2. Finite Element Method
FEM: Method for numerical solution of field problems.
Description - FEM cuts a structure into several elements
(pieces of the structure).
- Then reconnects elements at “nodes” as if
nodes were pins or drops of glue that hold elements
- This process results in a set of simultaneous
Number of degrees-of-freedom (DOF)
FEM: Finite (This is the origin of the name, Finite
4. Approximate Numerical
In mechanics of solids, our problem is to determine the
displacement u of the body shown.
It satisfying the equilibrium equations.
stresses are related to strains, which, in turn, are related
Solution of this set of equations is generally referred to as
an exact solution which are available for simple
geometries and loading conditions.
For problems of complex geometries and general
boundary and loading conditions, obtaining such
solutions is an almost impossible task.
Approximate solution methods usually employ potential
energy or variational methods, which place less stringent
conditions on the functions.
6. The stresses acting on the elemental volume dV are
shown in Fig. When the volume dV shrinks to a point, the
stress tensor is represented by placing its components in
a (3 x 3) symmetric matrix. However, we represent stress
by the six independent components as in
where σx, σy, σz are normal stresses and τyz, τxz, τzy
are shear stresses. Let us consider equilibrium of the
elemental volume shown in Fig. First we get forces on
faces by multiplying the stresses by the corresponding
areas. Writing ΣFx=0, ΣFy = 0, ΣFz = 0 and recognizing
dV = dx dy dz, we get the equilibrium equations
7. BOUNDARY CONDITIONS
Referring to Fig., we find that there are displacement boundary
conditions and surface-loading conditions. If u is specified on part of
the boundary denoted by Su, we have
U= 0 OR Sx
We can also consider boundary conditions such as u = a, where a
is a given displacement.
We now consider the equilibrium of an elemental tetrahedron ABCD,
shown in Fig., where DA, DB, and DC are parallel to the X-, y-, and
z-axes, respectively, and area ABC, denoted by dA, lies on the
surface. If n = [nx, ny, nz]T is the unit normal to dA, Consideration of
equilibrium along the three axes directions gives
These conditions must be satisfied on the boundary, St, where the
tractions are applied. In this description, the point loads must be
treated as loads distributed over small, but finite areas.
8. STRESS-STRAIN RELATIONS
For linear elastic materials, the stress-strain relations come from the
generalized Hooke's law. For isotropic materials, the two material
properties are Young's modulus (or modulus of elasticity) E and
Poisson's ratio v. Considering an elemental cube inside the body,
Hooke's law gives
The shear modulus (or modulus of rigidity), G, is given by
From Hooke's law relationships note that
Substituting for and so on into we get inverse relations.
9. D is the symmetric (6 x 6) material matrix given by
One dimension:- In one dimension, we have normal stress u along x and the
corresponding normal strain ϵ, Stress-strain relations are simply
Two dimensions:- In two dimensions, the problems are modeled as plane
stress and plane strain.
Plane Stress:- A thin planar body subjected to in-plane loading on its edge
surface is said to be in plane stress. A ring press fitted on a shaft, Here
stresses in z direction are set as zero. The Hooke's law relations then give
which is used as σ= Dϵ.
10. Plane Strain: If a long body of uniform cross section is
subjected to transverse loading along its length, a
small thickness in the loaded area, as shown in Fig,
can be treated as subjected to plane strain, Here ϵz ,
ϒzx, ϒ yz are taken as zero. Stress σ z may not be
zero in this case. The stress-strain relations can be
obtained directly from
D here is a (3 X 3) matrix, which relates three
stresses and three strains.
12. The total potential energy Π (Capital Pi) of an elastic
body, is defined as the sum of total strain energy (U) and
the work potential:
Π = Strain energy + Work potential
For linear elastic materials., the strain energy per unit
volume in the body is ½*σTϵ For
the elastic body shown in Fig. 1.1, the total strain energy
U is given by
The work potential WP is given by
The total potential for the general elastic body shown in
13. Rayleigh-Ritz Method(Potential Energy
For continua, the total potential energy II in previous Eq.
can be used for finding an approximate solution. The
Rayleigh-Ritz method involves the construction of an
assumed displacement field, say,
The functions Øi are usually taken as polynomials.
Displacements u, v, w must be kinematically admissible.
That is, u, v, w must satisfy specified boundary conditions.
Introducing stress-strain and strain-displacement relations,
and substituting above equations into potential energy
14. where r = number of independent unknowns. Now,
the extremum with respect to ai (i = 1 to r) yields the
set of r equations-
Principle of Minimum Potential Energy:- For
conservative systems, of all the kinematically
admissible displacement fields, those corresponding
to equilibrium extremize the total potential energy. If
the extremum condition is a minimum. the
equilibrium state is stable.
15. GALERKIN'S METHOD
Galerkin's method uses the set of governing equations in the
development of an integral form. it is usually presented as one of the
weighted residual methods. For our discussion, let us consider a
general representation of a governing equation on a region V:
Lu = P
For the one-dimensional rod considered, the governing equation is
the differential equation is
We may consider L as the operator operating on u.
