Slab on masonry walls, 5x6 m

Data and assumptions

• Clear spans: $l_x = 5$ m (short), $l_y = 6$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 3 kN/m², floor finish = 1.5 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: four edges discontinuous (corners held down); moment coefficients from IS 456 Annex D, Table 26.

Depth and effective span

Basic span/depth ratio = 20 (IS 456 cl. 23.2.1). Trial overall depth $D = 265$ mm, so $d = 265 - 20 - 5 = 240$ mm for the short-span bars and $d_y = 240 - 10 = 230$ mm for the long-span bars.

$$l_x = \min(5000 + 240,\ 5000 + 230) = 5230\ \text{mm (c/c governs)},\quad l_y = \min(6000 + 240,\ 6000 + 230) = 6230\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{6230}{5230} = 1.191 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.265 = 6.625 kN/m²
• Total working load $w$ = 6.625 + 1.5 + 3 = 11.125 kN/m²
• Factored load $w_u = 1.5 \times 11.125 = 16.688$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 16.688 \times 5.230^2 = 456.45$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.19$.

Check for depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 240^2 \times 10^{-6} = 158.98$ kN·m/m $>$ $M_{max} = 32.54$ kN·m/m. 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 $bD$ = 318 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends $0.1 l$ into the span: along the discontinuous long edge(s) (short-span direction): 50% of mid-span steel = 201 mm²/m, provide 8 mm @ 245 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 162 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 16.688 \times 5.230/2 = 43.64$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{43638}{1000 \times 240} = 0.182$ MPa Steel at support: $A_{st} = 201$ mm²/m, $p_t = 0.084\%$; $\tau_c = 0.28$ MPa (IS 456 Table 19, M20). For $D = 265$ mm, $k = 1.07$ (cl. 40.2.1.1), so $k\tau_c = 0.300$ MPa. $\tau_v = 0.182 < k\tau_c = 0.300$ MPa, so the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 402}{1000 \times 240} = 0.168\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{389}{402} = 233$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.17\%$ and $f_s = 233$ N/mm²): $k_t = 2.0$ (at such a low steel percentage the curve reaches its upper limit of 2.0). Permissible $l/d = 20 \times 2.0 = 40.0$; actual $l_x/d = 5230/240 = 21.8$. 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 Alternate bottom bars (201 mm²/m) run into the support: $M_1 = 17.12$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 240$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{17119556}{43638} + 240 = 750\ \text{mm}$$

Since 750 mm $>$ $L_d = 376$ mm, development length is satisfied.

Torsion reinforcement at corners

Corner size $l_x/5 = 5230/5 = 1046$ mm (say 1050 mm) in both directions.

• Corner with both edges discontinuous (4 corners): top and bottom meshes, each layer 3/4 of the mid-span $A_{st}$ = 0.75 × 389 = 292 mm²/m. Provide 8 mm @ 155 c/c in both directions, top and bottom, over 1050 mm × 1050 mm.
• No torsion steel is needed at corners where both edges are continuous (IS 456 D-1.10).

Detailing

PLAN  (long span horizontal, short span vertical)
  +------------------------------+
  |T                            T|
  | bot: 8mm@125 short span      |
  | bot: 8mm@155 long span       |
  |T                            T|
  +------------------------------+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars at edges (0.5 x mid-span steel)
   ~~~                      ~~~
  |------ slab D = 265 mm ------|
   bottom 8mm @ 125 c/c, alt. bars bent up
  [support]            [support]

• 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.
• All bars have 20 mm cover.
• Distribution/minimum steel $\ge$ 318 mm²/m is already satisfied by the bars provided.

Answer: $D = 265$ mm; short-span bottom 8 mm @ 125 c/c; long-span bottom 8 mm @ 155 c/c. Shear, deflection and development length are satisfied.