Four Linear Continuum Theories - Continuum Mechanics of Anisotropic Materials - page 148

The displacement u 2 represents the displacement curve. In the case of isotropy,
and in the notation customary of mechanics of materials, the deflection curve is
given by

Mx 2
2 EI :

y ð x Þ¼

Example 6.3.4
This problem concerns the propagation of elastic plane waves. The problem is to
determine the wave speed of a plane wave propagating in a symmetry direction in
an orthotropic material. Two kinds of plane waves are to be considered, one in
which the oscillating displacement component varies in the direction of propagation
and the other in which it varies in a direction perpendicular to the direction of
propagation. Let the direction of propagation be the x 1 direction and let the
direction perpendicular to the direction of propagation be the x 2 direction. Both
the x 1 and the x 2 directions are directions of the symmetry axes of the orthotropic
material. The displacement

that varies in the direction of its propagation is

u 1 ¼ u 1 ( x 1 , t ) and represents an time varying longitudinal motion of axial compres-

sion or tension or some combination of the two. The displacement that varies in the
perpendicular direction is u 1 ¼

u 1 ( x 2 , t ) and represents a time varying shearing

motion. The motion u 1 ¼ u 1 ( x 1 , t ) is called the longitudinal ( L ) motion and

u 1 ¼ u 1 ( x 2 , t ), the shear ( S ) motion. Neglect the action-at-a-distance force.

Solution : The differential equations governing these two motions are obtained from
the governing set of elasticity equations, the first three equations in this section.
First, from the strain-displacement relations (2.49) it follows that all the strain
components but one is zero for both the longitudinal and shearing motions and that
the nonzero components are given by

E 11 ¼ @

u 1 1

2 @

u 1

for ð L Þ and E 12 ¼ for ð S Þ:

@ x 1 @ x 2

Second, using these two results in the stress-strain relations for orthotropic
materials (6.22), it follows that

c 11 @

u 1

c 12 @

u 1

c 13 @

u 1

c 66 @

u 1

T 11 ¼

x 1 ;

T 22 ¼

x 1 ;

T 33 ¼

x 1 ;

for ð L Þ and T 12 ¼ for ð S Þ:

@ @ @ @ x 2

Third, upon substitution of these stresses and the functional form of the two
motions, u 1 ¼

u 1 ( x 2 , t ), in the stress equations of motion ( 6.18 )
one obtains the differential equations

u 1 ( x 1 , t ) and u 1 ¼

2 u 1
@

2 u 1
@

2 u 1
@

2 u 1
@

@

c L @ and @ c S @

t 2 ¼

for ð L Þ

t 2 ¼

for ð S Þ;

x 1 x 2