Fig.1 shows a part plan of the building. Slab (S1) has thickness of 120 mm & slab (S2) has thickness of 140 mm.

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written 6.0 years ago by

teamques10 ★ 69k

• modified 6.0 years 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 $B_1$

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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 $B_1$ = $72.135 + 46.25 = 118.39 KN/m$

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