The maximum shear stress is half difference between major and minor principal stresses.
\(?_{max}=\frac{1}{2}(?_1-?_2).\)

Boussinesq in 1885 studied and solved problems of stress distribution in soils due to a concentrated loads acting at ground surface. Darcy gave law of flow of water through soils. The stress function was introduced by G.B Airy in 1862.

Given,
Load q=80 kN/m²
Size of rectangle=2m*4m
Depth z=5m
Therefore m=2/5=0.4
n=4/5=0.8
? vertical stress at corner of rectangle is given by,
\(?_z=\frac{q}{4?}\left[\frac{2mn\sqrt{(m^2+n^2+1)}}{m^2+n^2+m^2 n^2+1}* \frac{m^2+n^2+2}{m^2+n^2+1}+tan^{-1}?\frac{2mn\sqrt{(m^2+n^2+1)}}{m^2+n^2+m^2 n^2+1} \right] \)
? \(?_z=\frac{80}{4?}\left[\frac{2*0.4*0.8\sqrt{(0.4^2+0.8^2+1)}}{0.4^2+0.8^2+0.4^2 0.8^2+1}* \frac{0.4^2+0.8^2+2}{0.4^2+0.8^2+1}+tan^{-1}\frac{2*0.4*0.8\sqrt{(0.4^2+0.8^2+1)}}{0.4^2+0.8^2-0.4^2 0.8^2+1} \right] \)
? ?_z=7.45 kN/m².

An isobar is a spatial, curved surface of shape of bulb. This is because vertical pressure on a given horizontal plane is same in all directions at all points located at equal radial distances around axis of loading.

Since in Terzaghi?s Theory of one dimensional consolidation, time lag in consolidation is due entirely to permeability of soil, re is very slow drainage of water that leads to neglecting of secondary consolidation.

Given,
Z=4m
Q=20 kN
r=0
The Boussinesq?s vertical stress ?_z is given by,
\(?_z=\frac{3Q}{2?z^2}\left[\frac{1}{1+(\frac{r}{z})^2}\right]^{\frac{5}{2}}\)
? \(?_z=\frac{3*20}{2?4^2}\)
? ?_z=0.597 kN/m².

The equilibrium equations in polar coordinates are given by,
1. \(\frac{??_r}{?r}+\frac{1}{r} \frac{??_{r?}}{??}+\frac{?_r-?_?}{r}=0 \)
2. \(\frac{1}{r}\frac{??_?}{??}+\frac{??_{r?}}{?r}+\frac{2?_{r?}}{r}=0.\)

Given,
(?_z)_max=0.235 kN/m²
r=4m
since maximum vertical stress is
\((?_z)_{max}=\frac{0.0888Q}{r^2} \)
? \(Q=\frac{(?_z)_{max}r^2}{0.0888} \)
? \(Q=\frac{0.235*4^2}{0.0888} \)
Q=42.34 kN.

let load intensity be q at a horizontal distance a and load intensity at distance x is qx=q*x/a.
? for any point P load is given by, \(?_z=\frac{q}{a?} \left[x?+z\right]. \)

Given,
uniformly loaded circular area of radius a=10m
depth z=1000m
vertical pressure ?_z = 80 kN/m²
The Boussinesq?s vertical pressure ?_z under a uniformly loaded circular area is given by,
\(?_z=q\left[1-[\frac{1}{1+(\frac{a}{z})^2}]^\frac{3}{2}\right]\)
? \(80=q\left[1-[\frac{1}{1+(\frac{1000}{10})^2}]^\frac{3}{2}\right]\)
? q=80 kN/m².