Question: The layout of the columns of a building is shown below. The outer columns are $400 mm \times 400$ mm in size and carry load of 800 kN each.
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The inner columns are $500 mm ×500 mm$ in size and carry load of 1200 kN each. Design a suitable raft foundation consider SBS of soil as $100 kN|m^2$. Adopt $M_{20}$ concrete and $Fe_{ 415}$ steel. Design only slab and main beam and draw the reinforcement details.

Mumbai University > Civil Engineering > Sem 8 > Design and Drawing of Reinforced Concrete Structure

Marks : 20

Year : MAY 2013

mumbai university ddrcs • 866 views
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Area of footing reqd. $=\dfrac {\text {Total ca}l^m load +10%\text {self wt.}}{SBS}\\ =\dfrac {(2×1100+2×1500)×1.1}{110}\\ =52m^2 $

Assume 0.5 projections from the column.

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Provide 12 m× 8 m raft footing

Factored upward soil pressure (w)

$w=\dfrac{1.5× \text {column load}}{\text {Area of footing}}\\ =\dfrac {1.5×(4×800+4×1200)}{12×8} \\ =87.5 \lt SBS$

∴ safe Introduce secondary beam as shown in figure to break the panel load and make it one way slab.

$W=87.5KN$

Cantilever slab $:-M= \dfrac {wl^2}2 = \dfrac {87.5\times2^2}2=175 KNm. $

Mid-span of continuous slab:-

$M(+ve)= \dfrac {wl^2}{10} = \dfrac {87.5\times2^2}{10}=35 KNm $

Support of secondary beams :-

$M(-ve)=\dfrac {wl^2}{12} = \dfrac {87.5\times3^2}{12}=29.16 KNm \\ Mu_{max} =175 KNm\\ ∴Mu_{max} =0.138 \space fck \space l\space d^2 \\ 175\times106=0.138\times 20\times 1000\times dr^2 \\ dr=251.8 mm $

Provide $D=350mm$

$$D=350-50-20/2$$

$=290mm\\ \text {Astmin} = \dfrac {0.12bD}{100}=420 mm^2$

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Dist. Steel:-

8 mm ∅ 150 mm c/c as astmin = 420 mm2 ( ∴ same steel to be provide as Astmi

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Design of beam $B_1$

Total upward load $=(2×3.6)×87.5 \\ =630 KN\\ udl = \dfrac {630}{3.6}=175 KN|m$

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$M_u=\dfrac {175×3.6^2}8=283.5 KNm$

By T- beam method:

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From IS code 456:2000 page.36

$l_f (codal)= \dfrac {l_0}6+6D_f+l_w \\ =\dfrac { 3600}6+6×350+400 \\ =3100 mm$

Assume $x_4=D_f=350 mm\\ M_{uf}=0.36\space fck \space l_f \space x_u (d-0.42x_u )\\ =0.36×20×3100×350(1090-0.42×350) \\ =7367 KNm ≫ M_u $

N.A. lies in flange

$Astr=\dfrac {0.5×20×3100×1090}{415} × (1-\sqrt{\dfrac {1-4.6×515×10^6}{20×3100×1090^2 }})\\ = 724 mm^2 \\ \text {Astmin} = \dfrac {0.85 lwd}{fy} =\dfrac {0.85×400×1090}{415}=894 mm^2 \\ ∴Astp=894 mm^2$

provide 5-16 mm∅

$pt.=\dfrac {100×1005}{400×1090}=0.23\% \\ Z_{uc} (page 73:IS:456:2000)=0.32 \\ V_{uc}=Z_{uc} bd=0.32×400×1090=139.52 kN \\ V_{u\space min}=0.4bd=0.4×400×1090=\dfrac {174.4kN }{313.92KN \lt 315 kN} $

Hence design & provide shear R|F

$V_{us}=V_{UD}-V_{UC} \\ =315-314 \\ =1 KN$

Provide 8 mm ∅ & 2 LG stirrups

$a_{sv} = 2×π⁄4 ×8^2 =100 mm^2 $

Spacing:-

$S_1=\dfrac {0.87 fy asvd}{V_{us}} = \dfrac {0.87×415×100×1090}{1×10^3 }=39355 mm \\ S_2=0.75×d=817.5 \\ S_3=300 mm$

Provide 8 mm ∅ 2 LG @ 300 mm c/c.

Primary beam $B_2$

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Hence design and provide shear R|F

$V_{us}=903-436=467 KN $

Assume 12 mm ∅ 4 LG stirrups

$As_v = 4×113=452 mm^2 $

Spacing $= s_1=\dfrac {0.87 f_y as_{vd}}{V_{us}} = \dfrac {0.87×415×452×1090}{467×10^3} =380.9 mm ≈380 mm \\ S_2=0.75×1090=817.5 mm \\ S_3=300 mm $

∴Provide 12 mm ∅ 4 LG stirrups @ 125 mm c/c.

Front view from plan

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Side view from plan

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modified 3.2 years ago  • written 3.2 years ago by gravatar for Aksh_31 Aksh_31630
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