KHOA HOC & C O N G N G H E
Cong thik van toe song Rayleigh trong ban khong gian dan hoi nen chju dUqrc dieu kien bien trcr khang
On the fomulas for Rayleigh wave velocities in a compressible elastic half-space with impedance boundary condition
Pham Thj Ha Giang
Tom tat
Su lan truyen cua song Rayleigh trong ban khong gian dan hoi dang hu6ng nen di/tfc v&i dieu liien bien t r d khang ditcrc nghien cuti gan day bdi cac tac gia Vinh va Xuan [1]. Cac tac gia da difa ra cong thii'c van toe song, dieu kien ton tai va duy nhat cua song. Tuy nhien, trong [1], dieu kien bien t r d khang ciii anh hifdng den u'ng suat tiep. Muc dich chinh cua bai bao la thiet lap cong thii'c van toe song trong trUcfng hdp ca Ung suat phap va illng suat tiep deu bi anh hudng bdi dieu kien bien t r d khang b^ng phifOng phap ham bien phtiTc.
Tif khoa: Song Rayleigh; cong thi^c van toe song Rayleigh; vdt lieu dan hoi nen difffc; dieu liien bien trd khdng; phuong phdp hdm bien phdc
Abstract
Tlie spread of Rayieigti waves in a compressible isotropic elastic lialf-space with impedance boundary conditionswas investigated recently by Vmh and Xuan [1]. The authors have provided thethefomularofvelocity, existence and uniqueness ofthe wave. However, in [1], tliere is only tangential stress which is affected by the impedance boundary condition. The mam purpose of this paper is to find such a formula for both tangential stress and normal stress which are affected by the impedance boundary condition case. By using the complex function method, an analytical exact formula for the velocity of Rayleigh waves has been derived.
Key words: Rayleigh waves, Rayleigh wave velocity;
the compressible elastic material, impedance boundary conditions; complex function method
TS. Phffm TliiHd Giang Bo man Ca hQc ly thuyet, Khoa Xdy dinig Email- hagiang8l3@gmail com DT 0945164695
Ngay nhan bai 25/02/2019 Ngay sira bai. 18/3/2019 Ngay duyet dang: 8/01/2020
1. Gio'i thieu
Trong phin Id-n cac nghien cuu vd sdng Rayleigh, ban khdng gian dan hoi duge gia thiit la tu do doi vdi irng suat, ti>c la ung suat blng khdng tren mat bidn cua nd. Sdng mat tuang ung dugc goi la "Sdng Rayleigh tu do ung suit" Mac diJ vay, trong nhiiu bai toan thuc t l cua am hoc, dien tCr hoc,..., diiu kien bien trd khang {impedance boundary cgndilions), lien he tuyln tinh cac ham cln lim va cac dao ham ciJa chijng tren bien, xult hien thuang xuydn, tham khao eae bai bao [2, 3] dii vdi ITnh vuc am hoc, [4. 5] ddi vo'i ITnh vuc dien-tf> hge, va cac tai iieu tham khao trong dd.
Bai loan truyin sdng Rayleigh trong ban khdng gian dan hoi dang hUo'ng ehiu diiu kien bien trd khang duac Godoy va edng su [6] nghien euu g i n day. Trong [6], cac tac gia da tim ra phuang trinh tan sac va ehung minh dugc sir tin tai va duy nhlt cua sdng P^ac du vay, cdng thuc cua van t i c sdng vin chua dugc lim ra, Bai loan nay da dugc giai quyit g i n day bd"i cac tac gia Vinh va Xuan [1].
Trong [1] va [6], eae tae gia chi xet trudng hgp diiu kien bien trd khang chi anh hUo'ng din ung suat tiep.
Mue dich cua bai bao la thiit lap cdng thuc van t i c sdng Rayleigh trong truang hgp ea ung suit phap va ung suit tiip d i u bj anh huang bdi diiu kien bien tra khang bang phuang phap ham biln phiFC
2. Cong thuc van tdc sdng
Trong tai iieu [7] lac gia Nguydn Quynh Xuan da thilt lap phuang trinh tan sac cho sdng Rayleigh truyin trong ban khdng gia dan hoi ddng hudng chiu dieu kien bien trd khang nhu sau,
+5^s,x{^fi^yl\-rx -1) = 0
(1)
Trong dd x^c^/cf, J^CT/CI vai CL. CJ lan lugt la van t i c sdng dgc va sdng ngang cua sdng Rayleigh, d^ (5, la cac tham sd trd khang vd huang, ChCi y ring tu diiu kien t i l d i n cua song Rayleigh chung ta chung minh duac 0<C-<CT . do vay 0 < J: < 1,
Elua vao bien mdi sau:
^ „ , ,H.1>1 2x-\
D I I vai biln mai phuang trinh tan sde trd thanh:
7r(n,) = , i ; ^ ( 9 - 2 ^ i ^ ; ) - 2 w ( 3 + V 2 ) + l
Phep biln doi (2) la anh xa 1-1 l u miin 0<x<'i
| w | < i Xet phuang trinh phirc sau, F{z)=C_l(z)+D(z))=0
(3) miln
3 8 TAP CHi KHOA HOC KIEN TRUC - XAY DUNG
Trong 66:
C(z) = z=(9-2<S,,S,)-2z(3 + *,*,) + l; (5)
• i j z ^ ( ( z + l M ^ , - 8 z )
V 2-r
+*,(2 + l ) ^ / ^ l ^ / z - l + , 5 , J 2 - 7 ( 2 + l)^/z + l 2-r (6) Trong bieu thi>c tren chung ta chon cac gia tn chinh cua cac can bac hai V z —1 ,
2-r
Khi dd phuang trinh (4) se Irting vai phuang trinh (3) vdi
|z| > l . Do do ta CO the ggi (4) la phuang trinh dang phuc cua (3). De tim w^, ta se lim nghiem thuc z, cua (4) thoa man | z t | > l . KJ' hieu L = l,uU vai L^=[-i,rH2-r)] va L-[yn2-r),\]. 5 - { 2 e C , z g i } , JV(z„) = )zsS 0|z-z„|<e) vai E la s l duang nhd iuy y, ZQ la mdl dllm luy ^ nlm trong mat phang C, Neu ham <f>(z) ehinh hinh trong miln QczC, chung ta viit ^(z) s H{£1).
Chu y ring vi 0<y<l nen 0<y/(2-y)<l. Ti> (4) la cd menh d i sau:
Menh di 1:
W) f(z)sH{S)
{fl) /(z) la ham bi ehan trong ^(-1) va A'(l) (X) /(z)-O(z^) khi | 2 K « ) ,
ifdA^) l3 fnpt h^Ti lien tue tCj' ben trai va ben phai eua L, vai cac gia trj bien l u ben phai ciJa J{z) l a / * ( 0 va cac gia trj bien tir ben phai eday^z) l a / " ( / ) duac xac djnh nhu sau'
f / ; ' ( / ) = C(r) + D*(f),reZ., F'{i) =
[F;(i) = C(i) + D^(ilteL, (7)
D^(t) = ±l(S,(l + \)yJt + ]-J]-l + <5,V2-?'(/ + l)V/ + l J ^ f /)
^2-r^\-ll-^-l((l + \)S,S,-Sl)
D;(/)-±/{V2^Vr7.
2-r
((r + l),5,^,-8/+^,(/ + l)^/7+T^/^^) + ^ , J 2 ^ { / + l)^/Ml ll ^
Dua vao ham g(t)(t e I ) duac xae i^nh nhu sau, F (I)
Hien nhien ta thay Xiy phu'ang trinh tren
Xet ham r(z) dipac dinh nghTa nhu' sau:
(11)
2m ^^ I dl.
{12) Chu y rang r(z) ehinh id tich phan dang Cauchy. D l i vdi
ham nay ta cd menh de sau;
Menh d l 2;
ir,) r(z)e//(S) ir,} r(») = o ir,) r(z) = n,(z)
vai z € A ' ( - l ) , r ( z ) = n , ( z ) v a i z e A ' ( l ) . fi„(z)(n,(z)) ia ham b| Chan trong N(-lXNil)) va co cac gia tri hiru han tai z - - l ( z = 1).
De chirng minh du-ac y, ta phai chu y rang {8]
logg{-])^logg(l) = 0.
Xet ham'
a ) ( z ) - e x p r ( z ) . (^3j Tir (jf) - (7^3) ta cd menh d l sau ddi vai ham 0(z):
Menh d§ 3:
' W 0 ( z ) E / f ( 5 ) ( ^ ) O ( z ) ^ 0 V z e 5 ((^3) O(z) = 0{\) k h i | 2 | ^ a ) {^4) 0 ( z ) = expQoCz)
vdi z e i V ( - l ) , 0 ( z ) = exp a j ( z ) v a i zeA^(l) Theo cdng thue Plemelj [8] la cd I h l suy ra ngay ham iD(z) thoa man cae dilu kien bien,
0 * ( / ) = g ( / ) 0 - ( / ) , / e L . (14) Bay gig la xet ham.
7(z) = F(z)/cD(2). (15) TCr ( f i H f , ) , ((*,)-(!*4), (11) va (15) taco ngay menh
d l sau:
Menh dS 4-
(y,) Y(Z)BH(S) (yi) 7 ( z ) = 0 ( z " ) khi [ z l ^ o D (.y,) 7 ( z ) bi Chan trong N(-l) va N(l)
(y,) Y\t) = Y-{t),t^L
Menh dS 5:
Y(z) la mdl da thue bac hai
ChCrng minh' Tu cac tinh chit yi va y^ cua ham Y(z), Chung ta cd the thiy rang Y(z) la mgt hamehinh hinh tren loan bg mat phang phuc ngeai trii' cac diem z=-l va z=\.
Nhung l u iyj), ia ed t h i suy ra rang eae diem nay la eae diem ky dj khu dugc. Nhu vay ed t h i ndi rang Y(z) chfnh hinh trdn loan bg mat phang phue C [8].Theo Djnh ly Liouviile [8] va ly^ thi Y(z) la mdl da Ihuc bac hai. Hoan ihanh ehCrng minh.
Dat:
P{z)^Y{z) = A^z^^A^z^-A^ ^^gj Theo (15) va (16) thi F(z) - P{z)<D(z). Tu {^^ va (^^i
ta thiy 0 ( z ) ?i 0 Vz e 5. Tu day ta cd ngay menh 6e sau:
Menh d4 6:
Phuang trinh F(z) ^ 0 o P(z) = 0 trong miin 5 u { - l } u { l } .
Chuy:
(i) Do linh khong lien tyc cua ham F(z) trong khoang (-1,1) nen phuang trinh F(z)=0 khdng ed nghiem trong
S O 3 7 - 2020
39
KHOA HOC & CONG NGHE
khoang nay. Dilu nay cd nghTa la nghiem cua phuang trinh Ff'zJ=f) n l m trong miln 5 u ( - l ) u { l } ,
(ii) Vi 0 < O ( 0 " < = o V / e { - l , l ) nen, iheo (I), hai nghiem cua phuang trinh P(z)=0 cung n l m trong miin 5 u { - l } u { l } .
(ill) Theo menh d l 6 thi thay vi lim nghiem cua phuang trinh sieu viet F(z)^0 ta se tim nghiem phuang trinh bac hai Pf'z)=0 trong miln 5 ' u { - i } u { l } . Phuang trinh nay dan gian ban r l i nhilu phuang trinh ban d l u .
Menh de 7:
Phuang trinh F(z)=0 cd hai nghiem phan biet z,=l va Z2=-A,/A2-J.
Chung minh
- Tu {1 )-(4) d i thay z / = / ehinh la mdl nghiem cua phuang tnnh F(z)=0.
- Tu msnh d l 6, phuang trinh bac hai P(z)^0 eung phai ed mgt nghiem z,^l. Theo Bjnh 1^ Vieta va (li). nghiem thu hai cua phuang trinh P(z)=0 la z ^ ^ - ^ / ^ ^ - ^ va nd ndm trong miln 5 V J { - 1 } U { 1 } . TLP day, eOng theo menh d l 6, Z2=-Ai/A2-1 la nghiem lhi> hai cua phuang trinh F(z)=0.
Hoan thanh chirng minh menh d l 7.
Nhu vay, d i cd the xac dinh dugc nghiem z^ eua phuang trinh F(z)=OXa phai xac djnh cac he s6Ai, A2-
Tu (11)tacd
loggil) - i6{t), 9{t) = Argg{t) (17}
Tu{12)[xem 1 ] t a c d
I = — \ ' " t " 9 { t ) d t , n = tl,\,2,i:
(18) Trong do:
271'
SO' dung (19) chung ta co the bieu diln e"^
^ - " • U I + £ L + . ^ + O ( Z - ' )
z ' (20)
^2,^:13 iacac hlng s6 dugc xac djnh tu d6ng nhat
{e''^^={-V{l))e'''''
Thay {18), {20) va (21) ddng nhat he sdddn d i n : Vdi fl,
thi>e sau.
(19) ntlu sau:
(21)
= — + -', 2
Khai triln . ^ ^ va J ( z + 1) thanh chuSi Laurent tai vo cijng, sau do thay vao bieu thuc ham F(z} ta dugc:
F(2) = B , z ' l-B,2 + £ , + 0 ( z " ' ) (23) Voi:
V 2 - ) ' ( ^ j - 8 + rS,^,) - ^ , ( 2 y - 3 ) + 8 + <5,J,-•Sfy
(24) Thay (22), (23) vao bieu thuc xac dinh P(z) ta du'oc:
(25)
^•iri
(22)
,a,V'
Si^ dyng (24, 25) IVlenh d l 7 cong thu'c nghiem cua (1) tuong ung vai song Rayleigh ia:
' " ' (26) vai: Z j ^ l - A . / A ^ . A, A^ duge xac dinh nhu (25),
Bay gia chung ta se gia sir ton tai hai sdng Rayleigh, nghTa la (1) tdn tai hai nghiem 0<x,i^Xy.<l. Tu (2) va (4) suy ra (4) cung seed 2 nghiem z'j^^z^ thda man jz'^j |z2|</, Tu menh de 6 va 7, ta cd phuang trinh P(z)=0 co 3 nghiem phan biet. Dilu nay khdng t h i xay ra vi P(z} la mgt da thirc bde 2. Vay n i u sdng Rayleigh t i n tai thi nd la duy nhlt.
Djnh ly
Nlu sdng Rayleigh ton tai thi nd la duy nhlt va binh phuang van t i c khdng ihir nguyen cua nd J:,=C^/C/' dugc tinh bdi edng thirc sau:
^' 2 z ,
vdi. Z2 = l - A | / A j 3. Ket Iuan
Su dung phuang phap ham biin phuc lac gia da din ra duge cdng thire van tie sdng Rayleigh trong ban khdng gian dan hdi ding hudng chiu dieu kien bien trd khang, va dieu kien bien tra khang anh hudng d i n ca ung suit phap lin irng suit tilp. Khi cho ^2=0, nghTa la diiu kien bien trd khang chl anh hudng d i n irng suat tilp thi cae k i t qua Irung vdi kel qua trong bai bao [1]./.
(27) A, A , duge xac dinh nhu (25).
Tai l i e u t h a m k h a o
I. Vmh PC, XiianNQ. (2016), Rayleigh waves with impedance boundary condition • Formula for the velocity, existence and uniqueness, European Journal of Mechanics A/Solids, 61, pp 180- 185
imperfectly conducting surfa Section B8, pp. 418-436.
Hiplmair, R. Lopei-Femandes:, M , Paganim convolution quadrature based impedance bo, Journal of Compulalional and Applied Maihe:
517.
existence of .-rface waves in an elastic half-space ^ilh itnpcd,.:,ce ho„„darv conditions". Wave Motion, 49. pp. 585-594.
, Anlipov,Y.A. (2002), "Diffraction ofa plane wave by a circular , ^ , c n •• . , * f j - . , r- , . , „ , - . , co,L.lh an impednancebomdaiy condition'', Joumal on Applied ^- f^*::if;:^!^^'^„„^:,fft!^/f „f ^^f:
Mathematics. 62.pp. 1122-1152 " " " ' " ' " " " "
. Qin, H-H.. Collon. D. (2012). "The inverse scattering problemfo. ' n • i.-^ • ^m-?, c - . „ , . lavilies with.mpadance boundary condition". Adv. Comput Math. ^ N^yenQm'nh Xuan gOi V.Jong .ndt Raykigh .
, , _ , _ , tra khang, Luan van Thae si Khoa hoc.
36. pp. 157-174
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(^"14), -Fast - pp.500-'' -.\
