For a W14 with KL = 15 ft m = 1.0 and U= 1.5. Substituting in Eq. [8.2], we obtainPUJB =^8 °0 + 200 x 2.0 + O = 1200 kips (5338 kN)In the AISC Column Load Tables (p. 2-19 of the LRFD Manual) if Fy = 36 ksi (248
mPa) and KL = 15 ft (4.57 m), Qfn = 1280 kips (>Pu>eff = 1200 kips) for a W14 x 159.
- Analyze the braced frame
Try a W14 x 159. To determine M11x (the second-order moment), use Eq. (H1-2).
M 14 = B 1 Mn^B 2 M 11Because the frame is braced, Mlt = O.
Mv = B 1 Mn; or Kx = ,B 1 X 200 kip-ftAccording to Eq. (Hl-3)
5l=
(l^)aLO
where Cm = 0.6 - 0.4(M^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*
=_^200 =_1Q
M 2 " 200
Cw = 0.6-0.4(-1.O) =1.0For a W14 x 159, Ix = 1900 in^4 (79,084 cm^4 )n Ti(^2) EI
x TJ
(^2) x 29,000 kips/in (^2) x 1900 in (^4) , , -
0 ,.. ~, ^cc , XT,
Pl= Ho? = (1.0 X
15 ft x^12 in/ft)
2 = (^16) ' 784 klpS (?4' (^655) ^
In Eq. (Hl-3)
BI = 1-800 kips/16,784 kips ~ L° 5
Here, M14x = 1.05 x 200 kip-ft = 210 kip-ft (284.6 kNm) the second-order required flexur-
al strength. (Substituting M14x = 210 kip-ft in preliminary design, Eq. [8.2] still leads to a
W14 x 159 as the trial section.)
Selecting the appropriate beam-column interaction formula, (Hl-Id) or (Hl-Ib) 9 we
have
P^= 800kips =Q63>Q2
Q 0 Pn 1280 kips
Use formula (Hl-Io), which, for Muy = O, reduces to
^ +! J^sio
$fn 9 Wn*