Download pdf - 36041 01 online appA p01-73 - · PDF file2 is uniformly distributed around the cap plate ... A circular aluminum tube of length L 650 mm is loaded in ... Ignore friction between the

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

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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


Strain gage

L

PP

35°35°

Clevis

Cable sling

P

Steel plate


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




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


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


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



d PP

d2d1

Steeltube

Polyethylenebar


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



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


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



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


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



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


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



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


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



P

Steelwire

P

Copper wire


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


L

A B45° 45°

P

P

C

a b c

B

P1 P2P3

A C D




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





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


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



L

B

A


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



P

Monel shellBrass core

d3

d1

d2

L


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



rod, dr

pin, dp

Clevis

ΔT



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:


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



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


P P

a

b

T d T



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:


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:


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



PPd


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:


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



H

P

P

PC

ABL

σθ/2

θτθ τθ

τθ τθ

σθ


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



L

dT T


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



L

dT T

L

dT T


Solution



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


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:


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



d2

d1 d

L1

d1

T1T2

L2

B C

d2

A


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



L1 L2

T

A B C

d1 d2T


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



C

D

A

B

TA

TB

TC

TD


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:


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



L

d2d1

d2T T

d2d1


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:


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)



df

P


Solution

f � 10�Hz P � 35�kW

�a � 45�MPa


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


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



df

P

d n

d2

d1


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:


Solution

RB �1

L�[P�a 2�P�(a b)] � 6.724�kN



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


A-4.3: A cantilever beam is loaded as shown in the figure. The bending momentat 0.5 m from the support is approximately:


Solution



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:


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


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:


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



A

E P

C DB

Cable4 m

2 m 3 m 2 m



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:


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



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


d

R

A

L

B

q

h

b



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:


Solution



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


known that reactions at A and B are 4.5 kN and 13.5 kN, respectively. Themaximum bending stress in the beam is approximately:


Solution



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


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



6b

2b

A B

D CP

2b


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:


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



L

AB

P

d


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



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



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


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



AB

q

L

q2—


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:


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



1.2 m

235 mm

20 mm

W1 = 4300 N

W2 = 700 N

P1 = 1500 N

7.5 m

y

xy

x

z


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



b

t

P P


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



z

y

C

200 mm12 mm12 mm

300

mm


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:


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



160 mm

8mm

90 mm

z

y

O


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


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



z

y

Cd1 d2 d3


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:


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



b t

t

z

y

O C

A


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



ha

b

hs

Aluminum

Steel

z

y

O


�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



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:


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)


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


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



y

xO

τxy

σy

σx


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:


` s1

s2 ` � 0.15

s2 � min(suP1, suP2) � �117.268�MPas1 � max(suP1, suP2) � 17.268�MPa

suP2 �sx sy

2

sx � sy


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



100 MPa

45 MPa


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:


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



100 MPa

45 MPa

y

xO

τxy

σy

σx


Solution

P � 160�kN b � 95�mm h � 300�mm

c � 0.8�m d � 200�mm


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



P

c

A

b d

h


Solution

L � 4.5�m

b � 95�mm h � 280�mm

a � 1.0�m d � 100�mm

q � 25�kNm



h

b

q

a

L


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


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:


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



Weld

Weld


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




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:


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



d

h

strain gage


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:


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:


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:


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



Welded seams

Welded seams

Welded seams


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:


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 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



Welded seams

Welded seams

Helical weld

α


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:


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



Helical weld

α

Helical weld

α



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


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:


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



P

T

T

P


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



F F

q = 1.5 kN/m

L = 2.5 m


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



L

q


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



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


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



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�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


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



BCA I1

I2

P

L2— L

2—


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�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



L

A B

CD

Pa


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



B

D

A C

L

Q

d 2d

B

D

A C

L Q

d 2d


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



B

L

A

ΔT


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



LΔT

gapfrictionlesssurface

d

t


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



W

A

B

C40° 50°

L

d


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



a L-a

Qcr

1EI

h

EI

h2


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)



d1 d2

BC

A

d � 200 mm

Q = 600 kN

1.5 m 1.0 m

2.5 m


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



EI

Pcr

L1

2E/32I

L2

Pcr

EI1

Pcr

L

2E/3I2

L

Pcr


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


