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and similarly for the other two pairs of faces. The total force acting on the parallelepiped in
the x direction thus is
∂σ xx
x + ∂σ xy
y + ∂σ xz
x
y
z
z
and similarly for the force in the y and z directions. Using Newton's second law of motion
(force = mass ×
acceleration), we can write
∂σ
2 u
x + ∂σ
y + ∂σ
z
xx
xy
xz
(A2.33)
x
y
z
= ρ
x
y
z
t 2
where
is the density of the parallelepiped and u is the x component of the displacement. (We
assume that all other body forces are zero; that is, gravity does not vary significantly across
the parallelepiped.) This equation of motion, which relates the second differential of the
displacement to the stress, can be simplified by expressing stress in terms of strain from
Eqs. (A2.19) and strain in terms of displacement from Eqs. (A2.3), (A2.6) and (A2.7).
Substituting for the stress from Eqs. (A2.19) into Eqs. (A2.33)gives
ρ
2 u
ρ
=
x (
λ +
2
µ
e xx )
+
y (2
µ
e xy )
+
z (2
µ
e xz )
(A2.34)
t 2
Substituting for the strains from Eqs. (A2.3), (A2.6) and (A2.7) into Eq. (A2.34)gives
µ
∂v
ρ
2 u
µ
u
x +
u
=
λ +
2
+
t 2
x
x
y
y
µ
w
z
x + u
(A2.35)
+
z
Assuming
λ
and
µ
to be constants, we can write
2 u
2 u
2 u
2 v
2 w
2 u
ρ
= λ
µ
+ µ
y + µ
z + µ
x +
2
+ µ
t 2
x 2
y 2
x
x
z 2
z 2
2 u
x 2
2 u
y 2
2 u
= λ
x
u
x +
y +
v
w
z
+
+
x + µ
+ µ
= λ
x + µ
2 u
x + µ
(A2.36a)
z 2 (see Appendix 1).
Likewise, the y and z components of the forces are used to yield equations for v and w :
2
2
x 2
2
y 2
2
where
is the Laplacian operator
/∂
+
/∂
+
/∂
2 v
ρ
)
2 v
=
(
λ + µ
y + µ
(A2.36b)
t 2
2 w
ρ
= ( λ + µ )
2 w
(A2.36c)
z + µ
t 2
These three equations are the equations of motion for a general disturbance transmitted
through a homogeneous, isotropic, perfectly elastic medium, assuming that we have
infinitesimal strain and no body forces. We can now manipulate these equations to put them
into a more useful form.
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