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If a solid shaft (diameter 20 cm, length 400 cm, N = 0.8 × 105 N/mm2) when subjected to a twisting moment, produces maximum shear stress of 50 N/mm2, the angle of twist in radians, is

A. 0.001

B. 0.002

C. 0.0025

D. 0.003

Answer: Option C


This Question Belongs to Civil Engineering >> Theory Of Structures

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Comments ( 2 )

  1. Sabireen Afridi
    Sabireen Afridi :
    3 years ago

    By Torsion Equation:

    T/Ip = Fs/ymax = G$/L.

    Where T = Torsion
    Ip = Polar moment of inertia
    Ymax = Max distance from nuetral axis
    G = MODULUS of rigidity
    $ = Angle of twist in Radian
    Fs = Max shear force.

    Hence Fs/Ymax = G$/L So 50/100 = 0.8 * 100000/4000 (All diamension in MM) = 0.0025 Answer.

  2. Aziz Panna
    Aziz Panna :
    4 years ago

    τ = Gθ/L
    =》 θ = τ L / G = (50 * 4000) /( 0.8 * 10^5) =2.5

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