Two-way slab simply supported, 6x4 m

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 6$ m (long); support (wall) width 230 mm; $f_{ck} = 25$ MPa, $f_y = 415$ MPa.
• Superimposed working load = 4 kN/m² (live load and finishes taken together as one load). 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: simply supported on four sides, corners not held down; moment coefficients from IS 456 Annex D, Table 27 (simply supported, corners free to lift).

Depth and effective span

Basic span/depth ratio = 20 (IS 456 cl. 23.2.1). Trial overall depth $D = 210$ mm, so $d = 210 - 20 - 5 = 185$ mm for the short-span bars and $d_y = 185 - 10 = 175$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 185 = 4185\ \text{mm},\quad l_y = 6000 + 185 = 6185\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{6185}{4185} = 1.478 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.210 = 5.250 kN/m²
• Total working load $w$ = 5.250 + 4 = 9.250 kN/m²
• Factored load $w_u = 1.5 \times 9.250 = 13.875$ kN/m²

Bending moments

From Table 27: $\alpha_x = 0.099$ at 1.4 and $0.104$ at 1.5; $\alpha_y = 0.051$ at 1.4 and $0.046$ at 1.5. Interpolating for $l_y/l_x = 1.478$:

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 13.875 \times 4.185^2 = 243.01$ kN·m/m.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 25 \times 1000 \times 185^2 \times 10^{-6} = 118.08$ kN·m/m $>$ $M_{max} = 25.00$ 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 $bD$ = 252 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm in both directions.

Since the corners are free to lift, no torsion steel is needed. Nominal top steel (8 mm @ 250 c/c, extending $0.1 l$ into the span) is provided along the supported edges to control cracks due to unintended fixity.

Check for shear

$V_u = w_u l_x/2 = 13.875 \times 4.185/2 = 29.03$ kN/m

$\tau_v = \dfrac{V_u}{bd} = \dfrac{29033}{1000 \times 185} = 0.157$ MPa

Steel at the support (alternate bars bent up): $A_{st} = 201$ mm²/m, $p_t = 0.109\%$, so $\tau_c = 0.29$ MPa (IS 456 Table 19, M25). For $D = 210$ mm, $k = 1.18$ (cl. 40.2.1.1), so $k\tau_c = 0.34$ MPa.

$\tau_v = 0.157 < k\tau_c = 0.34$ MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 402}{1000 \times 185} = 0.217\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{388}{402} = 232$ N/mm².

Modification factor from IS 456 Fig. 4 (for $p_t = 0.22\%$ and $f_s = 232$ N/mm²): $k_t \approx 1.8$.

Permissible $l/d = 20 \times 1.8 = 36.0$; actual $l_x/d = 4185/185 = 22.6$. Since actual < permissible, deflection is satisfied.

Check for development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 2.24$ MPa (M25, deformed bars, ×1.6): $L_d = \dfrac{0.87 \times 415 \times 8}{4 \times 2.24} = 322$ mm

At the simple support, $L_d \le 1.3\dfrac{M_1}{V} + L_0$ (IS 456 cl. 26.2.3.3(c)). Alternate bottom bars (201 mm²/m) run into the support: $M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 13.19$ kN·m/m, and $L_0$ = straight length beyond the centre of the wall $= 230/2 - 25$ (end cover) $= 90$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{13.19 \times 10^6}{29033} + 90 = 680\ \text{mm}$$

Since 680 mm $>$ $L_d = 322$ mm, development length is satisfied even with straight bars.

Detailing

PLAN  (long span horizontal, short span vertical)
  +------------------------------+
  |                              |
  | bot: 8mm@125 short span      |
  | bot: 8mm@195 long span       |
  |                              |
  +------------------------------+
  - / | : simply supported edge (nominal top bars 8mm@250)
  corners free to lift: no torsion steel

SECTION ALONG SHORT SPAN
  nominal top bars at edges (8mm@250, 0.1 l)
   ~~~                      ~~~
  |------ slab D = 210 mm ------|
   bottom 8mm @ 125 c/c, alt. bars bent up
  [wall]                  [wall]

• 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 $\ge$ 252 mm²/m is already satisfied by the bars provided.

Answer: $D = 210$ mm; short-span bottom 8 mm @ 125 c/c; long-span bottom 8 mm @ 195 c/c; nominal top bars 8 mm @ 250 c/c at the edges. Shear, deflection and development length are satisfied.