Nonlinear finite elements/Total Lagrangian approach

Total Lagrangian Approach edit

In the total Lagrangian approach, the discrete equations are formulated with respect to the reference configuration. The independent variables are   and  . The dependent variable is the displacement  .

Total Lagrangian Stress and Strain Measures edit

The strain is


For the reference configuration,


The Cauchy stress (force/current area) is


The engineering stress (force/initial area) is


The two stresses are related by


For the reference configuration,


Governing Equations in Total Lagrangian Form edit

The Total Lagrangian governing equations are:

  • Conservation of Mass:
For the axially loaded bar,
  • Conservation of Momentum:
  • For the bar   is constant. Therefore,
In short form,
If  , we get the equilibrium equation
  • Conservation of Energy:
For the bar:
In short form:
  • Constitutive Equations:
    • Total Form:
For a linear elastic material:
The superscript   refers to the fact that this function relates   and  .
For small strains:
    • Rate Form:
For a linear elastic material:

The Wave Equation edit

The momentum equation is


The total form constitutive equation is


Plug constitutive equation into momentum equation to get


If   is constant in the bar and the body force is zero,


This is the wave equation (hyperbolic PDE). The wave speed is  . If acceleration is zero, the equation becomes elliptic.

Initial and Boundary Conditions edit

The governing equation for the rod is second-order in time. So two initial conditions are needed.


The rod is initially at rest. Therefore, the initial conditions are


Since the problem is one-dimensional, there are two boundary points. Also, the momentum equation is second-order in the displacements. Therefore, at each end, either   or   must be prescribed. In mechanics, instead of  , the traction is prescribed.

Let   be the part of the boundary where displacements are prescribed. Let   be the part of the boundary where tractions (force vector per unit area) are prescribed.

Then the displacement boundary conditions are


The traction boundary conditions are


The unit normal to the boundary in the reference configuration is  .

For the axially loaded bar, the displacement boundary condition is


The unit normal to the boundary is


Therefore, the traction boundary condition is


For shock problems or fracture problems, an addition interior continuity or jump condition may be needed. This condition is written as




Weak Form for Total Lagrangian edit

The momentum equation (in its full form) is


To get the weak form over an element, we multiply the equation by a weighting function and integrate over the length of the element (from   to  ).


Integrate the first term by parts to get


Plug the above into the weak form and rearrange to get


If we think of the weighting function   as a variation of   that satisfies the kinematic admissibility conditions, we get


Recall that




Taking a derivative of this variation, we have




Therefore, we can alternatively write the weak form as


Comparing with the energy equation, we see that the first term above is the internal virtual work and the weak form is the principle of virtual work for the 1-D problem. The weak form may also be written as




Finite Element Discretization for Total Lagrangian edit

The trial solution is


where   is the number of nodes. In matrix form,


The test (weighting) function is


The derivatives of the test functions with respect to   are


We will derive the finite element system of equations after substituting these into the approximate weak form


Let us proceed term by term.

First LHS Term edit

The first term represents the virtual internal work


Plugging in the derivative of the test function, we get


The last substitution is made because the virtual internal work is the work done by internal forces in moving through a virtual displacement. The matrix form of the expression for virtual internal work is


The internal force is


Second LHS Term edit

The second term represents the virtual kinetic work


Plugging in the test function, we get


The inertial (kinetic) force is


Now, plugging in the trial function into the expression for the inertial force, we get






Note that the mass matrix is independent of time!

In matrix form,


The consistent mass matrix in matrix notation is


Plugging the expression for the inertial force into the expression for virtual kinetic work, we get


In matrix form,


RHS Terms edit

The right hand side terms represent the virtual external work


Plugging in the test function into the above expression gives


In matrix notation,


The external force is given by


In matrix notation,


Collecting the terms, we get the finite element system of equations


Now, the weighting function is arbitrary except at nodes where displacement BCs are prescribed. At these nodes the weighting function is zero.

For the bar, let us assume that a displacement is prescribed at node  . Then, the finite element system of equations becomes




Since the displacement at node   is known, the acceleration at node   is also known. Note that we have to differentiate the displacement twice to get the acceleration and hence the prescribed displacement must be a   function.

We can take the known acceleration   to the RHS to get


The above equation shows that prescribed boundary accelerations (and hence prescribed boundary displacements) contribute to nodes which are not on the boundary.

We can avoid that issue by making the mass matrix diagonal (using ad-hoc procedures that conserve momentum). If the mass matrix is diagonal, we get


In matrix form,


One way of generating a diagonal mass matrix or lumped mass matrix is the row-sum technique. The rows of the mass matrix are summed at lumped at the diagonal of the matrix. Thus,


Discrete Equations for Total Lagrangian edit

The finite element equations in total Lagrangian form are: