SSC JE Civil Engineering 23-Sep-2019 Shift 2 Solved Paper

Gradually applied axial load on a bar is given as σ = (F/A) (∵ F is the gradually applied load)
Here, work done is given as (F δ L) / 2 and strain energy stored = $\left({\mathit{\sigma }}^2∕2\mathit{E}\right)\mathit{A}\mathit{L}$
Work done is equal to the strain energy stored.
$\left(\mathit{F}\mathit{\delta }\mathit{L}\right)∕2=\left({\mathit{\sigma }}^2∕2\mathit{E}\right)\mathit{A}\mathit{L}$
Therefore, σ = (F/A) .....(1)
Suddenly applied axial load on a bar is given as σ = (2F/A), here work done = (F δ L)
$\left(\mathit{F}\mathit{\delta }\mathit{L}\right)=\left({\mathit{\sigma }}^2∕2\mathit{E}\right)\mathit{A}\mathit{L}$
Therefore, $\mathit{\sigma }=\left(2\mathit{F}∕\mathit{A}\right)$.....(2)
From (1) and (2), it can be concluded that
Hence, suddenly applied load is twice the gradually applied load.