Assuming a superimposed load of $3.5kN/m^2$ and a floor finish of $1kN/m^2$ design the slab. Use $M20$ concrete and HYSD steel of grade $Fe415.$ Assume that the slab corners are prevented from lifting. Draw the neat sketches of the reinforcement layout.

$$\dfrac {l_y}{lx}=\dfrac {5.75+023}{4+0.23}=\dfrac {5.98}{4.23}=1.41 < 2 $$

$\text { It is two way slab }\\ d= \dfrac {\text {short effective span}} {\text {S⁄D ratio ×M.F}}=\dfrac {4230}{20×1.4}=151.07 ≈160 \\ \text {D=d+effective cover } \\ =160+25=185mm \\ \text { Load Calculation: } \\ D.L = D×25=0.185×25=4.625kN/m^2 \\ L.L =3.5kN/m^2 \\ F.F = 1 kN/m^2 \\ Total =9.125kN/m^2 \\ \text {Factored load }(w_d) =9.125×1.5=12=13.68kN/m \\ \dfrac {l_y}{lx}=1.57 $

$$M_{ux}=α_x wdlx^2$$

$ =0.0998×13.68×4.23^2 \\ =24.42kNm \\ M_{uy}= α_y wdlx^2 \\ =0.138=0.0509×13.68×4.23^2 \\ =12.45kNm \\ M_{u\space max}=0.138f_{ck} bd^2\\ =0.138×20×1000×160^2 \\ =70.65kNm \gt M_{ux} \space \& \space M_{uy}\\ Astx= \dfrac {0.5×20×1000×160}{415}×[1-\sqrt {1-\dfrac {4.6×24.42×10^6}{20×1000×160^2 }}]\\ =449.09 mm^2 \\ \text { Assume 8mm∅ bars } ∅_x=∅_y=8d_y=152 \\ Asty= \dfrac {0.5×20×1000×152}{415}×[1-\sqrt {1-\dfrac {4.6×12.45×10^6}{20×1000×152^2 }}]\\ =234.47mm^2 \\ Astmin=0.12/100 b×D=222mm^2 \lt Astx \space \&\space Asty \\ \text { * Main steel in x-direction(8mm∅) } spacing =\dfrac {b×c/s\space area}{Asty}=\dfrac {1000×\pi/4 ×10^2}{449.09}=111.92 ≈100mm \\ \text { * Main steel in y-direction(8mm∅) } spacing =\dfrac {b×c/s\space area}{Asty}=\dfrac {1000×\pi/4 ×10^2}{234.47} =214.37 ≈200mm $