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Unformatted text preview: ll—l CHAPTER 11 REFLECTION AND REFRACTION AT A PLANE INTERFACE 111 INTRODUCTION When a dilantionalor shear wave impinges on a boundary between two
media, both ncﬁﬂeciion and haﬁaaciion take place. In general, an incident
dilatational or shear wave will produce both reflected and refracted
dilatational and shear waves, i.e., the boundary produces coupling between
the two types of waves. The amplitude of these reflected and refracted waves depends, among other things, on the incident angle with the boundary. 11.2 REFLECTION OF A PLANE DILATATIONAL WAVE FROM A PLANE FREE
BOUNDARY Consider first the case of an incident plane harmonic dilatational wave impinging on a plane free boundanu see Figure. Ty A1 (dilatational) free
surface wﬁﬂﬁﬂﬂ////M/”M?dilatational) 11—2 It can be shown that in order to satisfy all boundary conditions on
the free surface, the incident dilatational wave with amplitude Al’ must
give rise to 1W0 acﬁicctcd waves with amplitudes A2 and A3. One of these reflected waves must be another dilatational wave, while the second must be a shear wave. Denote the incident and reflected dilatational waves by ¢1 ¢2 A2 sin(wt +f2x + gzy + 61) , Al sin(wt4~flx + gly) (11.1) where $1 2 is the dibpﬂacemcnt moiion normal to the wave front, and
, l
l
 I
f = w cos a1 _ w Sln a1
1 c1 ’ gl‘ cl 1
(11.2) f
f — w cos a2 g _ w Sln d2
— ’ 2" 2
2 c1 c1 ‘
with 61 being the phase change.
Denote the reflected shear wave by
E
1
$3 = A3 s1n(wt — f3x + g3y + 62) , (11.3) where $3 is the displacement motion parallel to the wave front in the x—y *
plane (no motion in z—direction) , w cos 52 w sin 82
f3 = ——7;—— , g3 = ———— , (11.4)
2 2 and 62 is the phase change. This type of wave is sometimes called an SV—wave, i.e., ventécaﬂﬁy
pokahizcd Ahead wave. This nomenclature comes from geophysics and is
used to describe a variety of wave types; e.g., a paimany or P—wavc is
a dilatational wave, a Accondang or S—wavc is a shear wave, and an SH—
wave is a honizontaﬂﬁy poﬁahizcd Ahcah wave. The displacements corresponding to (11.1) and (11.3) are given by, ul = $1 cos a1
Vl = $1 sin a1
u2 = —¢2 cos a2
v2 = $2 sin a2
u3 = ¢3 sin 82 <
ll 3 $3 C08 82
The total displacements are then u = ul + u2 + u3 vl + v2 + v <1
[I 3 The normal stress OX follows from Hooke's law to be Q
II
E
+
I _ ‘ I — [)x(fl cos a1 + g1 s1n a1) + 2p fl cos a1]¢l . V 2 + g2 Sln a2) + 2p f2 cos uz]¢2
_ . _ a y + [A( f3 Sin a2 + g3 cos 82) Zn f3 Sln 82]¢3 , + [Mf2 cos a where
l _.
$1 — Al cos(wt + le + gly)
' — .—
¢2 — A2 cos(wt fzx + gzy + 61)
v __ _
$3 — A3 cos(mt f3X + g3y + 62) Hence it follows that EL. 2 G = [A1(A + 2p cos a1)cos(wt + gly) 
X x=0 Cl w 2
+ A2 c; [(A + 2p cos a2)cos(wt + gzy + 61)] BL. c2 [2U sin 82 cos 82 A3 cos(wt + g3y + 52)] (11.5) (11.6) (11.7) (11.8) 11—3 Now the stress UK at X = 0 must vanish to satisfy our boundary conditions. The only way (11.8) can vanish for all time and y is to have 0‘i = 0‘2
g1 = g2 ‘ g3 (11.9)
61 = 62 = 0
Under these conditions, (11.8) being equal to zero implies that
2 C1 2
(Al + A2)(A + Zu cos a1) = A3 E; u Sln 82 . (11.10)
Note that (11.9)2 yields that
Sln a1 _ Sln a2 _ Sln 82
" "' :
C1 C1 C2
and using (11.9), we get
sin a c
ﬁ=zl . (11.11)
2 2
Using the fact that
c 2 31 2a
A + 2p _ 1 _ n 1
_' _ s
U c22 sinZB2
(11.10) may be written as
(Al + A2) cos 282 Sln a1 = A3 Sln 82 sin 282 . (11.12)
Next we also must have Txy = 0 at x = 0; this condition implies that
Al
— sin 2a cos(mt + g y)
cl 1 1
A
— sin 2&2 cos(wt + gzy + 61) (11.13) 01 > o
N m +4 N cos 282 cos(wt + g3y + 62) = 0 11—4 11—5 Using the previous results (11.9) in relation (11.13) yields 2(Al — A2)cos 0L1 sin {32 — A3 cos 282 = O . (11.14) Equations (11.12) and (11.14) represent two equations for the two unknown amplitudes A and A (assuming Al is given). Notice that these amplitudes 2 3
are independent of the frequency. The following figure gives the solutions
A2 A3
K and X for various incident angles and values of Poisson's ratio.
