Answer:
b.2 times
Explanation:
Work done is equal to the strain energy stored.
(F δL) / 2 = (σ2 / 2E) AL
Therefore, σ = (F/A) ----------- (1)
- Suddenly applied load is given as σ = (2F/A), here work done = (F δL)
(F δL) = (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.
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Answers & Comments
Answer:
b.2 times
Explanation:
Work done is equal to the strain energy stored.
(F δL) / 2 = (σ2 / 2E) AL
Therefore, σ = (F/A) ----------- (1)
- Suddenly applied load is given as σ = (2F/A), here work done = (F δL)
(F δL) = (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.