--- title: "Two-way slab simply supported, 6x4 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:two-way-slab-simply-supported-6x4" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/two-way-slab-simply-supported-6x4 site: BCT Engineering --- # Two-way slab simply supported, 6x4 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2080 Bhadra, 14 marks - Video: [Design of a Simply Supported Two Way Slab using IS - 456 & SP -16](https://www.youtube.com/watch?v=_Gz2v3zvfMY) — Stan Academy - 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/two-way-slab-simply-supported-6x4 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a two-way slab simply supported on all four edges for a room 6 m x 4 m clear in size supported on 230 mm thick walls. The superimposed working load is 4 kN/m$^2$ and corners are not held down. Use M25 mix and Fe 415 grade steel. ## Short answer Adopt total depth 210 mm and effective depth 185 mm; effective short span is 4185 mm. Factored load is 13.88 kN/m2 giving short-span moment 25.00 and long-span moment 11.45 kNm/m. Provide 8 mm bars at 125 c/c short span and 195 c/c long span at bottom, with nominal 8 mm at 250 c/c top at edges; shear and deflection are safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 6$ m (long); support (wall) width 230 mm; $f_{ck} = 25$ MPa, $f_y = 415$ MPa. - Superimposed working load = 4 kN/m² (live load and finishes taken together as one load). 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: simply supported on four sides, corners not held down; moment coefficients from IS 456 Annex D, Table 27 (simply supported, corners free to lift). #### Depth and effective span Basic span/depth ratio = 20 (IS 456 cl. 23.2.1). Trial overall depth $D = 210$ mm, so $d = 210 - 20 - 5 = 185$ mm for the short-span bars and $d_y = 185 - 10 = 175$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 4000 + 185 = 4185\ \text{mm},\quad l_y = 6000 + 185 = 6185\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{6185}{4185} = 1.478 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.210 = 5.250 kN/m² - Total working load $w$ = 5.250 + 4 = 9.250 kN/m² - Factored load $w_u = 1.5 \times 9.250 = 13.875$ kN/m² #### Bending moments From Table 27: $\alpha_x = 0.099$ at 1.4 and $0.104$ at 1.5; $\alpha_y = 0.051$ at 1.4 and $0.046$ at 1.5. Interpolating for $l_y/l_x = 1.478$: | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ positive (mid-span, short) | 0.1029 | 25.00 | | $M_y$ positive (mid-span, long) | 0.0471 | 11.45 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 13.875 \times 4.185^2 = 243.01$ kN·m/m. #### Check of depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 25 \times 1000 \times 185^2 \times 10^{-6} = 118.08$ kN·m/m $>$ $M_{max} = 25.00$ 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 $bD$ = 252 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) | 25.00 | 388 | 8 mm @ 125 c/c | 402 | | Long span, bottom (mid-span) | 11.45 | 185 → 252 (min) | 8 mm @ 195 c/c | 258 | Since the corners are free to lift, no torsion steel is needed. Nominal top steel (8 mm @ 250 c/c, extending $0.1 l$ into the span) is provided along the supported edges to control cracks due to unintended fixity. #### Check for shear $V_u = w_u l_x/2 = 13.875 \times 4.185/2 = 29.03$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{29033}{1000 \times 185} = 0.157$ MPa Steel at the support (alternate bars bent up): $A_{st} = 201$ mm²/m, $p_t = 0.109\%$, so $\tau_c = 0.29$ MPa (IS 456 Table 19, M25). For $D = 210$ mm, $k = 1.18$ (cl. 40.2.1.1), so $k\tau_c = 0.34$ MPa. $\tau_v = 0.157 < k\tau_c = 0.34$ MPa, hence the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 402}{1000 \times 185} = 0.217\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{388}{402} = 232$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.22\%$ and $f_s = 232$ N/mm²): $k_t \approx 1.8$. Permissible $l/d = 20 \times 1.8 = 36.0$; actual $l_x/d = 4185/185 = 22.6$. Since actual < permissible, **deflection is satisfied**. #### Check for development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 2.24$ MPa (M25, deformed bars, ×1.6): $L_d = \dfrac{0.87 \times 415 \times 8}{4 \times 2.24} = 322$ mm At the simple support, $L_d \le 1.3\dfrac{M_1}{V} + L_0$ (IS 456 cl. 26.2.3.3(c)). Alternate bottom bars (201 mm²/m) run into the support: $M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 