basu-fem notes me623

8/11/2019 Basu-fem Notes Me623

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ME623: Finite Element Methods in Engineering Mechanics

Instructor: Sumit Basu

Email: [email protected]

Phone office: (0512 259) 7506

Office: NL211A

Phone residence: (0512 259) 8511


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Send assignments to

Ramkumar S [email protected] (odd numbered assignments)and

Deependra Singh Sekhawat [email protected] (even numberedassignments)

Evaluation

1. Assignments: 20%

2. 2 nd mid-semester exam: 30%

3. End-semester exam: 30%

4. ANSYS project: 20%


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Experiments versus simulations?

Qs. Sophisticated experiments can tell everything. Why do we need the FE method?

1: Experimental results are subject to interpretation. Interpretations are as good as thecompetence of the experimenter.

2. Experiments, especially sophisticated ones, can be expensive3. There are regimes of mechanical material behaviour that experiments cannot probe.

4. Generality of behaviour is often not apparent from experiments.

Experiments and simulations are like two legs of a human being. You need both to walk and it does notmatter which you use first!


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1943 : Richard Courant, a mathematician described a piecewise polynomialsolution for the torsion problem of a shaft of arbitrary cross section. Even holes.The early ideas of FEA date back to a 1922 bookby Hurwitz and Courant.

His work was not noticed by engineersand the procedure was impractical at the time due to the lack of digital computers.

A short history of FEA

1888-1972: b in Lublintz Germany

Student of Hilbert and Minkowski in Gottingen Germany

Ph.D in 1910 under Hilberts supervision.

1934: moved to New York University, founded the Courant Institute


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In the 1950 s: work in the aircraft industry introduced FE to practicing engineers.

A classic paper described FE work that was prompted by a need to analyzedelta wings, which are too short for beam theory to be reliable.

1960 : The name "finite element" was coined by structural engineer Ray Cloughof the University of California

By 1963 the mathematical validity of FE was recognized and themethod was expanded from its structural beginnings to include heat transfer,groundwater flow, magnetic fields, and other areas.

Large general-purpose FE software began to appear in the 1970s.By the late 1980 s the software was available on microcomputers,complete with color graphics and pre- and post-processors.By the mid 1990 s roughly 40,000 papers and books aboutFE and its applications had been published.

Professor emeritus of Structural Engineering at UC Berkley

Ph.D from MIT

Well known earthquake engineer


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Books

Concepts and applications of Finite element analysis: Cook, Malkus andPlesha, John Wiley and Sons, 2003.

T.R. Chandrupatla and A.D. Belegundu, Introduction to Finite Elements inEngineering, Second Edition, Prentice-Hall, 1997.

O. C. Zienkiewicz and R. L. Taylor, The Finite element method, vols 1 and2, Butterworth Heinemann, 2000

Klaus-Jurgen Bathe, Finite Element Procedures (Part 1-2), Prentice Hall,1995.


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Solving an engineering problem

Mathematical model: an

equation of motion

Eulers explicit scheme or first order Runge Kuttascheme


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Write a MATLAB code tointegrate the discretisedequations of motion with different

timesteps. Use l=1, g=10, initialvelocity=0, position=45 o.

Compare with the exact solution.


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x


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x

Alternately, the FEM idea.


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A simple example: another look at FEM

A 2-d truss with elements that canonly withstand tension.


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x

y

1 2 3

45

2

4

Direction of stretch

2 degrees of freedom (dof) per node


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2

4

X

Y

2

2/14/2

x

y


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Local stiffness matrix in local coordinatesystem (ndof X nnod, ndof X nnod)

Local force vector (ndof X nnod,1) Displacement vector (ndof X nnod, 1)

ndof : no. of dofs/node

nnod: no. ofnodes/element


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2

4

x

y

Transformation matrix


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3/1 4/2 7/3 8/4

3/1

4/2

7/3

8/4

Eg: 2,2 4,4

3,1 7,3

2

4

x

y

1 2 3

45


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104

93

42

312

104

93

22

111

DestinationLocal dof Elem no

1 2 3

45

1 2

3

4

5

6

7


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After assembling elements 1 and 2


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X+X+XX+X+XXXXXXX

X+X+XX+X+XXXXXXX

XXX+X+XX+X+XX+XX+XX+XX+X

XXX+X+XX+X+XX+XX+XX+XX+X

XXX+XX+XXX

XXX+XX+XXX

XXXXX+X+XX+X+XXX

XXXXX+X+XX+X+XXX

XXXXX+XX+XXXXXX+XX+X

Stiffness matrix is symmetric, diagonally dominant, sparse and banded.


