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Slab design, 4x6 m, Fe500

Short answer

Slab thickness 240 mm. Bottom reinforcement in both directions: 8 mm at 170 mm c/c. Top steel at discontinuous edges: 8 mm at 300 mm c/c extending 0.1 span. Designed as simply supported two-way slab with corners held down. Deflection satisfied.

Design a simply supported RC slab of 4 m x 6 m at limit state of collapse in flexure. Arrange the designed reinforcing bars and draw a neat sketch of slab showing arrangement of top and bottom reinforcements. Take, Live load = 4 kN/m2^2, Floor finish = 1 kN/m2^2, Grade of concrete = M20, Grade of steel reinforcement = Fe500.

Data and assumptions

  • Clear spans: lx=4l_x = 4 m (short), ly=6l_y = 6 m (long); support width 230 mm; fck=20f_{ck} = 20 MPa, fy=500f_y = 500 MPa.
  • Live load = 4 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.
  • 'Simply supported' slab with corners prevented from lifting (so top steel is provided at the corners); support width 230 mm assumed.
  • Panel type: four edges discontinuous (corners held down); moment coefficients from IS 456 Annex D, Table 26 (case 9: four edges discontinuous).

Depth and effective span

Basic span/depth ratio = 20 (IS 456 cl. 23.2.1). Trial overall depth D=240D = 240 mm, so d=240−20−5=215d = 240 - 20 - 5 = 215 mm for the short-span bars and dy=215−10=205d_y = 215 - 10 = 205 mm for the long-span bars.

lx=min⁡(lclear+d, c/c)=4000+215=4215 mm,ly=6000+215=6215 mml_x = \min(l_{clear} + d,\ c/c) = 4000 + 215 = 4215\ \text{mm},\quad l_y = 6000 + 215 = 6215\ \text{mm} lylx=62154215=1.474<2⇒two-way slab\frac{l_y}{l_x} = \frac{6215}{4215} = 1.474 < 2 \Rightarrow \text{two-way slab}

Loads (per m²)

  • Self-weight = 25 × 0.240 = 6.000 kN/m²
  • Total working load ww = 6.000 + 1 + 4 = 11.000 kN/m²
  • Factored load wu=1.5×11.000=16.500w_u = 1.5 \times 11.000 = 16.500 kN/m²

Bending moments

MomentCoefficientValue (kN·m/m)
MxM_x positive (mid-span, short)0.088025.79
MyM_y positive (mid-span, long)0.056016.42

M=α wu lx2M = \alpha\, w_u\, l_x^2 with wulx2=16.500×4.2152=293.14w_u l_x^2 = 16.500 \times 4.215^2 = 293.14 kN·m/m. Coefficients are interpolated for ly/lx=1.47l_y/l_x = 1.47.

Check of depth

Mu,lim=0.133 fckbd2=0.133×20×1000×2152×10−6=122.96M_{u,lim} = 0.133\, f_{ck} b d^2 = 0.133 \times 20 \times 1000 \times 215^2 \times 10^{-6} = 122.96 kN·m/m >> Mmax=25.79M_{max} = 25.79 kN·m/m. The 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 = 288 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)25.792888 mm @ 170 c/c296
Long span, bottom (mid-span)16.422888 mm @ 170 c/c296

Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 Annex D) extends 0.1l0.1 l into the span: along the discontinuous long edges (short-span direction), 50% of mid-span steel = 148 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edges (long-span direction), 50% of mid-span steel = 148 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

Vu=wulx/2=16.500×4.215/2=34.77V_u = w_u l_x/2 = 16.500 \times 4.215/2 = 34.77 kN/m τv=Vubd=347741000×215=0.162\tau_v = \dfrac{V_u}{bd} = \dfrac{34774}{1000 \times 215} = 0.162 MPa Steel at support: Ast=148A_{st} = 148 mm²/m, pt=0.069%p_t = 0.069\%; τc=0.28\tau_c = 0.28 MPa (IS 456 Table 19, M20, pt≤0.15%p_t \le 0.15\%). For D=240D = 240 mm, k=1.12k = 1.12 (cl. 40.2.1.1), so kτc=0.314k\tau_c = 0.314 MPa. τv=0.162<kτc=0.314\tau_v = 0.162 < k\tau_c = 0.314 MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: pt=100×2961000×215=0.138%p_t = \dfrac{100 \times 296}{1000 \times 215} = 0.138\%, fs=0.58fyAst,reqAst,prov=279.9f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 279.9 N/mm². Modification factor from IS 456 Fig. 4 (for pt=0.14%p_t = 0.14\% and fs=280f_s = 280 N/mm²): kt≈1.7k_t \approx 1.7 Permissible l/d=20×1.7=34l/d = 20 \times 1.7 = 34; actual lx/d=4215/215=19.6l_x/d = 4215/215 = 19.6. Since actual < permissible, deflection is satisfied (with a large margin, so a thinner slab could also be used).

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=0.87×500×84×1.92=453L_d = \dfrac{0.87 \times 500 \times 8}{4 \times 1.92} = 453 mm Alternate bottom bars (148 mm²/m) run into the support: M1=13.59M_1 = 13.59 kN·m/m, L0=max⁡(d,12ϕ)=215L_0 = \max(d, 12\phi) = 215 mm.

1.3M1V+L0=1.3×1358901434774+215=723 mm1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{13589014}{34774} + 215 = 723\ \text{mm}

Since 723 mm >> Ld=453L_d = 453 mm, development length is satisfied.

Torsion reinforcement at corners

Corner size lx/5=4215/5=843l_x/5 = 4215/5 = 843 mm (say 850 mm) in both directions.

  • All 4 corners have both edges discontinuous: provide top and bottom meshes, each layer 3/4 of the mid-span AstA_{st} = 0.75 × 288 = 216 mm²/m. Provide 8 mm @ 170 c/c in both directions, top and bottom, over 850 mm × 850 mm.

Detailing

PLAN  (long span horizontal, short span vertical)
  +------------------------------+
  |T                            T|
  | bot: 8mm@170 short span      |
  | bot: 8mm@170 long span       |
  |T                            T|
  +------------------------------+
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top + bottom (850 x 850)

SECTION ALONG SHORT SPAN
  top bars at edges (0.5 x mid-span steel)
   ~~~                      ~~~
  |------ slab D = 240 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 0.1l0.1 l from the support.
  • All bars have 20 mm cover.
  • Distribution/minimum steel ≥\ge 288 mm²/m is already satisfied by the bars provided.

Answer: D=240D = 240 mm; short-span bottom 8 mm @ 170 c/c; long-span bottom 8 mm @ 170 c/c; top bars 8 mm @ 300 c/c at the edges; torsion mesh 8 mm @ 170 c/c at the corners. Shear, deflection and development length are satisfied.

✓ Answer checked by Claude Opus 5.5 (high effort)

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