--- title: "Roof slab restrained on four sides, 6x3.5 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:roof-slab-restrained-on-four-sides" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/roof-slab-restrained-on-four-sides site: BCT Engineering --- # Roof slab restrained on four sides, 6x3.5 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2065 Shrawan (old course), 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/roof-slab-restrained-on-four-sides - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a roof slab for a room 6 m x 3.5 m restrained on all four sides by beams. It has to support super imposed load of 4 kN/m$^2$. Take M20 concrete and Fe250 steel. ## Short answer Slab thickness 125 mm. Short-span bottom bars 8 mm at 170 mm c/c, top bars over continuous long edges 8 mm at 125 mm c/c. Long-span bottom bars 8 mm at 265 mm c/c, top bars over continuous short edges 8 mm at 210 mm c/c. Designed as interior panel using IS 456 coefficients. Deflection satisfied. ## Answer #### Data and assumptions - Clear spans: $l_x = 3.5$ m (short), $l_y = 6$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 250$ MPa. - Live load = 4 kN/m², floor finish = 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 $d$. - The 'super-imposed load' of 4 kN/m² is taken as the total live load plus finishes; beam width 230 mm assumed; slab treated as an interior (fully restrained) panel. - 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 $D = 125$ mm, so $d = 125 - 20 - 5 = 100$ mm for the short-span bars and $d_y = 100 - 10 = 90$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 3500 + 100 = 3600\ \text{mm},\quad l_y = 6000 + 100 = 6100\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{6100}{3600} = 1.694 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.125 = 3.125 kN/m² - Total working load $w$ = 3.125 + 0 + 4 = 7.125 kN/m² - Factored load $w_u = 1.5 \times 7.125 = 10.688$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0584 | 8.10 | | $M_x$ positive (mid-span, short) | 0.0441 | 6.11 | | $M_y$ negative (continuous short edge) | 0.0320 | 4.43 | | $M_y$ positive (mid-span, long) | 0.0240 | 3.32 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 10.688 \times 3.600^2 = 138.51$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.69$. #### Check for depth $M_{u,lim} = 0.148\, f_{ck} b d^2 = 0.148 \times 20 \times 1000 \times 100^2 \times 10^{-6} = 29.60$ kN·m/m $>$ $M_{max} = 8.10$ 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 = 0.15% of $bD$ = 188 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm (short), 270 mm (long). | Location | $M_u$ (kN·m/m) | $A_{st}$ req (mm²/m) | Provide | $A_{st}$ prov | |---|---|---|---|---| | Short span, bottom (mid-span) | 6.11 | 292 | 8 mm @ 170 c/c | 296 | | Long span, bottom (mid-span) | 3.32 | 188 | 8 mm @ 265 c/c | 190 | | Short span, top over continuous long edge | 8.10 | 392 | 8 mm @ 125 c/c | 402 | | Long span, top over continuous short edge | 4.43 | 234 | 8 mm @ 210 c/c | 239 | #### Check for shear $V_u = w_u l_x/2 = 10.688 \times 3.600/2 = 19.24$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{19238}{1000 \times 100} = 0.192$ MPa Steel at support: $A_{st} = 402$ mm²/m, $p_t = 0.402\%$; $\tau_c = 0.43$ MPa (IS 456 Table 19, M20). For $D = 125$ mm, $k = 1.30$ (cl. 40.2.1.1), so $k\tau_c = 0.559$ MPa. $\tau_v = 0.192 < k\tau_c = 0.559$ MPa, so the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 296}{1000 \times 100} = 0.296\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 250 \times \dfrac{292}{296} = 143$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.30\%$ and $f_s = 143$ 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 $l_x/d = 3600/100 = 36.0$. Since actual < permissible, **deflection is satisfied**. #### Check for development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.2$ MPa (M20, plain mild-steel bars, so no 60% increase): $L_d = \dfrac{0.87 \times 250 \times 8}{4 \times 1.2} = 363$ mm Alternate bottom bars (148 mm²/m) run into the support: $M_1 = 3.16$ kN·m/m, $L_0 = \max(d,\ 12\phi) = 100$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{3156090}{19238} + 100 = 313\ \text{mm} $$ Since 313 mm $<$ $L_d = 363$ mm, straight bar ends are not enough. Plain bars must end in standard U-hooks anyway (IS 456 cl. 26.2.2.1), and a standard hook has an anchorage value of $16\phi$ = 128 mm. The available length becomes 313 + 128 = 441 mm $> 363$ mm, so **development length is satisfied** with hooked ends. #### Detailing ```text PLAN (long span horizontal, short span vertical) +==============================+ # # # bot: 8mm@170 short span # # bot: 8mm@265 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 = 125 mm ------| bottom 8mm @ 170 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 $\ge$ 188 mm²/m is already satisfied by the bars provided. **Answer:** $D = 125$ mm; short-span bottom 8 mm @ 170 c/c; long-span bottom 8 mm @ 265 c/c; top steel over continuous edge 8 mm @ 125 c/c (short direction); 8 mm @ 210 c/c (long direction). Shear, deflection and development length (with standard hooks at the bar ends) are satisfied. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The roof slab is restrained on all four sides by beams. Which moment coefficient case is used? A. Four edges discontinuous, corners held down B. Interior panel, all four edges continuous C. Two adjacent edges discontinuous D. One long edge discontinuous Answer: B. The answer treats the slab as an interior fully restrained panel, so Table 26 case: all four edges continuous, with negative and positive moments. (Key point: Panel type) 2. What minimum steel is taken for this slab and why is it different from Fe415 slabs? A. 0.12% of bD, for all grades B. 0.15% of bD, because mild steel Fe250 is used C. 0.30% of bD for roofs D. 0.85/fy of bD only Answer: B. IS 456 specifies 0.15% of the gross area for mild steel (Fe250) and 0.12% for high yield strength bars, so 0.15% x 1000 x 125 = 188 mm2/m. (Key point: Minimum steel for Fe250) 3. For Fe250 steel, which Mu,lim coefficient is used in the depth check? A. 0.138 fck b d^2 B. 0.133 fck b d^2 C. 0.148 fck b d^2 D. 0.36 fck b d^2 Answer: C. Mu,lim = 0.148 fck b d^2 for Fe250, 0.138 for Fe415 and 0.133 for Fe500. (Key point: Limiting moment constant) 4. With D = 125 mm, self-weight is 3.125 kN/m2 and super-imposed load is 4 kN/m2. What is wu? A. 7.125 kN/m2 B. 10.688 kN/m2 C. 6.0 kN/m2 D. 12.0 kN/m2 Answer: B. w = 3.125 + 4 = 7.125 kN/m2 and wu = 1.5 x 7.125 = 10.688 kN/m2. (Key point: Factored load) 5. Which moment governs the design of the short-span steel, and where is it? A. Positive mid-span moment in the long direction B. Negative moment over the continuous long edge, 8.10 kN·m/m C. Zero moment at the supports D. Torsion moment at corners Answer: B. The negative moment in the short direction, alpha = 0.0584, is largest (8.10 kN·m/m) and needs the most top steel. 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