chapter 2_finite element procedure

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Chapter2, M.Sc. Lecture Note

by Professor S.L. Chan 56

Chapter 2

Finite Element Procedure

First-order Linear and Second-order Nonlinear analysis

2.1 INTRODUCTION

In the previous chapter we discussed the formulation of an element matrix. We also noted

that in the finite element approach we need not assume a correct displacement function

satisfying all the boundary conditions. Instead, we express the displacement function in

term of the degree of freedom. After the assembling process, we obtain the stiffness

matrix for the complete structure. The displacements are then solved and from these we

can calculate the member forces for design. In this chapter, we discuss the process for

assembling and also the solution technique for a large stiffness matrix.

2.2 Additional example on element formulation - the simplest Constant Strain

Triangle (CST)

The formulation of a constant triangle is given to further standardise the finite element

procedure in this section. The element has three nodes at the corners of a triangle as

follows.

y+x+=u 21o α α α

y+x+=v 543 α α α

Since we have three nodes, we can at most assume three coefficients as for the

displacement as,

1

2

3





Equation 1 can be expressed in compact form as,

Put the conditions for u as,

u=u1 at x=x1, y=y1

u=u2 at x=x2, y=y2 (4)

u=u3 at x=x3, y=y3

we obtain the expression for nodal displacement u as,

[u] = [A] [α] (5)

[α] = [A-1

] [u] (6)

[u] = [ f(x,y) ]T [A]

-1 u

= [N] [u]

= [N1 N2 N3] [u1 u2 u3]T (7)

Similarly,

[v] = [N1 N2 N3] [v1 v2 v3]T (8)

][]

y)f(x,[=uTα

[ ]vuvuvu N0 N0 N0

0 N0 N0 N =

v

u332211

T

321

321

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂

∂

∂

∂

∂

∂

∂

∂

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂

∂

∂

∂

∂

∂

∂

∂

v

u

xy

y

0

0

x

=

x

v +

y

u

y

v

x

u

=][ ε





Substituting 5 into 6, we have in compact form,

(11)

For plane stress problem where stress perpendicular to the element is zero (σz = 0),

(12)

or,

[σ] = [D] [ε] (13)

The element matrix is then given by,

Can you derive a cubic element using the same procedure and in the last chapter ?

]u[]B[=][ε

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡

σ

σ

ν

ν

ε

ε

y

x

y

x

1-

-1

E

1 =

[B][C]][BTA=dv[B][D]][B=dv][[D]][=]k [TT

v

T

ve ∫∫ ε ε (14)





2.3 Theoretical Background

The element stiffness matrix, as its name implies, means the stiffness of the element. The

stiffness or resistance of a structure will then be equal to the sum of all its composing elements.Physically, the elements share the loads according to the ratio of their stiffness. As shown in the

Figure 2.1, the stiffness of the structure under a moment at joint A is equal to the sum of the four

elements.

In the case of a structure with more than 1 degree of freedom, the stiffness of the structure is still

equal to the sum of its constituent elements. However, the contributions of these elements must

be added to the corresponding degrees of freedom of the stiffness for the whole structure. For

example, if element A is connected to nodes i and j, its stiffness must be added to node i and j of

the global structure.





A revision on the stiffness matrix method of structural analysis

EA=1500 for all members

EI=100 EI=200

5

8 10

6

12

16

1

2 3

A

B

Order of reference = x y θ

x

y





Analyse the plane frame above. The linear element stiffness is as follows.

The transformation matrix for member "A" is (equation 26 in chapter 2):

⎥

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎢

⎣

⎡

I4

L

I6-

L

I12.SYM

00A

I2

L

I6-0I4

L

I6

L

I12-0

L

I6

L

I12

00A-00A

L

E =]k [

2

22

L

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

100|000

00.80.6-|000

00.60.8|000

--------------

000|100

000|00.80.6-

000|00.60.8

=[T]





