A solid steel bar of circular cross section has diameter d 5 2.5 in., L 5 60 in., and shear modulus of elasticity G 5 11.5 3 106 psi. The ba

Question

A solid steel bar of circular cross section has diameter d 5 2.5 in., L 5 60 in., and shear modulus of elasticity G 5 11.5 3 106 psi. The bar is subjected to torques T 5 300 lb-ft at the ends. Calculate the angle of twist between the ends. What is the maxi- mum shear stress and the shear stress at a distance rA 5 1.0 in. measured from the center of the bar

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niczorrrr 3 months 2021-08-02T15:14:51+00:00 1 Answers 10 views 0

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    2021-08-02T15:16:02+00:00

    Answer:

    A) θ = 4.9 x 10^(-3) rad

    B) τ_max = 1.173 ksi

    C) τ_a = 4.786 ksi

    Explanation:

    We are given;

    diameter; d = 2.5 inches = 0.2083 ft

    Length; L = 60 inches = 5 ft

    Torque; T = 300 lb.ft

    Shear modulus; G = 11.5 x 10^(6) psi = 11.5 x 144 x 10^(6) lb/ft² = 1.656 x 10^(9) lb/ft²

    A) Now, formula to determine angle of twist is given as;

    T/I_p = Gθ/L

    Where I_p is polar moment of inertia

    θ is angle of twist.

    Now I_p = πd⁴/32 = π(0.2083)⁴/32 = 1.85 x 10^(-4) ft⁴

    Thus, making θ the subject, we have;

    TL/GI_p = θ

    θ = (300 x 5)/(1.656 x 10^(9) x 1.85 x 10^(-4))

    θ = 4.9 x 10^(-3) rad

    B) Maximum shear stress is given by the formula ;

    τ_max = (Gθ/L)(d/2)

    From earlier, (Gθ/L) = T/I_p

    Thus, (Gθ/L) = 300/1.85 x 10^(-4) = 1621621.6216

    Thus,

    τ_max = 1621621.6216 x (0.2083/2)

    τ_max = 168891.89 lbf/ft²

    Converting to ksi = 168891.89/144000 ksi = 1.173 ksi

    C) Shear stress at radial distance is given as;

    τ_a = (Gθ/L)•r_a

    r_a is given as 5.1 inches = 0.425m

    τ_a = 1621621.6216 x 0.425 = 689189.189 lbf/ft²

    Converting to ksi = 689189.189/144000 ksi = 4.786 ksi

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