36041 01 online appA p01-73 - Cengage · A-1.9: A titanium bar (E 100 GPa, v 0.33) with square...
Transcript of 36041 01 online appA p01-73 - Cengage · A-1.9: A titanium bar (E 100 GPa, v 0.33) with square...
APPENDIX A:
FE EXAM REVIEW PROBLEMS & SOLUTIONS
JAMES M. GERE BARRY J. GOODNO
A
B
C
P1
dAB
tAB
dBC
tBC
P2
A-1.1: A hollow circular post ABC (see figure) supports a load P1 � 16 kNacting at the top. A second load P2 is uniformly distributed around the cap plateat B. The diameters and thicknesses of the upper and lower parts of the post aredAB � 30 mm, tAB � 12 mm, dBC � 60 mm, and tBC � 9 mm, respectively. Thelower 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
Solution
P1 � 16�kN dAB � 30�mm tAB � 12�mm
dBC � 60�mm tBC � 9�mm
Stress in AB:
Stress in BC: � must equal �AB
Solve for P2 P2 � �AB�ABC � P1 � 18.00�kN
Check: � same as in AB
A-1.2: A circular aluminum tube of length L � 650 mm is loaded in com-pression 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 strainin the longitudinal direction of 400 � 10�6. The shortening of the bar isapproximately:
(A) 0.12 mm(B) 0.26 mm(C) 0.36 mm(D) 0.52 mm
Solution
� � 400�(10�6) L � 650�mm
� ��L � 0.260�mm
sBC �P1 P2
ABC� 23.6�MPa
sBC �P1 P2
ABC
sAB �P1
AAB� 23.6�MPa
ABC �p
4�[dBC
2 � (dBC � 2�tBC)2] � 1442�mm2
AAB �p
4�[dAB
2 � (dAB � 2�tAB)2] � 679�mm2
APPENDIX A FE Exam Review Problems 1
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 1
A-1.3: A steel plate weighing 27 kN is hoisted by a cable sling that has a clevisat each end. The pins through the clevises are 22 mm in diameter. Each half ofthe cable is at an angle of 35° to the vertical. The average shear stress in each pinis approximately:
(A) 22 MPa(B) 28 MPa(C) 40 MPa(D) 48 MPa
Solution
W � 27�kN dp � 22�mm � � 35�deg
Cross sectional area of each pin:
Tensile force in cable:
Shear stress in each clevis pin (double shear):
A-1.4: A steel wire hangs from a high-altitude balloon. The steel has unit weight77kN/m3 and yield stress of 280 MPa. The required factor of safety against yieldis 2.0. The maximum permissible length of the wire is approximately:
(A) 1800 m(B) 2200 m(C) 2600 m(D) 3000 m
t �T
2�AP� 21.7�MPa
T �
aW
2b
cos(u)� 16.48�kN
Ap �p
4�d2
p � 380�mm2
2 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Strain gage
L
PP
35°35°
Clevis
Cable sling
P
Steel plate
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 2
Solution
�Y � 280�MPa FSY � 2
Allowable stress:
Weight of wire of length L: W � ��A�L
Max. axial stress in wire of length L: �max � ��L
Max. length of wire:
A-1.5: An aluminum bar (E � 72 GPa, v � 0.33) of diameter 50 mm cannotexceed a diameter of 50.1 mm when compressed by axial force P. The maximumacceptable 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 � 0.33
Lateral strain: �L � 0.002
Axial strain:
Axial stress: � � E��a � �436.4�MPa � below yield stress of 480 MPaso Hooke’s Law applies
Max. acceptable compressive load:
A-1.6: An aluminum bar (E � 70 GPa, v � 0.33) of diameter 20 mm is stretchedby axial forces P, causing its diameter to decrease by 0.022 mm. The maximumacceptable compressive load P is approximately:
(A) 73 kN(B) 100 kN(C) 140 kN(D) 339 kN
Pmax � s�ap
4�dinit
2b � 857 kN
�a ���L
n� �0.006
�L �dfinal � dinit
dinit
Lmax �sallow
g� 1818 m
smax �W
A
sallow �sY
FSY� 140.0�MPa
g � 77�kN
m3
APPENDIX A FE Exam Review Problems 3
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Solution
E � 70�GPa dinit � 20�mm �d � �0.022�mm v � 0.33
Lateral strain:
�L��0.001
Axial strain:
Axial stress: � � E��a � 233.3�MPa � below yield stress of 270 MPaso Hooke’s Law applies
Max. acceptable compressive load:
A-1.7: A polyethylene bar (E � 1.4 GPa, v � 0.4) of diameter 80 mm is insertedin 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 compressiveload P is approximately:
(A) 18 kN(B) 25 kN(C) 44 kN(D) 60 kN
Solution
E � 1.4�GPa d1 � 80�mm �d1 � 0.2�mm v � 0.4
Lateral strain:
�L �0.003
Axial strain:
Axial stress: � � E��a � �8.8�MPa � well below ultimate stress of28 MPa so Hooke’s Law applies
Max. acceptable compressive load:
Pmax � s�ap
4�d1
2b � 44.0�kN
�a ���L
v� �6.250 � 10�3
�L ��d1
d1
Pmax � s�ap
4�dinit
2b � 73.3�kN
�a ���L
v� 3.333 � 10�3
�L ��d
dinit
4 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
d PP
d2d1
Steeltube
Polyethylenebar
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 4
A-1.8: A pipe (E � 110 GPa) carries a load P1 � 120 kN at A and a uniformlydistributed load P2 � 100 kN on the cap plate at B. Initial pipe diameters andthicknesses 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’sratio 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 dBC � 70�mm
tBC � 10�mm P1 � 120�kN P2 � 100�kN
ABC �p
4�[dBC
2 � (dBC � 2�tBC)2] � 1885�mm2
APPENDIX A FE Exam Review Problems 5
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P2
dABtAB
dBCtBC
ABC Cap plate
P1
Axial strain of 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: �tBC � 0.0036�mm
Poisson’s ratio: � confirms value for brassgiven in properties table(also agrees with givenmodulus E)
v ���L
�BC� 0.34
�L ��tBC
tBC� 3.600 � 10�4
�BC ��(P1 P2)
E�ABC� �1.061 � 10�3
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 5
A-1.9: 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. Theincrease 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 P � 900�kN v � 0.33
6 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
bb
P
L
P
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
Final lateral dimension of bar: bf � b �L�b � 74.96040�mm
Final volume of bar: Vfinal � bf2�Lf � 1.68841562 � 107�mm3
Increase in volume of bar: �V � Vfinal � Vinit � 9156�mm3
A-1.10: An elastomeric bearing pad is subjected to a shear force V during a staticloading test. The pad has dimensions a � 150 mm and b � 225 mm, and thick-ness t � 55 mm. The lateral displacement of the top plate with respect to thebottom plate is 14 mm under a load V � 16 kN. The shear modulus of elasticityG 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
�V
Vinit� 0.000543
� �P
E�b2 � 1.60000 � 10�3
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 6
Ave. shear stress:
Ave. shear strain:
Shear modulus of elastomer:
A-1.11: A bar of diameter d � 18 mm and length L � 0.75 m is loaded in ten-sion by forces P. The bar has modulus E � 45 GPa and allowable normal stressof 180 MPa. The elongation of the bar must not exceed 2.7 mm. The allowablevalue of forces P is approximately:
(A) 41 kN(B) 46 kN(C) 56 kN(D) 63 kN
Solution
d � 18�mm L � 0.75�m E � 45�GPa �a � 180�MPaa � 2.7�mm
G �t
g� 1.902�MPa
g � atanad
tb � 0.249
t �V
a�b� 0.474�MPa
APPENDIX A FE Exam Review Problems 7
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
a
b
V
t
dPP
L
(1) allowable value of P based on elongation
�max � E��a � 162.0�MPa
� elongation governs
(2) allowable load P based on tensile stress
A-1.12: Two flanged shafts are connected by eight 18 mm bolts. The diameter ofthe 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 isapproximately:
(A) 19 kN�m(B) 22 kN�m(C) 29 kN�m(D) 37 kN�m
Pa2 � sa�ap
4 �d2b � 45.8�kN
Pa1 � smax�ap
4�d2b � 41.2�kN
�a �da
L� 3.600 � 10�3
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 7
Solution
db � 18�mm d � 240�mm �a � 90�MPa n � 8
Bolt shear area:
Max. torque:
A-1.13: A copper tube with wall thickness of 8 mm must carry an axial tensileforce of 175 kN. The allowable tensile stress is 90 MPa. The minimum requiredouter diameter is approximately:
(A) 60 mm(B) 72 mm(C) 85 mm(D) 93 mm
Solution
t � 8�mm P � 175�kN �a � 90�MPa
Tmax � n�(ta�As)�d
2� 22.0�kN�m
As �p � db
2
4� 254.5�mm2
8 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
T0
T0
P P
d
Required area based on allowable stress:
Area of tube of thickness t but unknown outer diameter d:
A � ��t�(d � t)
Solving for dmin:
so dinner � dmin � 2�t � 69.4�mmdmin �
Psa
p�t t � 85.4�mm
A �p
4�[d2 � (d � 2�t)2]
Areqd �Psa
� 1944�mm2
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 8
A-2.1: Two wires, one copper and the other steel, of equal length stretch the sameamount 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 wireis approximately:
(A) 1.00(B) 1.08(C) 1.19(D) 1.32
Solution
Es � 210�GPa Ec � 120�GPa
Displacements are equal: s � c
or
so Es�As � Ec�Ac
and
Express areas in terms of wire diameters then find ratio:
so
A-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 Bcannot 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 max � 2.7�mm
Statics: sum moments about A to find reaction at B
RB � PRB �
P�L
2 P�
L
2
L
dc
ds� B
Es
Ec� 1.323
p�dc2
4
ap�ds2
4b
�Es
Ec
Ac
As�
Es
Ec
P�L
