Find a) and b)
Extracted text: 2. Find Pa for the built-up column shown. Use A36 steel ( F, 36 ksi ). The length is 19 ft-6 in., and K= 1.0. ( E=29000 ksi ) For the W14X 257: A = 75.6 in.? I, = 3400 in.4 ly = 1290 in.4 bç = 15.995 in. (flange width) %3D d = 16.38 in. (section depth) (a) Find P, for the core W14×257. (b) Find Pa for the built-up column. W14 X 257 gosi b plate o batlo y
![Appendix<br>Expressions for axially loaded compression member:<br>KI<br>For-< C. =<br>2T²E<br>[1 – (Kl/r)²/(2C? )]Fy<br>F, =<br>5 3(Kl/r) _ (Kl/r)³<br>8Cc<br>Fy<br>|21²E<br>1212E<br>KI<br>For-> Cc<br>Fa<br>Fy<br>23(KI/r)2<br>Pa =<br>FaA<br>Expressions for beam:<br>MR = F,S, in which S =-<br>Expressions for tension member:<br>Pt = A,F , in which F = 0.6Fy<br>P; = A,Ft , in which F = 0.5Fu<br>P = A.Ft , in which Ae = U An, F = 0.5Fu<br>P = A,F, + A,F; , in which Fy = 0.30Fµ , F; = 0.50Fu<br>P = nr,, in which r, = ApFv, ry = d t F,<br>TABLE 2-1 Values for Reduction Coefficient, U<br>U = 0.90<br>W, M, S shapes or their tees.<br>Connection is to the flanges.<br>Minimum of three bolts per line in the direction of<br>Case I<br>stress.<br>(min.)<br>(min.)<br>U = 0,85<br>All shapes and built-up cross sections not meeting<br>the requirements of case I. Minimum of three bolts<br>per line in the direction of stress.<br>Case II<br>U = 0.75<br>All members whose connections have only two bolts<br>per line in the direction of stress.<br>Case III<br>Units:<br>ksi : kips per square inch<br>kips: kilo pounds<br>lb/ft: pounds per foot<br>kips/ft: kilo pounds per foot<br>](https://s3.us-east-1.amazonaws.com/storage.unifolks.com/qimg-008/008_dmeuzp0-o1gfgq5m.jpeg)
Extracted text: Appendix Expressions for axially loaded compression member: KI For-< c.="2T²E" [1="" –="" (kl/r)²/(2c?="" )]fy="" f,="5" 3(kl/r)="" _="" (kl/r)³="" 8cc="" fy="" |21²e="" 1212e="" ki="" for-=""> Cc Fa Fy 23(KI/r)2 Pa = FaA Expressions for beam: MR = F,S, in which S =- Expressions for tension member: Pt = A,F , in which F = 0.6Fy P; = A,Ft , in which F = 0.5Fu P = A.Ft , in which Ae = U An, F = 0.5Fu P = A,F, + A,F; , in which Fy = 0.30Fµ , F; = 0.50Fu P = nr,, in which r, = ApFv, ry = d t F, TABLE 2-1 Values for Reduction Coefficient, U U = 0.90 W, M, S shapes or their tees. Connection is to the flanges. Minimum of three bolts per line in the direction of Case I stress. (min.) (min.) U = 0,85 All shapes and built-up cross sections not meeting the requirements of case I. Minimum of three bolts per line in the direction of stress. Case II U = 0.75 All members whose connections have only two bolts per line in the direction of stress. Case III Units: ksi : kips per square inch kips: kilo pounds lb/ft: pounds per foot kips/ft: kilo pounds per foot