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Slab prevented from uplifting, 3.6x4.2 m

Short answer

Provide 200 mm slab, d=175 mm, lx=3.775 m c/c, ly=4.375 m, ly/lx=1.16, four edges discontinuous. Factored load 13.5 kN/m². Bottom: short span 8@205, long span 8@205. Top at all edges 8@300 (50% of positive). Shear, deflection, and development length checks are satisfied.

Design a slab for a room of size 3.6 m x 4.2 m prevented uplifting by walls (230 mm thick) loads for a intermediate storey of a residential building. Use M20 grade of concrete and Fe 415 grade of steel. Sketch the reinforcements. Carry out all necessary checks require in slab design. Take live load = 3 kN/m2^2, floor finish = 1 kN/m2^2.

Data and assumptions

  • Clear spans: lx=3.6l_x = 3.6 m (short), ly=4.2l_y = 4.2 m (long); support width 230 mm; fck=20f_{ck} = 20 MPa, fy=415f_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 dd.
  • Panel type: four edges discontinuous (corners held down); moment coefficients from IS 456 Annex D, Table 26.

Depth and effective span

Basic span/depth ratio = 20 (IS 456 cl. 23.2.1). Trial overall depth D=200D = 200 mm, so d=200−20−5=175d = 200 - 20 - 5 = 175 mm for the short-span bars and dy=175−10=165d_y = 175 - 10 = 165 mm for the long-span bars.

lx=min⁡(lclear+d, c/c)=3600+175=3775 mm,ly=4200+175=4375 mml_x = \min(l_{clear} + d,\ c/c) = 3600 + 175 = 3775\ \text{mm},\quad l_y = 4200 + 175 = 4375\ \text{mm} lylx=43753775=1.159<2⇒two-way slab\frac{l_y}{l_x} = \frac{4375}{3775} = 1.159 < 2 \Rightarrow \text{two-way slab}

Loads (per m²)

  • Self-weight = 25 × 0.200 = 5.000 kN/m²
  • Total working load ww = 5.000 + 1 + 3 = 9.000 kN/m²
  • Factored load wu=1.5×9.000=13.500w_u = 1.5 \times 9.000 = 13.500 kN/m²

Bending moments

MomentCoefficientValue (kN·m/m)
MxM_x positive (mid-span, short)0.068713.22
MyM_y positive (mid-span, long)0.056010.77

M=α wu lx2M = \alpha\, w_u\, l_x^2 with wulx2=13.500×3.7752=192.38w_u l_x^2 = 13.500 \times 3.775^2 = 192.38 kN·m/m. Coefficients are interpolated for ly/lx=1.16l_y/l_x = 1.16.

Check for depth

Mu,lim=0.138 fckbd2=0.138×20×1000×1752×10−6=84.53M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 175^2 \times 10^{-6} = 84.53 kN·m/m >> Mmax=13.22M_{max} = 13.22 kN·m/m. Section is under-reinforced; dd is adequate for flexure.

Reinforcement

Ast=0.5fckfy[1−1−4.6Mufckbd2]bdA_{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 bDbD = 240 mm²/m. Maximum spacing = min(3d3d, 300 mm) = 300 mm (short), 300 mm (long).

LocationMuM_u (kN·m/m)AstA_{st} req (mm²/m)ProvideAstA_{st} prov
Short span, bottom (mid-span)13.222408 mm @ 205 c/c245
Long span, bottom (mid-span)10.772408 mm @ 205 c/c245

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends 0.1l0.1 l into the span: along the discontinuous long edge(s) (short-span direction): 50% of mid-span steel = 123 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 123 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

Vu=wulx/2=13.500×3.775/2=25.48V_u = w_u l_x/2 = 13.500 \times 3.775/2 = 25.48 kN/m τv=Vubd=254811000×175=0.146\tau_v = \dfrac{V_u}{bd} = \dfrac{25481}{1000 \times 175} = 0.146 MPa Steel at support: Ast=123A_{st} = 123 mm²/m, pt=0.070%p_t = 0.070\%; τc=0.28\tau_c = 0.28 MPa (IS 456 Table 19, M20). For D=200D = 200 mm, k=1.20k = 1.20 (cl. 40.2.1.1), so kτc=0.336k\tau_c = 0.336 MPa. τv=0.146<kτc=0.336\tau_v = 0.146 < k\tau_c = 0.336 MPa, so the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: pt=100×2451000×175=0.140%p_t = \dfrac{100 \times 245}{1000 \times 175} = 0.140\%, fs=0.58fyAst,reqAst,prov=0.58×415×215245=211f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{215}{245} = 211 N/mm² (here Ast,req=215A_{st,req} = 215 mm²/m is the steel needed for bending; the minimum steel governs the bars provided). Modification factor from IS 456 Fig. 4 (for pt=0.14%p_t = 0.14\% and fs=211f_s = 211 N/mm²): kt=2.0k_t = 2.0 (at such a low steel percentage the curve reaches its upper limit of 2.0). Permissible l/d=20×2.0=40.0l/d = 20 \times 2.0 = 40.0; actual lx/d=3775/175=21.6l_x/d = 3775/175 = 21.6. Since actual < permissible, deflection is satisfied.

Check for development length

Ld=0.87fyϕ4τbdL_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}} with τbd=1.92\tau_{bd} = 1.92 MPa (M20, deformed bars, ×1.6): Ld=0.87×415×84×1.92=376L_d = \dfrac{0.87 \times 415 \times 8}{4 \times 1.92} = 376 mm Alternate bottom bars (123 mm²/m) run into the support: M1=7.63M_1 = 7.63 kN·m/m, L0=max⁡(d, 12ϕ)=175L_0 = \max(d,\ 12\phi) = 175 mm.

1.3M1V+L0=1.3×763364325481+175=564 mm1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{7633643}{25481} + 175 = 564\ \text{mm}

Since 564 mm >> Ld=376L_d = 376 mm, development length is satisfied.

Torsion reinforcement at corners

Corner size lx/5=3775/5=755l_x/5 = 3775/5 = 755 mm (say 760 mm) in both directions.

  • Corner with both edges discontinuous (4 corners): top and bottom meshes, each layer 3/4 of the mid-span AstA_{st} = 0.75 × 240 = 180 mm²/m. Provide 8 mm @ 205 c/c in both directions, top and bottom, over 760 mm × 760 mm.
  • 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@205 short span      |
  | bot: 8mm@205 long span       |
  |T                            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 at edges (0.5 x mid-span steel)
   ~~~                      ~~~
  |------ slab D = 200 mm ------|
   bottom 8mm @ 205 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 0.1l0.1 l from the support.
  • All bars have 20 mm cover.
  • Distribution/minimum steel ≥\ge 240 mm²/m is already satisfied by the bars provided.

Answer: D=200D = 200 mm; short-span bottom 8 mm @ 205 c/c; long-span bottom 8 mm @ 205 c/c. Shear, deflection and development length are satisfied.

✓ Answer checked by Claude Opus 5.5 (high effort)

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