Es�As�
P�L
Ec�Ac
APPENDIX A FE Exam Review Problems 9
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P
Steelwire
P
Copper wire
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 9
Method of Joints at B:
FAB � P (tension)
Force-displ. relation:
Check normal stress in bar AB:
� well below yield stress of290 MPa in tension
A-2.3: A brass rod (E � 110 GPa) with cross sectional area of 250 mm2 is loadedby forces P1 � 15 kN, P2 � 10 kN, and P3 � 8 kN. Segment lengths of the barare a � 2.0 m, b � 0.75 m, and c � 1.2 m. The change in length of the bar isapproximately:
(A) 0.9 mm(B) 1.6 mm(C) 2.1 mm(D) 3.4 mm
Solution
E � 110�GPa A � 250�mm2
a � 2�m b � 0.75�m
c � 1.2�m
P1 � 15�kN 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
s �Pmax
A� 102.0�MPa
Pmax �E�A
L�dmax � 459�kN
10 APPENDIX A FE Exam Review Problems
L
A B45° 45°
P
P
C
a b c
B
P1 P2P3
A C D
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems 11
d1
P
d2
L/2 L/2
P
Change in length:
positive so elongation
Check max. stress:
� well below yield stress for brass so OK
A-2.4: A brass bar (E � 110 MPa) of length L � 2.5 m has diameter d1 � 18 mmover 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 materialwith constant diameter d and length L. The elongation of the prismatic bar underthe 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
Volume of nonprismatic bar:
Diameter of prismatic bar of same volume:
Elongation of prismatic bar:
� less than for nonprismatic bard � P�L
E�Aprismatic� 3.09�mm
Vprismatic � Aprismatic�L � 459458�mm3
Aprismatic �p
4�d2 � 184�mm2
d � HVolnonprismatic
p
4�L
� 15.30�mm
Volnonprismatic � (A1 A2)�L
2� 459458�mm3
A2 �p
4�d2
2 � 113.097�mm2
A1 �p
4�d1
2 � 254.469�mm2
NAB
A� 68.0�MPa
�
dD
a b c� 2.384 � 10�4
dD �1
E�A�(NAB�a NBC�b NCD�c) � 0.942�mm
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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12 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Elongation of nonprismatic bar shown in fig. above:
A-2.5: A nonprismatic cantilever bar has an internal cylindrical hole of diameterd/2 from 0 to x, so the net area of the cross section for Segment 1 is (3/4)A. LoadP is applied at x, and load -P/2 is applied at x � L. Assume that E is constant. Thelength of the hollow segment, x, required to obtain axial displacement � 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:
Displacement at free end:
Set 3 equal to PL/EA and solve for x
or
So x � 3L/5
�P�(L � 5�x)
2�A�E �
P�L
E�A� 0 simplify S �
P�(3�L � 5�x)
2�A�E� 0
�P�(L � 5�x)
2�A�E�
P�L
E�A
d3 �
3�p
2� x
E�a3
4�Ab
�P
2�(L � x)
E�A simplify S �
P�(L � 5�x)
2�A�E
d3 �N1�x
E�a3
4�Ab
N2�(L � x)
E�A
N2 ��P
2N1 �
3�P
2
� �P�L
2�E�a 1
A1
1
A2b � 3.63�mm
2 3
dA
Segment 1 Segment 2
d2—
P2—
A34—
L – xx
P
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 12
A-2.6: A nylon bar (E � 2.1 GPa) with diameter 12 mm, length 4.5 m, andweight 5.6 N hangs vertically under its own weight. The elongation of the bar atits free end is approximately:
(A) 0.05 mm
(B) 0.07 mm
(C) 0.11 mm
(D) 0.17 mm
Solution
E � 2.1GPa L � 4.5�m d � 12�mm
W � ��L�A � 5.598 N
or
so
Check max. normal stress at top of bar
� ok - well below ult.stress for nylon
A-2.7: A monel shell (Em � 170 GPa, d3 � 12 mm, d2 � 8 mm) encloses a brasscore (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 Prequired 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
smax �W
A� 0.050�MPa
dB �g�L2
2�E� 0.053�mm
dB �(g�L�A)�L
2�E�AdB �
W�L
2�E�A
g � 11�kN
m3
A �p�d2
4� 113.097�mm2
APPENDIX A FE Exam Review Problems 13
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L
B
A
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 13
Compatibility: m � b
Statics: Pm Pb � P so
Set B equal to 0.10 mm and solve for load P:
so with b � 0.10�mm
and then
A-2.8: A steel rod (Es � 210 GPa, dr � 12 mm, ctes � 12 � 10�6/degreeCelsius) 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 theallowable normal stress in the rod is 70 MPa, the maximum permissible tem-perature drop �T is approximately:
(A) 14 degrees Celsius(B) 20 degrees Celsius(C) 28 degrees Celsius(D) 40 degrees Celsius
P �Eb�Ab
L�db�a1
Em�Am
Eb�Abb � 13.40�kN
Pb �Eb�Ab
L �dbdb �
Pb�L
Eb�Ab
Pb �P
a1 Em�Am
Eb�Abb
Pm �Em�Am
Eb�Ab �Pb
Pm�L
Em�Am�
Pb�L
Eb�Ab
Ab �p
4�d1
2 � 28.274�mm2
Am �p
4�(d3
2 � d22) � 62.832�mm2
14 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P
Monel shellBrass core
d3
d1
d2
L
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 14
Solution
Es � 210�GPa
dr � 12�mm dp � 15�mm
ctes � 12�(10�6)
�a � 45�MPa �a � 70�MPa
Force in rod due to temperature drop �T: and normal stress in rod:
Fr � Es�Ar�(ctes)��T
So �Tmax associated with normal stress in rod
degrees Celsius (decrease) � Controls
Now check �T based on shear stress in pin (in double shear):
A-2.9: A threaded steel rod (Es � 210 GPa, dr � 15 mm, ctes � 12 � 10�6/degree Celsius) is held stress free between rigid walls by a nut and washer(dw � 22 mm) assembly at each end. If the allowable bearing stress betweenthe washer and wall is 55 MPa and the allowable normal stress in the rod is90 MPa, the maximum permissible temperature drop �T is approximately:
(A) 25 degrees Celsius(B) 30 degrees Celsius(C) 38 degrees Celsius(D) 46 degrees Celsius
Solution
Es � 210�GPa dr � 15�mm dw � 22�mm
ctes � 12�(10�6)
�ba � 55�MPa �a � 90�MPa
Aw �p
4�(dw
2 � dr2) � 203.4�mm2
Ar �p
4�d2
r � 176.7�mm2
�Tmaxpin �ta�(2�Ap)
Es�Ar�ctes� 55.8
tpin �Fr
2�Ap
�Tmaxrod �sa
Es�ctes� 27.8
sr �Fr
Ar
Ap �p
4�dp
2 � 176.715�mm2
Ar �p
4�dr
2 � 113.097�mm2
APPENDIX A FE Exam Review Problems 15
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
rod, dr
pin, dp
Clevis
ΔT
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 15
16 APPENDIX A FE Exam Review Problems
rod, dr washer, dwΔT
Steel bolt
Copper tube
Force in rod due to temperature drop �T: and normal stress in rod:
Fr � Es�Ar�(ctes)��T
So �Tmax associated with normal stress in rod
degrees Celsius (decrease)
Now check �T based on bearing stress beneath washer:
degrees Celsius (decrease)
� Controls
A-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 untilit is just snug. The pitch of the bolt threads is 1.25 mm. The bolt is now tightenedby 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 Ec � 110�GPa L � 0.5�m
Ac � 400�mm2 As � 130�mm2
n � 0.25 p � 1.25�mm
Compatibility: shortening of tube and elongation of bolt � applied displacement of n � p
Statics: Pc � Ps
Solve for Ps
or Ps �n�p
L�a 1
Ec�Ac
1
Es�Asb
� 10.529�kNPs�L
Ec�Ac
Ps�L
Es�As� n�p
Pc�L
Ec�Ac
Ps�L
Es�As� n�p
�Tmaxwasher �sba�(Aw)
Es�Ar�ctes� 25.1
sb �Fr
Aw
�Tmaxrod �sa
Es�ctes� 35.7
sr �Fr
Ar
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 16
Stress in steel bolt:
� tension
Stress in copper tube:
� compression
A-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 are100 MPa and 48 MPa respectively. The maximum permissible load Pmax isapproximately:
(A) 56 kN(B) 62 kN(C) 74 kN(D) 91 kN
Solution
a � 38�mm b � 50�mm
A � a�b � 1900�mm2
�a � 100�MPa
�a � 48�MPa
Bar is in uniaxial tension so Tmax � �max/2; since 2 Ta � �a, shear stressgoverns
Pmax � �a�A � 91.2�kN
A-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 tem-perature change at which the wire goes slack is approximately:
(A) 5.7 degrees Celsius(B) �12.6 degrees Celsius(C) 12.6 degrees Celsius(D) �18.2 degrees Celsius
Solution
E � 110�GPa d � 2.0�mm
cte � 19.5�(10�6) T � 85�N
A �p
4�d2 � 3.14�mm2
sc �Ps
Ac� 26.3�MPa
ss �Ps
As� 81.0�MPa
APPENDIX A FE Exam Review Problems 17
P P
a
b
T d T
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 17
Normal tensile stress in wire due to pretension T and temperature increase �T:
Wire goes slack when normal stress goes to zero; solve for �T
degrees Celsius (increase in temperature)
A-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 � 11.5�kN
Normal stress in bar:
For bar in uniaxial stress, max. shear stress is on a plane at 45 deg. to axis ofbar and equals 1/2 of normal stress:
A-2.14: A steel plane truss is loaded at B and C by forces P � 200 kN. The crosssectional area of each member is A � 3970 mm2. Truss dimensions are H � 3 mand 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 � 3�m L � 4�m
Statics: sum moments about A to find vertical reaction at B
(downward)
Bvert ��P�H
L� �150.000�kN
tmax �s
2� 73.2�MPa
s �p
A� 146.4�MPa
A �p
4�d2 � 78.54�mm2
�T �
T
A
E�cte� 12.61
s �T
A � E�cte��T
18 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
PPd
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 18
Method of Joints at B:
CBvert � �Bvert
