## Q. 2.12

A short piece of steel pipe is to carry an axial compressive load of 1200 kN with a factor of safety of 1.8 against yielding (Figure 2.23). If the thickness of the pipe is to be 1/8th of outside diameter, find the minimum required outside diameter. Yield stress of steel is 270 GPa.

## Verified Solution

The steel pipe is to be designed for allowable stress level.

\begin{aligned} \text { Allowable stress } & =\frac{\text { Yield stress }}{\text { Factor of safety }} \\ & =\frac{270}{1.8} \\ & =150 MPa \end{aligned}

Inside diameter of pipe = d – (2d/8) = 0.75 d
Area of cross section of the pipe

$=\frac{\pi}{4}\left(d^2-0.75^2 d^2\right)=0.344 d^2$

The axial stress = σ= P/A

$150=\frac{1200 \times 10^3}{A} \text {, solving for } A \text {, we get, } A=8 \times 10^3 mm ^2$

∴        $0.344 d^2=8 \times 10^3$

$d=\sqrt{\frac{8 \times 10^3}{0.344}}=152.5 mm$