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Footing 300x400 column, factored moment 120 kNm

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.

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/m2^2 at 1.25 m below GL.

Data and assumptions

  • Column 400 mm × 300 mm (long side a=400a=400 mm along the footing length); footing concrete M20, steel Fe415; column concrete M25; column bars 8-20 mm.
  • Given loads are factored: Pu=1200P_u = 1200 kN, Mu=120M_u = 120 kN·m. Service loads = factored / 1.5: P=800.0P = 800.0 kN, M=80.0M = 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−60d = 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.

Areq=1.1 Pqa=1.1×800.0200=4.40 m2A_{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=0.100e = M/P = 80.0/800.0 = 0.100 m, which is less than L/6L/6, so there is no tension under the base. The size is chosen so that

qmax=1.1PBL(1+6eL)≤qaq_{max} = 1.1\frac{P}{BL}\left(1 + \frac{6e}{L}\right) \le q_a

Try L×B=3.1×1.7L \times B = 3.1 \times 1.7 m (A=5.27A = 5.27 m²): qmax=199.3q_{max} = 199.3 kN/m² ≤200.0\le 200.0 and qmin=134.7q_{min} = 134.7 kN/m² >0> 0. Safe.

Factored soil pressure

Pu=1.5×800.0=1200.0P_u = 1.5 \times 800.0 = 1200.0 kN, Mu=120.0M_u = 120.0 kN·m. pu=PuBL(1±6eL)p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right): pu,max=271.8p_{u,max} = 271.8 kN/m², pu,min=183.6p_{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 LL: (3100−400)/2=1350(3100 - 400)/2 = 1350 mm; along BB: (1700−300)/2=700(1700 - 300)/2 = 700 mm. Along LL (critical section at the column face on the side of maximum pressure, linear pressure): Mu,L=B[pfx22+(pmax−pf)x23]M_{u,L} = B\left[p_f\dfrac{x^2}{2} + (p_{max}-p_f)\dfrac{x^2}{3}\right], with pf=233.4p_f = 233.4 kN/m² at the face and x=1.35x = 1.35 m: Mu,L=401.2M_{u,L} = 401.2 kN·m. Along BB (average pressure): Mu,B=172.9M_{u,B} = 172.9 kN·m. Required depth for flexure: Mu,lim=0.138fckbd2M_{u,lim} = 0.138 f_{ck} b d^2 is far larger than the moment for d=640d = 640 mm, so the depth is governed by shear (one-way or punching). Trial D=700D = 700 mm, d=640d = 640 mm, found by the shear checks below.

Ast=0.5fckfy[1−1−4.6Mufckbd2]bdA_{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 bDbD.

DirectionMuM_u (kN·m)AstA_{st} req (mm²)ProvideAstA_{st} prov (mm²)
Along LL (width BB)401.2179916 mm @ 185 c/c1848
Along BB (width LL)172.9260412 mm @ 130 c/c2697

For the rectangular footing, the short-direction steel is placed in a central band of width B=1700B = 1700 mm in the proportion 2β+1=21.82+1=0.708\dfrac{2}{\beta+1} = \dfrac{2}{1.82+1} = 0.708 (IS 456 cl. 34.3.1), where β=L/B\beta = L/B.

  • Central band: 0.708×2604=18440.708 \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 dd from the column face)

  • Along LL side: Vu=315.9V_u = 315.9 kN, τv=VuBd=0.290\tau_v = \dfrac{V_u}{B d} = 0.290 MPa; pt=0.170%p_t = 0.170\%, τc=0.304\tau_c = 0.304 MPa (IS 456 Table 19, SP 16 formula). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=42.4V_u = 42.4 kN, τv=0.021\tau_v = 0.021 MPa; pt=0.136%p_t = 0.136\%, τc=0.288\tau_c = 0.288 MPa. Safe.

Check for two-way (punching) shear (at d/2d/2 from the column face)

  • Perimeter bo=2[(a+d)+(c+d)]=2[(400+640)+(300+640)]=3960b_o = 2[(a + d) + (c + d)] = 2[(400+640) + (300+640)] = 3960 mm
  • Vu=Pu−pu(a+d)(c+d)=1200.0−227.7×1.040×0.940=977.4V_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).
  • τv=Vubod=0.386\tau_v = \dfrac{V_u}{b_o d} = 0.386 MPa; ks=0.5+βc=1.00k_s = 0.5 + \beta_c = 1.00, τc′=ks×0.25fck=1.118\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118 MPa. τv<τc′\tau_v < \tau_c', safe.

Development length

Ld=0.87fyϕ4τbdL_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}, τbd=1.92\tau_{bd} = 1.92 MPa (M20, deformed bars): 16 mm bars → Ld=752L_d = 752 mm; 12 mm bars → 564564 mm. Available length from column face (less 50 mm end cover): 1300 mm along LL, 650 mm along BB. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area A2=400×300=0.120A_2 = 400\times300 = 0.120 m². Bearing stress =Pu/A2=10.00= P_u/A_2 = 10.00 MPa. Permissible bearing stress: column concrete 0.45fck=11.250.45 f_{ck} = 11.25 MPa; footing concrete 0.45fckA1/A20.45 f_{ck}\sqrt{A_1/A_2} with A1/A2≤2\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 Pu=1200P_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 Ldc=0.87fyϕ4×1.25 τbd=752L_{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 <Ldc< L_{dc}; dowels are bent at the bottom with 90° bends.

Detailing

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 LL (lower layer); along BB (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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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