--- title: "Corner slab design, 5x4 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:corner-slab-design-5x4-m" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/corner-slab-design-5x4-m site: BCT Engineering --- # Corner slab design, 5x4 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2082 Baisakh, 13 marks - Video: [Complete RCC Slab design as per IS 456-2000 code | manual calculation | two way slab design | online](https://www.youtube.com/watch?v=7WTLHg2gaUI) — Civil Engineering by Shravan - 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/corner-slab-design-5x4-m - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a corner slab (with two adjacent edges discontinuous) of clear dimension 5 m x 4 m in order to carry a live load of 3 kN/m$^2$ and floor finish of 1 kN/m$^2$. Use M20 concrete and Fe 415 steel. Carryout all the necessary checks and draw a neat sketch of plan and section of slab. ## Short answer Corner slab (two adjacent edges discontinuous), D = 175 mm, d = 150 mm, M20/Fe415. Factored load 12.56 kN/m²; max moment 13.42 kN·m/m. Reinforcement: short span bottom 8 mm @ 235 c/c, top 8 mm @ 195 c/c; long span both 8 mm @ 235 c/c; 8 mm @ 300 c/c at discontinuous edges. Shear and deflection (permissible l/d = 52) 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 = 1 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 is not given; 230 mm is assumed. - Panel type: two adjacent edges discontinuous; moment coefficients from IS 456 Annex D, Table 26 (case: two adjacent edges discontinuous). #### 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 + 3 = 8.375 kN/m² - Factored load $w_u = 1.5 \times 8.375 = 12.562$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0620 | 13.42 | | $M_x$ positive (mid-span, short) | 0.0466 | 10.09 | | $M_y$ negative (continuous short edge) | 0.0470 | 10.17 | | $M_y$ positive (mid-span, long) | 0.0350 | 7.57 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 12.562 \times 4.150^2 = 216.36$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$. #### Check of depth $M_{u,lim} = 0.138\, f_{ck} b d^2 = 0.138 \times 20 \times 1000 \times 150^2 \times 10^{-6} = 62.10$ kN·m/m $>$ $M_{max} = 13.42$ 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 $bD$ = 210 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) | 10.09 | 191 → 210 (min) | 8 mm @ 235 c/c | 214 | | Long span, bottom (mid-span) | 7.57 | 153 → 210 (min) | 8 mm @ 235 c/c | 214 | | Short span, top over continuous long edge | 13.42 | 257 | 8 mm @ 195 c/c | 258 | | Long span, top over continuous short edge | 10.17 | 208 → 210 (min) | 8 mm @ 235 c/c | 214 | Top steel at the discontinuous edges (for the negative moment due to partial fixity, IS 456 Annex D) extends $0.1 l$ into the span. Along the discontinuous long edge (short-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c; along the discontinuous short edge (long-span direction): 50% of mid-span steel = 107 mm²/m, provide 8 mm @ 300 c/c. #### Check for shear $V_u = w_u l_x/2 = 12.562 \times 4.150/2 = 26.07$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{26067}{1000 \times 150} = 0.174$ MPa Steel at the support: $A_{st} = 258$ mm²/m, $p_t = 0.172\%$, so $\tau_c = 0.30$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.37$ MPa. $\tau_v = 0.174 < k\tau_c = 0.37$ MPa, hence the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 214}{1000 \times 150} = 0.143\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{191}{214} = 215$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 215$ 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 $l_x/d = 4150/150 = 27.7$. 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 The check $L_d \le 1.3\dfrac{M_1}{V} + L_0$ (IS 456 cl. 26.2.3.3(c)) is made at the discontinuous edges. $L_0$ = straight length beyond the centre of the support $= 230/2 - 25$ (end cover) $= 90$ mm, plus the anchorage value of any bend. - Short-span bars at the discontinuous long edge (alternate bars bent up, 107 mm²/m, $d = 150$ mm): $M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 5.71$ kN·m/m, $1.3\dfrac{M_1}{V} = 1.3 \times \dfrac{5.71 \times 10^6}{26067} = 285$ mm. Straight: $285 + 90 = 375$ mm, just below 376 mm; with a 90° bend ($8\phi = 64$ mm): $439$ mm $> 376$ mm. - Long-span bars at the discontinuous short edge (107 mm²/m, $d = 140$ mm, same $V$ taken conservatively): $M_1 = 5.32$ kN·m/m, $1.3M_1/V = 265$ mm; with a 90° bend: $265 + 90 + 64 = 419$ mm $> 376$ mm. With standard 90° bends at the discontinuous edges, the **development length is satisfied**. #### Torsion reinforcement at corners Corner size $l_x/5 = 4150/5 = 830$ 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 × 210 = 158 mm²/m. Provide 8 mm @ 235 c/c in both directions, top and bottom, over 830 mm × 830 mm. - Corners with one edge discontinuous (2 corners): half of this = 0.375 × 210 = 79 mm²/m. Provide 8 mm @ 300 c/c in each direction, top and bottom. - No torsion steel is needed at the corner where both edges are continuous (IS 456 D-1.10). #### Detailing ```text PLAN (long span horizontal, short span vertical) +==============================+ # t| # bot: 8mm@235 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 ~~~~~~~ ~~~ |------ slab D = 175 mm ------| bottom 8mm @ 235 c/c, alt. bars bent up, 90 deg bend at the discontinuous edge [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 @ 235 c/c; long-span bottom 8 mm @ 235 c/c; top steel over the continuous long edge 8 mm @ 195 c/c (short direction) and over the continuous short edge 8 mm @ 235 c/c (long direction); 8 mm @ 300 c/c top bars at the discontinuous edges. Shear, deflection and development length (with 90° bends at the discontinuous edges) are satisfied. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the 5 m x 4 m panel designed as a two-way slab? A. Because ly/lx = 1.24 is less than 2, so bending occurs in both directions B. Because the live load is more than 3 kN/m2 C. Because two adjacent edges are discontinuous D. Because the slab is thicker than 150 mm Answer: A. A slab supported on four sides with ly/lx < 2 bends significantly in both directions, so it is designed as two-way using IS 456 Annex D coefficients. (Key point: Two-way slab classification) 2. How are the design moments in the two directions obtained for this panel? A. M = wu lx^2/8 for both directions B. M = alpha x wu x lx^2, with alpha from IS 456 Table 26 for two adjacent edges discontinuous, interpolated for ly/lx C. M = wu ly^2/8 for the long direction only D. M = 0.138 fck b d^2 in both directions Answer: B. IS 456 Annex D gives coefficients alpha for each edge condition; moment = alpha wu lx^2 with wu = 1.5 x (self-weight + finish + live load). (Key point: Moment coefficients and factored load) 3. For the corner where both edges are discontinuous, what torsion reinforcement is provided? A. None, because the corner is free B. Half of mid-span steel in the top layer only C. Top and bottom meshes, each layer 3/4 of mid-span Ast, over a corner of size lx/5 in both directions D. 3/4 of mid-span steel in the bottom layer only, over the whole slab Answer: C. Corners where two discontinuous edges meet tend to lift and twist, so IS 456 requires top and bottom meshes of 3/4 Ast over lx/5 x lx/5 (half of that at corners with one discontinuous edge). (Key point: Torsion steel at corners) 4. How does the answer verify the deflection criterion? A. Actual lx/d is compared with 26 x modification factor (from pt and fs); actual must be smaller B. Total deflection is computed by integration of the curvature C. lx/d must be greater than 26 D. Deflection is not checked for two-way slabs Answer: A. IS 456 cl. 23.2 limits span/depth to basic ratio times the tension-steel modification factor; 27.7 < 28.6 here, so the slab passes. 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