Basic Continuum Kinematics - Continuum Mechanics of Anisotropic Materials - page 45

dx 1

dx 1 + du 1

Fig. 2.8 An illustration for the geometric interpretation of the normal strain component E 11 .
The left and right illustrations of this figure represent the undeformed and deformed
configurations, respectively. The heavy black line represents the same material filament in the
two configurations. E 11 is equal to the change in length per unit length of the filament between
the two configurations. The original length is dx 1 and the change in length due to the deformation
is du 1 , thus E 11 ¼ du 1 / dx 1

dx 2 dx 2 + du 2

dx 3 + du 3

dx 3

dx 1 dx 1 + du 1

Fig. 2.9 An illustration for the geometric interpretation of the trace of the strain tensor, tr E , or the
divergence of the displacement field, u . 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. The volume element in the
undeformed configuration is dv o

∇ u ,tr E ¼ ∇

¼ dx 1 dx 2 dx 3 and the deformed volume is given by dv ¼ ( dx 1 þ

du 1 )( dx 2 þ du 2 )( dx 3 þ du 3 ). It may be shown (see text) that dv ¼ (1 þ tr E ) dv o . Thus the tr E

¼ ∇ u represents the change in volume per unit volume, ( dv dv o )/ dv o

may be neglected, it follows that dv ¼ (1 þ tr E ) dv o . Thus the tr E represents the

change in volume per unit volume, ( dv dv o )/ dv o .

The off-diagonal components of the strain tensor, for example E 12 , represent the
shearing strains. 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. To construct this geometric result algebrai-
cally, the unit vectors e 1 and e 2 are considered, see Fig. 2.10 . After deformation
these vectors are Fe 1 and Fe 2 , respectively, or, since F ¼ 1 þ E , the deformed

vectors are given by e 1 þ Ee 1 and e 2 þ Ee 2 , respectively. The dot product of the

vectors e 1 þ Ee 1 and e 2 þ Ee 2 is ( e 1 þ Ee 1 ) ( e 2 þ Ee 2 ) ¼ e 1 e 2 þ e 1 Ee 2 þ

e 2 Ee 1 þ Ee 1 Ee 2 , but since the unit vectors e 1 and e 2 are orthogonal, e 1 e 2 ¼ 0,

and also since Ee 1 Ee 2 is a higher order term because it contains the squares of the

displacement gradients (2.37), this expression reduces to ( e 1 þ Ee 1 ) ( e 2 þ Ee 2 ) ¼

e 1 Ee 2 þ e 2 Ee 1 . This result is further reduced by noting that e 1 Ee 2 ¼ e 2 Ee 1 ¼ E 12 ,