Chapter 7
Finite Element Analysis: An Introduction

 

 

 

 

Most practical design calculations involve components with a complicated three-dimensional geometry, and may also need to account for inherently nonlinear phenomena such as contact, large shape changes, or nonlinear material behavior.   These problems can only be solved using computer simulations.  The finite element method is by far the most widely used and versatile technique for simulating deformable solids.  This chapter gives a brief overview of the finite element method, with a view to providing the background needed to run simple simulations using a commercial finite element program.  More advanced analysis requires a deeper understanding of the theory and implementation of finite element codes, which will be addressed in the next chapter.

 

HEALTH WARNING: It is deceptively easy to use commercial finite element software: most programs come with a nice user-interface that allows you to define the geometry of the solid, choose a material model, generate a finite element mesh and apply loads to the solid with a few mouse clicks.   If all goes well, the program will magically turn out animations showing the deformation; contours showing stress distributions; and much more besides.  It is all too easy, however, to produce meaningless results, by attempting to solve a problem that does not have a well defined solution; by using an inappropriate numerical scheme; or simply using incorrect settings for internal tolerances in the code.  In addition, even high quality software can contain bugs.  Always treat the results of a finite element computations with skepticism!

 

 

 

7.1 A guide to using finite element software

 

The finite element method (FEM) is a computer technique for solving partial differential equations.  One application is to predict the deformation and stress fields within solid bodies subjected to external forces.  However, FEM can also be used to solve problems involving fluid flow, heat transfer, electromagnetic fields, diffusion, and many other phenomena. 

 

The principle objective of the displacement based finite element method is to compute the displacement field within a solid subjected to external forces.

 

 

To make this precise, visualize a solid deforming under external loads, as shown above. Every point in the solid moves as the load is applied.  The displacement vector u(x) specifies the motion of  the point at position x in the undeformed solid.  Our objective is to determine u(x).  Once u(x) is known, the strain and stress fields in the solid can be deduced.

 

There are two general types of finite element analysis in solid mechanics.  In most cases, we are interested in determining the behavior of a solid body that is in static equilibrium.  This means that both external and internal forces acting on the solid sum to zero.  In some cases, we may be interested in the dynamic behavior of a solid body.  Examples include modeling vibrations in structures, problems involving wave propagation, blast loading, and crash analysis. 

 

For Dynamic Problems the finite element method solves the equations of motion for a continuum $–$ essentially a more complicated version of ${\displaystyle \sum F}=ma$.  Naturally, in this case it must  calculate the motion of the solid as a function of time.

 

For Static Problems the finite element method solves the equilibrium equations ${\displaystyle \sum F}=0$.  In this case, it may not be necessary to calculate the time variation of motion.  However, some materials are history dependent (e.g metals deformed in the plastic regime).  In addition, a static equilibrium problem may have more than one solution, depending on the load history.  In this case the time variation of the solution must be computed.

 

For some applications, you may also need to solve additional field equations.  For example, you may be interested in calculating the temperature distribution in the solid, or calculating electric or magnetic fields.  In addition, special finite element procedures are available to calculate buckling loads and their modes, as well as natural frequencies of vibration and the corresponding mode shapes for a deformable solid. 

 

To set up a  finite element calculation, you will need to specify

 

1. The geometry of the solid.  This is done by generating a finite element mesh for the solid.  The mesh can usually be generated automatically from a CAD representation of the solid.

 

2. The properties of the material.  This is done by specifying a constitutive law for the solid.

 

3. The nature of the loading applied to the solid.  This is done by specifying the boundary conditions for the problem.

 

4. If your analysis involves contact between two more more solids, you will need to specify the surfaces that are likely to come into contact, and the properties (e.g. friction coefficient) of the contact.

 

5. For a dynamic analysis, it is necessary to specify initial conditions for the problem.  This is not necessary for a static analysis.

 

6. For problems involving additional fields, you may need to specify initial values for these field variables (e.g. you would need to specify the initial temperature distribution in a thermal analysis).

 

 

You will also need to specify some additional aspects of the problem you are solving and the solution procedure to be used:

 

1. You will need to specify whether the computation should take into account finite changes in the geometry of the solid.

 

2. For a dynamic analysis,  you will need to specify the time period of the analysis (or the number of time increments)

 

3.  For a static analysis you will need to decide whether the problem is linear, or nonlinear.   Linear problems are very easy to solve.  Nonlinear problems may need special procedures.

 

4.  For a static analysis with history dependent materials you will need to specify the time period of the analysis, and the time step size (or number of steps)

 

5. If you are interested in calculating natural frequencies and mode shapes for the system, you must specify how many modes to extract.

 

6. Finally, you will need to specify what the finite element method must compute.

 

The steps in running a finite element computation are discussed in more detail in the following sections.

 

 

 

7.1.1 The Finite Element Mesh for a 2D or 3D component

 

The finite element mesh is used to specify the geometry of the solid, and is also used to describe the displacement field within the solid.  A typical mesh (generated in the commercial FEA code ABAQUS) is shown below.

 

A finite element mesh may be three dimensional, like the example shown.  Two dimensional finite element meshes are also used to model simpler modes of deformation.  There are three main types of two dimensional finite element mesh, illustrated in the figure below

 

 

 

Plane Stress Finite Element Mesh: A plane stress finite element mesh is used to model a plate-like solid which is loaded in its own plane.  The solid must have uniform thickness, and the thickness must be much less than any representative cross sectional dimension.  A plane stress finite element mesh for a thin plate containing a hole is shown on the right. Only one quadrant of the specimen is modeled, since symmetry boundary conditions will be enforced during the analysis.

 

Plane Strain finite element mesh: A plane strain finite element mesh is used to model a long cylindrical solid that is prevented from stretching parallel to its axis.  For example, a plane strain finite element mesh for a cylinder which is in contact with a rigid floor is shown on the right. Away from the ends of the cylinder, we expect it to deform so that the out of plane component of displacement $u_3(x_1,x_2)=0$.  There is no need to solve for $u_3$, therefore, so a two dimensional mesh is sufficient to calculate $u_1(x_1,x_2)$ and $u_2(x_1,x_2)$.

 

As before, only one quadrant of the specimen is meshed: symmetry boundary conditions will be enforced during the analysis.

 

Axisymmetric finite element mesh: An axisymmetric mesh is used to model a solids that has rotational symmetry, which is subjected to axisymmetric loading.  An example is in the figure on the right, which compares a three dimensional mesh of an axisymmetric bushing to an axisymmetric mesh.  Note that the half the bushing has been cut away in the 3D view, to show the geometry more clearly.  In an axisymmetric analysis, the origin for the (x,y) coordinate system is always on the axis of  rotational symmetry.  Note also that to run an axisymmetric finite element analysis, both the geometry of the solid, and also the loading applied to the solid, must have rotational symmetry about the y axis.

 

In addition, special types of finite element can be used to model the overall behavior of a 3D solid, without needing to solve for the full 3D fields inside the solid.  Examples are shell elements; plate elements; beam elements and truss elements.  These will be discussed in a separate section below.

 

 

7.1.2 Nodes and Elements in a Mesh

 

A finite element mesh is defined by a set of nodes together with a set of finite elements, as shown below.

 

 

Nodes: The nodes are a set of discrete points within the solid body.  Nodes have the following properties:

 

1. A node number.  Every node is assigned an integer number, which is used to identify the node.  Any convenient numbering scheme may be selected $–$ the nodes do not need to be numbered in order, and numbers may be omitted.  For example, one could number a set of n nodes as 100, 200, 300… 100n, instead of 1,2,3…n.

 

2. Nodal coordinates.  For a three dimensional finite element analysis, each node is assigned a set of  $(x_1,x_2,x_3)$ coordinates, which specifies the position of the node in the undeformed solid.  For a two dimensional analysis, each node is assigned a pair of  $(x_1,x_2)$ coordinates.  For an axisymmetric analysis, the $x_2$ axis must coincide with the axis of rotational symmetry.

 

3. Nodal displacements.  When the solid deforms, each node moves to a new position.  For a three dimensional finite element analysis, the nodal displacements specify the three components of the displacement field u(x) at each node: $(u_1,u_2,u_3)$.  For a two dimensional analysis, each node has two displacement components $(u_1,u_2)$.  The nodal displacements are unknown at the start of the analysis, and are computed by the finite element program.

 

4. Other nodal degrees of freedom. For more complex analyses, we may wish to calculate a temperature distribution in the solid, or a voltage distribution, for example.  In this case, each node is also assigned a temperature, voltage, or similar quantity of interest.  There are also some finite element procedures which use more than just displacements to describe shape changes in a solid.  For example, when analyzing two dimensional beams, we use the displacements and rotations of the beam at each nodal point to describe the deformation.  In this case, each node has a rotation, as well as two displacement components.  The collection of all unknown quantities (including displacements) at each node are known as degrees of freedom.  A finite element program will compute values for these unknown degrees of freedom.

 

 

Elements are used to partition the solid into discrete regions.  Elements have the following properties.

 

1. An element number.  Every element is assigned an integer number, which is used to identify the element.  Just as when numbering nodes, any convenient scheme may be selected to number elements.

 

2. A geometry.  There are many possible shapes for an element.  A few of the more common element types are shown below.  Nodes attached to the element are shown in red.  In two dimensions, elements are generally either triangular or rectangular. In three dimensions, the elements are generally tetrahedra, hexahedra or bricks.  There are other types of element that are used for special purposes: examples include truss elements (which are simply axial members), beam elements, and shell elements. 

 

 

3. A set of faces.  These are simply the sides of the element.

 

4. A set of nodes attached to the element.  The figure below shows a typical finite element mesh.  Element numbers are shown in the center of each element; node numbers are shown next to each node (some node and element numbers are omitted for clarity). 

 

 

All the elements are 8 noded quadrilaterals.  Note that each element is connected to a set of nodes: element 1 has nodes (41, 45, 5, 1, 43, 25, 3, 21), element 2 has nodes (45, 49, 9, 5, 47, 29, 7, 25), and so on.  It is conventional to list the nodes the nodes in the order given, with corner nodes first in order going counterclockwise around the element, followed by the midside nodes.  The set of nodes attached to the element is known as the element connectivity.

 

5. An element interpolation scheme.  The purpose of a finite element is to interpolate the displacement field u(x) between values defined at the nodes.  This is best illustrated using an example.  Consider the two dimensional, rectangular 4 noded element shown on the right. Let $(u_1{}^{(a)},u_2{}^{(a)})$, $(u_1{}^{(b)},u_2{}^{(b)})$, $(u_1{}^{(c)},u_2{}^{(c)})$, $(u_1{}^{(d)},u_2{}^{(d)})$ denote the components of displacement at nodes a, b, c, d.  The displacement at an arbitrary point within the element can be interpolated between values at the corners, as follows

$\begin{array}{l}{u}_1=(1-\xi )(1-\eta )u_1{}^{(a)}+\xi (1-\eta )u_1{}^{(b)}+\xi \eta u_1{}^{(c)}+(1-\xi )\eta u_1{}^{(d)}\\ u_2=(1-\xi )(1-\eta )u_2{}^{(a)}+\xi (1-\eta )u_2{}^{(b)}+\xi \eta u_2{}^{(c)}+(1-\xi )\eta u_2{}^{(d)}\end{array}$

where

$\xi =x_1/B,\eta =x_2/H$

You can verify for yourself that the displacements have the correct values at the corners of the element, and the displacements evidently vary linearly with position within the element.

 

Different types of element interpolation scheme exist.  The simple example described above is known as a linear element.  Six noded triangles and 8 noded triangles are examples of quadratic elements: the displacement field varies quadratically with position within the element.  In three dimensions, the 4 noded tetrahedron and the 8 noded brick are linear elements, while the 10 noded tet and 20 noded brick are quadratic.   Other special elements, such as beam elements or shell elements, use a more complex procedure to interpolate the displacement field.

 

Some special types of element interpolate both the displacement field and some or all components of the stress field within an element separately. (Usually, the displacement interpolation is sufficient to determine the stress, since one can compute the strains at any point in the element from the displacement, and then use the stress$—$strain relation for the material to find the stress). This type of element is known as a hybrid element.  Hybrid elements are usually used to model incompressible, or nearly incompressible, materials.

 

6. Integration points.  One objective of a finite element analysis is to determine the distribution of stress within a solid.  This is done as follows.  First, the displacements at each node are computed (the technique used to do this will be discussed in Section 7.2 and Chapter 8.)  Then, the element interpolation functions are used to determine the displacement at arbitrary points within each element.  The displacement field can be differentiated to determine the strains.  Once the strains are known, the stress$—$strain relations for the element are used to compute the stresses.

 

In principle, this procedure could be used to determine the stress at any point within an element.  However, it turns out to work better at some points than others.  The special points within an element where stresses are computed most accurately are known as integration points.  (Stresses are sampled at these points in the finite element program to evaluate certain volume and area integrals, hence they are known as integration points).

 

For a detailed description of the locations of integration points within an element, you should consult an appropriate user manual.  The approximate locations of integration points for a few two dimensional elements are shown above on the right.

 

There are some special types of element that use fewer integration points than those shown in the figure.  These are known as reduced integration elements. This type of element must be used to analyze deformation of near-incompressible materials (e.g. rubbers or metals that experience large plastic strains).

 

7. A stress$—$strain relation and material properties.  Each element is occupied by solid material.  The type of material within each element (steel, concrete, soil, rubber, etc) must be specified, together with values for the appropriate material properties (mass density, Young’s modulus, Poisson’s ratio, etc).

 

 

 

7.1.3 Special Elements $–$ Beams, Plates, Shells and Truss elements

 

If you need to analyze a solid with a special geometry (e.g. a simple truss, a structure made of one or more slender beams, or plates) it is not efficient to try to generate a full 3D finite element mesh for each member in the structure.   Instead, you can take advantage of the geometry to simplify the analysis.

 

The idea is quite simple.  Instead of trying to calculate the full 3D displacement field in each member, the deformation is characterized by a reduced set of degrees of freedom.  Specifically:

 

1. For a pin jointed truss like the example shown on the right, we simply calculate the displacement of each joint in the structure. The members are assumed to be in a state of uniaxial tension or compression, so the full displacement field within each member can be calculated in terms of joint displacements.

 

2. For a beam, we calculate the displacement and rotation of the cross section along the beam.  These can be used to determine the internal shear forces bending moments, and therefore the stresses in the beam. A three dimensional beam has 3 displacement and 3 rotational degrees of freedom at each node.

 

3. For a plate, or shell, we again calculate the displacement and rotation of the plate section. A three dimensional plate or shell has 3 displacement and two rotational degrees of freedom at each node (these rotations characterize the rotation of a unit vector normal to the plate).  In some finite element codes, nodes on plates and shells have a fictitious third rotational degree of freedom which is added for convenience $–$ but you will find that attempting to impose boundary conditions on this fictitious degree of freedom has no effect on the deformation of the structure.

 

In an analysis using truss, beam or plate elements, some additional information must be specified to set up the problem:

 

1. For a truss analysis, each member in the truss is a single element.  The area of the member’s cross section must be specified.

 

2. For a beam analysis, the shape and orientation of the cross section must be specified (or, for an elastic analysis, you could specify the area moments of inertia directly).  There are also several versions of beam theory, which account differently for shape changes within the beam.  Euler-Bernoulli beam theory is the simple version covered in introductory courses.  Timoshenko beam theory is a more complex version, which is more accurate for  thicker beams.

 

3. For plates and shells, the thickness of the plate must be given.  In addition, the deformation of the plate can be approximated in various ways $–$ for example, some versions only account for transverse deflections, and neglect in-plane shearing and stretching of the plate; more complex theories account for this behavior.

 

 

Calculations using beam and plate theory also differ from full 3D or 2D calculations in that both the deflection and rotation of the beam or plate must be calculated.  This means that:

 

1. Nodes on beam elements have 6 degrees of freedom  $–$ the three displacement components, together with three angles representing rotation of the cross-section about three axes (see Fig. 17.11). <Fig 7.11 near here>  Nodes on plate or shell elements have 5 (or in some FEA codes 6) degrees of freedom.  The 6 degrees of freedom represent 3 displacement components, and two angles that characterize rotation of the normal to the plate about two axes (if the nodes have 6 degrees of freedom a third, fictitious rotation component has been introduced $–$ you will have to read the manual for the code to see what this rotation represents).

 

 

2. Boundary conditions may constrain both displacement and rotational degrees of freedom.  For example, to model a fully clamped boundary condition at the end of a beam (or the edge of a plate), you must set all displacements and all rotations to zero.

 

3. You can apply both forces and moments to nodes in an analysis.

 

Finally, in an analysis involving several beams connected together, you can connect the beams in two ways:

 

1. You can connect them with a pin joint, which forces the beams to move together at the connection, but allows relative rotation

 

2. You can connect them with a clamped joint, which forces the beams to rotate together at the connection.

 

 

In most FEA codes, you can create the joints by adding constraints, as discussed in Section 1.2.6 below. Occasionally, you may also wish to connect beam elements to solid, continuum elements in a model: this can also be done with constraints. 

 

 

 

7.1.4 Material Behavior

 

A good finite element code contains a huge library of different types of material behavior that may be assigned to elements.  A few examples are described below.

 

Linear Elasticity.  You are probably alreadly be familiar with the behavior of an isotropic linear elastic material (especially if you have read chapter 3!).  It has a uniaxial stress$—$strain response (valid only for small strains) as shown below

 

 

The stress--strain law for the material may be expressed in matrix form as

$\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ {\epsilon }_{33}\\ {\epsilon }_{12}\\ {\epsilon }_{13}\\ {\epsilon }_{23}\end{array}\right]={\displaystyle \frac{1}{E}\left[\begin{array}{cccccc}1& -\nu & -\nu & 0& 0& 0\\ -\nu & 1& -\nu & 0& 0& 0\\ -\nu & -\nu & 1& 0& 0& 0\\ 0& 0& 0& 1+\nu & 0& 0\\ 0& 0& 0& 0& 1+\nu & 0\\ 0& 0& 0& 0& 0& 1+\nu \end{array}\right]\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{33}\\ {\sigma }_{12}\\ {\sigma }_{13}\\ {\sigma }_{23}\end{array}\right]+\alpha \Delta T\left[\begin{array}{c}1\\ 1\\ 1\\ 0\\ 0\\ 0\end{array}\right]}$

Here, E and v are Young’s modulus and Poisson’s ratio for the material, while $\alpha$ denotes the thermal expansion coefficient.  Typical values (for steel) are E=210 GN/ ${\text{m}}^{\text{2}}$,  v=0.33, $\alpha =5\times {10}^{-6}{\text{K}}^{\text{-1}}$.  More extensive tables of values are given in Section 3.1.5, as well as the stress-strain relations for anisotropic elastic materials.

 

 

Elastic$—$plastic material behavior. Recall that the uniaxial stress$—$strain curve for an elastic$—$plastic solid looks something like the figure on the right.

 

The material behaves elastically until a critical stress (known as the yield stress) is reached.  If yield is exceeded, the material deforms permanently.  The yield stress of the matierial generally increases with plastic strain: this behavior is known as strain hardening.

 

The conditions necessary to initiate yielding under multiaxial loading are specified by a yield criterion, such as the Von-Mises or Tresca criteria.  These yield criteria are built into the finite element code.

 

The strain hardening behavior of a material is approximated by allowing the yield stress to increase with plastic strain.  The variation of yield stress with plastic strain for a material is usually specified by representing it as a series of straight lines, as shown in the figure.

 

 

 

7.1.5 Boundary conditions

 

Boundary conditions are used to specify the loading applied to a solid.  There are several ways to apply loads to a finite element mesh:

 

Displacement boundary conditions.  The displacements at any node on the boundary or within the solid can be specified.  One may prescribe $u_1$, $u_2$, $u_3$, or all three.  For a two dimensional analysis, it is only necessary to prescribe $u_1$ and/or $u_2$.

 

Various symbols are used to denote displacement boundary conditions applied to a finite element mesh: a few of these are illustrated below. Some user-interfaces use small conical arrowheads to indicate constrained displacement components.

 

 

For example, to stretch a 2D block of material vertically, while allowing it to expand or contract freely horizontally, we would apply boundary constraints to the top and bottom surface as shown on the right below.

 

Observe that one of the nodes on the bottom of the block has been prevented from moving horizontally, as well as vertically.  It is important to do this: the finite element program will be unable to find a unique solution for the displacement fields if the solid is free to slide horizontally. 

 

During the analysis, the finite element program will apply forces to the nodes with prescribed displacements, so as to cause them to move to their required positions.  If only the $u_1$ component of displacement is prescribed, then the force will act in the $e_1$ direction.  If  $u_2$ is prescribed, the force acts in direction $e_2$, and so on. Notice that you cannot directly apply a rotation to a node attached to a 2D or 3D solid.  Rotations can, however, be applied to the nodes attached to certain special types of element, such as beams, plates and shells, as well as rigid surfaces.

 

Symmetry conditions: Most finite element codes can automatically enforce symmetry and anti-symmetry boundary conditions. 

 

Prescribed forces. Any node in a finite element mesh may be subjected to a prescribed force.  The nodal force is a vector, and is specified by its three (or two for 2D) components, $(F_1,F_2,F_3)$.  Notice there is no direct way to apply a moment to a 3D solid $–$ you would need to do this by applying two point forces a small distance apart, or by applying contact loading, as outlined below.  Moments can be applied to some special types of element, such as shells, plates or beams. 

 

Distributed loads.  A solid may be subjected to distributed pressure or traction acting on ints boundary.  Examples include aerodynamic loading, or hydrostatic fluid pressure. Distributed traction is a  vector quantity, with physical dimensions of force per unit area in 3D, and force per unit length in 2D.  To model this type of loading in a finite element program, distributed loads may be applied to the the face of any element.

 

Default boundary condition at boundary nodes.  If no displacements or forces are prescribed at a boundary node, and no distributed loads act on any element faces connected to that node, then the node is assumed to be free of external force.

 

Body forces.  External body forces may act on the interior of a solid.  Examples of body forces include gravitational loading, or electromagnetic forces.  Body force is a vector quantity, with physical dimensions of force per unit volume.  To model this type of loading in a finite element program, body forces may be applied to the interior of any element. 

 

Contact.  This is probably the most common way to load a solid is through contact with another solid.   Special procedures are available for modeling contact between solids $–$ these will be discussed in a separate section below.

 

Load history.  In some cases, one may wish to apply a cycle of load to a solid.  In this case, the prescribed loads and displacements must be specified as a function of time.

 

 

General guidelines concerning boundary conditions.  When performing a static analysis, it is very important to make sure that boundary conditions are applied properly.  A finite element program can only solve a problem if a static equilibrium solution to the problem exists.

 

Difficulties arise if the user does not specify sufficient boundary constraints to prevent rigid body motion of a solid.  This is best illustrated by example.  Suppose we wish to model stretching a 2D solid, as described earlier.  The examples below show two correct ways to do this.

 

 

The examples below show various incorrect ways to apply boundary conditions.  In each case, one or more rigid body mode is unconstrained.

 

 

 

7.1.6 Constraints

 

You may sometimes need to use more complicated boundary conditions than simply constraining the motion or loads applied to a solid.  Some examples might be

 

1. Connecting different element types, e.g. beam elements to solid elements;

 

2. Enforcing periodic boundary conditions

 

3. Constraining a boundary to remain flat

 

4. Approximating the behavior of mechanical components such as welds, bushings, bolted joints, etc.

 

 

You can do this by defining constraints in an analysis.   At the most basic level, constraints can simply be used to enforce prescribed relationships between the displacements or velocities of individual nodes in the mesh.  You can also specify relationships between motion of groups of nodes.