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1 2 3

45

1 2

3

4

5

6

7

Sparsity pattern


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1 3 5

42

1 2

3

4

5

6

7

Half bandwidth


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Global force vector Global stiffness matrix

Global displacement vector

Notes:

Global stiffness matrix is singular i.e. it has zero eigenvalues

Hence it cannot be inverted!


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Nave approach for imposing displacement boundary conditions

L=a very large number. Also replace the corresponding dofs in the rhs vector byLXspecified displacement value


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The proper way of imposing displacement constraints


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Suppose dof k is specified

Traspose negative of the specified value X kth column to the right

Replace k th row and columns in the stiffness matrix to zero

Replace K(k,k) by 1

Set F(k)=specified value

Repeat above steps for all specified dofs.


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Solving the equations

K: requires huge storage, largest component of a FE code

Strategies:

Use sparsity: K_sparse=sparse(K);

Matlab command U=inv(K_sparse)*F;

Karl Friedrich Gauss(1777-1855)

Wilhelm Jordan (1842-1899)


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1 st elimination:

Row2- Row 1*(-6/18)

Row3-Row1*(-6/18)

2nd eliminationRow3-Row2*(-2/10)

Row4-Row2*(-6/10)

3 rd elimination:

Row4-Row3*(-7.2/9.6)

Original problem

Direct methods: Gauss Jordan elimination


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Write a code in Matlab to solve a nXn system of linear equations by Gaussianelimination with pivoting. Check the code on a small system. Take bigger and biggersystems and plot the time taken by your computer (use the Matlab tic function) againstthe system size n. Fit a curve to this plot.


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How does the solution procedure work?


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A digression: a little bit of linear algebra

Linear independence


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Boundary conditions

Force specified: eg. dof 9 and 10 in our example

Displacement specified: eg. dof 1,2,and 6 in our example

Both forces and displacements cannot be specified at the same dof.


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Eigenvalues and eigenvectors


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Change of basis


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Eigenvalues are real if A is real


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Eigenvalues are orthogonal


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Spectral Decomposition


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Norms


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A little more about solving systems of linear equations


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Iterative method of solving a system of linear equations:


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Iterative method of solving a system of linear equations:Jacobi method


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An example


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An example

Method of conjugate gradients for solving a system


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Method of conjugate gradients for solving a systemof linear simultaneous equations

All examples are sourced from An introduction to conjugate gradient method without theagonizing pain by J R Shewchuk of Carnegie Mellon University. Available on the web.


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Convergence of the example problem starting from (2,-2)


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A brief description of the conjugate gradient method


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Using coordinate axes assearch directions..


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\newcommand{\bm}[1]{\mbox{\boldmath $#1$}}


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Note that all previous searchvectors need to be stored. O(N 3)

operations are needed for theconjugation


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Thus conjugate gradient method works in the following way:


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Thus conjugate gradient method works in the following way:

Write a Matlab code to perform the conjugate gradient method on a 2X2 square,positive definite system of linear simultaneous equations.

The FEM scheme of things


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Read in data mainly nodalcoordinates, element connectivity,force and displacement boundaryconditions and material properties

For each element

Form local stiffness matrix

Assemble into global stiffness matrix & rhs vector

End

Form local rhs vector

Form destination array

Incorporate boundary conditions into stiffness and rhs

Solve

Stop


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Write a matlab code that

I.reads in the nodal coordinates and element connectivities of any 2-d trussstructure and displays the structure

2. Reads in the applied forces3. Reads in the displacement boundary condition information

4. Prepares the destination array

5. Computes the local stiffness in the global coordinate system for each elementand using the destination array puts them into appropriate locations in the global

stiffness

6. Applies boundary conditions

7. Solves the system and displays the deformed truss

Method of steepest gradient in action.


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Stiffness matrix from basics!


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1 2

L

x


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Intuitive but not easy


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2212

13

How do we solve a beam problem withdistributed loads?


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wL/4 wL/4wL/2

wL/8wL/4 wL/4 wL/4

Introduction to the theory of elasticity


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Number of independent elastic constants = 21

Number of elastic constants when there exists one plane of symmetry = 13

Number of elastic constants when there exists three mutually perpendicular planes ofsymmetry = 9

Number of independent elastic constants for an Isotropic material = 2


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X1,X1

X2,X2

X3

X3


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Strain displacement relationships


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x


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L/2

w(x)

2h y

0


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x

y

0

w

H

a

0


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x

y

0

0

w

H

a


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L

a

2h

P

x

x

Plane strain


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Plane stress


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Axisymmetry


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Axisymmetry

Geometry as well as loading should be axisymmetric!


