0
571views
Design beam B of typical flooring plan of steel building as shown following data for design 1) beam supports a brick wall of 230mm thick and 4m high having unit wt 20kN/m$^{3}$ 2)thickness of Rcc
1 Answer
0
0views

is 200mm top flange embedded in concrete unit weight of concrete is 25 kN/m$^{3}$

3) L.L on slab is 3kN/m$^{2}$ and F.F=1kN/m$^{2}$

enter image description here

Given

L.L=3kN/m$^{2}$

F.F=1kN/m$^{2}$

$\frac{ly}{lx}=\frac{6.5}{3}=2.167\gt$2 one way slab

load on beam=slab load+wall load+weight of beam

Slab load

Concrete load=25$\times$3$\times$0.2=15kN/m

Floor finish load=1$\times$3=3kN/m

Live load=3$\times$3=9kN/m

factor load=27$\times$1.5=40.5kN/m

Wall load

20$\times$0.230$\times$4=18.4kN/m

factored wall load=18.4$\times$1.5=27.6kN/m

Weight of beam

1kN/m

factored weight of beam=1$\times$1.5=1.5kN/m

load on beam=40.5+27.6+1.5

=69.6kN/m

enter image description here

load on beam B=Slab load+wall load+self weight

=0+27.6+1.5

=29.1 kN/m

enter image description here

1) V=shear force=$\frac{w1}{2}+\frac{w2\times l}{2}$

=$\frac{226.2}{2}+\frac{29.1\times6}{2}$

v=200.4kN

m$\frac{w_{1}L}{4}+\frac{w_{2}l^{2}}{8}$

m=470.25kN.m

2) Trial section

Zpreq=$\frac{m\ast \gamma mo}{fy}$

=$\frac{470.25\times 1.1}{250}$

=2069.1$\times 10^{3}mm^{3}$

select ISMB 500 having Zp=2074.67$\times 10^{3}mm^{3}$

fwc=(75+85.5)$\ast 10.2\ast\frac{250}{1.1}$

=372.67kN$\gt$ 200.4kN safe

Deflection

permissible =$\frac{span}{300}=\frac{4000}{300}$=20mm

act=$\frac{w_{1}l^{3}}{48EI}+\frac{5w_{2}l^{4}}{384 EI}$

$\int$act =36.2

$\int$act=3.2$\lt$ sper safe

Please log in to add an answer.