Question 10.SP.3: Column AB consists of a W10 × 39 rolled-steel shape made of ......
Column AB consists of a W10 × 39 rolled-steel shape made of a grade of steel for which \sigma_Y=36 ksi \text { and } E=29 \times 10^6 psi. Determine the allowable centric load P (a) if the effective length of the column is 24 ft in all directions, (b) if bracing is provided to prevent the movement of the midpoint C in the xz plane. (Assume that the movement of point C in the yz plane is not affected by the bracing.)
STRATEGY: The allowable centric load for part a is determined from the governing allowable stress design equation for steel, Eq. (10.38) or Eq. (10.40), based on buckling associated with the axis with a smaller radius of gyration since the effective lengths are the same. In part b, it is necessary to determine the effective slenderness ratios for both axes, including the reduced effective length due to the bracing. The larger slenderness ratio governs the design.
\sigma_{ cr }=\left[0.658^{\left(\sigma_Y / \sigma_{ c }\right)}\right] \sigma_Y (10.38)
\sigma_{ cr }=0.877 \sigma_e (10.40)
MODELING: First compute the slenderness ratio from Eq. (10.41) corresponding to the given yield strength \sigma_Y=36 ksi.
\frac{L}{r}=4.71 \sqrt{\frac{E}{\sigma_Y}} (10.41)
\frac{L}{r}=4.71 \sqrt{\frac{29 \times 10^6}{36 \times 10^3}}=133.7
ANALYSIS:
a. Effective Length = 24 ft. The column is shown in Fig. 1a. Knowing that r_y<r_x , buckling takes place in the xz plane (Fig. 2). For L = 24 ft and r=r_y= 1.98 in., the slenderness ratio is
\frac{L}{r_y}=\frac{(24 \times 12) \text { in. }}{1.98 \text { in. }}=\frac{288 \text { in. }}{1.98 \text { in. }}=145.5
Since L/r > 133.7, Eq. (10.39) in Eq. (10.40) is used to determine
\sigma_e=\frac{\pi^2 E}{(L / r)^2} (10.39)
\sigma_{ cr }=0.877 \sigma_e=0.877 \frac{\pi^2 E}{(L / r)^2}=0.877 \frac{\pi^2\left(29 \times 10^3 ksi \right)}{(145.5)^2}=11.86 ksi
The allowable stress is determined using Eq. (10.42)
\sigma_{ all }=\frac{\sigma_{ cr }}{1.67} (10.42)
=\frac{11.86 ksi }{1.67}=7.10 ksi
and
P_{\text {all }}=\sigma_{\text {all }} A=(7.10 ksi )\left(11.5 in ^2\right)=81.7 kips
b. Bracing at Midpoint C. The column is shown in Fig. 1b. Since bracing prevents movement of point C in the xz plane but not in the yz plane, the slenderness ratio corresponding to buckling in each plane (Fig. 3) is computed to determine which is larger.
xz Plane: Effective length = 12 ft = 144 in., r = r_y = 1.98 in.
L/r = (144 in.)∕(1.98 in.) = 72.7
yz Plane: Effective length = 24 ft = 288 in., r = r_x = 4.27 in.
L/r = (288 in.)∕(4.27 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, Eqs. (10.39) and (10.38) are used to determine \sigma_{ cr } :
\begin{aligned} & \sigma_e=\frac{\pi^2 E}{(L / r)^2}=\frac{\pi^2\left(29 \times 10^3 ksi \right)}{(72.7)^2}=54.1 ksi \\ & \sigma_{ cr }=\left[0.658^{\left(\sigma_\gamma / \sigma_e\right)}\right] F_Y=\left[0.658^{(36 ksi / 54.1 ksi )}\right] 36 ksi =27.3 ksi \end{aligned}
The allowable stress using Eq. (10.42) and the allowable load are
\begin{aligned} \sigma_{\text {all }} & =\frac{\sigma_{ cr }}{1.67}=\frac{27.3 ksi }{1.67}=16.32 ksi \\ P_{\text {all }} & =\sigma_{\text {all }} A=(16.32 ksi )\left(11.5 in ^2\right) \end{aligned}
P_{\text {all }}=187.7 kips
REFLECT and THINK: This sample problem shows the benefit of using bracing to reduce the effective length for buckling about the weak axis when a column has significantly different radii of gyration, which is typical for steel wide-flange columns.
