Betti's theorem

Betti's theorem is used in the design of compliant mechanisms by topology optimization approach.

Proof

Consider a solid body subjected to a pair of external force systems, referred to as ${\displaystyle F_i^P}$ and ${\displaystyle F_i^Q}$. Consider that each force system causes a displacement field, with the displacements measured at the external force's point of application referred to as ${\displaystyle d_i^P}$ and ${\displaystyle d_i^Q}$.

When the ${\displaystyle F_i^P}$ force system is applied to the structure, the balance between the work performed by the external force system and the strain energy is:

${\displaystyle \frac{1}{2}\sum _{i=1}^n{F}_i^P{d}_i^P=\frac{1}{2}{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^P{ϵ}_{ij}^{P}d\mathrm{\Omega }}$

The work-energy balance associated with the ${\displaystyle F_i^Q}$ force system is as follows:

${\displaystyle \frac{1}{2}\sum _{i=1}^n{F}_i^Q{d}_i^Q=\frac{1}{2}{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^Q{ϵ}_{ij}^{Q}d\mathrm{\Omega }}$

Now, consider that with the ${\displaystyle F_i^P}$ force system applied, the ${\displaystyle F_i^Q}$ force system is applied subsequently. As the ${\displaystyle F_i^P}$ is already applied and therefore won't cause any extra displacement, the work-energy balance assumes the following expression:

${\displaystyle \frac{1}{2}\sum _{i=1}^n{F}_i^P{d}_i^P+\frac{1}{2}\sum _{i=1}^n{F}_i^Q{d}_i^Q+\sum _{i=1}^n{F}_i^P{d}_i^Q=\frac{1}{2}{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^P{ϵ}_{ij}^{P}d\mathrm{\Omega }+\frac{1}{2}{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^Q{ϵ}_{ij}^{Q}d\mathrm{\Omega }+{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^P{ϵ}_{ij}^{Q}d\mathrm{\Omega }}$

Conversely, if we consider the ${\displaystyle F_i^Q}$ force system already applied and the ${\displaystyle F_i^P}$ external force system applied subsequently, the work-energy balance will assume the following expression:

${\displaystyle \frac{1}{2}\sum _{i=1}^n{F}_i^Q{d}_i^Q+\frac{1}{2}\sum _{i=1}^n{F}_i^P{d}_i^P+\sum _{i=1}^n{F}_i^Q{d}_i^P=\frac{1}{2}{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^Q{ϵ}_{ij}^{Q}d\mathrm{\Omega }+\frac{1}{2}{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^P{ϵ}_{ij}^{P}d\mathrm{\Omega }+{\int }_{\mathrm{\Omega }}{\sigma }_{ij}^Q{ϵ}_{ij}^{P}d\mathrm{\Omega }}$

If the work-energy balance for the cases where the external force systems are applied in isolation are respectively subtracted from the cases where the force systems are applied simultaneously, we arrive at the following equations:

${\displaystyle \sum _{i=1}^n{F}_i^P{d}_i^Q={\int }_{\mathrm{\Omega }}{\sigma }_{ij}^P{ϵ}_{ij}^{Q}d\mathrm{\Omega }}$

${\displaystyle \sum _{i=1}^n{F}_i^Q{d}_i^P={\int }_{\mathrm{\Omega }}{\sigma }_{ij}^Q{ϵ}_{ij}^{P}d\mathrm{\Omega }}$

If the solid body where the force systems are applied is formed by a linear elastic material and if the force systems are such that only infinitesimal strains are observed in the body, then the body's constitutive equation, which may follow Hooke's law, can be expressed in the following manner:

${\displaystyle {\sigma }_{ij}=D_{ijkl}{ϵ}_{kl}}$

Replacing this result in the previous set of equations leads us to the following result:

${\displaystyle \sum _{i=1}^n{F}_i^P{d}_i^Q={\int }_{\mathrm{\Omega }}{D}_{ijkl}{ϵ}_{ij}^P{ϵ}_{kl}^{Q}d\mathrm{\Omega }}$

${\displaystyle \sum _{i=1}^n{F}_i^Q{d}_i^P={\int }_{\mathrm{\Omega }}{D}_{ijkl}{ϵ}_{ij}^Q{ϵ}_{kl}^{P}d\mathrm{\Omega }}$

If we subtract both equations then we obtain the following result:

${\displaystyle \sum _{i=1}^n{F}_i^P{d}_i^Q=\sum _{i=1}^n{F}_i^Q{d}_i^P}$