Question

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

Answer 1

Identify the quantities in the given options. Also, try to recall the relationships connecting them. Check whether the given options match the standard expressions containing these quantities. Also, try rearranging the standard expressions and cross-check again to find the answer.

Complete step-by-step solution:
The quantities mentioned in the given options are nothing but the elastic constants that determine the deformation produced by a given stress system acting on the material. The elastic constants are:
(a) Young’s modulus or the modulus of Elasticity (E)
(b) Bulk modulus (K)
(c) Poisson’s ratio $\left( \sigma \right)$
(d) Shear modulus or the modulus of rigidity $\left( \eta \right)$
Now let us discuss each of these constants one by one.
(a) From Hooke’s law we know that stress applied on a body that is subjected to tensile or compressive stress is directly proportional to the strain within the elastic limits of that body. The ratio of this applied stress to the resultant strain is called Young’s modulus of elasticity (Y).
$Y=\dfrac{Fl}{A\Delta l}$
(b) In case if the applied stress was mutually perpendicular and equal, then the ratio of the stress applied to the volumetric strain is called Bulk modulus (K).
$K=\dfrac{FV}{A\Delta V}$
(c) Poisson’s ratio $\left( \sigma \right)$ is the ratio of lateral strain to the longitudinal strain. For an ideal elastic incompressible material it has a maximum value of 0.5.
$\sigma =\left| \dfrac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}} \right|$
Where, ${{\varepsilon }_{1}}$ and ${{\varepsilon }_{2}}$ are longitudinal and transverse strains respectively.
(d) When we subject a body to shear stress, the shape of the body is changed. The ratio of this applied shear stress to resultant shear strain is called shear modulus or the modulus of rigidity$\left( \eta \right)$.
$\eta =\dfrac{F}{A\theta }$
Where $\theta$ is the longitudinal strain.
Now, we have the relationship between Young’s modulus (Y), rigidity modulus $\left( \eta \right)$ and Poisson’s ratio $\left( \sigma \right)$ given by,
$Y=2\eta \left( 1+\sigma \right)$
Rearranging,
$1+\sigma =\dfrac{Y}{2\eta }$
$\Rightarrow \sigma =\dfrac{0.5Y}{\eta }-1$
$\therefore \sigma =\dfrac{0.5Y-\eta }{\eta }$
Therefore, we see that the correct relation of the modulus of elasticity is given by,
$\sigma =\dfrac{0.5Y-\eta }{\eta }$
Hence, option D $\sigma =\dfrac{0.5Y-\eta }{\eta }$ is correct.

Note: We have also had relations between the other elastic constants. The relation between $Y$, $K$
and $\sigma$ given by,
$Y=3K\left( 1-2\sigma \right)$
The relation between Y, K and $\eta$ given by,
$Y=\dfrac{9K\eta }{\left( 3K+\eta \right)}$
The relation between $\sigma$, $K$ and $\eta$ given by,
$\sigma =\dfrac{\left( 3K-2\eta \right)}{\left( 6K+2\eta \right)}$