REV IEW PRO BLEM S & SO LUTIO N S
JAMES M. G ERE
BARRY J. G O O DNO
Appendix A
FE Exam Review Problems
R-1.1: A plane truss has downward applied load P at joint 2 and another load P
applied leftward at joint 5. The force in member 3–5 is:
(A) 0
(B) ⫺P/2
(C) ⫺P
(D) ⫹1.5 P
Solution
⌺M1 ⫽ 0
V6 (3 L) ⫹ P L ⫺ P L ⫽ 0
so
3
5
P
V6 ⫽ 0
Method of sections
L
Cut through members 3-5,
2-5 and 2-4; use right hand
FBD
1
2
L
L
P
⌺M 2 ⫽ 0
6
4
L
5
P
F35 L ⫹ P L ⫽ 0
3
F35 ⫽ ⫺P
L
1
H1
V1
2
L
P
6
4
L
5
L
V6
P
F35
F25
L
2
F24
4
6
L
V6
1083
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1084
APPENDIX A FE Exam Review Problems
R-1.2: The force in member FE of the plane truss below is approximately:
(A) ⫺1.5 kN
(B) ⫺2.2 kN
(C) 3.9 kN
(D) 4.7 kN
Solution
3m
A
15 kN
3m
B
10 kN
3m
C
5 kN
D
3m
E
4.5 m
F
G
1m
Statics
⌺MA ⫽ 0
Ey (6 m) ⫺ 15 kN (3 m) ⫺ 10 kN (6 m) ⫺ 5 kN (9 m) ⫽ 0
E y ⫽ 25 kN
Ax
A
Ay
4.5 m
G
3m
B
15 kN
3m
C
10 kN
3m
5 kN
D
3m
E
F
1m
Ey
Method of sections: cut through BC, BE and FE; use right-hand FBD;
sum moments about B
1
⫺3
FFE (3 m) ⫺
FFE (3 m) ⫺ 10 kN (3 m) ⫺ 5 kN (6 m) ⫹Ey (3 m) ⫽ 0
110
110
Solving
FFE ⫽
5
110 kN
4
5 kN
10 kN
B
FFE ⫽ 3.95 kN
3
·FFE
10
FFE
3m
C
D
3m
E
1m
⫺
1
·FFE
10
Ey
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1085
APPENDIX A FE Exam Review Problems
R-1.3: The moment reaction at A in the plane frame below is approximately:
(A) ⫹1400 N⭈m
(B) ⫺2280 N⭈m
(C) ⫺3600 N⭈m
(D) ⫹6400 N⭈m
Solution
900 N
1.2 m
1200 N/m
900 N
1.2 m
C
B
3m
4m
Pin
connection
Bx
B
C
3m
By
Cy
A
Statics: use FBD of member BC to find reaction C yy
⌺MB ⫽ 0
Cy ⫽
Cy (3 m) ⫺ 900 N (1.2 m) ⫽ 0
900 N (1.2 m)
⫽ 360 N
3m
Sum moments about A for entire structure
⌺M A ⫽ 0
2
N
1
MA ⫹ Cy (3 m) ⫺ 900 N (1.2 m) ⫺ a1200 b 4 m a 4 mb ⫽ 0
m
2
3
Solving for MA
M A ⫽ 6400 N⭈m
900 N
1.2 m
1200 N/m
B
C
3m
Cy
4m
MA
A
Ax
Ay
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1086
APPENDIX A FE Exam Review Problems
R-1.4: A hollow circular post ABC (see figure) supports a load P1 ⫽ 16 kN
acting at the top. A second load P2 is uniformly distributed around the cap plate
at B. The diameters and thicknesses of the upper and lower parts of the post are
dAB ⫽ 30 mm, tAB ⫽ 12 mm, dBC ⫽ 60 mm, and tBC ⫽ 9 mm, respectively. The
lower part of the post must have the same compressive stress as the upper part.
The required magnitude of the load P2 is approximately:
(A) 18 kN
(B) 22 kN
(C) 28 kN
(D) 46 kN
P1
Solution
P1 ⫽ 16 kN
dAB ⫽ 30 mm
tAB ⫽ 12 mm
dBC ⫽ 60 mm
tBC ⫽ 9 mm
p
AAB ⫽ [dAB2 ⫺ (dAB ⫺ 2 tAB)2] ⫽ 679 mm2
4
A
tAB
dAB
P2
B
p
ABC ⫽ [dBC2 ⫺ (dBC ⫺ 2 tBC)2] ⫽ 1442 mm2
4
Stress in AB:
tBC
P1
sAB ⫽
⫽ 23.6 MPa
AAB
C
P1 ⫹ P2
⬍ must equal sAB
ABC
Stress in BC:
sBC ⫽
Solve for P2
P2 ⫽ ␴AB ABC ⫺ P1 ⫽ 18.00 kN
Check:
sBC ⫽
dBC
P1 ⫹ P2
⫽ 23.6 MPa
ABC
⬍ same as in AB
R-1.5: A circular aluminum tube of length L ⫽ 650 mm is loaded in compression by forces P. The outside and inside diameters are 80 mm and 68 mm,
respectively. A strain gage on the outside of the bar records a normal strain
in the longitudinal direction of 400 ⫻ 10⫺6. The shortening of the bar is
approximately:
(A) 0.12 mm
(B) 0.26 mm
(C) 0.36 mm
(D) 0.52 mm
Solution
␧ ⫽ 400 (10⫺6)
L ⫽ 650 mm
d ⫽ ␧ L ⫽ 0.260 mm
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APPENDIX A FE Exam Review Problems
1087
Strain gage
P
P
L
R-1.6: A steel plate weighing 27 kN is hoisted by a cable sling that has a clevis
at each end. The pins through the clevises are 22 mm in diameter. Each half of
the cable is at an angle of 35⬚ to the vertical. The average shear stress in each pin
is approximately:
(A) 22 MPa
(B) 28 MPa
(C) 40 MPa
(D) 48 MPa
Solution
W ⫽ 27 kN
dp ⫽ 22 mm
␪ ⫽ 35⬚
Cross sectional area of each pin:
p
Ap ⫽ d p2 ⫽ 380 mm2
4
P
Cable sling
Tensile force in cable:
W
a b
2
T⫽
⫽ 16.48 kN
cos(u)
Shear stress in each clevis
pin (double shear):
T
⫽ 21.7 MPa
t⫽
2 AP
35°
35°
Clevis
Steel plate
R-1.7: A steel wire hangs from a high-altitude balloon. The steel has unit weight
77kN/m3 and yield stress of 280 MPa. The required factor of safety against yield
is 2.0. The maximum permissible length of the wire is approximately:
(A) 1800 m
(B) 2200 m
(C) 2600 m
(D) 3000 m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1088
APPENDIX A FE Exam Review Problems
Solution
g ⫽ 77
kN
m3
␴Y ⫽ 280 MPa
sallow ⫽
Allowable stress:
Weight of wire of length L:
sY
⫽ 140.0 MPa
FSY
W ⫽ ␥ AL
Max. axial stress in wire of length L:
Lmax ⫽
Max. length of wire:
FSY ⫽ 2
smax ⫽
W
A
␴max ⫽ ␥ L
sallow
⫽ 1818 m
g
R-1.8: An aluminum bar (E ⫽ 72 GPa, ␯ ⫽ 0.33) of diameter 50 mm cannot
exceed a diameter of 50.1 mm when compressed by axial force P. The maximum
acceptable compressive load P is approximately:
(A) 190 kN
(B) 200 kN
(C) 470 kN
(D) 860 kN
Solution
E ⫽ 72 GPa
dinit ⫽ 50 mm
dfinal ⫽ 50.1 mm
dfinal ⫺ dinit
dinit
␧L ⫽ 0.002
Lateral strain:
␧L ⫽
Axial strain:
␧a ⫽
Axial stress:
␴ ⫽ E␧a ⫽ ⫺436.4 MPa
␯ ⫽ 0.33
⫺␧L
⫽ ⫺0.006
n
⬍ below yield stress of 480 MPa
so Hooke’s Law applies
Max. acceptable compressive load:R
p
Pmax ⫽ s a dinit2b ⫽ 857 kN
4
R-1.9: An aluminum bar (E ⫽ 70 GPa, ␯ ⫽ 0.33) of diameter 20 mm is stretched
by axial forces P, causing its diameter to decrease by 0.022 mm. The load P is
approximately:
(A) 73 kN
(B) 100 kN
(C) 140 kN
(D) 339 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1089
Solution
E ⫽ 70 GPa
dinit ⫽ 20 mm
␧L ⫽
Lateral strain:
⌬d
dinit
⌬d ⫽ ⫺0.022 mm
␯ ⫽ 0.33
d
P
P
␧L ⫽ ⫺0.001
⫺␧L
⫽ 3.333 ⫻ 10⫺3
v
Axial strain:
␧a ⫽
Axial stress:
␴ ⫽ E␧a ⫽ 233.3 MPa
⬍ below yield stress of 270 MPa
so Hooke’s Law applies
Max. acceptable load:
p
Pmax ⫽ s a dinit 2b ⫽ 73.3 kN
4
R-1.10: A polyethylene bar (E ⫽ 1.4 GPa, ␯ ⫽ 0.4) of diameter 80 mm is
inserted in a steel tube of inside diameter 80.2 mm and then compressed by axial
force P. The gap between steel tube and polyethylene bar will close when compressive load P is approximately:
(A) 18 kN
(B) 25 kN
(C) 44 kN
(D) 60 kN
Solution
E ⫽ 1.4 GPa
Lateral strain:
d1 ⫽ 80 mm
␧L ⫽
⌬d1 ⫽ 0.2 mm
⌬d1
d1
␯ ⫽ 0.4
Steel
tube
d1 d2
␧L ⫽ 0.003
Polyethylene
bar
⫺␧L
⫽ ⫺6.250 ⫻ 10⫺3
v
Axial strain:
␧a ⫽
Axial stress:
␴ ⫽ E␧a ⫽ ⫺8.8 MPa
⬍ well below ultimate stress of 28
MPa so Hooke’s Law applies
Max. acceptable compressive load:
p
Pmax ⫽ s a d1 2 b ⫽ 44.0 kN
4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1090
APPENDIX A FE Exam Review Problems
R-1.11: A pipe (E ⫽ 110 GPa) carries a load P1 ⫽ 120 kN at A and a uniformly
distributed load P2 ⫽ 100 kN on the cap plate at B. Initial pipe diameters and
thicknesses are: dAB ⫽ 38 mm, tAB ⫽ 12 mm, dBC ⫽ 70 mm, tBC ⫽ 10 mm.
Under loads P1 and P2, wall thickness tBC increases by 0.0036 mm. Poisson’s
ratio v for the pipe material is approximately:
(A) 0.27
(B) 0.30
(C) 0.31
(D) 0.34
Solution
E ⫽ 110 GPa
dAB ⫽ 38 mm
tAB ⫽ 12 mm
tBC ⫽ 10 mm
P1 ⫽ 120 kN
P2 ⫽ 100 kN
ABC ⫽
dBC ⫽ 70 mm
p
[dBC2 ⫺ (dBC ⫺ 2 tBC)2] ⫽ 1885 mm2
4
tBC dBC
Cap plate
C
tAB dAB
B
A
P1
P2
⫺(P1 ⫹ P2)
⫽ ⫺1.061 ⫻ 10⫺3
E ABC
Axial strain of BC:
␧BC ⫽
Axial stress in BC:
␴BC ⫽ E ␧BC ⫽ ⫺116.7 MPa
(well below yield stress of 550 MPa so Hooke’s
Law applies)
Lateral strain of BC:
␧L ⫽
⌬tBC ⫽ 0.0036 mm
⌬tBC
⫽ 3.600 ⫻ 10⫺4
tBC
Poisson’s ratio:
v⫽
⫺␧L
⫽ 0.34
␧BC
⬍ confirms value for brass
given in properties table
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1091
R-1.12: A titanium bar (E ⫽ 100 GPa, v ⫽ 0.33) with square cross section
(b ⫽ 75 mm) and length L ⫽ 3.0 m is subjected to tensile load P ⫽ 900 kN. The
increase in volume of the bar is approximately:
(A) 1400 mm3
(B) 3500 mm3
(C) 4800 mm3
(D) 9200 mm3
Solution
E ⫽ 100 GPa
b ⫽ 75 mm
L ⫽ 3.0 m
b
P ⫽ 900 kN
v ⫽ 0.33
b
P
P
L
Initial volume of bar:
Vinit ⫽ b2 L ⫽ 1.6875000 ⫻ 107 mm3
Normal strain in bar:
␧⫽
Lateral strain in bar:
␧L ⫽ ⫺v␧ ⫽ ⫺5.28000 ⫻ 10⫺4
Final length of bar:
Lf ⫽ L ⫹ ␧ L ⫽ 3004.800 mm
P
⫽ 1.60000 ⫻ 10⫺3
E b2
Final lateral dimension of bar:
Final volume of bar:
bf ⫽ b ⫹ ␧L b ⫽ 74.96040 mm
Vfinal ⫽ bf2 Lf ⫽ 1.68841562 ⫻ 107 mm3
Increase in volume of bar:
⌬V ⫽ Vfinal ⫺ Vinit ⫽ 9156 mm3
⌬V
⫽ 0.000543
Vinit
R-1.13: An elastomeric bearing pad is subjected to a shear force V during a static
loading test. The pad has dimensions a ⫽ 150 mm and b ⫽ 225 mm, and thickness t ⫽ 55 mm. The lateral displacement of the top plate with respect to the
bottom plate is 14 mm under a load V ⫽ 16 kN. The shear modulus of elasticity
G of the elastomer is approximately:
(A) 1.0 MPa
(B) 1.5 MPa
(C) 1.7 MPa
(D) 1.9 MPa
Solution
V ⫽ 16 kN
a ⫽ 150 mm
b ⫽ 225 mm
d ⫽ 14 mm
t ⫽ 55 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1092
APPENDIX A FE Exam Review Problems
Ave. shear stress:
t⫽
b
V
⫽ 0.474 MPa
ab
a
V
Ave. shear strain:
d
g ⫽ arctana b ⫽ 0.249
t
t
Shear modulus of elastomer:
t
G ⫽ ⫽ 1.902 MPa
g
R-1.14: A bar of diameter d ⫽ 18 mm and length L ⫽ 0.75 m is loaded in tension by forces P. The bar has modulus E ⫽ 45 GPa and allowable normal stress
of 180 MPa. The elongation of the bar must not exceed 2.7 mm. The allowable
value of forces P is approximately:
(A) 41 kN
(B) 46 kN
(C) 56 kN
(D) 63 kN
Solution
d ⫽ 18 mm
da ⫽ 2.7 mm
L ⫽ 0.75 m
E ⫽ 45 GPa
sa ⫽ 180 MPa
d
P
P
L
(1) allowable value of P based on elongation
da
⫽ 3.600 ⫻ 10⫺3
smax ⫽ E␧a ⫽ 162.0 MPa
L
p
⬍ elongation governs
Pa1 ⫽ smax a d 2 b ⫽ 41.2 kN
4
␧a ⫽
(2) allowable load P based on tensile stress
p
Pa2 ⫽ sa a d 2 b ⫽ 45.8 kN
4
R-1.15: Two flanged shafts are connected by eight 18 mm bolts. The diameter of
the bolt circle is 240 mm. The allowable shear stress in the bolts is 90 MPa.
