--- title: "Slab with two adjacent edges discontinuous, 4x5 m" subject: "Design of RCC Structure" chapter: 9 key: "semester_7/design_of_rcc_structure:slab-with-two-adjacent-edges-discontinuous" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/slab-with-two-adjacent-edges-discontinuous site: BCT Engineering --- # Slab with two adjacent edges discontinuous, 4x5 m Design of RCC Structure · Chapter 9: Design of slabs and staircase - Asked in: 2080 Baisakh, 12 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/slab-with-two-adjacent-edges-discontinuous - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-9 ## Question Design a reinforced concrete rectangular slab of size 4.0 m x 5.0 m to support an imposed load of 4 kN/m$^2$ and floor finish of 1 kN/m$^2$. The slab has two adjacent edges discontinuous with the slab resting on 275 mm wide beam. Check the safety of slab against shear and deflection. Use M20 grade concrete and Fe415 grade steel. (Design of torsional reinforcements is not required) ## Short answer Adopt total depth 175 mm and effective depth 150 mm; factored load is 14.06 kN/m2 giving maximum moment 15.03 kNm/m. Provide bottom 8 mm at 230 c/c short span and 235 c/c long span, top 8 mm at 170 c/c over continuous long edge, 215 c/c over continuous short edge, and 300 c/c at discontinuous edges. Shear 0.195 MPa is below 0.40 MPa and deflection is safe. ## Answer #### Data and assumptions - Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 275 mm; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. - Live load = 4 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 $d$. - 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 + 4 = 9.375 kN/m² - Factored load $w_u = 1.5 \times 9.375 = 14.062$ kN/m² #### Bending moments | Moment | Coefficient | Value (kN·m/m) | |---|---|---| | $M_x$ negative (continuous long edge) | 0.0620 | 15.03 | | $M_x$ positive (mid-span, short) | 0.0466 | 11.30 | | $M_y$ negative (continuous short edge) | 0.0470 | 11.38 | | $M_y$ positive (mid-span, long) | 0.0350 | 8.48 | $M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 14.062 \times 4.150^2 = 242.19$ 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 $>$ $M_{max} = 15.03$ 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) | 11.30 | 215 | 8 mm @ 230 c/c | 219 | | Long span, bottom (mid-span) | 8.48 | 172 → 210 (min) | 8 mm @ 235 c/c | 214 | | Short span, top over continuous long edge | 15.03 | 289 | 8 mm @ 170 c/c | 296 | | Long span, top over continuous short edge | 11.38 | 233 | 8 mm @ 215 c/c | 234 | 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 = 109 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 $V_u = w_u l_x/2 = 14.062 \times 4.150/2 = 29.18$ kN/m $\tau_v = \dfrac{V_u}{bd} = \dfrac{29180}{1000 \times 150} = 0.195$ MPa Steel at the support: $A_{st} = 296$ mm²/m, $p_t = 0.197\%$, so $\tau_c = 0.32$ MPa (IS 456 Table 19, M20). For $D = 175$ mm, $k = 1.25$ (cl. 40.2.1.1), so $k\tau_c = 0.40$ MPa. $\tau_v = 0.195 < k\tau_c = 0.40$ MPa, hence the slab is **safe in shear** and no shear reinforcement is needed. #### Check for deflection For the short-span bars: $p_t = \dfrac{100 \times 219}{1000 \times 150} = 0.146\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 415 \times \dfrac{215}{219} = 236$ N/mm². Modification factor from IS 456 Fig. 4 (for $p_t = 0.15\%$ and $f_s = 236$ 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 $= 275/2 - 25$ (end cover) $= 112.5$ mm, plus the anchorage value of any bend. - Short-span bars at the discontinuous long edge (alternate bars bent up, 109 mm²/m, $d = 150$ mm): $M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 5.84$ kN·m/m, $1.3\dfrac{M_1}{V} = 1.3 \times \dfrac{5.84 \times 10^6}{29180} = 260$ mm. Straight: $260 + 112.5 = 373$ mm, just below 376 mm; with a 90° bend ($8\phi = 64$ mm): $437$ 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, $1.3M_1/V = 237$ mm; with a 90° bend: $237 + 112.5 + 64 = 414$ mm $> 376$ mm. With standard 90° bends at the discontinuous edges, the **development length is satisfied**. #### Torsion reinforcement at corners (Not required by the question; given for completeness.) 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 × 215 = 161 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 × 215 = 81 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@230 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 8mm@170 8mm@300 ~~~~~~~ ~~~ |------ slab D = 175 mm ------| bottom 8mm @ 230 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 @ 230 c/c; long-span bottom 8 mm @ 235 c/c; top steel over the continuous long edge 8 mm @ 170 c/c (short direction) and over the continuous short edge 8 mm @ 215 c/c (long direction); 8 mm @ 300 c/c top bars at the discontinuous edges. The slab is safe in shear and deflection; development length is satisfied with 90° bends at the discontinuous edges. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the top steel at discontinuous edges determined? A. No top steel at all B. Full negative-moment steel as at continuous edges C. Same as bottom steel doubled D. 50% of the mid-span steel, extending 0.1 l into the span, for partial fixity Answer: D. A discontinuous edge on a wall or beam has some unintended fixity, so IS 456 requires negative steel of half the mid-span steel for 0.1 l. (Key point: Top steel at continuous vs discontinuous edges) 2. What shear criterion is used? A. tau_v must exceed tau_c B. tau_v = Vu/(bd) must be less than k tau_c, here 0.195 < 0.405 MPa C. Vu must exceed 0.87 fy Ast D. Shear is checked only with stirrups Answer: B. If tau_v is below k tau_c (tau_c from pt, k from slab depth), no shear reinforcement is needed. (Key point: Shear check) 3. Why does the deflection check use fs = 0.58 fy (Ast req / Ast prov)? A. Because deflection depends on concrete grade only B. Because steel stress at service load sets the modification factor kt; more steel provided lowers fs and increases the allowed l/d C. Because fs is the shear stress D. Because it is the ultimate strength Answer: B. IS 456 Fig. 4 gives kt in terms of service stress in steel and pt; excess steel reduces stress and allows a higher permissible l/d. (Key point: Effect of provided steel on deflection) 4. What is the development-length check at the support? A. Ld alone must exceed 1000 mm B. Bars need no anchorage in slabs C. 1.3 M1/V + L0 must be at least Ld = 0.87 fy phi/(4 tau_bd) D. Ld must be less than d Answer: C. At simple supports, positive-moment bars must satisfy Ld <= 1.3 M1/V + L0 (IS 456 cl. 26.2.3.3) so the bars anchor properly. 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