So bar force in AB is: AB � P CBhoriz � 400.0�kN (compression)
Max. normal stress in AB:
Max. shear stress is 1/2 of max. normal stress for bar in uniaxial stress and ison plane at 45 deg. to axis of bar:
A-2.15: A plane stress element on a bar in uniaxial stress has tensile stress of�� � 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
�� � 78�MPa
Plane stress transformation formulas for uniaxial stress:
and
� on element face � on element face at angle � at angle �90
sx �
su
2
sin(u)2sx �su
cos(u)2
tmax �sAB
2� 50.4�MPa
sAB �AB
A� 100.8�MPa
CBhoriz �L
H�CBvert � 200.0�kN
APPENDIX A FE Exam Review Problems 19
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
H
P
P
PC
ABL
σθ/2
θτθ τθ
τθ τθ
σθ
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 19
Equate above formulas and solve for �x
so
also �� � ��x�sin(�)�cos(�) � �55.154�MPa
Max. shear stress is 1/2 of max. normal stress for bar in uniaxial stress and ison plane at 45 deg. to axis of bar:
A-3.1: A brass rod of length L � 0.75 m is twisted by torques T until the angleof rotation between the ends of the rod is 3.5°. The allowable shear strain inthe copper is 0.0005 rad. The maximum permissible diameter of the rod isapproximately:
(A) 6.5 mm(B) 8.6 mm(C) 9.7 mm(D) 12.3 mm
Solution
L � 0.75�m
� � 3.5�deg
�a � 0.0005
Max. shear strain:
so
A-3.2: The angle of rotation between the ends of a nylon bar is 3.5°. The bardiameter is 70 mm and the allowable shear strain is 0.014 rad. The minimumpermissible length of the bar is approximately:
(A) 0.15 m(B) 0.27 m(C) 0.40 m(D) 0.55 m
dmax �2�ga�L
f� 12.28�mmgmax �
ad
2�fbL
tmax �sx
2� 58.5�MPa
sx �su
cos(u)2 � 117.0�MPa
u � atana 1
12b � 35.264�deg
tan(u)2 �1
2
20 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L
dT T
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 20
Solution
d � 70�mm
� � 3.5�deg
�a � 0.014
Max. shear strain:
so
A-3.3: A brass bar twisted by torques T acting at the ends has the followingproperties: L � 2.1 m, d � 38 mm, and G � 41 GPa. The torsional stiffness ofthe 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 � 38�mm
Polar moment of inertia, Ip:
Torsional stiffness, kT:
A-3.4: A brass pipe is twisted by torques T � 800 N�m acting at the ends causingan 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 isapproximately:
(A) 36.1 GPa(B) 37.3 GPa(C) 38.7 GPa(D) 40.6 GPa
kT �G�Ip
L� 3997�N�m
Ip �p
32�d4 � 2.047 � 105�mm4
Lmin �d
2�
f
ga� 0.15 mg �
r�f
L
APPENDIX A FE Exam Review Problems 21
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L
dT T
L
dT T
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 21
Solution
22 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
d2
d1
L
T T
T1d
T1
L � 2.1�m d1 � 38�mm d2 � 56�mm � � 3.5�deg T � 800�N�m
Polar moment of inertia:
Solving torque-displacement relation for shear modulus G:
A-3.5: An aluminum bar of diameter d � 52 mm is twisted by torques T1 at theends. 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
�a � 65�MPa
From shear formula:
A-3.6: A steel tube with diameters d2 � 86 mm and d1 � 52 mm is twisted bytorques at the ends. The diameter of a solid steel shaft that resists the same torqueat the same maximum shear stress is approximately:
(A) 56 mm(B) 62 mm(C) 75 mm(D) 82 mm
T1 max ta�Ip
ad
2b
� 1795�N�m
Ip �p
32�d4 � 7.178 � 105
�mm4
G �T�L
f�Ip� 36.1�GPa
Ip �p
32�(d2
4 � d14) � 7.608 � 105�mm4
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 22
Solution
d2 � 86�mm d1 � 52�mm
Shear formula for hollow pipe:
Shear formula for solid shaft:
Equate and solve for d of solid shaft:
A-3.7: A stepped steel shaft with diameters d1 � 56 mm and d2 � 52 mm istwisted by torques T1 � 3.5 kN�m and T2 � 1.5 kN�m acting in opposite direc-tions. 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
d � a32p
�
IPpipe
d2b1
3 � 82.0 �mm
d � D
16�Tp
T�ad2
2b
IPpipe
T13
tmax �
T�ad
2b
p
32�d4
simplify S 16�T
p�d3
tmax �
T�ad2
2b
IPpipe
IPpipe�
p
32�(d2
4 � d14) � 4.652 � 106�mm4
APPENDIX A FE Exam Review Problems 23
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
d2
d1 d
L1
d1
T1T2
L2
B C
d2
A
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 23
Polar moments of inertia:
Shear formula - max. shear stresses in segments 1 & 2:
A-3.8: A stepped steel shaft (G � 75 GPa) with diameters d1 � 36 mm andd2 � 32 mm is twisted by torques T at each end. Segment lengths are L1 � 0.9 mand L2 � 0.75 m. If the allowable shear stress is 28 MPa and maximum allow-able 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
d2 � 32�mm
G � 75�GPa
�a � 28�MPa
L1 � 0.9�m L2 � 0.75�m
�a � 1.8�deg
Polar moments of inertia:
Max torque based on allowable shear stress - use shear formula:
� controlsTmax2 � ta�a2�Ip2
d2b � 180�N�mTmax1 � ta�a2�Ip1
d1b � 257�N�m
tmax2 �
T�ad2
2b
Ip2tmax1 �
T�d1
2
Ip1
Ip2 �p
32�d2
4 � 1.029 � 105�mm4
Ip1 �p
32�d1
4 � 1.649 � 105�mm4
tmax2 �
T2�ad2
2b
Ip2� 54.3�MPatmax1 �
(T1 � T2)�d1
2
Ip1� 58.0�MPa
Ip2 �p
32�d2
4 � 7.178 � 105�mm4
Ip1 �p
32�d1
4 � 9.655 � 105�mm4
24 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L1 L2
T
A B C
d1 d2T
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 24
Max. torque based on max. rotation & torque-displacement relation:
A-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, therequired shaft diameter is approximately:
(A) 38 mm(B) 44 mm(C) 46 mm(D) 48 mm
Solution
�a � 50�MPa
TA � 975�N�m
TB � 1500�N�m
TC � 650�N�m
TD � 825�N�m
Find torque in each segment of shaft:
TAB � TA � 975.0�N�m TBC � TA � TB � �525.0�N�m
TCD � TD � 825.0�N�m
Shear formula:
Set T to Tallowable and T to torque in each segment; solve for required diameter d(largest controls)
Segment AB:
Segment BC:
Segment CD: d � a16 �|TCD|p�ta
b13
� 43.8�mm
d � a16 �|TBC|p�ta
b13
� 37.7�mm
d � a16 �|TAB|p�ta
b13
� 46.3�mm
t �
T�ad
2b
p
32�d4
simplify S 16�T
p�d3
Tmax �G�fa
aL1
Ip1
L2
Ip2b
� 185�N�m
f �T
G�aL1
Ip1
L2
Ip2b
APPENDIX A FE Exam Review Problems 25
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C
D
A
B
TA
TB
TC
TD
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 25
A-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 resultingmaximum tensile stress in the shaft is approximately:
(A) 38 MPa(B) 41 MPa(C) 49 MPa(D) 58 MPa
Solution
G � 27�GPa
d2 � 96�mm
d1 � 52�mm
Max. shear strain due to twist per unit length:
radians
Max. shear stress: �max � G��max � 40.7�MPa
Max. tensile stress on plane at 45 degrees & equal to max. shear stress:
�max � �max � 40.7�MPa
A-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 allowablenormal strain is 8.0 � 10�4. The required outside diameter d2 of the shaft isapproximately:
(A) 38 mm(B) 56 mm(C) 87 mm(D) 91 mm
Solution
T � 5.7�kN�m G � 27�GPa d1 � 52�mm
�a1 � 45�MPa �a � 8.0�(10�4)
Allowable shear strain based on allowable normal strain for pure shear
�a � 2��a � 1.600 � 10�3 so resulting allow. shear stress is:
�a2 � G��a � 43.2�MPa
gmax � ad2
2b �u � 1.508 � 10�3
u � 1.8�degm
26 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L
d2d1
d2T T
d2d1
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 26
So allowable shear stress based on normal strain governs �a � �a2
Use torsion formula to relate required d2 to allowable shear stress:
so rearrange equation to get
Solve resulting 4th order equation numerically, or use a calculator and trial & error
T � 5700000 N�mm d1 � 52 �a � 43.2
gives d2 � 91�mm
A-3.12: A motor drives a shaft with diameter d � 46 mm at f � 5.25 Hzand delivers P � 25 kW of power. The maximum shear stress in the shaft isapproximately:
(A) 32 MPa(B) 40 MPa(C) 83 MPa(D) 91 MPa
Solution
f � 5.25�Hz d � 46�mm
P � 25�kW
Power in terms of torque T:
P � 2� f T
Solve for torque T:
Max. shear stress using torsion formula:
A-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 theshaft is approximately:
(A) 35 mm(B) 40 mm(C) 47 mm(D) 61 mm
tmax �
T�ad
2b
Ip� 39.7�MPa
T �P
2�p�f� 757.9�N�m
Ip �p
32�d4 � 4.396 � 105
�mm4
f(d2) � d24
� a16p
�Ttab �d2 � d1
4
d24
� d14 �
16p
�Tta
�d2tmax �
T�ad2
2b
p
32�(d2
4 � d14)
APPENDIX A FE Exam Review Problems 27
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
df
P
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 27
Solution
f � 10�Hz P � 35�kW
�a � 45�MPa
Power in terms of torque T:
P � 2� f T
Solve for torque T:
Shear formula: or
Solve for diameter d:
A-3.14: A drive shaft running at 2500 rpm has outer diameter 60 mm and innerdiameter 40 mm. The allowable shear stress in the shaft is 35 MPa. The maximumpower that can be transmitted is approximately:
(A) 220 kW(B) 240 kW(C) 288 kW(D) 312 kW
Solution
n � 2500 rpm
d2 � 0.060 m
d1 � 0.040 m
Shear formula:
or
Power in terms of torque T:
