of the bar, there is no shear stress. In the first
case, the applied load in a uniform normal traction, so the normal stress
MUST be:
ANS: σ = 100 Pa
As stated above, since the applied load is in the axial direction and the
plane is normal to the axis of the bar:
ANS: τ = 0
(b) In the second case, the normal stress distribution is:
σ = P/A = (10 N)/(0.1 m2 )
ANS: σ = 100 Pa
As stated above, since the applied load is in the axial direction:
ANS: τ = 0
Problem 10.2 A prismatic bar with cross-sectional
area A = 4 in2 is subjected to tensile axial loads P . It
consists of a material that will safely support a tensile
normal stress of 60 ksi. Based on this criterion, what is
the largest safe value of P ?
Diagram:
Solution:
Using the definition of normal stress:
PALLOW = σALLOW A = (60, 000 lb/in2 )(4 in2 )
ANS:
PALLOW = 240, 000 lb = 240 kip
Problem 10.3 A prismatic bar has a solid circular
cross section with 20-mm diameter. It consists of a material that will safely support a tensile normal stress of
300 MPa. Based on this criterion, what is the largest
tensile load P to which the bar can be subjected?
Free Body Diagram:
Solution:
The cross-sectional area of the bar is:
A = πr 2 = π(0.01 m)2 = 3.142 × 10−4 m2
For a maximum normal stress of 300 × 106 Pa:
300 × 106 N/m2 =
ANS:
PM AX
3.142 × 10−4 m2
PMAX = 94.2 kN
Problem 10.4 The cross-sectional area of bar AB is
0.5 in2 . If the force F = 3 kips, what is the normal
stress on a plane perpendicular to the axis of bar AB?
Free Body Diagrams:
Solution:
Summing moments about point C:
ΣMC = 0 = (3000 lb)(6 ft) − FAB (4 ft)
FAB = 4500 lb (T)
The normal stress in member AB is:
σAB = FAB /AAB = (4500 lb)/(0.5 in2 )
ANS:
σAB = 9000 psi = 9 ksi
Problem 10.5 Bar AB of the frame in Problem 10.4 Free Body Diagram:
consists of material that will safely support a tensile normal stress of 20 ksi. If you want to design the frame to
support forces F as large as 8 kip, what is the minimum
required cross-sectional area of bar AB?
Solution:
The maximum safe load which can be supported by member AB is:
[1]
(FAB )MAX = σALLOW AAB = (20, 000 lb/in2 )(AAB )
Summing moments about point C:
ΣMC = 0 = F (6 ft)−(FAB )MAX (4 ft) = (8000 lb)(6 ft)−(FAB )MAX (4 ft)
(FAB )MAX = 12, 000 lb
Using Equation [1] to find the cross-sectional area of member AB:
12, 000 lb = (20, 000 lb/in2 )(AAB )
ANS:
AAB = 0.6 in2
Problem 10.6
Free Body Diagram:
The mass of the suspended box is 800 kg. The mass
of the crane’s arm (not including the hydraulic actuator
BC) is 200 kg, and its center of mass is 2 m to the right
of A. The cross-sectional area of the upper part of the
hydraulic actuator is 0.004 m2 . What is the normal stress
on a plane perpendicular to the axis of the upper part of
the actuator?
Solution:
The weight of the box is:
WB = mg = (800 kg)(9.81 m/sec2 ) = 7848 N
The weight of the crane’s arm is:
WARM = mg = (200 kg)(9.81 m/sec2 ) = 1962 N
To get the axial load in the actuator, sum moments about point A.
ΣMA = 0 = −WB (7 m)−WARM (2 m)+FACT (sin 63.4◦ )(3 m)
= −(7848 N)(7 m)−(1962 N)(2 m)+FACT (sin 63.4◦ )(3 m)−FACT cos 63.4(1.4 m)
FACT = 28, 634 N (C)
The normal stress on the upper portion of the actuator is:
σACT = FACT /AACT = (28, 634 N)/(0.004 m2 )
ANS:
σACT = 7.15 MPa
Problem 10.7 The cross-sectional area of the lower
part of the hydraulic actuator in Problem 10.6 is
0.010 m2 . What is the normal stress on a plane perpendicular to the axis of the lower part of the actuator?
Free Body Diagram:
Solution:
Summing moments about point A:
ΣMA = 0 = −(800kg)(9.81 m/sec2 )(7 m)−(200kg)(9.81 m/sec2 )(2 m)+FBC (cos 63.4◦ )(3 m)−FBC (cos 63.4◦ )(1.4 m)
FBC = 28, 634 N
The normal stress on the lower part of the actuator is:
σ=
FBC
28, 634 N
=
Note: The negative sign indicates compression.
A
0.010 m2
ANS:
σ = −2.86 MPa
Problem 10.8 The cross-sectional area of each bar is
A. What is the normal stress on a plane perpendicular
to the axis of one of the bars?
Free Body Diagram:
Solution:
Because the support is symmetrical, the axial load is the same in each
member. Draw the FBD where the load is applied and sum vertical
forces.
ΣFy = 0 = −F + 2[FAXIAL (sin β)]
FAXIAL = F/(2)(sin β)
The normal stress in the support members is:
σAXIAL = FAXIAL /A = [F/(2)(sin β)]/A
ANS:
σAXIAL = F/[2A sin β]
Problem 10.9 The angle β of the system in Prob- Free Body Diagram:
lem 10.8 is 60◦ . The bars are made of a material that will
safely support a tensile normal stress of 8 ksi. Based on
this criterion, if you want to design the system so that it
will support a force F = 3 kip, what is the minimum
necessary value of the cross-sectional area A?
Solution:
The maximum load which can be safely supported by EACH of the
support members is:
FMAX = σMAX (A) = (8000 lb/in2 )(A)
Summing vertical forces on the FBD:
ΣFy = 0 = −3000 lb + 2(8000 lb/in2 )(A)(sin 60◦ )
ANS:
A = 0.217 in2
Problem 10.10 Suppose that the horizontal distance
between the supports of the system in Problem 10.8 and
the load F are specified, and the prismatic bars are made
of a material that will safely a tensile normal stress σ0 .
You want to choose the angle β and the cross-sectional
area A of the bars so that the total volume of material
used is a minimum. What are β and A?
Free Body Diagram:
L = d/(cos β)
[1]
Solution:
Summing vertical forces on the FBD to find the force supported by the
two prismatic bars:
ΣFy = 0 = −F +2 [σ0 A(sin β)] where σ0 is the maximum allowable average normal stress.
From the above equation, the cross-sectional area of one of the prismatic bars may be expressed as:
A=
F
2σ0 (sin β)
[2]
Recall: sin 2β = 2 sin β cos β
Using Equations [1] and [2], the volume of a single bar is:
V = AL =
Fd
Fd
=
2σ0 sin β cos β
σ0 sin 2β
We see that the volume will be minimum when sin2β is maximum,
which is when sin 2β = 1, so:
ANS: β = 45◦ [3]
Using this value for β in Equation [2] to find A:
A=
ANS:
A = 0.707 σF
0
F
2σ0 (0.707)
Problem 10.11 The cross-sectional area of each bar
is 60 mm2 . If F = 4 kN, what are the normal stresses
on planes perpendicular to the axes of the bars?
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD:
ΣFx = 0 = −FAB (cos 60◦ ) + FAC (cos 45◦ )
FAB = 1.414FAC
[1]
Summing vertical forces on the FBD:
ΣFy = 0 = −40, 000 N + FAB (sin 60◦ ) + FAC (sin 45◦ )
Substituting Equation [1] into Equation [2]:
40, 000 N = [1.414FAC ](sin 60◦ ) + FAC (sin 45◦ )
FAC = 20, 708 lb
Therefore:
FAB = 29, 281 lb
The normal stresses in the two supporting members are:
σAB = FAB /AAB = (29, 281 lb)/(60 × 10−6 m2 )
ANS:
σAB = 488 MPa
σAC = FAC /AAC = (20, 708 lb)/(60 × 10−6 m2 )
ANS:
σAC = 345 MPa
[2]
Problem 10.12 The bars of the truss in Problem 10.11
are made of material that will safely support a tensile
normal stress of 600 MPa. Based on this criterion, what
is the largest safe value of the force F ?
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD:
ΣFx = 0 = −FAB (cos 60◦ ) + FAC (cos 45◦ )
FAB = 1.414FAC
[1]
Summing vertical forces on the FBD:
ΣFy = 0 = −F + FAB (sin 60◦ ) + FAC (sin 45◦ )
[2]
Substituting Equation [1] into Equation [2]:
F = [1.414FAC ](sin 60◦ ) + FAC (sin 45◦ )
FAC = 0.5177F
Therefore:
FAB = 0.732F
The normal stresses in the two supporting members are:
σAB = FAB /AAB = (0.732F )/(60×10−6 m2 ) = 600×106 N/ m2
(FMAX )AB = 49, 180 N
σAC = FAC /AAC = 0.5177F )/(60×10−6 m2 ) = 600×106 N/ m2
(FMAX )AC = 69, 538 N
We see that member AB will fail if subjected to (FMAX )AC , so the
maximum allowable load is:
ANS: FMAX = 49, 180 N = 49.2 kN
Problem 10.13 The cross-sectional area of each bar of
the truss is 400 mm2 . If F = 30 kN, what is the normal
stress on a plane perpendicular to the axis of member
BE?
Free Body Diagram:
Solution:
To find the axial load in member BE directly, sum vertical forces on
the FBD.
ΣFy = 0 = −30, 0000 N + FBE (sin 45◦ )
FBE = 42, 400 N (T)
The normal stress in member BE is:
σBE = FBE /ABE = (42, 400 N)/(400 × 10−6 )
ANS:
σBE = 106 MPa
Problem 10.14
In Problem 10.13, what is the normal stress on a plane
perpendicular to the axis of member BC?
Free Body Diagram:
Fig (A)
Fig (B)
Solution:
Summing vertical forces at point A:
ΣFY = 0 = −30, 000 N + PAC (sin 45◦ )
PAC = 42, 426 N (T)
Summing vertical forces at point C:
ΣFY = 0 = −PAC (sin 45◦ )+PBC = −(42, 426 N)(sin 45◦ )+PBC
PBC = 30, 000 N(C)
The normal stress in the lower part of the actuator is:
σ=
ANS:
30, 000 N
PBC
=
ABC
400 × 10−6 m2
σ = −75 MPa NOTE: (−) indicates compressive stress.
Problem 10.15 The truss in Problem 10.13 is made of
a material that will safely support a normal stress (tension
or compression) of 340 MPa. Based on this criterion,
what is the largest safe value of the force F ?
Solution:
We must first determine the axial load in EACH member of the truss.
Summing vertical forces at joint A:
ΣFy = 0 = −F + FAC (sin 45◦ )
FAC = 1.414F (T )
Summing horizontal forces at joint A:
ΣFx = 0 = FAB − FAC (cos 45◦ ) = FAB − (1.414F )(cos 45◦ )
Fig (A)
FAB = F (C)
Summing vertical forces at joint C:
ΣFy = 0 = FBC − FAC (sin 45◦ ) = FBC − (1.414F )(sin 45◦ )
FBC = F (C)
Summing horizontal forces at joint C:
ΣFx = −FCE + FAC (cos 45◦ ) = −FCE + (1.414F )(cos 45◦ )
FCE = F (T)
Fig (B)
Summing horizontal forces at joint E:
ΣFx = 0 = FCE + FBE (cos 45◦ ) = F + FBE (cos 45◦ )
FBE = 1.414F (C)
Summing vertical forces at joint E:
+ ↓ ΣFy = 0 = −FDE +FBE (sin 45◦ ) = −FDE −1.414F sin 45◦
FDE = F (T)
Summing moments about joint E:
Fig (C)
ΣME = 0 = −F (0.5 m) + Dx (0.25 m)
Ey
Dx = 2F
Therefore:
FBD = 2F (C)
Ex
We see that member BD is carrying the greatest load (2F ), and so we
recognize that this member controls the safe load which can be carried
by the truss. The largest load which can be safely carried by member
BD is:
FBD = σALLOW A = (340×106 N/ m2 )(400×10−6 m2 ) = 2F
ANS:
FMAX = 68, 000 N = 68 kN
Dx
F
Problem 10.16 The cross-sectional area of the pris- Free Body Diagram:
matic bar is A = 2 in2 and the axial force P = 20 kip.
Determine the normal and shear stresses on the plane P .
Draw a diagram isolating the part of the bar to the right
of plane P and show the stresses.
Solution:
The area of plane P is:
AP = (2 in2 )/(cos 70◦ )
AP = 5.848 in2
Summing vertical forces on the FBD:
ΣFy = 0 = −σAP (sin 70◦ ) + τ AP (cos 70◦ )
[1]
τ = 2.75σ
Summing horizontal forces on the FBD:
[2]
ΣFx = 0 = 20, 000 lb − σAP (cos 70◦ ) − τ AP (sin 70◦ )
Substituting Equation [1] into Equation [2]:
20, 000 lb = σAP (cos 70◦ )+(2.75σ)AP (sin 70◦ ) = σ(5.848 in2 )(cos 70◦ )+(2.75σ)(5.848 in2 )(sin 70◦ )
ANS: σ = 1169 psi
Using the determined value of σ in Equation [1]:
τ = 2.75σ = −3215 psi
τ = 2.75σ = −3215 psi Note: The negative sign conforms to the sign convention given in Figure 3.15
ANS:
Problem 10.17
Free Body Diagram:
If the normal stress on the plane P in Problem 10.16 is
6000 psi, what is the axial force P ?
Solution:
The area of plane P is:
AP = (2 in2 )/(cos 70◦ )
AP = 5.848 in2
Summing vertical forces on the FBD:
ΣFy = 0 = −σAP (sin 70◦ ) + τ AP (cos 70◦ )
[1]
τ = 2.75σ
Summing horizontal forces on the FBD:
[2]
ΣFx = 0 = P − σAP (cos 70◦ ) − τ AP (sin 70◦ )
Subtsituting Equation [1] into Equation [2]:
P = σAP (cos 70◦ )+(2.75σ)(AP )(sin 70◦ ) = (6000 lb)(5.848 in2 )(cos 70◦ )+(2.75)(6000 lb)(5.848 in2 )(sin 70◦ )
ANS:
P = 102, 600 lb = 102.6 kip
Problem 10.18 The cross-sectional area of the prismatic bar is 0.02 m2 . If the normal and shear stresses
on the plane P are σθ
=
1.25 MPa and τθ =
−1.5 MPa, what are the angle θ and the axial force P ?
Free Body Diagram:
θ
Solution:
The area of the slanted plane is:
AP = A/ cos θ
Summing vertical forces on the FBD:
ΣFy = 0 = τθ AP (cos θ) − σθ AP (sin θ)
τ0
=σ
0 (tan θ)
We can determine the magnitude of θ:
τθ
tan θ =
= (1.5 × 106 P a)/(1.25 × 106 Pa) = 1.2
σθ
ANS:
θ = 50.2◦
So the area of the plane is:
AP = A/ cos θ = (0.02 m2 )/(cos 50.2◦ )
AP = 0.0312 m2
Summing horizontal forces on the FBD:
ΣFx = 0 = P − τθ AP (sin θ) − σθ AP (cos θ)
P = (1.5×106 N/ m2 )(0.0312 m2 )(sin 50.2◦ )+(1.25×106 N/ m2 )(0.0312 m2 )(cos 50.2◦ )
ANS:
P = 61 kN
Problem 10.19 The cross-sectional area of the bar is
A = 0.5 in2 and the force F = 3000 lb. Determine the
normal stresses and the magnitudes of the shear stresses
on the planes (a) and (b).
Solution:
(a) The area of the plane is:
AP = A/ cos 30◦ = 0.5 in2 / cos 30◦
AP = 0.577 in2
To find the axial load in each member (note the symmetry), sum vertical
forces at the point where the load is applied.
ΣFy = 0 = −3000 lb + 2[P (sin 60◦ )
Fig (A)
P = 1732 lb
Summing forces in the y-direction (note the direction of the x- and
y-axes) on the FBD:
ΣFx = 0 = σAP (sin θ)−τ AP (cos θ) = σ(0.577 in2 )(sin 30◦ )−τ (0.577 in2 )(cos 30◦ )
[1]
τ = 0.577σ
Summing forces in the x-direction on the FBD:
ΣFx = 0 = 1732 lb − σAP (cos 30◦ ) − τ AP (sin 30◦ )
2
[2]
Fig (B)
2
1732 lb = σ(0.577 in )(0.866) + τ (0.577 in )(0.5)
Substituting Equation [1] into Equation [2]:
1732 lb = σ(0.577 in2 )(0.866) + (0.577σ)(0.577 in2 )(0.5)
ANS: σ = 2600 psi
Using this result in Equation [1]:
τ = 0.577σ
ANS: τ = 1500 psi
(b) The area of the plane is:
AP = A/ cos 60◦ = 0.5 in2 / cos 60◦
AP = 1 in2
In part (a) above, the axial load in the member was found to be P =
1732 lb. Summing forces in the y-direction (note the directions of the
x- and y-axes) on the FBD:
ΣFy = 0 = σAP (sin 60◦ ) − τ AP (cos 60◦ )
[1]
τ = 1.732σ
Summing forces in the x-direction:
ΣFx = 0 = −1732 lb + σAP (cos 60◦ ) + τ AP (sin 60◦ )
[2]
1732 lb = σ(1 in2 )(0.5) + τ (1 in2 )(0.866)
Substituting Equation [1] into Equation [2]:
1732 lb = σ(1 in2 )(0.5) + (1.732σ)(1 in2 )(0.866)
ANS: σ = 866 psi
Using this result in Equation [1]:
τ = 1.732σ
ANS:
τ = 1500 psi
Fig (C)
Problem 10.20 The truss in Problem 10.19 is constructed of a material that will safely support a normal
stress of 8 ksi and a shear stress of 3 ksi. Based on these
criteria, what is the largest force F that can safely be
applied
Solution:
We have seen from the text that maximum normal stress occurs when
θ = 0 and that maximum shear stress occurs when θ = 45◦ . Determining the axial load in each member of the truss while supporting the
load F :
ΣFy = 0 = −F + 2[P (sin 60◦ )]
P = F/(2 sin 60◦ )
[1]
P = 0.577F
Using the maximum normal stress as the design criterion:
Fig (A)
8000 lb/in2 = PMAX /0.5 in2
PMAX = 4000 lb
Using this value of PMAX in Equation [1]:
FMAX = PMAX /0.577 = 4000 lb/0.577
FMAX = 6932 lbThis is the load which produces the limiting normal stress.
Using the maximum shear stress as the design criteria, we pass a plane
through the cross-section at an angle of 45◦ .
The area of the plane is:
AP = A/ sin 45◦ = (0.5 in2 )/(0.707)
AP = 0.707 in2
Summing forces in the x-direction on the FBD:
ΣFx = 0 = σAP (cos 45◦ ) − τ AP (sin 45◦ )
[2]
σ=τ
Summing forces in the y-direction on the FBD:
[3]
Fig (B)
ΣFy = 0 = −P + σAP (sin 45◦ ) + τ AP (cos 45◦ )
Substituting Equation [1] into Equation [2]:
P = τ AP (0.707)+τ AP (0.707) = τ (0.707 in2 )(0.707)+τ (0.707 in2 )(0.707)
P = τ lb
Using the limiting shear stress as the design criterion:
[4]
P = 3000 lb
Now using Equation [1] in Equation [4]:
FM AX = P/0.577 = 3000 lb/0.577
ANS:
FMAX = 5196 lb
Fig (C)
Problem 10.21 Two marks are made 2 inches apart
on an unloaded bar. When the bar is subjected to axial
forces P , the marks are 2.004 inches apart. What is the
axial strain of the loaded bar?
Solution:
From the definition of strain:
ε=
L −L
L
=
2.004 in−2.000 in
2.000 in
ε = 0.002
Problem 10.22 The total length of the unloaded bar in
Problem 10.21 is 10 in. Use the result of Problem 10.21
to determine the total length of the loaded bar. What
assumption are you making when you do so?
Solution:
From the definition of strain:
ε=
L − L
2.004 in − 2.000 in
=
L
2.000 in
ε = 0.002
Total length of the deformed bar will be:
L = L(1 + ε) = (10 in)(1 + 0.002)
ANS:
L = 10.02 in
Problem 10.23 If the forces exerted on the bar in Problem 10.21 are P = 20 kip and the bar’s cross-sectional
area is A = 1.5 in2 , what is the modulus of elasticity of
the material?
Solution:
From Problem 10.21 we determined that the observed strain is 0.002.
The normal stress in the bar is:
σ=
P
20, 000 lb
=
= 13, 333 psi
A
1.5 in2
From the definition of the modulus of elasticity:
E=
σ
ε
=
13,333 lb/in2
0.002
E = 6.67 × 106 psi
Problem 10.24 A prismatic bar with length L = 6 m
and a circular cross section with diameter D = 0.02 m
is subjected to 20-kN compressive forces at its ends.
The length and diameter of the deformed bar are measured and determined to be L = 5.940 m and D =
0.02006 m. What are the modulus of elasticity and Poisson’s ration of the material?
Solution:
The strain in the bar is:
ε=
L − L
5.94 m − 6.0 m
=
= −0.01
L
6.0 m
The compressive stress in the bar is:
P
−20, 000 N
=
= −63.7 MPa
A
π(0.01 m)2
σ=
The modulus of elasticity for the material is:
E=
σ
63.7 MPa
=
ε
0.01
ANS: E = 6.37 GPa
Poison’s ration for the material is:
υ=
−εLAT
ε
=
(D −D)/D
ε
=
−(0.02006 m−0.02 m)/0.02 m
−0.01
υ = 0.3
Problem 10.25
The bar has modulus of elasticity E = 30 × 106 psi and
Poisson’s ration ν = 0.32. It has a circular cross section
with diameter D = 0.75 in. What compressive force
would have to be exerted on the right end of the bar to
increase its diameter to 0.752 in?
Solution:
From the definition of Poisson’s ratio:
0.32 =
−(0.752 in − 0.75 in)/0.75 in
ε
The strain which will be produced by the applied load is:
σ
P
1
ε=
=
2
2
6
E
π(0.375 in)
30 × 10 lb/in
Substituting the above expression for the strain into the expression for
Poisson’s ratio:
0.32 =
ANS:
−(0.752 in − 0.75 in)/0.75 in
(P/(π(0.375 in)2 )) 30×1061 lb/in2
P = 110.4 kip
Problem 10.26 What tensile force would have to be
exerted on the right end of the bar in Problem 10.25 to
increase its length to 9.02 in.? What is the bar’s diameter
after this load is applied?
Solution:
The strain in the bar will be:
ε=
L − L
9.02 in − 9.00 in
=
= 0.00222
L
9.00 in
The stress required to produce this strain is:
σ = Eε = (30 × 106 lb/in2 )(0.00222) = 66, 667 lb/in2
The load required to produce the stress is:
π(0.75 in)2
P = σA = (66, 667 lb/in2 )
4
ANS: P = 29.5 kip
The radial strain in the deformed bar is:
εLAT = εν = (0.00222)(−0.32) = 7.104 × 10−4
The diameter of the deformed bar is:
D = D(1 − εLAT ) = (0.75 in)(1 − 7.104 × 10−4 )
ANS:
D = 0.7495 in
Problem 10.27 A prismatic bar is 300 mm long and
has a circular cross section with 20-mm diameter. Its
modulus of elasticity is 120 Gpa and its Poisson’s ratio
is 0.33. Axial forces P are applied to the ends of the
bar which cause its diameter to decrease to 19.948 mm.
(a) What is the length of the loaded bar? (b) What is the
value of P ?
Solution:
We can use Poisson’s ratio to determine the extensional strain in the
material.
0.33 =
−(19.948 mm − 20 mm)/(20 mm)
→
ε
ε = 0.0079
(a) The length of the loaded bar is:
L = L(1 + ε) = (300 mm)(1 + 0.0079)
L = 302.37 mm
(b) The value of P is determined using the definition of the modulus
of elasticity.
ANS:
P = εEA = (0.0079)(120 × 109 N/m2 )(π)(0.01 m)2
ANS:
P = 297.8 kN
Problem 10.28 When unloaded, bars AB and AC are
each 36 in. in length and have a cross sectional area of
2 in2 . Their modulus of elasticity is E = 1.6 × 106 psi.
When the weight W is suspended at A, bar AB increases
its length by 0.01 in. What is the change in length of bar
AC?
FAB
30˚
60˚
20˚
FAC
W
Given the strain in member AB, the load W can be determined.
Solution:
We start by establishing the relationship between the loads in member
AB and AC.
Summing vertical forces at point A:
ΣFy = 0 = −W + FAB (cos 30◦ ) + FAC (sin 20◦ )
W = 0.866FAB + 0.342FAC
1.6 × 106 psi =
W = 9317 lb
FAC = 4733 lb (C)
The strain in member AC will be:
◦
ΣFx = 0 = FAC (cos 20 ) − FAB (sin 30 )
δAC =
[2]
(4733 lb)(36 in)
PL
=
= −0.532
AE
(2 in2 )(1.6 × 106 psi)
εAC = −0.532 Note: The negative strain results from the compressive load.
Solving Equations [1] and [2] together:
FAB = 0.954W (T)
FAC = 0.508W (C)
Problem 10.29 If a weight W = 12, 000 lb is suspended from the truss in Problem 10.28, what are the
changes in length of the two bars?
FAB
30˚
Solution:
Establish the relationship between the loads in member AB and AC.
Summing vertical forces at point A:
ΣFy = 0 = −12, 000 lb + FAB (cos 30◦ ) + FAC (sin 20◦ )
12, 000 lb = 0.866FAB + 0.342FAC
20˚
W = 12,000
[1]
Summing horizontal forces at point A:
ΣFx = 0 = FAC (cos 20◦ ) − FAB (sin 30◦ )
[2]
Solving Equations [1] and [2] together:
FAB = 0.954(12, 000 lb) = 11, 448 lb (T)
FAC = 0.508(12, 000 lb) = 6, 096 lb (C)
The stresses in the two bars are:
σAB =
FaB
11, 448 lb
=
= 5, 724 lb/in2
A
2 in2
σAC =
FAC
6, 096 lb
=
= 3, 048 lb/in2
A
2 in2
Strains in the two bars are:
εAB =
σAB
5, 724 lb/in2
=
= 0.00358
E
1.6 × 106 lb/in2
εAC =
σAC
−3, 048 lb/in2
=
= −0.00191
E
1.6 × 106 lb/in2
Changes in length for the two bars are:
δAB = εAB LAB = (0.00358)(36 in)
ANS:
δAB = 0.1288 in
0.954W/(2 in2 )
(0.1 in/36 in)
FAC = 0.508W = (0.508)(9317 lb)
[1]
◦
FAC = 0.532FAB
=
The load in member AC is:
Summing horizontal forces at point A:
FAC = 0.532FAB
FAB /A
ε
δAC = εAC LAC = (−0.00191)(36 in)
δAC = −0.0688 in
Problem 10.30 Bars AB and AC are each 300 mm
in length, have a cross-sectional area of 500 mm2 , and
have modulus of elasticity E = 72 Gpa. If a 24 kN
downward force is applied at A, what is the resulting
Solution:
displacement of point A?
FAB = P
FAC = P
30˚
Because the truss is symmetrical, the displacement will be ALL VERTICAL. Summing vertical forces at point A:
30˚
24,000 N
ΣFy = 0 = −24, 000 N + 2 [(P ) (sin 30◦ )]
P = 24, 000 N
Knowing the load in each member, we can calculate the strain in each
member.
24,000 N/(500×10−6 m2 )
72×109 N/m2
P/A
ε= E =
ε = 66.7 × 10−5
The new length for each of the two truss members is:
L = (300 mm)(1 + 66.7 × 10−5 ) = 300.2 mm
The new vertical distance from the overhead to point A is:
d = (300.2 mm)2 − [(300 mm) (cos 30◦ )]2 = 150.4 mm
The original vertical distance from the overhead to point A was:
d0 = (300 mm)2 − [(300 mm) (cos 30◦ )]2
d0 = 150 mm
The vertical displacement is:
D = d − d0 = 150.4 mm − 150 mm
D = 0.4 mm ↓
ANS:
Problem 10.31 Bars AB and AC of the truss shown in
Problem 10.30 are each 300 mm in length, have a crosssectional area of 500 mm2 , and are made of the same
material. When a 30-kN downward force is applied at
Solution:
point A, it deflects downward 0.4 mm. What is the
Summing vertical forces at point A:
modulus of elasticity of the material?
FAB = P
ΣFy = 0 = −30, 000 N + 2 [(P ) (sin 30◦ )]
P = 30, 000 N
The stress in each of the members is:
σ=
P
30, 000 N
=
= 60 × 106 N/m2 = 60 MPa
A
500 × 10−6 m2
The deformed length of each of the bars is:
L = [(0.3 m) (cos 30◦ )]2 + [(0.3 m) (sin 30◦ ) + 0.0004 m]2 = 0.3002 m
The strain in each of the bars is:
ε=
L − L
.3002 m − 0.3 m
=
= 6.667 × 10−4
L
0.3 m
The modulus of elasticity is:
E=
ANS:
E = 90 Gpa
σ
60 × 106 N/m2
=
ε
6.667 × 10−4
FAC = P
30,000 N
Problem 10.32
Bar AB has cross-sectional area A = 100 mm2 and
modulus of elasticity E = 102 Gpa. The distance H =
400 mm. If a 200-kN downward force is applied to
bar CD at D, through what angle in degrees does bar
CD rotate? (You can neglect the deformation of bar
CD.) Strategy: Because the bar’s change in length is
small, you can assume that the downward displacement
v of point B is vertical, and that the angle (in radians)
through which bar CD rotates is v/H.
Free Body Diagram:
Solution:
We can determine the angle of rotation by finding the vertical displacement at point B. Find the axial load in member AB by summing
moments about point C:
ΣMc = 0 = −(200 kN)(0.6 m) + FAB (sin 60◦ )(0.4 m)
FAB = 346.4 kN (C)
The strain in member AB will be:
P/A
ε= E =
ε = 0.034
(346,400 N)/(100×10−6 m)
102×109 N/m2
The original length of member AB is:
L = (300 mm)/(sin 60◦ ) = 346.3 mm
The new length of member AB is:
L = L(1 + ε) = (346.3 mm)(1 − 0.034) = 334.5 mm
The original height of point B is h = 300 mm.
The deformed height of point B is:
h = (L )2 − [300 mm/tan 60◦ ]2 = (334.5 mm)2 − (173.2 mm)2 = 286.2 mm
The change in vertical height at point B is:
v = h − h = 300 mm − 286.2 mm = 13.8 mm ↓
The angle, in radians, through which the bar CD rotates is:
ANS: θ = 13.8 mm/400 mm = 0.0345 radians = 1.98◦ clockwise
Problem 10.33 Bar AB in Problem 10.32 is made of
a material that will safely support a normal stress (in
tension or compression) of 5 GPa. Based on this criterion, through what angle in degrees can bar CD safely
be rotated relative to the position shown.
Free Body Diagram:
Solution:
Maximum allowable strain in the material is:
ε=
σ
5 × 109 N/m2
=
= 0.049
E
102 × 109 N/m2
Maximum allowable change in length for H is:
∆H = Lε = (0.4 m) (.049) = 0.0196 m
In the diagram, maximum allowable distance d is:
d = (∆H)(cos 30◦ ) = (0.0196 m)(cos 30◦ )
d = 0.01697 m
The angle through which bar AB may rotate is:
θ=
ANS:
d
±0.01697 m
=
= ±0.049 rad
LAB
(0.3/ sin 60◦ )
θ = ±2.8◦
Problem 10.34 If an upward force is applied at H
that causes bar GH to rotate 0.02 degrees in the counterclockwise direction, what are the axial strains in bars
AB, DC, and EF ? (You can neglect the deformation
of bar GH.)
Solution:
The vertical displacement at point H is:
θ = (0.02◦ /180◦ )(3.14159 radians) = 349 × 10−6 radians
The vertical displacement of points B, D and F are:
VB = (349 × 10−6 radians)(400 mm) = 0.14 mm
VD = (349 × 10−6 radians)(800 mm) = 0.28 mm
VF = (349 × 10−6 radians)(1200 mm) = 0.42 mm
The strains in each of the vertical bars is:
mm
= 0.00035 εCD =
ANS: εAB = 0.14
400 mm
0.28 mm
400 mm
= 0.00070
εEF =
0.42 mm
400 mm
= 0.00105
Problem 10.35 The bar has cross-sectional area A and
modulus of elasticity E. The left end of the bar is fixed.
There is initially a gap b between the right end of the bar
and the rigid wall (Figure 1). The bar is stretched until
it comes into contact with the rigid wall and is welded
to it (Figure 2). Notice that this problem is statically
indeterminate because the axial force in the bar after it
is welded to the wall cannot be determined from statics
alone. (a) What is the compatibility condition in this
problem? (b) What is the axial force in the bar after it is
welded to the wall?
Solution:
The compatibility condition requires that the bar’s change in length
must be limited to the amount of the gap, b. Using the relationship
δ = P L/AE to find the axial load:
= bAE
ANS: P = δAE
L
L
Problem 10.36 The bar has cross-sectional area A and
modulus of elasticity E. If an axial force F directed toward the right is applied at C, what is the normal stress in
the part of the bar to the left of C? (Strategy: Draw the
free-body diagram of the entire bar and write the equilibrium equation. Then apply the compatibility condition
that the increase in length of the part of the bar to the left
of C must equal the decrease in length of the part to the
right of C.)
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD:
[1]
ΣFx = 0 = F − RL − RR
The compatibility condition requires that the change in length of the
left portion of the bar must equal the change in length for the right
portion of the bar.
RL LL
RR LR
=
AL EL
AR ER
Since the denominators of the above equation are identical, we need
only consider the numerators.
[2]
RR = (LL /LR )(RL ) = [(L/3)/(2L/3)](RL ) = RL /2
Substituting Equation [2] into Equation [1]:
F = RL + RR = RL + (RL /2) = 3RL /2
or RL = (2/3)F
The stress in the left-hand portion of the bar is:
σ=
ANS:
σ = 2F/3A
RL
P
=
A
A
Problem 10.37 In Problem 10.36, what is the resulting Free Body Diagram:
displacement of point C?
Solution:
Substituting Equation [2] into Equation [1]:
We can use either the left-hand or right-hand portion of the bar to
determine the displacement of point C. The left-hand portion of the
bar is chosen.
Summing horizontal forces on the FBD:
[1]
ΣFx = 0 = F − RL − RR
F = RL + RR = RL + (RL /2) = 3RL /2
or RL = (2/3)F
The displacement of point C is:
The compatibility condition requires that the change in length of the
left portion of the bar must equal the change in length for the right
portion of the bar.
RL LL
RR LR
=
AL EL
AR ER
Since the denominators of the above equation are identical, we need
only consider the numerators.
[2]
RR = (LL /LR )(RL ) = [(L/3)/(2L/3)](RL ) = RL /2
Problem 10.38 The bar in Problem 10.36 has crosssectional area A = 0.005 m2 , modulus of elasticity
E = 72 GPa, and L = 1 m. It is made of a material that will safely support a normal stress (in tension
and compression) of 120 MPa. Based on this criterion,
what is the largest axial force that can be applied at C?
Free Body Diagram:
Solution:
We can use either the left-hand or right-hand portion of the bar to
determine the displacement of point C. The left-hand portion of the
bar is chosen.
Summing horizontal forces on the FBD:
[1]
ΣFx = 0 = F − RL − RR
The compatibility condition requires that the change in length of the
left portion of the bar must equal the change in length for the right
portion of the bar.
RL LL
RR LR
=
AL EL
AR ER
Since the denominators of the above equation are identical, we need
only consider the numerators.
[2]
RR = (LL /LR )(RL ) = [(L/3)/(2L/3)](RL ) = RL /2
Substituting Equation [2] into Equation [1]:
F = RL + RR = RL + (RL /2) = 3RL /2
or RL = (2/3)F
We see that the greater strain is in the left-hand portion of the bar
(RL > RR ). Using the given maximum allowable stress:
120 × 106 N/m2 =
ANS:
F = 900 kN
(2/3) F
(2/3) F
=
A
0.005 m2
PL
δ=
=
AE
ANS:
δ = 2F L/9AE
2 L
F
3
3
AE
Problem 10.39 In Example 3-7, determine the normal
stresses in parts A and B of the bar if the force applied at
the joint between parts A and B is (a) 40 kip; (b) 200 kip.
Free Body Diagram:
Solution:
(a) Assuming no reaction from the right-hand wall, total displacement
of the right-hand end of the bar is:
δ=
FL
(40, 000 lb) (10 in)
= 0.0106 in
= (2 in)2
AE
π 4
12 × 106 lb/in2
We see that the displacement is not sufficient to close the 0.02-in gap,
so there can be no reaction from the wall at the right.
The stress in the left-hand portion of the bar is:
σ=
P
40, 000 lb
=
(2 in)2
A
π 4
ANS: σ = 12, 732 lb/in2
(b) Again assuming no reaction from the right wall, we calculate the
displacement of the right-hand end of the bar.
δ=
PL
(200, 000 lb) (10 in)
= 0.053 in
= (2 in)2
AE
π 4
12 × 106 lb/in2
We see that this displacement is MORE than enough to close the 0.02in gap. The reaction at the right-hand wall must be sufficient to limit
the displacement of the right end of the bar to 0.02 in. This means that
the reaction at the right wall must be sufficient to displace the right end
of the bar a distance of:
δ = 0.053 in − 0.02 in = 0.033 in
The reaction at the right-hand wall is:
δ=
π
(R
R )(10 in)
(12×106 lb/in2 )
(2 in)2
4
+
(R
R )(8 in)
(4 in)2
π
(12×106 lb/in2 )
4
= 0.033 in
RR = 103, 672 lb
The stress in the right-hand portion of the bar is:
σR =
P
A
=
103,672 lb
π
σR = 8250psi
(4 in)2
4
Summing horizontal forces on the bar:
ΣFx = 0 = 200, 000 lb−RR −RL = 200, 000 lb−103, 672 lb−RL
RL = 96, 328 lb
The stress in the left hand portion of the bar is:
σL =
ANS:
σL = 30, 600 psi
P
96, 238 lb
=
(2 in)2
A
π 4
Problem 10.40 The bar has a circular cross section
and modulus of elasticity E = 70 GPa. Parts A and C
are 40 mm in diameter and part B is 80 mm in diameter.
If F1 = 60 kN and F2 = 30 kN, what is the normal
stress in part B?
Free Body Diagram:
Solution:
We must first determine the reactions at the left and right walls. We
allow the right-hand side of the bar to “float.”The displacement of the
“free” right-hand side of the bar is:
δR =
δR =
F1 L A
F2 L A
− AF2 LEB − A
AA EA
B B
A EA
(60,000N)(0.2 m)
π
(0.04 m)2
4
(70×106
N/m2 )
−
(30,000N)(0.4 m)
(0.08 m)2
π
(70×106 N/m2 )
4
−
(30,000N)(0.2 m)
(0.04 m)2
π
(70×106 N/m2 )
4
δR = 0.0341 m = 34.1 mm
The reaction at the right wall must be sufficient to prevent ANY displacement. Its magnitude is:
 
RR
0.2
m
0.4 m
 +
0.0341 m = 70×106 lb/in2  +
2
(0.08 m)2
π
(0.04 m)
4
π
4
0.2 m
π
(0.04 m)2
4


RR = 6000 N
The reaction at the left wall is:
ΣFx = 0 = 60, 000 N − 6, 000 N − 30, 000 N − RL
RL = 24, 000 N ←
The stress in section A is:
σB =
P
A
=
−24,000 N+60,000 N
π
(0.08 m)2
4
Note: The negative sign indicates a compressive stress.
ANS: σA = −7.16 MPa
Problem 10.41 In Problem 10.40, if F1 = 60 kN,
what force F2 will cause the normal stress in part C to
be zero?
Free Body Diagram:
Solution:
For the normal stress in section C to be zero, we see that the displacement of the intersection of sections B and C must be zero.
The equation for the displacement of the intersection of sections B and
C is:
δBC = 0 =
ANS:
(60, 000 N)(0.2 m)
F2 (0.2 m)
F2 (0.4 m)
−
−
2
2
2
9
2
9
2
π ((0.04 m) /4) 70 × 10 N/m
π ((0.04 m) /4) 70 × 10 N/m
π ((0.08 m) /4) 70 × 109 N/m2
F2 = 40, 000 N = 40 kN
Problem 10.42 The bar in Problem 10.40 consists of Free Body Diagram:
a material that will safely support a normal stress of
40 MPa. If F2 = 20 kN, what is the largest safe value
of F1 ?
Solution:
We see that section B has a cross-sectional area which is four times
that of sections A and C.
The cross-sectional areas are:
AA = AC = π (0.02 m)2 = 0.00126 m2
AB = π (0.04 m)2 = 0.00503 m2
With both ends of the bar restrained, total displacement must be zero.
R (0.2 m)
(F −R )(0.4 m)
(F −R −20,000 N)(0.2 m)
1
1
L
L
L
−
−
(0.00126 m2 )(70×109 n/m2 )
(0.00503 m2 )(70×109 N/m2 )
(0.00126 m2 )(70×109 N/m2 )
2.268 × 10−9 RL + 1.136 × 10−9 RL − 1.137 × 10−9 F1 + 2.268 × 10−9 RL − 2.268 × 10−9 F1 + 4.535 × 10−5 RL = 0
δR = 0 =
RL = 0.6F1 − 8, 000 N
Summing horizontal forces to find RR :
ΣFx = 0 = −RL +F1 −20, 000 N−RR = −0.6F1 +8004 N+F1 −20, 000 N−RR
RR = 0.4F1 − 12, 000 N
The axial loads in sections A, B and C are:
σa = 40 × 106 =
F1 = 97.3 kN
σb = 40 × 106 =
F1 = 523 kN
σc = 40 × 106 =
F1 = 156 kN
0.6f1 −8,000
0.00126
0.6f1 −8,000−F1
0.00503
0.4f1 −12,000
0.00126
The smallest of these three values for F1 is the highest allowable value.
ANS: F1 = 97.3 kN
Problem 10.43 Two aluminum bars (EAL = 10.0 × Free Body Diagram:
106 psi) are attached to a rigid support at the left and
a cross-bar at the right. An iron bar (EF E = 28.5 ×
106 psi) is attached to the rigid support at the left and
there is a gap b between the right end of the iron bar and
the cross-bar. The cross-sectional area of each bar is A =
0.5 in2 and L = 10 in. The iron bar is stretched until it
contacts the cross-bar and welded to it. Afterward, the
axial strain of the iron bar is measured and determined
to be εF E = 0.002. What was the size of the gap b?
Solution:
The foreshortening of the two aluminum bars plus the lengthening of
the steel bar must equal the gap, b. The lengthening of the steel bar is
(approximately):
δF E = (0.002)(10 in) = 0.02 in
Calculating the force in the steel bar:
PF E = εEA = (0.002)(28.5×106 lb/in2 )(0.5 in2 ) = 28, 500 lb
This same force is compressing the TWO aluminum bars. The aluminum bars are shortened by an amount of:
δAL =
PL
(28, 500 lb)(10 in))
= 0.0285 in
= AE
2 0.5 in2 10 × 106 lb/in2
The total original gap is:
b = δAL + δF E = 0.0285 in + 0.02 in
ANS:
b = 0.0485 in
Problem 10.44 In Problem 10.43, the iron will safely Free Body Diagram:
support a tensile stress of 100 ksi and the aluminum will
safely support a compressive stress of 40 ksi. What is
the largest safe value of the gap b?
Solution:
We see from the FBD that FF e = 2FAl .
Maximum allowable load in the aluminum bars is:
(σM AX )Al = 40, 000 lb/in2 =
FAl
→ (FAl )MAX = 20, 000 lb
0.5 in2
Maximum allowable load in the iron bar is:
(σM AX )F e = 100, 000 lb/in2 =
(FF e )M AX
→ (FF e )MAX = 50, 000 lb
0.5 in2
The maximum allowable stresses show that the allowable load in the
aluminum is the controlling criterion.
Using a load of 20, 000 lb in the aluminum and 40,000 in the iron:
b=
FAl L FF e L
(20, 000 lb)(10 in)
(40, 000 lb)(10 in)
+
=
+
AEAl AEF e
(0.5 in2 )(10 × 106 lb/in2 ) (0.5 in2 )(28.5 × 106 lb/in2 )
ANS:
b = 0.068 in
Problem 10.45 Bars AB and AC each have cross- Free Body Diagram:
sectional area A and modulus of elasticity E. If a downward force F is applied at A, show that the resulting
Fh
1
downward displacement of point A is EA
1+cos3 θ .
Solution:
Summing vertical forces at point A;
[1]
ΣFy = 0 = −F + PAB (cos θ) + PAC
We see that the VERTICAL movement of the A-end of member AB
will be:
[2]
vA = δAB (1/ cos θ)
From Equation [2], we see that the vertical movement of point A must
be the same for BOTH members of the truss, so:
h
1
PAB cos
PAC h
cos θ
=
AE
AE
[3]
PAB = PAC (cos2 θ)
Substituting Equation [3] into Equation [1]:
F = PAB (cos θ) + PAC
= PAC cos2 θ (cos θ) + PAC
F = PAC 1 + cos3 θ
F
PAC = 1+cos
3θ
Substituting this value for PAC into the expression for change in length
for member AC:
F
h
PAC h
1+cos3 θ
δAB =
=
AE
AE
Fh
1
ANS: δAB = AE 1+cos3 θ
Problem 10.46 If a downward force F is applied at
point A of the system shown in Problem 10.45, what are
the resulting normal stresses in bars AB and AC?
Free Body Diagram:
Solution:
Summing vertical forces at point A;
[1]
ΣFy = 0 = −F + PAB (cos θ) + PAC
We see that the VERTICAL movement of the A-end of member AB
will be:
[2]
vA = δAB (1/ cos θ)
From Equation [2], we see that the vertical movement of point A must
be the same for BOTH members of the truss, so:
h
1
PAB cos
PAC h
cos θ
=
AE
AE
[3]
PAB = PAC (cos2 θ)
Substituting Equation [3] into Equation [1]:
2
F = PAB (cos θ) + PAC
= PAC cos θ (cos θ) + PAC
3
F = PAC 1 + cos θ
[4]
PAC =
F
1 + cos3 θ
Using Equation [4] to determine the stress in member AC:
ANS:
[5]
σAC =
PAC
A
=
F
A(1+cos3 θ )
Combining Equations [3] and [4] to calculate the normal stress in
member AB:
F
cos2 θ
PAB
1+cos3 θ
σAB =
=
A
A
ANS:
σAB =
F cos2 θ
A[1+cos3 θ]
Problem 10.47 Each bar has a 500 − m m2 crosssectional area and modulus of elasticity E = 72 GPa.
If a 160-kN downward force is applied at A, what is the
resulting displacement of point A?
Free Body Diagram:
Solution:
The symmetry of the truss dictates that ALL of the displacement will
be vertical. We see that the axial loads in the two angled members will
be equal. In other words, δB = δC .
Summing vertical forces where the load is applied:
[1]
ΣFy = 0 = −160, 000 N + PA + 2PB (sin 60◦ )
The vertical displacement for each of the two angled members is:
vangle = δangle (1/ sin 60◦ )
Using the relationship between displacements for the straight member
and the two angled members:
δA = δB sin160◦ = δC sin160◦
0.3 m
1 PB ( sin
PA (0.3 m)
60◦ )
=
AE
AE
sin 60◦
[2]
PA = 1.333PB = 1.333PC
Substituting Equation [2] into Equation [1]:
0 = −160, 000 N + 1.333PB + 2PB (sin 60◦ )
PB = PC = 52, 201 N
From Equation [2] we get the axial load in member A:
PA = 1.333(52, 201 N) = 69, 584 N
The displacement of point A is:
(69,584 N)(0.3 m)
LA
=
= 0.580 mm ↓
ANS: δA = PAAE
(500×10−6 m2 )(72×109 N/m2 )
Problem 10.48 The bars in Problem 10.47 are made
of material that will safely support a tensile stress of
270 MPa. Based on this criterion, what is the largest
downward force that can safely be applied at A?
Free Body Diagram:
Solution:
The symmetry of the truss dictates that ALL of the displacement will
be vertical. We see that the axial loads in the two angled members will
be equal. In other words, δB = δC . Summing vertical forces where
the load is applied:
ΣFy = 0 = −F + PA + 2PB (sin 60◦ )
[1]
The vertical displacement for each of the two angled members is:
vangle = δangle (1/ sin 60◦ )
Using the relationship between displacements for the straight member
and the two angled members:
δA = δB sin160◦ = δC sin160◦
0.3 m
1 PB ( sin
PA (0.3 m)
60◦ )
=
AE
AE
sin 60◦
PA = 1.333PB = 1.333PC → PB = PC = 0.75PA
[2]
We see that the largest load is supported by the central member. Calculating the maximum allowable load in the members of the truss:
PA = PMAX = σMAX A = (270×106 N/m2 )(500×10−6 m2 ) = 135, 000 N
Using Equations [2] and [3] in Equation [1]:
F = 135, 000 N + 2[(0.75)(135, 000 N)](sin 60◦ )
ANS:
F = 310 kN
[3]
Problem 10.49 Each bar has a 500 − m m2 crosssectional area and modulus of elasticity E = 72 GPa.
If there is a gap h = 2 mm between the hole in the
vertical bar and the pin A connecting bars AB and AD,
what are the normal stresses in the three bars after the
vertical bar is connected to the pin at A?
Free Body Diagram:
Solution:
The downward displacement of bar C plus the upward displacement
of bars B and D equal 2 mm.
[1]
δA + δB / sin 60◦ = 2 × 10−3 m
The symmetry of the structure indicates that PB = PD . We also see
that the sum of vertical forces on point A must be zero after the bars
are connected.
ΣFy = 0 = −PA + 2(PB )(sin 60◦ )
[2]
PA = 1.732PB
Combining Equations [1] and [2]:
(1.732PB )(0.3 m)
(500×10−6 m2 )(72×109
PBC = 78, 295 N
N/m2 )
+
0.3 m
PB ( sin
60◦ )
(500×10−6 m2 )(72×109
Therefore: PA = (1.732)(78, 295 N)
PA = 135, 600 N
The normal stresses in the truss members are:
σA =
PA
A
=
135,600 N
500×10−6 m2
σA = 271 MPa
σB =
ANS:
PB
A
σB = 157 MPa
=
78295 N
500×10−6 m2
N/m2 )
1
sin 60◦
= 2 × 10−3 m
Problem 10.50 The bars in Problem 10.49 are made of
material that will safely support a normal stress (tension
or compression) of 400 MPa. Based on this criterion,
what is the largest safe value of the gap h?
Free Body Diagram:
Solution:
Summing horizontal forces at point A:
ΣFx = 0 = FB (cos 60◦ ) − FD (cos 60◦ ) → FB = FD
Summing vertical forces at point A:
ΣFy = 0 = FC +FB (sin 60◦ )+FD (sin 60◦ ) = FC +2FB (sin 60◦ )
Fc = −1.732FB = −1.732FD
[1]
From Equation [1] we see that the greatest load exists in member AC.
The magnitude of the maximum allowable force in member AC is:
(FC )MAX = σM AX A = (400 × 106 N/m2 )(500 × 10−6 m2 )
(FC )MAX = 200, 000 N (T)
From Equation [1] we see that the load in members AB and AD is:
FB = FD =
(FC )MAX
200, 000 N
=
1.732
1.732
FB = FD = 115, 473 N (C)
From the diagram we see that:
vAD = δAD / cos 30◦
[2]
The change in length for member AC is:
δAC =
δAC =
FC (0.3 m)
(500 × 10−6 m2 )(72 × 109 N/m2 )
(200, 000 N)(0.3 m)
(500 × 10−6 m2 )(72 × 109 N/m2 )
δAC = 0.00167 m = 1.67 mm ↑
The change in length for members AB and AD is:
δAB =
(115, 473 N) (0.3 m/sin 60◦ )
(500 × 10−6 m2 )(72 × 109 N/m2 )
δAB = δAD = 0.00111 m = 1.11 mm
The maximum allowable gap h is:
h = δAC +
ANS:
δAB
1.11 mm
= 1.67 mm +
cos 30◦
cos 30◦
h = 2.95 mm
Problem 10.51 The bar’s cross-sectional area is A −
(1 + 0.1x) in2 and the modulus of elasticity of the material is E = 12 × 106 psi. If the bar is subjected to tensile
axial forces P = 20 kip at its ends, what is the normal
stress at x = 6 in?
Free Body Diagram:
Solution:
At x = 6 in, the cross-sectional area of the bar is:
A = (1 + 0.1(6)) in2 = 1.6 in2
The normal stress in the bar at x = 6 in is:
σ = P/A = (20, 000 lb/(1.6 in2 )
ANS:
σ = 12, 500 psi = 12.5 ksi
Problem 10.52 What is the change in length of the bar
in Problem 10.51?
Free Body Diagram:
Solution:
The normal stress at any point in the bar is:
σ = P/A = (20, 000 lb)/(1 + 0.1x) in2
The strain at any point in the bar is:
ε=
σ
(20, 000 lb)/(1 + 0.1x) in2
0.001667 in2
=
=
2
6
E
(1
+ 0.1x) in2
12 × 10 lb/in
The change in length of the bar is:
10
10
10
0.001667 in2
1
0.001667 in 10
0.1
δ=
ε=
dx = 0.001667 in3
dx =
dx
2
2
0.1
(1 + 0.1x)
(1 + 0.1x) in
(1 + 0.1x)in
0
0
0
0
δ = 0.01667 [ln (1 + 0.1(10)) − ln (1 + 0.1(0))]
ANS:
δ = 0.01155 in
Problem 10.53 The cross-sectional area fo the bar in
Problem 10.51 is A = (1+az) in2 , where a is a constant,
and the modulus of elasticity of the material is E =
8 × 106 psi. When the bar is subjected to tensile axial
forces P = 14 kip at its ends, its change in length is δ =
0.01 in. What is the value of the constant a? (Strategy:
Estimate the value of a by drawing a graph of δ as a
function of a.
Free Body Diagram:
Solution:
The normal stress at any point in the bar is:
σ=
14, 000 lb
P
=
A
(1 + ax) in2
The strain at any point in the bar is:
ε=
σ
(14, 000/1 + ax) lb/in2
0.00175
=
=
2
6
E
1 + ax
8 × 10 lb/in
Total change in length of the bar may be expressed as:
10
10
0.00175
0.00175 10 a dx
0.00175
δ=
dx =
=
ln(1 + ax) +C
1
+
ax
a
1
+
ax
a
0
0
0
δ=
0.00175
[ln(1 + 10a)] + C
a
At x = 0, we see that δ = 0, so C = 0.
For the given change in length:
0.01 in =
0.00175
[ln(1 + 10a)]
a
Rearranging the terms in the equation and using a graphing calculator
to find the value of a:
ln(1 + 10a) − 5.714a = 0
ANS:
a = 0.1805 in−1
Problem 10.54 From x = 0 to x = 100 mm, the bar’s
height is 20 mm. From x = 100 mm to x = 200 mm,
its height varies linearly from 20 mm to 40 mm. From
x = 200 mm to x = 300 mm, its height is 40 mm. The
flat var’s thickness is 20 mm. The modulus of elasticity
of the material is E = 70 GPa. If the bar is subjected
to tensile axial forces P = 50 kN at its ends, what is its
change in length?
Free Body Diagrams:
Solution:
The problem is solved by considering each of the three sections of the
bar separately.
Elongation of the left-hand section of the bar is:
δL =
PL
(50, 000 N)(0.1 m)
=
= 0.1785 mm
AE
(0.02 m)(0.02 m)(70 × 109 N/m2
Elongation of the right-hand section of the bar is:
δR =
PL
(50, 000 N)(0.1 m)
=
= 0.0893 mm
AE
(0.02 m)(0.04 m)(70 × 109 N/m2
Determining the elongation of the center section of the bar will require
integration. The cross-sectional area at any point in the center section
is:
AC = (0.02 m)[0.02 m+((0.02 m)/(0.1 m)(x)] = 0.0004+0.004x
The stress at any point in the center section of the bar is:
σ=
50, 000
P
=
n/m2
A
0.0004 + 0.004x
Total elongation of the center section of the bar is:
σ
50, 000 N/(0.0004 + 0.004x) m2
dδC =
dx =
dx
2
E
70 × 109 N/m
δC =
0.1
0
7.143 × 10−7
7.143 × 10−7
dx =
0.0004 + 0.004x
0.004
0.1
0
0.004
dx
0.0004 + 0.004x
δC = 1.786×10−4 [ln (0.0004 + 0.004(0.1)) − ln (0.0004 + 0.004(0))]
δC = 0.1238 mm
Total change in length of the bar is:
δ = δL + δC + δR = 0.1785 mm + 0.1238 mm + 0.0893 mm
ANS:
δ = 0.392 mm
Problem 10.55 Fro x = 0 to x = 10in, the bar’s
cross-sectional area is A = 1 in2 . Fro x = 10 in to
x = 20 in, A = (0.1x) in2 . The modulus of elasticity
of the material is 12×106 psi. There is a gap b = 0.02 in
between the right end of the bar and the rigid wall. If the
bar is stretched so that is contacts the rigid wall and is
welded to it, what is the axial force in the bar afterward?
Free Body Diagrams:
Solution:
The deformation of the right-hand portion of the bar is:
dδR =
δR =
σ
P/A
P/0.1x in2
10P lb/in2
dx =
dx =
dx =
dx in
E
E
12 × 106 lb/in2
(12 × 106 lb/in2 )x
20
10
10P
10P
dx =
(12 × 106 )x
12 × 106
20
10
dx
P
=
[ln 20−ln 10] = (5.776×10−7 )P in
x
12 × 105
The expression for the total deformation of the bar is:
P
0.02 in = δL +δR =
+(5.77×10−7 )P
2
2
(1 in )(12 × 106 lb/in
ANS:
P = 3.0288 × 104 lbs
Problem 10.56 From x = 0 to x = 10 in, the crosssectional area of the bar in Problem 10.55 is A = 1 in2 .
The modulus of elasticity of the material is E = 12 ×
106 psi. There is a gap b = 0.02 in between the right
end of the bar and the rigid wall. If a 40 kip axial force
toward the right is applied to the bar at x = 10 in, what
is the resulting normal stress in the left half of the bar?
Free Body Diagram:
Solution:
We first determine whether the 40, 000−lb load is sufficient to close
the 0.02−in gap.
δ=
(40, 000 lb)(10 in)
(1 in2 )(12 × 106 lb/in2 )
= 0.0333 in,
which is larger than the 0.02−in gap, so there will be reaction at the
right-hand end of the bar.
Deformation of the left-hand portion of the bar is:
δL =
(40, 000 lb − RR )(10 in)
(1 in2 )(12 × 106 lb/in2 )
= 0.0333 in − 8.333 × 10−7 RR
Deformation of the right-hand portion of the bar is:
20
20
−RR dx
−RR
dx
−RR
δR =
=
=
[ln(20) − ln(10)] = −5.776×10−7 RR in
2
5
2
6
12 × 10 10 x
12 × 105
10 (0.1x in )(12 × 10 lb/in )
The expression for total deformation of the bar is:
δL + δR = 0.02 in
(0.0333 in − 8.333 × 10−7 RR ) − 5.776 × 10−7 RR = 0.02 in
RR = +9, 430 lb
RL = +30, 570 lb
Normal stress in the left-hand portion of the bar is:
σ=
ANS:
RL
30, 570 lb
=
1 in2
1 in2
σ = 30, 570lb/in2
Problem 10.57 The diameter of the bar’s circular
cross-section varies linearly from 10 mm at its left end
to 20 mm at its right end. The modulus of elasticity of
the material is E = 45 GPa. If the bar is subjected to
tensile axial forces P = 6 kN at its ends, what is the
normal stress at x = 80 mm?
Free Body Diagram:
Solution:
The function which describes the diameter of the bar is:
0.02 m − 0.01 m
D = 0.01 m+
(x) m = 0.01 m+(0.0667x) m
0.150 m
The diameter of the bar at x = 80 mm is:
D80 = 0.01 m + (0.0667)(0.08 m) = 0.01534 m
The cross-sectional area of the bar at x = 80 mm is:
A80 =
2
πD80
π(0.01534 m)2
=
= 0.0001847 m2
4
4
The normal stress in the bar at x = 80 mm is:
σ=
ANS:
P
6, 000 N
=
A
0.0001847 m2
σ = 32.5 MPa
Problem 10.58 What is the change in length of the bar
in Problem 10.57?
Free Body Diagram:
Solution:
The radius of the cross-sectional area at any point along the length of
the bar is:
0.02 − 0.01
d = 0.01 +
x m = (0.01 + 0.0667x) m
0.15
The cross-sectional area of the bar at any point along its length is:
A=π
d2
(0.01 + 0.0667x)2
=π
4
4
The change in length of the bar is:
0.15
0.15
0.15
σ
P
6, 000 N
δ=
dx =
dx =
dx
(0.01+0.0667x)2
E
AE
0
0
0
π
m2 (45 × 109 N/m2 )
4
Using Mathcad to evaluate the itegral:
ANS: δ = 0.127 mm
Problem 10.59 The bar is fixed at the left and is subjected to a uniformly distributed axial force. It has crosssectional area A and modulus of elasticity E. (a) Determine the internal axial force P in the bar as a function
of x. (b) What is the bar’s change in length?
Free Body Diagram:
Solution:
(a) The internal axial force at any point in the bar is:
P (x) = qL − qx = q(L − x)
ANS:
(b) The change in length of the bar is:
L
L
L
L
σ
P (x)
q(L − x)
qLx
qx2 δ=
dx =
dx =
dx =
−
AE
AE
AE
2AE 0
0 E
0
0
δ=
ANS:
qL2
2AE
Problem 10.60 The bar shown in Problem 10.59 has
length L = 2 m, cross-sectional area A = 0.03 m2 , and
modulus of elasticity E = 200 GPa. It is subjected to a
distributed axial force q = 12(1 + 0.4x) MN/m. What
is the bar’s change in length?
Free Body Diagram:
Solution:
The reaction at the wall is:
2
2
R=
12(1 + 0.4x) dx = (12x + 2.4x2 )0 = 33.6 MN ←
0
Total change in length for the bar is:
2
(33.6 − 12 − 4.8x) MN dx
δ=
2
2
9
0 (0.03 m )(200 × 10 N/m )
ANS:
δ = 0.0056 m = 5.6 mm
Problem 10.61 A cylindrical bar with 1−in diameter
fits tightly into a circular hole in a 5−in thick plate. The
modulus of elasticity of the marerial is E = 14×106 psi.
A 1000−lb tensile force is applied at the left end of the
bar, causing it to begin slipping out of the hole. At the
instant slipping begins, determine (a) the magnitude of
the uniformly distributed axial force exerted on the bar
by the plate; (b) the total change in the bar’s length.
Free Body Diagram:
Solution:
The distributed load along the 5−inch section of the bar is:
lb
ANS: q = 1,000
= 200 lb/in
5 in
The elongation of the 10−inch section of the bar is:
δ10 =
PL
(1, 000 lb)(10 in)
=
= 0.0009095 in
AE
π(0.5) in)2 (14 × 106 lb/in2 )
The elongation of the 5−inch section of the bar is:
5
(1000 − 200x) lb/π(0.5 in2 )2
σ
δ5 = εL = L =
dx = 0.0002273 in
E
14 × 106 lb/in2
0
Total change in length of the bar is:
δ = δ10 + δ5 = 0.0009095 in + 0.0002273 in
ANS:
δ = 0.00114 in
Problem 10.62 The bar has a circular cross section
with 0.002−m diameter and its modulus of elasticity is
E = 86.6 GPa. The bar is fixed at both ends and is
subjected to a distributed axial force q = 75 kN/m and
an axial force F = 15 kN. What is its change in length?
Free Body Diagram:
Solution:
Cross-sectional area of the bar is:
0.002 m 2
A=π
= 3.142 × 10−6 m2
2
Summing horizontal forces on the FBD to find the magnitude of R:
ΣFx = 0 = −R − 15, 000 N + (75, 000 N/m)(0.8 m)
R = 45, 000 N ←
Total change in length of the bar is:
0.8
(45, 000 N − 75, 000(x) N) dx
δ=
(3.142 × 10−6 m2 )(86.6 × 109 m2 )
0
ANS:
δ = 0.0441 m = 44.1 mm
Problem 10.63 In Problem 10.62, what axial force F
would cause the bar’s change in length to be zero?
Free Body Diagram:
Solution:
Summing horizontal forces on the FBD to find the reaction at the wall:
ΣFx = 0 = −R − F + (75, 000 N/m)(0.8 m)
R = (F − 60, 000 N
Setting the expression for total change in length equal to zero:
0.8
(F − 75, 000(x)) dx
δ=0=
(3.142 × 10−6 m2 )(86.6 × 109 N/m2
0
0.8
0 = F x − 37, 500x2 0 0.8F − 24, 000 N
ANS:
F = 30, 000 N
Problem 10.64 If the bar in Problem 10.62 is subjected
to a distributed force q = 75(1 + 0.2x) kN/m and an
axial force F = 15 kN, what is its change in length?
Free Body Diagram:
Solution:
The reaction at the wall can be found by summing horizontal forces:
0.8
ΣFx = 0 = −15, 000 N +
(75, 000)(1 + 0.2x) dx − RW
0
RW = 49, 800 N ←
Starting at the left-hand end of the bar, the function which describes
the axial force in the bar is:
x
P = 49, 800 N−
(75, 000)(1+0.2x) dx = 49, 800 N−75, 000x−7, 500x2
0
The cross-sectional area of the bar is:
A = π(0.001 m)2 = 3.1416 × 10−6 m2
Stress at any point in the bar is:
σ=
P
(49, 800 − 75, 000x − 7, 500x2 )
N/m2
=
A
3.1416 × 10−6
The change in length for the bar is:
0.8
σ
49, 800 − 75, 000x − 7, 500x2 ) N
δ= L=
dx
E
(3.1416 × 10−6 m2 )(86.6 × 109 N/m2 )
0
ANS:
δ = 0.0535m
Problem 10.65 The bar is fixed at A and B and is subjected to a uniformly distributed axial force. It has crosssectional area A and modulus of elasticity E. What are
the reactions at A and B?
Free Body Diagram:
Solution:
We recognize that the sum of horizontal forces on the bar must be zero.
ΣFx = 0 = −RA − RB + qL [1]
Removing the fixed structure at the right-hand end of the bar, the distributed load will produce a change in length of:
L
L
qx
qx2 qL2
δ=
dx =
=
2AE 0
2AE
0 AE
RB =
qL
2
Substituting this value for RB in Equation [1]:
RA = −RB + qL = −(qL/2) + qL
ANS:
RA = qL/2
Problem 10.66 What point of the bar in Problem 10.65
undergoes the largest displacement, and what is the displacement?
Free Body Diagram:
Solution:
We regognize that the sum of horizontal forces on the bar must be zero.
L
L
qx
qx2 qL2
Σ=
dx =
=
[1]
2AE 0
2AE
0 AE
The reaction at the right-hand support must be sufficient to produce a
reduction in length of −(qL2 )/(2AE).
−qL2
RB L
=
2AE
AE
RB =
qL
←
2
Substituting this value for RB in Equation [1]:
RA = −RB + qL = −(ql/2) + qL
RA = qL/2 ←
The expression for displacement of any point on the bar is:
L
(qL/2) − qx
δ=
dx [2]
AE
0
We know that maximum deflection occurs where the expression
dδ/dx = 0. Setting the derivative of Equation [2] equal to zero:
(qL/2) − qx
=0
AE
ANS: x = L/2 [3]
Using the value of x from Equation [3] in Equation [2] and evaluating:
L/2
L/2
(qL/2) − qx
(qLx/2) − qx2 /2
dx =
δ=
AE
AE
0
0
ANS:
δMAX =
qL2
8AE
Problem 10.67 A line L within an unconstrained sample of material is 200 mm long. The coefficient of thermal expansion of the material is α = 22 × 10−6◦ C−1 .
If the temperature of the material is increased by 30◦ C,
what is the length of the line?
Solution:
The new length of the line will be:
δ = Lα(∆T ) = (200 mm)(22 × 10−6◦ C−1 )(30◦ C) = 0.132
ANS:
L = 200.132 mm
Problem 10.68 The length of the line L within the
unconstrained sample of material shown in Problem 67 is
2 in. The coefficient of thermal expansion of the material
is α = 8 × 10−6◦ F−1 . After the temperature of the
material is increased, the length of the line is 2.002 in.
How much was the temperature increased?
Solution:
The thermal strain produced by the change in temperature is:
εT =
L − L
2.002 in − 2.000 in
=
= 0.001
L
2.000 in
The temperature increase required to produce the above thermal strain
is:
εT = α(∆T )
0.001 = (8 × 10−6◦ F−1 )(∆T )
ANS:
∆T = 125◦ F
Problem 10.69 Consider a 1 in. × 1 in. × 1 in. cube
within an unconstrained sample of material. The coefficient of thermal expansion of the material is α =
14 × 10−6◦ F−1 . If the temperature of the material is
decreased by 40◦ F, what is the volume of the cube?
Solution:
As the material warms, EVERY dimension (length, width and breadth)
is increased. The new length of one side of the cube will be:
L = L+∆L = L+(L)(α)(∆T ) = 1 in+(1 in)(14×10−6◦ C−1 )(−40◦ C)
L = 0.99944 in
The new volume of the cube will be:
V = (L )3 = (0.99944 in)3
ANS:
V = 0.998 in3
Problem 10.70 A prismatic bar is 200 mm long and
has a circular cross section with 30-mm diameter. After
the temperature of the unconstrained bar is increased, its
length is measured and determined to be 200.160 mm.
What is the bar’s diameter after the increase in temperature?
Solution:
Strain in the material will be the same in every direction.
The thermal strain in the direction of the length of the bar is:
εT =
L − L
200.160 mm − 200 mm
=
= 0.0008
L
200 mm
The same thermal strain in the direction of the radius will produce a
diameter of:
D = εD = (0.0008)(30 mm)
ANS:
D = 30.024 mm
Problem 10.71 If the increase in temperature in Problem 10.70 is 20◦ C, what is the coefficient of thermal
expansion of the bar?
Solution:
The original length of the bar is 200 mm, and the length of the bar after
thermal strain is 200.160 mm. The coefficient of thermal expansion
is:
∆L = L − L = L(α)(∆T )
200.160 mm − 200 mm = (200 mm)(α)(20◦ C)
ANS:
α = 40 × 10−6◦ C−1
Problem 10.72 If the increase in temperature in Problem 10.70 is 20◦ C and the modulus of elasticity of the
material is E = 72 GPa, what is the normal stress on a
plane perpendicular to the bar’s axis after the increase in
temperature? Strategy: Obtain a free-body diagram
by passing a plane through the bar.
Solution:
Since the bar is unconstrained, there will be no normal stress in the bar
regardless of the change in temperature.
ANS:
σ=0
Problem 10.73 The prismatic bar is made of material with modulus of elasticity E = 28 × 106 psi and
coefficient of thermal expansion α = 8 × 10−6◦ F−1 .
The temperature of the unconstrained bar is increased
by 50◦ F above its initial temperature T . (a) What is the
change in the bar’s length? (b) What is the change in
the bar’s diameter? (c) What is the normal stress on a
plane perpendicular to the bar’s axis after the increase in
temperature?
Strategy: To answer part (c), obtain a free-body diagram by passing a plane through the bar.
Solution:
(a) The new (heated) length of the bar will be:
∆L(α)(∆T ) = (15 in.)(8 × 10−6◦ F−1 )(50◦ F)
ANS:
L = 0.006 in.
(b) The change in the diameter of the bar is:
∆D = D(α)(∆T ) = (2 in.)(8 × 10−6◦ F−1 )(50◦ F)
ANS:
∆D = 0.0008 in.
(c) Because the bar is unconstrained, no forces are exerted on the ends
of the bar. The normal stress on EVERY plane passed through the bar
is
ANS:
σ=0
Problem 10.74 Suppose that the temperature of the
unconstrained bar in Problem 10.73 is increased by 50◦ F
above its initial temperature T and the bar is also subjected to 30,000-lb tensile axial forces at the ends. What
is the resulting change in the bar’s length?
Determine the change in length assuming that (a) the
temperature is first increased and then the axial forces
are applied; (b) the axial forces are applied first and then
the temperature is increased.
Solution:
Considering first the case in which the temperature changes first, then
the load is applied:
(∆L)T = Lα(∆T ) = (15 in.)(8×10−6◦ F−1 )(50◦ F) = 0.006 in.
(∆L)F =
PL
(30, 000 lb)(15.006 in)
=
= 0.0051 in.
AE
π(1 in)2 (28 × 106 lb/in2 )
Total change in length in this scenario is:
∆L = 0.006 in. + 0.0051 in.
ANS: ∆L = 0.0111 in.
Considering now the case in which the load is applied first, then the
temperature is increased:
(∆L)F =
PL
(30, 000 lb)(15 in)
=
= 0.0051 in.
AE
π(1 in)2 (28 × 106 lb/in2 )
(∆L)T = Lα(∆T ) = (15.005 in.)(8×10−6◦ F−1 )(50◦ F) = 0.006 in.
Total change in length in this scenario is:
∆L = 0.0051 in. + 0.006 in.
ANS:
∆L = 0.0111 in.
Problem 10.75 The prismatic bar is made of material
with modulus of elasticity E = 28 × 106 psi and coefficient of thermal expansion α = 8 × 10−6◦ F−1 . It is
constrained between rigid walls. If the temperature is
increased by 50◦ F above the bar’s initial temperature T ,
what is the normal stress on a plane perpendicular to the
bar’s axis?
Solution:
Because the bar is constrained, its length cannot change (∆L = 0).
∆L = Lα(∆T )−
ANS:
PL
P (15 in)
= (15 in)(8×10−6◦ F−1 )(50◦ F)−
AE
A(28 × 106 lb/in2 )
P/A = σ = −11, 200 lb/in2
NOTE: The (−) sign indicates a compressive stress.
Problem 10.76 The walls between which the prismatic bar in Problem 10.75 is constrained will safely support a compressive normal stress of 30, 000 psi. Based
on this criterion, what is the largest safe temperature increase to which the bar can be subjected?
Solution:
The maximum stress to which the wall can be subjected is also the
maximum stress to which the bar may be subjected.
∆L = 0 = Lα(∆T )−
ANS:
PL
15 in
= (15 in)(8×10−6◦ F−1 )(∆T )−(30, 000 lb/in2 )
2
AE
28 × 106 lb/in
∆T = 134◦ F
Problem 10.77 The prismatic bar in Problem 10.75
has a cross-sectional area A = 3 in2 and is made of
material with modulus of elasticity E = 28 × 106 psi
and coefficient of thermal expansion α = 8×10−6◦ F−1 .
It is constrained between rigid walls. The temperature is
increased by 50◦ F above the bar’s initial temperature T
and a 20,000-lb axial force to the right is applied midway
between the two walls. What is the normal stress on a
plane perpendicular to the bar’s axis to the right of the
point where the force is applied?
Solution:
Because the bar is constrained, the total change in length must be zero.
The change in length due to the applied load would be:
(∆L)F =
(20, 000 lb)(7.5 in)
(3 in2 )(28 × 106 lb/in2 )
= 0.0018 in.
The change in length due to the temperature change would be:
(∆L)T = Lα(∆T ) = (15in.)(8×10−6◦ F−1 )(50◦ F) = 0.006 in.
The reaction at the right-hand end of the bar will be sufficient to hold
the change in length to zero.
∆L = 0 = 0.0018 in. + 0.006 in. −
RR = 43, 680 lb ←
The stress in the right-hand portion of the bar is:
σR =
ANS:
RR
−43, 680 lb
=
A
3 in2
σR = 14, 560 lb/in2
RR (15 in)
(3 in2 )(28 × 106 lb/in2 )
Problem 10.78 The prismatic bar is made of material
with modulus of elasticity E = 28 × 106 psi and coefficient of thermal expansion α = 8×10−6◦ F−1 . It is fixed
to a rigid wall at the left. There is a gap B = 0.0002 in.
between the bar’s right end and the rigid wall.
If the temperature is increased by 50◦ F above the bar’s
initial temperature T , what is the normal stress on a plane
perpendicular to the bar’s axis?
Solution:
The rigid wall will exert an axial force on the bar sufficient to restrict
the bar’s elongation to 0.002 inches.
∆L = 0.002 in. = Lα(∆T )−
ANS:
RR L
15 in
= (15 in)(8×10−6◦ F−1 )(50◦ F)−σ
2
AE
28 × 106 lb/in
σ = −7, 467 lb/in2
NOTE: The (−) sign indicates a compressive stress.
Problem 10.79 Bar A has a cross-sectional area of
0.04 m2 , modulus of elasticity E = 70 GPa, and coefficient of thermal expansion α = 14 × 10−6◦ C−1 . Bar B
has a cross-sectional area of 0.01 m2 , modulus of elasticity E = 120 GPa, and coefficient of thermal expansion
α = 16 × 10−6◦ C−1 . There is a gap B = 0.4 mm
between the ends of the bars. What minimum increase
in the temperature of the bars above their initial temperature T is necessary to cause them to come into contact?
Solution:
The sum of the expansions of the two bars will be 0.0004 m.
εT = LA αA (∆T )+LB αB (∆T ) = (1 m)(14×10−6 /◦ C)(∆T )+(1 m)(16×10−6 /◦ C)(∆T )
ANS:
∆T = 13.3◦ C
Problem 10.80 If the temperature of the bars in Problem 10.79 is increased by 40◦ C above their initial temperature T , what are the normal stresses in the bars?
Solution:
The reaction at the walls will be sufficient to restrict the total elongation
of the bars to 0.0004 m.
∆L = LA αA (∆T ) + LB αB (∆T ) − σA
LA
LB
− σB
EA
EB
[1]
The relationship between the stress in the two bars is:
PA = PB → σB = 4σA
[2]
Substituting Equation [2] into Equation [1]:
0.0004 m = (1 m)(14×10−6 /◦ C)(40◦ C)+(1 m)(16×10−6 /◦ C)(40◦ C)−σA
ANS:
σA = −16.8 MPa σB = −67.2 MPa
NOTE: The (−) indicates a compressive stress.
1m
70 ×
109
N/m
2
−4σA
1m
120 ×
109
N/m
2
Problem 10.81 Each bar has a 2 − in2 cross-sectional
area, modulus of elasticity E = 14 × 106 psi, and coefficient of thermal expansion α = 11 × 10−6◦ F−1 . The
normal stresses in the bars are initially zero. If their
temperature is increased by 40◦ F from their initial temperature T , what is the resulting displacement of point
A.
Solution:
The original length of each bar is:
L = (36 in)2 + (18 in)2 = 40.25 in
The change in length for each bar due to the temperature increase is:
(∆L)T = Lα(∆T ) = (41.25 in)(11×10−6 /◦ F)(40◦ F) = 0.0177in.
The new vertical distance from the fixed surface to point A is:
y = (40.25 in + 0.0177 in)2 − (18 in)2 = 36.021 in
The vertical displacement of point A is:
v = y − y = 36.021 in − 36 in
ANS:
v = 0.021 in
Problem 10.82 If the temperature of the bars in Problem 10.81 is decreased by 30◦ F from their initial temperature T , what force would need to be applied at A
so that the total displacement of point A caused by the
temperature change and the force is zero?
Solution:
The original length of each bar is:
L = (36 in)2 + (18 in)2 = 40.25 in
The amount by which the bars’ length decreases due to the decrease
in temperature is:
(∆L)T = Lα(∆T ) = (40.25 in)(11×10−6 /◦ F)(−30◦ ) = −0.0133 in
The force required to return the bars to their original length is:
0.0133 in =
P (40.25 in)
(2 in2 )(14 × 106 lb/in2 )
→ P = 9, 252 lb
The load, F , required to produce this axial load in BOTH bars is:
ΣFy = 0 = −F + 2P (sin 60◦ ) = −F + 2(9.252 lb)(sin 60◦ )
ANS:
F = 16, 000 lb
Problem 10.83 You are designing a bar with a solid
circular cross section that is to support a 4-kN tensile
axial load. You have decided to use 6061-T6 aluminum
alloy (See Appendix D), and you want the factor of safety
to be S = 2. Based on this criterion, what should the
bar’s diameter be?
Solution:
From Appendix B, the yield stress of the material is σy = 270 ×
106 N/ m2 .
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
ANS:
D = 6.14 mm
270 × 106 N/m2
4, 000 N
= 2
2
π D
2
Problem 10.84 You are designing a bar with a solid
circular cross section with 5-mm diameter that is to support a 4-kN tensile axial load, and you want the factor of
safety to be at least S = 2. Choose an aluminum alloy
from Appendix D that satisfies this requirement.
Solution:
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
σy
4, 000 N
=
2
π(0.0025 m)2
From Appendix D, the yield stress for the material is σy = 407 ×
106 N/ m2 .
ANS:
σy = 407 × 106 N/m2
Either of aluminum alloys 7075-T6 or 2014-T6 will support the load.
Problem 10.85 You are designing a bar with a solid
circular cross section that is to support a 4000-lb tensile
axial load. You have decided to use ASTM-A572 structural steel (See Appendix D), and you want the factor
of safety to be S = 1.5. Based on this criterion, what
should the bar’s diameter be?
Solution:
From Appendix, the yield stress for the material is σy
50, 000 lb/in2 .
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
ANS:
=
50, 000 N/m2
4, 000 lb
= 2
1.5
π D
2
D = 0.391 in
Problem 10.86 You are designing a bar with a solid
circular cross section with 1/2-in. diameter that is to
support a 4000-lb tensile axial load, and you want the
factor of safety to be at least S = 3. Choose a structural
steel from Appendix D that satisfies this requirement.
Solution:
Calculating the required cross-sectional area for the circular bar:
σy
P
=
→
S
A
σy
4, 000 lb
=
3
π(1/4 in)2
σy = 61, 115 lb/in2
ANS:
ASTM - A 514 will support the load.
Problem 10.87 The horizontal beam of length L =
2 m supports a load F = 30-kN. The beam is supported
by a pin support and the brace BC. The dimension
h = 0.54 m. Suppose that you want to make the brace
out of existing stock that has cross-sectional area A =
0.0016 m2 and yield stress σY = 400 MPa. If you want
the brace to have a factor of safety S = 1.5, what should
the angle θ be?
Free Body Diagram:
Solution:
The maximum allowable force in the brace is:
σy
400 × 106 N/m2
(FBC )M AX =
A=
(0.0016 m2 ) = 426.7×103 N
S
1.5
Summing moments about the pin connection at the wall (at D):
ΣMD = 0 = (30, 000 N)(2 m)−FBC (sin θ)(0.54 m) = (30, 000 N)(2 m)−(426.7×103 N)(cos θ)(0.54 m)
cos θ = 0.2604
ANS:
θ=
74.9◦
Problem 10.88 Consider the system shown in Problem 10.87. The horizontal beam of length L = 4 ft
supports a load F = 20 kip. The beam is supported
by a pin support and the brace BC. The dimension
h = 1 ft and the angle θ = 60◦ . Suppose that you want
to make the brace out of existing stock that has yield
stress σy = 50 ksi. If you want to design the brace
BC to have a factor of safety S = 2, what should its
cross-sectional area be?
Free Body Diagram:
Solution:
Summing moments about the pin connection at the wall (at D):
ΣMD = 0 = (20, 000 lb)(4 ft) − FBC (cos 60◦ )(1.732 ft)
FBC = 92, 380 lb
Calculating the cross-sectional area of the brace required to support
the load:
FBC
σy
=
→ :
A
S
ANS:
A = 3.69 in2
92, 380 lb
50, 000 lb/in2
=
A
2
Problem 10.89 The horizontal beam shown in Fig- Free Body Diagram:
ure P10-87 is of length L and supports a load F . The
beam is supported by a pin support and the brace BC.
Suppose that the brace is to consist of a specified material
for which you have chosen an allowable stress σALLOW ,
and you want to design the brace so that its weight is a
minimum. You can do this by assuming that the brace is
subjected to the allowable stress and choosing the angle
θ so that the volume of the brace is a minimum. What is
the necessary angle θ?
Solution:
Let P be the compressive axial load in the bar. Summing moments
about the pin connection at the wall (at D):
ΣMD = 0 = LF − (P )(cos θ)(b) → P = LF/(b cos θ)
[1]
The maximum allowable value of P is:
P = σALLOW A [2]
We see that the volume of the bar is:
V = LBAR A = (LBAR P )/(σALLOW )
[3]
Using Equation [1] in Equation [3]:
V σALLOW =
LF
lBAR
LF
1
·
=
·
[4]
cos θ
b
cos θ sin θ
V σALLOW
1
=
LF
cos θ sin θ
We see from Equation [4] that V is minimum where the product
(cos θ)(sin θ) is maximum. Using a graphing calculator to find the
angle at which the product is a maximum:
ANS:
θ = 45◦
Problem 10.90 In Problem 10.89, draw a graph showing the dependence of the volume of the brace on the
angle θ for 5◦ < θ < 85◦ . Notice that the graph is
relatively flat near the optimum angle, meaning that the
designer can choose θ within a range of angles near the
optimum value and still obtain a near-optimum design.
Solution:
Using Equation [4] from the solution to Problem 10.89:
V σALLOW =
LF
lBAR
LF
1
·
=
·
[4]
cos θ
b
cos θ sin θ
V σALLOW
1
=
LF
cos θ sin θ
We note that the quantities L and F are constant. We draw the graph
of the function
f (θ) =
1
sin(θ) cos(θ)
We see that the graph is practically flat from approximately θ = 39◦
to approximately θ = 51◦ , giving a designer considerable latitude in
choosing the angle for member BC.
Problem 10.91 The truss is a preliminary design for a
structure to attach one end of a stretcher to a rescue helicopter. Based on dynamic simulations, the design engineer estimates that the downward forces the stretcher
will exert will be no greater than 360 lb. At A and
at B. Assume that the members of the truss have the
same cross-sectional area. (Choose a material from Appendix D) and determine the cross-sectional area so that
the structure has a factor of safety S = 2.5.
Free Body Diagrams:
Solution:
The method of joints is used to determine the axial loads in each member of the truss.
At joint B:
ΣFy = 0 = −360 lb + FBD → FBD = 360 lb (T)
ΣFx = 0 = FBC (cos 18.4◦ ) → FBC = 0
At joint A:
ΣFy = 0 = −360 lb + FAC → FAC = 360 lb (T)
At joint C:
ΣFy = 0 = −360 lb + FCF (sin 63.4◦ ) → FCF = 402.6 lb (T)
ΣFx = 0 = −(402.6 lb)(cos 63.4◦ )+FCD → FCD = 180.3 lb (C)
At joint F : From symmetry we see that
FDF = FCF → FDF = 402.6 lb (C)
ΣFx = 0 = (402.6 lb)(cos 63.4◦ )+FDF (cos 63.4◦ )−FF G → FF G = 360.5 lb (T)
The largest axial load is 402.6 lb.
From Appendix D we see that σy for 2014-T6 Al is 60, 000 lb/in2 .
Calculating the required cross-sectional area for the truss members:
σy
PM AX
=
→
S
A
ANS:
A = 0.017 in2
60, 000 lb/in2
402.6 lb
=
2.5
A
Problem 10.92 Upon learning of an upgrade in the helicopter’s engine, the engineer designing the truss shown
is Problem 10.91 does new simulations and concludes
that the downward forces on the stretcher will exert at A
and B may be as large as 400 lb. He also decides the
truss will be made of existing stock with cross-sectional
area A = 0.1 in2 . Choose an aluminum alloy from Appendix B so that the structure will have a factor of safety
of at least S = 5.
Free Body Diagrams:
Solution:
The method of joints is used to determine the axial loads in each member of the truss.
At joint A:
ΣFy = 0 = −400 lb + FAC → FAC = 400 lb (T)
At joint B:
ΣFy = 0 = −400 lb + FBD → FBD = 400 lb (T)
ΣFx = 0 → FBC = 0
At joint C:
ΣFy = 0 = −400 lb + FCF (sin 63.4◦ ) → FCF = 447.4 lb (T)
ΣFx = 0 = −(447.4 lb)(cos 63.4◦ )+FCD → FCD = 200.3 lb (C)
At joint F : From symmetry we see that
FDF = FCF → FDF = 447.4 lb (C)
ΣFx = 0 = (447.4 lb)(cos 63.4◦ )+(447.4(cos 63.4◦ )−FF G → FF G = 400.7 lb (T)
The largest axial load is 447.4 lb.
Calculating the required cross-sectional area for the truss members:
σy
PM AX
=
→
S
A
σy
447.4 lb
=
5
0.1 in2
σy = 22, 370 lb/in2
The aluminum alloys in Appendix D which exceed this minimum yield
stress are:
ANS: 2014-T6, 6061-T6 and 7075-T6.
Problem 10.93 Two candidate truss designs to sup- Free Body Diagrams:
port the load F are shown. Members of a given crosssectional area A and yield stress σY = are to be used.
Ay
Compare the factors of safety and weights of the two designs and discuss reasons that might lead you to choose
Ax
one design over the other. (The weights can be compared
by calculating the total lengths of their members.)
Bx
Solution:
C
Considering candidate truss (a);
At joint C:
(a)
ΣFy = 0 = −F + FAC (sin 26.6◦ ) → FAC = 2.23F (T)
ΣFx = 0 = −(2.23F )(cos 26.6◦ )+FBC → FBC = 1.994F (C)
Summing vertical forces at point B:
Ay
D
Ax
→ FAB = 0
So the largest load in truss (a) is:
FMAX = 2.23F
ANS:
The factor of safety for truss (a) is:
σy
σy
S=
=
σMAX
2.23F/A
ANS:
The length of truss (a) is:
La = 2h + h + (2h)2 + h2 = 5.24h
Considering candidate truss (b):
At joint C:
ΣFy = 0 = −F + FCD (sin 45◦ ) → FCD = 1.414F
ΣFx = 0 = −(1.414F )(sin 45◦ ) + FBC → FBC = F
At joint D:
ΣFy = 0 = FBD (sin 45◦ ) − FCD (sin 45◦ ) → FBD = 1.414F
ΣFx = 0 = −2(1.414F )(cos 45◦ ) + FAD → FAD = 2F
At joint B:
ΣFy = 0 = FBD (sin 45◦ ) − FAB → FAB = F
The largest load in truss (b) is:
FMAX = 2F
ANS:
S=
The factor of safety for truss (b) is:
σy
σMAX
=
σy
2F/A
ANS:
The length of truss (b) is:
√
Lb = h + h + 2h + 2( 2h) = 6.83h
If the design constraint is cost, truss (a) contains less material and fewer
joints. If the design constraint is strength, truss (b) is the better choice.
Bx
C
(b)
Problem 10.94 The cross-sectional area of bar AB is
0.015 m2 . If the force F = 20 kN, what is the normal
stress on a plane perpendicular to the axis of bar AB?
Diagram:
Solution:
Sum moments about point C to find the load in member AB.
ΣMC = 0 = (20, 000 N)(3 m) − FAB (sin 60◦ )(2 m)
FAB = 34, 641 N
The normal stress in member AB is:
σAB = FAB /AAB = (34, 641 N)/(0.015 m2 )
ANS:
σAB = 2.31 MPa
Problem 10.95 Bar AB of the frame in Problem 10.94
consists of a material that will safely support a tensile
normal stress of 20 MPa. Based on this criterion, what
is the largest safe value of the force F ?
Solution:
The maximum safe load which can be supported by member AB is:
(FAB )MAX = (20 × 106 N/ m2 )(0.015 m2 )
(FAB )MAX = 300, 000 N = 300 kN
Summing moments about point C to find FMAX :
ΣMC = 0 = −(FAB )MAX (sin 60◦ )(2 m)+FMAX (3 m) = −(300, 000 N)(sin 60◦ )(2 m)+FMAX (3 m)
ANS:
FMAX = 173, 200 N = 173.2 kN
Problem 10.96 The system shown supports half of the
weight of the 680-kg excavator. The cross-sectional area
of member AB is 0.0012 m2 . If the system is stationary,
what normal stress acts on a plane perpendicular to the
axis of member AC?
Free Body Diagrams:
Solution:
The weight of the excavator is:
W = mg = (680 kg)(9.81 m/sec2 )
W = 6671 N
The weight which must be supported by EACH SIDE of the system is
W/2 = 3335 N.
The axial load in member AB can be determined directly by summing
moments about point D.
ΣMD = 0 = (3335 N)(0.2 m)−FAB (sin 41.2◦ )(0.45 m)−FAB (cos 41.2◦ )(0.1 m)
FAB = 1795 N
The normal stress in member AB is:
σAB = FAB /AAB = (1795 N)/(0.0012 m2 )
ANS:
σAB = 1.496 MPa = 1.5 MPa
Problem 10.97
Free Body Diagram:
Member AC in Problem 10.96 has a cross-sectional area
of 0.0014 m2 . If the system is stationary, what normal
stress acts on a plane perpendicular to the axis of member
AC?
Solution:
The load in member AB was (1795 N) was calculated in the solution
for Problem 10.96. The axial load in member AC can be determined
directly by summing vertical forces at joint A.
ΣFy = 0 = −FAB (sin 41.2◦ )+FAC (sin 48.4◦ ) = −(1795 N)(sin 41.2◦ )+FAC (sin 48.4◦ )
FAC = 1581 N (C)
The normal stress in member AC is:
σAC = FAC /AAC = (1581 N)/(0.0014 m2 )
ANS: σAC = −1.13 MPa Note: The negative sign indicates a
compressive stress.
Problem 10.98 The bar has modulus of elasticity E =
30 × 106 psi, Poisson’s ration ν = 0.32, and a circular
cross section with diameter D = 0.75 in. There is a gap
b = 0.02 in. between the right end of the bar and the
rigid wall. If the bar is stretched so that it contacts the
rigid wall and is welded to it, what is the bar’s diameter
afterward?
Solution:
The strain which must be produced in order to close the gap is: ε =
L −L
in−9.00 in
= 9.02 9.00
= 0.0022
L
in
We can use the definition of Poisson’s ratio to establish the change in
diameter.
−(D −D)/D
0.32 =
ε
D = 0.7495 in
=
−(D−0.75 in)/(0.75 in)
0.0022
Problem 10.99 After the bar in Problem 10.98 is
welded to the rigid wall, what is the normal stress on
a plane perpendicular to the bar’s axis?
Solution:
A strain of ε = 0.0022 was determined in the solution to Problem 10.98.
The stress required to produce this strain is found using the modulus
of elasticity.
E=
σ
ε
σ = εE = (0.00222) 30 × 106 lb/in2
σ = 66.7 ksi
Problem 10.100 The link AB of the pliers has a
cross-sectional area of 40 mm2 and elastic modulus
E = 210 GPa. If forces F = 150 N are applied to
the pliers, what is the change in length of link AB?
Free Body Diagram:
Solution:
Summing moments about point D on the lower handle allows us to
solve directly for PAB .
ΣMD = 0 = −(150 N)(0.13 m) + (PAB )(sin 23.2◦ )(0.03 m)
PAB = 1650 N (C)
The length of link AB is:
LAB = (70 mm)2 + (30 mm)2 = 76.16 mm
The change in length for link AB is:
δ=
ANS:
PL
AE
=
(−1650 N)(76.16 mm)
N/m2 )
(40×10−6 m2 )(210×109
δ = −0.015 mm
Problem 10.101
Suppose that you want to design the pliers in Problem 10.100 so that forces F as large as 450 N can be
applied. The link AB is to be made of a material that
will support a compressive normal stress of 200 MPa.
Based on this criterion, what minimum cross-sectional
area must link AB have?
Free Body Diagram:
Solution:
With an applied load of 450 N to the lower handle, we can solve
directly for the load in link AB by summing moments about point D.
ΣMD = 0 = PAB (sin 23.2◦ ) (0.03 m) − (450 N) (0.13 m)
PAB = 4950 N
Determining the minimal safe cross-sectional area for link AB:
A=
ANS:
P
4950 N
=
σALLOW
200 × 106 N/m2
A = 24.8 × 10−6 m2 = 24.8 mm2
Problem 10.102 Each bar has a cross-sectional area
2
of 3 in2 and modulus of elasticity E = 12 × 106 lb/in .
If a 40-kip horizontal force directed toward the right is
applied at A, what are the normal stresses in the bars?
Free Body Diagram:
Solution:
Step 1—Equilibrium:
Members AB and AD are assumed to be in tension. Member AC is
assumed to be in compression. Summing vertical forces at point A:
ΣFy = 0 = PAB (sin 40◦ ) − PAC (sin 50◦ ) + PAD (sin 70◦ )
PAC = +0.839PAB + 1.227PAD
[1]
Summing horizontal forces at point A:
ΣFx = 0 = 40, 000 lb−PAB (cos 40◦ )−PAD (cos 70◦ )−PAC (cos 50◦ )
PAC = +62, 229 lb − 1.192PAB − 0.532PAD
[2]
Step 2—Lengths:
LAB =
60
= 93.343
sin 40◦
LAC =
60
= 78.324
sin 50◦
LAD =
60
= 63.857
sin 70◦
Step 3—Force–Deformation:
δAB
=
=
PAB LAB
AE
PAB (93.343)
(3)(12 × 106 )
δAB = 2.59286 × 106 PAB
δAC
=
=
PAC LAC
AE
PAC (78.324)
(3)(12 × 106 )
δAC = 2.17567 × 106 PAC
δAD
=
=
[3]
[4]
PAD LAD
AE
PAD (63.857)
(3)(12 × 106 )
δAD = 1.77364 × 106 PAD
[5]
Step 4—Compability: Changes in length related to the horizontal (u)
and vertical (v) displacements
δAB = u cos 40◦ + v sin 40◦
δAB = 0.76604u + 0.64288v
[6]
δAC = u cos 50◦ − v sin 50◦
δAC = 0.64278u − 0.76604v
[7]
δAD = u cos 70◦ + v sin 70◦
δAD = 0.34202u + 0.93969v
[8]
Step 5—Unknowns:
PAB , PAC , PAC , δAB , δAC , δAC , u, v
There are eight unknowns and eight equations. Solve the system simultaneously to find
Loads:
PAB = 22, 336.227
PAC = 30, 495.968
PAD = 9581.85
Horizontal displacement:
u = 0.087
Vertical displacement:
v = 0.01358
Normal stresses:
σAB =
ANS:
σAB = 7.445 ksi
σAC =
ANS:
22, 336.227
= 7445.409
3
30, 495.968
= 10, 165.32
3
σAC = 10.17 ksi
σAD =
ANS:
σAD = 3.19 ksi
9581.85
= 3193.95
3
Problem 10.103 The bars of the system in Problem 10.102 consist of a material that will safely support
a tensile normal stress of 20 ksi. Based on this criterion,
what is the largest downward force that can safely be
applied at A?
Free Body Diagram:
Solution:
Summing horizontal forces at point A:
ΣFx = 0 = FAC (cos 50◦ )−FAD (cos 70◦ )−FAB (cos 40◦ )
[1]
Summing vertical forces at point A:
ΣFy = 0 = −F +FAB (sin 40◦ )+FAD (sin 70◦ )+FAC (sin 50◦ )
The lengths of the bars are:
LAB = (60 in)/(sin 40◦ ) = 93.3 in.
[3]
LAC = (60 in)/(sin 50◦ ) = 78.3 in
[4]
LAD = (60 in)/(sin 70◦ ) = 63.9 in
[5]
The stresses in the three bars are:
σAB = FAB /3 in2
[6]
σAC = FAC /3 in2
[7]
σAD = FAD /3 in2
[8]
The horizontal displacement may be expressed in three different ways.
u = (LAB + δAB )(cos θAB ) − LAB (cos 40◦ )
[9]
u = (LAC + δAC )(cos θAC ) − LAC (cos 50◦ )
[10]
u = (LAD + δAD )(cos θAD ) − LAD (cos 70◦ )
[11]
The vertical displacement may be expressed in three different ways.
v = (LAB + δAB )(cos θAB ) − LAB (sin 40◦ )
[12]
v = (LAC + δAC )(cos θAC ) − LAC (sin 50◦ )
[13]
v = (LAD + δAD )(cos θAD ) − LAD (sin 70◦ )
[14]
Assuming that bar AD carries the largest load among the three bars
and will thus be the limiting stress consideration, we solve this system
of 14 equations together.
ANS:
F = 112.3 kip
[2]
Problem 11.1 A cube of material is subjected to a pure
shear stress τ = 9 MPa. The angle β is measured ad
determined to be 89.98◦ . What is the shear modulus G
of the material?
Diagram:
Solution:
Converting the shear strain angle into radians:
γ=
(90◦ − 89.98◦ )
π = 3.49 × 10−4 radians
180◦
Using the definition of the shear modulus:
G=
ANS:
τ
9 × 106 N/m2
=
γ
3.49 × 10−4
G = 25.8 GPa
Problem 11.2 If the cube in Problem 11.1 consists of
material with shear modulus G = 4.6 × 106 psi and the
shear stress τ = 8000 psi, what is the angle β in degrees?
Free Body Diagram:
Solution:
The shear strain will be:
γ=
τ
8, 000 lb/in2
=
= 1.739×10−3 radians = 0.0996◦
G
4.6 × 106 lb/in2
The angle β is:
β = 90◦ − γ = 90◦ − 0.0996◦
ANS:
β = 89.9◦
Problem 11.3 If the cube in Problem 11.1 consists of
aluminum alloy that will safely support a pure stress of
270 MPa and G = 26.3 GPa, what is the largest shear
strain to which the cube can safely be subjected?
Solution:
The shear strain will be:
γ=
ANS:
τ
270 × 106 N/m2
= 0.010266
=
G
26.3 × 109 N/m2
γ = 0.0103
Problem 11.4 The cube of material is subjected to a
pure shear stress τ = 12 MPa. What are the normal
stress and the magnitude of the shear stress on the plane
P?
Free Body Diagram:
Solution:
Summing vertical forces on the free body diagram:
ΣFY = 0 = −(12×106 N/m2 )A(cos 30◦ )+τP A(cos 30◦ )+σA(sin 30◦ )
[1]
0.866τP + 0.5σP = 10.39 × 106 N/m2
Summing horizontal forces on the free body diagram:
ΣFX = 0 = −(12×106 N/m2 )A(sin 30◦ )−τP A(sin 30◦ )+σP A(cos 30◦ )
[2]
− 0.5τP + 0.866σP = 6 × 106 N/m2
Solving Equations [1] and [2] together:
ANS:
ANS:
τP = 6 MPa
σP = 10.4 MPa
Problem 11.5 In Problem 11.4, what are the magnitudes of the maximum tensile, compressive, and shear
stresses to which the material is subjected?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = σA A(cos θ)−(12×106 N/m2 )A(sin θ)−τ A(sin θ)[1]
Solving Equation [1] for σA :
τ =
σA (cos θ) − (12 × 106 N/m2 )(sin θ)
= σA (cot θ)−12×106 N/m2
sin θ
We see that σA is maximum when cot θ is minimum (θ = 0◦ ), or:
ANS:
σMAX = τ = 12 MPa
Summing forces in the y-direction on the element:
ΣFy = 0 = −(12×106 N/m2 )(A)(cos θ)+τ A(cos θ)+σA(sin θ)
Solving Equation [2] for τ
τ = 12 MPa − σ(tan θ)
We see that τ is maximum when tan θ is minimum (θ = 0), or:
ANS:
τMAX = τ = 12 MPa
[2]
Problem 11.6 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . If the
normal stress on the plane P is 14 MPa, what is τ ?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = (14×106 N/m2 )A(cos 30◦ )−τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 24.25×106 N/m2 −τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = (14×106 N/m2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −8.083×106 N/m2 +τ
Solving Equations [1] and [2] together:
24.25 × 106 N/m2 − τ = −8.083 × 106 N/m2 + τ
ANS:
τ = 16.17 MPa
Problem 11.7 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . The shear
modulus of the material is G = 28 GPa. If the normal
stress on the plane P is 80 MPa, what is the shear strain
of the cube.
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = (80×106 N/m2 )A(cos 30◦ )−τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 138.6 MPa−τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = (80×106 N/m2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −46.19 MPa+τ
Solving Equations [1] and [2] together:
τ = 92.4 MPa
The shear strain is:
γ=
ANS:
τ
92.4 × 106 N/m2
=
= 0.003299
G
28 × 109 N/m2
γ = 0.0033
[2]
[2]
Problem 11.8 The cube of material is subjected to a
pure shear stress τ = 20 ksi.
(a) What are the normal stress and the magnitude of the
shear stress on the plane P ? (b) What are the magnitudes
of the maximum tensile, compressive, and shear stresses
to which the material is subjected?
Free Body Diagram:
Solution:
Summing vertical forces on the free body diagram:
ΣFY = 0 = (20, 000 lb/in2 )A(cos 30◦ )−τθ A(cos 30◦ )−σθ A(sin 30◦ )
[1] τθ (0.866) + σθ (0.5) = 17, 320 lb/in2
ΣFX = 0 = −(20, 000 lb/in2 )A(sin 30◦ )−τθ A(sin 30◦ )+σθ A(cos 30◦ )
[2] τθ (0.5) − σθ (0.866) = −10, 000 lb/in2
Solving Equations [1] and [2] together:
σθ = −17, 316 lb/in2
(a) ANS:
ANS: τθ = 10, 000 lb/in2
(b) ANS: 20 ksi
Problem 11.9 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . If the
normal stress on the plane P is -20 ksi, what is τ ?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = (20, 000 lb/in2 )A(cos 30◦ )+τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 34, 641 lb/in2 +τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = −(20, 000 lb/in2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −11, 547 lb/in2 −τ
Solving Equations [1] and [2] together:
−34, 641 lb/in2 − τ = 11, 547 lb/in2 + τ
ANS:
τ = −23, 094 lb/in2
[2]
Problem 11.10 The cube of material shown in Problem 11.4 is subjected to a pure shear stress τ . The shear
modulus of the material is G = 4 × 106 psi. If the normal stress on the plane P is -12 ksi, what is the shear
strain of the cube?
Free Body Diagram:
Solution:
Summing forces in the x-direction on the element:
ΣFx = 0 = −(12, 000 lb/in2 )A(cos 30◦ )+τ A(sin 30◦ )−τ A(sin 30◦ ) → τ = 20, 784 lb/in2 +τ
[1]
Summing forces in the y-direction on the element:
ΣFy = 0 = −(12, 000 lb/in2 )A(sin 30◦ )+τ A(cos 30◦ )−τ A(cos 30◦ ) → τ = −6, 928 lb/in2 −τ
Solving Equations [1] and [2] together:
20, 784 lb/in2 − τ = −6, 928 lb/in2 + τ
τ = 13, 856 lb/in2
The shear strain is:
γ=
ANS:
τ
13, 856 lb/in2
=
G
4 × 106 lb/in2
γ = 0.00346
Problem 11.11 If a bar has a solid circular cross section with 15-mm diameter, what is the polar moment of
inertia of its cross section in m4 ?
Solution:
The polar moment of inertia for the cross section is:
π
π 0.015 m 4
J = r4 =
2
2
2
ANS:
J= 4.97 × 10−9 m4
Problem 11.12 If a bar has a hollow circular cross
section with 2-in. outer radius and 1-in. inner radius,
what is the polar moment of inertia of its cross section?
Solution:
The polar moment of inertia for the cross section is:
π
π
J=
(ro )4 − (ri )4 =
(2in)4 − (1in)4
2
2
ANS:
J= 23.56 in4
[2]
Problem 11.13 The bar has a circular cross section
with 15-mm diameter and the shear modulus of the material is G = 26 GPa. If the torque T = 10 N − m, determine (a) the magnitude of the maximum shear stress
in the bar; (b) the angle of twist of the end of the bar in
degrees.
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 c4 = π2 (0.0075m)4
J = 4.97 × 10−9 m4
Maximum shear stress in the bar is:
τMAX =
ANS:
Tρ
(10 N − m) (0.0075 m)
=
J
4.97 × 10−9
(a) τ MAX = 15.1 × 106 N/m2
The angle of twist for the bar is:
φ=
ANS:
LT
(0.8m)(10 N − m)
=
JG
(4.97 × 10−9 m4 )(26 × 109 N/m2 )
φ = 0.0619 rad = 3.547◦
Problem 11.14 If the bar in Problem 11.13 is subjected Free Body Diagram:
to a torque T that causes the end of the bar to rotate 4◦ ,
what is the magnitude of the maximum shear stress in
the bar?
Solution:
The polar moment of inertia for the cross section is:
π
π
J = c4 = (0.0075)4 = 4.97 × 10−9
2
2
Using the angle of rotation at the end of the bar to determine the applied
torque:
◦ 4
TL
(0.8 m)(T )
(π) rad =
=
→ T = 11.27 N − m
180◦
JG
(4.97 × 10−9 m4 )(26 × 109 N/m2 )
Maximum shear stress in the cross section is:
τMAX =
ANS:
Tc
(11.27 N − m)(0.0075 m)
=
J
4.97 × 10−9 m4
τ MAX = 17.01 MPa
Problem 11.15 The bar in Problem 11.13 is to be used
in an application that requires that it be subjected to an
angle of twist no greater than 1◦ . What is the maximum
allowable value of the torque T ?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 c4 = π2 (0.0075 m)4
J = 4.97 × 10−9 m4
Converting the angle of twist into radians:
φ=
1◦
π = 0.0175 radians
180◦
The torque which will produce this angle of twist is:
(0.0175) 4.97 × 10−9 m4 26 × 109 N/m2
φJG
T =
=
L
0.8 m
ANS:
T = 2.82 N − m
Problem 11.16 The solid circular shaft that connects
the turbine blades of the hydroelectric power unit to the
generator has a 0.4-m radius and supports a torque T =
2 MN − m. What is the maximum shear stress in the
shaft?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
π
π
J = c4 = (0.4 m)4 = 0.0402 m4
2
2
Maximum shear stress in the shaft is:
τMAX =
ANS:
Tρ
(2 × 106 N − m)(0.4 m)
=
J
0.0402 m4
τ MAX = 19.9 MPa
Problem 11.17 Consider the solid circular shaft in
Problem 11.16. The shear modulus of the material is
G = 80 GPa. What angle of twist per unit meter of
length is caused by the 2-MN-m torque?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
π
π
J = c4 = (0.4 m)4 = 0.0402 m4
2
2
The angle of twist per meter of length is:
φ
T
2 × 106 N − m
=
=
L
JG
(0.0402 m4 )(80 × 109 N/m2 )
ANS:
φ
L
= 0.00062 rad/m = 0.0356 degrees/m
Problem 11.18 If the shaft in Problem 11.16 has a
hollow circular cross section with 0.5-m outer radius and
0.3-m inner radius, what is the maximum shear stress?
Free Body Diagram:
Solution:
The polar moment of inertia for the hollow shaft is:
π
π
J = (ro − ri ) =
(0.5 m)4 − (0.3 m)4
2
2
J = 0.0855 m4
Maximum shear stress in the shaft is:
2 × 106 N − m (0.5 m)
Tρ
τMAX =
=
J
0.0855 m4
ANS:
τ MAX = 11.7 MPa
Problem 11.19 The propeller of the wind generator is
supported by a hollow circular shaft with 0.4-m outer
radius and 0.3-m inner radius. The shear modulus of the
material is G = 80 GPa. If the propeller exerts an 840kN-m torque on the shaft, what is the resulting maximum
shear stress?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 (0.4 m)4 − (0.3 m)4
J = 0.0275 m4
Maximum shear stress in the shaft is:
τMAX =
ANS:
Tc
(840, 000 N − m) (0.4 m)
=
J
0.0275 m4
τ MAX = 12.2 MPa
Problem 11.20 In Problem 11.19, what is the angle of
twist of the propeller shaft per meter of length?
Free Body Diagram:
Solution:
The polar moment of inertia for the shaft is:
J = π2 (0.4 m)4 − (0.3 m)4
J = 0.0275 m4
Angle of twist for the shaft is:
φ=
ANS:
LT
(1 m) (840, 000 N − m)
=
JG
(0.0275 m4 ) (80 × 109 N/m2 )
φ = 0.000382 rad = 0.0219◦
Problem 11.21 In designing a new shaft for the wind
generator in Problem 11.19, the engineer wants to limit
the maximum shear stress in the shaft to 10 MPa, but design constraints require retaining the 0.4-m outer radius.
What new inner radius should she use?
Solution:
The polar moment of inertia for the shaft is:
J = π2 (0.4 m)4 − (ri )4
J = 0.0402 − 1.571ri4
Maximum shear stress in the shaft is:
τMAX =
ANS:
Tc
(840, 000 N − m) (0.4 m)
=
= 10 × 106 N/m2
J
(0.0402 − 1.571ri4 )
ri = 0.2546 m
Problem 11.22 The bar has a circular cross section
with 1-in. diameter and the shear modulus of the material
is G = 5.8 × 106 psi. If the torque T = 1000 in − lb,
determine (a) the magnitude of the maximum shear stress
in the bar; (b) the magnitude of the angle of twist of the
right end of the bar relative to the wall in degrees.
Free Body Diagram:
Solution:
Maximum torque in the shaft is 1,000 in-lb.
The polar moment of inertia for the shaft is:
J = π2 (0.5 in)4
J = 0.0982 in4
(a) Maximum shear stress in the shaft is:
TC
(1, 000 in − lb) (0.5 in)
=
J
0.0982 in4
τMAX =
ANS:
τMAX = 5092.958 lb/in2
(b) The angle of twist in the 8-inch section of the bar is:
φ8 in =
φ8 in
(8 in)(500 in−lb))
(0.0982 in4 )(5.8×106 in4 )
= 0.00702 rad = 0.402◦
The angle of twist in the 6-inch section of the bar is:
φ6 in =
φ6 in
(6 in)(1,000 in−lb)
(0.0982 in4 )(5.8×106 lb/in2 )
= 0.0105 rad = 0.604◦
Total angle of twist for the bar is:
φ = φ8 in + φ6 in = 0.402◦ + 0.604◦
ANS:
φ = 1.006◦
Problem 11.23 For the bar in Problem 11.22, what
value of the torque T would cause the angle of twist of
the end of the bar to be zero?
Free Body Diagram:
Solution:
The torque in the 8-inch section of the bar is (T − 500 in − lb).
The torque in the 6-inch section of the bar is T .
The equation for total angle of twist for the bar is:
0=
L8 in T8 in L6 in T6 in
(8 in) (T − 500 in − lb) (6 in) T
+
=
+
JG
JG
JG
JG
Solving the equation for T :
ANS:
T = 286 in − lb
Problem 11.24 Part A of the bar has a solid circular cross section and Part B has a hollow circular
cross section. The shear modulus of the material is
G = 3.8 × 106 psi. Determine the magnitudes of the
maximum shear stresses in parts A and B of the bar.
Free Body Diagram:
Solution:
The torque in the solid section of the bar is 250,000 in-lb. The torque
in the hollow section of the bar is 100,000 in-lb. Polar moment of
inertia for the solid section of the bar is:
π
JS = (2 in)4 = 25.13 in4
2
Polar moment of inertia for the hollow section of the shaft is:
π
JH =
(2 in)4 − (1 in)4 = 23.56 in4
2
Maximum shear stress in the solid section of the bar is:
(τMAX )S =
ANS:
TS cS
(250, 000 in − lb) (2 in)
=
JS
25.13 in − lb
(τMAX )S = 19, 896.54 lb/in2 = 19.89 ksi
Maximum shear stress in the hollow section of the bar is:
(τMAX )H =
ANS:
TH cH
(100, 000 in − lb) (2 in)
=
JH
23.56 in4
(τMAX )H = 8, 488.96 lb/in2 = 8.49 ksi
Problem 11.25 For the bar in Problem 11.24, deter- Free Body Diagram:
mine the magnitude of the angle of twist of the end of
the bar in degrees.
Solution:
The torque in the solid section of the bar is 250,000 in-lb. The torque
in the hollow section of the bar is 100,000 in-lb. Polar moment of
inertia for the solid section of the bar is:
π
JS = (2 in)4 = 25.13 in4
2
Polar moment of inertia for the hollow section of the shaft is:
π
JH =
(2 in)4 − (1 in)4 = 23.56 in4
2
The angle of twist for the solid section of the shaft is:
φ=
LS TS
(7 in) (250, 000 in − lb)
= JS GS
25.13 in4 3.8 × 106 lb/in2
φS = 0.0183 rad = 1.05◦
The angle of twist for the hollow section of the shaft is:
φ=
LH TH
(14 in) (100, 000 in − lb)
= JH GH
23.56 in4 3.8 × 106 lb/in2
φH = 0.0156 rad = 0.896◦
Total angle of twist for the shaft is:
φt = φS + φH = 1.05◦ + 0.896◦
φ = 1.95◦
ANS:
Problem 11.26 For the bar in Problem 11.24, deter- Free Body Diagram:
mine the magnitude of the maximum shear stresses in
parts A and B of the bar and the magnitude of the angle
of twist of the end of the bar in degrees if the 150 in-kip
couple acts in the opposite direction.
Solution:
Polar moments of inertia for the two sections of the bar are:
π
π
JA = (2in)4 = 25.13 in4 JB =
(2in)4 − (1in)4 = 23.56 in4
2
2
From the FBD we see that the torque in the sections of the bar is:
TA = −50, 000 in − lb
TB = 100, 000 in − lb
Maximum shear stresses in the sections of the bar are:
(τA )MAX =
(50, 000 in − lb)(2 in)
25.13 in4
(τB )MAX =
(100, 000 in − lb)(2 in)
23.56 in4
(τA )MAX = 3, 980 lb/in2 (τB )MAX = 8, 488.96 lb/in2 ≈ 8.49 ksi
ANS:
The angles of twist in each of the sections of the bar are:
φA = −
(50, 000 in − lb)(7 in)
(25.13 in4 )(3.8 × 106 lb/in2 )
φA = −0.00367 radians
φB =
(100, 000 in − lb)(14 in)
(23.56 in4 )(3.8 × 106 lb/in2 )
φB = 0.01564 radians
Total angle of twist is:
φ = φA + φB = −0.00367 rad + 0.01564 rad
ANS:
φ = 0.01197 rad = 0.686◦
Problem 11.27 The lengths LA = LB = 200 mm and
LC = 240 mm. The diameter of parts A and C of the bar
is 25 mm and the diameter of part B is 50 mm. The shear
modulus of the material is G = 80 GPa.If thetorqueT
= 2.2 kN-m, determine the magnitude of the angle of
twist of the right end of the bar relative to the wall.
Free Body Diagram:
Solution:
The torques in the three sections of the blade are:
TA = 2, 200 N − m−8, 000 N − m+4, 000 N − m = −1, 800 N − m
TB = 2, 200 N − m − 8, 000 N − m = −5, 800 N − m
TC = 2, 200 N − m
The polar moment of inertia for sections A ad C is:
π
JA = JC = (0.0125 m)4 = 38.35 × 10−9 m4
2
The polar moment of inertia for section B of the bar is:
π
JB = (0.025 m)4 = 614 × 10−9 m4
2
The angle of twist in each section of the bar is:
φA =
L A TA
JA GA
=
φB =
L B TB
JB GB
=
φC =
L C TC
JC GC
=
(0.2 m)(−1,800 N−m)
m4 )(80×109 N/m2 )
(38.35×10−9
(0.2 m)(−5,800 N−m)
m4 )(80×109 N/m2 )
(614×10−9
= −0.0236 rad = −1.353◦
(0.24 m)(2,200 N−m)
m4 )(80×109 N/m2 )
(38.35×10−9
= −0.1173 rad = −6.723◦
= 0.172 rad = 9.861◦
Total angle of twist I the bar is:
φ = φA + φB + φC = −6.723◦ − 1.353◦ + 9.861◦
ANS:
φ = 1.78◦
Problem 11.28 For the bar in Problem 11.27, what
value of the torque T would cause the angle of twist of
the right end of the bar relative to the wall to be zero?
Free Body Diagram:
Solution:
The torques in the three sections of the blade are:
TA = T − 8, 000 N − m + 4, 000 N − m = T − 4, 000 N − m
TB = T − 8, 000 N − m
TC = T
The equation which expresses the angle of twist in the bar is:
0=
LA TA LB TB LC TC
(0.2 m) (T − 4, 000 N − m) (0.2 m) (T − 8, 000 N − m) (0.24 m) (T )
+
+
=
+
+
JA GA JB GB JC GC
JG
JG
JG
Solving the above equation for T :
ANS:
T = 1, 988.84 N − m = 1.99 kN − m
Problem 11.29 The bar in Problem 11.27 is made of
a material that can safely support a pure shear stress of
1.1 GPa. Based on this criterion, what is the range of
positive values of the torque T that can safely be applied?
Free Body Diagram:
Solution:
We see from the FBD that the torques in the three sections of the bar
are:
TA = −4, 000 N − m−T
TB = −8, 000 N − m+T
TC = T
Note: As the torque T increases, the torque in section A initially
decreases. We can see that the center section, B, will NOT be
factor in determining the limiting torque (large J and relatively small
T ).
The torque, T , which will result in maximum shear stress in section A
is:
τA = 1.1×109 N/m2 =
(−4, 000 N − m + T )(0.0125 m)
→ T = 7, 375 N − m
π
(0.0125 m)4
2
The torque, T , which will result in maximum shear stress in section C
is:
τC = 1.1 × 109 N/m2 =
T (0.0125 m)
→ T = 3, 375 N − m
m)4
π
(0.0125
2
The range of allowable torques, T , is:
ANS:
3, 375 N − m ≤ T ≤ 7, 375 N − m
Problem 11.30 The bars AB and CD each have a solid
circular cross section with 30-mm diameter and consist
of a material with a shear modulus G = 28 GPa. The
ratio of the gears are rB = 120 mm and rC = 90 mm.
If the torque TA = 200 N − m, what are the maximum
shear stresses in the bars?
Diagram:
Solution:
The polar moment of inertia for the two shafts is:
π
J = (0.015 m)4 = 79.52 × 10−9 m4
2
Maximum shear stress in bar AB is:
(τMAX )AB =
ANS:
TAB cAB
(200 N − m) (0.015 m)
=
JAB
79.52 × 10−9 m4
(τMAX )AB = 37.7 MPa
The torque in shaft CD is:
rc
0.045 m
TCD =
TAB =
200 N − m = 150 N − m
rb
0.060 m
Maximum shear stress in bar CD is:
(τMAX )CD =
ANS:
TCD cCD
(150 N − m) (0.015 m)
=
JCD
79.52 × 10−9 m4
(τMAX )CD = 28.3 MPa
Problem 11.31 In Problem 11.30, what is the angle of
twist at A? (Assume that the deformations of the gears
are negligible.)
Diagram:
Solution:
The polar moment of inertia for the two shafts is:
π
J = (0.015m)4 = 79.52 × 10−9 m4
2
The torques on the two shafts are:
90 mm
TA = 200 N − m TC =
TA = (0.75)(200 N − m) = 150 N − m
120 mm
Total angle of twist at A is:
φ = φCD +φAB =
ANS:
(1m)(150 N − m)
(1m)(200 N − m)
+
(79.52 × 10−9 m4 )(28 × 109 N/m2 ) (79.52 × 10−9 m4 )(28 × 109 N/m2 )
φ = 0.157 rad = 9◦
Problem 11.32 Consider the system shown in Prob- Diagram:
lem 11.30. The bars AB and CD each have a solid
circular cross section with 30-mm diameter. The radii
of the gears must satisfy the relation rB +rC = 210 mm.