 

 

 

7.1.7 Contacting Surfaces and Interfaces

 

In addition to being subject to forces or prescribed displacements, solid objects are often loaded by coming into contact with another solid. An example is shown on the right

 

Modern finite element codes contain sophisticated capabilities for modeling contact.  Unfortunately, contact can make a computation much more difficult, because the region where the two solids come into contact is generally not known a priori, and must be determined as part of the solution.  This almost always makes the problem nonlinear $–$ even if both contacting solids are linear elastic materials.  In addition, if there is friction between the contacting solids, the solution is history dependent.

 

For this reason, many options are available in finite element packages to control the way contacting surfaces behave.

 

There are three general cases of contact that you may need to deal with:

 

1. A deformable solid contacts a stiff, hard solid whose deformation may be neglected.  In this case the hard solid is modeled as a rigid surface, as outlined below.

 

2. You may need to model contact between two deformable solids

 

3. The solid comes into contact with itself during the course of deformation (this is common in components made from rubber, for example, and also occurs during some metal forming operations).

 

 

Whenever you model contact, you will need to

 

1. Specify pairs of surfaces that might come into contact.  One of these must be designated as the master surface and the other must be designated as the slave surface. (In these more enlightened times the surfaces are also referred to as ‘Primary’ and ‘Secondary’ surfaces, but this terminology does not capture the roles of the two surfaces in quite the same way.   Perhaps the best terminology would be ‘leader’ and ‘follower’ surfaces).

 

2. Define the way the two surfaces interact, e.g. by specifying the coefficient of friction between them.

 

Modeling a stiff solid as a rigid surface:  In many cases of practical interest one of the two contacting solids is much more compliant than the other.  Examples include a rubber in contact with metal, or a metal with low yield stress in contact with a hard material like a ceramic.  As long as the stresses in the stiff or hard solid are not important, its deformation can be neglected.

 

In this case the stiffer of the two solids may be idealized as a rigid surface.  Both 2D and 3D rigid surfaces can be created, as shown on the right.

 

A rigid surface (obviously) can’t change its shape, but it can move about and rotate.  Its motion is defined using a reference point on the solid, which behaves like a node.  To move the solid around during an analyisis, you can define displacement and rotational degrees of freedom at this node.  Alternatively, you could apply forces and moments to the reference point.   Finally, in a dynamic analysis, you can give the rigid solid appropriate inertial properties (so as to create a rigid projectile, for example).

 

Defining a contact pair $–$ master/slave surfaces: (or ‘primary’ and ‘secondary’ surfaces) Whenever you set up a finite element calculation that involves contact, you need to specify pairs of surfaces that may come into contact during the analysis.  One of each pair must be designated the master surface, the other must be designated the slave surface.

 

This rather obscure finite element terminology refers to the way that contact constraints are actually applied during a computation.  The geometry of the master surface will be interpolated as a smooth curve in some way (usually by interpolating between nodes).  The slave surface is not interpolated.  Instead, each individual node on the slave surface is constrained so as not to penetrate into the master surface.  For example, the nodes on the slave surface shown on the right would be forced to remain outside the boundary of the master surface.

 

For a sufficiently fine mesh, the results should not be affected by your choice of master and slave surface.  However, it improves convergence (see below to learn what this means) if you choose the more rigid of the two surfaces to be the master surface.  If you don’t know which surface is more rigid, just make a random choice.  If you run into convergence problems later, try switching them over.

 

Contact parameters You can define several parameters that control the behavior of two contacting surfaces:

 

1. The contact formulation - `finite sliding’ or `small sliding’ $–$ specifies the expected relative tangential displacement of the two surfaces.  `Finite sliding’ is the most general, but is computationally more demanding.  `Small sliding’ should be used if the relative tangential displacement is likely to be less than the distance between two adjacent nodes.

 

2. You can specify the relationship between the contact pressure and separation between the contacting surfaces. Alternatively, you can assume the contact is `hard’ $–$ this means the interface can’t withstand any tension, and the two contacting surfaces cannot inter-penetrate. 

 

3. You can specify the tangential behavior of the interface $–$ for example by giving the friction coefficient.

 

 

 

7.1.8 Initial Conditions and external fields

 

For a dynamic analysis, it is necessary to specify the initial velocity and displacement of each node in the solid.  The default value is zero velocity and displacement.

 

In addition, if you are solving a coupled problem $–$ one involving coupled deformation and heat flow, for example - you may need to specify initial values for the additional field variables (e.g. the temperature distribution)

 

 

 

7.1.9 Solution procedures and time increments

 

The finite element method calculates the displacement $u(x^a)$ at each node in the finite element mesh by solving the equations of static equilibrium or the equations of motion.  In this section, we briefly outline some of the solution procedures, and the options available to control them.

 

Linear or Nonlinear Geometry As you know, you can simplify the calculation of internal forces in a structures by neglecting shape changes when solving the equations of equilibrium.  For example, when you solve a truss problem, you usually calculate forces in each member based on the undeformed shape of the structure.

 

You can use the same idea to simplify calculations involving deformable solids.  In general, you should do so whenever possible.  However, if either

 

1. You anticipate that material might stretch by more than 10% or so or

 

2. You expect that some part of the solid might rotate by more than about 10 degrees

 

3. You wish to calculate buckling loads for your structure

 

you should account for finite geometry changes in the computation.  This will automatically make your calculation nonlinear (and so more difficult), even if all the materials have linear stress-strain relations.

 

 

Time stepping for dynamic problems:  For a dynamic problem, the nodal displacements $u(x^a,t)$ must be calculated as a function of time.  The displacements are calculated by solving the equations of motion for the system, which look something like

$M{\displaystyle \frac{d^{2}u}{d{t}^2}+Ku=F(t)}$

where M and K are called mass and stiffness matrices.  Both M and K can be functions of u.  There are 2.5 ways to integrate this equation.

