Transient Response of Two-Dimensional Cantilevered Semi-Infinite and Finite Elastic Plates, Subjected to Base Motions Citation Garrott, W. Riley (1977) Transient Response of Two-Dimensional Cantilevered Semi-Infinite and Finite Elastic Plates, Subjected to Base Motions. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/e3aq-9d16. https://resolver.caltech.edu/CaltechTHESIS:11182025-210824229 Abstract This research is concerned with the response of a two-dimensional, isotropic, homogeneous, elastic, cantilevered plate subjected to a step transverse velocity at the base. The investigation uses a method by Miklowitz which is based on a double Laplace transform and a boundedness condition on the solution. The case of a semi-infinite plate is solved, for long-time, to find the shear and normal stresses at the base. The solution in the interior of the plate is shown to agree with that obtained by the Bernoulli-Euler approximate theory. The solution is then extended to the case of the finite length plate, with traveling wave and vibrational forms of the solution being found for the interior of the plate. At the base of the plate the investigation shows that the normal stress is singular at the corners while the shear stress is non-singular. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Applied Mechanics and Physics) Degree Grantor: California Institute of Technology Division: Engineering and Applied Science Major Option: Applied Mechanics Minor Option: Physics Thesis Availability: Public (worldwide access) Research Advisor(s): Miklowitz, Julius Thesis Committee: Unknown, Unknown Defense Date: 9 May 1977 Funders: Funding Agency Grant Number Earle C. Anthony Fellowship UNSPECIFIED Record Number: CaltechTHESIS:11182025-210824229 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:11182025-210824229 DOI: 10.7907/e3aq-9d16 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 17767 Collection: CaltechTHESIS Deposited By: Benjamin Perez Deposited On: 19 Nov 2025 18:12 Last Modified: 19 Nov 2025 18:28 Thesis Files PDF - Final Version See Usage Policy. 19MB Repository Staff Only: item control page

Transient Response of Two-Dimensional Cantilevered Semi-Infinite and Finite Elastic Plates, Subjected to Base Motions Thesis by W. Riley Garrott In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California ( 1977) (Submitted May 9, 1977) ii ACKNOWLEDGMENTS The author wishes to express his sincere gratitude and apprecia tion to Professor Julius Miklowitz for suggesting the problem and for his valuable guidance, suggestions, and assistance given during the research and preparation of this thesis. My thanks also go to Marijan Dravinski, Sigrid Kite and to the other faculty and students in Applied Mechanics for their help. The author is grateful for the Earle C. Anthony Fellowship, the teaching assistantships, and the tuition scholarships granted him by the California Institute of Technology and the financial support received from the National Science Foundation during the course of this work. iii ABSTRACT This research is concerned with the response of a two-dimensional, isotropic, homogeneous, elastic, cantilevered plate subjected to a step transverse velocity at the base . The investigation uses a method by Miklowitz which is based on a double Laplace transform and a boundedness condition on the solution. The case of a semi-infinite plate is solved, for long-time, to find the shear and normal stresses at the base. The solution in the in- terior of the plate is shown to agree with that obtained by the Bernoulli-Euler approximate theory. The solution is then extended to the case of the finite length plate, with traveling wave and vibrational forms of the solution being found for the interior of the plate. At the base of the plate the investigation shows that the normal stress is singular at the corners while the shear stress is non-singular. iv TABLE OF CONTENTS Title I. Page INTRODUCTION 1 THE SEMI-INFINITE PLATE 3 1. Statement of the Problem 3 2. Formal Solution 5 3. Solution of the Boundedness Equations 11 4. Forms for the Edge Unknowns 12 Long-time Approximation to the Boundedness Equations 16 Numerical Solution of the Bo.undedness Equations 23 Derivation of the Formal Long-time Solution 25 a. Near-field Asymptotic Solution for x .... 0. 32 b. Inversion by Residue Theory for the Domain h < x << c t s Lowest Mode Contribution 42 44 Contribution of the Complex Segments of the Higher Branches 50 Inversion of the Time Transform 54 II. THE FINITE PLATE 59 1. Formulation, Formal Solution and Entirety Condition 59 2. Forms for the Edge Unknowns from Euler-Bernoulli Approximate Theory 63 3. Solution of the Entirety Equations 68 4. Inversion of the Time Transform-Traveling Wave Form 83 5. Inversion of the Time Transform- Vibrational Form REFERENCES 91 101 1 INTRODUCTION Analysis of semi-infinite waveguides based on the equations of motion for a linear elastic , homogeneous , isotropic medium is a subject of long standing interest . Recently a method has been developed by Miklowitz for solving nonseparable elastic wave g uide problems involving nonmixed edge or end conditions. For the semi - infinite wave - guide, the method uses a Laplace transform on the propagation coordi nate, and a related boundedness condition on the solution which generates integral equations for the edge unknowns (displacements and strains). Solution of these integral equations determines the formal solution to a problem. The first problem solved by Miklowitz [ 1] was a symmetrically loaded waveguide with nonmixed displacement end con ditions, i.e. a cantilevered semi-infinite plate. be found in Miklowitz [z], Further details may [3]. Sinclair and Miklowitz [4] extended the method to non-mixed symmetric stress end conditions. More recently , they have also extended the technique to antisymmetric stress end con ditions [ 5] . They found the solution to the problem of the semi-infinite plate under a sudden end moment and zero end shear stress . information for the near and far field was obtained. Lo ng-time References for oth- er work on plates with non-mixed edge conditions are given by Miklowitz in [ 1] . In the current work, the foregoing general ideas have been extended to the finite waveguide. Here the essential differences are that a finite Laplace transform on the propagation coordinate replaces the onesided Laplace transform for the semi -infinite waveguide, and a related 2 entirety condition on the transformed solution replaces the above-mentioned boundedness condition. To solve the problem of the cantilevered finite length plate, the solution of the problem for a similar semi-infinite plate is needed. So the first case solved here is the problem of a cantilevered semi-infinite plate, subjected to a step transverse velocity at the base where the normal displacement is assumed to be zero. The integral equations re- sulting from the boundedness condition were solved for long-time to yield the shear and normal strains at the base, with the latter becoming singular at the corners. The exact theory solution and the Euler- Bernoulli approximate theory solution are shown to agree for the longtime-near-field region away from the base. For the finite length cantilevered plate, the solution obtained from the Euler- Bernoulli approximate theory is used to reduce the entirety condition to the same set of equations that resulted from the boundedness condition for the semi-infinite plate. The strains at the base are shown to be the strains at the base for the semi-infinite plate multiplied by a reflection function. The