Infinitesimal Elasticity, Variational Principles, Castigliano Theorems

Gauge Institute H. Vic Dannon and Vadim Komkov

Infinitesimal Elasticity, Variational Principles, and

Castigliano Theorems [email protected] 1918-2008

H. Vic Dannon and Vadim Komkov

February, 2012

Abstract The Equilibrium Equations of Elasticity are

never derived as Euler’s Variational Equations. A

Hamiltonian is not identified in Elasticity, and Hamilton’s

Equations are absent from Elasticity Theory.

Elasticity Theory includes Variational Principles known as

Castigliano Theorems. Do the Castigliano Theorems of

Elasticity Theory stem from Variational Principles that are

unknown in Mechanics?

We show that Castigliano’s Theorems are equivalent to

Euler’s Variational Equations, and to Hamilton’s Variational

Equations.

To that end we had to set up the theory of elasticity with

infinitesimals, and give proofs to fundamental theorems of

elasticity we found unproven.

1


The incomplete analysis that left Elasticity Theorems

unproven, was not helped by concern about tensors in

several Elasticity texts. For the sake of clarity, we avoided

proofs with tensor indices here.

Keywords: Infinitesimal, Variation, Euler’s Variational

Equation, Variational Calculus, Variational Derivative,

Legendre Sufficient Condition, Elasticity, Stress, Strain,

Energy, Castigliano, Beam Bending, Shear.

2000 Mathematics Subject Classification; 49-02;26E30;

26E15; 26E20; 26A12; 46S20; 97I40;

Physics and Astronomy Classification Scheme

;45.20.Jj; 4510Db;

2


Contents

Introduction 1. Hyper-real Line

2. Strain and Stress on a Bar

3. Strain Energy of a Bar

4. Variational principles of a Bar

5. Strain on a Plate

6. Stress on a Plate

7. Strain Energy of a Plate

8. Variational principles on a Plate

9. Strain on a Box

10. Stress on a Box

11. Strain Energy of a Box

12. Variational Principles on a Box

13. Castigliano Theorems

14. Erroneous Derivations of Castigliano Theorems

15. Castigliano Theorems and the Principle of Virtual Work

16. Castigliano Theorems and the Hamiltonian

17. Castigliano and Beam Bending without Shear

References

3


Introduction

0.1 Virtuals, and Infinitesimals

Elasticity Theory uses terms such as “Virtual Work”,

“Virtual displacement”, “Virtual Strain”, and “Virtual

Stress” to describe Infinitesimal functions that in Calculus

are known as Variations, and are not virtual.

The statements about infinitesimals are unclear, if not plain

wrong. For instance, in [Chou, p.154] we have,

“ Notice that in

12

( )x x x xδ σ ε σ δε= ,

no factor of 12

is present, since the stress is constant

during the virtual displacement.”

In fact, the stress

x xEσ ε= varies according to

x xEδσ δε= , and we have

1 1 12 2 2

( )

x

x x x x x x

Eδε

δ σ ε σ δε ε δσ= +

1 12 2

x

x x x xE

σ

σ δε ε δε= +

4


. x xσ δε=

0.2 Infinitesimals and Limits

In terms of the Calculus of Limits, Stress is defined by

0limA

PAΔ →

ΔΔ

,

where is the force on the area . PΔ AΔ

It is difficult to imagine what this limit amounts to when

. 0AΔ →

It is easier to comprehend the quotient

dPdA

,

where the infinitesimal dA is always positive, and never

vanishes.

To avoid the fogginess of limits in Calculus, we have to use

infinitesimals.

0.3 The Strain Energy

The reason why the infinitesimal strain energy density for a

bar is

x xd dμ σ ε= ,

is beyond all texts. The text by [Timoshenko] supplies only a

5


plausibility argument for it.

It is even less obvious why the infinitesimal strain energy

density of a plate is

2xx xx xy xy yy yyd d d dμ σ ε σ ε σ ε= + + .

Then, without any explanation, texts conclude that

1 12 2xx xx xy xy yy yyμ σ ε σ ε σ ε= + + ,

leaving unexplained the fact that the strain in the x

direction on a plate depends on the stress in the y direction.

0.4 Variational Principles of Elasticity

The Principle of Virtual Work, and the Principle of Minimal

Potential Energy at Equilibrium, are stand-alones,

disconnected from mainstream Variational Principles of

Mechanics.

The Principle of Virtual Work is equivalent to Elasticity’s

Equilibrium equations, which are hinted to be Euler’s

Variational Equations. But are never derived from Euler’s

Variational Theory.

The Principle of Minimal Potential Energy can be

established only with the Calculus of Variations. But

Variations are alien to Elasticity Theory.

6


Weinstock [Weinstock, pp. 199-260] derives the equations of

Equilibrium in order to discuss transverse vibrations of bars,

and plates. But once the kinematics is eliminated from his

derivation, the Elasticity of structures remains, where

Hooke’s law is replaced by ( )f x k= x , and a body is replaced

by its center of mass.

0.5 Castigliano Theorems, and Hamilton's Equations

A Hamiltonian is not identified in Elasticity, and Hamilton’s

Equations are absent from Elasticity Theory.

Elasticity Theory of Structures includes Variational

Principles known as Castigliano Theorems.

Do the Castigliano Theorems stem from Variational

Principles that are unknown in Mechanics?

We show that Castigliano’s Theorems are equivalent to

Euler’s Variational Equations, and to Hamilton’s Variational

Equations.

7


1.

Hyper-real Line Each real number α can be represented by a Cauchy

sequence of rational numbers, so that . 1 2 3( , , ,...)r r r nr α→

The constant sequence ( is a constant hyper-real. , , ,...)α α α

In [Dan1] we established that,

1. Any totally ordered set of positive, monotonically

decreasing to zero sequences constitutes a

family of infinitesimal hyper-reals.

1 2 3( , , ,...)ι ι ι

2. The infinitesimals are smaller than any real number,

yet strictly greater than zero.

3. Their reciprocals (1 2 3

1 1 1, , ,...ι ι ι ) are the infinite hyper-

reals.

4. The infinite hyper-reals are greater than any real

number, yet strictly smaller than infinity.

5. The infinite hyper-reals with negative signs are

smaller than any real number, yet strictly greater than

. −∞

6. The sum of a real number with an infinitesimal is a

8


non-constant hyper-real.

7. The Hyper-reals are the totality of constant hyper-

reals, a family of infinitesimals, a family of

infinitesimals with negative sign, a family of infinite

hyper-reals, a family of infinite hyper-reals with

negative sign, and non-constant hyper-reals.

8. The hyper-reals are totally ordered, and aligned along

a line: the Hyper-real Line.

9. That line includes the real numbers separated by the

non-constant hyper-reals. Each real number is the

center of an interval of hyper-reals, that includes no

other real number.

10. In particular, zero is separated from any positive

real by the infinitesimals, and from any negative real

by the infinitesimals with negative signs, . dx−

11. Zero is not an infinitesimal, because zero is not

strictly greater than zero.

12. We do not add infinity to the hyper-real line.

13. The infinitesimals, the infinitesimals with

negative signs, the infinite hyper-reals, and the infinite

hyper-reals with negative signs are semi-groups with

9


respect to addition. Neither set includes zero.

14. The hyper-real line is embedded in , and is

not homeomorphic to the real line. There is no bi-

continuous one-one mapping from the hyper-real onto

the real line.

∞

15. In particular, there are no points on the real line

that can be assigned uniquely to the infinitesimal

hyper-reals, or to the infinite hyper-reals, or to the non-

constant hyper-reals.

16. No neighbourhood of a hyper-real is

homeomorphic to an ball. Therefore, the hyper-

real line is not a manifold.

n

17. The hyper-real line is totally ordered like a line,

but it is not spanned by one element, and it is not one-

dimensional.

10


2.

Strain, and Stress on a Bar

2.1 Strain-Displacement Relation for a Bar

2 2( ') ( ) 2dx dx dxdu− ≈

2 '( )

x

dx u x dxε

= .

The x -gradient of the displacement is the Axial Strain

elongation per unit length inx direction. ,x xuε ≡ =

Proof: Elastic Deformation of a bar, transforms

the point x to , 'x

and the infinitesimally close x , to . dx+ ' 'x dx+

'x x→ ,

' 'x dx x dx+ → + .

The Displacement of x to 'x

( ) 'u x x x≡ − , satisfies

'du dx dx≡ − . Therefore,

11


2 2' ' ( ) ( )dx dx dxdx dx du dx− = + −

2dxdu dudu= +

Keeping first order in du , we have,

2dxdu≈

. ,

2( ) '( )( )

xu

dx u x dx=

And is approximated by . 2( ) (dx du dx+ − 2) 2( ) ( )xdx dxε

2.2 The Axial Stress xσ

The Axial Stress is the force stretching the bar per

unit area of the bar’s cross section.

( )x xσ

2.3 Hook’s Law, Stress-Strain Relation in a Bar

The strain and the stress satisfy Hooke’s Law

x xEσ ε≡ ,

where the Young Modulus is a constant that

characterizes the Elastic material.

E

2.4 Body Force on a Bar ( )xf x

12


A body force is an exterior force ( )xf x , such as gravitational,

or magnetic, per unit length.

2.5 The Equilibrium Equation of a Bar

, ( ) 0x x xf xσ + = .

Proof: If a body force ( )xf x per unit length applies to the bar,

the static equilibrium over dx satisfies the equation

,

( ) ( ) ( ) 0

x x

x x x

dx

x dx x f x dx

σ

σ σ+ − + = .

Hence, the Equilibrium Equation is

, ( ) 0x x xf xσ + = .

13


3.

Strain Energy of a Bar

3.1 Infinitesimal Strain Energy of a Bar is

x xd dμ σ ε= ,

xx

ddμ

σε

= .

Proof: The infinitesimal work done by the axial stress ,

along the displacement is

( )x xσ

xdu

( ) ( ) ( )x x x x x x xx dx xdu du du dxσ σ σ

+− = ∂

, ,

[ ( ) ]

x x x x

x x x x x x

du

du du dx

σ

σ σ= ∂ + ∂

,[ ]x x x x xdu d dxσ σ ε= +

The infinitesimal work done by the body force ( )xf x is

( )x xf x dxdu

Thus, the total infinitesimal work is the sum

, ,

0

[ ] ( ) [ ( )]x x x x x x x x x x x x xdu d dx f x dxdu f x du dx d dxσ σ ε σ σ ε

=

+ + = + + ,

14


x xd dxσ ε=

The Infinitesimal work is stored in the bar as Strain Energy.

The Infinitesimal Axial Strain Energy per unit length is

x xd dμ σ ε= . That is,

xx

ddμ

σε

= .

3.2 Axial Strain Energy density of a Bar is

212 xEμ ε= .

Proof:

212

( )x x x x xd d E d Edμ σ ε ε ε ε= = =

212 xEμ ε= .

3.3 Complementary Axial Strain Energy of a Bar is

212 xE

μ σ∗ = ,

xx

ddμ

εσ

∗= .

Proof: The complementary axial strain energy density is the

axial strain energy density expressed in terms of the stress.

15


by substituting 1x xEε = σ into μ , we obtain

212 xE

μ σ∗ = .

Although equals , it is denoted by μ to specify its

dependence on the stress.

μ∗ μ ∗

Differentiating both sides,

1

x

x xEd d

ε

μ σ∗ = σ .

xx

ddμ

εσ

∗= .

16


4.

Variational Principles of a Bar

4.1 Euler’s Variational Equation

The deflection u that minimize or maximizes the Bar’s

energy per unit length,

( , , ')F F x u u=

satisfies Euler’s Variational Equation

( , , ') 0F x u uδ = ,

which is,

0'

F d Fu

u dx uδ

⎛ ⎞∂ ∂ ⎟⎜ − =⎟⎜ ⎟⎟⎜ ∂ ∂⎝ ⎠.

or,

0'

F d Fu dx u

∂ ∂− =

∂ ∂.

Proof: The Bar’s infinitesimal energy is 12

12

( , ( ), '( )

x dx

x dx

F u u d

ξ

ξ

ξ ξ ξ= +

= −∫ ξ .

The deflection u that minimizes or maximizes the bar’s

energy satisfies Euler’s Variational Equation

( , , ') 0F x u uδ = .

17


That is,

' 0'x

xx D u

F Fu u

u u δ

δ δ∂ ∂

+ =∂ ∂

.

Integration by parts, where vanishes on the boundary,

[Dan3] leads to the variational equation

uδ

0'

F d Fu

u dx uδ

⎛ ⎞∂ ∂ ⎟⎜ − =⎟⎜ ⎟⎟⎜ ∂ ∂⎝ ⎠.

Since this holds for any variation , the deflection u , that

minimizes or maximizes the bar’s energy satisfies Euler’s

uδ

Equation

'F d Fu dx u

∂ ∂−

∂ ∂.

4.2 If the body forces originate from a Potential per ( )uυ

unit length, so that

xfuυ∂

− =∂

Then, Euler’s Equation is the Equilibrium Equation

, 0x x xfσ + = .

Proof: The Energy per unit length is

( , ( ), '( ))F x u x u x μ υ= +

18


12 x xσ ε υ= + .

Here,

xF

fu u

υ∂ ∂= = −

∂ ∂

' xd F ddx u dx

σ∂

=∂

Euler’s Equation is

0'

x xf

F d Fu dx u

σ−

∂ ∂− =

∂ ∂

That is, Euler’s Equation is the equilibrium Equation

0x x xf σ+ ∂ = .

4.3 Principle of Virtual Work for a Bar

At Equilibrium, U Wδ δ=

Proof: The equilibrium deflection u that minimizes or

maximizes the bar’s energy satisfies the Euler Variational

Equation

0

F

F d Fu

u dx uδ

δ⎛ ⎞∂ ∂ ⎟⎜ − =⎟⎜ ⎟⎟⎜ ∂ ∂⎝ ⎠

.

Thus, at equilibrium, the Variation of F , [Dan3], satisfies

0.Fδ =

19


Therefore,

( )U W U Wδ δ δ− = −

0

( , , ')x l

x

F x u u dxδ=

=

= ∫

0

( , , ') 0x l

x

F x u u dxδ=

=

= =∫

That is, . U Wδ δ=

4.4 Principle of Minimal Energy of a Bar

The Bar’s Energy is Minimal at Equilibrium

Proof: In [Dan3], we established Legendre’s claim that a

Sufficient Condition for to be Minimal is that ( , , ')x b

x a

F x u u dx=

=∫

0

2

20

( ') u

F

u

∂>

∂.

Here,

{ }2 2

2122 2

[ '] 0( ') ( ')

FE u E

u u

∂ ∂= =

∂ ∂> ,

since E , Young’s Elastic Modulus, is always positive.

Therefore, the Energy of the bar is minimal at the

equilibrium.

20


5.

Strain on a Plate

5.1 Strain-Displacement Relations

2 2 2 2 2( ') ( ') ( ) ) 2[ ( ) 2 ( )xx xy yydx dy dx dy dx dxdy dyε ε ε+ − − ≈ + + 2 ]

2 , xx xy

yx yy

dxdx dy

dy

ε εε ε

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎡ ⎤= ⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

.

Proof: Elastic Deformation of a plate, transforms the point

to ( ' , and the infinitesimally close point

, to ( ' .

( , )x y , ')x y

( ,x dx y dy+ + ) ', ' ')x dx y dy+ +

'x x→ , , 'y y→

' 'x dx x dx+ → + , . ' 'y dy y dy+ → +

The Displacements

( , ) 'xu x y x x≡ − ,

21


( , ) 'yu x y y y≡ −

satisfy 'xdu dx dx= − , 'ydu dy dy= − .

Therefore,

2 2 2 2 2 2 2( ') ( ') [( ) ( ) ] ( ) ( ) ( ) ( )x ydx dy dx dy dx du dx dy du dy+ − + = + − + + − 2

)y

,

⎤⎥⎥⎥⎦

2 22 2 ( ) (x y xdxdu dydu du du= + + +

Keeping first order in , and in , xdu ydu

2 2x ydxdu dydu≈ +

{ }, , , ,2 [ ] [ ]x x x y y x y ydx u dx u dy dy u dx u dy= + + + .

, , ,

2 2

2[ ( ) ( ) ( ) ]

xx yyxy yx

x x x y y x y ydx u dx dx u u dy dy u dy

ε εε ε=

= + + +

. 2 , xx xy

yx yy

dxdx dy

dy

ε εε ε

⎡ ⎤ ⎡⎢ ⎥ ⎢⎡ ⎤= ⎣ ⎦ ⎢ ⎥ ⎢⎢ ⎥ ⎢⎣ ⎦ ⎣

5.2 The Axial Strains on a Plate are

,xx x xuε ≡ = elongation per unit length inx direction,

,yy y yuε ≡ = elongation per unit length in y direction.

5.3 Shear Strains on a Plate are

1, ,2

( )xy yx x y y xu uε ε= ≡ + .

22


As the drawing shows,

,y xu dx = deflection at length dx angle at radius dx ⇒ ,y xu =

,x yu dy = deflection at length dy angle at radius dy ⇒ ,x yu =

23


6.

Stress on a Plate

6.1 The Axial Stresses , xxσ yyσ

The Axial Stress is the force per unit length of the

side, stretching the plate on both sides in the x direction.

( , )xx x yσ

y

The Axial Stress is the force per unit length of the x

side , stretching the plate on both sides in the y direction.

( , )yy x yσ

6.2 Hook’s Law for Axial Stresses on a Plate

Young’s Elastic Modulus E =

Poisson Ratio ν =

24


2

1

11xx xx

yy yy

Eσ νσ νν

⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢=⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

εε

⎤⎥⎥⎥⎦

Proof: The stretching along y , by 1yyE

σ , contracts the plate

along x by 1yyE

ν σ− . Then,

1 1xx xx yyE Eε σ ν σ= − ⇒ E . xx xx yyε σ νσ= −

Similarly,

1 1yy yy xxE Eε σ ν σ= − ⇒ E . yy xx yyε νσ σ= − +

σσ

⎤⎥⎥⎥⎦

That is, 1

1xx xx

yy yy

E

E

ε νε ν

⎡ ⎤ ⎡ ⎤ ⎡−⎢ ⎥ ⎢ ⎥ ⎢=⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣.

Hence,

21

1( )

1

1

xx

yy Exx xx yy

E

E

ν

ε νε

σ εν

ν

−

−

= = +−

−

νε ,

21

1

( )1

1

xx

yy Eyy xx yy

E

E

ν

εν ε

σ νε + εν

ν

−

−= =

−−

.

