The relation between Young's modulus (Y),modu | vedclass

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(D) The relations between elastic constants are given by:$Y = 3K(1 - 2\sigma)$  --- $(1)$$Y = 2\eta(1 + \sigma)$  --- $(2)$From $(1)$,we have: $\frac{

Vedclass · Accepted Answer

$\frac{1}{Y} = \frac{1}{3\eta} + \frac{1}{9K}$

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(D) The relations between elastic constants are given by:$Y = 3K(1 - 2\sigma)$  --- $(1)$$Y = 2\eta(1 + \sigma)$  --- $(2)$From $(1)$,we have: $\frac{Y}{3K} = 1 - 2\sigma$  --- $(3)$From $(2)$,we have: $\frac{Y}{2\eta} = 1 + \sigma$  --- $(4)$To eliminate $\sigma$,multiply equation $(4)$ by $2$:$\frac{Y}{\eta} = 2 + 2\sigma$  --- $(5)$Adding equations $(3)$ and $(5)$:$\frac{Y}{3K} + \frac{Y}{\eta} = (1 - 2\sigma) + (2 + 2\sigma)$$\frac{Y}{3K} + \frac{Y}{\eta} = 3$Dividing both sides by $3Y$:$\frac{1}{9K} + \frac{1}{3\eta} = \frac{1}{Y}$Thus,the correct relation is $\frac{1}{Y} = \frac{1}{3\eta} + \frac{1}{9K}$.

The relation between Young's modulus $(Y)$,modulus of rigidity $(\eta)$,and bulk modulus $(K)$ for an elastic material is:

Similar Questions

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