Use $Fe415$ steel and M20 grade concrete. Use following BM coefficient. Negative moment at continuous

Positive moment at midspan

Assume beam width (b) = 230mm

$$ \frac{l_y}{l_x} = \frac{5+0.23}{4+0.23} = \frac{5.23}{4.23} = 1.23 \lt 2$$

Therefore, it is a two way slab

$$ d = \frac{short \space effective \space span}{(S/D \space ratio \times M.F)} $$

$$ dr = \frac{4230}{26 \times 1.4} = 116.2 = 125$$

$$ D = dr + d' = 125 + 25 = 150mm $$

Load Calculation

$$ D.L = D \times 25 = 0.15 \times 25 = 3.75 kN/m^2 $$

$$ L.L = 1 kN/m^2 $$

$$ F.F = 1kN/m^2 $$

$$ Total = 8.25 kN/m^2 $$

$$ Factored load (w_d) = 8.25 \times 1.5 = 12.375 = 12.375 kN/m $$

$$ +ve \space BM(M_{ux} = a_x w d l x^2 = 0.037 \times 12.7 \times 4.23^2 $$

$$ (M_{ux}) +ve = 8.19lNm $$

$$ (M_{ux}) -ve = a_x w d (l x)^2 = 0.0625 \times 12.37 \times 4.23^2 $$

$$ (M_{ux}) -ve = 13.83 kNm $$

$$ (M_{uy}) -ve = a_y w d (l x)^2 = 0.047 \times 13.37 \times 4.23^2 $$

$$ (M_{uy}) -ve = 10.41 kNm $$

$$ (M_{umax} = 0.138 f_{ck} bd^2 = 0.138 \times 20 \times 1000 \times 125^2 $$

$$ = 43.125 kNm \gt [(M_{ux}) + ve(M_{uy} + ve(M_{ux} - ve (M_{uy} -ve $$

Therefore Safe.

$$ (Astx) + ve = \frac{0.5 \times 20 \times 1000 \times 125}{415} \times [ 1 - \sqrt{1 - \frac{4.6 \times 8.19 \times 10^6}{20 \times 1000 \times 125^2}} ]$$

$$ =187.39 mm^2 \approx 188mm^2 $$

$$ (Astx) - ve = \frac{0.5 \times 20 \times 1000 \times 125}{415} \times [ 1 - \sqrt{1 - \frac{4.6 \times 13.83 \times 10^6}{20 \times 1000 \times 125^2}} ]$$

$$ =324 mm^2 $$

Assume 8mm $\phi$ bars $\phi_x = \phi_y = 8d_y = d - \frac{d_x}{2} - \frac{d_y}{2} $

$ = 125 - \frac{8}{2} - \frac{8}{2} = 117 mm$

$$ (Asty) + ve = \frac{0.5 \times 20 \times 1000 \times 117}{415} \times [ 1 - \sqrt{1-\frac{4.6 \times 8.19 \times 10^6}{20 \times 1000 \times 117^2}} $$

$$= 201 mm^2$$

$$ (Asty) - ve = \frac{0.5 \times 20 \times 1000 \times 117}{415} \times [ 1 - \sqrt{1-\frac{4.6 \times 10.41 \times 10^6}{20 \times 1000 \times 117^2}} $$

$$= 258 mm^2$$

$$ Astmin = \frac{0.12bd}{100} = \frac{0.12 \times 1000 \times 150}{100}$$

$$ 180 mm^2 \lt [(Astx) = ve, (Astx) - ve, (Asty) + ve, (Asty) - ve] $$

Therefore Safe.

• Main steel in x-direction (8mm$\phi$) spacing +ve = $\frac{1000 \times \pi / 4 \times 8^2}{188} = 267.37 \approx 250 mm$

Spacing -ve = $\frac{1000 \times \pi / 4 \times 8^2}{324} = 155.14 \approx 150mm$

• Main steel in y-direction (8mm$\phi$) spacing +ve = $\frac{1000 \times \pi / 4 \times 8^2}{201} = 250.07 \approx 225 mm$

Spacing -ve = $\frac{1000 \times \pi / 4 \times 8^2}{258} = 194.82 \approx 175mm$