Principles of Virtual Work: Integral Form of the Basic Equations - Mechanics of Structures: Variational and Computational Methods - page 78

The work for the whole body of volume V then amounts to

( x , y , ... , γ yz )

W i =− (σ

x d

+ σ

y d

+···+ τ

yz d

γ ) dV

x y yz

V ( 0 , 0 , ... 0 )

ij

σ T d dV

=− σ =−

ij d ij dV

(2.6)

V 0 V 0

with [Chapter 1, Eqs. (1.3) and (1.6)]
σ =

τ yz ] T σ 23 ] T

[ σ x σ y σ z τ xy τ xz = [ σ 11 σ 22 σ 33 σ 12 σ 13

(2.7)

11 22 33 γ 12 γ 13 γ 23 ] T
The generalized Hooke's law takes the form [Chapter 1, Eqs. (1.34) and (1.35)]

yz ] T

= [ xx yy zz 2 xy 2 xz 2 = [

σ ij = E ijkl kl

or σ = E , so that for elastic bodies

1
2

1
2

T E dV W i =− E ijkl

kl dV

=−

ij

V V

1
2

1
2

σ T dV =−

V σ

ij dV

=− (2.8)

ij

V

where σ T E T .

For a three-dimensional linear elastic body for which the strain energy is equal to but
opposite in sign to the work of the internal forces,

T

= T E T

= ( E ) and, because E is symmetric, E =

1
2

1
2

T E dV U i =− W i = U 0 () dV = E ijkl

kl dV

= (2.9)

ij

V V V

2 T E is the specific potential of a unit volume or the strain energy density .
With a symmetric matrix E , U 0 has the quadratic form

1

in which U 0 () =

1
2 T E

U 0 () =

G

23

2 γ

ν 1

2
11

2
22

2
33

2 2

12

2
13

2

= + + +

ν ( 11 + 22 + 33 )

+ + γ + γ (2.10)

1 − 2

Here, it has been assumed that the material is isotropic, and E is taken from Eq. (1.34). A
quadratic function which is never negative for arbitrary values of the variables is called a
positive definite (quadratic) function. Such a function is zero only when each variable is zero.
As is shown in introductory mechanics of solids textbooks for isotropic materials, the shear
modulus of elasticity G must be positive, and Poisson's ratio

ν can vary between zero and

one half, i.e., G 5. Thus, the strain energy density of Eq. (2.10) is a positive

definite function in the strain components.
It is useful to visualize again the strain energy in terms of a stress-strain curve of an
element, such as an extension bar. From Eq. (2.4)
U 0
A =

> 0 , 0 ≤ ν ≤ 0 .

x
0 σ

x d

(2.11)

x

is the area under the curves of Fig. 2.2. The area above the curve would be x d

σ x . Using

integration by parts, it follows that

U 0
A

σ = σ − σ = σ −

x d x d

(2.12)

x x x x x x

The area above the curve is referred to as the complement of the area U 0 / A . Set

σ x

U 0
A

x d

σ = (2.13)

x

0