6.3 The Shear Stresses, , xyσ yxσ

25


The Shear Stress is the force per unit length, along

the y sides,

( , )xy x yσ

The Shear Stress is the force per unit length, along

the sides,

( , )yx x yσ

x

6.4 Hook’s Law for a Shear Stresses on a Plate

The shear stress satisfies Hooke’s Law. xyσ

2xy xyGσ ε= ,

where G Shear Modulus. =

Similarly, in isotropic material,

. 2yx yxGσ ε=

Since , we have xy yxε ε=

26


6.5 . xy yxσ σ=

6.6 Hook’s Law for a Plate


Poisson Ratio ν =

G Shear Modulus=21

Eg

ν=

−

2

1 0

1 01

0 0 2

xx xx

yy yy

xy xy

E

g

σ νσ ν

νσ ε

⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢=⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

εε

⎤⎥⎥⎥⎥⎥⎦

6.7 21E

Gν

=+

21xy xy xyE

Gσ εν

= =+

ε

Proof: By 6.2

2

1

11xx xx

yy yy

Eσ νσ νν

⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢=⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

εε

⎤⎥⎥⎥⎦.

The matrix 1

1

νν

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

27


has the eigen values 1 , and 1 , and can be rotated so

that

ν− ν+

' ' 11

' ' 121

01 0

0 1 01xx xx xx

yy yy yy

EE ν

ν

σ ν ενσ εν

+

−

⎡ ⎤ ⎡ ⎤ ⎡ ⎤'

'

ε

ε

⎡ ⎤⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥+− ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦,

That rotation diagonalizes also the matrix

1 0

1 0

0 0 2g

νν

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

.

The equation for its eigen values yields

2gλ = ; ; . 1λ ν= + 1λ ν= −

For , 1λ ν= −

2 1g ν= − ,

22 (1 )

11

E EG ν

νν= − =

+−,

1xy xyE

σ εν

=+

.

6.8 Body Forces ( , )xf x y , ( , )yf x y

Body forces are exterior forces ( , )xf x y , and ( , )yf x y such as

gravitational, or magnetic, per unit area.

28


6.9 The Equilibrium Equations for a Plate

, , 0xx x xy y xfσ σ+ + = ,

, , 0yx x yy y yfσ σ+ + = .

Proof: If body forces ( , )xf x y , and ( , )yf x y per unit area apply

to the plate,

In the x direction,

the force due to is xxσ

,

,[ ( , ) ( , )]

xx x

xx xx xx x

dx

x dx y x y dy dxdy

σ

σ σ σ=

+ − = ,

the force due to is yxσ

, ,

,[ ( , ) ( , )]

yx y xy y

yx yx xy y

dy dy

x y dy x y dx dxdy

σ σ

σ σ σ

= =

+ − = ,

and the force due to xf is ( , )xf x y dxdy ,

Therefore, the Equilibrium of forces in the x direction is,

29


, , 0xx x xy y xfσ σ+ + = .

Similarly, the Equilibrium of forces in the y direction is,

, , 0yy y yx x yfσ σ+ + = .

30


7.

Strain Energy of a Plate 7.1 Infinitesimal Strain Energy per unit area

2xx xx xy xy yy yyd d d dμ σ ε σ ε σ ε= + +

. (summation over ; ) ij ijdσ ε= 1,2i = 1,2j =

xxxx

μσ

ε∂

=∂

; 2 xyxy

μσ

ε∂

=∂

; yyyy

μσ

ε∂

=∂

Proof: In the x direction,

The infinitesimal work done by , along is xxdyσ xdu

( , ) ( , )( ) ( ) ( )xx x xx x x xx xx dx y x ydydu dydu du dxdyσ σ σ

+− = ∂

, ,

[ ( ) ]

xx x x x

x xx x xx x x

du

du du dxdy

σ

σ σ= ∂ + ∂

The infinitesimal work done by , along is yxdxσ xdu

31


( , ) ( , )( ) ( ) ( )yx x yx x y yx xx y dy x ydxdu dxdu du dydxσ σ σ

+− = ∂

,,

[ ( ) ]

x yxy y

y yx x yx y x

du

du du dxdy

σ

σ σ= ∂ + ∂

The infinitesimal work by xf dxdy is ( , )x xf x y dxdydu

In the y direction,

The infinitesimal work done by , along is yydxσ ydu

( , ) ( , )( ) ( ) ( )yy y yy y y yy yx y dy x ydxdu dxdu du dxdyσ σ σ

+− = ∂

,,

[ ( ) ]

y yyy y

y yy y yy y y

du

du du dxdy

σ

σ σ= ∂ + ∂

The infinitesimal work done by , along is xydyσ ydu

( , ) ( , )( ) ( ) ( )xy y xy y x xy yx dx y x ydydu dydu du dxdyσ σ σ

+− = ∂

,,

[ ( ) ]

y xxy x

x yx y yx x y

du

du du dxdy

σ

σ σ= ∂ + ∂

The infinitesimal work by yf is ( , )y yf x y dxdydu

32


Thus, the total infinitesimal work is the sum

+ , , , ,

0

{[ ( , )] }

xx

xx x xy y x x xx x x xy x yf x y du d u du dxdy

ε

σ σ σ σ+ + + +

, , , ,

0

{[ ( , )] }

yy

yx x yy y y y yy y y xy y xf x y du d u du dxdy

ε

σ σ σ σ+ + + + +

, ,

2

{ }

xy xy

xx xx xy x y yx y x yy yy

d

d du du d d

σ ε

σ ε σ σ σ ε= + + + xdy

The Infinitesimal work is stored in the plate as Strain

Energy. The Infinitesimal Strain Energy per unit area is

2xx xx xy xy yy yyd d d dμ σ ε σ ε σ ε= + + .

Hence,

xxxx

μσ

ε∂

=∂

; 2 xyxy

μσ

ε∂

=∂

; yyyy

μσ

ε∂

=∂

.

7.2 Strain Energy density of a Plate

1 12 2xx xx xy xy yy yyμ σ ε σ ε σ ε= + +

12 ij ijσ ε= (summation over ) , 1,i j = 2

2

2 212 1

[ 2 ] 2Exx yy xx yy xyG

νε νε ε ε ε

−= + + + 2

Proof:

2xx xx xy xy yy yyd d d dμ σ ε σ ε σ ε= + +

Substituting

33


21( )E

xx xx yyνσ ε

−= + νε ,

21( )E

yy xx yyνσ νε

−= + ε ,

1E

xy xyνσ ε

+= ,

2 211 1( ) 2 (E E Exx yy xx xy xy xx yy yyd d d

νν νμ ε νε ε ε ε νε ε

+− −= + + + + )dε

2 2 2 22 21 1

2 11 1 1 1( )E E E E E

xx yy xx xy xx yy yyd d d dνν ν ν ν

ε νε ε ε νε ε+− − − −

= + + + + 22

dε

2 2

2 2 212 11 1

( ) [ ] (E Exx yy yy xx xx yy xyd d d

νν νε ε ν ε ε ε ε ε

+− −= + + + + )E d

2 22 2 21

2 11 1

2

( ) ( ) (E Exx yy yy xx xy

G

d dνν ν

ε ε ν ε ε ε+− −

= + + + )E d

Integrating,

2 22 21

2 1 1( ) 2E Exx yy xx yy xyG

ν νμ ε ε νε ε

− −= + + + 2ε

2

2 212 1

[ 2 ] 2Exx yy xx yy xyG

νε νε ε ε ε

−= + + + 2

2 2

12 1 1{ [ ] [ ] }

xx yy

E Exx yy xx xx yy yy xy xyν ν

σ σ

ε νε ε νε ε ε σ ε− −

= + + + + .

7.3 Complementary Strain Energy of a Plate

1 12 2xx xx xy xy yy yyμ σ ε σ ε σ∗ = + + ε

2 21 12 2

( 2 )xx xx yy yy xyE Gσ νσ σ σ σ= − + + 2

34


xxxx

με

σ

∗∂=

∂; 2 xy

xy

με

σ

∗∂=

∂; yy

yy

με

σ

∗∂=

∂

Proof: The complementary strain energy density is the

strain energy density expressed in terms of the stresses.

Substituting 1 1

xx xx yyE Eε σ ν= − σ

1 1yy yy xxE Eε σ ν= − σ

12xy xyG

ε σ=

into μ ,

1 12 2xx xx xy xy yy yyμ σ ε σ ε σ ε= + +

1 1 12 2 2

( ) (xx xx yy xy xy yy xx yyE G Eσ σ νσ σ σ σ νσ σ= − + + − + )

2 21 12 2

( 2 )xx xx yy yy xyE Gσ νσ σ σ σ= − + + 2



μ∗ μ ∗

1 1xx yy xxE E

xx

μσ ν σ ε

σ

∗∂= − =

∂.

1 1yy xx yyE E

yy

μσ ν σ ε

σ

∗∂= − =

∂

1 2xy xyGxy

μσ ε

σ

∗∂= =

∂.

35


8.

