Triangular Foundation Slab Soil Bearing Pressure (Civil & Construction Engineering) Solution

A geotechnical structural engineer at the UNLV Howard R. Hughes College of Engineering is designing a reinforced concrete mat foundation for a heavy construction crane tower on an alluvial Las Vegas soil subgrade. The footprint of the triangular foundation slab forms a closed and bounded region $D$ in the horizontal plane with vertices at:

where spatial dimensions $x$ and $y$ are measured in meters ($\text{m}$).

Under severe eccentric overturning wind loads and tower deadweight, the contact soil bearing pressure distribution beneath the slab is modeled by the multivariable quadratic function:

To prevent differential settlement and localized punch-through failure, the engineer must find the global maximum and global minimum soil bearing pressures across the entire slab footprint.

1. State why the Extreme Value Theorem guarantees the existence of absolute extrema for $P(x, y)$ on this domain.
2. Perform Tier 1 (Interior Search): Find all critical points of $P(x, y)$ and determine whether they lie strictly within the open interior of the triangular slab.
3. Perform Tier 2 (Boundary Edge Search): Parameterize each of the three boundary edges of the triangle and locate all 1D critical points along the perimeter.
4. Perform Tier 3 (Vertex Check): Evaluate the bearing pressure at each corner vertex of the foundation.
5. Assemble the Master Comparison Table and state the absolute minimum and absolute maximum soil bearing pressures, identifying their precise physical coordinates.

Step-by-Step Solution:

1. Extreme Value Theorem Justification: State the Extreme Value Theorem for Functions of Two Variables (defined in Extrema on Closed and Bounded Regions):

   • A continuous function defined on a closed and bounded domain in $\mathbb{R}^2$ is mathematically guaranteed to attain an absolute maximum and an absolute minimum value on that domain.
   • Here, $P(x, y)$ is a polynomial and hence continuous everywhere on $\mathbb{R}^2$. The domain $D$ is a solid triangle including all perimeter edges and vertices (closed) and is contained entirely within a disk of radius $5\text{ m}$ (bounded). Therefore, absolute extrema are guaranteed to exist.

2. Tier 1: Interior Search: Compute the first partial derivatives of $P(x, y)$:

   $$P_x = \frac{\partial}{\partial x}(x^2 + y^2 - 2x - 4y + 20) = 2x - 2$$
   $$P_y = \frac{\partial}{\partial y}(x^2 + y^2 - 2x - 4y + 20) = 2y - 4$$
   Set both derivatives to zero:
   $$2x - 2 = 0 \implies x = 1$$
   $$2y - 4 = 0 \implies y = 2$$
   This yields candidate point $(1, 2)$. Check Region Interior: The boundary hypotenuse line connecting $(4, 0)$ to $(0, 3)$ has equation:
   $$y = -\frac{3}{4}x + 3$$
   At $x = 1$, the hypotenuse height is $y_{\text{edge}} = -\frac{3}{4}(1) + 3 = 2.25$. Because $y = 2 < 2.25$ and $x = 1 > 0$, the point $(1, 2)$ lies strictly inside the open interior of the triangular slab. Evaluate the pressure at $(1, 2)$:
   $$P(1, 2) = 1^2 + 2^2 - 2(1) - 4(2) + 20 = 1 + 4 - 2 - 8 + 20 = 15\text{ kPa}$$

3. Tier 2: Boundary Edge Search:

   • Bottom Edge ($y = 0, 0 \le x \le 4$):
     $$g_1(x) = P(x, 0) = x^2 - 2x + 20$$
     Set $g_1'(x) = 2x - 2 = 0 \implies x = 1$. Since $1 \in [0, 4]$, candidate is $(1, 0)$. Evaluate: $P(1, 0) = 1^2 - 2(1) + 20 = 19\text{ kPa}$.
   • Left Edge ($x = 0, 0 \le y \le 3$):
     $$g_2(y) = P(0, y) = y^2 - 4y + 20$$
     Set $g_2'(y) = 2y - 4 = 0 \implies y = 2$. Since $2 \in [0, 3]$, candidate is $(0, 2)$. Evaluate: $P(0, 2) = 2^2 - 4(2) + 20 = 4 - 8 + 20 = 16\text{ kPa}$.
   • Hypotenuse Edge ($y = -\frac{3}{4}x + 3, 0 \le x \le 4$): Substitute $y = -\frac{3}{4}x + 3$ into $P(x, y)$:
     $$g_3(x) = x^2 + \left(-\frac{3}{4}x + 3\right)^2 - 2x - 4\left(-\frac{3}{4}x + 3\right) + 20$$
     $$g_3(x) = x^2 + \frac{9}{16}x^2 - \frac{9}{2}x + 9 - 2x + 3x - 12 + 20 = \frac{25}{16}x^2 - \frac{7}{2}x + 17$$
     Differentiate and set to zero:
     $$g_3'(x) = \frac{50}{16}x - \frac{7}{2} = \frac{25}{8}x - \frac{7}{2} = 0 \implies x = \frac{7/2}{25/8} = \frac{7}{2} \cdot \frac{8}{25} = \frac{28}{25} = 1.12\text{ m}$$
     Since $1.12 \in [0, 4]$, find corresponding $y$:
     $$y = -\frac{3}{4}(1.12) + 3 = -0.84 + 3 = 2.16\text{ m}$$
     Evaluate pressure at $(1.12, 2.16)$:
     $$P(1.12, 2.16) = (1.12)^2 + (2.16)^2 - 2(1.12) - 4(2.16) + 20 = 1.2544 + 4.6656 - 2.24 - 8.64 + 20 = 15.04\text{ kPa}$$

4. Tier 3: Vertex Check: Evaluate at the three corner vertices:

   • Origin $V_1(0, 0)$: $P(0, 0) = 20\text{ kPa}$.
   • East Vertex $V_2(4, 0)$: $P(4, 0) = 4^2 - 2(4) + 20 = 16 - 8 + 20 = 28\text{ kPa}$.
   • North Vertex $V_3(0, 3)$: $P(0, 3) = 3^2 - 4(3) + 20 = 9 - 12 + 20 = 17\text{ kPa}$.

5. Master Comparison & Geotechnical Conclusion:

   Candidate Point $(x, y)$	Location / Tier	Bearing Pressure $P(x, y)$
   $(1, 2)$	Interior Critical Point	$15.00\text{ kPa}$ (Absolute Minimum)
   $(1.12, 2.16)$	Hypotenuse Edge	$15.04\text{ kPa}$
   $(0, 2)$	Left Vertical Edge	$16.00\text{ kPa}$
   $(0, 3)$	North Vertex $V_3$	$17.00\text{ kPa}$
   $(1, 0)$	Bottom Horizontal Edge	$19.00\text{ kPa}$
   $(0, 0)$	Corner Vertex $V_1$	$20.00\text{ kPa}$
   $(4, 0)$	East Vertex $V_2$	$28.00\text{ kPa}$ (Absolute Maximum)

   • Absolute Minimum Soil Pressure: $P_{\min} = 15.00\text{ kPa}$, occurring at the interior coordinates $(1\text{ m}, 2\text{ m})$.
   • Absolute Maximum Soil Pressure: $P_{\max} = 28.00\text{ kPa}$, occurring at the eastern footing toe $(4\text{ m}, 0\text{ m})$.
   • Geotechnical engineers will size the footing thickness and subgrade soil compaction against the peak edge stress of $28.00\text{ kPa}$ to prevent localized punch-through failure.