Minimum Potential Energy Principle

8/12/2019 Minimum Potential Energy Principle

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MANE 4240 & CIVL 4240

Introduction to Finite Elements

Principles of minimumpotential energy and

Rayleigh-Ritz

Prof. Suvranu De


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Reading assignment:

Section 2.6 + Lecture notes

Summary: Potential energy of a system

Elastic barString in tension

Principle of Minimum Potential EnergyRayleigh-Ritz Principle


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A generic problem in 1D

11

00

10;022

xat u

xat u

x xdx

ud

Approximate solution strategy:GuessWhere j o(x), j 1(x), are known functions and a o, a1, etc areconstants chosen such that the approximate solution

1. Satisfies the boundary conditions2. Satisfies the differential equation

Too difficult to satisfy for general problems!!

...)()()()( 22110 xa xa xa xu o j j j


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Potential energy

Wloadingof energy potential(U)energyStrain

The potential energy of an elastic body is defined as


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F

uF

x k

k u

k

1

Hookes Law F = ku

Strain energy of a linear spring

F = Force in the springu = deflection of the springk = stiffness of the spring


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Strain energy of a linear spring

F

u u+du

Differential strain energy of the springfor a small change in displacement(du) of the spring

FdudU For a linear spring

kududU

The total strain energy of the spring2u

0uk

21

duuk U

dU


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Strain energy of a nonlinear spring

F

u u+du

FdudU The total strain energy of the spring

curventdispalcemeforcetheunder AreaduFUu

0

dU


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Potential energy of the loading (for a single spring as in the figure)

FuW

Potential energy of a linear spring

Wloadingof energy potential(U)energyStrain Fuku

21

2

F

x k

k u

Example of how to obtain the equlibr


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Principle of minimum potential energy for a system of springs

For this system of spring, first write down the total potentialenergy of the system as:

3x2

2322

21 Fd)dd(k 21

)d(k 21

x x x system

Obtain the equilibrium equations by minimizing the potential energy

1k F x

2k

1xd 2xd 3xd

)2(0)dd(k

d

)1(0)dd(k dk d

232

3

232212

Equation F

Equation

x x

x

system

x x x x

system


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Principle of minimum potential energy for a system of springs

In matrix form, equations 1 and 2 look like

F x

x 0

d

d

k k

k k k

3

2

22

221

Does this equation look familiar?

Also look at example problem worked out in class


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Axially loaded elastic bar

x

y

x=0 x=L

A(x) = cross section at x b(x) = body force distribution(force per unit length)E(x) = Youngs modulus u(x) = displacement of the bar

at x

x

F

dxdu

Axial strain

Axial stressdx

duEE

Strain energy per unit volume of the bar2

dxduE

21

21dU

Strain energy of the bar

Adx2

1dV

2

1dUU

L

0x

since dV=Adx


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Axially loaded elastic bar

Strain energy of the bar

L L

0

2

0dx

dxdu

EA21

dxA21

U

Potential energy of the loading

L)Fu(xdx buW0

L

Potential energy of the axially loaded bar

L)Fu(xdx budxdxdu

EA21

00

2

L L


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Principle of Minimum Potential EnergyAmong all admissible displacements that a body can have, the onethat minimizes the total potential energy of the body satisfies thestrong formulation

Admissible displacements : these are any reasonable displacementthat you can think of that satisfy the displacement boundaryconditions of the or iginal problem ( and of course certain minimumcontinuity requir ements ). Example:

Exact solution for thedisplacement field u exact (x)

Any other admissibledisplacement field w(x)

L0 x


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Lets see what this means for an axially loaded elastic bar

A(x) = cross section at x b(x) = body force distribution(force per unit length)E(x) = Youngs modulus x

y

x=0 x=L

x

F

Potential energy of the axially loaded bar corresponding to theexact solution u exact (x)

L)(xFudx budxdx

duEA

21

)(u exact0 exact0

2exact

exact

L L


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Potential energy of the axially loaded bar corresponding to theadmissible displacement w(x)

L)Fw(xdx bwdxdxdw

EA21

(w)00

2

L L

Exact solution for thedisplacement field u exact (x)

Any other admissibledisplacement field w(x)

L0x


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Example:

L xbdx

ud AE 0;0

2

2

L xat F dxdu

EA

xat u 00

Assume EA=1; b=1; L=1; F=1Analytical solution is

22

2 x xuexact

67

1)(xudxudxdx

du21

)(u exact1

0 exact

1

0

2exact

exact

Potential energy corresponding to this analytical solution


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Now assume an admissible displacement

xw

Why is this an admissible displacement? This displacement isquite arbitrary. But, it satisfies the given displacement boundarycondition w(x=0)=0. Also, its first derivate does not blow up.

