Question 10.SP.5: Using the allowable-stress method, determine the largest loa...
Using the allowable-stress method, determine the largest load \mathbf{P} that can be safely carried by a W310 \times 74 steel column of 4.5-\mathrm{m} effective length. Use E=200 \mathrm{GPa} and \sigma_{Y}=250 \mathrm{MPa}.
The largest slenderness ratio of the column is L / r_{y}=(4.5 \mathrm{~m}) /(0.0498 \mathrm{~m})= 90.4. Using Eq. (10.41)
{\frac{L}{r}}=4.71{\sqrt{\frac{E}{\sigma_{Y}}}} (10.41)
with E=200 \mathrm{GPa} and \sigma_{Y}=250 \mathrm{MPa}, we find that the slenderness ratio at the junction between the two equations for \sigma_{c r} is L / r=133.2. Thus, we use Eqs. (10.38) and (10.39)
\sigma_{\mathrm{cr}}=\mathrm{[0.658^{(\sigma_Y/ \sigma_e)}\sigma_{Y}} (10.38)
\sigma_{e}=\frac{\pi^{2}E}{(L/r)^{2}} (10.39)
and find that \sigma_{c r}= 162.2 MPa. Using Eq. (10.42),
\sigma_{\mathrm{all}}={\frac{\sigma_{\mathrm{cr}}}{1.67}} (10.42)
the allowable stress is
\left(\sigma_{\text {all }}\right)_{\text {centric }}=162.2 / 1.67=97.1 \mathrm{MPa}
For the given column and loading, we have
\frac{P}{A}=\frac{P}{9.42 \times 10^{-3} \mathrm{~m}^{2}} \quad \frac{M c}{I}=\frac{M}{S}=\frac{P(0.200 \mathrm{~m})}{1.050 \times 10^{-3} \mathrm{~m}^{3}}
Substituting into Eq. (10.58), we write
\begin{gathered} \frac{P}{A}+\frac{M c}{I} \leq \sigma_{\text {all }} \\ \frac{P}{9.42 \times 10^{-3} \mathrm{~m}^{2}}+\frac{P(0.200 \mathrm{~m})}{1.050 \times 10^{-3} \mathrm{~m}^{3}} \leq 97.1 \mathrm{MPa} \quad P \leq 327 \mathrm{kN} \end{gathered}
The largest allowable load \mathbf{P} is thus \mathbf{P}=327 \mathrm{kN} \downarrow