Variational Principles on a

Plate

8.1 Euler’s Variational Equations

The deflections , and that minimize or maximize xu yu

the plate’s energy per unit area,

( , , , , , , )x y xx xy yyF F x y u u ε ε ε=

satisfy Euler’s Variational Equation

( , , , , , , ) 0x y xx xy yyF x y u uδ ε ε ε = ,

which is,

1 12 2

0x y x x y yx xx xy y xy yy

F F F F F Fu u

u uδ δ

ε ε ε ε

⎛ ⎞ ⎛∂ ∂ ∂ ∂ ∂ ∂⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜− ∂ − ∂ + − ∂ − ∂ =⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝

⎞

⎠.

or, 12

0x yx xx xy

F F Fu ε ε

∂ ∂ ∂− ∂ − ∂ =

∂ ∂ ∂,

12

0x yy xy yy

F F Fu ε ε

∂ ∂ ∂− ∂ − ∂ =

∂ ∂ ∂.

Proof: The Plate’s infinitesimal energy is

36


1 12 2

1 12 2

( , , , , , , )

x dx y dy

x dx y dy

F u u d d

ξ η

ξ η ξξ ξη ηηξ η

ξ η ε ε ε ξ η= + = +

= − = −∫ ∫ .

The deflections , and that minimizes or maximize the

plate’s energy, satisfy Euler’s Variational Equation

xu yu

( , , , , , , ) 0x y xx xy yyF x y u uδ ε ε ε = .

That is,

1 12 2

0

x x y yy x x y

x xx xy y yyx xx xy y yyu uu u

F F F F Fu u

u uδ δδ δ

δ δε δε δ δεε ε ε

∂ ∂∂ + ∂

∂ ∂ ∂ ∂ ∂+ + + +

∂ ∂ ∂ ∂ ∂=

1 12 2

0x x x y x y x y y yx xx xy y xy yy

F F F F F Fu u u u u u

u uδ δ δ δ δ δ

ε ε ε ε∂ ∂ ∂ ∂ ∂ ∂

+ ∂ + ∂ + + ∂ + ∂ =∂ ∂ ∂ ∂ ∂ ∂

Integration by parts, where , and vanish on the

boundary, [Dan3] leads to the variational equation

xuδ yuδ

1 12 2


F F F F F Fu u

u uδ δ

ε ε ε ε

⎛ ⎞ ⎛∂ ∂ ∂ ∂ ∂ ∂⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜− ∂ − ∂ + − ∂ − ∂ =⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝

⎞

⎠.

Since the variations , and are independent, the

deflections , and that minimizes or maximize the

plates’s energy satisfy Euler’s Variational Equations

xuδ yuδ

xu yu

37


12

0x yx xx xy

F F Fu ε ε∂ ∂ ∂

− ∂ − ∂ =∂ ∂ ∂

12

0x yy xy yy

F F Fu ε ε

∂ ∂ ∂− ∂ − ∂ =

∂ ∂ ∂

8.2 If the body forces originate from a Potential per ( , )x yu uυ

unit area, so that

xx

fuυ∂

− =∂

yy

fuυ∂

− =∂

Then, Euler’s Equations are the Equilibrium Equations

, , 0xx x xy y xfσ σ+ + =

, , 0yx x yy y yfσ σ+ + = .

Proof: The Energy per unit area is

( , , , , , , )x y xx xy yyF x y u u ε ε ε μ υ= +

1 12 2xx xx xy xy yy yyσ ε σ ε σ ε υ= + + + .

The Euler’s Variational Equations are

38


12

2

0

x xx xy

x yx xx xy

f

F F Fu

σ σ

ε ε−

∂ ∂ ∂− ∂ − ∂ =

∂ ∂ ∂

12

0

y yx yy

x yy xy yy

f

F F Fu

σ σ

ε ε

−

∂ ∂ ∂− ∂ − ∂ =

∂ ∂ ∂

That is, Euler’s Equations are the equilibrium Equations

0x x xx y xyf σ σ+ ∂ + ∂ = ,

0y x yx y yyf σ σ+ ∂ + ∂ = .

8.3 Principle of Virtual Work


Proof: The equilibrium deflections , and that minimize

the bar’s energy satisfy the Euler Variational Equation

xu yu

1 12 2


F

F F F F F Fu u

u u

δ

δ δε ε ε ε

⎛ ⎞ ⎛∂ ∂ ∂ ∂ ∂ ∂⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜− ∂ − ∂ + − ∂ − ∂ =⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝

⎞

⎠.


0.Fδ =

Therefore,

( )U W U Wδ δ δ− = −

39


0 0

( , , , , , , )yx

y lx l

x y xx xy yyx y

F x y u u dxdyδ ε ε==

= =

= ∫ ∫ ε

0 0

( , , , , , , ) 0yx

y lx l

x y xx xy yyx y

F x y u u dxdyδ ε ε ε==

= =

= =∫ ∫


8.4 Principle of Minimal Energy for a Plate

The Plate’s Energy is Minimal at Equilibrium

Proof: Legendre’s Sufficient Condition for

0 0

( , , , , , , )yx

y lx l

x y xx xy yyx y

F x y u u dxdyε ε ε==

= =∫ ∫

to be Minimal at the equilibrium generalizes to

2

2equilibrium ;

0( )

x yij u u

F

ε

∂>

∂.

By 7.2,

22 21

2 1[ 2 ] 2Exx yy xx yy xyF G

νε νε ε ε ε

−= + + + 2 υ+ ,

where is Young’s Elastic Modulus, 0E >

ν Poisson’s Ratio, and = 0 1ν< <

1EGν+

= =Shear Modulus.

40


Hence,

2

2

2 10

( )E

xx

Fνε −

∂= >

∂,

2

2

2 10

( )E

yy

Fνε −

∂= >

∂,

2

124 2

( )E

xy

FG

νε +

∂= = >

∂0

Therefore, the Total Energy of the plate is minimal at the

equilibrium.

41


9.

Strain on a Box

5.1 Strain-Displacement Relations for a box

2 2 2 2 2 2( ') ( ') ( ') ( ) ) ( )dx dy dz dx dy dz+ + − − + ≈

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

)

=

2 2 22[ ( ) ( ) ( ) 2 2 2 ]xx yy zz xy yz zxdx dy dz dxdy dydz dzdxε ε ε ε ε ε≈ + + + + +

. 2 , ,xx xy xz

yx yy yz

zx zy zz

dx

dx dy dz dy

dz

ε ε εε ε εε ε ε

⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤= ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

Proof: Elastic Deformation of a body, transforms the point

to ( ' , and ( , , to

.

( , , )x y z , ', ')x y z ,x dx y dy z dz+ + +

( ' ', ' ', ')x dx y dy z dz+ + +

The Displacements ( , , ) 'xu x y z x x≡ − , ( , , ) 'yu x y z y y≡ − ,

( , , ) 'zu x y z z z≡ − satisfy

'xdu dx dx= − , 'ydu dy dy= − ,

'zdu dz dz= − Therefore,

2 2 2 2 2 2( ') ( ') ( ') [( ) ( ) ( ) ]dx dy dz dx dy dz+ + − + +

42


2 22 2 2 ( ) ( ) (x y z x ydxdu dydu dzdu du du du= + + + + + 2)z

z

+

+

}

,

Keeping first order in , , and xdu ydu zdu

2 2 2x ydxdu dydu dzdu≈ + +

, , ,2{ [ ]x x x y x zdx u dx u dy u dz= + +

, , ,[ ]y x y y y zdy u dx u dy u dz+ + +

. , , ,[ ]z x z y z zdz u dx u dy u dz+ + +

, ,2[ ( ) ( ) ( ) ]

xx yy zz

x x y y z zdx u dx dy u dy dz u dz

ε ε ε

= + +

, , , , , ,

2 2 2 2 2 2

2[ ( ) ( ) ( ) ]

xy yx yz zy zx xz

x y y x y z z y z x x zdx u u dy dy u u dz dz u u dx

ε ε ε ε ε ε= = =

+ + + + + +

2 , ,xx xy xz

yx yy yz

zx zy zz

dx

dx dy dz dy

dz

ε ε εε ε εε ε ε

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

.

5.2 The Axial Strains on a Box are

, ( , , )xx x xu x y zε ≡ =

=

=

elongation per unit length inx direction,

elongation per unit length in y direction, , ( , , )yy y yu x y zε ≡

elongation per unit length in y direction. , ( , , )zz z zu x y zε ≡

5.3 Shear Strains on the faces of a Box are

1, ,2

[ ( , , ) ( , ,xy yx x y y xu x y z u x y zε ε= ≡ + )],

43


1, ,2

[ ( , , ) ( , , )yz zy y z z yu x y z u x y zε ε= ≡ + ],

1, ,2

[ ( , , ) ( , , )zx xz z x x zu x y z u x y zε ε= ≡ + ]

As the drawing shows,

,y xu dx = deflection at length dx angle at radius dx ⇒ ,y xu =

,x yu dy = deflection at length dy angle at radius dy ⇒ ,x yu =

44


10.

Stress on a Box

10.1 The Axial Stresses , , xxσ yyσ zzσ

The Axial Stress is the force per unit area of the

face, stretching the box in the x direction.

( , , )xx x y zσ

yz

The Axial Stress is the force per unit area of the xz

face , stretching the box in the y direction.

( , , )yy x y zσ

The Axial Stress is the force per unit area of the xy

face , stretching the box in the z direction.

( , , )zz x y zσ

45


10.2 Hooke’s Law for Axial Stresses on a Box


Poisson Ratio ν =

1

1(1 )(1 2 )

1

xx xx

yy yy

zz zz

Eσ ν νσ ν ν

ν νσ ν ν

⎡ ⎤ ⎡ ⎤ ⎡−⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢= −⎢ ⎥ ⎢ ⎥ ⎢+ −⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

ν εν εν ε

⎤⎥⎥⎥⎥⎥⎦

Proof: Stretching along y , by 1yyE

σ , and along by z 1zzE

σ ,

contracts the box along x by 1 1yy zzE E

ν σ ν σ− − . Then,

1 1 1xx xx yy zzE E Eε σ ν σ ν σ= − − ⇒ . xx xx yy zzEε σ νσ νσ= − −

Similarly,

1 1 1yy yy xx zzE E Eε σ ν σ ν σ= − − ⇒ E . yy xx yy zzε νσ σ νσ= − + −

1 1 1zz zz xx yyE E Eε σ ν σ ν σ= − − ⇒ E . yy xx yy zzε νσ νσ σ= − − +

σσσ

⎤⎥⎥⎥⎥⎥⎦

That is, 1

1

1

xx xx

yy yy

zz zz

E

E

E

ε ν νε ν νε ν ν

⎡ ⎤ ⎡ ⎤ ⎡− −⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢= − −⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢− −⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

.

46


Since

2

1

1 (1 ) (1

1

ν νν ν νν ν

− −− − = + −

− −

2 )ν ,

(1 )(1 2 )

1

1[(1 ) ]

1

1

1

xx

yy

zz Exx xx yy zz

E

E

Eν ν

ε ν νε νε ν

σ νν ν

ν νν ν

+ −

− −−

−= = − +

− −− −− −

ε νε νε+ ,

(1 )(1 2 )[ (1 )E

yy xx yy zzν νσ νε ν ε

+ −= + − ]νε+ ,

(1 )(1 2 )[ (1 )ν− ]E

zz xx yy zzν νσ νε νε ε

+ −= + + .

10.3 The Shear Stresses, , , on box faces xyσ yxσ

47


The Shear Stress is the force per unit area in the

plane, along y ,

( , , )xy x y zσ

xy

The Shear Stress is the force per unit area in yx

plane, along x ,

( , , )yx x y zσ

10.4 Hooke’s Law for a Shear Stresses on box faces

The shear stress satisfies Hooke’s Law. xyσ

2xy xyGσ ε= ,

2yz yzGσ ε= ,

2zx zxGσ ε= .

where G Shear Modulus. =

Similarly, in isotropic material,

2yx yxGσ ε= ,

2zy zyGσ ε= ,

2xz xzGσ ε= .

10.5 ; ; . xy yxσ σ= yz zyσ σ= xy yxσ σ=

Proof: Since , , . xy yxε ε= yz zyε ε= zx xzε ε=

48


10.6 Hook’s Law for a Box


Poisson Ratio ν =

G Shear Modulus=21

Eg

ν=

−

(1 )(1 2 )

1

1

0 1

2

2

2

xx xx

yy yy

zz zzE

xy xy

yz yz

zx zx

g

g

g

ν ν

σ ν ν νσ ν ν νσ ν νσ εσ εσ ε

+ −

⎡ ⎤ ⎡ ⎤ ⎡−⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢=⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

εεε

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

10.7 21E

Gν

=+

21xy xy xyE

Gσ εν

= =+

ε

21yz yz yzE

Gσ εν

= =+

ε

21zx zx zxE

Gσ εν

= =+

ε

Proof: By 10.2

49


(1 )(1 2 )

1

1

1

xx xxE

yy yy

zz zz

ν ν

σ ν νσ ν νσ ν ν

+ −

⎡ ⎤ ⎡ ⎤ ⎡−⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢= −⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

ν εν εν ε

⎤⎥⎥⎥⎥⎥⎦

ν

.

The matrix 1

1

1

ν ν νν ν νν ν

⎡ ⎤−⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

has the eigen values 1 , and 1 , and can be rotated so

that

2ν− ν+

' ' 11

' ' 1(1 )(1 2 ) 1

' ' 11 2

0 01 2 0 0

0 1 2 0 0 0

0 0 1 0 0

xx xx xxE

yy yy yy

zz zz zz

Eν

ν ν ν

ν

σ ενσ ν ε

νσ ε

+

+ − +

−

⎡ ⎤ ⎡ ⎤ ⎡ ⎤'

'

'

ε

ε

ε

⎡ ⎤⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥= − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥+⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

.

That rotation diagonalizes also the matrix

1

1

0 1

2

2

2

g

g

g

ν ν νν ν νν ν

⎡ ⎤−⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

.

The equation for its eigen values yields

2gλ = ; ; . 1λ ν= + 1 2λ ν= −

For , 1 2λ ν= −

50


2 1 2g ν= − ,

(1 )(1 2 ) (1 )2 (1 2 )E EG

ν νν

+ − += − =

ν,

(1 )E

xy xyνσ ε

+=

(1 )E

yz yzνσ ε

+=

(1 )E

zx zxνσ ε

+= .

10.8 Body Forces ( , , )xf x y z , ( , , )yf x y z , ( , , )zf x y z

Body forces are exterior forces such as gravitational, or

magnetic, per unit volume.

10.9 The Equilibrium Equations for a Box

, , , 0xx x xy y xz z xz fσ σ σ+ + + = ,

, , , 0yx x yy y yz z yfσ σ σ+ + + = ,

, , , 0zx x zy y zz z zfσ σ σ+ + + = .

Proof: If body forces ( , , )xf x y z , ( , , )yf x y z , and ( , , )zf x y z per

unit volume apply to the box,

51


In the x direction,

the force due to is xxσ

,

,[ ( , , ) ( , , )]

xx x

xx xx xx x

dx

x dx y z x y z dydz dxdydz

σ

σ σ σ=

+ − = ,

the force due to is yxσ

, ,

,[ ( , , ) ( , , )]

yx y xy y

yx yx xy y

dy dy

x y dy z x y z dxdz dxdydz

σ σ

σ σ σ

= =

+ − = ,

the force due to is zxσ

, ,

,[ ( , , ) ( , , )]

zx z xz z

zx zx xz z

dy dz

x y z dz x y z dxdy dxdydz

σ σ

σ σ σ= =

+ − = ,

and the force due to xf is ( , , )xf x y z dxdydz .

52


Therefore, the Equilibrium of forces in the x direction is,

, , , 0xx x xy y xz z xz fσ σ σ+ + + = .

Similarly,

in the y direction

the Equilibrium of forces is

, , , 0yx x yy y yz z yfσ σ σ+ + + = ,

and in the z direction

the Equilibrium of forces is

, , , 0zx x zy y zz z zfσ σ σ+ + + = .

53


11.

Strain Energy of a Box 11.1 Infinitesimal Strain Energy of a Box

2 2 2xx xx yy yy zz zz xy xy zy zy zx zxd d d d d d dμ σ ε σ ε σ ε σ ε σ ε σ ε= + + + + +

. (summation over ; ). ij ijdσ ε= 1,2,3i = 1,2,3j =

xxxx

μσ

ε∂

=∂

; yyyy

μσ

ε∂

=∂

; zzzz

μσ

ε∂

=∂

; 2 xyxy

μσ

ε∂

=∂

; 2 yzyz

μσ

ε∂

=∂

; 2 xzzx

μσ

ε∂

=∂

Proof: In the x direction,

54


The infinitesimal work done by , along is xxdydzσ xdu

( , , ) ( , , )( ) ( ) ( )xx x xx x x xx xx dx y z x y zdydzdu dydzdu du dxdydzσ σ σ

+− = ∂

, ,

[ ( ) ]

xx x x x

x xx x xx x x

du

du du dxdydz

σ

σ σ= ∂ + ∂

The infinitesimal work done by , along is yxdydzσ xdu

( , , ) ( , , )( ) ( ) ( )yx x yx x y yx xx y dy z x y zdydzdu dxdzdu du dydxdzσ σ σ

+− = ∂

,,

[ ( ) ]

x yxy y

y yx x yx y x

du

du du dxdydz

σ

σ σ= ∂ + ∂

The infinitesimal work done by , along is zxdxdyσ xdu

( , , ) ( , , )( ) ( ) ( )zx x zx x z zx xx y z dz x y zdxdydu dxdydu du dydxdzσ σ σ

+− = ∂

,,

[ ( ) ]

x zxz z

z zx x zx z x

du

du du dxdydz

σ

σ σ= ∂ + ∂

The infinitesimal work done by ( , , )xf x y z dxdydz is

( , , )x xf x y z dxdydzdu .

In the y direction,

55


The infinitesimal work done by , along is yydxdzσ ydu

( , , ) ( , , )( ) ( ) ( )yy y yy y y yy yx y dy z x y zdxdzdu dxdzdu du dxdydzσ σ σ

+− = ∂

,,

[ ( ) ]

y yyy y

y yy y yy y y

du

du du dxdydz

σ

σ σ= ∂ + ∂

The infinitesimal work done by , along is xydydzσ ydu

( , , ) ( , , )( ) ( ) ( )xy y xy y x xy yx dx y z x y zdydzu dydzdu du dxdydzσ σ σ

+− = ∂

,,

[ ( ) ]

y xyx x

x xy y yx x y

du

du du dxdydz

σ

σ σ= ∂ + ∂

56


The infinitesimal work done by , along is zydxdyσ ydu

( , , ) ( , , )( ) ( ) ( )zy y zy y z zy yx y z dz x y zdxdydu dxdydu du dxdydzσ σ σ

+− = ∂

,,

[ ( ) ]

y zyz z

z zy y yz z y

du

du du dxdydz

σ

σ σ= ∂ + ∂

The work by ( , , )yf x y z dxdydz is ( , , )y yf x y z dxdydzdu .

