CONTINUUM MECHANICS for ENGINEERS

which, upon factoring the left-hand side and dividing by (dX) 2 , becomes
dx − dX dx + dX dX dX
i j
= 2εij
dX dX dX dX
But dX i /dX = N i, a unit vector in the direction of dX, and for small deformations
we may assume (dx + dX)/dX ≈ 2, so that
dx − dX
= εijNN i j = Nˆ ⋅ε⋅Nˆ dX
(4.7-9)
The scalar ratio on the left-hand side of this equation is clearly the change
in length per unit original length for the element in the direction of . It
is known as the longitudinal strain, or the normal strain and we denote it by
. If, for example, is taken in the X1 direction so that , then
ˆN
ˆ N
Nˆ = Iˆ 1
e ( ˆ N )
ˆ ˆ
ˆ ˆ
e
( Iˆ
1)
= Iˆ ⋅ε⋅ Iˆ
= ε
1 1 11
Likewise, for N = I , or the normal strains are found to be ε22 and
2 N = I3
ε33, respectively. Thus, the diagonal elements of the small (infinitesimal)
strain tensor represent normal strains in the coordinate directions.
To gain an insight into the physical meaning of the off-diagonal elements
of the infinitesimal strain tensor we consider differential vectors dX (1) and
dX (2) at position P which are deformed into vectors dx (1) and dx (2) , respectively.
In this case, Eq 4.6-19 may be written,
dx (1) ⋅ dx (2) = dX (1) ⋅ dX (2) + dX (1) ⋅ 2ε ⋅ dX (2) (4.7-10)
which, if we choose dX (1) and dX (2) perpendicular to one another, reduces to
dx (1) ⋅ dx (2) = dx (1) dx (2) cosθ = dX (1) ⋅ 2ε ⋅ dX (2) (4.7-11)
where θ is the angle between the deformed vectors as shown in Figure 4.3.
If now we let , the angle measures the small change in the original
right angle between dX (1) and dX (2) π
θ = −γ
γ
2
and also
π
cosθ = cos γ = sin ≈
2 −
⎛ ⎞
γ γ
⎝ ⎠
since is very small for infinitesimal deformations. Therefore, assuming
as before that dx (1) ≈ dX (1) and dx (2) ≈ dX (2) γ
because of small deformations
γ ≈ θ = ⋅ ⋅ ≈ ⋅ ⋅
cos ( 1)
( 2)
dX
dX
2ε Nˆ 2εNˆ
( 1)
dx dx
( 2) ( 1) ( 2)
(4.7-12)