JN Reddy - 1 Lecture Notes on ENERGY PRINCIPLES AND ...mechanics.tamu.edu/wp-content/uploads/2017/03/04-07_Work...2017/03/04 · =05. x r r T R A A L x • d = diameter of the shaft

MEEN 618: ENERGY AND VARIATIONAL METHODS

Read: Chapter 4

WORK, ENERGY , AND VARIATIONAL CALCULUSCONTENTS

Work done External and internal work done Strain energy and strain energy

density Complementary strain energy Strain energy and comple-

mentary strain energy ofTrusses, Torsional members, and beams

The principle of minimum totalpotential energy

Virtual work done Elements of variational calculus

JN Reddy - 1 Lecture Notes on ENERGY PRINCIPLES AND VARIATIONAL METHODS

Work done Magnitude of the force multiplied by the magnitude of the displacement in the direction of the force:

F

u

F uW

B

●

●

●

A

FudF u

F uB

A

dW d

W d


Energy is the capacity to do work

002 21 1

02 200

,s se

es

dE F de F ke

E F de ke ke

●

Fe

m

k

0W FeF

F

0e e

Work done

F

0e

sF ke=

sF ke=

kk

m●

●

●

fsFsF

e0ede

sFNote that

0e e

dE Wde

(strain energy)


Work and Energy 4

EXTERNAL AND INTERNAL WORK IN A DEFORMABLE BODY

JN Reddy

Work done by external forces

( ) ( ) ( ) ( )EW d s s d

f x u x t u

t d

f d

d

d

In calculating the external work done, the applied (external) forces (or moments) are assumed to be independent of the displacements (or rotations) they cause in a body.


Work and Energy 5JN Reddy

Work done by internal forces (1D)

1dx1x

A11σ 11σ

1u 1u1

1 1 11 11

udu dx dxx∂

= =∂

d

0U

0*U

11

11 11 1 11 11 1

0 1

0 11 0 00

( )

( )

( ,)

A d dx d Adx

dU Adx

U dU U U d

STRAIN ENERGY DENSITY AND STRAIN ENERGY



Strain energy of a 3D solid

12 ( ) : ( )dU

x x

●

0 00

0

( ,)ij

ij

ij

i

ij ij

ij j

U d U U d

U d d

If the only energy stored in the body is the strain energy, we write

For a linear elastic body, we have

IU W

STRAIN ENERGY DENSITY AND STRAIN ENERGY



11

11 11 1 11 11 1

0 1

0 11 0 00

*

* * * *

( )

)

( )

( ,

A d dx d AdxdU Adx

U dU U U d

Complementary strain energy for 1D

COMPLEMENTARY STRAIN ENERGY DENSITY AND COMPLEMENTRY STRAIN ENERGY

d 0U

0*U

Complementary strain energy for 3D

0 00

* * *( ,)ij

ijij ijU d U U d


STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TRUSSES

0 0 01 1

0 0 01 01

0

( ) ( ) ( )

* *( ) *( ) *( )

, ( )

, ( )

i

i

i

i

i i

N Ni i i

i ii i

N

i

i i i

Ni i i

i ii i

U U d A LU U

U U d A LU

d

dU

A truss is a collection of uniaxial members, each of which can only carry axial force (compressive or tensile). The members are connected through pins that allow relative rotation.

The strain energy and complementary strain energy of a truss with N members each having ist own length, area of cross section, and modulus are


AN EXAMPLE

OB

C

vO

a u−

b v+

u

(a)(b)

O1

K = material constantAi = cross-sectional area of the ith

memberPin connections

B

C30 =

2b

aP

OP

2F

1F

2

10

0

, ,,

KK

1 1 12 2 2 2 22 2 21

2 2

21 1 1 2 1BO BOBO

( ) ( )a u v a u v au u ua a a a

Find the (1) strain energy and (2) complementary strain energy of the truss

1 12 2 2 2 2 22 2

22 2 2 2

12

2 2 2 2

21 1

3 31 2 14

CO COCO

( ) ( ) ( ) ( )a u b v a b u v bv aua b a b

bv au bv au v ua b a b a

(1) From Fig. (b), we have the following strains:

STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TRUSSES - AN EXAMPLE


(1) continued [note that the stress in member 1 is compressive and it is tensile in member 2; see part (b) to confirm this]

The strain energy densities of each member is

1 1

2 2

13 3 2

1 1 1 1 1 1 20 30 0

33 2

2 2 2 2 2 2 20 0 0

2 23 3

2 2 3 33 3 4

( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

,K K uU d K da

K K v uU d K da

1 2

1 2 1 20 0 1 1 0 2 2 0

313 22

1 23

2 4 3 33 43 3

( ) ( ) ( ) ( )

V VU U d U d A LU A L U

KA a u KA a v ua a

The total strain energy of the truss is




(2) From Fig. (a), we have

2 1 1 22 0 30 3 2sin , cos , , ,F F PF PF P F

1 1

2 2

21 331 1 1 1 10 2 3 20 0

1

22 332 2 2 2 20 2 2 3 20 0

2

1 33

1 1 83 3

( ) ( )

( ) ( )

( )*( ) ( ) ( ) ( ) ( )

( )*( ) ( ) ( ) ( ) ( )

,PU d dK K A K

PU d dK K A K

The complementary strain energy densities of each member are

3 31 2

0 1 1 0 2 2 2 2 2 21 2

1 3 3 163 3

* *( ) *( ) P a P aU U A L U A LA K A K

The total complementary strain energy of the truss is


STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF TORSION OF CIRCULAR SHAFTS

The shear stress in a shaft subjected to torque

( )xTrrJ

dr+

r

0 5.R d=

( )x rr

T

RA

A

Lx

•●●

d = diameter of the shaftJ = polar moment of inertia

2 20 0 0

2 4 22

x x

x x x x x xGU d G d G

The strain energy density and strain energy are

2 ,x xr TLL GJ

The complementary strain energy density is

2 22 22

20 0 0 0

1 12 2 2

/L d L

xG rU d G dr rd dx GJ dx

L L

20 0

2 22 22

0 0 0 0

122

1 1 1 12 2 2

*

/*

x

x x x

L d L

xv

U dG

Tr TU d dr rd dx dxG G J GJ


STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF STRAIGHT E-B BEAMS

The strain energy density and strain energy of the Euler-Bernoulli (E-B) beam are (linear elastic material)

222

0 20 0

2 222 2

0 2 20 0

2 2

12 2

( )xx xx

xx xx xx xx xx

L L

xx xxv A

E E du d wU d E x d zdx dx

E du d w du d wU U d z dAdx A D dxdx dx dx dx

2( ) , ( )xx xxA A

A E x dA EA D z E x dA EI

2 22 2

0 0 0

22 22

0 2 20 0

122 2 2 2

12

1*

* * , ( )

xx xz xx xzxx xx xz xz

L Ls

sA A

N VQU d dE G E A G Ib

f VN M AU U dAdx dx f Q z dAEA EI GA I b

The complementary strain energy density and complementary strain energy of the Euler-Bernoulli beam are


STRAIN ENERGY AND COMPLEMENTRY STRAIN ENERGY OF E-B BEAMS - AN EXAMPLE

A

B D

b

vP

hP

a

xvV P=

hN P= hP

vPvM P x=hP

vP

hP

vP

v hP b P a+

vP bhP

vPhP vP

vP b

x

x

Ph

vP

x

hV P=

vN P=

v hM P b P x= +

vP b

Determine the complementary strain energy of the (determinate) frame structure.

AN EXAMPLE:

BA BA

BA

DB DB DB

, ,

, ,

v h

v h

h v v

N P V P

M P b P x

N P V P M P

2 2 2 2 3 22

0

2 22

0

2 22 2 2 2 3

12 2 2 2 6 2

12 2 2

1 12 2 3 2

DB

BA

*

*

,b

h s v h v s vv

av s h

v h

v s hv v h h

P f P P b P b f P bU P x dxEA EI GA EA EI GA

P f PU P b P x dxEA EI GA

P a f P aP ab P P a b P aEA EI GA

All members have the same E, A, and I


TOTAL POTENTIAL ENERGY AND COMPLEMENTRY ENERGY

12

( ) ( ) :EU V d d d

u σ ε ε f u t u

The potential energy of a 3D solid ( , )I E EW U W V

For an E-B beam, they are

12

*( ) : ( )d d d

σ σ ε σ f u t u

The complementary energy of a 3D solid

22 2

20 0

12

( , ) ...L Ldu d wu w EA EI dx fu qw dx

dx dx

22 2

0 0

12

*( , ) ...L L

sxx xz

f VN M dx fu qw dxEA EI GA


THE PRINCIPLE OF MINIMUM TOTAL POTENTIAL ENERGY

Minimum nature of PE:

22 2

20 0

22 2

20 0

22 20 0

20

12

12

12

( , )L L

L L

L

du d wu w EA EI dx fu qw dxdx dx

du d wEA EI dx fu qw dxdx dx

du d wEA EI ddx dx

0 00

220 0

2 20

22 20 0

20

1 0 02

( , ) ( ) ( )

( , ) ( , )

L

L

L

x fu qw dx

du d wdu d wEA EI dxdx dx dx dx

du d wu w EA EI dxdx dx

u w u w

because u and w satisfy the equilibrium equations


VIRTUAL WORKVirtual displacements are those which satisfy the homogeneous form of the specified kinematic boundary conditions, but otherwise arbitrary.

,x u

Fq

w

,z w

00 0 0 ˆˆ( ) , ( ) , ˆ

x

dwu u w wdx

=

= = = − =

21 1 1 1

ˆˆˆ( ) , ( )u x u a x w x w x b x

2 2 2 31 1 1 1 2 1 2 2 1 2 , ; ,u a x w b x u a x a x w b x b x

Set of admissible displacements

Set of admissible virtual displacements

Virtual work done by actual forces in moving through their respective displacements is

( , )q F

0

( ) ( ) cos ( ) sin ( )L

W q s w s ds F u L F w L


COMPLEMENTARY VIRTUAL WORK

Virtual forces are those which satisfy the self-equilibrium conditions, but otherwise arbitrary.

0 0 0* ( ) ( ) ( ) ( ) ( )

( ) ( )

W P u L F w L P u F w M

P u L F w L

L

F( )q x

P

( )f x,x u

,z w

0

0 00 0

0

( )( )

x

uw

dwdx =

==

=

P

F

P

FM F L = ⋅

,z w

,x uL

The set is clearly is in self-equilibrium.

( , )P F

The virtual work doneby virtual forces in moving through actual Displacements is


VIRTUAL WORK DONE

0 ( ) : ( )I ij kl ijW U d d d

δ σ ε δε

Internal virtual work done in a 3D body

Internal complementary virtual work done

0* * ( )( ) :I ij kl ijW U d d d

δ ε σ δσ

tf u uEW d d

External virtual work done in a 3D body

External complementary virtual work done

u tf u*

uEW d d

* * *;I E I EW W W W W W Total virtual virtual work done


VIRTUAL WORK DONE FOR E-B BEAMS

2

20

L

Id u d wW N M dxdx dx

Internal virtual work done

Internal complementary virtual work done

0 1 1

02* ( ) ( ) ( )L

I xx xx xzW N M V dx

( ) ( ) q( ) ( ) VW pointforcesb d

Ea c

W f x u x dx x w x dx

External virtual work done in a 3D body

External complementary virtual work done* ( ) ( ) ( ) ( ) VW

b d

Ea c

W f x u x dx q x w x dx


COMPLEMENTARY VIRTUAL WORK DONE: AN EXAMPLE

a

b

A

B DhPδ

vPδ

v hP b P aδ δ+hPδ

vPδ

x

xhPδ

vPδ

hN Pδ δ=

vV Pδ δ=

vM P xδ δ=vPδ

hPδvP bδ

vN Pδ δ=hV Pδ δ= h vM P x P bδ δ δ= +

0

* *

* *

,

E

L

Is

E v h

N M VW U N M V dxEA EI K GA

W V v P u P

DB BA DB BA

DB BA

, , ,

,h v v h

v v h

N P N P V P V P

M P x M P b P x

0

3

0

22

2

3

12

12

DB DBDB DB DB DB

BA BABA BA BA BA

*

*

,

bDB

s

h v vh v

s

bBA

s

vv h v

v h

N M VU N M V dxEA EI K GA

P b P b P bP PEA EI K GA

N M VU N M V dxEA EI K GA

P a aP b a P b PEA EI

a aP b PEI

3

3h

hs

P a PK GA


FIRST VARIATION and VARIATIONAL SYMBOLThe delta operator used in conjunction with virtual quantities has special importance in variational calculus. The operator is called the variational operator because it is used to denote a variation (or change) in a given quantity.

u u, ; ( ) ( ) ( ) ( ) u v v u v v u

2 2 2

2

2

22 2

O

( , , ) ( , , )

( , , )

( ) ( )( ) ( , , )! !

( )

F F x u v u v F x u uF FF x u u v vu u

v F v v F F x u uu u u

F Fv vu u


0

0

( , )

( ) ( )( ) ( )

dF u v u vFd

F u v F u vu v u v

F F F Fv v u uu u u u

0lim F F FF v v

u u

F F F Fv v u uu u u u

FIRST VARIATION OF A FUNCTION OF A DEPENDENT VARIABLE

Define

Alternatively,


ANALOGY BETWEEN TOTAL DIFFERENTIALAND VARIATIONAL OPERTOR

F F FdF dx du dux u u

F Fx FF u ux u u

0

1 2 1 2 1 2 1 2 1 2

11 1 2 1 21 1 12

2 2

3 4

21( ) ( ) (, ( )

(

)

, ( ) ( ) () )n n

F F F F F F F F F FF F F F F F n F FF F

Analogy

Properties

0 0 0 0

(1)

(2)

( ) ( )( )( )a a a a

du dv d dv udx dx dx dx

udx vdx vdx udx


FIRST VARIATION OF A FUNCTIONAL

A functional is a mapping (or operator) from a vector spaceU into the real number field R. Thus, if (i.e. u is an element of U), then is a real number.

Fu U

( )F u

The First Variation of a Functional

A functional

( ) ( , , ) ,b

a

duI u F x u u dx udx

( ; ) ( , , )b b b

a a a

x bb

a x a

F FI u u F x u u dx F dx u u dxu u

F d F Fudx uu dx u u


FUNDAMENTAL LEMMA OF VARIATIONAL CALCULUS AND EULER EQUATIONS

0( ) ( ) ( ) ( )b

aG x x dx B a a

Lemma: If G is an integrable function and is arbitraryin a< x <b and is arbitrary, then the statement

( )x( )a

0 0( ) and ( )G x a x b B a implies that

which are called the Euler equations. If

0( ; )x bb

a x a

F d F FI u u udx uu dx u u

0 andI

0 0;x b

x a

F d F Fa x bu dx u u

is arbitraryin( , ) and at and , thenu a b x a x b


EULER EQUATIONS OF FUNCTIONAL IN 2D,INVOLVING TWO DEPENDENT VARIABLES

( , ) ( , , , , , , , ) , , .x x y y xuI u v F x y u v u v u v dxdy u etcx

Given the functional

find the Euler equations if . 0I

0 x y x yx y x y

x y x y

F F F F F Fu u u v v v dxdyu u u v v v

F F F F F Fu vu x u y u v x v y v

x y x yx y x y

dxdy

F F F Fn n u n n v dxdyu u v v

We have


EULER EQUATIONS OF FUNCTIONAL IN 2D,INVOLVING TWO DEPENDENT VARIABLES

If is arbitrary in and on , then the Euler Equations are as follows:

u

0

0

0

0

in

on

x y

x y

x yx y

x yx y

F F Fu x u y u

F F Fv x v y v

F Fn nu uF Fn nv v

See the textbook for examples (some will be discussed in the class)


Problems with Constraints-1

: Find the minimum of the function ( , ) with no constraints.

Necessary condition

0

Sin

P

ce and are arbitrary and independent, we have

0 a

robl

nd 0

em 1 F x y

F FdF dx dyx y

dx dy

F Fx y

¶ ¶= + =

¶ ¶

¶ ¶= =

¶ ¶

: Find the minimum of the function ( , ) subjected to the constraint ( , ) 0

Necessary condition

0

But and are not inde

Problem 2

pendent of each otherbecause of the

F x yG x y

F FdF dx dyx y

dx dy

=

¶ ¶= + =

¶ ¶

cannotconstraint, we set

0 and 0F Fx y

¶ ¶= =

¶ ¶



Introduce new function( , , ) ( , ) ( , )

where is the Lagrange multiplier. Set

0

Now , , and are independent of each other; so

Lagrange M

,we can se

ultiplier Method:L

L L LL

F x y F x y G x y

F F FdF dx dy dx y

dx dy d

l l

l

ll

l

= +

¶ ¶ ¶= + + =

¶ ¶ ¶

t

0, 0,

( , ) 0.

L L

L

F FF G F Gx x x y y y

F G x y

l l

l

¶ ¶¶ ¶ ¶ ¶= + = = + =

¶ ¶ ¶ ¶ ¶ ¶¶

= =¶


[ ]2Introduce new function

( , , ) ( , ) ( , )2

where is the penalty parameter. Set

0

Since and are independent of each other, we c

Penalty Functi

an set

on Method:

P

P PP

P

F x y F x y G x y

F FdF dx dyx y

dx dy

F F GGx x

gl

g

g

= +

¶ ¶= + =

¶ ¶

¶ ¶ ¶= +

¶ ¶ ¶0, 0.PF F GG

x y y yg

¶ ¶ ¶= = + =

¶ ¶ ¶

Comparison of the two methods

Lagrange Multiplier M

0,

et :

0

hod

.L LF FF G F Gx x x y y y

l l¶ ¶¶ ¶ ¶ ¶

= + = = + =¶ ¶ ¶ ¶ ¶ ¶

Penalty Function

0, 0.

Method:P PF FF G F GG Gx x x y y y

g g¶ ¶¶ ¶ ¶ ¶

= + = = + =¶ ¶ ¶ ¶ ¶ ¶

We find that in the penalty function method we can compute( , )G x yg gl g=

AN EXAMPLE: ALGEBRAIC PROBLEM

F (x, y) = 2x2 + y2 − 8x+ y + 1, G(x, y) ≡ 2x− y = 0Lagrange Multiplier Method

4x− 8 + 2λ = 0, 2y + 1− λ = 0, 2x− y = 0x = 0.5, y = 1.0, λ = 3.0

Penalty Function Method

4x− 8 + 2γ(2x− y) = 0, 2y + 1− γ(2x− y) = 0

xγ =8 + 3γ

4 + 6γ, yγ =

3γ − 12 + 3γ

Clearly, as γ →∞, we havelimγ→∞xγ = 0.5 = x, lim

γ→∞ yγ = 1.0 = y

Table: A comparison of the penalty solution with the exact forvarious values of the penalty parameter γ.

γ 1.0 10.0 25.0 50.0 100.0 1000.0

xγ 1.1 0.5938 0.5390 0.5197 0.5099 0.5010yγ 0.4 0.9063 0.9610 0.9803 0.9901 0.9990G(xγ , yγ) 1.8 0.2813 0.1169 0.0592 0.0298 0.0030λγ 1.8 2.8125 2.9221 2.9605 2.9801 2.9980

Penalty Function Method (Example -continued)

),( γγγ γλ yxG=

AN EXAMPLE: CONTINUUM PROBLEM-1

: Find the minimum of the functional ( , ) subjected to the constraint ( , ) 0.Minimize

subjected to(

Lagrange Multi

Pro

( , ) ( , ,

plie

ble

, , , )r Metho

, , )