In the z direction,

The infinitesimal work done by , along is zzdxdyσ zdu

( , , ) ( , , )( ) ( ) ( )zz z zz z z zz zx y z dz x y zdxdydu dxdzdu du dxdydzσ σ σ

+− = ∂

57


,,

[ ( ) ]

z zzz z

z zz z zz z z

du

du du dxdydz

σ

σ σ= ∂ + ∂

The infinitesimal work done by , along is zxdydzσ zdu

( , , ) ( , , )( ) ( ) ( )zx z zx z x zx zx dx y z x y zdydzu dydzdu du dxdydzσ σ σ

+− = ∂

,,

[ ( ) ]

z xzx x

x zx z zx x z

du

du du dxdydz

σ

σ σ= ∂ + ∂

The infinitesimal work done by , along is zydxdzσ zdu

( , , ) ( , , )( ) ( ) ( )zy z zy z y zy zx y dy z x y zdxdzdu dxdzdu du dxdydzσ σ σ

+− = ∂

,,

[ ( ) ]

z yyz z

y zy y yz y z

du

du du dxdydz

σ

σ σ= ∂ + ∂

The work by ( , , )zf x y z dxdydz is ( , , )z zf x y z dxdydzdu .

The total infinitesimal work is the sum

, , ,

0

{[ ( , , )] }xx x xy y xz z x xf x y z du dxdydzσ σ σ+ + + +

, , ,

0

{[ ( , , )] }yx x yy y yz z y yf x y z du dxdydzσ σ σ+ + + + +

, , ,

0

{[ ( , , )] }zx x zy y zz z z zf x y z du dxdydzσ σ σ+ + + + +

+

, , ,{ }

xx

xx x x xy x y xz x zd u du du dxdydz

ε

σ σ σ+ + +

58


, , ,{ }

yy

yy y y yx y x yz y zd u du du dxdydz

ε

σ σ σ+ + + +

ydz

, , , ,{ }

zz

zz z z zy z y z x z xd u du du dxdydz

ε

σ σ σ+ + +

{ }xx xx yy yy zz zzd d d dxdσ ε σ ε σ ε= + +

, , , , , ,

2 2 2

{ }

xy xy yz yz zx zx

xy x y yx y x yz y z zy z y zx z x xz x z

d d d

du du du du du du dxdydz

σ ε σ ε σ ε

σ σ σ σ σ σ+ + + + + +

The Infinitesimal work is stored in the box as Strain Energy.

The Infinitesimal Strain Energy per unit volume is

2 2 2xx xx yy yy zz zz xy xy yz yz zx zxd d d d d d dμ σ ε σ ε σ ε σ ε σ ε σ ε= + + + + + .

Hence,

xxxx

μσ

ε∂

=∂

; yyyy

μσ

ε∂

=∂

; zzzz

μσ

ε∂

=∂

;

2 xyxy

μσ

ε∂

=∂

; 2 yzyz

μσ

ε∂

=∂

; 2 xzzx

μσ

ε∂

=∂

.

11.2 Strain Energy density of a Box

1 1 12 2 2xx xx yy yy zz zz xy xy yz yz zx zxμ σ ε σ ε σ ε σ ε σ ε σ ε= + + + + + .

12 ij ijσ ε= . (summation over ). , 1,2,i j = 3

(1 ) 2 2 212 (1 )(1 2 ) (1 )(1 2 )

( ) {E Exx yy zz xx yy yy zz zz xx

ν νν ν ν ν

ε ε ε ε ε ε ε ε ε−+ − + −

= + + + + + } +

. 2 2 22 ( )xy yz zxG ε ε ε+ + +

59


Proof:

2 2 2xx xx yy yy zz zz xy xy yz yz zx zxd d d d d d dμ σ ε σ ε σ ε σ ε σ ε σ ε= + + + + +

Substituting

(1 )(1 2 )[(1 ) ]E

xx xx yy zzν νσ ν ε νε

+ −= − + + νε ,

(1 )(1 2 )[ (1 )E

yy xx yy zzν νσ νε ν ε

+ −= + − ]νε+ ,

(1 )(1 2 )[ (1E

zz xx yy zzν νσ νε νε

+ −= + + ) ]ν ε− ,

1E

xy xyνσ ε

+= ,

1E

yz yzνσ ε

+= ,

1E

zx zxνσ ε

+= .

(1 )(1 2 )[(1 ) ]E

xx yy zz xxd dν ν

μ ν ε νε νε+ −

= − + + ε +

(1 )(1 2 )

[ (1 ) ]Exx yy zz yydν ν

νε ν ε νε ε+ −

+ + − + +

(1 )(1 2 )

[ (1 ) ]Exx yy zz zzdν ν

νε νε ν ε ε+ −

+ + + − +

1 1 1

2 2 2E E Exy xy yz yz zx zxd d

ν ν νε ε ε ε ε ε

+ + ++ + + d .

(1 ) 2 2 212 (1 )(1 2 )

( )Exx yy zzdν

ν νε ε ε−

+ −= + + +

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

{[ ] [ ] [ ]}

zz xxxx yy yy zz

Eyy xx xx yy yy zz zz yy zz xx xx zz

dd d

d d d d d dνν ν

ε εε ε ε ε

ε ε ε ε ε ε ε ε ε ε ε ε+ −

+ + + + + + +

2 2 21

2

( )Exy yz zx

G

dν

ε ε ε+

+ + + .

60


Integrating,

(1 ) 2 2 212 (1 )(1 2 ) (1 )(1 2 )

( ) {E Exx yy zz xx yy yy zz zz xx

ν νν ν ν ν

μ ε ε ε ε ε ε ε ε−+ − + −

= + + + + + }ε +

. 2 2 22 ( )xy yz zxG ε ε ε+ + +

12 (1 )(1 2 )

[ (1 ) ]

yy

Exx yy zz yyν ν

σ

νε ν ε νε ε+ −

= + − + +

12 (1 )(1 2 )

[ (1 )

zz

Exx yy zz zzν ν

σ

νε νε ν ε ε+ −

+ + + − ] +

2 2 2

zxxy yz

xy xy yz yz zx zxG G G

σσ σ

ε ε ε ε ε ε+ + +

12{ }xx xx yy yy zz zz xy xy yz yz zx zxσ ε σ ε σ ε σ ε σ ε σ ε= + + + + + .

11.3 Complementary Strain Energy of a Box

1 1 12 2 2xx xx yy yy zz zz xy xy yz yz zx zxμ σ ε σ ε σ ε σ ε σ ε σ ε∗ = + + + + +

2 2 212

( ) (xx yy zz xx yy yy zz zz xxE Eνσ σ σ σ σ σ σ σ σ= + + − + + )+

2 2 212

( )xy yz zxGσ σ σ+ + +

xxxx

με

σ

∗∂=

∂; yy

yy

με

σ

∗∂=

∂; zz

zz

με

σ

∗∂=

∂;

2 xyxy

με

σ

∗∂=

∂; 2 yz

yz

με

σ

∗∂=

∂; 2 zx

zx

με

σ

∗∂=

∂

61


Proof: The complementary strain energy density is the

strain energy density expressed in terms of the stresses.

Substituting 1 1 1

xx xx yy zzE E Eε σ ν σ ν σ= − − ,

1 1 1yy xx yy zzE E Eε ν σ σ ν= − + − σ ,

1 1 1zz xx yy zzE E Eε ν σ ν σ= − − + σ ,

12xy xyG

ε σ= ,

12yz yzG

ε σ= ,

12zx zxG

ε σ= ,

into μ ,

1 1 12 2 2xx xx yy yy zz zz xy xy yz yz zx zxμ σ ε σ ε σ ε σ ε σ ε σ ε= + + + + + ,

12

( )xx xx yy zzEσ σ νσ νσ= − − +

12

( )yy xx yy zzEσ νσ σ νσ+ − + − +

12

( )zz xx yy zzEσ νσ νσ σ+ − − + +

2 21 1 12 2 2xy yz zxG G G

σ σ+ + + 2σ ,

2 2 212

( ) (xx yy zz xx yy yy zz zz xxE Eνσ σ σ σ σ σ σ σ σ= + + − + + )+

2 2 212

( )xy yz zxGσ σ σ+ + + .

62




μ∗ μ ∗

1 1 1xx yy zz xxE E E

xx

μσ ν σ ν σ ε

σ

∗∂= − − =

∂,

1 1 1yy xx zz yyE E E

yy

μσ ν σ ν σ ε

σ

∗∂= − − =

∂,

1 1 1zz xx yy zzE E E

zz

μσ ν σ ν σ ε

σ

∗∂= − − =

∂,

1 2xy xyGxy

μσ ε

σ

∗∂= =

∂,

1 2yz yzGyz

μσ ε

σ

∗∂= =

∂,

1 2zx zxGzx

μσ ε

σ

∗∂= =

∂.

63


12.

