Sheet1d1/15.L720mmb0,3 .d216mmsayap6tf1/40 .b18mmbadan97.7142857143tw1/200 .d7mmhd-(2.tf)684mmLuas BrutoA2.(bf .tf) + h .tw12564.000Ix2.bf.tf.((h/2)+(tf/2))^2+((tw*h^3)/12)+(((2*bf)*tf^3)/12)1144895472.00000

Sheet2Data IWFd1/15.L720mmL=10800b0,3 .d216mmtf1/40 .b18mmtw1/200 .d7mmhd-(2.tf)684mmqu41.5N/mmfy240MpaE200000N/mm2G77000N/mm2PenyelesaianLuas BrutoA=2.(bf .tf) + h .tw12564InersiaIy=2((tf.bf^3)/12)+((h.tw^3)/12)30252639.000Ix=2.bf.tf.(((h/2)+(tf/2))^2)+((tw.h^3)/12)+(((2.bf).tf^3)/12)1144895472

ry=(Iy/A)^0.549.0701819873rx=(Ix/A)^0.5301.8693051039Modulus penampang arah x dan ySx=(Ix/cy)3180265.2dimana cy=360Sy=(Iy/cx)280117.027777778cx=108Momen tahanan plastisZx=((tw.h^2)/4)+h.tf.bf3478140Zy=(h.tw^2+2.tf.b^2)/4428283Inersia RotasiJ=(2.bf.tf^3+h.tw^3)/3918012Tekuk torsi CwCw=(Iy.ho^2)/414908621509756dimanaho=702A. Kuat LenturPengaruh tekuk torsi lateral (penampang I)Lp=1,76 .ry (E/fy)^0.52493.1000846009X1=(/sx)*((E*G*J*A)/2)^0.59309.3646341231X2=4.((sx/(G*J))^2)*(cw/Iy)0.0039900903Lr=ry(X1/fL)*(1+(1+X2*(fL^2))^0.5)^0.59224.6808059669fL=170Bentang diantara pengaku lateral === >3Cb= 1.01L=L/33600maka :Lp < L < Lr.=== > rumus 4.4Mn=Cb.(Mr+(Mp-Mr)*((Lr-L)/(Lr-Lp))704466747.623845dimanaMr=Sx.(fy-70)540645084Mp=Zx.fy834753600Cb=Tabel 3.11.01Pengaruh Tekuk LokalBatsan kelangsingan penampang tabel 3.1 hal 11Sayap=pf = 170/(fy)^0.510.9734528143rf = 370/(fy-70)^0.528.3777045874Badan=pw = 1680/(fy)^0.5108.4435336938rw = 2550/(fy)^0.5164.6017922138Kelangsingan Komponen PenampangSayap=(bf/2)/tf6rwlangsing

Badan LangsingJika h/t < rmaka Mn=Kg.S.fcrdimana S=200000Mpaar=(bf.tf)/(h.tw)0.8120300752makaKg = 1-(ar/(1200+(300.ar))*((h/tw)-(2550/((fy)^0.5))1.0376242224h=684tw=7fy=240Cb=1.01tf=18b=216L= 3600Mencari tegangan kritis fcrFaktor kelangsingan pelat sayap (g)A.Berdasar panjang bentangg = L/rt63.376840363dimanaIytl = ((tf*(b^3))/12)+((h/6)*(tw^3))/12)15119802.5Atl = b.tf+(h/6).tw4686makart = ((Iytl/Atl)^0.5)56.8030842084p = 1.76*((E/fy)^0.5)50.8068236887r = 4.40*((E/fy)^0.5)127.0170592217B.Berdasar Tebal pelat sayapg = bf/(2.tf)6p = 0.38*((E/fy)^0.5)10.9696551146r = 1.35*(((kc*E)/fy)^0.5)24.7904052186dimanaKc = 4/((h/tw)^0.5)0.4046513191Menghitung tegangan Kritis fcrA. p < g < rmaka fcr = Cb.fy(1-((g-p)/(r-p))194.5219113245lihat kondisi syaratnyaB. g < p < rmaka fcr = fy240maka diambil nilai fcr terkecil 194.5219113245pilih nilai nyamakaMn=Kg.S.fcr40368129.3958102