Steps of rectangular footing design

Steps for design of a rectangular RC footing (isolated)

A rectangular footing is used when the column is rectangular, when there is a moment on the column, or when there is a space restriction on one side.

1. Load and area: take the column service load $P$ (add about 10% for the weight of footing and backfill) and the safe bearing capacity $q_a$. $$A_{req} = \frac{1.1P}{q_a}$$
2. Dimensions: choose $L \times B$ so that the projections beyond the column faces are equal in both directions, i.e. $L - a = B - c$ (where $a \times c$ is the column size), with $LB \ge A_{req}$. This gives about equal cantilever moments per metre width in both directions. If the column carries a moment $M$, check $q_{max} = \dfrac{1.1P}{LB}\left(1 + \dfrac{6e}{L}\right) \le q_a$ and $q_{min} \ge 0$ with $e = M/P$.
3. Factored upward soil pressure: $p_u = \dfrac{P_u}{LB}$ ($P_u = 1.5P$) (or the maximum pressure for a footing with moment). The self-weight is not included in bending and shear design.
4. Depth from the greater of: (a) one-way shear at a distance $d$ from the column face, and (b) punching shear at $d/2$ from the column face (permissible stress $k_s \cdot 0.25\sqrt{f_{ck}}$), and the bending requirement. Adopt $D$ and $d$.
5. Bending moment at the face of the column in each direction: $$M_{u,L} = p_u B\frac{(L-a)^2}{8}, \qquad M_{u,B} = p_u L\frac{(B-c)^2}{8}$$ Find $A_{st}$ in each direction (minimum 0.12% of $bD$).
6. Distribution of steel: long-direction bars are spread uniformly over the width $B$. For short-direction bars, a fraction $\dfrac{2}{\beta + 1}$ of the total ($\beta = L/B$) is concentrated in a central band of width $B$; the rest is spread equally in the two outer portions (IS 456 cl. 34.3.1).
7. Checks: one-way shear $\tau_v \le \tau_c$ in both directions, punching shear, development length of the bars ($L_d \le$ available length from the face of the column), and bearing at the column-footing interface; provide dowels.
8. Detail: bars at the bottom, with 50 mm cover, plan and sections.

Why shear reinforcement is not provided in a footing

• A footing is a thick member, and its depth is made large enough that the shear stress is less than the permissible shear stress of concrete (for both one-way and punching shear). The concrete alone carries the shear, so stirrups are not needed.
• Stirrups in a thick, heavy slab cast against soil are difficult to place and hold, and would cause congestion.
• Bent-up bars or shear reinforcement are not efficient against punching in a thick footing, and the depth is easily increased at little cost.
• Increasing the depth also gives rigidity, so the assumption of a uniform soil pressure is better satisfied.