36041 01 online appA p01-73 - Cengage...the cable is at an angle of to the vertical. The average shear stress in each pin is approximately: (A) 22 MPa (B) 28 MPa (C) 40 MPa (D) 48

APPENDIX A:

FE EXAM REVIEW PROBLEMS & SOLUTIONS

JAMES M. GERE BARRY J. GOODNO

© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A-1.1: A plane truss has downward applied load P at joint 2 and another load Papplied leftward at joint 5. The force in member 3–5 is:

(A) 0(B)(C)(D) �1.5 P

�P�P/2

Appendix A

P

PL

L

LL1 6

2 4

53

Solution

so

Method of sections

Cut through members 3-5, 2-5 and 2-4; use right hand FBD

F35 � �P

F35 L � P L � 0

�M2 � 0

V6 � 0

V6 (3

L) � P

L � P

L � 0

�M1 � 0

P

P

L

LLL

16

2 4

53

H1 V1 V6

F35

F25

F24

P

L

L

62 4

5

V6

1083

FE Exam Review Problems

78572_app_ptg01_hr_1083-1168.qxd 1/24/12 12:15 PM Page 1083

A-1.2: The force in member FE of the plane truss below is approximately:

(A)(B)(C) 3.9 kN(D) 4.7 kN

Solution

�2.2 kN�1.5 kN

1084 APPENDIX A FE Exam Review Problems


A B C D

G

E

F

3 m

4.5 m

3 m15 kN 5 kN

3 m

1 m

10 kN3 m

A B C D

G

E

Ey

Ay

Ax

F

3 m

4.5 m

3 m15 kN 5 kN

3 m

1 m

10 kN3 m

CB D5 kN10 kN

3 m

Ey

FFE

E3 m

1 m

·FFE310

·FFE�110

Statics

Ey � 25 kN

�MA � 0 Ey (6 m) � 15 kN

(3 m) � 10 kN

(6 m) � 5 kN

(9 m) � 0

Method of sections: cut through BC, BE and FE; use right-hand FBD; sum moments about B

Solving

FFE � 3.95 kN

FFE � 5

4 110 kN

�3

110 FFE (3 m) �

1

110 FFE

(3 m) � 10 kN (3 m) � 5 kN (6 m) �Ey (3 m) � 0


A-1.3: The moment reaction at A in the plane frame below is approximately:

(A)(B)(C)(D)

Solution

�6400 N�m�3600 N�m�2280 N�m�1400 N�m

APPENDIX A FE Exam Review Problems 1085


B

C

A

900 N

3 m

4 mPinconnection

1200 N/m1.2 m

Statics: use FBD of member BC to find reaction Cyy

Sum moments about A for entire structure

Solving for MA

MA � 6400 N�m

MA � Cy (3 m) � 900 N (1.2 m) �

1

2 a1200

Nmb 4 m a2

3 4 mb � 0

�MA � 0

Cy � 900 N (1.2 m)

3 m � 360 N

�MB � 0 Cy (3 m) � 900 N (1.2 m) � 0

B

C

900 N

3 m

1.2 m

CyBy

Bx

B

C

A

900 N

3 m

4 m

1200 N/m1.2 m

MA

Cy

Ax

Ay


A-1.4: 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 sAB

Solve for P2 P2 � �AB ABC � P1 � 18.00 kN

Check: � same as in AB

A-1.5: 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

d � 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



A

B

C

P1

dAB

tAB

dBC

tBC

P2




A-1.6: 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 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 � �

Cross sectional area of each pin:

Tensile force in cable:

Shear stress in each clevis pin (double shear):

A-1.7: 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 d 2

p � 380 mm2

35�

35�

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 � AL

Max. axial stress in wire of length L: �max � L

Max. length of wire:

A-1.8: An aluminum bar (E � 72 GPa, � � 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: � � Ea � �436.4 MPa � below yield stress of 480 MPaso Hooke’s Law applies

Max. acceptable compressive load:

A-1.9: An aluminum bar (E � 70 GPa, � � 0.33) of diameter 20 mm is stretchedby axial forces P, causing its diameter to decrease by 0.022 mm. The load P isapproximately:

(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 � � 0.33

Lateral strain:

�0.001

Axial strain:

Axial stress: � � Ea � 233.3 MPa � below yield stress of 270 MPaso Hooke’s Law applies

Max. acceptable load:

A-1.10: A polyethylene bar (E � 1.4 GPa, � � 0.4) of diameter 80 mm isinserted in a steel tube of inside diameter 80.2 mm and then compressed by axialforce P. The gap between steel tube and polyethylene bar will close when com-pressive load 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 � � 0.4

Lateral strain:

L �0.003

Axial strain:

Axial stress: � � Ea � �8.8 MPa � well below ultimate stress of 28MPa 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 �

L ��d

dinit

d PP

d2d1

Steeltube

Polyethylenebar




P2

dABtAB

dBCtBC

ABC Cap plate

P1

Axial strain of BC:

Axial stress in BC: �BC � EBC � �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

v ��L

BC� 0.34

L ��tBC

tBC� 3.600 10�4

BC ��(P1 � P2)

E ABC� �1.061 10�3

A-1.11: 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




A-1.12: A titanium bar (E � 100 GPa, v � 0.33) with square cross section (b � 75 mm) and length L � 3.0 m is subjected to tensile load P � 900 kN. 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 � b2L � 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 � Lb � 74.96040 mm

Final volume of bar: Vfinal � bf2Lf � 1.68841562 107 mm3

Increase in volume of bar: �V � Vfinal � Vinit � 9156 mm3

A-1.13: 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




a

b

V

t

dPP

L

(1) allowable value of P based on elongation

smax � Ea � 162.0 MPa

� elongation governs

(2) allowable load P based on tensile stress

A-1.15: 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

Ave. shear stress:

Ave. shear strain:

Shear modulus of elastomer:

A-1.14: 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 sa � 180 MPada � 2.7 mm

G �t

g� 1.902 MPa

g � arctanad

tb � 0.249

t �V

a b� 0.474 MPa




Solution

db � 18 mm d � 240 mm ta � 90 MPa n � 8

Bolt shear area:

Max. torque:

A-1.16: 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 sa � 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 [d

2 � (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: ds � dc

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 dmax � 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. Comparethis nonprismatic bar to a prismatic bar of the same volume of material with constant diameter d and length L. The elongation of the prismatic bar under thesame 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 d 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 d � PL/EA at the free end is:

(A) x � L/5(B) x � L/4(C) x � L/3(D) x � 3L/5

Solution

Forces in Segments 1 & 2:

Displacement at free end:

Set d3 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� �

P (L � 5 x)

2 AE

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 � 2.1 GPa 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: dm � db

Statics: Pm � Pb � P so

Set dB equal to 0.10 mm and solve for load P:

so with db � 0.10 mm

and then

A-2.8: A steel rod (Es � 210 GPa, dr � 12 mm, ) is heldstress free between rigid walls by a clevis and pin (dp 15 mm) assembly at eachend. If the allowable shear stress in the pin is 45 MPa and the allowable normalstress in the rod is 70 MPa, the maximum permissible temperature drop �T isapproximately:

(A) 14 (B) 20 (C) 28 (D) 40 �C

�C�C�C

�as � 12 10�6> �C

P �Eb

Ab

L db

a1 � Em

Am

Eb Ab

b � 13.40 kN

Pb �Eb

Ab

L dbdb �

Pb L

Eb Ab

Pb �P

a1 � Em

Am

Eb Ab

b

Pm �Em Am

Eb Ab Pb

Pm L

EmAm�

PbL

EbAb

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




rod, dr

pin, dp

Clevis

ΔT

Solution

Es � 210 GPa

dr � 12 mm dp � 15 mm

�s � 12(10�6)

ta � 45 MPa sa � 70 MPa

Force in rod due to temperature drop �T: and normal stress in rod:

Fr � EsAr (�s)�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, �s � 12 10�6 )is held stress free between rigid walls by a nut and washer (dw 22 mm)assembly at each end. If the allowable bearing stress between the washer andwall is 55 MPa and the allowable normal stress in the rod is 90 MPa, themaximum permissible temperature drop �T is approximately:

(A) 25 (B) 30 (C) 38 (D) 46

Solution

Es � 210 GPa dr � 15 mm dw � 22 mm

�s � 12(10�6)

sba � 55 MPa sa � 90 MPa

> �C

Aw �p

4 (dw

2 � dr2) � 203.4 mm2

Ar �p

4 d2

r � 176.7 mm2

�C�C�C�C

�> �C

�Tmaxpin �ta

(2 Ap)

Es Ar

as� 55.8

tpin �Fr

2 Ap

�Tmaxrod �sa

Es as

� 27.8

sr �Fr

Ar

> �C

Ap �p

4 dp

2 � 176.715 mm2

Ar �p

4 dr

2 � 113.097 mm2




rod, dr washer, dwΔT

Steel bolt

Copper tube

Force in rod due to temperature drop �T: and normal stress in rod:

Fr � Es Ar (�s)�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

as� 25.1

sb �Fr

Aw

�Tmaxrod �sa

Es as� 35.7

sr �Fr

Ar




P P

a

b

T d T

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

sa � 100 MPa

ta � 48 MPa

Bar is in uniaxial tension so Tmax � smax/2; since 2ta � sa, shear stress gov-erns

