The ratio of stress induced due to sudden applied axial load to s
A. 2
B. 1.5
C. 2.5
D. 3
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
Strain energy (U)
U = Strain energy per unit volume (u) × Volume of the Member (V)
U = Area under load-deformation curve
Strain energy per unit volume ⇒ \(u=\frac{1}{2}\times stress(\sigma)\times strain (\epsilon) =\frac{1}{2}\times \sigma \times\frac{\sigma}{E}=\frac{\sigma^2}{2E}\)
Also,
Load deformation curve for gradual and sudden loading is shown below:
In gradual loading, the loading starts from zero and increases gradually till the body is fully loaded, while in sudden loading, the load is suddenly applied on the body.
Calculation:
1. For Gradual Loading:
Strain Energy (U):
\(U=\frac{1}{2}\times P × δ L=\frac{1}{2}\times P ×\frac{\sigma\times L}{E}\) ......(i)
\(U = u \times V=\frac{\sigma^2}{2E}\times A\times L\) ......(ii)
From equation (i) and (ii)
\(\therefore \frac{1}{2}\times P ×\frac{\sigma\times L}{E} =\frac{\sigma^2}{2E}\times A\times L\)
\({\sigma _{gradual}} = \frac{P}{A}\)
2. For sudden loading
Strain Energy (U):
\(U=P × δ L=P ×\frac{\sigma\times L}{E}\) ......(i)
\(U = u \times V=\frac{\sigma^2}{2E}\times A\times L\) ......(ii)
From equation (i) and (ii)
\(\therefore P ×\frac{\sigma\times L}{E} =\frac{\sigma^2}{2E}\times A\times L\)
\({\sigma _{sudden}} = \frac{{2P}}{A}\)
∴The ratio of stress-induced due to sudden applied axial load to stress-induced due to gradually applied axial load on a bar is 2.
