SSC JE Civil Engineering 2012 Solved Paper

Question : 133 of 200

Marks: +1, -0

The angle of twist of a closely helical spring under an axial torque is given by

Solution:

Torsion equation;

T/r = T/{Ip} =E. θ/L

Angle of twist;

θ= {TL}/{EI_p} = {T.2πDn}/ ({E πd^4}/32)

={64TDn}/{Ed^4}

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