Question 10.SP.3: Column AB consists of a W10 × 39 rolled-steel shape made of ...
Column A B consists of a W 10 \times 39 rolled-steel shape made of a grade of steel for which \sigma_{Y}=36 \mathrm{ksi} and E=29 \times 10^{6} \mathrm{psi}. Determine the allowable centric load \mathbf{P}(a) if the effective length of the column is 24 \mathrm{ft} in all directions, (b) if bracing is provided to prevent the movement of the midpoint C in the x z plane. (Assume that the movement of point C in the y z plane is not affected by the bracing.)
We first compute the value of the slenderness ratio from Eq. 10.41 corresponding to the given yield strength \sigma_{Y}=36 \mathrm{ksi}.
\frac{L}{r}=4.71 \sqrt{\frac{29 \times 10^{6}}{36 \times 10^{3}}}=133.7
a. Effective Length =\mathbf{2 4} \mathrm{ft}. Since r_{y}<r_{x}, buckling will take place in the x z plane. For L=24 \mathrm{ft} and r=r_{y}=1.98 in., the slenderness ratio is
\frac{L}{r_{y}}=\frac{(24 \times 12) \mathrm{in.}}{1.98 \mathrm{in.}}=\frac{288 \mathrm{in} .}{1.98 \mathrm{in.}}=145.5
Since L / r>133.7, we use Eq. (10.39)
\sigma_{e}={\frac{\pi^{2}E}{(L/r)^{2}}} (10.39)
in Eq. (10.40)
\sigma_{c r}=0.877\sigma_{e} (10.40)
to determine \sigma_{c r}
\sigma_{c r}=0.877 \sigma_{e}=0.877 \frac{\pi^{2} E}{(L / r)^{2}}=0.877 \frac{\pi^{2}\left(29 \times 10^{3} \mathrm{ksi}\right)}{(145.5)^{2}}=11.86 \mathrm{ksi}
The allowable stress, determined using Eq. (10.42), and P_{\text {all }} are
\begin{aligned} & \sigma_{\text {all }}=\frac{\sigma_{c r}}{1.67}=\frac{11.86 \mathrm{ksi}}{1.67}=7.10 \mathrm{ksi} \\ & P_{\text {all }}=\sigma_{\text {all }} A=(7.10 \mathrm{ksi})\left(11.5 \mathrm{in}^{2}\right)=81.7 \mathrm{kips} \end{aligned}
b. Bracing at Midpoint C. Since bracing prevents movement of point C in the x z plane but not in the y z plane, we must compute the slenderness ratio correspoinding to buckling in each plane and determine which is larger.
x z \text { Plane: } \quad \text { Effective length }=12 \mathrm{ft}=144 \text { in., } r=r_{y}=1.98 \text { in. }
L / r=(144 \text { in. }) /(1.98 \text { in. })=72.7
yz Plane: Effective length =24 \mathrm{ft}=288 in., r=r_{x}=4.27 \mathrm{in}.
L / r=(288 \text { in. }) /(4.27 \text { in. })=67.4
Since the larger slenderness ratio corresponds to a smaller allowable load, we choose L / r=72.7. Since this is smaller than L / r=133.7, we use Eqs. (10.39) and (10.38)
\sigma_{\mathrm{cr}}=[0.658^{\sigma_Y/\sigma_e}]\sigma_{Y} (10.38)
to determine \sigma_{c r}
\sigma_{e}=\frac{\pi^{2} E}{(L / r)^{2}}=\frac{\pi^{2}\left(29 \times 10^{3} \mathrm{ksi}\right)}{(72.7)^{2}}=54.1 \mathrm{ksi}
\sigma_{c r}=\left[0.658^{\left(\sigma_{\gamma} / \sigma_{e}\right)}\right] F_{Y}=\left[0.658^{(36 \mathrm{ksi} / 54.1 \mathrm{ksi})}\right] 36 \mathrm{ksi}=27.3 \mathrm{ksi}
We now calculate the allowable stress using Eq. (10.42) and the allowable load.
\begin{aligned} \sigma_{\text {all }} & =\frac{\sigma_{c r}}{1.67}=\frac{27.3 \mathrm{ksi}}{1.67}=16.32 \mathrm{ksi} \\ P_{\text {all }} & =\sigma_{\text {all }} A=(16.32 \mathrm{ksi})\left(11.5 \mathrm{in}^{2}\right) \quad P_{\text {all }}=187.7 \mathrm{ksi} \end{aligned}
