04 Energy Method

8/8/2019 04 Energy Method

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Kashi Nath Saha, Mech. Engg. Dept., Jadavpur Univ., Kol-32 12/05/10 04:04

Advantages of energy method:

When the exact solution of the governing differential equation is not available,it may yield a good approximate solution.

The method has the inherent ability to converge to the exact solution if a largenumber of terms are taken to represent the unknown field (say, deflectionshape).

The numerical scheme of the solution technique is quite simple in view of the

advancement in numerical methods and computational fields. The method is suitable to be used in advanced techniques like FEM, which can

take up irregularities in geometry, material, boundary condition, etc.

It can be shown that for stable equilibrium, any virtual displacement will cause apositive change in the total potential energy of the system, which means that for asystem in stable equilibrium is minimum. p


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Conservation of energy:work done = energy stored

Kinetic energy T:

== P k U 21212

Tenergykineticof increment

doneworkof Increment

=

====dT )r r md( dt)r ( )r (m

r d F

21

r r mT =21

Strain energy U:energy in spring = work done

Conservation of energy for conservative systems

E = total energy = T + U = constant


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Energy methods Example

0

0

=+=

x xm xkx

) E ( dt d

0=+ kx xm

Same as vector mechanics

Work-energy principles have manyuses, but one of the most useful isto derive the equations of motion.

Conservation of energy: E = const.

22

2

2

21

21

21

21

xmkxT U E

xmT

kxU

+=+=

=

=


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Type of boundary conditions:

1. Geometric boundary condition they refer to the kinematic conditions,like, deflection, slope, curve, etc., of the boundary.

2. Forced boundary condition refers to the forced conditions of the boundary.

Type of functions:1. eigen-functions 2. admissible functions 3. coordinate functions

Success or failure in applying the assumed series solution for the assumed field(say deflection w ), largely depends on the choice of proper coordinate functions.In the majority of cases a few terms of the series give a sufficiently accurate resultif the coordinate functions are the eigen-functions of the problem. In other cases,satisfactory result can be obtained when the coordinate functions are chosen froma set of orthogonal functions.

Generation of orthogonal set: In case of vectors, Gram-Schmidt orthogonalisation procedure is well known.However, the scheme is not readily available for generating functions. In thiscase, it is implemented with the help of a starting function, an eigen-function or

an admissible function or at least a coordinate function.


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Linear independence:

The set of n-vectors v 1, v2, , v n are linearly independent if their linear combination,

c1v1 + c2v2 + + c nvn ( = 0 ) is zero only when all the arbitraryscalars (c i s ) are zero. In this case, the vectors will form anorthogonal set, such that

=

It is known in vector algebra that to form a basis of a n -dimensional vector space, the basis vectors must come from a set

of linearly independent vectors, which again forms an orthogonalset. Similarly the solution space for the assumed unknown field(say deflection w ) is complete, when they are formed through a setof orthogonal functions.

( )i jv v d

1,

0,

for i j

for i j

=


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Expression of strain energy due to bending of a beam / column: Assumption The beam / column is slender, i.e., the effect of shear deformationcan be neglected.


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Expression of potential energy of a column:

Expression of total potential


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Energy approach for stability problem


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Buckling mode shapes:


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Energy approach for dynamic problem

( )

( ) ( )

+

+

=

=

L t t

t

t

L L

t

t

L

dxuu Adt u xu

E Adxu xu

E A x

uudx At

dt dxu x x

u E Auudx A

00 0

0

2

1

2

1

2

1

0

0

221

21

21

21

um

xu

xu

E )(E x x

=====

====

energykinetic T

UenergystrainenergypotentialV

densityenergystrainUo

Hamiltons principle:

( ) dt u xu

E Adxu xu

E A x

u At

t

t

L L

+

= 21 0 0

0

= 21

)(0t

t dt V T


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Fixed-free bar Free vibration

E

L

in

2=

are the eigenfunctions

L xi

2sin

For free vibration:

General solution:

Hence

)cos()(),( t x At xu n =

are the frequencies (eigenvalues)

2

2

22

2

xut u =

),5,3,1( =i

E

=

= wave speed


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Ritz method Free vibration

Start with Hamiltons principle after I.B.P. in time:

Seek an approximate solution to u(x, t):

In time: harmonic function cos( t) ( = n)In space: X(x) = a 1 1(x)

where: a 1 = constant to be determined

1(x) = known function of position

( ) ( ) dt dxu x x

u E Auu A

t

t

t

L

= 21 0

0

1(x) must satisfy the following:Satisfy the homogeneous form of the EBC.

u(0) = 0 in this case.Be sufficiently differentiable as required by HP.


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0cos0 = = L p D ii orsolution)(trivialEither

= wave speed

E

=

For any time dependent problem:

=

+

=

,5,3,1 2sin

2cos

2sin),(

iii L

t i B

Lt i

A L xi

t xu

Free vibration:

[ ] ( ) ( )[ ]

=

++=1

sincos)sin()cos(),(i

iiiiiiii x p D x pC t p Bt p At xu

EBC:

NBC:

0)0( =u

00 =

=

== L x L x

x

u

x

u E A

General solution:

EBC [ ]

==+=

1

0)sin()cos(),0(i

iiiii t p Bt p AC t u

( )[ ]

== =+=

1

0)sin()cos(cosi

iiiiiii

L x t p Bt p A L p p D

xu

0=iC

25

23

2

oror= L p i

),5,3,1(2

== i L

i p i

NBC

Fixed-free bar General solution


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One-term Ritz approximation

Ritz estimate is higher than the exactOnly get one frequencyIf we pick a different basis/trial/approximation function 1, we would

get a different result.

)cos()cos()(

)cos()cos()(),()(

1

1111

t xt xu

t xat xat xu x x

=====

:eapproximatAlso:Pick

[ ] dt t dx E A x x Aat t

L)(cos)1)(1())((0 2

0

21

2

1

=Substituting:

222

23

2 333

L

E L

L E A L A =

==

L L RITZ

732.13 ==

L L EXACT 571.1

2 ==

1010

22 adx E Aadx x A L L =

Hence

[ ]{ } [ ]{ }( )a K aM =2 :formmatrixin

= L x EXACT

2sin1

x RITZ =1


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One-term Ritz approximation (cont.)

Both mode shape and natural frequency are exact.But all other functions we pick will never give us a frequency lower

than the exact.

= L x x

2sin)(1 :pickweif What

( ) ( )

dt t dx L x

L E A

L x

Aa

dt dxu x x

u E Auu At

t

t

L

t t

L

)(cos2

cos22

sin0

0

2

0

2

2

221

0

2

1

2

1

=

=Substituting:

EXACT RITZ L E

L

===

22Hence

( )( ) )cos(2sin)cos()(

)cos(2sin)cos()(),(

1

111

t L xt xu

t L xat xat xu

====

:eapproximatAlso

= L x

Ldxd

2cos

21


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Two-term Ritz approximation (cont.)

E == 22 and

==+ Laa La L 4526.00)3785.01713.0( 2212

=

0

0

)534()4(

)4()3(

2

1

532422

42232

a

a

L L L L

L L L L

leads to

Solving characteristic polynomial (for det[ ]=0 ) yields 2 frequencies:

L L RITZ RITZ 67.5)(5767.1)( 21 == and

Substitution of:

L L EXACT EXACT 7123.4)(5708.1)( 21 == and

Mode 1:

Let a 1 = 1:

L x x x X 21 4526.0)( =:1shapeMode

== Laa La L 38.10)10.5043.7( 22122

Mode 2:

L x x x X 22 38.1)( =:2shapeMode


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MODE SHAPES OF BEAM VIBRATION


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Modeshape plots forSFF

FandCCSSboundary

conditions


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Modeshape plots forSFF

FandCCSSboundary

conditions

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04 Energy Method

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