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 .
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,
The Total Lagrangian governing equations are:
- For the axially loaded bar,
- Conservation of Momentum:
- For the bar is constant. Therefore,
- In short form,
- If , we get the equilibrium equation
- For the bar:
- In short form:
- For a linear elastic material:
- The superscript refers to the fact that this function relates and .
- For small strains:
- For a linear elastic material:
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.
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
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,
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.
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
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,
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,
The finite element equations in total Lagrangian form are: