Page 171 - Handbook of Civil Engineering Calculations, Second Edition

1.154 STRUCTURAL STEEL ENGINEERING AND DESIGN

For a W14 with KL 15 ft m 1.0 and U 1.5. Substituting in Eq. [8.2], we obtain
P u,eff 800 + 200 2.0 + 0 1200 kips (5338 kN)
In the AISC Column Load Tables (p. 2-19 of the LRFD Manual) if F y 36 ksi (248
mPa) and KL 15 ft (4.57 m), c P n 1280 kips (>P u,eff 1200 kips) for a Wl4 159.
2. Analyze the braced frame
Try a W14 159. To determine M ux (the second-order moment), use Eq. (H1-2).

M u B 1 M nt + B 2 M lt
Because the frame is braced, M lt 0.
or M ux B 1 200 kip-ft
M u B 1 M nt
According to Eq. (H1-3)
C m
B 1 1.0
(1 – P u /P e )
where C m 0.6 – 0.4(M 1 /M 2 ) for beam-columns not subjected to lateral loads between
supports.
For M 1 M 2 200 kip-ft (271 kNm) in single curvature bending (i.e., end moments
in opposite directions)

M 1 200
– –1.0
M 2 200
C m 0.6 – 0.4(–1.0) 1.0

4
4
For a W14 159, I x 1900 in. (79,084 cm )
2 × 29,000 kips/sq.in. × 1900 in. 4
2
EI x
P l 16,784 kips (74,655 kN)
(Kl) 2 (1.0 × 15 ft × 12 in./ft) 2
In Eq. (H1-3)
1.0
B 1 1.05
1 – 800 kips/16,784 kips
Here, M ux 1.05 200 kip-ft 210 kip-ft (284.6 kNm) the second-order required flex-
ural strength. (Substituting M ux 210 kip-ft in preliminary design, Eq. [8.2] still leads to
a W14 159 as the trial section.)
Selecting the appropriate beam-column interaction formula, (H1-1a) or (H1-1b), we
have

P u 800 kips
0.63 > 0.2
c P n 1280 kips
Use formula (H1-1a), which, for M uy 0, reduces to
P u 8 M ux
+ 1.0
c P n 9 b M nx