A -INTRODUCTION AND OVERVIEW - Universiti …taminmn/MMJ1153 Computational Method in Solid... · A –INTRODUCTION AND OVERVIEW ... MMJ1153 – COMPUTATIONAL METHOD IN SOLID MECHANICS

MMJ1153 – COMPUTATIONAL METHOD IN SOLID MECHANICS

A - INTRODUCTION AND OVERVIEW

INTRODUCTION AND OVERVIEW M.N. Tamin, CSMLab, UTM 1


Course Content:

A – INTRODUCTION AND OVERVIEW

Numerical method and Computer-Aided Engineering; Physical

problems; Mathematical models; Finite element method;.

B – REVIEW OF 1-D FORMULATIONS

Elements and nodes, natural coordinates, interpolation function, bar

elements, constitutive equations, stiffness matrix, boundary


elements, constitutive equations, stiffness matrix, boundary

conditions, applied loads, theory of minimum potential energy; Plane

truss elements; Examples.

C – PLANE ELASTICITY PROBLEM FORMULATIONS

Constant-strain triangular (CST) elements; Plane stress, plane

strain; Axisymmetric elements; Stress calculations; Programming

structure; Numerical examples.


COMPUTER-AIDED ENGINEERING (CAE)

The use of computers to analyze and simulate the

function (structural, motion, etc.) of mechanical,

electronic or electromechanical systems.

• Computer-Aided Design (CAD)

- Drafting


- Drafting

- Solid modeling

- Animation & visualization

- Dimensioning & tolerancing

• Engineering Analyses

- Analytical & numerical methods


The Process of FE analysis

INTRODUCTION AND OVERVIEW M.N. Tamin, CSMLab, UTM

Ref: K.J. Bathe, Finite Element Procedures, Prentice-Hall, 1996

4


02

2

=+ yEI

P

dx

yd

Buckling of Euler Column

PHYSICAL PROBLEM AND MATHEMATICAL MODEL


=

−

−+

2

1

2

12

2

1

22

221

0

0

x

x

m

m

x

x

kk

kkkω

Eigenvalue problem


EXAMPLE OF A MODEL

Steady-state heat conduction through a thick wall


)(:

0

xTSolution

Qdx

dTk

dx

d=+

+

=

+ ∞hTR

R

R

T

T

T

hkkk

kkk

kkk

LLLLLL

L

L

MM

L

M

L

L

2

1

2

1

21

22221

11211


REVIEW OF MATRIX ALGEBRA



Matrix Algebra

In this course, we need to solve system of linear equations in the

form

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

=+++

=+++

=+++

...

............................................

...

...

22222121

11212111

(2-1)


nnnnnn bxaxaxa =+++ ...2211

where x1, x2, …, xn are the unknowns.

Eqn. (2-1) can be written in a matrix form as

[ ]{ } { }bxA = (2-2)

where [A] is a (n x n) square matrix, {x} and {b} are (n x 1) vectors.

8


[ ] { } { }

=

=

= n

n

b

b

b

x

x

x

aaa

aaa

aaa

:b and ,

:x ,

...

::::

...

...

A2

1

2

1

22221

11211

The square matrix [A] and the {x} and {b} vectors are is given by,


… (2-3)

Note:

Element located at ith row and jth column of matrix [A] is denoted by aij. For

example, element at the 2nd row and 2nd column is a22.

nnnnnn bxaaa ...21

9


Matrix Multiplication

The product of matrix [A] of size (m x n) and matrix [B] of size (n x

p) will results in matrix [C], with size (m x p).

Note: The (ij)th component of [C], i.e. cij, is obtained by taking the DOT

product,

(2-4)[ ] [ ] [ ] ) x ( ) x ( ) x (

pmpnnm

CBA =


product,

])[ ofcolumn th ( ])[ of rowth ( BjAicij ⋅= (2-5)

Example:

2) x (2 2) x (3 )3 x 2(

710-

157

30

25

41

120

312

=

−

−

10


Matrix Transposition

If matrix [A] = [aij], then transpose of [A], denoted by [A]T, is given by

[A]T = [aji]. Thus, the rows of [A] becomes the columns of [A]T.

Example:

−

−

=32

60

51

][A


Note: In general, if [A] is of dimension (m x n), then [A]T has the dimension

of (n x m).

−

24

32

Then,

−

−=

2365

4201][ TA

11


Transpose of a Product

The transpose of a product of matrices is given by the product of the

transposes of each matrices, in reverse order, i.e.

TTTT ABCCBA ][][][])][][([ = (2-6)

Determinant of a Matrix


Consider a 2 x 2 square matrix [x],

[ ]

=

2221

1211

xx

xxx

(2-7)

The determinant of this matrix is give by,

[ ] 12212211 xxxxxdet −=

12


EXAMPLE

Given that:

=

=

−=

−=

2

1

}{ 041

213

][

301

013][

41

01][

ED

CA


3302

Find the product for the following cases:

a) [A][C]

b) [D]{E}

c) [C]T[A]

13


Solution of System of Linear Equations

System of linear algebraic equations can be solved for the unknown using

the following methods:

a) Cramer’s Rule

b) Inversion of Coefficient Matrix

c) Gaussian Elimination

d) Gauss-Seidel Iteration


Example: Solve the following SLEs using Gauss elimination method.

(iii) 6 122

(ii) 8324

(i) 11312

321

321

321

−=−+−

=+−

=−+

xxx

xxx

xxx

14


Gauss Elimination Method

Reducing a set of n equations in n unknowns to an equivalent triangular

form (forward elimination). The solution is determined by back substitution

process.

