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1 Chapter 1 Foundation of Solid Mechanics and Variational Methods
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Page 1: 1- Some Fundamental Concepts

1

Chapter 1

Foundation of Solid Mechanics

and Variational Methods

Page 2: 1- Some Fundamental Concepts

2

1- Some Fundamental Concepts 1-1- Physical Problems, Mathematical Models, Solutions 1-2- Continuum Mechanics 1-3- Boundary value problem solution 1-4- Approximate solution of a boundary value problem 2- Concepts of Stress, Strain, Constitutive Relations and Various Form of Energy 2-1- STRESS

2.1.1- Force Distributions 2.1.2- Stress 2.1.3- Equations of Motion

2.2- STRAIN

2.2.1- Physical Interpretation of Strain Terms 2.2.2- The Rotation Tensor 2.2.3- Compatibility Equations

2.3- HOOKE’S LAW 3- Boundary-Value Problems for Linear Elasticity 4- Energy Consideration 5- Principles of Virtual Work 6- The Method of Total Potential Energy 7- Differential Equations VS Functional for Continuous Systems 7.1- Formulation of Continuous Systems

7.1.1- Examples of Differential Approach 7.1.2- Example of Variational Approach 8- No. of Rigid Body Modes in a System 9- Sample Problems

Page 3: 1- Some Fundamental Concepts

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1- Some Fundamental Concepts 1-1- Physical Problems, Mathematical Models, Solutions The main objective of this section is describing the concepts of body and mathematical modeling. Procedures for formulation and solution of a Mathematical model of a physical problem are discussed. The following diagram shows a general view of the modeling from body to model to solution. Influence of environment Applied Mechanics State variables involve displacements, velocities, pressure, temperature, stress, strain, charge, position, etc. Influence of environment can be due to forces, temperature changes, etc. Properties are determined from laboratory testing. 1-2- Continuum Mechanics Things that we can perceive, see, hear, or build can be explained by using certain principles and laws of natures: conservation of mass, energy, linear

BODY STATE 1

State Variables at STATE 1

MA

TER

IAL

MODEL IF

FEASIBLE

State Variables at STATE 2

SOLUTION IF

POSSIBLE

BODY STATE 2

PRO

PER

TIES

Page 4: 1- Some Fundamental Concepts

4

and angular momenta, the laws of electromagnetic flux, and the concept of thermodynamic irreversibility. These are among the fundamental principles on which the subject of mechanics is based. The subject of continuum mechanics is based on the foregoing governing principles, which are independent of the internal constitution of material. However, the response of a system or a medium subjected to (external) forces can not be determined uniquely only with the governing field equations derived from the basic principles. The internal constitution of material plays an important role in the subject of continuum mechanics. Study of the response of a substance or body under external excitation constitutes the major endeavor in engineering. In engineering applications, the response behavior can be studied at macroscopic level without considering atomic and molecular structure. The subject of studying material behavior at the macroscopic level can be called continuum mechanics.

By invoking physical principles and constitutive behavior, we obtain equations governing the behavior of continuous system (a boundary value problem). A solution to a boundary value problem in continuum mechanics requires constitutive equations in addition to the governing field equations. The basic principles governing Newtonian mechanics are a) conservation of mass, b) conservation of momentum, c) conservation of moment of

P1

P2

Continuous body

External forces

Physical principles

Constitutive behavior

Solution procedure

Response

Page 5: 1- Some Fundamental Concepts

5

momentum (or angular momentum), d)conservation of energy, and e)laws of thermodynamics; these principles are considered to be valid for all materials irrespective of their internal constitution. Therefore, a unique solution to a boundary value problem in continuum mechanics cannot be obtained only with the application of governing field equations. Hence a unique determination of the response require additional consideration that account for the nature of different materials. The equations that model the behavior of a material are called ‘constitutive equations’ or ‘constitutive laws’ or ‘constitutive model’. 1-3- Boundary value problem solution A solution to a BVP can be obtained using different approaches. The following diagram shows a schematic view of the problem. ERROR?

ERROR? MATERIAL SCIENCE

ERROR? Model studies or direct experiment include checking of the approximate solution with the state variables in laboratory which involves dimensional analysis and similitude.

MODEL

BODY

APPROXIMATE SOLUTION OF THE MATH. MODEL

SOLUTION OF THE MATH. MODEL

OBSERVED SOULTION

Page 6: 1- Some Fundamental Concepts

6

1-4- Approximate solution of a boundary value problem A mathematical model (BVP) of a real-life problem is often difficult to obtain an exact solution. The finite element method (FEM) can be viewed as a method of finding approximate solutions for the BVP problems. Two approaches of Weighted Residual Method (WRM) and Energy Methods (EM) are used for finding approximate solution of BVP. A number of schemes are employed under the WRM, among which are collocation, subdomain, least squares, and Galerkin’s methods. Galerkin’s method has been the most commonly used residual method for finite element applications. This method is based on minimization of the residual left after an approximate or trial solution is substituted into the differential equation governing a problem. The EM procedures are based on the idea of finding consistent states of bodies or structures associated with stationary values of a scalar quantity assumed by the loaded bodies. In engineering, usually this quantity is a measure of energy or work. The process of finding stationary values of energy requires use of mathematical disciplines called variational calculus involving use of variational principles. For many problems, both approaches yield exactly the same results. The following diagram shows a schematic view of use of these two approaches. Primitives are those involve physical quantities associated with the state variables, e.g. Time, Length, Force, etc. Based on the primitives we establish the axioms for problem solving, i.e. try to obtain a solution for the assumed model. Here the primary objective is to make sure that the mathematical model represents the real body. Choice of axioms depends on the type of problem, form of geometry and the physical quantities involved. There are two kinds of axioms in applied mechanics.

i) Newtonian Axiom (Newton’s Axiom) It defines force as momentum change, vector forces act on each particle of the body and an equilibrium differential equation (or momentum balance) governs the solution throughout the body.

ii) Leibnitz Axiom (Work Axiom) It defines work as the effect of forces acting on the body from which a work function is obtained, e.g. potential energy, complementary energy, kinetic energy, etc. A solution is

Page 7: 1- Some Fundamental Concepts

7

obtained as an extremum problem, e.g., a minimum or a maximum or a saddle point problem.

