b) both ends are hinged.

Take $\sigma_{c} = 550 N/mm^{2}, a = (1/1600)$

$$ \begin{array}{l}{\text { Data: } \mathrm{L}=4.2 \mathrm{m}, \text { a) Both ends are fixed. } \mathrm{b} \text { ) Both ends are hinged }} \ {\sigma \mathrm{c}=550 \mathrm{MPa}, \quad a=\frac{1}{1600}} \ {\text { Find: } \mathrm{P}_{\mathrm{R}}}\end{array} $$ $$ \begin{array}{l}{\mathrm{A}=\frac{\pi}{4}\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right)=\frac{\pi}{4}\left(300^{2}-200^{2}\right)=39269.91 \mathrm{mm}^{2}} \ {\text { Imin }=\frac{\pi}{64}\left(\mathrm{D}^{4}-\mathrm{d}^{4}\right)=\frac{\pi}{64}\left(300^{4}-200^{4}\right)=319068003.9 \mathrm{mm}^{4}} \ {\mathrm{K}_{\min }=\sqrt{\frac{\operatorname{Imin}}{\mathrm{A}}}=\sqrt{\frac{319068003.9}{39269.91}}=90.138 \mathrm{mm}}\end{array} $$ $$ \begin{array}{l}{\lambda=\frac{\mathrm{Le}}{\mathrm{K}_{\min }}} \ {\text { Case a }) \quad \mathrm{Le}=\frac{\mathrm{L}}{2}=\frac{4200}{2}=2100 \mathrm{mm}} \ {\text { Case b }) \quad \mathrm{Le}=\mathrm{L}=4200 \mathrm{mm}}\end{array} $$ $$ \begin{array}{l}{\text { Case a) } \lambda=\frac{2100}{90.138}=23.297} \ {\text { Case b) } \lambda=\frac{4200}{90.138}=46.595}\end{array} $$ By using Rankine's Formula, $$ \begin{array}{l}{\mathrm{P}_{\mathrm{R}}=\frac{\sigma_{\mathrm{c}} \mathrm{A}}{1+\mathrm{a} \lambda^{2}}} \ {\text { Case a) } \mathrm{P}_{\mathrm{R}}=\frac{550 \times 39269.91}{1+\left(\frac{1}{1600}\right) \times 23.297^{2}}=16127647.85 \mathrm{N}=16.13 \times 10^{3} \mathrm{kN}} \ {}\end{array} $$ $$ \text { Case b ) } \mathrm{P}_{\mathrm{R}}=\frac{550 \times 39269.91}{1+\left(\frac{1}{1600}\right) \times 46.595^{2}}=9163741.91 \mathrm{N}=9.16 \times 10^{3} \mathrm{kN} $$