Slab panel 2 for BM and SF (floor plan)

Data and assumptions

• Clear spans: $l_x = 3$ m (short), $l_y = 5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa.
• Live load = 3 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 $d$.
• Panel 2 is the top-right panel: its left edge (common with panel 1) and bottom edge (common with panel 3) are continuous; the right edge and top edge are on the outer wall (discontinuous). Bay dimensions 3 m × 5 m are taken as clear spans; wall thickness 230 mm.
• Panel type: two adjacent edges discontinuous; 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 $D = 140$ mm, so $d = 140 - 20 - 5 = 115$ mm for the short-span bars and $d_y = 115 - 10 = 105$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 3000 + 115 = 3115\ \text{mm},\quad l_y = 5000 + 115 = 5115\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5115}{3115} = 1.642 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.140 = 3.500 kN/m²
• Total working load $w$ = 3.500 + 1 + 3 = 7.500 kN/m²
• Factored load $w_u = 1.5 \times 7.500 = 11.250$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 11.250 \times 3.115^2 = 109.16$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.64$.

Check for depth

$M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 115^2 \times 10^{-6} = 36.50$ kN·m/m $>$ $M_{max} = 8.75$ 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 $bD$ = 168 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 = 85 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 85 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 11.250 \times 3.115/2 = 17.52$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{17522}{1000 \times 115} = 0.152$ MPa Steel at support: $A_{st} = 223$ mm²/m, $p_t = 0.194\%$; $\tau_c = 0.32$ MPa (IS 456 Table 19, M20). For $D = 140$ mm, $k = 1.30$ (cl. 40.2.1.1), so $k\tau_c = 0.416$ MPa. $\tau_v = 0.152 < k\tau_c = 0.416$ MPa, so the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 170}{1000 \times 115} = 0.148\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{163}{170} = 231$ N/mm² (here $A_{st,req} = 163$ mm²/m is the steel needed for bending; the minimum steel governs the bars provided). Modification factor from IS 456 Fig. 4 (for $p_t = 0.15\%$ and $f_s = 231$ 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 $l_x/d = 3115/115 = 27.1$. 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 (85 mm²/m) run into the support: $M_1 = 3.48$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 115$ mm.

$$1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{3483013}{17522} + 115 = 373\ \text{mm}$$

Since 373 mm $<$ $L_d = 376$ mm, the bars are given a standard 90° bend at the support (anchorage value $8\phi$ = 64 mm). The available length then becomes 373 + 64 = 437 mm $> L_d$, so development length is satisfied with bent ends.

Torsion reinforcement at corners

Corner size $l_x/5 = 3115/5 = 623$ mm (say 630 mm) in both directions.

• Corner with both edges discontinuous (1 corner): top and bottom meshes, each layer 3/4 of the mid-span $A_{st}$ = 0.75 × 168 = 126 mm²/m. Provide 8 mm @ 295 c/c in both directions, top and bottom, over 630 mm × 630 mm.
• Corner with only one discontinuous edge (2 corners): half of 3/4 of the mid-span $A_{st}$ = 0.375 × 168 = 63 mm²/m. Provide 8 mm @ 300 c/c in each direction, top and bottom.
• 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@295 short span      |
  # bot: 8mm@295 long span       |
  #                             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 over      top bars at
  continuous edge    discontinuous edge (0.5 x)
   ~~~~~~~                       ~~~
  |------ slab D = 140 mm ------|
   bottom 8mm @ 295 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 $\ge$ 168 mm²/m is already satisfied by the bars provided.

Answer: $D = 140$ mm; short-span bottom 8 mm @ 295 c/c; long-span bottom 8 mm @ 295 c/c; top steel over continuous edge 8 mm @ 225 c/c (short direction); 8 mm @ 295 c/c (long direction). Shear, deflection and development length (with bent bar ends) are satisfied.