Nonlinear finite elements/Homework 5/Solutions/Problem 1

Problem 1: Nonlinear Beam BendingEdit

Given:

The differential equations governing the bending of straight beams are

 

SolutionEdit

Part 1Edit

Show that the weak forms of these equations can be written as

 


First we get rid of the shear force term by combining the second and third equations to get

 

Let   be the weighting function for equation (1) and let   be the weighting function for equation (2).

Then the weak forms of the two equations are

 

To get the symmetric weak forms, we integrate by parts (even though the symmetry is not obvious at this stage) to get

 

Integrate equation (6) again by parts, and get

 

Collect terms and rearrange equations (5) and (7) to get

 

Rewriting the equations, we get

 

Hence shown.

Part 2Edit

The von Karman strains are related to the displacements by

 

The stress and moment resultants are defined as

 

For a linear elastic material, the stiffnesses of the beam in extension and bending are defined as

 

where   is the Young's modulus of the material.

Derive expressions for   and   in terms of the displacements   and   and the extensional and bending stiffnesses of the beam assuming a linear elastic material.


The stress-strain relations for an isotropic linear elastic material are

 

Since all strains other than   are zero, the above equations reduce to

 

If we ignore the stresses   and  , the only allowable value of   is zero. Then the stress-strain relations become

 

Plugging this relation into the stress and moment of stress resultant equations, we get

 

Plugging in the relations for the strain we get

 

Since both   and   are independent of   and  , we can take these quantities outside the integrals and get

 

Using the definitions of the extensional, extensional-bending, and bending stiffness, we can then write

 

To write these relations in terms of   and   we substitute the expressions for the von Karman strains to get

 

These are the expressions of the resultants in terms of the displacements.

Part 3Edit

Express the weak forms in terms of the displacements and the extensional and bending stiffnesses.

The weak form equations are

 

At this stage we make two more assumptions:

  1. The elastic modulus is constant throughout the cross-section.
  2. The  -axis passes through the centroid of the cross-section.

From the first assumption, we have

 

From the second assumption, we get

 

Then the relations for   and   reduce to

 

Let us first consider equation (11). Plugging in the expression for   we get

 

We can also write the above in terms of virtual displacements by defining  ,  , and  . Then we get

 

Next, we do the same for equation (12). Plugging in the expressions for   and  , we get

 

We can write the above in terms of the virtual displacements and the generalized forces by defining

 

to get

 

Part 4Edit

Assume that the approximate solutions for the axial displacement and the transverse deflection over a two noded element are given by

 

where  .

Compute the element stiffness matrix for the element.

The weak forms of the governing equations are

 

Let us first write the approximate solutions as

 

where   are generalized displacements and

 

To formulate the finite element system of equations, we substitute the expressions from   and   from equations (19) and (20) into the weak form, and substitute the shape functions   for  ,   for  .

For the first equation (17) we get

 

After reorganizing, we have

 

We can write the above as

 

where   and

 

For the second equation (18) we get

 

After rearranging we get

 

We can write the above as

 

where   and

 

In matrix form, we can write

 

or

 

The finite element system of equations can then be written as

 

or

 

Part 5Edit

Show the alternate procedure by which the element stiffness matrix can be made symmetric.

The stiffness matrix is unsymmetric because   contains a factor of   while   does not. The expressions of these terms are

 

To get a symmetric stiffness matrix, we write equation (18) as

 

The quantity is green is assumed to be known from a previous iteration and adds to the   terms.

Repeating the procedure used in the previous question

 

After rearranging we get

 

We can write the above as

 

where   and

 

This gives us a symmetric stiffness matrix.

Part 6Edit

Derive the element tangent stiffness matrix for the element.

Equation (21) can be written as

 

where

 

The residual is

 

For Newton iterations, we use the algorithm

 

where the tangent stiffness matrix is given by

 

The coefficients of the tangent stiffness matrix are given by

 

Recall that the finite element system of equations can be written as

 

where the subscripts have been changed to avoid confusion.

Therefore, the residuals are

 

The derivatives of the residuals with respect to the generalized displacements are

 

Differentiating, we get

 

These equations can therefore be written as

 

Now, the coefficients of  ,  , and   of the symmetric stiffness matrix are independent of   and  . Also, the terms of   are independent of the all the generalized displacements. Therefore, the above equations reduce to

 

Consider the coefficients of  :

 

From our previous derivation, we have

 

Therefore,

 

The tangent stiffness matrix coefficients are therefore

 

Next, {consider the coefficients of  :

 

The coefficients of   are

 

Therefore, the derivatives are

 

Therefore the coefficients of   are

 

Finally, for the   coefficients, we start with

 

and plug in the derivatives of the stiffness matrix coefficients