Balance of Mechanical Energy

✨ 19.9.2024, 21:27

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✏️ 20.8.2025, 17:07

This is not a new balance, since it is derived from the balance of linear momentum by scalar multiplication with the velocity,

\begin{align} \rho\,\dot{\textbf{v}}\cdot \textbf{v} = \textrm{div}\textbf{T}\cdot \textbf{v} + \textbf{b}\cdot \textbf{v} \end{align}

Using Gauss's Theorem and integration by parts

\begin{align} \label{eq:c9q45cowqt} \rho\,\dot{\textbf{v}}\cdot \textbf{v} = \textrm{div} \left( \textbf{v} \,\textbf{T}\right) - \textbf{T} \cdot \textrm{grad}\; \textbf{v} + \textbf{b}\cdot \textbf{v} \end{align}

1. Symmetric Stress Tensor - Angular momentum preserved

In case of a symmetric stress tensor, \(\textbf{T} = \textbf{T}^{ T}\) (non-polar media), we can use the split of the velocity gradient into symmetric and skew-symmetric parts, \(\textbf{L} = \textbf{D} + \textbf{W}\) and the fact that \(\textbf{T} \cdot \textbf{W} = 0\)

\begin{align} \textbf{T} \cdot \textbf{L} = \textbf{T} \cdot \left( \textbf{D} + \textbf{W} \right) = \textbf{T} \cdot \textbf{D} \end{align}

so that \eqref{eq:c9q45cowqt} reads

\begin{align} \rho\,\dot{\textbf{v}}\cdot \textbf{v} = \textrm{div} \left( \textbf{v} \,\textbf{T}\right) - \textbf{T} \cdot \textbf{D} + \textbf{b}\cdot \textbf{v} \end{align}

With respect to the current configuration, the solution is obtained by taking the integral over the system volume

\begin{align} \int_{ \mathcal{R}_t}\rho\;\dot{\textbf{v}}\cdot \textbf{v} \;\textrm{d}v &= \int_{ \mathcal{R}_t} \textrm{div} \left( \textbf{v} \,\textbf{T}\right) \;\textrm{d}v - \int_{ \mathcal{R}_t}\textbf{T} \cdot \textbf{D}\;\textrm{d}v +\int_{ \mathcal{R}_t} \textbf{b}\cdot \textbf{v}\;\textrm{d}v \end{align}

where

\begin{align} \int_{ \mathcal{R}_t} \textrm{div} \left( \textbf{v} \,\textbf{T}\right) \;\textrm{d}v = \int_{ \partial\mathcal{R}_t} \left( \textbf{v} \,\textbf{T}\right) \cdot \textbf{n} \;\textrm{d}a =\int_{ \partial\mathcal{R}_t} \textbf{v} \cdot\textbf{t} \;\textrm{d}a \end{align}

we arrive at the balance of mechanical power in the form of

\begin{align} \int_{ \mathcal{R}_t}\rho\;\dot{\textbf{v}}\cdot \textbf{v} \;\textrm{d}v &= - \int_{ \mathcal{R}_t}\textbf{T} \cdot \textbf{D}\;\textrm{d}v + \int_{ \partial\mathcal{R}_t} \textbf{v} \cdot\textbf{t} \;\textrm{d}a +\int_{ \mathcal{R}_t} \textbf{b}\cdot \textbf{v}\;\textrm{d}v \end{align}

1.1. Power of external forces and power of internal forces

Here we define power contributions

\begin{align} L_a &:= \int_{ \partial\mathcal{R}_t} \textbf{v} \cdot\textbf{t} \;\textrm{d}a + \int_{ \mathcal{R}_t} \textbf{b}\cdot \textbf{v} \;\textrm{d}v \\ L_i &:= \int_{ \mathcal{R}_t}\textbf{T} \cdot \textbf{D} \;\textrm{d}v \end{align}

and together with the rate of change of the kinetic energy (power) Kinetic Energy

\begin{align} L_a = \dot{K} + L_i \end{align}

meaning that the power of the external forces is responsible for the change of the kinetic energy of the body (system) and the deformation of the body.

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Balance of Energy
One of the most important laws in physics and engineering
✨ 19.9.2024, 22:51
✏️ 28.8.2025, 10:31

Balance of Mechanical Energy

1. Symmetric Stress Tensor - Angular momentum preserved

1.1. Power of external forces and power of internal forces

Links to this note

Balance of Energy

One of the most important laws in physics and engineering