Ch13 - CHAPTER 13 HALF—SPACE PROBLEMS 13-1 lNIRODUCTIoN...

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Unformatted text preview: CHAPTER 13 HALF—SPACE PROBLEMS 13-1 lNIRODUCTIoN We now consider a couple of specific boundary value problems dealing with the wave motion in a semi—infinite or half space of elastic material. These problems have application to geophysical phenomena and have been studied in quite some detail: see e.g., Ewing, Jardetzky and Press; Fung; or Graff. 13-2 LAMB’S PROBLEM: LINE LOAD SUDDENLY APPLIED ON AN ELASTIC HALF—SPACE*‘ Consider a semi~infinite body of homogeneous isotropic linear elastic material occupying the space 2.: O. The medium is at rest for t < 0; however, at t = O, a uniform concentrated line load Q is suddenly applied normal to the free surface 2 = 0, along the entire y—axis. * For additional details see: Mooney, H., “Some Numerical Solutions for Lamb’s Problem”, Bull. Seism. Soc. Amer, Vol 64, 473, 1974; “Elementary Solutions to Lamb’s Problem for a Point Source and Their Relevance to Three Dimensional Studies of Spontaneous Crack Propagation”, Bull. Seism. Soc. Amer, Vol 69, 947, 1975. i ! i We assume that the deformation state is plane strain; hence, _,§9 _ E! u = u(x,z,t) — 8X 32 V = 0 (13.1) 29.+ QT W : W<X3Z9t) _ az 8X ’ where we are using the displacement potentials ¢ and ii: {0,¢,0}. The stress components therefore follow to be 2 2 2 2 3x X 3X 32 2 3X32 32 32 32 82 g =)\(..$+_$)+2u(___qi+_.lp._) z 2 2 2 BXBZ 3x Bz 8z (13.2) oy=v(ox+oz) zx 3x8z a 2 2 X Bz T =T =0 Zy XY The displacement potentials must satisfy the wave equations 2 l on V¢=—§¢ C1 (13.3) 2 — l n vw———C2w 2 The boundary conditions on this problem are TZX(X,0,t) = O sz(x,0,t) = O (13.4) OZ(X’O’t) = ‘Q 6(X)H(t) : Where 6(X) is the vitae delta fianction defined by 0 , X<0 6(x) = w , x=0 (13.5) 0 , X>O , 13—2 such that (also see equation (1.79)) f 6(x)dx = 1 , (13.6) and H(t) is the unit 0& Heauiéide étep fianctian defined by equation (1.73). Note the following property M) = 6(2) dg . (13.7) We also place certain conditions at infinity, and require that: 1. A11 displacements and stresses remain fiiniie at infinity. 2. At large distances from the point of application of the load, the disturbances are outgoing waves, i.e., no waves coming from infinity. These are called the fitniteneéb and nadiaiion conditiané. To start the solution, take the Laplace transform with respect to time of (13.3) and (13.4); this gives 2— 2- 2 M+M=L$ 3x2 322 c 2 1 (13.8) 32$ + BEE _ S2 __ ———— ——w 3x 822 c 2 2 a%' a}' f" 2 ——9—-+ ——$-— ——$ = o , @ z = 0 3x82 2 2 3x 32 (13.9) 2— 2— 2— 2— A(§_g +.§*$) + ZUC§_9.+ §_ELO = _.g§£§l @ Z = 0. 8X2 BZ2 az2 3x82 3 ’ An elementary particular solution to equations (13.8) is given by __ _- iksx — 2 + cl_2 sz ¢ = ¢(Xazas) = e _. _' iksx ~ sz + c2—2 sz W = W(Xaz’s) = e 13-3 13-4 Therefore by the principle of superposition, the general solution to (13.8) follows to be __ w —‘s[\/cl”2 + k2 z — ikx] ¢ = f Pl(k)e dk _m .1.1__1- (13.10) _' m —s[¢é2“2 + k2 z — ikx] w = f P2(k)e dk , where P1 and P2 are arbitrary functions of k. The quantities Vél_2 + k2 and c2 so that E and E'+ O as z + m. Substituting (13.10) into the boundary conditions (13.9) gives 00 f [—21k/cl“2 + k2 Pl(k) — (2k2 + c2"2)P2(k)]s2eikadk = o _m [:szZ + c2_2)Pl(k) — 2ik/c2_2 + k2 P2(k)]uszeiksxdk = — Qaéx) . To satisfy (13.11)l we must have Pl(k) = (2k2 + c2‘2)R(k) (13.12) P2(k) —2ik/Cl—2 + k2 R(k) for some arbitrary function R(k). Now it can be shown that one can represent the Dirac delta by 6(X) = 00 lim 1 sin k8 ikx 8+0 2W8 I k e dk 00 1 ikx =fife dk; —00 so by letting k + ks in the above equation, we get 00 s iksx 2“ f e dk . (13.13) —00 6(x) = Relations (13.12) and (13.13) along with (13.11)2 yields + k must have positive real parts along the path of integration 13—5 00 _ _ — 'k f [(21.2 + c2 2)2 — 4k2/<cl 2 + k2)<c2 2 + k2)]R(.k)uszel SX dk = _00 °° 1k — %f e SX dk , which is satisfied by R(k) = — J—z [(2k2 + c2’2)2 — 4k2 (k2 + cl_2)(k2 + (124)]? (13.14) 2flus Hence equations (13.10), (13.12) and (13.14) give 2 _2 —s(./cl'2 + k2 z -— ikx) __ Q m (2k + c2 )e ¢ = — 2 f dk 2mm —00 Mk) ( .12.... 13.15) 00 —S(\/C2 + Z _ __ i cl + k e w = 2 I dk ’ nus —°° Mk) where A(k) = (21:2 + (:2’"2)2 — 4k2/(k2 + cl”2)(k2 + (22‘2) (13.16) and is called the Rag/£21911 gum/um aqua/Lon. The stresses follow from (13.2) to be _ Q co (2k2 + c2"2)2 g oz = — E; [00 ————A—-———— e dk so 2 W +39% 1‘— /(c '2 + k2)(c ‘2 + k2) endk TI _00 A l 2 ‘ °° k 2 2 /————2 2 3; __.— _ —:—l; — -— _ TZX — 7T [00 A (2k + c2 ) (cl + k )e dk (13.17) g; °° k 2 2 2 2 g l __ _ fT—m n W [ooAQk +c2 )(cl +k)edk Q 0° 1 2 2 2 2 2 a ex = — ELK (c2 — 2C1 — 2k )(2k + c2 )e dk 2 °° k2 1 2 2 2 WT f TVCC12+k)(c2 +k) endk , E I z i E g E 1 $ 13—6 where E = —s/él_2 + k2 z + iskx Jf—:§’""“§‘ . n = —s c + k z + 1skx . 2 To find the inverse transform of (13.15) or (13.17) is quite difficult. A method of inversion due to: Carniard, L., "Reflextion et Refraction des Ones Seismigues Progressives”, Paris, Gauthiers—Villars, 1935, and later modified by DeHoop, A. T., "Representation Theorems for the Displacement in an Elastic Solid and Their Application to Elastodynamic Diffraction Theory”, Thesis, Technische Hogeschool Te Delft, 1958, consists of considering k as a complex number and deforming the path of integration (—w to 00) so that the integrals in (13.15) or (13.17) can be recognized as the Laplace transforms of certain explicit functions of time. This would allow one to write down the inverse transform by inspection. Consider for this case the transformation from k to t such that _ 2 . t = cl,2 + k z — 1kx , and deform the original path of integration on to another path, say F1, in such a way that the corresponding path Tl on the t—plane is the positive real axis from 0 to m. Hence integrals like those in (13.15) or (13.17) take the form 00 f f(x,z,t)e_Stdt , 0 so that the inverse Laplace transform would be simply f(x,z,t). Graphically we have, 1 1 i i l i \\2 Pl (hyperbola: "\~ Re(k) Im(k) = constant5\\\\ fer...“ Original path of integration ra—iclfil Branch ///fli‘ . —1 points a“, 2—1C2 \\\\j k — plane f-ik e r kr: roots of A(k) = 0. Hence by using this inversion technique, the stress OZ is given by (l) (1) 8k 3k _ _ 9_. _ E;_ (l) + _ (l) - Oz ‘ 2w H(t cl)[M1(k+ ) at M1(k— ) 8t 1 (2) (2) 8k 6k 2Q :_ (2) +- _ (2) — +1T H(t— c2)[M2(k+ )at M2(k- ) at ] 3 (13.18) 2 <1) <1) <3) <3) Ski) Q +37% ft [M2<k+ )_M2(k~ H at 2 <2) <2) <4) <4) 3km _g + +7r f6 ft [M2(k+ )_M2(k— H at where r = V; + 22 G = tan—1 Cg , 0 :_e :_W 13—7 Mle) (1) ft f<2) (2) II I! II II II II (2k2 + c2'2)2[(2k (2k2 + c k2V(c _2 + k2)(c2— Ff A(k) l 2 —2 2) A(k) + c "2)2 — 413/03 + c [Zn—1 2 +k) 2 2 / —2 . .5 i 2 — cl Sine + i r cosG —2 2 l )(k + c 2 2 sinG + i-% cose r . i(— c -2 — E'---sin6 +-E cose)i5 2 r2 r -2 t t 1( C2 — ~§ sine + E~cose)i6 ,(6-+O) r c l , O < 6 < (308—1 wg —- - c l O , otherwise #:2”‘_.":M l , E—‘> t > E-—~cos6 + r c — c 2 sine c —- —-c 2 l 2 l O , otherwise c l , W - COS_l —g-< e < N C .r l O , otherwise —73T'———:" l , 'E— > t >~£- Icosel + rVé — c 2 sinG c —- —-c 2 l 2 2 0 otherwise (13.19) 13—8 The other stress components are obtained in a similar manner and have similar forms to the previous equations. Examining the wave pattern of the stress response for this problem, we find the wave motion shown in the following figure§ * Also see: Dally, J .W. and Thau, S.A. , “Observations of Stress Wave Propagation in a Half- Plane with Boundary Loading”, Int. J. Solids Struct, V01. 3, 293, 1967. 13—9 head wave or Von~Schmidt wave front Rayleigh wave front \ I l I . dilatational ' shear ‘ P—wave front \\\\\ S—wave front N,\\Nfl ‘ MWMWMWWWWWW,» ' z i l Region_lz (r > clt) Undisturbed, all terms in (13.18) are zero. Region II: (clt > r > czt) Dilatational wave field from the first term of (13.18). Region III: (r < c2t) Shear wave field from the second term of (13.18), and dilatational wave field. c c 2 l(—-2-)<6<Tr;0<6<cos—l(-2-)) cl —- —- *— cl Head wave field from the third and fourth terms of (13.18) ; Region IV: (clt > r > c t; W — cos— and dilatational wave field. In addition, the terms Ml(k£l)) and M2(k£2)) in (13.18), also will reveal Rayleigh surface waves. 13—10 13-3 S_IEADY_ SiAIt M_o\/ING LINE LOAD P3031 EM* We now investigate the wave motion problem of a line load of magnitude P moving with constant speed V across the face surface of an isotropic elastic half space. We assume Ataady Atate condiiionb so that the load has been applied and moving for an infinitely long time. As before the problem is modeled by plane strain theory; hence, the displacements and stresses are again given by equations (13.1) and (13.2). The displacement potentials must, as usual, satisfy the wave equations (13.3). The boundary conditions for this problem are written as O (x,0,t) = —P6(X + Vt) Z (13. 20) TZX(X,O,t) = sz(x,0,t) = O . Because of the steady nature of this problem, the solution is neatly handled by using the moving coondinateé (€,z,t) defined by 1 E = X + Vt j z = 2 (13.21) t=t . Using this change of coordinates, the wave equations (13.3) become * see Cole, J. and Huth, J. , “Stresses Produced in a Half-Plane by Moving Loads”, J. Appl. Mech., pp. 433-436, 1958. 2252+§32=Xiizi 3&2 az2 cl2 agz (13.22) fl+fl=fi§fl agz 3Z2 c22 352 Next by introducing the Mach numbcnb M = V/cl and M = V/c2, we can 1 2 write equations (13.22), for the following three cases, as: 1 > M > M : Subsonic 2 l 2 2 21E M_ 81 + — 0 agz 322 2 2 (13.23) B28w LIL 27+ 2—0 , 35 32 > > : ' M2 M1 1‘ Supersonic 'E 2 33$ _ gig = 0 l 3.52 32 (13.24) ~2 222222-.) 2 2 2 _ ’ BE Bz M2 > 1 > Ml: Transonic 2 2 2M M_ B + — O 1 8‘52 322 2 2 (13.25) —22_12 32- B2 2 ‘ 2 “ 0 ’ 3E 82 where 2 2 —-2 Bl — 1 — Ml — — B1 2 2 2 m ll [—1 I a: II | ml It is important to note that equations (13.23) and (13.25)l are ekfiipiic while equations (13.24) and (13.25)2 are hypcabOZic. Only the hyperbolic equations will produce wave propagation; so let us therefore consider only * this supersonic case. * See Fung for further details concerning the other cases. 13—11 The general solutions to the wave equations (13.24) follow, from our D'Alembert travelling wave solution (1.8), to be ¢ = fl(€ - Biz) + gl(€ + Eiz) w = f2<€ —'Ezz) + g2(€ + Ezz). However the radiation condition implies that gl(€ + Eiz) = g2(§ + 322).: O, i.e., no waves coming from infinity. Our solution is then ¢ = flu: - E12) 11) = — (13.26) In order to use the boundary conditions (13.20), we rewrite the stress equations (13.2) in terms of our moving coordinates, as 2 2 _ 2 _ 2 §11._ §_i_ ox — MM2 2M1 + 2) 2 2p BEBZ 3E o=(M2—2)fl+2fi— (1327) z u 2 2 BEBX ' 3S 2 2 _ 3 ¢ _ 2 _ .§J£ sz — 2“ BEBX “(M2 2) 2 3E The boundary conditions (13.20) become OZ(E,09t) = “P6(€) (13.28) TZX(€909t) = 0 Using (13.27)l 3 in (13.28) yields 2 gig 82m P (M2 ' 2) 8g2 + 2 agaz = "fi 5(a) 2 2 (l3.29) ii __ 2 _ 3.1 = _ 2 agaz (M2 2) 2 0 , @ Z — 0. 31: Integrating the above equation with respect to E and using (13.26) produces 13—12 13—13 2 (M2 1 “v1 _3 - 2>fl (a) — zszfz (a) — — U H<€> (13.30) _ I 2 v _ _ —261fl (a) — (M2 — 2)f2 (E) — o , @ z — 0 where we have taken the constant (function) of integration to be zero. Solving equations (13.30) for f' and g' yields . _ 2 EL_ fl (a) - -<M2 - 2) “A H<a> W P (13.31) ; fZ'CE) = 281 m HOE) , where 2 2 —-— A = (2 — M2 ) + 48182 # 0 Finally using the solution (13.31) with (13.26) in (13.1) gives the following displacements u=1[m—M2me-Ez>+fifi ma—EaJ MA 2 l l 2 2 (13.32) w=§3fl5%2—mma—Qm+2§ma-Qa1 Likewise the stresses would follow from (13.27). The motion specified by equations (13.32), indicates that the displacements are marked by two Mach waveé defined by the lines a — E12 = o and g — Eéz = o. The medium will be undisturbed in front of these waves as shown in the following figure. «m«——4>.E undisturbed g — Biz = 0 region __ E - Blz = 0 —wave .. P—wave ...
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Ch13 - CHAPTER 13 HALF—SPACE PROBLEMS 13-1 lNIRODUCTIoN...

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