Feo Dpp Sample

Michael R. Lindeburg, PE

Professional Publications, Inc. Belmont, California

FEPRACTICEPROBLEMS for the Other Disciplines Fundamentals of Engineering Exam

OTHERDISCIPLINES

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17 Thermal, Hoop, andTorsional StressPRACTICE PROBLEMS

1. The maximum torque on a 0.15 m diameter solidshaft is 13 500 Nm. What is most nearly the maximumshear stress in the shaft?

(A) 20 MPa

(B) 23 MPa

(C) 28 MPa

(D) 34 MPa

2. The unrestrained glass window shown is subjected toa temperature change from 0C to 50C. The coefficientof thermal expansion for the glass is 8.8 106 1/C.

1.2 m

2 m

What is most nearly the change in area of the glass?

(A) 0.00040 m2

(B) 0.0013 m2

(C) 0.0021 m2

(D) 0.0028 m2

3. The cylindrical steel tank shown is 3.5 m in diameter,5 m high, and filled to the top with a brine solution.Brine has a density of 1198 kg/m3. The thickness of thesteel shell is 12.5 mm. Neglect the weight of the tank.

3.5 m

5 m

0.65 m

What is the approximate hoop stress in the steel 0.65 mabove the rigid concrete pad?

(A) 1.2 MPa

(B) 1.4 MPa

(C) 7.2 MPa

(D) 11 MPa

4. A steel shaft is shown. The shear modulus is 80 GPa.

outsidediameter= 50 mm

insidediameter= 30 mm

L = 1.0 m

r1

r2

Most nearly, what torque should be applied to the endof the shaft in order to produce a twist of 1.5?

(A) 420 Nm(B) 560 Nm(C) 830 Nm(D) 1100 Nm

P P I * w w w . p p i 2 p a s s . c o m

Strength

of

Materials

5. For the shaft shown, the shear stress is not to exceed110 MPa.

r1

r2

r1 = 0.015 mr2 = 0.025 m

T

What is most nearly the largest torque that can beapplied?

(A) 1700 Nm(B) 1900 Nm(C) 2300 Nm(D) 3400 Nm

6. An aluminum (shear modulus = 2.8 1010 Pa) rod is25 mm in diameter and 50 cm long. One end is rigidlyfixed to a support. Most nearly, what torque must beapplied at the free end to twist the rod 4.5 about itslongitudinal axis?

(A) 26 Nm(B) 84 Nm(C) 110 Nm(D) 170 Nm

7. A circular bar at 10C is constrained by rigid concretewalls at both ends. The bar is 1000mm long and has a cross-sectional area of 2600 mm2.

1000 mm

A = 2600 mm2

E = modulus of elasticity = 200 GPa = coefficient of thermal expansion = 9.4 106 1/C

What is most nearly the axial force in the bar if thetemperature is raised to 40C?

(A) 92 kN

(B) 110 kN

(C) 130 kN

(D) 150 kN

8. A 3 m diameter solid bar experiences opposing torquesof 280 Nm at each end.

T

T

d

What is most nearly the maximum shear stress in thebar?

(A) 2.2 Pa

(B) 31 Pa

(C) 42 Pa

(D) 53 Pa

9. A 12.5 mm diameter steel rod is pinned between tworigid walls. The rod is initially unstressed. The rodstemperature subsequently increases 50C. The rod isadequately stiffened and supported such that bucklingdoes not occur. The coefficient of linear thermal expan-sion for steel is 11.7 106 1/C. The modulus of elas-ticity for steel is 210 GPa.

d 12.5 mm

3.5 m

What is the approximate axial force in the rod?

(A) 2.8 kN

(B) 15 kN

(C) 19 kN

(D) 58 kN


17-2 F E O T H E R D I S C I P L I N E S P R A C T I C E P R O B L E M S

Stre

ngth

of

Materia

ls

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .SOLUTIONS

1. The polar moment of inertia is

J pr4

2 p

2

0:15 m

2

4

4:97 105 m4

The shear stress is

TrJ

13 500 Nm 0:15 m

2

4:97 105 m4 20:37 106 Pa 20 MPa

The answer is (A).

