LECTURE 2

The main topics in this session will include the following: Review of matrix algebra Types of finite elements Concept of Stiffness Matrix 1-D Spring Element. Derivation of the stiffness matrix for a Spring Element and

a Spring Assemblage Example of solution of spring assemblage using Direct Stiffness Method

Learning objectives: At the end of this session, you should be: Able to perform the matrix operations Familiar with the concept of stiffness matrix Familiar with the types of finite elements Able to develop the stiffness matrix for a Spring Assemblage

2.1 REVIEW OF LINEAR ALGEBRA Matrix methods are a necessary tool used in the finite element method for purposes of simplifying the formulation of the element stiffness equations, for purposes of long hand solutions of various problems, and, most important, for use in programming the method for computer applications. Hence, matrix notation represents a simple and easy-to-use notation for writing and solving sets of simultaneous algebraic equations. Definition of a Matrix

A matrix is a square or rectangular array of numbers set out in rows and columns.

nmnn

m

m

aaa

aaa

aaa

A

21

22221

11211

Matrix is represented by a single symbol [A], ij

a designates an individual element of the matrix. The first subscript i always designates the number of the row in which the element lies. The second subscript j designates the column.

For example, element 21a is in row 2 and column 1.

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The matrix with n rows and m columns is said to have a dimension of n by m (or n x m). It is referred to as an n by m matrix.

Matrices with row dimension n = 1 are called row vectors.

Matrices with only one column (m = 1) are referred to as column vectors and are

denoted by , i.e.

1

2

m

Matrices where n = m are called square matrices. The diagonal consisting of the

elements nn

aaaa ,,,,332211

is termed the main diagonal. Square matrices are particularly important when solving sets of simultaneous

linear equations. Matrix Operating Rules

Addition (subtraction) of two matrices, say, [A] and [B], is accomplished by adding (subtracting) corresponding terms in each matrix. The elements of the resulting matrix [C] are computed as

ijijijbac

for ni ,,2,1 and .,,2,1 mj Addition and subtraction can be performed only between matrices having the same dimensions.

The multiplication of a matrix [A] by a scalar g is obtained by multiplying every element of [A] by g:

nmnn

m

m

gagaga

gagaga

gagaga

AgD

21

22221

11211

The product of two matrices is represented as [C]=[A][B], where the elements of [C] are defined as follows:

n

kkjikij

bac1

3

where n = the column dimension of [A] and the row dimension of [B]. Multiplication of two matrices can be performed only if the first matrix has as many columns as the number of rows in the second matrix.

lnlmmn CBA

For matrix [A] of order 2 x 2 and matrix [B] of order 2 x 2, after multiplying the two matrices, we have

11 11 12 21 11 12 12 22

21 11 22 21 21 12 22 22

[ ]ija b a b a b a b

ca b a b a b a b

For example, let

2 1[ ]

3 2A

1 1[ ]

2 0B

The product [A] [B] is then

2(1) 1(2) 2( 1) 1(0) 4 2[ ][ ]

3(1) 2(2) 3( 1) 2(0) 7 3A B

Transpose of a Matrix Any matrix can be transposed. This operation is frequently used in finite element equation formulations. The transpose of a matrix [A] is commonly denoted by [A]T . The superscript T is used to denote the transpose of a matrix [A]. The transpose of a matrix is obtained by interchanging rows and columns; that is, the first row becomes the first column, the second row becomes the second column, and so on. For example, if we let

2 1

[ ] 3 2

4 5

A

then

2 3 4[ ]

1 2 5

TA

where we have interchanged the rows and columns of [A] to obtain the transpose.

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Special Types of a Square Matrix A diagonal matrix is a square matrix where all elements off the main diagonal are equal to zero, as in

44

33

22

11

0

0

a

a

a

a

A

Note that where large blocks of elements are zero, they are left blank.

An identity (unit) matrix is a diagonal matrix where all elements on the main diagonal are equal to 1 and all other elements equal to zero, as in

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

I

The symbol [I] is used to denote the identity matrix. The identity matrix has properties similar to unity. Inverse of a Matrix The inverse of a matrix is a matrix such that

1 1[ ] [ ] [ ][ ] [ ]A A A A I where the superscript, -1, denotes the inverse of [A] as [A]-1. Note that the inverse of a matrix may only be calculated for a non-singular square matrix (non-zero determinant) and may be obtained by several methods (e.g., the Gauss-Jordan method).