P � 2� f T � 2�(n/60) T � (� n/30) T
Pmax � 312 kWPmax �2�p�n
60�Tmax � 3.119 � 105
W
Tmax �2�ta�Ip
d2� 1191.2 N�mt �
T�ad2
2b
Ip
Ip �p
32�(d2
4 � d14) � 1.021 � 10�6
m4
ta � 35�(106) N
m2
d � a16�Tp�ta
b13
� 39.8�mm
t �16.T
p�d3t �
T�ad
2b
p
32�d 4
T �P
2�p�f� 557.0�N�m
28 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
df
P
d n
d2
d1
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 28
A-4.1: A simply supported beam with proportional loading (P � 4.1 kN) hasspan length L � 5 m. Load P is 1.2 m from support A and load 2P is 1.5 m fromsupport 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 c � 1.5�m
L � a b c � 5.00 m
P � 4.1�kN
Statics to find reaction force at B:
Moment just left of load 2P:
M � RB�c � 10.1�kN�m � compression on top of beam
A-4.2: A simply-supported beam is loaded as shown in the figure. The bendingmoment 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
Solution
RB �1
L�[P�a 2�P�(a b)] � 6.724�kN
APPENDIX A FE Exam Review Problems 29
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AC
B
1.8 kN/m7.5 kN
1.0 m 1.0 m0.5 m
5.0 m3.0 m
A B
2PP
La b c
Statics to find reaction force at A:
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 beam
RA �1
5�m� cc1.8�
kNm
�(3�m � 0.5�m)2
2 d7.5�kN�(3�m 1�m)d�7.125�kN
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 29
A-4.3: A cantilever beam is loaded as shown in the figure. The bending momentat 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
30 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AB
1.8 kN/m4.5 kN
1.0 m1.0 m 3.0 m
5.0 m 1.0 m
BA C
9 kN4.5 kN
1.0 m
Cut beam at 0.5 m from support; use statics and right-hand FBD to findinternal moment at that point
(tension on top of beam)
A-4.4: An L-shaped beam is loaded as shown in the figure. The bending momentat 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
Solution
� 18.5�kN�m
M � 0.5�m�(4.5�kN) a0.5�m 1.0�m 3.0�m
2b �1.8�
kNm
�(3.0�m)
Use statics to find reaction at B; sum moments about A
RB �1
5�m�[9�kN�(6�m) � 4.5�kN�(1.�m)] � 9.90�kN
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 30
Cut beam at midpoint of AB; use right hand FBD, sum moments
� tension ontop of beam
A-4.5: A T-shaped simple beam has a cable with force P anchored at B andpassing over a pulley at E as shown in the figure. The bending moment just leftof 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:
(downward)
Now cut beam & cable just left of CE & use left FBD; 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:
Solve for P:
A-4.6: A simple beam (L � 9 m) with attached bracket BDE has force P � 5 kNapplied 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
P � �35
16�(1.25) � 2.73 kN
MC �4
5�P�(3) a�4
7 �Pb �(2 3) simplify S �
16�P
35
MC �4
5�P�(3) VA�(2 3)
VA ��4
7�P
VA ��1
7�m�[P�(4�m)]
M � RB�a5�m
2b � 9�kN�a5�m
2 1�mb � 6.75�kN�m
APPENDIX A FE Exam Review Problems 31
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
A
E P
C DB
Cable4 m
2 m 3 m 2 m
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 31
32 APPENDIX A FE Exam Review Problems
A CB
1.6 m 1.6 m 1.6 m
4.5 kN • m15 kN/m
A C
L
DE
P
B
L—6
L—3
L—2
Sum moments about B to get reaction at A:
Cut beam at midspan, use left FBD & sum moments to find moment atmidspan:
Mmspan � RA�s1.6) � 15�s1.6)�a1.6
2b S 11.85 kN�m
RA �1
3.2� c15�s1.6)�a1.6
1.6
2b 4.5d S 19.40625 kN
Solution
Sum moments about A to find reaction at C:
Cut through beam just right of B, then use FBD of BCto find moment at B:
Substitute numbers for L and P:
L � 9�m P � 5�kN
A-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
MB S 5�L�P
12� 18.8�kN� m
MB � RC�aL
2
L
3b S
5�L�P
12
RC �1
L� cp�aL
6
L
3b d S
P
2
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 32
A-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
d � 1.5�mm R � 0.6�m
A-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 includesthe 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 � 5�m b � 200�mm h � 280�mm
Section modulus:
Max. moment at midspan:
Max. flexural stress at midspan:
smax �Mmax
S� 11.4�MPa
Mmax �q�L2
8� 29.7�kN�m
S �b�h2
6� 2.613 � 106�m3
q � 9.5�kNm
�max �d
2�aR d
2b
� 1.248 � 10�3
�max �
d
2
R d
2
S d
2�aR d
2b
APPENDIX A FE Exam Review Problems 33
d
R
A
L
B
q
h
b
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 33
A-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 maximumbending stress in the pipe is approximately:
(A) 28 MPa(B) 33 MPa(C) 47 MPa(D) 59 MPa
Solution
34 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
s
L
L � 12�m s � 4�m d2 � 100�mm d1 � 75�mm
Pipe cross sectional properties:
Uniformly distributed weight of pipe, q:
Vertical force at each lift point:
Max. moment is either at lift points (M1) or at midspan (M2):
� controls, tension on top
� tension on top
Max. bending stress at lift point:
A-5.4: A beam with an overhang is loaded by a uniform load of 3 kN/m over itsentire length. Moment of inertia Iz � 3.36 � 106 mm4 and distances to top andbottom of the beam cross section are 20 mm and 66.4 mm, respectively. It is
smax �
u M1 u �ad2
2b
I� 59.0�MPa
M2 � F�s
2 � q�
L
2�aL
4b � �1.484�kN�m
M1 � �q�aL � s
2b �aL � s
2b � �3.958�kN�m
F �q�L
2� 1.484�kN
q � gCI�A � 0.247�kNm
I �p
64�sd2
4 � d14) � 3.356 � 106�mm4A �
p
4�sd2
2 � d12) � 3436�mm2
gCI � 72�kN
m3
d2
d1
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 34
known that reactions at A and B are 4.5 kN and 13.5 kN, respectively. Themaximum bending stress in the beam is approximately:
(A) 36 MPa(B) 67 MPa(C) 102 MPa(D) 119 MPa
Solution
APPENDIX A FE Exam Review Problems 35
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AB
C
4 m 2 m
3kN/m
z
y
C
20 mm
66.4 mm
RA � 4.5�kN Iz � 3.36�(106)�mm4
Location of max. positive moment in AB (cut beam at location of zero shear &use left FBD):
� compression on top ofbeam
Compressive stress on top of beam at xmax:
Tensile stress at bottom of beam at xmax:
Max. negative moment at B (use FBD of BC to find moment; compression onbottom of beam):
st2 �Mneg�(20�mm)
Iz� 35.7�MPa
sc2 �Mneg�(66.4�mm)
Iz� 118.6�MPa
Mneg � a3�kNm
b �(2�m)2
2� 6.000�kN�m
st1 �Mpos�(66.4�mm)
Iz� 66.696�MPa
sc1 �Mpos�(20�mm)
Iz� 20.1�MPa
Mpos � RA�xmax � 3�kNm
�xmax
2
2� 3.375�kN�mxmax
RA
q S 1.5�m
q � 3�kNm
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 35
A-5.5: A steel hanger with solid cross section has horizontal force P � 5.5 kNapplied at free end D. Dimension variable b � 175 mm and allowable normalstress is 150 MPa. Neglect self weight of the hanger. The required diameter ofthe 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:
NA � P
(leftward)
MA � P�(2�b) � 1.9�kN�m
(tension on bottom)
Max. normal stress at bottom of cross section at A:
Set �max � �a and solve for required diameter d:
(���a)�d3 � (4�P)�d � 64�P�b � 0 � solve numerically or by trial &error to find
dreqd � 5.11 cm
A-5.6: A cantilever wood pole carries force P � 300 N applied at its free end, aswell as its own weight (weight density � 6 kN/m3). The length of the pole isL � 0.75 m and the allowable bending stress is 14 MPa. The required diameterof 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
�a � 14�MPa gw � 6�kN
m3
smax �4�P�(16�b d)
p�d 3smax �P
ap�d2
4b
(2�P.b)�ad
2b
ap�d 4
64b
36 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6b
2b
A B
D CP
2b
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 36
Uniformly distributed weight of pole:
Max. moment at support:
Section modulus of pole cross section:
Set Mmax equal to �a � S and solve for required min. diameter d:
Or
� solve numerically or by trial
& error to find
dreqd � 5.50 cm
Since wood pole is light, try simpler solution which ignores self weight:
P�L � �a�S Or
A-5.7: A simply supported steel beam of length L � 1.5 m and rectangular crosssection (h � 75 mm, b � 20 mm) carries a uniform load of q � 48 kN/m, whichincludes its own weight. The maximum transverse shear stress on the cross sec-tion at 0.25 m from the left support is approximately:
(A) 20 MPa(B) 24 MPa(C) 30 MPa(D) 36 MPa
Solution
L � 1.5�m
h � 75�mm b � 20�mm
q � 48�kNm
dreqd � cP�L�a 32p�sa
b d13
� 5.47�cm
ap�sa
32b �d3 � P�L
ap�sa
32b �d3 � ap�gw�L2
8b �d2 � P�L � 0
P�L cgw�ap�d2
4b d �L�
L
2 � saap�d3
32b � 0
S �
p�d4
64
ad
2b
S p�d3
32S �
I
ad
2b
Mmax � P�L w�L�L
2
w � gw�ap�d2
4b
APPENDIX A FE Exam Review Problems 37
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L
AB
P
d
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 37
Cross section properties:
A � b�h � 1500�mm2
Support reactions:
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:
Or more simply . . .
