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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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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

For linear elastic materials , the strain energy per unit volume in the body is “ ”

For elastic body total strain energy (U) is

12

Tσ ε

12

TU dvσ ε= ∫

The work potential is given by

The total potential energy for the general elastic body is

. .T T Ti iV S

iWP u fdV u Tds u P= − − −∑∫ ∫

. .12

T T T Ti iV S

i

dv u fdV u Tds u Pπ σ ε= − − −∑∫ ∫ ∫

Principal of minimum potential energy

For conservative systems, of all the kinematically admissible displacement

fields, those corresponding to equilibrium extremize the total potential energy . If the

extremum condition is a minimum , the equilibrium state is stable

2

1

3 4

2

1

3

K2`K1`

K3K4

q1

q3

q2

Example

Figure-1

Figure 1 ,shows a system of spring .

The total potential energy is given by

where δ1 , δ2 , δ3 , and δ4 are extensions of four spring . since

δ1 = q 1 - q 2

δ2 = q 2

δ3 = q 3 - q 2

δ4 = - q 3

2 2 2 21 1 2 2 3 3 4 4 1 1 3 3

1 1 1 12 2 2 2

k k k k F q F qπ δ δ δ δ= + + + − −

we have

where q 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 3

1 1 1 12 2 2 2

k q q k q k q q k q F q F qπ = − + + − + − −

• For equilibrium of this 3- DOF system , we need to minimize to ∏ with respect to q 1 , q 2 , and q 3 the three equations are given by

i = 1 ,2 ,3

which are

= k1 (q 1 - q 2) - F1 = 0

0iq

π∂ =∂

1qπ∂∂

= -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

2qπ∂∂

3qπ∂∂

If on the other hand , we proceed to write the equilibrium of the system by considering the equilibrium of each separate node as shown in figure 2We can write

K1δ1 = F1

K2δ2 - K1δ1 - K3δ3 = 0

K3δ3 - K4δ4 = F3

Which is precisely the set of equations represented in Eq-1

We see clearly that the set of equation “1”is obtained in a routine manner using the potential energy approach, without any reference to the free body diagrams .

This make the potential energy approach attractive for large and complex problems .

RAYLEIGH-RITZ METHOD

Rayleigh-Ritz method involves the construction of an assumed displacement field, say

u = ∑ai Φi ( x, y, z) i = 1 to L

v = ∑ajΦj( x, y, z) j = L + 1 to M

w = ∑ak Φk( x, y, z) k = M + 1 to N

N > M > L

Eq-1

The functions Φi 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)

∏= ∏ ( a1, a2, a3, ….…….. ar )

where r = no of independent unknowns the extremum with respect to ai,( i = 1 to r) yields the set of r equation

i = 1, 2, 3……… ,r0ia

π∂ =∂

Example

The potential energy of for the linear 1-Drod with body force is neglected , is

where u1= u (x = 1)

( )2

10

1 22

ld uE A d x ud xπ ⎡ ⎤= −⎢ ⎥⎣ ⎦∫

1 1

E = 1, A= 1

X

Y

21

let as consider a polynomial function u = a1 + a2x + a3x3 this must satisfy

u = 0, at x = 0u = 0 at x = 2

thus 0 = a1

0 = a1 +2 a2+ 4a3

Figure- 2

Hence a2 = -2a3u = a3 (-2x + x2)u1 = -a3

then , and

( )2

10

1 22

lduEA dx udxπ ⎡ ⎤= −⎢ ⎥⎣ ⎦∫

23 3

22 23

a a⎛ ⎞= +⎜ ⎟⎝ ⎠

( )32 1du a xdx

= − +

we set

Resulting in a3 = -0.75u1 = - a3 = 0.75

the stress in bar given by

exact solution is obtained if piecewise polynomial interpolation is used in the construction of u .

30a

π∂ =∂

( )1.5 1duE xdx

σ = = −

GALERKIN’S METHOD• Galerkin’s method uses the set of governing

equations in the development of an integral form.

• It is usually presented as one of the weighted residual methods.

• Let us consider a general representation of a governing equation on a region “V”

Lu = PWhere , “L” as operator operating on “u”

For the one-dimensional rid considered in previous example

Governing equation

( ) 0d duEAdx dx =

( )dd EAdx dx⎛ ⎞⎜ ⎟⎝ ⎠

We may consider L as operator, operating on “u”

• The exact solution needs to satisfy “Lu=P”at every point x .• If we seek an approximate solution , if

introduces an error , called the residual

• The approximate methods revolve around setting the residual relative to a weighting function Wi ,i = 0 to n

( )xεu

( )x L u Pε = −

( ) 0iW Lu P dV− =∫

The weighting function Wi are chosen from the 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 φ

1

n

i ii

u Q G=

= ∑

Where the coefficient are arbitrary , except for requiring that satisfy boundary conditions were is prescribed.

1

n

i ii

Gφ φ=

= ∑

Given by

For elastic materials

∫ =+Φ+∂∂

+∂

∂+

∂∂

Vxx

xzxyx dVfzyx

0......])[( ττσ

∫ ∫ ∫+∂∂

−=∂∂

V V Sx dSndV

xdV

xαθθαθα

∫ ∫ ∫ ∑−−−V V S i

TTTT PTdSfdVdV φφφφεσ )(

Example let us consider the problem of the previous example and solve it by Galerkin’s approach.The equilibrium equation is

• u=0 at x=0• u=0 at x=0Multiplying this differential equation by Integrating by pars, we get

0d duEAdx dx =

Figure- 2

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

( ) ( )2 1 2

0 10

0ddu du duEA EA EAdx dx dx dxφ φ φ− + + =∫

φ

duEA dx

2

10

2 0dduEA dx dxφ φ− + =∫

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

φφ

( )212u x x u= −

( )212x xφ φ= −

( )2

21

0

2 2 2 0u x d xφ⎡ ⎤− − + =⎢ ⎥

⎣ ⎦∫

This is to be satisfied for every .We get

( )1 18 2 03 uφ − + =

1 0.75u =