--- title: "Interior slab panel 3.5x4.5 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:interior-slab-panel-3-5x4-5" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/interior-slab-panel-3-5x4-5 site: BCT Engineering --- # Interior slab panel 3.5x4.5 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2079 Baisakh, 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/interior-slab-panel-3-5x4-5 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design an interior panel of a slab for a room having clear dimension 3.5 m x 4.5 m. The slab rests on 250 mm wide beam. Assume live load of 4 kN/m$^2$ and floor finish of 0.6 kN/m$^2$. Use M20 mix and Fe415 grade of steel. Check for shear and deflection is also required. Draw reinforcement detailing in plan and sections. ## Short answer Two-way slab with D=150 mm, d=125 mm. Factored load 12.525 kN/m². Provide 8 mm bars at 275 mm c/c both ways for all locations. Shear (τ_v=0.182 MPa < kτ_c=0.36 MPa) and deflection (l/d=29 < 52) are 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. 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: interior panel (all four edges continuous); moment coefficients from IS 456 Annex D, Table 26 (case: interior panel). #### Depth and effective span Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 150$ mm, so $d = 150 - 20 - 5 = 125$ mm for the short-span bars and $d_y = 125 - 10 = 115$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 3500 + 125 = 3625\ \text{mm},\quad l_y = 4500 + 125 = 4625\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{4625}{3625} = 1.276 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.150 = 3.750 kN/m² - Total working load $w$ = 3.750 + 0.6 + 4 = 8.350 kN/m² - Factored load $w_u = 1.5 \times 8.350 = 12.525$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0460 | 7.58 | | $M_x$ positive (mid-span, short) | 0.0350 | 5.77 | | $M_y$ negative (continuous short edge) | 0.0320 | 5.27 | | $M_y$ positive (mid-span, long) | 0.0240 | 3.95 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 12.525 \times 3.625^2 = 164.59$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.28$. #### Check of depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 125^2 \times 10^{-6} = 43.13$ kN·m/m $>$ $M_{max} = 7.58$ 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 $bD$ = 180 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) | 5.77 | 131 → 180 (min) | 8 mm @ 275 c/c | 183 | | Long span, bottom (mid-span) | 3.95 | 97 → 180 (min) | 8 mm @ 275 c/c | 183 | | Short span, top over continuous long edges | 7.58 | 173 → 180 (min) | 8 mm @ 275 c/c | 183 | | Long span, top over continuous short edges | 5.27 | 130 → 180 (min) | 8 mm @ 275 c/c | 183 | #### Check for shear $V_u = w_u l_x/2 = 12.525 \times 3.625/2 = 22.70$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{22702}{1000 \times 125} = 0.182$ MPa Steel at the support: $A_{st} = 183$ mm²/m, $p_t = 0.146\%$ ($< 0.15\%$), so $\tau_c = 0.28$ MPa (IS 456 Table 19, M20). For $D = 150$ mm, $k = 1.30$ (cl. 40.2.1.1), so $k\tau_c = 0.36$ MPa. $\tau_v = 0.182 < k\tau_c = 0.36$ MPa, hence the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 183}{1000 \times 125} = 0.146\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{131}{183} = 172$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.15\%$ and $f_s = 172$ 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 $l_x/d = 3625/125 = 29.0$. 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 All four edges are continuous, so there is no simple support where the check $L_d \le 1.3M_1/V + L_0$ governs: the bottom bars run across the beams into the adjoining panels. The top bars over the supports extend $0.3 l_x = 0.3 \times 3500 = 1050$ mm from the face of the support, which is much more than $L_d = 376$ mm, so the **development length is satisfied**. #### Detailing ```text PLAN (long span horizontal, short span vertical) +==============================+ # # # bot: 8mm@275 short span # # bot: 8mm@275 long span # # # +==============================+ = / # : continuous edge (top bars 8mm@275) (all four edges continuous: no torsion steel needed) SECTION ALONG SHORT SPAN top bars over top bars over continuous edge continuous edge 8mm@275 8mm@275 ~~~~~~~ ~~~~~~~ |------ slab D = 150 mm ------| bottom 8mm @ 275 c/c, alt. bars bent up [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 $\ge$ 180 mm²/m is already satisfied by the bars provided. - No torsion reinforcement is needed, since all corners are formed by continuous edges (IS 456 D-1.10). **Answer:** $D = 150$ mm; short-span bottom 8 mm @ 275 c/c; long-span bottom 8 mm @ 275 c/c; top steel over the continuous edges 8 mm @ 275 c/c in both directions. 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 the 3.5 m x 4.5 m panel designed as a two-way slab? A. Because ly/lx = 1.276 is less than 2 B. Because the slab is supported on 250 mm wide beams C. Because the live load is 4 kN/m^2 D. Because all four edges are continuous, regardless of the span ratio Answer: A. A slab supported on all sides is two-way when ly/lx < 2; here ly/lx = 4625/3625 = 1.276. (Key point: One-way or two-way) 2. How is the design load obtained for this panel (D = 150 mm)? A. Only live load 4 kN/m^2 multiplied by 1.5 B. Self-weight 3.75 + finish 0.6 + live 4 = 8.35 kN/m^2, then multiplied by 1.5 to get 12.525 kN/m^2 C. Self-weight + live load, multiplied by 1.2 D. Total 8.35 kN/m^2 used directly without any factor Answer: B. Dead load (self-weight + finish) plus live load gives the working load, which is multiplied by the load factor 1.5 for limit state design. (Key point: Factored load) 3. Why is no shear reinforcement needed? A. Because tau_v is greater than tau_c so concrete alone fails B. Because slabs never need a shear check C. Because tau_v = 0.182 MPa is less than k*tau_c = 0.374 MPa D. Because the steel percentage is above 2% Answer: C. The nominal shear stress is below the permissible shear strength of the slab (k times tau_c), so concrete alone resists the shear. (Key point: Shear check) 4. Bottom bars give 1.3M1/V + L0 = 358 mm, less than Ld = 376 mm. What is done? A. Slab depth is doubled B. The check is ignored because slabs need no anchorage C. Bar diameter is increased to 16 mm D. Bars are given a standard 90 degree bend at the support, raising the available length to 422 mm Answer: D. A bend adds an anchorage value (8 phi = 64 mm), making the available length 422 mm which exceeds Ld. 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