Roof slab restrained on four sides, 6x3.5 m

Data and assumptions

• Clear spans: $l_x = 3.5$ m (short), $l_y = 6$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 250$ MPa.
• Live load = 4 kN/m², floor finish = 0 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 $d$.
• The 'super-imposed load' of 4 kN/m² is taken as the total live load plus finishes; beam width 230 mm assumed; slab treated as an interior (fully restrained) panel.
• 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 $D = 125$ mm, so $d = 125 - 20 - 5 = 100$ mm for the short-span bars and $d_y = 100 - 10 = 90$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 3500 + 100 = 3600\ \text{mm},\quad l_y = 6000 + 100 = 6100\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{6100}{3600} = 1.694 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.125 = 3.125 kN/m²
• Total working load $w$ = 3.125 + 0 + 4 = 7.125 kN/m²
• Factored load $w_u = 1.5 \times 7.125 = 10.688$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 10.688 \times 3.600^2 = 138.51$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.69$.

Check for depth

$M_{u,lim} = 0.148\, f_{ck} b d^2 = 0.148 \times 20 \times 1000 \times 100^2 \times 10^{-6} = 29.60$ kN·m/m $>$ $M_{max} = 8.10$ 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 = 0.15% of $bD$ = 188 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 270 mm (long).

Check for shear

$V_u = w_u l_x/2 = 10.688 \times 3.600/2 = 19.24$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{19238}{1000 \times 100} = 0.192$ MPa Steel at support: $A_{st} = 402$ mm²/m, $p_t = 0.402\%$; $\tau_c = 0.43$ MPa (IS 456 Table 19, M20). For $D = 125$ mm, $k = 1.30$ (cl. 40.2.1.1), so $k\tau_c = 0.559$ MPa. $\tau_v = 0.192 < k\tau_c = 0.559$ MPa, so the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 296}{1000 \times 100} = 0.296\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 250 \times \dfrac{292}{296} = 143$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.30\%$ and $f_s = 143$ 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 $l_x/d = 3600/100 = 36.0$. Since actual < permissible, deflection is satisfied.

Check for development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.2$ MPa (M20, plain mild-steel bars, so no 60% increase): $L_d = \dfrac{0.87 \times 250 \times 8}{4 \times 1.2} = 363$ mm Alternate bottom bars (148 mm²/m) run into the support: $M_1 = 3.16$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 100$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{3156090}{19238} + 100 = 313\ \text{mm}$$

Since 313 mm $<$ $L_d = 363$ mm, straight bar ends are not enough. Plain bars must end in standard U-hooks anyway (IS 456 cl. 26.2.2.1), and a standard hook has an anchorage value of $16\phi$ = 128 mm. The available length becomes 313 + 128 = 441 mm $> 363$ mm, so development length is satisfied with hooked ends.

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@170 short span      #
  # bot: 8mm@265 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 = 125 mm ------|
   bottom 8mm @ 170 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 $\ge$ 188 mm²/m is already satisfied by the bars provided.

Answer: $D = 125$ mm; short-span bottom 8 mm @ 170 c/c; long-span bottom 8 mm @ 265 c/c; top steel over continuous edge 8 mm @ 125 c/c (short direction); 8 mm @ 210 c/c (long direction). Shear, deflection and development length (with standard hooks at the bar ends) are satisfied.