Page 125 - Handbook of Civil Engineering Calculations, Second Edition

1.108 STRUCTURAL STEEL ENGINEERING AND DESIGN

2. 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 clas-
sified as a secondary member. To
compute the transverse shear, assume
that the column will be loaded to its
capacity of 432 kips (1921.5 N).
Then force per bar /2(0.02)(432)
1
(16.1/14) 5.0 kips (22.24 N).
Also, L/r 140; therefore, r
16.1/140 0.115 in. (2.9210 mm).
For a rectangular section of thick-
ness t, r 0.289t. Then t 0.115/
0.289 0.40 in. (10.160 mm). Set t
7 /16 in. (11.11 mm); r 0.127 in.
(3.226 mm); L/r 16.1/0.127 127; f
9.59 kips/sq.in.(66. 123 MPa); A
2
5.0/9.59 0.52 sq.in. (3.355 cm ).
From the Manual, the minimum width
required for /2-in. (12.7 mm) rivets
1
1
1 /2 in. (38.1 mm). Therefore, use a flat
FIGURE 15. Lacing and tie plates.
1
bar 1 /2 /16 in. (38.1 11.11 mm); A
7
2
0.66 sq.in. (4.258 cm ).
3. 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
5 /16 in. (355.6 7.94 mm). The rivet pitch is limited to six diameters, or 3 in. (76.2 mm).
SELECTION OF A COLUMN WITH A LOAD
AT AN 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 K x L 30 ft (9.2 m) and K y L 15 ft (4.6 m).

Calculation Procedure:
1. 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
0.5
factor K
defined by K
K(1 – m/2) . Thus, for this column, m 56/186 0.30. And
K
x L 30(1 – 0.15) 0.5 27.6 ft (8.41 m).
2. Select a trial section on the basis of the K y L value
From the AISC Manual for a W8 40, capacity 186 kips (827.3 kN) when K y L 16.2
ft (4.94 m) and r x /r y 1.73.