Question 21.P.9: A tubular column has an effective length of 2.5 m and is to ...

The second moment of area of the column is given by

I=\frac{\pi\left[D^{4}-\left(\frac{7 D}{8}\right)^{4}\right]}{64}=0.0203 D^{4}

The area of cross section is given by

A=\frac{\pi\left[D^{2}-\left(\frac{7 D}{8}\right)^{2}\right]}{4}=0.1841 D^{2}

Then

r^{2}=\frac{0.0203 D^{4}}{0.1841 D^{2}}=0.11 D^{2}

Substituting in Eq. (21.27) P=\frac{\sigma_{S} A}{1+k\left(L_{e} / r\right)^{2}}

300 \times 3 \times 10^{3}=\frac{0.1841 D^{2}}{\left[1+\left(\frac{1}{7500}\right) \frac{\left(2.5 \times 10^{3}\right)^{2}}{\left(0.11 D^{2}\right)}\right]}

which simplifies to

D^{4}-14.85 \times 10^{3} D^{2}-0.125 \times 10^{3}=0

Solving

D = 122 mm

Say

D = 128 mm, t = 8 mm.