Explanation Given information: The length of the tube is 2 m. Calculation: Refer the shear stress-strain diagram. The value of yield shear stress is 350 MPa and the value of yield shear strain is 0.007 rad. The expression to find the plastic torque ( T P ) in the shaft with outer radius ( c 0 ) and inner radius ( c i ) is shown below: T P = 2 π τ Y ∫ c i c 0 ρ 2 d ρ = 2 π τ Y [ ρ 3 3 ] c i c 0 = 2 π τ Y ( ( c 0 ) 3 3 − ( c i ) 3 3 ) (1) The value of outer radius is 100 mm and the value of inner radius is 60 mm. Substitute 350 MPa for τ Y , 100 mm for c 0 , and 60 mm for c i in Equation (1). T P = 2 π ( 350 MPa ) ( ( 100 mm ) 3 3 − ( 60 mm ) 3 3 ) = 2 π ( 350 MPa × 1 N mm 2 1 MPa ) ( ( 100 mm ) 3 3 − ( 60 mm ) 3 3 ) = 574.702 × 10 6 N ⋅ mm = 574.7 kN ⋅ m Therefore, the torque required for the material to become fully plastic is 574.7 kN ⋅ m _ . The radius of the tube at yield ( ρ Y ) is 60 mm . Find the angle of twist ( ϕ P ) of the tube using the relation: ϕ P = γ Y ρ Y L (2) Here, L is the length of the tube. Substitute 0.007 rad for γ Y , 60 mm for ρ Y , and 2 m for L in Equation (2). ϕ P = 0.007 rad 60 mm ( 2 m ) = 0.007 rad 60 mm ( 2 m × 10 3 mm 1 m ) = 0.2333 rad Find the value of the shear modulus of elasticity of the material ( G ) using the relation: G = τ Y γ Y (3) Substitute 350 MPa for τ Y and 0.007 rad for γ Y in Equation (3). G = 350 MPa 0.007 rad = 50 × 10 3 MPa = 50 × 10 3 MPa ( 10 6 Pa 1 MPa ) = 50 × 10 9 Pa Find the angle of twist ( ϕ ' P ) of the tube when the reverse torque of 574

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ISBN: 9780137605521

Author: HIBBELER

Publisher: RENT PEARS

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Chapter 5.10, Problem 140P

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The torque required for the material to become fully plastic.

The permanent angle of twist of the tube.

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Chapter 5.10, Problem 139P

Chapter 5.10, Problem 141P

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The torque required for the material to become fully plastic. The permanent angle of twist of the tube.

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