Boussinesq's Equation for Vertical Stress

The question asks us to find the ratio of vertical stresses at two different depths beneath a concentrated load applied on the soil surface. We will use Boussinesq's equation, which describes the stress distribution in a semi-infinite elastic soil mass subjected to surface loads.

For a concentrated vertical load Q applied on the surface of a homogeneous, elastic, and semi-infinite medium, the vertical stress ($\sigma_z$) at a depth z directly below the point of load application is given by the formula:

$$ \sigma_z = \frac{3Q}{2\pi z^2} $$

Here, Q represents the magnitude of the concentrated load, and z is the vertical depth from the surface.

Calculating Stress Ratio

We need to calculate the ratio of the vertical stress at a depth of $z_1 = 3$ m to the vertical stress at a depth of $z_2 = 5$ m.

Let $\sigma_{z1}$ be the vertical stress at depth $z_1 = 3$ m. Using the formula:

$$ \sigma_{z1} = \frac{3Q}{2\pi (3)^2} = \frac{3Q}{2\pi \times 9} $$

Let $\sigma_{z2}$ be the vertical stress at depth $z_2 = 5$ m. Using the formula:

$$ \sigma_{z2} = \frac{3Q}{2\pi (5)^2} = \frac{3Q}{2\pi \times 25} $$

The question asks for the ratio $\frac{\sigma_{z1}}{\sigma_{z2}}$. Let's calculate this ratio:

$$ \frac{\sigma_{z1}}{\sigma_{z2}} = \frac{\left( \frac{3Q}{2\pi \times 9} \right)}{\left( \frac{3Q}{2\pi \times 25} \right)} $$

We can simplify this expression by cancelling out the common terms ($3Q$ and $2\pi$):

$$ \frac{\sigma_{z1}}{\sigma_{z2}} = \frac{\frac{1}{9}}{\frac{1}{25}} $$

To find the ratio, we invert the denominator and multiply:

$$ \frac{\sigma_{z1}}{\sigma_{z2}} = \frac{1}{9} \times \frac{25}{1} = \frac{25}{9} $$

Now, we convert the fraction to a decimal:

$$ \frac{25}{9} \approx 2.777... $$

Rounding to two decimal places, the ratio is approximately 2.77.

Therefore, the ratio of vertical stresses at depths of 3 m and 5 m vertically below the point of application of the load is approximately 2.77.