Question: Effective Column Length Lyy = 8 . 0 m Effective Column Length Lzz = 8 . 0 m Calculate: Myy = Mzz = Using these

Effective Column Length
Lyy =8.0 m
Effective Column Length
Lzz =8.0 m
Calculate:
Myy =
Mzz =
Using these formulas:
My,Ed =((h/2)+100) x R2=
Mz,Ed =((tw/2)+100) x (R1 R3)=
My,Ed = My,Ed X L2-3/(L2-3+ L1-2)=
Mz,Ed = Mz,Ed X L2-3/(L2-3+ L1-2)=
Note:
Beam 1=((30 x 1.35)+(15 x 1.50) x (4/2)=126
Beam 2=(20 x 1.35) x (6/2)=81
Beam 3=((30 x 1.35)+(15 x 1.50)) x (8/2)=252
Total Design Load =459.00 kN
Column:-305x305x97
Steel Grade fy=275 N/mm2
Allowable Design Stress =275N/mm2
Properties:
h=307.9 mm
b=305.3 mm
tw=9.9 mm
tf=15.4 mm
r=20 mm
Iyy=22250 cm4
Izz=7308 cm4
iyy =13.4 cm
izz =7.69 cm
Wpl,yy=1592 cm3
Wpl,zz=726 cm3
Iw=1.56 dm6
It=91.2 cm4
A=123 cm2
E=210000 N/mm2
Calculate:
Web Buckling:
(Equation 6.22) hw/tw =(72/\eta =66.56)
Flexural Buckling about "zz" Length A:
Effective Unrestrained Length =8.0 m
(Equation 6.50)\lambda zz =
(See - Table 6.2) h/b =
(Table 6.1)\alpha zz =
(Clause 6.3.1.2)\Phi zz =
(Clause 6.3.1.2)\chi zz =1, & 1/(\lambda LT^2)
(Equation 6.47) Nb,z, Rd =
Calculate:
Lateral Torsional Buckling: Length A:
Effective Unrestrained Length =8.0 m
NCCI SN002a Table 1.1\lambda LT =
(See NA - Table 6.5) h/b =
(Table 6.3)\alpha LT =(curve)
(Clause 6.3.2.3-(1))\Phi LT =
(Clause 6.3.2.3-(1))\chi LT =1, & 1/(\lambda LT^2)
(Equation 6.55) My,b,Rd =
Mz,cb,Rd =
Interaction equation =
Simplified Interaction criterion =
(NEd/Nmin,b,Rd)+(My,Ed/My,b,Rd)+1.5(Mz,Ed/Mz,cb,Rd)=
Effective Column Length Lyy = 8 . 0 m Effective

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