--- title: "Slab panel 2 for BM and SF (floor plan)" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-panel-2-for-bm-and" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-panel-2-for-bm-and site: BCT Engineering --- # Slab panel 2 for BM and SF (floor plan) Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2069 Chaitra, 14 marks - Video: [HOW TO CONSIDER TWO WAY SLAB BENDING MOMENT COEFFICIENTS AS PER IS 456 TABLE 26](https://www.youtube.com/watch?v=XC48mZ2OJBE) — Structural Design Mentor - 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-panel-2-for-bm-and - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question RC slab of the floor of a residential building is subjected to live load of 3 kN/m$^2$ and floor finishes of 1 kN/m$^2$. Design the slab panel 2 for BM and SF. Draw neat sketches of slab showing top and bottom arrangements of reinforcing bars. [Figure: floor plan of four panels numbered 1, 2 (top row) and 4, 3 (bottom row); 2 bays of 5 m horizontally and 2 bays of 3 m vertically; one-brick-thick wall around] ## Short answer Slab thickness 140 mm. Short span bottom bars 8 mm at 295 mm c/c, top bars over continuous edge 8 mm at 225 mm c/c. Long span bottom bars 8 mm at 295 mm c/c, top bars over continuous edge 8 mm at 295 mm c/c; discontinuous edges 8 mm at 300 mm c/c. Shear is safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 3$ m (short), $l_y = 5$ 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 2 is the top-right panel: its left edge (common with panel 1) and bottom edge (common with panel 3) are continuous; the right edge and top edge are on the outer wall (discontinuous). Bay dimensions 3 m × 5 m are taken as clear spans; wall thickness 230 mm. - Panel type: two adjacent edges 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 $D = 140$ mm, so $d = 140 - 20 - 5 = 115$ mm for the short-span bars and $d_y = 115 - 10 = 105$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 3000 + 115 = 3115\ \text{mm},\quad l_y = 5000 + 115 = 5115\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{5115}{3115} = 1.642 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.140 = 3.500 kN/m² - Total working load $w$ = 3.500 + 1 + 3 = 7.500 kN/m² - Factored load $w_u = 1.5 \times 7.500 = 11.250$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0801 | 8.75 | | $M_x$ positive (mid-span, short) | 0.0600 | 6.55 | | $M_y$ negative (continuous short edge) | 0.0470 | 5.13 | | $M_y$ positive (mid-span, long) | 0.0350 | 3.82 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 11.250 \times 3.115^2 = 109.16$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.64$. #### Check for depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 115^2 \times 10^{-6} = 36.50$ kN·m/m $>$ $M_{max} = 8.75$ 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 $bD$ = 168 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) | 6.55 | 168 | 8 mm @ 295 c/c | 170 | | Long span, bottom (mid-span) | 3.82 | 168 | 8 mm @ 295 c/c | 170 | | Short span, top over continuous long edge | 8.75 | 219 | 8 mm @ 225 c/c | 223 | | Long span, top over continuous short edge | 5.13 | 168 | 8 mm @ 295 c/c | 170 | 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 = 85 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 85 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 11.250 \times 3.115/2 = 17.52$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{17522}{1000 \times 115} = 0.152$ MPa Steel at support: $A_{st} = 223$ mm²/m, $p_t = 0.194\%$; $\tau_c = 0.32$ MPa (IS 456 Table 19, M20). For $D = 140$ mm, $k = 1.30$ (cl. 40.2.1.1), so $k\tau_c = 0.416$ MPa. $\tau_v = 0.152 < k\tau_c = 0.416$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 170}{1000 \times 115} = 0.148\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{163}{170} = 231$ N/mm² (here $A_{st,req} = 163$ mm²/m is the steel needed for bending; the minimum steel governs the bars provided). Modification factor from IS 456 Fig. 4 (for $p_t = 0.15\%$ and $f_s = 231$ 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 $l_x/d = 3115/115 = 27.1$. 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 (85 mm²/m) run into the support: $M_1 = 3.48$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 115$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{3483013}{17522} + 115 = 373\ \text{mm} $$ Since 373 mm $<$ $L_d = 376$ mm, the bars are given a standard 90° bend at the support (anchorage value $8\phi$ = 64 mm). The available length then becomes 373 + 64 = 437 mm $> L_d$, so **development length is satisfied** with bent ends. #### Torsion reinforcement at corners Corner size $l_x/5 = 3115/5 = 623$ mm (say 630 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 × 168 = 126 mm²/m. Provide 8 mm @ 295 c/c in both directions, top and bottom, over 630 mm × 630 mm. - Corner with only one discontinuous edge (2 corners): half of 3/4 of the mid-span $A_{st}$ = 0.375 × 168 = 63 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 T| # bot: 8mm@295 short span | # bot: 8mm@295 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 at continuous edge discontinuous edge (0.5 x) ~~~~~~~ ~~~ |------ slab D = 140 mm ------| bottom 8mm @ 295 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 $\ge$ 168 mm²/m is already satisfied by the bars provided. **Answer:** $D = 140$ mm; short-span bottom 8 mm @ 295 c/c; long-span bottom 8 mm @ 295 c/c; top steel over continuous edge 8 mm @ 225 c/c (short direction); 8 mm @ 295 c/c (long direction). Shear, deflection and development length (with bent bar ends) are satisfied. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Panel 2 has effective spans lx = 3115 mm and ly = 5115 mm. How is it classified and why? A. One-way slab, because ly is larger than lx B. Two-way slab, because ly/lx = 1.64 is less than 2 C. Two-way slab, because ly/lx is greater than 2 D. One-way slab, because the edges are discontinuous Answer: B. A slab supported on four sides is two-way when ly/lx < 2; here 5115/3115 = 1.64. (Key point: One-way or two-way) 2. Which edges of panel 2 are treated as continuous when choosing the moment coefficients? A. Right and top edges, which lie on the outer wall B. All four edges C. Left edge (with panel 1) and bottom edge (with panel 3) D. None, since the slab rests on walls Answer: C. Edges shared with neighbouring panels are continuous; the right and top edges are on the outer wall and are discontinuous, giving the case of two adjacent edges discontinuous. (Key point: Edge conditions of panel 2) 3. The total working load is 7.5 kN/m2 (self-weight 3.5 + finish 1 + live 3). What factored load is used for design? A. 7.5 kN/m2 B. 9.0 kN/m2 C. 11.25 kN/m2 D. 12.0 kN/m2 Answer: C. The limit state design uses the partial safety factor 1.5, so wu = 1.5 x 7.5 = 11.25 kN/m2. (Key point: Factored load) 4. What torsion reinforcement is provided at a corner of panel 2 where both edges are discontinuous? A. None, as torsion is ignored in slabs B. Top and bottom meshes, each layer 3/4 of mid-span steel, over lx/5 x lx/5 C. Top mesh only, equal to the full mid-span steel D. Bottom mesh only, 1/2 of mid-span steel Answer: B. IS 456 D-1.8 requires top and bottom meshes of 3/4 of the mid-span steel over a corner of size lx/5 (about 630 mm) in both directions where two edges are discontinuous. 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