Area of footing required $= \dfrac {(\text {total column load})+10%\text {self wt.}}{SBS} \\ = \dfrac {(2×500+2×1500+600+1200+2×2000+2×550+2×1200)×1.1}{110} \\ =133m^2 $

Area provided $=21.6×12.6=272.16m^2 \gt 133m^2 $

Factored upward soil pressure (w)

$w=\dfrac {1.5×\text {column load}}{\text {Area of footing}} \\ = \dfrac {1.5×(1000+3000+600+4000+1200+1100+2400)}{21.6×12.6} \\ =73.3kN|m^2 \lt SBS$

∴ safe

Slab

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

$W=73.3kN|m^2$

Cantilever slab $:-M=\dfrac {wl^2}2=\dfrac {73.3×0.3^2}2=3.29KNm. $

Mid-span of continuous slab:-

$M(+ve) = \dfrac {wl^2}{10} = \dfrac {73.3ⅹ3^2}{10}=65.97KNm $

Support of secondary beams :-

$M(-ve)=\dfrac {wl^2}{12}= \dfrac {73.3\times3^2}{12}=54.97KNm \\ Mu_{max} =65.97 KNm \\ ∴Mu _{max} =0.138\space fck\space l\space d^2\\ 48.85\times106=0.138\times20\times1000\times dr^2 \\dr=154mm$

Provide $D=250 mm$

$$D=250-50-20/2$$

$=190mm \\ \text {Astmin }= \dfrac {0.12bD}{100} = \dfrac {0.12}{100}×1000×250=300mm^2 $

Dist. Steel:-

Astmin=$300mm^2$ ,Provide 8 mm ∅

Spacing $= \dfrac {1000×50}{300}=166.67 mm \\ ≈150 mm ∅$

Provide $8 mm ∅ 150 $ mm c/c

Design of beam $B_1$

Total upward load $=(6.7×3)×73.3 \\ =1474 KN\\ udl= \dfrac {1474}{6.7}=220 KN|m$

$M_u=\dfrac {wl^2}8=\dfrac {218×4.05^2}8=446.96 KNm$

By T-beam method:-

From IS code 456:2000 page.36

$l_f (codal)= \dfrac {l_0}6+6D_f+l_w \\ =\dfrac { 6700}6+6×250+300 \\ =2917 mm$

Assume $x_4=D_f=250 mm$

$$M_{uf}=0.36\space fck \space l_f \space x_u \space(d-0.42x_u ) $$

$ =0.36×20×2917×250(990-0.42×250) \\ =4647 KNm\gt 1235 kNm$

N.A. lies in flange

$Astr=\dfrac {0.5×20×2917×990}{415} × (1-\sqrt{\dfrac {1-4.6×1235×10^6}{20×2917×990^2}}) \\ = 3548 mm^2$

Provide $8-25 mm ∅$

Astmin $=\dfrac {0.85 lwd}{fy} = \dfrac {0.85×300×990}{415 } =609mm^2 \lt Astr \\ Astp=3927 mm^2 \\ pt.=\dfrac {100×3927}{300×990}=1.32 \% \\ Z_{uc} (page 73:IS:456:2000)=0.684 \\ V_{uc}=Z_{uc} bd=0.684×300×990=203.15kN \\ V_{u\space min}=0.4bd=0.4×300×990=\dfrac {118.8kN }{321.95KN \lt V_{uD}}$

Hence design & provide shear R|F

$V_{us}=V_{UD}-V_{UC} \\ =737-322 \\ =415 KN$

Provide $8 mm ∅ $& logged stirrups

$$a_{sv}=4×π⁄4×8^2=200 mm^2$$

Spacing :-

$S_1=\dfrac {0.87 \space fy\space asvd}{V_{us}} =\dfrac {0.87×415×200×990}{415×10^3 }=172.26 mm≈150mm \\ S_2=0.75×990=742.5mm \\ S_3=300mm$

Provide 8 mm ∅ 4 LG @ 300 mm c/c.

Primary beam $B_1$

Hence design and provide shear R|F

$V_{us}=1800-229=1571 KN$

Assume 12 mm ∅ 4 LG stirrups

$As_v = 4×113=452 mm2 $

Spacing $= s_1=\dfrac {0.87\space f_y\space as_vd}{V_{us}} =\dfrac {0.87×415×452×990}{1571×10^3}=102.84 mm ≈100 mm \\ S_2=0.75×990=742.5 mm \\ S_3=300 mm $

Provide 12 mm ∅ 4 LG stirrups @ 100 mm c/c.

Side view in plan