Ignore friction between the flange plates. The maximum value of torque T0 is
approximately:
(A) 19 kN⭈m
(B) 22 kN⭈m
(C) 29 kN⭈m
(D) 37 kN⭈m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1093
Solution
db ⫽ 18 mm
d ⫽ 240 mm
ta ⫽ 90 MPa
Bolt shear area:
n⫽8
T0
2
As ⫽
p db
⫽ 254.5 mm2
4
T0
Max. torque:
Tmax ⫽ n (ta As)
d
⫽ 22.0 kN⭈m
2
R-1.16: A copper tube with wall thickness of 8 mm must carry an axial tensile
force of 175 kN. The allowable tensile stress is 90 MPa. The minimum required
outer diameter is approximately:
(A) 60 mm
(B) 72 mm
(C) 85 mm
(D) 93 mm
Solution
t ⫽ 8 mm
P ⫽ 175 kN
sa ⫽ 90 MPa
d
P
P
Required area based on allowable stress:
Areqd ⫽
P
⫽ 1944 mm2
sa
Area of tube of thickness t but unknown outer diameter d:
A⫽
p 2
[d ⫺ (d ⫺ 2 t)2]
4
A ⫽ ␲t (d ⫺ t)
Solving for dmin:
P
sa
dmin ⫽
⫹ t ⫽ 85.4 mm
pt
so
dinner ⫽ dmin ⫺ 2t ⫽ 69.4 mm
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1094
APPENDIX A FE Exam Review Problems
R-2.1: Two wires, one copper and the other steel, of equal length stretch the same
amount under an applied load P. The moduli of elasticity for each is: Es ⫽ 210 GPa,
Ec ⫽ 120 GPa. The ratio of the diameter of the copper wire to that of the steel wire
is approximately:
(A) 1.00
(B) 1.08
(C) 1.19
(D) 1.32
Solution
Es ⫽ 210 GPa
Copper
wire
Ec ⫽ 120 GPa
ds ⫽ dc
Displacements are equal:
or
PL
PL
⫽
Es As Ec Ac
so
Es As ⫽ Ec Ac
and
Ac Es
⫽
As Ec
Steel
wire
P
P
Express areas in terms of wire diameters then find ratio:
p dc2
4
Es
⫽
2
Ec
p ds
a
b
4
so
dc
Es
⫽
⫽ 1.323
ds B Ec
R-2.2: A plane truss with span length L ⫽ 4.5 m is constructed using cast iron pipes
(E ⫽ 170 GPa) with cross sectional area of 4500 mm2. The displacement of joint B
cannot exceed 2.7 mm. The maximum value of loads P is approximately:
(A) 340 kN
(B) 460 kN
(C) 510 kN
(D) 600 kN
Solution
L ⫽ 4.5 m
E ⫽ 170 GPa
A ⫽ 4500 mm2
dmax ⫽ 2.7 mm
Statics: sum moments about A to find reaction at B
P
RB ⫽
L
L
⫹P
2
2
L
RB ⫽ P
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1095
APPENDIX A FE Exam Review Problems
P
P
C
45°
A
45°
B
L
Method of Joints at B:
FAB ⫽ P (tension)
Force-displ. relation:
Pmax ⫽
EA
dmax ⫽ 459 kN
L
s⫽
Check normal stress in bar AB:
Pmax
⫽ 102.0 MPa
A
⬍ well below yield stress of
290 MPa in tension
R-2.3: A brass rod (E ⫽ 110 GPa) with cross sectional area of 250 mm2 is loaded
by forces P1 ⫽ 15 kN, P2 ⫽ 10 kN, and P3 ⫽ 8 kN. Segment lengths of the bar
are a ⫽ 2.0 m, b ⫽ 0.75 m, and c ⫽ 1.2 m. The change in length of the bar is
approximately:
(A) 0.9 mm
(B) 1.6 mm
(C) 2.1 mm
(D) 3.4 mm
Solution
E ⫽ 110 GPa
a⫽2m
A ⫽ 250 mm2
A
c ⫽ 1.2 m
P1 ⫽ 15 kN
P2
P1
b ⫽ 0.75 m
C
B
a
b
D
P3
c
P2 ⫽ 10 kN
P3 ⫽ 8 kN
Segment forces (tension is positive):
NAB ⫽ P1 ⫹ P2 ⫺ P3 ⫽ 17.00 kN
NBC ⫽ P2 ⫺ P3 ⫽ 2.00 kN
NCD ⫽ ⫺P3 ⫽ ⫺8.00 kN
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1096
APPENDIX A FE Exam Review Problems
Change in length:
dD ⫽
1
(NAB a ⫹ NBC b ⫹ NCD c) ⫽ 0.942 mm
EA
dD
⫽ 2.384 ⫻ 10⫺4
a⫹b⫹c
⬍
positive so elongation
Check max. stress:
NAB
⫽ 68.0 MPa
A
⬍ well below yield stress for brass so OK
R-2.4: A brass bar (E ⫽ 110 MPa) of length L ⫽ 2.5 m has diameter d1 ⫽ 18 mm
over one-half of its length and diameter d2 ⫽12 mm over the other half. Compare
this nonprismatic bar to a prismatic bar of the same volume of material with
constant diameter d and length L. The elongation of the prismatic bar under the
same load P ⫽ 25 kN is approximately:
(A) 3 mm
(B) 4 mm
(C) 5 mm
(D) 6 mm
Solution
L ⫽ 2.5 m
P ⫽ 25 kN
d1 ⫽ 18 mm
d2 ⫽ 12 mm
E ⫽ 110 GPa
d2
P
P
p
A1 ⫽ d12 ⫽ 254.469 mm2
4
A2 ⫽
d1
L/2
L/2
p 2
d2 ⫽ 113.097 mm2
4
Volume of nonprismatic bar:
Vol nonprismatic ⫽ (A1 ⫹ A2)
L
⫽ 459458 mm3
2
Diameter of prismatic bar of same volume: d ⫽
Aprismatic ⫽
p 2
d ⫽ 184 mm2
4
Volnonprismatic
⫽ 15.30 mm
p
L
H
4
Vprismatic ⫽ Aprismatic L ⫽ 459458 mm3
Elongation of prismatic bar:
d⫽
PL
⫽ 3.09 mm
E Aprismatic
⬍ less than d for nonprismatic bar
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1097
Elongation of nonprismatic bar shown in fig. above:
⌬⫽
PL 1
1
a
⫹
b ⫽ 3.63 mm
2 E A1
A2
R-2.5: A nonprismatic cantilever bar has an internal cylindrical hole of diameter
d/2 from 0 to x, so the net area of the cross section for Segment 1 is (3/4)A. Load
P is applied at x, and load -P/2 is applied at x ⫽ L. Assume that E is constant. The
length of the hollow segment, x, required to obtain axial displacement
d ⫽ PL/EA at the free end is:
(A) x ⫽ L/5
(B) x ⫽ L/4
(C) x ⫽ L/3
(D) x ⫽ 3L/5
Solution
Forces in Segments 1 & 2:
N1 ⫽
3P
2
N2 ⫽
⫺P
2
Segment 1
Segment 2
3
—A
4
d
A
P
—
2
P
Displacement at free end:
d3 ⫽
d
—
2
x
3
2
L–x
N1 x
N2 (L ⫺ x)
⫹
EA
3
E a Ab
4
3P
⫺P
(L ⫺ x)
x
2
2
P (L ⫺ 5 x)
⫽⫺
d3 ⫽
⫹
EA
2 AE
3
E a Ab
4
Set d3 equal to PL/EA and solve for x
⫺
P (L ⫺ 5 x) P L
or
⫽
2AE
EA
⫺
P (L ⫺ 5 x)
PL
P (3 L ⫺ 5 x)
⫺
⫽ 0 simplify S ⫺
⫽0
2AE
EA
2AE
So x ⫽ 3L/5
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1098
APPENDIX A FE Exam Review Problems
R-2.6: A nylon bar (E ⫽ 2.1 GPa) with diameter 12 mm, length 4.5 m, and
weight 5.6 N hangs vertically under its own weight. The elongation of the bar at
its free end is approximately:
(A)
0.05 mm
(B)
0.07 mm
(C)
0.11 mm
(D)
0.17 mm
Solution
E ⫽ 2.1 GPa
L ⫽ 4.5 m
d ⫽ 12 mm
A
2
pd
A⫽
⫽ 113.097 mm2
4
g ⫽ 11
L
kN
m3
W ⫽ ␥ L A ⫽ 5.598 N
dB ⫽
WL
2EA
B
dB ⫽
or
(g L A) L
2 EA
2
so
dB ⫽
gL
⫽ 0.053 mm
2E
Check max. normal stress at top of bar smax ⫽
W
⫽ 0.050 MPa
A
⬍ ok - well below ult.
stress for nylon
R-2.7: A monel shell (Em ⫽ 170 GPa, d3 ⫽ 12 mm, d2 ⫽ 8 mm) encloses a brass
core (Eb ⫽ 96 GPa, d1 ⫽ 6 mm). Initially, both shell and core are of length 100 mm.
A load P is applied to both shell and core through a cap plate. The load P
required to compress both shell and core by 0.10 mm is approximately:
(A) 10.2 kN
(B) 13.4 kN
(C) 18.5 kN
(D) 21.0 kN
Solution
Em ⫽ 170 GPa
Eb ⫽ 96 GPa
d1 ⫽ 6 mm
d2 ⫽ 8 mm
d3 ⫽ 12 mm
L ⫽ 100 mm
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APPENDIX A FE Exam Review Problems
1099
P
Monel shell
Brass core
L
d1
d2
d3
Am ⫽
p 2
(d3 ⫺ d22) ⫽ 62.832 mm2
4
Ab ⫽
p 2
d1 ⫽ 28.274 mm2
4
dm ⫽ db
Compatibility:
Pm L
Pb L
⫽
Em Am Eb Ab
Pm ⫽
Pm ⫹ Pb ⫽ P
Statics:
Em Am
Pb
Eb Ab
so
Pb ⫽
P
Em Am
a1 ⫹
b
Eb Ab
Set dB equal to 0.10 mm and solve for load P:
db ⫽
Pb L
Eb Ab
and then
so
P⫽
Pb ⫽
Eb Ab
db
L
with
db ⫽ 0.10 mm
Eb Ab
Em Am
db a1 ⫹
b ⫽ 13.40 kN
L
Eb Ab
R-2.8: A steel rod (Es ⫽ 210 GPa, dr ⫽ 12 mm, as ⫽ 12 ⫻ 10⫺6 > ⬚C ) is held
stress free between rigid walls by a clevis and pin (dp ⫽ 15 mm) assembly at each
end. If the allowable shear stress in the pin is 45 MPa and the allowable normal
stress in the rod is 70 MPa, the maximum permissible temperature drop ⌬T is
approximately:
(A) 14 ⬚C
(B) 20 ⬚C
(C) 28 ⬚C
(D) 40 ⬚C
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1100
APPENDIX A FE Exam Review Problems
Solution
Es ⫽ 210 GPa
dp ⫽ 15 mm
dr ⫽ 12 mm
p 2
Ar ⫽ dr ⫽ 113.097 mm2
4
p
Ap ⫽ dp2 ⫽ 176.715 mm2
4
pin, dp
∆T
rod, dr
Clevis
␣s ⫽ 12(10 ) > ⬚C
⫺6
ta ⫽ 45 MPa
sa ⫽ 70 MPa
Force in rod due to temperature drop ⌬T:
and normal stress in rod:
sr ⫽
Fr ⫽ Es Ar (␣s) ⌬T
Fr
Ar
So ⌬Tmax associated with normal stress in rod
⌬Tmaxrod ⫽
sa
⫽ 27.8
Es as
degrees Celsius (decrease) ⬍ Controls
Now check ⌬T based on shear stress in pin (in double shear):
⌬Tmaxpin ⫽
ta (2 Ap)
Es Ar as
tpin ⫽
Fr
2 Ap
⫽ 55.8
R-2.9: A threaded steel rod (Es ⫽ 210 GPa, dr ⫽ 15 mm, ␣s ⫽ 12 ⫻ 10⫺6 > ⬚C )
is held stress free between rigid walls by a nut and washer (dw ⫽ 22 mm)
assembly at each end. If the allowable bearing stress between the washer and
wall is 55 MPa and the allowable normal stress in the rod is 90 MPa, the
maximum permissible temperature drop ⌬T is approximately:
(A) 25 ⬚C
(B) 30 ⬚C
(C) 38 ⬚C
(D) 46 ⬚C
Solution
Es ⫽ 210 GPa
dr ⫽ 15 mm
Ar ⫽
p 2
dr ⫽ 176.7 mm2
4
Aw ⫽
p 2
(dw ⫺ dr2) ⫽ 203.4 mm2
4
␣s ⫽ 12(10⫺6) > ⬚C
sba ⫽ 55 MPa
dw ⫽ 22 mm
sa ⫽ 90 MPa
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APPENDIX A FE Exam Review Problems
rod, dr
washer, dw
∆T
Force in rod due to temperature drop ⌬T:
and normal stress in rod:
sr ⫽
Fr ⫽ Es Ar (␣ s)⌬T
1101
Fr
Ar
So ⌬Tmax associated with normal stress in rod
⌬Tmaxrod ⫽
sa
⫽ 35.7
Es as
degrees Celsius (decrease)
Now check ⌬T based on bearing stress beneath washer:
sba (Aw)
⫽ 25.1
Es Ar as
⌬Tmaxwasher ⫽
sb ⫽
Fr
Aw
degrees Celsius (decrease)
⬍ Controls
R-2.10: A steel bolt (area ⫽ 130 mm2, Es ⫽ 210 GPa) is enclosed by a copper tube
(length ⫽ 0.5 m, area ⫽ 400 mm2, Ec ⫽ 110 GPa) and the end nut is turned until
it is just snug. The pitch of the bolt threads is 1.25 mm. The bolt is now tightened
by a quarter turn of the nut. The resulting stress in the bolt is approximately:
(A) 56 MPa
(B) 62 MPa
(C) 74 MPa
(D) 81 MPa
Solution
Es ⫽ 210 GPa
2
Ac ⫽ 400 mm
n ⫽ 0.25
Ec ⫽ 110 GPa
L ⫽ 0.5 m
2
As ⫽ 130 mm
Copper tube
p ⫽ 1.25 mm
Compatibility: shortening of tube
and elongation of bolt ⫽ applied
displacement of n ⫻ p
Steel bolt
Ps L
Pc L
⫹
⫽ np
Ec Ac
Es As
Statics:
Pc ⫽ Ps
Solve for Ps
Ps L
Ps L
⫹
⫽ np
Ec Ac
Es As
or
Ps ⫽
np
⫽ 10.529 kN
1
1
La
⫹
b
Ec Ac Es As
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1102
APPENDIX A FE Exam Review Problems
Stress in steel bolt:
ss ⫽
Ps
⫽ 81.0 MPa
As
⬍ tension
Stress in copper tube:
sc ⫽
Ps
⫽ 26.3 MPa
Ac
⬍ compression
R-2.11: A steel bar of rectangular cross section (a ⫽ 38 mm, b ⫽ 50 mm)
carries a tensile load P. The allowable stresses in tension and shear are
100 MPa and 48 MPa respectively. The maximum permissible load Pmax is
approximately:
(A) 56 kN
(B) 62 kN
(C) 74 kN
(D) 91 kN
Solution
a ⫽ 38 mm
b ⫽ 50 mm
A ⫽ a b ⫽ 1900 mm2
b
sa ⫽ 100 MPa
P
P
ta ⫽ 48 MPa
a
Bar is in uniaxial tension so Tmax ⫽ smax/2; since 2 ta ⬍ sa, shear stress governs
Pmax ⫽ ta A ⫽ 91.2 kN
R-2.12: A brass wire (d ⫽ 2.0 mm, E ⫽ 110 GPa) is pretensioned to T ⫽ 85 N.
The coefficient of thermal expansion for the wire is 19.5 ⫻ 10⫺6 > ⬚C. The temperature change at which the wire goes slack is approximately:
(A) ⫹5.7 ⬚C
(B) ⫺12.6 ⬚C
(C) ⫹12.6 ⬚C
(D) ⫺18.2 ⬚C
Solution
E ⫽ 110 GPa
d ⫽ 2.0 mm
␣b ⫽ 19.5 (10 ) > ⬚C
⫺6
p
A ⫽ d 2 ⫽ 3.14 mm2
4
T ⫽ 85 N
T
d
T
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APPENDIX A FE Exam Review Problems
1103
Normal tensile stress in wire due to pretension T and temperature increase ⌬T:
s⫽
T
⫺ E ab ⌬T
A
Wire goes slack when normal stress goes to zero; solve for ⌬T
T
A
⌬T ⫽
⫽ ⫹12.61
E ab
degrees Celsius (increase in temperature)
R-2.13: A copper bar (d ⫽ 10 mm, E ⫽ 110 GPa) is loaded by tensile load
P ⫽ 11.5 kN. The maximum shear stress in the bar is approximately:
(A) 73 MPa
(B) 87 MPa
(C) 145 MPa
(D) 150 MPa
Solution
E ⫽ 110 GPa
d ⫽ 10 mm
p 2
A ⫽ d ⫽ 78.54 mm2
4
P ⫽ 11.5 kN
d
P
P
Normal stress in bar:
p
s ⫽ ⫽ 146.4 MPa
A
For bar in uniaxial stress, max. shear stress is on a plane at 45 deg. to axis of
bar and equals 1/2 of normal stress:
s
tmax ⫽ ⫽ 73.2 MPa
2
R-2.14: A steel plane truss is loaded at B and C by forces P ⫽ 200 kN. The cross
sectional area of each member is A ⫽ 3970 mm2. Truss dimensions are H ⫽ 3 m
and L ⫽ 4 m. The maximum shear stress in bar AB is approximately:
(A) 27 MPa
(B) 33 MPa
(C) 50 MPa
(D) 69 MPa
Solution
P ⫽ 200 kN
A ⫽ 3970 mm2
H⫽3m
L⫽4m
Statics: sum moments about A to find vertical reaction at B
Bvert ⫽
⫺P H
⫽ ⫺150.000 kN
L
(downward)
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1104
APPENDIX A FE Exam Review Problems
P
P
C
H
B
L
A
P
Method of Joints at B:
CBvert ⫽ ⫺Bvert
CBhoriz ⫽
L
CBvert ⫽ 200.0 kN
H
AB ⫽ P ⫹ CBhoriz ⫽ 400.0 kN (compression)
So bar force in AB is:
Max. normal stress in AB:
sAB ⫽
AB
⫽ 100.8 MPa
A
Max. shear stress is 1/2 of max. normal stress for bar in uniaxial stress and is
on plane at 45 deg. to axis of bar:
tmax ⫽
sAB
⫽ 50.4 MPa
2
R-2.15: A plane stress element on a bar in uniaxial stress has tensile stress of
s␪ ⫽ 78 MPa (see fig.). The maximum shear stress in the bar is approximately:
(A) 29 MPa
(B) 37 MPa
(C) 50 MPa
(D) 59 MPa
Solution
␴u ⫽ 78 MPa
σθ /2
Plane stress transformation formulas for
uniaxial stress:
sx ⫽
su
cos(u)2
and
⵩ on element face
at angle ␪
τθ τθ
σθ
θ
su
2
sx ⫽
sin(u)2
⵩ on element face
at angle ␪⫹90
τθ τθ
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1105
APPENDIX A FE Exam Review Problems
Equate above formulas and solve for sx
tan(u)2 ⫽
u ⫽ atana
so
sx ⫽
1
2
1
b ⫽ 35.264⬚
12
su
⫽ 117.0 MPa
cos(u)2
also ␶u ⫽ ⫺sx sin(u) cos(u) ⫽ ⫺55.154 MPa
Max. shear stress is 1/2 of max. normal stress for bar in uniaxial stress and is
on plane at 45 deg. to axis of bar:
tmax ⫽
sx
⫽ 58.5 MPa
2
R-2.16: A prismatic bar (diameter
d0 ⫽ 18 mm) is loaded by force P1.
