--- title: "Slab rigidly fixed with beams, 6.5x4 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-rigidly-fixed-with-beams-6" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-rigidly-fixed-with-beams-6 site: BCT Engineering --- # Slab rigidly fixed with beams, 6.5x4 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2071 Chaitra, 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/slab-rigidly-fixed-with-beams-6 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design slab of a room of size 6.5 m x 4 m for a live load of 4.5 kN/m$^2$ and floor finish of 1 kN/m$^2$ of slab are rigidly fixed with beam. Take width of beam 230 mm. Use M20 concrete and TMT bars. Draw top and bottom reinforcement detailing with sections. Carry out all checks required for slab design. ## Short answer Provide 175 mm slab, d=150 mm, lx=4.15 m c/c, ly=6.65 m, ly/lx=1.60, treated as interior panel. Factored load 14.81 kN/m². Bottom: short 8@235, long 8@235. Top: short negative 14.25 kNm/m with 8@180, long negative 8.16 with 8@235. Shear and deflection checks pass. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 6.5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 4.5 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 $d$. - TMT bars taken as Fe415; slab treated as an interior panel (all four edges continuous/fixed with beams). - Panel type: interior panel (all four edges continuous); 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. $$ l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 6500 + 150 = 6650\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{6650}{4150} = 1.602 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.175 = 4.375 kN/m² - Total working load $w$ = 4.375 + 1 + 4.5 = 9.875 kN/m² - Factored load $w_u = 1.5 \times 9.875 = 14.812$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0559 | 14.25 | | $M_x$ positive (mid-span, short) | 0.0426 | 10.88 | | $M_y$ negative (continuous short edge) | 0.0320 | 8.16 | | $M_y$ positive (mid-span, long) | 0.0240 | 6.12 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 14.812 \times 4.150^2 = 255.11$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.60$. #### 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 $>$ $M_{max} = 14.25$ 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) | 10.88 | 210 | 8 mm @ 235 c/c | 214 | | Long span, bottom (mid-span) | 6.12 | 210 | 8 mm @ 235 c/c | 214 | | Short span, top over continuous long edge | 14.25 | 274 | 8 mm @ 180 c/c | 279 | | Long span, top over continuous short edge | 8.16 | 210 | 8 mm @ 235 c/c | 214 | #### Check for shear $V_u = w_u l_x/2 = 14.812 \times 4.150/2 = 30.74$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{30736}{1000 \times 150} = 0.205$ MPa Steel at support: $A_{st} = 279$ mm²/m, $p_t = 0.186\%$; $\tau_c = 0.31$ 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. $\tau_v = 0.205 < k\tau_c = 0.388$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{207}{214} = 233$ N/mm² (here $A_{st,req} = 207$ 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 $f_s = 233$ N/mm²): $k_t = 2.0$ (at such a low steel percentage the curve reaches its upper limit of 2.0). Permissible $l/d = 26 \times 2.0 = 52.0$; 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. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{5706338}{30736} + 150 = 391\ \text{mm} $$ Since 391 mm $>$ $L_d = 376$ mm, **development length is satisfied**. #### Detailing ```text PLAN (long span horizontal, short span vertical) +==============================+ # # # bot: 8mm@235 short span # # bot: 8mm@235 long span # # # +==============================+ = / # : 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 @ 180 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. Why is this slab treated as an interior panel? A. Because all four edges are fixed or continuous with beams B. Because the slab is 6.5 m long C. Because TMT bars are used D. Because it has no live load Answer: A. A slab rigidly fixed to beams on all sides has continuous edges, so the interior panel coefficients of Table 26 apply. (Key point: Interpreting rigidly fixed) 2. TMT bars are not given a grade in the question. What is assumed? A. Fe250 B. Fe415 C. Fe550 with fy = 550 D. Plain bars with fy = 250 Answer: B. TMT bars are taken as Fe415 with fy = 415 MPa, as stated in the assumptions. (Key point: Reinforcement grade) 3. Where are top bars needed in this slab? A. Only at mid-span B. Over the continuous edges where negative moments occur C. Nowhere, since bottom steel is enough D. Only at the corners Answer: B. Negative (hogging) moments over the supports are resisted by top bars, here 8 mm at 180 c/c in the short direction. (Key point: Top steel) 4. Vu = 30.74 kN/m gives tau_v = 0.205 MPa against k tau_c = 0.395 MPa. What follows? A. Shear reinforcement is required B. Depth must be doubled C. Slab is safe in shear without shear reinforcement D. Use stirrups at 100 mm Answer: C. When tau_v is below k tau_c the concrete alone carries the shear, so no shear reinforcement is needed. 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