Nonlinear finite elements/Homework 5/Solutions/Problem 1

Problem 1: Nonlinear Beam BendingEdit


The differential equations governing the bending of straight beams are



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




The finite element system of equations can then be written as




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




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




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