For member 'B', [T]=[I], an identity matrix. Thus it is equal to the original

stiffness matrix. The final assembled stiffness matrix is :-

⎥⎥

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∑

8012-2.4

00150.SYM

4012-0120

122.4-07.257.57

00150-3.671.42246.4

000204.8-3.640

0004.854.77-71.42-4.854.77

0003.6-71.42-96.43-3.6-71.4296.43

=]k [=[K] L

2

=1i

Load vector :

For member "A", it is equal to :-

⎥⎥⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢⎢

⎣

⎡

406-02060

6-1.206-1.2-00015000150-

206-04060

61.2-061.20

00150-00150

=[T]]k [][T eT





In global axis, it is equal to :-

Adding the loads at global node, the final load vector is :-

Solving,

⎥⎥⎥⎥⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎢⎢⎢

⎣

⎡

20

8-

020-

8-

0

=][F a

⎥⎥⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎢

⎣

⎡

20

6.4-

4.8

20-

6.4-

4.8

=][F][Te

T

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

0

0

0

20

18.4-

4.8

20-

6.4-

4.8

=[F]





The displacement in local axis can then be obtained by transforming the

displacement to local coordinate as,

⎥⎥⎥

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

0

0

0

0.3320

0.7228-

0.2354

0.7739-

0

0

=

[F]][K =[u]-1

⎥⎥⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎢⎢

⎣

⎡

17.88-

9.79

36.8-

0

6.21

36.80

=

]u[[T]]k [=]u[]k [=]F[ geaeaeaeaI





Are you confident that the bracing system in the

above structure is adequate against instability ?

⎥⎥⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎢

⎣

⎡

4.60

2.25-

35.31-

17.88

2.25

35.31

=

]u[[T]]k [=]u[]k [=]F[ ge be be be bI





Procedure for a nonlinear analysis

1. Read in data such as the material and sectional properties, the coordinate of the

nodes, the connectivity of the elements, the boundary conditions and the

loadings.

2. Form the element stiffness matrix and transform it form the local to the global

axes as,

in which [L] is the local to global transformation matrix given by Reference 1

][L]k [[L]=]k [T

localeglobale

⎥⎥⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢⎢

⎣

⎡

′

Q

Q

ll-l

0Ql

Q

l-

Q

ll-l

=]L[

zyz

y

zyxx

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

′

′

′

′

]L[000

0]L[00

00]L[0

000]L[

=[L]





where lx, ly and lz are respectively the direction cosine of the element with respect to the

x, y and z axes and Q is equal to (lx2 + lz

2)½.

3. Partition it into 4 sub-matrices blocks (only for element with 2 nodes) as,

4. Let the element is connected to nodes i and j, add these sub-matrices to the

block of matrix corresponding to nodes i and j of the whole structure. That is,

5. Repeat the process from (2) and (3) for all elements and the global stiffness

matrix is then assembled.

⎥⎥⎦

⎤

⎢⎢⎣

⎡

k k

k k =k][

2221

1211

e

k +K =K 22e jj jj

k +K =K 11eiiii

k +K =K 12eijij





6. Solve for the displacements of the whole structure as,

7. Select the nodal displacements for each element from its connectivity and then

transformation the displacements at the two nodes of the element from global to

local coordinates as,

8. Calculate the resisting forces for each element as,

9. Transforming the elemental forces from local to global coordinates and

assembling, the resisting forces for the complete structure are obtained as,

10. For linear analysis, the analysis is completed and the total resisting forces for

the whole structure can be used to check the numerical error.

For present bifurcation analysis, the element forces determined in step 8 will be

written on a data file and will be re-used for next process described below.

11. After calculating the internal forces for the elements, the computer program will

be directed to go back to step 2. However, in the second iterative process, the

geometric stiffness matrix with the calculated forces in the previous iteration

[F]][K =][r -1

global

][r ][L=][r global

T

local

][r k][=][Fglobalelocal

]F[[L]=[F] local

elementsof .no=i

1=i

∑





will be added to the linear element stiffness matrix before transformation and

assembling. Thus,

12. The analysis process is continued until step 8 so that the determinant of the

global structural stiffness matrix can be determined. If this determinant is very

small or, in practical computer implementation, the change for two successive

loads is negligibly small, the load assumed in the first iteration is the bifurcation

load. Otherwise, the load factor, λ, and the determinant will be stored in the

program for next bifurcation load prediction.

