--- title: "Footing 300x400 column, factored moment 120 kNm" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:footing-300x400-column-factored-moment-120" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-300x400-column-factored-moment-120 site: BCT Engineering --- # Footing 300x400 column, factored moment 120 kNm Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2072 Chaitra, 14 marks - Video: [Design of a Rectangular Footing](https://www.youtube.com/watch?v=LwDIS8RGQ3s) — Stan Academy - 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-300x400-column-factored-moment-120 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated rectangular footing for a column of size 300 mm x 400 mm. The column is carrying 8-20 mm dia bars with M25 concrete. The column is carrying a factored axial load of 1200 kN and the factored moment of 120 kN-m. Sketch the details of designed reinforcements in plan and sections. Also check the bearing stress and development length required. Adopt M20 grade concrete for footing. Grade of steel used is Fe415 [text partly cut off in scan]. Assume bearing capacity of soil = 200 kN/m$^2$ at 1.25 m below GL. ## Short answer Provide a 3.1 m × 1.7 m footing, 700 mm deep. Longitudinal steel 16 mm @ 185 mm c/c; transverse steel central band 12 mm @ 100 mm c/c, outer strips 12 mm @ 115 mm c/c. Bearing stress and development length are satisfactory. ## Answer #### Data and assumptions - Column 400 mm × 300 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M25; column bars 8-20 mm. - Given loads are factored: $P_u = 1200$ kN, $M_u = 120$ kN·m. Service loads = factored / 1.5: $P = 800.0$ kN, $M = 80.0$ kN·m. - Allowable (safe) bearing capacity of soil = 200 kN/m² at 1.25 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 800.0}{200} = 4.40\ \text{m}^2 $$ Eccentricity $e = M/P = 80.0/800.0 = 0.100$ 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 $L \times B = 3.1 \times 1.7$ m ($A = 5.27$ m²): $q_{max} = 199.3$ kN/m² $\le 200.0$ and $q_{min} = 134.7$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 800.0 = 1200.0$ kN, $M_u = 120.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 271.8$ kN/m², $p_{u,min} = 183.6$ kN/m² (average 227.7 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 $L$: $(3100 - 400)/2 = 1350$ mm; along $B$: $(1700 - 300)/2 = 700$ 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 $p_f = 233.4$ kN/m² at the face and $x = 1.35$ m: $M_{u,L} = 401.2$ kN·m. Along $B$ (average pressure): $M_{u,B} = 172.9$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 640$ 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 $B$) | 401.2 | 1799 | 16 mm @ 185 c/c | 1848 | | Along $B$ (width $L$) | 172.9 | 2604 | 12 mm @ 130 c/c | 2697 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 1700$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.82+1} = 0.708$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.708 \times 2604 = 1844$ mm² → 12 mm @ 100 mm c/c (17 bars). - Each outer strip (700 mm wide): remaining steel 380 mm² or minimum 0.12% = 588 mm², whichever is more → 12 mm @ 115 mm c/c (6 bars). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 315.9$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.290$ MPa; $p_t = 0.170\%$, $\tau_c = 0.304$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 42.4$ kN, $\tau_v = 0.021$ 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 $b_o = 2[(a + d) + (c + d)] = 2[(400+640) + (300+640)] = 3960$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1200.0 - 227.7 \times 1.040 \times 0.940 = 977.4$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.386$ 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): 1300 mm along $L$, 650 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 400\times300 = 0.120$ m². Bearing stress $= P_u/A_2 = 10.00$ 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 18.00 MPa. The lesser, 11.25 MPa, governs. Capacity of bearing = 1350 kN $\ge$ $P_u = 1200$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 600 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 ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 700 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 3.1 m, B = 1.7 m ``` - Bottom mesh: 16 mm @ 185 c/c along $L$ (lower layer); along $B$ (upper layer), 12 mm @ 100 c/c in the central 1.7 m band and 12 mm @ 115 c/c in each outer strip, 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 3.1 m × 1.7 m × 700 mm thick; 16 mm @ 185 c/c along the length; along the width, 12 mm @ 100 c/c in the central 1.7 m band and 12 mm @ 115 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With e = M/P = 0.100 m < L/6 = 0.517 m, how does soil pressure vary? A. Uniform over the base B. Trapezoidal, p_max = 271.8 and p_min = 183.6 kN/m^2, both compressive C. Triangular with tension on one side D. Zero at the centre Answer: B. Small eccentricity gives P/A(1 +- 6e/L) with both ends in compression; q_max is limited to the safe bearing capacity. (Key point: Eccentricity and pressure) 2. How is steel placed in the short direction (B = 1700 mm)? A. Uniformly over the length B. 2/(beta+1) = 0.708 of it in a central band of width B, the rest in outer strips (min 0.12%) C. All in the outer strips D. Half in each strip Answer: B. IS 456 cl. 34.3.1 concentrates a fraction 2/(beta+1) of the short-direction steel in the central band under the column. (Key point: Short-direction steel in rectangular footing) 3. Column is M25, footing M20. Which limit applies and what follows? A. 0.45 f_ck = 11.25 MPa (column) governs; bearing capacity 1350 kN >= 1200 kN so minimum dowels 0.5% of column area (600 mm^2) suffice B. 9 MPa governs and dowels carry everything C. 18 MPa governs D. No bearing check required Answer: A. Lesser of column 11.25 MPa and footing up to 18 MPa governs; bearing capacity exceeds P_u, so only minimum dowels are needed. (Key point: Bearing at column base) 4. Short-direction 12 mm bars (L_d = 564 mm) have only 650 mm available from column face. What is concluded? A. Not satisfied B. 564 mm is less than 650 mm, satisfied C. 12 mm bars need 752 mm D. Check is skipped Answer: B. L_d = 0.87 f_y phi/(4 tau_bd) = 564 mm for 12 mm bars, which fits within the available 650 mm. 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