Pmax � ta 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 . The tem-perature change at which the wire goes slack is approximately:

(A) �5.7 (B) �12.6 (C) �12.6 (D) �18.2

Solution

E � 110 GPa d � 2.0 mm

�b � 19.5 (10�6) T � 85 N

A �p

4 d2 � 3.14 mm2

> �C

�C�C�C

�C

> �C

sc �Ps

Ac� 26.3 MPa

ss �Ps

As� 81.0 MPa




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 ab� �12.61

s �T

A � E ab �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 ofs� � 78 MPa (see fig.). The maximum shear stress in the bar is approximately:


Solution

�u � 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 sx

so

also �u � �sx sin(u) cos(u) � �55.154 MPa

Max. shear stress is 1/2 of max. normal stress for bar in uniaxial stress and ison plane at 45 deg. to axis of bar:

A-2.16: A prismatic bar (diameter) is loaded by force P1.

A stepped bar (diameters with radius R of fillets

) is loaded by force P2. Theallowable axial stress in the material is75 MPa. The ratio of the max-imum permissible loads that can beapplied to the bars, considering stressconcentration effects in the stepped bar, is:

(A) 0.9(B) 1.2(C) 1.4(D) 2.1

P1/P2

� 2 mmd2 � 25 mm,

d1 � 20 mm,d0 � 18 mm

tmax �sx

2� 58.5 MPa

sx �su

cos(u)2 � 117.0 MPa

u � atana 1

12b � 35.264�

tan(u)2 �1

2

P1

P2

d1

d0d1

d2

P2

P1

FIG. 2-66 Stress-concentration factor K for round bars with shoulder fillets.The dashed line is for a full quater-circular fillet.

K

0 0.05 0.10

1.11.2

1.5

0.15 0.20 0.25 0.30

3.0

2.5

2.0

1.5

RD1

K = P

1/4 D2

max nom

nom =

D2 – D12

R =

= 2 P

R

PD1

D2D1 D2

ss s

p


Solution

Prismatic bar

Stepped bar

from stress conc. Fig. 2-66

R

d1�

2 mm

20 mm� 0.100

d2

d1�

25 mm

20 mm� 1.250 so K � 1.75

P1 max � sallow ap d0

2

4b � (75 MPa) cp (18 mm)2

4d � 19.1 kN



K

0 0.05 0.10

1.11.2

1.5

0.15 0.20 0.25 0.30

3.0

2.5

2.0

1.5

RD1

K = P

1/4 D2

max nom

nom =

D2 – D12

R =K = 1.75

= 2 P

R

PD1

D2D1 D2

ss s

p

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

f � 3.5°

ga � 0.0005

Max. shear strain:

so dmax �2 ga

L

f� 12.28 mmgmax �

ad

2 fbL

P1 max

P2 max�

19.1 kN

13.5 kN� 1.41

P2 max � sallow

K ap d1

2

4b � a75 MPa

Kb cp(20 mm)2

4d � 13.5 kN

L

dT T




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

Solution

d � 70 mm

f �

ga � 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 of thebar 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

3.5�

L

dT T

L

dT T


Solution



d2

d1

L

T T

T1d

T1

L � 2.1 m d1 � 38 mm d2 � 56 mm f � 3.5° 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

ta � 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

T1

3

tmax �

T ad

2b

p

32 d4

�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

ta � 28 MPa

L1 � 0.9 m L2 � 0.75 m

fa �

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

1.8�

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

ta � 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

b1

3� 43.8 mm

d � a16 |TBC|p ta

b1

3� 37.7 mm

d � a16 |TAB|p ta

b1

3� 46.3 mm

t �

T ad

2b

p

32 d4

�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: tmax � Ggmax � 40.7 MPa

Max. tensile stress on plane at 45 degrees & equal to max. shear stress:

smax � tmax � 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

ta1 � 45 MPa a � 8.0(10�4)

Allowable shear strain based on allowable normal strain for pure shear

ga � 2a � 1.600 10�3 so resulting allow. shear stress is:

ta2 � Gga � 43.2 MPa

gmax � ad2

2b u � 1.508 10�3

u � 1.8�>m



L

d2d1

d2T T

d2d1


So allowable shear stress based on normal strain governs ta � ta2

Use torsion formula to relate required d2 to allowable shear stress:

and rearrange equation to get

Solve resulting 4th order equation numerically, or use a calculator and trial & error

T � 5700000 N�mm d1 � 52 mm ta � 43.2 MPa

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 � 2p 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 d 4 � 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

ta � 45 MPa


P � 2p 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 � 2p f T � 2p(n/60) 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-3.15: A prismatic shaft (diameter ) is loaded by torque T1. A stepped shaft (diameters radius R of fillets

) is loaded by torque T2. The allowable shear stress in the material is 42MPa. The ratio of the maximum permissible torques that can be applied tothe shafts, considering stress concentration effects in the stepped shaft is:

(A) 0.9(B) 1.2(C) 1.4(D) 2.1

T1/T2

� 2 mmd2 � 25 mm,d1 � 20 mm,

d0 � 19 mm



T1

T1

d0

R

T2 T2

D2D1

FIG. 3-59 Stress-concentration factor K for a stepped shaft in torsion. (Thedashed line is for a full quarter-circular fillet.)