CALCULUS OF VARIATION FUNCTIONAL ANALYSIS

PRIMITIVES

NEWTONIAN AXIOM

MOMENTUM

LEIBNITZ AXIOM

ENERGY OR WORK

CALCULUS

DIFF.EQUATIONS, BOUNDARY VALUE

PROBLEMS

VARIATIONAL PRINCIPLES, WORK FUNCTIONAL

GALERKIN

FINITE ELEMENT METHOD

EXTREMUM

Page 8: 1- Some Fundamental Concepts

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2- Concepts of Stress, Strain, Constitutive Relations and Various Form of Energy Ref : Energy and Finite Element Methods in Structural Mechanics By: I.H. Shames 1985 2-1- STRESS 2-1-1- Force Distributions In study of continuous media 2 classes of forces exist:

a. body force distribution It acts directly on the distribution of matter in the domain of specification.

⎩⎨⎧

volumeunitPermassunitPer

notationIndextxxxBor

notationVectortzyxB i ),,,(),,,( 321

b. Surface Tractions

In discussing a continuum there may be some boundary with force distributed on the boundary. The force is applied to such boundary directly from material outside the domain.

⎩⎨⎧

elementareathetonormalbenotneedareaunitPer

txxxTortzyxT i ),,,(),,,( 321

ν T

3x dAtonormalν 2x Superscript referring to the direction of the area element at the point of application of the 1x surface traction ),,,(),,,( 321

)()( txxxTortzyxT iνν

If the area element has the unit normal in the xj direction then We would express the traction vector on this element as: ji

ji

j TTorT =)()(

dA

Page 9: 1- Some Fundamental Concepts

9

2-1-2- STRESS 3x )3(T

)(

,,

,,

,,

333231)3(

232221)2(

131211)1(

StressT

T

T

T

ijij

i

i

i

τ

τττ

τττ

τττ

=⎪⎭

⎪⎬

=

=

=

)1(T )2(T 2x Traction forces on orthogonal faces 1x Sign convention: Normal stress directed outward from interface (+) tensile stress Normal stress directed toward surface (-) compressive stress

Shear stress is (+) if both stress itself and unit normal point in +ive coordinates directions or both points in –ive coordinate directions.

3x 2x 1x Knowing ijT for a set of axes, i.e. for three orthogonal interfaces at a point, we can determine a stress vector )(νT for an interface at the point having any direction whatever.

)( stressshearjitermsdiagonaloffstressnormal

ij

ii

≠==

ττ

Page 10: 1- Some Fundamental Concepts

10

Consider a point P in a continuum (any point in the domain) 3x 2x P Newton’s law for the mass center in 1x direction Cauchy’s Formula jijjjiiiiii TTTorTTTT ννννν

νν ==++=)(

332211)(

Knowing ijT we can get the traction vector for any interface at the point. This formula can be used to relate tractions on the boundary to stresses directly next to the boundary. Prove Cauchy’s Formula.

1x

1x

3x

A

Tv

B

C ν

P

νvectornormalhasABC

1xΔ

2xΔ 3xΔ

2x Tv

1 Tv

2

Tv3

Page 11: 1- Some Fundamental Concepts

11

2-1-3- Equations of Motion Consider an element of the body of mass dm at P Newton’s 2nd law:

Vdmfd &= ∫ ∫ ∫ ∫ ∫∫∫∫ =+

D DiiiS dVVdVBdAT ρυ &)(

jij vτ

Gauss’ Theorem Consider a continuous, differentiable nth order tensor field Tjk….over a volume V with its boundary surface defined by S. The Gauss’s Theorem in a generalized form is given by:

( ) dATx

TijkS

i

jk

V

ν...dV...

∫∫∫ ∫ ∫ =∂

or: ( ) dATT ijkSijk

V

ν...dV)( ,... ∫∫∫ ∫ ∫ =

where νi are the direction cosines of the unit outward normal. For Tjk….a zero order Tensor, say a scalar function φ,

dAiSiV

νϕϕ ∫∫∫ ∫ ∫ =dV,

or: −−

∫∫∫ ∫ ∫ =∇ AdS

V

ϕϕ dV

where the differential area dAAd−−

=ν . The above equation is generally referred to as Gauss’ Law.

∫ ∫ ∫ =−+ 0)( , dVVB iijij

D

ρτ &

D is arbitrary

Using moment of momentum equation will also result in:

kjjkHM ττ =⇒= 0&

Further investigations: - Transformation Equations for stress - Principal stresses (given a system of stresses for an orthogonal set of interfaces at a point , we can associate a stress vector for interfaces having any direction in space

jντ ijV

iT = Now: is there a direction ν such that stress vector is collinear withν ?

iijij VB &ρτ =+,

Page 12: 1- Some Fundamental Concepts

12

2-2- STRAIN Means of expressing the deformation of a body

Line segment in the undeformed geometry If body is given a rigid body motion each line segment in the body under goes no change in length. Change in length of line segments in the body, (or distance between points) can serve as a measure of deformation (change of shape and size) of the body.

iidxdxdsAB == 22)( distance between points When forces are applied, body will deform. It is convenient now to consider that the xi reference is labeled the iξ reference when considering deformed state:

3ξ *ξd *B *A 2ξ 1ξ

Deformation can be depicted by mapping of each point from coordinate ix to coordinate iξ . We can say then for a deformation:

),,( 321 xxxii ξξ = Since mapping is one-to-one:

),,( 321 ξξξii xx = :

1x

2x

3x

)( ixA

Bidx

Page 13: 1- Some Fundamental Concepts

13

We can express:

jj

ii d

xdx ξ

ξ ⎟⎟⎠

⎞⎜⎜⎝

∂∂

= jj

ii dx

xd ⎟

⎟⎠

⎞⎜⎜⎝

∂∂

ξ

kmk

i

m

iii dd

xxdxdxds ξξ

ξξ ∂∂

∂∂

==2

lkl

i

k

iii dxdx

xxdddsBA

∂∂

∂∂

===ξξ

ξξ2*** 2

jiijjiijj

k

i

k dxdxdxdxxx

dsds εδξξ

22*2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

∂∂

∂∂

=−

jiijjij

k

i

kij dddd

xxdsds ξξηξξ

ξξδ 22*

2

=⎟⎟⎠

⎞⎜⎜⎝

∂∂

∂∂

−=−

Strain terms: Green strain

⎟⎟⎠

⎞⎜⎜⎝

⎛−

∂∂

∂∂

= ijj

k

i

kij xx

δξξ

ε21

Lagrange coordinates ( ijε expressed as function of coordinates in the undeformed state)

Almansi measure of strain

⎟⎟⎠

⎞⎜⎜⎝

∂∂

∂∂

−=j

k

i

kijij

xxξξ

δη21

Eulerian coordintes ( ijη formulated as function of coordinate for deformed state) displacement field iu 33 ,ξx

iii xu −= ξ iu *A 22 ,ξx A

11 ,ξx

Page 14: 1- Some Fundamental Concepts

14

We may express iu as a function of Lagrange coordinate ix , in which case it expresses the displacement from the position ix in the undeformed state to the deformed position iξ . On the other hands iu can equally well be expressed in terms of iξ , the Eulerian coordinates; in which case it expresses the displacement that must have taken place to get to the position iξ from some undeformed configuration.

