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A solid circular shaft of 3m length and 200mm diameter is subjected to a torque of 100knm. Find relative rotation between end x section of the shaft. Take g = 1×10^5 nmm^2-1×105-n-mm2

Question-Answer › Category: Strength of Materials › A solid circular shaft of 3m length and 200mm diameter is subjected to a torque of 100knm. Find relative rotation between end x section of the shaft. Take g = 1×10^5 nmm^2-1×105-n-mm2
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Sareyam verma asked 1 year ago

A solid circular shaft of 3m length and 200mm diameter is subjected to a torque of 100knm. Find relative rotation between end x section of the shaft. Take g = 1×10^5 nmm^2

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Arpit Raj answered 1 year ago

We have given the followings in the question,   Diameter of shaft, d= 200 mm           
Length of shaft L = 3 m = 3000 mm
Torque , T = 100 KNm = 100*10^6 Nmm          Modulus of Rigidity, G = 1*105 N/mm2 = 105 MPa       ( since 1 MPa = 1 N/mm2)          
We know that rotation of shaft relative to cross section is given by angle of twist,            θ = TL / GJ           
Where J= Polar moment of Inertia           
for solid shaft, J= πd4 / 32 = π*(200)4 / 32 = 1.57 *108 mm4        
So    θ = (100*106*3000) / 105 * 1.57*108
= 0.0191 rad = 0.0191*180/π                
 θ = 1.09 degree        
Hence relative rotation between end x section is 1.09 degree.

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