Desain batang tekan dan lentur

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

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