## Chapter 2

## Q. 2.21

A steel bar 3 m long has a circular section of diameter 1.5 cm over one-half of its length and diameter of 1cm over the other half as in Figure 2.34.

(a) How much will the bar elongate under a tensile load of 50 kN?

(b) If the same volume of the material is rolled into a bar of constant diameter d and length

3 m, what will be elongation under the same load? Take E = 200 GPa.

## Step-by-Step

## Verified Solution

Areas of cross sections:

A_{ AB }=\frac{\pi}{4} \times 15^2=176.7 mm ^2, A_{ AB }=\frac{\pi}{4} \times 10^2=78.53 mm ^2

(a) Elongation of the bar ABC = =d l_{ ABC }=\frac{50 \times 1000 \times 1500}{176.7 \times 200 \times 10^3}+\frac{50 \times 1000 \times 1500}{18.53 \times 200 \times 10^3}

= 2.122 + 4.775 = 6.897 mm

(b) Elongation of the bar if diameter is made constant. Since volume is same, we have

\frac{\pi d^2}{4} \times 3000=176.7 \times 1500+78.53 \times 1500

On simplification, d² = 162.48, ∴ d = 12.75 mm.

With this uniform diameter, the change in length of the bar would be

d l_{ AC }=\frac{50 \times 10^3 \times 3000}{\left(\pi \times 12.75^2 \times 200 \times 10^3\right) / 4}=5.878 mm