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determine capacity of ISHB 400 82.2 kg/m with 1 cover plate of 400$\times$20 mm on each flange is used as column of effective height 6.5m design suitable slab base take m20 concrete and sbc of soil
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180kN/m$^{2}$ Design gusseted base of above column

Given ISHB 400 at 82.2 kg/m

cover plate on each flange =400$\times$20

effective length of column=6500mm

SBC of soil=150$\frac{kN}{m^{2}}$=180Kpa

Step I

Requirement

1.design column
2 design base

3.design gusset base

Pd=Ae$\ast$ fcd

Pd=264.66$\times 10^{2}\ast fcd$

$\gamma$ xx=193.8

$\gamma$yy=95.5

$\frac{Kl}{\gamma min}=\frac{6500}{65.5}$=68.06

60 $ \ \ \ \ \ \ $ 168

68.06 $\ \ $ fcd

70 $ \ \ \ \ \ \ $152

$\frac{70-60}{70-68.06}=\frac{152-168}{152-x(fcd)}$

fcd=155.10 mpa

Pd=(264.66$\times 10^{2})\ast$155.10

=4104.88$\times10^{3}$N

Pd=4104.88$\times 10^{3}N$

Slab Base

Step I Area Required

AP=$\frac{Pu}{0.6\ast fck}$

=$\frac{4104.88\times10^{3}}{0.6\times 20}$

AP=342.08$\times 10^{3}mm^{2}$

Length of pole=$(\frac{Lc-Bc}{2})+\sqrt{(\frac{Lc-Bc}{2}){2}^{2}+Ap}$

Lc=440mm

Bc=400mm

=$(\frac{440-400}{2})+\sqrt{(\frac{440-400}{2})^{2}+342.08\times 10^{3}}$

LP=605.22mm

provide =610mm

width of plate=$\frac{Ap}{Lp}=\frac{342.08\times 10^{3}}{610}$

Bp=560.79mm

provide width of plate=570mm

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thickness of plate fp=$\sqrt{2.5\ast w(a^{2}-0.3b^{2})\ast \frac{\gamma mo}{fy}}$

w=$\frac{pu}{Ae}=\frac{4104.88\times 10^{3}}{(610\times 570)}$

w=11.80 $\lt 0.6\ast fck$

w=11.80 $\lt$12 hence ok

a=85 mm

b=85 mm

tp=$\sqrt{2.5\ast 1.8(85^{2}-0.3\times 85^{2})+\frac{1.1}{250}}$

=25.62 mm

tp=provide 26mm

Step II design concrete block

Area of concrete block =$\frac{1.1\times fu}{1.5\times 180}$

=$\frac{1.1\times 4105}{1.5\times 180}$

Af=16.72m$^{2}$

for equal projection

Lf=$(\frac{Lp-Bp}{2})+\sqrt{(\frac{Lp-Bp}{2})^{2}+Ap}$

=$(\frac{610-570}{2})+\sqrt{(\frac{610-570}{2})^{2}+16.72\times 10^{6}}$

Lf=4109.06mm$^{2}$

provide 4110mm

width of footing=$\frac{Af}{Lf}$

=$\frac{16.72\times 10^{6}}{4110}$

Bf=4068.19mm

provide Bf=4100mm

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