Question 10.SP.5: Using the Allowable Stress Design method, determine the larg......
Using the Allowable Stress Design method, determine the largest load P that can be safely carried by a W310 × 74 steel column of 4.5-m effective length. Use E = 200 GPa and \sigma_Y = 250 MPa.
STRATEGY: Determine the allowable centric stress for the column, using the governing Allowable Stress Design equation for steel, Eq. (10.38) or Eq. (10.40) with Eq. (10.42). Then use Eq. (10.55) to calculate the load P.
\sigma_{ cr }=\left[0.658^{\left(\sigma_\gamma / \sigma_c\right)}\right] \sigma_Y (10.38)
\sigma_{ cr }=0.877 \sigma_e (10.40)
\sigma_{ all }=\frac{\sigma_{ cr }}{1.67} (10.42)
\frac{P}{A}+\frac{M c}{I} \leq \sigma_{\text {all }} (10.55)
MODELING and ANALYSIS:
The largest slenderness ratio of the column is L / r_y = (4.5 m)/(0.0498 m) = 90.4. Using Eq. (10.41) with E = 200 GPa and \sigma_Y= = 250 MPa, the slenderness ratio at the junction between the two equations for \sigma_{ cr } is L/r = 133.2. Thus, Eqs. (10.38) and (10.39) are used, and \sigma_{ cr } = 162.2 MPa. Using Eq. (10.42), the allowable stress is
\sigma_e=\frac{\pi^2 E}{(L / r)^2} (10.39)
\frac{L}{r}=4.71 \sqrt{\frac{E}{\sigma_Y}} (10.41)
\left(\sigma_{\text {all }}\right)_{\text {centric }}=162.2 / 1.67=97.1 MPa
For the given column, replacing the eccentric loading with a centric force-couple system acting at the centroid (Fig. 1), we write
\frac{P}{A}=\frac{P}{9.42 \times 10^{-3} m ^2} \quad \frac{M c}{I}=\frac{M}{S}=\frac{P(0.200 m )}{1.050 \times 10^{-3} m ^3}
Substituting into Eq. (10.55), we obtain
\begin{gathered} \frac{P}{A}+\frac{M c}{I} \leq \sigma_{\text {all }} \\ \frac{P}{9.42 \times 10^{-3} m ^2}+\frac{P(0.200 m )}{1.050 \times 10^{-3} m ^3} \leq 97.1 MPa \quad P \leq 327 kN \end{gathered}
The largest allowable load P is P = 327 kN ↓