The exact solution needs to satisfy governing eq. at every point x. If
we seek an approximate solution ũ it introduces an error ϵ(x), called
The approximate methods revolve around setting the residual
relative to a weighting function wi to zero:
16. GALERKIN'S METHOD in
Let us turn our attention to the equilibrium
equations In elasticity. Galerkin’s method requires
is an arbitrary displacement consistent with the
boundary conditions of u. if is a unit
normal at a point x on the surface.
17. The integration by parts formula is
where a and (J are functions of (x, y, z). For
multidimensional problems, above Eq is usually referred to
as the Green-Gauss theorem or the divergence theorem.
Using this formula, integrating Eq. by parts, and
rearranging terms, we get
18. One-Dimensional Problems
The total potential energy and the stress-strain and
strain-displacement relationships are now used in
developing the finite element method for a one-
For the one-dimensional problem, the stress, strain,
displacement, and loading depend only on the variable x.
so variables are as
Furthermore, the stress-strain and strain-displacement
19. For one-dimensional problems, the differential volume dV
can be written as dV = Adx
The loading consists of three types: the body force F, the
traction forte T, and the point load Pi. These forces are
shown acting on a body in Fig.
Body Force:- A body force is a distributed force acting on every
elemental volume of the body and has the units of force per unit
traction Force:- A traction force is a distributed load acting on the
surface of the body. For the one-dimensional problem
considered here, however, the traction force is defined as force
per unit length.
Point Load:- Finally, Pi is a force acting at a point i and ui is the
x displacement at that point.
21. 1-D FINITE ELEMENT
The finite element modeling of a one-dimensional
body is considered in Section.
The basic idea is to discretize the region and
express the displacement field in terms of values
at discrete points.
Linear elements are introduced first.
Stiffness and load concepts are developed using
potential energy and Galerkin approaches.
Boundary conditions are then considered.
22. The first step is to model the bar as a stepped shaft, consisting of a
discrete number of elements, each having a uniform cross section.
Specifically, let us model the bar using four finite elements. A simple
scheme for doing this is to divide the bar into four regions, as shown in
In the finite element model, every element connects to two nodes.
However, cross-sectional area, traction, and body forces can differ in
magnitude from element to element.
Better approximations are obtained by increasing the number of
It is convenient to define a node at each location where a point load is
In a one-dimensional problem, every node is permitted to displace only
in the ±x direction. This, each node has only one degree of freedom
The five-node finite element model in Fig. has five dofs.
The displacements along each dof are denoted by Q1, Q2, Q3... Q5. In
fact, the column vector Q = [Q1. Q2 .. 'Q5]T is called the global
24. SHAPE FUNCTIONS
Consider a typical finite element e in Fig.
In the local number scheme, the first node will be numbered 1 and
the second node 2.
The notation X1 = x-coordinate of node 1
X2 =x-coordinate of node 2 is used.
We define a natural or intrinsic coordinate system, denoted by ξ as
25. From Fig. we see that ξ = -1 at node 1 and ξ = 1 at node 2.
The length of an element is covered when ξ changes from -1
We use this system of coordinates in defining shape
functions, which are used in interpolating the displacement
Now the unknown displacement field within an element will be
interpolated by a linear distribution. Fig.3.6
To implement this linear interpolation, linear shape functions
will be introduced as
27. The shape functions N1 and N2 are shown in Figs. Respectively.
The graph of the shape function N1 in Fig.(a) is obtained from equation
N1(ξ) by noting that N1 = 1 at ξ = -1, N1 = 0 at ξ = 1.
and N1 is a straight line between the two points.
Similarly, the graph of N2 in Fig. 3.7(b) is obtained from Eq. N2(ξ).
Once the shape functions are defined, the linear displacement field within
the element can be written in terms of the nodal displacements q1 and q2
u= N1q1 + N2q2
or, in matrix notation, as u=Nq
Where N = [N1 N2] and q=[q1 q2]T
In these equations, q is referred to as the element displacement
It is readily verified from equation of u that u=q1 at node 1 and u=q2 at
node 2, and that u varies linearly.
It may be noted that the transformation from x to ξ in eq of u can be
written in terms of N1 and N2 as
x= N1x1 + N2x2
28. Strain - displacement matrix
The strain--displacement relation as
Upon using the chain rule of differentiation, we obtain
From the relation between x and ξ, we have
So, can be written as
29. where the (1 x 2) matrix B, called the element
strain-displacement matrix, is given by
Note: Use of linear shape functions results in a
constant B matrix and, hence, in a constant strain
within the element. The stress, from Hooke's law,
30. Element stiffness matrix
Consider the strain energy term
Substituting for and in
In the finite element model, the cross-
sectional area of element e, denoted by Ae, is
31. Also, B is a constant matrix. Further, the transformation
from x to ξ in shape function equation.
where -1 <ξ<1, and le = IX2 – X1l is the length of the
The element strain energy Ue is now written as
where Ee is Young's modulus of element e.
Noting that and substituting for B from Eq.
32. This equation is of the form
where the element stiffness matrix ke is given by
observe that Ke is linearly proportional to the
product Ae Ee and inversely proportional to the
33. PROPERTIES OF K (Stiffness
1. The dimension of the global stiffness K is (N x N),
where, N is the number of nodes. This follows
from the fact that each node has only one degree
1. K is symmetric.
2. K is a banded matrix. That is. all elements outside of
the band are zero.