--- title: "Isolated footing 450x450, service moment 110 kNm" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-footing-450x450-service-moment-110" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-footing-450x450-service-moment-110 site: BCT Engineering --- # Isolated footing 450x450, service moment 110 kNm Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2082 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/isolated-footing-450x450-service-moment-110 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing for a column of size 450 mm x 450 mm with 8-20 mm diameter bars as longitudinal reinforcement in order to carry a service axial load of 1000 kN and service moment of 110 kNm. Take the safe bearing capacity of soil as 140 kN/m$^2$. Use M20 concrete and Fe 415 steel. ## Short answer Use a 4.0 m x 2.3 m footing, overall depth 700 mm (d=640 mm). Provide 16 mm bars @150 mm c/c along the longer direction; in the short direction, place a central band of 12 mm bars @100 mm c/c and outer strips of 12 mm @120 mm c/c. One-way and two-way shear are safe. ## Answer #### Data and assumptions - Column 450 mm × 450 mm (long side $a=450$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm. - Service load $P = 1000$ kN, service moment $M = 110$ kN·m. - Allowable (safe) bearing capacity of soil = 140 kN/m². - The service moment acts about one axis. - 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 Weight of footing and backfill is taken as 10% of the column load. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1000.0}{140} = 7.86\ \text{m}^2 $$ Eccentricity $e = M/P = 110.0/1000.0 = 0.110$ 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 $L \times B = 4.0 \times 2.3$ m ($A = 9.20$ m²): $q_{max} = 139.3$ kN/m² $\le 140.0$ and $q_{min} = 99.8$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 1000.0 = 1500.0$ kN, $M_u = 165.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 189.9$ kN/m², $p_{u,min} = 136.1$ kN/m² (average 163.0 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 $L$: $(4000 - 450)/2 = 1775$ mm; along $B$: $(2300 - 450)/2 = 925$ mm. Along $L$ (critical section at the column face on the side of maximum pressure): the pressure is 166.1 kN/m² at the face and 189.9 kN/m² at the edge, so per metre width $M = 166.1 \times \dfrac{1.775^2}{2} + (189.9 - 166.1) \times \dfrac{1.775^2}{3} = 261.6 + 25.0 = 286.6$ kN·m/m, and $M_{u,L} = 286.6 \times 2.3 = 659.3$ kN·m. Along $B$ (average pressure): $M_{u,B} = 279.0$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 640$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 700$ mm, $d = 640$ mm**, found 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 $bD$. | Direction | $M_u$ (kN·m) | $A_{st}$ req (mm²) | Provide | $A_{st}$ prov (mm²) | |---|---|---|---|---| | Along $L$ (width $B$) | 659.3 | 2980 | 16 mm @ 150 c/c | 3083 | | Along $B$ (width $L$) | 279.0 | 3360 (minimum) | 12 mm, banded (below) | 4184 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 2300$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.74+1} = 0.730$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.730 \times 3360 = 2453$ mm² → 12 mm @ 100 mm c/c (23 bars, 2601 mm²). - Each outer strip (850 mm wide): remaining steel 453 mm² or minimum 0.12% = 714 mm², whichever is more → 12 mm @ 120 mm c/c (7 bars, 792 mm²). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 475.9$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.323$ MPa; $p_t = 0.209\%$, $\tau_c = 0.327$ MPa (Table 19). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 185.9$ kN, $\tau_v = 0.073$ MPa; $p_t = 0.16\%$, $\tau_c \approx 0.29$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(450+640) + (450+640)] = 4360$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1500.0 - 163.0 \times 1.090 \times 1.090 = 1306.3$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.468$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118$ 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): 16 mm bars → $L_d = 752$ mm; 12 mm bars → $564$ mm. Available length from column face (less 50 mm end cover): 1725 mm along $L$, 875 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 450\times450 = 0.203$ m². Bearing stress $= P_u/A_2 = 7.41$ 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. Capacity of bearing = 1822 kN $\ge$ $P_u = 1500$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 1012 mm²: use 8 bars of 20 mm (2513 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 752$ mm (and the same lap with the column bars). Available depth in footing = 700 - 50 - 28 = 622 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN (4.0 m x 2.3 m; L horizontal) +--------------------------------------------+ | | | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | | | |--+---+---+---+-+----------+--+---+---+---+-| | | | | | | column | | | | | | |--+---+---+---+-+----------+--+---+---+---+-| | | | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | | | +--------------------------------------------+ <-0.85 m><---central band 2.3 m---><-0.85 m> - = 16 mm @ 150 c/c (along L) | = 12 mm along B: @ 100 c/c in central band, @ 120 c/c in outer strips ELEVATION | column | | | +----------------+----------+----------------+ | (dowels) | D = 700 mm | o o o o o o o o o o o o o o | bottom mesh, 50 mm cover +--------------------------------------------+ <----------------- 4000 mm ------------------> ``` - Bottom mesh: 16 mm @ 150 c/c along $L$ (lower layer); along $B$ (upper layer), 12 mm @ 100 c/c in the central 2.3 m band and 12 mm @ 120 c/c in each 0.85 m outer strip; 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 4.0 m × 2.3 m × 700 mm thick; 16 mm @ 150 c/c along the length; along the width, 12 mm @ 100 c/c in the central 2.3 m band and 12 mm @ 120 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the required footing area taken as 1.1P/qa rather than P/qa? A. To allow 10% extra for the weight of footing and backfill B. To apply a load factor of 1.1 for safety C. To account for the moment of 110 kNm D. To reduce the bearing capacity of soil by 10% Answer: A. The weight of footing and backfill is taken as 10% of the column load, so the soil must carry 1.1P. (Key point: Required base area) 2. With e = M/P = 0.110 m and L = 4.0 m, what does e < L/6 tell us? A. The footing must be made deeper B. The base pressure is uniform C. Part of the base is in tension D. There is no tension under the base, so q_min is positive Answer: D. When e is inside the middle third (L/6), the whole base stays in compression and the pressure varies linearly with q_min > 0. (Key point: Eccentricity check) 3. Why is the footing's self-weight excluded when finding the factored soil pressure p_u used for bending? A. It is negligible compared with the column load B. It is already included in the safe bearing capacity C. It acts directly over the soil and causes no bending in the footing D. It is carried by the dowels Answer: C. Self-weight is supported by the soil directly beneath it, so it does not bend the footing slab. (Key point: Factored soil pressure) 4. Why is the depth D = 700 mm decided by the shear checks and not by flexure? A. Flexural capacity 0.138 fck b d^2 is far larger than the moment, so shear controls B. The moment is larger than the limiting moment C. Minimum steel needs a thick footing D. IS 456 fixes footing depth at 700 mm Answer: A. The section could resist the bending moment with much less depth, so one-way or punching shear sets the thickness. (Key point: Depth governed by shear) ## Close variants set in other exams - [Isolated footing 400x400, service moment 120 kNm](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-footing-400x400-service-moment-120) — Design of RCC Structure, chapter 11, asked 2× ## Related in Design of RCC Structure - [Isolated footing 450x450, 1600 kN service](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-footing-450x450-1600-kn-service) — Design of RCC Structure, chapter 11 - [Rectangular footing, 450x450, uniaxial moment 240 kNm](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Arectangular-footing-450x450-uniaxial-moment-240) — Design of RCC Structure, chapter 11 - [Footing with load transfer check, 1000 kN](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Afooting-with-load-transfer-check-1000) — Design of RCC Structure, chapter 11 - [Isolated square footing, 600 kN and 50 kNm](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-square-footing-600-kn-and) — Design of RCC Structure, chapter 11, asked 2× - [Footing, 450x500 column, 6#20, uniaxial moment](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Afooting-450x500-column-6-20-uniaxial) — Design of RCC Structure, chapter 11 - [Footing 300x400 column, factored moment 120 kNm](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Afooting-300x400-column-factored-moment-120) — Design of RCC Structure, chapter 11 ## Related in other subjects - [Brick column design for 80 kN axial load](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Abrick-column-80kn-roof) — Concrete Technology and Masonry Structure, chapter 8 - [Check 25 cm x 25 cm Sal column, 200 kN + 20 kNm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-column-sal-25x25-200kn-20knm) — Design of Steel and Timber Structure, chapter 11 - [Timber column under axial load and bending](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Atimber-column-combined-bending-axial) — Design of Timber and Masonry Structures, chapter 3 - [Select Sal column, 200 kN + 30 kNm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-column-sal-select-bending-3-2m) — Design of Steel and Timber Structure, chapter 11