3.1 Galerkin’s Residual
Let \(\delta v\) be virtual velocity based on the current configuration of a solid body. The virtual work \(\delta w\) (per unit volume and time) done by residual (r) during the motion of body is \(\delta w = r \cdot \delta v\). By choosing the linear momentum equation (derived here) as residual, from Galerkin’s Residual Method:
\(\displaystyle \int\limits_{}(\rho b + div(\sigma )) \cdot \delta v\:dV = 0\)
We need to find an alternative expression to \(\delta v \cdot div(\sigma )\) by product rule of derivation in indicial notation:
\((\delta v_i\:\sigma_{ij})_{,j} = \delta v_{i,j}\:\sigma_{ij} + \delta v_i\:\sigma_{ij,j}\)
\(div(\sigma \:\delta v) = \bigtriangledown (\delta v):\sigma + \delta v \cdot div(\sigma )\)
Moreover, application of Gauss Integral Theorem on \(div(\sigma \delta v)\):
\(\displaystyle \int\limits_{}div(\sigma \:\delta v)\:dV = \int\limits_{}div(\delta v\:\sigma )\:dV = \int\limits_{}(\delta v\:\sigma ) \cdot n\:dA\)
Insertion of these two into Galerkin’s equation yields:
\(\displaystyle \int\limits_{}(\rho b \cdot\delta v \:-\: \nabla (\delta v):\sigma )\:dV + \int\limits_{}\delta v\:(\sigma \cdot n)\:dA = 0\)
Note that \(t = \sigma \cdot n\). The gradient of \(v\) is velocity gradient, so \(\nabla(v) = L\). It is possible to write L in terms of symmetric and skew-symmetric parts:
\(L = d + w\)
\(\displaystyle \int\limits_{}L:\sigma\:dV = \int\limits_{}(d + w):\sigma\:dV\)
Here, \(w:\sigma = 0\) because w is skew-symmetric and \(\sigma \) is symmetric \((\sigma_{ij} = \sigma_{ji})\):
\(\displaystyle \frac{1}{2}(L_{ij} \:-\: L_{ji}):\sigma_{ij} = \frac{1}{2}(L_{ij}:\sigma_{ij}\:-\: L_{ji}:\sigma_{ji}) = 0\)
Finally, after substituting, the equation becomes:
\(\boxed{\displaystyle \int\limits_{}(\sigma :\delta d)\:dV \:- \int\limits_{}\delta v \cdot \rho b\:dV \:- \int\limits_{}\delta v \cdot t\:dA = 0}\)
In this equation, the first term represents the internal work while the others external work. It’s so important that \(\delta d\) and \(\delta v\) is compatible in the equation because:
\(\begin{aligned}
\delta d = \frac{1}{2}(\delta L + \delta L^T) &= \frac{1}{2}(\nabla \left(\delta v\right) + \nabla \left(\delta v\right)^T)\\
&=\frac{1}{2}\left[\frac{\partial \delta v}{\partial x} + \left(\frac{\partial \delta v}{\partial x}\right)^T\right]
\end{aligned}\)
It becomes very similar to small strain \(\varepsilon\) if we replace v with u:
\(\displaystyle \delta\varepsilon = \frac{1}{2}\left[\frac{\partial \delta u}{\partial x} + \left(\frac{\partial \delta u}{\partial x}\right)^T\right]\)
3.2 Energetic Conjugation
Since Cauchy’s stress (σ) represents current force per unit deformed area, we cannot calculate the integral over the current volume because it is still unknown in the incremental stage. Therefore, we need another type of stress measure according to undeformed area. Knowing that (d) is the symmetric part of velocity gradient, substitution into the expression below:
\(\begin{aligned}
\int\limits_V\sigma :d\:dV&= \int\limits_{V_0}\sigma :(F^{- T} \cdot \dot{E} \cdot F^{- 1})\:J\:dV_0\\
&=\int\limits_{V_0}J\:(F^{- 1} \cdot \sigma \cdot F^{- T}):\dot{E}\:dV_0\\
&= \int\limits_{V_0}S:\dot{E}\:dV_0
\end{aligned}\)
“Second Piola-Kirchoff” stress tensor (S) is used here. It is expressed as the transformed current force per unit undeformed area. It’s a symmetric tensor like Cauchy’s tensor. (E) is nothing but Green-Lagrange strain tensor.
\(\boxed{S = J(F^{- 1} \cdot \sigma \cdot F^{- T})}\)
Above, the Jacobian transformation is used, which transforms the volume or area in current configuration to that of initial configuration. Also, the expression inside the integral is re-arranged with the help of indicial form in the second line. The derivations are given below.
The virtual work equation is written in terms of Piola-Kirchoff stress tensor and Green-Lagrange strain tensor. Also, to be consistent, the terms related to external work can be transformed into the undeformed configuration.
\(\displaystyle \int\limits_{}(S:\delta \dot{E})\:dV_0- \int\limits_{}\delta v_0 \cdot \rho b\:dV_0- \int\limits_{}\delta v_0 \cdot t\:dA_0 = 0\)
Finally, due to compatibility of \(\delta u\) and \(\delta\varepsilon\), the final form can be written in terms of virtual displacement instead of virtual velocity. Thus, the virtual work done unit time:
\(\boxed{\int\limits_{}(S:\delta E)\:dV_0- \int\limits_{}\delta u \cdot \rho b\:dV_0- \int\limits_{}\delta u \cdot t\:dA_0 = 0}\)
References:
[1] Reddy, J.N. (2004) An Introduction to Nonlinear Finite Element Analysis.
[2] Belytschko, T., Liu, W. K., Moran, B., & Elkhodary, K. (2014). Nonlinear finite elements for continua and structures.