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CL


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Plane strain

Axisymmetric


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Axisymmetric geometry, Non-axisymmetric loading

Digression: A little bit of variational calculusu(x)


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ab

u(x)

u(x)


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ab

y(x)

(x)

E l i


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Euler equation

Some more interesting problems


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Euler equation for several independent variables


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This is zero on the boundary


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Necessary condition for a function v(x,y) minimising or maximising thefunctional J

v(x,y)


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u(x,y)

v(x,y)

(x,y)

Weak anchoring


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What about the other way round?


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Linear, differential operator

Example :


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Laplace equation

Solving Laplace equation with given boundary conditions is equivalent to minimisingthe variational statement

The other way round: strong form to weak form


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Example


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The route to the variational principle starting from an self-adjoint pde is outlined. The routefor non-self-adjoint cases is not known.

Principle of minimum potential energy

We assume the existence of W


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Potential of the loads

Internal energy

Finding a displacement field satisfying the boundary conditions is equivalent tominimising the above variational statement.


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Bar of c/s A, modulus E loadedby a body force cx where c hasunits of force/area


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Assignment: This is the Rayleigh-Ritz technique where the problem is reduced tothat of determining a few constant coefficients. Carry out the solution with threeundetermined constants.

Principle of virtual work: weak form of the eqns of equilibrium


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Internal virtual workExternal virtual work


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Virtual work equation

Strong form to weak formanother example (from Reddy,pp69)


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Shear force

Bending moment

A 1-d problem with exact solution

x


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L

Piecewise linear approximation: the assumeddisplacement method

x


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L

Strain displacement matrix

Shape function matrix

VS

2-d domain, triangular elements

Linear triangle, 3 nodes/element, 2dofs/node


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Shape functions


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Strain-displacement matrix


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Strains are constant within an element: Constant strain triangle (CST)


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f

Calculation of nodal forces due to surface tractions:example


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x

y

f S3x

2

31

f S3y

f S2x

f S2y

(2,0)

(2,2)


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t

A


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t=thickness of the element

A=surface area of the element

A

Assignment 7: Write a Matlab function to generate the plane strain CSTstiffness matrix given the elastic constants and the coordinates of the three

nodes


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Example

y

El 2

23300 psi

1000 lb

2 iThickness (t) = 0.5 in


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x

El 11

4 3 in

2 in E= 3010 6 psi

=0.25

(a) Compute the unknown nodal displacements.(b) Compute the stresses in the two elements.

Realize that this is a plane stress problem and therefore we need to use

psi E

D 72 10

2100

02.38.0

08.02.3

100

01

01

1=

=


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2.1002

00

Step 1: Node-element connectivity chart

24

Node 3

33

Area(sqin)

42

Node 2

3211

Node 1ELEMENT

004

203

232031

yx Node

Nodal coordinates

Step 3: Compute element stiffness matrices

)1(T)1()1(T)1()1(

3.05333.02.045.05.09833.0

BDB)5.0)(3(BDB

== At k


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710

2.0

05333.0

02.02.1

3.00045.0

2.02.02.13.04.1

=

u1 u2 u4 v4v2v1

Step 2: Compute strain-displacement matrices for the elements

=332211

321

321

000

000

21

bcbcbc

ccc

bbb

A B

Recall

123312231

213132321

x xc x xc x xc y yb y yb y yb

=== ===

with


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For Element #1:

1(1)

2(2)

4(3)(local numbers within brackets)

0;3;3

0;2;0

321

321

======

x x x

y y y

Hence

033

202

321

321

======

ccc

bbb

=

200323

003030020002

61)1( B

Therefore

For Element #2:

=

200323

003030

020002

61)2( B

)2(T)2()2(T)2()2(

3.05333.02.045.05.09833.0

BDB)5.0)(3(BDB

== At k


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710

2.0

05333.0

02.02.1

3.00045.0

2.02.02.13.04.1

=

u3 u4 u2 v2v4v3

Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees offreedom

014433 ===== vvuvu Notice that

Hence we need to calculate only a small (3x3) stiffness matrix


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710

4.102.0

0983.045.0

2.045.0983.0

=K

u1 u2 v2

u1u2

v2

Step 5: Compute consistent nodal loads

= y

x

x

f

f

f

f

2

2

1

= y f 2

0

0


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yS y f f 210002 +=

The consistent nodal load due to traction on the edge 3-2

lb x

dx x

dx N

tdx N f

x

x

xS y

22529

502

50

3150

)5.0)(300(

)300(

3

0

2

3

0

3

0 233

3

0 2332

=

==

=

==

=

=

=

3 2

3232 x

N =

lb

f f yS y

1225

100022

=

+=Hence

Step 6: Solve the system equations to obtain the unknown nodal loads


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f d K =

=

1225

0

0

4.102.0

0983.045.0

2.045.0983.0

10

2

2

17

v

u

u

Solve to get

=

in

inin

v

uu

4

4

4

2

2

1

109084.0

101069.0102337.0

Step 7: Compute the stresses in the elements

)1()1()1( d BD=

With

In Element #1


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With

[ ][ ]00109084.0101069.00102337.0

d 444

442211)1(

== vuvuvuT

Calculate

psi

=

1.76

1.1391

1.114)1(

)2()2()2( d BD=

With

[ ])2(d =T

In Element #2


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[ ]

[ ]44224433

)(

109084.0101069.00000

d

=

= vuvuvu

Calculate

psi

=

35.363

52.281.114

)2(

Notice that the stresses are constant in each element

Quadratic traingle (Linear strain triangle)


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Can geometry be simplified (plane stress,strain, axisymm, 3d, ymmetries?)

Material properties? Material? Elastic orelasto plastic. Model for constitutive

behaviour?


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Simplified geometry.

What elements to use?

Mesh. Insert boundaryconditions appropriately.

Solve


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Solve.

Post process, check theresults.

Solution involves formingelement stiffnesses and loads,assembling, applying bcs and

solving.

Interpolation and shape functions


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The intepolation functions are same as the shape functions we got for the 1-d example.

12

3

Quadratic interpolation in 1-d


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Bilinear rectangle (through interpolation formulas)

x

y


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2a2b

This is equivalent to assuming a displacement variation with terms

An example: Classical Kirchoff Plate


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Completeness, compatibility etc.

Requirements of a FE modelMonotonic convergence

Completeness Element must represent rigid body modes + constant strain states


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Element must rotate and translateas a rigid body.


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orthonormality

Generalised eigenproblem


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Parasitic shear gets worse with (a/b) ratio. This condition is known as shear locking.

For completeness we require:

Rigid body modes to be represented (in 2-d two translations and one rotation)

Constant strain states must be represented (in 2-d two normal and one shear strain)

Interelement continuity of displacements must be satisfied.

Rigid body modes for a plane stress quad

Flexural modes, 0.495, 0.495 Shear mode, 0.769


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Stretch mode,0.769

Uniform

extension mode1.43


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Parasitic shear makes the Q4 element unusable in bending


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More on convergence


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Starting from a trial function and always adding more terms to it (equivalent to h 0 inFEM) will always make the potential go towards a lower value.

Decrease in the total potential implies increase in the strain energy. Thus predictedstiffness is always higher than actual in FE analysis.

Spatially anisotropic elements depend on coordinate

orientations.

Rate of convergence depends on the completenessof the polynomial used.

Pascals triangle


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Isoparametric element formulation


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First proposed by: B.M. Irons, Engineering applications of numerical integration instiffness methods, AIAA Journal, v4, n11, 1966, 2035-2037.

The Q4 element revisited


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34

r=-1

r=1

s=1/2

s=-1/2

3

4

s=1

r

s

y

1

1

11


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1 2r=1/2

s=-12r=-1/2

x

Parent element Physical element

Calculation of gradients in 2-d


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J Jacobian matrix


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Follows from the Q4 element doneearlier. Put a=b=1.


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Stiffness matrix needs to be integrated numerically.


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Numerical integration: Gauss quadrature-1d

r=1r=-1


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Sampling points between -1 and 1Weights

r=-1 r=1

The idea behind a quadrature scheme.


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10 577352

201

weightsSampling pointsn

Gauss quadrature: sample points and weights


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0.5555

0.55550.8888

0.77459,

-0.774590

3

0.347850.34785

0.652140.65214

0.86114-0.86114

0.33998-0.33998

4

11

0.57735,-0.57735

2


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Important rule: n point Gauss quadrature integrates a polynomial of order 2n-1 exactly

GI in 2-d


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For the Q4 element, 2X2 quadrature is required

1

r s

r 2 rs s 2

r 3 r 2s rs 2 s 3

2X2


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Write a MATLAB code to formulate the stiffness matrix of a 4 noded quadrilateralisoparametric element given the nodasl coordinates. Use Gauss quadrature.

Assume material properties as E = 200 Gpa and Poissons ratio=0.3.

Term paper topics

Group 1 : Model the problem of a line load on a half space. Compare with the theoreticalsolution of stresses near the load point. Refine the mesh to see if you converge to the exactsolution.

