To determine The torque required to form an elastic core. The angle of twist. The permanent angle of twist. To sketch: The residual stress distribution in the shaft. Explanation Given information: Length of the shaft is 2 m. The radius of the elastic core is ρ Y = 23 mm . Calculation: The expression to find torque required to form an elastic core ( T ) in the shaft is shown below: T = π τ Y 6 ( 4 c 3 − ρ Y 3 ) (1) Here, τ Y is the yield stress, c is the outer radius of the shaft, and ρ Y is the radius of the elastic core. Outer radius value of the shaft is 40 mm. Refer the shear stress-strain diagram. The value of yield shear stress is 150 MPa. The value of yield shear strain is γ Y = 0.005 rad . Substitute 150 MPa for τ Y , 40 mm for c , and 23 mm for ρ Y in Equation (1). T = π ( 150 MPa ) 6 ( 4 ( 40 mm ) 3 − ( 23 mm ) 3 ) = 19.151 × 10 6 N ⋅ mm = 19.151 × 10 6 N ⋅ mm × 10 − 3 kN 1 N × 10 − 3 m 1 mm ≈ 19.2 kN ⋅ m Therefore, the torque required to form an elastic core is 19.2 kN ⋅ m _ . Find the angle of twist ( ϕ ) of the shaft using the relation: ϕ = γ Y ρ Y L (2) Here, L is the length of the shaft. Substitute 0.005 rad for γ Y , 23 mm for ρ Y , and 2 m for L in Equation (2). ϕ = 0.005 rad 23 mm ( 2 m ) = 0.005 rad 23 mm ( 2 m × 10 3 mm 1 m ) = 0.4348 rad = 0.4348 rad ( 180 ° π rad ) ϕ = 24.9 ° Therefore, the angle of twist is 24.9 ° _ . Find the value of the shear modulus of elasticity of the material ( G ) using the relation: G = τ Y γ Y (3) Substitute 150 MPa for τ Y and 0.005 rad for γ Y in Equation (3). G = 150 MPa 0

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ISBN: 2810014920922

Author: HIBBELER

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Chapter 5.10, Problem 5.134P

To determine

The torque required to form an elastic core.

The angle of twist.

The permanent angle of twist.

To sketch:

The residual stress distribution in the shaft.

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Chapter 5.10, Problem 5.133P

Chapter 5.10, Problem 5.135P

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