   # -7 A steel tube (G = 11.5 x 10 6 psi) has an outer diameter d 2 = 2 . 0 in. and an inner diameter d t =1 ,5 in. When twisted by a torque 7", the tube develops a maximum normal strain of 170 x 10-6. What is the magnitude of the applied torque T? ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347 ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 3, Problem 3.5.7P
Textbook Problem
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## -7 A steel tube (G = 11.5 x 106 psi) has an outer diameter d2= 2.0 in. and an inner diameter dt=1,5 in. When twisted by a torque 7", the tube develops a maximum normal strain of 170 x 10-6.What is the magnitude of the applied torque T?

To determine

The magnitude of the applied torque.

### Explanation of Solution

Given information:

The shear modulus is 11.5×106psi , the outside diameter is 2in , the inside diameter is 1.5in and the maximum normal strain in the bar is 170×106 .

Write the expression for polar moment of inertia from torsion equation.

Ip=π32(d24d14) ..... (I)

Here, the polar moment of inertia is IP , the outside diameter is d2 and inside diameter is d1

Write the expression for maximum shear strain.

γmax=2εmax ...... (II)

Here, maximum shear strain is γmax and the maximum normal strain is εmax .

Write the expression for applied torque from torsion Equation.

T=2Ipτmaxd2...... (III)

Here, the applied torque is T , polar moment of inertia is Ip , maximum shear stress of the shaft is τmax , and outer diameter of the shaft is d2 .

Write the expression for maximum shear stress.

τmax=Gγmax...... (IV)

Here, the maximum shear stress is τmax , the shear modulus of the shaft is G and maximum shear strain is γmax .

Substitute Gγmax for τmax in Equation (III).

T=2Ip(Gγmax)d2...... (V)

Calculation:

Substitute 1.5in for d2 and 2in for d2 in Equation (I).

Ip=π32(( 2in)4( 1

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