--- title: "Isolated footing 450x450, 1600 kN service" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-footing-450x450-1600-kn-service" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-footing-450x450-1600-kn-service site: BCT Engineering --- # Isolated footing 450x450, 1600 kN service Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2073 Shrawan, 14 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-450x450-1600-kn-service - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing for a square column 450 mm x 450 mm, reinforcement with 8-20 dia bars and carrying a service load of 1600 kN. Assume bearing capacity of soil as 250 kN/m$^2$ and depth of foundation as 1.5 m. Adopt M20 concrete and Fe 500 steel. Also check the development length and bearing stress in concrete. ## Short answer Provide a 2.7 m × 2.7 m footing, 700 mm deep, with 12 mm bars at 130 mm c/c both ways. Development length 680 mm is satisfied. Bearing stress requires dowels; column bars extended into footing are adequate. ## Answer #### Data and assumptions - Column 450 mm × 450 mm (long side $a=450$ mm along the footing length); footing concrete M20, steel Fe500; column concrete M20; column bars 8-20 mm. - Service load $P = 1600$ kN. - Allowable (safe) bearing capacity of soil = 250 kN/m² at 1.5 m depth. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1600.0}{250} = 7.04\ \text{m}^2 $$ Provide a square footing $L \times B = 2.7 \times 2.7$ m, area = 7.29 m² $\ge$ 7.04 m². #### Factored soil pressure $P_u = 1.5 \times 1600.0 = 2400.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{2400.0}{2.7 \times 2.7} = 329.2$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(2700 - 450)/2 = 1125$ mm; along $B$: $(2700 - 450)/2 = 1125$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 562.5$ kN·m, $M_{u,B} = 562.5$ kN·m. Required depth for flexure: $M_{u,lim} = 0.133 f_{ck} b d^2$ 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 $B$) | 562.5 | 2268 | 12 mm @ 130 c/c | 2349 | | Along $B$ (width $L$) | 562.5 | 2268 | 12 mm @ 130 c/c | 2349 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 431.1$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.249$ MPa; $p_t = 0.136\%$, $\tau_c = 0.288$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 431.1$ kN, $\tau_v = 0.249$ MPa; $p_t = 0.136\%$, $\tau_c = 0.288$ 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 - $V_u = P_u - p_u (a+d)(c+d) = 2400.0 - 329.2 \times 1.090 \times 1.090 = 2008.9$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.720$ 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): 12 mm bars (both directions) → $L_d = 680$ mm. Available length from column face (less 50 mm end cover): 1075 mm along $L$, 1075 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 $= P_u/A_2 = 11.85$ 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 $<$ $P_u = 2400$ kN, so the excess 578 kN must be carried by dowels: $A_{s,dow} = \dfrac{577500}{0.87 f_y} = 1328$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels, which is adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 906$ mm (and the same lap with the column bars). Available depth in footing = 700 - 50 - 24 = 626 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 700 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 2.7 m, B = 2.7 m ``` - Bottom mesh: 12 mm @ 130 c/c along $L$ (lower layer), 12 mm @ 130 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 2.7 m × 2.7 m × 700 mm thick; 12 mm @ 130 c/c along the length and 12 mm @ 130 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Bearing stress P_u/A_2 = 11.85 MPa exceeds the permissible 9.00 MPa. What is the consequence? A. The column must be enlarged B. Load beyond bearing capacity (1822 kN) must be carried by dowels C. The footing must be redesigned in M30 D. Bearing need not be checked Answer: B. Bearing capacity is 0.45 f_ck x A_2 = 1822 kN < 2400 kN, so the excess 578 kN passes through dowels (1328 mm^2 needed, 2513 mm^2 available). (Key point: Bearing stress check) 2. Which value governs the permissible bearing stress at the column-footing interface? A. The greater of column and footing values B. The lesser: column 0.45 f_ck = 9.00 MPa versus footing up to 18.00 MPa C. Always 18 MPa D. 0.87 f_y Answer: B. The lesser of the two permissible stresses is used, as the weaker side would crush first. (Key point: Permissible bearing stress) 3. Which expression gives development length of footing bars and what is the check? A. L_d = 0.87 f_y phi/(4 tau_bd) = 680 mm for 12 mm bars, which is less than the 1075 mm available from column face B. L_d = 0.87 f_y/phi, 12 mm C. L_d = 40 phi always, compared with D D. L_d is not needed for footings Answer: A. Bars must be anchored beyond the critical section at column face; available 1075 mm exceeds required 680 mm, so satisfied. (Key point: Development length) 4. Why does L_dc = 0.87 f_y phi/(4 x 1.25 tau_bd) = 906 mm contain the factor 1.25? A. Bond stress in compression is increased by 25%, making the length shorter B. Steel stress is raised 25% C. It is a safety factor increasing length D. It allows for cover Answer: A. IS 456 allows 25% higher design bond stress for bars in compression, so the compression length is less than in tension. 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