--- title: "Slab prevented from uplifting, 3.6x4.2 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-prevented-from-uplifting-3-6x4" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-prevented-from-uplifting-3-6x4 site: BCT Engineering --- # Slab prevented from uplifting, 3.6x4.2 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2072 Kartik, 16 marks - Video: [Design of a Two Way Slab - Corners Held Down / Corners Prevented from Lifting / Restrained Slab](https://www.youtube.com/watch?v=QIdl-6NaKlw) — 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-prevented-from-uplifting-3-6x4 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a slab for a room of size 3.6 m x 4.2 m prevented uplifting by walls (230 mm thick) loads for a intermediate storey of a residential building. Use M20 grade of concrete and Fe 415 grade of steel. Sketch the reinforcements. Carry out all necessary checks require in slab design. Take live load = 3 kN/m$^2$, floor finish = 1 kN/m$^2$. ## Short answer Provide 200 mm slab, d=175 mm, lx=3.775 m c/c, ly=4.375 m, ly/lx=1.16, four edges discontinuous. Factored load 13.5 kN/m². Bottom: short span 8@205, long span 8@205. Top at all edges 8@300 (50% of positive). Shear, deflection, and development length checks are satisfied. ## Answer #### Data and assumptions - Clear spans: $l_x = 3.6$ m (short), $l_y = 4.2$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 3 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 $d$. - Panel type: four edges discontinuous (corners held down); moment coefficients from IS 456 Annex D, Table 26. #### Depth and effective span Basic span/depth ratio = 20 (IS 456 cl. 23.2.1). Trial overall depth $D = 200$ mm, so $d = 200 - 20 - 5 = 175$ mm for the short-span bars and $d_y = 175 - 10 = 165$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 3600 + 175 = 3775\ \text{mm},\quad l_y = 4200 + 175 = 4375\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{4375}{3775} = 1.159 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.200 = 5.000 kN/m² - Total working load $w$ = 5.000 + 1 + 3 = 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) | |---|---|---| | $M_x$ positive (mid-span, short) | 0.0687 | 13.22 | | $M_y$ positive (mid-span, long) | 0.0560 | 10.77 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 13.500 \times 3.775^2 = 192.38$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.16$. #### Check for depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 175^2 \times 10^{-6} = 84.53$ kN·m/m $>$ $M_{max} = 13.22$ 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 $bD$ = 240 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) | 13.22 | 240 | 8 mm @ 205 c/c | 245 | | Long span, bottom (mid-span) | 10.77 | 240 | 8 mm @ 205 c/c | 245 | 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 long edge(s) (short-span direction): 50% of mid-span steel = 123 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 123 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 13.500 \times 3.775/2 = 25.48$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{25481}{1000 \times 175} = 0.146$ MPa Steel at support: $A_{st} = 123$ mm²/m, $p_t = 0.070\%$; $\tau_c = 0.28$ MPa (IS 456 Table 19, M20). For $D = 200$ mm, $k = 1.20$ (cl. 40.2.1.1), so $k\tau_c = 0.336$ MPa. $\tau_v = 0.146 < k\tau_c = 0.336$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 245}{1000 \times 175} = 0.140\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{215}{245} = 211$ N/mm² (here $A_{st,req} = 215$ 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 $f_s = 211$ N/mm²): $k_t = 2.0$ (at such a low steel percentage the curve reaches its upper limit of 2.0). Permissible $l/d = 20 \times 2.0 = 40.0$; actual $l_x/d = 3775/175 = 21.6$. 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 (123 mm²/m) run into the support: $M_1 = 7.63$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 175$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{7633643}{25481} + 175 = 564\ \text{mm} $$ Since 564 mm $>$ $L_d = 376$ mm, **development length is satisfied**. #### Torsion reinforcement at corners Corner size $l_x/5 = 3775/5 = 755$ mm (say 760 mm) in both directions. - Corner with both edges discontinuous (4 corners): top and bottom meshes, each layer 3/4 of the mid-span $A_{st}$ = 0.75 × 240 = 180 mm²/m. Provide 8 mm @ 205 c/c in both directions, top and bottom, over 760 mm × 760 mm. - 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 T| | bot: 8mm@205 short span | | bot: 8mm@205 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 at edges (0.5 x mid-span steel) ~~~ ~~~ |------ slab D = 200 mm ------| bottom 8mm @ 205 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. - All bars have 20 mm cover. - Distribution/minimum steel $\ge$ 240 mm²/m is already satisfied by the bars provided. **Answer:** $D = 200$ mm; short-span bottom 8 mm @ 205 c/c; long-span bottom 8 mm @ 205 c/c. 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. Why is basic span/depth taken as 20 rather than 26? A. Because the concrete is M20 B. Because the slab is simply supported (all four edges discontinuous) C. Because it is a cantilever D. Because the span exceeds 10 m Answer: B. IS 456 gives 20 for simply supported spans and 26 for continuous spans. (Key point: Basic l/d for this slab) 2. Which moments are calculated for a panel with four discontinuous edges and corners held down? A. Only positive moments, Mx and My, at mid-span B. Negative moments at all edges C. Only the moment in the long direction D. Only torsion moments Answer: A. No edges are continuous, so Table 26 gives only mid-span sagging coefficients, 0.0687 and 0.0560 here. (Key point: Moments for corners held down) 3. Because the corners are prevented from lifting, what is provided at all four corners? A. Nothing B. Top and bottom meshes of 3/4 mid-span Ast over lx/5 x lx/5 C. Stirrups of 8 mm at 100 c/c D. Extra bottom bars only in the short direction Answer: B. Both edges at each corner are discontinuous, so the full torsion mesh (3/4 of mid-span steel in each layer) is placed at the four corners. (Key point: Torsion steel) 4. Ld = 0.87 fy phi / (4 tau_bd) = 376 mm. The bar anchorage check uses 1.3 M1/V + L0 = 564 mm. Why is it satisfied? A. Because 564 mm is greater than Ld B. Because 564 mm is less than the slab depth C. Because L0 is zero D. Because tau_bd is 1.92 MPa Answer: A. The available anchorage must be at least Ld; 564 mm exceeds 376 mm so the bars are adequately anchored. 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