Desain batang tekan dan lentur
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Transcript of Desain batang tekan dan lentur
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Pu = 100,000.00 kg= 1,000.00 kN
coba pakai 400 200 8 13 = 1,000,000.0 NMuZ = 20 kNm
= 20,000,000.0 NmmAg = 8192 mm2
Fy = 240 MPaE = 200000 MPaLb = 4000 mmbf = 200 mmtf = 13 mmh = 400 mmtw = 8 mm
400 8 374
Ix = 229648682.7 mm4
Iy = 17349290.67 mm4
Sx = Wx = 1148243.413 mm3
Sy = Wy = 173492.9067 mm3
rx = 167.4314381 mm
imin = ry = 46.01992322 mmk = 1
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periksa kelangsingan penampangFlens Web
T=d-2(t+k1) 42.25 < 42.93< h1=h- 2(tf+r0) OK-Profil KOMPAK
OK-Profil KOMPAK
Lp = mm
1 < 1.5 MyMn1 = 308,628,480 Nmm < 413,367,629
ok
X1 = 13270.63818
X2 = 0.000255779
Mp=1.5My = 413,367,629 NmmLr = mm Mr = 195,201,380 Nmm
Mmax = 20,000,000 MA = 5,000,000 MB = 20,000,000 MC = 5,000,000 Cb = 1.56
2
Mn2 = 526,710,945 > 413,367,629 pakai Mp= 1.5 My
3
Mn3=Mcr = 772,953,080 Nmm > 413,367,629 Nmmpakai Mp= 1.5 My
Mn-final = 413,367,629 Nmm < 413,367,629 Nmmok
Mn = 413,367,629 Nmm > 20,000,000 NmmProfil Ok
dipakai bila Pu 0.893 > 0.2 Pn pakai persamaan 1
= 0.94 < 1profil OK
Lp