K

01.00

1.50

2.00

0.10

1.5

1.21.1

0.20

D2 D1T

R

T

tmax = Ktnom tnom = 16T D1

3——

= 2D1

D2—–

D1

R—–

= + 2RD1D2

p


A B

2PP

La b c


Solution

Prismatic shaft

Stepped shaft

so from graph (see Fig. 3-59)

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 � RBc � 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:


RB �1

L [P a � 2 P (a � b)] � 6.724 kN

T1 max

T2 max�

56.6

48.9� 1.16

T2 max �tallow

K ap d1

3

16b �

42 MPa

1.35 c p

16 (20 mm)3d � 48.9 N�m

K � 1.35

d2

d1�

25 mm

20 mm� 1.250

R

d1�

2 mm

20 mm� 0.100

T1max �tallow IP

d0

2

� tallow ap d0

3

16b � 42 MPa cp

16 (19 mm)3d � 56.6 N�m



Solution

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

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

RA �1

5 m cc1.8

kNm

(3 m � 0.5 m)2

2 d�7.5 kN (3 m �1 m)d�7.125 kN


AB

1.8 kN/m4.5 kN

1.0 m1.0 m 3.0 m

AC

B

1.8 kN/m7.5 kN

1.0 m 1.0 m0.5 m

5.0 m3.0 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:


� 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)




Solution


5.0 m 1.0 m

BA C

9 kN4.5 kN

1.0 m

Use statics to find reaction at B; sum moments about A

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:

MC �4

5 P (3) � a�4

7 Pb (2 � 3) � �

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

RB �1

5 m [9 kN (6 m) � 4.5 kN (1. m)] � 9.90 kN

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

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:


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 �5 L P

12� 18.8 kN� m

MB � RC aL

2 �

L

3b �

5 L P

12

RC �1

L cP aL

6 �

L

3b d �

P

2

P �35

16 (1.25) � 2.73 kN




Sum moments about B to get reaction at A:

Cut beam at midspan, use left FBD & sum moments to find moment atmidspan:

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

q � 9.5 kNm

max �d

2 aR � d

2b

� 1.248 10�3

max �

d

2

R � d

2

�d

2 aR � d

2b

Mmspan � RA s1.6) � 15 s1.6) a1.6

2b � 11.85 kN�m

RA �1

3.2 c15 s1.6) a1.6 �

1.6

2b � 4.5d � 19.40625 kN



d

R


Section modulus:

Max. moment at midspan:

Max. flexural stress at midspan:

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

smax �Mmax

S� 11.4 MPa

Mmax �q L2

8� 29.7 kN�m

S �b h2

6� 2.613 106 m3



s

L

d2

d1

A

L

B

q

h

b

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: 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


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


Solution

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



AB

C

4 m 2 m

3 kN/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:

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� 1.5 m

q � 3 kNm


Max. negative moment at B (use FBD of BC to find moment; compression onbottom of beam):

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 (2b) � 1.9 kN�m

(tension on bottom)

Max. normal stress at bottom of cross section at A:

Set smax � sa and solve for required diameter d:

(�sa)d3 � (4P)d � 64Pb � 0 � solve numerically or by trial &error to find

dreqd � 5.11 cm

smax �4 P (16 b � d)

p d 3smax �P

ap d2

4b

�

(2 P b) ad

2b

ap d 4

64b

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



6b

2b

A B

D CP

2b


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

sa � 14 MPa

Uniformly distributed weight of pole:

Max. moment at support:

Section modulus of pole cross section:

Set Mmax equal to sa 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:

PL � sa S Or

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

�p d3

32S �

I

ad

2b

Mmax � P L � w L L

2

w � gw ap d2

4b

gw � 6 kN

m3



L

AB

P

d


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

Cross section properties:

A � bh � 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

tmax �3 V0.25

2 A� 24.0 MPa

tmax �V0.25

Q

I 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 � 48 kNm



q

b

h

L


Solution



36 mm

P at L/3

36 mm

12 mm12 mm12 mm

q

L

L � 0.5 m W � 4.8 N

h � 36 mm b � 36 mm ta � 0.3 MPa


A � bh � 1296 mm2

Max. shear force at left support:

Shear stress on glued joint at left support; set � a then solve for Pmax:

Or Or

so for ta � 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

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 h

t �

Vmax ab h2

9b

ab h3

12b b

t �Vmax

Qjoint

I b

tt

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

q �W

L� 9.60

Nm


Solution

L � 0.65 m b � 50 mm h � 170 mm

sa � 95 MPa ta � 12 MPa


A � bh � 8500 mm2

Reaction force and moment at A:

Compare max. permissible values of q based on shear and moment allowablestresses; smaller value controls

So, since ta � 12 MPa

So, since sa � 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:


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

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



AB

q

L

q2—


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 � P1H � �10.410 kN�m � results in compression at right

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:

st �P

A �

P ad

2 �

t

2b a t

2b

I�

9 P

2 b t

d �t

6I �

b t 3

12

sc �N

A �

|M| ad2

2b

I� 15.9 MPa

I �p

64 (d2

4 � d14) � 7.873 107 mm4

A �p

4 (d2

2 � d12) � 13509 mm2



1.2 m

235 mm

20 mm

W1 = 4300 N

W2 = 700 N

P1 = 1500 N

7.5 m

y

xy

x

z

b

t

P P


Compressive stress at bottom of block:

Ratio of max. tensile to compressive stress in block:

A-5.12: A rectangular beam with semicircular notches has dimen-sions and . The maximum allowable bending stressin the plastic beam is , and the bending moment is

. The minimum permissible width of the beam is:

(A) 12 mm(B) 20 mm(C) 28 mm(D) 32 mm

M � 185 N�ms max � 6.5 MPa

h1 � 140 mmh � 160 mm

9

5� 1.8ratio � ` st

sc ` �

9

5

sc �P

A �

P ad

2 �

t

2b a t

2b

I � �

5 P

2 b t



h = h1 + 2R

R—h1

3.0

2.5

2.0

1.50 0.05 0.10 0.15 0.20 0.25 0.30

K

1.05

= 1.2

1.1

b = thickness

K = s maxs nom

= 6M bh 2 1

s nom

h1h

M M

2Rhh1—

MM

h h1

2R

FIG. 5-50 Stress-concentration factor K for a notched beam of rectangularcross section in pure bending ( height of beam; thickness of beam,perpendicular to the plane of the figure). The dashed line is for semicircularnotches ( )h � h1 � 2R

b �h �


Solution

From Fig 5-50:

sallow

K�

6 M

b h12 so b min �

6 M K

sallow h12 �

6 (185 N�m) (2.25)

6.5 MPa C(140 mm)2D � 19.6 mm

K � 2.25

R

h1�

10

140� 0.071

h

h1�

160

140� 1.143

R �1

2 (h � h1) �

1

2 (160 mm � 140 mm) � 10.000 mm



z

y

C

200 mm12 mm12 mm

300

mm

h = h1 + 2R

R—h1

3.0

2.5

2.0

1.50 0.05 0.100.071 0.15 0.20 0.25 0.30

K

1.05

= 1.2

1.1

b = thickness

K = s maxs nom

= 6M bh 2 1

s nom

h1h

M M

2Rhh1—

K = 2.25

A-6.1: A composite beam is made up of a 200 mm 300 mm core (Ec � 14 GPa)and an exterior cover sheet (300 mm 12 mm, Ee � 100 GPa) on each side.Allowable stresses in core and exterior sheets are 9.5 MPa and 140 MPa, 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

Solution

b � 200 mm t � 12 mm

h � 300 mm

Ec � 14 GPa Ee � 100 GPa

sac � 9.5 MPa

sae � 140 MPa


Composite beam is symmetric about both axes so each NA is an axis ofsymmetry

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, notexterior cover sheet, controlsmoments about both axes

A-6.2: A composite beam is made up of a 90 mm 160 mm wood beam (Ew �11 GPa) and a steel bottom cover plate (90 mm 8 mm, Es � 190 GPa).Allowable stresses in wood and steel are 6.5 MPa and 110 MPa, respectively.The allowable bending moment about the z-axis of the composite beam is mostnearly:

(A) 2.9 kN?m(B) 3.5 kN?m(C) 4.3 kN?m(D) 9.9 kN?m

ratioz_to_y �Mmax_cz

Mmax_cy� 0.72

Mmax_ey � sae (Ec

Icy � Ee Iey)

ab

2 � tbEe

� 136.2kN�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




Solution

b � 90 mm t � 8 mm

h � 160 mm

Ew � 11 GPa Es � 190 GPa

saw � 6.5 MPa

sas � 110 MPa

Aw � bh � 14400 mm2

As � bt � 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)

