- Compute the value of KxL
associated with a uniform-
strength column, and compare
this with the actual value
Thus, KxL = 1.73(15.3) = 26.5 ft (8.1 m) <
30 ft (9.2 m). The section is therefore inade-
quate. - Try a specific column section
of larger size FIGURE 13
Trying W8 x 48, the capacity - 200 kips
(889.6 kN) when KyL 17.7 ft (5.39 m). For
uniform strength, KxL = 1.74(17.7) = 30.8 >
30 ft (9.39 m > 9.2 m). The W8 x 48 there-
fore appears to be satisfactory. - Verify the design
To verify the design, record the properties of this section and compute the slenderness ra-
tios. For this grade of steel and thickness of member, the yield-point stress is 50 kips/in^2
(344.8 MPa), as given in the Manual. Thus, A = 14.11 in^2 (91038 cm^2 ); rx = 3.61 in
(91.694 mm); ry = 2.08 in (52.832 mm). Then KJLIrx = 30(12)73.61 - 100; KyLlry =
15(12)72.08-87. - Determine the allowable stress and member capacity
From the Manual, /= 14.71 kips/in^2 (101.425 MPa) with a slenderness ratio of 100. Then
P = 14.11(14.71) = 208 kips (925.2 kN). Therefore, use W8 x 48 because the capacity of
the column exceeds the intended load.
STRESS IN COLUMN WITH PARTIAL
RESTRAINT AGAINST ROTATION
The beams shown in Fig. \4a are rigidly connected to a W14 x 95 column of 28-ft (8.5-
m) height that is pinned at its foundation. The column is held at its upper end by cross
bracing lying in a plane normal to the web. Compute the allowable axial stress in the col-
umn in the absence of bending stress.
Major (x)
axis- Minor (y)
axis
column Minor axis(a) Framing plan at top•Major axis(b) Restraint conditionsFIGURE 14