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Design Zp=M×8mofy=3.34×106×1.10250=14696mm3
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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+MyMDy1

3.34×1068.66×106+1.47×1061.92×106 1

1.15 < unsafe

check for shera capacity (Z directions)

Vd=Fy×h.tw3×γmo250×100×43×1.10=52.49kN

Also

0.6Vd=31.49kN > 2.54Kn sfe

Shear capacity in y direction

Vd=fy×bf×tf3×γmo=250×(2×50)×6.43×1.10

=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

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