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Module: 2 Finite Element Formulation Techniques Lecture 4: Stiffness Matrix and Boundary Conditions

2.4.1Element Stiffness Matrix The stiffness matrix of a structural system can be derived by various methods like variationalprinciple, Galerkin method etc. The derivation of an element stiffness matrix has already been discussed in earlier lecture. The stiffness matrix is an inherent property of the structure. Element stiffness is obtained with respect to its axes and then transformed this stiffness to structure axes.The properties of stiffness matrix are as follows:

Stiffness matrix issymmetric and square. In stiffness matrix, all diagonal elements are positive. Stiffness matrix is positive definite

Forexample,ifKisasymmetricnnrealmatrixandxisnonzerocolumnvector,thenKwillbepositivedefinitewhilexTKxispositive. 2.4.2Global Stiffness Matrix A structural system is an assemblage of number of elements. These elements are interconnected together to form the whole structure. Therefore, the element stiffness of all the elementsarefirst need to be calculated and then assembled together in systematic manner. It may be noted that the stiffness at a joint is obtained by adding the stiffness of all elements meeting at that joint. To start with, the degrees of freedom of the structure are numberedfirst. This numbering will start from 1 to n where n is the total degrees of freedom. These numberings are referred to as degrees of freedom corresponding to global degrees of freedom. The element stiffness matrix of each element should be placed in their proper position in the overall stiffness matrix. The following steps may be performed to calculate the global stiffness matrix of the whole structure.

a. Initialize global stiffness matrix K as zero. The size of global stiffness matrix will be equal to the total degrees of freedom of the structure.

b. Compute individual element properties and calculate local stiffness matrix k of that element.

c. Add local stiffness matrix k to global stiffness matrix K using proper locations d. Repeat the Step b. and c. till all local stiffness matrices are placed globally.

The steps to be followed in the computer program are shown in the form of flow chart in Fig. 2.4.1 for assembling the local stiffness matrix to global stiffness matrix.

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Fig. 2.4.1Assembly of stiffness matrix from local to global

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2.4.3Boundary Conditions Under this section, procedure to include the effect of boundarycondition in the stiffness matrix for the finite element analysis will be discussed. The solution cannot be obtained unless support conditions are included in the stiffness matrix. This is because, if all the nodes of the structure are included in displacement vector, the stiffness matrix becomes singular and cannot be solved if the structure is not supported amply, and it cannot resist the applied loads.A solution cannot be achieved until the boundary conditions i.e., the known displacements are introduced.

In finite element analysis, the partitioning of the global matrix is carried out in a systematic way for the hand calculation as well as for the development of computer codes. In partitioning, normally the equilibrium equations can be partitioned by rearranging corresponding rows and columns, so that prescribed displacements are grouped together. For example, let considerthe equation of equilibrium is expressed in compact form as:

F K d (2.4.1) Where, [K] is the global stiffness matrix, {d} is the displacement vector consisting of global degrees of freedom, and {F} is the load vector corresponding to degrees of freedom. By the method of partitioning the above equation can be partitioned in the following manner.

K KF d

F dK K

(2.4.2)

Where,subscripts refers to the displacements free to move and refers to the prescribed support displacements.As the prescribed displacements {d} are known,eq. (2.4.2) may be written in expanded form as: F = K d + K d (2.4.3) Thus it is possible to obtain the free displacement of the structure{d} as

-1d = K F - K d (2.4.4) If the displacements at supports {d}are zero, then the above equation can be simplified to the following expression.

-1d = K F (2.4.5) Thus, by rearranging assembled matrix, the portion corresponding to the unknown displacements in eq.(2.4.4) can be taken out for the solution purpose. This is possible as the known displacements {d}are restrained, i.e., displacementsare zero. If the support has some known displacements, then eq. (2.4.4) can be used to find the solution. If the few supports of the structures yield, then the above method may be modified by partitioning the stiffness matrix into three parts as shown below:

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{ }{ }{ }

[ ] { }{ }{ }

K K KF d

F K K K d

F dK K K

aa ab aga a

b ba bb bg b

g gga gb gg

= (2.4.6)

Here, refers to unknown displacement; refers to known displacement (0) and refers to zero displacement. Thus, the above equation can be separated and solved independently to find required unknown results as shown below.

{ } [ ]{ } { } { }[ ]{ } { } { } { } { }{ } [ ] { } { }{ }1

F K d K d K d

or, K d F K d as d 0

Thus, d K F K d

a aa a ab b ag g

aa a a ab b g

-a aa a ab b

= + + = - =

= - (2.4.7)

For computer programming, several techniques are available for handling boundary conditions. One of the approachesis to make the diagonal element of stiffness matrix corresponding to zero displacement as unity and corresponding all off-diagonal elements as zero. For example,let consider a 33 stiffness matrix with following force-displacement relationship.

1 11 12 13 1

2 21 22 23 2

3 31 32 33 3

F k k k dF k k k dF k k k d

(2.4.8)

Now, if the third node has zero displacement (i.e., d3= 0) then the matrix will be modified as follows to incorporate the boundary condition.

1 11 12 1

2 21 22 2

3

00

0 0 0 1

F k k dF k k d

d

(2.4.9) Thus, while inverting whole matrix, d3 will become zero automatically.

To incorporate known support displacement in computer programming following procedure may be adopted. Considering the displacement d2has known value of , 1st row of eq. (2.4.8) can be written as:

1 11 1 12 2 13 3F k d k d k d (2.4.10)

Or

1 12 11 1 13 3F k k d k d (2.4.11)

Now the 2nd row of eq. (2.4.8) has to become:

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2d (2.4.12) Similarly 3rd row will be:

3 32 31 1 33 3F k k d k d (2.4.13) Thus above three equations can be written in a combined form as

1 12 11 13 1

2

3 32 31 33 3

00 1 0

0

F k k k dd

F k k k d

(2.4.14) Another approach may also be followed to take care the known restrained displacementsby assigning a higher value (say =1020) in the diagonal element corresponding to that displacement.

1 11 12 13 120 20

22 21 22 23 2

3 31 32 33 3

10 10F k k k d

k k k k dF k k k d

(2.4.15)

20 2022 21 1 22 2 23 310 k k d k 10 d k d\d = + +

As d3 is corresponding to zero displacement, the above equation can be simplified to the following. 20 20

22 21 1 22 220 20

22 22 2

2

10 k k d k 10 d

or 10 k k 10 dd known displacement is ensured

\d = + d =

=d If the overall stiffness matrix is to be formed in half band form then the numbering of nodes should be such that the bandwidth is minimum. For this the labels are put in a systematic manner irrespective of whether the joint displacements are unknowns or restraints. However, if the unknown displacements are labeled first then the matrix operations can be restricted up to unknown displacement labels and beyond that the overall stiffness matrix may be ignored.