Basic Continuum Kinematics - Continuum Mechanics of Anisotropic Materials - page 46

dx 2 + du 2

φ

dx 2

(π/2 - φ)

dx 1 + du 1

dx 1

Fig. 2.10 Illustration for the geometric interpretation of the shearing strain E 12 . The left and right
illustrations of this figure represent the undeformed and deformed configurations, respectively.
The heavy black lines represent the same material filaments in the two configurations. E 12 is equal
to one-half the change in angle that was originally a right angle between the x 1 and x 2 axes, f /2 in

this figure

thus it follows that ( e 1 þ 2 E 12 . Recalling the formula (A61) for

the dot product of two vectors, say u and v , as equal to the magnitude of the first times
the magnitude of the second times the cosine of the angle (say

Ee 1 ) ( e 2 þ Ee 2 ) ¼

z ) between them, u v

¼ u i v i ¼j u jj v j cos z , it follows that 2 E 12 ¼j e 1 þ Ee 1 jj e 2 þ Ee 2 j cos ( p /2 f ),

where the angle ( p /2 f ) is illustrated in Fig. 2.10 . The magnitude of j e 1 þ Ee 1 j

is the square root of ( e 1 þ Ee 1 ) ( e 1 þ Ee 1 ) ¼ e 1 e 1 þ e 1 Ee 1 þ e 1 Ee 1 þ Ee 1 Ee 1 ,

but since e 1 e 1 ¼ 1and Ee 1 Ee 1 is a higher order term, this reduces to the square root

of 1 2 E 11 , by a parallel of the arguments used above to obtain the formula for 2 E 12 .

At this point a classical approximation is used. This approximation is that 1

þ

þ e

2 may be neglected; the proof of this
approximation follows easily if one squares it. Then, since the square of E 11
is a higher order term, the square root of 1

√ (1 þ 2 e ) if summands of the order e

þ

2 E 11 is given by 1

þ E 11 ,thus2 E 12

¼ þ E 11 )(1 þ E 22 )cos( p f

) r E 12 ¼

þ E 11 )(1 þ E 22 )sin f

(1 /2 (1 or

expanding; 2 E 12 ¼ sin f þ ( E 11 þ E 22 )sin f þ E 11 E 22 sin f . Finally, since the

angle is small as are E 11 and E 22 , thus the neglect of higher order

terms gives 2 E 12 ¼ f , and the interpretation of E 12 as one-half the change in an angle
that was originally a right angle between the x 1 and x 2 axes (Fig. 2.10 ). These geometric
interpretations of the components of E as the change in the length of filaments, the
change in angles and the change in volume deformation between the undeformed
and the deformed configurations are the rationale for calling E the strain tensor.

f is small, sin f

Example 2.3.1
The deformation gradient and the inverse deformation gradient for the motion given
by ( 2.12 ) were computed in Example 2.1.4. Determine the restriction on the motion
given by ( 2.12 ) so the motion is infinitesimal. Find the strain tensor E and the
rotation tensor Y for the infinitesimal motion.