A-5.8: A simply supported laminated beam of length L � 0.5 m and square crosssection weighs 4.8 N. Three strips are glued together to form the beam, with theallowable shear stress in the glued joint equal to 0.3 MPa. Considering also theweight of the beam, the maximum load P that can be applied at L/3 from the leftsupport is approximately:
(A) 240 N(B) 360 N(C) 434 N(D) 510 N
Solution
tmax �3�V0.25
2�A� 24.0�MPa
tmax �V0.25�Q
1�b� 24.0�MPa
R �q�L
2� 36.0�kN
I �b�h3
12� 7.031 � 105�mm4
Q � ab� h
2b �
h
4� 14062�mm3
38 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
q
b
h
L
L � 0.5�m W � 4.8�N
h � 36�mm b � 36�mm �a � 0.3�MPa
q �W
L� 9.60�
Nm
36 mm
P at L/3
36 mm
12 mm12 mm12 mm
q
L
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 38
Cross section properties:
A � b�h � 1296�mm2
Max. shear force at left support:
Shear stress on glued joint at left support; set T � Ta then solve for Pmax:
Or Or
so for �a � 0.3�MPa
A-5.9: An aluminum cantilever beam of length L � 0.65 m carries a distributedload, which includes its own weight, of intensity q/2 at A and q at B. The beamcross section has width 50 mm and height 170 mm. Allowable bending stress is95 MPa and allowable shear stress is 12 MPa. The permissible value of loadintensity q is approximately:
(A) 110 kN/m(B) 122 kN/m(C) 130 kN/m(D) 139 kN/m
Solution
L � 0.65�m b � 50�mm h � 170�mm
�a � 95�MPa �a � 12�MPa
Cross section properties:
A � b�h � 8500�mm2
Reaction force and moment at A:
MA �5
12�q�L2
MA �q
2 �L�
L
2
1
2 �
q
2 �L�
2�L
3RA �
3
4 �q�LRA �
1
2�aq
2 qb �L
S �b�h2
6� 2.408 � 105�mm3I �
b�h3
12� 2.047 � 107�mm4
Pmax �3
2�a3�b�h�ta
4 �
q�L
2b � 434 N
ta �4
3�b�h� c q�L
2 P�a2
3b d
ta �4�Vmax
3�b�ht �
Vmax�ab�h2
9b
ab�h3
12b �b
t �Vmax�Qjoint
I�b
Vmax �q�L
2 P�a2
3b
I �b�h3
12� 1.400 � 105�mm4
Qjoint � ab� h
3b �ah
2 �
h
6b � 5184�mm3
APPENDIX A FE Exam Review Problems 39
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AB
q
L
q2—
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 39
Compare max. permissible values of q based on shear and moment allowablestresses; smaller value controls
So, since �a � 12�MPa
So, since �a � 95�MPa
A-5.10: An aluminum light pole weighs 4300 N and supports an arm of weight700 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 crosssection at the base has outside diameter 235 mm and thickness 20 mm. Themaximum compressive stress at the base is approximately:
(A) 16 MPa(B) 18 MPa(C) 21 MPa(D) 24 MPa
Solution
H � 7.5�m B � 1.2�m
W1 � 4300�N W2 � 700�N
P1 � 1500�N
d2 � 235�mm t � 20�mm
d1 � d2 � 2�t � 195�mm
Pole cross sectional properties at base:
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
I �p
64�(d2
4 � d14) � 7.873 � 107�mm4
A �p
4�(d2
2 � d12) � 13509�mm2
qmax2 �12
5�sa�S
L2 � 130.0�kNm
sa �
5
12�q�L2
Ssmax �
MA
S
qmax1 �8
9�ta�A
L� 139�
kNm
ta �3
2� ±
3
4 �q�L
A ≤tmax �
3
2�RA
A
40 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.2 m
235 mm
20 mm
W1 = 4300 N
W2 = 700 N
P1 = 1500 N
7.5 m
y
xy
x
z
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 40
Compressive stress at right side at base of pole:
A-5.11: Two thin cables, each having diameter d � t/6 and carrying tensile loadsP, are bolted to the top of a rectangular steel block with cross section dimensionsb � t. The ratio of the maximum tensile to compressive stress in the block dueto loads P is:
(A) 1.5(B) 1.8(C) 2.0(D) 2.5
Solution
Cross section properties of block:
A � b�t
Tensile stress at top of block:
Compressive stress at bottom of block:
Ratio of max. tensile to compressive stress in block:
A-5.12: 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, respec-tively. The ratio of the maximum permissible bending moment about the z-axis tothat about the y-axis is most nearly:
(A) 0.5(B) 0.7(C) 1.2(D) 1.5
9
5� 1.8ratio � ` st
sc ` S
9
5
sc �P
A �
P�ad
2
t
2b �a t
2b
I S �
5�P
2�b�t
st �P
A
P�ad
2
t
2b �a t
2b
I S
9�P
2�b�t
d �t
6I �
b�t3
12
sc �N
A
|M|�ad2
2b
I� 15.9�MPa
APPENDIX A FE Exam Review Problems 41
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
b
t
P P
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 41
Solution
b � 200�mm t � 12�mm
h � 300�mm
Ec � 14�GPa Ee � 100�GPa
�ac � 9.5�MPa
�ae � 140�MPa
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:
Bending about z axis based on allowable stress in each material (lesser valuecontrols)
Bending about y axis based on allowable stress in each material (lesser valuecontrols)
� allowable stress in the core, not
exterior cover sheet, controls
moments about both axes
ratioz_to_y �Mmax_cz
Mmax_cy� 0.72
Mmax_ey � sae�(Ec�Icy Ee�Iey)
ab
2 tbEe
� 136.2�kN�m
Mmax_cy � sac�(Ec�Icy Ee�Iey)
b
2 �Ec
� 74.0�kN�m
Mmax_ez � sae�
aEc�Icz Ee�Iezbh
2 �Ee
� 109.2�kN�m
Mmax_cz � sac�
aEc�Icz Ee�Iezbh
2 �Ec
� 52.9�kN�m
Iey �2�h�t3
12� 2�(t�h)�ab
2
t
2b2
� 8.099 � 107�mm4
Iez �2�t�h3
12� 5.400 � 107�mm4
Icy �h�b3
12� 2.000 � 108�mm4Icz �
b�h3
12� 4.500 � 108�mm4
42 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
z
y
C
200 mm12 mm12 mm
300
mm
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 42
A-5.13: 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 mostnearly:
(A) 2.9 kN�m(B) 3.5 kN�m(C) 4.3 kN�m(D) 9.9 kN�m
Solution
b � 90�mm t � 8�mm
h � 160�mm
Ew � 11�GPa Es � 190�GPa
�aw � 6.5�MPa
�as � 110�MPa
Aw � b�h � 14400�mm2
As � b�t � 720�mm2
Locate NA (distance h2 above base) by summing 1st moments of EA aboutbase of beam; then find h1 � dist. from NA to top of beam:
h1 � h t � h2 � 118.93�mm
Moments of inertia of wood and steel about NA:
Allowable moment about z axis based on allowable stress in each material(lesser value controls)
Mmax_s � sas�(Ew�Iw Es�Is)
h2�Es� 10.11�kN�m
Mmax_w � saw�(Ew�Iw Es�Is)
h1�Ew� 4.26�kN�m
Iw �b�h3
12 Aw�ah1 �
h
2b2
� 5.254 � 107�mm4
Is �b�t3
12 As�ah2 �
t
2b2
� 1.467 � 106�mm4
h2 �
Es�As� t
2 Ew�Aw�at
h
2b
Es�As Ew�Aw� 49.07�mm
APPENDIX A FE Exam Review Problems 43
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
160 mm
8mm
90 mm
z
y
O
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 43
A-5.14: A steel pipe (d3 � 104 mm, d2 � 96 mm) has a plastic liner with innerdiameter d1 � 82 mm. The modulus of elasticity of the steel is 75 times that ofthe modulus of the plastic. Allowable stresses in steel and plastic are 40 MPa and550 kPa, respectively. The allowable bending moment for the composite pipe isapproximately:
(A) 1100 N�m(B) 1230 N�m(C) 1370 N�m(D) 1460 N�m
Solution
d3 � 104�mm d2 � 96�mm d1 � 82�mm
�as � 40�MPa
�ap � 550�kPa
Cross section properties:
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)
Modular ratio: n � 75
Divide through by Ep in moment expressions above
Mmax_ p � sap�(Ip n�Is)
ad2
2b
� 1374�N�m
Mmax_s � sas�(Ip n�Is)
ad3
2b �n
� 1230�N�m
n �Es
Ep
Mmax_p � sap�(Ep�Ip Es Is)
ad2
2b �Ep
Mmax_s � sas�(Ep�Ip Es�Is)
ad3
2b �Es
Ip �p
64�(d2
4� d1
4) � 1.950 � 106�mm4
Is �p
64�(d3
4 � d24) � 1.573 � 106�mm4
Ap �p
4�(d2
2 � d12) � 1957.2�mm2
As �p
4�(d3
2 � d22) � 1256.6�mm2
44 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
z
y
Cd1 d2 d3
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 44
A-5.15: 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 bendingmoment of 1.75 N�m is applied about the z axis. The ratio of the maximum stress inthe aluminum to that in the copper is approximately:
(A) 0.6(B) 0.8(C) 1.0(D) 1.5
Solution
b � 25�mm t � 1.5�mm Aa � b�t � 37.5�mm2 Ac � Aa � 37.500�mm2
M � 1.75�N�m
Ea � 70�GPa Ec � 110�GPa
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:
h1 � 2�t � h2 � 1.667�mm h1 h2 � 3.000�mm 2�t � 3.000�mm
Moments of inertia of aluminum and copper strips about NA:
Bending stresses in aluminum and copper:
Ratio of the stress in the aluminum to that of the copper:
A-5.16: 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 momentis applied about the z axis resulting in a maximum stress in the aluminum of55 MPa. The maximum stress in the steel is approximately:
(A) 86 MPa(B) 90 MPa(C) 94 MPa(D) 98 MPa
sa
sc� 0.795
sc �M�h2�Ec
Ea�Ia Ec�Ic� 52.6�MPasa �
M�h1�Ea
Ea�Ia Ec�Ic� 41.9�MPa
Ia �b�t3
12 Aa�ah1 �
t
2b2
� 38.542�mm4
Ic �b�t3
12 Ac�ah2 �
t
2b2
� 19.792�mm4
h2 �
Ec�Ac�t
2 Ea�Aa�at
t
2b
Ec�Ac Ea�Aa� 1.333�mm
APPENDIX A FE Exam Review Problems 45
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
b t
t
z
y
O C
A
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 45
Solution
b � 25�mm ha � 42�mm hs � 68�mm
Ea � 72�GPa Es � 190�GPa �a � 55�MPa
Aa � b�ha � 1050.0�mm2
As � b�hs � 1700.0�mm2
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:
h1 � ha hs � h2 � 65.57�mm
h1 h2 � 110.00�mm
Moments of inertia of aluminum and steel parts about NA:
Set max. bending stress in aluminum to given value then solve for moment M:
Use M to find max. bending stress in steel:
A-6.1: A rectangular plate (a � 120 mm, b � 160 mm) is subjected to com-pressive stress �x � � 4.5 MPa and tensile stress �y � 15 MPa. The ratio of thenormal stress acting perpendicular to the weld to the shear stress acting along theweld is approximately:
(A) 0.27(B) 0.54(C) 0.85(D) 1.22
Solution
a � 120�mm b � 160�mm
u � atanaa
bb � 36.87�deg
ss �M�h2�Es
Ea�Ia Es�Is� 98.4�MPa
M �sa�(Ea�Ia Es�Is)
h1�Ea� 3.738�kN�m
Ia �b�ha
3
12 Aa�ah1 �
ha
2b2
� 2.240 � 106�mm4
Is �b�hs
3
12 As�ah2 �
hs
2b2
� 8.401 � 105�mm4
h2 �
Es�As� hs
2 Ea�Aa�ahs
ha
2b
Ea�Aa Es�As� 44.43�mm
46 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
ha
b
hs
Aluminum
Steel
z
y
O
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 46
�x � �4.5�MPa �y � 15�MPa
�xy � 0
Plane stress transformation: normal and shear stresses on y-face of element rotated through angle �(perpendicular to & along weld seam):
` su
tu ` � 0.85
tu � �sx � sy
2�sinc2�au
p
2b d txy�cosc2�au
p
2b d � �9.36�MPa
� 7.98�MPasu �sx sy
2
sx � sy
2�cosc2�au
p
2b d txy�sinc2�au
p
2b d
APPENDIX A FE Exam Review Problems 47
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
σy
Weld σxab
y
xO
τxy
σy
σx
A-6.2: A rectangular plate in plane stress is subjected to normal stresses �x and�y and shear stress �xy. Stress �x is known to be 15 MPa but �y and �xy areunknown. However, the normal stress is known to be 33 MPa at counterclock-wise angles of 35° and 75° from the x axis. Based on this, the normal stress �y
on the element below is approximately:
(A) 14 MPa(B) 21 MPa(C) 26 MPa(D) 43 MPa
Solution
�x � 15 �35 � 33 �75 � �35
Plane stress transformations for 35 deg & 75 deg:
su �sx sy
2
sx � sy
2�cos(2�u) txy�sin(2�u)
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 47
For � � 35 deg:
Or �y 2.8563��xy � 69.713
And for � � 75 deg:
Or �y 0.5359��xy � 34.292
Solving above two equations for �y and Txy gives:
so �y � 26.1 MPa
A-6.3: A rectangular plate in plane stress is subjected to normal stresses �x �35 MPa, �y � 26 MPa, and shear stress �xy � 14 MPa. The ratio of the magni-tudes of the principal stresses (�1/�2) is approximately:
(A) 0.8(B) 1.5(C) 2.1(D) 2.9
Solution
�x � 35�MPa �y � 26�MPa �xy � 14�MPa
Principal angles:
Plane stress transformations:
s2 �sx sy
2
sx � sy
2�cos(2�uP2) txy�sin(2�uP2) � 15.79�MPa
s1 �sx sy
2
sx sy
2�cos(2�uP1) txy�sin(2�uP1) � 45.21�MPa
uP2 � uP1 p
2� 126.091�deg
uP1 �1
2�atana 2�txy
sx � syb � 36.091�deg
asy
txyb � a1 2.8563
1 0.5359b�1
�a69.713
34.292b � a26.1
15.3b MPa
sx sy
2
sx � sy
2�cos[2�(u75)] txy�sin[2�(u75)] � s75
u75 � 75� p
180
sx sy
2
sx � sy
2�cos[2�(u35)] txy�sin[2�(u35)] � s35
u35 � 35� p
180
48 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
y
xO
τxy
σy
σx
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 48
Ratio of principal stresses:
A-6.4: A drive shaft resists torsional shear stress of 45 MPa and axial compres-sive stress of 100 MPa. The ratio of the magnitudes of the principal stresses(�1/�2) is approximately:
(A) 0.15(B) 0.55(C) 1.2(D) 1.9
Solution
�x � �100�MPa �y � 0
�xy � �45�MPa
Principal angles:
Plane stress transformations:
� actually �2
� this is �1
So
Ratio of principal stresses:
A-6.5: A drive shaft resists torsional shear stress of 45 MPa and axial compres-sive stress of 100 MPa. The maximum shear stress is approximately:
(A) 42 MPa(B) 67 MPa(C) 71 MPa(D) 93 MPa
` s1
s2 ` � 0.15
s2 � min(suP1, suP2) � �117.268�MPas1 � max(suP1, suP2) � 17.268�MPa
suP2 �sx sy
2
sx � sy
2�cos(2�uP2) txy�sin(2�uP2) � 17.27�MPa
suP1 �sx sy
2
sx � sy
2�cos(2�uP1) txy�sin(2�uP1) � �117.27�MPa
uP2 � uP1 p
2� 110.994�deg
uP1 �1
2�atana 2�txy
sx � syb � 20.994�deg
s1
s2� 2.86
APPENDIX A FE Exam Review Problems 49
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
100 MPa
45 MPa
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 49
Solution
�x � �100�MPa �y � 0
�xy � �45�MPa
Max. shear stress:
A-6.6: A drive shaft resists torsional shear stress of �xy � 40 MPa and axial com-pressive stress �x � �70 MPa. One principal normal stress is known to be 38 MPa(tensile). The stress �y is approximately:
(A) 23 MPa(B) 35 MPa(C) 62 MPa(D) 75 MPa
Solution
�x � �70�MPa �xy � 40�MPa
�y � unknown �prin � 38�MPa
Stresses �x and �y must be smaller than the given principal stress so:
�1 � �prin
Substitute into stress transformation equation and solve for sy:
A-6.7: 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 principalstresses (�1/�2) 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
� 23.2�MPa
sx sy
2 B asx � sy
2b2
txy2 � s1 solve, sy S
626�MPa
27
tmax � B asx � sy
2b2
txy2 � 67.3�MPa
50 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
100 MPa
45 MPa
y
xO
τxy
σy
σx
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 50
Solution
P � 160�kN b � 95�mm h � 300�mm
c � 0.8�m d � 200�mm
Cross section properties:
A � b�h � 28500�mm2
Moment, shear force and normal and shear stresses at A:
MA � �P�c � �1.280 � 105�kN�mm VA � P
Plane stress state at A: �x � �A �xy � �A �y � 0
Principal stresses:
Ratio of principal stresses (�1 / �2):
A-6.8: 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 ofthe magnitudes of the principal stresses (�1/�2) at a point a � 1.0 m from theleft support and distance d � 100 mm up from the bottom of the beam isapproximately:
(A) 9(B) 17(C) 31(D) 41
` s1
s2 ` � 17.9
s2 �sx sy
2 B asx � sy
2b2
txy2 � �1.767�MPa
s1 �sx sy
2 B asx � sy
2b2
txy2 � 31.709�MPa
uP �1
2�atana 2�txy
sx � syb � 13.283�deg
sA �
�MA�ad � h
2b
I� 29.942�MPatA �
VA�QA
I�b� 7.485�MPa
QA � [b�(h � d)]� c h2 �
(h � d)
2 d � 9.500 � 105�mm3
I �b�h3
12� 2.138 � 108�mm4
d
h� 0.667
APPENDIX A FE Exam Review Problems 51
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P
c
A
b d
h
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 51
Solution
L � 4.5�m
b � 95�mm h � 280�mm
a � 1.0�m d � 100�mm
q � 25�kNm
52 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
h
b
q
a
L
Cross section properties:
A � b�h � 26600�mm2
Moment, shear force and normal and shear stresses at distance a from left support:
Plane stress state: �x � � �xy � � �y � 0�MPa
Principal stresses:
Ratio of principal stresses (�1 / �2): ` s1
s2 ` � 40.7
s2 �sx sy
2 B asx � sy
2b2
txy2 � �0.254�MPa
s1 �sx sy
2 B asx � sy
2b2
txy2 � 10.324�MPa
uP �1
2�atana 2�txy
sx � syb � 8.909�deg
s �
�Ma�ad � h
2b
I� 10.070�MPat �
Va�Q
I�b� 1.618�MPa
Ma �q�L
2 �a �
q�a2
2� 4.375 � 104�kN�mmVa �
q�L
2 � q�a � 31.250�kN
Q � [b�(h � d)]� c h2 �
(h � d)
2 d � 8.550 � 105�mm3
I �b�h3
12� 1.738 � 108�mm4
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 52
A-7.1: A thin wall spherical tank of diameter 1.5 m and wall thickness 65 mmhas internal pressure of 20 MPa. The maximum shear stress in the wall of thetank is approximately:
(A) 58 MPa(B) 67 MPa(C) 115 MPa(D) 127 MPa
Solution
d � 1.5�m t � 65�mm p � 20�MPa
Thin wall tank since:
Biaxial stress:
� � 115.4�MPa
Max. shear stress at 45 deg. rotation is 1/2 of �
A-7.2: A thin wall spherical tank of diameter 0.75 m has internal pressureof 20 MPa. The yield stress in tension is 920 MPa, the yield stress in shear is475 MPa, and the factor of safety is 2.5. The modulus of elasticity is210 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 E � 210�GPa
�Y � 920�MPa �Y � 475�MPa FSY � 2.5
� 0.28 �a � 1220�(10�6)
Thickness based on tensile stress:
t1 �
p�ad
2b
2�a sY
FSYb
� 10.190�mm
tmax �s
2� 57.7�MPa
s �
p�ad
2b
2�t
t
ad
2b
� 0.087
APPENDIX A FE Exam Review Problems 53
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Weld
Weld
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 53
Thickness based on shear stress:
Thickness based on normal strain:
t3 � 10.54�mm � largest value controls
A-7.3: A thin wall cylindrical tank of diameter 200 mm has internal pressure of11 MPa. The yield stress in tension is 250 MPa, the yield stress in shear is140 MPa, and the factor of safety is 2.5. The minimum permissible thickness ofthe tank is approximately:
(A) 8.2 mm(B) 9.1 mm(C) 9.8 mm(D) 11.0 mm
Solution
d � 200�mm p � 11�MPa
�Y � 250�MPa �Y � 140�MPa FSY � 2.5
Wall thickness based on tensile stress:
Wall thickness based on shear stress:
A-7.4: A thin wall cylindrical tank of diameter 2.0 m and wall thickness 18 mmis open at the top. The height h of water (weight density � 9.81 kN/m3) in thetank at which the circumferential stress reaches 10 MPa in the tank wall isapproximately:
(A) 14 m(B) 18 m(C) 20 m(D) 24 m
t2 �
p�ad
2b
2�a tY
FSYb
� 9.821�mm t2
ad
2b
� 0.098
t1 �
p�ad
2b
sY
FSY
� 11.00�mm � larger value governs t1
ad
2b
� 0.110
t3 �
p�ad
2b
2��a�E �(1 � v)
t2 �
p�ad
2b
4�a tY
FSYb
� 9.868�mm
54 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 54
Solution
d � 2�m t � 18�mm �a � 10�MPa
Pressure at height h: ph � �w�h
Circumferential stress:
Set �c equal to �a and solve for h:
A-7.5: The pressure relief valve is opened on a thin wall cylindrical tank, withradius to wall thickness ratio of 128, thereby decreasing the longitudinal strainby 150 � 10�6. Assume E � 73 GPa and v � 0.33. The original internal pres-sure in the tank was approximately:
(A) 370 kPa(B) 450 kPa(C) 500 kPa(D) 590 kPa
Solution
�L � 148�(10� 6)
E � 73�GPa � 0.33
Longitudinal strain:
Set � to �L and solve for pressure p:
A-7.6: A cylindrical tank is assembled by welding steel sections circumferen-tially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is2.0 MPa. The maximum stress in the heads of the tank is approximately:
(A) 38 MPa(B) 45 MPa(C) 50 MPa(D) 59 MPa
p �2�E��L
rt�(1 � 2�n)� 497�kPa
� �p
2�E�ar
tb �(1 � 2� )
rt �r
t rt � 128
h �sa�t
(gw)�ad
2b
� 18.3 m
sc �
ph�ad
2b
t sc �
(gw�h)�ad
