balok baja, sama berat namun beda kekuatan lebih

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LTB

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Balok baja dua tumpuan jepit, beban merata, Mtump = 2*Mlap, optimasi dgn haunched, sliced section. kriteria kekuatan: tegangan von mises maksimum, stabilitas dan serviceability:lendutan. Questionable, LTB for haunced? Cb values? elastic or plastic, residual stress effect? etc. Studying advanced problem using Code_Aster as FE solver.

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Validation – Calculation pad.

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Lb = 1200
w = 1.00
q = w*20 = 20.00
M1=(1/12)*q*Lb^2 = 2400000.00
Z = 2480.62
Sigma1=M1/Z = 967.50
‘ Bot = 1351.9 kg/cm2
‘ Top = 1408.7 kg/cm2
M2=(1/24)*q*Lb^2 =1200000.00
Sigma2=M2/Z = 483.75
‘ Bot = 708.9 kg/cm2
‘ Top = 672.3 kg/cm2
Es=2000000
nu = 0.30
G = Es/(2*(1+nu)) = 769230.77
h = 60.00
b = 20.00
tf = 1.70
tw = 1.10
Iy = ((1/6)*tf*b^3)+((1/12)*(h-tf)*tw^3)  = 2273.13
d = h – 2*tf  = 56.60
Ix = ((1/12)*b*(d+2*tf)^3) – ((1/12)*(b-tw)*d^3) = 74418.64
J = (1/3)*((2*b*tf^3)+(h*tw^3)) = 92.13
Cw = (h^2*Iy)/4 = 2045819.80
w = 1.00
q = w*b = 20.00
As = h*tw = 66.00
delta = ((q * Lb^4) / (384*Es*Ix)) + ((q*Lb^2)/(8*G*As)) = 0.80
K = 1.00
Ms = Min(M1,M2)*(-1) = -1200000.00
Ml = Max(M1,M2) = 2400000.00
Cb = 1.75+1.05*(Ms/Ml)+0.3*(Ms/Ml)^2 = 1.30  < 2.3
Mcra = Cb*(Pi/(K*Lb))*Sqrt((Es*Iy*G*J)+((Pi*Es)/Lb)^2*Iy*Cw)  = 2282234.25
Sigmacra = Mcra / Z  = 920.03
Lamda = 1.21034
wcr = w * Lamda  = 1.21
qcr = wcr * b=24.21
Mcrfe = 1/24 * qcr * Lb^2=1452408.00
Rltb = Mcra/Mcrfe=1.57

Mmax =  2400000.00
rx = 0.25
x = rx * Lb= 300.00
Ma = (q/12)*(6*Lb*x – Lb^2 – 6*x^2) = 300000.00
rx = 0.50
x = rx * Lb= 600.00
Mb = (q/12)*(6*Lb*x – Lb^2 – 6*x^2) = 1200000.00
rx = 0.75
x = rx * Lb = 900.00
Mc = (q/12)*(6*Lb*x – Lb^2 – 6*x^2) = 300000.00

Cb = (12.5*Mmax) / (2.5*Mmax + 3*Ma + 4*Mb + 3*Mc) = 2.38
‘ Conservatively takes Cb = 1.00

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all units are in (Kgf-cm)

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Keterangan tambahan, dilain kesempatan :)

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