Chapter 2
Q. 2.9
A steel pipe of length 6 m, outside diameter 45 mm and wall thickness 10 mm as shown in Figure 2.20 is subjected to an axial compressive load of 400 kN. Assuming that E = 200 GPa and poisson’s ratio of 0.3 find: (a) shortening of the pipe, (b) the increase in outside diameter, (c) the increase in wall thickness.
Step-by-Step
Verified Solution
The steel pipe is shown in Figure 2.20.
(a) To calculate the shortening of the pipe we make use of the formula:
\begin{gathered} d l=\frac{P L}{A E}, A=\frac{\pi}{4}\left(45^2-25^2\right)=1099.55 mm ^2 \\ L=6000 mm , E=200 \times 10^3 N / mm ^2 \\ d l=\frac{400 \times 10^3 \times 6000}{1099.55 \times 200 \times 10^3}=10.91 mm \end{gathered}
(b) Increase in outside diameter
\begin{aligned} \text { Linear strain } & =E_{\text {long }}=\frac{10.91}{6000} \\ & =0.0018 \\ \text { Lateral strain } & =0.3 \times 0.0018 \\ & =0.00054 \end{aligned}
∴ Increase in outside diameter
dd = d × 0.00054
= 45 × 0.00054
= 0.0243 mm
(c) Increase in wall thickness
dt = 10 × 0.00054
= 0.0054 mm