A stepped bar (diameters d1 ⫽ 20 mm,
d2 ⫽ 25 mm, with radius R of fillets
⫽ 2 mm) is loaded by force P2. The
allowable axial stress in the material is
75 MPa. The ratio P1/P2 of the maximum permissible loads that can be
applied to the bars, considering stress
concentration effects in the stepped
bar, is:
(A) 0.9
(B) 1.2
(C) 1.4
(D) 2.1
P1
P2
d0
d1
P1
d2
d1
P2
FIG. 2-66 Stress-concentration factor K for round bars with shoulder fillets.
The dashed line is for a full quater-circular fillet.
3.0
R
D2
=2
D1
P
D2
1.5
2.5
s
K = s max
1.2
K
D1
nom
1.1
s nom =
P
P
p D21/4
2.0
R=
1.5
0
D2 – D1
2
0.05
0.10
0.15
R
D1
0.20
0.25
0.30
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1106
APPENDIX A FE Exam Review Problems
Solution
Prismatic bar P1 max ⫽ sallow a
p d02
p (18 mm)2
b ⫽ (75 MPa) c
d ⫽ 19.1 kN
4
4
2 mm
d2 25 mm
R
⫽
⫽ 0.100
⫽
⫽ 1.250 so K ⫽ 1.75
d1 20 mm
d1 20 mm
Stepped bar
from stress conc. Fig. 2-66
3.0
R
D2
=2
D1
P
D2
1.5
2.5
s
K = s max
1.2
K
D1
s nom =
nom
1.1
P
P
p D21/4
2.0
K = 1.75
R=
1.5
0
P2 max ⫽
D2 – D1
2
0.05
0.10
0.15
R
D1
0.20
0.25
0.30
75 MPa p(20 mm)2
sallow p d12
a
b⫽a
bc
d ⫽ 13.5 kN
K
4
K
4
P1 max
19.1 kN
⫽
⫽ 1.41
P2 max
13.5 kN
R-3.1: A brass rod of length L ⫽ 0.75 m is twisted by torques T until the angle
of rotation between the ends of the rod is 3.5°. The allowable shear strain in
the copper is 0.0005 rad. The maximum permissible diameter of the rod is
approximately:
(A) 6.5 mm
(B) 8.6 mm
(C) 9.7 mm
(D) 12.3 mm
Solution
L ⫽ 0.75 m
d
T
T
f ⫽ 3.5°
ga ⫽ 0.0005
L
Max. shear strain:
d
a fb
2
gmax ⫽
so
L
dmax ⫽
2 ga L
⫽ 12.28 mm
f
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APPENDIX A FE Exam Review Problems
1107
R-3.2: The angle of rotation between the ends of a nylon bar is 3.5°. The bar
diameter is 70 mm and the allowable shear strain is 0.014 rad. The minimum
permissible length of the bar is approximately:
(A) 0.15 m
(B) 0.27 m
(C) 0.40 m
(D) 0.55 m
Solution
d ⫽ 70 mm
d
T
f ⫽ 3.5⬚
T
ga ⫽ 0.014
L
Max. shear strain:
g⫽
rf
L
so
L min ⫽
df
⫽ 0.15 m
2 ga
R-3.3: A brass bar twisted by torques T acting at the ends has the following
properties: L ⫽ 2.1 m, d ⫽ 38 mm, and G ⫽ 41 GPa. The torsional stiffness of the
bar is approximately:
(A) 1200 N⭈m
(B) 2600 N⭈m
(C) 4000 N⭈m
(D) 4800 N⭈m
Solution
G ⫽ 41 GPa
L ⫽ 2.1 m
d
T
T
d ⫽ 38 mm
L
Polar moment of inertia, Ip:
Ip ⫽
p 4
d ⫽ 2.047 ⫻ 105 mm4
32
Torsional stiffness, kT:
kT ⫽
G Ip
L
⫽ 3997 N⭈m
R-3.4: A brass pipe is twisted by torques T ⫽ 800 N⭈m acting at the ends causing
an angle of twist of 3.5 degrees. The pipe has the following properties: L ⫽ 2.1 m,
d1 ⫽ 38 mm, and d2 ⫽ 56 mm. The shear modulus of elasticity G of the pipe is
approximately:
(A) 36.1 GPa
(B) 37.3 GPa
(C) 38.7 GPa
(D) 40.6 GPa
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1108
APPENDIX A FE Exam Review Problems
Solution
T
T
d1
L
L ⫽ 2.1 m d1 ⫽ 38 mm
d2
d2 ⫽ 56 mm
Polar moment of inertia: Ip ⫽
f ⫽ 3.5° T ⫽ 800 N ⭈ m
p 4
(d2 ⫺ d14) ⫽ 7.608 ⫻ 105 mm4
32
Solving torque-displacement relation for shear modulus G:
G⫽
TL
⫽ 36.1 GPa
f Ip
R-3.5: An aluminum bar of diameter d ⫽ 52 mm is twisted by torques T1 at the
ends. The allowable shear stress is 65 MPa. The maximum permissible torque T1
is approximately:
(A) 1450 N⭈m
(B) 1675 N⭈m
(C) 1710 N⭈m
(D) 1800 N⭈m
Solution
d ⫽ 52 mm
T1
d
T1
ta ⫽ 65 MPa
Ip ⫽
p 4
4
d ⫽ 7.178 ⫻ 105 mm
32
From shear formula:
T1 max
ta Ip
d
a b
2
⫽ 1795 N⭈m
R-3.6: A steel tube with diameters d2 ⫽ 86 mm and d1 ⫽ 52 mm is twisted by
torques at the ends. The diameter of a solid steel shaft that resists the same torque
at the same maximum shear stress is approximately:
(A) 56 mm
(B) 62 mm
(C) 75 mm
(D) 82 mm
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APPENDIX A FE Exam Review Problems
1109
Solution
d2 ⫽ 86 mm
IPpipe ⫽
d1 ⫽ 52 mm
p 4
(d2 ⫺ d14 ) ⫽ 4.652 ⫻ 106 mm4
32
Shear formula for hollow pipe:
d2
Ta b
2
tmax ⫽
IPpipe
d
Ta b
2
16 T
tmax ⫽
⫽
p 4
p
d3
d
32
Equate and solve for d of solid shaft:
d
d1
d2
Shear formula for solid shaft:
d⫽D
1
16 T
p
3
d2
Ta b
2
IPpipe
T
1
32 IPpipe 3
d⫽a
b ⫽ 82.0 mm
p d2
R-3.7: A stepped steel shaft with diameters d1 ⫽ 56 mm and d2 ⫽ 52 mm is
twisted by torques T1 ⫽ 3.5 kN⭈m and T2 ⫽ 1.5 kN⭈m acting in opposite directions. The maximum shear stress is approximately:
(A) 54 MPa
(B) 58 MPa
(C) 62 MPa
(D) 79 MPa
Solution
d1 ⫽ 56 mm
d2 ⫽ 52 mm
T1 ⫽ 3.5 kN⭈m T2 ⫽ 1.5 kN⭈m
T1
d1
d2
B
A
L1
T2
C
L2
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1110
APPENDIX A FE Exam Review Problems
Polar moments of inertia:
p 4
d1 ⫽ 9.655 ⫻ 105 mm4
32
p
Ip2 ⫽ d24 ⫽ 7.178 ⫻ 105 mm4
32
Ip1 ⫽
Shear formula - max. shear stresses in segments 1 & 2:
(T1 ⫺ T2)
tmax1 ⫽
Ip1
d1
2
d2
T2 a b
2
tmax2 ⫽
⫽ 54.3 MPa
Ip2
⫽ 58.0 MPa
R-3.8: A stepped steel shaft (G ⫽ 75 GPa) with diameters d1 ⫽ 36 mm and
d2 ⫽ 32 mm is twisted by torques T at each end. Segment lengths are L1 ⫽ 0.9 m
and L2 ⫽ 0.75 m. If the allowable shear stress is 28 MPa and maximum allowable twist is 1.8 degrees, the maximum permissible torque is approximately:
(A) 142 N⭈m
(B) 180 N⭈m
(C) 185 N⭈m
(D) 257 N⭈m
Solution
d1 ⫽ 36 mm
d1
d2
T
d2 ⫽ 32 mm
G ⫽ 75 GPa
A
ta ⫽ 28 MPa
C
B
L1
T
L2
L1 ⫽ 0.9 m L2 ⫽ 0.75 m
fa ⫽ 1.8⬚
Polar moments of inertia:
p 4
d1 ⫽ 1.649 ⫻ 105 mm4
32
p
Ip2 ⫽ d24 ⫽ 1.029 ⫻ 105 mm4
32
Ip1 ⫽
Max torque based on allowable shear stress - use shear formula:
T
tmax1 ⫽
d1
2
Ip1
Tmax1 ⫽ ta a
d2
Ta b
2
tmax2 ⫽
Ip2
2 Ip1
2 Ip2
b ⫽ 257 N⭈m Tmax2 ⫽ ta a
b ⫽ 180 N⭈m ⬍ controls
d1
d2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1111
Max. torque based on max. rotation & torque-displacement relation:
f⫽
L2
T L1
a
⫹
b
G Ip1
Ip2
Tmax ⫽
G fa
⫽ 185 N⭈m
L1
L2
a
⫹
b
Ip1
Ip2
R-3.9: A gear shaft transmits torques TA ⫽ 975 N⭈m, TB ⫽ 1500 N⭈m, TC ⫽
650 N⭈m and TD ⫽ 825 N⭈m. If the allowable shear stress is 50 MPa, the
required shaft diameter is approximately:
(A) 38 mm
(B) 44 mm
(C) 46 mm
(D) 48 mm
Solution
ta ⫽ 50 MPa
TA
TA ⫽ 975 N⭈m
TB
TB ⫽ 1500 N⭈m
TC
TC ⫽ 650 N⭈m
A
TD ⫽ 825 N⭈m
TD
B
C
Find torque in each
segment of shaft:
D
TAB ⫽ TA ⫽ 975.0 N⭈m
TBC ⫽ TA ⫺ TB ⫽ ⫺525.0 N⭈m
TCD ⫽ TD ⫽ 825.0 N⭈m
Shear formula:
d
Ta b
2
16 T
⫽
t⫽
p 4
p d3
d
32
Set t to tallowable and T to torque in each segment; solve for required diameter d
(largest controls)
1
Segment AB:
16 |TAB| 3
d⫽a
b ⫽ 46.3 mm
p ta
1
16 |TBC| 3
b ⫽ 37.7 mm
Segment BC: d ⫽ a
p ta
1
Segment CD:
16 |TCD| 3
b ⫽ 43.8 mm
d⫽a
p ta
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1112
APPENDIX A FE Exam Review Problems
R-3.10: A hollow aluminum shaft (G ⫽ 27 GPa, d2 ⫽ 96 mm, d1 ⫽ 52 mm)
has an angle of twist per unit length of 1.8°/m due to torques T. The resulting
maximum tensile stress in the shaft is approximately:
(A) 38 MPa
(B) 41 MPa
(C) 49 MPa
(D) 58 MPa
Solution
G ⫽ 27 GPa
d2
T
T
d2 ⫽ 96 mm
d1 ⫽ 52 mm
L
u ⫽ 1.8⬚>m
Max. shear strain due to
twist per unit length:
d1 d2
d2
gmax ⫽ a b u ⫽ 1.508 ⫻ 10⫺3
2
Max. shear stress:
radians
tmax ⫽ Ggmax ⫽ 40.7 MPa
Max. tensile stress on plane at 45 degrees & equal to max. shear stress:
smax ⫽ tmax ⫽ 40.7 MPa
R-3.11: Torques T ⫽ 5.7 kN⭈m are applied to a hollow aluminum shaft (G ⫽
27 GPa, d1 ⫽ 52 mm). The allowable shear stress is 45 MPa and the allowable
normal strain is 8.0 ⫻ 10⫺4. The required outside diameter d2 of the shaft is
approximately:
(A) 38 mm
(B) 56 mm
(C) 87 mm
(D) 91 mm
Solution
T ⫽ 5.7 kN⭈m
G ⫽ 27 GPa
ta1 ⫽ 45 MPa
␧a ⫽ 8.0(10⫺4)
d1 ⫽ 52 mm
d1 d2
Allowable shear strain based on allowable normal strain for pure shear
ga ⫽ 2␧a ⫽ 1.600 ⫻ 10⫺3
so resulting allow. shear stress is:
ta2 ⫽ Gga ⫽ 43.2 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1113
APPENDIX A FE Exam Review Problems
So allowable shear stress based on normal strain governs
ta ⫽ ta2
Use torsion formula to relate required d2 to allowable shear stress:
tmax ⫽
d2
Ta b
2
p 4
(d2 ⫺ d14)
32
16 T
4
4
and rearrange equation to get d 2 ⫺ d 1 ⫽ p t d2
a
Solve resulting 4th order equation numerically, or use a calculator and trial & error
T ⫽ 5700000 N⭈mm
d1 ⫽ 52 mm
16 T
f(d2) ⫽ d2 4 ⫺ a
b d ⫺ d1 4
p ta 2
ta ⫽ 43.2 MPa
gives
d2 ⫽ 91 mm
R-3.12: A motor drives a shaft with diameter d ⫽ 46 mm at f ⫽ 5.25 Hz
and delivers P ⫽ 25 kW of power. The maximum shear stress in the shaft is
approximately:
(A) 32 MPa
(B) 40 MPa
(C) 83 MPa
(D) 91 MPa
Solution
f ⫽ 5.25 Hz
d ⫽ 46 mm
P ⫽ 25 kW
p 4
5
4
Ip ⫽
d ⫽ 4.396 ⫻ 10 mm
32
Power in terms of torque T:
P ⫽ 2p f T
f
d
Solve for torque T:
T⫽
P
⫽ 757.9 N⭈m
2pf
P
Max. shear stress using torsion formula:
d
Ta b
2
tmax ⫽
⫽ 39.7 MPa
Ip
R-3.13: A motor drives a shaft at f ⫽ 10 Hz and delivers P ⫽ 35 kW of power.
The allowable shear stress in the shaft is 45 MPa. The minimum diameter of the
shaft is approximately:
(A) 35 mm
(B) 40 mm
(C) 47 mm
(D) 61 mm
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1114
APPENDIX A FE Exam Review Problems
Solution
f ⫽ 10 Hz
f
P ⫽ 35 kW
d
ta ⫽ 45 MPa
P
Power in terms of torque T:
P ⫽ 2pf T
Solve for torque T:
T⫽
P ⫽ 557.0 N⭈m
2pf
Shear formula:
d
Ta b
2
t⫽
p 4
d
32
t⫽
or
16 T
pd
3
1
Solve for diameter d:
16 T 3
d⫽a
b ⫽ 39.8 mm
p ta
R-3.14: A drive shaft running at 2500 rpm has outer diameter 60 mm and inner
diameter 40 mm. The allowable shear stress in the shaft is 35 MPa. The maximum
power that can be transmitted is approximately:
(A) 220 kW
(B) 240 kW
(C) 288 kW
(D) 312 kW
Solution
n ⫽ 2500 rpm
ta ⫽ 35 (106)
N
m2
d
n
d2 ⫽ 0.060 m
d1
d1 ⫽ 0.040 m
Ip ⫽
d2
p 4
(d2 ⫺ d14) ⫽ 1.021 ⫻ 10⫺6 m4
32
Shear formula:
d2
Ta b
2
t⫽
Ip
or
Tmax ⫽
2 ta Ip
d2
⫽ 1191.2 N⭈m
Power in terms of torque T:
P ⫽ 2p f T ⫽ 2p(n/60) T
2pn
5
Pmax ⫽ 60 Tmax ⫽ 3.119 ⫻ 10 W
Pmax ⫽ 312 kW
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1115
R-3.15: A prismatic shaft (diameter d0 ⫽ 19 mm ) is loaded by torque T1.
A stepped shaft (diameters d1 ⫽ 20 mm, d2 ⫽ 25 mm, radius R of fillets
⫽ 2 mm) is loaded by torque T2. The allowable shear stress in the material is 42
MPa. The ratio T1/T2 of the maximum permissible torques that can be applied to
the shafts, considering stress concentration effects in the stepped shaft is:
(A) 0.9
(B) 1.2
(C) 1.4
(D) 2.1
T1
d0
T1
D2
R
D1
T2
T2
FIG. 3-59 Stress-concentration factor K for a stepped shaft in torsion. (The
dashed line is for a full quarter-circular fillet.)
2.00
R
T
1.2
K
D2
1.1
tmax = Ktnom
1.5
1.50
D1
T
16T
tnom = ——
p D13
D2
—–
=
D1 2
D2 = D1 + 2R
1.00
0
0.20
0.10
R–
—
D1
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1116
APPENDIX A FE Exam Review Problems
Solution
Prismatic shaft
T1max ⫽
tallow IP
p d0 3
p
⫽ tallow a
b ⫽ 42 MPa c (19 mm)3 d ⫽ 56.6 N⭈m
d0
16
16
2
Stepped shaft
d2 25 mm
R
2 mm
⫽
⫽
⫽ 1.250
⫽ 0.100 so from graph (see Fig. 3-59)
d1 20 mm
d1 20 mm
K ⫽ 1.35
T2 max ⫽
tallow p d13
a
b
K
16
⫽
T1 max 56.6
⫽ 1.16
⫽
T2 max 48.9
42 MPa p
c (20 mm)3 d ⫽ 48.9 N⭈m
1.35
16
R-4.1: A simply supported beam with proportional loading (P ⫽ 4.1 kN) has
span length L ⫽ 5 m. Load P is 1.2 m from support A and load 2P is 1.5 m from
support B. The bending moment just left of load 2P is approximately:
(A) 5.7 kN⭈m
(B) 6.2 kN⭈m
(C) 9.1 kN⭈m
(D) 10.1 kN⭈m
Solution
a ⫽ 1.2 m
b ⫽ 2.3 m
L ⫽ a ⫹ b ⫹ c ⫽ 5.00 m
2P
P
c ⫽ 1.5 m
A
B
P ⫽ 4.1 kN
a
b
L
c
Statics to find reaction force at B:
RB ⫽
1
[P a ⫹ 2 P (a ⫹ b)] ⫽ 6.724 kN
L
Moment just left of load 2P:
M ⫽ RB c ⫽ 10.1 kN⭈m
⬍ compression on top of beam
R-4.2: A simply-supported beam is loaded as shown in the figure. The bending
moment at point C is approximately:
(A) 5.7 kN⭈m
(B) 6.1 kN⭈m
(C) 6.8 kN⭈m
(D) 9.7 kN⭈m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1117
Solution
7.5 kN
1.8 kN/m
C
A
1.0 m
B
0.5 m
1.0 m
3.0 m
5.0 m
Statics to find reaction force at A:
RA ⫽
1
kN (3 m ⫺ 0.5 m)2
cc1.8
d ⫹7.5 kN (3 m ⫹1 m)d ⫽7.125 kN
m
5m
2
Moment at point C, 2 m from A:
M ⫽ RA (2 m) ⫺ 7.5 kN (1.0⭈m) ⫽ 6.75 kN⭈m
⬍ compression
on top of beamR
R-4.3: A cantilever beam is loaded as shown in the figure. The bending moment
at 0.5 m from the support is approximately:
(A) 12.7 kN⭈m
(B) 14.2 kN⭈m
(C) 16.1 kN⭈m
(D) 18.5 kN⭈m
Solution
4.5 kN
1.8 kN/m
A
1.0 m
B
1.0 m
3.0 m
Cut beam at 0.5 m from support; use statics and right-hand FBD to find
internal moment at that point
M ⫽ 0.5 m (4.5 kN) ⫹ a0.5 m ⫹ 1.0 m ⫹
⫽ 18.5 kN⭈m
3.0 m
kN
b 1.8
(3.0 m)
m
2
(tension on top of beam)
R-4.4: An L-shaped beam is loaded as shown in the figure. The bending moment
at the midpoint of span AB is approximately:
(A) 6.8 kN⭈m
(B) 10.1 kN⭈m
(C) 12.3 kN⭈m
(D) 15.5 kN⭈m
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1118
APPENDIX A FE Exam Review Problems
Solution
4.5 kN
9 kN
1.0 m
A
B
5.0 m
C
1.0 m
Use statics to find reaction at B; sum moments about A
RB ⫽
1
[9 kN (6 m) ⫺ 4.5 kN (1. m)] ⫽ 9.90 kN
5m
Cut beam at midpoint of AB; use right hand FBD, sum moments
M ⫽ RB a
5m
5m
b ⫺ 9 kN a
⫹ 1 mb ⫽ 6.75 kN⭈m
2
2
⬍ tension on
top of beam
R-4.5: A T-shaped simple beam has a cable with force P anchored at B and
passing over a pulley at E as shown in the figure. The bending moment just left
of C is 1.25 kN⭈m. The cable force P is approximately:
(A) 2.7 kN
(B) 3.9 kN
(C) 4.5 kN
(D) 6.2 kN
Solution
MC ⫽ 1.25 kN⭈m
Sum moments about D to
find vertical reaction at A:
VA ⫽
⫺1
[P (4 m)]
7m
VA ⫽
⫺4
P
7
E
P
Cable
4m
A
B
C
D
(downward)
Now cut beam & cable just
left of CE & use left FBD;
2m
show VA downward & show
vertical cable force component
of (4/5)P upward at B; sum moments
at C to get MC and equate to given
numerical value of MC to find P:
3m
2m
4
MC ⫽ P (3) ⫹ VA (2 ⫹ 3)
5
4
⫺4
16 P
MC ⫽ P (3) ⫹ a Pb (2 ⫹ 3) ⫽ ⫺
5
7
35
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1119
Solve for P:
P⫽
35
(1.25) ⫽ 2.73 kN
16
R-4.6: A simple beam (L ⫽ 9 m) with attached bracket BDE has force P ⫽ 5 kN
applied downward at E. The bending moment just right of B is approximately:
(A) 6 kN⭈m
(B) 10 kN⭈m
(C) 19 kN⭈m
(D) 22 kN⭈m
Solution
Sum moments about A to find
reaction at C:
RC ⫽
B
A
C
1
L
L
P
cP a ⫹ b d ⫽
L
6
3
2
D
E
P
Cut through beam just
right of B, then use FBD of BC
to find moment at B:
L
—
6
L
—
3
L
—
2
L
L
5LP
L
MB ⫽ RC a ⫹ b ⫽
2
3
12
Substitute numbers for L and P:
L⫽9m
MB ⫽
P ⫽ 5 kN
5LP
⫽ 18.8 kN⭈ m
12
R-4.7: A simple beam AB with an overhang BC is loaded as shown in the figure.
The bending moment at the midspan of AB is approximately:
(A) 8 kN⭈m
(B) 12 kN⭈m
(C) 17 kN⭈m
(D) 21 kN⭈m
Solution
4.5 kN · m
15 kN/m
A
C
B
1.6 m
1.6 m
1.6 m
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1120
APPENDIX A FE Exam Review Problems
Sum moments about B to
get reaction at A:
RA ⫽
1
1.6
b ⫹ 4.5d ⫽ 19.40625 kN
c15 s1.6) a1.6 ⫹
3.2
2
Cut beam at midspan, use left FBD & sum moments to find moment at
midspan:
1.6
Mmspan ⫽ RA s1.6) ⫺ 15 s1.6) a b ⫽ 11.85 kN⭈m
2
R-5.1: A copper wire (d ⫽ 1.5 mm) is bent around a tube of radius R ⫽ 0.6 m.
The maximum normal strain in the wire is approximately:
(A) 1.25 ⫻ 10⫺3
(B) 1.55 ⫻ 10⫺3
(C) 1.76 ⫻ 10⫺3
(D) 1.92 ⫻ 10⫺3
Solution
␧max ⫽
d
2
d
d
R⫹
2
d ⫽ 1.5 mm
␧max ⫽
d
⫽
2 aR ⫹
d
b
2
R
R ⫽ 0.6 m
d
d
2 aR ⫹ b
2
⫽ 1.248 ⫻ 10⫺3
R-5.2: A simply supported wood beam (L ⫽ 5 m) with rectangular cross section
(b ⫽ 200 mm, h ⫽ 280 mm) carries uniform load q ⫽ 6.5 kN/m which includes
the weight of the beam. The maximum flexural stress is approximately:
(A) 8.7 MPa
(B) 10.1 MPa
(C) 11.4 MPa
(D) 14.3 MPa
Solution
L⫽5m
q ⫽ 9.5
b ⫽ 200 mm
h ⫽ 280 mm
kN
m
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1121
APPENDIX A FE Exam Review Problems
Section modulus:
q
2
S⫽
bh
⫽ 2.613 ⫻ 106 m3
6
A
h
B
Max. moment at midspan:
Mmax ⫽
q L2
⫽ 29.7 kN⭈m
8
b
L
Max. flexural stress at midspan:
smax ⫽
M max
⫽ 11.4 MPa
S
R-5.3: A cast iron pipe (L ⫽ 12 m, weight density ⫽ 72 kN/m3, d2 ⫽ 100 mm,
d1 ⫽ 75 mm) is lifted by a hoist. The lift points are 6 m apart. The maximum
bending stress in the pipe is approximately:
(A) 28 MPa
(B) 33 MPa
(C) 47 MPa
(D) 59 MPa
Solution
d1
d2
s
L
L ⫽ 12 m
s⫽4m
d2 ⫽ 100 mm
d1 ⫽ 75 mm
gCI ⫽ 72
kN
m3
Pipe cross sectional properties:
A⫽
p
p 2
sd2 ⫺ d12) ⫽ 3436 mm2 I ⫽ sd24 ⫺ d14) ⫽ 3.356 ⫻ 106 mm4
4
64
Uniformly distributed weight of pipe, q:
Vertical force at each lift point:
F⫽
q ⫽ gCI A ⫽ 0.247
kN
m
qL
⫽ 1.484 kN
2
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1122
APPENDIX A FE Exam Review Problems
Max. moment is either at lift points (M1) or at midspan (M2):
L⫺s L⫺s
ba
b ⫽ ⫺1.979 kN⭈m
M1 ⫽ ⫺q a
2
4
M2 ⫽ F
s
L L
⫺ q a b ⫽ ⫺1.484 kN⭈m
2
2 4
⬍ controls, tension on top
⬍ tension on top
d2
u M1 u a b
2
Max. bending stress at lift point: smax ⫽
⫽ 29.5 MPa
I
R-5.4: A beam with an overhang is loaded by a uniform load of 3 kN/m over its
entire length. Moment of inertia Iz ⫽ 3.36 ⫻ 106 mm4 and distances to top and
bottom of the beam cross section are 20 mm and 66.4 mm, respectively. It is
known that reactions at A and B are 4.5 kN and 13.5 kN, respectively. The
maximum bending stress in the beam is approximately:
(A) 36 MPa
(B) 67 MPa
(C) 102 MPa
(D) 119 MPa
Solution
3 kN/m
A
y
C
B
20 mm
z
4m
C
2m
Iz ⫽ 3.36 (106) mm4
RA ⫽ 4.5 kN
q⫽3
66.4 mm
kN
m
Location of max. positive moment in AB (cut beam at location of zero shear &
use left FBD):
x max
RA
⫽ 1.5 m
q
M pos ⫽ RA x max ⫺ 3
kN x max 2
⫽ 3.375 kN⭈m
m 2
⬍ compression on top of
beam
Compressive stress on top of beam at xmax:
sc1 ⫽
M pos (20 mm)
Iz
⫽ 20.1 MPa
Tensile stress at bottom of beam at xmax:
st1 ⫽
M pos (66.4 mm)
Iz
⫽ 66.696 MPa
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APPENDIX A FE Exam Review Problems
1123
Max. negative moment at B (use FBD of BC to find moment; compression on
bottom of beam):
M neg ⫽ a3
sc2 ⫽
st2 ⫽
kN (2 m)2
⫽ 6.000 kN⭈m
b
m
2
M neg (66.4 mm)
Iz
M neg (20 mm)
Iz
⫽ 118.6 MPa
⫽ 35.7 MPa
R-5.5: A steel hanger with solid cross section has horizontal force P ⫽ 5.5 kN
applied at free end D. Dimension variable b ⫽ 175 mm and allowable normal
stress is 150 MPa. Neglect self weight of the hanger. The required diameter of
the hanger is approximately:
(A) 5 cm
(B) 7 cm
(C) 10 cm
(D) 13 cm
Solution
P ⫽ 5.5 kN
b ⫽ 175 mm
␴a ⫽ 150 MPa
Reactions at support:
6b
A
B
NA ⫽ P
(leftward)
2b
D
MA ⫽ P (2 b) ⫽ 1.9 kN⭈m
(tension on bottom)
C
P
2b
Max. normal stress at bottom of cross section at A:
d
(2 P b) a b
2
P
⫹
smax ⫽
4
2
pd
pd
b
b
a
a
64
4
smax ⫽
4 P (16 b ⫹ d)
pd3
Set smax ⫽ sa and solve for required diameter d:
(␲sa)d3 ⫺ (4 P)d ⫺ 64Pb ⫽ 0
⬍ solve numerically or by trial &
error to find
dreqd ⫽ 5.11 cm
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1124
APPENDIX A FE Exam Review Problems
R-5.6: A cantilever wood pole carries force P ⫽ 300 N applied at its free end, as
well as its own weight (weight density ⫽ 6 kN/m3). The length of the pole is
L ⫽ 0.75 m and the allowable bending stress is 14 MPa. The required diameter
of the pole is approximately:
(A) 4.2 cm
(B) 5.5 cm
(C) 6.1 cm
(D) 8.5 cm
Solution
P ⫽ 300 N
L ⫽ 0.75 m
sa ⫽ 14 MPa
gw ⫽ 6
kN
m3
Uniformly distributed weight of pole:
p d2
b
w ⫽ gw a
4
A
B
d
Max. moment at support:
L
Mmax ⫽ P L ⫹ w L
2
P
L
Section modulus of pole cross section:
S⫽
I
d
a b
2
p d4
64
p d3
S⫽
⫽
32
d
a b
2
Set Mmax equal to sa ⫻ S and solve for required min. diameter d:
P L ⫹ cgw a
p d2
L
p d3
b d L ⫺ sa a
b⫽0
4
2
32
Or
a
p sa 3
p gw L2 2
bd ⫺a
b d ⫺ P L ⫽ 0 ⬍ solve numerically or by trial
32
8
& error to find
dreqd ⫽ 5.50 cm
Since wood pole is light, try simpler solution which ignores self weight:
PL ⫽ sa S
Or
p sa 3
b d ⫽ PL
a
32
1
32 3
dreqd ⫽ cP L a
b d ⫽ 5.47 cm
p sa
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APPENDIX A FE Exam Review Problems
1125
R-5.7: A simply supported steel beam of length L ⫽ 1.5 m and rectangular cross
section (h ⫽ 75 mm, b ⫽ 20 mm) carries a uniform load of q ⫽ 48 kN/m, which
includes its own weight. The maximum transverse shear stress on the cross section at 0.25 m from the left support is approximately:
(A) 20 MPa
(B) 24 MPa
(C) 30 MPa
(D) 36 MPa
Solution
kN
m
L ⫽ 1.5 m
q ⫽ 48
h ⫽ 75 mm
b ⫽ 20 mm
q
Cross section properties:
A ⫽ bh ⫽ 1500 mm2
h
h h
Q ⫽ ab b ⫽ 14062 mm3
2 4
L
b h3
I⫽
⫽ 7.031 ⫻ 105 mm4
12
b
Support reactions:
R⫽
qL
⫽ 36.0 kN
2
Transverse shear force at 0.25 m from left support:
V0.25 ⫽ R ⫺ q (0.25 m) ⫽ 24.0 kN
Max. shear stress at NA at 0.25 m from left support:
V0.25 Q
⫽ 24.0 MPa
Ib
3 V0.25
tmax ⫽
⫽ 24.0 MPa
2A
tmax ⫽
Or more simply . . .
