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: Length of the tube is 2 m. Maximum shear strain γ max = 0.006 rad . Calculation: The outer radius of the tube is 45 mm. The inner radius of the tube is 40 mm. Refer the shear stress-strain diagram: The value of yield shear stress τ Y = 240 MPa . The value of yield shear strain γ Y = 0.003 rad . Sketch the shear strain diagram as shown in Figure 1. Refer Figure 1. Find the angle of twist of the tube as shown below: ϕ = γ max L c = 0.008 rad × 2 m 45 mm = 0.008 rad × 2 m 45 mm × 10 − 3 m 1 mm = 0.3556 rad Find the angle of twist ( ϕ ) of the tube using the relation: ϕ = γ Y ρ Y L (1) Here, ρ Y is the radius of elastic core and L is the length of the tube. Substitute 0.3556 rad for ϕ , 0.003 rad for γ Y , and 2 m for L in Equation (1). 0.3556 rad = 0.003 rad ρ Y ( 2 m ) ρ Y = 0.016875 m The radius of elastic core is less than the inner radius of the tube. Hence 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 (2) Substitute τ Y for τ , 45 mm for c i , and 40 mm for c 0 in Equation (2). T P = 2 π τ Y ∫ 40 mm 45 mm ρ 2 d ρ = 2 π τ Y [ ρ 3 3 ] 40 mm 45 mm = 2 π ( 240 MPa ) ( ( 45 mm ) 3 3 − ( 40 mm ) 3 3 ) = 13.6345 × 10 6 N ⋅ mm T P = 13.6345 × 10 3 N ⋅ m ≈ 13.6 kN ⋅ m Therefore, the applied torque T is 13 .6 kN ⋅ m _ . Find the value of the shear modulus of elasticity of the material ( G ) using the relation: G = τ Y γ Y (3) Substitute 240 MPa for τ Y and 0.003 rad for γ Y in Equation (3). G = 240 MPa 0.003 rad = 80 × 10 3 MPa = 80 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 ( ( 45 mm ) 4 − ( 40 mm ) 4 ) = 2.42 × 10 6 mm 4 . Substitute 13.6345 × 10 3 N ⋅ m for T ' P , 2 m for L , 2.42 × 10 6 mm 4 for J , and 70 × 10 9 Pa for G in Equation (4). ϕ ' P = ( 13.6345 × 10 3 N ⋅ m ) ( 2 m ) ( 2.42 × 10 6 mm 4 ) ( 80 × 10 9 Pa ) = ( 13

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Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

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Chapter 5.10, Problem 5.142P

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 5.141P

Chapter 5, Problem 5.143RP

Chapter 5 Solutions

Mechanics of Materials

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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.