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Interior slab panel 4.5x6.5 m

Short answer

Provide 195 mm slab, effective depth 170 mm, clear short span 4.5 m, c/c 4.67 m, ly/lx=1.43. Factored load 17.06 kN/m². Bending moments: short positive 14.72 kNm/m (8@200), negative 19.19 (8@150); long positive 8.93 (8@210), negative 11.91 (8@210). Shear and deflection are satisfactory.

Design and detail an interior panel of a slab resting on RCC beams on all sides for a room having clear dimensions of 4.5 m x 6.5 m. The slab is subjected to a super-imposed live load of 4 kN/m2^2 and floor finishes load of 2.5 kN/m2^2. Take M20 concrete and Fe415 steel.

Data and assumptions

  • Clear spans: lx=4.5l_x = 4.5 m (short), ly=6.5l_y = 6.5 m (long); support width 230 mm; fck=20f_{ck} = 20 MPa, fy=415f_y = 415 MPa.
  • Live load = 4 kN/m², floor finish = 2.5 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: 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=195D = 195 mm, so d=195−20−5=170d = 195 - 20 - 5 = 170 mm for the short-span bars and dy=170−10=160d_y = 170 - 10 = 160 mm for the long-span bars.

lx=min⁡(lclear+d, c/c)=4500+170=4670 mm,ly=6500+170=6670 mml_x = \min(l_{clear} + d,\ c/c) = 4500 + 170 = 4670\ \text{mm},\quad l_y = 6500 + 170 = 6670\ \text{mm} lylx=66704670=1.428<2⇒two-way slab\frac{l_y}{l_x} = \frac{6670}{4670} = 1.428 < 2 \Rightarrow \text{two-way slab}

Loads (per m²)

  • Self-weight = 25 × 0.195 = 4.875 kN/m²
  • Total working load ww = 4.875 + 2.5 + 4 = 11.375 kN/m²
  • Factored load wu=1.5×11.375=17.062w_u = 1.5 \times 11.375 = 17.062 kN/m²

Bending moments

MomentCoefficientValue (kN·m/m)
MxM_x negative (continuous long edge)0.051619.19
MxM_x positive (mid-span, short)0.039614.72
MyM_y negative (continuous short edge)0.032011.91
MyM_y positive (mid-span, long)0.02408.93

M=α wu lx2M = \alpha\, w_u\, l_x^2 with wulx2=17.062×4.6702=372.11w_u l_x^2 = 17.062 \times 4.670^2 = 372.11 kN·m/m. Coefficients are interpolated for ly/lx=1.43l_y/l_x = 1.43.

Check for depth

Mu,lim=0.138 fckbd2=0.138×20×1000×1702×10−6=79.76M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 170^2 \times 10^{-6} = 79.76 kN·m/m >> Mmax=19.19M_{max} = 19.19 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 = 234 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)14.722478 mm @ 200 c/c251
Long span, bottom (mid-span)8.932348 mm @ 210 c/c239
Short span, top over continuous long edge19.193268 mm @ 150 c/c335
Long span, top over continuous short edge11.912348 mm @ 210 c/c239

Check for shear

Vu=wulx/2=17.062×4.670/2=39.84V_u = w_u l_x/2 = 17.062 \times 4.670/2 = 39.84 kN/m τv=Vubd=398411000×170=0.234\tau_v = \dfrac{V_u}{bd} = \dfrac{39841}{1000 \times 170} = 0.234 MPa Steel at support: Ast=335A_{st} = 335 mm²/m, pt=0.197%p_t = 0.197\%; τc=0.32\tau_c = 0.32 MPa (IS 456 Table 19, M20). For D=195D = 195 mm, k=1.21k = 1.21 (cl. 40.2.1.1), so kτc=0.387k\tau_c = 0.387 MPa. τv=0.234<kτc=0.387\tau_v = 0.234 < k\tau_c = 0.387 MPa, so the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: pt=100×2511000×170=0.148%p_t = \dfrac{100 \times 251}{1000 \times 170} = 0.148\%, fs=0.58fyAst,reqAst,prov=0.58×415×248251=238f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{248}{251} = 238 N/mm². Modification factor from IS 456 Fig. 4 (for pt=0.15%p_t = 0.15\% and fs=238f_s = 238 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=26×2.0=52.0l/d = 26 \times 2.0 = 52.0; actual lx/d=4670/170=27.5l_x/d = 4670/170 = 27.5. 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 (126 mm²/m) run into the support: M1=7.59M_1 = 7.59 kN·m/m, L0=max⁡(d, 12ϕ)=170L_0 = \max(d,\ 12\phi) = 170 mm.

1.3M1V+L0=1.3×759474439841+170=418 mm1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{7594744}{39841} + 170 = 418\ \text{mm}

Since 418 mm >> Ld=376L_d = 376 mm, development length is satisfied.

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@200 short span      #
  # bot: 8mm@210 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 = 195 mm ------|
   bottom 8mm @ 200 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.
  • Top bars over continuous supports extend 0.3lx0.3 l_x from the face of the support; all bars have 20 mm cover.
  • Distribution/minimum steel ≥\ge 234 mm²/m is already satisfied by the bars provided.

Answer: D=195D = 195 mm; short-span bottom 8 mm @ 200 c/c; long-span bottom 8 mm @ 210 c/c; top steel over continuous edge 8 mm @ 150 c/c (short direction); 8 mm @ 210 c/c (long direction). Shear, deflection and development length are satisfied.

✓ Answer checked by Claude Opus 5.5 (high effort)

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