SSC JE Mechanical Engineering 2018 Set 1 Paper

Question : 130 of 200

Marks: +1, -0

If the diameter of the column is reduced by 30%, then what will be the change in the Euler’s buckling load (in %)?

Solution:

Euler’s Buckling load

P_e={π^2EI}/{l_e^2}

I={πd^4}/64

P_e∝d^4.

Diameter reduced to 30%, so new diameter 0.7d
Change of Euler’s load

={d^4-(0.7d)^4}/d^4×100=75%

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