To determine The applied torque T . The permanent angle of twist of the tube. To sketch: The residual shear-stress distribution in the tube. Explanation Given information: The Length of the tube is 2 m. The Maximum shear strain is γ max = 0.006 rad . Calculation: The outer radius of the tube is 35 mm. The inner radius of the tube is 30 mm. Refer the shear stress-strain diagram: The value of yield shear stress τ Y = 210 MPa . The value of yield shear strain γ Y = 0.003 rad . Sketch the shear strain in the shaft as shown in Figure 1. Refer Figure 1. Find the value of shear strain in the inner radius of the tube as shown below: γ 30 mm = 0.006 rad 35 mm = 0.005143 rad The tube behaves fully plastic. The expression to find the plastic torque ( T P ) in the tube with inner radius ( c i ) and outer radius ( c 0 ) is shown below: T P = 2 π ∫ c i c 0 τ ρ 2 d ρ (1) Substitute τ Y for τ , 35 mm for c i , and 30 mm for c 0 in Equation (1). T P = 2 π τ Y ∫ 30 mm 35 mm ρ 2 d ρ = 2 π τ Y [ ρ 3 3 ] 30 mm 35 mm = 2 π ( 210 MPa ) ( ( 35 mm ) 3 3 − ( 30 mm ) 3 3 ) = 6.98219 × 10 6 N ⋅ mm T P = 6.98219 × 10 3 N ⋅ m ≈ 6.98 kN ⋅ m Therefore, the applied torque T is 6.98 kN ⋅ m _ . Find the angle of twist ( ϕ P ) of the tube using the relation: ϕ P = γ max c 0 L (2) Here, L is the length of the tube. Substitute 0.006 rad for γ max , 35 mm for c 0 , and 2 m for L in Equation (2). ϕ P = 0.006 rad 35 mm ( 2 m ) = 0.006 rad 35 mm ( 2 m × 10 3 mm 1 m ) = 0.34286 rad Find the value of the shear modulus of elasticity of the material ( G ) using the relation: G = τ Y γ Y (3) Substitute 210 MPa for τ Y and 0.003 rad for γ Y in Equation (3). G = 210 MPa 0.003 rad = 70 × 10 3 MPa = 70 GPa Find the angle of twist ( ϕ ' P ) of the tube when the reverse torque of T P = 6.98219 × 10 3 N ⋅ m is applied using the relation: ϕ ' P = T ' P L J G (4) The polar moment of inertia of the hollow tube of outer radius ( c 0 ) and inner radius ( c i ) is, J = π 2 ( c 0 4 − c i 4 ) = π 2 ( ( 35 mm ) 4 − ( 30 mm ) 4 ) = 1.0848 × 10 6 mm 4 . Substitute 6.98219 × 10 3 N ⋅ m for T ' P , 2 m for L , 1.0848 × 10 6 mm 4 for J , and 70 × 10 9 Pa for G in Equation (4). ϕ ' P = ( 6.98219 × 10 3 N ⋅ m ) ( 2 m ) ( 1.0848 × 10 6 mm 4 ) ( 70 × 10 9 Pa ) = ( 6

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 5.10, Problem 5.140P

To determine

The applied torque T.

The permanent angle of twist of the tube.

To sketch:

The residual shear-stress distribution in the tube.

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Chapter 5.10, Problem 5.139P

Chapter 5.10, Problem 5.141P

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The applied torque T . The permanent angle of twist of the tube. To sketch: The residual shear-stress distribution in the tube.

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The applied torque T. The permanent angle of twist of the tube. To sketch: The residual shear-stress distribution in the tube.

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Chapter 5 Solutions

The applied torque T.

The permanent angle of twist of the tube.

To sketch:

The residual shear-stress distribution in the tube.