LectureACM_2

8/13/2019 LectureACM_2

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Variational principles in engineeringmechanics

Nguyn Xun Hng

University of Science

School of Math & Computer Science


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Contents

2. Vectors and Tensors

4. Introduction to Variational Principles

6. Finite element method in deformable solids

5. Overview of some advanced variational forms

1. Introduction

7. Finite element method in plates and shells structures

8. Introduction to Advanced finite element methods

9. Practical applications

3. Overview of Partial Differential Equations (PDE)


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4. Introduction to variational principles


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Introduction to variational principles

Strong form

Due to difficulty in finding the exact solution of most

differential equations

Alternative (weak) form

Simpler to find the exact solution of differential

equations based on Variational Principles

Relaxed some strict requirements

Background


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Consider the functional

with a single variablex ( , , , ), ( )x xxF F x u u u u u x

u is called theprimary variable of functionalF

u

xa b

u=vu

u+vAssumed that u is changed tou+u u+v, where=constant.

The operator iscalledthe

variational symbol.

Note that v is restricted by the

condition v(a)=v(b)=0, for all .

Therefore, we have the following conditions:

u(a)+v(a)=u(a), and u(b)+v(b)=u(b).


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Family of curves passing through x=1

and x=3


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General form of variational formulations

Discuss on the variation u=v: u describes an admissible displacement in the function

u(x)at a fixedvalue of the independent variablex.

u

xa b

u=vu

u+vIf u is specified at points onboundary, u = 0 at these

points because the known

value of u is can not be

varied.

How is admissible displacementin the function u(x)at a

fixedvalue of the independent variablex ?


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These are a set of u (orv)that satisfy constraint conditionsenforced on the whole problem. Reader should be revisited

in the courses of Analytical Mechanicsfor more details.

u is known as a virtual displacement (it is arbitrary

elsewhere and vanishes on boundary of problem).

Now we describe the change of u to the change in the

function F. We define

( , , ') ''

''

F F F

F x u u x u ux u u

F Fu u

u u

is called the first variation ofF.


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Anactual displacement u changed to u +du, we obtain thetotal differential ofF:

( , , ') ''

F F FdF x u u dx du du

x u u

Question: What is the difference between F and dF in

both mathematic and physical means ?


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Some properties of the variation operator:( )

1 2 1 2F F F F

( )1 2 2 1 1 2

F F F F F F

( ) ( )1 2 1 1 22

2 2

F F F F F

F F

( ) ( )d d dv duu vdx dx dx dx

....

( )b b

a a

udx udx


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Example: Boundary-value problem (BVP) of bar (Poisson

equation in one dimension (1D)):

21 ( )2

L L

0 0duk dx - qudxdx

Prove that = 0 leads to boundary-value above problem

2

20 (0, )

(0) 0, | 0x L

d uk q in L

dx

duu kdx

(strong form)

(potential energy)

( )0

L L

0 0

du d uk dx - q udx

dx dx

(weak form)


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Variational formulation of BVP

- Assume that we need to find the variational form of the

following partial differential equation (1D):

( ) 0 (0, ),where ''

x

F F uin L u u

u x u x

( ) 0'

L

0

F Fv dx

u x u

(Note that uor v

can be considered

as test function)

D 0 Non 0, and on

Fu u g L

u'

subjected to the boundary condition

Three basic steps to obtain the variational form of BVP

Step 1: BVP is multiplied by a test function or variation v


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0( | | ) 0' ' '

L

x L x0

F v F F Fv dx v v

u x u u u

Step 2: perform the part integral on the second term:

Step 3: inserting boundary conditions

00

'

L

0

F v Fv dx vg u x u

If is a linear functional in u, i.e, ( )F

q x

u

Finally, we obtain a weak form (variational form):

00

0

'

L L

0

v Fdx qvdx vg

x u

(weak form)


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- Find the variational form of the following partial

differential equation (2D) (strong form (SF)):

( ) ( ) 0x y

F F Fin

u x u y u

( ) ( ) 0x y

F F Fv d

u x u y u

D N

D N

on , and onx y n

x y

F Fu u n n q

u u



(2.1)

(2.2)


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( ) ( )D N

x y x y

x y x y x y

F v F v F F F F Fv d v n n d v n n d

u x u y u u u u u



( ) 0N

x y

x y x y

F v F v F F Fv d v n n d

u x u y u u u

If is a linear functional in u, i.e, ( , )F

q x y

u

Finally, we obtain a weak form (WF)):

( ) 0N

n

x y

v F v F d qvd q vd

x u y u


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The total potential energy of problem is now given by

int( , ) ( ) ext1 = a u u f u U W2

Let a(u,v)be bilinear form and symmetric and f(v) is linear.

