--- title: "Slab three edges continuous, 5.5x4 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-three-edges-continuous-5-5x4" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-three-edges-continuous-5-5x4 site: BCT Engineering --- # Slab three edges continuous, 5.5x4 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2072 Chaitra, 15 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-three-edges-continuous-5-5x4 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question A rectangular slab panel 5.5 m x 4.0 m (clear span) is continuous over three edges and discontinuous over one short edge. The slab is to rest on 250 mm wide beam. The slab is subjected to live load of 5 kN/m$^2$ and floor finishes load of 1.0 kN/m$^2$. Design the slab. Sketch the arrangement of reinforcement bars at support and mid span separately with torsional re-bars. Check whether the section satisfies the deflection criteria. (Check for shear and development length not necessary) ## Short answer Provide 175 mm slab, d=150 mm, lx=4.15 m c/c, ly=5.65 m, ly/lx=1.36. Factored load 15.56 kN/m². Bottom bars: short 8@235, long 8@235; top over continuous edges: short 8@180, long 8@235; discontinuous short edge top 8@300 (50%). Deflection check: actual l/d=27.7, permissible=52, safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 5.5$ m (long); support width 250 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 5 kN/m², floor finish = 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 $d$. - Panel type: one short edge discontinuous; 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 $D = 175$ mm, so $d = 175 - 20 - 5 = 150$ mm for the short-span bars and $d_y = 150 - 10 = 140$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 5500 + 150 = 5650\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{5650}{4150} = 1.361 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.175 = 4.375 kN/m² - Total working load $w$ = 4.375 + 1.0 + 5 = 10.375 kN/m² - Factored load $w_u = 1.5 \times 10.375 = 15.562$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0535 | 14.33 | | $M_x$ positive (mid-span, short) | 0.0402 | 10.78 | | $M_y$ negative (continuous short edge) | 0.0370 | 9.92 | | $M_y$ positive (mid-span, long) | 0.0280 | 7.50 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 15.562 \times 4.150^2 = 268.03$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.36$. #### Check for depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10$ kN·m/m $>$ $M_{max} = 14.33$ kN·m/m. Section is under-reinforced; $d$ is adequate for flexure. #### Reinforcement $$ A_{st} = \frac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \frac{4.6 M_u}{f_{ck} b d^2}}\right] b d $$ Minimum steel = 0.12% of $bD$ = 210 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 300 mm (long). | Location | $M_u$ (kN·m/m) | $A_{st}$ req (mm²/m) | Provide | $A_{st}$ prov | |---|---|---|---|---| | Short span, bottom (mid-span) | 10.78 | 210 | 8 mm @ 235 c/c | 214 | | Long span, bottom (mid-span) | 7.50 | 210 | 8 mm @ 235 c/c | 214 | | Short span, top over continuous long edge | 14.33 | 275 | 8 mm @ 180 c/c | 279 | | Long span, top over continuous short edge | 9.92 | 210 | 8 mm @ 235 c/c | 214 | Top steel at discontinuous edges (negative moment due to partial fixity, IS 456 D-1.6) extends $0.1 l$ into the span: along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 15.562 \times 4.150/2 = 32.29$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{32292}{1000 \times 150} = 0.215$ MPa Steel at support: $A_{st} = 279$ mm²/m, $p_t = 0.186\%$; $\tau_c = 0.31$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.388$ MPa. $\tau_v = 0.215 < k\tau_c = 0.388$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{205}{214} = 231$ N/mm² (here $A_{st,req} = 205$ mm²/m is the steel needed for bending; the minimum steel governs the bars provided). Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 231$ N/mm²): $k_t = 2.0$ (at such a low steel percentage the curve reaches its upper limit of 2.0). Permissible $l/d = 26 \times 2.0 = 52.0$; actual $l_x/d = 4150/150 = 27.7$. Since actual < permissible, **deflection is satisfied**. #### Check for development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.92$ MPa (M20, deformed bars, ×1.6): $L_d = \dfrac{0.87 \times 415 \times 8}{4 \times 1.92} = 376$ mm Alternate bottom bars (107 mm²/m) run into the support: $M_1 = 5.71$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 150$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{32292} + 150 = 380\ \text{mm} $$ Since 380 mm $>$ $L_d = 376$ mm, **development length is satisfied**. #### Torsion reinforcement at corners Corner size $l_x/5 = 4150/5 = 830$ mm (say 830 mm) in both directions. - Corner with only one discontinuous edge (2 corners): half of 3/4 of the mid-span $A_{st}$ = 0.375 × 210 = 79 mm²/m. Provide 8 mm @ 300 c/c in each direction, top and bottom. - No torsion steel is needed at corners where both edges are continuous (IS 456 D-1.10). #### Detailing ```text PLAN (long span horizontal, short span vertical) +==============================+ # t| # bot: 8mm@235 short span | # bot: 8mm@235 long span | # 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 over continuous edge continuous edge ~~~~~~~ ~~~~~~~ |------ slab D = 175 mm ------| bottom 8mm @ 235 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 $0.1 l$ from the support. - Top bars over continuous supports extend $0.3 l_x$ 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:** $D = 175$ mm; short-span bottom 8 mm @ 235 c/c; long-span bottom 8 mm @ 235 c/c; top steel over continuous edge 8 mm @ 180 c/c (short direction); 8 mm @ 235 c/c (long direction). Shear, deflection and development length are satisfied. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which Table 26 case applies to a panel continuous over three edges and discontinuous over one short edge? A. Interior panel B. One short edge discontinuous C. One long edge discontinuous D. Four edges discontinuous Answer: B. The coefficient case is named by the discontinuous edge, here one short edge. (Key point: Panel type) 2. With lx = 4150 mm and ly = 5650 mm, ly/lx = 1.36. How is the slab treated? A. One-way, since ly > lx B. Two-way, since ly/lx < 2 C. Two-way, since ly/lx > 1 D. Flat slab Answer: B. A rectangular panel supported on four sides is two-way when ly/lx < 2. (Key point: Aspect ratio) 3. Short-span bars have pt = 0.143% and fs about 231 N/mm^2, giving kt = 1.07. What is checked? A. Actual lx/d = 27.7 against 26 x 1.07 = 27.8 B. Actual lx/d against 20 x 1.07 C. Actual ly/d against 26 D. Ast against minimum steel Answer: A. The permissible l/d is the basic ratio 26 modified by kt, and actual lx/d must not exceed it. (Key point: Deflection check) 4. Which corners need torsion steel in this panel? A. Two corners where one edge is discontinuous, using half the full corner mesh over lx/5 B. All four corners with full mesh C. No corners at all D. Only corners where both edges are continuous Answer: A. Corners meeting one discontinuous and one continuous edge get half the mesh (0.375 x mid-span Ast); corners with both edges continuous need none. 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