◀ Back to Note

Tony's Intertwingled Memex • Reference Manual

Structural Column Strain Energy Stability (Civil & Mechanical Engineering) Solution

Formulas & Definitions

A load-bearing structural steel support column in a commercial building framework experiences internal strain energy UUU (in Joules, J\text{J}J) due to combined axial mechanical compression loads and solar thermal expansion. The total strain energy is modeled as a function of the axial force load xxx (in kilonewtons, kN\text{kN}kN) and the surface temperature shift yyy (in degrees Celsius, ∘C^\circ\text{C}∘C):

U(x,y)=0.5x2+0.04x2y+3y2U(x, y) = 0.5x^2 + 0.04x^2 y + 3y^2U(x,y)=0.5x2+0.04x2y+3y2

  1. Compute the first-order partial derivatives UxU_xUx​ and UyU_yUy​.
  2. Compute all four second-order partial derivatives (Uxx,Uyy,Uxy,UyxU_{xx}, U_{yy}, U_{xy}, U_{yx}Uxx​,Uyy​,Uxy​,Uyx​).
  3. Verify Clairaut's Theorem for the mixed partial derivatives.
  4. Construct the 2×22 \times 22×2 Hessian matrix HU(x,y)\mathbf{H}_U(x, y)HU​(x,y) and evaluate its determinant D=det⁡(HU)D = \det(\mathbf{H}_U)D=det(HU​) at nominal service load (x,y)=(10 kN,25∘C)(x, y) = (10\text{ kN}, 25^\circ\text{C})(x,y)=(10 kN,25∘C).
  5. Interpret the physical structural meaning of the mixed partial derivative UxyU_{xy}Uxy​ and explain why Uxx>0U_{xx} > 0Uxx​>0 confirms structural stability against buckling.

Step-by-Step Solution:

  1. Compute First-Order Partial Derivatives: State the First-Order Partial Derivative Definitions (see Partial Derivatives):

    Ux=∂∂x[0.5x2+0.04x2y+3y2]=1.0x+0.08xyU_x = \frac{\partial}{\partial x}[0.5x^2 + 0.04x^2 y + 3y^2] = 1.0x + 0.08xyUx​=∂x∂​[0.5x2+0.04x2y+3y2]=1.0x+0.08xy
    Uy=∂∂y[0.5x2+0.04x2y+3y2]=0.04x2+6yU_y = \frac{\partial}{\partial y}[0.5x^2 + 0.04x^2 y + 3y^2] = 0.04x^2 + 6yUy​=∂y∂​[0.5x2+0.04x2y+3y2]=0.04x2+6y

  2. Compute Pure and Mixed Second-Order Partials: State the Second-Order Partial Derivative Formulas (see Second-Order Partial Derivatives):

    • Pure Partials:
      Uxx=∂∂x[x+0.08xy]=1.0+0.08yU_{xx} = \frac{\partial}{\partial x}[x + 0.08xy] = 1.0 + 0.08yUxx​=∂x∂​[x+0.08xy]=1.0+0.08y
      Uyy=∂∂y[0.04x2+6y]=6U_{yy} = \frac{\partial}{\partial y}[0.04x^2 + 6y] = 6Uyy​=∂y∂​[0.04x2+6y]=6
    • Mixed Partials:
      Uxy=∂∂y[x+0.08xy]=0.08xU_{xy} = \frac{\partial}{\partial y}[x + 0.08xy] = 0.08xUxy​=∂y∂​[x+0.08xy]=0.08x
      Uyx=∂∂x[0.04x2+6y]=0.08xU_{yx} = \frac{\partial}{\partial x}[0.04x^2 + 6y] = 0.08xUyx​=∂x∂​[0.04x2+6y]=0.08x
  3. Verify Clairaut's Theorem:

    Uxy=Uyx=0.08xU_{xy} = U_{yx} = 0.08xUxy​=Uyx​=0.08x
    Since U(x,y)U(x, y)U(x,y) is a smooth polynomial, its second partials are continuous everywhere on R2\mathbb{R}^2R2, verifying Clairaut's Theorem.

  4. Construct Hessian Matrix and Evaluate Determinant: State the Hessian Matrix definition:

    HU(x,y)=[UxxUxyUyxUyy]=[1.0+0.08y0.08x0.08x6]\mathbf{H}_U(x, y) = \begin{bmatrix} U_{xx} & U_{xy} \\ U_{yx} & U_{yy} \end{bmatrix} = \begin{bmatrix} 1.0 + 0.08y & 0.08x \\ 0.08x & 6 \end{bmatrix}HU​(x,y)=[Uxx​Uyx​​Uxy​Uyy​​]=[1.0+0.08y0.08x​0.08x6​]

    • Evaluate at (x,y)=(10,25)(x, y) = (10, 25)(x,y)=(10,25):
      • Uxx(10,25)=1.0+0.08(25)=1.0+2.0=3.0 J/kN2U_{xx}(10, 25) = 1.0 + 0.08(25) = 1.0 + 2.0 = 3.0\text{ J/kN}^2Uxx​(10,25)=1.0+0.08(25)=1.0+2.0=3.0 J/kN2
      • Uyy(10,25)=6.0 J/(∘C)2U_{yy}(10, 25) = 6.0\text{ J/}(^\circ\text{C})^2Uyy​(10,25)=6.0 J/(∘C)2
      • Uxy(10,25)=0.08(10)=0.8 J/(kN⋅∘C)U_{xy}(10, 25) = 0.08(10) = 0.8\text{ J/(kN}\cdot^\circ\text{C)}Uxy​(10,25)=0.08(10)=0.8 J/(kN⋅∘C)
        HU(10,25)=[3.00.80.86.0]\mathbf{H}_U(10, 25) = \begin{bmatrix} 3.0 & 0.8 \\ 0.8 & 6.0 \end{bmatrix}HU​(10,25)=[3.00.8​0.86.0​]
    • Determinant:
      D=det⁡(HU)=(3.0)(6.0)−(0.8)2=18.0−0.64=17.36>0D = \det(\mathbf{H}_U) = (3.0)(6.0) - (0.8)^2 = 18.0 - 0.64 = 17.36 > 0D=det(HU​)=(3.0)(6.0)−(0.8)2=18.0−0.64=17.36>0
  5. Physical & Structural Interpretation:

    • Mixed Partial UxyU_{xy}Uxy​: Physically represents the thermal-mechanical coupling stiffness. It measures how thermal expansion increases the column's mechanical stiffness response to compressive loading.
    • Stability & Convexity: Because D=17.36>0D = 17.36 > 0D=17.36>0 and Uxx=3.0>0U_{xx} = 3.0 > 0Uxx​=3.0>0, the strain energy function is strictly convex (concave upward) in all directions. In structural mechanics, positive curvature of the strain energy potential proves that the column is in a stable equilibrium state, guaranteeing resistance against sudden elastic buckling under these combined thermal-load service conditions.