Kinematics and Mechanics of Large Elastic Deformations - Continuum Mechanics of Anisotropic Materials - page 310

On the other hand, the relationship between d A and d a is constructed by twice
substituting d X F 1 ¼ d x into d A ¼ d X ( r ) d X ( s ),

F 1 F 1 dA ¼f dx ð r Þg f dx ð s Þg: (11.40)

The vector d A is then dotted with the inverse deformation gradient F 1 from the
left, thus

F 1 F 1 F 1 F 1

dA ¼f dx ð r Þg f dx ð s Þg : (11.41)

The right-hand side of this equation may be expanded, as a similar one for d a
was above, and, using the fact that a determinant of a product of matrices is the
product of the determinants (Section A.8, page 372), rewritten as

F 1 J 1 da

dA ¼ : (11.42)

These formulas relating d A and d a are called the formulas of Nanson . These
formulas will be employed in relating the various definitions of stress associated
with large deformations.

Example 11.5.1
Consider the plane area that forms the right-hand face of the unit cube illustrated in
Fig. 11.9 . Use the formulas of Nanson to determine the magnitude and orientation
of the deformed area as a result of the deformation specified in Example Problem
11.3.1.
Solution : The undeformed area is represented by d A T ¼

(1, 0, 0). The value of J is

2 √ 3 and the tensor of inverse deformation gradients is given by

2 3

1 1

3

0

p

p

2

4 5 :

F 1 ¼

1

0

2 0
001

F 1 , thus From ( 11.42 ) it follows that d a ¼ J d A

8
<

2 3 9

=
; ¼

2 3

1
2

3

1

0

2

p

p

2 p

4 5 4 5 :

da ¼ ½ 100 1

0

1

0

2 0
001

:

The deformed representation of the plane area that forms the right-hand face of
the unit cube illustrated in Fig. 11.7 is shown as the right sloping right-hand face in
Fig. 11.8 . From that figure it is seen that the face associated with this area is two
units high and one unit wide, and that the unit normal to this face is indeed (2/

√ 3,

1/ √ 3, 0).