 

1. The most direct method is called explicit time integration, or explicit dynamics and works something like this.  Remember that for a dynamic calculation, the values of u and $v=du/dt$ are known at t=0.  We can therefore compute M and K at time t=0, and then use them to calculate the acceleration $a=d^{2}u/d{t}^2$ at t=0, as

$a=M^{-1}\left(F(t)-Ku\right)$

The acceleration can then be used to find the velocity $v(\Delta t)$ and displacement $u(\Delta t)$ at time $\Delta t$ as

$v(\Delta t)=v(0)+a\Delta tu(\Delta t)=u(0)+v(0)\Delta t+{\displaystyle \frac{1}{2}a\Delta t^2}$

This procedure can then be applied repeatedly to march the solution through time.

 

2. The second procedure is called implicit time integration or implicit dynamics.  The procedure is very similar to explicit time integration, except that instead of calculating the mass and stiffness matrices at time t=0, and using them to calculate acceleration at t=0, these quantities are calculated at time $\Delta t$ instead.  This is a bit more time consuming to do, however, because it involves more equation solving.

 

3. The 2.5^th method is called Modal Dynamics and only works if M and K are constant. In this case one can take the Fourier transform of the governing equation and integrate it exactly.  This method is used to solve linear vibration problems.

 

 

The following guidelines will help you to choose the most appropriate method for your application:

 

1. For explicit dynamics each time step can be calculated very fast.  However, the method is stable only if $\Delta t$ is very small $–$ specifically, the time interval must be smaller than the time taken for an elastic wave to propagate from one side of an element to the other.  This gives $\Delta t<L\sqrt{\rho (1-2\nu )/2\mu (1-\nu )}$, where $\rho$ is the mass density of the solid, $\mu$ is its shear modulus and $\nu$ its Poisson’s ratio.   Explicit dynamics works best for rapid, transient problems like crash dynamics or impact.  It is not good for modeling processes that take place over a long time.  If elastic wave propagation is not the main focus of your computation, you can sometime speed up the calculations by increasing the density $\rho$ (but you have to be careful to make sure this does not affect the results).  This is called mass scaling.

 

2. For implicit dynamics the cost of computing each time step is much greater.  The algorithm is unconditionally stable, however, and will always converge even for very large $\Delta t$.  This is the method of choice for problems where inertial loading is important, but rapid transients are not the focus of the analysis.

 

3. Modal Dynamics only works for linear elastic problems.  It is used for vibration analysis.

 

 

Nonlinear Solution Procedures for Static Problems: If a problem involves contact, plastically deforming materials, or large geometry changes it is nonlinear.  This means that the equations of static equilibrium for the finite element mesh have the general form

$F^{(b)}(u^{(a)})=0$

where F() denotes a set of b=1,2…N vector functions of the nodal displacements $u(x^a)$, a=1,2…N, and N is the number of nodes in the mesh.

 

The nonlinear equations are solved using the Newton-Raphson method, which works like this.  You first guess the solution to the equations $–$ say $u^a=w^{(a)}$.  Of course (unless you are a genius) w won’t satisfy the equations, so you try to improve the solution by adding a small correction $dw$.  Ideally, the correction should be chosen so that

$F^{(b)}(w^{(a)}+d{w}^a)=0$

but of course it’s not possible to do this.  So instead, we take a Taylor expansion to get

$F^{(b)}(w^{(a)}+d{w}^a)\approx F^{(b)}(w^{(a)})+{\displaystyle \frac{d{F}^{(b)}(u)}{d{u}^{(a)}}d{w}^{(a)}=0}$

The result is a system of linear equations of the form $F^{(b)}+Kd{w}^{(a)}=0$, where $K=d{F}^{(b)}/d{u}^{(a)}$ is a constant matrix called the stiffness matrix.  The equations can now be solved for $dw$; the guess for w can be corrected, and the procedure applied over again.  The iteration is repeated until $\left|F^{(b)}(u^{(a)})\right|<\epsilon$, where $\epsilon$ is a small tolerance.

 

In problems involving hard contact, an additional iterative method is used to decide which nodes on the slave surface contact the master surface.  This is just a brute-force method $–$ it starts with some guess for contacting nodes; gets a solution, and checks it.  If any nodes are found to penetrate the master surface, these are added to the list of nodes in contact.  If any nodes are experiencing forces attracting them to the master surface, they are removed from the list of nodes in contact.

 

The problem with any iterative procedure is that it may not converge $–$ that is, repeated corrections $dw$ either take the solution further and further away from the solution, or else just spiral around the solution without ever reaching it.

 

The solution is (naturally) more likely to converge if the guess $w^{(a)}$ is close to the correct solution.  Consequently, it is best to apply the loads to a nonlinear solid gradually, so that at each load step the displacements are small.  The solution to one load increment can then be used as the initial guess for the next.

 

Convergence problems are the curse of FEM analysts.  They are very common and can be exceedingly difficult to resolve.  Here are some suggestions for things to try if you run into convergence problems:

 

1. Try applying the load in smaller increments.  most commercial codes will do this this automatically $–$ but will stop the computation if the increment size falls below a minimum value.  You can try reducing the minimum step size..

 

2. Convergence problems are sometimes caused by ill conditioning in the stiffness matrix.  This means that the equations $F^{(b)}+Kd{w}^{(a)}=0$ cannot be solved accurately.  Ill conditioning can arise because of (i) severely distorted elements in the mesh; (ii) material behavior is incompressible, or nearly incompressible; and (iii) The boundary conditions in the analysis do not properly constrain the solid. You can fix (i) by modifying the mesh $–$ some FEM codes contain capabilities to automatically remove element distorsion during large deformation. You can avoid problems with incompressibility (ii) by using reduced integration elements or hybrid elements.  Problems with boundary conditions (iii) can usually be corrected by adding more constraints.  There is one common problem where this is hard to do $–$ if the motion of a body in your analysis is constrained only by contacts with other solids (e.g. a roller between two surfaces) the stiffness matrix is always singular at the start of the analysis.  Some finite element codes contain special procedures to deal with this problem.

