Question

The ultimate stress for a hollow steel column which carries an axial load of 2 MN is 500 N/mm If the external diameter of the column 250 mm, determine the internal diameter. Take the factor of safety as 4.0. 2​

Answer 1

Given:

Given,

Yield stress = 500 N/mm2

Factor of safety = 4.0

External diameter of the tube = 250 mm

Load P = 2MN = $2\times {{10}^{6}}$ N

To find:

We need to determine the internal diameter.

Solution:

We know that, Area A = load/stress = $\[2\times {{10}^{6}}/500\text{ }=\text{ 4}000m{{m}^{2}}\]$

For hollow steel column,

Area of the hollow $\[A=\text{ }\frac{\pi }{4}\text{ }\left( {{D}^{2}}~-{{d}^{2}} \right)\text{ }=4000m{{m}^{2}}\]$

                                   $=\[\text{ }\frac{\pi }{4}\text{ }\left( {{250}^{2}}~-{{d}^{2}} \right)\text{ }=4000m{{m}^{2}}\]$

                                   $=\[\text{3}\text{.14(62500-}{{\text{d}}^{2}})$

                                   $=4000m{{m}^{2}}$

Now,

$& \text{(62500-}{{\text{d}}^{2}})=\frac{4000}{3.14} \\ \\ & \text{(62500-}{{\text{d}}^{2}})=1273.9 \\ \\ & \text{-}{{\text{d}}^{2}}=1273.9-62500 \\\\ & \text{-}{{\text{d}}^{2}}=-61226.1 \\ \\ & {{\text{d}}^{2}}=61226.1$

To find the worth of internal diameter taking root on the each sides of above the equation and simplify it.

$\begin{align} & d=\sqrt{61226.1} \\ & \\ & d=247.4mm \\ \end{align}$

Therefore, the internal diameter of the column is $d=247.4mm$.