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05 Solutions 46060 5/25/10 3:53 PM Page 214 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writ...

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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

A shaft is made of a steel alloy having an allowable shear stress of tallow = 12 ksi. If the diameter of the shaft is 1.5 in., determine the maximum torque T that can be transmitted. What would be the maximum torque T¿ if a 1-in.-diameter hole is bored through the shaft? Sketch the shear-stress distribution along a radial line in each case. •5–1.

T TÀ

Allowable Shear Stress: Applying the torsion formula tmax = tallow = 12 =

Tc J T (0.75) p 2

(0.754)

T = 7.95 kip # in.

Ans.

Allowable Shear Stress: Applying the torsion formula tmax = tallow = 12 =

T¿c J T¿ (0.75) p 2

(0.754 - 0.54)

T¿ = 6.381 kip # in. = 6.38 kip # in. tr = 0.5 in =

T¿r = J

6.381(0.5) p 2

(0.754 - 0.54)

Ans.

= 8.00 ksi

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5–2. The solid shaft of radius r is subjected to a torque T. Determine the radius r¿ of the inner core of the shaft that resists one-half of the applied torque 1T>22. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.

r r

T

a)

tmax = t =

Since t =

r¿ =

b)

L0 L0 L0

Tc Tr 2T = p 4 = J p r3 2 r

(T2 )r¿ p 2

(r¿)4

=

T p(r¿)3 T r¿ 2T a = b r pr3 p(r¿)3

r¿ t ; r max r 1

24

= 0.841 r

r 2

dT = 2p r 2

L0

Ans.

r¿

tr2 dr r¿

dT = 2p r 2

r tmax r2 dr L0 r r¿

dT = 2p

r 2T 2 a 3 br dr L0 r pr

r¿

T 4T r3 dr = 4 2 r L0 r¿ =

r 1

24

= 0.841r

Ans.

215

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5–3. The solid shaft is fixed to the support at C and subjected to the torsional loadings shown. Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points.

10 kN⭈m C A

50 mm

The internal torques developed at Cross-sections pass through point B and A are shown in Fig. a and b, respectively. The polar moment of inertia of the shaft is J =

p (0.0754) = 49.70(10 - 6) m4. For 2

point B, rB = C = 0.075 Thus, tB =

4(103)(0.075) TB c = 6.036(106) Pa = 6.04 MPa = J 49.70(10 - 6)

Ans.

From point A, rA = 0.05 m. tA =

6(103)(0.05) TArA = 6.036(106) Pa = 6.04 MPa. = J 49.70 (10 - 6)

216

Ans.

B

75 mm 4 kN⭈m 75 mm

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*5–4. The tube is subjected to a torque of 750 N # m. Determine the amount of this torque that is resisted by the gray shaded section. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.

75 mm 100 mm 750 N⭈m 25 mm

a) Applying Torsion Formula: tmax =

Tc = J

750(0.1) p 2

(0.14 - 0.0254)

tmax = 0.4793 A 106 B =

= 0.4793 MPa

T¿(0.1) p 2

(0.14 - 0.0754)

T¿ = 515 N # m

Ans.

b) Integration Method: r t = a b tmax c

dA = 2pr dr

and

dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr T¿ =

L

0.1m

2ptr2 dr = 2p

=

=

r tmax a br2 dr c L0.075m

0.1m 2ptmax r3 dr c L0.075m

2p(0.4793)(106) r4 0.1 m c d2 0.1 4 0.075 m

= 515 N # m

Ans.

5–5. The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.

tmax =

Tmax c = J

A

30 N⭈m

90(0.02) p 2

4

4

(0.02 - 0.0185 )

20 N⭈m

= 26.7 MPa

Ans.. 80 N⭈m

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5–6. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and F allow free rotation of the shaft.

F E D C B

(tBC)max = (tDE)max =

35(12)(0.375) TBC c = 5070 psi = 5.07 ksi = p 4 J 2 (0.375)

Ans.

25(12)(0.375) TDE c = p = 3621 psi = 3.62 ksi 4 J 2 (0.375)

Ans.

A

35 lb⭈ft

5–7. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress developed in regions CD and EF of the shaft. The bearings at A and F allow free rotation of the shaft.

