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Two-way slab on RCC beams, 4x6 m

Short answer

Design as four-edge discontinuous panel (IS 456 case 9). Trial depth 175 mm, d=150 mm; factored load 14.44 kN/m². Short-span steel 10 mm @ 180 c/c, long-span 10 mm @ 270 c/c; shear and deflection safe.

Design a two-way slab resting on RCC beams on all sides for a room having clear dimensions of 4 m x 6 m. The slab is subjected to a super-imposed live load of 2.5 kN/m2^2 and floor finishes (screeds and flooring) load of 2.75 kN/m2^2. Take M20 concrete grade and Fe415 steel grade.

Data and assumptions

  • Clear spans: lx=4l_x = 4 m (short), ly=6l_y = 6 m (long); fck=20f_{ck} = 20 MPa, fy=415f_y = 415 MPa.
  • Live load = 2.5 kN/m², floor finish = 2.75 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.
  • Beam width not given; 230 mm assumed.
  • Panel type: the slab covers a single room, so it is not continuous over any of the beams: four edges discontinuous, with corners held down by the monolithic beams. Moment coefficients from IS 456 Annex D, Table 26 (case 9), and torsion steel is provided at the corners.

Depth and effective span

Basic span/depth ratio = 20 (IS 456 cl. 23.2.1, simply supported). Trial overall depth D=175D = 175 mm, so d=175−20−5=150d = 175 - 20 - 5 = 150 mm for the short-span bars and dy=150−10=140d_y = 150 - 10 = 140 mm for the long-span bars.

lx=min⁡(lclear+d, c/c)=4000+150=4150 mm,ly=6000+150=6150 mml_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 6000 + 150 = 6150\ \text{mm} lylx=61504150=1.482<2⇒two-way slab\frac{l_y}{l_x} = \frac{6150}{4150} = 1.482 < 2 \Rightarrow \text{two-way slab}

Loads (per m²)

  • Self-weight = 25 × 0.175 = 4.375 kN/m²
  • Total working load ww = 4.375 + 2.75 + 2.5 = 9.625 kN/m²
  • Factored load wu=1.5×9.625=14.438w_u = 1.5 \times 9.625 = 14.438 kN/m²

Bending moments

MomentCoefficientValue (kN·m/m)
MxM_x positive (mid-span, short)0.088321.95
MyM_y positive (mid-span, long)0.056013.92

M=α wu lx2M = \alpha\, w_u\, l_x^2 with wulx2=14.438×4.1502=248.65w_u l_x^2 = 14.438 \times 4.150^2 = 248.65 kN·m/m. αx\alpha_x is interpolated for ly/lx=1.48l_y/l_x = 1.48 (0.085 at 1.4, 0.089 at 1.5).

Check of depth

Mu,lim=0.138 fckbd2=0.138×20×1000×1502×10−6=62.10M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10 kN·m/m >> Mmax=21.95M_{max} = 21.95 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 = 210 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)21.9543110 mm @ 180 c/c436
Long span, bottom (mid-span)13.9228810 mm @ 270 c/c291

Top steel at the discontinuous edges (negative moment due to partial fixity by the beams, IS 456 Annex D): 50% of the mid-span steel, extending 0.1l0.1 l into the span. Along the long edges (short-span direction): 218 mm²/m, provide 8 mm @ 230 c/c (219 mm²/m). Along the short edges (long-span direction): 145 mm²/m, provide 8 mm @ 300 c/c (168 mm²/m).

Check for shear

Vu=wulx/2=14.438×4.150/2=29.96V_u = w_u l_x/2 = 14.438 \times 4.150/2 = 29.96 kN/m τv=Vubd=299581000×150=0.200\tau_v = \dfrac{V_u}{bd} = \dfrac{29958}{1000 \times 150} = 0.200 MPa Steel at support (alternate bars bent up): Ast=218A_{st} = 218 mm²/m, pt=0.145%p_t = 0.145\%; τc=0.28\tau_c = 0.28 MPa (IS 456 Table 19, M20). For D=175D = 175 mm, k=1.25k = 1.25 (cl. 40.2.1.1), so kτc=0.35k\tau_c = 0.35 MPa. τv=0.200<kτc=0.35\tau_v = 0.200 < k\tau_c = 0.35 MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: pt=100×4361000×150=0.291%p_t = \dfrac{100 \times 436}{1000 \times 150} = 0.291\%, fs=0.58fyAst,reqAst,prov=0.58×415×431436=237.9f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{431}{436} = 237.9 N/mm². Modification factor from IS 456 Fig. 4 (for pt=0.29%p_t = 0.29\% and fs=238f_s = 238 N/mm²): kt≈1.5k_t \approx 1.5 Permissible l/d=20×1.5=30l/d = 20 \times 1.5 = 30; actual lx/d=4150/150=27.7l_x/d = 4150/150 = 27.7. 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=0.87×415×104×1.92=470L_d = \dfrac{0.87 \times 415 \times 10}{4 \times 1.92} = 470 mm Alternate bottom bars (218 mm²/m) run into the support: xu=0.87×415×2180.36×20×1000=10.9x_u = \dfrac{0.87 \times 415 \times 218}{0.36 \times 20 \times 1000} = 10.9 mm, M1=0.87×415×218×(150−0.42×10.9)=11.45M_1 = 0.87 \times 415 \times 218 \times (150 - 0.42 \times 10.9) = 11.45 kN·m/m, L0=max⁡(d,12ϕ)=150L_0 = \max(d, 12\phi) = 150 mm.

1.3M1V+L0=1.3×1145300029958+150=647 mm1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{11453000}{29958} + 150 = 647\ \text{mm}

Since 647 mm >> Ld=470L_d = 470 mm, development length is satisfied.

Torsion reinforcement at corners

Corner size lx/5=4150/5=830l_x/5 = 4150/5 = 830 mm (say 850 mm) in both directions. All 4 corners have both edges discontinuous, so each of the four layers (top and bottom, both directions) gets 3/4 of the short-span mid-span steel = 0.75 × 431 = 323 mm²/m. Provide 8 mm @ 150 c/c (335 mm²/m) over 850 mm × 850 mm at each corner.

Detailing

PLAN  (long span horizontal, short span vertical)
  +------------------------------+
  |T                            T|
  | bot: 10mm@180 short span     |
  | bot: 10mm@270 long span      |
  |T                            T|
  +------------------------------+
  - / | : discontinuous edge over beam
          (top bars 8mm@230 / 8mm@300, 0.1 l long)
  T : torsion mesh top + bottom, 8mm@150 (850 x 850)

SECTION ALONG SHORT SPAN
  top bars at edges (0.5 x mid-span steel)
   ~~~                      ~~~
  |------ slab D = 175 mm ------|
   bottom 10mm @ 180 c/c, alt. bars bent up
  [beam]                  [beam]
  • 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 210 mm²/m is already satisfied by the bars provided.

Answer: D=175D = 175 mm; short-span bottom 10 mm @ 180 c/c; long-span bottom 10 mm @ 270 c/c; top bars at the edges 8 mm @ 230 c/c (short direction) and 8 mm @ 300 c/c (long direction); torsion mesh 8 mm @ 150 c/c at all four corners. Shear, deflection and development length are satisfied.

✓ Answer checked by Claude Opus 5.5 (high effort)

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