2. Changes in temperature affect each linear dimension.

width LT To 8:8 106 1C

1:2 m50C 0C 0:000528 m

height 8:8 106 1C

2 m50C 0C 0:00088 m

Ainitial 2 m1:2 m 2:4 m2Afinal 2 m 0:00088 m

1:2 m 0:000528 m 2:40211 m2

DA Afinal Ainitial 2:40211 m2 2:4 m2 0:00211 m2 0:0021 m2

Alternative Solution

The area coefficient of thermal expansion is, for allpractical purposes, equal to 2.

The change in area is

DA 2AoDT 2 8:8 106 1C

2:4 m250C 0C

0:00211 m2 0:0021 m2

The answer is (C).

3. Determine whether the tank is thin-walled or thick-walled.

tr 12:5 mm

3:5 m2

1000

mmm

0:007 < 0:1

Use formulas for thin-walled cylindrical tanks. The pres-sure is

p gh

1198 kgm3

9:81

ms2

5 m 0:65 m

51 123 PaThe hoop stress is

t prt pd

2t 51 123 Pa3:5 m

2 12:5 mm1000

mmm

0@

1A

7:157 106 Pa 7:2 MPaThe answer is (C).

4. Convert the twist angle to radians.

1:5 2p rad360

0:026 rad

Calculate the polar moment of inertia, J.

r1 15 mm 0:015 mr2 25 mm 0:025 mJ p

2r4o r4i p2

0:025 m4 0:015 m4

5:34 107 m4

The torque is

T GJL

0:026 rad 80 GPa 109 PaGPa

5:34 107 m41:0 m

1119 Nm 1100 Nm

The answer is (D).

5. Since the shear stress is largest at the outer diameter,the maximum torque is found using this radius. For anannular region,

J p2r4o r4i p2

0:025 m4 0:015 m4

5:34 107 m4


17-4 F E O T H E R D I S C I P L I N E S P R A C T I C E P R O B L E M S

Stre

ngth

of

Materia

ls

The torque is

Tmax Jr2

110 MPa 106 Pa

MPa

5:34 107 m4

0:025 m

2349:9 Nm 2300 Nm

The answer is (C).

6. Convert degrees to radians.

4:5 2p rad360

7:854 102 rad

The polar moment of inertia is

J pr4

2 p

2

25 mm

2 1000 mmm

0B@

1CA

4

3:83 108 m4

Rearrange the twist angle equation to solve for torque.

T GJL

7:854 102 rad2:8 1010 Pa3:83 108 m4

50 cm

100cmm

168:7 Nm 170 Nm

The answer is (D).

7. The elongation due to temperature change is

LT2 T1 9:4 106 1C

1000 mm40C 10C 0:282 mm

Rearrange the elongation equation to solve for force.

F EAL

0:282 mm 200 GPa 106 kPa

GPa

2600 mm2

1 m 1000 mmm

3

146:6 kN 150 kN

The answer is (D).

8. Maximum shear stress occurs at the outer surface.The shear is

TrJ

T

d2

p32

d4280 Nm 3 m

2

p32

3 m4

52:8 Pa 53 Pa

The answer is (D).

9. The thermal strain is

"t DT 11:7 106 1C

50C 0:000585 m=m

The thermal stress is

t E"t 210 GPa 109 PaGPa

0:000585

mm

1:2285 108 Pa

(This is less than the yield strength of steel.)

The compressive force in the rod is

F A

1:2285 108 Pap 12:5 mm2 1000 mm

m

0B@

1CA

2

15 076 N 15 kN

The answer is (B).

10. The total change in length is

t LinitialT To 11 106 1C

10 km50C 20C 0:0033 km

Add the change in length to the initial length.

L Linitial t 10 km 0:0033 km 10:0033 km

The answer is (C).

11. Tanks under external pressure fail by buckling (i.e.,collapse), not by yielding. They should not be designedusing the simplistic formulas commonly used for thin-walled tanks under internal pressure.

The answer is (D).


T H E R M A L , H O O P , A N D T O R S I O N A L S T R E S S 17-5

Strength

of

Materials

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