Question 2.9: A steel pipe of length 6 m, outside diameter 45 mm and wall ......
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.
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