F E D C B

(tEF)max =

TEF c = 0 J (tCD)max =

25 lb⭈ft 40 lb⭈ft 20 lb⭈ft

Ans.

A

25 lb⭈ft 40 lb⭈ft 20 lb⭈ft

35 lb⭈ft

15(12)(0.375) TCD c = p 4 J 2 (0.375)

= 2173 psi = 2.17 ksi

Ans.

218

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*5–8. The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress on the shaft.

300 N⭈m

500 N⭈m

A 200 N⭈m

Internal Torque: As shown on torque diagram. Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying torsion Formula.

C 400 N⭈m 300 mm

abs = tmax

=

Tmax c J 400(0.015) p 2

(0.0154)

D B

400 mm

= 75.5 MPa

Ans.

The shaft consists of three concentric tubes, each made from the same material and having the inner and outer radii shown. If a torque of T = 800 N # m is applied to the rigid disk fixed to its end, determine the maximum shear stress in the shaft.

500 mm

•5–9.

J =

T ⫽ 800 N⭈m

ri ⫽ 20 mm ro ⫽ 25 mm

2m

p p p ((0.038)4 - (0.032)4) + ((0.030)4 - (0.026)4) + ((0.025)4 - (0.020)4) 2 2 2 -6

ri ⫽ 26 mm ro ⫽ 30 mm

4

J = 2.545(10 ) m tmax =

800(0.038) Tc = 11.9 MPa = J 2.545(10 - 6)

Ans.

219

ri ⫽ 32 mm ro ⫽ 38 mm

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5–10. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d.

T R

r

n is the number of bolts and F is the shear force in each bolt. T

T F = nR

T - nFR = 0; T

tavg =

4T F nR = p 2 = A ( 4 )d nRpd2

Maximum shear stress for the shaft: tmax =

Tr 2T Tc = p 4 = J pr3 2r 4T 2T = nRpd2 p r3

tavg = tmax ; n =

2 r3 Rd2

Ans.

5–11. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench.

C

B

A

15 lb 6 in.

tAB =

tBC

Tc = J

Tc = = J

8 in.

210(0.375) p 2

(0.3754 - 0.344)

Ans. 15 lb

210(0.5) p 2

= 7.82 ksi

(0.54 - 0.434)

= 2.36 ksi

Ans.

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*5–12. The motor delivers a torque of 50 N # m to the shaft AB. This torque is transmitted to shaft CD using the gears at E and F. Determine the equilibrium torque Tⴕ on shaft CD and the maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts.

A 50 mm 30 mm

Equilibrium:

B

a + ©ME = 0; a + ©MF = 0;

50 - F(0.05) = 0

F = 1000 N

35 mm T


T¿ - 1000(0.125) = 0 T¿ = 125 N # m

C E 125 mm D

F

Ans.

Internal Torque: As shown on FBD. Maximum Shear Stress: Applying torsion Formula. (tAB)max =

50.0(0.015) TAB c = 9.43 MPa = p 4 J 2 (0.015 )

Ans.

(tCD)max =

125(0.0175) TCDc = 14.8 MPa = p 4 J 2 (0.0175 )

Ans.

•5–13. If the applied torque on shaft CD is T¿ = 75 N # m, determine the absolute maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating.

A 50 mm

Equilibrium:

30 mm

a + ©MF = 0;

75 - F(0.125) = 0;

a + ©ME = 0;

600(0.05) - TA = 0

B

F = 600 N

35 mm T


TA = 30.0 N # m Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula (tEA)max =

30.0(0.015) TEA c = 5.66 MPa = p 4 J 2 (0.015 )

Ans.

(tCD)max =

75.0(0.0175) TCDc = 8.91 MPa = p 4 J 2 (0.0175 )

Ans.

221

C E 125 mm D

F

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5–14. The solid 50-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress in the shaft.

250 N⭈m

75 N⭈m

A

325 N⭈m 150 N⭈ m

B 500 mm

C D

The internal torque developed in segments AB , BC and CD of the shaft are shown in Figs. a, b and c.