R-5.8: A simply supported laminated beam of length L ⫽ 0.5 m and square cross
section weighs 4.8 N. Three strips are glued together to form the beam, with the
allowable shear stress in the glued joint equal to 0.3 MPa. Considering also the
weight of the beam, the maximum load P that can be applied at L/3 from the left
support is approximately:
(A) 240 N
(B) 360 N
(C) 434 N
(D) 510 N
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1126
APPENDIX A FE Exam Review Problems
Solution
q P at L/3
12 mm
12 mm 36 mm
12 mm
L
36 mm
L ⫽ 0.5 m
W ⫽ 4.8 N
h ⫽ 36 mm
b ⫽ 36 mm
q⫽
N
W
⫽ 9.60
m
L
ta ⫽ 0.3 MPa
Cross section properties:
h h h
Qjoint ⫽ ab b a ⫺ b ⫽ 5184 mm3
3 2 6
A ⫽ bh ⫽ 1296 mm2
I⫽
b h3
⫽ 1.400 ⫻ 105 mm4
12
Max. shear force at left support:
Vmax ⫽
2
qL
⫹ Pa b
2
3
Shear stress on glued joint at left support; set t ⫽ ta then solve for Pmax:
t⫽
Vmax Qjoint
ta ⫽
Ib
Or
t⫽
4 qL
2
c
⫹ P a bd
3bh 2
3
Vmax a
b h2
b
9
b h3
a bb
12
Or
ta ⫽
4 Vmax
3 bh
so for ta ⫽ 0.3 MPa
qL
3 3 b h ta
Pmax ⫽ a
⫺
b ⫽ 434 N
2
4
2
R-5.9: An aluminum cantilever beam of length L ⫽ 0.65 m carries a distributed
load, which includes its own weight, of intensity q/2 at A and q at B. The beam
cross section has width 50 mm and height 170 mm. Allowable bending stress is
95 MPa and allowable shear stress is 12 MPa. The permissible value of load
intensity q is approximately:
(A) 110 kN/m
(B) 122 kN/m
(C) 130 kN/m
(D) 139 kN/m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1127
Solution
L ⫽ 0.65 m
b ⫽ 50 mm
sa ⫽ 95 MPa
h ⫽ 170 mm
ta ⫽ 12 MPa
Cross section properties:
q
—
2
B
A
L
A ⫽ bh ⫽ 8500 mm2
b h3
⫽ 2.047 ⫻ 107 mm4
12
Reaction force and moment at A:
I⫽
1 q
RA ⫽ a ⫹ qb L
2 2
5
M A ⫽ q L2
12
q
S⫽
3
RA ⫽ q L
4
b h2
⫽ 2.408 ⫻ 105 mm3
6
q L 1 q 2L
MA ⫽ L ⫹
L
2 2 22 3
Compare max. permissible values of q based on shear and moment allowable
stresses; smaller value controls
tmax ⫽
3 RA
2 A
3
qL
3 4
ta ⫽ ±
≤
2
A
kN
8 ta A
⫽ 139
m
9 L
5
q L2
MA
12
sa ⫽
smax ⫽
S
S
12 sa S
kN
qmax2 ⫽
⫽ 130.0
m
5 L2
So, since ta ⫽ 12 MPa
qmax1 ⫽
So, since sa ⫽ 95 MPa
R-5.10: An aluminum light pole weighs 4300 N and supports an arm of weight
700 N, with arm center of gravity at 1.2 m left of the centroidal axis of the pole.
A wind force of 1500 N acts to the right at 7.5 m above the base. The pole cross
section at the base has outside diameter 235 mm and thickness 20 mm. The
maximum compressive stress at the base is approximately:
(A) 16 MPa
(B) 18 MPa
(C) 21 MPa
(D) 24 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1128
APPENDIX A FE Exam Review Problems
Solution
W2 = 700 N
H ⫽ 7.5 m
B ⫽ 1.2 m
W1 ⫽ 4300 N
W2 ⫽ 700 N
1.2 m
P1 ⫽ 1500 N
d2 ⫽ 235 mm
P1 = 1500 N
W1 = 4300 N
t ⫽ 20 mm
d1 ⫽ d2 ⫺ 2t ⫽ 195 mm
7.5 m
20 mm
z
y
Pole cross sectional properties at base:
x
p
A ⫽ (d2 2 ⫺ d1 2) ⫽ 13509 mm2
4
y
235 mm
x
p
I⫽
(d2 4 ⫺ d1 4) ⫽ 7.873 ⫻ 107 mm4
64
Compressive (downward) force at base of pole:
N ⫽ W1 ⫹ W2 ⫽ 5.0 kN
Bending moment at base of pole:
M ⫽ W2 B ⫺ P1 H ⫽ ⫺10.410 kN⭈m
⬍ results in compression at right
Compressive stress at right side at base of pole:
N
sc ⫽
⫹
A
d2
|M| a b
2
⫽ 15.9 MPa
I
R-5.11: Two thin cables, each having diameter d ⫽ t/6 and carrying tensile loads
P, are bolted to the top of a rectangular steel block with cross section dimensions
b ⫻ t. The ratio of the maximum tensile to compressive stress in the block due
to loads P is:
(A) 1.5
(B) 1.8
(C) 2.0
(D) 2.5
Solution
Cross section properties of block:
A ⫽ bt
bt3
I⫽
12
t
d⫽
6
b
P
P
t
Tensile stress at top of block:
P
st ⫽ ⫹
A
t
t
d
Pa ⫹ b a b
2
2 2
9P
⫽
I
2bt
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1129
Compressive stress at bottom of block:
P
sc ⫽ ⫺
A
t
t
d
Pa ⫹ b a b
2
2 2
5P
⫽ ⫺
I
2bt
Ratio of max. tensile to compressive stress in block:
ratio ⫽ `
st
9
`⫽
sc
5
9
⫽ 1.8
5
R-5.12:
A rectangular beam with semicircular notches has dimensions h ⫽ 160 mm and h 1 ⫽ 140 mm. The maximum allowable bending stress
in the plastic beam is s max ⫽ 6.5 MPa, and the bending moment is
M ⫽ 185 N⭈m. The minimum permissible width of the beam is:
(A) 12 mm
(B) 20 mm
(C) 28 mm
(D) 32 mm
2R
M
M
h
h1
3.0
2R
h
— = 1.2
h1
K
M
M
h
h1
2.5
h = h1 + 2R
1.1
2.0
s
K = s max s nom= 6M2
nom
bh 1
b = thickness
1.05
1.5
0
0.05
0.10
0.15
0.20
0.25
0.30
R
—
h1
FIG. 5-50 Stress-concentration factor K for a notched beam of rectangular
cross section in pure bending (h ⫽ height of beam; b ⫽ thickness of beam,
perpendicular to the plane of the figure). The dashed line is for semicircular
notches (h ⫽ h1 ⫹ 2R)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1130
APPENDIX A FE Exam Review Problems
Solution
R⫽
1
1
(h ⫺ h1) ⫽ (160 mm ⫺ 140 mm) ⫽ 10.000 mm
2
2
10
R
⫽
⫽ 0.071
h 1 140
h
160
⫽
⫽ 1.143
h 1 140
From Fig 5-50:
K ⫽ 2.25
sallow 6 M
⫽ 2
K
b h1
b min ⫽
so
3.0
6MK
6 (185 N⭈m) (2.25)
⫽
⫽ 19.6 mm
sallow h21
6.5 MPa C (140 mm)2D
2R
h
— = 1.2
h1
K
M
M
h
h1
2.5
h = h1 + 2R
K = 2.25
1.1
2.0
s
K = s max s nom= 6M2
nom
bh 1
b = thickness
1.05
1.5
0
0.05 0.071 0.10
0.15
0.20
0.25
0.30
R
—
h1
R-6.1: A composite beam is made up of a 200 mm ⫻ 300 mm core (Ec ⫽ 14 GPa)
and an exterior cover sheet (300 mm ⫻ 12 mm, Ee ⫽ 100 GPa) on each side.
Allowable stresses in core and exterior sheets are 9.5 MPa and 140 MPa, respectively. The ratio of the maximum permissible bending moment about the z-axis to
that about the y-axis is most nearly:
(A) 0.5
(B) 0.7
(C) 1.2
y
(D) 1.5
b ⫽ 200 mm
t ⫽ 12 mm
z
h ⫽ 300 mm
Ec ⫽ 14 GPa
Ee ⫽ 100 GPa
sac ⫽ 9.5 MPa
sae ⫽ 140 MPa
C
300 mm
Solution
200 mm
12 mm
12 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1131
Composite beam is symmetric about both axes so each NA is an axis of
symmetry
Moments of inertia of cross section about z and y axes:
b h3
h b3
⫽ 4.500 ⫻ 108 mm4
⫽ 2.000 ⫻ 108 mm4
Icy ⫽
12
12
2 t h3
Iez ⫽
⫽ 5.400 ⫻ 107 mm4
12
t 2
2 h t3
b
Iey ⫽
⫽ 2 (t h) a ⫹ b ⫽ 8.099 ⫻ 107 mm4
12
2
2
Icz ⫽
Bending about z axis based on allowable stress in each material (lesser value
controls)
Mmax_cz ⫽ sac
aEc Icz ⫹ Ee Iez b
⫽ 52.9 kN⭈m
Mmax_ez ⫽ sae
aEc Icz ⫹ Ee Iez b
⫽ 109.2 kN⭈m
h
Ec
2
h
Ee
2
Bending about y axis based on allowable stress in each material (lesser value
controls)
Mmax_cy ⫽ sac
Mmax_ey ⫽ sae
ratioz_to_y ⫽
(Ec Icy ⫹ Ee Iey)
b
Ec
2
(Ec Icy ⫹ Ee Iey)
b
a ⫹ tbEe
2
Mmax_cz
⫽ 0.72
Mmax_cy
⫽ 74.0 kN⭈m
⫽ 136.2kN⭈m
⬍ allowable stress in the core, not
exterior cover sheet, controls
moments about both axes
R-6.2: A composite beam is made up of a 90 mm ⫻ 160 mm wood beam (Ew ⫽
11 GPa) and a steel bottom cover plate (90 mm ⫻ 8 mm, Es ⫽ 190 GPa).
Allowable stresses in wood and steel are 6.5 MPa and 110 MPa, respectively.
The allowable bending moment about the z-axis of the composite beam is most
nearly:
(A) 2.9 kN?m
(B) 3.5 kN?m
(C) 4.3 kN?m
(D) 9.9 kN?m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1132
APPENDIX A FE Exam Review Problems
Solution
b ⫽ 90 mm
y
t ⫽ 8 mm
h ⫽ 160 mm
Ew ⫽ 11 GPa
Es ⫽ 190 GPa
160 mm
saw ⫽ 6.5 MPa
sas ⫽ 110 MPa
z
O
8
mm
Aw ⫽ bh ⫽ 14400 mm2
As ⫽ bt ⫽ 720 mm2
90 mm
Locate NA (distance h2 above base) by summing 1st moments of EA about
base of beam; then find h1 ⫽ dist. from NA to top of beam:
Es As
h2 ⫽
t
h
⫹ Ew Aw at ⫹ b
2
2
⫽ 49.07 mm
Es As ⫹ Ew Aw
h1 ⫽ h ⫹ t ⫺ h2 ⫽ 118.93 mm
Moments of inertia of wood and steel about NA:
Is ⫽
Iw ⫽
b t3
t 2
⫹ As ah2 ⫺ b ⫽ 1.467 ⫻ 106 mm4
12
2
b h3
h 2
⫹ Aw ah1 ⫺ b ⫽ 5.254 ⫻ 107 mm4
12
2
Allowable moment about z axis based on allowable stress in each material
(lesser value controls)
Mmax_w ⫽ saw
Mmax_s ⫽ sas
(Ew Iw ⫹ Es Is)
⫽ 4.26 kN⭈m
h1 Ew
(Ew Iw ⫹ Es Is)
⫽ 10.11 kN⭈m
h2 Es
R-6.3: A steel pipe (d3 ⫽ 104 mm, d2 ⫽ 96 mm) has a plastic liner with inner
diameter d1 ⫽ 82 mm. The modulus of elasticity of the steel is 75 times that of
the modulus of the plastic. Allowable stresses in steel and plastic are 40 MPa and
550 kPa, respectively. The allowable bending moment for the composite pipe is
approximately:
(A) 1100 N?m
(B) 1230 N?m
(C) 1370 N?m
(D) 1460 N?m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1133
Solution
d3 ⫽ 104 mm
d2 ⫽ 96 mm
d1 ⫽ 82 mm
sas ⫽ 40 MPa
y
sap ⫽ 550 kPa
Cross section properties:
z
p
(d3 2 ⫺ d2 2) ⫽ 1256.6 mm2
4
p
Ap ⫽ (d2 2 ⫺ d1 2) ⫽ 1957.2 mm2
4
p
Is ⫽
(d3 4 ⫺ d2 4) ⫽ 1.573 ⫻ 106 mm4
64
p
Ip ⫽
(d2 4 ⫺ d1 4) ⫽ 1.950 ⫻ 106 mm4
64
C
As ⫽
d1
d2 d3
Due to symmetry, NA of composite beam is the z axis
Allowable moment about z axis based on allowable stress in each material
(lesser value controls)
Mmax_s ⫽ sas
(Ep Ip ⫹ Es Is)
Modular ratio:
Mmax_p ⫽ sap
d3
a b Es
2
n⫽
Es
n ⫽ 75
Ep
(Ep Ip ⫹ Es Is)
d2
a b Ep
2
Divide through by Ep in moment expressions above
Mmax_s ⫽ sas
(Ip ⫹ nIs)
Mmax_ p ⫽ sap
d3
a bn
2
⫽ 1230 N⭈m
(Ip ⫹ nIs)
d2
a b
2
⫽ 1374 N⭈m
R-6.4: A bimetallic beam of aluminum (Ea ⫽ 70 GPa) and copper (Ec ⫽ 110 GPa)
strips has width b ⫽ 25 mm; each strip has thickness t ⫽ 1.5 mm. A bending
moment of 1.75 N?m is applied about the z axis. The ratio of the maximum stress in
the aluminum to that in the copper is approximately:
(A) 0.6
(B) 0.8
(C) 1.0
(D) 1.5
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1134
APPENDIX A FE Exam Review Problems
Solution
b ⫽ 25 mm t ⫽ 1.5 mm Aa ⫽ b t ⫽ 37.5 mm2 Ac ⫽ Aa ⫽ 37.5 mm2
M ⫽ 1.75 N⭈m
Ea ⫽ 70 GPa
y
Ec ⫽ 110 GPa
t
Equate 1st moments of EA
about bottom of beam to locate NA
(distance h2 above base); then find
h1 ⫽ dist. from NA to top of beam:
Ec Ac
h2 ⫽
A
z
O
C
b
t
t
t
⫹ Ea Aa at ⫹ b
2
2
⫽ 1.333 mm
Ec Ac ⫹ Ea Aa
h1 ⫽ 2t ⫺ h2 ⫽ 1.667 mm
h1 ⫹ h2 ⫽ 3.000 mm
2 t ⫽ 3.000 mm
Moments of inertia of aluminum and copper strips about NA:
Ic ⫽
Ia ⫽
bt 3
t 2
⫹ Ac ah2 ⫺ b ⫽ 19.792 mm4
12
2
bt3
t 2
⫹ Aa ah1 ⫺ b ⫽ 38.542 mm4
12
2
Bending stresses in aluminum and copper:
sa ⫽
Mh1 Ea
⫽ 41.9 MPa
Ea Ia ⫹ Ec Ic
sc ⫽
Mh2 Ec
⫽ 52.6 MPa
Ea Ia ⫹ Ec Ic
sa
⫽ 0.795
sc
Ratio of the stress in the aluminum to that of the copper:
R-6.5: A composite beam of aluminum (Ea ⫽ 72 GPa) and steel (Es ⫽ 190 GPa)
has width b ⫽ 25 mm and heights ha ⫽ 42 mm, hs ⫽ 68 mm. A bending moment
is applied about the z axis resulting in a maximum stress in the aluminum of
55 MPa. The maximum stress in the steel is approximately:
(A) 86 MPa
(B) 90 MPa
(C) 94 MPa
(D) 98 MPa
y
Aluminum
Solution
b ⫽ 25 mm
ha ⫽ 42 mm
Ea ⫽ 72 GPa
Es ⫽ 190 GPa
ha
hs ⫽ 68 mm
sa ⫽ 55 MPa
Steel
z
O hs
Aa ⫽ bha ⫽ 1050.0 mm2
As ⫽ bhs ⫽ 1700.0 mm2
b
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APPENDIX A FE Exam Review Problems
1135
Locate NA (distance h2 above base) by summing 1st moments of EA about
base of beam; then find h1 ⫽ dist. from NA to top of beam:
Es As
h2 ⫽
hs
ha
⫹ Ea Aa ahs ⫹ b
2
2
⫽ 44.43 mm
Ea Aa ⫹ Es As
h1 ⫽ ha ⫹ hs ⫺ h2 ⫽ 65.57 mm
h1 ⫹ h2 ⫽ 110.00 mm
Moments of inertia of aluminum and steel parts about NA:
Is ⫽
Ia ⫽
b hs3
hs 2
⫹ As ah2 ⫺ b ⫽ 8.401 ⫻ 105 mm4
12
2
b ha3
ha 2
⫹ Aa ah1 ⫺ b ⫽ 2.240 ⫻ 106 mm4
12
2
Set max. bending stress in aluminum to given value then solve for moment M:
M⫽
sa (Ea Ia ⫹ Es Is)
