FEM Introduction Mohammed Bakkesin

Introduction to the Finite Element Method Spring 2010

Introduction to the Finite Element Method Dr. Mohammad Mohamm ad Tawfik Tawfik

Course Objectives • The student should be capable of writing simple programs to solve different problems using finite element method.


Assessment • 10%

Assignments (1 per week)

• 20%

Quizzes (best 2 out of 3)

– Week of 12/11/2006 – Week of 20/12/2006 – Week of 17/1/2006

• 20%

Course Project

• 25%

Midterm exam (Week of 2/12/2006)

• 25%

Final exam (starting 3/2/2007) Introduction to the Finite Element Method Dr. Mohammad Mohamm ad Tawfik Tawfik

Fundamental Course Agreement • Homework is sent in electronic format (No hardcopies are accepted) • Computer programs have to written in MATLAB or Mathematica script • No late homework is accepted • No excuses are accepted for missing a quiz • Best two out of three quizzes are counted Introduction to the Finite Element Method Dr. Mohammad Tawfik

References • J.N. Reddy, An Introduction to the Finite Element Method 3rd ed., McGraw Hill, ISBN 007-124473-5 • D.V. Hutton, “Fundamentals of Finite Element Analysis” 1st ed., McGraw Hill, ISBN 007121857-2 • K. Bathe, “Finite Element Procedures,” Prentice Hall, 1996. (in library) • T. Hughes, “The finite Element Method: Linear Static and Dynamic Finite Element analysis, ” Dover Publications, 2000. (in library) “

”

Introduction to the Finite Element Method Dr. Mohammad Tawfik

Numerical Solution of Boundary Value Problems Weighted Residual Methods


Objectives • In this section we will be introduced to the general classification of approximate methods • Special attention will be paid for the weighted residual method • Derivation of a system of linear equations to approximate the solution of an ODE will be presented using different techniques Introduction to the Finite Element Method Dr. Mohammad Tawfik

Why Approximate? • Ignorance • Readily Available Packages • Need to Develop New Techniques • Good use of your computer! • In general, the problem does not have an analytical solution!


Classification of Approximate Solutions of D.E.’s • Discrete Coordinate Method – Finite difference Methods – Stepwise integration methods • Euler method • Runge-Kutta methods • Etc…

• Distributed Coordinate Method


Distributed Coordinate Methods • Weighted Residual Methods – Interior Residual • Collocation • Galrekin • Finite Element

– Boundary Residual • Boundary Element Method

• Stationary Functional Methods – Reyligh-Ritz methods – Finite Element method Introduction to the Finite Element Method Dr. Mohammad Tawfik

Basic Concepts • A linear differential equation may be written in the form:

L f  x  g  x 

• Where L(.) is a linear differential operator. • An approximate solution maybe of the form: n

f  x    ai i x  i 1


Basic Concepts • Applying the differential operator on the approximate solution, you get: n

  L f  x   g  x   L  ai i  x   g  x   i   1

n

  ai L i  x   g  x   0 i 1 n

 ai L i  x   g  x  R x i 1

Residue


Handling the Residue • The weighted residual methods are all based on minimizing the value of the residue. • Since the residue can not be zero over the whole domain, different techniques were introduced.


Collocation Method • The idea behind the collocation method is similar to that behind the buttons of your shirt! • Assume a solution, then force the residue to be zero at the collocation points


Collocation Method R x j



0

R x j  n

 ai L i  x j   F  x j   0 i 1


Example Problem


The bar tensile problem

2

EA

 u  x

2

 F  x   0


Bar application 2

EA

 u  x

 

 F x  0

2

n

    a  x 

u x

i

i

i 1

n

a

EA

i

i 1

2

 

d  i x dx

2

 F  x   R x  Applying the collocation method n

EA ai i 1

d 2 i x j dx

2

 

 F x j  0


In Matrix Form  k k    k n

11

k 21

...

12

k 22

...

1

 k 2 n

 ...

 F  x      F  x  k n a                  k nn  an   F  xn  k n1   a1  2

2

1

2

k ij



EA

d 2 i  x  dx 2

x x j 

Solve the above system for the “generalized coordinates” ai to get the solution for u(x)


Notes on the trial functions • They should be at least twice differentiable! • They should satisfy all boundary conditions! • Those are called the “ Admissibility Conditions”.


Using Admissible Functions • For a constant forcing function, F(x)=f • The strain at the free end of the bar should be zero (slope of displacement is zero). We may use:   x 

   Sin

 x

  2l 


Using the function into the DE: EA

2

 

dx

2

d  x

     EA   2l 

2

  x  Sin   2l 

• Since we only have one term in the series, we will select one collocation point! • The midpoint is a reasonable choice!         EA  Sin a    f   2l   4   2

1


Solving: a1



2

f EA 2l  Sin 2

4





4 2l 

• Then, the approximate solution for this problem is: • Which gives the maximum displacement to be:

2

f EA

2



0.57

l f EA

  x  u x   0.57 Sin  EA  2l  l 2 f

ul 



• And maximum strain to be: u x 0

l 2 f exact 0.57 EA



0.9

lf

EA

exact







0.5

1.0




The Subdomain Method (free reading) • The idea behind the subdomain method is to force the integral of the residue to be equal to zero on an subinterval of the domain


The Subdomain Method x j 1

 R x dx  0

x j x j 1

n

x j 1

 a  L  x dx   g  x dx  0 i

i 1

i

x j

x j


Bar application 2

EA

 u  x

 

 F x  0

2

n

    a  x 

u x

i

i

i 1

n

a

EA

i

i 1

2

 

d  i x dx

2

 F  x   R x  Applying the subdomain method x j1

n

a 

EA

i

i 1

x j

d  i  x  2

dx

2

x j 1



dx   F  x dx x j


In Matrix Form  x  d   x    x   i EA  dxai     F  x dx  x dx   x  j

1

2

j

1

2

j

j



The Galerkin Method • Galerkin suggested that the residue should be multiplied by a weighting function that is a part of the suggested solution then the integration is performed over the whole domain!!! • Actually, it turned out to be a VERY GOOD idea


The Galerkin Method

 R x   x dx  0 j

Domain

n

 a   x L  xdx    x g  xdx  0 i

i 1

j

Domain

i

j

Domain


Bar application 2

EA

 u  x

 

 F x  0

2

n

    a  x 

u x

i

i

i 1

n

a

EA

i

i 1

 

2

d  i x dx

2

 F  x   R x  Applying Galerkin method n

a 

EA

i

i 1

 j  x 

Domain

d 2 i  x 

dx

2

dx  

  x F  x dx j

Domain


In Matrix Form    d  i  x   EA   j  x  dxai     j  x  F  x dx dx  Domain   Domain  2

2



Same conditions on the functions are applied • They should be at least twice differentiable! • They should satisfy all boundary conditions! • Let’s use the same function as in the collocation method:   x 

   Sin

 x

  2l 


Substituting with the approximate solution: n

a 

EA

 j

i

i 1

 x 

Domain

d 2 i  x  dx

2

2

  x F  xdx

dx  

j

Domain

l

     x    x   EA  a  Sin Sin dx  2l   2l   2l  1

0

l

  x    Sin  fdx  2l  0

    EA   2l 

2

a1

l 2



2l



a1



f EA

2

16l 3





0.52

fl 2 EA


Substituting with the approximate solution: (Int. by Parts) n

a 

EA

 j

i

i 1



 x 

Domain

 j  x 

Dom ain

    EA   2l 

d 2 i  x  dx

  j  x 

Zero! 2

a1

l 2

d 2 i  x 



2

dx

2

dx



j

Domain

dx

d  i  x 

2l

  x F  xdx

dx  

l

 0



d  j  x  d  i  x  dx

Dom ain

a1



f EA

dx

dx

2

16l 3





0.52

fl 2 EA


What did we gain? • The functions are required to be less differentiable • Not all boundary conditions need to be satisfied • The matrix became symmetric!


Summary • We may solve differential equations using a series of functions with different weights. • When those functions are used, Residue appears in the differential equation • The weights of the functions may be determined to minimize the residue by different techniques • One very important technique is the Galerkin method. Introduction to the Finite Element Method Dr. Mohammad Tawfik

NOTE • Next Sunday 5/11 (No lecture) • Following week 12/11, Quiz #1 will be held covering all the material up-to this lecture • Homework #1 is due next week (Electronic submission of report and code is mandatory.


Report Should Include … • Cover page • Introduction section indicating the procedure you used with the equations as implemented in your code • Results section • Observations and Conclusions if any according to the output of your program. Introduction to the Finite Element Method Dr. Mohammad Tawfik

Homework #1 •

Solve the beam bending problem, for beam displacement, for a simply supported beam with a load placed at the center of the beam using – Collocation Method – Subdomain Method – Galerkin Method

• •

Use three term Sin series that satisfies all BC’s Write a program that produces the results for n-term solution.

4

d w dx

4

w(0)



F ( x)



w(l )

2

dx

0

2

d w(0) 2



d w(l ) 

dx

2



0


Exact Solution

( )

w x

x



3

12



x



13 x 60

3

12

0  x  1 / 2

x



2

4



7 x 15



3 10

1 / 2  x  1


The Finite Element Method 2nd order DE’s in 1-D


Objectives • Understand the basic steps of the finite element analysis • Apply the finite element method to second order differential equations in 1-D


The Mathematical Model • Solve: d  du    a   cu  f  0 dx  dx  0  x  L

• Subject to:

 du  u0  u0 ,  a   Q0  dx  x L Introduction to the Finite Element Method Dr. Mohammad Tawfik

Step #1: Discretization • At this step, we divide the domain into elements. • The elements are connected at nodes. • All properties of the domain are defined at those nodes.


Step #2: Element Equations • Let’s concentrate our attention to a single element. • The same DE applies on the element level, hence, we may follow the procedure for weighted residual methods on the element level!

d  du    a   cu  f  0 dx  dx  x1  x  x2 u x1   u1 , u x2   u2 ,

 du   du   Q1 ,  a   Q2 a   dx  x x  dx  x x 1

2


Polynomial Approximation • Now, we may propose an approximate solution for the primary variable, u(x), within that element. • The simplest proposition would be a polynomial!


Polynomial Approximation • Interpolating the values of displacement knowing the nodal displacements, we may write:

   b x  b

u x

1

  u u x   u u x1 2

0

1

 b1 x1  b0

2

 b1 x2  b0

 x  x   x  x  u   u u x     x  x   x  x  2

1

1

2

1

2

2

1


Polynomial Approximation  

u x

 x2  x   x  x1  u1   u2    x2  x1   x2  x1  u1    1u1  2u2   1  2      x u e  u2 


Step #2: Element Equations (cont’d) • Assuming constant domain properties:

a

d 2u dx

2

 cu  f  0 x1  x  x2

• Applying the Galerkin method:   d  i  x    a j  x  ui  c j  x  i  x ui  j  x  f dx  0  dx  Domain 2

2


Step #2: Element Equations (cont’d) • Note that:

 d  i  x     a j  x  dx  dx  Dom ain 2

2

 a j  x 

• And:

d  i  x  dx

 

d  1 x dx

x2

a x1

1  

he

d  j  x  d  i  x 



Dom ain

,

dx

 

d  2 x dx

dx

dx

1 

he


Step #2: Element Equations (cont’d) • For i=j=1: (and ignoring boundary terms)  1         x x x x a u  f     c   x  he  he    he dx  0    2

x2

2

2

1

2

1

• Which gives:  a che  fh   u  e  0 2  he 3  1


Step #2: Element Equations (cont’d) • Repeating for all terms:  a  h  e

 1 1 che 2 1 1   6 1   

1 u1 

     u 2    2

fhe 1 2

 1

• The above equation is called the e le me n t . equation


What happens for adjacent elements?


Homework #2 • Derive the element equation without ignoring the boundary terms. • What are differences in the element equation. • The solution should be handed using the same report format (use equation editor to write your report). Introduction to the Finite Element Method Dr. Mohammad Tawfik

Finite Element Procedure 1. Connecting Elements 2. Boundary Conditions 3. Solving Equations Introduction to the Finite Element Method Dr. Mohammad Tawfik

Objectives • Learn how the finite element model for the whole domain is assembled • Learn how to apply boundary conditions • Solving the system of linear equations


Recall • In the previous lecture, we obtained the element equation that relates the element degrees of freedom to the externally applied fields  a  h  e

 1  1 che 2 1 u1  fhe 1  1 1   6 1 2 u   2 1      2    k k  u   f  • Which maybe written:      k k  u   f  1

2

1

1

3

4

2

2


Two –Element example k  k

k 21  u11 

k  k

k 22  u12 

1

1 1

3

2

1 2

3

 f       k  u   f  1

1

4

2

1

1

1

2

 f       k  u   f  2

1

2

2

4

2

2

k  k 0 

k 21

1

1 1

3

2

k 41  k 12 k 32

 u  k  u k   u 0

1

2

2

2

2

4

3

  f  Q        f   Q   f  Q     1

1

2

2

3

3

    


Illustration: Bar application 2

EA

 u  x

2

 

 F x  0

1. Discretization: Divide the bar into N number of elements. The length of each element will be (L/N) 2. Derive the element equation from the differential equation for constant properties an externally applied force: x  EA  d  j d  i   x  he  dx dx u j  f  i dx  0   2

2

1


Performing Integration:  EA  d  j d  i   x  he 2  dx dx u j  f  i dx  0  

x2

1

EA  1



he  1

 1 u e  fhe 1  e    1  u  2 1 1

2

Note that if the integration is evaluated from 0 to h e, where he is the element length, the same results will be obtained.


Two –Element bar example EA 



1

he 1 EA  1

 he 1

1u   f      1  u   f  1

1

1

1

1

1

2

2

1 u   f      1  u   f  2

1

2

2

2

1

2

2

 1  1 0  u1  EA    fhe   1 2  1 u2    2 he    0  1 1  u3 

1 R     2   0  1  0     


Applying Boundary Conditions


Applying BC’s • For the bar with fixed left side and free right side, we may force the value of the left-displacement to be equal to zero:  1 1 0  0  EA    fhe   1 2  1 u2    2 he    0  1 1  u3 

1  R     2   0  1  0     


Solving • Removing the first row and column of the system of equations: EA  2



he  1

• Solving:

 1 u  fhe 2      1  u  2 1  2

3

u  fhe 3     u  2 EA 4 2

2

3


Secondary Variables • Using the values of the displacements obtained, we may get the value of the reaction force:    1 1 0  0   1 2  1  3 fhe   fhe   2  2  0  1 1   4 fhe     2 

1 R     2   0  1  0     


Secondary Variables • Using the first equation, we get: 

3 fhe 2

R



fhe 2



R

2 fhe

 

• Which is the exact value of the reaction force. Introduction to the Finite Element Method Dr. Mohammad Tawfik

Summary • In this lecture, we learned how to a s s e m b l e the g l o b a l matrices of the finite element model; how to apply the boundary conditions, and solve the system of equations obtained. • And finally, how to obtain the secondary variables.


Homework #3 • Problems #3.9 & 3.13 from the text book • Write down a computer code that solves the problem for N elements.


Bars and Trusses


Objectives


Bar Example (Ex. 4.5.2, p. 187)

• Consider the bar shown in the above figure. • It is composed of two different parts. One steel tapered part, and uniform Aluminum part. • Calculate the displacement field using finite element method. Introduction to the Finite Element Method Dr. Mohammad Tawfik

Bar Example • The bar may be represented by two elements. • The stiffness matrices of the two elements may be obtained using the following integration:  d   dx   EA x  d    dx x2

K e

x1

1

2

  d    dx 

1

d  2 dx

 1  1 h  h   dx   EA x   1 1  dx    h h   x2

x1

2

2

e

e

2

2

e

e


Bar Example • For the Aluminum bar: E=10 7 psi, and A=1 in2. we get: K Al



10

7

120

2

 1  1  1 10  1  1 1 dx  120 1 1 

x2

7

x1

• For the Steel bar: E=3810 7 psi, and A=(1.5-0.5x/96) in2. we get: K Fe 

3.10 96

2

7

x2

  x  1.5  1

0.5 x   1 96

  1

 1  1 4.75.10  1 dx      1 96  1 1 7


Bar Example • Assembling the Stiffness matrix and utilizing the external forces, we get: 0  u1   0   R   49.5  49.5       4 5  8.33 u2   2.10    0  10  49.5 57.8   0  8.33 8.33  u3   105   0 

• The boundary conditions may be applied and the system of equations solved. Introduction to the Finite Element Method Dr. Mohammad Tawfik

Bar Example • Solving, we get:

u  0.061   in u  0.181 2

3

• For the secondary variables:

R

30000 lb

 


Reading Task • Please read and understand examples, 4.5.1 & 4.5.3.


Trusses • A truss is a set of bars that are connected at frictionless joints. • The Truss bars are generally oriented in the plain.


Trusses • Now, the problem lies in the transformation of the local displacements of the bar, which are always in the direction of the bar, to the global degrees of freedom that are generally oriented in the plain.


Equation of Motion 1 0 EA  h  1  0

0  1 0 u1  0

0

0

1

0

0

 F 1       0  v1   0       0 u 2   F 2      0 v 2   0  Introduction to the Finite Element Method Dr. Mohammad Tawfik

Transformation Matrix u  v  u v 

1

1

2

2

 0 0  Cos  Sin   u    Sin  Cos   0 0  v       0 Cos  Sin   u 0     0  Sin  Cos  v  Local DOF  0

1

1

2

2

 Local DOF 

     Transformed DOF

d T Transforme DOF


The Equation of Motion Becomes • Substituting into the FEM: • Transforming the forces: • Finally:

  

K T   F

T 

T

    

K T   T

T

F

 K    F  Introduction to the Finite Element Method Dr. Mohammad Tawfik

Recall  K  T  

T

Where:  Cos   Sin  

T  

  

 

K T

  Cos  Sin 

0

0

0

0

  0 0  Cos  Sin     Sin  Cos  0

0

1 0 EA  K   h 1  0

0

1

0

0

0

0

0

1

0

0

  0  0


Element Stiffness Matrix in Global Coordinates 0 0 Cos   Sin   1  Sin  Cos   0 0 0 EA   K    Cos   Sin   1 0 0 h    Sin  Cos    0 0  0 0 0  Cos  Sin    Sin  Cos   0 0    0 Cos  Sin   0   0  Sin  Cos   0

0

1

0

0

0

0

0

1

0

0

  0  0


Element Stiffness Matrix in Global Coordinates 1 1    Cos    Sin2  Sin2   Cos   2 2  1  1  Sin2  Sin   Sin2  Sin      EA 2 2   K    1 1 h   Sin2   Cos    Sin2  Cos     2 2  1  1 Sin2  Sin     Sin2   Sin   2  2  2

2

2

2

2

2

2

2


Example: 4.6.1 pp. 196-201 • Use the finite element analysis to find the displacements of node C.


Element Equations 1  EA 0  K    L  1  0 1

0

1

0

0

0

0

0

1

0

0

  0  0

0 0 0 1 EA  K 2    L 0 0  0  1

0

0

 1 0 0  0 1 0

0.3536  0.3536  0.3536   0.3536  0.3536  0 . 3536  0 . 3536  0 . 3536 EA 3   K    0.3536 0.3536  L  0.3536  0.3536   0.3536   0.3536  0.3536 0.3536


Assembly Procedure Procedure 0.3536  1 0  0.3536  0.3536   1.3536  0.3536  0.3536 0 0 0.3536  0.3536     0 1 0 0 0 1 EA   K     0 0 0 1 0 1  L    0.3536  0.3536 0 0 0.3536 0.3536    1.3536   0.3536  0.3536 0  1 0.3536 


Global Force Vector  F x   F x   F    F y    y  Remember!  F x   F x   F        NO distributed load F F y    y  is applied to a truss  F x   P       F y   2 P  1

1

1

1

2

2

2

2

3

3


Boundary Conditions U 1



V 1



U 2



V 2



0

Remove the corresponding rows and columns

EA 0.3536 0.3536  U 3   P       L 0.3536 1.3536  V 3   2 P  Continue! (as before) Introduction to the Finite Element Method Dr. Mohammad Mohamm ad Tawfik Tawfik

Results U 3  5.828

F 1 x F 2 x

PL EA

, V 3  

P , F 1 y

 



0, F 2 y

3 PL

EA

P ,

 



3 P


Postcomputation e

e





P 1 Ae

u  v  u v 

1

1

2

2

e



P 2 Ae

 P  A E   L  P  e

1 e

2

e

e

1  1 

 1 u   1  u 

  Cos  Sin  0 0  u     0 0   Sin  Cos  v    0 Cos  Sin   u   0    0  Sin  Cos  v 0 

1

1

2

2

e 1 e 2

  

     


Postcomputation



(1)



0,



( 2)

 

3 P A

,



( 3)



2

P A


Summary • In this lecture we learned how to apply the finite element modeling technique to bar problems with general orientation in a plain.


Homework #5 • Problem 4.27, – Due 13/12/2006 before 9:00am

• Problem 4.44, – Due 20/12/2006 before 9:00am


Announcements • Compensation Tutorial for E15: – Next Sunday 17/12/2006 3 rd Period in H6

• Next Lecture: – Wednesday 20/12/2006 3 rd Period in H6

• Next Quiz: – Wednesday 20/12/2006 3 rd Period in H6 – (This Lecture is included) Introduction to the Finite Element Method Dr. Mohammad Tawfik

Term Projects • A problem has got to be solved using the finite element method • A report is going to be presented by each group presenting the problem and its solution


The Report should contain: • Cover page – Project Title – Names of team members

• Table of contents • Introduction and literature survey – – – –

Introduction to the problem Historical background and relevance of the problem Papers and books that presented the problem Latest achievements in the problem


The Report should contain: • The finite element derivation – Governing equation – Derivation of the element matrices • Using Glerkin method • Application of Symbolic manipulator to derive the matrix equations will be appreciated

– Solution procedure


The Report should contain: • The numerical results and verification – Program results – Verification of results compared to published results – Parametric study

• Discussion – Observations of the results – Further work that may be performed with the problem – Future developments of the model

• References Introduction to the Finite Element Method Dr. Mohammad Tawfik

Evaluation • Report (50%) • Code (30%) – Structured: Functions built, easily modified – Readability: Organization, remarks – Length: The shorter the better

• Results (20%)


Projects • Heat transfer in a 2-D heat sink • 2-D flow around a blunt body in a wind tunnel • Vibration characteristics of a pipe with internal fluid flow • Panel flutter of a beam • Rotating Timoshenko beam/blade Introduction to the Finite Element Method Dr. Mohammad Tawfik

Heat transfer in a 2-D heat sink • The heat sink will have heat flowing from one side • Convection transfer on the surfaces • Different boundary conditions on the other three sides • Plot contours of temperature distribution with different boundary conditions Introduction to the Finite Element Method Dr. Mohammad Tawfik

2-D flow around a blunt body in a wind tunnel • Potential flow in a duct • Rectangular body with different Dimensions • Study the effect of the body size on the flow speed on both sides • Plot contours of potential function, pressure, and velocity potential Introduction to the Finite Element Method Dr. Mohammad Tawfik

Vibration characteristics of a pipe with internal fluid flow • Study the change of the natural frequencies with the flow speed under different boundary conditions and fluid density • Indicate the flow speeds at which instabilities occur


Panel flutter of a beam • A fixed-fixed beam is subjected to flow over its surface • Plot the effect of the flow speed on the natural frequencies of the beam • Indicate the speed at which instability occurs


Rotating Timoshenko beam/blade • Rotating beams undergo centrifugal tension that results in the change of its natural frequencies • Study the effect of rotation speed on the beam natural frequencies and frequency response to excitations at the root


Teams • 2-3 Students teams • Names and selected projects should be submitted before 4PM on Thursday 21/12/2006


Work Progress • A report should be submitted By 4PM every Wednesday • 27/12/2006 – The report should contain a preliminary literature survey – Problem statement – Governing equations

