Question 21.P.10: A short length of hollow tube 32 mm external diameter and 25...

In Eq. (21.27) P=\frac{\sigma_{S} A}{1+k\left(L_{e} / r\right)^{2}} \sigma_{\mathrm{S}} A = 70 kN. Also the area of the cross section of the column is

A=\frac{\pi}{4}\left(32^{2}-25^{2}\right)=313.4 \mathrm{~mm}^{2}

and its second moment of area is

I=\frac{\pi}{64}\left(32^{4}-25^{4}\right)=32297.1 \mathrm{~mm}^{4}

The radius of gyration of the column cross section is then

r=\sqrt{(32297.1 / 313.4)}=10.15 \mathrm{~mm}

The equivalent length, L_{e} , of the column is equal to 0.5 × 2.5 × 10³ = 1250 mm. Now substituting the above in Eq. (21.27)

24.1 \times 10^{3}=\frac{70 \times 10^{3}}{1+k(1250 / 10.15)^{2}}

which gives

k = 0.000126

For a column with pinned ends L_{e} is equal to the actual length of the column which in this case is 1500 mm. Therefore, from Eq. (21.27) the crippling load is

P=\frac{70 \times 10^{3}}{1+0.000126(1500 / 10.15)^{2}}=18700 \mathrm{~N}

or

P = 18.7 kN