5-1

MEG 741 Energy and Variational Methods in Mechanics I

Brendan J. O’Toole, Ph.D.Associate Professor of Mechanical Engineering

Howard R. Hughes College of EngineeringUniversity of Nevada Las Vegas

TBE B-122(702) 895 - 3885bj@me.unlv.edu

Chapter 2: Principles of Virtual Work: Integral Form of the Basic Equations

5-2

Virtual Work and Variational Methods

Energy principles provide an alternative to Newtonian methodsas a means of deriving and solving governing equations.

5-3

Work & Energy• Applied forces, moments, and torques do work

on a structure, e.g., W=F.L

• This work changes the potential energy state of the internal forces– This is referred to as internal energy or strain energy

• The strain energy can be defined in terms of stresses & strains

• Applied Work (external) = Change of internal energy

direction in the force ofcomponent theis where , dsFs sdsFdW =

5-4

Work and Energy (continued)

ii WU −=

The strain energy in an elastic body, Ui, is equal to and has the opposite sign of the work done by the internal forces, Wi.

( )dzFdyFdWd zy +−=−=Π

Let’s say that Π is the potential energy stored in a 2-D structure and it is going to do some work, W. And d Π is the exact differential of Π.

where Fy and Fz are the forces in the y and z directions.

• The work done by internal energy is conservative.

• The change in energy when a force moves from point A to point B is independent of the path taken.

• If a force starts at point A and moves around and ends up back at point A, no net energy has been stored in the structure.

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Work and Potential Energy of Internal ForcesExternal forces, N, are applied at the ends of the rod shown below.

What is the work done during the deformation of this bar?

NN

L

( )[ ]

AN

dxddxdu

x

xxxx

σ

εεεε

=

=−+=stress of in terms written becan force internal The

iselement aldifferenti theoflength in change The

First examine the work done by the internal forces, N, over the differential element of length, dx.

dx

N N

dx

N N

x1 2

xx dεε +xε

The net work over the differential element is N•u :

AdxddxAdworknet xxxx εσεσ ==

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Internal Work due to Axial LoadThe internal work over the length of the bar is found by integrating the previous expression: NN

L

dx

N N

xxxi

xi

x

L

i

L

xxi

L

xxi

VVEW

LAEW

dxAEW

AdxdEW

AdxdW

x

x

εσε

ε

ε

εε

εσε

ε

212

21

221

2

0 21

0 0

0 0

−=−=

−=

−=

−=

−=

∫∫ ∫∫ ∫

Since σx = Eεx

If εx is constant over L

The capacity of the internal forces to do work is called strain energy, Ui. The strain energy is considered a positive quantity. The work done by the internal forces is negative:

xxxi

xxxi

VVEW

VVEU

εσε

εσε

212

21

212

21

−=−=

==ii WU −=

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General LoadingThe internal work over the length of the bar is found by integrating the previous expression:

dVddd

dddW

Vyzyzxzxzxyxy

zzyyxxi yzxzxy

zyx

∫∫∫∫∫∫∫

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

++

+++= γγγ

εεε

γτγτγτ

εσεσεσ

000

000

∫∫∫∫

==

−=−=

VVi

VVi

dVdVU

dVdVW

or

Eεεσ

EεεσTT

TT

ε

ε

21

21

21

21

The strain energy density is defined as the strain energy per unit volume.

Eεεσ TT ε21

21 ==

=

o

io

UV

UU

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Strain Energy Density and Complementary Strain Energy Density

xσ

oU

*oU

xε

Uo ≡ strain energy density

Uo ≡ complementary strain energy density*

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Example 2.1: Find the Strain Energy in a Beam with an Internal Axial Force and Bending Moment

∫ ∫

∫

∫

⎥⎦

⎤⎢⎣

⎡++=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +=

=+=

=

L

Ai

Vi

xxx

V xxi

dAdxI

zMAINMz

AN

EU

dVI

MzAN

IMz

AN

EU

EIMz

AN

dxU

0 2

22

2

2 2121

121

,

21

σεσ

εσ

∫∫

=

=

A

A

IdAz

zdA2

0

thatrecall

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

L

i dxEI

MEANU

0

22

22The Strain Energy for Beam Problems:

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In Some Cases, the Internal Work is not Simply Equal to a Potential Function

If the internal work depends on the loading history (and not just the end states) then it is not equal to a potential function.

