Question: A hall in a building has a floor consisting of a One way slab Continous slab cast with simply supported beams of width 250mm spaced at 4m c/c (fig3).
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The clear span of beam is 9m. Assuming live load=$3kN/m^2$, partition load $1kN/m^2$ and load due to finishes 0.6 $kN/m^2$.

Design the slab with Grade of concrete is M20 and steel is Fe 415. Use LSM method. Draw neat reinforcement diagram.

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

ddrcs mumbai university • 182 views
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modified 6 weeks ago by gravatar for Yashbeer Yashbeer160 written 4 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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Solution:

$l_y = 9m$

$l_x = 4m$

$\frac{l_y}{l_x} = \frac{9}{4} = 2.25 \gt 2$

One way continuous slab

Given:

Live load = $3 KN/m^2$

Partition load = $1 KN/m^2$

F.F = $0.6 KN/m^2$

M20 Fe415

beam = 250 mm

brick wall = 230 mm

1] Calculation of depth

$\begin{aligned} d_{req} &= \frac{l_x}{\frac{l}{d} \times MF} \\ &= \frac{4000}{26 \times 1.4} \\ &= 109.89 \sim 120 mm \\ \text{} \\ D &= 120 + 20 + \frac{10}{2} \\ &= 145 mm \end{aligned}$


2] Load calculation

i) Dead load

$\begin{aligned} \text{a) s/w of slab } &= 25 \times D \\ &= 25 \times 0.145 \\ &= 3.625 KN/m^2 \\ \text{} \\ \text{Partition load} &= 1 KN/m^2 \\ \text{Floor finish} &= 0.6 KN/m^2 \\ &\quad \text{----------------} \\ &\quad \quad 5.225 KN/m^2 \\ \text{} \\ W_u D_L &= 1.5 \times 5.225 \\ &= 7.84 KN/m \end{aligned}$

ii) Live load

$LL = 3 KN/m^2$

$W_{ULL} = 4.5 KN/m^2$

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$D.L \quad \frac{+1}{12} \quad \frac{-1}{16} \quad \frac{1}{16} \quad \frac{-1}{12}$

$L.L \quad \frac{+1}{10} \quad \frac{-1}{9} \quad \frac{1}{12} \quad \frac{-1}{9}$

$\begin{aligned} M_{u1} &= \frac{1}{12} \times 7.84 \times 4^2 + \frac{1}{10} \times 4.5 \times 4^2 \\ &= 17.65 KNm \\ \text{} \\ M_{uB} &= \frac{-1}{10} \times 7.84 \times 4^2 - \frac{1}{9} \times 4.5 \times 4^2 \\ &= -20.54 KNm \\ \text{} \\ M_{u2} &= \frac{1}{16} \times 7.84 \times 4^2 + \frac{1}{12} \times 4.5 \times 4^2 \\ &= 13.84 KNm \\ \text{} \\ M_{uc} &= \frac{-1}{12} \times 7.84 \times 4^2 + (\frac{-1}{9}) \times 4.5 \times 4^2 \\ &= -18.45 KNm \end{aligned}$


4] Check for depth

$\begin{aligned} mm_{max} &= Ru_{max} \ bd^2 \\ d &= \sqrt{\frac{20.54 \times 100}{0.138 \times 20 \times 1000}} \\ d &= 86.26 mm \lt 120 mm \\ \therefore \quad & \text{Safe in depth} \end{aligned}$


5] Calculation of steel

a) Main steel

$\begin{aligned} \text{1) } Ast_1 &= \frac{0.5 \times 20 \times 1000 \times 120}{415} \left[ 1- \sqrt{1- \frac{46 \times 176.5 \times 10^6}{20 \times 1000 \times 120^2}} \right] \\ &= 441.24 mm^2 \\ \text{} \\ Ast_{min} &= 174 mm^2 \\ \text{} \\ &\text{Use comm} \phi \text{} \\ Spacing &= \frac{\pi / 4 \times 10^2}{441.24} \times 1000 \\ &= 177.99 mm \sim 150 mm \\ & \text{Use 10mm } \phi \text{ @ 150 mm} \\ \end{aligned}$

2) $Ast_B = 521.30 mm^2$

Provide 10mm $\phi$ 125 mm c/c

3) $Ast_2 = 339.53 mm^2$

Provide 10mm $\phi$ @ 200 mm c/c

4) $Ast_c = 463.14 mm^2$

Provide 10mm $\phi$ @ 150 mm c/c

b) Distribution steel

$Ast_{min} = 174 mm^2$

Provide 8 mm $\phi$ @ 275 mm c/c

$Ast_p = \frac{\pi/4 \times 10^2}{150} \times 1000 = 523.6 mm^2$


6] Check for deflection Page no 38 IS456

$\begin{aligned} Fs &= 0.58 \times 415 \times \frac{441.29}{53.6} \\ &= 202.83 N/mm^2 \sim 24 N/mm^2 \\ \text{} \\ Pt \% &= \frac{523.6}{1000 \times 120} \times 100 \\ &= 0.436 \% \\ \text{} \\ MF &= 1.3 \\ \text{} \\ d &= \frac{4000}{26 \times 1.3} \\ &= 118.34 mm \lt 120 mm \\ \therefore \quad & \text{Safe in deflection} \end{aligned}$


7] Check for shear

At B

$\begin{aligned} V_{uB} &= \alpha_{DL} \times W_{UD} \times l_x + \alpha_u W_{ULL} \times l_x \\ &= 0.6 \times 7.84 \times 4 + 0.6 \times 7.84 \times 4 \\ &= 37.63 KN \\ \text{} \\ V_{uD} &= V_{uB} = 37.63 KN \\ \text{} \\ V_{uc} &= k \ \tau_{UC} \ bd \\ k &= 1.3 \ (for \ d \le 150) \\ \text{} \\ Pt \% &= \frac{Ast}{bd} \times 100 \\ &= \frac{521.30}{1000 \times 120} \times 100 \\ &= 0.43 \\ \text{} \\ 0.25 &- 0.36 \\ 0.43 &- 1 \\ 0.50 &- 0.48 \\ \text{} \\ \tau_{uc} &= 0.4120 N/mm^2 \\ \text{} \\ V_{uc} &= k \ \tau_{uc} \ bd \\ &= 1.3 \times 0.421 \times 1000 \times 120 \\ &= 65.67 KN \gt V_{UD} = 37.63 KN\\ \therefore \quad & \text{Safe in shear} \end{aligned}$


8] Check for development length

$\begin{aligned} Ld &= \frac{0.87 f_y \phi}{4 \tau_{bd}} \\ &= \frac{0.87 \times 415 \times 10}{4 \times 1.2 \times 1.6} \\ &= 470.11 mm \\ \text{} \\ M_1 &= \frac{17.65}{2} = 8.825 KN \\ \text{} \\ V_A &= 21.9 KN \\ \text{} \\ l_o &= 12 \phi \text{ or d}\\ &= 120 mm \\ &= 644.03 mm \\ \text{} \\ L_d & \lt \frac{1}{3} \frac{M_1}{V} + l_s \end{aligned}$

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modified 6 weeks ago  • written 6 weeks ago by gravatar for Yashbeer Yashbeer160
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