Stress Increment Calculation Under Rectangular Load

This problem focuses on calculating the stress increment, which is the increase in stress at a specific point within the soil caused by a surface load. This is a key calculation in foundation design to ensure the soil can withstand the applied loads without excessive settlement or failure.

Analyzing the Problem Parameters

We need to find the stress increment at a depth of 10 m below a corner of a rectangular load. The provided details are:

• Depth ($z$): 10 m
• Load Intensity ($q$): Given as 250 kN. Load intensity typically means pressure (force per unit area), such as kN/m² or kPa. The options are given in kPa.
• Influence Factor ($I$): 0.0179

A calculation using $q = 250$ kN/m² would yield $250 \times 0.0179 = 4.475$ kPa. This result does not align with any of the provided options (0.0448 kPa, 0.0525 kPa, 0.0425 kPa, 0.061 kPa).

Assumption Clarification: To match the given options, it's highly probable that the intended load intensity was 2.5 kN/m² (or 2.5 kPa). The value '250 kN' might be missing its units (like /m²) or could be a typo. We will proceed using the assumed load intensity of 2.5 kN/m².

• Assumed Load Intensity ($q$): 2.5 kN/m²

Calculating Stress Increment Using Influence Factor

The stress increase ($\Delta \sigma$) at a certain depth beneath a loaded area, particularly under the corner of a rectangular load, is often calculated using an influence factor ($I$). The formula used is:

$$ \Delta \sigma = q \times I $$

Where:

• $\Delta \sigma$ represents the stress increment (in units like kPa).
• $q$ is the load intensity or pressure applied at the surface (in units like kPa).
• $I$ is the influence factor, a dimensionless value that depends on the geometry of the load (length, width) and the location (depth, horizontal position) relative to the load.

Step-by-Step Calculation

1. Identify Given Values:
   • Assumed Load Intensity ($q$): 2.5 kN/m²
   • Influence Factor ($I$): 0.0179
2. Apply the Formula: Insert the values into the stress increment equation.

   $$ \Delta \sigma = 2.5 \text{ kN/m}^2 \times 0.0179 $$

3. Perform the Multiplication: Calculate the product.

   $$ \Delta \sigma = 0.04475 \text{ kN/m}^2 $$

4. Unit Conversion: Convert the result from kN/m² to kPa. Note that 1 kN/m² = 1 kPa.

   $$ \Delta \sigma = 0.04475 \text{ kPa} $$

5. Rounding: Round the calculated value to match the precision of the options.

   $$ \Delta \sigma \approx 0.0448 \text{ kPa} $$

Conclusion on Stress Increment

Following the calculation with the assumed load intensity of 2.5 kN/m², the stress increment is found to be approximately 0.0448 kPa. This result corresponds to one of the multiple-choice options.