--- title: "Slab with one short edge discontinuous, 6x4 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-with-one-short-edge-discontinuous" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-with-one-short-edge-discontinuous site: BCT Engineering --- # Slab with one short edge discontinuous, 6x4 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2075 Chaitra, 16 marks - Video: [Design of a Simply Supported Two Way Slab using IS - 456 & SP -16](https://www.youtube.com/watch?v=_Gz2v3zvfMY) — Stan Academy - 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 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a RCC slab for a room of clear dimensions 6 m x 4 m whose one short edge is discontinuous and corners are restrained at supports. The live load on the slab is 4 kN/m$^2$ and superimposed load of 1.20 kN/m$^2$. Adopt M20 grade concrete and Fe415 grade steel. Check the slab for deflection, and development length. Give the detail sketches, sectional view along short span with reinforcement details along with torsional reinforcements. ## Short answer Slab thickness 175 mm, effective depth 150 mm (short) and 140 mm (long). Short bottom: 8 mm @ 235 c/c; long bottom: 8 mm @ 235 c/c; top over continuous long edge: 8 mm @ 185 c/c; top over continuous short edge: 8 mm @ 235 c/c; discontinuous edge top: 8 mm @ 300 c/c. Shear, deflection and development length are safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 6$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 4 kN/m², floor finish = 1.2 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 $d$. - Support width not given; 230 mm assumed. Superimposed load 1.2 kN/m² is taken as floor finish. - 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. $$ l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 6000 + 150 = 6150\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{6150}{4150} = 1.482 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.175 = 4.375 kN/m² - Total working load $w$ = 4.375 + 1.2 + 4 = 9.575 kN/m² - Factored load $w_u = 1.5 \times 9.575 = 14.362$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0566 | 14.01 | | $M_x$ positive (mid-span, short) | 0.0435 | 10.75 | | $M_y$ negative (continuous short edge) | 0.0370 | 9.15 | | $M_y$ positive (mid-span, long) | 0.0280 | 6.93 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 14.362 \times 4.150^2 = 247.36$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.48$. #### 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 $>$ $M_{max} = 14.01$ 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) | 10.75 | 210 | 8 mm @ 235 c/c | 214 | | Long span, bottom (mid-span) | 6.93 | 210 | 8 mm @ 235 c/c | 214 | | Short span, top over continuous long edge | 14.01 | 269 | 8 mm @ 185 c/c | 272 | | Long span, top over continuous short edge | 9.15 | 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 $V_u = w_u l_x/2 = 14.362 \times 4.150/2 = 29.80$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{29802}{1000 \times 150} = 0.199$ MPa Steel at support: $A_{st} = 272$ mm²/m, $p_t = 0.181\%$; $\tau_c = 0.31$ 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. $\tau_v = 0.199 < k\tau_c = 0.388$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{204}{214} = 229$ N/mm² (here $A_{st,req} = 204$ 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 $f_s = 229$ 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. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{29802} + 150 = 399\ \text{mm} $$ Since 399 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 @ 185 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 discontinuity is assumed in this panel? A. One short edge discontinuous, other three edges continuous B. One long edge discontinuous C. All four edges discontinuous D. Two adjacent edges discontinuous Answer: A. The problem states one short edge discontinuous, so that case of IS 456 Table 26 is used. (Key point: Panel case) 2. What top steel runs along the discontinuous short edge? A. None B. 50% of mid-span long-span steel, extending 0.1 l into the span C. Full mid-span steel D. Torsion mesh only Answer: B. Partial fixity at the discontinuous edge produces negative moment, handled by half the mid-span steel. (Key point: Top steel at discontinuous edge) 3. How is the development length checked at supports? A. Ld must exceed 1.3M1/V + L0 B. Ld is taken as 12 phi C. Ld must not exceed 1.3M1/V + L0, here 399 mm > 376 mm D. It is ignored for slabs Answer: C. Bars must satisfy Ld <= 1.3M1/V + L0; the available 399 mm exceeds the 376 mm required. (Key point: Development length) 4. Where is torsion reinforcement provided in this slab? A. At all four corners with full strength B. At the two continuous-edge corners C. Nowhere in the slab D. At the two corners meeting the discontinuous edge, half-strength mesh Answer: D. Only corners where a discontinuous edge meets a continuous one need (half) torsion steel; none where both edges are continuous. 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