--- title: "Slab with one short edge discontinuous, 4x5 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-with-one-short-edge-discontinuous-2" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-with-one-short-edge-discontinuous-2 site: BCT Engineering --- # Slab with one short edge discontinuous, 4x5 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2074 Asoj, 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-with-one-short-edge-discontinuous-2 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a slab panel having one short edge discontinuous for a room size of 4 m x 5 m. The edges of slab is supported on walls of width 250 mm. The slab is carrying a live load of 4 kN/m$^2$ and floor finish of 0.75 kN/m$^2$. Use M20 Concrete and Fe415 steel. Sketch the reinforcement detailing in plan and sections. Check for deflection and development length are necessary. ## Short answer Slab depth 170 mm, effective depth 145 mm (short) and 135 mm (long). Bottom: short 8 mm @ 245 c/c, long 8 mm @ 245 c/c. Top over continuous long edge: 8 mm @ 220 c/c; over continuous short edge: 8 mm @ 245 c/c. Discontinuous short edge top: 8 mm @ 300 c/c. Deflection, shear, and development length are adequate. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 250 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 4 kN/m², floor finish = 0.75 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 $d$. - 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 $D = 170$ mm, so $d = 170 - 20 - 5 = 145$ mm for the short-span bars and $d_y = 145 - 10 = 135$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 4000 + 145 = 4145\ \text{mm},\quad l_y = 5000 + 145 = 5145\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{5145}{4145} = 1.241 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.170 = 4.250 kN/m² - Total working load $w$ = 4.250 + 0.75 + 4 = 9.000 kN/m² - Factored load $w_u = 1.5 \times 9.000 = 13.500$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0492 | 11.42 | | $M_x$ positive (mid-span, short) | 0.0372 | 8.64 | | $M_y$ negative (continuous short edge) | 0.0370 | 8.58 | | $M_y$ positive (mid-span, long) | 0.0280 | 6.49 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 13.500 \times 4.145^2 = 231.94$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$. #### Check for depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 145^2 \times 10^{-6} = 58.03$ kN·m/m $>$ $M_{max} = 11.42$ 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 $bD$ = 204 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) | 8.64 | 204 | 8 mm @ 245 c/c | 205 | | Long span, bottom (mid-span) | 6.49 | 204 | 8 mm @ 245 c/c | 205 | | Short span, top over continuous long edge | 11.42 | 226 | 8 mm @ 220 c/c | 228 | | Long span, top over continuous short edge | 8.58 | 204 | 8 mm @ 245 c/c | 205 | 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 = 103 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 13.500 \times 4.145/2 = 27.98$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{27979}{1000 \times 145} = 0.193$ MPa Steel at support: $A_{st} = 228$ mm²/m, $p_t = 0.158\%$; $\tau_c = 0.29$ MPa (IS 456 Table 19, M20). For $D = 170$ mm, $k = 1.26$ (cl. 40.2.1.1), so $k\tau_c = 0.365$ MPa. $\tau_v = 0.193 < k\tau_c = 0.365$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 205}{1000 \times 145} = 0.141\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{169}{205} = 198$ N/mm² (here $A_{st,req} = 169$ 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 $f_s = 198$ 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 $l_x/d = 4145/145 = 28.6$. 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 (103 mm²/m) run into the support: $M_1 = 5.29$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 145$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5291593}{27979} + 145 = 391\ \text{mm} $$ Since 391 mm $>$ $L_d = 376$ mm, **development length is satisfied**. #### Torsion reinforcement at corners Corner size $l_x/5 = 4145/5 = 829$ 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 × 204 = 76 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@245 short span | # bot: 8mm@245 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 = 170 mm ------| bottom 8mm @ 245 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 $\ge$ 204 mm²/m is already satisfied by the bars provided. **Answer:** $D = 170$ mm; short-span bottom 8 mm @ 245 c/c; long-span bottom 8 mm @ 245 c/c; top steel over continuous edge 8 mm @ 220 c/c (short direction); 8 mm @ 245 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. With clear span 4000 mm and d = 145 mm, how is the effective short span obtained? A. Clear span + d = 4145 mm (not more than c/c) B. Clear span only = 4000 mm C. Clear span + support width = 4250 mm always D. Clear span - d = 3855 mm Answer: A. The effective span is the smaller of clear span + d and centre-to-centre distance. (Key point: Effective span) 2. Why are moment coefficients taken from the 'one short edge discontinuous' case of Table 26? A. Because the short span is 4 m B. Because the panel has three continuous edges and one discontinuous short edge C. Because the slab has one long edge discontinuous D. Because the panel is simply supported Answer: B. Table 26 coefficients depend on edge conditions, and this panel is continuous on three edges and discontinuous on one short edge. (Key point: Panel type and coefficients) 3. Calculated Ast for several locations was below 204 mm^2/m. What is provided there? A. Less steel, as calculation allows it B. Minimum steel 0.12% of bD = 204 mm^2/m C. Minimum steel 0.85 bd / fy D. Steel for 1% reinforcement Answer: B. For Fe415 slabs the minimum steel is 0.12% of the gross section (bD), so 8 mm at 245 c/c giving 205 mm^2/m is provided. (Key point: Minimum steel governs) 4. Basic l/d is 26 and modification factor kt = 1.13, so permissible l/d = 29.4. Actual lx/d = 28.6. What is the conclusion? A. Fails, depth must be increased B. Fails, steel must be increased C. Satisfied, since actual l/d is less than permissible D. Cannot be decided without shear check Answer: C. Deflection control is met when actual span/depth does not exceed basic ratio x modification factor. 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