--- title: "Footing for 30x35 cm column, M25" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:footing-for-30x35-cm-column-m25" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-for-30x35-cm-column-m25 site: BCT Engineering --- # Footing for 30x35 cm column, M25 Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2079 Bhadra, 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/footing-for-30x35-cm-column-m25 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design a footing for a rectangular column of size 30 cm x 35 cm reinforced with 8-20 mm dia. bars. The column is subjected to a factored axial load and moment of 1000 kN and 80 kN-m, respectively. The allowable bearing capacity of soil is 140 kN/m$^2$. At a depth of 1.6 m. Use M25 concrete and TMT bars for column and footing both. Sketch all of the reinforcement required. ## Short answer Provide a 3.2 m × 1.9 m footing, 650 mm thick. Reinforcement: along length, 12 mm bars @ 140 mm c/c; along width, central band with 17 bars of 12 mm @ 110 mm c/c and each outer strip with 5 bars of 12 mm @ 130 mm c/c. ## Answer #### Data and assumptions - Column 350 mm × 300 mm (long side $a=350$ mm along the footing length); footing concrete M25, steel Fe500; column concrete M25; column bars 8-20 mm. - Given loads are factored: $P_u = 1000$ kN, $M_u = 80$ kN·m. Service loads = factored / 1.5: $P = 666.7$ kN, $M = 53.3$ kN·m. - Allowable (safe) bearing capacity of soil = 140 kN/m² at 1.6 m depth. - TMT bars taken as Fe500. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 666.7}{140} = 5.24\ \text{m}^2 $$ Eccentricity $e = M/P = 53.3/666.7 = 0.080$ 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 $L \times B = 3.2 \times 1.9$ m ($A = 6.08$ m²): $q_{max} = 138.7$ kN/m² $\le 140.0$ and $q_{min} = 102.5$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 666.7 = 1000.0$ kN, $M_u = 80.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 189.1$ kN/m², $p_{u,min} = 139.8$ kN/m² (average 164.5 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 $L$: $(3200 - 350)/2 = 1425$ mm; along $B$: $(1900 - 300)/2 = 800$ mm. Along $L$ (critical section at the column face on the side of maximum pressure, linear pressure): $M_{u,L} = B\left[p_f\dfrac{x^2}{2} + (p_{max}-p_f)\dfrac{x^2}{3}\right]$, with $p_f = 167.2$ kN/m² at the face and $x = 1.425$ m: $M_{u,L} = 350.7$ kN·m. Along $B$ (average pressure): $M_{u,B} = 168.4$ kN·m. Required depth for flexure: $M_{u,lim} = 0.133 f_{ck} b d^2$ is far larger than the moment for $d = 590$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 650$ mm, $d = 590$ 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 $B$) | 350.7 | 1482 (minimum) | 12 mm @ 140 c/c | 1535 | | Along $B$ (width $L$) | 168.4 | 2496 | 12 mm @ 140 c/c | 2585 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 1900$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.68+1} = 0.745$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.745 \times 2496 = 1860$ mm² → 12 mm @ 110 mm c/c (17 bars). - Each outer strip (650 mm wide): remaining steel 318 mm² or minimum 0.12% = 507 mm², whichever is more → 12 mm @ 130 mm c/c (5 bars). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 289.9$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.259$ MPa; $p_t = 0.137\%$, $\tau_c = 0.291$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 110.5$ kN, $\tau_v = 0.059$ MPa; $p_t = 0.137\%$, $\tau_c = 0.291$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(350+590) + (300+590)] = 3660$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1000.0 - 164.5 \times 0.940 \times 0.890 = 862.4$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.399$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.250$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 2.24$ MPa (M25, deformed bars): 12 mm bars (both directions) → $L_d = 583$ mm. Available length from column face (less 50 mm end cover): 1375 mm along $L$, 750 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 350\times300 = 0.105$ m². Bearing stress $= P_u/A_2 = 9.52$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 11.25$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 22.50 MPa. The lesser, 11.25 MPa, governs. Capacity of bearing = 1181 kN $\ge$ $P_u = 1000$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 525 mm²: use 8 bars of 20 mm (2513 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 777$ mm (and the same lap with the column bars). Available depth in footing = 650 - 50 - 24 = 576 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 650 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 3.2 m, B = 1.9 m ``` - Bottom mesh: 12 mm @ 140 c/c along $L$ (lower layer); along $B$ (upper layer), 12 mm @ 110 c/c in the central 1.9 m band and 12 mm @ 130 c/c in each outer strip, 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 3.2 m × 1.9 m × 650 mm thick; 12 mm @ 140 c/c along the length; along the width, 12 mm @ 110 c/c in the central 1.9 m band and 12 mm @ 130 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Given factored P_u = 1000 kN and M_u = 80 kNm, the soil pressure check uses: A. Service P = 666.7 kN and M = 53.3 kNm B. Factored loads directly C. P = 1500 kN D. Only the axial load Answer: A. Dividing by the load factor 1.5 gives service loads for comparison with the allowable bearing capacity. (Key point: Load conversion) 2. Why are only minimum dowels (0.5% of column area, 525 mm^2) needed? A. The column bars are not continued B. Bearing capacity 1181 kN is at least Pu = 1000 kN, so bearing transfers the whole load C. The footing is very deep D. Steel is Fe500 Answer: B. With 0.45 fck = 11.25 MPa governing, the capacity exceeds Pu, so dowels just need to meet the minimum. (Key point: Bearing transfer) 3. For the 3.2 m x 1.9 m footing, how much short-direction steel lies in the central band of width B? A. Half of it B. 1/beta of it C. All of it D. 2/(beta+1) = 0.745 of it, the rest in the outer strips Answer: D. IS 456 cl. 34.3.1 puts 2/(beta+1) of the steel in the central band, with beta = L/B = 1.68; outer strips get the remainder or minimum steel. 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