Question 16.S-P.2: 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 σ_{Y} = 36 ksi and E = 29 × 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.)
We first compute the value of the slenderness ratio from Eq. 16.25 corresponding to the given yield strength σ_{Y} = 36 ksi.
\frac{L}{r} = 4.71 \sqrt{\frac{E}{σ_{Y}}} (16.25)
\frac{L}{r} = 4.71 \sqrt{\frac{29 × 10^{6}}{36 × 10^{3}}} =133.7
a. Effective Length = 24 ft. Since r_{y} < r_{x}, buckling will take place in the xz plane. For L = 24 ft and r = r_{y} = 1.98 in., the slenderness ratio is
\frac{L}{r_{y}} = \frac{(24 × 12) in.}{1.98 in.} = \frac{288 in.}{1.98 in.} = 145.5
Since L/r > 133.7, we use Eq. (16.23) in Eq. (16.24) to determine σ_{cr}
σ_{e} = \frac{π^{2}E}{(L/r)^{2}} (16.23)
σ_{cr} = 0.877σ_{e} (16.24)
σ_{cr} = 0.877σ_{e} = 0.877 \frac{π^{2}E}{(L/r)^{2}} = 0.877 \frac{π^{2}(29 × 10^{3} ksi)}{(145.5)^{2}} = 11.86 ksi
The allowable stress, determined using Eq. (16.26), and P_{all} are
σ_{all} = \frac{σ_{cr}}{1.67} (16.26)
σ_{all} = \frac{σ_{cr}}{1.67} = \frac{11.86 ksi}{1.67} = 7.10 ksi
P_{all} = σ_{all} A = (7.10 ksi) (11.5 in^{2}) = 81.7 kips
b. Bracing at Midpoint C. Since bracing prevents movement of point C in the xz plane but not in the yz plane, we must compute the slenderness ratio corresponding to buckling in each plane and 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 = 145.5, we use Eqs.(16.23) and (16.22) to determine σ_{cr}
σ_{e} = \frac{π^{2}E} {(L/r)^{2}} = \frac{π^{2}(29 × 10^{3} ksi)}{(72.7)^{2}} = 54.1 ksi
σ_{cr} = [ 0.658^{(σ_{Y}/σ_{e})} ] F_{Y} = [ 0.658^{(36 ksi/54.1 ksi)} ] 36 ksi = 27.3 ksi
We now calculate the allowable stress using Eq. (16.26) and the allowable load.
σ_{all} = \frac{σ_{cr}}{1.67} = \frac{27.3 ksi}{1.67} 5=16.32 ksi
P_{all} = σ_{all}A = (16.32 ksi) (11.5 in^{2}) P_{all} = 187.7 ksi
