Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 3, Problem 3.8.14P

The composite shaft shown in the figure is manufactured by shrink-Fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are dY= 40 mm for the brass core and d2= 50 mm for the steel sleeve. The shear moduli of elasticity are Gb= 36 GPa for the brass and Gs= 80 GPa for the steel.

(a)

Assuming that the allowable shear stresses

in the brass and steel are rb= 48 MPa and

ts= 80 MPa, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft.

(b)

If the applied torque T = 2500 kN · m, find

the required diameter d2so that allowable shear

stress t3is reached in the steel.

  Chapter 3, Problem 3.8.14P, The composite shaft shown in the figure is manufactured by shrink-Fitting a steel sleeve over a , example  1

  Chapter 3, Problem 3.8.14P, The composite shaft shown in the figure is manufactured by shrink-Fitting a steel sleeve over a , example  2

(a)

Expert Solution
Check Mark
To determine

The maximum permissible torque applied to the shaft.

Answer to Problem 3.8.14P

The maximum permissible torque applied to the shaft is 1520.21Nm .

Explanation of Solution

Given information:

The outer diameter of steel sleeve is 50mm , the outside diameter of the brass core is 40mm . is 2.25in , the shear modulus of the steel is 80GPa , the shear modulus for the brass is 36GPa , allowable shear stress in brass is 48MPa and allowable shear stress in steel is 80MPa .

Write the expression for total torque of the system.

  T=Tb+Ts ...... (I)

Here, the total torque in the system is T , torque in the brass core is Tb and torque in the steel sleeve is Ts .

Write the expression for polar moment of inertia of the brass core.

  Ipb=π32(d14)...... (I)

Here, the polar moment of inertia of brass core is IPb , the diameter of the brass core is d1 .

Write the expression for polar moment of inertia of steel sleeve.

  Ips=π32(d24d14)...... (II)

Here, the polar moment of inertia of steel sleeve is IPs , the diameter of the steel sleeve is d2 .

Write the expression for torque on the brass core

  Tb=T(GbIpbGbIpb+GsIps)...... (III)

Here, the torque in the bar is Tb , shear modulus of steel sleeve is Gs , the shear modulus of brass core is Gb and total twisted torque is T .

Write the expression for torque in the steel sleeve.

  Ts=T(GsIpsGbIpb+GsIps) ...... (IV)

Here, the torque in the bar is Ts .

Write the expression for maximum shear stress in brass core.

  τb=Tb(d12)Ipb...... (V)

Here, the maximum shear stress in the brass core is τb .

Write the expression for maximum shear stress in the steel sleeve.

  τs=Ts(d22)Ips...... (VI)

Here, the maximum shear stress in the bar is τs .

Calculation:

Substitute 40mm for d1 in Equation (I).

  Ipb=π32(40mm)4=π32(40mm( 1m 1000mm))4=π32(0.04m)4=2.51×107m4

Substitute 50mm for d2 and 40mm for d1in Equation (II).

  Ips=π32(( 50mm)4( 40mm)4)=π32(( 50mm( 1m1000mm ))4( 40mm( 1m1000mm ))4)=π32(6.25×106m42.56×106m4)=3.62×107m4

Substitute 3.62×107m4 for Ips , 2.51×107m4 for Ipb , 80GPa for Gs and 36GPa for Gb in Equation (III).

  Tb=T(36GPa( 2.51× 10 7 m 4 )36GPa( 2.51× 10 7 m 4 )+80GPa( 3.62× 10 7 m 4 ))=T(36GPa( 109Pa 1GPa )( 2.51× 10 7 m 4 )36GPa( 109Pa 1GPa )( 2.51× 10 7 m 4 )+80GPa( 109Pa 1GPa )( 3.62× 10 7 m 4 ))=T(9036Pam49036Pam4+28960Pam4)=T(0.237)

Substitute 3.62×107m4 for Ips , 2.51×107m4 for Ipb , 80GPa for Gs and 36GPa for Gb in Equation (IV).

  Ts=T(80GPa( 3.62× 10 7 m 4 )36GPa( 2.51× 10 7 m 4 )+80GPa( 3.62× 10 7 m 4 ))=T(80GPa( 109Pa 1GPa )( 3.62× 10 7 m 4 )36GPa( 109Pa 1GPa )( 2.51× 10 7 m 4 )+80GPa( 109Pa 1GPa )( 3.62× 10 7 m 4 ))=T(28960Pam49036Pam4+28960Pam4)=T(0.762)

Substitute 3.62×107m4 for Ips , T(0.237) for Tb , 48MPa for τb and 40mm for d1 in Equation (V).

  48MPa=(T( 0.237))( ( 40mm )2)2.51×107m448MPa( 106Pa1MPa)=(T( 0.237))(40mm( 1m 1000mm ))2(2.51× 10 7m4)48×106Pa=(T(0.237))(79681.2)48×106Pa(79681.2)(0.237)=T

  T=2541.77Nm

Substitute 3.62×107m4 for Ipb , T(0.762) for Ts , 80MPa for τs and 50mm for d2 in Equation (VI).

  80MPa=(T( 0.762))( ( 50mm )2)3.62×107m480MPa( 106Pa1MPa)=(T( 0.762))(50mm( 1m 1000mm ))2(3.62× 10 7m4)80×106Pa=(T(0.762))(69060.77)80×106Pa(69060.77)(0.762)=T

  T=1520.21Nm

Conclusion:

The maximum permissible torque applied to the shaft is 1520.21Nm .

(b)

Expert Solution
Check Mark
To determine

The diameter of the steel sleeve.

Answer to Problem 3.8.14P

The diameter of the steel sleeve is 0.0569m .

Explanation of Solution

Given information:

The applied torque is 2500kNm .

Substitute, 40mm for d1 in Equation (I)

  Ips=π32(d24( 40mm)4)=π32(d24( 40mm( 1m1000mm ))4)=(0.098)(d242.56×106m4)

Substitute T(GsIpsGbIpb+GsIps) for Ts in Equation (VI).

  τs=(T( GsIps GbIpb+GsIps ))( d 2 2)Ips=(( G s T I ps G b I pb + G s I ps ))(d22)(1I ps)=GsTd22(GbI pb+GsI ps) ...... (VII)

Calculation:

Substitute 2500kNm for T , 80GPa for Gs , 36GPa for Gb and 2.51×107m4 for Ipb , (0.098)(d242.56×106m4) for Ips and 80MPa for τs in Equation (VII).

80MPa= ( 80GPa )( 2500kNm ) d 2 2( ( 36GPa )( 2.51× 10 7 m 4 ) +( 36GPa )( ( 0.098 )( d 2 4 2.56× 10 6 m 4 ) ) )

80MPa( 10 6 Pa 1MPa )= ( 80GPa( 10 9 Pa 1GPa ) )( 2500Nm ) d 2 2( ( 36GPa( 10 9 Pa 1GPa ) )( 2.51× 10 7 m 4 ) +( 80GPa( 10 9 Pa 1GPa ) )( ( 0.098 )( d 2 4 2.56× 10 6 m 4 ) ) )

80MPa( 10 6 Pa 1MPa )= ( 2× 10 14 Nm ) d 2 2( 9036Pa m 4 +( 80GPa( 10 9 Pa 1GPa ) ×( ( 0.098 )( d 2 4 2.56× 10 6 m 4 ) ) ) )

6283.2 d 2 4 d 2 8.8467× 10 3 =0

  d2=0.0569m

Conclusion:

The diameter of the steel sleeve is 0.0569m .

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Chapter 3 Solutions

Mechanics of Materials (MindTap Course List)

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