Finite Element Method
Many practical problems in engineering are either extremely difficult or impossible to solve by conventional analytical methods. Such methods involve finding mathematical equations which define the required variables. For example, the distribution of stresses and displacements in a solid component, or of pressure and velocities in the flow of a fluid, might be required.
In the past, the common practice was to simplify such problems to the point where an analytical solution could be obtained which, it was hoped, bore some resemblance to the solution of the real problem. Because of the uncertainties involved in such a procedure (over-simplification of the initial problem), large factors of safety would be introduced.
What is FEM (FEA)?
The basis of the finite element method (also commonly known as Finite Element Analysis) is that a body or structure may be divided into smaller elements of finite dimensions called Finite Elements. The original body or structure is then considered as an assemblage of these elements connected at a finite number of joints called Nodes or Nodal Points.
Steps involved in a Finite Element Analysis procedure
1. Discretization (subdivision of the body or structure into smaller elements)
2. Application of mathematical model to each element (formulation of the properties of the elements)
3. Combination of the elements to obtain the properties of the whole body/structure (assembly)
Thus, instead of solving the problem for the entire body/structure in one operation, in the method of finite elements, attention is mainly devoted to the formulation of properties of the constituent elements. A common procedure is then adopted for combining the elements.
Single Spring Element
When a pin-jointed truss element is part of a structure, its ends will be able to move due to (i). displacement of the structure (lump displacement) (ii). deformation of the member
This may be modelled by a spring element.
The nodes of the spring element are indicated by points i and j in Fig. 1.
Let if = nodal force at node i iu = nodal displacement at node i
For this particular case of the spring element, we will consider only one degree of freedom (DOF) for each node of the spring. We will assume that displacements are possible only along the axis of the spring.
Spring element loaded by oppositely directed forces
Let us first evaluate the response of a spring element to which oppositely directed forces of magnitude N are applied.
If we now consider that a spring element is used to model a truss member of length L and cross-sectional area A, the stiffness 1k of the spring will be given by
1AEk L= 
where E is Young's Modulus for the material.
Using the sign convention that forces and displacements are positive in the positive x-direction, the forces may be related to the displacements by the following equations:
1 1 1 2 1 1 1 2f k u u k u k u = =  2 1 2 1 1 1 1 2f k u u k u k u + = = 
In matrix form, the system of equations can be written as
1 1 1 122 1 1
f k k uuf k k
if = (element) nodal force vector ju = (element) nodal displacement vector
ijk = (element) "stiffness" matrix
local stiffness matrix
We shall denote by eiK
or eiK the element stiffness matrix for
the i numbered element. Therefore,
1 11 1
e ek k
K Kk k
An important property of the stiffness matrix for an element (and as will be seen later, for a complete structure) is that it is symmetrical.
Assembly of Spring Elements
Consider a system of combined spring elements. For this particular configuration of linked spring elements, the assembly process is equivalent to finding the stiffness matrix for the whole system.
As a n example, let us consider a system consisting of two spring elements (Fig. 2). Note that 1F , 2F and 3F are externally applied loads.
The element nodal force components are denoted by ijf , where i is the element number, and j is the node number for each element.
Drawing free body diagrams for each spring element allows for the representation of internal nodal forces ( ijf ).
Using equation , the force-displacement relationship for each element may be written as
11 1 1 11 21 12
f k k uuk kf
221 2 2
2 32 22
uf k kuk kf
Assembly Method 1
We expand the two matrix equations so that they are in equivalent form
11 1 11
1 21 123
0 0 00
k k ufuk kfu
1 2 2 22 32 22
0 0 0 000
f k k uuk kf
1 1 1
0 0 0e
k kK k k
2 2 2
0 0 0 0
0eK k k
The matrix equation for the whole system is obtained by adding the matrix equation for each system (superposition)
(assembly of spring elements addition)
11 1 1 1
1 22 1 1 1 2 2 2
2 32 22
f k k uf f k k k k u
1 2+e eK K K=
1) It is imperative that the element stiffness matrices are first expanded before proceeding with the addition operation.
2) The global stiffness matrix will have dimensions n n , where n is the number of degrees of freedom (DOF) of the system.
Assembly Method 2
An alternative way of assembling the stiffness matrix for the system is to consider the equilibrium of forces for each node of the system.
global stiffness matrix
Equilibrium of forces at node 1 For node 1, the element nodal force is equal to the externally applied force 1F :
11 1f F=
1 1 1 1 2 F k u k u =
Equilibrium of forces at node 2 The sum of element nodal forces applied to node 2 is equal to the externally applied force 2F :
1 22 1 2f f F+ =
( )2 1 1 1 2 2 2 3 F k u k k u k u+ = +
Equilibrium of forces at node 3 The element nodal force applied to node 3 is equal to the externally applied force 3F :
22 3f F=
3 2 2 2 3 F k u k u = +
Writing the equations for nodal equilibrium in matrix form, we obtain the same global stiffness as when we expand the element stiffness matrices and added them together.
Assembly Method 3
A 3rd method for performing the assembly of elementary stiffness matrices is to use a table of connectivity.
As an introduction to the use of the method, a sample connectivity table is given below. Details of the method will be introduced through a sample problem.
Element no. Local Element
Node 1 Node 2
1 1 2
2 2 3
Boundary and Load Conditions
Let us assume that:
1 0u = (node 1 is fixed) 2 3F F P= =
11 1 11 1
1 22 1 2 1 1 2 2 2
2 33 2 22
f k kF uf f F k k k k u
uF k kf
global node numbers
1 1 2 2 232 2
k kFP k k k k uP uk k
which reduces to
21 2 232 2
uk k kPuP k k
and 1 1 2.F k u=
The unknowns are 23
= and the reaction force 1F .
Solving the equations, we obtain the nodal displacements and the reaction force.
Pu ku P P
+ 1 2F P=
For the spring system above,
1 100 /k N mm= , 2 200 /k N mm= , 3 100 /k N mm= 500P N= , 1 4 0u u= =
Find: a) the global stiffness matrix b) the displacements of nodes 2 and 3 c) the reaction forces at nodes 1 and 4 d) the force in the spring 2
For the above spring system, with arbitrarily numbered nodes and elements as shown, find the global stiffness matrix.
Note that the analysis is based on the fact that the elements used sustain only axial forcesrod or truss member
For this particular configurationa) each node of a b) the bar is no longer aligned with the global
Local and Global Coordinates
The spring element can be used important engineering problemexample. Structures by frictionless hinges which imply that only tension or compression exists in the bars.shown below.
russ element having 2 DOF / NODE
Note that the analysis is based on the fact that the elements used sustain only axial forces (bars or rods). The terms
member can be used indifferently in what follows.
configuration, we consider the following:each node of a bar has 2 degrees of freedom
no longer aligned with the global x coordinate
Local and Global Coordinates
The spring element can be used as a simplified modelimportant engineering problems. A structure is one such
Structures normally consist of trusses, i.e. bars joined by frictionless hinges which imply that only tension or
ts in the bars. A typical truss structure is
2 DOF / NODE
Note that the analysis is based on the fact that the elements The terms bar,
can be used indifferently in what follows.
model in many A structure is one such
normally consist of trusses, i.e. bars joined by frictionless hinges which imply that only tension or