- Design the lacing bars
The lacing system must be capable of
transmitting an assumed transverse shear
equal to 2 percent of the axial load; this
shear is carried by two bars, one on each
side. A lacing bar is classified as a sec-
ondary member. To compute the trans-
verse shear, assume that the column will
be loaded to its capacity of 432 kips
(1921.5 N).
Then force per bar =
1
/ 2 (0.02)(432)
x (16.1/14) = 5.0 kips (22.24 N). Also, LIr
< 140; therefore, r = 16.1/140 = 0.115 in
(2.9210mm).
For a rectangular section of thickness
t, r ='0.289/. Then t = 0.115/0.289 = 0.40
in (10.160 mm). Set t =^7 Ae in (11.11
mm); r = 0.127 in (3.226 mm); LIr =
16.1/0.127 = 127; / = 9.59 kips/in^2
(66. 123 MPa); A = 5.0/9.59 = 0.52 in^2
(3.355 cm^2 ). From the Manual, the mini- FIGURE 15. Lacing and tie plates,
mum width required for !/2-in (12.7 mm)
rivets = !!/2 in (38.1 mm). Therefore, use a
flat bar W 2 x V 16 in (38.1 x U.U mm);
A = 0.66 in^2 (4.258 cm^2 ). - Design the end tie plates in accordance with the Specification
The minimum length = 14 in (355.6 mm); t = 14/50 = 0.28. Therefore, use plates 14 x^5 / 16
in (355.6 x 7.94 mm). The rivet pitch is limited to six diameters, or 3 in (76.2 mm).
SELECTION OFA COLUMN WITHA LOAD
ATAN INTERMEDIATE LEVEL
A column of 30-ft (9.2-m) length carries a load of 130 kips (578.2 kN) applied at the top
and a load of 56 kips (249.1 kN) applied to the web at midheight. Select an 8-in (203.2-
mm) column of A242 steel, using KxL = 30 ft (9.2 m) and KyL = 15 ft (4.6 m).
Calculation Procedure:- Compute the effective length of the column with respect
to the major axis
The following procedure affords a rational method of designing a column subjected to a
load applied at the top and another load applied approximately at the center. Let m = load
at intermediate level, kips per total load, kips (kilonewtons). Replace the factor K with a
factor K' defined by K' = K(I - w/2)
05
. Thus, for this column, m = 56/186 = 0.30. And
KxL = 30(1 - 0.15) °^5 = 27.6 ft (8.41 m). - Select a trial section on the basis of the KyL value
From the AISC Manual for a W8 x 40, capacity = 186 kips (827.3 kN) when KjL = 16.2 ft
(4.94m) and rjry =1.73.