--- title: "Interior slab design, 4x5 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:interior-slab-design-4x5-m" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/interior-slab-design-4x5-m site: BCT Engineering --- # Interior slab design, 4x5 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2081 Baisakh, 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/interior-slab-design-4x5-m - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a RC slab (interior panel) resting on RCC beams on all sides for a room having clear dimensions of 4 m x 5 m. The slab is subjected to superimposed live load of 3 kN/m$^2$ and floor finish load of 2 kN/m$^2$. Use M20 grade concrete and Fe415 grade steel. Perform necessary checks and provide structural drawings. ## Short answer Adopt total depth 165 mm and effective depth 140 mm; effective spans are 4140 mm and 5140 mm, ratio 1.24. Factored load is 13.69 kN/m2 giving maximum moment 10.48 kNm/m. Provide 8 mm bars at 250 c/c bottom both ways, 230 c/c top over long edges, 250 c/c top over short edges; shear and deflection are safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 3 kN/m², floor finish = 2 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 (beam) width is not given; 230 mm is assumed. - 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 $D = 165$ mm, so $d = 165 - 20 - 5 = 140$ mm for the short-span bars and $d_y = 140 - 10 = 130$ mm for the long-span bars. $$ l_x = \min(l_{clear} + d,\ c/c) = 4000 + 140 = 4140\ \text{mm},\quad l_y = 5000 + 140 = 5140\ \text{mm} $$ $$ \frac{l_y}{l_x} = \frac{5140}{4140} = 1.242 < 2 \Rightarrow \text{two-way slab} $$ #### Loads (per m²) - Self-weight = 25 × 0.165 = 4.125 kN/m² - Total working load $w$ = 4.125 + 2 + 3 = 9.125 kN/m² - Factored load $w_u = 1.5 \times 9.125 = 13.688$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0447 | 10.48 | | $M_x$ positive (mid-span, short) | 0.0337 | 7.90 | | $M_y$ negative (continuous short edge) | 0.0320 | 7.51 | | $M_y$ positive (mid-span, long) | 0.0240 | 5.63 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 13.688 \times 4.140^2 = 234.60$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$. #### Check of depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 140^2 \times 10^{-6} = 54.10$ kN·m/m $>$ $M_{max} = 10.48$ 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 $bD$ = 198 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) | 7.90 | 160 → 198 (min) | 8 mm @ 250 c/c | 201 | | Long span, bottom (mid-span) | 5.63 | 122 → 198 (min) | 8 mm @ 250 c/c | 201 | | Short span, top over continuous long edges | 10.48 | 214 | 8 mm @ 230 c/c | 219 | | Long span, top over continuous short edges | 7.51 | 164 → 198 (min) | 8 mm @ 250 c/c | 201 | #### Check for shear $V_u = w_u l_x/2 = 13.688 \times 4.140/2 = 28.33$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{28333}{1000 \times 140} = 0.202$ MPa Steel at the support: $A_{st} = 219$ mm²/m, $p_t = 0.156\%$, so $\tau_c = 0.29$ MPa (IS 456 Table 19, M20). For $D = 165$ mm, $k = 1.27$ (cl. 40.2.1.1), so $k\tau_c = 0.37$ MPa. $\tau_v = 0.202 < k\tau_c = 0.37$ MPa, hence the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 201}{1000 \times 140} = 0.144\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{160}{201} = 192$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 192$ 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 $l_x/d = 4140/140 = 29.6$. Since actual < permissible, **deflection is satisfied**. #### Check for development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.92$ 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 $0.3 l_x = 0.3 \times 4000 = 1200$ 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@250 short span # # bot: 8mm@250 long span # # # +==============================+ = / # : continuous edge (top bars) (all four edges continuous: no torsion steel needed) SECTION ALONG SHORT SPAN top bars over top bars over continuous edge continuous edge 8mm@230 8mm@230 ~~~~~~~ ~~~~~~~ |------ slab D = 165 mm ------| bottom 8mm @ 250 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 $\ge$ 198 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:** $D = 165$ mm; short-span bottom 8 mm @ 250 c/c; long-span bottom 8 mm @ 250 c/c; top steel over the continuous long edges 8 mm @ 230 c/c (short direction) and over the continuous short edges 8 mm @ 250 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. What does 'interior panel' mean for choosing the moment coefficients? A. All four edges are continuous, so negative moments occur at all supports and the Table 26 interior-panel row is used B. All four edges are simply supported C. Only the short edges are continuous D. Only the long edges are continuous Answer: A. An interior panel is continuous over all four edges, giving hogging moments at all supports and the smallest positive moments. (Key point: Interior panel edge condition) 2. How is the effective span lx obtained for the slab supported on beams? A. Clear span only, 4000 mm B. Centre-to-centre of beams always C. Clear span + d (= 4140 mm), limited to the c/c distance D. Clear span + 230 mm always Answer: C. IS 456 cl. 22.2: effective span is the lesser of clear span + d and centre-to-centre distance of supports. (Key point: Effective span and depth) 3. Which limits govern the choice of bars here? A. Minimum 0.12% of bD (Fe 415) and maximum spacing the lesser of 3d and 300 mm B. Minimum 1% of bD and spacing at most 100 mm C. Minimum 0.85/fy and spacing at most d D. No minimum steel is required when moment is small Answer: A. IS 456 cl. 26.5.2 sets 0.12% minimum for HYSD bars and a maximum spacing of 3d or 300 mm, so low-moment locations get the minimum steel. (Key point: Minimum steel and spacing limits) 4. What is the shear check conclusion and why? A. tau_v = 0.202 MPa exceeds tau_c, so stirrups are needed B. tau_v = Vu/(bd) = 0.202 MPa is less than k tau_c = 0.372 MPa, so no shear reinforcement is needed C. Shear is ignored for slabs D. tau_v must equal tau_c exactly Answer: B. Slabs are normally safe in shear when tau_v < k tau_c, where k = 1.27 for D = 165 mm and tau_c depends on pt and concrete grade. 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