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2.2 CONCEPT OF STIFFNESS MATRIX The elastic spring is one of the simplest structural components to analyse, which has a direct relationship between the force in the spring and the displacement of its free end:

Figure 2.1 Simple elastic spring

The relationship between the internal force and displacement of the free end is given by

F k

where k is termed the stiffness of the spring and corresponds to the slope of the force-displacement curve. Knowing the applied force allows this equation to be inverted to give the displacement as

F

k

It is possible to extend this approach to find the stiffness of a complete structure, which has numerous points that can deflect. Using matrix notation

F k

The quantity k is termed the stiffness matrix for the structure, which relates the

applied nodal forces, F , to the unknown nodal displacements, . The unknown nodal displacements are then found by solving the set of linear simultaneous equations, giving

1

k F

To obtain a clearer understanding of elements ijk of the stiffness matrix, we write out

the expanded form of the stiffness equation as

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1 11 12 1 1

1 21 22 2 1

1 2

x n x

y n y

nz n n nn nz

F k k k

F k k k

F k k k

where the force components of the vector 1 1 1 2 2 2T

x y z x y z nx ny nzF F F F F F F F F F act

at various nodes (1, 2, , n) on a structure and the corresponding set of nodal

displacements 1 1 1 2 2 2T

x y z x y z nx ny nz . The subscripts to the right of F

and identify the node and the direction of the force or displacement. For instance, 1xF

denotes the force at node 1 applied in the x direction. Now assume a structure to be forced into a displaced configuration defined by

1 1 11, 0x y z nz . Then from the above stiffness equation, we have

1 11 1 21 1, ,x y nz nF k F k F k

It can be concluded that the element stiffness coefficients kij represent the force Fi in

the direction of the ith degree of freedom due to a unit displacement j in the jth degree of freedom while all other displacements are zero. 2.3 TYPES OF FINITE ELEMENTS As it has been discussed earlier, the first step of the FE method involves dividing the body into an equivalent system of finite elements with associated nodes and choosing the most appropriate element type to model the actual physical behaviour of a system. The discretised body or mesh is often created with mesh-generation programs or pre-processor programs available to the user. The choice of element types used in a finite element analysis depends on physical composition of the body under actual loading conditions. Judgement concerning the appropriateness of one-, two-, or three-dimensional elements is necessary. It should be emphasised that the choice of the most appropriate element for a particular problem is one of the major tasks that must be carried out by the designer/analyst. The primary line elements consist of spring, bar (or truss) and beam elements. The spring elements are characterised by a value of the axial spring stiffness coefficient. The bar and beam elements have a cross-sectional area. In general, the cross-sectional area within the element is kept constant. Theses elements are often used to model trusses and frame structures. The simplest line element (called a linear element) has two nodes, one at each end, although higher-order elements having three nodes or more also exist. The line elements are the simplest of elements to consider and will be discussed to illustrate many of the basic concepts of the finite element method.

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The basic two-dimensional (or plane) elements are loaded by forces in their own plane (plane stress or plane strain conditions). They are triangular or quadrilateral elements. The simplest two-dimensional elements have corner nodes only (linear elements) with straight sides or boundaries. There are also higher-order elements, typically with midside nodes (called quadratic elements) and curved sides. They are often used to model a wide range of engineering problems. The most common three-dimensional elements are tetrahedral and hexahedral (or brick); they are used when it becomes necessary to perform a three-dimensional stress analysis. The basic three-dimensional elements have corner nodes only and straight edges. Higher-order three-dimensional elements also exist, typically with midedge nodes and curved surfaces for their sides. The axisymmetric element is developed by rotating a triangle or quadrilateral about a fixed axis located in the plane of the element through 360 deg. This element can be used when the geometry and loading of the problem are axisymmetric. The following table illustrates some of the elements currently used by finite element software for the solution of structural/mechanical problems:

Name Shape Nodes DoF/node DoF

Spring/Truss (Displacement)

2 1 2

Simple Beam (Bending)

2 2 4

Triangular (Plane Stress/ Plane Strain)

3 2 6

Quadrilateral (Plane Stress/ Plane Strain)

4 2 8

Tetrahedron (Three-

dimensional stress)

4 3 12

Hexahedron (Three-

dimensional stress)