If the torque TA = 200 kN − m and the bars are made
of a material that will safely support a pure shear stress
of 40 MPa, what is the largest safe value of the radius
rC ?
Solution:
The torque in bar AB is TAB = 200 N − m. The shear stress in bar
AB is:
τAB =
(200 N − m)(0.015 m)
= 37.73 MPa
π
(0.015 m)4
2
The shear stress in bar CD, limited to 40 MPa, can be expressed as:
rC
rC
τAB =
(37.73 MPa) [1]
τCD = 40×106 N/m2 =
rB
rB
A second equation we can use is:
rB + rC = 0.21 m
[2]
Solving equations [1] and [2] together:
ANS:
rC = 0.108 m = 108 mm
Problem 11.33 The bar has a circular cross section Free Body Diagram:
with 1-in. diameter. If the torque TO = 1000 in − lb,
determine the magnitudes of the maximum shear stresses
in parts A and B of the bar.
Solution:
The polar moment of inertia for the bar is:
π
π
J = c4 = (0.5 in)4 = 0.0982 in4
2
2
Since the sum of moments on the bar must be zero:
MA + MB = 1, 000 in − lb
[1]
We see that φA = φB , so we have:
(8 in) (MA )
(6 in) MB
=
JG
JG
MA = 0.75MB
[2]
Solving equations [1] and [2] together, we get:
MA = 428.6 in − lb
MB = 571.4 in − lb
Maximum shear stress in each section of the bar is:
(τMAX )A =
ANS:
(τMAX )A = 2180 lb/in2
(τMAX )B =
ANS:
M A cA
(428.6 in − lb) (0.5 in)
=
JA
0.0982 in4
M B cB
(571.4 in − lb) (0.5 in)
=
JB
0.0982 in4
(τMAX )B = 2910 lb/in2
Problem 11.34 Suppose that the bar in Problem 11.33 Free Body Diagram:
consists of a material that will safely support a maximum
shear stress of 40 ksi. Based on this criterion, what is
the maximum safe magnitude of the torque TO ?
Solution:
The polar moment of inertia for the bar is:
π
π
J = c4 = (0.5 in)4 = 0.0982 in4
2
2
We see that:
MA + MB = T
[1]
Since φA = φB, we also see that:
MA (8 in)
MB (6 in)
=
JG
JG
MA = 0.75MB
[2]
Solving equations [1] and [2] together, we get:
MA = 0.429T
MB = 0.571T
We see that the maximum torque is in section B. With a maximum
allowable shear stress of 40,000 psi:
40, 000 lb/in2 =
(0.571T ) (0.5 in)
0.0982 in4
Maximum allowable torque is:
ANS:
T = 13.8 in − kip
Problem 11.35 Suppose that the bar in Problem 11.33 Free Body Diagram:
is subjected to a torque T0 = 10, 000 in − lb and consists
of a material that will safely support a maximum shear
stress of 40 ksi. Based on this criterion, what is the
largest distance from the left end of the bar at which the
torque can safely be applied?
Solution:
The polar moment of inertia for the bar is:
π
π
J = c4 = (0.5 in)4 = 0.0982 in4
2
2
To find the maximum allowable moment:
40, 000 lb/in2 =
MMAX (0.5 in)
0.0982 in4
MMAX = 7856 in − lb
We see that:
MA + MB = 10, 000 in = lb
As the applied moment moves from left-to-right, the moment at the
right-hand end increases. Knowing that φA = φB :
((10, 000 in − lb) − MMAX )(14in − LR )
MMAX LR
=
JG
JG
((10, 000 in − lb) − 7856 in − lb)(14in − LR )
(7856 in − lb)(LR )
=
JG
JG
LR = 3 in
ANS:
LL = 11 in
Problem 11.36 The bar is fixed at both ends. It consists Free Body Diagram:
of material with shear modulus G = 28 GPa and has a
solid circular cross section. Part A is 40 mm in diameter
and part B is 20 mm in diameter. Determine the torques
exerted on the bar by the walls.
Solution:
Polar moments of inertia for the two sections are:
π
π
π
π
JA = c4 = (0.02 m)4 = 251×10−9 m4 JB = c4 = (0.01 m)4 = 15.7×10−9 m4
2
2
2
2
We see that:
MA + MB = 1, 200 N − m
[1]
Since φA = φB , we also see that:
(0.16 m) (MA )
(0.12 m) (MB )
=
(251 × 10−9 m4 ) G
(15.7 × 10−9 m4 ) G
MA = 12MB
[2]
Solving equations [1] and [2] together:
MA = 1107.7 N − m
ANS:
ANS:
MB = 92.3 N − m
Problem 11.37 Determine the magnitudes of the max- Free Body Diagram:
imum shear stresses in parts A and B of the bar in Problem 11.36.
Solution:
JA =
π 4
c
2
=
π
2
(0.02 m)4 = 251 × 10−9 m4
JB =
We see that:
MA + MB = 1, 200 N − m
[1]
Since φA = φB , we also see that:
(0.16 m) (MA )
(0.12 m) (MB )
=
(251 × 10−9 m4 ) G
(15.7 × 10−9 m4 ) G
MA = 12MB
[2]
Solving equations [1] and [2] together:
MA = 1107.7 N − m
MB = 92.3 N − m
Maximum shear stress in the two sections is:
(τMAX )A =
ANS:
M A cA
(1107.7 N − m) (0.02 m)
=
JA
251 × 10−9 m4
(τMAX )A = 88.3 MPa
(τMAX )B =
ANS:
M B cB
(92.3 N − m) (0.01 m)
=
JB
15.7 × 10−9 m4
(τMAX )B = 58.8 MPa
π 4
π
c = (0.01 m)4 = 15.7×10−9 m4
2
2
Problem 11.38 Each bar is 10 in. long and has a solid Free Body Diagram:
circular cross section. Bar A has a diameter of 1 in. and
its shear modulus is 6 × 106 psi. Bar B has a diameter
of 2 in. and its shear modulus is 3.8 × 106 psi. The ends
of the bars are separated by a small gap. The free end of
bar A is rotated 2◦ about the bar’s axis and the bars are
welded together. What are the magnitudes of the angles
of twist (in degrees) of the two bars afterward?
Solution:
Polar moments of inertia for the two bars are:
π
π
π
π
JA = r 4 = (0.5 in)4 = 0.0982 in4 JB = r 4 = (1 in)4 = 1.571 in4
2
2
2
2
The moment required to produce an angle of twist of two degrees in
bar A is:
(0.0349 rad) 0.0982 in4 6 × 106 lb/in2
φJG
M =
=
= 2056 in − lb
L
10 in
After the two bars are welded together, each of the welded ends will
rotate until equilibrium is achieved. The total of the two deflection
angles will be 2◦ . Because the bars, after welding, are in contact with
each other, the moments exerted by each of the bars are equal. We
have φA = φB and MA = MB , so:
MA (10 in)
MA (10 in)
+
= (2◦ )
0.0982 in4 6 × 106 lb/in2
1.571 in4 3.8 × 106 lb/in2
3.14159 rad
180◦
MA = MB = 1872 in − lb
The angle of twist in each bar is:
ANS:
φA =
MA LA
JA GA
=
ANS:
φB =
MB LB
JB GB
=
(1872 in−lb)(10 in)
lb/in2 )
= 0.03177 rad = 1.82◦
(0.0982 in4 )(6×106
(1872 in−lb)(10 in)
lb/in2 )
(1.571 in4 )(3.8×106
= 0.00314 rad = 0.18◦
Problem 11.39 In Problem 11.38, the ends of the bars
are separated by a small gap. Suppose that the free end
of bar A is rotated 2◦ about its axis. The fee end of bar B
is rotated 2◦ about its axis in the opposite direction, and
the bars are welded together. What are the magnitudes of
the maximum shear stresses in the two bars afterward?
Solution:
Polar moments of inertia for the two bars:
π
π
JA = (0.5 in)4 = 0.0982 in4 JB = (1 in)4 = 1.571 in4
2
2
We see that the resulting moments in the two bars will have the same
magnitude. We also see that the total angle of twist for the two bars
will be 4◦ (0.0698 rad) when the bars achieve equilibrium.
MA = MB
[1]
φA + φB = 0.0698 rad
[2]
MA (10 in)
MA (10 in)
+
= 0.0698
(0.0982 in4 )(6 × 106 lb/in2 ) (1.571 in4 )(3.8 × 106 lb/in2 )
MA = MB = 3743 in − lb
Calculating maximum shear stress in each bar:
(3743 in−lb)(0.5 in)
0.0982 in4
ANS:
(τMAX )A =
MA rA
JA
ANS:
(τMAX )B =
(3742 in−lb)(1 in)
1.571 in4
=
= 2.38 ksi
= 19.06 ksi
Problem 11.40 The lengths LA = LB = 200 mm Free Body Diagram:
and LC = 240 mm. The diameter of parts A and C is
25 mm and the diameter of part B is 50 mm. The shear
modulus of the material is G = 80 GPa. What is the
magnitude of the maximum shear stress in the bar?
Solution:
Polar moments of inertia for the sections of the bar are:
π
π
JA = JC = (0.0125 m)4 = 3.835×10−8 m4 JB = (0.025 m)4 = 6.136×10−7 m4
2
2
We see that the sum of the three angles of twist from A to C must be
zero, so we have:
φA + φB + φC = 0
(0.2 m)(−MC + (8000 N − m) − (4000 N − m)) (0.2 m)(−MC + (8000 N − m)
(−MC )(0.24 m)
+
+
=0
(3.835 × 10−8 m4 )(G)
(6.136 × 10−7 m4 )(G)
(3.835 × 10−8 m4 )(G)
MC = 1989 N − m
Because the sum of moments on the bar is zero:
−4000 N − m + 8000 N − m − 1989 N − m − MA = 0
MA = 2011 N − m
Maximum shear stresses in each of the three sections are:
N−m)(0.0125 m)
= 655 MPa
ANS: (τMAX )A = (2011
3.835×10−8 m4
(τMAX )B =
((8000 N − m) − (1989 N − m))(0.025 m)
= 245 MPa
6.136 × 10−7 m4
(τMAX )C =
(1989 N − m)(0.0125 m)
= 648 MPa
3.835 × 10−8 m4
Problem 11.41 In Problem 11.40, through what angle
does the bar rotate at the position where the 8 kN-m
couple is applied?
Solution:
Polar moments of inertia for the sections of the bar are:
π
π
JA = JC = (0.0125 m)4 = 3.835×10−8 m4 JB = (0.025 m)4 = 6.136×10−7 m4
2
2
We see that the sum of the three angles of twist from A to C must be
zero, so we have:
φA + φB + φC = 0
(0.2 m)(−MC + (8000 N − m) − (4000 N − m)) (0.2 m)(−MC + (8000 N − m)
(−MC )(0.24 m)
+
+
=0
(3.835 × 10−8 m4 )(G)
(6.136 × 10−7 m4 )(G)
(3.835 × 10−8 m4 )(G)
MC = 1989 N − m
Because the sum of moments on the bar is zero
−4000 N − m + 8000 N − m − 1989 N − m − MA = 0
MA = 2011 N − m
The simplest means of determining the angle of twist is to start from
the right-hand end.
φ=
ANS:
LC TC
(0.24 m)(1989 N − m)
=
JC GC
(3.835 × 10−8 m4 )(80 × 109 N/m2 )
φ = 0.1556 rad = 8.91◦
Problem 11.42 The collar is rigidly attached to bar A.
The cylindrical bar A is 80 mm in diameter and its shear
modulus is G = 66 GPa. There are gaps b = 2 mm
between the arms of the collar and the ends of the identical bars B and C. Bars B and C are 30 mm in diameter
and their modulus of elasticity is E = 170 GPa. If the
bars B and C are extended so that they come into contact
with the arms of the collar and are welded to them, what
is the magnitude of the maximum shear stress in bar A
afterward?
Free Body Diagram:
Solution:
The polar moment of inertia for bar A is:
π
π
J = r4 = (0.04 m)4 = 4.021 × 10−6 m4
2
2
The torque exerted upon bar A by the bars B and C is:
T = 2[P (0.3 m)]
[1]
The compatibility condition for the gap between the collar and bars B
and C is:
PL
0.002 m = rφ + AE
0.002 m = (0.3 m) φ +
P (0.4 m)
π(0.015 m)2 (66×109 N/m2 )
0.002 m = 0.3φ + 8.57 × 10−9 P
[2]
Combining Equations [1] and [2]:
0.002 m = (0.3 m)
(0.6P N − m) (0.8 m)
P (0.4 m)
+
(4.021 × 10−6 m4 ) (66 × 109 N/m2 )
π (0.015 m)2 (170 × 109 N/m2 )
P = 3663 N
Using Equation [1] to determine the torque:
T = 2198 N − m
Maximum shear stress in bar A is:
(τMAX )A =
ANS:
Tr
(2198 N − m) (0.04 m)
=
J
4.021 × 10−6 m4
(τMAX )A = 21.9 MPa
Problem 11.43 In Example 11.2, what is the magnitude of the maximum shear stress in the bar?
Free Body Diagram:
Solution:
Maximum shear stress occurs at the wall (smallest cross-section for
the bar). From the given function for J, the polar moment of inertia at
the wall (x = 0) is:
J = 0.00016 m4
The radius of the bar at the wall is needed.
J = (π/2)(r)4 → r =
2J
π
1/4
=
1/4
2(0.00016 m4 )
π
= 0.1005 m
Maximum shear stress in the bar at the wall is:
τMAX =
(200, 000 N − m)(0.1005m)
0.00016 m4
τMAX = 125.6 MPa
ANS:
Problem 11.44 In Example 4.2, suppose that the
torque T is applied to the bar at x = 1 m. What is
the magnitude of the angle of twist of the entire bar?
Free Body Diagram:
Solution:
From the given function for J, the polar moment of inertia at x = 1 m
is:
J = 0.00016 + 0.0006(1) m4 = 0.00076 m4
The radius of the bar at x = 1 m is:
r=
2J
π
1/4
=
2(0.00076 m4 )
π
1/4
= 0.148 m
The angle of twist for the entire bar is:
1
φ=
0
ANS:
(200, 000 N − m)dx
=
(0.00016 + 0.0006x2 )(47 × 109 N/m2 )
φ = 0.015207 radians =
0.8609◦
≈ 0.861
200, 000 N − m
47 × 109 N/m2
1
0
dx
(0.00016 + 0.0006x2 )
Problem 11.45 The bar has a solid circular crosssection. Its polar moment of inertia is given by J =
(0.1 + 0.15x) in4 , where x is the axial position in
inches, and the shear modulus of the material is G =
4.6 × 106 psi. If the bar is subjected to an axial torque
T = 20 in − kip, what is the magnitude of the maximum
shear stress at x = 6 in?
Free Body Diagram:
Solution:
At x = 6 in., the polar moment of inertia is:
J = (0.1 + 0.15(6)) in4 = 1 in4
The radius of the bar at x = 6 in. is:
2 1 in4
2J 1/4
r=
=
π
π
1/4
= 0.893 in.
Maximum shear stress in the bar at x = 6 in. is:
τMAX =
ANS:
Tr
(20, 000 in − lb) (0.893 in)
=
J
1 in4
τMAX = 17, 860 lb/in2 = 17.86 ksi
Problem 11.46 What is the angle of twist (in degrees)
of the entire bar in Problem 11.45?
Free Body Diagram:
Solution:
The angle of twist is determine by integrating over the length of the
bar.
10
φ=
0
(20, 000 in − lb)dx
=
(0.1 + 0.15x)(4.6 × 106 lb/in2 )
20, 000 in − lb
4.6 × 106 lb/in2
φ = 0.029 [ln(1.6) − ln(0.1)]
ANS:
φ = 0.080365 rad = 4.6045◦
10
0
dx
20, 000 in − lb
=
(0.1 + 0.15x)
4.6 × 106 lb/in2
1
0.15
1
0
0.15dx
(0.1 + 0.15x)
Problem 11.47 Suppose that an axial hole is drilled
through the bar in Problem 11.45 so that it has a hollow
circular cross section with inner radius ri = 0.3 in. What
is the angle of twist (in degrees) of the entire bar due to
the 20-in-kip torque?
Free Body Diagram:
Solution:
The new expression for the polar moment of inertia is:
π
J = (0.1 + 0.15x) − (0.3 in)4 = (0.08728 + 0.15x) in4
2
The angle of twist is determined by integrating from x = 0 to x = 10.
10
φ=
0
φ=
T dx
=
JG
10
0
(20, 000 in − lb)dx
=
(0.08728 + 0.15x)(4.6 × 106 lb/in2 )
20, 000 in − lb
4.6 × 106 lb/in2
ANS:
1
0.15
10
0
20, 000 in − lb
4.6 × 106 lb/in2
10
0
dx
(0.0873 + 0.15x)
0.15dx
= 0.02899 [ln(1.587) − ln(0.0873)]
(0.0873 + 0.15x)
φ = 0.084077 rad = 4.817◦
Problem 11.48 The radius of the bar’s circular cross Free Body Diagram:
section varies linearly from 10 mm at x = 0 to 5 mm
at x = 150 mm. The shear modulus of the material is
G = 17 GPa. What torque T would cause a maximum
shear stress of 10 MPa at x = 80 mm?
Solution:
An expression for the radius of the bar for any value of x is:
5 mm
r = 0.010 m −
x = (0.010 − 0.0333x) m
150 mm
At x = 80 mm, the radius of the bar is:
r = 0.010 m − (0.0333)(0.08 m) = 0.0073333 m
The polar moment of inertia at x = 80 mm is:
π
J = (0.007336 m)4 = 4.5428 × 10−9 m4
2
The torque required to produce a maximum shear stress of 10 MPa at
x = 80 mm is:
10 × 106 N/m2 4.549 × 10−9 m4
τJ
T =
=
r
0.007336 m
ANS:
T = 6.19 N − m
Problem 11.49 In Problem 11.48, what torque T
would cause the end of the bar to rotate one degree?
Free Body Diagram:
Solution:
An expression for the radius of the bar for any value of x is:
5 mm
r = 0.010 m −
x = (0.010 − 0.0333x) m
150 mm
The expression for the polar moment of inertia at any point on the bar
is:
π
J = (0.010 − 0.0333x)4 m4
2
The integral expression for the angle of twist in the bar is:
0.15
φ=
0
T dx
=
JG
0.15
0
T dx
π
2
(0.010 − 0.0333x)4 m4 (17 × 109 N/m2 )
= 0.01745 rad
Recognizing that T , φ, π, and G are constant, the expression for the
angle of twist reduces to:
π(0.01745 rad)(17×109 N/m2 )
2T
4.66×108
T
ANS:
(−0.0333) =
0.15
0
=
0.15
0
dx
(0.010−0.0333x)4
(−0.0333)dx
(0.010−0.0333x)4
=
(0.010−0.0333x)−3
−3
T = 6.67 N − m
Problem 11.50 In Problem 11.48, suppose that the Free Body Diagram:
torque T at the end of the bar is 20 N-m and you want to
apply a torque in the opposite direction at x = 75 mm
so that the angle through which the end of the bar rotates
is zero. What is the magnitude of the torque you must
apply?
Solution:
An expression for the radius of the bar for any value of x is:
5 mm
r = 0.010 m −
x = (0.010 − 0.0333x) m
150 mm
The expression for the polar moment of inertia at any point on the bar
is:
π
J = (0.010 − 0.0333x)4 m4
2
The angle of twist is calculated by integrating over each of the two
sections if the bar.
0.075
0=
0
0=
π
2
((20 N − m) − T )dx
+
[(0.01 − 0.0333x)4 ] m4 (17 × 109 N/m2 )
((20 N − m) − T )
π
(17 × 109 N/m2 )
2
ANS:
1
−0.033
T = 101.902 N − m
0.075
0
0.15
π
0.075 2
(20 N − m)dx
[(0.01 − 0.0333x)4 ] m4 (17 × 109 N/m2 )
(−0.033)dx
+
(0.01 − 0.0333x)4
20 N − m
π
(17
× 109 N/m2 )
2
1
−0.033
0.15
0.075
dx
(0.01 − 0.0333x)4
Problem 11.51 Bars A and B have solid circular cross
sections and consist of material with shear modulus G =
17 GPa. Bar A is 150 mm long and its radius varies
linearly from 10 mm at its left to 5 mm at its right end.
The prismatic bar B is 100 mm long and its radius is
5 mm. There is a small gap between the bars. The end
of bar A is given an axial rotation of one degree and the
bars are welded together. What is the torque in the bars
afterward?
