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Box 1.56 (The canonical representation of left angular shear and right angular shear. Special
case: G 12 =0, c 12 =0and g 12 =0, C 12 =0).
Left and right angular parameters Ψ, and Ψ r :
cos Ψ l = C 11 d u
d s
versus c 11 d U
d S
d s
d S
d S
d s =cos Ψ r ,
cos Ψ l = G 11 d u
1
λ
c 11 d U
1
Λ =cos Ψ r ,
versus
d s
d S
(1.331)
sin Ψ l = C 22 d v
d s
versus c 22 d V
d S
d s
d S
d S
d s =sin Ψ r ,
sin Ψ l = C 22 d v
1
Λ =sin Ψ r .
Left and right stretches, left and right principal stretches:
1
λ
versus c 22 d V
d s
d S
λ 2 ;= d S 2
d s 2
versus Λ 2 := d s 2
d S 2 ,
λ 1 = C 11
versus Λ 1 = c 11
G 11 .
(1.332)
g 11
λ 2 = C 22
versus Λ 2 = c 22
G 22 ,
g 22
G 11 U =cos Ψ 1
cos Ψ r = g 11 u
C 11
g 11
cos Ψ l c 11
G 11
1
λ
1
Λ =cos Ψ r ,
cos Ψ l =cos Ψ r
versus
(1.333)
C 22
g 22
sin Ψ l c 22
G 22
1
λ
1
Λ =sin Ψ r ,
sin Ψ l =sin Ψ r
versus
cos Ψ l =cos Ψ r λ 1
λ
cos Ψ l Λ 1
versus
Λ =cos Ψ r ,
sin Ψ l =sin Ψ r λ 2
sin Ψ l Λ 2
versus
Λ =sin Ψ r ,
(1.334)
λ
tan Ψ l = λ 2
Λ 2
Λ 1 tan Ψ l =tan Ψ r .
λ 1 tan Ψ r
versus
Left and right angular shear, left and right angulardistortion:
tan( Ψ l − Ψ r )=tan l
versus tan r =tan( Ψ r − Ψ l ) ,
tan l =
versus tan r =
tan Ψ r
1+tan Ψ l tan Ψ r
tan Ψ l
tan Ψ l
1+tan Ψ r tan Ψ l ,
tan Ψ r
tan l = tan Ψ l
versus tan r = tan Ψ r
Λ 2 Λ 1 tan Ψ l
1+ Λ 2 Λ 1 tan 2 Ψ l
λ 2 λ 1 tan Ψ r
1+ λ 2 λ 1 tan 2 Ψ r
,
(1.335)
 
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