Slab rigidly fixed with beams, 6.5x4 m

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 6.5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 4.5 kN/m², floor finish = 1 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 $d$.
• TMT bars taken as Fe415; slab treated as an interior panel (all four edges continuous/fixed with beams).
• Panel type: interior panel (all four edges continuous); moment coefficients from IS 456 Annex D, Table 26.

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 175$ mm, so $d = 175 - 20 - 5 = 150$ mm for the short-span bars and $d_y = 150 - 10 = 140$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 6500 + 150 = 6650\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{6650}{4150} = 1.602 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.175 = 4.375 kN/m²
• Total working load $w$ = 4.375 + 1 + 4.5 = 9.875 kN/m²
• Factored load $w_u = 1.5 \times 9.875 = 14.812$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 14.812 \times 4.150^2 = 255.11$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.60$.

Check for depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10$ kN·m/m $>$ $M_{max} = 14.25$ 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 $bD$ = 210 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long).

Check for shear

$V_u = w_u l_x/2 = 14.812 \times 4.150/2 = 30.74$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{30736}{1000 \times 150} = 0.205$ MPa Steel at support: $A_{st} = 279$ mm²/m, $p_t = 0.186\%$; $\tau_c = 0.31$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.388$ MPa. $\tau_v = 0.205 < k\tau_c = 0.388$ MPa, so the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{207}{214} = 233$ N/mm² (here $A_{st,req} = 207$ mm²/m is the steel needed for bending; the minimum steel governs the bars provided). Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ 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 $l/d = 26 \times 2.0 = 52.0$; actual $l_x/d = 4150/150 = 27.7$. 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 (107 mm²/m) run into the support: $M_1 = 5.71$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 150$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{30736} + 150 = 391\ \text{mm}$$

Since 391 mm $>$ $L_d = 376$ mm, development length is satisfied.

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@235 short span      #
  # bot: 8mm@235 long span       #
  #                              #
  +==============================+
  = / # : 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 over      top bars over
  continuous edge    continuous edge
   ~~~~~~~            ~~~~~~~
  |------ slab D = 175 mm ------|
   bottom 8mm @ 235 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.
• 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 $\ge$ 210 mm²/m is already satisfied by the bars provided.

Answer: $D = 175$ mm; short-span bottom 8 mm @ 235 c/c; long-span bottom 8 mm @ 235 c/c; top steel over continuous edge 8 mm @ 180 c/c (short direction); 8 mm @ 235 c/c (long direction). Shear, deflection and development length are satisfied.