⫽ 3.738 kN⭈m
h1Ea
Use M to find max. bending stress in steel: ss ⫽
M h2 Es
⫽ 98.4 MPa
Ea Ia ⫹ Es Is
R-7.1: A rectangular plate (a ⫽ 120 mm, b ⫽ 160 mm) is subjected to compressive stress sx ⫽ ⫺ 4.5 MPa and tensile stress sy ⫽ 15 MPa. The ratio of the
normal stress acting perpendicular to the weld to the shear stress acting along the
weld is approximately:
(A) 0.27
(B) 0.54
(C) 0.85
(D) 1.22
Solution
a ⫽ 120 mm
a
u ⫽ arctana b ⫽ 36.87⬚
b
sx ⫽ ⫺4.5 MPa
σy
b ⫽ 160 mm
sy ⫽ 15 MPa
ld
We
a
b
σx
txy ⫽ 0
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1136
APPENDIX A FE Exam Review Problems
Plane stress transformation: normal and shear stresses on y-face of element
rotated through angle u (perpendicular to & along weld seam):
su ⫽
sx ⫹ sy
2
tu ⫽ ⫺
`
⫹
sx ⫺ sy
2
sx ⫺ sy
2
cosc2 au ⫹
sinc2 au ⫹
su
` ⫽ 0.85
tu
p
p
b d ⫹ txy sinc2 au ⫹ b d ⫽ 7.98 MPa
2
2
p
p
b d ⫹ txy cosc2 au ⫹ b d ⫽ ⫺9.36 MPa
2
2
R-7.2: A rectangular plate in plane stress is subjected to normal stresses sx and
sy and shear stress txy. Stress sx is known to be 15 MPa but sy and txy are
unknown. However, the normal stress is known to be 33 MPa at counterclockwise angles of 35° and 75° from the x axis. Based on this, the normal stress sy
on the element below is approximately:
(A) 14 MPa
(B) 21 MPa
(C) 26 MPa
(D) 43 MPa
Solution
sx ⫽ 15 MPa
s35 ⫽ 33 MPa
s75 ⫽ s35
Plane stress transformations for 35⬚ & 75⬚:
su ⫽
sx ⫹ sy
2
sx ⫺ sy
⫹
2
cos(2 u) ⫹ txy sin(2 u)
y
σy
τxy
σx
O
x
For u ⫽ 35⬚:
sx ⫹ sy
2
Or
⫹
u35 ⫽ 35⬚
sx ⫺ sy
2
cos[2 (u35)] ⫹ txy sin[2 (u35)] ⫽ s35
sy ⫹ 2.8563txy ⫽ 69.713
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
And for u ⫽ 75⬚:
sx ⫹ sy
2
⫹
1137
u75 ⫽ 75⬚
sx ⫺ sy
2
cos[2 (u75)] ⫹ txy sin[2 (u75)] ⫽ s75
sy ⫹ 0.5359 txy ⫽ 34.292
Or
Solving above two equations for sy and txy gives:
sy
1 2.8563 ⫺1 69.713
26.1
b⫽c
d a
b⫽a
b MPa
txy
1 0.5359
34.292
15.3
a
so ␴y ⫽ 26.1 MPa
R-7.3: A rectangular plate in plane stress is subjected to normal stresses sx ⫽
35 MPa, sy ⫽ 26 MPa, and shear stress txy ⫽ 14 MPa. The ratio of the magnitudes of the principal stresses (s1/s2) is approximately:
(A) 0.8
(B) 1.5
(C) 2.1
(D) 2.9
Solution
y
sx ⫽ 35 MPa
sy ⫽ 26 MPa
txy ⫽ 14 MPa
σy
Principal angles:
uP1 ⫽
2txy
1
arctana
b ⫽ 36.091⬚
s
2
x ⫺ sy
uP2 ⫽ uP1 ⫹
τxy
O
σx
p
⫽ 126.091⬚
2
x
Plane stress transformations:
s1 ⫽
s2 ⫽
sx ⫹ sy
2
sx ⫹ sy
2
⫹
⫹
sx ⫹ sy
2
sx ⫺ sy
2
cos(2uP1) ⫹ txy sin(2uP1) ⫽ 45.21 MPa
cos(2uP2) ⫹ txy sin(2uP2) ⫽ 15.79 MPa
Ratio of principal stresses:
s1
⫽ 2.86
s2
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1138
APPENDIX A FE Exam Review Problems
R-7.4: A drive shaft resists torsional shear stress of 45 MPa and axial compressive stress of 100 MPa. The ratio of the magnitudes of the principal stresses
(s1/s2) is approximately:
(A) 0.15
(B) 0.55
(C) 1.2
(D) 1.9
Solution
sx ⫽ ⫺100 MPa
sy ⫽ 0
txy ⫽ ⫺45 MPa
Principal angles:
2txy
1
arctana
b ⫽ 20.994⬚
s
2
x ⫺ sy
p
uP2 ⫽ uP1 ⫹ ⫽ 110.994⬚
2
uP1 ⫽
100 MPa
45 MPa
Plane stress transformations:
suP1 ⫽
sx ⫹ sy
2
⬍ actually s2
suP2 ⫽
sx ⫹ sy
2
⫹
⫹
sx ⫺ sy
2
sx ⫺ sy
2
cos(2uP1) ⫹ txy sin(2uP1) ⫽ ⫺117.27 MPa
cos(2uP2) ⫹ txy sin(2uP2) ⫽ 17.27 MPa
⬍ this is s1
So
s1 ⫽ max(suP1, suP2) ⫽ 17.268 MPa s2 ⫽ min(suP1, suP2) ⫽ ⫺117.268 MPa
Ratio of principal stresses:
`
s1
` ⫽ 0.15
s2
R-7.5: A drive shaft resists torsional shear stress of 45 MPa and axial compressive stress of 100 MPa. The maximum shear stress is approximately:
(A) 42 MPa
(B) 67 MPa
(C) 71 MPa
(D) 93 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1139
Solution
sx ⫽ ⫺100 MPa
sy ⫽ 0
txy ⫽ ⫺45 MPa
Max. shear stress:
tmax ⫽
sx ⫺ sy 2
a
b ⫹ txy2 ⫽ 67.3 MPa
B
2
100 MPa
45 MPa
R-7.6: A drive shaft resists torsional shear stress of txy ⫽ 40 MPa and axial compressive stress sx ⫽ ⫺70 MPa. One principal normal stress is known to be 38 MPa
(tensile). The stress sy is approximately:
(A) 23 MPa
(B) 35 MPa
(C) 62 MPa
(D) 75 MPa
Solution
sx ⫽ ⫺70 MPa
txy ⫽ 40 MPa
sy is unknown
sprin ⫽ 38 MPa
y
σy
τxy
Stresses sx and sy must be smaller than
the given principal stress so:
σx
O
x
s1 ⫽ sprin
Substitute into stress transformation equation
and solve for sy:
sx ⫹ sy
2
⫹
sx ⫺ sy 2
626 MPa
b ⫹ txy2 ⫽ s1 solve, sy ⫽
B
2
27
⫽ 23.2 MPa
a
R-7.7: A cantilever beam with rectangular cross section (b ⫽ 95 mm, h ⫽ 300 mm)
supports load P ⫽ 160 kN at its free end. The ratio of the magnitudes of the principal
stresses (s1/s2) at point A (at distance c ⫽ 0.8 m from the free end and distance
d ⫽ 200 mm up from the bottom) is approximately:
(A) 5
(B) 12
(C) 18
(D) 25
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1140
APPENDIX A FE Exam Review Problems
Solution
P ⫽ 160 kN
b ⫽ 95 mm
c ⫽ 0.8 m
d ⫽ 200 mm
h ⫽ 300 mm
d
⫽ 0.667
h
P
Cross section properties:
A ⫽ bh ⫽ 28500 mm2
I⫽
b h3
⫽ 2.138 ⫻ 108 mm4
12
QA ⫽ [b (h ⫺ d)] c
A
h
c
b
d
h (h ⫺ d)
⫺
d ⫽ 9.500 ⫻ 105 mm3
2
2
Moment, shear force and normal and shear stresses at A:
MA ⫽ ⫺Pc ⫽ ⫺1.280 ⫻ 105 kN⭈mm
VA QA
tA ⫽
⫽ 7.485 MPa
Ib
h
⫺MA ad ⫺ b
2
⫽ 29.942 MPa
sA ⫽
I
sx ⫽ sA
Plane stress state at A:
VA ⫽ P
txy ⫽ tA
sy ⫽ 0
Principal stresses:
uP ⫽
s1 ⫽
s2 ⫽
2 txy
1
b ⫽ 13.283⬚
arctana
sx ⫺ sy
2
sx ⫹ sy
2
sx ⫹ sy
2
⫹
⫹
sx ⫺ sy 2
b ⫹ txy2 ⫽ 31.709 MPa
B
2
a
sx ⫺ sy 2
b ⫹ txy2 ⫽ ⫺1.767 MPa
B
2
a
Ratio of principal stresses (s1 / s2):
s1
` s ` ⫽ 17.9
2
R-7.8: A simply supported beam (L ⫽ 4.5 m) with rectangular cross section
(b ⫽ 95 mm, h ⫽ 280 mm) supports uniform load q ⫽ 25 kN/m. The ratio of
the magnitudes of the principal stresses (s1/s2) at a point a ⫽ 1.0 m from the
left support and distance d ⫽ 100 mm up from the bottom of the beam is
approximately:
(A) 9
(B) 17
(C) 31
(D) 41
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APPENDIX A FE Exam Review Problems
1141
Solution
kN
m
L ⫽ 4.5 m
b ⫽ 95 mm
h ⫽ 280 mm
a ⫽ 1.0 m
d ⫽ 100 mm
q ⫽ 25
q
h
b
a
L
Cross section properties:
A ⫽ bh ⫽ 26600 mm2
I⫽
b h3
⫽ 1.738 ⫻ 108 mm4
12
Q ⫽ [b (h ⫺ d)] c
h (h ⫺ d)
⫺
d ⫽ 8.550 ⫻ 105 mm3
2
2
Moment, shear force and normal and shear stresses at distance a from left
support:
Va ⫽
q a2
qL
qL
⫺ q a ⫽ 31.250 kN Ma ⫽
a⫺
⫽ 4.375 ⫻ 104 kN⭈mm
2
2
2
Va Q
t⫽
⫽ 1.618 MPa
Ib
Plane stress state: sx ⫽ s
h
⫺Ma ad ⫺ b
2
s⫽
⫽ 10.070 MPa
I
txy ⫽ t
sy ⫽ 0
Principal stresses:
uP ⫽
s1 ⫽
s2 ⫽
2 txy
1
arctana
b ⫽ 8.909⬚
s
2
x ⫺ sy
sx ⫹ sy
2
sx ⫹ sy
2
⫹
⫹
sx ⫺ sy 2
b ⫹ txy2 ⫽ 10.324 MPa
B
2
a
sx ⫺ sy 2
b ⫹ txy2 ⫽ ⫺0.254 MPa
B
2
a
Ratio of principal stresses (s1 / s2):
`
s1
` ⫽ 40.7
s2
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1142
APPENDIX A FE Exam Review Problems
R-8.1: A thin wall spherical tank of diameter 1.5 m and wall thickness 65 mm
has internal pressure of 20 MPa. The maximum shear stress in the wall of the
tank is approximately:
(A) 58 MPa
(B) 67 MPa
(C) 115 MPa
(D) 127 MPa
Solution
d ⫽ 1.5 m
t ⫽ 65 mm
Weld
t
⫽ 0.087
d
a b
2
Thin wall tank since:
Biaxial stress:
d
pa b
2
s⫽
2t
p ⫽ 20 MPa
␴ ⫽ 115.4 MPa
Max. shear stress at 45 deg. rotation is 1/2 of s
tmax ⫽
s
⫽ 57.7 MPa
2
R-8.2: A thin wall spherical tank of diameter 0.75 m has internal pressure
of 20 MPa. The yield stress in tension is 920 MPa, the yield stress in shear is
475 MPa, and the factor of safety is 2.5. The modulus of elasticity is
210 GPa, Poisson’s ratio is 0.28, and maximum normal strain is 1220 ⫻
10⫺6. The minimum permissible thickness of the tank is approximately:
(A) 8.6 mm
(B) 9.9 mm
(C) 10.5 mm
(D) 11.1 mm
Solution
d ⫽ 0.75 m
p ⫽ 20 MPa
sY ⫽ 920 MPa
␯ ⫽ 0.28
E ⫽ 210 GPa
tY ⫽ 475 MPa
FSY ⫽ 2.5
⫺6
␧a ⫽ 1220(10 )
Weld
Thickness based on tensile stress:
d
pa b
2
⫽ 10.190 mm
t1 ⫽
sY
2a
b
FSY
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APPENDIX A FE Exam Review Problems
1143
Thickness based on shear stress:
d
pa b
2
t2 ⫽
⫽ 9.868 mm
tY
4a
b
FSY
Thickness based on normal strain:
d
pa b
2
t3 ⫽
(1 ⫺ n)
2 ␧a E
t3 ⫽ 10.54 mm
⬍ largest value controls
R-8.3: A thin wall cylindrical tank of diameter 200 mm has internal pressure of
11 MPa. The yield stress in tension is 250 MPa, the yield stress in shear is
140 MPa, and the factor of safety is 2.5. The minimum permissible thickness of
the tank is approximately:
(A) 8.2 mm
(B) 9.1 mm
(C) 9.8 mm
(D) 11.0 mm
Solution
d ⫽ 200 mm
sY ⫽ 250 MPa
p ⫽ 11 MPa
tY ⫽ 140 MPa
FSY ⫽ 2.5
Wall thickness based on tensile stress:
d
pa b
2
⫽ 11.00 mm
t1 ⫽
sY
FSY
⬍ larger value governs
Wall thickness based on shear stress:
d
pa b
2
⫽ 9.821 mm
t2 ⫽
tY
2a
b
FSY
t1
⫽ 0.110
d
a b
2
t2
⫽ 0.098
d
a b
2
R-8.4: A thin wall cylindrical tank of diameter 2.0 m and wall thickness 18 mm
is open at the top. The height h of water (weight density ⫽ 9.81 kN/m3) in the
tank at which the circumferential stress reaches 10 MPa in the tank wall is
approximately:
(A) 14 m
(B) 18 m
(C) 20 m
(D) 24 m
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1144
APPENDIX A FE Exam Review Problems
Solution
d⫽2m
t ⫽ 18 mm
Pressure at height h:
sa ⫽ 10 MPa
gw ⫽ 9.81
kN
m3
d
ph ⫽ gw h
d
ph a b
2
Circumferential stress: sc ⫽
t
d
(gw h) a b
2
sc ⫽
t
h
Set sc equal to sa and solve for h:
h⫽
sa t
⫽ 18.3 m
d
(gw) a b
2
R-8.5: The pressure relief valve is opened on a thin wall cylindrical tank, with
radius to wall thickness ratio of 128, thereby decreasing the longitudinal strain
by 150 ⫻ 10⫺6. Assume E ⫽ 73 GPa and v ⫽ 0.33. The original internal pressure in the tank was approximately:
(A) 370 kPa
(B) 450 kPa
(C) 500 kPa
(D) 590 kPa
Solution
rt ⫽
r
t
rt ⫽ 128
␧L ⫽ 148 (10⫺ 6)
E ⫽ 73 GPa
n ⫽ 0.33
strain gage
Longitudinal strain:
␧⫽
p r
a b (1 ⫺ 2␯)
2E t
Set ␧ to ␧L and solve for pressure p:
p⫽
2 E ␧L
⫽ 497 kPa
rt (1 ⫺ 2n)
R-8.6: A cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is
2.0 MPa. The maximum stress in the heads of the tank is approximately:
(A) 38 MPa
(B) 45 MPa
(C) 50 MPa
(D) 59 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1145
Solution
d ⫽ 1.5 m
t ⫽ 20 mm
p ⫽ 2.0 MPa
Welded seams
d
pa b
2
sh ⫽
⫽ 37.5 MPa
2t
R-8.7: A cylindrical tank is assembled by welding steel sections circumferentially.
Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa. The
maximum tensile stress in the cylindrical part of the tank is approximately:
(A) 45 MPa
(B) 57 MPa
(C) 62 MPa
(D) 75 MPa
Solution
d ⫽ 1.5 m
t ⫽ 20 mm
p ⫽ 2.0 MPa
Welded seams
d
pa b
2
sc ⫽
⫽ 75.0 MPa
t
R-8.8: A cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa.
The maximum tensile stress perpendicular to the welds is approximately:
(A) 22 MPa
(B) 29 MPa
(C) 33 MPa
(D) 37 MPa
Solution
d ⫽ 1.5 m
t ⫽ 20 mm
d
pa b
2
sw ⫽
⫽ 37.5 MPa
2t
p ⫽ 2.0 MPa
Welded seams
R-8.9: A cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is
2.0 MPa. The maximum shear stress in the heads is approximately:
(A) 19 MPa
(B) 23 MPa
(C) 33 MPa
(D) 35 MPa
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1146
APPENDIX A FE Exam Review Problems
Solution
d ⫽ 1.5 m
t ⫽ 20 mm
p ⫽ 2.0 MPa
Welded seams
d
pa b
2
th ⫽
⫽ 18.8 MPa
4t
R-8.10: A cylindrical tank is assembled by welding steel sections circumferentially.
Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa. The
maximum shear stress in the cylindrical part of the tank is approximately:
(A) 17 MPa
(B) 26 MPa
(C) 34 MPa
(D) 38 MPa
Solution
d ⫽ 1.5 m
t ⫽ 20 mm
p ⫽ 2.0 MPa
Welded seams
d
pa b
2
tmax ⫽
⫽ 37.5 MPa
2t
R-8.11: A cylindrical tank is assembled by welding steel sections in a helical
pattern with angle a ⫽ 50 degrees. Tank diameter is 1.6 m, thickness is 20 mm,
and internal pressure is 2.75 MPa. Modulus E ⫽ 210 GPa and Poisson’s ratio
n ⫽ 0.28. The circumferential strain in the wall of the tank is approximately:
(A) 1.9 ⫻ 10⫺4
(B) 3.2 ⫻ 10⫺4
(C) 3.9 ⫻ 10⫺4
(D) 4.5 ⫻ 10⫺4
Solution
d ⫽ 1.6 m
E ⫽ 210 GPa
t ⫽ 20 mm
n ⫽ 0.28
p ⫽ 2.75 MPa
a ⫽ 50⬚
Circumferential stress:
d
pa b
2
sc ⫽
⫽ 110.0 MPa
t
Helical weld
α
Circumferential strain:
␧c ⫽
sc
(2 ⫺ n) ⫽ 4.50 ⫻ 10⫺4
2E
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1147
R-8.12: A cylindrical tank is assembled by welding steel sections in a helical
pattern with angle a ⫽ 50 degrees. Tank diameter is 1.6 m, thickness is
20 mm, and internal pressure is 2.75 MPa. Modulus E ⫽ 210 GPa and
Poisson’s ratio n ⫽ 0.28. The longitudinal strain in the the wall of the tank is
approximately:
(A) 1.2 ⫻ 10⫺4
(B) 2.4 ⫻ 10⫺4
(C) 3.1 ⫻ 10⫺4
(D) 4.3 ⫻ 10⫺4
Solution
d ⫽ 1.6 m
t ⫽ 20 mm
E ⫽ 210 GPa
n ⫽ 0.28
p ⫽ 2.75 MPa
a ⫽ 50⬚
Longitudinal stress:
Helical weld
d
pa b
2
sL ⫽
⫽ 55.0 MPa
2t
α
Longitudinal strain:
␧L ⫽
sL
(1 ⫺ 2n) ⫽ 1.15 ⫻ 10⫺4
E
R-8.13: A cylindrical tank is assembled by welding steel sections in a helical
pattern with angle a ⫽ 50 degrees. Tank diameter is 1.6 m, thickness is
20 mm, and internal pressure is 2.75 MPa. Modulus E ⫽ 210 GPa and
Poisson’s ratio n ⫽ 0.28. The normal stress acting perpendicular to the weld
is approximately:
(A) 39 MPa
(B) 48 MPa
(C) 78 MPa
(D) 84 MPa
Solution
d ⫽ 1.6 m
t ⫽ 20 mm
p ⫽ 2.75 MPa
E ⫽ 210 GPa
␯ ⫽ 0.28 a ⫽ 50⬚
Helical weld
α
Longitudinal stress:
d
pa b
2
⫽ 55.0 MPa So sx ⫽ sL
sL ⫽
2t
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1148
APPENDIX A FE Exam Review Problems
Circumferential stress:
d
pa b
2
sc ⫽
⫽ 110.0 MPa so
t
sy ⫽ sc
u ⫽ 90⬚ ⫺ a ⫽ 40.000⬚
Angle perpendicular to the weld:
Normal stress perpendicular to the weld:
s40 ⫽
sx ⫹ sy
2
⫹
sx ⫺ sy
2
cos (2 u) ⫽ 77.7 MPa
R-8.14: A segment of a drive shaft (d2 ⫽ 200 mm, d1 ⫽ 160 mm) is subjected
to a torque T ⫽ 30 kN?m. The allowable shear stress in the shaft is 45 MPa. The
maximum permissible compressive load P is approximately:
(A) 200 kN
(B) 286 kN
(C) 328 kN
(D) 442 kN
Solution
d2 ⫽ 200 mm
d1 ⫽ 160 mm
ta ⫽ 45 MPa
P
T ⫽ 30 kN⭈m
Cross section properties:
T
p 2
(d2 ⫺ d12) ⫽ 11310 mm2
4
p 4
Ip ⫽
(d2 ⫺ d14) ⫽ 9.274 ⫻ 107 mm4
32
A⫽
T
Normal and in-plane shear stresses:
sx ⫽ 0
sy ⫽
⫺P
A
txy ⫽
d2
Ta b
2
IP
⫽ 32.349 MPa
P
Maximum in-plane shear stress: set ␶max ⫽ ␶allow then solve for sy
t max ⫽
sx ⫺ sy 2
a
b ⫹ txy2
2
B
Finally solve for P ⫽ sy A:
so sy ⫽ #4 (ta ⫺ txy)2 ⫽ 25.303 MPa
Pmax ⫽ ␴y A ⫽ 286 kN
R-8.15: A thin walled cylindrical tank, under internal pressure p, is compressed
by a force F ⫽ 75 kN. Cylinder diameter is d ⫽ 90 mm and wall thickness
t ⫽ 5.5 mm. Allowable normal stress is 110 MPa and allowable shear stress is
60 MPa. The maximum allowable internal pressure pmax is approximately:
(A) 5 MPa
(B) 10 MPa
(C) 13 MPa
(D) 17 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1149
Solution
d ⫽ 90 mm
t ⫽ 5.5 mm
F ⫽ 75 kN
A ⫽ 2p
sa ⫽ 110 MPa
d
t ⫽ 1555 mm2
2
d
pmax a b
2
sc ⫽
t
F
F
Circumferential normal stress:
and setting sc ⫽ sa and solving for pmax:
2t
pmaxc ⫽ sa a b ⫽ 13.4 MPa
d
⬍ controls
Longitudinal normal stress:
d
pmax a b
2
F
⫺
sL ⫽
2t
A
or
sL ⫽
pmax d F
⫺
4t
A
So set sL ⫽ sa and solve for pmax:
F 4t
pmaxL ⫽ asa ⫹ b
⫽ 38.7 MPa
A d
Check also in-plane & out-of-plane shear stresses: all are below allowable
shear stress so circumferential normal stress controls as noted above.
R-9.1: An aluminum beam (E ⫽ 72 GPa) with a square cross section and
span length L ⫽ 2.5 m is subjected to uniform load q ⫽ 1.5 kN/m. The allowable bending stress is 60 MPa. The maximum deflection of the beam is
approximately:
(A) 10 mm
(B) 16 mm
(C) 22 mm
(D) 26 mm
Solution
E ⫽ 72 (103)MPa
q ⫽ 1.5
sa ⫽ 60 MPa
q = 1.5 kN/m
N
mm
L ⫽ 2500 mm
Max. moment and deflection at L/2:
Mmax ⫽
q L2
8
dmax ⫽
L = 2.5 m
5 q L4
384 E I
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1150
APPENDIX A FE Exam Review Problems
Moment of inertia and section modulus for square cross section (height ⫽
width ⫽ b)
I⫽
b4
12
S⫽
Flexure formula
I
b3
⫽
6
b
a b
2
qL2
8
smax ⫽ 3
b
a b
6
Mmax
S
smax ⫽
so
b3 ⫽
3 qL2
4 smax
Max. deflection formula
dmax ⫽
dmax ⫽
5q L4
so
b4
384 E a b
12
5q L4
2
3 qL
ca
smax
4
384 E ≥
12
solve for dmax if smax ⫽ sa
⫽ 22.2 mm
1
b 3d
4
¥
R-9.2: An aluminum cantilever beam (E ⫽ 72 GPa) with a square cross section
and span length L ⫽ 2.5 m is subjected to uniform load q ⫽ 1.5 kN/m. The allowable bending stress is 55 MPa. The maximum deflection of the beam is approximately:
(A) 10 mm
(B) 20 mm
(C) 30 mm
(D) 40 mm
Solution
E ⫽ 72(103) MPa
q ⫽ 1.5
sa ⫽ 55 MPa
q
N
mm
L ⫽ 2500 mm
L
Max. moment at support & max. deflection at L:
Mmax ⫽
q L2
q L4
dmax ⫽
2
8EI
Moment of inertia and section modulus for square cross section (height ⫽
width ⫽ b)
I⫽
b4
12
S⫽
b3
I
⫽
6
b
a b
2
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APPENDIX A FE Exam Review Problems
1151
Flexure formula
q L2
2
smax ⫽ 3
b
a b
6
Mmax
S
smax ⫽
so
b3 ⫽ 3
q L2
smax
Max. deflection formula
dmax ⫽
dmax ⫽
q L4
b4
8E a b
12
so solve for dmax if smax ⫽ sa
q L4
⫽ 29.9 mm
q L2 1 4
b 3d
ca3
smax
8E ≥
¥
12
R-9.3: A steel beam (E ⫽ 210 GPa) with I ⫽ 119 ⫻ 106 mm4 and span length
L ⫽ 3.5 m is subjected to uniform load q ⫽ 9.5 kN/m. The maximum deflection
of the beam is approximately:
(A) 10 mm
(B) 13 mm
(C) 17 mm
(D) 19 mm
Solution
E ⫽ 210(103) MPa
q ⫽ 9.5
I ⫽ 119(106) mm4
⬍ strong axis I for W310⫻52
N
mm
L ⫽ 3500 mm
y
MA
q
A
L
x
B
k = 48EI/L3
RB = kδB
Max. deflection at A by superposition of SS beam mid-span deflection & RB/k:
dmax ⫽
5 q (2 L)4
(q L)
⫹
⫽ 13.07 mm
384 E I
48 E I
a 3 b
L
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1152
APPENDIX A FE Exam Review Problems
R-9.4: A steel bracket ABC (EI ⫽ 4.2 ⫻ 106 N?m2) with span length L ⫽ 4.5 m
and height H ⫽ 2 m is subjected to load P ⫽ 15 kN at C. The maximum rotation of joint B is approximately:
(A) 0.1⬚
(B) 0.3⬚
(C) 0.6⬚
(D) 0.9⬚
Solution
I ⫽ 20 ⫻ 106 mm4
E ⫽ 210 GPa
6
⬍ strong axis I for W200 ⫻ 22.5
2
EI ⫽ 4.20 ⫻ 10 N⭈m
C
P ⫽ 15 kN
L ⫽ 4.5 m
P
H⫽2m
H
B
A
L
Max. rotation at B: apply statically-equivalent moment P⫻H at B on SS beam
uBmax ⫽
(P H) L
⫽ 0.614⬚
3EI
uBmax ⫽ 0.011 rad
R-9.5: A steel bracket ABC (EI ⫽ 4.2 ⫻ 106 N?m2) with span length L ⫽ 4.5 m
and height H ⫽ 2 m is subjected to load P ⫽ 15 kN at C. The maximum horizontal displacement of joint C is approximately:
(A) 22 mm
(B) 31 mm
(C) 38 mm
(D) 40 mm
Solution
E ⫽ 210 GPa
I ⫽ 20 ⫻ 106 mm4
⬍ strong axis I for W200 ⫻ 22.5
EI ⫽ 4.20 ⫻ 106 N⭈m2
C
P
P ⫽ 15 kN
L ⫽ 4.5 m
H
H⫽2m
B
A
L
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APPENDIX A FE Exam Review Problems
1153
Max. rotation at B: apply statically-equivalent moment P ⫻ H at B on SS beam
uBmax ⫽
(P H) L
⫽ 0.614⬚
3EI
uBmax ⫽ 0.011 rad
Horizontal deflection of vertical cantilever BC:
dBC ⫽
P H3
⫽ 9.524 mm
3EI
Finally, superpose uB ⫻ H and dBC
dC ⫽ uBmax H ⫹ dBC ⫽ 31.0 mm
R-9.6: A nonprismatic cantilever beam of one material is subjected to load P at
its free end. Moment of inertia I2 ⫽ 2 I1. The ratio r of the deflection dB to the
deflection d1 at the free end of a prismatic cantilever with moment of inertia I1
carrying the same load is approximately:
(A) 0.25
(B) 0.40
(C) 0.56
(D) 0.78
Solution
Max. deflection of prismatic cantilever
(constant I1)
A
P L3
d1 ⫽
3 E I1
I2
L
—
2
P
C
I1
B
L
—
2
Rotation at C due to both load P & moment
PL/2 at C for nonprismatic beam:
L 2
L L
Pa b
aP b
2
2 2 3 L2 P
uC ⫽
⫹
⫽
2 E I2
E I2
8 E I2
Deflection at C due to both load P & moment
PL/2 at C for nonprismatic beam:
L 3
L L 2
Pa b
aP b a b
2
2 2
5 L3 P
dCl ⫽
⫹
⫽
3 E I2
2 E I2
48 E I2
Total deflection at B:
L 3
Pa b
2
L
dB ⫽ dCl ⫹ uC ⫹
2
3 E I1
L 3
P
b
a
2
5 L3 P
3 L2 P L
L3 P (7 I1 ⫹ I2)
dB ⫽
⫹a
b ⫹
⫽
48 E I2
8 E I2 2
3 E I1
24 E I1 I2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1154
APPENDIX A FE Exam Review Problems
Ratio dB/d1:
L3 P (7 I1 ⫹ I2)
24 E I1 I2
7 I1 1
⫽
⫹
r⫽
8 I2 8
P L3
3 E I1
7 1
1
so r ⫽ a b ⫹ ⫽ 0.563
8 2
8
R-9.7: A steel bracket ABCD (EI ⫽ 4.2 ⫻ 106 N?m2), with span length L ⫽ 4.5 m
and dimension a ⫽ 2 m, is subjected to load P ⫽ 10 kN at D. The maximum
deflection at B is approximately:
(A) 10 mm
(B) 14 mm
(C) 19 mm
(D) 24 mm
Solution
E ⫽ 210 GPa
I ⫽ 20 ⫻ 106 mm4
⬍ strong axis I for W200⫻22.5
EI ⫽ 4.20 ⫻ 106 N⭈m2
L
A
P ⫽ 10 kN
B
D
L ⫽ 4.5 m a ⫽ 2 m
a
C
P
h ⫽ 206 mm
Statically-equivalent loads at end of cantilever AB:
• downward load P
• CCW moment P ⫻ a
Downward deflection at B by superposition:
dB ⫽
P L3 (P a) L2
⫺
⫽ 24.1 mm
3EI
2EI
dB
⫽ 0.005
L
R-10.1: Propped cantilever beam AB has moment M1 applied at joint B. Framework
ABC has moment M2 applied at C. Both structures have constant flexural rigidity EI.
If the ratio of the applied moments M 1/M 2 ⫽ 3/2, the ratio of the reactive moments
M A1/M A2 at clamped support A is:
(A) 1
(B) 3/2
(C) 2
(D) 5/2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1155
APPENDIX A FE Exam Review Problems
Solution
M1
L/2
B
A
C
M2
y
y
A
B
x
x
MA2
MA1
L
L
From Prob. 10.3-1:
MA1 ⫽
M1
2
M2
2
Statically-equivalent moment at B is M2
so
MA2 ⫽
MA1 M1
⫽
MA2 M2
so
MA1 3
⫽
MA2 2
Ratio of reactive moments is
R-10.2: Propped cantilever beam AB has moment M1 applied at joint B. Framework
ABC has moment M2 applied at C. Both structures have constant flexural rigidity EI.
If the ratio of the applied moments M 1/M 2 ⫽ 3/2, the ratio of the joint rotations at
B, uB1/uB2, is:
(A) 1
(B) 3/2
(C) 2
(D) 5/2
Solution
M2
y
y
M1
L/2
B
A
A
x
MA1
θB1
L
C
B
MA2
x
θB2
L
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1156
APPENDIX A FE Exam Review Problems
From Prob 10.3-1:
uB1 ⫽
M1 L
4 EI
Statically-equivalent moment at B is M 2
Ratio of joint rotations is
u B1
M1
⫽
u B2
M2
uB2 ⫽
so
M2 L
4 EI
uB1
3
⫽
u B2
2
so
R-10.3: Structure 1 with member BC of length L/2 has force P1 applied at joint C.
