Question: Following figure shows the layout plan of columns of a building. Design a raft foundation for the building. Take net bearing capacity = 80kn/m3. Each corner column carries a load of 700 kN.
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Central columns carry load of 1000 kN. Draw neat sketch showing reinforcement details.

enter image description here

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

ddrcs mumbai university • 391 views
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modified 3 months ago by gravatar for Yashbeer Yashbeer160 written 7 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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Solution:

1) Calculation of upward pressure

$w =\frac {P}{A}$

= $\frac {Total \ Load}{Area \ of \ raft}$

= $\frac{700 \times 4 + 1000 \times 2}{8 \times 4}$

= $150 \ kN /m^2 $ $\gt 80 \ kN /m^2$

$\therefore$ Rt is not safe

2) Area of raft

Area of raft read = $\frac {Total \ Load}{SBC}$

= $\frac{700 \times 4 + 1000 \times 2}{80}$

= $60 m^2$

Area of raft available = $4 \times 8$

= $32 \ m^2$

Area of raft read > Area of raft available

$\therefore$ Provide nominal offset of 1m from Face of Column

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$Column \ Size - 300 \times 300\ mm $

$L = 4+4+1+1+\frac{0.3}{2} +\frac{0.3}{2}$

= $10.3 m$

$B = 4+1+1+\frac{0.3}{2} +\frac{0.3}{2}$

= $6.3 m$

$Actual \ Area = 10.3 \times 6.3 = 64.89 \ m^2$

3) Calculation of upward Pressure

$w = \frac{Actual \ load}{Actual \ Area}$

$\frac{4800}{64.89}$

= $73.79 \lt 80 \ kN \ m^2$

Limiting upward pressure

$w_u - 1.5 \times 73.79$

= $110.685 kN /m^2$

4) Calculation of B.M

$\frac{l_y}{l_n} = \frac{4}{4} = 1$

For Interior Pandel (Case )

$-\alpha_x = 0.032$ and $-\alpha_y = 0.032$

$\alpha_x = 0.024$ and $\alpha_y = 0.024$

$-m_x = -m_y = 0.032 \times 110.685 \times 4^2$

= $56.67 \ kN.m$

$m_x = m_y = 0.024 \times 110.685 \times 4^2$

= $42.50 \ kN.m$

5) Calculation of depth of raft

a) B.M

$d =\sqrt{\frac{56.67 \times 10^3}{0.138 \times 20 \times 1000}}$

$d = 143.29 \ mm$

b) Considering Punching shear, Critical section acts @ distance $\frac{d}{2}$ from face of coloumn.

$\tau_v = k \tau_c$

$k \tau_c = R.H.S$

$k = 0.5 + \frac{b}{d}$

$k = 0.5 + \frac{0.3}{0.3}$

= $1.5 \gt 1$

$\therefore k=1$

but

$\tau_c = 0.25 \sqrt{FCk}$

= $0.25 \sqrt {20}$

$\tau_c = 1.118 N/mm^2$

$k\tau_c = 1.118 N/mm^2$

enter image description here

$\tau_c = 1.118 \times 10^3 kN/m^2$

$\tau_v = \frac{v_u}{pd} = \frac{1500-110.685(0.3 + d)^2}{4(0.3 + d) \times d}$

$\tau_v = k\tau_c$

$1.118 \times 10^3= \frac{1500-110.685(0.3 + d)^2}{4(0.3 + d) \times d}$

$d = 0.436 m$

$d = 450 \ mm$

$D = 500 \ mm$

6) Calculation of steel

a)

$-m_x = -m_y = 56.67 kNm$

$Ast_x = Ast_y = \frac{0.5 \times 20 \times1000 \times 450}{415} [1-{\sqrt{1-\frac{4.6 \times 56.67 \times 10^6}{20 \times 1000 \times 450}}}]$

$Ast_x = Ast_y = 354.77 \ mm^2$

$Ast_{min} =\frac{0.12}{100} \times 1000 \times 500 = 600 \ mm^2$

Use $Ast_{min} = 600 \ mm^2$

Use $16 \ mm \phi$

Spacing $300 \ mm$

Provide 16mm $\phi$ @ 300 mm c/c

b)

$m_x = m_y = 42.50 \ kN.m$

$Ast_x = Ast_y = 264.95 \ mm^2$

Provide $Ast_{min} = 600 \ mm^2$

Use 16 mm $\phi$ @ 300 mm c/c

enter image description here

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modified 3 months ago  • written 3 months ago by gravatar for Ankit Pandey Ankit Pandey70
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