Using the interaction method, solve Sample Prob. 10.5 . Assume [latex]\left(\sigma_{\text {all }}\right)_{\text {bending }}=[/latex] [latex]150 \mathrm{MPa}[/latex].

Using Eq. (10.60), we write [latex]\frac{P / A}{\left(\sigma_{\text {all }}\right)_{\text {centric }}}+\frac{M c / I}{\left(\sigma_{\text {all }}\right)_{\text {bending }}} \leq 1[/latex] Substituting the given allowable bending stress and the allowable centric stress found in Sample Prob. 10.5 , as well as the other given data, we have [latex] \begin{gathered} \frac{P /\left(9.42 \times 10^{-3} \mathrm{~m}^{2}\right)}{97.1 \times 10^{6} \mathrm{~Pa}}+\frac{P(0.200 \mathrm{~m}) /\left(1.050 \times 10^{-3} \mathrm{~m}^{3}\right)}{150 \times 10^{6} \mathrm{~Pa}} \leq 1 \\ P \leq 423 \mathrm{kN} \end{gathered} [/latex] The largest allowable load [latex]\mathbf{P}[/latex] is thus [latex]\mathbf{P}=423 \mathrm{kN} \downarrow[/latex]

Question 10.SP.6: Using the interaction method, solve Sample Prob. 10.5. Assum...

Mechanics of Materials

Using the interaction method, solve Sample Prob. 10.5. Assume $\left(\sigma_{\text {all }}\right)_{\text {bending }}=$ $150 \mathrm{MPa}$ .

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Using Eq. (10.60), we write

$\frac{P / A}{\left(\sigma_{\text {all }}\right)_{\text {centric }}}+\frac{M c / I}{\left(\sigma_{\text {all }}\right)_{\text {bending }}} \leq 1$

Substituting the given allowable bending stress and the allowable centric stress found in Sample Prob. 10.5, as well as the other given data, we have

$\begin{gathered} \frac{P /\left(9.42 \times 10^{-3} \mathrm{~m}^{2}\right)}{97.1 \times 10^{6} \mathrm{~Pa}}+\frac{P(0.200 \mathrm{~m}) /\left(1.050 \times 10^{-3} \mathrm{~m}^{3}\right)}{150 \times 10^{6} \mathrm{~Pa}} \leq 1 \\ P \leq 423 \mathrm{kN} \end{gathered}$

The largest allowable load $\mathbf{P}$ is thus $\mathbf{P}=423 \mathrm{kN} \downarrow$