Question 10.SP.6: Using the interaction method, solve Sample Prob. 10.5. Assum......
Using the interaction method, solve Sample Prob. 10.5. Assume \left(\sigma_{\text {all }}\right)_{\text {bending }}=150 MPa.
STRATEGY: Use the allowable centric stress found in Sample Prob. 10.5 to calculate the load P.
MODELING and ANALYSIS:
Using Eq. (10.57),
\frac{P / A}{\left(\sigma_{\text {all }}\right)_{\text {centric }}}+\frac{M c / I}{\left(\sigma_{\text {all }}\right)_{\text {bending }}} \leq 1 (10.57)
Substituting the allowable bending stress and the allowable centric stress found in Sample Prob. 10.5, as well as the other given data, we obtain
\frac{P /\left(9.42 \times 10^{-3} m ^2\right)}{97.1 \times 10^6 Pa }+\frac{P(0.200 m ) /\left(1.050 \times 10^{-3} m ^3\right)}{150 \times 10^6 Pa } \leq 1
P ≤ 423 kN
The largest allowable load P is P = 423 kN ↓