Footing for 400x300 column, reversible moment

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 8-16 mm.
• Service load $P = 1000$ kN, service moment $M = 100$ kN·m.
• Allowable (safe) bearing capacity of soil = 200 kN/m² at 1.5 m depth.
• Loads are service loads. The moment is reversible, so the same steel is provided on both sides.
• 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.

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1000.0}{200} = 5.50\ \text{m}^2$$

Eccentricity $e = M/P = 100.0/1000.0 = 0.100$ 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 $L \times B = 3.6 \times 1.8$ m ($A = 6.48$ m²): $q_{max} = 198.0$ kN/m² $\le 200.0$ and $q_{min} = 141.5$ kN/m² $> 0$. Safe.

Factored soil pressure

$P_u = 1.5 \times 1000.0 = 1500.0$ kN, $M_u = 150.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 270.1$ kN/m², $p_{u,min} = 192.9$ kN/m² (average 231.5 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 $L$: $(3600 - 400)/2 = 1600$ mm; along $B$: $(1800 - 300)/2 = 750$ 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 $p_f = 235.8$ kN/m² at the face and $x = 1.6$ m: $M_{u,L} = 595.9$ kN·m. Along $B$ (average pressure): $M_{u,B} = 234.4$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 740$ mm, so the depth is governed by shear (one-way or punching). Trial $D = 800$ mm, $d = 740$ 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 $bD$.

For the rectangular footing, the short-direction steel is placed in a central band of width $B = 1800$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{2.00+1} = 0.667$ (IS 456 cl. 34.3.1), where $\beta = L/B$.

• Central band: $0.667 \times 3456 = 2304$ mm² → 16 mm @ 150 mm c/c (12 bars).
• Each outer strip (900 mm wide): remaining steel 576 mm² or minimum 0.12% = 864 mm², whichever is more → 16 mm @ 180 mm c/c (5 bars).

Check for one-way shear (at distance $d$ from the column face)

• Along $L$ side: $V_u = 403.8$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.303$ MPa; $p_t = 0.181\%$, $\tau_c = 0.312$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 8.3$ kN, $\tau_v = 0.003$ MPa; $p_t = 0.133\%$, $\tau_c = 0.288$ MPa. Safe.

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

• Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+740) + (300+740)] = 4360$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 1500.0 - 231.5 \times 1.140 \times 1.040 = 1225.6$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.380$ 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): 1550 mm along $L$ (adequate), 700 mm along $B$. Along $B$ the available 700 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 $A_2 = 400\times300 = 0.120$ m². Bearing stress $= P_u/A_2 = 12.50$ 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 $<$ $P_u = 1500$ kN, so the excess 150 kN must be carried by dowels: $A_{s,dow} = \dfrac{150000}{0.87 f_y} = 415$ mm²; the column bars (8-16 mm, 1608 mm²) are continued into the footing as dowels, which is adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 602$ mm (and the same lap with the column bars). Available depth in footing = 800 - 50 - 32 = 718 mm $\ge L_{dc}$, adequate.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 800 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 3.6 m, B = 1.8 m

• Bottom mesh: 16 mm @ 150 c/c along $L$ (lower layer); along $B$ (upper layer), 16 mm @ 150 c/c in the central 1.8 m band and 16 mm @ 180 c/c in each outer strip, 50 mm clear cover; dowels 8-16 mm.

Answer: Footing 3.6 m × 1.8 m × 800 mm thick; 16 mm @ 150 c/c along the length; along the width, 16 mm @ 150 c/c in the central 1.8 m band and 16 mm @ 180 c/c in the outer strips.