design internal beam b1 refer floor system

Given

Span of beam=8m

Live load=4kN/m$^{2}$

Flour finish load=1.5kN/m$^{2}$

Fy=250

fe=410

Required=design internal beam B1

$\frac{Ly}{Lx}=\frac{8}{4}=2\geq$2 one way slab

$\frac{ly}{lx}\leq$2 two way slab

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

Concrete load=25$\frac{(kN}{ms})\times(4m)\times 0.140m$

=14kN/m

F.F.L=1.5$\times$4=6kNm

L.L =4$\times$4=16kN/m

slab load=concrete load+EFL+L.L

=14+6+16

=36kN/m

Assume weight of beam=1kN/m

weight of beam =1$\times$4=1kNm

Total load no of beam=36+1=37kN.m

factored w=37$\times$1.5

w=55.5kN/m

Step I factor shear and factor movement

V=$\frac{wl}{2}=\frac{55.5\times 8}{2}$

V=22.2kN

m=$\frac{wl^{2}}{8}=\frac{55.5\times8^{2}}{8}$=444kN/M

Step II Trial section

Zpreq=$\frac{M\times \gamma ma}{fy}$

=$\frac{444\times10^{6}\times1.1}{250}$

Zfreq=1953.6$\times10^{3}mm^{3}$

ISMB 500having Zp= 2074.67$\times10^{3}mm^{3}$

h=500.bf= 180mm,tf=17.2mm

tw=10.2mm, $\gamma_{1}$=17mm

IZ=45218.3$\times10^{4}mm^{4}$

Ze=1808.7$\times10^{3}mm^{3}$

Step III check for section classification

$\epsilon=\sqrt{\frac{250}{fy}}$=1

$\frac{b}{tf}=\frac{Df/2}{tf}=\frac{180/2}{17.2}$=5.23$\lt$9.4$\epsilon$

$\frac{d}{tw}=\frac{D-2tf-2\gamma _{1}}{tw}=\frac{500-(2\times17.2)\times(2\times 17)}{10.2}$

=4.22$\lt$ 84$\epsilon$ section is plastic

Step IV

Vd=$\frac{Av\times fy}{\sqrt{3}\times \gamma mo}$ Av=500$\times$10.2

=$\frac{5100\times 250}{\sqrt{3}\times 1.1}$

Vd=669.20kN $\gt$ V=222kN

Step V

0.6 Vd=0.6$\times$66.920$\times 10^{3}$

V=222$\times 10^{3}$N $\lt 40152\times 10^{3}$N (low shear

md=$\frac{Bd\times(Zp)\times fy}{\gamma mo}$

=$\frac{1\times(2074.67\times10^{3})\times 250}{11}$

md=471.52kNm$\gt$m=444kNm

$\leq1.2\times ze\times\frac{fy}{\gamma mo}$ safe

$\leq(1.2\times 1808.7\times 10^{3})\times\frac{250}{1.1}$

$\lt$ 493.28kN safe

Step VI $\frac{d}{tw}+42.2 \lt 97 \epsilon$ no need to check shear buckling

Web crippling

Fwc=(b1+n2)$\ast tw\ast \frac{fy}{\gamma mo}$

n2=2.5(tf+$\gamma_{1}$)

=2.5(17.2+17)

n$_{2}$=85.5

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

=372.67kN $\gt$ 222kN safe

Deflection

$\int permissible =\frac{span}{300}=\frac{8000}{300}$=26.66mmm

$\int act=\frac{5(55.5/1.5)\times(8000)^{4}}{384(2\times10^{5})\ast 45218.3\times10^{4}}$ $\int$ act=21.82mm

$\int act$(21.82)mm$\lt$Sper(26.66mm) safe