2b
t
gw � 9.81�kN
m3
APPENDIX A FE Exam Review Problems 55
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
d
h
strain gage
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 55
Solution
d � 1.5�m t � 20�mm p � 2.0�MPa
A-7.7: 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. Themaximum 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
A-7.8: A cylindrical tank is assembled by welding steel sections circumferen-tially. 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 p � 2.0�MPa
A-7.9: A cylindrical tank is assembled by welding steel sections circumferen-tially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is2.0 MPa. The maximum shear stress in the heads is approximately:
(A) 19 MPa(B) 23 MPa(C) 33 MPa(D) 35 MPa
sw �
p�ad
2b
2�t� 37.5�MPa
sc �
p�ad
2b
t� 75.0�MPa
sh �
p�ad
2b
2�t� 37.5�MPa
56 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Welded seams
Welded seams
Welded seams
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 56
Solution
d � 1.5�m t � 20�mm p � 2.0�MPa
A-7.10: 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. Themaximum 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
A-7.11: A cylindrical tank is assembled by welding steel sections in a helicalpattern with angle � � 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 � 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 t � 20�mm p � 2.75�MPa
E � 210�GPa � 0.28 � � 50�deg
Circumferential stress:
Circumferential strain:
�c �sc
2�E �(2 � n) � 4.50 � 10�4
sc �
p�ad
2b
t� 110.000�MPa
tmax �
p�ad
2b
2�t� 37.5�MPa
th �
p�ad
2b
4�t� 18.8�MPa
APPENDIX A FE Exam Review Problems 57
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Welded seams
Welded seams
Helical weld
α
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 57
A-7.12: A cylindrical tank is assembled by welding steel sections in a helicalpattern with angle � � 50 degrees. Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus E � 210 GPa andPoisson’s ratio � 0.28. The longitudinal strain in the the wall of the tank isapproximately:
(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 p � 2.75�MPa
E � 210�GPa � 0.28 � � 50�deg
Longitudinal stress:
Longitudinal strain:
A-7.13: A cylindrical tank is assembled by welding steel sections in a helicalpattern with angle � � 50 degrees. Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus E � 210 GPa andPoisson’s ratio � 0.28. The normal stress acting perpendicular to the weldis 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 � � 50�deg
Longitudinal stress:
sL �
p�ad
2b
2�t� 55.000�MPa So sx � sL
�L �sL
E �(1 � 2�n) � 1.15 � 10�4
sL �
p�ad
2b
2�t� 55.000�MPa
58 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Helical weld
α
Helical weld
α
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 58
Circumferential stress:
Angle perpendicular to the weld: � � 90�deg � � � 40.000�deg
Normal stress perpendicular to the weld:
A-7.14: A segment of a drive shaft (d2 � 200 mm, d1 � 160 mm) is subjectedto a torque T � 30 kN�m. The allowable shear stress in the shaft is 45 MPa. Themaximum 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 �a � 45�MPa
T � 30�kN�m
Cross section properties:
Normal and in-plane shear stresses:
�x � 0
Maximum in-plane shear stress: set �max � �allow then solve for �y
Finally solve for P � �y A: Pmax � �y�A � 286�kN
A-7.15: A thin walled cylindrical tank, under internal pressure p, is compressedby 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 is60 MPa. The maximum allowable internal pressure pmax is approximately:
(A) 5 MPa(B) 10 MPa(C) 13 MPa(D) 17 MPa
t max � B asx � sy
2b2
txy2 So sy � #4�(ta � txy)
2 � 25.303�MPa
txy �
T�ad2
2b
IP� 32.349�MPasy �
�P
A
Ip �p
32�(d2
4 � d14) � 9.274 � 107�mm4
A �p
4�(d2
2 � d12) � 11310�mm2
s40 �sx sy
2
sx � sy
2�cos (2�u) � 77.7�MPa
sc �
p�ad
2b
t� 110.000�MPa So sy � sc
APPENDIX A FE Exam Review Problems 59
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P
T
T
P
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 59
Solution
d � 90�mm t � 5.5�mm �a � 110�MPa
F � 75�kN
Circumferential normal stress:
and setting �c � �a and solving for pmax:
Longitudinal normal stress:
So set �L � �a and solve for pmax:
Check also in-plane & out-of-plane shear stresses: all are below allowableshear stress so circumferential normal stress controls as noted above.
A-8.1: An aluminum beam (E � 72 GPa) with a square cross section andspan length L � 2.5 m is subjected to uniform load q � 1.5 kN/m. The allow-able bending stress is 60 MPa. The maximum deflection of the beam isapproximately:
(A) 10 mm(B) 16 mm(C) 22 mm(D) 26 mm
Solution
E � 72�(103)MPa �a � 60 MPa
L � 2500 mm
Max. moment and deflection at L/2:
Mmax �q�L2
8 dmax �
5�q�L4
384�E�I
q � 1.5 N
mm MPa �
N
mm2
pmaxL � asa F
Ab �
4�t
d� 38.7�MPa
sL �
pmax�ad
2b
2�t �
F
A Or sL �
pmax�d
4�t �
F
A
pmaxc � sa�a2�t
db � 13.4�MPa � controls
sc �
pmax�ad
2b
t
A � 2�p�d
2 �t � 1555�mm2
60 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F F
q = 1.5 kN/m
L = 2.5 m
36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 60
Moment of inertia and section modulus for square cross section (height �width � b)
Flexure formula
Max. deflection formula
A-8.2: An aluminum cantilever beam (E � 72 GPa) with a square cross sectionand span length L � 2.5 m is subjected to uniform load q � 1.5 kN/m. The allow-able bending stress is 55 MPa. The maximum deflection of the beam is approx-imately:
(A) 10 mm(B) 20 mm(C) 30 mm(D) 40 mm
Solution
E � 72�(103) MPa �a � 55 MPa
L � 2500 mm
Max. moment at support & max. deflection at L:
Moment of inertia and section modulus for square cross section (height �width � b)
S �I
ab
2b
S b3
6I �
b4
12
d max �q�L4
8�E�IM max �
q�L2
2
MPa �N
mm2q � 1.5 N
mm
d max �5�q�L4
384�E� ≥ ca3
4�q�L2
s max b1
3d4
12 ¥
� 22.2 mm
dmax �5�q�L4
384�E�ab4
12b
so solve for d max if s max � sa s max � sa
smax �Mmax
S smax �
q�L2
8
ab3
6b
so b3 �3
4�q�L2
smax
I �b4
12 S �
1
ab
2b
S b3
6
APPENDIX A FE Exam Review Problems 61
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L
q
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 61
Flexure formula
so
Max. deflection formula
so solve for max if �max � �a �max � �a
A-8.3: A steel beam (E � 210 GPa) with I � 119 � 106 mm4 and span lengthL � 3.5 m is subjected to uniform load q � 9.5 kN/m. The maximum deflectionof the beam is approximately:
(A) 10 mm(B) 13 mm(C) 17 mm(D) 19 mm
Solution
E � 210�(103) MPa I � 119�(106) mm4 � strong axis I for W310�52
L � 3500 mm
MPa �N
mm2q � 9.5 N
mm
dmax �q�L4
8�E� ≥ ca3�
q�L2
s max b1
3d4
12 ¥
� 29.9 mm
dmax �q�L4
8�E�ab4
12b
b3 � 3�q�L2
smax smax �
q�L2
2
ab3
6b
smax �Mmax
S
62 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
RB = kδB
MA
k = 48EI/L3A B
qy
xL
Max. deflection at A by superposition of SS beam mid-span deflection & RB/k:
d max �5�q�(2�L)4
384�E�I
(q�L)
a48�E�I
L3 b� 13.07 mm
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 62
A-8.4: A steel bracket ABC (EI � 4.2 � 106 N�m2) with span length L � 4.5 mand height H � 2 m is subjected to load P � 15 kN at C. The maximum rota-tion of joint B is approximately:
(A) 0.1 degrees(B) 0.3 degrees(C) 0.6 degrees(D) 0.9 degrees
Solution
E � 210�GPa I � 20�106�mm4 � strong axis I for W200�22.5
E�I � 4.20 � 106�N�m2
P � 15�kN
L � 4.5�m H � 2�m
APPENDIX A FE Exam Review Problems 63
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
A
C
BH
P
L
A
C
BH
P
L
Max. rotation at B: apply statically-equivalent moment P�H at B on SS beam
�Bmax � 0.011�rad
A-8.5: A steel bracket ABC (EI � 4.2 � 106 N�m2) with span length L � 4.5 mand height H � 2 m is subjected to load P � 15 kN at C. The maximum hori-zontal 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
E�I � 4.20 � 106�N�m2
P � 15�kN
L � 4.5�m H � 2�m
uBmax �(P�H)�L
3�E�I� 0.614�deg
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 63
Max. rotation at B: apply statically-equivalent moment P � H at B on SS beam
�Bmax � 0.011�rad
Horizontal deflection of vertical cantilever BC:
Finally, superpose �B � H and BC
C � �Bmax�H BC � 31.0�mm
A-8.6: A nonprismatic cantilever beam of one material is subjected to load P atits free end. Moment of inertia I2 � 2 I1. The ratio r of the deflection B to thedeflection 1 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)
Rotation at C due to both load P & moment PL/2 at C for nonprismatic beam:
Deflection at C due to both load P & moment PL/2 at C for nonprismatic beam:
Total deflection at B:
dB �5�L3�p
48�E�I2 a3�L2�P
8�E�I2b �
L
2
P�aL
2b3
3�E�I1 simplify S
L3�P�(7.I1 I2)
24�E�I1�I2
dB � dCl uC�L
2
P�aL
2b3
3�E�I1
dCl �
P�aL
2b3
3�E�I2
aP�L
2b �aL
2b2
2�E�I2 simplify S
5�L3�P
48�E�I2
uC �
P�aL
2b2
2�E�I2
aP�L
2b �
L
2
E�I2 simplify S
3�L2�P
8�E�I2
d1 �P�L3
3�E�I1
dBC �P�H3
3�E�I� 9.524�mm
uBmax �(P�H)�L
3�E�I� 0.614�deg
64 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
BCA I1
I2
P
L2— L
2—
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 64
Ratio B/1:
so
A-8.7: A steel bracket ABCD (EI � 4.2 � 106 N�m2), with span length L � 4.5 mand dimension a � 2 m, is subjected to load P � 10 kN at D. The maximumdeflection 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