A-6.3: 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

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


Solution

d3 � 104 mm d2 � 96 mm d1 � 82 mm

sas � 40 MPa

sap � 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

A-6.4: A bimetallic beam of aluminum (Ea � 70 GPa) and copper (Ec � 110 GPa)strips has width b � 25 mm; each strip has thickness t � 1.5 mm. A 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

Mmax_ p � sap (Ip � nIs)

ad2

2b

� 1374 N�m

Mmax_s � sas (Ip � nIs)

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


Solution

b � 25 mm t � 1.5 mm Aa � b t � 37.5 mm2 Ac � Aa � 37.5 mm2

M � 1.75 N�m

Ea � 70 GPa 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 � 2t � h2 � 1.667 mm h1 � h2 � 3.000 mm 2 t � 3.000 mm

Moments of inertia of aluminum and copper strips about NA:

Bending stresses in aluminum and copper:

Ratio of the stress in the aluminum to that of the copper:

A-6.5: A composite beam of aluminum (Ea � 72 GPa) and steel (Es � 190 GPa)has width b � 25 mm and heights ha � 42 mm, hs � 68 mm. A bending momentis applied about the z axis resulting in a maximum stress in the aluminum of55 MPa. The maximum stress in the steel is approximately:


Solution

b � 25 mm ha � 42 mm hs � 68 mm

Ea � 72 GPa Es � 190 GPa sa � 55 MPa

Aa � bha � 1050.0 mm2

As � bhs � 1700.0 mm2

sa

sc� 0.795

sc �Mh2

Ec

Ea Ia � Ec Ic

� 52.6 MPasa �Mh1

Ea

Ea Ia � Ec

Ic� 41.9 MPa

Ia �b t

3

12� Aa

ah1 �t

2b2

� 38.542 mm4

Ic �b t

3

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

ha

b

hs

Aluminum

Steel

z

y

O




σy

Weld σxab

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 � 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-7.1: A rectangular plate (a � 120 mm, b � 160 mm) is subjected to com-pressive stress sx � � 4.5 MPa and tensile stress sy � 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

sx � �4.5 MPa sy � 15 MPa

txy � 0

u � arctanaa

bb � 36.87�

ss �M h2

Es

Ea Ia � Es

Is� 98.4 MPa

M �sa (Ea

Ia � Es Is)

h1Ea� 3.738 kN�m

Ia �b ha

3

12� Aaah1 �

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




Plane stress transformation: normal and shear stresses on y-face of elementrotated through angle u (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

su �sx � sy

2�

sx � sy

2 cosc2 au �

p

2b d � txy sinc2 au �

p

2b d � 7.98 MPa

y

xO

τxy

σy

σx

A-7.2: A rectangular plate in plane stress is subjected to normal stresses sx andsy and shear stress txy. Stress sx is known to be 15 MPa but sy and txy 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 sy

on the element below is approximately:


Solution

sx � 15 MPa s35 � 33 MPa s75 � s35

Plane stress transformations for & :

su �sx � sy

2 �

sx � sy

2 cos(2 u) � txy sin(2 u)

75�35�

For :

Or sy � 2.8563txy � 69.713

sx � sy

2�

sx � sy

2 cos[2 (u35)] � txy sin[2 (u35)] � s35

u35 � 35�u � 35�


y

xO

τxy

σy

σx



And for :

Or sy � 0.5359 txy � 34.292

Solving above two equations for sy and txy gives:

so �y � 26.1 MPa

A-7.3: A rectangular plate in plane stress is subjected to normal stresses sx �35 MPa, sy � 26 MPa, and shear stress txy � 14 MPa. The ratio of the magni-tudes of the principal stresses (s1/s2) is approximately:

(A) 0.8(B) 1.5(C) 2.1(D) 2.9

Solution

sx � 35 MPa sy � 26 MPa txy � 14 MPa

Principal angles:

Plane stress transformations:

Ratio of principal stresses:s1

s2� 2.86

s2 �sx � sy

2�

sx � sy

2 cos(2uP2) � txy sin(2uP2) � 15.79 MPa

s1 �sx � sy

2�

sx � sy


uP2 � uP1 � p

2� 126.091�

uP1 �1

2 arctana 2txy

sx � syb � 36.091�

asy

txyb � c1 2.8563

1 0.5359d�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�u � 75�




100 MPa

45 MPa

A-7.4: 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(s1/s2) is approximately:

(A) 0.15(B) 0.55(C) 1.2(D) 1.9

Solution

sx � �100 MPa sy � 0

txy � �45 MPa

Principal angles:

Plane stress transformations:

� actually s2

� this is s1

So

Ratio of principal stresses:

A-7.5: 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(2uP1) � txy sin(2uP1) � �117.27 MPa

uP2 � uP1 �p

2� 110.994�

uP1 �1

2 arctana 2txy

sx � syb � 20.994�




y

xO

τxy

σy

σx

100 MPa

45 MPa

Solution

sx � �100 MPa sy � 0

txy � �45 MPa

Max. shear stress:

A-7.6: A drive shaft resists torsional shear stress of txy � 40 MPa and axial com-pressive stress sx � �70 MPa. One principal normal stress is known to be 38 MPa(tensile). The stress sy is approximately:


Solution

sx � �70 MPa txy � 40 MPa

sy is unknown sprin � 38 MPa

Stresses sx and sy must be smaller than the given principal stress so:

s1 � sprin

Substitute into stress transformation equation and solve for sy:

A-7.7: A cantilever beam with rectangular cross section (b � 95 mm, h � 300 mm)supports load P � 160 kN at its free end. The ratio of the magnitudes of the principalstresses (s1/s2) at point A (at distance c � 0.8 m from the free end and distance d � 200 mm up from the bottom) is approximately:

(A) 5(B) 12(C) 18(D) 25

� 23.2 MPa

sx � sy

2� B asx � sy

2b2

� txy2 � s1 solve, sy �

626 MPa

27

tmax � B asx � sy

2b2

� txy2 � 67.3 MPa


Solution

P � 160 kN b � 95 mm h � 300 mm

c � 0.8 m d � 200 mm


A � bh � 28500 mm2

Moment, shear force and normal and shear stresses at A:

MA � �Pc � �1.280 105 kN�mm VA � P

Plane stress state at A: sx � sA txy � tA sy � 0

Principal stresses:

Ratio of principal stresses (s1 / s2):

A-7.8: A simply supported beam (L � 4.5 m) with rectangular cross section(b � 95 mm, h � 280 mm) supports uniform load q � 25 kN/m. The ratio ofthe magnitudes of the principal stresses (s1/s2) 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 arctana 2 txy

sx � syb � 13.283�

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 � bh � 26600 mm2

Moment, shear force and normal and shear stresses at distance a from left support:

Plane stress state: sx � s txy � t sy � 0

Principal stresses:

Ratio of principal stresses (s1 / s2): ` 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 arctana 2 txy

sx � syb � 8.909�

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




Weld

Weld

A-8.1: 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 s

A-8.2: 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

sY � 920 MPa tY � 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




Thickness based on shear stress:

Thickness based on normal strain:

t3 � 10.54 mm � largest value controls

A-8.3: 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

sY � 250 MPa tY � 140 MPa FSY � 2.5

Wall thickness based on tensile stress:

Wall thickness based on shear stress:

A-8.4: 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 � n)

t2 �

p ad

2b

4 a tY

FSYb

� 9.868 mm


Solution

d � 2 m t � 18 mm sa � 10 MPa

Pressure at height h: ph � gwh

Circumferential stress:

Set sc equal to sa and solve for h:

A-8.5: 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 n � 0.33

Longitudinal strain:

Set to L and solve for pressure p:

A-8.6: 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 � 2n)

� 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-8.7: A cylindrical tank is assembled by welding steel sections circumferentially.Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa. 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-8.8: 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-8.9: 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-8.10: A cylindrical tank is assembled by welding steel sections circumferentially.Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa. 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-8.11: A cylindrical tank is assembled by welding steel sections in a helicalpattern with angle a � 50 degrees. Tank diameter is 1.6 m, thickness is 20 mm,and internal pressure is 2.75 MPa. Modulus E � 210 GPa and Poisson’s ratio n � 0.28. The circumferential strain in the wall of the tank is approximately:

(A) 1.9 10�4

(B) 3.2 10�4

(C) 3.9 10�4

(D) 4.5 10�4

Solution

d � 1.6 m t � 20 mm p � 2.75 MPa

E � 210 GPa n � 0.28


Circumferential strain:

c �sc

2 E (2 � n) � 4.50 10�4

sc �

p ad

2b

t� 110.000 MPa

a � 50�

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-8.12: A cylindrical tank is assembled by welding steel sections in a helicalpattern with angle a � 50 degrees. Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus E � 210 GPa andPoisson’s ratio n � 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 n � 0.28

Longitudinal stress:

Longitudinal strain:

A-8.13: A cylindrical tank is assembled by welding steel sections in a helicalpattern with angle a � 50 degrees. Tank diameter is 1.6 m, thickness is 20 mm, and internal pressure is 2.75 MPa. Modulus E � 210 GPa andPoisson’s ratio n � 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

Longitudinal stress:

sL �

p ad

2b

2 t� 55.000 MPa So sx � sL

a � 50�

L �sL

E (1 � 2n) � 1.15 10�4

sL �

p ad

2b

2 t� 55.000 MPa

a � 50�

Helical weld

α

Helical weld

α



Angle perpendicular to the weld:

Normal stress perpendicular to the weld:

A-8.14: 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 ta � 45 MPa

T � 30 kN�m


Normal and in-plane shear stresses:

sx � 0

Maximum in-plane shear stress: set �max � �allow then solve for sy

Finally solve for P � sy A: Pmax � �y A � 286 kN

A-8.15: 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

u � 90� � a � 40.000�

sc �

p ad

2b

t� 110.000 MPa So sy � sc



P

T

T

P




Solution

d � 90 mm t � 5.5 mm sa � 110 MPa

F � 75 kN

Circumferential normal stress:

and setting sc � sa and solving for pmax:

Longitudinal normal stress:

So set sL � sa 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-9.1: 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 sa � 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

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

� controlspmaxc � sa a2 t

db � 13.4 MPa

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-9.2: 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 sa � 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

�b3

6I �

b4

12

d max �q L4

8 E IM max �

q L2

2

q � 1.5 N

mm

d max �5q L4

384 E ≥ ca3

4 q L2

s max b1

3d4

12 ¥

� 22.2 mm

dmax �5q L4

384 E ab4

12b

so solve for d max if s max � sa

smax �Mmax

S smax �

qL2

8

ab3

6b

so b3 �3

4

qL2

smax

I �b4

12 S �

I

ab

2b

�b3

6



L

q




Flexure formula

so

Max. deflection formula

so solve for dmax if smax � sa

A-9.3: 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 W31052

L � 3500 mm

q � 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-9.4: 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(B) 0.3(C) 0.6(D) 0.9

Solution

E � 210 GPa I � 20 106 mm4 � strong axis I for W200 22.5

EI � 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 PH at B on SS beam

uBmax � 0.011 rad

A-9.5: 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

EI � 4.20 106 N�m2

P � 15 kN

L � 4.5 m H � 2 m

uBmax �(P H) L

3 E I� 0.614�




Max. rotation at B: apply statically-equivalent moment P H at B on SS beam

uBmax � 0.011 rad

Horizontal deflection of vertical cantilever BC:

Finally, superpose uB H and dBC

dC � uBmax H � dBC � 31.0 mm

A-9.6: 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 dB to thedeflection d1 at the free end of a prismatic cantilever with moment of inertia I1

carrying the same load is approximately:

(A) 0.25(B) 0.40(C) 0.56(D) 0.78

Solution

Max. deflection of prismatic cantilever (constant I1)

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�

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�

5 L3 P

48 E I2

uC �

P aL

2b2

2 E I2�

aP

L

2b

L

2

E I2�

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�

BCA I1

I2

P

L2— L

2—


Ratio dB/d1:

so

A-9.7: 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 � strong axis I for W20022.5

EI � 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-10.1: Propped cantilever beam AB has moment M1 applied at joint B. FrameworkABC has moment M2 applied at C. Both structures have constant flexural rigidity EI.If the ratio of the applied moments , the ratio of the reactive moments

at clamped support A is:

(A) 1(B) 3/2(C) 2(D) 5/2

MA1/MA2

M1/M2 � 3/2

dB

L� 0.005dB �

P L3

3 E I�

(P a) L2

2 E I� 24.1 mm

I � 20 106 mm4

r �7

8 a1

2b �

1

8� 0.563

r �

L3 P (7 I1 � I2)

24 E I1 I2

P L3

3 E I1

�7 I1

8 I2�

1

8



L

A B

CD

Pa




From Prob. 10.3-1:

Statically-equivalent moment at B is M2 so

Ratio of reactive moments is

A-10.2: Propped cantilever beam AB has moment M1 applied at joint B. FrameworkABC has moment M2 applied at C. Both structures have constant flexural rigidity EI.If the ratio of the applied moments , the ratio of the joint rotations atB, , is:

(A) 1(B) 3/2(C) 2(D) 5/2

Solution

uB1/uB2

M1/M2 � 3/2

MA1

MA2�

M1

M2 so

MA1

MA2�

3

2

MA2 �M2

2

MA1 �M1

2

Ax

B

L

M1

MA1

y

Ax

B

L

M2 C

L/2

MA2

y

A B

L

M1

θB1

x

y

θB2

Ax

B

L

M2 C

L/2

y

MA1 MA2

Solution


Documents

36041 01 online appA p01-73 - Cengage...the cable is at an angle of to the vertical. The average shear stress in each pin is approximately: (A) 22 MPa (B) 28 MPa (C) 40 MPa (D) 48