Structure 2 with member BC of length L has force P2 applied at C. Both structures
have constant flexural rigidity EI. If the ratio of the applied forces P1/P2 ⫽ 5/2, the
ratio of the joint B rotations uB1/uB2 is:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
Solution
P2
C
L
P1
C
y
y
L/2
A
B
θB1
MA1
B
A
x
x
θB2
MA2
L
L
Statically-equivalent moment at B is M1 ⫽ P1 ⫻ L/2
From Prob. 10.3-1:
u B1 ⫽
L
bL
2
L2 P1
⫽
4 EI
8 EI
aP1
Statically-equivalent moment at B is M2 ⫽ P2 ⫻ L
AP 2 LB L L2 P 2
⫽
4 EI
4 EI
u B1
P1
⫽
Ratio of joint rotations is
u B2 2 P2
so
u B2 ⫽
so
u B1 5
⫽
u B2 4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1157
APPENDIX A FE Exam Review Problems
R-10.4: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. The required ratio of the applied
forces P1/P2 so that joint B rotations uB1 and uB2 are equal is approximately:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
Solution
P2
C
L
P1
C
y
y
L/2
A
B
MA1
B
A
x
x
MA2
θB1
θB2
L
L
Statically-equivalent moment at B is M1 ⫽ P1 ⫻ L/2
From Prob. 10.3-1:
uB1 ⫽
L
bL
2
L2 P 1
⫽
4 EI
8 EI
aP1
Statically-equivalent moment at B is M 2 ⫽ P2 ⫻ L
so
uB2 ⫽
(P2 L) L L2 P 2
⫽
4 EI
4 EI
Ratio of joint rotations is
uB1
P1
⫽
uB2 2 P2
and
uB1
⫽ 1 so
uB2
P1
⫽2
P2
R-10.5: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. If the ratio of the applied forces
P1/P2 ⫽ 5/2, the ratio of the joint B reactions RB1/RB2 is:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1158
APPENDIX A FE Exam Review Problems
Solution
P2
C
L
P1
C
y
y
L/2
A
B
B
A
x
x
RB1
RB2
L
L
Statically-equivalent moment at B is M 1 ⫽ P1 ⫻ L/2
From Prob. 10.3-1:
RB1 ⫽
⫺3 M1
2 L
L
aP1 b
2
⫺3
3P1
or RB1 ⫽
⫽⫺
2
L
4
Statically-equivalent moment at B is M 2 ⫽ P2 ⫻ L
so
RB2 ⫽
⫺3 AP2 LB
3 P2
⫽⫺
2
L
2
Ratio of reactions at B is
RB1
P1
⫽
RB2 2 P 2
so
RB1 1 5
5
⫽ a b⫽
RB2 2 2
4
R-10.6: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. The required ratio of the applied
forces P1/P2 so that joint B reactions RB1 and RB2 are equal is:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1159
APPENDIX A FE Exam Review Problems
Solution
P2
C
L
P1
C
y
y
L/2
A
B
B
A
x
x
RB2
RB1
L
L
Statically-equivalent moment at B is M 1 ⫽ P1 ⫻ L/2
From Prob. 10.3-1: RB1 ⫽
⫺3 M1
2 L
RB1 ⫽
or
⫺3
2
L
b
2
3 P1
⫽⫺
L
4
aP1
Statically-equivalent moment at B is M 2 ⫽ P2 ⫻ L
so
RB2 ⫽
⫺3 AP2 LB
3 P2
⫽⫺
2
L
2
Ratio of reactions at B is
RB1
P1
⫽
RB2 2 P2
and
RB1
⫽ 1 so
RB2
P1
⫽2
P2
R-10.7: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. If the ratio of the applied forces
P1/P2 ⫽ 5/2, the ratio of the joint C lateral deflections dC1/dC2 is approximately:
(A) 1/2
(B) 4/5
(C) 3/2
(D) 2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1160
APPENDIX A FE Exam Review Problems
Solution
P1
δC1
C
y
L/2
A
B
x
L
L 3
L
P1 a b
P1 L
2
2 L 5 L3 P 1
⫹
⫽
d C1 ⫽
3 EI
4 EI 2
48 EI
d C2 ⫽
P 2 L3 P 2 L L
7 L3 P 2
⫹
L⫽
3 EI
4 EI
12 EI
dC1
5P1
⫽
dC2 28 P 2
5 5 25
⫽
28 2 56
25
⫽ 0.446
56
R-11.1: Beam ACB has a sliding support at A and is supported at C by a pinned end
steel column with square cross section (E ⫽ 200 GPa, b ⫽ 40 mm) and height
L ⫽ 3.75 m. The column must resist a load Q at B with a factor of safety 2.0 with
respect to the critical load. The maximum permissible value of Q is approximately:
(A) 10.5 kN
(B) 11.8 kN
(C) 13.2 kN
(D) 15.0 kN
Solution
E ⫽ 200 GPa
b ⫽ 40 mm
I⫽
n ⫽ 2.0
L ⫽ 3.75 m
b4
⫽ 2.133 ⫻ 105 mm4
12
Statics: sum vertical forces to
find reaction at D:
RD ⫽ Q
So force in pin-pin column is Q
p2 E I
Qcr ⫽ 2 ⫽ 29.9 kN
Pcr ⫽ Qcr
L
A
C
B
2d
d
Q
L
D
Allowable value of Q:
Qcr
Qallow ⫽
⫽ 15.0 kN
n
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1161
R-11.2: Beam ACB has a pin support at A and is supported at C by a steel column
with square cross section (E ⫽ 190 GPa, b ⫽ 42 mm) and height L ⫽ 5.25 m.
The column is pinned at C and fixed at D. The column must resist a load Q at
B with a factor of safety 2.0 with respect to the critical load. The maximum permissible value of Q is approximately:
(A) 3.0 kN
(B) 6.0 kN
(C) 9.4 kN
(D) 10.1 kN
Solution
E ⫽ 190 GPa
n ⫽ 2.0
b ⫽ 42 mm
L ⫽ 5.25 m
A
B
C
2d
d
Effective length of pinned-fixed column:
Le ⫽ 0.699 L ⫽ 3.670 m
Q
L
b4
I⫽
⫽ 2.593 ⫻ 105 mm4
12
D
Statics: use FBD of ACB and sum moments
about A to find force in column as a multiple of Q:
FCD ⫽
Q (3 d)
⫽ 3Q
d
So force in pin-fixed column is 3Q
Pcr ⫽ 3 Qcr Pcr ⫽
p2 E I
⫽ 36.1 kN
Le2
So
Qcr ⫽
Pcr
⫽ 12.0 kN
3
Allowable value of Q:
Qallow ⫽
Qcr
⫽ 6.0 kN
n
R-11.3: A steel pipe column (E ⫽ 190 GPa, a ⫽ 14 ⫻ 1026 per degree Celsius,
d2 ⫽ 82 mm, d1 ⫽ 70 mm) of length L ⫽ 4.25 m is subjected to a temperature
increase ⌬T. The column is pinned at the top and fixed at the bottom. The temperature increase at which the column will buckle is approximately:
(A) 36 °C
(B) 42 °C
(C) 54 °C
(D) 58 °C
Solution
E ⫽ 190 GPa
d2 ⫽ 82 mm
L ⫽ 4.25 m
d1 ⫽ 70 mm
a ⫽ [14 (10⫺6)]/ ⬚C
A⫽
p 2
(d2 ⫺ d12) ⫽ 1432.57 mm2
4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1162
APPENDIX A FE Exam Review Problems
Effective length of pinned-fixed column:
Le ⫽ 0.699L ⫽ 3.0 m
B
p
(d24 ⫺ d14) ⫽ 1.04076 ⫻ 106 mm4
I⫽
64
∆T
L
Axial compressive load in bar: P ⫽ EA a(⌬T)
Equate to Euler buckling load and solve for ⌬T:
p2 E I
L2
⌬T ⫽ e
EAa
A
Or
⌬T ⫽
p2 I
⫽ 58.0⬚C
a A L2e
R-11.4: A steel pipe (E ⫽ 190 GPa, a ⫽ 14 ⫻ 10⫺6/ ⬚C , d2 ⫽ 82 mm, d1 ⫽ 70 mm)
of length L ⫽ 4.25 m hangs from a rigid surface and is subjected
to a temperature increase ⌬T ⫽ 50 °C. The column is fixed at the top and has a
small gap at the bottom. To avoid buckling, the minimum clearance at the bottom
should be approximately:
(A) 2.55 mm
(B) 3.24 mm
(C) 4.17 mm
(D) 5.23 mm
Solution
E ⫽ 190 GPa
L ⫽ 4250 mm
a ⫽ [14(10⫺6)] / °C
d2 ⫽ 82 mm
A⫽
⌬T ⫽ 50 °C
d1 ⫽ 70 mm
p 2
(d2 ⫺ d12) ⫽ 1433 mm2
4
∆T
L
p
I ⫽ (d24 ⫺ d14) ⫽ 1.041 ⫻ 106 mm4
64
Effective length of fixed-roller support column:
gap
frictionless
surface
Le ⫽ 2.0L ⫽ 8500.0 mm
Column elongation due to temperature increase:
d1 ⫽ a⌬TL ⫽ 2.975 mm
Euler buckling load for fixed-roller column:
Pcr ⫽
p2 E I
⫽ 27.013 kN
L2e
Column shortening under load of P ⫽ Pcr:
d2 ⫽
Pcr L
⫽ 0.422 mm
EA
Mimimum required gap size to avoid buckling: gap ⫽ d1 ⫺ d2 ⫽ 2.55 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1163
R-11.5: A pinned-end copper strut (E ⫽ 110 GPa) with length L ⫽ 1.6 m is constructed of circular tubing with outside diameter d ⫽ 38 mm. The strut must
resist an axial load P ⫽ 14 kN with a factor of safety 2.0 with respect to the critical load. The required thickness t of the tube is:
(A) 2.75 mm
(B) 3.15 mm
(C) 3.89 mm
(D) 4.33 mm
Solution
E ⫽ 110 GPa
Pcr ⫽ nP
L ⫽ 1.6 m
d ⫽ 38 mm
n ⫽ 2.0
P ⫽ 14 kN
Pcr ⫽ 28.0 kN
t
Solve for required moment of inertia I in
terms of Pcr then find tube thickness
Pcr ⫽
p2 E I
L2
I⫽
Pcr L2
p2 E
d
4
I ⫽ 66025 mm
Moment of inertia
I⫽
Solve numerically for min. thickness t:
p 4
[d ⫺ (d ⫺ 2 t)4]
64
tmin ⫽ 4.33 mm
d 4 ⫺ (d ⫺ 2 t)4 ⫽ I
64
p
d ⫺ 2tmin ⫽ 29.3 mm ⫽ inner diameter
R-11.6: A plane truss composed of two steel pipes (E ⫽ 210 GPa, d ⫽ 100 mm,
wall thickness ⫽ 6.5 mm) is subjected to vertical load W at joint B. Joints A and
C are L ⫽ 7 m apart. The critical value of load W for buckling in the plane of the
truss is nearly:
(A) 138 kN
(B) 146 kN
(C) 153 kN
(D) 164 kN
Solution
E ⫽ 210 GPa
L⫽7m
d ⫽ 100 mm
t ⫽ 6.5 mm
B
Moment of inertia
p 4
[d ⫺ (d ⫺ 2 t)4] ⫽ 2.097 ⫻ 106 mm4
I⫽
64
Member lengths:
LBA ⫽ L cos(40⬚) ⫽ 5.362 m
LBC ⫽ L cos(50⬚) ⫽ 4.500 m
d
W
40°
50°
A
C
L
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1164
APPENDIX A FE Exam Review Problems
Statics at joint B to find member forces FBA and FBC:
• Sum horizontal forces at joint B:
FBA cos(40⬚) ⫽ FBC cos(50⬚)
cos(50⬚)
cos(40⬚)
FBA ⫽ FBC
cos(50⬚)
⫽ 0.839
cos(40⬚)
• Sum vertical forces at joint B:
a⫽
where
W ⫽ FBA sin(40⬚) ⫹ FBC sin(50⬚)
cos(50⬚)
W ⫽ FBC
sin(40⬚) ⫹ FBC sin(50⬚)
cos(40⬚)
1
FBC ⫽ W b where b ⫽
cos(50⬚)
a
sin(40⬚) ⫹ sin(50⬚)b
cos(40⬚)
⫽ 0.766
So member forces in terms of W are:
FBC ⫽ Wb and FBA ⫽ FBC a
with ab ⫽ 0.643
or
FBA ⫽ W(ab)
Euler buckling loads in BA & BC:
p2 E I
⫽ 151.118 kN
FBA_cr ⫽
LBA2
b
so WBA_cr ⫽ FBA_cr ⫽ 138 kN
a
FBC_cr ⫽
p2 E I
⫽ 214.630 kN
LBC2
⬍ lower value controls
so
WBC_cr ⫽ b FBC_cr ⫽ 164 kN
R-11.7: A beam is pin-connected to the tops of two identical pipe columns, each
of height h, in a frame. The frame is restrained against sidesway at the top of
column 1. Only buckling of columns 1 and 2 in the plane of the frame is of
interest here. The ratio (a/L) defining the placement of load Qcr, which causes
both columns to buckle simultaneously, is approximately:
(A) 0.25
(B) 0.33
(C) 0.67
(D) 0.75
Qcr
Solution
a
Draw FBD of beam only; use
statics to show that Qcr causes
forces P1 and P2 in columns 1 & 2
respectively:
P1 ⫽ a
P2 ⫽
L⫺a
b Qcr
L
EI
h
L-a
EI
1
h
2
a
Qcr
L
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1165
Buckling loads for columns 1 & 2:
Pcr1 ⫽
Pcr2 ⫽
p2 EI
L⫺a
b Qcr
⫽a
L
(0.699 h)2
p2 EI
a
⫽ a b Qcr
L
h2
Solve above expressions for Qcr, then solve for required a/L so that columns
buckle at the same time:
p2 EI
L
p2 EI L
a
b
⫽
a b
(0.699 h)2 L ⫺ a
h2 a
L
p2 EI L
p2 EI
b⫺ 2 a b⫽0
2 a
a
(0.699 h) L ⫺ a
h
Or
L
L
⫺ ⫽0
a
0.6992 (L ⫺ a)
Or
Or
a
⫽ 0.6992
L⫺a
a
0.6992
⫽
⫽ 0.328
L (1 ⫹ 0.6992)
So
R-11.8: A steel pipe column (E ⫽ 210 GPa) with length L ⫽ 4.25 m is
constructed of circular tubing with outside diameter d2 ⫽ 90 mm and inner
diameter d1 ⫽ 64 mm. The pipe column is fixed at the base and pinned at the
top and may buckle in any direction. The Euler buckling load of the column
is most nearly:
(A) 303 kN
(B) 560 kN
(C) 690 kN
(D) 720 kN
Solution
E ⫽ 210 GPa
L ⫽ 4.25 mm
d2 ⫽ 90 mm
d1 ⫽ 64 mm
Moment of inertia
I⫽
p
(d24 ⫺ d14)
64
d1
d2
I ⫽ 2.397 ⫻ 106 mm4
Effective length of column for fixed-pinned case:
Le ⫽ 0.699 L ⫽ 2.971 m
Pcr ⫽
p2 E I
⫽ 563 kN
L2e
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1166
APPENDIX A FE Exam Review Problems
R-11.9: An aluminum tube (E ⫽ 72 GPa) AB of circular cross section has a
pinned support at the base and is pin-connected at the top to a horizontal beam
supporting a load Q ⫽ 600 kN. The outside diameter of the tube is 200 mm and
the desired factor of safety with respect to Euler buckling is 3.0. The required
thickness t of the tube is most nearly:
(A) 8 mm
(B) 10 mm
(C) 12 mm
(D) 14 mm
Solution
E ⫽ 72 GPa
SMc ⫽ 0
L ⫽ 2.5 m
n ⫽ 3.0
2.5 Q
1.5
P ⫽ 1000 kN
P⫽
Find required I based on critical
buckling load
Critical load
Q ⫽ 600 kN
d ⫽ 200 mm
Q = 600 kN
C
B
Pcr ⫽ Pn
Pcr ⫽ 3000 kN
Pcr ⫽
1.5 m
1.0 m
p2 E I
L2
2.5 m
Pcr L2
I⫽ 2
p E
d ⫽ 200 mm
6
4
I ⫽ 26.386 ⫻ 10 mm
Moment of inertia
p 4
I⫽
[d ⫺ (d ⫺ 2 t)4]
64
B
d ⫺ 4 d4 ⫺ I
t min ⫽
2
64
p
A
tmin ⫽ 9.73 mm
R-11.10: Two pipe columns are required to have the same Euler buckling load
Pcr. Column 1 has flexural rigidity E I and height L1; column 2 has flexural
rigidity (4/3)E I and height L2. The ratio (L2/L1) at which both columns will
buckle under the same load is approximately:
(A) 0.55
(B) 0.72
(C) 0.81
(D) 1.10
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
APPENDIX A FE Exam Review Problems
1167
Solution
Equate Euler buckling load expressions
for the two columns considering their
different properties, base fixity conditions
and lengths:
2
p2 a Eb (2 I)
3
p EI
⫽
(0.699 L1)2
L22
2
Pcr
Pcr
E
I
2E/3
2I
L1
L2
Simplify then solve for L2 /L1:
2
L2
4
a
b ⫽
0.699 L1
3
L2
4
⫽
0.6992 ⫽ 0.807
L1 B 3
R-11.11: Two pipe columns are required to have the same Euler buckling load
Pcr. Column 1 has flexural rigidity E I1 and height L; column 2 has flexural
rigidity (2/3)E I2 and height L. The ratio (I2/I1) at which both columns will
buckle under the same load is approximately:
(A) 0.8
(B) 1.0
(C) 2.2
(D) 3.1
Solution
Equate Euler buckling load expressions
for the two columns considering their
different properties, base fixity conditions
and lengths:
2
p2 a Eb (I2)
3
p E I1
⫽
(0.699 L)2
L2
2
Pcr
Pcr
E
I1
2E/3
I2
L
L
Simplify then solve for I2/I1:
L2
E
I2
⫽
I1 (0.699 L)2 2
E
3
3
I2
2
⫽
⫽ 3.07
I1 (0.699)2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.