E�I � 4.20 � 106�N�m2
P � 10�kN
L � 4.5�m a � 2�m
h � 206�mm
Statically-equivalent loads at end of cantilever AB:
• downward load P• CCW moment P � a
Downward deflection at B by superposition:
A-9.1: Beam ACB has a sliding support at A and is supported at C by a pinnedend steel column with square cross section (E � 200 GPa, b � 40 mm) andheight L � 3.75 m. The column must resist a load Q at B with a factor of safety2.0 with respect to the critical load. The maximum permissible value of Q isapproximately:
(A) 10.5 kN(B) 11.8 kN(C) 13.2 kN(D) 15.0 kN
dB
L� 0.005dB �
P�L3
3�E�I �
(P�a)�L2
2�E�I� 24.1�mm
r �7
8�a1
2b
1
8� 0.563
r �
L3�P�(7�I1 I2)
24�E�I1�I2
P�L3
3�E�I1
simplify S 7�I1
8�I2
1
8
APPENDIX A FE Exam Review Problems 65
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L
A B
CD
Pa
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 65
Solution
E � 200�GPa n � 2.0
b � 40�mm L � 3.75�m
Statics: sum vertical forces to find reaction at D:
RD � Q
So force in pin-pin column is Q
Allowable value of Q:
A-9.2: Beam ACB has a pin support at A and is supported at C by a steel columnwith 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 atB with a factor of safety 2.0 with respect to the critical load. The maximum per-missible 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
Effective length of pinned-fixed column:
Le � 0.699�L � 3.670 m
Statics: use FBD of ACB and sum moments about A to find force in column as a multiple of Q:
So force in pin-fixed column is 3Q
So Qcr �Pcr
3� 12.0�kNPcr � 3�Qcr Pcr �
p2�E�I
Le2 � 36.1�kN
FCD �Q�(3�d)
d S 3�Q
I �b4
12� 2.593 � 105�mm4
Qallow �Qcr
n� 15.0�kN
Pcr � Qcr Qcr �p2�E�I
L2 � 29.9�kN
I �b4
12� 2.133 � 105�mm4
66 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
B
D
A C
L
Q
d 2d
B
D
A C
L Q
d 2d
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 66
Allowable value of Q:
A-9.3: A steel pipe column (E � 190 GPa, � � 14 � 10�6 per degree Celsius,d2 � 82 mm, d1 � 70 mm) of length L � 4.25 m is subjected to a temperatureincrease �T. The column is pinned at the top and fixed at the bottom. The tem-perature 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 L � 4.25�m � � [14�(10�6)]
Effective length of pinned-fixed column:
Le � 0.699�L � 3.0 m
Axial compressive load in bar: P � EA �(�T)
Equate to Euler buckling load and solve for �T:
A-9.4: A steel pipe (E � 190 GPa, � � 14 � 10�6 per degree Celsius, d2 � 82 mm,d1 � 70 mm) of length L � 4.25 m hangs from a rigid surface and is subjectedto a temperature increase �T � 50 °C. The column is fixed at the top and has asmall gap at the bottom. To avoid buckling, the minimum clearance at the bottomshould 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
� � [14�(10�6)] / °C �T � 50 °C
�T �
p2�E�I
Le2
E�A�a Or �T �
p2�I
a�A�Le2 � 58.0�C
I �p
64�(d2
4 � d14) � 1.04076 � 106�mm4
d2 � 82�mm d1 � 70�mm A �p
4 �(d2
2 � d12) � 1432.57�mm2
Qallow �Qcr
n� 6.0�kN
APPENDIX A FE Exam Review Problems 67
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
B
L
A
ΔT
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 67
d2 � 82 mm d1 � 70 mm
Effective length of fixed-roller support column:
Le � 2.0�L � 8500.0 mm
Column elongation due to temperature increase:
1 � ���T�L � 2.975 mm
Euler buckling load for fixed-roller column:
Column shortening under load of P � Pcr:
Mimimum required gap size to avoid buckling: gap � 1 � 2 � 2.55 mm
A-9.5: A pinned-end copper strut (E � 110 GPa) with length L � 1.6 m is con-structed of circular tubing with outside diameter d � 38 mm. The strut mustresist an axial load P � 14 kN with a factor of safety 2.0 with respect to the crit-ical 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 L � 1.6�m d � 38�mm n � 2.0 P � 14�kN
Pcr � n�P Pcr � 28.0�kN
Solve for required moment of inertia I in terms of Pcr then find tube thickness
I � 66025�mm4
Moment of inertia Solve numerically for min. thickness t:
tmin � 4.33�mm d � 2�tmin � 29.3�mm � inner diameter
d 4 � (d � 2�t)4 � I� 64p
I �p
64�[d 4 � (d � 2�t)4]
Pcr �p2�E�I
L2 I �Pcr�L2
p2�E
d2 �Pcr�L
E�A� 0.422 mm
Pcr �p2�E�I
Le2 � 27.013 kN
I �p
64�(d2
4 � d14) � 1.041 � 106 mm4
A �p
4�(d2
2 � d12) � 1433 mm2
68 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
LΔT
gapfrictionlesssurface
d
t
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 68
A-9.6: 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 andC are L � 7 m apart. The critical value of load W for buckling in the plane of thetruss is nearly:
(A) 138 kN(B) 146 kN(C) 153 kN(D) 164 kN
Solution
E � 210�GPa L � 7�m
d � 100�mm t � 6.5�mm
Moment of inertia
Member lengths:
LBA � L�cos(40�deg) � 5.362 m
LBC � L�cos(50�deg) � 4.500 m
Statics at joint B to find member forces FBA and FBC:
• Sum horizontal forces at joint B:
FBA�cos(40�deg) � FBC�cos(50�deg)
where
• Sum vertical forces at joint B:
W � FBA�sin(40�deg) FBC�sin(50�deg)
So member forces in terms of W are:
FBC � W�� and FBA � FBC�� or FBA � W�(���)with ��� � 0.643
Euler buckling loads in BA & BC:
so � lower value controls
so WBC_cr � b�FBC_cr � 164�kNFBC_cr �p2�E�I
LBC2 � 214.630�kN
WBA_cr �b
a �FBA_cr � 138�kN
FBA_cr �p2�E�I
LBA2 � 151.118�kN
� 0.766
FBC � W�b where b �1
a cos(50�deg)
cos(40�deg)�sin(40�deg) sin(50�deg)b
W � FBC� cos(50�deg)
cos(40�deg)�sin(40�deg) FBC�sin(50�deg)
a � cos(50�deg)
cos(40�deg)� 0.839
FBA � FBC� cos(50�deg)
cos(40�deg)
I �p
64�[d4 � (d � 2�t)4] � 2.097 � 106�mm4
APPENDIX A FE Exam Review Problems 69
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
W
A
B
C40° 50°
L
d
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 69
A-9.7: A beam is pin-connected to the tops of two identical pipe columns, eachof height h, in a frame. The frame is restrained against sidesway at the top ofcolumn 1. Only buckling of columns 1 and 2 in the plane of the frame is ofinterest here. The ratio (a/L) defining the placement of load Qcr, which causesboth columns to buckle simultaneously, is approximately:
(A) 0.25(B) 0.33(C) 0.67(D) 0.75
Solution
Draw FBD of beam only; use statics to show that Qcr causes forces P1 and P2 in columns 1 & 2 respectively:
Buckling loads for columns 1 & 2:
Solve above expressions for Qcr, then solve for required a/L so that columnsbuckle at the same time:
Or
Or Or
So
A-9.8: A steel pipe column (E � 210 GPa) with length L � 4.25 m isconstructed of circular tubing with outside diameter d2 � 90 mm and innerdiameter d1 � 64 mm. The pipe column is fixed at the base and pinned at thetop and may buckle in any direction. The Euler buckling load of the columnis most nearly:
(A) 303 kN(B) 560 kN(C) 690 kN(D) 720 kN
a
L�
0.6992
(1 0.6992)� 0.328
a
L � a� 0.6992L
0.6992�(L � a) �
La
� 0
p2�El
(0.699�h)2 a L
L � ab �
p2�El
h2 �aLab � 0
p2�El
(0.699�h)2 a L
L � ab �
p2�El
h2 �aLab
Pcr2 �p2�El
h2 � aa
Lb �Qcr
Pcr1 �p2�El
(0.699�h)2 � aL � a
Lb �Qcr
P2 �a
L�Qcr
P1 � aL � a
Lb �Qcr
70 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
a L-a
Qcr
1EI
h
EI
h2
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 70
Solution
E � 210�GPa L � 4.25�mm
d2 � 90�mm d1 � 64�mm
Moment of inertia
I � 2.397 � 106�mm4
Effective length of column for fixed-pinned case:
Le � 0.699�L � 2.971 m
A-9.9: An aluminum tube (E � 72 GPa) AB of circular cross section has a pinnedsupport at the base and is pin-connected at the top to a horizontal beam sup-porting a load Q � 600 kN. The outside diameter of the tube is 200 mm and thedesired factor of safety with respect to Euler buckling is 3.0. The required thick-ness t of the tube is most nearly:
(A) 8 mm(B) 10 mm(C) 12 mm(D) 14 mm
Solution
E � 72 GPa L � 2.5 m n � 3.0 Q � 600 kN d � 200 mm
�Mc � 0 P � 1000�kN
Find required I based on critical buckling loadCritical load
Pcr � P�n
Pcr � 3000�kN
I � 26.386 � 106�mm4
Moment of inertia
tmin � 9.73�mmt min �
d � B4 d4 � I� 64p
2
I �p
64�[d4 � (d � 2�t)4]
I �Pcr�L2
p2�E
Pcr �p2�E�I
L2
P �2.5�Q
1.5
Pcr �p2�E�I
Le2 � 563�kN
I �p
64�(d2
4 � d14)
APPENDIX A FE Exam Review Problems 71
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
d1 d2
BC
A
d � 200 mm
Q = 600 kN
1.5 m 1.0 m
2.5 m
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 71
A-9.10: 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 underthe same load is approximately:
(A) 0.55(B) 0.72(C) 0.81(D) 1.10
Solution
Equate Euler buckling load expressions for the two columns considering their different properties, base fixity conditions and lengths:
Simplify then solve for L2/L1:
A-9.11: 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 underthe 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:
p2�E�I1
(0.699�L)2 �
p2�a2
3�Eb �(I2)
L2
L2
L1� B
4
3�0.6992 � 0.807
a L2
0.699�L1b2
�4
3
p2�E�I
(0.699�L1)2 �
p2�a2
3�Eb �(2�I)
L22
72 APPENDIX A FE Exam Review Problems
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
EI
Pcr
L1
2E/32I
L2
Pcr
EI1
Pcr
L
2E/3I2
L
Pcr
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 72
Simplify then solve for I2/I1:
I2
I1�
3
2
(0.699)2 � 3.07
I2
I1�
L2
(0.699�L)2�E
2
3�E
APPENDIX A FE Exam Review Problems 73
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36041_01_online_appA_p01-73.qxd 3/22/11 11:30 AM Page 73