Skip to main content

Two-adjacent-edges discontinuous slab, 3.5x4.5 m

Short answer

Two adjacent edges discontinuous, D=155 mm, d=130 mm. Factored load 14.212 kN/m². Provide 8 mm bars: short span bottom @250, long bottom @270, short top @185, long top @235, and 8 mm @300 at discontinuous edges. Shear and deflection safe.

Design a two adjacent sides (edges) discontinuous reinforced concrete slab for room having clear dimensions of 3.5 m x 4.5 m. The slab rest on 250 mm wide beam. Consider 25 mm thick PCC floor finish and live load on slab as 4.0 kN/m2^2 and partition wall load on slab as 1.0 kN/m2^2. Use M20 concrete and Fe415 steel. Check also the slab safe in shear deflection or not. Show the reinforcement and arrangement in plan and section (along short span only). (Design of torsional reinforcements in slab not required).

Data and assumptions

  • Clear spans: lx=3.5l_x = 3.5 m (short), ly=4.5l_y = 4.5 m (long); support width 250 mm; fck=20f_{ck} = 20 MPa, fy=415f_y = 415 MPa.
  • Live load = 4 kN/m², floor finish = 0.6 kN/m², partition = 1.0 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.
  • 25 mm PCC floor finish: 0.025 × 24 kN/m³ = 0.6 kN/m².
  • 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=155D = 155 mm, so d=155−20−5=130d = 155 - 20 - 5 = 130 mm for the short-span bars and dy=130−10=120d_y = 130 - 10 = 120 mm for the long-span bars.

lx=min⁡(lclear+d, c/c)=3500+130=3630 mm,ly=4500+130=4630 mml_x = \min(l_{clear} + d,\ c/c) = 3500 + 130 = 3630\ \text{mm},\quad l_y = 4500 + 130 = 4630\ \text{mm} lylx=46303630=1.275<2⇒two-way slab\frac{l_y}{l_x} = \frac{4630}{3630} = 1.275 < 2 \Rightarrow \text{two-way slab}

Loads (per m²)

  • Self-weight = 25 × 0.155 = 3.875 kN/m²
  • Total working load ww = 3.875 + 0.6 + 4 + 1.0 = 9.475 kN/m²
  • Factored load wu=1.5×9.475=14.212w_u = 1.5 \times 9.475 = 14.212 kN/m²

Bending moments

MomentCoefficientValue (kN·m/m)
MxM_x negative (continuous long edge)0.063811.94
MxM_x positive (mid-span, short)0.04808.99
MyM_y negative (continuous short edge)0.04708.80
MyM_y positive (mid-span, long)0.03506.55

M=α wu lx2M = \alpha\, w_u\, l_x^2 with wulx2=14.212×3.6302=187.28w_u l_x^2 = 14.212 \times 3.630^2 = 187.28 kN·m/m. Coefficients are interpolated for ly/lx=1.28l_y/l_x = 1.28.

Check of depth

Mu,lim=0.138 fckbd2=0.138×20×1000×1302×10−6=46.64M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 130^2 \times 10^{-6} = 46.64 kN·m/m >> Mmax=11.94M_{max} = 11.94 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 = 186 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)8.991988 mm @ 250 c/c201
Long span, bottom (mid-span)6.55156 → 186 (min)8 mm @ 270 c/c186
Short span, top over continuous long edge11.942668 mm @ 185 c/c272
Long span, top over continuous short edge8.802118 mm @ 235 c/c214

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 = 101 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge (long-span direction): 50% of mid-span steel = 93 mm²/m, provide 8 mm @ 300 c/c.

Check for shear

Vu=wulx/2=14.212×3.630/2=25.80V_u = w_u l_x/2 = 14.212 \times 3.630/2 = 25.80 kN/m

τv=Vubd=257961000×130=0.198\tau_v = \dfrac{V_u}{bd} = \dfrac{25796}{1000 \times 130} = 0.198 MPa

Steel at the support: Ast=272A_{st} = 272 mm²/m, pt=0.209%p_t = 0.209\%, so τc=0.33\tau_c = 0.33 MPa (IS 456 Table 19, M20). For D=155D = 155 mm, k=1.29k = 1.29 (cl. 40.2.1.1), so kτc=0.42k\tau_c = 0.42 MPa.

τv=0.198<kτc=0.42\tau_v = 0.198 < k\tau_c = 0.42 MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: pt=100×2011000×130=0.155%p_t = \dfrac{100 \times 201}{1000 \times 130} = 0.155\%, fs=0.58fyAst,reqAst,prov=0.58×415×198201=237f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{198}{201} = 237 N/mm².

Modification factor from IS 456 Fig. 4 (for pt=0.15%p_t = 0.15\% and fs=237f_s = 237 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=3630/130=27.9l_x/d = 3630/130 = 27.9. Since actual < permissible, the slab is safe in deflection.

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 =250/2−25= 250/2 - 25 (end cover) =100= 100 mm, plus the anchorage value of any hook or bend.

  • Short-span bars at the discontinuous long edge (alternate bars bent up, 101 mm²/m, d=130d = 130 mm): M1=0.87fyAstd(1−Astfybdfck)=4.64M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 4.64 kN·m/m, 1.3M1V=1.3×4.64×10625796=2341.3\dfrac{M_1}{V} = 1.3 \times \dfrac{4.64 \times 10^6}{25796} = 234 mm. Straight: 234+100=334234 + 100 = 334 mm <376< 376 mm; with a 90° bend (8ϕ=648\phi = 64 mm): 398398 mm >376> 376 mm.
  • Long-span bars at the discontinuous short edge (93 mm²/m, d=120d = 120 mm, same VV taken conservatively): M1=3.96M_1 = 3.96 kN·m/m, 1.3M1/V=2001.3M_1/V = 200 mm; with a 90° bend: 364364 mm <376< 376 mm; with a standard U-hook (16ϕ=12816\phi = 128 mm): 200+100+128=428200 + 100 + 128 = 428 mm >376> 376 mm.

So the short-span bars are given 90° bends and the long-span bars standard hooks at the discontinuous edges; the development length is then satisfied.

Torsion reinforcement at corners

(Not required by the question; given for completeness.) Corner size lx/5=3630/5=726l_x/5 = 3630/5 = 726 mm (say 730 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 × 198 = 148 mm²/m. Provide 8 mm @ 270 c/c in both directions, top and bottom, over 730 mm × 730 mm.
  • Corners with one edge discontinuous (2 corners): half of this = 0.375 × 198 = 74 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@250 short span      |
  # bot: 8mm@270 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@185            8mm@300
   ~~~~~~~            ~~~
  |------ slab D = 155 mm ------|
   bottom 8mm @ 250 c/c, alt. bars bent up,
   90 deg bend at the discontinuous edge
  [beam 250]          [beam 250]
  • 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 186 mm²/m is already satisfied by the bars provided.

Answer: D=155D = 155 mm; short-span bottom 8 mm @ 250 c/c; long-span bottom 8 mm @ 270 c/c; top steel over the continuous long edge 8 mm @ 185 c/c (short direction) and over the continuous short edge 8 mm @ 235 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 bends/hooks at the discontinuous edges.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Design of RCC Structure chapter 9 questions →