Nonlinear finite elements/Total Lagrangian approach

Total Lagrangian ApproachEdit

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 MeasuresEdit

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 FormEdit

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 EquationEdit

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 ConditionsEdit

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

 

where

 

Weak Form for Total LagrangianEdit

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

 

Therefore,

 

Taking a derivative of this variation, we have

 

Also,

 

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

 

where,

 

Finite Element Discretization for Total LagrangianEdit

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 TermEdit

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 TermEdit

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

 

where

 

and

 

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 TermsEdit

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

 

or,

 

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 LagrangianEdit

The finite element equations in total Lagrangian form are: