This solution explains how to calculate the vertical stress at a specific depth below the center of a ring foundation, based on the principles of soil mechanics and Boussinesq's theory for stress distribution.

Ring Foundation Vertical Stress Calculation

We need to determine the increase in vertical stress in the soil at a depth of 4 m directly below the center of a ring foundation. The foundation has an outer diameter of 8 m and a thickness of 1 m, carrying a uniform load intensity of 200 kN/m².

Key Parameters for Stress Calculation

First, let's list the important values given in the problem:

• Uniform load intensity, q = 200 kN/m²
• Depth below the foundation, z = 4 m
• Outer diameter of the ring foundation, D_o = 8 m
• Thickness of the ring foundation, B = 1 m

Determining Inner and Outer Radii

The foundation is a ring (annulus). To find the stress caused by this ring, we consider it as the difference between two solid circular areas:

• Outer radius, R_o = D_o / 2 = 8 m / 2 = 4 m
• Inner diameter, D_i = D_o - 2 * B = 8 m - 2 * 1 m = 6 m
• Inner radius, R_i = D_i / 2 = 6 m / 2 = 3 m

Boussinesq Formula for Vertical Stress

For a uniformly loaded circular area, the vertical stress increase ($\Delta \sigma_z$) at a depth z directly below the center is given by:

$$ \Delta \sigma_z = q \left[ 1 - \frac{z^3}{(R^2 + z^2)^{3/2}} \right] $$

Where:

• q is the uniform load intensity.
• z is the depth below the load.
• R is the radius of the circular loaded area.

Calculating Stress Under the Ring Foundation

The stress caused by the ring foundation is the stress caused by the outer circle minus the stress caused by the inner circle.

Step 1: Stress from Outer Circular Load ($\Delta \sigma_{zo}$)

Using the formula with R_o = 4 m and z = 4 m:

$$ \Delta \sigma_{zo} = 200 \text{ kN/m}^2 \left[ 1 - \frac{(4 \text{ m})^3}{((4 \text{ m})^2 + (4 \text{ m})^2)^{3/2}} \right] $$

$$ \Delta \sigma_{zo} = 200 \left[ 1 - \frac{64}{(16 + 16)^{3/2}} \right] $$

$$ \Delta \sigma_{zo} = 200 \left[ 1 - \frac{64}{(32)^{3/2}} \right] $$

Calculate $(32)^{3/2}$: $(32)^{3/2} = (\sqrt{32})^3 = (4\sqrt{2})^3 = 64 \times 2\sqrt{2} = 128\sqrt{2} \approx 181.019 $$

$$ \Delta \sigma_{zo} = 200 \left[ 1 - \frac{64}{181.019} \right] $$

$$ \Delta \sigma_{zo} = 200 \left[ 1 - 0.3535 \right] $$

$$ \Delta \sigma_{zo} = 200 \times 0.6465 \approx 129.3 \text{ kN/m}^2 $$

Step 2: Stress from Inner Circular Load ($\Delta \sigma_{zi}$)

Using the formula with R_i = 3 m and z = 4 m:

$$ \Delta \sigma_{zi} = 200 \text{ kN/m}^2 \left[ 1 - \frac{(4 \text{ m})^3}{((3 \text{ m})^2 + (4 \text{ m})^2)^{3/2}} \right] $$

$$ \Delta \sigma_{zi} = 200 \left[ 1 - \frac{64}{(9 + 16)^{3/2}} \right] $$

$$ \Delta \sigma_{zi} = 200 \left[ 1 - \frac{64}{(25)^{3/2}} \right] $$

Calculate $(25)^{3/2}$: $(25)^{3/2} = (\sqrt{25})^3 = 5^3 = 125$$

$$ \Delta \sigma_{zi} = 200 \left[ 1 - \frac{64}{125} \right] $$

$$ \Delta \sigma_{zi} = 200 \left[ 1 - 0.512 \right] $$

$$ \Delta \sigma_{zi} = 200 \times 0.488 = 97.6 \text{ kN/m}^2 $$

Step 3: Total Stress Due to Ring Foundation ($\Delta \sigma_{z\_ring}$)

Subtract the stress from the inner circle from the stress from the outer circle:

$$ \Delta \sigma_{z\_ring} = \Delta \sigma_{zo} - \Delta \sigma_{zi} $$

$$ \Delta \sigma_{z\_ring} \approx 129.3 \text{ kN/m}^2 - 97.6 \text{ kN/m}^2 $$

$$ \Delta \sigma_{z\_ring} \approx 31.7 \text{ kN/m}^2 $$

Final Result Verification

The calculated vertical stress caused by the ring foundation at a depth of 4 m below the center is approximately 31.7 kN/m². This matches one of the provided options.