•

10/1/2007 – The report should contain a deeper literature survey – The preliminary derivations of the finite element model

•

17/1/2007 – A more mature version of the report should be presented – Preliminary results of the code – List of the program script should be included

•

24/1/2007 – Final version of the report should be presented together with the code


Beams and Frames


Beams and Frames • Beams are the most-used structural elements. • Many real structures may be approximated as beam elements • Two main beam theories: – Euler-Bernoulli beam theory – Timoshenko beam theory Introduction to the Finite Element Method Dr. Mohammad Tawfik

Euler-Bernoulli Beam Theory • The main assumption in the EulerBernoulli beam theory is that the beam’s thickness is too small compared to the beam length • That assumption resulted in that the sheer deformation of the beam may be neglected without much error in the analysis Introduction to the Finite Element Method Dr. Mohammad Tawfik

Governing Equation • The equation governing the deformation of and E-B beam under transverse loading may be written in the form: 2  d d w   EI  x  2   F ( x) 2  dx  dx  2


The Thin-Beam Elements • The thin beam element has a special feature, namely, the two degrees of freedom at each node are related.


Beam Interpolation Function ( )  a1  a2 x  a3 x

w x

 

w x

1

x

x

2

2

x

 a4 x 3

3

a

   H  xa

w x

   dH  x  a  

dw x dx



dx





  a  0 H x x

1

2 x

3 x

2

a


Beam Interpolation Function 

w0



w' 0





w1



w'1



 H 0a  H 0a x



w l



w2

w' l  w'2 





 H l a  H x l a

 w1   H 0   a1   w'   H 0 a   1   x   2       T a  w2   H l   a3    w'2   H x l  a4  Introduction to the Finite Element Method Dr. Mohammad Tawfik

Beam Interpolation Function  w1  1  w'  0  1     w2  1 w'2  0

0

0

1

0

l

l

1

2l

2

  a1    0 a2    3 l  a3    2 3l   a4  0

a a   a a

1

2

3

4

1     0  3    l   2  l 2

3

0

0

1 2

0 3

l 1

2

2

l

l

2 3

l

 w  0    1  w'  l   w 1  w'   l  0

1

1

2

2

2

     


Beam Interpolation Function a T 

1





w  e

w x   H  xa   H  x T  w   N  x w 1

e

e



4

    N  x w

w x

i

i

i 1


Beam Interpolation Function  3 x 2 x 1  l  l   x  2 x  x  T l l       N x N x       3 x  2 x  l l  x x    l l  2

2

3

3

2

3

3

2

3

2

3

2

3

2

          


Interpolation Functions 1 N1

N3

0.8

0.6

) x ( N

0.4

0.2 N2

0 0

0.2

0.4

0.6

0.8

N4

1

-0.2

-0.4 X


Beam Stiffness Matrix • The governing equation is: 2  d d w   EI  x  2   F ( x) 2  dx  dx  2

• Using the series solution 4

    N  xw

w x

i

i

i 1


Beam Stiffness Matrix • The governing equation becomes 2  d d N i  EI  x  2  2  dx i 1 dx  4

2

 wi   F ( x)  R( x)  

• Applying Galerkin Galerkin method: method: 2  4 d 2   d N i        R x N dx EI x w F x N dx ( ) ( )    j i j 2 0 0   i1 dx2     dx  

l e

l e


Beam Stiffness Matrix • Using integration by parts, twice, and ignoring the boundary terms, we get: l e 2 2  4    d N d N j i   EI  x    w F x N dx  0  ( ) i j 2 2     dx dx i 1  0     • In matrix form:

 l e

 EI  x N

xx

0

l e

 N dxw   F ( x) N dx e

xx

xx

0 Introduction to the Finite Element Method Dr. Mohammad Tawfik

Use of Symbolic Manipulator Beam Example


O p t i o n a l Homework #6 • Derive the expression for the interpolation function for a beam in terms of nodal displacements and slopes. • Try to use a symbolic manipulator to generate the expressions. 2

 A

d w 2

dt

4

 EI

d w dx

4

 F ( x)


Two Dimensional Elements


2-D Elements • In this section, we will be introduced to two dimensional elements with single degree of freedom per node. • Detailed attention will be paid to rectangular elements.


