Compare the crippling loads calculated by Rankine’s theory and Euler’s theory for a 4 m long hollow steel column having inner and outer diameter of 50 mm and 46 mm respectively.

Consider both the ends of the column are pin jointed, yield strength = 310 MPa, Rankine’s constant= 1/7500 and E = 210 GPa.

\begin{array}{l} \text { Given Data: } \ D_{0}=50 \mathrm{~mm}, \quad D_{\text {in }}=46 \mathrm{~mm} \ L_{e}=4 \mathrm{~m}=4000 \mathrm{~mm} \ f_{c}=310 \mathrm{MPa} \ \alpha=\frac{1}{7500}, E=210 \times 10^{3} \mathrm{MPa} \end{array} \begin{array}{l} \text { As per Euler's theory, }\ P=\frac{\pi^{2} E I}{Le^{2}};\ P= \frac{π^{2} \times 210 \times 10^{3} \times \frac{\pi}{64}\left[50^{4}-46^{4}\right]}{(4000)^2};\ P=11.27 \mathrm{kN} \text { ,Euler Load. } \end{array} \begin{array}{l} \text { As per Rainkine's Theory,} \text { }\ P=\frac{f_{c} A}{1+\alpha \lambda^{2}};\ \lambda=\frac{l}{\gamma};\ I=A r^{2}\ \Rightarrow \frac{π}{64 }\left(50^{4}-46^{4}\right)=\frac{\pi}{4}\left(50^{2}-46^{2}\right)\ \Rightarrow \frac{\left(50^{2}-46^{2}\right)\left(50^{2}+46^{2}\right)}{16}=\left(50^{2}-46\right) \times r^{2}\ \Rightarrow \quad r=16.985 \mathrm{~mm} \end{array} $P=\frac{310 \times \frac{\pi}{4}\left(50^{2}-46^{2}\right)}{1+\frac{1}{7500} \times(16.985)^{2}}\\P=90.03 \mathrm{kN}$