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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
Rayleigh-Ritz Principle
...)()(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
Rayleigh-Ritz Principle
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: