--- title: "Rectangular footing, 600x600 column, 1150 kN" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:rectangular-footing-600x600-column-1150-kn" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/rectangular-footing-600x600-column-1150-kn site: BCT Engineering --- # Rectangular footing, 600x600 column, 1150 kN Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2072 Kartik, 16 marks - Video: [How to decide the size of footing? | Area of footing | Design of RCC footing | Civil Tutor](https://www.youtube.com/watch?v=HAyyvpl0vlY) — Civil Tutor - 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/rectangular-footing-600x600-column-1150-kn - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design a rectangle footing to carry a column load of 1150 kN and BM of 250 kN-m from 600 x 600 mm square column with the 20 mm diameter steel. The bearing capacity of soil is 200 kN/m$^2$. Consider depth of foundation as 1.5 m. Take unit weight of earth is 17 kN/m$^3$. Use M20 concrete and Fe 415 steel. ## Short answer Footing size 4.0 m × 2.2 m, 850 mm deep. Longitudinal steel 16 mm @ 155 mm c/c; transverse steel central band 16 mm @ 145 mm c/c and outer strips 16 mm @ 175 mm c/c. ## Answer #### Data and assumptions - Column 600 mm × 600 mm (long side $a=600$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm. - Service load $P = 1150$ kN, service moment $M = 250$ kN·m. - Allowable (safe) bearing capacity of soil = 200 kN/m² at 1.5 m depth, soil unit weight 17 kN/m³. - Loads are service loads. - 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 Net safe bearing capacity = $q - \gamma D_f = 200 - 17 \times 1.5 = 174.50$ kN/m² (this takes care of the weight of footing and backfill). $$ A_{req} = \frac{P}{q_{net}} = \frac{1150.0}{174.50} = 6.59\ \text{m}^2 $$ Eccentricity $e = M/P = 250.0/1150.0 = 0.217$ m, which is less than $L/6$, so there is no tension under the base. The size is chosen so that $$ q_{max} = \frac{P}{BL}\left(1 + \frac{6e}{L}\right) \le q_a $$ Try $L \times B = 4.0 \times 2.2$ m ($A = 8.80$ m²): $q_{max} = 173.3$ kN/m² $\le 174.5$ and $q_{min} = 88.1$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 1150.0 = 1725.0$ kN, $M_u = 375.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 259.9$ kN/m², $p_{u,min} = 132.1$ kN/m² (average 196.0 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 $L$: $(4000 - 600)/2 = 1700$ mm; along $B$: $(2200 - 600)/2 = 800$ 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 $p_f = 205.6$ kN/m² at the face and $x = 1.7$ m: $M_{u,L} = 768.8$ kN·m. Along $B$ (average pressure): $M_{u,B} = 250.9$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 790$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 850$ mm, $d = 790$ 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 $B$) | 768.8 | 2790 | 16 mm @ 155 c/c | 2854 | | Along $B$ (width $L$) | 250.9 | 4080 | 16 mm @ 195 c/c | 4124 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 2200$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.82+1} = 0.710$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.710 \times 4080 = 2895$ mm² → 16 mm @ 145 mm c/c (15 bars). - Each outer strip (900 mm wide): remaining steel 592 mm² or minimum 0.12% = 918 mm², whichever is more → 16 mm @ 175 mm c/c (5 bars). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 491.3$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.283$ MPa; $p_t = 0.164\%$, $\tau_c = 0.299$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 7.8$ kN, $\tau_v = 0.002$ MPa; $p_t = 0.131\%$, $\tau_c = 0.288$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(600+790) + (600+790)] = 5560$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1725.0 - 196.0 \times 1.390 \times 1.390 = 1346.3$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.306$ 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 (both directions) → $L_d = 752$ mm. Available length from column face (less 50 mm end cover): 1650 mm along $L$ (adequate), 750 mm along $B$. Along $B$ the available 750 mm is slightly less than $L_d = 752$ mm, so the bars are given 90° bends at the ends (each bend adds an anchorage value of $8\phi = 128$ mm), which is adequate. #### Bearing stress and transfer of load from column Column area $A_2 = 600\times600 = 0.360$ m². Bearing stress $= P_u/A_2 = 4.79$ 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 = 3240 kN $\ge$ $P_u = 1725$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 1800 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 = 850 - 50 - 32 = 768 mm $\ge L_{dc}$, **adequate**. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 850 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 4.0 m, B = 2.2 m ``` - Bottom mesh: 16 mm @ 155 c/c along $L$ (lower layer); along $B$ (upper layer), 16 mm @ 145 c/c in the central 2.2 m band and 16 mm @ 175 c/c in each outer strip, 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 4.0 m × 2.2 m × 850 mm thick; 16 mm @ 155 c/c along the length; along the width, 16 mm @ 145 c/c in the central 2.2 m band and 16 mm @ 175 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In sizing the footing, the net safe bearing capacity is 200 - 17 x 1.5 = 174.5 kN/m^2. Why is the term 17 x 1.5 subtracted? A. It accounts for the weight of footing and backfill above the base B. It is the factor of safety applied to the soil C. It accounts for the uplift due to the moment D. It converts service pressure to factored pressure Answer: A. The overburden (footing plus backfill) is already carried by the soil at foundation level, so it is deducted from the bearing capacity. (Key point: Net safe bearing capacity) 2. The eccentricity e = M/P = 0.217 m is less than L/6 for L = 4.0 m. What does this tell us? A. Shear reinforcement is required B. The footing must be made square C. There is no tension under the base (q_min > 0) D. The footing will overturn Answer: C. When e < L/6 the resultant lies within the middle third, so the whole base stays in compression. (Key point: Eccentricity check) 3. Why are factored pressures (p_u) used for bending and shear design, and what governs the depth here? A. Service loads are used for flexure; depth is governed by minimum steel B. Load factor 1.5 is applied for limit state design; depth is governed by shear, not flexure C. Factored loads include footing self-weight; depth is governed by bearing D. Factored loads are used only for the soil check; depth is governed by development length Answer: B. Structural design is by limit state with 1.5 x service load, and M_u,lim is far larger than the moment, so one-way or punching shear fixes the depth. (Key point: Factored loads and depth governing) 4. For the short-direction steel in a rectangular footing, what fraction of the total steel is placed in the central band of width B? A. 0.5 B. L/B C. 2/(beta+1), where beta = L/B (about 0.71 here) D. 1/(beta+1) Answer: C. IS 456 cl. 34.3.1 puts 2/(beta+1) of the short-direction steel in the central band; the rest goes in the outer strips. 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