CONTINUUM MECHANICS for ENGINEERS

with analogous expressions for and in the other coordinate directions.
Consider next the unit extension (longitudinal strain) in any direction
at P. This may be expressed in terms of the stretch as
ˆN ˆn
ˆN
(4.8-10)
Notice that the unit extension is zero when the stretch is unity, as occurs
with a rigid body displacement. If Nˆ Iˆ
,
or, solving for E 11,
(4.8-11)
1 2
E11 = e + e
(4.8-12)
( Iˆ ) ( Iˆ
2 )
For small deformation theory where E11 → ε11, and for which may be
neglected in comparison to , the above equation asserts that E11 = ε11 = .
The change in angle between any two line elements may also be given in
terms of stretch. Let dX (1) and dX (2) be arbitrary vectors which become dx (1)
and dx (2) , respectively, during a deformation. By the dot product, dx (1) ⋅ dx (2) =
dx (1) dx (2) cos θ, we may compute the angle θ between dx (1) and dx (2) N
e e
( Nˆ
)
( Iˆ
)
from its
cosine, which with the help of Eq 4.8-3 takes the form
cos θ =
e
dx − dX
= = Λ − 1= NCN ˆ ⋅ ⋅ ˆ −1
dX
( ˆ ) ( ˆ N N)
or upon dividing the numerator and denominator by the scalar product
dX (1) dX (2) and making use of Eqs 4.8-4 and 4.6-12 we obtain
Nˆ ˆ
1⋅C⋅N2 cos θ = (4.8-13)
Λ Λ
( Nˆ Nˆ
1) ( 2)
Thus, for elements originally in the Î and directions, the angle between
1 Î2 them in the deformed configuration may be determined from
= 1
e = Iˆ ⋅C⋅Iˆ − 1= C − 1= 1+ 2E −1
( ˆ I
1
)
1 1 11
1 1
( 1)
( 2)
( 1)
dx
dx
F⋅dX ⋅ =
( 1)
( 2)
dx dx dX ⋅C⋅dX cos
θ 12
( 1) ( 1)
C12 C12
= =
Λ Λ C C
( ˆ ) ( ˆ I1 I2)
⋅
11 22
11
( 2)
F⋅dX dX ⋅C⋅dX ( 2) ( 2)
2
e
( ˆ )
(4.8-14)