To determine the locus of the maximum vertical stress $$\sigma_z$$ at any given radial distance $ r $, we need to find the value of $ z $ that maximizes the expression given for vertical stress:

$$\sigma_z = \frac{3P}{2\pi} \frac{z^3}{(r^2 + z^2)^{5/2}}$$

We will do this by differentiating $ \sigma_z $ with respect to $ z $ and then setting the derivative to zero to find the critical points.

Let's denote the vertical stress by $ f(z) $:

$$ f(z) = \frac{3P}{2\pi} \frac{z^3}{(r^2 + z^2)^{5/2}} $$

To find the critical points, we need to take the derivative of $ f(z) $ with respect to $ z $ and set it equal to zero:

$$ \frac{d}{dz} \left( \frac{z^3}{(r^2 + z^2)^{5/2}} \right) = 0 $$

Let's use the quotient rule for differentiation:

$$ \frac{d}{dz} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dz} - u \frac{dv}{dz}}{v^2} $$

Here, $ u = z^3 $ and $ v = (r^2 + z^2)^{5/2} $. Now calculate the derivatives $ \frac{du}{dz} $ and $ \frac{dv}{dz} $:

- $ \frac{du}{dz} = 3z^2 $

- $ \frac{dv}{dz} = \frac{5}{2} (r^2 + z^2)^{3/2} \cdot 2z = 5z (r^2 + z^2)^{3/2} $

Plugging these back into the quotient rule:

$$ \frac{d}{dz} \left( \frac{z^3}{(r^2 + z^2)^{5/2}} \right) = \frac{(r^2 + z^2)^{5/2} \cdot 3z^2 - z^3 \cdot 5z (r^2 + z^2)^{3/2}}{(r^2 + z^2)^5} $$

Simplifying the numerator:

$$ = \frac{(r^2 + z^2)^{3/2} [3z^2(r^2 + z^2) - 5z^4]}{(r^2 + z^2)^5} $$

$$ = \frac{3z^2 (r^2 + z^2) - 5z^4}{(r^2 + z^2)^{5/2}} $$

Setting the derivative equal to zero:

$$ 3z^2 (r^2 + z^2) - 5z^4 = 0 $$

Factor the equation:

$$ z^2 (3r^2 + 3z^2 - 5z^2) = 0 $$

$$ z^2 (3r^2 - 2z^2) = 0 $$

Thus, either $ z^2 = 0 $ or $ 3r^2 = 2z^2 $. Since $ z = 0 $ does not provide the maximum for non-zero depth $ z $, we have:

$$ 3r^2 = 2z^2 $$

This simplifies to:

$$ z^2 = \frac{3}{2} r^2 $$

Therefore, the locus of the maximum vertical stress $$\sigma_z$$ is given by:

$$ z^2 = \frac{3}{2} r^2 $$

Hence, the correct option is:

Option A: $$ z^2 = \frac{3}{2} r^2 $$

Stress Distribution of Soil · Geotechnical Engineering · GATE CE

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