Conventional design of mat foundation

The conventional (rigid) method treats the mat as an infinitely rigid body, so the soil pressure follows a straight line (planar) distribution.

Assumptions

• The mat is rigid, so its deflection is small compared with the soil's.
• The soil pressure is linearly distributed, with its centroid on the line of action of the resultant of the column loads.

Steps

1. Total load: $Q = Q_1 + Q_2 + \dots + Q_n$ (column loads, plus mat weight if not included).
2. Eccentricities: the resultant lies at $(\bar x, \bar y)$ from the centre; $e_x = \dfrac{\sum Q_ix_i}{Q}$ and $e_y = \dfrac{\sum Q_iy_i}{Q}$. So $M_x = Qe_y$ and $M_y = Qe_x$.
3. Soil pressure at any point $(x, y)$:

$$q = \frac{Q}{A} \pm \frac{M_y\,x}{I_y} \pm \frac{M_x\,y}{I_x}$$

with $A = BL$, $I_x = \dfrac{BL^3}{12}$ and $I_y = \dfrac{LB^3}{12}$. Check that $q_{max} \le q_{all}$ and $q_{min} > 0$ (no tension). 4. Divide into strips: separate the mat into strips of width $B_1$ in each direction, each strip bounded by the centre lines between columns. 5. Strip loads: for a strip of length $L_1$, find the average soil pressure $q_{av}$ (from the pressures at its two ends) and the total soil reaction $q_{av}B_1L_1$. The sum of the column loads on the strip, $\Sigma Q$, is generally not equal to this reaction. 6. Modify the loads so the strip is in equilibrium. Take the average load $Q_{av} = \dfrac{\Sigma Q + q_{av}B_1L_1}{2}$. Multiply the column loads by $F_c = \dfrac{Q_{av}}{\Sigma Q}$ and the soil pressure by $F_s = \dfrac{Q_{av}}{q_{av}B_1L_1}$. 7. Shear force and bending moment diagrams of the strip loaded with the modified column loads and the soil pressure, treated as a continuous beam. 8. Design: reinforcement for the moments, and check for punching (two-way) shear at columns and beam shear, as in IS 456.

 Column loads Q1  Q2  Q3
              v   v   v
        ______________________
        ^^^^^^^^^^^^^^^^^^^^^^   <- q (soil pressure)