0

0d

m 3

b

a

I u vG u v

G u

I u v F x u u v v dx

u v v

Id

ò ¢ ¢=

¢ ¢

=

=

=

¶=

0

ba

F F F Fu u v v dxu u v v

G G G GG u u v vu u v v

d d d d

d d d d d

òæ ö¶ ¶ ¶ ÷ç ¢ ¢+ + + ÷ç ÷ç ¢ ¢è ø¶ ¶ ¶ ¶

¶ ¶ ¶ ¶¢ ¢= = + + +¢ ¢¶ ¶ ¶ ¶

0 ba

ba

F F F Fu u v vu u v vG G G Gu u v v dxu u v v

F G d F G uu u dx u u

F G d F G vv v dx v v

d d d d

l d d d d

l l d

l l d

ò

ò

é¶ ¶ ¶ ¶¢ ¢ê= + + +¢ ¢ê ¶ ¶ ¶ ¶ë

ùæ ö¶ ¶ ¶ ¶ ÷ç ¢ ¢ ú+ + + + ÷ç ÷ç ú¢ ¢è ø¶ ¶ ¶ ¶ ûìé ùæ öï ¶ ¶ ¶ ¶ï ÷çê ú= + - +í ÷ç ÷çê ú¢ ¢ï è ø¶ ¶ ¶ ¶ë ûïî

üé ùæ ö ï¶ ¶ ¶ ¶ ï÷çê ú+ + - + ý÷ç ÷çê ú¢ ¢ ïè ø¶ ¶ ¶ ¶ë û ïþSelect such that the coefficient of is zero.Then the coefficient of is zero because isindependent.

dx

vu u

l dd d



( , , ) ( , ) ( , , , )( )

baL

ba

baL

ba

I u v I u v G u u v v dxF G dx

F G F GI u uu u u uF G F Gv v G dxv v v vF G d Fu u dx u

l ll

d l d l d

l d l d dl

l

ò

ò

ò

ò

¢ ¢º += +éæ ö æ ö¶ ¶ ¶ ¶÷ ÷ç ç ¢ê= + + +÷ ÷ç ç÷ ÷ç çê ¢ ¢è ø è ø¶ ¶ ¶ ¶ë

ùæ ö æ ö¶ ¶ ¶ ¶÷ ÷ç ç ¢ ú+ + + + +÷ ÷ç ç÷ ÷ç ç ú¢ ¢è ø è ø¶ ¶ ¶ ¶ ûìé¶ ¶ ¶ïïê= + - +í ¢ïê ¶ ¶ ¶ëïî

G uu

F F d F G v G dxv v dx v v

l d

l l d dl

ùæ ö¶ ÷ç ú÷ç ÷ç ú¢è ø¶ ûüùé æ ö ï¶ ¶ ¶ ¶ ï÷ç úê+ + - + + ý÷ç ÷ç ú¢ ¢ ïê è ø¶ ¶ ¶ ¶ë û ïþ


212( , ) (

Penalty, )

Function Meth( , , , )

odbaP

baP

ba

I u v I u v G u u v v dxF G F GI G u G uu u u uF G F GG v G v dxv v v vFu

g

d g d g d

g d g d

ò

ò

ò

é ù¢ ¢º + ë ûéæ ö æ ö¶ ¶ ¶ ¶÷ ÷ç ç ¢ê= + + +÷ ÷ç ç÷ ÷ç çê ¢ ¢è ø è ø¶ ¶ ¶ ¶ë

ùæ ö æ ö¶ ¶ ¶ ¶÷ ÷ç ç ¢ú+ + + +÷ ÷ç ç÷ ÷ç ç ú¢ ¢è ø è ø¶ ¶ ¶ ¶ ûìé¶ïïê= íïê ¶ëïî

( , , , )

G d F GG G uu dx u u

F G d F GG G v dxv v dx v v

G u u v vg g g g g

g g d

g g d

l g

ùæ ö¶ ¶ ¶ ÷ç ú+ - + ÷ç ÷ç ú¢ ¢è ø¶ ¶ ¶ ûüùé æ ö ï¶ ¶ ¶ ¶ ï÷ç úê+ + - + ý÷ç ÷ç ú¢ ¢ ïê è ø¶ ¶ ¶ ¶ë û ïþ

¢ ¢=

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JN Reddy - 1 Lecture Notes on ENERGY PRINCIPLES AND ...mechanics.tamu.edu/wp-content/uploads/2017/03/04-07_Work...2017/03/04 · =05. x r r T R A A L x • d = diameter of the shaft