⎪⎪⎭

⎪⎪⎬

+∂∂

=∂∂

∂∂

−=∂∂

ijj

i

j

i

j

iij

j

i

xu

x

ux

δξ

ξδ

ξ

substitution in previous equation for ijε

⎟⎟⎠

⎞⎜⎜⎝

∂∂

∂∂

+∂

∂+

∂∂

=j

k

i

k

i

j

j

iij x

uxu

xu

xu

21ε Initial undeformed geometry

Indicate what must occur during a given deformation.

⎟⎟⎠

⎞⎜⎜⎝

∂∂

∂∂

−∂

∂+

∂∂

=j

k

i

k

i

j

j

iij

uuuuξξξξ

η21 Deformed instantaneous

geometry of body Indicate what must have occurred to reach this geometry from an earlier undeformed state. So far no restriction on magnitude of deformation, Infinitesimal strain:

11 <<∂∂

<<∂∂

j

i

j

i uxu

ξ

( )ji

jijjj

iji

j

ji

i

xJu

uxJx

xJxJ

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂−=⎥

⎤⎢⎣

⎡−

∂∂

∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂∂

=∂

∂ξ

δξξξξ

)(

for infinitesimal strain i

juξ∂∂

can be dropped

no need to distinguish between Eulerian and lograngian coordinates in expressing strains

( )ijjii

j

j

iijij uu

xu

xu

,,21

21

+=⎟⎟⎠

⎞⎜⎜⎝

∂+

∂∂

== ηε

ii x∂∂

=∂∂ξ

Page 15: 1- Some Fundamental Concepts

15

xyyx

xyx

xx xu

yu

xu

γεε21

21

=⎟⎟⎟

⎜⎜⎜

∂+

∂∂

=∂∂

= =ijγ engineering shear strain

yzzy

yzy

yy yu

zu

yu

γεε21

21

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂

∂=

∂=

nzzu

xxz

zz xu

zu

zu

γεε51

21

=⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

=∂∂

=

2-2-1- Physical interpretation of strain terms A small rectangular parallelepiped at P. We have also placed a Cartesian reference at P. Imagine the body has some deformation:

z Let’s focus on line yPQ Δ=

Projection of **QP in the y direction yQP )( **

PyQyy uuyQP )()()( ** −+Δ= Taylor series for Qyu )( in terms of Pyu )( :

pyP

ypy uy

yu

uy )(...)( −⎥⎥⎦

⎢⎢⎣

⎡+Δ⎟⎟

⎞⎜⎜⎝

⎛∂

∂++Δ=

...+Δ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+Δ= yyu

yP

y

Net y component of elongation of segment PQ

...)( ** +Δ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=Δ− yyu

yQPP

yy

y

x

G zΔ

N

yΔ Q P xΔ

R

*P *Q

Qu

P yΔ Q

Pu

Deformed geometry

*

*

QQPP

Page 16: 1- Some Fundamental Concepts

16

Where with coalescence of QP & , we may drop subscript P :

yyyy

yu

yyQP

ε=∂∂

Δ−)( **

=∴ yyε change in length in the y direction per uint original length of vanishingly small line segment originally in the y direction.

Now consider PR of xΔ and PQ of yΔ

We are interested in the projection of **RP and **QP on to the xy place (on to the place the line segments were in undeformed state)

yu

tgyy

yyu

tg xp

x

∂∂

==⇒+Δ

+Δ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

= θθδ

θ 2

...

0→Δy

similarly : x

uy

∂=β

x

y

z

S

R

Q PxΔ

*P

*Q *R

Line segments in initial and deformed geometry

P

x

y

R

Q

Pxu )( ...)( +Δ⎟⎟

⎞⎜⎜⎝

⎛∂∂

+

⎪⎪⎭

⎪⎪⎬

yyuu

p

xPx

of Q in the x direction

*Q

*P

*R

θ

Pxu )( = displacement component of

P in x direction

= displacement component

2yδ = second order increment for a small deformation

β

44 844 762yy δ+Δ

Page 17: 1- Some Fundamental Concepts

17

xyxyyx

xu

yu γεβθ ==

∂∂

+∂∂

=+ 2

ijγ = change from a right angle of vanishingly small line segments originally in the ji& directions at a point

Now effect of strain on a infinitesimal rect. Parallelepiped in the undeformed geometry.

Zero shear stress means side will remain orthogonal on deformation. However position and orientation of the element may change as length of the sides and volume.

Existence of shear stress means sides may lose they mutual perpendicularity, (parallelograms instead of rectangles) ∴Size of the rectangular parallelepiped is changed by normal strain while the basic shape is changed by shear strain.

Prove : ⎟⎠⎞

⎜⎝⎛ =Δ

iiVV ε

2-2-2- The Rotation Tensor Previously, we considered stretching of a line element to generate ijε and then used the deformation of a vanishingly small rectangular parallelepiped to give physical interpretation to the component of strain tensor. We now introduce rotation tensor. This time rather than considering just the stretch of a vanishingly small line element, we consider the complete mutual relative motion of the end points of line element. (include rotation as well as stretching)

xd

zd

yd

Page 18: 1- Some Fundamental Concepts

18

Consider PN the relative movement of end points can be given by using disp. field.

Pj

pjPPN Ux

xuUUU −

⎥⎥

⎢⎢

⎡+Δ⎟

⎟⎠

⎞⎜⎜⎝

∂∂

+=− ...

Expand NU as a Taylor series about P

In limit ⎟⎟⎠

⎞⎜⎜⎝

∂∂

=∂∂

=→Δ jj

iij

jj dx

xu

dunotationindexdxxudux 0

Thus the relative movement idu between the two adjacent point idx a part is

44344214434421)(

21)(

21

,,,,, ijjiijjiji uuuuu −++=

ijε ijw (rotation tensor skew-symetric)

Assume Rigid body motion PGPQ , ; same xδφ

( ) ( )

y

uyyu

uSin

PzP

zPz

x ′Δ

−⎥⎥⎦

⎢⎢⎣

⎡+Δ⎟⎟

⎞⎜⎜⎝

⎛∂∂

+

=

...δφ

( )yyy ′Δ=Δ→Δ 0

yuz

x ∂∂

=δφ

( )

z

zz

uuu

SinPG P

yPyPy

x ′Δ⎥⎥⎦

⎢⎢⎣

⎡+Δ⎟⎟

⎞⎜⎜⎝

⎛∂

∂+−

=→

...)(

: δφ

z

u yx ∂

∂−=δφ

y

z

Q yΔ P

G

xδφ

*P

y′Δ xδφ

zΔ ( ) ...+Δ⎟⎟

⎞⎜⎜⎝

⎛∂∂

+

⎪⎪⎭

⎪⎪⎬

yyuu

P

zPz

}

( ) ...+Δ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+ zzu

uP

yPy

Page 19: 1- Some Fundamental Concepts

19

233221 ww

zu

yu yz

x −==⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂−

∂∂

=→ δφ

For other 2 components:

311321 ww

xu

zu zx

y −==⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

=δφ

122121 ww

yu

xu xy

z −==⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂

∂=δφ

For rigid body movement, the nonzero components of the rotation tensor give the infinitesimal rotation components of the element. What does ijw represent when the rectangular parallelepiped is undergoing a movement including deformation of the element and not just BR. rotation? Each line segment in the rectangular volume has its own angle of rotation and we can show that ijw for such situation gives the average rotation components of all the line segments in the body. However we shall term the component of ijw the rigid body rotation components.

From experiment ijε portion of equation ijw related to the stress ijτ

Further investigation: Transformation equation for strain.

2-2-3- Compatibility equations Strain-displacement relations

( )ijjiij uu ,,21

+=ε (*)

If sui ' are know, ijε can be obtained. The inverse problem of finding the displacement field from a strain field is not so simple.

Three functions iu must be determined by integral of 6 partial differential equations (*) to ensure single-valued continuous solution iu , we must impose certain restriction of ijε

can not set forth any ijε . to expect unique solution, the following equations are to be satisfied:

Page 20: 1- Some Fundamental Concepts

20

⎪⎪⎩

⎪⎪⎨

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

−∂∂

=∂∂

∂∂∂

=∂∂

+∂∂

)2(2

)2(

2

2

2

2

2

2

equationsmorezyxxzy

equationsmoreyxxy

xyxzyzxx

xyyyxx

γγγε

γεε

2.3 HOOKE’S Law Linear elastic behavior klijklij C ετ = generalized Hook law tensororderCtensororder th

ijklnd

ijij 42, ⇒ετ jiklijklijklij CCsymmetricCsymmetric =⇒τ ijlkijklkl CCsymmetric =⇒ε It can be shown that Cijkl =Cklij ( Using Energy Concept It can be proved.) Thus, starting with 81 terms for Cijkl (=34), we may show, using the three aforementioned symmetry relations for Cijkl , that only 21 of these terms are independent. We will assume now that the material is homogeneous (which has same composition throughout) so we may consider Cijkl to be a set of constants for a given reference. For an isotropic material, in which the material properties at a point are not dependent on direction, we have:

ijeeijij Gεεδλτ 2+=

This is the general form of Hooke’s law giving stress components in terms of strain components for isotropic materials. The constant λ and G are the so-called Lame constants. It can be seen that as a result of isotropy the number of independent elastic moduli has been reduced fron 21 to 2. The inverse of Hooke’s law yielding: ijkkijij T

EEδντνε −

+=

1

equationstotal 6

Page 21: 1- Some Fundamental Concepts

21

E and ν are Young’s modulus and the poisson ratio stemming from one-dimensional test data. ( )[ ] xyxyxyzzyyxxxx GEE

ττνεττντε2111

=+

=+−=

( )[ ]zzxxyyyy E

ττντε +−=1

GGGE

GVVEV

VEG

++

=+

=−+

=+

=λλ

λλνλ )23(

)(2)21()1()1(2

3- Boundary-value problems for linear elasticity

The complete system of equations for linear elasticity for homogeneous, isotropic solid includes the equilibrium equations:

)3(0, equationsB ijij =+τ

The stress-strain law: )6(2 equationsG ijijllij ⊕+= εδελτ Strain displacement relations:

)6(*)(21

,, equationsvu ijjiij +=ε

We have 15 equations and 15 unknowns. When explicit use of the displacement field is not made, we must be sure that the compatibility equations are satisfied. It must be understood that iB and )(v

iT have resultants that satisfy equilibrium equations for the body as dictated by Rigid body mechanics. In this regard that iB and )(v

iT must be statically compatible.

Page 22: 1- Some Fundamental Concepts

22

We may pose three classes of boundary values problems:

1st kind B.V. problem: determine the distribution of stresses and displacements in the interior of the body under a given body force distribution and a given surface traction over the boundary.

2nd kind B.V. problem: determine the distribution of stresses and displacements in the interior of the body under the action of a given body force distribution and a prescribed displacement distribution over the entire boundary.

Mixed B.V. problem: determine the distribution of stresses and displacements in the interior of the body under the action of a given body force distribution with a given traction distribution over part of the boundary )( 1s and a prescribed displacement distribution over the remaining part of the boundary 2s .

Note : on the surfaces where the )(v

iT are prescribed, Cauchy’s formula jij

vi TT ν=)( must apply.

1st kind: convenient to express basic equations in terms of stresses. To do this:

ijkkijij EE

δτντνε −+

=1 in compatibility euations

Using equilibrium equations, we can arrive at the Beltrami-Michell system of equations:

)(11

1,,

2,

2ijjiijijij BBKK +−=∇

+−

++∇ δ

νν

ντ

where kkK τ=

The solution of these equations, subject to the satisfaction of Cauch’s formula on the boundary for simply connected domains, will lead to a set of stress components that both satisfy the equilibrium equations and are derivable from a single-valued continuous displacement field.

substitutide

Page 23: 1- Some Fundamental Concepts

23

2nd kind: Substitute equations * and ☺ in the equilibrium equations to yield differential equations with the displacement field as the dependent variable. Then we get Navier equations of elasticity: 0)( ,

2 =+++∇ ijiji BuGuG λ

For dynamic conditions we need only employ the following equations in place of the equilibrium equations.

iijij uB &ρτ =+, The results are the addition of the term tup && on the right side of the above equations. If the above equation can be solved in conjunction with the prescribed displacements on the surface and if the resulting solution is singled –valued and continuous the problem may be considered solved. Solution for mixed BV problems will be investigated using different techniques introduced partly in this notes such as variational approach.

