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APPENDIX A:

FE EXAM REVIEW PROBLEMS & SOLUTIONS

JAMES M. GERE BARRY J. GOODNO

A

B

C

P1

dABtAB

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 16kN dAB 30mm tAB 12mm

dBC 60mm tBC 9mm

Stress in AB:

Stress in BC: must equal AB

Solve for P2 P2 ABABC P1 18.00kN

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 400 106. The shortening of the bar isapproximately:

(A) 0.12 mm(B) 0.26 mm(C) 0.36 mm(D) 0.52 mm

Solution

400(106) L 650mm

L 0.260mm

sBC P1 P2

ABC 23.6MPa

sBC P1 P2

ABC

sAB P1

AAB 23.6MPa

ABC p

4[dBC

2 (dBC 2tBC)2] 1442mm2

AAB p

4[dAB

2 (dAB 2tAB)2] 679mm2

APPENDIX A FE Exam Review Problems 1

2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 1

A-1.3: A steel plate weighing 27 kN is hoisted by a cable sling that has a clevisat each end. The pins through the clevises are 22 mm in diameter. Each half ofthe cable is at an angle of 35 to the vertical. The average shear stress in each pinis approximately:

(A) 22 MPa(B) 28 MPa(C) 40 MPa(D) 48 MPa

Solution

W 27kN dp 22mm 35deg

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

2AP 21.7MPa

T

aW2b

cos(u) 16.48kN

Ap p

4d2p 380mm

2

2 APPENDIX A FE Exam Review Problems

2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Strain gage

L

PP

3535

Clevis

Cable sling

P

Steel plate

36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 2

Solution

Y 280MPa 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-1.5: 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 72GPa dinit 50mm dfinal 50.1mm 0.33

Lateral strain: L 0.002

Axial strain:

Axial stress: Ea 436.4MPa below yield stress of 480 MPaso Hookes 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 sap4 dinit2b 857 kN

a L

n 0.006

L dfinal dinit

dinit

Lmax sallow

g 1818 m

smax W

A

sallow sYFSY

140.0MPa

g 77kN

m3

APPENDIX A FE Exam Review Problems 3

2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 3

Solution

E 70GPa dinit 20mm d 0.022mm v 0.33

Lateral strain:

L0.001

Axial strain:

Axial stress: Ea 233.3MPa below yield stress of 270 MPaso Hookes Law applies

Max. acceptable compressive load:

A-1.7: A polyethylene bar (E 1.4 GPa, v 0.4) of diameter 80 mm is insertedin a steel tube of inside diameter 80.2 mm and then compressed by axial force P.The gap between steel tube and polyethylene bar will close when compressiveload P is approximately:

(A) 18 kN(B) 25 kN(C) 44 kN(D) 60 kN

Solution

E 1.4GPa d1 80mm d1 0.2mm v 0.4

Lateral strain:

L 0.003

Axial strain:

Axial stress: Ea 8.8MPa well below ultimate stress of28 MPa so Hookes Law applies

Max. acceptable compressive load:

Pmax sap4 d12b 44.0kN

a L

v 6.250 103

L d1d1

Pmax sap4 dinit2b 73.3kN

a L

v 3.333 103

L d

dinit

4 APPENDIX A FE Exam Review Problems

d PP

d2d1

Steeltube

Polyethylenebar

36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 4

A-1.8: A pipe (E 110 GPa) carries a load P1 120 kN at A and a uniformlydistributed load P2 100 kN on the cap plate at B. Initial pipe diameters andthicknesses are: dAB 38 mm, tAB 12 mm, dBC 70 mm, tBC 10 mm.Under loads P1 and P2, wall thickness tBC increases by 0.0036 mm. Poissonsratio v for the pipe material is approximately:

(A) 0.27(B) 0.30(C) 0.31(D) 0.34

Solution

E 110GPa dAB 38mm tAB 12mm dBC 70mm

tBC 10mm P1 120kN P2 100kN

ABC p

4[dBC

2 (dBC 2tBC)2] 1885mm2

APPENDIX A FE Exam Review Problems 5

P2

dABtABdBCtBC

ABC Cap plate

P1

Axial strain of BC:

Axial stress in BC: BC EBC 116.7MPa

(well below yield stress of 550 MPa so Hookes Law applies)

Lateral strain of BC: tBC 0.0036mm

Poissons ratio: confirms value for brassgiven in properties table(also agrees with givenmodulus E)

v LBC

0.34

L tBCtBC

3.600 104

BC (P1 P2)

EABC 1.061 103

36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 5

A-1.9: A titanium bar (E 100 GPa, v 0.33) with square cross section (b 75 mm) and length L 3.0 m is subjected to tensile load P 900 kN. Theincrease in volume of the bar is approximately:

(A) 1400 mm3

(B) 3500 mm3

(C) 4800 mm3

(D) 9200 mm3

Solution

E 100GPa b 75mm L 3.0m P 900kN v 0.33

6 APPENDIX A FE Exam Review Problems

bb

P

L

P

Initial volume of bar: Vinit b2L 1.6875000 107mm3

Normal strain in bar:

Lateral strain in bar: L v 5.28000 104

Final length of bar: Lf L L 3004.800mm

Final lateral dimension of bar: bf b Lb 74.96040mm

Final volume of bar: Vfinal bf2Lf 1.68841562 10

7mm3

Increase in volume of bar: V Vfinal Vinit 9156mm3

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 16kN a 150mm b 225mm d 14mm t 55mm

V

Vinit 0.000543

P

Eb2 1.60000 103

36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 6

Ave. shear stress:

Ave. shear strain:

Shear modulus of elastomer:

A-1.11: A bar of diameter d 18 mm and length L 0.75 m is loaded in ten-sion by forces P. The bar has modulus E 45 GPa and allowable normal stressof 180 MPa. The elongation of the bar must not exceed 2.7 mm. The allowablevalue of forces P is approximately:

(A) 41 kN(B) 46 kN(C) 56 kN(D) 63 kN

Solution

d 18mm L 0.75m E 45GPa a 180MPaa 2.7mm

G t

g 1.902MPa

g atanadtb 0.249

t V

ab 0.474MPa

APPENDIX A FE Exam Review Problems 7

a

b

V

t

dPP

L

(1) allowable value of P based on elongation

max Ea 162.0MPa

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 kNm(B) 22 kNm(C) 29 kNm(D) 37 kNm

Pa2 saap4 d2b 45.8kN

Pa1 smaxap4 d2b 41.2kNa

daL

3.600 103

36041_01_online_appA_p01-73.qxd 3/22/11 11:29 AM Page 7

Solution

db 18mm d 240mm a 90MPa n 8

Bolt shear area:

Max. torque: