Potential Fem

7/29/2019 Potential Fem

1/32

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


2/32

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 =


3/32

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 =


4/32

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


5/32

2

1

3 4

2

1

3

K2`K1`

K3

K4

q1

q3

q2

Example

Figure-1


6/32

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 = + + +


7/32

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 = + + +


8/32


9/32

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


10/32

= -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


11/32

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


12/32

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 .


13/32

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


14/32

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)


15/32


16/32

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 =


17/32

1 1

E = 1, A= 1

X

Y

2

1


18/32

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


19/32

Figure- 2


20/32

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 = +


21/32

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 = =


22/32

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


23/32

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


24/32

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 =


25/32

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=

=


26/32

Where the coefficient are arbitrary , except

for requiring that satisfy boundary

conditions were is prescribed.

iu

1

n

i ii G =

=

Given by


27/32

For elastic materials

=+++

+V

xxxzxyx dVf

zyx0......])[(

+=

V V S

x dSndVx

dVx

V V S i

TTTTPTdSfdVdV )(


28/32

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

=


29/32

Figure- 2


30/32

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 + =


31/32

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

+ =


32/32

This is to be satisfied for every .

We get

)1 18 2 03 u + =

1 0.75u =

1

Date post:	14-Apr-2018
Category:	Documents
Upload:	deval-shah
View:	222 times
Download:	0 times

Potential Fem

Documents