--- title: "Slab on masonry walls, 5x6 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-on-masonry-walls-5x6-m" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-on-masonry-walls-5x6-m site: BCT Engineering --- # Slab on masonry walls, 5x6 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2076 Chaitra, 14 marks - Video: [How to Design Reinforced Concrete Slabs (Part 1)](https://www.youtube.com/watch?v=SH9Y7gilklk) — Light Bearers - 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-on-masonry-walls-5x6-m - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a RC slab over a room 5 m x 6 m. The slab is supported on masonry walls all round with adequate restraint and corners are held down. The live load on slab is 3 kN/m$^2$ and floor finish 1.5 kN/m$^2$. The thickness of supporting wall is 230 mm. Use M20 concrete mix and Fe415 grade steel. Also draw the top and bottom reinforcement detailing with their section and plan. Check for deflection and development length is necessary. ## Short answer Four edges discontinuous with corners held down, D=265 mm, d=240 mm. Factored load 16.688 kN/m². Provide 8 mm bars: short span bottom @125, long bottom @155, top at edges @245 (short) and @300 (long). Shear and deflection safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 5$ m (short), $l_y = 6$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 3 kN/m², floor finish = 1.5 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 $D = 265$ mm, so $d = 265 - 20 - 5 = 240$ mm for the short-span bars and $d_y = 240 - 10 = 230$ mm for the long-span bars. $$ l_x = \min(5000 + 240,\ 5000 + 230) = 5230\ \text{mm (c/c governs)},\quad l_y = \min(6000 + 240,\ 6000 + 230) = 6230\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{6230}{5230} = 1.191 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.265 = 6.625 kN/m² - Total working load $w$ = 6.625 + 1.5 + 3 = 11.125 kN/m² - Factored load $w_u = 1.5 \times 11.125 = 16.688$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ positive (mid-span, short) | 0.0713 | 32.54 | | $M_y$ positive (mid-span, long) | 0.0560 | 25.56 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 16.688 \times 5.230^2 = 456.45$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.19$. #### Check for depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 240^2 \times 10^{-6} = 158.98$ kN·m/m $>$ $M_{max} = 32.54$ 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 $bD$ = 318 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) | 32.54 | 389 | 8 mm @ 125 c/c | 402 | | Long span, bottom (mid-span) | 25.56 | 318 | 8 mm @ 155 c/c | 324 | 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 = 201 mm²/m, provide 8 mm @ 245 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 162 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 16.688 \times 5.230/2 = 43.64$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{43638}{1000 \times 240} = 0.182$ MPa Steel at support: $A_{st} = 201$ mm²/m, $p_t = 0.084\%$; $\tau_c = 0.28$ MPa (IS 456 Table 19, M20). For $D = 265$ mm, $k = 1.07$ (cl. 40.2.1.1), so $k\tau_c = 0.300$ MPa. $\tau_v = 0.182 < k\tau_c = 0.300$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 402}{1000 \times 240} = 0.168\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{389}{402} = 233$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.17\%$ and $f_s = 233$ 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 $l_x/d = 5230/240 = 21.8$. 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 (201 mm²/m) run into the support: $M_1 = 17.12$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 240$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{17119556}{43638} + 240 = 750\ \text{mm} $$ Since 750 mm $>$ $L_d = 376$ mm, **development length is satisfied**. #### Torsion reinforcement at corners Corner size $l_x/5 = 5230/5 = 1046$ mm (say 1050 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 × 389 = 292 mm²/m. Provide 8 mm @ 155 c/c in both directions, top and bottom, over 1050 mm × 1050 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@125 short span | | bot: 8mm@155 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 = 265 mm ------| bottom 8mm @ 125 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 $\ge$ 318 mm²/m is already satisfied by the bars provided. **Answer:** $D = 265$ mm; short-span bottom 8 mm @ 125 c/c; long-span bottom 8 mm @ 155 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. Which moment coefficient case applies to this slab? A. Four edges discontinuous with corners held down B. Interior panel with all edges continuous C. One long edge discontinuous D. Two adjacent edges discontinuous Answer: A. The slab rests on masonry walls all round with corners held down, so Table 26 case for four discontinuous edges is used, giving only positive moments. (Key point: Support condition) 2. Why is the basic span/depth ratio taken as 20 here? A. The slab is continuous B. The slab is simply supported-type (edges discontinuous) C. It is a cantilever D. Because Fe415 steel is used Answer: B. IS 456 gives 20 for simply supported spans and 26 for continuous ones; this panel is not continuous. (Key point: Basic l/d ratio) 3. Why is the development length satisfied? A. Because 750 mm is less than 376 mm B. Because bars have hooks of 500 mm C. Because the cover is 20 mm D. 1.3M1/V + L0 = 750 mm exceeds Ld = 376 mm Answer: D. The check Ld <= 1.3M1/V + L0 passes since the available length is greater than the required Ld. 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