Data: A=300×300 mm2 , d = 20 mm$\phi$

No. of steel bar = 8, P = 360kN, m =15

Find: $\sigma_{c}, \sigma_{s}$

$$ \begin{array}{l}{\mathrm{A}_{\mathrm{s}}=\mathrm{n} \times \frac{\pi}{4} \mathrm{d}^{2}=8 \times \frac{\pi}{4} 20^{2}=2513.27 \mathrm{mm}^{2}} \\ {\mathrm{A}_{\mathrm{c}}=\mathrm{A}_{\mathrm{g}}-\mathrm{A}_{\mathrm{s}}} \\ {\mathrm{A}_{\mathrm{c}}=300 \times 300-2513.27} \\ {\mathrm{A}_{\mathrm{c}}=87486.72 \mathrm{mm}^{2}} \\ {\frac{\sigma_{\mathrm{s}}}{\sigma_{\mathrm{c}}}=m} \\ {\sigma_{\mathrm{s}}=\mathrm{m} \times \sigma_{\mathrm{c}}} \\ {\sigma_{\mathrm{s}}=15 \sigma_{\mathrm{c}}}\end{array} $$

\begin{array}{l}{\mathrm{P}=\mathrm{P}_{\mathrm{s}}+\mathrm{P}_{\mathrm{c}}} \ {\mathrm{P}=\sigma_{\mathrm{s}} \mathrm{A}_{\mathrm{s}}+\sigma_{\mathrm{c}} \mathrm{A}_{\mathrm{c}}} \ {360 \times 10^{3}=\left(15 \sigma_{\mathrm{c}}\right) 2513.27+\sigma_{\mathrm{c}} 87486.72} \ {360 \times 10^{3}=(37699.11+87486.72) \sigma_{\mathrm{c}}} \ \sigma_{\mathrm{c}=2.876 \mathrm{N} / \mathrm{mm}^{2}}\end{array}

$$ \begin{array}{l}{\sigma_{\mathrm{s}}=15 \sigma_{\mathrm{c}}} \\ {\sigma_{\mathrm{s}}=15 \times 2.876} \\ {\sigma_{\mathrm{s}}=43.136 \mathrm{N} / \mathrm{mm}^{2}}\end{array} $$