0
Fig.1 shows a part plan of the building. Slab (S1) has thickness of 120 mm & slab (S2) has thickness of 140 mm.

For the building, live load=4kN/m2, floor finish=1kN/m2. Beam B1 is 8.2m long, 200mm wide & 450 mm deep. Also beam B1 carries a masonry wall 115mm thick & 3m high with masonry unit wt=18kn/m3/ Calculate Factored(design) UDL carried by the beam B1. Include the self-weight of the beam itself.

enter image description here

Civil Eng > Sem 8 > Design and Drawing of Reinforced Concrete structure

ddrcs mumbai university • 333  views
0  upvotes
0

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 $B_1$

enter image description here

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

0  upvotes
Please log in to add an answer.

Next up

Read More Questions

If you are looking for answer to specific questions, you can search them here. We'll find the best answer for you.

Search

Study Full Subject

If you are looking for good study material, you can checkout our subjects. Hundreds of important topics are covered in them.

Know More