400 mm 500 mm

The maximum torque occurs in segment AB. Thus, the absolute maximum shear stress occurs in this segment. The polar moment of inertia of the shaft is p J = (0.0254) = 0.1953p(10 - 6)m4. Thus, 2

A tmax B abs =

250(0.025) TAB c = 10.19(106)Pa = 10.2 MPa = J 0.1953p(10 - 6)

222

Ans.

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5–15. The solid shaft is made of material that has an allowable shear stress of tallow = 10 MPa. Determine the required diameter of the shaft to the nearest mm.

15 N⭈m 25 N⭈m A

30 N⭈m B

60 N⭈m C

70 N⭈m D E

The internal torques developed in each segment of the shaft are shown in the torque diagram, Fig. a. Segment DE is critical since it is subjected to the greatest internal torque. The polar p d 4 p 4 moment of inertia of the shaft is J = a b = d . Thus, 2 2 32

tallow

TDE c = ; J

d 70 a b 2 10(106) = p 4 d 32 d = 0.03291 m = 32.91 mm = 33 mm

223

Ans.

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*5–16. The solid shaft has a diameter of 40 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stress is maximum.

15 N⭈m 25 N⭈m A

30 N⭈m B

The internal torque developed in each segment of the shaft are shown in the torque diagram, Fig. a.

60 N⭈m C

70 N⭈m D E

Since segment DE subjected to the greatest torque, the absolute maximum shear p stress occurs here. The polar moment of inertia of the shaft is J = (0.024) 2 = 80(10 - 9)p m4. Thus,

tmax =

70(0.02) TDE c = 5.57(106) Pa = 5.57 MPa = J 80(10 - 9)p

Ans.

The shear stress distribution along the radial line is shown in Fig. b.

The rod has a diameter of 1 in. and a weight of 10 lb/ft. Determine the maximum torsional stress in the rod at a section located at A due to the rod’s weight.

•5–17.

4.5 ft B

Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;

TA - 10(4)(2) = 0

TA = 80 lb # ft a

The polar moment of inertia of the cross section at A is J =

12in b = 960 lb # in. 1ft

p (0.54) = 0.03125p in4. 2

Thus

tmax =

960 (0.5) TA c = = 4889.24 psi = 4.89 ksi J 0.03125p

Ans.

224

4 ft

A 1.5 ft 1.5 ft

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5–18. The rod has a diameter of 1 in. and a weight of 15 lb/ft. Determine the maximum torsional stress in the rod at a section located at B due to the rod’s weight.

4.5 ft B

4 ft

Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;

TB - 15(4)(2) = 0

TB = 120 lb # ft a

12 in b = 1440 lb # in. 1ft

p The polar moment of inertia of the cross-section at B is J = (0.54) 2 = 0.03125p in4. Thus,

tmax =

1440(0.5) TB c = = 7333.86 psi = 7.33 ksi J 0.03125p

Ans.

225

A 1.5 ft 1.5 ft

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5–19. Two wrenches are used to tighten the pipe. If P = 300 N is applied to each wrench, determine the maximum torsional shear stress developed within regions AB and BC. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm. Sketch the shear stress distribution for both cases.

P

B

Internal Loadings: The internal torque developed in segments AB and BC of the pipe can be determined by writing the moment equation of equilibrium about the x axis by referring to their respective free - body diagrams shown in Figs. a and b. ©Mx = 0; TAB - 300(0.25) = 0

TAB = 75 N # m

TBC = 150 N # m

Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254 - 0.014 B = 22.642(10 - 9)m4. 2

A tmax B AB =

75(0.0125) TAB c = 41.4 MPa = J 22.642(10 - 9)

A tAB B r = 0.01 m = A tmax B BC =

Ans.

TAB r 75(0.01) = 33.1 MPa = J 22.642(10 - 9)

150(0.0125) TBC c = 82.8 MPa = J 22.642(10 - 9)

A tBC B r = 0.01 m =

Ans.

TBC r 150(0.01) = 66.2 MPa = J 22.642(10 - 9)

The shear stress distribution along the radial line of segments AB and BC of the pipe is shown in Figs. c and d, respectively.

226

A

250 mm P

And ©Mx = 0; TBC - 300(0.25) - 300(0.25) = 0

250 mm

C

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