--- title: "Two-adjacent-edges discontinuous slab, 3.5x4.5 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:two-adjacent-edges-discontinuous-slab-3" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/two-adjacent-edges-discontinuous-slab-3 site: BCT Engineering --- # Two-adjacent-edges discontinuous slab, 3.5x4.5 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2078 Bhadra, 16 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/two-adjacent-edges-discontinuous-slab-3 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a two adjacent sides (edges) discontinuous reinforced concrete slab for room having clear dimensions of 3.5 m x 4.5 m. The slab rest on 250 mm wide beam. Consider 25 mm thick PCC floor finish and live load on slab as 4.0 kN/m$^2$ and partition wall load on slab as 1.0 kN/m$^2$. Use M20 concrete and Fe415 steel. Check also the slab safe in shear deflection or not. Show the reinforcement and arrangement in plan and section (along short span only). (Design of torsional reinforcements in slab not required). ## Short answer Two adjacent edges discontinuous, D=155 mm, d=130 mm. Factored load 14.212 kN/m². Provide 8 mm bars: short span bottom @250, long bottom @270, short top @185, long top @235, and 8 mm @300 at discontinuous edges. Shear and deflection safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 3.5$ m (short), $l_y = 4.5$ m (long); support width 250 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 4 kN/m², floor finish = 0.6 kN/m², partition = 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 $d$. - 25 mm PCC floor finish: 0.025 × 24 kN/m³ = 0.6 kN/m². - Panel type: two adjacent edges discontinuous; moment coefficients from IS 456 Annex D, Table 26 (case: two adjacent edges discontinuous). #### Depth and effective span Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 155$ mm, so $d = 155 - 20 - 5 = 130$ mm for the short-span bars and $d_y = 130 - 10 = 120$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 3500 + 130 = 3630\ \text{mm},\quad l_y = 4500 + 130 = 4630\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{4630}{3630} = 1.275 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.155 = 3.875 kN/m² - Total working load $w$ = 3.875 + 0.6 + 4 + 1.0 = 9.475 kN/m² - Factored load $w_u = 1.5 \times 9.475 = 14.212$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0638 | 11.94 | | $M_x$ positive (mid-span, short) | 0.0480 | 8.99 | | $M_y$ negative (continuous short edge) | 0.0470 | 8.80 | | $M_y$ positive (mid-span, long) | 0.0350 | 6.55 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 14.212 \times 3.630^2 = 187.28$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.28$. #### Check of depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 130^2 \times 10^{-6} = 46.64$ kN·m/m $>$ $M_{max} = 11.94$ kN·m/m. The 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 $bD$ = 186 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm in both directions. | Location | $M_u$ (kN·m/m) | $A_{st}$ req (mm²/m) | Provide | $A_{st}$ prov | |---|---|---|---|---| | Short span, bottom (mid-span) | 8.99 | 198 | 8 mm @ 250 c/c | 201 | | Long span, bottom (mid-span) | 6.55 | 156 → 186 (min) | 8 mm @ 270 c/c | 186 | | Short span, top over continuous long edge | 11.94 | 266 | 8 mm @ 185 c/c | 272 | | Long span, top over continuous short edge | 8.80 | 211 | 8 mm @ 235 c/c | 214 | Top steel at the discontinuous edges (for the negative moment due to partial fixity, IS 456 Annex D) extends $0.1 l$ into the span. Along the discontinuous long edge (short-span direction): 50% of mid-span steel = 101 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge (long-span direction): 50% of mid-span steel = 93 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 14.212 \times 3.630/2 = 25.80$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{25796}{1000 \times 130} = 0.198$ MPa Steel at the support: $A_{st} = 272$ mm²/m, $p_t = 0.209\%$, so $\tau_c = 0.33$ MPa (IS 456 Table 19, M20). For $D = 155$ mm, $k = 1.29$ (cl. 40.2.1.1), so $k\tau_c = 0.42$ MPa. $\tau_v = 0.198 < k\tau_c = 0.42$ MPa, hence the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 201}{1000 \times 130} = 0.155\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{198}{201} = 237$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.15\%$ and $f_s = 237$ N/mm²): at such a low steel percentage the curve reaches its upper limit, so $k_t = 2.0$. Permissible $l/d = 26 \times 2.0 = 52.0$; actual $l_x/d = 3630/130 = 27.9$. Since actual < permissible, the slab is **safe in deflection**. #### 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 The check $L_d \le 1.3\dfrac{M_1}{V} + L_0$ (IS 456 cl. 26.2.3.3(c)) is made at the discontinuous edges. $L_0$ = straight length beyond the centre of the support $= 250/2 - 25$ (end cover) $= 100$ mm, plus the anchorage value of any hook or bend. - Short-span bars at the discontinuous long edge (alternate bars bent up, 101 mm²/m, $d = 130$ mm): $M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 4.64$ kN·m/m, $1.3\dfrac{M_1}{V} = 1.3 \times \dfrac{4.64 \times 10^6}{25796} = 234$ mm. Straight: $234 + 100 = 334$ mm $< 376$ mm; with a 90° bend ($8\phi = 64$ mm): $398$ mm $> 376$ mm. - Long-span bars at the discontinuous short edge (93 mm²/m, $d = 120$ mm, same $V$ taken conservatively): $M_1 = 3.96$ kN·m/m, $1.3M_1/V = 200$ mm; with a 90° bend: $364$ mm $< 376$ mm; with a standard U-hook ($16\phi = 128$ mm): $200 + 100 + 128 = 428$ mm $> 376$ mm. So the short-span bars are given 90° bends and the long-span bars standard hooks at the discontinuous edges; the **development length is then satisfied**. #### Torsion reinforcement at corners (Not required by the question; given for completeness.) Corner size $l_x/5 = 3630/5 = 726$ mm (say 730 mm) in both directions. - Corner with both edges discontinuous (1 corner): top and bottom meshes, each layer 3/4 of the mid-span $A_{st}$ = 0.75 × 198 = 148 mm²/m. Provide 8 mm @ 270 c/c in both directions, top and bottom, over 730 mm × 730 mm. - Corners with one edge discontinuous (2 corners): half of this = 0.375 × 198 = 74 mm²/m. Provide 8 mm @ 300 c/c in each direction, top and bottom. - No torsion steel is needed at the corner where both edges are continuous (IS 456 D-1.10). #### Detailing ```text PLAN (long span horizontal, short span vertical) +==============================+ # t| # bot: 8mm@250 short span | # bot: 8mm@270 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 over top bars at continuous edge discontinuous edge 8mm@185 8mm@300 ~~~~~~~ ~~~ |------ slab D = 155 mm ------| bottom 8mm @ 250 c/c, alt. bars bent up, 90 deg bend at the discontinuous edge [beam 250] [beam 250] ``` - 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 $\ge$ 186 mm²/m is already satisfied by the bars provided. **Answer:** $D = 155$ mm; short-span bottom 8 mm @ 250 c/c; long-span bottom 8 mm @ 270 c/c; top steel over the continuous long edge 8 mm @ 185 c/c (short direction) and over the continuous short edge 8 mm @ 235 c/c (long direction); 8 mm @ 300 c/c top bars at the discontinuous edges. The slab is safe in shear and deflection; development length is satisfied with bends/hooks at the discontinuous edges. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 25 mm PCC floor finish converted to a load? A. 0.025 m x 24 kN/m^3 = 0.6 kN/m^2 B. 25 mm x 25 kN/m^3 = 625 kN/m^2 C. It is ignored in design D. It is added to the live load and then factored by 1.2 Answer: A. Thickness times unit weight gives the load per m^2: 0.025 x 24 = 0.6 kN/m^2. (Key point: Floor finish load) 2. What top steel is provided along discontinuous edges? A. No steel at all B. 50% of the mid-span steel, extending 0.1 l into the span, for partial fixity C. Twice the mid-span steel D. Same as the steel over continuous edges Answer: B. A discontinuous edge develops some negative moment due to partial fixity, so half the mid-span steel is provided on top. (Key point: Top steel at discontinuous edges) 3. For the corner where both edges are discontinuous, what torsion steel is provided? A. No torsion steel B. Only a bottom mesh of full mid-span steel over the whole slab C. Top and bottom meshes of 3/4 of mid-span steel over lx/5 x lx/5 D. A mesh of 1/4 of mid-span steel over lx/2 Answer: C. IS 456 D-1.8: each layer has 3/4 of the mid-span steel over a corner of size lx/5; half of that for corners with one discontinuous edge. (Key point: Torsion at corners) 4. How is deflection checked? A. Deflection is computed by integrating the moment diagram B. Actual l/d must exceed the permissible l/d C. Only the self-weight is checked D. Actual lx/d = 27.9 is compared with permissible 26 x modification factor (1.08) = 28.1 Answer: D. The l/d limit method multiplies the basic ratio 26 by the modification factor kt from pt and fs; actual must be below this. 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