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 = (σ2∕2E)AL Work done is equal to the strain energy stored. (FδL)∕2=(σ2∕2E)AL Therefore, σ = (F/A) .....(1) Suddenly applied axial load on a bar is given as σ = (2F/A), here work done = (F δ L) (FδL)=(σ2∕2E)AL Therefore, σ=(2F∕A).....(2) From (1) and (2), it can be concluded that Hence, suddenly applied load is twice the gradually applied load.