SSC JE Mechanical Engineering 2013 Paper

Question : 120 of 200

Marks: +1, -0

For a key to be equally strong in shearing and crushing, the width of the key, assuming that the allowable crushing stress is twice the allowable shear stress, should be

2.5 times its thickness
2 times its thickness
1.5 times its thickness
equal to its thickness

Validate

Solution:

t → thickness of key
w → width of key
p → load

\Load/{\Crushing \area}=2×(\load/{\shearing \area})

1/(t/2×L)=2(1/(W×L)),[t=W]

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