Page 24: 1- Some Fundamental Concepts

24

4- Energy consideration

We have described the stress tensor arising from equilibrium consideration and the strain tensor from kinematics considerations. These tensors are related to each other by laws that are called constitutive laws. In general such relations include temperature and time as other variables. In addition, they often require knowledge of the history of deformation lending to the instantaneous condition of interest in order to properly relate stress and strain. We assume that the constitutive laws relate stress and strain directly and uniquely. That is,

.).(),........,,( 331211 LClawveConstitutiijij εεεττ =

Consider an infinitesimal rectangular element under the action of normal stresses only.

The displacements of faces 1 and 2 in the x direction are xu as ,dxxuu x

x ∂∂

+

Increment of mechanical work done by the stresses on the element during deformations is:

+⎟⎠⎞

⎜⎝⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

++− dzdydxxuuddxdzdydu x

xx

xxxxxxx

τττ

10 ⟨⟨⎟⎠

⎞⎜⎝

⎛∂∂

+ kdxu

kuddddB xx

xzyxx

Canceling terms and deleting the higher order expressions:

zyxxxxxx

xx dddduBx

Txu

d ⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ +∂∂

+⎟⎠

⎞⎜⎝

⎛∂∂

τ

zzτ dyyyy

yy ∂∂

τ

xxτ dxxxx

xx ∂∂

+ττ

yyτ

dzzzz

zz ∂∂

+ττ

(2)

(1)

0=mequilibriu

Page 25: 1- Some Fundamental Concepts

25

dVddddxu

d xxxxzyxx

xx εττ =⎟⎠⎞

⎜⎝⎛∂∂

Similar expression for y and z directions can be obtained. Thus for normal stresses on an element, the incremental of mechanical work for isotropic materials is:

( ) )( stressesnormaldVddd zzzzyyyyxxxx ετετετ ++

=w Mechanical work per unit volume zzzzyyyyxxxx ddddw ετετετ ++=

Now consider the case of pure shear:

The mechanical increment of work

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+ dydxx

ddzdxdyy

xyxy

xyxy

γβγ

ττ

( ) 1,,0 ≤≤⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂++ kdykdx

xddydzdxB xy

xyx ηβγ

ηγ

dVddxdydzd xyxyxyxy ετγτ 2=→

Thus, for pure shear stresses on all faces we get the following result for increments of mechanical work:

Mechanical work increment per unit volume at a point for a general state of stress is:

ijijddw ετ= (valid only for infinitesimal deformation)

Now integrating from 0 to some strain level ijε we get:

∫ === ij udW ijij

εετ

0 strain energy density function which is the mechanical

work performed on an element per unit volume at a point during a deformation.

xyτ xyγ dy

dx

dz

dxxxy

xy ∂∂

γ

dyyxy

xy ∂∂

τ

( ) dVddd yzyzxzxzxyxy ετετετ ++2

Page 26: 1- Some Fundamental Concepts

26

ijij

ijijuddu τε

ετ =∂∂

⇒=

( u is point function, integral independent of path then perfect differential )

Total strain energy dVdUv

ijijij

∫ ∫ ∫ ∫ ⎥⎦⎤

⎢⎣⎡=

εετ

0

)(20 yzyzxzxzxyxyzzzzyyyyxxxxijij dddddddw ij ετετετετετετετε

+++++== ∫∫

∫ ∫ ∫=v

dVwU

Examples of Calculating Total strain energy

Uniaxial stress dVdU xx

vxx

ij )(0

εσε

∫ ∫ ∫ ∫=

E

dE

dw xx

xxx

2

21 σσσεσ === ∫∫

Ew x

2

21 σ

=

dVE

U x2

21 σ

∫ ∫ ∫=

Pure bending

dyE

dw xxxxx

ij2

0 21 σεσ

ε== ∫

∫∫∫ ⎟⎠⎞

⎜⎝⎛= dV

IMY

EU

2

21

dxI

ME

dAYdxIM

E ∫∫∫∫ ==2

22

2

21

21

2xxEIWEIM ==

ρ ∫=

l

oxx dxWIEU 2

21

Shear stress

GG

ddW ijij

ijijij

2

21 ττ

τετ === ∫∫

xyxy Gγτ =

dVGU ij /21 2∫= τ

EddE σεεσ ==

IMY

εσ E=

x

y

z

Page 27: 1- Some Fundamental Concepts

27

5- Principles of virtual work Particle Mechanics : Virtual work is defined as the work done on a particle by all the forces acting on the particle as this particle is given a small hypothetical displacement , a virtual displacement , which is consistent with the constraints present. The applied forces are kept constant during the virtual displacement. Deformable body: Same as particle with specifying a continuous

displacement field with small deformation and constraint, applied force kept constant. We conveniently denote a virtual displacement by employing the variational operationδ .

In general situation we would have as load possibilities a body force distribution iB through out the body as well as surface tractions )(v

iT over part of the boundary , 1S , of the body. Over the remaining part of the boundary, 2S , we have prescribed the displacement field iu ,in which case, to avoid violating the constraints we must be sure that 0=iuδ on 2S . Virtual work for such a general solution would be:

∫ ∫ ∫ ∫ ∫+=v s

iv

iiivirt dsuTdvuBW δδδ )(

iB and )(viT must not depend on iuδ in computation of virtWδ . We can expand

the surface integral to cover entire surface since 20 Sonui =δ , thus 21 SSS +=

We now develop the principle of virtual work for a deformable body

dsudvuBW ijv s

ijiivirt δυτδδ ∫ ∫ ∫ ∫ ∫+=

∫ ∫ ∫ ∫ ∫ ∫+=v v

jiijii dVudVuB ,)( δτδ

∫ ∫ ∫ ∫ ∫ ∫++=v v

jiijijiji dVudVuB ,, )()( δτδτ

We now introduce a kinematically compatible strain field variation ijδε (it is because it is formed directly from the displacement field variation). ijijijijjiji WWuu δδεεδδδ +=+== )()()( ,,

Page 28: 1- Some Fundamental Concepts

28

Because of skew symmetry of the rotation tensor and the symmetry of the stress tensor, 0=ijij wδτ ijijjiij u δετδτ =,)(

∫ ∫ ∫∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ++=+⇒v

ijijv s v

iijijiv

iii dVdVuBdsuTdVuB δετδτδδ )( ,)(

We now impose the condition that we have static equilibrium. This means in the above equation that:

1. External load iB and )(viT are such that there is overall equilibrium

for the body from the point of view rigid body mechanics we say that iB and )(v

iT are statically compatible. 2. At any point in the body 0, =+ ijij BT

∫ ∫ ∫∫ ∫ ∫ ∫ ∫ =+⇒v

ijijv s

iv

iii dVTdsuTdVuB δεδδ )(

This is the principle of v.w. for a deformable body We can say that necessary condition for equilibrium is that for any kinematically compatible deformation field ),( ijiu δεδ , the external v.w. with statically compatible body forces and surface traction , must equal the internal v.w. This is sufficient for equilibrium. Another more useful interpretation of the principle of v.w. is as follows. The necessary requirements for equilibrium of a particular stress field ijτ are that :

1. iB and )(viT are statically compatible

2. The particular stress field ijτ satisfies the v.w. equilibrium for any kinematically compatible, admissible, deformation field.