 

3. Try to isolate the source of the problem.  Convergence issues can often be traced to one or more of the following: (i) Severe material nonlinearity; (ii) Contact and (iii) Geometric nonlinearity.  Try to change your model to remove as many of these as possible $–$ e.g. if you are doing a plasticity computation with contact and geometric nonlinearity, try doing an elastic calculation and see if it works.  If so, the problems are caused by material nonlinearity.  Similarly, try analyzing the two contacting solids separately, without the contact; or try the computation without nonlinear geometry.  Once you’ve traced the source of the problem, you might be able to fix it by changing the material properties, contact properties or loading conditions.

 

4. Convergence problems are often caused by some kind of mechanical or material failure in the solid, which involve a sudden release of energy.  In this case, the shape of the solid may suddenly jump from one static equilibrium configuration to another, quite different, equilibrium configuration.  There is a special type of loading procedure (called the Riks method) that can be used to stabilize this kind of problem.

 

5. Some boundary value problems have badly behaved governing equations.  For example, the equations governing plane strain deformation of a perfectly plastic solid become hyperbolic for sufficiently large strains.  Static FEM simply won’t work for these problems.  Your best bet is to try an explicit dynamic calculation instead, perhaps using mass scaling to speed up the calculation.

 

 

Load steps and increments: When you set up a finite element model, you usually apply the load in a series of steps.  You can define different boundary conditions in each step.  Unless you specify otherwise, the loads (or displacements) will vary linearly from their values at the start of the step to their values at the end of the step, as illustrated on the right.

 

In a nonlinear analysis, the solution may not converge if the load is applied in a single increment.  If this is the case, the load must be applied gradually, in a series of smaller increments.  Many finite element codes will automatically reduce the time increment if the solution fails to converge.

 

 

 

7.1.10 Output

 

The finite element method (in solid mechanics) always calculates the displacement of each node in the mesh $–$ these are the unknown variables in the computation.  However, these may not be the quantities you are really interested in.  A number of quantities can be computed from the displacement fields, including:

 

1.  Velocity and acceleration fields

2.  Strain components, principal strains, and strain invariants,or their rates

3.  Elastic and plastic strains or strain rates

4.  Stress components; principal stresses; stress invariants

5.  Forces applied to nodes or boundaries

6.  Contact pressures

7.  Values of material state variables (e.g. yield stresses)

8.  Material failure criteria

 

All these quantities can be computed as functions of time at selected points in the mesh (either at nodes, or at element integration points); as functions of position along paths connecting nodes; or as contour plots.

 

 

 

7.1.11 Units in finite element computations

 

A finite element code merely solves the equations of motion (or equilibrium), together with any equations governing material behavior.  Naturally, equations like F=ma and $\sigma =E\epsilon$ do not contain any units a priori.  Consequently, when entering geometric dimensions, material data and loads into an FEA code, you can use any system of units you like, but the units of all quantities must be consistent.  You have to be very careful with this.  When you sketch the part you are modeling, it is often convenient to enter dimensions in cm, inches, or mm.  This is fine $–$ but then cm, inches or mm must be used for any other material or load data that contain length dimensions.  For example, if you use cm to dimension your part, then you must enter data for Young’s modulus and yield stress in $N/c{m}^2$, and you must also specify pressures acting on the system in  $N/c{m}^2$.  In this case, the FEA code will report stresses in units of $N/c{m}^2$.

 

 

 

7.1.12 Using Dimensional Analysis to simplify FEA analysis.

 

You may have used dimensional analysis to find relationships between data measured in an experiment (especially in fluid mechanics).  The same idea can be used to relate variables you might compute in an FEA analysis (e.g. stress), to the material properties of your part (e.g. Young’s modulus) and the applied loading.

 

The basic idea is simple, and is best illustrated by example.  Suppose we wish to use FEA to calculate the deflection of the tip of a cantilever with length L, Young’s modulus E and area moment of inertia I, which is subjected to a force P as shown on the right. We would set this up as an FEA problem, entering data for L, E, I, and P in the code, and computing $\delta$.  We could express the functional relationship as

$$\delta =f(E,I,P,L)$$

If we were asked to calculate the function f numerically, we would have to run simulations where we vary E, I, L and P independently. This would be very painful. Fortunately, since the relationship must be independent of the system of units, we know we can re-write this expression so that both left and right hand side are dimensionless $–$ i.e. as combinations of variables that have no units.  Noting that $\delta$ and L have dimensions of length, I has dimensions of ${\text{length}}^{\text{4}}$, P has dimensions $\text{mass}\times {\text{length/time}}^{\text{2}}$ and E has dimensions of $\text{mass/(length}\times {\text{time}}^{\text{2}})$, we could put

${\displaystyle \frac{\delta }{L}=g({\displaystyle \frac{P}{E{L}^2},{\displaystyle \frac{L^4}{I})}}}$

Now, we only need to calculate the function g.  We could do this by keeping L and I fixed, and varying P to see the results of varying the first group; we could then keep P and L fixed and vary I to see the effect of varying the second group.  The results could be displayed graphically as shown on the right

 

If we had done a linear analysis (no nonlinear geometric effects) the curves would be straight lines.

 

There is often more than one choice of dimensionless group, and some are better than others.  For example, for the beam problem we could create a new dimensionless group by multiplying together the two groups in the function g $–$ this gives

${\displaystyle \frac{\delta }{L}=h({\displaystyle \frac{P{L}^2}{EI},{\displaystyle \frac{L^4}{I})}}}$

This turns out to be a much better choice. In fact, if we conducted a linear analysis we would find that the function h is independent of $L^4/I$.  In this case the data would collapse onto a single master-curve as shown on the right.

 

Unfortunately, dimensional analysis alone will not tell you the best dimensionless groups.  You have to use your physical intuition to identify them.  For the beam example, you might remember that E and I always appear as the product EI in the governing equations $–$ so it makes sense to try to find dimensionless groups that combine them in this way.   In other examples, you may see some physical significance of combinations of variables (they might look like a kinetic energy, or a pressure, for example) which might help you to choose the best set. 

 

The beauty of using dimensional analysis to simplify numerical simulations is that, unlike in experiments, you don’t need to guess what variables influence the results.  You know exactly what they are, because you typed them into the program!