We obtain the weak form:

( , ) ( )a u v f v

( , ) ( ) , ( )N

n

x y

v F v F a u v d f v qvd q vd x u y u

where

It is clear that = 0 leads to the strong form of boundary-value

above problem

(2.3)


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int ext = U - W =0 if only if u = u

The Minimum Potential Energy Principle

The Minimum Potential Energy (MPE) principle states that

the actual displacement solution that satisfies the

strong form is that which renders stationary

with respect to admissible variations of the exact

displacement field .

u(x)

u = u + v

u(x)

We say that produces the Minimum Potential Energy:u

(u) (u ), for all u V

smooth, and |D

V u u u

where

(2.4)

Variational Form (VF)


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An illustration of

energy functional

2

1 ( , , , )d

x

x xxx

F x u u u x

Variation of a minimizing function

u

x

u(x) + v(x)

u(x)

x

0

( )| 0

d

d

2

2

( , ) 2 ( )

( , ) ( , ) ( , ) ( , ) 2 ( ) 2 ( )

( , ) 2 [ ( , ) ( )] ( , ) 2 ( )

2 = a u v u v f u v

a u u a u v a v u a v v f u f v

a v v a u v f v a u u f u


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We rewrite the Minimum Potential Energy (Minimization

Problems (MP)):

0(u) (u v ), for all v V

where 0 smooth, and | 0DV v v

(2.4)

The equivalence of solution

SF WFEquivalence ?

VF

Equivalence ?


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Examples

( ) (0, )d du

k ru x in Ldx dx

1. Given the boundary-value problem


D N0 on 0, and | 1 onx Lduu k Ldx

Establish the weak form and the potential energy ?

Solve:

0( ) 0

L d duv k ru x dx

dx dx



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Examples


00

| 0L

x L

x

dv du duk rvu vx dx kv

dx dx dx

0 0( ) ( ) 0

L Ldv duk rvu dx vxdx v L

dx dx


Week form:

0 0( ) ( )

L Ldv duk rvu dx vxdx v L

dx dx

Energy functional:

2 2

0 0

[ ( ) ] ( )L L1 du

= k ru dx uxdx u L

2 dx


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Examples

2. The energy functional:

0

0 in

0

n N

D

u

ug u q on

n

u on

with

Hint:

2

0

1[ ]

2Nn

1 = u ud g u q u d

2

| 0D

u

. It is shown that the minimization of (u) leads

to the solution of the PDE

(u + v) (u) and set0( ) 0

d u + v |

d

Or we can use Eq. (2.1) to obtain the PDE


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The condition for minimization problem

1.L -ispositive definite:

0, 0uLud u ud u

2.Lis self-adjoint:

, , 0vLud uLvd u v

Example:

u q a minimization problem can be obtained for

0u v on

u u q U no minimization problem can be obtained

for 0u v on

Can one find a functional for a variational principle for any

differential equations ?

(symmetry)


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Variational formulation in solid mechanics

int ext = U - W =0 if only if u = u

The Principle of Minimum Potential Energy

The Minimum Potential Energy (MPE) principle states that

the actual displacement solution that satisfies the

strong form is that which renders stationary

with respect to admissible variations of the exact

displacement field .

u(x)

u = u + v

u(x)

We say that produces the Minimum Potential Energy:u

(u) (u ), for all u V

smooth, and |D

V u u u

where

(2.4)

Variational Form (VF)


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Tonti diagram for the continuum model of a bar member

Variational Formulation of Bar member

duE E

dx

'0F q

F EA

( )x

du

dx

The total potential energy (PTE) of the barU W

[ ( )] [ ( )] [ ( )]u x U u x W u x

u:primary variable


governing equations


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Variational Formulation of Bar member