Note: the mathematical relation between a deformation field and a stress field is independent of any constitutive law and applies to all materials within the limitations of small deformation.

external virtual work internal virtual work

Page 29: 1- Some Fundamental Concepts

29

We have shown that the satisfaction of the principle of v.w. is a necessary relation between the external loads and stresses in a body in equilibrium.

We can also show that satisfaction of the principle of v.w. is sufficient to satisfy the equilibrium requirement of a body.

Assume v.w. equilibrium is valid

dVu

dVu

dVuu

dVv

ijij

v

jiij

v v

ijjiijijij ∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫ ∫ ∫ ∫ +=⎟⎟

⎞⎜⎜⎝

⎛ +=

2)(

2)(

2,,,, δ

τδ

τδτδετ

∫ ∫ ∫=v

jiij dVu ,)(δτ

We made use of symmetry of ijτ . We can write the last expression as follows:

∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫−=

v v vijijjiijjiij dVudVudVu δτδτδτ ,,, )()(

Using divergence theorem

∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫−=

v s vijijjiijjiij dVudsudVu δτυδτδτ ,,)(

∫ ∫ ∫ ∫ ∫−=

1

,s v

ijijjiij dVudsu δτυδτ

We have made use of the fact that 20 Sonui =δ Now substituting these results for the last integral in the principle of virtual work, that was found previously and is as the following:

∫ ∫ ∫∫ ∫ ∫ ∫ ∫ =+v

ijijv s

iv

iii dVdsuTdVuB δετδδ )(

Results in the followings:

0)()( )(, =−++∫ ∫ ∫ ∫ ∫ dsuTdVuB

V sijij

viiijij δυτδτ

Since iuδ is arbitrary, we must conclude VinBiiij 0, =+τ By the same reasoning 1

)( SonT jijv

i υτ= We have generated Newton’s law for equilibrium at any point inside the body and Cauchy’s formula, which ensure equilibrium at the boundary. ⇒ Satisfaction of principle of v.w. is both necessary and sufficient for equilibrium.

Page 30: 1- Some Fundamental Concepts

30

6- The Method of Total Potential Energy Note: Calculus of Variations has to be reviewed.

We now develop from the virtual work idea, the concept of total potential energy which applies to elastic body (not necessary linear elastic):

∫ ∫ ∫∫ ∫ ∫ ∫ ∫ =+v

ijijv s

iv

iii dVdsuTdVuB δετδδ )(

uuuddu ij

ijij

ijijij

1δδεε

τε

ετ =∂∂

⇒=∂∂

⇒=

UudVdVudsuTdVuBv

vv si

viii

111)( δδδδδ ===+ ∫∫∫∫ ∫ ∫∫ ∫ ∫ ∫ ∫

Note: iuδ is virtual displacement field. A priori not related to stress field We define potential energy V of applied load as a functional of displacement field ui

dsuTdVuBVv s

iv

iii∫ ∫ ∫ ∫ ∫−−= )( Bi and Tiν prescribed

dsuuu

TdVuuu

BVv s

jj

ivij

j

ii∫ ∫ ∫ ∫ ∫ ∂

∂−

∂∂

−= δδδ )(1

ijj

i

uu

δ=∂∂

dsuTdVuBVv s

iv

iii∫ ∫ ∫ ∫ ∫−−= δδδ )(1

0)(1 =+VUδ (Principle of total potential energy)

VU +=π (Total Potential Energy)

dsuTdVuBUv s

iv

iii∫ ∫ ∫ ∫ ∫−−= )(π

0)(1 =πδ Principle of total potential energy

Interpretation: The necessary requirements for equilibrium of a particular stress field ijτ :

1. Bi and Tiν are statically compatible

Page 31: 1- Some Fundamental Concepts

31

2. The deformation field, to which the field ijτ is related through a constitutive law for elastic behavior, extremize TPE with respect to all other kinematically compatible admissible deformation fields.

Extremization of the TPE w.r.t admissible deformation fields is necessary for equilibrium to exist between the forces and the stresses in a body. Just in the method of virtual work, we can show it to be a sufficient condition for equilibrium. We can show that TPE is actually a local minimum for the equilibrium configuration under loads Bi and Ti

ν compared with the TPE corresponding to neighboring admissible configurations with the same Bi and Ti

ν.

Examine the difference between TPE of equilibrium state and an admissible neighboring state ii uu δ+ and ijij δεε + show that the second variation of TPE is positive.

The total potential energy theorem states that’ of all the admissible fields which satisfy compatibility and essential boundary conditions, the actual one which satisfies equilibrium and stress BC’s provide a minimum to π.

The total potential (π) is also called the functional of the problem.

Assume that in the functional (π) the highest derivative of a state variable (wrt a space coordinates) is of order m, i.e. the operator contains at most mth order derivatives. Such a problem we call Cm-1variational problem. Considering the boundary of the problem, we can identify two classes on bc’s: Essential bc’s (geometric): correspond to prescribed displacement and rotations. The order of the derivatives in the essential bc’s is in a Cm-1 Problem, at most m-1.

Natural boundary conditions (force bc’s): corresponds to prescribed boundary force and momentums. The highest derivative in this bc’s are of order m to 2m-1. By invoking the stationary of the functional a problem, the problem governing differential equation and natural and essential bc’s can be derived. In Cm-1 variational problem, the order of the highest derivative presented in the problem governing differential equation is 2m. Therefore, integration by parts is employed m times. Effect of bc’s are included implicitly in π.

Page 32: 1- Some Fundamental Concepts

32

7- Differential Equations VS functional for continuous systems

We can get a solution to a partial differential equation which is satisfied at each point in the body and also satisfy a set of boundary conditions. A solution obtained, maybe for displacements or stresses, etc. A functional represents a number (scalar) and for naturally occurring functional, it may represent work, energy or power or etc. In some instances, it may not represent any physical quantity. At extremum, it yields a solution to the differential equation (equilibrium or momentum balance or heat balance, etc.). ∫= )()(.. functionaldxyfIge Existence of a functional and solution obtained as extremum of this functional also helps to determine as to what kind of equilibrium is achieved. This leads to theory of stability, for example, if it is a minimum at extremum then the solution obtained is stable! To go from differential equation to variational problem we need to know operational algebra or calculus (functional analysis) and to go from variational problem to differential equation we need to know the calculus of variations. 7.1 Formulation of continuous systems We consider a typical differential element with the objective of obtaining differential equations that express the element equilibrium requirements, constitutive relations, and element interconnectivity requirements. These differential equations must hold throughout the domain of the system and before the solution can be obtained they must be supplemented by boundary conditions and, in dynamic analysis, also by initial conditions. Two different approaches can be followed to generate the system governing differential equations.

1. The direct method (differential equations) 2. The variational method

Page 33: 1- Some Fundamental Concepts

33

The direct method In this method, we establish the equilibrium and constitutive requirements of typical differential elements in terms of state variables. These considerations lead to a system of differential equations in the state variables. In general the equations must be supplemented by additional differential equations that impose appropriate constraints on the state variables in order that all compatibility requiremvents be satisfied. Finally to complete the formulation all the boundary conditions and in a dynamic analysis the initial conditions are stated in differential formulation for a continous system, a differential element with objective of obtaining differential equation that express element equilibrium is found. This differential equation must hold through the domain of the system. The D.E must be supplemented by B.C.’S and dynamic analysis, initial condition example. 7.1.1 Examples of differential approach Example 1- Beam element u a) Differential Ele.

b) Equilibrium in vertical direct

dxtumpdxdx

xV

2

2

∂∂

=+∂∂

or equating sum of the moment about the left hand face to zero

x

),( txu ),( txp

P(x)

M

V

dxxMM∂∂

+

dxxVV∂∂

+

2

2

tum

∂∂

02

2

=+∂∂

−∂∂ p

tum

xV

Page 34: 1- Some Fundamental Concepts

34

022 2

2

=+∂∂

−−∂∂

++⎟⎠⎞

⎜⎝⎛

∂∂

+ MdxxMMdxdx

tumdxpdxdxdx

xVV

*0=∂∂

+xMV

now xu∂∂

=θ also from elementary beam theory x

EIM∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

−=⇒∂∂

=⇒ 2

2*

2

2

xuEI

xV

xuEIMθ

⇒ Transverse vibrate of beam For a unique solution we must specify bc’s Note of the elementary beam theory Plain remain plain

Ptum

xuEI

x=

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

2

2

2

2

2

2

lxuxu====

@00@0

02

2

=∂∂

=xuEIM

00

==

VM

free

Simply supported.

0

0

=∂∂

=

=

xu

u

θ

xdEIM∂

Δ′ Δ0)(

0

0

max =−−

=⇒

∫dA

cy

ydA

F

σ

IMy

dAycyM

=

⎟⎠⎞

⎜⎝⎛−= ∫

σ

σ max

dy

y

θd

p

xΔ ρθε

ε

1/)/(

'(

==Δ

Δ=

ΔΔ==

=

dxd

xye

yxe

yEIM

sHookEI

My

⇒ Mρ/121 xd

=

Page 35: 1- Some Fundamental Concepts

35

Example 2- Dam’s Reservoir Example 3- Rod subjected to step load E young modulus p ρ mass density A cross section 1) Differential element

Equilibrium udxAAAdxx

&&ρσσσ =−⎟⎠⎞

⎜⎝⎛

∂∂

+ )(

Constitutive relation

xuE∂∂

Combining equations: ρEC

tu

Cxu

=∂∂

=∂∂

2

2

22

2 1

)3(bc

)4(bc

)2(bc

)1(bc pc

p &&22 1

=∇

x

),( txu Ro

Ro

R(t)

t

xA1σ

dx

Rod

dxxAdxx +∂

∂+ )( 1

σσ

Page 36: 1- Some Fundamental Concepts

36

b.c’s 0),( =tou RotlxuEA =∂∂ ),(

Initial load 0),( =oxu 0),( =∂∂ ox

tu

Page 37: 1- Some Fundamental Concepts

37

7-1-2 Examples of Variational approach Example1. Beam dxxwxpdxwEIw

LL

xx )()(2

)(00

2 ∫∫ −=Π

dxwPdxwwEI L

xxxx

Lδδδ ∫∫ −=Π

002

2

dxwwEIwEIw x

L

xxxL

xxx δδ ∫−=⇒00

dxwwEIwEIwwEIw

L

xxxxL

xxxL

xxx δδδ ∫+−=000

: ⇒−+−=Π ∫

Lxxx

Lxxx

L

xxxx wEIwwEIwdxwPEIw000

)( δδδδ 00 =−≤≤ PEIwlx xxxx bc’s 0

0=L

xxx wEIw δ 0

0=L

xxx wEIW δ In general on the bc’s at Lxorx ==0

444 3444 21loadingofenergypotential

43421

energystrainwU )(

L

11

2CC

mm =

=−

xwwbcessential ,⇒

1 1

Page 38: 1- Some Fundamental Concepts

38

Example2: 2

0

22

0 21

2)(

21

L

L

xxx

LwkdxwPdxwEI +−=Π ∫∫

* dxwEIU xx

L 2

021∫=

LLPw ′−== δδ ∫= )( wtoduelengthinchangenodsL

)0(110

22

22 LdxwLdxdxdwdxdwds

L′→′+=⇒⎟

⎠⎞

⎜⎝⎛+=+= ∫

.211

0

2 dispsmallfordxwLL

∫′

⎟⎠⎞

⎜⎝⎛ ′+≈

dxwLdxwdxL

LLL

∫∫∫′′′′+′=′+≈

0

2

0

2

0 21

21

LLsmallisdxwLL

L′≈′=′−= ∫

′δδ

0

2

21

w

x k

p

Buckling of the column

prove it *

dx dw ds

x

wL′

δ

L

11

2CC

mm =

=−

xwwcbess ,.. ⇒

dxwPWL

∫ ′=0

2

2

Page 39: 1- Some Fundamental Concepts

39

Example3. Rod subjected to STEP load

RudxfudxuEA LBL

x

L−−=Π ∫∫ 0

2

0 21

00

0=−−==Π ∫ ∫ RudxfudxuuEA L

BL

xx δδδδ RudxfudxuuAEuuEA L

BL

xx

LLx δδδδ −−−= ∫∫ 000

( ) [ ] 0000

=−−++−==∫ uuEAuRuEAdxufuEA

xxLLx

L Bxx δδδ

arbitraryisuδ 0=+⇒ B

xx fuEA 0=== Lx uorRuEALx δ

0

000 0

====

xuoruEAx x δ

problemiationalCm var1 0=

xRo

),( txu

lengthunitforforcebodf

tLuutuusbc

B

L

=

===

),(),0(0' 0

Page 40: 1- Some Fundamental Concepts

40

Example 4. 2-D Variational Principle

∫∫ΓΩ

Γ−Ω⎥⎦⎤

⎢⎣⎡ −+=

q

dqdQkkJ yx φφφφ 22

22

onlypositionsoffunctionsareQK & )(0 boundarytheofparton φδφ Γ= qonspecifiedq Γ [ ] ∫∫