8 3 24

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2.4 LINEAR SPRING ELEMENT This section introduces some of the basic concepts on which the direct stiffness method is founded. The linear spring is introduced first because it provides a simple tool to illustrate the basic concepts. Step 1 Stiffness Matrix for a Spring Element We will now derive the stiffness matrix for a one-dimensional linear spring that is, a spring that obeys Hookes law and resists forces only in the direction of the spring. Consider the linear spring element shown in Figure 2.2. Reference nodes are located at the ends of the element. These reference points are called the nodes of the spring element. The positive nodal forces are f1 and f2. The local x axis acts in the direction of the spring so that we can directly measure displacements and forces along the spring. The local nodal displacements are d1 and d2 for the spring element. These nodal displacements are called the degrees of freedom at each node. Positive direction for the forces and displacements at each node are taken in the positive x direction as shown from node 1 to node 2 in the figure. The symbol k is called the spring constant or stiffness of the spring. We now want to develop a relationship between nodal forces and nodal displacements for a spring element. This relationship will be the stiffness matrix. Step 2 Define the Stress-Strain Relationship

The tensile forces T produce a total elongation (deformation) of the spring. The typical total elongation of the spring is shown in Figure 2.3. Here d1 is a negative value

x

d1 d2 L

T T

Figure 2.2 Spring element with positive nodal displacement and force conventions

Figure 2.3 Deformed spring

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because the direction of displacement is opposite the positive x direction, whereas d2 is a positive value. The deformation of the spring is then represented by

2 1d d (2.1)

The stress-strain relationship can be expressed in terms of the force-deformation relationship as

T k (2.2) Now, using Eq. (2.1) in Eq. (2.2), we obtain

2 1T k d d (2.3) Step 3 Derive the Element Stiffness Matrix and Equations We now derive the spring element stiffness matrix. By sign convention for nodal forces and equilibrium, we have

1 2f T f T (2.4)

Using Eqs. (2.3) and (2.4), we have

1 1 2

2 2 1

f k d d

f k d d

(2.5)

Now expressing Eqs. (2.5) in a single matrix equation yields

1 1

2 2

f dk k

f dk k

(2.6)

This relationship holds for the spring along the x axis. From our basic definition of a stiffness matrix, we obtain

k k

kk k

(2.7)

as the stiffness matrix for a linear spring element. Here k is called the local stiffness

matrix for the element. We can observe from Eq. (2.7) that k is a symmetric square matrix.

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Step 4 Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions

The global stiffness matrix and global force matrix for the whole structure are assembled using force equilibrium equations, force-deformation, compatibility equations, and the direct stiffness method described later. This step applies for structures composed of more than one element such that

( ) ( )

1 1

and N N

e e

e e

K k F f

(2.8)

where k and f are now element stiffness and force matrices expressed in a global coordinate system. (Note: the sign used in this context does not imply a simple summation of element matrices but rather denotes that these element matrices must be assembled according to the direct stiffness method.) Step 5 Solve for the Nodal Displacements The displacements are then determined by imposing boundary conditions, such as

support conditions, and solving a system of equations, F K d , simultaneously. Step 6 Solve for the Element Forces Finally, the element forces are determined by back-substitution, applied to each element, into equations similar to Eqs. (2.5). 2.4 ASSEMBLY OF SPRING ELEMENTS Structures such as trusses, frames, and bridges comprise basic structural components connected together to form the overall structural systems. To analyse these structures, we must determine the total structure stiffness matrix for a system of elements. Before considering the whole structure, we will determine the total structure stiffness matrix for a spring assemblage by using the force-displacement matrix relations derived in Section 2.3 for the spring element. We will then illustrate the procedure outlined in Step 4 above. We will consider the example of the two-spring assemblage shown in Figure 2.4. Here we fix node 1 and node 3 and apply axial force F2 at node 2. The stiffness of spring elements 1 and 2 are k1 and k2, respectively. The nodes of the assemblage have been numbered 1, 2, and 3. The x axis is the global axis of the assemblage. The local axis of each element coincides with the global axis of the assemblage. For element 1, using Eq. (2.6), we have

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1 1 1 1

2 1 1 2

f k k d

f k k d

(2.9)

and for element 2, we have

2 22 2

3 32 2

f dk k

f dk k

(2.10)

Let us consider the forces acting on each node. The free-body diagram of nodes, which shows the force acting on nodes 1 through 3 of this model, is depicted in Figure 2.5. Static equilibrium requires that the sum of the forces acting on each node be zero. This requirement creates the following three equations:

1 1 1 2 1

1 1 1 2 2 2 3 2

2 2 2 3 3

( )

k d k d F

k d k k d k d F

k d k d F

(2.11)

In matrix form, Eqs. (2.11) are expressed by

Figure 2.4 Two-spring assemblage

F1 k1(d2 d1) k2(d3 d2)

k1(d2 d1)

F3

k2(d3 d2)

F2

Node 1 Node 2 Node 3

Figure 2.5 Free body diagram of the nodes

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1 1 1 1

2 1 1 2 2 2

3 2 2 3

0

0

F k k d

F k k k k d

F k k d

(2.12)

Equation (2.12) is now written as the single matrix equation

F K d (2.13)

where 1

2

3

F

F F

F

is called the global nodal force matrix, 1

2

3

d

d d

d

is called the

global nodal displacement matrix, and

1 1

1 1 2 2

2 2

0

0

k k

K k k k k

k k

(2.14)

is called the global stiffness matrix. We will consider the complete solution to this example problem after considering a more practical method of assembling the total stiffness matrix and discussing the support boundary conditions in Section 2.5. 2.5 ASSEMBLING THE GLOBAL STIFFNESS MATRIX BY DIRECT STIFFNESS

METHOD We will now consider a more convenient method for constructing the global stiffness matrix. This method is based on proper superposition of the individual element stiffness matrices making up a structure. Referring to the two-spring assemblage of Section 2.4, the element stiffness matrices are given in Eqs. (2.9) and (2.10) as

1 1 2 2

1 2 2 3

1 1 2 2

k k k k

k kk k k k

(2.15)

Here the dis written above the columns and next to the rows indicate the degrees of freedom associated with each element row and column. The two element stiffness matrices are not associated with the same degrees of freedom; that is, element 1 is associated with axial displacement at nodes 1 and 2,

d1 d2

d1

d2

d2 d3

d2

d3

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whereas element 2 is associated with axial displacements at nodes 2 and 3. Therefore, the element stiffness matrices cannot be added together in their present form. To superimpose the element matrices, we must expand them to the size of the total structure (spring assemblage) stiffness matrix so that each element stiffness matrix is associated with all the degrees of freedom of the structure. To expand each element stiffness matrix to the order of the global stiffness matrix, we simply add rows and columns of zeros for those displacements not associated with that particular element. For element 1, we rewrite the stiffness matrix in expanded form as

1 1

1 2 1 1

0

0

0 0 0

k k

k k k

(2.16)

Similarly, for element 2, we have

2 3 2 22 2

0 0 0

0

0

k k k

k k

(2.17)

The global stiffness matrix is then obtain as

1 1

1 2 2 3 1 1 2 2

2 2

0

0

k k

K k k k k k k

k k

(2.18)

and the total set of stiffness equations is written as

1 1 1 1

1 1 2 2 2 2

2 2 3 3

0

0

k k d F

k k k k d F

k k d F

(2.19)

This method of directly assembling individual element stiffness matrices to form the total structure stiffness matrix and the total set of stiffness equations is called the direct stiffness method. It is the most important step in the finite element method. 2.6 Boundary Conditions We must specify boundary (support) conditions for structure such as the spring

assemblage, or K will be singular; that is, the determinant of K will be zero, and its inverse will not exist. This means the structural system is unstable. Without specifying

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adequate support conditions, the structure will be free to move as a rigid body and not resist any applied load. As indicated in Figure 2-4, all boundary conditions are such that the displacements are zero at node 1 (d1 = 0) and node 3 (d3 = 0) because they are fixed. Therefore, Eq. (2.19) can be written as

1 1 1

1 1 2 2 2 2

2 2 3

0 0

0 0

k k F

k k k k d F

k k F

(2.20)

The final equation is obtained by deleting the rows and columns corresponding to the zero-displacement degrees of freedom. Here row 1 and column 1 are deleted because d1 = 0. However, F1 and F3 are not necessarily zero and can be determined once d2 is solved for. After solving Eq. (2.20) for d2, we have

2

2

1 2

Fd

k k

(2.21)

Now we can determine reaction forces F1 and F3 by substituting d2 in the first and third equations of (2.20) as

11 1 2 2

1 2

23 2 2 2

1 2

kF k d F

k k

kF k d F

k k

(2.22)

As the last step in the FEA analysis, the internal forces can be obtained by making use of the local element equations Eq. (2.6).