principles of minimum potential energy
TRANSCRIPT
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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
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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
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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
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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
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
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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 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
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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
Rayleigh-Ritz Principle
...)()(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
Rayleigh-Ritz Principle
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.
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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
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
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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: