Principles of Minimum Potential Energy

8/12/2019 Principles of Minimum Potential Energy

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A generic problem in 1D

11

00

10;02

2

xatu

xatu

xxdx

ud

Approximate solution strategy:

Guess

Wherej o(x), j 1(x), are known functions and ao, a1, etc areconstants chosen such that the approximate solution

1. Satisfies the boundary conditions

2. Satisfies the differential equation

Too difficult to satisfy for general problems!!

...)()()()( 22110 xaxaxaxu o


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Potential energy

Wloadingofenergypotential(U)energyStrain

The potential energy of an elastic body is defined as


3/27

F

uF

x

k

ku

k

1

Hookes Law

F = ku

Strain energy of a linear spring

F = Force in the spring

u = deflection of the spring

k = stiffness of the spring


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Strain energy of a linear spring

F

u u+du

Differential strain energy of the spring

for a small change in displacement

(du) of the spring

FdudU

For a linear spring

kududU

The total strain energy of the spring

2u

0uk

21duukU

dU


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Strain energy of a nonlinear spring

F

u u+du

FdudU

The total strain energy of the spring

curventdispalcemeforcetheunderAreaduFUu

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

Wloadingofenergypotential(U)energyStrain

Fuku21 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 potential

energy of the system as:

3x

2

232

2

21 Fd)dd(k21)d(k

21

xxxsystem

Obtain the equilibrium equations by minimizing the potential energy

1k F

x

2k

1xd 2xd 3xd

)2(0)dd(kd

)1(0)dd(kdkd

232

3

232212

EquationF

Equation

xx

x

system

xxxx

system


8/27

Principle of minimum potential energy for a system of springs

In matrix form, equations 1 and 2 look like

Fx

x 0

d

d

kk

kkk

3

2

22

221

Does this equation look familiar?

Also look at example problem worked out in class


9/27

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

dx

duAxial strain

Axial stressdx

duEE

Strain energy per unit volume of the bar

2

dx

duE

2

1

2

1dU

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

LL

0

2

0dx

dx

duEA

2

1dxA

2

1U

Potential energy of the loading

L)Fu(xdxbuW0

L

Potential energy of the axially loaded bar

L)Fu(xdxbudxdxduEA

21

00

2

LL


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Principle of Minimum Potential Energy

Among all admissible displacements that a body can have, the one

that minimizes the total potential energy of the body satisfies the

strong formulation

Admissible displacements: these are any reasonable displacement

that you can think of that satisfy thedisplacement boundary

conditions of the original problem(and of course certain minimumcontinuity requirements). Example:

Exact solution for the

displacement field uexact(x)

Any other admissible

displacement field w(x)

L0x


12/27

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 modulusx

y

x=0 x=L

x

F

Potential energy of the axially loaded bar corresponding to the

exact solution uexact(x)

L)(xFudxbudxdx

duEA2

1)(u exact0 exact0

2

exactexact

LL


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Potential energy of the axially loaded bar corresponding to the

admissible displacement w(x)

L)Fw(xdxbwdxdx

dwEA

2

1(w)

00

2

LL

Exact solution for the

displacement field uexact(x)

Any other admissible

displacement field w(x)

L0x


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Example:

Lxbdx

ud

AE 0;02

2

LxatFdx

duEA

xatu

00

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

22

2x

xuexact

6

71)(xudxudx

dx

du

2

1)(u exact

1

0 exact

1

0

2

exactexact

Potential energy corresponding to this analytical solution


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Now assume an admissible displacement

w

Why is this an admissible displacement? This displacement is

quite arbitrary. But, it satisfies the given displacement boundary

condition w(x=0)=0. Also, its first derivate does not blow up.

11)w(xdxwdxdx

dw

2

1(w)

1

0

1

0

2

Potential energy corresponding to this admissible displacement

Notice

(w))(u

16

7

exact since


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Principle of Minimum Potential Energy

Among 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 boundary

conditions), and w is an admissible displacement (that is quite

arbitrary except for the fact that it satisfies the displacement

boundary conditions and its first derivative does not blow up),

then(w))(uexact

unless w=uexact (i.e. the exact solution minimizes the potential

energy)


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The Principle of Minimum Potential Energy and the strong

formulation are exactly equivalent statements of the same

problem.

The exact solution (uexact) that satisfies the strong form, renders

the 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 strong

formulation. The long answer is, it

1. requires only the first derivative to be finite

2. incorporates the force boundary condition automatically. The

admissible displacement (which is the function that you need tochoose) needs to satisfy only the displacement boundary condition


18/27

Finite element formulation, takes as its starting point, not the

strong formulation, but the Principle of Minimum PotentialEnergy.

Task is to find the function w that minimizes the potential energy

of the system

From the Principle of Minimum Potential Energy, that function w

is the exact solution.

L)Fw(xdxbwdxdx

dwEA

2

1(w)

00

2

LL


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Step 1. Assume a solution

...)()()()( 22110 xaxaxaxw o

Wherej o(x), j 1(x), are admissible functions and ao, 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(xdxbwdxdx

dwEA

2

1(w)

00

2

LL


...)()(F

dx...b

dx...dx

d

dx

dEA

2

1,...)a,(a

1100

0 1100

0

2

11

0010

LxaLxaaa

aa

L

L


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Step 3. Obtain the coefficients ao, a1, etc by setting

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

i

ai


The approximate solution is

...)()()()( 22110 xaxaxaxu o

Where the coefficients have been obtained from step 3


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Example of application of Rayleigh Ritz Principle

x

x=0 x=2

x=1

F

E=A=1

F=2

1

2

0

2

1)Fu(xdxdx

du

2

1(u)

xatappliedFloadofEnergyPotential

EnergyStrain

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

Let us assume a polynomial admissible displacement field2

210u xaxaa

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


23/27

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

2

210

2

u

xxa

xaxaa


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Now we apply Rayleigh Ritz principle, which says that if I plug

this approximation into the expression for the potential energy P , Ican obtain the unknown (in this case a2) by minimizing P

4

3

023

8

0

23

4

2Fdx2dx

d

2

1

1)Fu(xdxdxdu

21(u)

2

2

2

2

2

2

1

2

2

2

0

2

2

2

2

0

2

a

a

a

aa

xxaxxaxat

evaluated


25/27

2

2

2

2

210

2

4

3

2

u

xx

xxa

xaxaa

Hence the approximate solution to this problem, using the

Rayleigh-Ritz principle is

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

212

10uexactxforx

xforx


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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

Approximate

solution

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

St

ress

Approximate

stress

Exact Stress

The stress within the bar:

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Principles of Minimum Potential Energy

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