SongDo Partners
Plate-Thickness ratio for Flexure
λffp = 0.38 √(E / Fy)
=
9.24
λffr = 1.00 √(E / Fy)
=
24.32
λff1 = B1 / 2 / tf1
=
7.14
< λffp = 9.24
→ Compact
λfwp = 3.76 √(E / Fy)
=
91.45
λfwr = 5.70 √(E / Fy)
=
138.63
λfw1 = (H1-2(tf1+r1)-tw1)/2/tw1
=
20.94
< λfwp = 91.45
→ Compact
Section Yielding Strength
Mpx = min[Fy×Zx, 1.6×Fy×Sx]
=
675.03
kN·m
Lateral-Torsional Buckling
rtsx = √[√(Iy×Cw) / Sx]
=
221.47
mm
c = 1 (2-axis symmetry H Section)
=
1.0
ho1 = (H1 - tf1) / 2
=
218.0
mm
Lp = 1.76×ry×√(E/Fy)
=
5.794
m
Lr1 = 1.95rtsx(E/0.7Fy)×√{(Jc)/(Sx×ho1)}×√[1+√{1+6.76(...)²}]
=
30.22
m
Lb
=
9.050
m
< Lr1 = 30.22
→ No Problem
Fcr,mbx = (Cb×π²E)/(Lb/rtsx)²×√{1+0.078(Jc)/(Sx×ho1)×(Lb/rtsx)²}
=
1,445.77
MPa
Mnx = Mpx
=
675.03
kN·m
φb
=
0.9
φbMnx
=
607.53
kN·m
> Mux = 0.00
→ O.K
Plate-Thickness ratio for Flexure
λff2 = B2 / 2 / tf2
=
7.14
< λffp = 9.24
→ Compact
λfw2 = (H2-2(tf2+r2)-tw2)/2/tw2
=
20.94
< λfwp = 91.45
→ Compact
Section Yielding Strength
Mpy = min[Fy×Zy, 1.6×Fy×Sy]
=
675.03
kN·m
Lateral-Torsional Buckling
rtsy = √[√(Iy×Cw) / Sy]
=
221.47
mm
ho2 = (H2 - tf2) / 2
=
218.0
mm
Mny = Mpy
=
675.03
kN·m
φbMny
=
607.53
kN·m
> Muy = 0.00
→ O.K
Pu / φcPn
=
0.888
> 0.20
→ O.K
Pu/φPn + 8/9×(Mux/φMnx + Muy/φMny)
=
0.888
< 1.000
→ O.K
Shear coefficient, Cv
kvw
=
5.00
(보강된 웹)
λvw = {H2-2(tf2+r2)-tw2}/2/tw2
=
20.94
< 2.24√(E/Fy) = 54.48
Cvw = 1
=
1.00
(약축방향 재하 2축대칭)
Shear Strength, φVn
As1 = Aw + 2Af
=
9,650.00
mm²
Vny = 0.6×Fy×As1×Cvw
=
2,055.45
kN
φvw
=
1.00
φvwVny
=
2,055.45
kN
> Vuy = 0.00
→ O.K
Shear Strength, φVn
As2 = Aw + 2Af
=
9,650.00
mm²
Vnx = 0.6×Fy×As2×Cvw
=
2,055.45
kN
φvwVnx
=
2,055.45
kN
> Vux = 0.00
→ O.K