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•5–1.

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.

T

T¿

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

214

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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¿ =

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

= 0.841 r

Ans.

24

r

2

b)

r¿

dT = 2p

L0

r

2

r¿

dT = 2p

L0

r

2

L0

tr2 dr

L0

r

tmax r2 dr

L0 r

r¿

dT = 2p

r 2T 2

a 3 br dr

L0 r pr

r¿

4T

T

= 4

r3 dr

2

r L0

r¿ =

r

1

Ans.

= 0.841r

24

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 kNиm

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 =

TArA

6(103)(0.05)

= 6.036(106) Pa = 6.04 MPa.

=

J

49.70 (10 - 6)

216

Ans.

B

75 mm

4 kNиm

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 Nиm

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

0.1m

T¿ =

L

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 Nиm

90(0.02)

p

2

4

4

(0.02 - 0.0185 )

20 Nиm

= 26.7 MPa

Ans..

80 Nиm

217

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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 =

35(12)(0.375)

TBC c

= 5070 psi = 5.07 ksi

= p

4

J

2 (0.375)

Ans.

(tDE)max =

25(12)(0.375)

TDE c

= 3621 psi = 3.62 ksi

= p

4

J

2 (0.375)

Ans.

A

35 lbиft

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 lbиft

40 lbиft

20 lbиft

Ans.

A

25 lbиft

40 lbиft

20 lbиft

35 lbиft

15(12)(0.375)

TCD c

= p

4

J

2 (0.375)

= 2173 psi = 2.17 ksi

Ans.

218

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300 Nиm

*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.

500 Nиm

A

200 Nиm

Internal Torque: As shown on torque diagram.

C

Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying

torsion Formula.

400 Nиm

300 mm

abs =

tmax

Tmax c

J

400(0.015)

=

p

2

(0.0154)

Ans.

= 75.5 MPa

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

T ϭ 800 Nиm

ri ϭ 20 mm

ro ϭ 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 ϭ 26 mm

ro ϭ 30 mm

4

J = 2.545(10 ) m

tmax =

B

400 mm

•5–9.

J =

D

800(0.038)

Tc

= 11.9 MPa

=

J

2.545(10 - 6)

Ans.

219

ri ϭ 32 mm

ro ϭ 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 =

F

4T

nR

= p 2 =

A

( 4 )d

nRpd2

Maximum shear stress for the shaft:

tmax =

Tc

Tr

2T

= 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.

220

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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 T 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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250 Nиm

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.

75 Nиm

A

325 Nиm

150 Nи m

B

500 mm

The internal torque developed in segments AB , BC and CD of the shaft are shown

in Figs. a, b and c.

C

D

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 Nиm

25 Nиm

A

30 Nиm

B

60 Nиm

C

70 Nиm

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

70a 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 Nиm

25 Nиm

A

30 Nиm

B

The internal torque developed in each segment of the shaft are shown in the torque

diagram, Fig. a.

60 Nиm

C

70 Nиm

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.

•5–17.

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.

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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*5–20. Two wrenches are used to tighten the pipe. If the

pipe is made from a material having an allowable shear stress

of tallow = 85 MPa, determine the allowable maximum force

P that can be applied to each wrench. The pipe has an outer

diameter of 25 mm and inner diameter of 20 mm.

P

250 mm

C

B

A

250 mm

Internal Loading: By observation, segment BC of the pipe is critical since it is

subjected to a greater internal torque than segment AB. Writing the moment

equation of equilibrium about the x axis by referring to the free-body diagram

shown in Fig. a, we have

©Mx = 0; TBC - P(0.25) - P(0.25) = 0

TBC = 0.5P

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

tallow =

TBC c

;

J

85(106) =

0.5P(0.0125)

22.642(10 - 9)

P = 307.93N = 308 N

Ans.

227

P

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•5–21.

The 60-mm-diameter solid shaft is subjected to the

distributed and concentrated torsional loadings shown.

Determine the absolute maximum and minimum shear

stresses on the outer surface of the shaft and specify their

locations, measured from the fixed end A.

A

2 kNиm/m

1.5 m

1200 Nиm

C

The internal torque for segment BC is Constant TBC = 1200 N # m, Fig. a. However,

the internal for segment AB varies with x, Fig. b.

TAB - 2000x + 1200 = 0

TAB = (2000x - 1200) N # m

The minimum shear stress occurs when the internal torque is zero in segment AB.

