ABSTRACT
A well-known criteria that we remember from engineering and represents deformation, is named strain. Defining engineering strain is very simple, if we imagine a rod with an initial length l0 (see Fig. 2.1), when subjected to tension its length increases to ln, then the engineering strain is defined by:
εE n l l l
= − 0 0
(2.1.1)
If we multiply the numerator and denominator of Eq. (2.1.1) into l ln0 + then we have:
nl l l l l l l
l l
l l l
l l
l =
−( ) +( ) +( )
= −
+
= −0 0
2 2
+ −
= − +( )l l
l
l l ln
Eε
In the above expression if εE << 2, then we can ignore it and define a new strain known as “Green strain” which will be:
εG n l l
l =
(2.1.2)
Comparing Eqs. (2.1.1) and (2.1.2) shows that ε ε εE G
= +( )1 2
or
alternatively:
ε ε εG E E= +( )1 2 It is obvious that if εE <<1 (small deformation), in Eq. (3) we can ignore
the term εE 2
2 and then we can say that Green strain εG and engineering
strain εE are close ε εG E≅ . Meanwhile we can rearrange Eq. (2.1.2) into
the following form:
l l ln G 2
0 22− = ε (2.1.3)
When the deformation is large and is not a uni-axial tension type, we can also define the Green strain tensor by relaying on a simple definition in Eq. (2.1.3). According to Fig. 2.2, vector r shows position of a point in un-deformed position and the metric tensor can be shown by g , the differential length element ds can be found from Eq. (1.3.1) written by:
ds g dx dx i jij i j2 1 2 3= =. , , (2.1.4)
In the deformed body vector
R is position of the same point and the metric tensor will change to G because the deformation may be too big that
causes change of the natural coordinates. Similarly the differential length element dS can be found from Eq. (1.3.1) written by:
dS G dx dx i jij i j2 1 2 3= =. , , (2.1.5)
General form of Eq. (2.1.3) gives Green strain tensor like this:
dS ds dx dxij i j2 2 2− = ε (2.1.6)
By substituting Eqs. (2.1.4) and (2.1.5) into Eq. (2.1.6) we have:
G dx dx g dx dx dx dxij i j
ij i j− = 2 ε
Simplify the above equation gives the Green strain tensor which is:
ε ij ij ijG g= −( )12 (2.1.7) Equation (2.1.7) shows that Green strain tensor has covariant components; therefore, it should be expressed by contra-variant base vectors as indicated in Eq. (1.2.11) ε ij can be expressed by the un-deformed contravariant base vectors like this:
(2.1.8)
According to Fig. 2.2, the displacement vector u is:
R r u= +
If we differentiate versus the coordinate αi, then we have ∂ ∂
= ∂ ∂
+ ∂ ∂
R r u
i i iα α α
since ∂ ∂
=
R G i
iα , ∂ ∂
= r g i
iα and ∂
∂ =
u u i
iα , then we can write:
(2.1.9)
By substituting Eq. (2.1.9) and also g g gij i j= ⋅ into Eq. (2.1.7) the Green strain tensor can be written in terms of displacement vector as follows:
ε ij i j j i i jg u g u u u= ⋅ + ⋅ + ⋅( )12 , , , , (2.1.10)
It is also possible that express the Green strain tensor in terms of the deformed covariant base vectors, that is, substitute and also G G Gij i j= ⋅ into Eq. (2.1.7) to have the following expression:
ε ij i j j i i jG u G u u u= ⋅ + ⋅ − ⋅( )12 , , , , (2.1.11)
It is also possible to use covariant differentiation of the displacement components from Eq. (1.3.16) to change the strain formulas for strain tensor, that is,
u u g u u gm m
u U G u U Gm m
It should be remembered that Eq. (1.3.16) is written in terms of covariant base vectors while Eqs. (2.1.12a) and (2.1.12b) are written in terms
of contra-variant base vectors. This is because Green strain is a covariant tensor and needs an expression with covariant components.