13. There are a number of methods to predict the next load. We now use an

implicit polynomial iteration process as follows,

14. The load vector is now multiplied by the load factor, λ , and then go to step 2.

15. The iterative process is repeated until condition at step 12 is satisfied.

k +k =k GeLee

DETDET-DET

- += i

1i-i

1i-ii1i+

λ λ λ λ





Procedure for a second order nonlinear analysis

Conventional Newton-Raphson Method

Equilibrium Path

Displacement, u

Modified Newton-Raphson Method

F K = Constant within a load cycle

Divergence Load

Equilibrium Path

Load, F

Displacement, u

K = Variable for every iteration

Divergence Load

T

T

KO T

KO T

0

F0

F1

F1

Load, F

u0

1u

u0

1u





For a second order nonlinear analysis, steps 1 to 6 are the same bit the

displacements must be small or incremental and the stiffness must be formed on the

basis of the updated geometry.Thus,

Note [L] must be formed on the basis of the updated coordinate (and therefore we can

see some books refer this approach as updated Lagrangian formulation). In general,

[F] is not equal to applied force [F]App. The unbalanced force must then be used to

calculate the unbalanced force and thus the strain and stress corresponding to this.

F][]K [=]r [-1

Tglobal ∆∆

]r [+][x=][xglobali1i+

∆

]r [][L=]r [global

T

local ∆∆

]r []k [=]F[localelocal

∆∆

]F[+][F=][Flocallocalilocal1i+ ∆

][F[L] =[F]local

ofElements. No=i

1=i

∑





and [∆F] will then be substituted into equation 28.

The steps from (1) to (15) represent a general procedure. In the following section,

important numerical techniques for these steps are described.

The Global Structure Stiffness

A general structure can normally have hundreds or even thousands degrees of freedom.

consequently the size of the stiffness matrix is very large. However, we can reduce

significantly its size by noting the following properties.

(a) The matrix is symmetric. It can be easily verified by the fact that the coefficient

k ij is equal to k ji and the transformation matrix on the two sides of the matrix is

transpose to each other. This property allows us to store nearly half of the

matrix in the program and thus storage is reduced by half approximately.

(b) The coefficients of the matrix are normally cluster around the diagonal of the

matrix and therefore only terms inside the bandwidth3 or the skyline profile

4

are needed to be saved. Also, the factorisation needs to be done only on the

non-zero terms inside the bandwidth or the skyline profile. This leads to a great

saving in computer storage and computer time. Alternatively, the frontal

solution solver5 can be used to reduce the in-core computer storage by

eliminating the nodes in the stiffness matrix. In the first two methods, the

numbering of nodes will affect the storage size whilst in the third method, the

numbering of elements will affect the memory size. In our computer program,

the skyline profile method is employed to store the stiffness matrix.

][F-[F]=F][App

∆





References

1. Gere, J.M. and Weaver, W.J., "Analysis of Framed Structures", Van Nostrand

Reinhold, N.Y. 1965.

2. Bathe, K.J., "Finite Element Procedures in Engineering Analysis", Prentice Hall,

Englewood Cliffs, New Jersey, U.S.A. 1982.

3. Rockey, K.C., Evans, H.R., Griffiths, D.W. and Nethercot, D.A., "The Finite

Element Method", 2nd edition, Granada, 1983.

4. Felippa, C.A., "Solution of Linear Equations with Skyline-stored Symmetric

Matrix", Computers and Structures, Vol. 5, 1975, pp. 13-29.

5. Irons, B. and Ahmad, S., "Techniques of Finite Elements", Ellis Horwood, 1990.

6. Zienkiewicz, O.C., "The Finite Element Method", 3rd Edition, McGraw-Hill,

1977. A State-of-the-art Good Book on the Finite Element Method.

7 Chan, S.L. and Chui, P.P.T., “Nonlinear static and cyclic analysis of steel

frames with semi-rigid connections”, Elservier, 2000, 336, pp.

chapter 2_finite element procedure

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