Question

When a uniform metallic wire is stretched the lateral strain produced in it is beta. If neu and y are the poisson's ratio and young's modulus for wire then what is elastic potential density?

Answer 1

Answer:

Elastic Potential energy density is given as

$u = \frac{1}{2}(Y)(\frac{\beta}{\nu})^2$

Explanation:

As we know that the ratio of lateral strain and longitudinal strain is known as poisson's ratio

So we will have

$\frac{\delta r/ r}{\delta L/L} = \nu$

so we have

$\frac{\beta}{\nu} = \frac{\delta L}{L}$

Also we know that stress and strain ratio is known as young's modulus

so we will have

$stress = Y strain$

$stress = Y(\frac{\beta}{\nu})$

now we know that elastic potential density is given as

$u = \frac{1}{2}(stress)(strain)$

$u = \frac{1}{2}(Y)(\frac{\beta}{\nu})^2$

#Learn

Topic : Elasticity

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