Page 25 - Mechanics of Asphalt Microstructure and Micromechanics

18 Ch a p t e r O n e

Therefore:
. . .
•
•+
dS F dS F = J dS = JtrLdS Multiply both sidees by F -1
o
o
. .
•
•
dS dS F F = JtrLdS • F = trLdS
+
o
−1
−1
.
L =• −1 One has:
F F
.
•
−
dS = trLdS dS L
(1-89)
1.6.2.6.3 Volume Change
In the reference configuration:
dV = dX 1 () • dX 2 ( ) × dX 3 ( ) = ξ dX dX dX 3 ( ) ) (1-90)
2 ( )
0
1 ()
ABC A B C
In the current configuration:
dV = dx • dx ( ) 2 × dx ( ) = ξ dx dx dx ( )
()
1
3
3
( )
2
()
1
ijk i j k
•
•
•
2
3
3
1
2
()
1
dx () = F dX dx ( ) = F dX ( ) dx ( ) = F dX ( )
1
3
2
2
dx () 1 = x dX () dx ( ) = x dX () dx ( ) 3 = x dX ()
X
,
iA A i B B i C C
,
,
dV = ξ x x x dX dX dX ( ) 3 = JdV 0
2
()
1
( )
,
,
,
ijk i A i B i C A B C
. .
dV = J dV = J trL dV = ( trL dV (1-91)
o
o
)
)
(

If the deformation is isochoric, then dV = 0 trL = 0
1.6.2.6.4 Rate of Stretch Ratio
Considering the identity:
dxn = x dXN (1-92)
,
i i A A
nα = x N
i i A A
,
or
nα = F N
•
So:
·
·
•
nα + nα = F N = • • L nα•
L F N =
·
·
nnαα = L n
•
+
/
· ·
•+
nn nnαα = nL n
•
••
/
·
·
nL
/
αα =• • n