Variational Principles on a Box

12.1 Euler’s Variational Equations

The deflections , , and th at minimize or

maximize

xu yu zu

the box energy per unit volume,

( , , , , , , , , , , , )x y z xx yy zz xy yz zxF F x y z u u u ε ε ε ε ε ε=

satisfy Euler’s Variational Equation

( , , , , , , , , , , , ) 0x y z xx yy zz xy yz zxF x y z u u uδ ε ε ε ε ε =ε ,

which is,

1 12 2x y z

x xx xy xz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜ − ∂ − ∂ − ∂ +⎟⎜ ⎟⎟⎜ ∂ ∂ ∂ ∂⎝ ⎠x

1 12 2x y z

y yx yy yz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜+ − ∂ − ∂ − ∂ ⎟⎜ ⎟⎟⎜∂ ∂ ∂ ∂⎝ ⎠y +

1 12 2

0x y z zz zx zy zz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜+ − ∂ − ∂ − ∂ =⎟⎜ ⎟⎟⎜∂ ∂ ∂ ∂⎝ ⎠.

or,

1 12 2

0x y zx xx xy xz

F F F Fu ε ε ε∂ ∂ ∂ ∂

− ∂ − ∂ − ∂ =∂ ∂ ∂ ∂

,

64


1 12 2

0x y zy yx yy yz

F F F Fu ε ε ε

∂ ∂ ∂ ∂− ∂ − ∂ − ∂ =

∂ ∂ ∂ ∂,

1 12 2

0x y zz zx zy zz

F F F Fu ε ε ε

∂ ∂ ∂ ∂− ∂ − ∂ − ∂ =

∂ ∂ ∂ ∂.

Proof: The Box infinitesimal energy is 1 1 12 2 2

1 1 12 2 2

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

x dx y dy z dz

x dx y dy z dz

F u u u d d

ξ η ζ

ξ η ζ ξξ ηη ζζ ξη ηζ ζξξ η ζ

ξ η ζ ε ε ε ε ε ε ξ η ζ= + = + = +

= − = − = −∫ ∫ ∫ d

ε

.

The deflections , , and that minimizes or maximize

the box energy, satisfy Euler’s Variational Equation

xu yu zu

( , , , , , , , , , , , ) 0x y z xx yy zz xy yz zxF x y z u u uδ ε ε ε ε ε = .

That is,

x x z zy y

x xx y yy zx xx y yy z zzu uu

F F F F F Fu u u

u u uδ δδ

δ δε δ δε δ δεε ε

∂ ∂∂

∂ ∂ ∂ ∂ ∂ ∂+ + + + + +

∂ ∂ ∂ ∂ ∂ ∂ zzε+

1 11 1 1 12 22 2 2 2

0

x z z xx y y x z y y z

xy yz zxxy yz xy

u uu u u u

F F F

δ δδ δ δ δ

δε δε δεε ε ε

∂ + ∂∂ + ∂ ∂ + ∂

∂ ∂ ∂+ + +∂ ∂ ∂

= .

1 12 2x x x y x z

x xx xy xz

F F F Fu u u

uδ δ δ

ε ε ε∂ ∂ ∂ ∂

+ ∂ + ∂ + ∂∂ ∂ ∂ ∂ xuδ +

65


1 12 2y y y x y z

y yy yx yz

F F F Fu u u

uδ δ δ

ε ε ε∂ ∂ ∂ ∂

+ + ∂ + ∂ + ∂∂ ∂ ∂ ∂ yuδ +

1 12 2

0z z z x z y zz zz zx zy

F F F Fu u u u

uδ δ δ δ

ε ε ε∂ ∂ ∂ ∂

+ + ∂ + ∂ + ∂∂ ∂ ∂ ∂

=

Integration by parts, where , , and vanish on the

boundary, [Dan3] leads to the variational equation

xuδ yuδ zuδ

1 12 2x y z

x xx xy xz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜ − ∂ − ∂ − ∂ +⎟⎜ ⎟⎟⎜ ∂ ∂ ∂ ∂⎝ ⎠x

1 12 2x y z

y yx yy yz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜+ − ∂ − ∂ − ∂ ⎟⎜ ⎟⎟⎜∂ ∂ ∂ ∂⎝ ⎠y +

1 12 2

0x y z zz zx zy zz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜+ − ∂ − ∂ − ∂ =⎟⎜ ⎟⎟⎜∂ ∂ ∂ ∂⎝ ⎠.

Since the variations , , and are independent, the

deflections , , and that minimizes or maximize the

box energy satisfy Euler’s Variational Equations

xuδ yuδ zuδ

xu yu zu

1 12 2

0x y zx xx xy xz

F F F Fu ε ε ε∂ ∂ ∂ ∂

− ∂ − ∂ − ∂ =∂ ∂ ∂ ∂

,

1 12 2

0x y zy yx yy yz

F F F Fu ε ε ε

∂ ∂ ∂ ∂− ∂ − ∂ − ∂ =

∂ ∂ ∂ ∂,

66


1 12 2

0x y zz zx zy zz

F F F Fu ε ε ε

∂ ∂ ∂ ∂− ∂ − ∂ − ∂ =

∂ ∂ ∂ ∂.

12.2 If the body forces originate from a Potential ( , , )x y zu u uυ

per unit volume, so that

xx

fuυ∂

− =∂

,

yy

fuυ∂

− =∂

,

zz

fuυ∂

− =∂

,

Then, Euler’s Equations are the Equilibrium Equations

, , , 0xx x xy y xz z xfσ σ σ+ + + = ,

, , , 0yx x yy y yz z yfσ σ σ+ + + = ,

, , , 0zx x zy y zz z zfσ σ σ+ + + = .

Proof: The Energy per unit volume is

( , , , , , , , , , , , )x y z xx yy zz xy yz zxF x y z u u u ε ε ε ε ε ε μ υ= +

1 1 12 2 2xx xx yy yy zz zz xy xy yz yz zx zxσ ε σ ε σ ε σ ε σ ε σ ε υ= + + + + + + .

The Euler’s Variational Equations are

67


1 12 2

22

0

xx xzxy

x y zx xx xy xz

F F F Fu

σ σσ

ε ε ε∂ ∂ ∂ ∂

− ∂ − ∂ − ∂ =∂ ∂ ∂ ∂

,

1 12 2

2 2

0

xy yy yz

x y zy yx yy yz

F F F Fu

σ σ σ

ε ε ε∂ ∂ ∂ ∂

− ∂ − ∂ − ∂ =∂ ∂ ∂ ∂

,

1 12 2

2 2

0

xz zzzy

x y zz zx zy zz

F F F Fu

σ σσ

ε ε ε∂ ∂ ∂ ∂

− ∂ − ∂ − ∂ =∂ ∂ ∂ ∂

.

That is, Euler’s Equations are the equilibrium Equations

, , , 0xx x xy y xz z xfσ σ σ+ + + = ,

, , , 0yx x yy y yz z yfσ σ σ+ + + = ,

, , , 0zx x zy y zz z zfσ σ σ+ + + = .

12.3 Principle of Virtual Work


Proof: The equilibrium deflections , and that minimize

the box energy per unit volume satisfy the Euler Variational

Equation

xu yu

68


1 12 2x y z

x xx xy xz

F F F FF u

uδ δ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜= − ∂ − ∂ − ∂ ⎟⎜ ⎟⎟⎜∂ ∂ ∂ ∂⎝ ⎠x +

1 12 2x y z

y yx yy yz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜+ − ∂ − ∂ − ∂ ⎟⎜ ⎟⎟⎜∂ ∂ ∂ ∂⎝ ⎠y +

1 12 2

0x y z zz zx zy zz

F F F Fu

uδ

ε ε ε

⎛ ⎞∂ ∂ ∂ ∂ ⎟⎜ ⎟⎜+ − ∂ − ∂ − ∂ =⎟⎜ ⎟⎟⎜∂ ∂ ∂ ∂⎝ ⎠.


0.Fδ =

Therefore,

( )U W U Wδ δ δ− = −

0 0 0

( , , , , , , , , , , , )yx z

y lx l z l

x y z xx yy zz xy yz zxx y z

F x y z u u u dxdydzδ ε ε ε ε ε ε== =

= = =

= ∫ ∫ ∫

0 0 0

( , , , , , , , , , , , )yx z

y lx l z l


F x y z u u u dxdydzδ ε ε ε ε ε ε== =

= = =

= ∫ ∫ ∫

. 0=


12.4 Principle of Minimal Energy for a Plate

The Box Energy is Minimal at Equilibrium

Proof: Legendre’s Sufficient Condition for

69


0 0 0

( , , , , , , , , , , , )yx z

y lx l z l


F x y z u u u dxdydzε ε ε ε ε ε== =

= = =∫ ∫ ∫

to be Minimal at the equilibrium generalizes to

2

2equilibrium ; ;

0( )

x y zij u u u

F

ε

∂>

∂.

By 11.2,

(1 ) 2 2 212 (1 )(1 2 )

( )Exx yy zzF ν

ν νε ε ε−

+ −= + + +

(1 )(1 2 )

{ }Exx yy yy zz zz xx

νν ν

ε ε ε ε ε ε+ −

+ + + +

υ

. 2 2 22 ( )xy yz zxG ε ε ε+ + + +

where is Young’s Elastic Modulus, 0E >

ν Poisson’s Ratio, and = 0 1ν< <

1EGν+

= =Shear Modulus.

Hence, 2

(1 )

(1 )(1 2 )20

( )E

xx

F νν νε

−+ −

∂= >

∂,

2(1 )

(1 )(1 2 )20

( )E

yy

F νν νε

−+ −

∂= >

∂,

2(1 )

(1 )(1 2 )20

( )E

zz

F νν νε

−+ −

∂= >

∂

70


2

124 2

( )E

xy

FG

νε +

∂= = >

∂0 ,

2

124 2

( )E

yz

FG

νε +

∂= = >

∂0 ,

2

124 2

( )E

zx

FG

νε +

∂= = >

∂0

Therefore, the Total Energy of the bar is minimal at the

equilibrium.

71


13.

Castigliano Theorems There are infinitely many body forces applying to the

infinitely many material particles of an elastic structure.

In Engineering, those many forces are replaced by several

forces concentrated and at their points of application in the

structure.

Then, Castigliano Theorem for the Displacements yields the

displacements as the derivatives of the complementary

strain energy with respect to the forces at the application

points.

And Castigliano Theorem for the Forces yields the forces as

the derivatives of the strain energy with respect to the

displacements at the application points.

13.1 Castigliano Theorem for the Displacements

If an elastic structure is at equilibrium under concentrated

Body Forces

1P , ,… 2P

applied at several points

72


Then the equilibrium displacements at the application

points are

11

Uu

P

∗∂=

∂, 2

2

Uu

P

∗∂=

∂,…,

where is the complementary strain energy of the

structure.

1( ,... )nU P P∗

Proof: in chapters 15, and 16.

13.2 Castigliano Theorem for the Forces

If an elastic structure is at equilibrium under concentrated

Body Forces that cause displacements

1u , ,… 2u

at the application points

Then the equilibrium forces at the application points are

11

UP

u∂

=∂

, 22

UP

u∂

=∂

,…,

where is the strain energy of the structure. 1,( ,... )nU u u

Proof: in chapters 15, and 16.

73


14.

Erroneous Derivations of

Castigliano Theorems

Castigliano Theorems follow from the Variational Principles

of Elasticity. They can be derived from the Principle of

Virtual Work, or from the Principle of Minimal Potential.

Attempting to prove them otherwise may fail.

The following erroneous derivation appears in [Timoshenko],

and is followed by [Mura].

[Mura, p.124] has

“The variation of the strain energy of an elastic

body of volume G is

,

,

( )j ij i ij j i

ij i jG u u

U u

σ δ σ δ

δ σ δ

∂ −

= ∫∫∫ dυ (12.7)

,( )

ij j iG

j ij i ij j iG G

n u ds

u d u d

σ δ

σ δ υ σ δ υ

∂

= ∂ −

∫∫

∫∫∫ ∫∫∫

where

is the surface that bounds G , G∂

74


is direction cosines so that jn j jn ds ds=

If

, 0ij jσ = , in G

ij j inσ = F

i

on G∂

we have Castigliano’s Theorem

[ ]i iG

U u F u dsδ δ∂

= ∫∫ .”

Clearly, there is no way to avoid the surface integration, and

pull from under the surface integral, in order to find iuδ

i

Uu

∂∂

.

But even more puzzling is the requirement , in G . , 0ij jσ =

By the equilibrium equations, this means that the body

forces, xf , yf , and zf vanish in the elastic body.

In other words, this requires bridges subject to no

gravitational forces.

Timoshenko’s derivation [Timoshenko, p.255] makes a

similarly puzzling claim, just wrapped in variational

notation, He has

0x xx y xy z xzδσ δσ δσ∂ + ∂ + ∂ = .

75


That means,

0x xx y xy z xzδ σ δ σ δ σ∂ + ∂ + ∂ = ,

( )

x

x xx y xy z xz

f

δ σ σ σ∂ + ∂ + ∂ = 0 ,

0xfδ = , in G

Similarly,

0yfδ = , in G

0zfδ = , in G .

Then, the partial derivatives

1

UP

∗∂∂

, 2

UP

∗∂∂

,…

are not defined.

76


15.

Castigliano Theorems and the

Principle of Virtual Work

15.1 Proof of Castigliano Theorem for Forces

If under several concentrated body forces applied at several

points, the strain energy U depends on the displacements,

1( ,... )nU U u u= ,

the infinitesimal work along ,…. is 1uδ nuδ

1 1 ... n nW P u P uδ δ= + + δ .

By the Principle of Virtual Work,

1 1 ... n nU W P u P uδ δ δ δ= = + +

Therefore,

11

UP

u∂

=∂

,… nn

UP

u∂

=∂

.

15.2 Proof of Castigliano Theorem for Displacements

77


If under several concentrated body forces applied at several

points, the Complementary Strain energy U depends on the

Forces,

∗

1( ,... )nU U P P∗ ∗= ,

the infinitesimal work by ,…. is 1Pδ nPδ

1 1 ... n nW u P uδ δ∗ = + + Pδ

P

.

By the Principle of Virtual Work,

1 1 ... n nU W u P uδ δ δ δ∗ ∗= = + +

Therefore,

11

Uu

P

∗∂=

∂,… n

n

Uu

P

∗∂=

∂.

78


16.

Castigliano Theorems and the

Hamiltonian

16.1 The Hamiltonian of Concentrated Forces

is the potential

1 1 1 1( ,... , ,... ) ....n n nP P u u U Pu P uΗ = − − n−

where

1 1( ,... ) ( ,... )n nU U u u U P P∗= =

16.2 Castigliano Theorem for the Forces is Hamilton’s

Equations

ii

UP

u∂

=∂

Proof: By the Principle of Minimal Potential, 0iu

∂Η=

∂.

79


16.3 Castigliano Theorem for the displacements is

Hamilton’s Equations

ii

UP

u∂

=∂

Proof: By the Principle of Minimal Potential, 0iP

∂Η=

∂.

80


17.

Castigliano and Beam Bending

without Shear

A beam of length l parallel to the ground, is attached to a

wall at one of the beam ends. A uniform force P pointing in

the ground direction, applies at each point along the beam,

and bends it down by y .

17.1 The x -Deflection due to the y -length Bending

( ) yx

uu y y

xθ

∂= − =

∂

81


Proof: ( ) xx

uu y y

yθ

⎛ ⎞∂ ⎟⎜ ⎟= − = −⎜ ⎟⎜ ⎟∂⎝ ⎠

Zero shear means that 0y xu u

x y

∂ ∂+ =

∂ ∂, Therefore,

yx

uu y

x

∂=

∂

17.2 The x -Strain due to the y -length Bending

2

20

yxx

y

uyx

ε=

⎛ ⎞⎟∂⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟∂⎜ ⎟⎟⎜⎝ ⎠.

17.3 The x -Stress due to the y -length Bending

2

20

yxx

y

uEy

xσ

=

⎛ ⎞⎟∂⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟∂⎜ ⎟⎟⎜⎝ ⎠.

17.4 The Bending Moment at , x

( )M x Px=2

22

0

y

AyI

uE y d

x=

∂=

∂ ∫∫ A

where I is the Inertia Moment.

Proof:

82


( ) xxA

M x ydAσ= ∫∫

2

22

0

y

A y

uEy dA

x=

∂=

∂∫∫

2

22

0

y

AyI

uE y d

x=

∂=

∂ ∫∫ A .

17.5 The x -Stress due to the y -length Bending

( )xx

M xy

Iσ = .

Proof: 2

20

( ) ( )y

y

u M x M xEy Ey y

EI Ix=

⎛ ⎞⎟∂⎜ ⎟⎜ ⎟ = =⎜ ⎟⎜ ⎟∂⎜ ⎟⎟⎜⎝ ⎠.

17.6 The Strain Energy of the Bending at x l , =

2

0

1( ) ( )

2

x l

x

U l M x dxEI

=

=

= ∫

Proof: 2

0

12

x l

xxx A

U dE

σ=

=

= ∫ ∫∫ Adx

83


2

0

1 ( )2

x l

x A

M xy dAdx

E I

=

=

⎛ ⎞⎟⎜= ⎟⎜ ⎟⎟⎜⎝ ⎠∫ ∫∫

2 22

0

1( )

2

x l

x A

I

M x y dAdxEI

=

=

= ∫ ∫∫

2

0

1( )

2

x l

x

M x dxEI

=

=

= ∫ .

17.7 The vertical deflection at x is l=3

3PlEI

Proof: By Castigliano, the vertical deflection at x is l=

2

0

( ) 1( )

2

x l

Px

U lM x dx

P EI

=

=

∂= ∂

∂ ∫

2

0

1( )

2

x l

Px

M x dxEI

=

=

= ∂∫

0

12 (

2

x l

PPxx x

M M x dEI

=

=

= ∂∫ ) x

3

2

03

x l

x

P Px dx

EI EI

=

=

= =∫l .

84


17.8 The deflection angel is 2

2PlEI

Proof: By Castigliano, the deflection angel is

2

0

( ) 1( )

2

x l

Mx

U lM x dx

M EI

=

=

∂= ∂

∂ ∫

2

0

1( )

2

x l

Mx

M x dxEI

=

=

= ∂∫

0

12

2

x l

Pxx

MdxEI

=

=

= ∫

2

02

x l

x

P Pxdx

EI EI

=

=

= =∫l .

85


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Documents

Infinitesimal Elasticity, Variational Principles, Castigliano Theorems