--- title: "Isolated square footing, 600 kN and 50 kNm" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-square-footing-600-kn-and" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-square-footing-600-kn-and site: BCT Engineering --- # Isolated square footing, 600 kN and 50 kNm Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2067 Asar (old course), 10 marks - Video: [Design of a Square Footing](https://www.youtube.com/watch?v=jYCc0QFKwR4) — 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/isolated-square-footing-600-kn-and - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated square footing foundation of uniform thickness for a 400 mm x 400 mm column subjected to an axial load of 600 kN and a moment of 50 kNm at service state. Consider bearing capacity of soil as 150 kN/m$^2$ and concrete grade M20 and steel grade Fe415. ## Short answer Square footing 2.4 m × 2.4 m, depth 450 mm (effective 390 mm). Reinforcement along moment direction: 12 mm @ 170 mm c/c, other direction: 12 mm @ 195 mm c/c; shear checks satisfactory. ## Answer #### Data and assumptions - Column 400 mm × 400 mm ($a = c = 400$ mm); footing and column concrete M20, steel Fe415; column bars assumed 8-16 mm. - Service load $P = 600$ kN, service moment $M = 50$ kN·m. - Allowable (safe) bearing capacity of soil = 150 kN/m². - Square footing, as required by the question. - Clear cover 50 mm for the footing (IS 456 cl. 26.4.2.2); effective depth $d = D - 60$ mm (cover + half bar). Load factor 1.5. #### Size of footing The weight of the footing and backfill is taken as 10% of the column load. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 600.0}{150} = 4.40\ \text{m}^2 $$ Eccentricity $e = M/P = 50.0/600.0 = 0.083$ m, which is less than $L/6$, so there is no tension under the base. The size is chosen so that $$ q_{max} = 1.1\frac{P}{BL}\left(1 + \frac{6e}{L}\right) \le q_a $$ Try $L \times B = 2.4 \times 2.4$ m ($A = 5.76$ m²): $q_{max} = 138.5$ kN/m² $\le 150.0$ and $q_{min} = 90.7$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 600.0 = 900.0$ kN, $M_u = 75.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 188.8$ kN/m², $p_{u,min} = 123.7$ kN/m² (average 156.2 kN/m²). Self-weight does not cause bending in the footing, so it is excluded. #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(2400 - 400)/2 = 1000$ mm; along $B$: $(2400 - 400)/2 = 1000$ mm. Along $L$ (critical section at the column face on the side of maximum pressure, linearly varying pressure): pressure at the column face $= 188.8 - \dfrac{188.8 - 123.7}{2.4}\times 1.0 = 161.7$ kN/m², so $$ M_{u,L} = B\left[\frac{161.7 \times 1.0^2}{2} + \frac{(188.8 - 161.7)\times 1.0^2}{3}\right] = 2.4\,(80.85 + 9.03) = 215.7\ \text{kN·m} $$ Along $B$ (average pressure): $M_{u,B} = 156.2 \times 2.4 \times 1.0^2/2 = 187.5$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2 = 1007$ kN·m for $d = 390$ mm, far larger than the moments, so the depth is governed by shear (one-way or punching). Trial **$D = 450$ mm, $d = 390$ mm**, confirmed by the shear checks below. $A_{st} = \dfrac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \dfrac{4.6 M_u}{f_{ck} b d^2}}\right] b d$; minimum steel = 0.12% of $bD$ = 1296 mm². | Direction | $M_u$ (kN·m) | $A_{st}$ req (mm²) | Provide | $A_{st}$ prov (mm²) | |---|---|---|---|---| | Along $L$ (width $B$) | 215.7 | 1589 | 12 mm @ 170 c/c | 1597 | | Along $B$ (width $L$) | 187.5 | 1374 | 12 mm @ 195 c/c | 1392 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 264.3$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.282$ MPa; $p_t = 0.171\%$, $\tau_c = 0.297$ MPa (Table 19). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 228.8$ kN, $\tau_v = 0.244$ MPa; $p_t = 0.149\%$, $\tau_c = 0.28$ MPa (lowest tabulated value, for $p_t \le 0.15\%$). Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+390) + (400+390)] = 3160$ mm - $V_u = P_u - p_u (a+d)(c+d) = 900.0 - 156.2 \times 0.790 \times 0.790 = 802.5$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.651$ MPa; $k_s = 0.5 + \beta_c = 1.5$, limited to $1.0$; $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 1.92$ MPa (M20, deformed bars): for 12 mm bars in both directions, $L_d = 564$ mm. Available length from the column face (less 50 mm end cover): 950 mm along $L$ and 950 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 5.62$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 9.00$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 18.00 MPa. The lesser, 9.00 MPa, governs. Bearing capacity = 1440 kN $\ge$ $P_u = 900$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of the column area = 800 mm²: use 8 bars of 16 mm (1608 mm²) as dowels, the same as the column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 602$ mm (and the same lap with the column bars). Available depth in the footing = 450 - 50 - 24 = 376 mm $< L_{dc}$; the dowels are bent at the bottom with 90° bends. (Bends do not count in compression, but this is acceptable here because the dowels are nominal: bearing alone transfers the load.) #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 450 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 2.4 m, B = 2.4 m ``` - Bottom mesh: 12 mm @ 170 c/c along $L$ (lower layer), 12 mm @ 195 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-16 mm. **Answer:** Footing 2.4 m × 2.4 m × 450 mm thick; 12 mm @ 170 c/c along the length and 12 mm @ 195 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the final 2.4 x 2.4 m footing larger than the A_req = 4.40 m^2 (about 2.1 x 2.1 m)? A. Code requires a minimum 2.4 m B. Eccentricity (M/P = 0.083 m) makes q_max larger than the average, so size is increased until q_max <= q_a C. To increase shear capacity D. To provide the minimum steel Answer: B. The area from 1.1P/q_a assumes uniform pressure; with moment q_max exceeds average and the size is checked and enlarged. (Key point: Area required with moment) 2. Why is the depth (D = 450 mm) chosen by shear checks and not by M_u,lim? A. Because M_u,lim = 0.138 f_ck b d^2 is far larger than the applied moment B. Because shear is always larger than flexure C. Because steel is Fe415 D. Because the column is square Answer: A. The section is lightly stressed in flexure, so shear (one-way and punching) controls the required depth. (Key point: Depth governed by shear) 3. Punching shear: tau_v = 0.651 MPa and permissible tau_c' = k_s x 0.25 sqrt(f_ck) = 1.118 MPa. What does this indicate? A. The footing is unsafe in punching B. The footing is safe; no shear steel is needed C. Depth must be reduced D. Moment transfer must be added Answer: B. tau_v < tau_c' so the footing is safe against punching shear. 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