Kolom lantai 1 sampai lantai 3

x r KL = 71 , 6 48 , 157 8 , 2 = 72,755 Cm = 0.6 - 0.4       2 1 M M = 0.6 - 0.4       004 , 12596 1274,004 = 0,559 Pe = 2 2 .       x r KL Ag E  = 2 2 755 , 72 147 29000  = 7940,52 kip  1 =        Pe Pu Cm 1 =        52 , 7940 118,048 1 0,559 = 0,567  1  1 = 1  2 =          L H h Pu . 1 1 Dimana : h = hasil SAP U 1 = 0,386 ft = 4,632 in H = W 1 = 4,36 kip  2 =        57,48 5,29071 4,632 118,048 1 1 = 2,2 Mlt = W 1 . L = 5,2907157.48 = 833,179 kip in Mux =  1 . Mnt +  2. Mlt = 1 5660,658 + 2,9833,179 = 8076,878 kip in  Efek Balok Lb = 157,48 in Lp = 222 in Lr = 3168 in  maka Lp  Lb  Lr Mencari nilai Cb : Cb = 1.75 + 1.05       2 1 M M + 0.3 2 2 1       M M = 1.75 + 1.05       004 , 12596 1274,688 + 0.3       004 , 12596 1274,688 = 1,887 2,3 memenuhi Cb =                004 , 12596 1274,688 4 . 6 . 1 4 . 6 , 1 2 1 M M = 1,787 2,3 memenuhi Karena kedua harga Cb memenuhi, maka diambil Cb = 1,887 Mp = Zx.Fy = 105036 = 37800 kip.in Mr = Fy – FrSx = 36 – 10838 = 21788 kip in Mn = Cb                   Lp Lr Lp Lb Mr Mp Mp = 1                   222 3168 222 157,48 21788 37800 37800 = 38150,677 kip in dipakai Mn = Mp Mn = 0,9 37800 = 34020 kip in  Efek Kolom y r KL = 43 , 4 48 . 157 1 , 3 = 110,201  c = E Fy r KL 2  = 110,201 29000 36 2  = 1,24 Fcr = 0,658 2 c  .Fy = 18,915 Pn = Ag. Fcr = 147 . 18,915 = 2780,507  c Pn =  c . Pn = 0,85 . 2780,507 = 2363,431 kip  maka  c Pn Pu 2363,431 kip 118,048 kip Pn Pu c  0.2 431 , 2363 048 , 118 0.2 0,0499 0,2 Maka : 1 Mn Mu n 2 Pu    P 1 34020 12596,004 363,431 2 . 2 118,048   0,370 1  Kontrol Penampang Kompak  Tekuk Badan 36 640 190 , 2 19,60 Fy 640 tw d    = 8,949  106,667  Penampang Kompak  Tekuk Sayap 36 65 5 , 3 2 010 , 17 65 . 2     Fy tf bf = 2,43  10,833  Penampang Kompak  Kontrol Geser Vu = 88,101 d = 19,60 in tw = 2,19 in 59 949 , 8   tw h ФVn = 0,6 x fy x d x tw = 0,6 x 36 x 19,60 x 2,19 = 927,158 kips ФVn Vu 519,656 kips 88,101 kips

b. Kolom lantai 4 dan 5

Data : Frame = 32 Matas = - 563,1112 Kip-ft = 6757,334 Kip-in Mbwh = 445,164 Kip-ft = 5341,968 Kip-in Pu = -78,188 Kip Vu = 76,85 Kip L = 13,1232 ft = 157,48 in W4 = 21,1628 Kip Mnt = 707,716 Kip-in Profil W 14 x 370 Dimana : Ag = 109 in 2 E = 29000 r x = 7,07 in G = 11200 r y = 4,27 in J = 222 in 4 Sx = 607 in 3 F R = 10 Ksi untuk baja gilig Cw = 116000 in 6 Zx = 736 in 3 Iy = 1990 in 4 P W 7 Mencari nilai K portal: G A = balok kolom L EI L EI   =     88 , 314 3400 48 . 157 2660 = 1,56 G B = 10 menurut LRFD gambar C-C2.2 Dari LRFD 6-186 unbraced frame didapat K = 1,5  Efek Kolom x r KL = 07 , 7 48 , 157 5 , 1 = 33,412 Cm = 0.6 - 0.4       2 1 M M = 0.6 - 0.4       968 , 5341 6757,334 = 0,094 Pe = 2 2 .       x r KL Ag E  = 412 , 33 109 29000 2 2  = 27917,649 kip  1 =        Pe Pu Cm 1 =        649 , 27917 78,188 1 0,094 = 0,09  1  1 = 1  2 =          L H h Pu . 1 1 Dimana : h = hasil SAP U 1 = 4,584 in H = W 4 = 21,1628 kip  2 =        157,48 21,1628 4,584 78,188 1 1 = 1,12 6757,334 kip.in 5341,968 kip.in