Example: A slender bar has an applied axial stress and a temperature change x σx, ∆T

xσ

xεA B

C D

oε ( ) finalxε

xσ

xεA B

C D

( ) finalxεoε

Case I: Temperature Change Applied First

• ∆T applied with no constraint (stress free expansion from A to B)

• Applied Stresses cause εx between B and C

• Arrive at final stress and strain at D

Case II: Stress Applied First

• Applied Stresses cause εx between A and C

• ∆T applied under constant stress from C to D

• Arrive at same final stress and strain at D

The internal work is equal to the area under the stress strain curves. It is not the same in both cases.

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Work of the Applied Forces (External Work)

x ΝConsider an axial bar with an applied force

Fe

Fu

e

u

e

uNW

kudu

kuW

Nku

duNW

F

F

21

21

that assume2

0

0

=

==

=

=

∫

∫

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General Equations for External Work

∫∫

∫ ∫∫ ∫

+=

+=

ViV

Siie

V

u

iVS

u

iie

dVupdsupW

dVdupdsdupW

i

p

i

i

p

i

21

21

00

responselinear a with materialsFor

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Some Fundamentals of Variational Calculus• A review of variational calculus is presented in Appendix I of

the textbook and summarized in the following pages.• Variational calculus involves finding extrema of functionals.

– A function is usually an expression involving independent variables; for example: f = f(x)

– A functional is a function of a function, or a function of dependent variables; for example, f = f(u,u´) , where u = g(x) and u´ = du/dx

• Variational calculus is used to derive energy concepts such as the principal of virtual work.

• The principle of virtual work requires an understanding of terms like – virtual displacements– virtual forces, and – virtual work

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General Variational Calculus Problem

• Find the function u(x) such that

– In other words, find the u(x) that makes Π an extreme value• Necessary Conditions

– u = u(x) in the interval a ≤ x ≤ b– F is a known function (like strain energy density for example)– Usually require that u(x) be twice differentiable with respect

to x, (u´ and u″ exist)– And F be twice differentiable with respect to x, u, and u´

( ) ) (where stationary rendered is ,, dxdu

b

audxuuxF =′′=Π ∫

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Virtual Displacements and the Variational Operator, δ

• Let u(x) be a family of neighboring paths of the extremizing path u(x)

uu ˆ,( )xu

( )xu

x

uδdu

dx

ax = bx =

)()()(ˆ xuxuxu δ+=Where u(x) is the extremizing path

and δu is a small variation away from the extremizing path. Note that:

endpoints at the 0ˆ

=−=

uuuu

δδ

• Notice that there is a big difference between du and δu• δ is called the variational operator and it has similar properties

as the differential operator d

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Properties of the Variational Operator δ• The variational operator behaves like the differential operator in many ways.• For example, assume F = F(u,u´), where u´=du/dx, Then:

( )( )

( )

( ) ( ) 11

11

22

2121

2

1

212121

2121

FFnF

FFFFF

FF

FFFFFFFFFF

uuFu

uFF

nn δδδ

δδδ

δδδδδδ

δδδ

−=

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛

+=+=+

′′∂

∂+

∂∂

=

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Functionals and δ• A functional is an integral expression whose integrand(s) are

functions of dependent variables and their derivatives.

