--- title: "Restrained floor slab, 4x5 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:restrained-floor-slab-4x5-m" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/restrained-floor-slab-4x5-m site: BCT Engineering --- # Restrained floor slab, 4x5 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2074 Chaitra, 15 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/restrained-floor-slab-4x5-m - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a restrained floor slab for a room 4 m x 5 m in size to support a live load of 5 kN/m$^2$, with two adjacent sides discontinuous. Use M20 concrete and Fe415 grade steel. Sketch the details of reinforcements. ## Short answer Slab thickness 175 mm, effective depth 150 mm (short) and 140 mm (long). Short bottom: 8 mm @ 210 c/c; long bottom: 8 mm @ 235 c/c; top over continuous long edge: 8 mm @ 155 c/c; top over continuous short edge: 8 mm @ 190 c/c; top at both discontinuous edges: 8 mm @ 300 c/c. Shear and deflection checks are satisfied. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 5 kN/m², floor finish = 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 $d$. - Support width (230 mm) and floor finish (1.0 kN/m²) are not given and are assumed. - 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 $D = 175$ 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. $$ l_x = \min(l_{clear} + d,\ c/c) = 4000 + 150 = 4150\ \text{mm},\quad l_y = 5000 + 150 = 5150\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{5150}{4150} = 1.241 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.175 = 4.375 kN/m² - Total working load $w$ = 4.375 + 1.0 + 5 = 10.375 kN/m² - Factored load $w_u = 1.5 \times 10.375 = 15.562$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0620 | 16.63 | | $M_x$ positive (mid-span, short) | 0.0466 | 12.50 | | $M_y$ negative (continuous short edge) | 0.0470 | 12.60 | | $M_y$ positive (mid-span, long) | 0.0350 | 9.38 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 15.562 \times 4.150^2 = 268.03$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$. #### 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 $>$ $M_{max} = 16.63$ 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) | 12.50 | 239 | 8 mm @ 210 c/c | 239 | | Long span, bottom (mid-span) | 9.38 | 210 | 8 mm @ 235 c/c | 214 | | Short span, top over continuous long edge | 16.63 | 322 | 8 mm @ 155 c/c | 324 | | Long span, top over continuous short edge | 12.60 | 259 | 8 mm @ 190 c/c | 265 | 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 = 120 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge(s) (long-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 15.562 \times 4.150/2 = 32.29$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{32292}{1000 \times 150} = 0.215$ MPa Steel at support: $A_{st} = 324$ mm²/m, $p_t = 0.216\%$; $\tau_c = 0.33$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.413$ MPa. $\tau_v = 0.215 < k\tau_c = 0.413$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 239}{1000 \times 150} = 0.160\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{239}{239} = 241$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.16\%$ and $f_s = 241$ 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 (120 mm²/m) run into the support: $M_1 = 6.37$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 150$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{6374248}{32292} + 150 = 407\ \text{mm} $$ Since 407 mm $>$ $L_d = 376$ mm, **development length is satisfied**. #### Torsion reinforcement at corners Corner size $l_x/5 = 4150/5 = 830$ mm (say 830 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 × 239 = 179 mm²/m. Provide 8 mm @ 235 c/c in both directions, top and bottom, over 830 mm × 830 mm. - Corner with only one discontinuous edge (2 corners): half of 3/4 of the mid-span $A_{st}$ = 0.375 × 239 = 90 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| # bot: 8mm@210 short span | # bot: 8mm@235 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 (0.5 x) ~~~~~~~ ~~~ |------ slab D = 175 mm ------| bottom 8mm @ 210 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 @ 210 c/c; long-span bottom 8 mm @ 235 c/c; top steel over continuous edge 8 mm @ 155 c/c (short direction); 8 mm @ 190 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. Self-weight 4.375, floor finish 1.0 and live load 5 kN/m^2 give w = 10.375 kN/m^2. What design load is used for the moments? A. 10.375 kN/m^2 B. 1.5 x 10.375 = 15.56 kN/m^2 C. 1.5 x 5 = 7.5 kN/m^2 D. 1.5 x 4.375 + 6 = 12.56 kN/m^2 Answer: B. Limit state design uses the factored load wu = 1.5 x total working load, applied in M = alpha x wu x lx^2. (Key point: Factored load) 2. How much top steel is provided along the discontinuous edges, and for what reason? A. None, since the moment there is zero B. Twice the mid-span steel to resist shear C. 50% of mid-span steel, for negative moment from partial fixity, extending 0.1 l into the span D. Same as the mid-span steel, extending 0.3 l Answer: C. A discontinuous edge on a wall has some fixity, causing hogging moment, so half of the mid-span steel is placed on top for 0.1 l. (Key point: Top steel at discontinuous edges) 3. For the corner where both edges are discontinuous, what torsion reinforcement is provided? A. No steel, as the corner is unrestrained B. Top and bottom meshes, each 3/4 of mid-span Ast, over lx/5 x lx/5 C. Half of mid-span Ast in the bottom layer only D. Top mesh only, equal to the continuous-edge steel Answer: B. Corners held down tend to lift, so IS 456 D-1.8 gives meshes in top and bottom, each layer 3/4 of mid-span steel, over a corner of size lx/5. 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