ΓΩ

Γ−Ω−+=q

dqdQkkJ yyxx δφφδδφφδφφδ

Note: xx xx

)()( δφδφφδφδ =∂∂

=∂∂

=

Integrate by part the first two terms Ω−== ∫∫ ∫ ∫

ΩΩ Ω Γ

dkdykdxdykdxdyk xxxxxxx δφφδφφδφφδφφ )()(

∫ ∫ ∫

Ω Ω Γ

+Ω−= dxkdkdxdyk yyyyy δφφδφφδφφ )(

[ ] [ ] Γ−Γ++Ω++−= ∫∫∫

ΓΓΩ

dqdvkvkdQkkJq

yyxxyyxx δφδφφφδφφφδ )()(

x

y

φΓ

*

*

*

Γdvx

Γdvy

dy

dx

θ

yv v

xv θ

θθ

sincos

Γ=Γ=

ddxddy

44 344 21

nk∂∂φ

Page 41: 1- Some Fundamental Concepts

41

∫ ∫ ∫ ∫Γ Γ Γ Γ Γ=

Γ+Γ+Γ=Γ++=Γ

q fon

dk qfn

φφδφ

φδφφ

0

Ω

−=⇒ Jδ ∫ ∫Γ Γ

=Γ+Γ−+Ωq f

dkdqkd nn 0)( δφφδφφδφ

Arbitrary=δφ ⇒equEuler. Ω= in0

fn

qn

kor

Lonqk

Γ==

=−

00

0

φδφ

φ

Heat Conduction : If Qk & constant =∇ φ2 const. Poisson’s equation If 0&1 == Qk 02 =∇ φ Laplace equation Also other form of equations such as Torsion problem (Poisson’s equation) or Irrotational flow (Laplace equation), seepage problem or flow through porous media are examples of the above equations. Example5. Transient 2-D Heat Equivalent steady state variational principle for any time t :

{ } Γ⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ −++Γ+⎥⎦

⎤⎢⎣⎡ +−+= ∫∫∫

ΓΓΩ

dqTdqTdxdyCQkkTJCA

CCAyyxx φφφαφφφφφφφ2

221)( 22 &

φ is a function of x,y and time t

t∂∂

=φφ&

0)( =φδ J at any time t

t∂∂φ must be considered fixed in the calculus of variation formulation

Page 42: 1- Some Fundamental Concepts

42

[ ]dACCQkkTJ yyyxxx∫Ω

++−+= δφφφφδδφδφφδφφφδ &&)(

[ ] Γ−++Γ+ ∫∫

ΓΓ

dqTdqTCA

CCA δφφαδφφαδφδφ

∫ ∫ ∫ ∫

Γ Ω Γ Ω

Ω−Γ+Ω−Γ= dkdvkdkdvkT yyyyyyxxxxxx δφφδφφδφφδφφ )()(

∫ ∫ ∫ ∫

Γ Γ

=Γ−+′+Γ+Ω+Ω−A C

dqTdqTdCdQ CCA 0)( δφφαφαδφδφφδφ &

0)()( =−++ φφφ &CQkk yyyxxx [ ] Ω+−−−= ∫

Ω

dCQkkTT yyyxxx δφφφφφδ &)()()(

∫ ∫

Γ Γ

Γ+Γ++A

dqTdvkvkT Ayyyxxx δφδφφφ )(

Γ

=Γ−++C

dqT CC 0)( δφφφα

On BB φφ =Γ 0=++Γ AyyyxxxA qvkvk φφ 0)( =−+++Γ CCyyyxxxC qvkvk φφαφφ

lyrespectiveandonareaunitperinputheatspecifiedqq

volumeunitperinputHeatQ

directionyintyConductiviThermalkdirectionxintyConductiviThermalketempreatur

CA

CA

y

x

ΓΓ=

=

===

,

φ

CBA Γ+Γ+Γ=Γ

y

x

Ω *

**

ThickncssTCBA

=Γ+Γ+Γ=Γ

Page 43: 1- Some Fundamental Concepts

43

* Problem: For a transient 2-D heat flow, the equivalent steady state variational principle at time t can be written as:

( )[ ] Γ⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ −++Γ++−+= ∫∫∫

ΓΓΩ

dqdqdxdyCQkkJCA

CCAyyxx φφφαφφφφφφφ2

221)( 22 &

qqQkk

ttyx Ayx ,,),,(∫ ∂

∂==

φφφ &

You are asked to find the Euler equation and the appropriate boundary condition

1- assumption about displacement field 2- sometime assumption about constitutive law 3- variational process as it relates to the T.P.E 4- it gives us proper equations of equilibrium and proper BC’S

(Certain internal constraints due to displacement assumptions) 8. No. of Rigid body modes in a system In a variational form we try to find the strain energy U. The rigid body motions are not accompanied by change in strain energy. The No. of non contributing terms (from the displacement model) to the strain energy are the No. of rigid body modes. “Bathe” P.173 If the structure is not supported, there will be a number of linearly independent vectors, qVUU ,.......,, 21 for which the expression i

Ti UKU is equal

to zero, i.e. zero strain energy is stored in the system when iU is the displacement vector. Such vector iU is said to represent a rigid body mode of the system.

Ω *

**

CBA Γ+Γ+Γ=Γ

Page 44: 1- Some Fundamental Concepts

44

9. Sample Problems 1- For the beam shown, write down the variational principle (Potential Energy) which also includes the boundary actions. Find out the Euler Lagrange equation and the associated boundary conditions.

21 , KK are translational spring constnts 21 , θθ KK are rotational spring constnts

Are there any rigid body modes present? 2- Figure 2 show a system of beam-column with transverse and tangential springs. a) Write down the functional (Total Potential Energy) for the system. Perform the first variation (Fig 2). b) Derive the Euler-Lagrange equations and the associated bc’s c) How many rigid body modes do exist? d) Perform the second variation Π2δ to show weather the problem is a minimum or maximum. Note: )(2 Π=Π δδδ

uy,

xu,

)(xp

V

1P 1M

1k x1θk uniform EI

constant 0P unit length

2k

2θk 2M

2P

NK

ux,

uy,

)(xq

L TK


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