Question

A steel rod 4 m long and 20 mm diameter is subjected to an axial tensile load of 45

kN. Find the change in length and diameter of the rod. Es = 2 X 105 N/mm2

. Poisson’s

ratio =1/4​

Answer 1

A steel rod 4 m long and 20 mm diameter is subjected to an axial tensile load of 45 kN.

We have to find the change in length and diameter of the rod.

length of steel rod , L = 4m

diameter of steel rod, d = 20 mm = 0.02 m

so, cross sectional area of rod, A = πd²/4

= $\frac{3.14 \times (0.02)^2}{4}$

= 3.14 × 10¯⁴ m²

axial tensile load, F = 45 kN = 45000 N

modulus of elasticity, E = 2 × 10⁵ N/mm² = 2 × 10¹¹ N/m²

to determine the elongation of rod, we use $\delta=\frac{FL}{AE}$

= $\frac{45000\times4}{3.14\times10^{-4}\times2\times10^{11}}$

= 2.86 × 10¯³ m

= 2.86 mm

Therefore the change in the length of the steel rod is 2.86 mm.

we know, $\mu=\frac{\sigma_y}{\sigma_x}$

where $\sigma_x$ is longitudinal strain and $\sigma_y$ is lateral strain.

here, $\sigma_x$ = (2.86 mm)/(4m)

and $\sigma_y$ = δ/(20mm)

μ = 1/4 = 0.25

∴ 0.25 = $\frac{\frac{\delta}{20mm}}{\frac{2.86mm}{4m}}$

⇒0.25 = $\frac{\delta\times 4}{20\times2.86\times10^{-6}}$

⇒3.575 × 10¯⁶ m = δ

Therefore the change in diameter of the steel rod is 3.575 μm