Question: Design a suitable raft slab connecting the columns of a building (fig 4). Design of the continuous beam over the raft slab is not needed.
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The size of the building is 12m x 12m with columns spaced at 4m intervals. Service load transmitted by the each column is 450kn column size is 300mm x 300mm. Soil SBC is 120 Mpa. Use M20 grade concrete & Fe415 steel. Use LSM.

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

ddrcs mumbai university • 259 views
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modified 3 months ago by gravatar for Yashbeer Yashbeer160 written 6 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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Given:

Column size: $(300 \times 300 ) mm$

Column load: $1450 KN$

$S_{BC}$ of soil: 120m pq

M20

Fe415

Solution:

1] Load

$\begin{aligned} \text{a) Total load on column } &= 4 \times 450 \\ &= 1800 KN \times 3 \\ &= 5400 KN \\ \text{} \\ \text{b) Self wt of footing } &= 10 \% \times 5406 \\ &= 540 KN \\ \text{} \\ \text{Total load } &= 5400 + 540 \\ &= 5940 KN \\ \end{aligned}$


2] Area of raft read

$\begin{aligned} &= \frac{\text{Total Load}}{\text{SBC of soil}} \\ A &= \frac{5940}{120} \\ &= 49.5 m^2 \\ A & = L \times B \\ \text{} \\ \text{length of raft} &= 12 \times 4 = 48 m \\ \text{width of raft} &= \frac{A}{L} = \frac{49.5}{48} = 1.03 m \\ \text{width of raft B } &= 1.05 m \\ \text{} \\ Overhang &= \frac{1.05 - 0.3}{0.375 m} \end{aligned}$


3] Calculation of upward pressure

$\begin{aligned} \text{a) } W &= \frac{P}{\text{Area of raft}} \\ &= \frac{5400}{48 \times 1.05} \\ &= 107.14 KN/m^2 \\ \text{} \\ w_u &= 1.5 \times 107.14 \\ &= 160.71 KN/m \quad \text{ (Assume 1 m of width) } \\ \text{} \\ \text{b) } & \text{Calculation of moment} \\ m_u &= \frac{w_u \ a^2}{2} \\ &= \frac{160.71 \times 0.375^2}{2} \\ &= 11.3 KNm \end{aligned}$


4] Calculation of depth

$\begin{aligned} m_u &= mu_{max} = Ru_{max} \ bd^2 \\ d &= \sqrt{\frac{11.3 \times 10^6}{0.13 \times 20 \times 1000}} \\ &= 63.98 mm \sim 100 mm \\ D &= 100 + 50 = 150 mm \end{aligned}$


5] Calculation of steel

$\begin{aligned} Ast &= \frac{0.5 \ fck \ bd}{f_y} \left[ 1- \sqrt{1 - \frac{4.6 m_u}{fck \ db^2} } \right] \\ &= \frac{0.5 \times 20 \times 1000 \times 100}{415} \left[ 1 - \sqrt{1 - \frac{4.6 \times 11.3 \times 10^6}{20 \times 1000 \times 100^2} } \right] \\ &= 336.65 mm^2 \\ \text{} \\ Ast_{min} &= \frac{0.12}{100} \times 1000 \times 150 \\ &= 180 mm^2 \\ \text{} \\ Ast_{read} & \gt Ast_{min} \quad \therefore Safe \\ \text{Provide } & Ast_{req} \text{ use 12 mm } \phi \\ Spacing &= \frac{\pi / 4 \times 12^2}{336.65} \times 1000 \\ &= 335.95 \sim 300 mm \\ \text{Provide } & \text{ 12mm } \phi \text{ @ 300 mm c/c } \\ \end{aligned}$


Design of Beam over raft

a) Load transmitted on beam parameter run $ = 107.14 \times 1.05 = 112.497 KN/m$

Maximum bending moment $ = \frac{wl^2}{10}$

$m = \frac{112.49 \times 4^2}{10} = 179.99 KNm = 180 KNm $

$m_u = 1.5 \times 180 = 270 KNm$

b) Calculation of Depth

$d = \sqrt{\frac{270 \times 10^6}{0.138 \times 20 \times 3000}} = 571.04 mm$

So as to take care of shear stress

$D = 1.5 \times 571.04 = 856.56 mm$

$d = 875 - 50 = 825 mm$


C] Calculation of steel

$Ast = \frac{0.5 \times 20 \times 300 \times 825}{415} \left[ 1 - \sqrt{1 - \frac{4.6 \times 270 \times 10^6}{20 \times 300 \times 825}} \right] \\ = 988.88 mm^2$

Use 16 mm $\phi$

No. of bars $ = \frac{988.88}{\pi / 4 \times 16^2} = 4.91$

Provide 5 - 16 mm $\phi$

$Ast_p = 5 \times \pi / 4 \times 162 = 1005.3 mm^2$

d) Design of shear

$\begin{aligned} \text{Shear coming on beam} &= 0.6 W_L \\ &= 0.6 \times 112.97 \times 4 \\ \text{} \\ V_D &= 271.12 \\ \text{} \\ V_{uD} &= 1.5 \times 271.12 \\ &= 406.68 KN \\ \text{} \\ V_{UC} &= \tau_{uc} \ bd \\ pt \% &= \frac{1005.3}{300 \times 825} \times 100 = 0.4 \end{aligned}$

by interpolating

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$\begin{aligned} \tau_{uc} &= 0.432 \\ V_{UC} &= 0.432 \times 300 \times 825 \\ &= 106.92 KN \lt V_{UD} \\ \therefore \quad & \text{S/F is required} \\ \text{} \\ V_{UD} = V_{UC} + V_{US} \\ V_{US} = 299.76 KN \\ \text{} \\ S_v &= \frac{0.87 \ fy \ Asv \ d}{V_{US}} \\ &= \frac{0.87 \times 415 \times 2 \times \pi / 4 \times 8^2 \times 825}{299.76 \times 103} \\ &= 99.89 mm \sim 90 mm \end{aligned}$

Provide 2L - 8mm @ 90 mm c/c

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modified 3 months ago  • written 3 months ago by gravatar for Yashbeer Yashbeer160
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