By setting TAB = 0,

0 = 2000x - 1200

x = 0.6 m

Ans.

And

d = 1.5 m - 0.6 m = 0.9 m

Ans.

tmin = 0

Ans.

The maximum shear stress occurs when the internal torque is the greatest. This

occurs at fixed support A where

d = 0

Ans.

At this location,

(TAB)max = 2000(1.5) - 1200 = 1800 N # m

The polar moment of inertia of the rod is J =

tmax =

p

(0.034) = 0.405(10 - 6)p. Thus,

2

(TAB)max c

1800(0.03)

= 42.44(106)Pa = 42.4 MPa

=

J

0.405(10 - 6)p

228

Ans.

B

0.8 m

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5–22. The solid shaft is subjected to the distributed and

concentrated torsional loadings shown. Determine the

required diameter d of the shaft to the nearest mm if the

allowable shear stress for the material is tallow = 50 MPa.

A

2 kNиm/m

1.5 m

1200 Nиm

C

The internal torque for segment BC is constant TBC = 1200 N # m, Fig. a. However,

the internal torque for segment AB varies with x, Fig. b.

TAB - 2000x + 1200 = 0 TAB = (2000x - 1200) N # m

For segment AB, the maximum internal torque occurs at fixed support A where

x = 1.5 m. Thus,

A TAB B max = 2000(1.5) - 1200 = 1800 N # m

Since A TAB B max 7 TBC, the critical cross-section is at A. The polar moment of inertia

p d 4

pd4

of the rod is J =

. Thus,

a b =

2 2

32

tallow =

Tc

;

J

50(106) =

1800(d>2)

pd4>32

d = 0.05681 m = 56.81 mm = 57 mm

229

Ans.

B

0.8 m

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*5–24. The copper pipe has an outer diameter of 2.50 in.

and an inner diameter of 2.30 in. If it is tightly secured to the

wall at C and a uniformly distributed torque is applied to it

as shown, determine the shear stress developed at points A

and B. These points lie on the pipe’s outer surface. Sketch

the shear stress on volume elements located at A and B.

B

A

C

125 lbиft/ft

4 in.

9 in.

12 in.

Internal Torque: As shown on FBD.

Maximum Shear Stress: Applying the torsion formula

tA =

TA c

J

125.0(12)(1.25)

=

tB =

p

2

(1.254 - 1.154)

Ans.

= 3.02 ksi

Ans.

TB c

J

218.75(12)(1.25)

=

= 1.72 ksi

p

2

(1.254 - 1.154)

•5–25.

The copper pipe has an outer diameter of 2.50 in.

and an inner diameter of 2.30 in. If it is tightly secured to

the wall at C and it is subjected to the uniformly distributed

torque along its entire length, determine the absolute

maximum shear stress in the pipe. Discuss the validity of

this result.

B

A

C

125 lbиft/ft

4 in.

9 in.

Internal Torque: The maximum torque occurs at the support C.

Tmax = (125 lb # ft>ft)a

12 in.

25 in.

b = 260.42 lb # ft

12 in.>ft

Maximum Shear Stress: Applying the torsion formula

abs =

tmax

Tmax c

J

260.42(12)(1.25)

=

p

2

(1.254 - 1.154)

Ans.

= 3.59 ksi

According to Saint-Venant’s principle, application of the torsion formula should be

as points sufficiently removed from the supports or points of concentrated loading.

230

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5–26. A cylindrical spring consists of a rubber annulus

bonded to a rigid ring and shaft. If the ring is held fixed and

a torque T is applied to the shaft, determine the maximum

shear stress in the rubber.

ro

ri

T

h

T

r

T

F

t =

=

=

A

2prh

2p r2 h

Shear stress is maximum when r is the smallest, i.e. r = ri. Hence,

tmax =

T

2p ri 2 h

Ans.

300 Nиm

5–27. The A-36 steel shaft is supported on smooth

bearings that allow it to rotate freely. If the gears are

subjected to the torques shown, determine the maximum

shear stress developed in the segments AB and BC. The

shaft has a diameter of 40 mm.

100 Nиm

A

The internal torque developed in segments AB and BC are shown in their

respective FBDs, Figs. a and b.

The polar moment of inertia of the shaft is J =

A tAB B max

200 Nиm

B

p

(0.024) = 80(10-9)p m4. Thus,

2

C

300(0.02)

TAB c

= 23.87(106)Pa = 23.9 MPa

=

=

J

80(10-9)p

A tBC B max =

200(0.02)

TBC c

= 15.92(106) Pa = 15.9 MPa

=

J

80(10-9)p

231

Ans.

