SSC JE Civil Engineering 2013 Solved Paper

Question : 167 of 200

Marks: +1, -0

The equivalent stiffness of two springs of stiffness

S_1

and

S_2

joined in series is given by S =

Solution:

Springs joined in series

1/S=1/S_1+1/S_2={S_1+S_2}/{S_1S_2}

S={ S_1 S_2} /(S_1 + S_2 )

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