Basic approach

-Any equation can be multiplied (or divided) by a nonzero scalar


-Any equation can be multiplied (or divided) by a nonzero scalar

-Any equation can be added to (or subtracted from) another equation

-The position of any two equations in the set can be interchanged

15


Example: Solve the following SLEs using Gaussian elimination.

(iii) 6 122

(ii) 8324

(i) 11312

321

321

321

−=−+−

=+−

=−+

xxx

xxx

xxx


Eliminate x1 from eq.(ii) and eq.(iii). Multiply eq.(ii) by 0.5 we get,

(iii) 6 122

*(ii) 45.112

(i) 11312

321

321

321

−=−+−

=+−

=−+

xxx

xxx

xxx

16


Subtract eq.(ii)* from eq.(i), we obtain

Add eq.(iii) with eq.(i), yields

(iii) 6 122

**(ii) 75.420

(i) 11312

321

321

321

−=−+−

=−+

=−+

xxx

xxx

xxx


Add eq.(iii) with eq.(i), yields

*(iii) 5 430

**(ii) 75.420

(i) 11312

321

321

321

=−+

=−+

=−+

xxx

xxx

xxx

17


Eliminate x2 from eq.(iii)*. Multiply eq.(ii)** by 3 and eq.(iii)* by 2

we get

Subtract eq.(iii)** from eq.(ii)***, we obtain

(i) 11312 =−+ xxx

**(iii) 10 860

***(ii) 215.1360

(i) 11312

321

321

321

=−+

=−+

=−+

xxx

xxx

xxx


***(iii) 11 5.500

**(ii) 75.420

(i) 11312

321

321

321

=−+

=−+

=−+

xxx

xxx

xxx

From eq.(iii)*** we determine the value of x3, i.e.

25.5

113 −=

−=x

18


Back substitute value of x3 into eq.(ii)** and solve for x2, we get

12

)2(5.472 −=

−+=x

Back substitute value of x2 and x3 into eq.(i) and solve for x1, we

get

6=x


61 =x

19


Example

Solve the following systems of linear equations by using the

Gaussian elimination method.

340

1242

223

321

321

321

=++

=+−

=−+−

xxx

xxx

xxxa)


6222

8324

11312

340

321

321

321

321

−=−+−

=+−

=−+

=++

xxx

xxx

xxx

xxx

b)

20


Steps in solving a continuum problem by FEM

� Identify and understand the problem(This essential step is not FEM)

� Select the solution domainSelect the solution region for analysis.

� Discretize the continuumDivide the solution region into finite number of elements,


Divide the solution region into finite number of elements,

connected to each other at specified points / nodes.

� Select interpolation functionsChoose the type of interpolation function to represent the

variation of the field variables over the element.

� Derive element characteristic matrices and vectorsEmploy direct, variational, weighted residual or energy

balance approach.

[k](e) {φ}(e) = {f}(e)


Steps > (Continued)

� Assemble the element characteristic matrices and vectors

� Combine the element matrix equations and form the matrix

equations expressing the behavior of the entire solution region /

system.

� Modify the system equations to account for the boundary conditions

of the problem.


of the problem.

� Solve the system equations

Solve the set of simultaneous equations to obtain the unknown

nodal values of the field variables.

� Make additional computations, if desired

Use the resulting nodal values to calculate other important

parameters.


What is the problem?

Solution region of

interest

P


Task:

To simulate stress and strain fields in the

vicinity of a sharp notch under tensile load

Other examples:

-Scratches on a tensile surface

-Oil groove on a shaft

-Threaded connections

-Rivet holes under tension

Stress concentration along a shaft


� Select the solution domain

� Discretize the continuum

� Choose inerpolation functions

� Derive element characteristic

matrices and vectors

� Draw the model geometry

� Mesh the model geometry

� Select element type

� (The FEA software was written to do this)

Input material properties

FE procedures: FE software user steps:


matrices and vectors

� Assemble element

characteristic matrices and

vectors

� Solve the system equations

� Make additional

computations, if desired

Input material properties

� (The computer will assemble it)

Input specified load and boundary conditions

� (The compiler will solve it)

Request for output

� Post-process the result file


� Draw the model geometry

� Mesh the model geometry

� Select element type

[20-node Hexahedral elements]

� Input material properties

EXAMPLE: Stresses in a C(T) specimen


� Input material properties

[Elastic modulus, Poisson’s ratio]

� Input specified load and BC

[Pin loading, displacement rate,

zero displacement at lower pin]

� Request for output

[Displacement, strain and stress

components]

� Post-process the result file


EXAMPLE: Stresses in a C(T) specimen


� Check for obvious mistakes/errors

� Extract useful information

� Explain the physics/ mechanics

� Validate the results


MODELING CAPABILITIES – Microelectronic device reliability

- Prediction of spatial distribution of physical parameters

Silicon Die

Heat sink

Solder


MT1

TC1

TD1

0.00

0.02

0.04

0.06

0.08

0.10

0 1 2 3 4 5

N

εin

-40

125

183

25

Re-flow

T

Time

- Prediction of damage evolution characteristics

Substrate

Motherboard PCB

Solder joints

Solder

mask


w

MODELING CAPABILITIES – Deflection of composite laminates beam


• 14 layers [0/45/90/-45/45/-45/02]s

• Layer thickness = 0.3571 mm

Al Comp-0º Comp-90º

E E11 E22

70 GPa 44.74 GPa 12.46 GPa

)L(6

EI3

−= xwx

y

Documents

A -INTRODUCTION AND OVERVIEW - Universiti …taminmn/MMJ1153 Computational Method in Solid... · A –INTRODUCTION AND OVERVIEW ... MMJ1153 – COMPUTATIONAL METHOD IN SOLID MECHANICS