Ans.

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300 Nиm

*5–28. The A-36 steel shaft is supported on smooth

bearings that allow it to rotate freely. If the gears are

subjected to the torques shown, determine the required

diameter of the shaft to the nearest mm if tallover = 60 MPa.

100 Nиm

The internal torque developed in segments AB and BC are shown in their

respective FBDs, Fig. a and b

A

200 Nиm

B

Here, segment AB is critical since its internal torque is the greatest. The polar

p d 4

pd4

moment of inertia of the shaft is J =

. Thus,

a b =

2 2

32

C

tallow

TC

=

;

J

60(106) =

300(d>2)

pd4>32

d = 0.02942 m = 30 mm

Ans.

•5–29.

When drilling a well at constant angular velocity,

the bottom end of the drill pipe encounters a torsional

resistance TA . Also, soil along the sides of the pipe creates a

distributed frictional torque along its length, varying

uniformly from zero at the surface B to tA at A. Determine

the minimum torque TB that must be supplied by the drive

unit to overcome the resisting torques, and compute

the maximum shear stress in the pipe. The pipe has an outer

radius ro and an inner radius ri .

TA +

TB

B

L

1

t L - TB = 0

2 A

tA

2TA + tAL

TB =

2

Ans.

Maximum shear stress: The maximum torque is within the region above the

distributed torque.

tmax =

tmax =

Tc

J

(2TA + tAL)

] (r0)

2

p 4

4

(r

r

i)

2 0

[

(2TA + tAL)r0

=

Ans.

p(r40 - r4i )

232

A

TA

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5–30. The shaft is subjected to a distributed torque along

its length of t = 110x22 N # m>m, where x is in meters. If the

maximum stress in the shaft is to remain constant at

80 MPa, determine the required variation of the radius c of

the shaft for 0 … x … 3 m.

x

x

T =

L

t dx =

Tc

t =

;

J

L0

10 x2dx =

6

80(10 ) =

3m

c

10 3

x

3

t ϭ (10x2) Nиm/m

3

(10

3 )x c

p

2

c4

c3 = 26.526(10-9) x3

c = (2.98 x) mm

Ans.

5–31. The solid steel shaft AC has a diameter of 25 mm and

is supported by smooth bearings at D and E. It is coupled to

a motor at C, which delivers 3 kW of power to the shaft

while it is turning at 50 rev>s. If gears A and B remove 1 kW

and 2 kW, respectively, determine the maximum shear stress

developed in the shaft within regions AB and BC. The shaft

is free to turn in its support bearings D and E.

TC =

3(103)

P

=

= 9.549 N # m

v

50(2p)

TA =

1

T = 3.183 N # m

3 C

3 kW

2 kW

25 mm

1 kW

A

D

(tAB)max =

3.183 (0.0125)

TC

= 1.04 MPa

= p

4

J

2 (0.0125 )

Ans.

(tBC)max =

9.549 (0.0125)

TC

= 3.11 MPa

= p

4

J

2 (0.0125 )

Ans.

233

B

E

C

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*5–32. The pump operates using the motor that has a

power of 85 W. If the impeller at B is turning at 150 rev>min,

determine the maximum shear stress developed in the

20-mm-diameter transmission shaft at A.

150 rev/min

A

Internal Torque:

v = 150

rev 2p rad 1 min

= 5.00p rad>s

¢

≤

min

rev

60 s

P = 85 W = 85 N # m>s

T =

P

85

=

= 5.411 N # m

v

5.00p

Maximum Shear Stress: Applying torsion formula

tmax =

Tc

J

5.411 (0.01)

=

p

4

2 (0.01 )

= 3.44 MPa

Ans.

•5–33.

The gear motor can develop 2 hp when it turns at

450 rev>min. If the shaft has a diameter of 1 in., determine

the maximum shear stress developed in the shaft.

The angular velocity of the shaft is

v = ¢ 450

rev

2p rad

1 min

≤ ¢

≤ ¢

≤ = 15p rad>s

min

1 rev

60 s

and the power is

P = 2 hp ¢

550 ft # lb>s

≤ = 1100 ft # lb>s

1 hp

Then

T =

P

1100

12 in

=

= 23.34 lb # ft a

b = 280.11 lb # in

v

15p

1ft

The polar moment of inertia of the shaft is J =

tmax =

p

(0.54) = 0.03125p in4. Thus,

2

280.11 (0.5)

Tc

=

= 1426.60 psi = 1.43 ksi

J

0.03125p

Ans.

234

B

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5–34. The gear motor can develop 3 hp when it turns at

150 rev>min. If the allowable shear stress for the shaft is

tallow = 12 ksi, determine the smallest diameter of the shaft

to the nearest 18 in. that can be used.

The angular velocity of the shaft is

v = a 150

rev

2p rad 1 min

ba

ba

b = 5p rad>s

min

1 rev

60 s

and the power is

P = (3 hp) a

550 ft # lb>s

b = 1650 ft # lb>s

1 hp

Then

T =

P

1650

12 in

=

= (105.04 lb # ft)a

b = 1260.51 lb # in

v

5p

1 ft

The polar moment of inertia of the shaft is J =

tallow =

Tc

;

J

12(103) =

p d 4

pd4

a b =

. Thus,

2 2

32

1260.51 (d>2)

pd4>32

d = 0.8118 in. =

7

in.

8

Ans.

5–35. The 25-mm-diameter shaft on the motor is made

of a material having an allowable shear stress of

tallow = 75 MPa . If the motor is operating at its maximum

power of 5 kW, determine the minimum allowable rotation

of the shaft.

Allowable Shear Stress: The polar moment of inertia of the shaft is

p

J =

A 0.01254 B = 38.3495(10-9) m4.

2

tallow =

Tc

;

J

75(106) =

T(0.0125)

38.3495(10-9)

T = 230.10 N # m

Internal Loading:

T =

P

;

v

230.10 =

5(103)

v

v = 21.7 rad>s

Ans.

235

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*5–36. The drive shaft of the motor is made of a material

having an allowable shear stress of tallow = 75 MPa. If the

outer diameter of the tubular shaft is 20 mm and the wall

thickness is 2.5 mm, determine the maximum allowable

power that can be supplied to the motor when the shaft is

operating at an angular velocity of 1500 rev>min.

Internal Loading: The angular velocity of the shaft is

v = a 1500

rev

2p rad 1 min

ba

ba

b = 50p rad>s

min

1 rev

60 s

We have

T =

P

P

=

v

50p

Allowable Shear Stress: The polar moment of inertia of the shaft is

p

J = A 0.014 - 0.00754 B = 10.7379(10-9) m4.

2

tallow =

Tc

;

J

75(106) =

a

P

b(0.01)

50p

10.7379(10-9)

P = 12 650.25 W = 12.7 kW

Ans.

•5–37.

A ship has a propeller drive shaft that is turning at

1500 rev>min while developing 1800 hp. If it is 8 ft long and

has a diameter of 4 in., determine the maximum shear stress

in the shaft caused by torsion.

Internal Torque:

v = 1500

rev 2p rad 1 min

a

b

= 50.0 p rad>s

min 1 rev

60 s

P = 1800 hpa

T =

550 ft # lb>s

b = 990 000 ft # lb>s

1 hp

990 000

P

=

= 6302.54 lb # ft

v

50.0p

Maximum Shear Stress: Applying torsion formula

tmax =

6302.54(12)(2)

Tc

=

p 4

J

2 (2 )

= 6018 psi = 6.02 ksi

Ans.

236

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5–38. The motor A develops a power of 300 W and turns

its connected pulley at 90 rev>min. Determine the required

diameters of the steel shafts on the pulleys at A and B if the

allowable shear stress is tallow = 85 MPa.

60 mm

90 rev/min

A

B

150 mm

Internal Torque: For shafts A and B

vA = 90

rev 2p rad 1 min

a

b

= 3.00p rad>s

min

rev

60 s

P = 300 W = 300 N # m>s

P

300

=

= 31.83 N # m

vA

3.00p

TA =

vB = vA a

rA

0.06

b = 3.00pa

b = 1.20p rad>s

rB

0.15

P = 300 W = 300 N # m>s

TB =

P

300

=

= 79.58 N # m

vB

1.20p

Allowable Shear Stress: For shaft A

tmax = tallow =

85 A 106 B =

TA c

J

31.83 A d2A B

A B

p dA 4

2 2

dA = 0.01240 m = 12.4 mm

Ans.

For shaft B

tmax = tallow =

85 A 106 B =

TB c

J

79.58 A d2B B

A B

p dB 4

2 2

dB = 0.01683 m = 16.8 mm

Ans.

237

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