The relationship between Young's modulus Y, B | vedclass

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Question

(A) The relationship between the elastic moduli is derived from the Poisson's ratio $\sigma$. We have the following standard relations: $Y = 3K(1 - 2\

Vedclass · Accepted Answer

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

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(A) The relationship between the elastic moduli is derived from the Poisson's ratio $\sigma$. We have the following standard relations: $Y = 3K(1 - 2\sigma)$ $Y = 2\eta(1 + \sigma)$ From the first equation,$1 - 2\sigma = \frac{Y}{3K} \implies 2\sigma = 1 - \frac{Y}{3K} \implies \sigma = \frac{1}{2} - \frac{Y}{6K}$. From the second equation,$1 + \sigma = \frac{Y}{2\eta} \implies \sigma = \frac{Y}{2\eta} - 1$. Equating the two expressions for $\sigma$: $\frac{Y}{2\eta} - 1 = \frac{1}{2} - \frac{Y}{6K}$ $\frac{Y}{2\eta} + \frac{Y}{6K} = \frac{3}{2}$ Multiplying by $6\eta K$: $3YK + Y\eta = 9\eta K$ $Y(3K + \eta) = 9\eta K$ $Y = \frac{9\eta K}{3K + \eta}$ Thus,the correct option is $A$.

The relationship between Young's modulus $Y,$ Bulk modulus $K,$ and modulus of rigidity $\eta$ is:

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