Load Decomposition and Energy Interaction in Linear Structural Analysis

How elastic energy is decomposed with displacements

In linear structural analysis, the response to multiple external load components can be decomposed thanks to the principle of superposition. Suppose the total external force vector is written as

\mathbf{F} = \mathbf{F}_A + \mathbf{F}_B + \mathbf{F}_C

For a linear system with stiffness matrix \mathbf{K}, the displacement satisfies

\mathbf{K}\mathbf{u} = \mathbf{F}

Because the system is linear, the displacement can be decomposed as

\mathbf{u} = \mathbf{u}_A + \mathbf{u}_B + \mathbf{u}_C

where

\mathbf{K}\mathbf{u}_A = \mathbf{F}_A, \qquad
\mathbf{K}\mathbf{u}_B = \mathbf{F}_B, \qquad
\mathbf{K}\mathbf{u}_C = \mathbf{F}_C

The total strain energy is

U = \frac{1}{2}\mathbf{u}^\top \mathbf{K}\mathbf{u}
= \frac{1}{2}\mathbf{F}^\top \mathbf{u}

Substituting the decomposed fields gives

U =
\frac{1}{2}
(\mathbf{u}_A + \mathbf{u}_B + \mathbf{u}_C)^\top
\mathbf{K}
(\mathbf{u}_A + \mathbf{u}_B + \mathbf{u}_C)

Expanding this expression yields

U =
\frac{1}{2}
\Big(
\mathbf{u}_A^\top \mathbf{K}\mathbf{u}_A
+
\mathbf{u}_B^\top \mathbf{K}\mathbf{u}_B
+
\mathbf{u}_C^\top \mathbf{K}\mathbf{u}_C
\Big)
+
\Big(
\mathbf{u}_A^\top \mathbf{K}\mathbf{u}_B
+
\mathbf{u}_A^\top \mathbf{K}\mathbf{u}_C
+
\mathbf{u}_B^\top \mathbf{K}\mathbf{u}_C
\Big)

The first three terms represent the “self-energy” contributions of each load component. The remaining terms are cross terms that arise from interactions between different load-induced displacement fields.

Therefore, although the displacement field can be fully decomposed due to linearity, the total strain energy is not simply the sum of independent energy contributions. The quadratic nature of the energy functional introduces coupling terms. Increasing \mathbf{F}_A​ does not reduce the contribution of \mathbf{F}_B rather, the total energy changes according to both the individual self-energies and the interaction terms.

How should we estimate energy contribution from a Force Vector

Marginal contribution

Define the contribution of load component F_A as the increase in total energy when it is added to the system:

\Delta U_A
= U(\mathbf{F})
     U(\mathbf{F} - \mathbf{F}_A)

This definition includes cross terms and measures the actual influence of \mathbf{F}_A​ on the full system response.

Pros: Physically intuitive (“what changes when A is added”).
Cons: Can depend on ordering if multiple loads are added sequentially.

Symmetric Energy Decomposition

To avoid ordering dependence, distribute cross terms equally among interacting loads.

First compute

M_{ij}=\mathbf{F}_i^\top \mathbf{K}^{-1}\mathbf{F}_j

The self-energy is

U_i^{\text{self}}=\frac{1}{2} M_{ii}

Then define the allocated energy as

U_i=
\frac{1}{2} M_{ii}
+
\frac{1}{2}
\sum_{j \ne i} M_{ij}

This equally splits each cross term between participating components. By construction,

\sum_i U_i = U

The contribution ratio is then

r_i = \frac{U_i}{U}

Pros: Order-independent and sums exactly to 100%
Cons: This is a mathematical allocation rule, not a physical separation