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" ( P )= ^ i~, (P) = 7T+X axoj/ dz dy ox dz axSubstitution of eqs. 3.143 and 3.144 into eq. 3.145 yields
3.145)
Since
we have f° ( 1
1 \ 92i?
f° 1
also (;
/° / i i \ a 2 i? P = 7 - ^ - -3- TTTT' 8?r \cf cjj dxdy
and 8vr \ C ,
cf/ oxaz
3.147)
156
CHAPTER 3. ELASTIC
WAVES IN A HOMOGENEOUS
MEDIUM
We see that the point force ffc causes displacement of particles in all directions, and, as was pointed out earlier, this field arises as a result, of superposition of the longitudinal and shear waves. It is also instructive to determine the dilatation and curl of field s. Performing differentiations, we obtain div s = - ^f°^
d 1
or diVS
= -
f°
d 1
4
^
^
and
f°
d 1
curias = 0,
curias = — ^2 — —, 4nc s dz R
curias = --r^ 2 ^—5 4irc s dy R
curl r s = 0,
curies = - ^ ^ ,
curl 2 s = l
or l
(3.150)
Thus, in general, the point force produces both deformation of an elementary volume and its rotation. It is easy to derive by analogy expressions for field s when the point force is directed along either the y— or z—axis.
3.4
Propagation of waves caused by the point force
As in the previous section, we assume that force F coordinate origin, and it has the x-component only:
is applied in the vicinity of the
F = (F x , 0, 0)
(3.151)
Here Fx is an arbitrary function of time. To begin with, we will use results derived from studying the displacement field in equilibrium. First of all, let us introduce the potentials of the body force f acting on unit mass, f = grad U + curl A,
(3.152)
where all three functions, f, U, and A, depend on time and coordinates of a point. Taking a divergence and after it curl from both sides of eq. 3.152, we obtain Poisson's equations for both potentials: V 2 £/ = div f,
V 2 A = -curl f
(3.153)
3.4. PROPAGATION OF WAVES CAUSED BY THE POINT FORCE
157
The last equality is valid, provided that div A = 0 These equations are exactly the same as in the case of equilibrium, and, correspondingly, for the point force /° at the origin, eq. 3.128, we have
Here r0 is the distance from the origin to an arbitrary point q(x', '', z'). Note that potentials U(q,t) and A(q,t) synchronously change with the force applied at the origin and this happens regardless of the position of point q. This fact vividly demonstrates the auxiliary character of functions U and A. Scalar and vector potentials of field s The relationship between the displacement s and its potentials, s = grad f + curl tp,
(3.155)
is always valid, since it follows from the system of equations curl s = W
and
div s = 0
Now, as in the case of equilibrium, we establish a linkage between the potential of fields t(q, t) and s(q, t). With this purpose in mind, we use the equation of motion (Chapter 2): d2s (X + /i)grad div s + /iV2s + pi = p— Substitution of eqs. 3.155 and 3.152 into eq. 3.156 yields (A + /i)grad div(grad<^ + curl •0) + /*V (grad<^ + curl ift) d2 +p(grad U + curl A) = p—-(giadip + curl ip) z at
(3.156)
158
CHAPTER 3. ELASTIC
WAVES IN A HOMOGENEOUS
MEDIUM
or (A + 2/i)gradV2w + p grad U + a curl V2t/> + p curl A —p grad—^- + p curl —— otz at1 This means that
that is,
W
^
=- ^
(3.157)
and
These arc inhomogencous wave equations, and their right sides arc represented by potentials of the body force. Earlier, we learned that longitudinal and shear waves caused by the point spherical source obey homogeneous wave equations at regular points. As we know, the solutions for scalar potential
0. In the same manner as for s i , field s 2 decays with distance as 1/R, provided that the argument t — R/cs remains constant. Correspondingly, we again cannot distinguish the near, intermediate, and far zones. From eq. 3.200, it follows that field s 2 is directly proportional to projection of the point force on the direction defined by unit vector i s , which is perpendicular to the radius-vector R . Now, using eq. 3.200, we determine div s2 and curl s 2 . Since s 2fl = 0,
s28 = — ^ / x ° (t - - j cos 9 cos
• 0
and
V; -> 0,
if
r -> oo
(6.22)
and the Sommerfeld condition of radiation has to be met (Part II). 4. Finally, at the free surface, normal and shear stresses must vanish TZZ = 0
and
TXZ = 0,
ft
z = -H
(6.23)
We assume that in the Cartesian system of coordinates, displacement s : s — grad if + curl ip
(6.24)
is characterized by only two components, u and w, s = ui + wk
(6.25)
For this reason we have chosen the y-component of vector potential ipy. Correspondingly, eq. 6.24 gives ~ u
dip
dip
= ^r-^r> ox
az
~ w=
dip
Oil)
^r + ^ az
ox
.
.
6 26
-
6.1 LINEAR SOURCE OF P and SV WAVES ...
281
Then, taking into account Hooke's law: ~ . , ~ dw TZZ = Adivs + 2^. — ,
~ ^
r
=
I du dw \ M ^ + a ^ ) '
. , (6-27)
cqs. 6.23 become
-x& + J*»+*±\=Bi I oz2
2J!!L+ ^ 2_ ^z=0
ox oz f
ox oz
ox
(r28)
oz
Note that we have already used these boundary conditions studying Rayleigh waves. Thus, we have formulated the boundary value problem, and in accordance with the theorem of uniqueness, only one wavefield obeys all four conditions. First, taking into account that the free surface is a plane, z = —H, it is natural to find a solution of the Helmholtz equations in the Cartesian system of coordinates. For instance, in the case of scalar potential, we have
(6 29)
a £ + a # + *«V = °.
'
which describes the potential at regular points. Applying the method of separation of variables (Part II), we represent the function
x< = ezx + (p ezy,
ez,z, = ezz,
(E-95)
ex,y, = exy + 2ip(eyy - exx)
It is clear that any plane in an isotropic medium is a plane of symmetry and elastic properties are independent on a direction. For this reason, we can use eq. E-93, which can be written in new system of coordinates: Mi = - y ( e £ v + eylyl + e'f v ) + cl2{ey>y<ezlzl + e zV e x - x / + e ^ e ^ y )
(E-96)
Now we represent u\ as a sum of UQ and terms, depending on (p in the first power. Substituting cqs. E-95 into cq. E-96 and preserving terms with