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Vertical Pressure below the corner of uniformly load rectangular area

Calculate the intensity of stress at a point 'P located as shown in figure at a depth of 2.5 m due to a load of 2 kN/m2 over the footing area of 4m * 5 m.

• For area $A_{1} and A_{3}$
• For area $A_{2} and A_{4}$
m = L/Z = 3/2.5 = 1.20, n = B/Z = 2/2.5 = 0.8 Using Boussinesq equation \begin{aligned} K &=\frac{1}{4 \pi}\left[\frac{2 m n\left(m^{2}+n^{2}+1\right)}{m^{2}+n^{2}+m^{2} n^{2}+1} \cdot \frac{\left(m^{2}+n^{2}+2\right)}{\left(m^{2}+n^{2}+1\right)}+\tan ^{-1}\left(\frac{2 m n\left(m^{2}+n^{2}+1\right)^{1 / 2}}{m^{2}+n^{2}-m^{2} n^{2}+1}\right)\right] \\ K_{1} &=K_{3}=\frac{1}{4 \pi}\left[\frac{2 \times 0.8 \times 0.8\left(0.8^{2}+0.8^{2}+1\right)^{1 / 2}}{0.8^{2}+0.8^{2}+0.8^{2} \times 0.8^{2}+1} \cdot \frac{\left(0.8^{2}+0.8^{2}+2\right)}{\left(0.8^{2}+0.8^{2}+1\right)}\right.\\ &\left.\quad+\tan ^{-1}\left(\frac{2 \times 0.8 \times 0.8\left(0.8^{2}+0.8^{2}+1\right)^{1 / 2}}{0.8^{2}+0.8^{2}-0.8^{2} \times 0.8^{2}+1}\right)\right] \\=& \frac{1}{4 \pi}\left(1.0329+45.94^{\circ}\right)=\frac{1}{4 \pi}(1.0329+0.8021 \mathrm{rad} .)=0.1461 \\=& K_{4}=\frac{1}{4 \pi}\left[\frac{2 \times 1.20 \times 0.8\left(1.20^{2}+0.8^{2}+1\right)^{1 / 2}}{1.20^{2}+0.8^{2}+1.20^{2} \times 0.8^{2}+1} \cdot \frac{\left(1.20^{2}+0.8^{2}+2\right)}{\left(1.20^{2}+0.8^{2}+1\right)}\right.\\ \therefore \quad &\left.\quad+\tan ^{-1}\left(\frac{2 \times 1.20 \times 0.8\left(1.20^{2}+0.8^{2}+1\right)^{1 / 2}}{1.20^{2}+0.8^{2}-1.20^{2} \times 0.8^{2}+1}\right)\right] \\=& \frac{1}{4 \pi}\left[1.1154+57.35^{\circ}\right]=\frac{1}{4 \pi}[1.1154+1.00 \mathrm{rad}]=0.1684 \\=&\left(K_{1}+K_{2}+K_{3}+K_{4}\right) \cdot q \\=&(0.1461+0.1684+0.1461+0.1684) \times 250 \\=& 0.629 \times 250=157.24 \mathrm{kN} / \mathrm{m}^{2} \end{aligned} Using Westergaard's equation \begin{aligned} K &=\frac{1}{2 \pi}\left[\cot ^{-1}\left(\frac{1}{2 m^{2}}+\frac{1}{2 n^{2}}+\frac{1}{4 m^{2} n^{2}}\right)^{1 / 2}\right] \\ K_{1} &=K_{3}=\frac{1}{2 \pi}\left[\cot ^{-1}\left(\frac{1}{2 \times 0.8^{2}}+\frac{1}{2 \times 0.8^{2}}+\frac{1}{4 \times 0.8^{2} \times 0.8^{2}}\right)^{1 / 2}\right]=0.0948 \\ K_{2} &=K_{4}=\frac{1}{2 \pi}\left[\cot ^{-1}\left(\frac{1}{2 \times 1.20^{2}}+\frac{1}{2 \times 0.8^{2}}+\frac{1}{4 \times 1.20^{2} \times 0.8^{2}}\right)^{1 / 2}\right]=0.1168 \\ \sigma_{z} &=\left(K_{1}+K_{2}+K_{3}+K_{4}\right) \cdot q \\ &=(2 \times 0.0948+2 \times 0.1168) \times 250=105.8 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}