Interior slab panel 3.5x4.5 m

Data and assumptions

• Clear spans: $l_x = 3.5$ m (short), $l_y = 4.5$ m (long); support width 250 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 4 kN/m², floor finish = 0.6 kN/m². Unit weight of concrete 25 kN/m³. Clear cover 20 mm (mild exposure, IS 456 Table 16); 10 mm bars assumed for working out $d$.
• Panel type: interior panel (all four edges continuous); moment coefficients from IS 456 Annex D, Table 26 (case: interior panel).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 150$ mm, so $d = 150 - 20 - 5 = 125$ mm for the short-span bars and $d_y = 125 - 10 = 115$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 3500 + 125 = 3625\ \text{mm},\quad l_y = 4500 + 125 = 4625\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{4625}{3625} = 1.276 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.150 = 3.750 kN/m²
• Total working load $w$ = 3.750 + 0.6 + 4 = 8.350 kN/m²
• Factored load $w_u = 1.5 \times 8.350 = 12.525$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 12.525 \times 3.625^2 = 164.59$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.28$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 125^2 \times 10^{-6} = 43.13$ kN·m/m $>$ $M_{max} = 7.58$ kN·m/m. The section is under-reinforced; $d$ is adequate for flexure.

Reinforcement

$$A_{st} = \frac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \frac{4.6 M_u}{f_{ck} b d^2}}\right] b d$$

Minimum steel = 0.12% of $bD$ = 180 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm in both directions.

Check for shear

$V_u = w_u l_x/2 = 12.525 \times 3.625/2 = 22.70$ kN/m

$\tau_v = \dfrac{V_u}{bd} = \dfrac{22702}{1000 \times 125} = 0.182$ MPa

Steel at the support: $A_{st} = 183$ mm²/m, $p_t = 0.146\%$ ($< 0.15\%$), so $\tau_c = 0.28$ MPa (IS 456 Table 19, M20). For $D = 150$ mm, $k = 1.30$ (cl. 40.2.1.1), so $k\tau_c = 0.36$ MPa.

$\tau_v = 0.182 < k\tau_c = 0.36$ MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 183}{1000 \times 125} = 0.146\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{131}{183} = 172$ N/mm².

Modification factor from IS 456 Fig. 4 (for $p_t = 0.15\%$ and $f_s = 172$ N/mm²): at such a low steel percentage the curve reaches its upper limit, so $k_t = 2.0$.

Permissible $l/d = 26 \times 2.0 = 52.0$; actual $l_x/d = 3625/125 = 29.0$. Since actual < permissible, deflection is satisfied.

Check for development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.92$ MPa (M20, deformed bars, ×1.6): $L_d = \dfrac{0.87 \times 415 \times 8}{4 \times 1.92} = 376$ mm

All four edges are continuous, so there is no simple support where the check $L_d \le 1.3M_1/V + L_0$ governs: the bottom bars run across the beams into the adjoining panels. The top bars over the supports extend $0.3 l_x = 0.3 \times 3500 = 1050$ mm from the face of the support, which is much more than $L_d = 376$ mm, so the development length is satisfied.

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@275 short span      #
  # bot: 8mm@275 long span       #
  #                              #
  +==============================+
  = / # : continuous edge (top bars 8mm@275)
  (all four edges continuous: no torsion steel needed)

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
  8mm@275            8mm@275
   ~~~~~~~            ~~~~~~~
  |------ slab D = 150 mm ------|
   bottom 8mm @ 275 c/c, alt. bars bent up
  [beam 250]          [beam 250]

• Bottom bars: short-span bars in the lower layer, long-span bars above them; alternate bars may be bent up at $0.1 l$ from the support.
• Top bars over continuous supports extend $0.3 l_x$ from the face of the support; all bars have 20 mm cover.
• Distribution/minimum steel $\ge$ 180 mm²/m is already satisfied by the bars provided.
• No torsion reinforcement is needed, since all corners are formed by continuous edges (IS 456 D-1.10).

Answer: $D = 150$ mm; short-span bottom 8 mm @ 275 c/c; long-span bottom 8 mm @ 275 c/c; top steel over the continuous edges 8 mm @ 275 c/c in both directions. Shear, deflection and development length are satisfied.