U W

Governing equations Virtual work principle Minimum TPE

Strong form: '' ( ) 0EAu q x

Weak form: ( , ) ( )a u u f u ( 0)or U W

TPEint( , ) ( ) ext

1 = a u u f u U W

2

0

0

( , ) ,

( )

l

l

du d ua u u EA dx

dx dx

f u q udx


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Variational Formulation of beam member


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z

x

w

x

xp zu

x

The Bernoulli-Euler Beam Theory

Strain

2

2

x

wz

x

up

x


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2

2

wE Ez Ez

x

2

A A

M z dA E z dA EI

Tonti diagram for the continuum model of a thin beam member

( )w x

( )x w

M EI

M q


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Variational Formulation of thin beam

U W


Strong form: '''' ( ) 0EIw q x

Weak form: ( , ) ( )a w w f w ( 0)or U W

TPEint( , ) ( ) ext

1 = a w w f w U W

2

2 2

2 20

0

( , ) ,

( )

l

l

d w d wa w w EI dx

dx dx

f w q wdx

(fixed beam)


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Variational Formulation of plane problem

Plane StressPlane Strain Axial symmetry

0z

0

z

0, 0

0

r z

r z

u


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Displacement field:( )

( )

u x

v x

u

Plane stress problem

x

y

xy

x

y

xy

Stress field:

Strain field:

Plane strain problem


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Compatibility equation

(kinematic):

0

0

x

y s

xy

x

u

vy

y x

u

u

Constitute equation : s u

D D u

0

0s

x

y

y x

symmetric-gradientoperator

Equilibrium

equations :

b = 0

0

0

xyxx

xy y

y

bx y

bx y

=

=

in

x x xy y x

xy x y y y

n n t

n n t

=

=on

or


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Tonti diagram for the continuum model of plane problem

u

t

s

u

u b = 0

Dij j i

n t

Displacement vector u is primary variables

We obtain a variational principle of single field as follows

u = u


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U W


Strong form:

Weak form: ( , ) ( )a f u u u ( 0)or U W

TPEint( , ) ( ) ext

1 = a f U W

2 u u u

( , ) d ,

( ) d dt

T

T T

a

f

u u D

u u b u t

b = 0


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The dynamic problem

mx F

Hamiltons Principle and Lagranges equation in dynamics

Dynamic motion of a rigid bodysystem can be described

in several approaches such as Newtons2nd law, energy

concept

Newtons equation of motion:

Lagranges equation of motion: ii i

d L LQ

dt q q

whereL =T- V being Lagrange functionT - kinetic energy,V -potential energy

Above Lagranges equation can is obtained via calculus

of variation from Hamiltons principle:

whereL =T- V being Lagrange functionT - kinetic energy,V -potential energy


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The dynamic problem

2

1

( ) 0t

t

T V dt

Or minimize

2

1

( )t

t

I T V dt

This implies that actual path followed by a dynamic process is

such as to make the integral of (T-V) a minimum.

Hamiltonsprinciple for deformable body can be expressed as

2

1

0t

t

Ldt whereL =T- being Lagrange function

T - kinetic energy,=U+V totalpotential energy


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The dynamic problem

.2T d

t t

u u

where

( . . )V d dS

f u t u

1

2

T

U d Strain energy

External work

Note: u(x,t1)=u(x,t2)


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The dynamic problem

2

0. ( )

2 2

L AT d dx

t t t

u u u

Example: Let us consider the axial motion of an elasticbar of length L, area of cross section A, modulus of

elasticityE, mass density , subjected to a distributed load

f, and an end loadP. Find the equations of motion of the

bar

0

( )L

V fudx Pu L

2

0

1 1( )

2 2

LT u

U d AE dxx


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5. Advanced variational principles


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Advanced variational principles

Variational principleTotal potential energyTotal complementary energy

Hellinger-Reissner (HR)

Hybrid principles

Veubeke-Hu-Washizu (VHW)

Approximate

field:

Displacements(one field)

Stresses (one field)

Displacements & stresses (two fields)

Displacements, stresses, strains(three fields)


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Restriction to solid problems

Strong form Tonti diagram for the continuum model of solid problems

u

t

s

u

u b = 0

Dij j i

n t

u = u

What is happened as 1)

s

u

u 2)

u u

3) uD

etc

=> impose a weak form


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The Hellinger-Reissner (HR) Principle

u

1D

The starting Weak Form for derivation of the HR principle

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