For the 2-D BV Problem • Let’s consider a problem with a single dependent variable • We may set one degree of freedom to each node; say f i. • Further, let’s only consider a rectangular element that is aligned with the physical coordinates Introduction to the Finite Element Method Dr. Mohammad Tawfik

A Rectangular Element • For the approximation of a general function f(x,y) over the element you need a 2-D interpolation function

f  x, y   a1  a2 x  a3 y  a4 xy


Let’s follow the same procedure!


2-D Interpolation Function f ( x, y)  a1  a2 x  a3 y  a4 xy

  f  H 0,0a f a,0 f a, b  f   H a, ba f 0, b f 0,0



1

3



f  x y  ,



f 2





f 4





 H  x y a ,

 H 0, aa  H 0, ba

 f 1   H 0,0  a1   f   H a,0 a   2   2       T a  f 3   H a, b  a3     f 4   H 0b  a4  Introduction to the Finite Element Method Dr. Mohammad Tawfik

2-D Interpolation Function  f  1 0 0 0   a   f  1 a 0 0  a               f  1 a b ab a   f  1 0 b 0  a  1

1

2

2

3

3

4

4

a a   a a

1

2

3

4

0 0 1   1 1 0   a a   1  0 0  b   1  1 1  ab ab ab

   f  0   f    1    f b    1  f  ab  0

1

2

3

4


2-D Interpolation Function f  x, y    H  x, y a   N  x, y  f e

 x y xy  1  a  b  ab   x xy      a ab  T  N  x, y    N  x, y     xy   ab    y xy   b  ab  Introduction to the Finite Element Method Dr. Mohammad Tawfik

How does this look like?


2-D Interpolation Functions

N1

1

1

0.9

0.9

0.8

0.8

0.7

0.7

0.6

0.6

N2

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2 0.9

0.1

0.6

0

0.3 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x

0.9 0.1

y

0.6

0

0.3 0

0 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x

y

0 1


2-D Interpolation Functions

N3

1

1

0.9

0.9

0.8

0.8

0.7

0.7

0.6

0.6

N4

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2 0.9

0.1

0.6

0

0.3 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x

0.9 0.1

y

0.6

0

0.3 0

0 1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x

y

0 1


Example: Laplace Equation 

 4

2

  0

2



 x

2





2



 y

2

0

 

   i N i  x, y    N  x, y  

e

i 1


Example: Laplace Equation 4

   i N i  x, y    N  x, y   e

i 1

Applying the Galerkin method and integrating by parts, the element equation becomes

  N  N    N  N dA  0 e

x

x

y

y

Area


The Element Equaiton  2a  b  a  2b a b  2a  b    2a  b   2a  b a b  e 1  a  2b    0 6ab   a  b  2a  b 2a  b  a  2b    a b a  2b 2a  b   2a  b  2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2


The Logistic Problem!


The Logistic Problem • In the 2-D problems, the numbering scheme, usually, is not as straight forward as the 1-D problem


1-D Example • Element #1 is associated with nodes 1&2 • Element #2 is associated with nodes 2&3, etc…


2-D Example


2-D Example


For Element #5 Local Node Number

Global Node Number

1

5

2

6

3

9

4

8 Introduction to the Finite Element Method Dr. Mohammad Tawfik

Contribution of element #5 to global matrix 1

2

3

4

5

6

5

1,1

6

7

8

9

1,2

1,4

1,3

2,1

2,2

2,4

2,3

8

4,1

4,2

4,4

4,3

9

3,1

3,2

3,4

3,3

10

11

12

1 2 3 4

7

10 11 12


A Solution for the Logistics ’ Problem • One solution of the logistic problem is to keep a record of elements and the mapping of the local numbering scheme to the global numbering scheme in a table!


Elements Register: Global Numbering Element Number

Node Number 1

2

3

4

1

1

2

5

4

2

4

5

8

7

3

7

8

11

10

4

2

3

6

5

5

5

6

9

8

6

8

9

12

11


FEM Introduction Mohammed Bakkesin

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