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If a solid shaft (diameter 20 cm length 400 cm N...

This question belongs to Civil Engineering Theory Of Structures
Theory of Structures
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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

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Correct Answer: Option C
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3 Comments
RAJKUMAR YADAV
RAJKUMAR YADAV 2 years ago
400*20/0.8*10^5*50
=0.0025
Sabireen Afridi
Sabireen Afridi 6 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.
Aziz Panna
Aziz Panna 6 years ago
τ = Gθ/L
=》 θ = τ L / G = (50 * 4000) /( 0.8 * 10^5) =2.5
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