2.1 Linear Momentum
Linear momentum and mass are introduced by definition:
\(\displaystyle I = \int\limits_P\rho {\mathcal{v}}\:dV\)
\(\displaystyle m = \int\limits_P\rho\:dV\)
Also, we need mechanical force as the summation of surface and body forces over the volume:
\(\displaystyle F = \int\limits_P\rho b\:dV + \int\limits_Pt\:dA\)
The relation between linear momentum and mechanical force is given:
\(\displaystyle \frac{dI}{dt} = F\)
\(\displaystyle \frac{d}{dt}\int\limits_P\rho v\:dV = \int\limits_P\rho b\:dV + \int\limits_Pt\:dA\)
Re-write the first term on the left hand side and the second term on the right hand side. In the equation (**) both Cauchy’s Theorem and Gauss Integral Theorem are used. Also, since the mass is conserved, the rate of density is zero.
\(\displaystyle \frac{d}{dt}\int\limits_P\rho v\:dV = \int\limits_P(\dot{\rho}v + \rho \dot{v})\:dV\) (*)
\(\displaystyle \int\limits_Pt\:dA = \int\limits_P\sigma n\:dA = \int\limits_Pdiv(\sigma )\:dV\) (**)
\(\dot{\rho} = 0\) (***)
Substitution of these 3 identities into equation above:
\(\displaystyle \int\limits_P\rho \dot{v}\:dV = \int\limits_P(\rho b + div(\sigma ))\:dV\)
\(\displaystyle \int\limits_P(\rho \dot{v} – \rho b – div(\sigma ))\:dV = 0\)
Localization theorem indicates that for a continuous function F(x) = 0 if
\(\displaystyle \int\limits_PF(x)\:dx = 0\)
\(\rho \dot{v} = \rho b + div(\sigma )\)
For a static equilibrium \(\dot{v} = 0\). Finally:
\(\boxed{\rho b + div(\sigma ) = 0}\)
References:
[1] Holzapfel, G.A. (2000) Nonlinear Solid Mechanics: A Continuum Approach for Engineering.
[2] Belytschko, T., Liu, W. K., Moran, B., & Elkhodary, K. (2014). Nonlinear finite elements for continua and structures.