written 10 months ago by
Yashbeer ★ 210

•
modified 10 months ago

Given:
$LL = 4 KN/m^2$
$FF = 1 KN/m^2$
$\begin{aligned}
Bean \ B_1 &= 8.2 m \ long \\
&= 200 mm \ wide \\
&= 450 mm \ deep
\end{aligned}$
load calculation for slab $CS_{2}$
1] Dead Load
$\begin{aligned} \text{s / w of slab} &=25 \times D \\ &=25 \times 0.450 \\ &=11.25 \mathrm{ KN/m^2} \\
FF &= 1 \ KN/m^2 \\
T.L &= 12.25 \ KN/m^2 \\
\text{} \\
W_u D_L &= 1.5 \times 12.25 \\
&= 18.375 KN/m \quad \text{...Assume Lm of strip}
\end{aligned}$
$\text{Live load} = 4 KN/m^2 \\ W_u L_L = 6 KN/m^2$
$\text{Total load} = 18.375 + 6 = 24.375 KN/m$
2] Load calculation of slab $S_1$
a) s/w of slab = $25 \times D = 25 \times 0.450 = 11.25 KN/m$
b) LL = $3 KN/m$
c) FF= $1 KN/m$
$Total = 15.25 KN/m$
$w_u = 1.5 \times 15.25 = 22.875 KN/m$
Load calculation on Beam $B_1$
a) Load transferred by
$\text{Slab } S_1 = \frac{W_u l_x}{2}\bigg[ 1  \frac{1}{3\beta^2}\bigg] + \frac{W_u l_x}{2}\bigg[ 1  \frac{1}{3 \beta^2} \bigg]$
$\beta = \frac{l_y}{l_x} = \frac{4.1}{3} = 1.36$
$= \frac{22.875 \times 3}{2} \bigg[ 1  \frac{1}{3 \times 1.362} \bigg] \times 2 = 56.26 KN/m$
b) Wall load = $1.5 \times 18 \times 0.115 \times 3 = 9.315 KN/m$
c) $\text{s/w of beam} = 10 \% (a+b) = 10\%(56.26 + 9.315) = 6.56 KN/m$
$W_u = 72.135 KN/m$
2] Calcuations
$\begin{aligned}
\text{ a) load transfered by } S_2 &= \frac{W_uL_x}{2} \left[ 1  \frac{1}{3 \beta^2} \right] \\
&= \frac{22.875 \times 3}{2} \left[ 1 \frac{1}{3 \times 2.7^2} \right] \\
&= 32.74 KN/m
\end{aligned}$
$\begin{aligned}
\text{ b) Wall load } &= 1.5 \times 18 \times 0.115 \times 3 \\
&= 9.315 KN/m
\end{aligned}$
$\begin{aligned}
\text{ b) S/w of beam } &= \frac{10}{100} (32.74 \times 9.315) \\
&= 4.20 KN/m
\end{aligned}$
$W_u = 46.25 KN/m$
Total load (UDL) on beam $B_1$ = $72.135 + 46.25 = 118.39 KN/m$