13.19$ kN·m/m, and $L_0$ = straight length beyond the centre of the wall $= 230/2 - 25$ (end cover) $= 90$ mm. $$ 1.3\frac{M_1}{V} + L_0 = 1.3 \times \frac{13.19 \times 10^6}{29033} + 90 = 680\ \text{mm} $$ Since 680 mm $>$ $L_d = 322$ mm, **development length is satisfied** even with straight bars. #### Detailing ```text PLAN (long span horizontal, short span vertical) +------------------------------+ | | | bot: 8mm@125 short span | | bot: 8mm@195 long span | | | +------------------------------+ - / | : simply supported edge (nominal top bars 8mm@250) corners free to lift: no torsion steel SECTION ALONG SHORT SPAN nominal top bars at edges (8mm@250, 0.1 l) ~~~ ~~~ |------ slab D = 210 mm ------| bottom 8mm @ 125 c/c, alt. bars bent up [wall] [wall] ``` - 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. - All bars have 20 mm cover. - Distribution/minimum steel $\ge$ 252 mm²/m is already satisfied by the bars provided. **Answer:** $D = 210$ mm; short-span bottom 8 mm @ 125 c/c; long-span bottom 8 mm @ 195 c/c; nominal top bars 8 mm @ 250 c/c at the edges. 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 no torsion reinforcement provided in this slab? A. Because the slab is thick B. Because the corners are free to lift, so torsional restraint is not developed C. Because M25 concrete is strong D. Because the slab is one-way Answer: B. When corners are not held down, they lift and no torsion arises at them; the Table 27 coefficients for this case already account for that and torsion steel is not needed. (Key point: No torsion steel when corners free) 2. What basic span/depth ratio is used for the simply supported slab? A. 26 B. 35 C. 20 D. 40 Answer: C. IS 456 cl. 23.2.1 gives 20 for simply supported spans (26 for continuous), then multiplies by the modification factor kt. (Key point: Basic span/depth ratio) 3. Which moments are calculated, and which direction is larger? A. Negative moments at all edges, long direction larger B. Only positive mid-span moments; Mx (short span, 25.0 kN m/m) exceeds My (long span, 11.45 kN m/m) C. Only the long-span moment D. Equal moments in both directions Answer: B. Simply supported edges have no hogging moments; the short span carries more load, so Mx > My, giving more steel in the short direction. (Key point: Moments in a simply supported slab) 4. How is the 4 kN/m2 superimposed load used in the factored load? A. Added to the self-weight (25 x 0.21 = 5.25) to get 9.25 kN/m2, then multiplied by 1.5 B. Multiplied by 1.5 and self-weight ignored C. Multiplied by 0.87 D. Used directly without factoring Answer: A. Total working load = self-weight + superimposed load, and wu = 1.5 x 9.25 = 13.875 kN/m2. (Key point: Using superimposed load) ## Related in Design of RCC Structure - [Slab, two adjacent edges discontinuous, 5x4 m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-two-adjacent-edges-discontinuous-5x4) — Design of RCC Structure, chapter 9 - [Two-adjacent-edges discontinuous slab, 3.5x4.5 m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Atwo-adjacent-edges-discontinuous-slab-3) — Design of RCC Structure, chapter 9 - [Slab on masonry walls, 5x6 m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-on-masonry-walls-5x6-m) — Design of RCC Structure, chapter 9 - [Two-way slab with torsion, panel 1](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Atwo-way-slab-with-torsion-panel) — Design of RCC Structure, chapter 9 - [Slab continuous over three edges, 5x4 m (225 mm beam)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-continuous-over-three-edges-5x4-2) — Design of RCC Structure, chapter 9 - [Slab design, three edges continuous, 5x4 m (Fe500)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-design-three-edges-continuous-5x4) — Design of RCC Structure, chapter 9 ## Related in other subjects - [Brick and mortar for 230 mm wall (100 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-230mm-100kn-5m) — Concrete Technology and Masonry Structure, chapter 8, asked 7× - [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7 - [Simply supported beam with two elements](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Asimply-supported-beam-two-elements) — Finite Element Method, chapter 3 - [Design of a simply supported timber beam](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Adesign-timber-floor-beam) — Design of Timber and Masonry Structures, chapter 3