 

The following steps (known as the Buckingham Pi theorem) will tell you how many dimensionless groups to look for:

 

1. List the variable you are computing, and also the variables you entered into the code to define the problem.  Count the total number of variables and call it n

 

2. List the dimensions, in terms of fundamental units (i.e. mass, length, time, electric current, and luminous intensity) of all the variables

 

3. Count the number of independent fundamental units that appear in the problem (e.g. if mass, length and time appear independently, then there are 3 different units) and call the number k.  Units are independent if they don’t always appear in the same combination.  For example, in our beam problem mass and time are not independent, because they appear together as ${\text{mass/time}}^{\text{2}}$ in both P and E.  The beam problem has length, and ${\text{mass/time}}^{\text{2}}$ as two independent combinations of fundamental units.

 

4. A total of n-k independent dimensionless groups must appear in the dimensionless relationship.

 

 

For the beam problem, we had 5 variables $\delta ,E,I,P,L$, and two independent combinations of fundamental units, so we expect to see three dimensionless groups $–$ which is precisely what we got. 

 

 

 

7.1.13 Simplifying FEA analysis by scaling the governing equations

 

An alternative approach to identifying the dimensionless parameters that control the solution to a problem is to express the governing equations themselves in dimensionless form.  This is a much more powerful technique, but is also somewhat more difficult to use.

 

We can illustrate the procedure using our beam problem again.  Let x measure distance along the beam, and let w denote its vertical deflection, as indicated on the right. You may remember that linear Euler-Bernoulli beam theory gives the following governing equation for w

$EI{\displaystyle \frac{d^{4}w}{d{x}^4}=0}$

(the right hand side vanishes because no forces act on 0<x<L) while the boundary conditions are

$\left.\begin{array}{c}w=0\text{(no deflection at wall)}\\ {\displaystyle \frac{dw}{dx}=0\text{ (no rotation at wall)}}\end{array}\right\}x=0$         

     $\left.\begin{array}{c}EI{\displaystyle \frac{d^{2}w}{d{x}^2}=0\text{(no moment at free end)}}\\ EI{\displaystyle \frac{d^{3}w}{d{x}^3}=P\text{ (Shear force equal to }P\text{)}}\end{array}\right\}x=L$

(If you don’t remember these it doesn’t matter $–$ it’s the scaling discussed below that’s important). 

 

We now re-write the equations so that they are dimensionless.  We aways start by replacing all field variables (in this case w and x) with dimensionless quantities.  In this case we could use $\xi =x/L\widehat{w}=w/L$. Substituting $x=\xi Lw=\widehat{w}L$ gives

${\displaystyle \frac{EI}{L^3}{\displaystyle \frac{d^4\widehat{w}}{d{\xi }^4}=0}}$

 

$\left.\begin{array}{c}L\widehat{w}=0\text{(no deflection at wall)}\\ {\displaystyle \frac{d\widehat{w}}{d\xi }=0\text{ (no rotation at wall)}}\end{array}\right\}\xi =0$ 

 

$\left.\begin{array}{c}{\displaystyle \frac{EI}{L}{\displaystyle \frac{d^2\widehat{w}}{d{\xi }^2}=0\text{(no moment at free end)}}}\\ {\displaystyle \frac{EI}{L^2}{\displaystyle \frac{d^3\widehat{w}}{d{\xi }^3}=P\text{ (Shear force equal to }P\text{)}}}\end{array}\right\}\xi =1$

 

We now look and see if we can make further simplifications.  Our objective is to remove as many material and geometric parameters from the equations as possible, by defining new dimensionless field variables or introducing dimensionless combinations of material or geometric variables.  In this case, we see that if we define a new dimensionless displacement W so that

$WP{L}^2/(EI)=\widehat{w}$

substitute, and cancel as many terms as possible, the governing equations become

${\displaystyle \frac{d^{4}W}{d{\xi }^4}=0}$

 

$\left.\begin{array}{c}W=0\text{(no deflection at wall)}\\ {\displaystyle \frac{dW}{d\xi }=0\text{ (no rotation at wall)}}\end{array}\right\}\xi =0$

 

$\left.\begin{array}{c}{\displaystyle \frac{d^{2}W}{d{\xi }^2}=0\text{(no moment at free end)}}\\ {\displaystyle \frac{d^{3}W}{d{\xi }^3}=1\text{ (Shear force equal to }P\text{)}}\end{array}\right\}\xi =1$

In this form, the governing equations contain absolutely no material or geometric parameters.  The solution for W must therefore be independent of L,E,I or P.  We can solve the equation just once, and then work out the tip deflection from the value of W at $\xi =1$.  Specifically,

$\delta =w(x=L)=L\widehat{w}(\xi =1)={\displaystyle \frac{P{L}^3}{EI}W(\xi =1)}$

 

This scaling procedure is the best way to simplify numerical computations.  It is more difficult to apply than dimensional analysis, however, and it is possible (although perhaps not a good idea) to run an FEA simulation of a problem where you don’t actually know the governing equations! In this case you should just use standard dimensional analysis to try to simplify the problem.

 

 

 

7.1.14 Dimensional analysis $–$ closing remarks

 

It is good practice to scale a mechanics problem as outlined in the preceding sections, and present results in dimensionless form. Not all practicing engineers and managers are really comfortable with it, however -   they don’t want to see data presented in dimensionless form $–$ they want to see real numbers (often for good reason $–$ after all, the numbers are needed to make a design decision at the end of the day). 

 

The best way to deal with this when presenting results in a report is to divide it into two sections $–$ one where you present `master curves’ in dimensionless form, and another where the master curves are converted to give real numbers for quantities of interest. 

 

You can make dimensional analysis work in your favor.  It is not uncommon for your boss to tell you to run a series of simulations where you vary a parameter that can be shown on dimensional grounds to have absolutely no effect.  Also, dimensional analysis often tells you that varying two parameters are equivalent $–$ for example, if you already had data for deflection as a function of load in the beam problem, you could use that to immediately calculate the deflection as a function of length.   Don’t tell your bosses this $–$ it will just confuse them.  Instead,  tell them that you will work night and day to get the simulations done by the end of the week.  Then take the week off, and plot up the data 5 mins before your next meeting.