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Hillside Foundation Slope Stability (Civil & Construction Engineering) Solution

Formulas & Definitions

A geotechnical and civil engineering firm assesses hillside terrain stability prior to constructing a reinforced concrete retaining wall and foundation slab for a commercial development in Henderson, Nevada.

A drone-based LiDAR survey generates a continuous digital elevation model E(x,y)E(x, y)E(x,y) (in meters, m\text{m}m) across the building pad:

E(x,y)=50−0.05x3+0.1xy+0.02y2E(x, y) = 50 - 0.05x^3 + 0.1xy + 0.02y^2E(x,y)=50−0.05x3+0.1xy+0.02y2
where xxx represents the distance eastward (in meters) and yyy represents the distance northward (in meters) relative to an anchored geodetic survey monument.

  1. State how cross-sectional vertical cuts y=y0y = y_0y=y0​ and x=x0x = x_0x=x0​ define directional slopes on the terrain surface.
  2. Determine the algebraic expressions for the directional slopes ∂E∂x\frac{\partial E}{\partial x}∂x∂E​ and ∂E∂y\frac{\partial E}{\partial y}∂y∂E​.
  3. Evaluate the slopes ∂E∂x\frac{\partial E}{\partial x}∂x∂E​ and ∂E∂y\frac{\partial E}{\partial y}∂y∂E​ at the proposed foundation center location (x,y)=(4,10) m(x, y) = (4, 10)\text{ m}(x,y)=(4,10) m.
  4. State whether the ground slopes uphill or downhill when walking due east (+x+x+x) and due north (+y+y+y) from the center location, and state the angle of inclination along each cardinal direction in degrees (θ=arctan⁡(m)\theta = \arctan(m)θ=arctan(m)).

Step-by-Step Solution:

  1. State the Cross-Sectional Trace Principle:

    • Slicing the 3D elevation surface with a plane y=y0y = y_0y=y0​ isolates an east-west cross-section where longitude yyy is constant; the derivative ∂E∂x\frac{\partial E}{\partial x}∂x∂E​ is the instantaneous slope ΔEΔx\frac{\Delta E}{\Delta x}ΔxΔE​ along this cut.
    • Slicing with a plane x=x0x = x_0x=x0​ isolates a north-south cross-section where latitude xxx is constant; the derivative ∂E∂y\frac{\partial E}{\partial y}∂y∂E​ is the instantaneous slope ΔEΔy\frac{\Delta E}{\Delta y}ΔyΔE​ along this cut. (Defined in Geometric Interpretation of Partial Derivatives.)
  2. Compute the Algebraic Slope Expressions:

    • Differentiating with respect to xxx (holding yyy constant):
      ∂E∂x=∂∂x(50)−0.05∂∂x(x3)+0.1y∂∂x(x)+0.02∂∂x(y2)\frac{\partial E}{\partial x} = \frac{\partial}{\partial x}(50) - 0.05 \frac{\partial}{\partial x}(x^3) + 0.1y \frac{\partial}{\partial x}(x) + 0.02 \frac{\partial}{\partial x}(y^2)∂x∂E​=∂x∂​(50)−0.05∂x∂​(x3)+0.1y∂x∂​(x)+0.02∂x∂​(y2)
      ∂E∂x=0−0.05(3x2)+0.1y(1)+0=−0.15x2+0.1y\frac{\partial E}{\partial x} = 0 - 0.05(3x^2) + 0.1y(1) + 0 = -0.15x^2 + 0.1y∂x∂E​=0−0.05(3x2)+0.1y(1)+0=−0.15x2+0.1y
    • Differentiating with respect to yyy (holding xxx constant):
      ∂E∂y=∂∂y(50)−0.05∂∂y(x3)+0.1x∂∂y(y)+0.02∂∂y(y2)\frac{\partial E}{\partial y} = \frac{\partial}{\partial y}(50) - 0.05 \frac{\partial}{\partial y}(x^3) + 0.1x \frac{\partial}{\partial y}(y) + 0.02 \frac{\partial}{\partial y}(y^2)∂y∂E​=∂y∂​(50)−0.05∂y∂​(x3)+0.1x∂y∂​(y)+0.02∂y∂​(y2)
      ∂E∂y=0−0+0.1x(1)+0.02(2y)=0.1x+0.04y\frac{\partial E}{\partial y} = 0 - 0 + 0.1x(1) + 0.02(2y) = 0.1x + 0.04y∂y∂E​=0−0+0.1x(1)+0.02(2y)=0.1x+0.04y
  3. Evaluate at (x,y)=(4,10) m(x, y) = (4, 10)\text{ m}(x,y)=(4,10) m:

    • For the east-west slope:
      ∂E∂x∣(4,10)=−0.15(4)2+0.1(10)=−0.15(16)+1.0=−2.40+1.0=−1.40 m/m\left.\frac{\partial E}{\partial x}\right|_{(4, 10)} = -0.15(4)^2 + 0.1(10) = -0.15(16) + 1.0 = -2.40 + 1.0 = -1.40\text{ m/m}∂x∂E​​(4,10)​=−0.15(4)2+0.1(10)=−0.15(16)+1.0=−2.40+1.0=−1.40 m/m
    • For the north-south slope:
      ∂E∂y∣(4,10)=0.1(4)+0.04(10)=0.40+0.40=0.80 m/m\left.\frac{\partial E}{\partial y}\right|_{(4, 10)} = 0.1(4) + 0.04(10) = 0.40 + 0.40 = 0.80\text{ m/m}∂y∂E​​(4,10)​=0.1(4)+0.04(10)=0.40+0.40=0.80 m/m
  4. Interpret Cardinal Behavior and Angles of Inclination:

    • Eastward Direction (+x+x+x): Since ∂E∂x=−1.40<0\frac{\partial E}{\partial x} = -1.40 < 0∂x∂E​=−1.40<0, walking due east is walking downhill at a −140%-140\%−140% grade.
      θx=arctan⁡(−1.40)≈−54.46∘\theta_x = \arctan(-1.40) \approx -54.46^\circθx​=arctan(−1.40)≈−54.46∘
      (A steep drop of 54.5∘54.5^\circ54.5∘ below the horizontal).
    • Northward Direction (+y+y+y): Since ∂E∂y=0.80>0\frac{\partial E}{\partial y} = 0.80 > 0∂y∂E​=0.80>0, walking due north is walking uphill at an +80%+80\%+80% grade.
      θy=arctan⁡(0.80)≈38.66∘\theta_y = \arctan(0.80) \approx 38.66^\circθy​=arctan(0.80)≈38.66∘
      (A climb of 38.7∘38.7^\circ38.7∘ above the horizontal).
    • Engineering Conclusion: The steep drop to the east mandates an engineered retaining wall system to prevent catastrophic landslide slippage along the cut slope.