Desain batang tekan dan lentur

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Download Desain batang tekan dan lentur
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  P u =100,00000 kg=1,00000 kNcoba pakai400200813=1,000,0000 NM uZ =20kNm=20,000,0000 NmmA g =8192mm 2 F y =240MPaE=200000MPaL b =4000mmb f =200mmt f =13mmh=400mmt w =8mm4008374I x =2296486827mmI y =1734929067mmS x = W x =1148243413mm 3 S y = W y =1734929067mm 3 r x =1674314381mmi min  = r y =4601992322mmk=1 λ <025λ c =0958 025<λ<12ω =1493 λ >12φ Pn=1,119,402 N > 1,000,000 N OK Z x =1285952mm 3 Z y =265984mm 3 J 1 =3589813333mm 5 J 2 =3567626667mm 5 C w1 =694E+11mmC w2 649E+11mm DESAIN BATANG TEKAN DAN LENTUR 1313200 λ ω  67061 431 −= 1 = ω  2 251  λ ω  = ω φ φ   ygn F  AP   = AIi  ymin = EFrLk   yc π λ  = 4  2 h I C   yw = ( ) 4t2ht 2btZ  f 2w2f f y −+= ( ) ( ) 2f wf f f x  t2h4tthtbZ  −+−= ∑ =  hb J   31  3  periksa kelangsingan penampangFlensWebT=d-2(t+k 1 )4225<4293<h 1 =h- 2(t f +r 0 )OK-Profil KOMPAKOK-Profil KOMPAKLp=mm 1 <15 M y M n1 =308,628,480 Nmm<413,367,629okX 1 =1327063818X 2 =0000255779M p =15M y =413,367,629 NmmLr=mmM r =195,201,380 NmmMmax=20,000,000 M A =5,000,000 M B =20,000,000 M C =5,000,000 C b =156  2 M n2 =526,710,945 >413,367,629pakai Mp= 15 My 3 M n3 =M cr =772,953,080 Nmm>413,367,629Nmmpakai Mp= 15 MyM n-final =413,367,629 Nmm<413,367,629Nmmok φ M n =413,367,629 Nmm>20,000,000 NmmProfil Okdipakai bila Pu0893>02 φ Pnpakai persamaan 1=094 <1profil OKLp<Lb<Lr maka Mn=Mn27691614233813709166 persamaan 1 yrf   F250t2b =<  λ   yrw1 F665th =<  λ     y y F  E r  Lp  761 = 2EGJASX x1 π  = 2xyw2 GJSIC4X       = ( ) ( ) 2ry2ry1yr  FFX11 FFXrL −++−= ( )  pprpbrppbn  MLLLLMMMCM <      −−−−= xy1n  ZFM = wy2bybbcr  CI LEGJEI LCM      +=  π π  198 ≤      ++ nybuynxbuxnu  M  M  M  M PP φ φ φ