Basic Equations: Differential Form - Mechanics of Structures: Variational and Computational Methods - page 6

notation, the stresses are gathered together as

σ x
σ y
σ z
τ xy
τ xz
τ

τ yz ] T

σ = = [ σ x σ y σ z τ xy τ xz (1.3a)

yz

It is sometimes convenient to write the stresses as

σ yz ] T

σ = [ σ xx σ yy σ zz σ xy σ xz (1.3b)

or

σ 23 ] T

σ = [ σ 11 σ 22 σ 33 σ 12 σ 13 (1.3c)

where x 3 have been used.

The deformed shape of a solid subjected to loading can be described by the three dis-
placement components

= 1, y = 2, z =

u x = u x ( x, y, z ) = u = u 1

u y = u y ( x, y, z ) = v = u 2 (1.4)

u z = u z ( x, y, z ) = w = u 3

or in vector notation

u x
u y
u z

=

u z ] T ] T u 3 ] T

u = [ u x u y = [ u vw = [ u 1 u 2 (1.5)

As indicated in Problem 1.35, rotary displacements can be derived from the translations
u x ,u y , and u z . The normal strains are designated by

x , y , and z , while the shear strains

are γ

xy ,

γ

xz , and

γ

yz . As in the case of stresses, the strains are symmetric, i.e.,

γ = γ

ji .In

ij

matrix notation,

x

y

z

yz ] T

= = [ γ γ γ

x y z xy xz

γ

xy

γ

xz

γ

yz

or

yz ] T

= [ xx yy zz 2 xy 2 xz 2 (1.6)

or

23 ] T = [ 2 2 2

11 22 33 12 13

where the relationship j, will be explained in Section 1.2.

Note that the strain components are placed in the matrix in the same order as the cor-
responding stress quantities. Then a measure of the internal work

γ = 2

ij ,i

=

ij

ik is simply equal
to the vector product σ T . Note that the definitions of Eq. (1.6) in which

σ

ik

γ = 2

yz , etc.,

yz

σ T . To show this, return to do not violate the equality σ ik ik = σ

ik , of Eq. (1.1). Since

ik

= σ

the stress and strain components are symmetric, e.g.,

31 , ik of Eq. (1.1) becomes 13 ik