Interior slab design, 4x5 m

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 3 kN/m², floor finish = 2 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 $d$.
• Support (beam) width is not given; 230 mm is assumed.
• 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 $D = 165$ mm, so $d = 165 - 20 - 5 = 140$ mm for the short-span bars and $d_y = 140 - 10 = 130$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 140 = 4140\ \text{mm},\quad l_y = 5000 + 140 = 5140\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5140}{4140} = 1.242 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.165 = 4.125 kN/m²
• Total working load $w$ = 4.125 + 2 + 3 = 9.125 kN/m²
• Factored load $w_u = 1.5 \times 9.125 = 13.688$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 13.688 \times 4.140^2 = 234.60$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$.

Check of depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 140^2 \times 10^{-6} = 54.10$ kN·m/m $>$ $M_{max} = 10.48$ 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 $bD$ = 198 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm in both directions.

Check for shear

$V_u = w_u l_x/2 = 13.688 \times 4.140/2 = 28.33$ kN/m

$\tau_v = \dfrac{V_u}{bd} = \dfrac{28333}{1000 \times 140} = 0.202$ MPa

Steel at the support: $A_{st} = 219$ mm²/m, $p_t = 0.156\%$, so $\tau_c = 0.29$ MPa (IS 456 Table 19, M20). For $D = 165$ mm, $k = 1.27$ (cl. 40.2.1.1), so $k\tau_c = 0.37$ MPa.

$\tau_v = 0.202 < k\tau_c = 0.37$ MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 201}{1000 \times 140} = 0.144\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{160}{201} = 192$ N/mm².

Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 192$ 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 $l_x/d = 4140/140 = 29.6$. Since actual < permissible, deflection is satisfied.

Check for development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.92$ 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 $0.3 l_x = 0.3 \times 4000 = 1200$ 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@250 short span      #
  # bot: 8mm@250 long span       #
  #                              #
  +==============================+
  = / # : continuous edge (top bars)
  (all four edges continuous: no torsion steel needed)

SECTION ALONG SHORT SPAN
  top bars over      top bars over
  continuous edge    continuous edge
  8mm@230            8mm@230
   ~~~~~~~            ~~~~~~~
  |------ slab D = 165 mm ------|
   bottom 8mm @ 250 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 $\ge$ 198 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: $D = 165$ mm; short-span bottom 8 mm @ 250 c/c; long-span bottom 8 mm @ 250 c/c; top steel over the continuous long edges 8 mm @ 230 c/c (short direction) and over the continuous short edges 8 mm @ 250 c/c (long direction). Shear, deflection and development length are satisfied.