Nonlinear finite elements/Homework 5/Solutions/Problem 1

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Problem 1: Nonlinear Beam Bending[edit | edit source]

Given:

The differential equations governing the bending of straight beams are

Solution[edit | edit source]

Part 1[edit | edit source]

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 2[edit | edit source]

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 3[edit | edit source]

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 4[edit | edit source]

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 5[edit | edit source]

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 6[edit | edit source]

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

\\

The derivatives are

and

Therefore, the coefficients of the tangent stiffness matrix can be written as