1 1 1
10
09
08
07
v=0‘25
z—‘~\
O \
"E 06 / ’1 1,593: \\\
a) l/ / , Vii \\\ \\
E 05 / ’/ , \\\ \\
E / / // \\\ \
a. / / / ‘__.l \\ \
E 04 ’ / // 3 \ \
< / / , \ \
/ // \ \
////// \\
03 [7,, . :\\\
/// —Az/,1,——— ‘=0'3 \\\\
0.2 III \ \
Ix, C \\
Z
27%“ is.
o;1 ,/ .»=o'25 \
0 ‘
i ‘ ,
0.0 J J j J L 1 0 10 20 30 40 50 60 ‘70 80 90
‘ Incidence a11g)c,01’,(deg)\—/ 11—6 113 REFLECTION OF A PLANE SHEAR WAVE FROM A PLANE FREE BOUNDARY Consider now the case an incident plane harmonic shear wave, reflecting off of a plane free boundary as shown in the figure. We consider separately the incident SH— and SV—waves. B3 (dilat a:::?é1) i
2
I
w” i
1
I
i
I
i
i Incident SH—Wave For a shear wave with its vibration direction parallel to the z—axis,
i.e., a SH—wave, there will be no motion in the x or y directions, so that
u = v = 0. One can then show for this case that the boundary conditions
OX = Txy = sz = O at X = 0 will be satisfied with simply a single reflected shear wave with reflected angle and amplitude equal to incident angle and amplitude but with a 180° phase shift of amplitude. Incident SV—Wave For a shear wave with motion in the x—y plane, the boundary conditions
at X = O can only be satisfied if the incident wave gives rise to both reflected shear, B2 and dilatational, B3 waves as shown. Going through a similar analysis as before one finds that Bi=8'2
(11.15)
1 C
C sin a5”
' 1
Sin Bl 2 11—7 The condition 0 l = 0 implies that
X x=0 (Bl + B2)sin 281 sin 81 = B3 sin a; cos ZBi , (11.16)
while the condition T I = 0 yields
xy x=0
(Bl — B2)cos 281 = 2B3 sin 81 cos aé . (11.17) 3 Again (11.16) and (11.17) represent two equations for the two unknown amplitudes B2 and B3. The following figure illustrates the solutions B2/Bl i and B3/B for various incident angles and Poisson's ratios. 1 Amplitude ratio 5 10 15 20 25 30 35
Incidence angle, 5"(deg) 11—8 114 REFLECTION AND REFRACTION OF A PLANE DILATATIONAL WAVE
FROM A PLANE INTERFACE BETWEEN Two MEDIA Consider the case of two media perfectly bonded together at a plane
interface, and pass an incident dilatational wave through this interface, as shown in the following figure. interface Since the two materials are perfectly bonded, no slipping occurs at the interface, and hence the following maiching condétéonA must hold at the interface.
ua = ub
Va = Vb
_ 7'c
wa — wb
_ (11.18)
(Gx)a — (OX)b
( xy)a = (Txy)b
(sz)a = (sz)b , at x = 0
“ wa = Wb = O for the incident dilatational case. 11—9 Relations (11.18) state that the displacements and stresses must be continuous at the interface. Conditions (11.18) imply that
Sln a1 = sin a2 = Sln 82 Sln a3 Sin 83
(cl)a (cl)a (C2)a = (11.19)
(cl)b (c2)b Specifically, (11.18)l implies (Al — A2)cos a1 + A3 sin 82 — A4 cos a3 — A5 sin 83 = 0 , (11.20) while (11.18)2 yields (Al + A2)sin a1 + A3 cos 82  A4 sin a3 + A5 cos 83 = O , (11.21) *
relation (11.18)4 gives ‘
. pb
(Al + A2)(cl)a cos 282  A3(C2)a Sln 282 — A4(Cl)b (5;) cos 283
pb (11.22)
— A5(C2)b (6;) Sln 283 = O ,
and (11.18)5 produces the result i
2 C1
pa(c2)a [(Al — A2)31n 2&1 — A3 (2;)a cos 282]
 (11.23)
_ pb(C2): [A4 :Cl:a sin 2&3 * A5 :Clia cos 283] = 0
C1 b “2, b
Relations (11.20) — (11.23) together with (11.19) may be solved for A2, A3, A4 and A given the incident amplutide Al. 53 115 BEFLECTION AND REFRACTION OF A PLANE SHEAR WAVE FROM A
ELANE INTERFACE BETWEEN Two MEDIA Next consider the incident shear wave case, as shown in the following figure. ll—lO Bl (shear)(SV) ‘ B2 (shear)(SV) A. (dilatational) 3 (dilatational) [if B4 Incident SH—Wave As before when the vibration direction of the incident shear wave is
parallel to the z—aXis, there is no motion normal to the interface, and hence no dilatational waves are reflected or refracted. For this case one finds
1:!
81 82 t
. . (11.24)
Sln 83 _ (c2)b
.1—
81n Bl (c2)a
Also relations (11.18)3 @ imply that
’
Bl + B2  BS = 0
(11.25)  v _ _  v =
pa Sln 281(Bl B2) Bspb Sln 283 0 Equations (11.25) may be solved for BZ/Bl and BS/Bl' Incident SV—Wave
For the incident SV wave case we find
sin 8' sin 8' Sin a' sin a' sin 8'
( ) l = ( 3___= ?__3_~.=.z—_y_— = ?——7—— , (11.26)
C2a C2a Cla Clb C2b and relations (11.18) imply (Bl — B2)sin 81 + B3cos mg + B4cos a; — Bssin 85 = 0
(Bl + Bz)cos Bi + B3sin aé — B4sin a3 — 35cos 85 = 0
(c2)a(Bl + B2)sin ZBi — B3(cl)acos ZBi
+ B4(C1)b g: cos 233  B5(c2)b Eisin 285 = 0 (11.27)
C2
pa(c2)a[(B1 — B2)cos ZBi — B3CEI)asin 2aé]
—pb(c2)b[(§f)bB4sin 2&5 + Bscos 285] = 0 Again the four equations (11.27) with (11.26) allows a solution for all amplitudes given the incident amplitude B1. 11.6 CRITICAL ANGLE OF INCIDENCE When the velocity of propagation for the reflected or refracted wave
is higher than that of the incident wave, there will be a Cﬂiiicaﬂ angﬂe 06
incidence which makes the angle of reflection or refraction equal to% .
For angles of incidence greater than this critical angle (measured from interface normal), the previous relationships break down. Consider again the following specific cases for the free surface reflection. Incident Dilatational Wave: Free Surface
The analysis for this case is valid for all angles of incidence; hence no critical angle exists. ll—ll 11—12 Incident Shear Wave (SV): Free Surface
For this case denote the displacements for each wave in complex form; so we can write i(wt + fix + giy) ¢i = BlRe{e } . . . incident shear wave
i(wt — fix + giy + 61)
¢é = BZRe{e } . . . reflected shear wave (11.28)
mm;  féx + g'zy + 62)
¢é = B3Re{e } . . . reflected dilatational wave §
and
I ' V
f' = w cos 81 ' = w Sin 81
1 c2 ’ g1 c2
. . , (11.24) 1
U) COS 062 U) Sln 01,2
f' = , g' =  r
2 cl 2 Cl
Recall :
sin a' C
sin 8'2 _ E; ' E
1 2
C2
Now if sin 8' > ——' =$> sin a' > 1 , (11.30)
1 cl 2 g
i
which means cos a5 = iv 1 — sinzaé , is a pure imaginary number. Hence 6
we may write I
v = i  5
f2 1m
§
where (11.31) 1
w C1 2 2 1/2 1
m = “C— [(:—) Sin Bi — 1] . E
1 2 E
Thus we get the form for the reflected dilatational wave to be 1
(19' = B e_mX cos(wt + g'y) , (11.32)
3 3 2 ‘
where we have dropped the unbounded form, emx.
plaﬂc Note that (11.32) corresponds to a wave, the amplitude of which decays A exponentially with distance from the free surface and whose phase is independent of X. We also find that B1 = B2 and that 61 can no longer be taken as 0
c
or‘l but must be a function of B and l'.
2 1 c2 Note a critical angle is defined by (11.30), i.e.,
c
sin e =~2~= /—il_ . (11.33)
c c
1
For Bi_3 6C we get $5 = g, and hence the reflected wave motion (11.32) is
actually a bunﬁace wave. We will discuss these types of waves in more
detail in the next chapter. Similar phenomena occur with reflected and refracted waves at an interface between two media(£§fone£ytdque$>. ll—l3 ...
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