--- title: "Footing, 450x500 column, 6#20, uniaxial moment" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:footing-450x500-column-6-20-uniaxial" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-450x500-column-6-20-uniaxial site: BCT Engineering --- # Footing, 450x500 column, 6#20, uniaxial moment Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2080 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/footing-450x500-column-6-20-uniaxial - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing for a 450 mm x 500 mm sized column, with 6#20 mm diameter bars. Carrying factored axial load of 1100 kN and factored uniaxial moment of 120 kN-m at the column base. Take the depth of footing 1.5 m and safe bearing capacity of soil 100 kN/m$^2$. Use M20 concrete and Fe500 grade steel. ## Short answer Provide a 4.1 m × 2.3 m footing, 650 mm deep. Along the 4.1 m side, use 12 mm bars @ 130 mm c/c; along the 2.3 m side, place a central band of 21 bars (12 mm @ 105 mm c/c) and outer strips of 7 bars each (12 mm @ 125 mm c/c). ## Answer #### Data and assumptions - Column 500 mm × 450 mm (long side $a=500$ mm along the footing length); footing concrete M20, steel Fe500; column concrete M20; column bars 6-20 mm. - Given loads are factored: $P_u = 1100$ kN, $M_u = 120$ kN·m. Service loads = factored / 1.5: $P = 733.3$ kN, $M = 80.0$ kN·m. - Allowable (safe) bearing capacity of soil = 100 kN/m² at 1.5 m depth. - Column concrete grade assumed M20 (same as footing). Moment acts about the minor axis so that it varies the pressure along the 500 mm side. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 733.3}{100} = 8.07\ \text{m}^2 $$ Eccentricity $e = M/P = 80.0/733.3 = 0.109$ 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 $L \times B = 4.1 \times 2.3$ m ($A = 9.43$ m²): $q_{max} = 99.2$ kN/m² $\le 100.0$ and $q_{min} = 71.9$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 733.3 = 1100.0$ kN, $M_u = 120.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 135.3$ kN/m², $p_{u,min} = 98.0$ kN/m² (average 116.6 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 $L$: $(4100 - 500)/2 = 1800$ mm; along $B$: $(2300 - 450)/2 = 925$ 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 $p_f = 118.9$ kN/m² at the face and $x = 1.8$ m: $M_{u,L} = 483.7$ kN·m. Along $B$ (average pressure): $M_{u,B} = 204.6$ 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 $B$) | 483.7 | 1956 | 12 mm @ 130 c/c | 2001 | | Along $B$ (width $L$) | 204.6 | 3198 | 12 mm @ 140 c/c | 3312 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 2300$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.78+1} = 0.719$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.719 \times 3198 = 2299$ mm² → 12 mm @ 105 mm c/c (21 bars). - Each outer strip (900 mm wide): remaining steel 450 mm² or minimum 0.12% = 702 mm², whichever is more → 12 mm @ 125 mm c/c (7 bars). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 361.2$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.266$ MPa; $p_t = 0.147\%$, $\tau_c = 0.288$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 160.2$ kN, $\tau_v = 0.066$ MPa; $p_t = 0.137\%$, $\tau_c = 0.288$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(500+590) + (450+590)] = 4260$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1100.0 - 116.6 \times 1.090 \times 1.040 = 967.8$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.385$ 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): 1750 mm along $L$, 875 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 500\times450 = 0.225$ m². Bearing stress $= P_u/A_2 = 4.89$ 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 = 2025 kN $\ge$ $P_u = 1100$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 1125 mm²: use 6 bars of 20 mm (1885 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 906$ 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 = 4.1 m, B = 2.3 m ``` - Bottom mesh: 12 mm @ 130 c/c along $L$ (lower layer); along $B$ (upper layer), 12 mm @ 105 c/c in the central 2.3 m band and 12 mm @ 125 c/c in each outer strip, 50 mm clear cover; dowels 6-20 mm. **Answer:** Footing 4.1 m × 2.3 m × 650 mm thick; 12 mm @ 130 c/c along the length; along the width, 12 mm @ 105 c/c in the central 2.3 m band and 12 mm @ 125 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The loads given are factored (1100 kN, 120 kNm). How is the footing size checked against the safe bearing capacity? A. Use factored loads directly with qa B. Divide by 1.5 to get service loads P = 733.3 kN and M = 80 kNm C. Multiply by 1.5 again D. Ignore the moment Answer: B. Safe bearing capacity relates to service loads, so factored values are divided by 1.5 before sizing. (Key point: Service vs factored loads) 2. The moment is taken about the minor axis, so the pressure varies along which side? A. The 450 mm side B. Both sides equally C. Neither side D. The 500 mm (long) side along footing length L Answer: D. The footing length L is aligned with the 500 mm side, and the pressure varies linearly along L, with e = M/P = 0.109 m < L/6. (Key point: Moment direction) 3. Along L, where is the critical section for bending and which pressure is used? A. At the column face on the side of maximum pressure, with linear pressure B. At d from the face with average pressure C. At the footing centre D. At the footing edge Answer: A. The cantilever moment is taken at the column face on the more heavily loaded side, using the linear pressure. (Key point: Bending moment on the long projection) 4. For Fe500 the limiting moment is written 0.133 fck b d^2 (instead of 0.138). Why? A. Concrete grade changed B. Higher-strength steel has a smaller limiting neutral axis depth, so lower Mu,lim C. Cover is larger D. Footing is rectangular Answer: B. Fe500 has a smaller xu,max/d than Fe415, which reduces the limiting moment coefficient. 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