Explanation Given information: The diameter of the solid steel is d = 4 in . . The angle of the strain gage parallel to the axis of the shaft is β = 25 ° . The modulus of rigidity of the material is G = 11.5 × 10 6 psi . The gage reading in x -axis is ε x ′ = 300 × 10 − 6 in ./in . . Calculation: Show the general stress element for the given condition as in Figure 1. Consider the torsion in the steel shaft. The normal stress in x -axis is σ x = 0 . The normal stress in y -axis is σ y = 0 . The shear stress in xy- plane is τ x y = τ 0 . Find the normal strain in x -axis ( ε x ) as follows; ε x = 1 E ( σ x − ν σ y ) Here, the modulus of elasticity is E and the Poisson’s ratio is ν . Substitute 0 for σ x and 0 for σ y . ε x = 1 E ( 0 − ν ( 0 ) ) = 0 Find the normal strain in y -axis ( ε y ) as follows; ε y = 1 E ( σ y − ν σ x ) Substitute 0 for σ x and 0 for σ y . ε y = 1 E ( 0 − ν ( 0 ) ) = 0 Find the shearing strain in xy -plane ( γ x y ) as follows; γ x y = τ x y G Substitute τ 0 for τ x y . γ x y = τ 0 G γ x y 2 = τ 0 2 G Find the polar moment of inertia ( J ) of the shaft using the equation. J = π d 4 32 Substitute 4 in. for d . J = π × 4 4 32 = 25

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ISBN: 9780100257061

Author: BEER

Publisher: YUZU

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Chapter 7.9, Problem 152P

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Find the torque T in the strain gage.

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Chapter 7.9, Problem 151P

Chapter 7.9, Problem 153P

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Find the torque T in the strain gage.

Solution Summary: The author shows the general stress element for the given condition as in Figure 1. The torque T indicated by a gage reading is underset_T=113.2.

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Find the torque T in the strain gage.

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