11)w(xdxwdxdxdw

21

(w)1

0

1

0

2

Potential energy corresponding to this admissible displacement

Notice

(w))(u

167

exact

since


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Principle of Minimum Potential EnergyAmong all admissible displacements that a body can have, the one

that minimizes the total potential energy of the body satisfies thestrong formulation

Mathematical statement: If uexact is the exact solution (which

satisfies the differential equation together with the boundaryconditions), and w is an admissible displacement (that is quitearbitrary except for the fact that it satisfies the displacementboundary conditions and its first derivative does not blow up ),

then (w))(u exact unless w=u exact (i.e. the exact solution minimizes the potentialenergy )


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he Principle of Minimum Potential Energy and the strongformulation are exactly equivalent statements of the same

problem.

The exact solution (u exact ) that satisfies the strong form, rendersthe potential energy of the system a minimum.

So, why use the Principle of Minimum Potential Energy?The short answer is that it is much less demanding than the strongformulation. The long answer is, it1. requires only the first derivative to be finite2. incorporates the force boundary condition automatically. Theadmissible displacement (which is the function that you need tochoose) needs to satisfy only the displacement boundary condition


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Finite element formulation, takes as its starting point, not thestrong formulation, but the Principle of Minimum PotentialEnergy .

Task is to find the function w that minimizes the potential energyof the system

From the Principle of Minimum Potential Energy, that function wis the exact solution.

L)Fw(xdx bwdxdxdw

EA21

(w)00

2

L L


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Step 1. Assume a solution

...)()()()( 22110 xa xa xa xw o j j j

Where j o(x), j 1(x), are admissible functions and a o, a1,etc are constants to be determined from the solution.

Rayleigh-Ritz Principle

The minimization of the potential energy is difficult to perform

exactly.The Rayleigh-Ritz principle is an approximate way of doing this.


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Step 2. Plug the approximate solution into the potential energy

L)Fw(xdx bwdxdx

dwEA

2

1(w)

00

2

L L


...)()(F

dx... b

dx...dxd

dxd

EA21

,...)a,(a

1100

0 1100

0

21

10

010

L xa L xa

aa

aa

L

L

j j

j j

j j


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Step 3. Obtain the coefficients a o, a1, etc by setting

,...2,1,0,0(w)

i

a i


The approximate solution is

...)()()()( 22110 xa xa xa xu o j j j

Where the coefficients have been obtained from step 3


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Example of application of Rayleigh Ritz Principle

x

x=0 x=2x=1

F

E=A=1

F=2

1

2

0

2

1)Fu(xdxdxdu

21

(u)

xat

applied F load of Energy Potential

EnergyStrain

The potential energy of this bar (of length 2) is

Let us assume a polynomial admissible displacement field 2

210u xa xaa

Note that this is NOT the analytical solution for this problem.


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Example of application of Rayleigh Ritz Principle

For this admissible displacement to satisfy the displacement

boundary conditions the following conditions must be satisfied:

0422)u(x

00)u(x

210

0

aaa

a

Hence, we obtain

21

0

2

0

aa

a

Hence, the admissible displacement simplifies to

22

2210

2

u

x xa

xa xaa


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Now we apply Rayleigh Ritz principle , which says that if I plugthis approximation into the expression for the potential energy , Ican obtain the unknown (in this case a 2) by minimizing

43

0238

0

234

2Fdx2

dx

d

2

1

1)Fu(xdxdxdu

21(u)

2

2

2

22

2

1

22

2

0

22

2

2

0

2

a

a

a

aa

x xa x xa xat

evaluated


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2

22

2210

24

3

2

u

x x

x xa

xa xaa

Hence the approximate solution to this problem, using theRayleigh-Ritz principle is

Notice that the exact answer to this problem (can you prove this?) is

212

10u exact x for x

x for x


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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

x

Exact solution

Approximatesolution

The displacement solution :

How can you improve the approximation?


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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

-1.5

-1

-0.5

0

0.5

1

1.5

x

S t r e s s

Approximatestress

Exact Stress

The stress within the bar:

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Minimum Potential Energy Principle