Modulus of Elasticity

Question

Consider a cube of unit dimensions. Let $\alpha$ and $\beta$ be the lateral and longitudinal strains. The expressions for moduli of elasticity on applying unit tension -

1) At one edge: Young's modulus $E=\frac{1}{\alpha}$

2) Normally to all the sides: Bulk modulus $K=\frac{1}{3(\alpha-2\beta)}$

3) Tangentially: Rigidity modulus $N=\frac{1}{2(\alpha+\beta)}$

Here, I was able to understand the first expression... But, How do we obtain the expressions for Bulk & Shear moduli..? Explanation please...

Answer 1

For instance, let us derive 2). Let us consider a unit cube and apply unit tension along $x$, unit tension along $y$, and unit tension along $z$ (to derive bulk modulus). For example, tension along $x$ causes strain $\alpha$ along $x$ and strains $-\beta$ along $y$ and along $z$ (the sign of lateral strain is typically opposite to that of longitudinal strain: when you extend a cube in one direction, it typically contracts in two other directions. We consider a linear theory, so strains from tension in all three directions are summed up. Therefore, the total strain in direction $x$ equals $\alpha$ (caused by tension in direction $x$) plus $-\beta$ (caused by tension in direction $y$) plus $-\beta$ (caused by tension in direction $z$) equals $\alpha-2\beta$. The same is true for total strain along directions $y$ and $z$. Therefore, the length of a side of the unit cube after unit tension is applied in all directions is $1+\alpha-2\beta$. Therefore, the volume of the unit cube after unit tension is applied in all directions is $(1+\alpha-2\beta)^3\approx1+3(\alpha-2\beta)$. Thus the relative volume change is 3(\alpha-2\beta), and the bulk modulus is $\frac{1}{3(\alpha-2\beta)}$.