Chapter 20, Problem 27P

The vertical stress ${\sigma }_z$ under the corner of a rectangular area subjected to a uniform load of intensity q is given by the solution of Boussinesq's equation:

$\sigma =\frac{q}{4\pi }\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2{n}^2}\frac{m^2+n^2+2}{m^2+n^2+1}+{\sin }^{-1}\left(\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2{n}^2}\right)\right]$

Because this equation is inconvenient to solve manually, it has been reformulated as

${\sigma }_z=q{f}_z\left(m,n\right)$

FIGURE P20.27

TABLE P20.27

m	n = 1.2	n = 1.4	n = 1.6
0.1	0.02926	0.03007	0.03058
0.2	0.05733	0.05894	0.05994
0.3	0.08323	0.08561	0.08709
0.4	0.10631	0.10941	0.11135
0.5	0.12626	0.13003	0.13241
0.6	0.14309	0.14749	0.15027
0.7	0.15703	0.16199	0.16515
0.8	0.168430	0.17389	0.17739

where $f_z\left(m,n\right)$ is called the influence value and m and n are dimensionless ratios, with $m=a/z$ and $n=b/z$ and a and b as defined in Fig. P20.27. The influence value is then tabulated, a portion of which is given in Table P20.27. If $a=4.6$ and $b=14$, use a third-order interpolating polynomial to compute ${\sigma }_z$ at a depth 10 m below the corner of a rectangular footing that is subject to a total load of 100 t (metric tons). Express your answer in tonnes per square meter. Note that q is equal to the load per area.