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Slab with two adjacent edges discontinuous, 4x5 m

Short answer

Adopt total depth 175 mm and effective depth 150 mm; factored load is 14.06 kN/m2 giving maximum moment 15.03 kNm/m. Provide bottom 8 mm at 230 c/c short span and 235 c/c long span, top 8 mm at 170 c/c over continuous long edge, 215 c/c over continuous short edge, and 300 c/c at discontinuous edges. Shear 0.195 MPa is below 0.40 MPa and deflection is safe.

Design a reinforced concrete rectangular slab of size 4.0 m x 5.0 m to support an imposed load of 4 kN/m2^2 and floor finish of 1 kN/m2^2. The slab has two adjacent edges discontinuous with the slab resting on 275 mm wide beam. Check the safety of slab against shear and deflection. Use M20 grade concrete and Fe415 grade steel. (Design of torsional reinforcements is not required)

Data and assumptions

  • Clear spans: lx=4l_x = 4 m (short), ly=5l_y = 5 m (long); support width 275 mm; fck=20f_{ck} = 20 MPa, fy=415f_y = 415 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.
  • Panel type: two adjacent edges discontinuous; moment coefficients from IS 456 Annex D, Table 26 (case: two adjacent edges discontinuous).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). 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=5000+150=5150 mml_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 5000 + 150 = 5150\ \text{mm} lylx=51504150=1.241<2⇒two-way slab\frac{l_y}{l_x} = \frac{5150}{4150} = 1.241 < 2 \Rightarrow \text{two-way slab}

Loads (per m²)

  • Self-weight = 25 × 0.175 = 4.375 kN/m²
  • Total working load ww = 4.375 + 1 + 4 = 9.375 kN/m²
  • Factored load wu=1.5×9.375=14.062w_u = 1.5 \times 9.375 = 14.062 kN/m²

Bending moments

MomentCoefficientValue (kN·m/m)
MxM_x negative (continuous long edge)0.062015.03
MxM_x positive (mid-span, short)0.046611.30
MyM_y negative (continuous short edge)0.047011.38
MyM_y positive (mid-span, long)0.03508.48

M=α wu lx2M = \alpha\, w_u\, l_x^2 with wulx2=14.062×4.1502=242.19w_u l_x^2 = 14.062 \times 4.150^2 = 242.19 kN·m/m. Coefficients are interpolated for ly/lx=1.24l_y/l_x = 1.24.

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=15.03M_{max} = 15.03 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 in both directions.

LocationMuM_u (kN·m/m)AstA_{st} req (mm²/m)ProvideAstA_{st} prov
Short span, bottom (mid-span)11.302158 mm @ 230 c/c219
Long span, bottom (mid-span)8.48172 → 210 (min)8 mm @ 235 c/c214
Short span, top over continuous long edge15.032898 mm @ 170 c/c296
Long span, top over continuous short edge11.382338 mm @ 215 c/c234

Top steel at the discontinuous edges (for the negative moment due to partial fixity, IS 456 Annex D) extends 0.1l0.1 l into the span. Along the discontinuous long edge (short-span direction): 50% of mid-span steel = 109 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge (long-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

Vu=wulx/2=14.062×4.150/2=29.18V_u = w_u l_x/2 = 14.062 \times 4.150/2 = 29.18 kN/m

τv=Vubd=291801000×150=0.195\tau_v = \dfrac{V_u}{bd} = \dfrac{29180}{1000 \times 150} = 0.195 MPa

Steel at the support: Ast=296A_{st} = 296 mm²/m, pt=0.197%p_t = 0.197\%, so τc=0.32\tau_c = 0.32 MPa (IS 456 Table 19, M20). For D=175D = 175 mm, k=1.25k = 1.25 (cl. 40.2.1.1), so kτc=0.40k\tau_c = 0.40 MPa.

τv=0.195<kτc=0.40\tau_v = 0.195 < k\tau_c = 0.40 MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: pt=100×2191000×150=0.146%p_t = \dfrac{100 \times 219}{1000 \times 150} = 0.146\%, fs=0.58fyAst,reqAst,prov=0.58×415×215219=236f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{215}{219} = 236 N/mm².

Modification factor from IS 456 Fig. 4 (for pt=0.15%p_t = 0.15\% and fs=236f_s = 236 N/mm²): at such a low steel percentage the curve reaches its upper limit, so kt=2.0k_t = 2.0.

Permissible l/d=26×2.0=52.0l/d = 26 \times 2.0 = 52.0; 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=0.87×415×84×1.92=376L_d = \dfrac{0.87 \times 415 \times 8}{4 \times 1.92} = 376 mm

The check Ld≤1.3M1V+L0L_d \le 1.3\dfrac{M_1}{V} + L_0 (IS 456 cl. 26.2.3.3(c)) is made at the discontinuous edges. L0L_0 = straight length beyond the centre of the support =275/2−25= 275/2 - 25 (end cover) =112.5= 112.5 mm, plus the anchorage value of any bend.

  • Short-span bars at the discontinuous long edge (alternate bars bent up, 109 mm²/m, d=150d = 150 mm): M1=0.87fyAstd(1−Astfybdfck)=5.84M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 5.84 kN·m/m, 1.3M1V=1.3×5.84×10629180=2601.3\dfrac{M_1}{V} = 1.3 \times \dfrac{5.84 \times 10^6}{29180} = 260 mm. Straight: 260+112.5=373260 + 112.5 = 373 mm, just below 376 mm; with a 90° bend (8ϕ=648\phi = 64 mm): 437437 mm >376> 376 mm.
  • Long-span bars at the discontinuous short edge (107 mm²/m, d=140d = 140 mm, same VV taken conservatively): M1=5.32M_1 = 5.32 kN·m/m, 1.3M1/V=2371.3M_1/V = 237 mm; with a 90° bend: 237+112.5+64=414237 + 112.5 + 64 = 414 mm >376> 376 mm.

With standard 90° bends at the discontinuous edges, the development length is satisfied.

Torsion reinforcement at corners

(Not required by the question; given for completeness.) Corner size lx/5=4150/5=830l_x/5 = 4150/5 = 830 mm in both directions.

  • Corner with both edges discontinuous (1 corner): top and bottom meshes, each layer 3/4 of the mid-span AstA_{st} = 0.75 × 215 = 161 mm²/m. Provide 8 mm @ 235 c/c in both directions, top and bottom, over 830 mm × 830 mm.
  • Corners with one edge discontinuous (2 corners): half of this = 0.375 × 215 = 81 mm²/m. Provide 8 mm @ 300 c/c in each direction, top and bottom.
  • No torsion steel is needed at the corner where both edges are continuous (IS 456 D-1.10).

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                             t|
  # bot: 8mm@230 short span      |
  # bot: 8mm@235 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 over      top bars at
  continuous edge    discontinuous edge
  8mm@170            8mm@300
   ~~~~~~~            ~~~
  |------ slab D = 175 mm ------|
   bottom 8mm @ 230 c/c, alt. bars bent up,
   90 deg bend at the discontinuous edge
  [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 210 mm²/m is already satisfied by the bars provided.

Answer: D=175D = 175 mm; short-span bottom 8 mm @ 230 c/c; long-span bottom 8 mm @ 235 c/c; top steel over the continuous long edge 8 mm @ 170 c/c (short direction) and over the continuous short edge 8 mm @ 215 c/c (long direction); 8 mm @ 300 c/c top bars at the discontinuous edges. The slab is safe in shear and deflection; development length is satisfied with 90° bends at the discontinuous edges.

✓ Answer checked by Claude Opus 5.5 (high effort)

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