A bar of size 3m long, 250 mm wide and 150 mm thick is subjected to pull of 50 kN. Find change in length, breadth and thickness. E= 2 X 10^5 N/mm^2 , Poisson’s ratio= 0.25. Calculate modulus of rigidity and bulk modulus. Also, find volumetric strain and change in volume.

Solution :

Given :

Length, $\quad L=3 ~m=3000~mm$

Width, $\quad b=250~mm$

Diameter or Thickness, $\quad d=150~mm$

Young Modulus, $\quad E=2×10^5~N/mm^2$

Poisson ratio, $\quad \mu =0.25$

$\begin{aligned}\sigma=\frac{P}{A} &=\frac{50 \times 10^{3}}{250 \times 150}\\ &=1.33 {~N} /{mm}^{2}\end{aligned}$

1) change in length (dL)

$\begin{aligned}dL=\frac{P L}{A E} &=\frac{50 \times 10^{3} \times 3000}{250 \times 150 \times 2 \times 10^{5}}\\ &=0.02 {~mm} \end{aligned}$

2) Change in breadth (db)

$\begin{aligned} \mu &=\frac{\text { lateral Strain }}{\text { longitudinal Strain }} \\ \mu &=\frac{\frac{db}{b}}{\frac{dL}{L}}\\ \frac{d b}{b} &= \mu\frac{d L}{L} \Rightarrow d b =\frac{\mu dL ~b}{L}\\ db &=\frac{0.25 \times 0.02 \times 250}{3000}\\ db &=4.16 \times 10^{-4} {~mm} \end{aligned}$

3) Change in thickness (dt)

$\begin{aligned} d t &=\mu \frac{d L}{L} \times t\\ &=\frac{0.25 \times 0.02 \times 150}{3000}\\ &=2.5 \times 10^{-4} {~mm} \end{aligned}$

4) Modulus of rigidity (G)

$E=2 G(1+\mu )$

$\begin{aligned} G &=\frac{E}{2(1+\mu )}\\ &=\frac{2 \times 10^{5}}{2(1+0.25)}\\ &=8 \times 10^{4} {~N} /{mm}^{2} \end{aligned}$

5) Bulk modulus (K)

$ \begin{aligned} E &=3 K(1-2 \mu) \\ K &=\frac{E}{3(1-2 \mu)}\\ &=\frac{2 \times 10^{5}}{3(1-2(0.25))}\\ &=13.33 \times 10^{4} {~N}/{mm}^{2} \end{aligned} $

6) Volumetric Strain (dV/V)

$\begin{aligned} \frac{d V}{V}=\frac{\sigma}{K}&=\frac{1.33}{13.33×10^6}\\ &= 9.97 \times 10^{-6} \end{aligned}$

7) Change in Volume (dV)

$\begin{aligned} V &=\frac{\pi }{4}d^2L\\ &=\frac{\pi }{4}(150)^2×3000\\ &=53.01×10^6 ~mm^3 \end{aligned}$

$\begin{aligned}{dV} &={9.97 \times 10^{-6}}×V\\ dV&=9.97 \times 10^{-6}×{53.01\times 10^{6}}\\ &=528.5 {~mm^3}\end{aligned}$