Group 2 : Model a hard elliptical inclusion in a softer matrix. The matrix is infinitely largecompared to the inclusion and is subjected to uniaxial stresses applied at infinity. Find outthe theoretical solution to this problem and check how the strains inside the inclusion varywith its ellipticity. Change the Poissons ratio of the matrix and the inclusion and see how thestress fields in the matrix and inclusion change.


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Group 3 : Model a sharp crack in a infinite body. Find out the theoretical solution to thisproblem. The stresses at the crack tip should be infinite. Compare with the theoreticalsolution and see whether with mesh refinement you can get close to the theoretical solution.

Group 4 : Model the problem of a series of periodically placed holes in a thin film. Thisproblem has already been discussed. Discuss further with Dr Ghatak.

Group 5 : Model the bending of a functionally graded beam where the gradation isexponential in the depth direction. Compare with the solution for a homogeneous beam tosee the differences due to the property variation.

Group 6 : Model an internally pressurised hollow cylinder in plane strain. Start with a thickcylinder and using theoretical results show how, as you move towards a thin cylinder thesolution changes.

Group 7: Analyse using you own code a deep beam using a structured mesh composed oflinear strain and constant strain triangles. Check all stresses with theoretical estimates.Check convergence with mesh refinement.

Group 8: Verify the Euler buckling load of a slender beam using Finite Elements. Yoursolutions may break down at the point of buckling. How good are your estimates comparedto the theoretical solutions?

Group 9: Analyse using your own code a internally pressurised plane strain thick cylinderand check the results with theoretical estimates.

Group 10: Model a infinite wedge with tractions on one of the boundaries. Verify the stresssolutions near the tip of the wedge with theoretical estimates. What happens in the case of


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p g ppa right angled wedge subject to shear tractions on one edge?

Group 11: A beam is rotating at a fixed angular velocity about its geometric center. Find thestress distribution in the beam and compare with theoretical results.

Group 12: Solve the stresses for a 90 degree curved beam of inner radius a and outerradius b subjected to a normal sinusoidal loading Asin( ). Find solutions to the problem as varies from 0 to 2. Comment on the solution at =1;Group 13: Solve the problem of a heavy beam on a elastic foundation. Compare yourresults with theoretical solutions for different beam aspect ratios. Investigate when the

theoretical solution breaks down.Group 14: An infinite plate subjected to a remote tensile load contains an elliptical hole.Using a FEM simulation determine the stress concentration at the tip of the ellipse as afunction of the ellipticity of the hole. Compare with theoretical results.

Step by step guide to formulating a problem in FEA

Eg. Heat conduction in a 2-d domain

Q1. What are the governing equations and boundary conditions for the problem?


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Q2. What is the variable to be solved for?

T(x,y) a scalar quantity 1 dof per node.

Q3. Does a variational statement exist?

Check for self-adjointness if self adjoint write variational principle

In this case self adjointness exists and hence


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is the variational principle. Thus the correct T(x,y) makes theabove a minimum and also satisfies all the boundary conditions.

If variational principle does not exist, use a weighted residual method (we will learnabout it later)

Q.4 Variational principle exists. Now what?


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If we use the Rayleigh Ritz principle, we should start the discretisation right away. Willdeal with that route later..

Q5. What element to choose?

Tricky. Eigenvalue analysis may help. Experience may too.

Let us choose a 4 noded iso-p quad in this case.Thus


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What if we used the Rayleigh Ritz technique?


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Define geometry

Mesh geometry

Loop over elementsForm local stiffness

Form surface force vector if required

Assemble stiffness

Assemble forces

d


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end

Apply essential boundary conditions

Solve

Post process results

Method of weighted residuals


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weights

Example: beam bending


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Example 2: Heat transfer in 2-d


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Some special elements: singular elements1

2

34

5

6

7

89

Include if node I is defined


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x

y


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Vanishes at x=0 Jacobian issingular at x=0

3


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2

5

7

1,4,8 6Singularity exists all over the element.

Eg. Show that singularity exists along x axis.What happens if 1,4,8 are given different nodenumbers?


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Errors in FE analysis

Round off errors in ill-conditioned

systems P


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Discretisation error


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From Cook, Malkaus

and Plesha (2002)

In a linear element, displacement error is proportional to the square of element size while


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, p p p qstrain error is linearly proportional to element size.

Displacements are more accurate near the nodes, strains are accurate inside elements.


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Eg. For a problem meshed with 3 noded triangles, p=1

Thus error in energy O(h 2)

Change to 6 noded triangle: p=2 and energy error O(h 4)

Reduce element size by half: error reduced by a factor of 4.

basu-fem notes me623

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