Rectangular foundation, stress at point P

Method

Divide the 4 m × 5 m rectangle into four rectangles that have P as a common corner. The stress under a corner of a rectangle is $\Delta\sigma_z = q\,I$ where $I = f(m, n)$, $m = B/z$ and $n = L/z$. Here $z = 2.5$ m and $q = 200$ kN/m².

   2 m       3 m
 +-------+-----------+
 | 2x2   |   3x2     |  2 m
 |   (1) |   (2)     |
 +-------P-----------+
 |   (3) |   (4)     |  2 m
 | 2x2   |   3x2     |
 +-------+-----------+

For $(m,n) = (0.8, 1.2)$ the table has no column for 1.2, so $I$ is interpolated between $n = 1.0$ (0.1598) and $n = 2$ (0.1812): $I = 0.1598 + 0.2\times(0.1812 - 0.1598) = 0.1641$.

Stress at P

$$\Delta\sigma_z = q\,(2\times0.1401 + 2\times0.1641) = 200\times0.6084 = 121.7\ \text{kN/m}^2$$

Answer: $\Delta\sigma_z \approx 122$ kN/m² at 2.5 m below P.

Using the exact formula for the influence factor (which gives $I = 0.1461$ for 0.8 × 0.8 and $0.1684$ for 0.8 × 1.2), the result is about 126 kN/m², so the table-based answer is within about 3%.