Stress Increase Under Circular Loaded Area

PhysicsSoil MechanicsMedium

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Lect. 8 - 5:

A flexible circular area is subjected to a uniformly distributed load of $150\text{ kN/m}^2$ shown in figure below. The diameter of the loaded area is $2\text{ m}$. Determine the stress increase $\Delta\sigma$ in a soil mass at a point located $3\text{ m}$ below the loaded area at $r = 0.0, 0.4\text{ m}, 0.8\text{ m},$ and $1\text{ m}.$

Use Boussinesq's solution

SOLUTION:

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Step 1

Hi arjan, let's solve this geotechnics problem where we calculate the stress increase beneath a flexible circular loaded area using Boussinesq's theory.

Determining Stress Increase under a Circular Area

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Step 2

Let's list the given parameters first. We have a uniformly distributed load, lowercase q, of one hundred fifty kilonewtons per square meter.

$$q = 150 \; \text{kN/m}^2$$
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Step 3

The diameter is two meters, which means the radius R of the circular area is one meter. We are looking for the stress increase delta sigma at a depth z of three meters.

$$D = 2 \; \text{m} \implies R = 1 \; \text{m}$$
$$z = 3 \; \text{m}$$
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Step 4

According to Boussinesq's solution for a circular area, the vertical stress increase depends on the ratios of radius over depth and radial distance over depth. We define two non-dimensional parameters, capital A and capital B.

Boussinesq Formula

$$\Delta \sigma = q \cdot I_z$$
$$I_z = 1 - \left[ \frac{1}{1 + (R/z)^2} \right]^{3/2} \text{ (for center point)}$$
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Step 5

For points not at the center, we use influence charts or updated formulas involving ratios. Here, r is the horizontal distance from the center. Let's calculate the common ratio r over z for each case.

$$r/z \text{ values needed for } r = 0, 0.4, 0.8, 1.0$$
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Step 6

We also need the ratio R over z, which is constant for all points in this problem.

$$R/z = 1 / 3 = 0.333$$

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About This Question

Subject
Physics
Topic
Soil Mechanics
Difficulty
Medium
Question Type
Open Ended

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