Question

For the cube of given dimensions which is fixed at the bottom and acted upon by the force as shown in figure, identify the correct statement(s).
[Poisson's ratio $=0.8$ and Young's modulus $=2\times {10}^{11}{\text{N/m}}^2$]

• A. Normal stress is $25\sqrt{3}{\text{N/m}}^2$
• B. Shear stress is $25{\text{N/m}}^2$
• C. Longitudinal strain is $12.5\sqrt{3}\times {10}^{-11}$
• D. Magnitude of lateral strain is $10\sqrt{3}\times {10}^{-11}$

Answer 1

Answer: (A), (B), (C), (D)

Magnitude of lateral strain is $10\sqrt{3}\times {10}^{-11}$

Given,
Poisson's ratio i.e $\nu =0.8$
Young's modulus $=2\times {10}^{11}{\text{N/m}}^2$
According to question


$\text{Normal stress}=\frac{\text{Normal force}}{\text{Area}}$

Normal force is $25\sqrt{3}\text{N}$ acting peprendicular to the surface of the cube.
$\Rightarrow {\sigma }_N=\frac{25\sqrt{3}}{1}=25\sqrt{3}{\text{N/m}}^2...\left(1\right)$
$\text{Shear stress}=\frac{\text{Shear force}}{\text{Area}}$
Similarly, $25\text{N}$ is the shear force, acting parallel to the surface of cube.
$\Rightarrow {\sigma }_S=\frac{25}{1}=25{\text{N/m}}^2...\left(2\right)$
Young's modulus has been defined as ratio of stress and strain for the normal direction, at the cross-sectional area where force is acting.
$Y=\frac{{\sigma }_N}{{\epsilon }_N}$
$\Rightarrow {\epsilon }_N=\frac{25\sqrt{3}}{2\times {10}^{11}}$
$\Rightarrow {\epsilon }_N=12.5\sqrt{3}\times {10}^{-11}$

${\epsilon }_N$ is the strain along normal direction, also known as longitudinal strain.

Poisson ratio is defined as,
$\nu =-\frac{\text{Lateral strain}}{\text{Longitudinal strain}}$
$\Rightarrow 0.8=\frac{-{\epsilon }_S}{12.5\sqrt{3}\times {10}^{-11}}$
$\Rightarrow {\epsilon }_S=-10\sqrt{3}\times {10}^{-11}$
Hence, magnitude of lateral strain is $10\sqrt{3}\times {10}^{-11}$

$\therefore \left(a\right),\left(b\right),\left(c\right),\left(d\right)$ are correct choices.