written 6.1 years ago by |
Zp=M×8mofy=3.34×106×1.10250=14696mm3
From shear table select ISLC 100 mm having h=100mm, bp=50mm , tf=6.4mm
tw=4mm γ=6mm, Ixx=164.7×104mm Iyy=24.8×104mm4, Zxx=32.9×103mm3 Zyy=7.3×103mm3
Zpz, prov=38.09×103mm3
Zpy=Zy× s
=7.3×103×1.1576=8.45×103mm3
h1=74.3mm
sectional classification:
btf=506.4=7.8<9.4ϵ
dtw=74.34=18.58<42ϵ
section is plastic
check for moment capacity of section
Mdz=Bd×ZPz×fyγyγmo....(Bb=1)
=1×38.09×1032501.10
=5654.55×103N.mm
=8.66×106N.mm
Mdy=Bd×Zpy×gyγyγmo=12×8.45×103×2501.10
=1.92×106 N.mm
Using Interaction Equation
MzMdz+MyMDy≤1
3.34×1068.66×106+1.47×1061.92×106≤ 1
1.15 < unsafe
check for shera capacity (Z directions)
Vd=Fy×h.tw√3×γmo250×100×4√3×1.10=52.49kN
Also
0.6Vd=31.49kN > 2.54Kn sfe
Shear capacity in y direction
Vd=fy×bf×tf√3×γmo=250×(2×50)×6.4√3×1.10
=83.98×103NAlso0.6Vd=50.39kN\gt1.47kNChceckfordeflection\intmax=\frac{span}{300}=\frac{4000}{300}=13.33mmworking\int=\frac{5}{384}\times \frac{wL^{4}}{EI}=\frac{5}{384}\times \frac{0.845\times 4000^{4}}{2\times 10^{5}\times 164.7\times 10^{4}}\int=8.55mm\lt \int$ max