Question

Which of the following relations is true ?
A. $3Y = K(1 + \sigma )$
B. $K = \dfrac{{9\eta Y}}{{Y + \eta }}$
C. $\sigma = (6K + \eta )Y$
D. $\sigma = \dfrac{{0.5Y - \eta }}{\eta }$

Answer 1

Remember all the equations of elasticity and apply them accordingly.If you have the equations memorised, then this will be easy, else we must derive the equations accordingly. Types of modulus of elasticity are : Young’s Modulus $Y$, Bulk Modulus $K$, Modulus of rigidity.

Complete step by step answer:
Value of Young’s modulus = $\dfrac{\text{longitudinal stress}}{\text{longitudinal strain}}$ .
Thus $Y=\dfrac{1}{\alpha } - - - - (1)$ , where $\alpha$ is the longitudinal strain per unit stress.
We also know that Bulk modulus $K=\dfrac{1}{{3(\alpha - 2\beta )}}- - - - (2)$
where $\beta$ is the lateral strain per unit stress.
We know that rigidity modulus $\eta = \dfrac{1}{{2(\alpha + \beta )}} - - - - (3)$ .
Also , Poisson’s ratio $\sigma = \dfrac{\beta }{\alpha } - - - - (4)$ .

Now we can derive the relation for the above question,
$\eta = \dfrac{1}{{2(\alpha + \beta )}}$
$\Rightarrow \eta = \dfrac{{1/\alpha }}{{2(1 + \beta /\alpha )}}$
But we know that $\beta /\alpha$ is Poisson's ratio. Hence,
$\eta = \dfrac{Y}{{2(1 + \sigma )}}$
$\Rightarrow 1 + \sigma = \dfrac{{0.5Y}}{\eta }$
Now moving that one to the right hand side , we get
$\therefore \sigma = \dfrac{{0.5Y - \eta }}{\eta }$

Hence, the correct answer is option D.

Note: Some important points which are to be remembered are as follows:-
-Coefficient of elasticity is always dependent on the material and its temperature and purity, and not on its stress and strain.
-The fundamental characteristic of elasticity of a material is its strain, not its stress.
-The values of the three coefficients of elasticity $Y,K{\text{ }}and{\text{ }}\eta$ are always different and are related to each other.
-The potential energy due to elasticity equals $\dfrac{1}{2} \times stress \times strain$ . -From this we can relate it to spring force and use it in required equations.