Free Body Diagram:
Solution:
An expression for the radius of bar A is:
.005
rA = 0.010 −
x m = ((0.010 − 0.0333x) m)
0.150
An expression for the polar moment of inertia for bar A is:
π
π
J = r 4 = (0.01 m − (0.0333x) m)4
2
2
After joining, the resulting bar will be subjected to a moment at end A
and a moment at end B, or:
MA − MB = 0
MA = MB
[1]
When the two bars are joined, the angle of twist for bar A will be
reduced and an angle of twist will be introduced into bar B. The sum
of these two angles of twist will be one degree (0.0175 radians), or:
φA + φB = 0.01745 radians
[2]
The angle of twist in bar A can be described by:
0.15
φA =
0
π
2
MA dx
(0.01m − (0.033x) m)4 (17 × 109 N/m2 )
=
2MA
π (17 × 109 N/m2 )
φA = 2.615 × 10−3 MA
Using this expression for φA and Equation [1] in Equation [2]:
2.615 × 10−3 MA +
ANS:
π
2
MA (0.1 m)
(0.005 m)4 (17 × 109 N/m2 )
MA = 2.027 N − m
= 0.01745 rad
0.15
0
dx
(0.01 m − (0.033x) m)4
Problem 11.52 The aluminum alloy bar has a circular Free Body Diagram:
cross section with 20-mm diameter, length L = 120 mm,
and a shear modulus of 28 GPa. If the distributed torque
is uniform and causes the end of the bar to rotate 0.5◦ ,
what is the magnitude of the maximum shear stress in
the bar?
Solution:
The expression for the angle of twist at the end of the bar is:
0.12
0.0087266 rad =
(T dx)x
π
2
0
(0.01 m)4 (28 × 109 N/m2 )
=
T
π
2
(0.01 m)4 (28 × 109 N/m2 )
0.12
xdx
0
T = 533.078 N − m/m
Maximum torque in the bar is at the wall and has a magnitude of:
TMAX = (533 N − m/m)(0.12 m) = 63.969 N − m
Maximum shear stress in the bar is:
τMAX =
ANS:
(64 N − m)(0.01 m)
π
(0.01 m)4
2
τMAX = 40.7 MPa
Problem 11.53 If the distributed torque in Prob- Free Body Diagram:
lem 11.48 is given by the equation c = co (x/L)3 and
causes the end of the bar to rotate 0.5◦ , what is the magnitude of the maximum shear stress in the bar?
Solution:
The polar moment of inertia for the bar is:
π
J = (0.01 m)4 = 1.571 × 10−8 m4
2
The moment applied at any point on the bar is:
x 3
M = C0
dx
0.12
The angle through which the end of the bar rotates is:
π φ = (0.5◦ )
= 0.00873 rad
180◦
The expression for the angle of twist is:
0.12
φ=
C0
3
x
0.12
xdx
JG
0
0.00873 rad =
→ 0.00873 rad =
(1.571 × 10−8
C0
JG(0.12 m)3
The total torque applied to the bar is:
0.12
0
x 3
dx = 39.99M − n
0.12
Maximum shear stress in the bar is:
τMAX =
ANS:
Tρ
(39.99 N − m)(0.01 m)
=
J
1.571 × 10−8 m4
τMAX = 25.5 MPa
x4 dx
0
C0
m4 )(28 × 109 N/m2 )(0.12)3
C0 = 1333 N − m/m
T = (1333 N − m)
0.12
0.12
x4 dx
0
Problem 11.54 A cylindrical bar with 1-in. diameter
fits tightly in a circular hole in a 5-in. thick plate. The
shear modulus of the material is G = 5.6 × 106 psi. A
12,000 in-lb axial torque is applied at the left end of the
bar. The distributed torque exerted on the bar by the plate
is given by the equation c = c0 [1−(x/5)1/2 ] in − lb/in,
where c0 is a constant and x is the axial position in inches
measured from the left side of the plate. Determine the
constant c0 and the magnitude of the maximum shear
stress in the bar at x = 2 in.
Free Body Diagram:
Solution:
The polar moment of inertia for the bar is:
π
J = (0.5 in)4 = 0.0982 in4
2
We know that the sum of moments on the bar must equal zero. That
is, the distributed torque must be equal and opposite to the 12,00 in-lb
applied torque.
5
12, 000 in − lb =
c0 1 −
0
ANS:
x 1/2
5
dx = c0 x −
x3/2
√
3
5
2
5
= c0 (1.667)
0
c0 = 7200 in − lb/in
Starting from the left-hand end of the bar, we calculate the torque at
x = 2 in:
2
T = (12, 000 in − lb) − (7200)
0
x1/2
1− √
5
T = 3669.6 in − lb
Maximum shear stress at x = 2 in is:
τ =
ANS:
Tr
(3669.6 in − lb)(0.5 in)
=
J
0.0982 in4
τ = 18.684 ksi
dx
Problem 11.55 In Problem 11.54, what is the magnitude of the angle of twist of the left end of the bar relative
to its right end?
Free Body Diagram:
Solution:
For the 10-inch section, the angle of twist is:
φ10 =
LT
(10in)(12, 000 in − lb)
=
= 0.2182 rad = 12.5◦
JG
(0.0982 in4 )(5.6 × 106 lb/in2 )
For the 5-inch section, the angle of twist is:
LT
φ5 =
=
JG
5
0
(7200 in − lb/in) 1 −
(0.0982 in4 )(5.6 ×
1/2
x√
dx(x)
5
6
10 lb/in2 )
φ5 = 0.02182 rad = 1.25◦
Total angle of twist for the bar is:
φ = φ10 + φ5 = 12.5◦ + 1.25◦
ANS:
φ = 13.75◦
Problem 11.56 The aluminum alloy bar has a circular Free Body Diagram:
cross section with 20-mm diameter and a shear modulus
of 28 GPa. What is the magnitude of the maximum shear
stress in the bar due to the uniformly distributed torque?
Solution:
The polar moment of inertia for the bar is:
π
J = (0.01 m)4 = 1.571 × 10−8 m4
2
From the symmetry of the loading, we see that the angle of twist will
be maximum at x = 60 mm. Total moment applied to the bar is:
T = (8, 500 N − m/m)(0.12 m) = 1, 020 N − m
As a result of the symmetric loading, the reactions at the ends of the
bar are:
ML = MR = T /2 = (1, 020 N − m)/2 = 510 N − m
Maximum shear stress is found at the extreme ends of the bar, where
the torque is maximum (510 N-m).
τMAX =
ANS:
Tr
(510 N − m)(0.01 m)
=
J
1.571 × 10−8 m4
τMAX = 324.6 MPa
Problem 11.57 One type of high-strength steel drill
pipe used in drilling oil wells has a 5-in. outside diameter
and 4.28-in. inside diameter. If the steel will safely
support a shear stress of 95 ksi, what is the largest torque
to which the pipe can safely be subjected?
Free Body Diagram:
Solution:
The polar moment of inertia for the pipe cross-section is:
π 4
π
J=
r − ri4 =
(2.5 in)4 − (2.14 in)4
2 o
2
J = 28.42 in4
Maximum permissible torque for the pipe is:
95, 000 lb/in2 28.4 in4
τMAX J
TMAX =
=
ρ
2.5 in
ANS:
T MAX = 1.08 × 106 in − lb
Problem 11.58 The drill pipe described in Problem 11.57 has a shear modulus G = 12 × 106 psi. If it
is used to drill an oil well 20,000 ft deep and the drilling
operation subjects the bottom of the pipe to a torque
T = 7500 in − lb, what is the resulting angle of twist
(in degrees) of the 20,000-ft pipe?
Free Body Diagram:
Solution:
The polar moment of inertia for the cross-section is:
J = π2 (2.5 in)4 − (2.14 in)4
J = 28.4 in4
Being careful of units, we apply the equation for the angle of twist:
(20, 000 ft) 12 in
(7500 in − lb)
LT
ft
φ=
= JG
28.4 in4 12 × 106 lb/in2
ANS:
φ= 5.28 rad = 302.5◦
Problem 11.59 The radius R = 200 mm. The in- Free Body Diagram:
finitesimal element is at the surface of the bar. What are
the normal stress and the magnitude of the shear stress
on the plane P ?
Solution:
The polar moment of inertia for the bar is:
π
J = (0.2 m)4 = 0.002513 m4
2
The shear stress at the surface of the bar (which is maximum shear
stress) is:
τ =
TR
(400 N − m) (0.2 m)
=
= 31, 834 N/m2
J
0.0025 m4
Summing vertical forces on the sectioned element:
ΣFy = 0 = (32, 000 N/m) (A) (sin 35◦ )−τ A (sin 35◦ )+σA (cos 35◦ )
[1]
τ − 0.7002σ = −32, 000 N/m2
Summing horizontal forces on the sectioned element:
ΣFX = 0 = − 32, 000 N/m2 A (sin 35◦ )−τ A (sin 35◦ )+σA (cos 35◦ )
[2]
τ − 1.428σ = −32, 000 N/m2
Solving equations [1] and [2] together:
ANS: σ = −30 kPa Note: The negative sign indicates a compressive stress.
ANS:
τ = 10.94 kPa
Problem 11.60 For the element in Problem 11.59, determine the normal stress and the magnitude of the shear
stress on the plane P shown.
Free Body Diagram:
Solution:
Summing horizontal forces on the sectioned element:
ΣFX = 0 = 32, 000 N/m2 A (cos 20◦ )−τ A (cos 20◦ )−σA (sin 20◦ )
[1]
τ + 0.364σ = 32, 000 N/m2
Summing vertical forces on the sectioned element:
ΣFY = 0 = 32, 000 N/m2 A (sin 20◦ )+τ A (sin 20◦ )−σA (cos 20◦ )
[2] − τ + 2.747σ = 32, 000 N/m2
Solving equations [1] and [2] together:
ANS:
ANS:
σ = 20.5 kPa
τ = 24.5 kPa
Problem 11.61 Part A of the bar has a solid circular Free Body Diagram:
cross section and part B has a hollow circular cross section. The bar is fixed at both ends and the shear modulus
of the material is G = 3.8 × 106 psi. Determine the
torques exerted on the bar by the walls.
Solution:
The polar moments of inertia are:
JA =
π 4
π
c = (2 in)4 = 25.13 in4
2
2
JB =
π 4
π
ro − ri4 =
(2 in)4 − (1 in)4 = 23.56 in4
2
2
We see that:
MA + MB = 150, 000 in − lb
[1]
Knowing that φA = φB , we also see that:
(7 in) (MA )
(14 in)(MB )
=
(23.56 in4 )G
25.13 in4 G
MA = 2.133MB
[2]
Solving Equations [1] and [2] together:
ANS:
ANS:
MA = 102.1 in − kip
MB = 47.9 in − kip
Problem 11.62 Determine the magnitudes of the max- Free Body Diagram:
imum shear stresses in parts A and B of the bar in Problem 11.61.
Solution:
The polar moments of inertia are:
π
π
JA = c4 = (2 in)4 = 25.13 in4
2
2
JB =
π 4
π
ro − ri4 =
(2 in)4 − (1 in)4 = 23.56 in4
2
2
We see that:
MA + MB = 150, 000 in − lb
We also see that:
(7 in)(MA )
(25.13 in4 )G
=
[1]
(14 in)(MB )
(23.56 in4 )G
MA = 2.133MB
[2]
Solving equations [1] and [2] together:
MA = 102.1 in − kipMB = 47.9 in − kip
Maximum torques in each section of the shaft:
(τMAX )A =
ANS:
(τMAX )A = 8.13 ksi
(τMAX )B =
ANS:
M A cA
(102.1 in − kip) (2 in)
=
JA
25.13 in4
M B cB
(47, 900 in − lb) (2 in)
=
JB
23.56 in4
(τMAX )B = 4.07 ksi
Problem 11.63 Suppose that you want to decrease the
weight of the bar in Problem 11.61 by increasing the
inside diameter of Part B. The bar is made of material
that will safely support a pure shear stress of 10 ksi.
Based on this criterion, what is the largest safe value of
the inside diameter?
Solution:
The polar moments of inertia for the two sections of the bar are:
π
π
π JA = (2 in)4 = 25.13 in4 JB =
(2 in)4 − ri4 = 25.13 in4 − ri4
2
2
2
We see that:
MA + MB = 150, 000 in − lb
[1]
Using the given limit of 10,000 ksi in shear stress:
10, 000 lb/in2 =
MB (2 in)
(25.13 in4 − π2 ri4 )
MB = 125, 650 in − lb − 7854ri4
[2]
Since the angle of twist must be the same for each section of the bar
(using equation [1]:
MA LA
MB LB
=
JA GA
JB GB
((150, 000 in − lb) − MB ) (7 in)
MB (14in)
= (25.13 in4 )(G)
25.13 in4 − π2 ri4 (G)
MB = 75, 000 in − lb−0.5MB −4688 in − lb+0.0313MB ri4
Solving Equations [2] and [3] together:
0 = −118, 163 in4 + 15, 714ri4 − 245.8ri8
Using the quadratic equation to solve for ri4 :
−15, 714 ± (15, 714)2 − 4(−245.8)(−118, 163)
ri4 =
2(−245.8)
ri4 = 8.704, 55.23
(clearly the 55,23 result is impossible, so it is discarded)
ri = 1.718
ANS:
di = 3.44 in
[2]
Problem 11.64 The bar has a circular cross section
with polar moment of inertia J and shear modulus G.
The distributed torque c = c0 (x/L)2 . What are the
magnitudes of the torques exerted on the bar by the left
and right walls?
Free Body Diagram:
Solution:
The total torque exerted by the distributed torque is:
L
T =
c0
x 2
L
0
c0
L2
dx =
L
x2 dx =
0
c0
L2
L3
3
=
c0 L
3
The equation of equilibrium for the bar is:
TA + TB − T = 0 → TA + TB = c0 L/3
[1]
We see that the angle of twist at each end of the bar must be zero. An
expression for the angle of twist in the bar is:
3
L
L
x 2
x
x c0 L
dx
c0 L
dx
2
TA L
TA L
0=
−
=
−
JG
JG
JG
JG
0
0
The equation can be simplified by multiplying both sides of the equation by the product JG.
0 = TA L −
ANS:
TA =
c0
L2
L
x3 dx = TA L −
0
c0 L
4
Using the above value of TA in Equation [1]:
ANS:
TB =
c0 L
12
c0
L2
L4
4
Problem 11.65 In Problem 11.64, at what axial position x is the bar’s angle of twist the greatest and what is
its magnitude?
Free Body Diagram:
Solution:
The total torque exerted by the distributed torque is:
L
T =
c0
0
x 2
L
c0
dx = 2
L
L
x2 dx =
0
c0
L2
L3
3
=
c0 L
3
The equation of equilibrium for the bar is:
TA + TB − T = 0 → TA + TB = c0 L/3
[1]
We see that the angle of twist at each end of the bar must be zero. An
expression for the angle of twist in the bar is:
3
L
L
x 2
x
x c0 L
dx
c0 L
dx
2
TA L
TA L
0=
−
=
−
JG
JG
JG
JG
0
0
The equation can be simplified by multiplying both sides of the equation by the product JG.
0 = TA L −
c0
L2
L
x3 dx = TA L −
0
TA =
c0
L2
L4
4
c0 L
4
Using the above value of TA in Equation [1]:
TB =
c0 L
12
Considering the loading from end A to end B, we see that the load at
any point on the bar is:
x2
L
C0 2 dx x − C0
x = M [2]
L
4
The angle of maximum twist occurs where the applied moment is
maximum. We find the location of maximum moment by setting the
differential of Equation [2] equal to zero.
x3
L
=0
C 0 2 − C0
L
4
The value of x where the maximum angle of twist occurs is:
ANS:
x=
L
41/3
= 0.63L
Calculating the magnitude of the maximum angle of twist:
0.63L
0.63L
x2
C0 L
2 dx x
C0
C0
φ=
=
x3 dx =
JG
JGL2
JGL2
0
ANS:
0
φ=
2
0L
0.0394 CJG
0.63L
x4 4 0
Problem 11.66 If the bar in Problem 11.64 is acted
upon by the distributed load C = C0 (x/L)2 and is free
of external torque from x = L/2 to x = L, what are the
magnitudes of the torque exerted on the bar by the left
and right walls?
Free Body Diagram:
Solution:
The total moment applied to the bar is:
L/2
M =
C0
0
L/2
x2
x3 L
dx
=
C
= C0
0
L2
3L2 0
24
The reaction at the right-hand end of the bar must be sufficient to off-set
the angle of twist which results from the applied moment.
L/2
x2
C0 L
2 dx x
MR L
−
=0
JG
JG
0
Multiplying through by JG:
L/2
C0
0
C0
ANS:
MR =
x3
dx − MR L = 0
L2
L2
− MR L = 0
64
C0 L
64
Since the sum of moments acting on the bar must equal zero:
ML + MR =
ML +
ANS:
ML =
5
C L
192 0
C0 L
24
C0 L
C0 L
=
64
24
Problem 12.1 The components of plain stress at a point
p of a material are σx = 20 MPa, σy = 0 and τxy = 0.
If θ = 45◦ , what are the stresses σx , σy and τxy
at point
p?
Solution:
Using Equation (12-7) to find σx :
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
σx == 20 MPa+0
+ 20 MPa−0
(cos 90◦ ) + 0
2
2
σx =
ANS:
σx = 10 MPa
Using Equation (12-9) to find σy :
σy =
σy =
ANS:
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
20 MPa+0
− 20 MPa−0
(cos 90◦ ) − 0
2
2
σy = 10 MPa
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
− 20 MPa−0
(sin 90◦ ) + 0
2
=−
τxy
=
τxy
ANS:
= −10 MPa
τxy
Problem 12.2 The components of plane stress at a
point p of a material are σx = 0, σy = 0 and τxy =
25 ksi. If θ = 45◦ , what are the stresses σx , σy and τxy
at point p?
Solution:
Using Equation (12-7) to find σx :
σx =
σx
ANS:
=
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
0+0
0−0
+ 2 (cos 90◦ ) + (25 ksi) (sin 90◦ )
2
σx = 25 ksi
Using Equation (12-9):
σy =
σy
ANS:
=
σx +σy
σ −σ
− x 2 y cos 2θ − τxy sin 2θ
2
0+0
0−0
− 2 cos(90◦ ) − 25 sin(90◦ )
2
σy = −25 ksi
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
− 0−0
(sin
90◦ ) + (25 ksi) (cos 90◦ )
2
=−
τxy
τxy
ANS:
=
=0
τxy
Problem 12.3 The components of plane stress at a
point p of a material are σx = −8 ksi, σy = 6 ksi
and τxy = −6 ksi. If θ = 30◦ , what are the stresses σx ,
σy and τxy
at point p?
Solution:
Using Equation (12-7) to find σx :
σx =
σx
=
ANS:
σx +σy
σ −σ
+ x 2 y (cos 2θ) + τxy (sin 2θ)
2
−8 ksi+6 ksi
ksi
+ −8 ksi−6
(cos 60◦ ) + (−6
2
2
ksi) (sin 60◦ )
σx = −9.7 ksi
Using Equation (12-9):
σy =
σy
=
ANS:
σx +σy
σ −σ
− x 2 y (cos 2θ) − τxy (sin 2θ)
2
−8 ksi+6 ksi
ksi
− −8 ksi−6
(cos 60◦ ) − (−6
2
2
ksi) (sin 60◦ )
σy = 7.7 ksi
:
Using Equation (12-8) to find τxy
σx −σy
(sin 2θ) + τxy (cos 2θ)
2
(−8−6)
− 2
sin 60◦ + (−6) cos 60◦
=−
τxy
=
τxy
ANS:
= 3.062 ksi
τxy
Problem 12.4 During liftoff, strain gauges attached to
one of the Space Shuttle main engine nozzles determine
that the components of plane stress σx = 66.46 MPa,
σy = 82.54 MPa, and τxy
= 6.75 MPa at θ = 20◦ .
What are the stresses σx , σy and τxy at that point?
Solution:
Adding Equations (12-7) and (12-9):
σx + σy = σx + σy → σx + σy = 149 MPa
[1]
Using Equation [1] in Equation (12-9):
82.54 MPa =
149 MPa σx − σy
−
[cos (40◦ )] − τxy [sin (40◦ )]
2
2
8.04 MPa = −(σx − σy )(0.383) − (0.643)τxy
τxy = −12.5 MPa − (σx − σy )(0.596)
[2]
Substituting the given information into Equation (12-8):
6.75 MPa = −
σx − σy
[sin (40◦ )] + τxy [cos (40◦ )]
2
6.75 MPa = −(0.321)(σx − σy ) + τxy (0.766)
τxy = 8.81 MPa + (0.419)(σx − σy )
[3]
Subtracting Equations [2] and [3]:
0 = −21.31 − 1.015(σx − σy ) → σx − σy = −20.99 MPa
Adding Equations [1] and [4]:
2σx = 128.01 MPa
ANS:
σx = 64 MPa
σy = 85 MPa
τxy = 0 MPa
[4]