• The first variation of a functional can be calculated as follows

( ) ( ) dxdu

b

audxuuxFu =′′=Π ∫ where,,,

∫

∫∫

⎟⎠⎞

⎜⎝⎛ ′

′∂∂

+∂∂

=Π

==Π

b

a

b

a

b

a

dxuuFu

uF

FdxFdx

δδδ

δδδ

therefore

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Extrema of Functionals• Suppose we wish to find the extemum of Π, where

• The necessary condition for the functional to have an extremum is that its first variation be zero, or

( ) ( ) ( ) ( ) ba

b

aubuuaudxuuxFu ==′=Π ∫ , ,,,

( ) 0

0

0

=⎟⎠⎞

⎜⎝⎛

′∂∂

+∂∂

=Π

=⎟⎠⎞

⎜⎝⎛ ′

′∂∂

+∂∂

=Π

=Π

∫

∫b

a

b

a

dxdx

uduFu

uF

dxuuFu

uF

δδδ

δδδ

δ

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Extrema of Functionals• Use integration by parts to evaluate the second term in the

previous integral. The general procedure for integration by parts:

[ ]( )

( )

0

becomes 0

then, , , Define

=⎥⎦⎤

⎢⎣⎡

′∂∂

+⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

′∂∂

−∂∂

=⎟⎠⎞

⎜⎝⎛

′∂∂

+∂∂

=∴==

−=

∫

∫

∫∫′∂

∂

b

a

b

a

b

a

dxud

uF

b

a

ba

b

a

uuFudx

uF

dxd

uF

dxdx

uduFu

uF

utdts

tdsstsdt

δδ

δδ

δδ

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0=⎥⎦⎤

⎢⎣⎡

′∂∂

+⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

′∂∂

−∂∂

∫b

a

b

au

uFudx

uF

dxd

uF δδ

Extrema of Functionals (continued)

• This equation can be simplified– The variation of u, (δu), at points a and b must be zero since u is

specified at those points; therefore:

bxauF

dxd

uF

u

udxuF

dxd

uF

uuF

b

a

b

a

<<=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

′∂∂

−∂∂

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

′∂∂

−∂∂

=⎥⎦⎤

⎢⎣⎡

′∂∂

∫

,0

then value,zero)-(nonarbitrary an is if and

0

leaving ,0

δ

δ

δEuler Lagrange

EquationThis is a necessary

condition for a function u(x) to extremize the functional Π.

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Example: Find the Path of Minimum Length Between Points 1 and 2

22 dydxds +=y

x

dy

dx

ax = bx =

ds

1

2

( )

dxy

dx

dydx

dydxds

dxdy

dxdx

dxdx

∫∫∫

∫∫

′+=Π

+=Π

+=Π

+==Π

2

1

2

2

1

2

1

22

2

1

222

1

1

2

2

2

2

21 yF ′+=In this case:

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Minimum Path Example (Continued)

Apply the Euler Lagrange Equation:

0=⎟⎟⎠

⎞⎜⎜⎝

⎛′∂

∂−

∂∂

=⎟⎠⎞

⎜⎝⎛

′∂∂

−∂∂

yF

dxd

yF

uF

dxd

uFLet u = y for this problem:

( ) ( )

01

0

:BecomesEquation LagrangeEuler The1

2121

0

2

2

2 21

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

′+

′−

′+

′=

′∂∂

′′+=′∂

∂

=∂∂

−

yy

dxd

yy

yF

yyyF

yF

),(1 2

yyFFyF′=

′+=

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Minimum Path Example (Completed)

Solve the Euler Lagrange Equation:

( )

BAxyAy

AC

Cy

Cyy

Cyy

Cy

yy

ydxd

+==′

=−

=′

′+=′

′+=′

=′+

′∴=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

′+

′

2

2

222

2

22

1

1

1

1 ,0

1

Apply boundary conditions at points 1 and 2 to find the

unknown constants A and B.

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Next Class

• Discuss some Problems from HW 1• Discuss Semester Projects• Define Virtual Work Using Variational Calculus