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POTENTIAL ENERGY, .The total potential energy of an elastic body , is
defined as the sum of total strain energy (U) and the
work potential (WP) .
= U + WP
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For linear elastic materials , the
strain energy per unit volume in
the body is
For elastic body total strain
energy (U) is
1
2
T
1
2
TU dv =
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The work potential is given by
The total potential energy for the generalelastic body is
. .T T T
i iV Si
WP u fdV u Tds u P=
. .
12
T T T T i i
V Si
dv u fdV u Tds u P =
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Principal of minimum potentialenergy
For conservative systems, of all the
kinematically admissible displacement
fields, those corresponding to equilibriumextremize the total potential energy . If the
extremum condition is a minimum , theequilibrium state is stable
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2
1
3 4
2
1
3
K2`K1`
K3
K4
q1
q3
q2
Example
Figure-1
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Figure 1 ,shows a system of spring .
The total potential energy is given by
where 1 , 2 , 3 , and 4 are extensions of four spring .since 1 = q 1 - q 2
2 = q 23 = q 3 - q 24 = - q 3
2 2 2 2
1 1 2 2 3 3 4 4 1 1 3 31 1 1 12 2 2 2
k k k k F q F q = + + +
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we have
whereq 1 , q 2 , and q 3 are the displacements of nodes
1 , 2 and 3 respectively
( ) ( )2 22 21 1 2 2 2 3 3 2 4 3 1 1 3 31 1 1 12 2 2 2
k q q k q k q q k q F q F q = + + +
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For equilibrium of this 3- DOF system , we need
to minimize to with respect to q 1 , q 2 , and q 3the three equations are given by
i = 1 ,2 ,3
which are
= k1 (q 1 - q 2) - F1 = 0
0iq
=
1q
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= -k1 (q 1 - q 2) + k2 q2 - k3 (q 3 - q 2) = 0
= k3 (q 3 - q 2) + k4 q3 F3 = 0
Equilibrium equation can be put in the form of
K q = F as follows
K1
-K3
-K1
-K1 K1+ K2+ K3 -K3
0
0 K3+ K4
=
q1
q2
q3
F1
0
F3
.. .. 1 1
2q
3q
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If on the other hand , we proceed to write theequilibrium of the system by considering the
equilibrium of each separate node as shown in
figure 2We can write
K11 = F1K22 - K11 - K33 = 0K33 - K44 = F3
Which is precisely the set of equations represented in
Eq-1
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We see clearly that the set of equation 1
is obtained in a routine manner using thepotential energy approach, without any
reference to the free body diagrams .
This make the potential energy approach
attractive for large and complex problems .
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RAYLEIGH-RITZ METHOD
Rayleigh-Ritz method involves the construction of
an assumed displacement field, say
u = ai i ( x, y, z) i = 1 to L
v = aj j( x, y, z) j = L + 1 to M
w = akk( x, y, z) k = M + 1 to N
N > M > L
Eq-1
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The functions i
are usually taken as
polynomials. Displacements u, v, w must
satisfy boundary conditions.
Introducing stress-strain and strain-
displacement relation Substituting
equation 1 in to (PE)
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Example
The potential energy of for the linear 1-D
rod with body force is neglected , is
where u1= u (x = 1)
( )2
1
0
12
2
ld uE A d x u
d x =
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1 1
E = 1, A= 1
X
Y
2
1
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let as consider a polynomial function
u = a1
+ a2x + a
3x3 this must satisfy
u = 0, at x = 0
u = 0 at x = 2
thus 0 = a10 = a
1+2 a
2+ 4a
3
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Figure- 2
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Hence
a2
= -2a3
u = a3 (-2x + x2
)u
1= -a
3
then , and
( )
2
10
12
2
l
duEA dx udx
= 2
3 3
22 2
3
a a
= +
( )32 1du a xdx = +
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we set
Resulting in a3
= -0.75
u1
= - a3
= 0.75
the stress in bar given by
exact solution is obtained if piecewise
polynomial interpolation is used in theconstruction of u .
3 0a
=
( )1.5 1du
E xdx = =
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GALERKINS METHOD Galerkins method uses the set of governing
equations in the development of an integral form.
It is usually presented as one of the weightedresidual methods.
Let us consider a general representation of agoverning equation on a region V
Lu = P
Where , L as operator operating on u
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For the one-dimensional rid considered inprevious example
Governing equation
) 0d duEAdx dx =
( )dd EAdx dx
We may consider L as operator, operating
on u
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The exact solution needs to satisfy Lu=Pat every point x .
If we seek an approximate solution , if
introduces an error , called the residual
The approximate methods revolve aroundsetting the residual relative to a weighting
function Wi ,i = 0 to n
( )x u
)x L u P =
) 0iW Lu P dV =
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The weighting function Wi are chosen fromthe basis functions used for constructing
Here ,we choose the weighting function to
be linear combination of the basis function
Gi . Specifically ,consider an arbitrary
function
1
n
i i
i
u Q G=
=
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Where the coefficient are arbitrary , except
for requiring that satisfy boundary
conditions were is prescribed.
iu
1
n
i ii G =
=
Given by
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For elastic materials
=+++
+V
xxxzxyx dVf
zyx0......])[(
+=
V V S
x dSndVx
dVx
V V S i
TTTTPTdSfdVdV )(
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Examplelet us consider the problem of the previousexample and solve it by Galerkins approach.
The equilibrium equation is
u=0 at x=0
u=0 at x=0
Multiplying this differential equation byIntegrating by pars, we get
0d duEAdx dx
=
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Figure- 2
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Where is zero at x = 0 and x = 2.
is the tension in the rod ,which
takes a jump of magnitude 2 at x = 1 , thus
( ) ( )
21 2
0 10
0ddu du duEA EA EAdx dx dx dx + + =
duEAdx
2
1
0
2 0dduEA dx dx + =
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Now we use the same polynomial (basis ) for u
and
if u1 and are the value at x = 1 ,thus
Substituting these and E = 1, A = 1 in the
previous integral yields
( )2 12u x x u=
( )2
12x x =
( )2
2
1
0
2 2 2 0u x d x
+ =
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This is to be satisfied for every .
We get
)1 18 2 03 u + =
1 0.75u =
1