traveling wave and vibrational forms of the solution are found for the interior of the plate, away from the base. 3 I. THE SEMI-INFINITE PLATE 1. Statement of the Problem To solve the finite cantilevered plate problem, it is first neces- sary to find the solution of the semi-infinite cantilevered plate. problem is shown in Fig. 1. This A homogeneous, isotropic, linear elastic plate in plane strain of width 2h is built into a rigid base, and suddenly this base is given a uniform velocity in the width direction. The problem is formulated as a standard plane strain elastodynamic boundary value problem. Displacements and y directions, respectively. 2 cdu xx u and v are taken to be in the x The governing equations are (x,y,t) + (2 cd-c 2)v (x,y,t) + c 2 u (x,y,t) = s xy s yy ( 1. 1) ( 1. 2) axy (x,y,t) = ulv (x,y,t)+u (x,y,t)J •L X y .. for x > O, -h < y<h, t > O. Here c 2 A+2µ = d p and c 2 =_g_ are, respecs p tively, the dilational and equivoluminal body wave speeds, A and µ 4 y O"yy = O"xy =O u = 0 h h Fig. 1 X Coordinates, Displacements and Boundary Conditions for the Semi-Infinite Plate in Plane Strain. 5 are the Lanie' constants, p is the mass density and k 2 2 2 = cd/ cs. Sub- scripts in this work, when associated with displacements, indicate differentiation; but when associated with stresses identify the component in the usual way. Initial and boundary conditions are u(x,y,O) = ut(x,y,O)=v(x,y,O)=vt(x,y,O) for x>O, -h:5:y:5:h, ( 1. 3) and u(O,y,t)=O -h :s;; y :s;; h, t ::: 0 ayy (x, ±h, t) = o xy (x, ±h, t) = 0 for x > 0, t 2: 0 ( 1. 4a) ( 1. 4b) The radiation conditions are limit {u, ux, etc .}-- 0 for v, v , etc. X .,.00 -h :s;; y :s;; h, t 2: 0 . (1. 5) X The problem is an antisymmetric (flexural) one with respect to the midplane y = 0. It models a very tall building whose rigid base (ground) suddenly moves horizontally but not vertically. The problem is one of wave radiation into the plate, and no interaction with the base , except for wave reflection there. 2. Formal Solution Tlw problem is dc•composed into a rigid body n1otion and a resid- ual probl<>m. Tlw rigid body motion is 6 u(.x, y, t) for x : :c: O, -h s y :5: h, t:::c:O ( 1. 6 ) v(x,y,t) The residual problem must satisfy the following initial and boundary conditions u(x, y, 0) = u/x, y, 0) = v(x, y, 0) = 0 for X > 0, -h :,:; y :,:; h ,(1.7) vt(x,y,O) = -VQ u(O,y,t) = v(O,y,t) = 0 for a yy (x,±h,t) = a xy -h ~ y sh, t::::: O (x,±h,t) = 0 for ( 1. Sa) x> 0, t z 0 ( 1. Sb) - h s y :,:; h, t :;, : 0 ( 1. 9) The radiation conditions now are limit { V }- -V t x-oo 0 u, u , etc. for X limit x-oo .- 0 etc . V x' The residual problem has the same form as the ones considered in [ 5). The formal solution can be obtained in exactly the same manner a s was don e in the fir s t p a rt o f that pape r. I s I) This give s (s ee Eq. (19) of 7 1 u(x, y, t) = Z1ri J - u(x, y, p)e Pt dp Br v{x,y,t) - p 1 r ~(x,y,p)eptdp 21ri ..i Br p (1. 10) u(x,y,p) - ~(x, y, p) - where Br p and Br s r 1 ""' sx u(s,y,p)e ds 21ri J Br s r 1 ""' sx v(s,y,p)e ds 21ri J Br s are the Bromwich contours in the p- and s- planes, respectively, and ""' u(s,y,p) ==c u (s,y,p) + ""'P u (s,y,p) C ""'P v( S, y, p) -- ""' V ( S, y, p) + V ( S, y, p) :::,. (1.lla) ""'C u (s, y, p) = c (s, p) sinh a, y + c (s, p) sinh j3 y 1 2 (l.llb) 8 ~(s,y,p) = :;J: {t: sinhc,,(y-y')+~ sinh ~(y-y'~g(s,y'. p) 2 +k s [ cosh Cl/ (y-y' )-cosh ~ (y-y')}(s, y', p)} dy' (l. llc ) ~p (s, y, p) = :! .r: {[" + sinh Cl/ (y-y')+ ~Z sinh ~ (y-yf (s, y'. p) :z [cosh a, (y-y')-cosh ~ (y-y'Dg(s, y', p} dy' , 2 2 2 C(s p) = - (s ) rk (zs -k )coshf3h• I(s,p)+2s f3sinhf3h•J(s,p~ 1 ' L s,p s L J, (1.lld) 2 2 Cz(s,p) = -L(-s~p) [zk sa,coshc,h• I(s,p)-(zs -k!)sinhc,h• J(s,p~ , 9 I(s, p) = l 2k s Jh{ s O 2k ~ 2s 2 -k 2) ---'---'-----~ sinh (h-y) + 21-' A srnh • 13 (h-y) ] g ( s, y, p ) ai s Ci 2 2 + [(zs - k!) cosh a, (h-y )- 2s cosh + ~ (h-y 1] h( s, y, pi} dy k 2 -2 .) 7 u(O,h,p) ( (1. lle) J(s, p) = :; (ff• 2 cosh a, (h-yl-( 2s 2 -k;) cosh ~ (h-y~ g(s, y, p) - v(O, h, p) 2r ] 2 g(s,y,p) = k Lsu(O,y,p)+u)O,y,p) +(k -l)vy(O,y,p) ( 1. 11£) VO 1 [ 2 ] h(~,y,p) = - + 2 sv(O,y,p)+v)O,y,p)+(k -l)uy<O,y,p) 2 scd k , (1.12) k S the transform paran1eters for the time and x =..E_ and p and s C are s Laplace transforms, 10 respectively. A bar over a quantity indicates that it has been Laplace transformed with respect to t while a tilda indicates that it has been Laplace transformed with respect to x. first term in h(s, y, p) It should be noted that the comes from the nonzero initial condition, which is the second of (1.7), and hence is not present in [5]. L(s, p) is the generalized Rayleigh- Lamb frequency equa- tion for antisymmetric harmonic waves. Define s. (p) J as the roots of L[s. (p),p] = 0 J Then, as shown in [1], [5], g(s,y,p) (1. 13) and h(s,y,p) must, for Re[s. (p)] > 0, satisfy the following boundedness condition (see pp. 8-12 J in [1] and (22) in [5]). 2 cosh f3 .y] - 2s. h A .h h(s ., y, p) JCOS t-'· J J s. ,_L_ - 2 k O!. J 2 7 _ } + ( k -2)-u(0,h,p) + Y/sj'p)v(0,h,p) (zi-k 2 ) YJ_(sJ_,p) = 2 s tanh CY.h = 2k s.0/. J J J ( 1. 14) , -2s.f3. 1 2( k 2s.-k J s k z) tanh {3.h J 11 The first of ( 1. 14) are two coupled integral equations fo r g(s,y,p) and h(s,y,p). Solving these equations completes the formal solution of the problem. 3. Solution of the Boundedness Equations Using the boundary conditions at x = 0, (1. 1 lf) reduces to 2- g(s,y,p) = k u (0,y,p) X ( 1. 15 ) h(s,y,p) The unknown Laplace transformed edge strains u (0, y, p) and X vX(0, y, p) are found by assuming for them representations with unknown coefficients. The representations consist of a singular term which corresponds to the behavior of the strains at the corners y = ±h plus a Fourier series. If the singularity at the corner is the same as the assumed singular form, then the Fourier series only has to represent a regular function of y. The unknown coefficients in the Fourier series will decay as or faster as n, 1 /n 2 the number of the term in the series, becomes large. It should be noted that calculation of the values of the edge unknowns, to a given level of accuracy, requires only a finite number of terms in the series because of this two part representation. In [ 1], Miklowitz found that a dynamically loaded elastic waveguide that was built-in at the base had the same types of singularities 12 at the corners as did a similar statically loaded waveguide. The types of singularities in the present problem should also be calculable from statics. Since the present problem does not have a static limit, the singularities of a static right-angled wedge with one edge built-in and the other edge free will be used. The possible stress (and strain) singularities of a static rightangled wedge are known from the work of Knein [ 6], Williams [ 7] and Uflyand [ 8]. As these works show, the dominant stresses are, near the corner, proportional to ner and q r -q where r is the distance from the cor- is a real positive number. · For a fixed corner angle, pends only on Pois son I s ratio. In the remainder of this work, be set equal to 0. 2433 which makes q de- v will q very close to 1/4. Forms for the Edge Unknowns It is necessary to assume forms for the unknown edge strains u {O,y,p) and; (O,y,p). From the antisymmetry of the problem, u will be odd in y and ; will be even. So u will be represented by X X X X X an antisymmetrized singular term plus a Fourier sine series while V X will be represented by a symmetrized singular term plus a Fourier cosine series. In order that the Fourier series converge rapidly, it is necessary to choose the correct singular forms for the edge unknowns. Based on the results of the right angled wedge problem, any singulari ities that are present are expected to be of the r-4 type. However, since the present problem has two corners which will interact, one or both of the strains may not be singular at the corners. Various 13 combinations of strain singularities were tried. These were, both strains singular, both strains nonsingular, and one strain singular and one nonsingular. For each assumed form of the edge strains, g(s, y, p) and h(s, y, p) were calculated from (1 . 15 ), substituted into the boundedness equations ( 1. 14) and the resulting simple integrations were performed. This gave an infinite set of algebraic equations which were then approximated for long-time. The equations were solved numeri- cally using the method of reduction to see if the unknown coefficients could be determined. This procedure will be shown in more detail for the case that solved the boundedness equations. The boundedness equations were solved by assuming that u was singular and that ~ was not. X X This gives the following forms for the unknown edge strains: 00 . n1ry b n ( P ) srn 2h u (O,y,p) = bo(p{(l-y/h)-¼-(l+y/h)-¼+z-¼y/h ]+; X n=2 neven (1. 16) 00 ~ (0, y, p) = ao(p) + ~ X n= 2 n1ry an ( p ) cos Zh n even a (p) , an(p), b (p) and bn(p) 0 0 are the unknown coefficients, functions of the Laplace transform parameter p. Substitute (L 16) into (1.15) to get g(s,y,p) in turn these into the boundedness equations, ( 1. 14). and h(s,y,p) and The latter equa- tions are then integrated with the aid of the following integrals: 14 h 1 f sin h a. y (1 +y / h) - dy = T J 0 4 J Q'. , J (1. 17) h J cosh ex .y cos~? dy =(-l) 0 J h r s inh ex .y sin n 1rhy dy = 2 Jo J n,r where Sn= 2 h, placed by !3 .. J n I2 a . s inh cv . h J 2 2J (ex.+ e ) J n n 6 sinh a .h n J 2 (a ~+e ) J n n even and another analogous set of integrals with even reJ Then the boundedness equations become an infinite set a,. of algebraic equations: 00 fRe} llm f a 0 (p)M?(s.,p)+ \ J J ,_ ~ n 0 a (p)M. (s ,p)+b (p)N. (s.,p) 0 n J j J J n =2 n even ( 1. 18) 00 + \.I b (p)tfl(s.,p)+Q.(s ., p)} = 0 n J J J J n-;, 2 neven where 15 2 0 M.(s.,p) = s.Y.(s . ,p)/13. J J J J J J n/2 (-1) s.Y . (s.,p) n 1_ M . (s.,p) - - _ _ ____.,_J___.1.._.... J J (13f + e!) 0 N.(s.,p) J J ( 1. 19) n N.(s.,p) J J 2 2 Q. ( S. , p) = v Y. (S,, p) / C j3. 0 J J J J s J and Y. ( s . , p) J J is as in ( 1 . 14) . Equations ( 1. 18) can be solved for the unknown coefficients a (p), an (p), b (p) 0 0 and b n (p). a (p),a (p),a (p), 2 4 0 . , of s . (p) J That is, for a certain number of unknowns, b (p),b (p),b (p), . . . , matching numbers 2 0 4 (which are infinite in number; see [ 1]) are available to give a sufficient numb e r of equations from (1. 18) to solve for these unknowns. To proceed further, a representation for the The lon g -time solution will be considered here . s. (p) is needed. J 16 Long-time Approximation to the Boundedness Equations The long-time solution can be derived from the first two of ( 1. 10) by using Watson's Lemma. This gives (see Sec. 5.10. 2.4 of [9]) u(x, y, t~ 1 21ri [ v(x, y, t~ t >> 1 For p small, the roots s.(p) J r r u(x, y' p ~ ept dp ( 1. 20) JBr l~(x,y,p}j p p ~ 1 of L{s,p) are,forthelowestmode, .. (1.21) s O( p ) = ± ( 1±i ) y y = / where ' 2 cP r , pg = c p ~ is the "plate" wave speed and ✓ p(l-v 2 ) h r = -- is the radius of gyration for the plate section. modes, s.(p) = s.,j ~ 1, J J g J3 where f(s.) = sin 2s.h J J For the higher s. 1s a complex constant satisfying J {1.22) 2s.h = 0 J Equation ( 1. 21) is the generalized frequency-wave number relation for the Euler-Bernoulli approximate theory. f(s) is a well known function in the analysis of two dimensional elastostatic layer problems, hence its occurrence here is not surprising. The zeros of f(s) are an or- dered infinite set, corresponding to the piercing points of L (s, p) in the plane p = O (cf.( 1], ing points). Fig. 8 for corresponding symmetric wave pierc- Hillman and Salzer, [ 10], give the first ten roots to six 17 decimal places. The Euler-Bernoulli approximate theory frequency-wave number relation is known to be a good approximation to the lowest mode of the antisymmetric Rayleigh- Lamb frequency equation for a range of p small but greater than zero. Furthermore, as Sinclair and Miklowitz show in (70) of [ 5], for the higher modes It follows that the zeros of f( s) limit p-t0 { -d s. (pJ ) = 0 dp - are a good approximation to the s .(p), j ::::>: 1, for p small. This shows that the long-tirne approximation J will be valid for t large but not neces __sarily infinite. It remains to approximate (1. 19) for (1.21) and (1. 22) and with Re[ s/p)] > 0. this gives, for p small with s.(p) as in J For the lowest mode, s (p), 0 p .... 0 h - - + O(p) k2 (1. 23) Nno(so,P) = (-It/2(l+i)2h2~ + O(p~) nTT 2c r • p g V h - (- l+i)-0- 2 Zed ~ ,f-~_g + 0(1) 18 Approximating ( 1. 19) for the higher modes, s. ( p), j ~ 1, J gives 12 (-1t s.1tan s.h l) 2s~ n J [(k2 M.(s.,p)=2 2 2 - 2- 2J 2 J J k ( 9 - s. ) k (8 - §. ) n J n J ( 1. 24) ( - 1 )n /2 9 tan s,., . h ~ k 2 l N~(S.,p) c ( 2 l J J -s. \ k n J e 2"'s . 2 ] 2 ,.,2)+ 1 + O(p ) en2J-s. J 1---y-)~ .nz) 2 2 Q.(s., p) = -v tan s.h/cd s~ + O(p ) 0 J J J J where h H,., = sj JO 1 l. ,., l. -4 h cos s.y(l- h) dy J , ( 1. 25) (cont.) 19 The coefficients in (1 . 23) and (1. 24) a r e split into their real and imaginary parts and the order in p of each of the terms is determined . Substituting the order of each term into (1. 18) gives 3 0( 1) O(p~) O(p½) O(p) O(p) O(p 2) 0( 1) 0( I) 0( 1) 0( 1) 1 O(p½) ao(p) O(p -2) O(p2) . . a2(p) O(p 2) . . . 0( 1) 0( 1) bO(p) 0( 1) 0( 1) b2(p) 1 1 _1-. = 0(1) (1. 26) 0( 1) Using Cramer's rule on (1. 26) gives that 1 ao(p) = O(p -2) ( 1. 27) a (p) = bo(p) = b (p) = O(p n n ( 1. 2 7) shows that the unknown edge strains, -1 ) u (0 ,y,t ) and v (0,y,t), X are constant for long-time. Define X 20 ( 1. 28) A n where A , An' B 0 0 B n p a (p) b (p) = n and Bn are independent of p. n p Substituting (1. 23), (1. 24) and (1. 28) into (1. 18), retaining only the lowest order terms in p, and simplifying gives the following set of simultaneous equations: l. (X) -2- 4 2 B + \ 7 0 /._ n (-1/ n= 2 2B n nTT = ( 1. 2 9) neven irRe } • m where j ~ fl \' • "n "n "0 + B 0 N. {z.)} [ A M . (z.)+B N. (z.) n{ 2 n J J n J J J J n even (X) = o goes from 1 to infinity and n/2 __ (-1) 4z.tanz. _.,_ M. {z.) _______.._ " n J J (cont.) 21 12 ,.. n N. (z.) J J 2mr tan z.~(k2_ 1) 8z~ + ] 1 (n 2TT 2 -4zj2) k 2 ( n 2 TT 2 -4zj2) (-1t = ' (1.30) 1 1 R~ = r cos z.r(l+r)- 4 dr J .io J r. rl 1 S'. = , sin z. r ( 1 - r) - 4 d r J "o J ,1 . 1 T~ = ' sin z.r(l+rf 4 dr J .io J where z . is obtained from J sin 2z. = 2z. J J The first two of ( 1. 29) have important physical meanings. net shear force at the base is given by The 22 rh Q(O,t) = 1 a (O,y,t)dy .,_h xy Using ( 1. 2) , substituting for (1.31) vX (O, y, p) fr om ( 1 . 16) a n d i n te g r a tin g gives Substituting for a (p) from (1. 28) and the first of (1 . 29) and inverting 0 gives ( 1. 3 2) which goes to zero for long-time. The net moment at the b a se is given by h M(O,t) =J-h axx (O,y,t)ydy ( 1. 33) Us,ing (1. 2), (1. 16), (l. 28) and integrating gives M(O , p) = 2µk 2 2 { _1,. 4 LO B h2 5 0 " -- ) 7 p n~2 n 2 (-1) neven Substituting from the second of (1 . 29) and inve r t i ng g i ves M(O , t) = 2vohp c r H(t ) p g which is constant for long-time. The problem can also be solved approximately using (1. 3 4 ) 23 Euler-Bernoulli approximate theory for the quantities Q(0, t) and When the shear force and moment are calculated, ( 1. 3 2) and M(0, t). (1. 34) result. So the approximate solution has the same net shear force and net moment at the base as does the exact solution for longtime. It should be emphasized however that the exact theory governs the important singularity in uX (0, y, t). The following section assesses this important contribution to the problem. Numerical Solution of the Boundedness Equations Equations (1. 29) were solved nymerically using Fortran IV. The method of reduction was used to calculate the solution. The values of the unknowns were calculated using more and more unknowns until convergence to the final value of each unknown was reached. Conver- gence should be obtained for a relatively small number of unknowns since the Fourier series in (1. 16) are not being called upon to represent the singularities at y = ±h. Values of the unknown coefficients for Table I. v = 0.2433 are shown in As can be seen, the coefficients converge for 30 through 38 unknowns. The coefficients also decay faster than 1 /n 2 for large n. Therefore, the coefficients in Table I are a solution to (1. 29) and hence ( 1 . 16) is a solution to the boundedness condition for long-time. A similar procedure was carried out for each of the other possible singular forms for the unknown base strains. For all of these cases, the unknown coefficients failed to converge by the time fifty unknowns were used. This indicates that the other singular terms do not 24 TABLE I Coefficients ( Coeffi cients V A n and = 0. 243 3 B n ) Number of Unknowns Used 20 24 28 30 32 34 38 42 A2 0.5165 0.5165 0.5166 0.5166 0.5166 0.5166 0.5167 0.5167 A4 -0.2397 -0.2399 -0.23 99 -0.2399 -0.2399 -0.2399 -0. 2400 A6 0. 1529 0. 153 2 0. 1533 0. 153 3 0.1533 0. 1534 o. 1534 0. 1534 AS -0.1111 -0.1118 -0.1121 -0.1122 -0.1123 -0.1123 -0.1124 -0.1124 AlO 0.0859 0.0875 0.0882 0 . 0883 0.0884 0.0885 0.0886 0.0887 Al2 -0.0674 -0.0710 -0.0722 -0.0725 -0.0728 Al4 0.0505 0.0586 0.0610 0.0616 0.0620 0.0623 Al6 -0.0271 -0.04 70 -0.0527 -0.053 5 -0.0539 BO 0.7151 B2 0.2936 B4 -0.0970 -0.0978 B6 0.0469 -0.2398 -0.0729 -0.0731 -0.0732 0. 06 26 0.0628 -0.0545 -0.0548 0.7136 0.7129 0 . 7127 0.7127 0.7125 0.7126 0.7128 0.2950 0.2956 0.2958 0.2958 0.2959 0.2958 0.2956 -0.0981 -0.0982 -0.0982 -0.0983 -0.0982 -0.0981 0.0475 0.0477 0.0478 0.0478 0 . 0478 0.0478 0.0477 BS -0.026 5 -0.02 71 -0.0273 -0.0274 -0.0274 -0.0274 -0.0274 -0.0273 BIO 0.0161 0.0168 0.0171 0.0171 0.0171 0.0172 0.0172 0.0171 Bl2 -0.0097 -0.0108 -0.0112 -0.0112 -0.0113 -0.0113 -0.0112 Bl4 0.0049 0.0069 0.0075 0.0076 0.0077 0.0077 0.0078 0.0077 Bl6 0.0011 -0.003 7 -0.0048 -0.0051 -0.0052 -0.0053 -0.0517 -0.0111 -0.0050 -0.0053 25 represent the base strains correctly. The coefficients from Table I were used to calculate the strains at the base. Graphs of these strains are shown in Figs. 2 and 3. The normal strain, u (0,y,t), X becomes infinite as the corners are approached indicating that there is restraint in the On the other hand, the shear strain, x y=±h direction. v (0,y,t), does not become inx finite for long-time, probably because the motion in the direction is restrained to a lesser degree since the motion the base of the plate wants to make in the thickness direction is the same as is given by the boundary condition. Note that the shear strain may be singular at the cor- ners for short-time. 4. Derivation of the Formal Long-Time Solution Once the transformed edge unknowns, ~ (0, y, p) X have been determined for small p, calculated from ( 1. 11). g(s, y, p} and -; (0, y, p), X the formal long-time solution can be and h( s, y, p) are found by substituting the hase strains from (1.16) into (1.15). The integrandsin(l.ll)arenow known and the indicated integrations are performed. The resulting forms for the doubly transformed solutions are 4 u(s,y,p) = ) u.(s,y,p) -J j=1 (1.35) 4 \ c.v. -,.- ( s,y,p ) v(s, y, p) = , .' 1 J J= where 26 TO 3 -- 2 >- 0 ---:::, >< u-01 >0 z - 0 <( a:: r- en -I _J <( ~ 0:: 0 z -2 -3 TO -ro 0 -I THICKNESS Fig. 2 COORDINATE , y/h Normal Strain at the Base. ro 27 3 -- 2 >. 0 ......., X > -01 0 u> 0 --- ---- z -<l: er rCf) -I er <l: w I Cf) -2 -3 -I 0 THICKNESS Fig. 3 COORDINATE , Shear Strain at the Base. y/h . 28 I (s,y,p) u L(s, p) ii 3 (s,y,p) = -:~ b 0 (p)r°'z sinh<>y(H"-R°') + ~ sinh~y(H~-R~)], s (1. 36) ::,. v (s,y,p) = fA(s,p) • 1 I (s,y,p) V L(s, p) where (1. 37a) 29 (1.37b) (1. 37c) k -1 ~ 2s s 2 2 ) ( 2 2 2) + 1 + 2 a (p) ~ 2} ~(---:-T" k CY +e k n CY +8 n n • ( 1. 37d) (1. 37e) + ~ ~osh ~y ~~-T~)-sinh ~y (H~-R~}z+r;(:~ + 1)} , 30 ( 1. 37 f) (1. 37g) +),lj , - sinh 13 y ( s;-T;)+ cosh 13 y (H;-R~)+ z-¼(+. a h 13 h fH(s,y,p) (l.37i) 2 2 I u ( s , y, p) = - k s [ ( 2 s - k !) cos h p h s inh c, y + 2 c, p cos h c, h s inh p y] (cont.) 31 I v(s, y, p) = -k 2 " ~ 2s 2 -k!) cosh Ph cosh "y - 2s 2 cosh "h cosh pj L(s, p) is the Rayleigh-Lamb frequency equation as in (1. 12 ), S T ry are as in the first two of ( 1. 17) with a. replaced by J 0t Ct (1.38) , and and ( 1. 39) 32 Eouations ( 1. 35) through ( 1. 39) have resulted from the form of the edge unknowns assumed in ( 1 . 16), i.e . they are not directly de- pendent on the assumption of small p. For use in the long-time solu- tion, however, they will have to be appro x imated for p small to be consistent with the approximations that were used to determine the unknown coefficients a (p), an (p) , b (p) and bn (p) t~at they contain. 0 0 The doubly transformed solution will now be inverted, using approx imations that g ive the solution for two different regions of the plate. First, the asymptotics of the Laplace transform will be used to find the solution as x -+ 0. In the following section the doubly transformed solu- tion is inverted by residue theory. away from the base, x = 0, This solution is valid in a region but behind the body wave fronts. Since p must be small, information about the wave fronts cannot be obtained. a. Near-field Asymptotic Solution for x - 0 The near-field asymptotic solution, valid as x .... 0, will be ob- tained in the same manner as it was by Miklowitz in [ 1] and [ 2]. Ap- plying Watson I s Lemma to the present case gives (see Sec. 5. 10. 2. 3 of = --4 u(x , y , PJ {_ 2TT1 v(x,y,p) X..,. 0 L s ~ (s, y, p)} e sxd s ( 1. 40) :=.. { v(s, y, p) ls\ ... ro Since p must be small , ( 1. 35) through ( 1.39) will be expanded with 2 2 I sh I>> 1 and ik s /s l<< l. This gives 33 (1.41) It should be noted that since large here is necessarily on Br , s (1.41) shows, assuming again that p and f3 is real, that the real parts of are large through the value of Im s remain constant. Note that for very large work that follows s while the imaginary parts Is I, et = f3 = Im s. is chosen on the upper half of Br s. since ~ and ~ are even in a approximation holds all over and Br s f3, et In the However, it follows that the large I sh I in the usual way. Because of the singular terms in (1. 16), y = h must be given special consideration. For the ''interior" solution, 0 ~ y < h, the following approximations are valid: L(s,p) = 2 1 s2k2 (k 2 s k2 I (s,y,p)= u -1) e-2ish i 4 s 2 k 2 (k 2 -l)(h-y)e-is(h+y) s (1. 42) 2 1) (h -y ) e -is(h+y) - -J ( s y p ) ""' -1 s 2k2 (k u ' ' 4 s k2 I ( s, y, p ) ·~ - -1 s 2 1<: 2 (I<: 2 - l)(I1-y ) e -is(h+y) V 4 S (cont.) 34 Now S Ct and T 0/ may be written in the form ( 1. 4 3) For large a these become Make the variable change e ah sCt = 2 y = h-r . h J Then 1 e - ar (h/ r )4 dr 0 ( 1. 44) Since a is a real , positive, large parameter in ( 1 . 44) , accordin g to (1. 41), these integrals may be approximated by Watson's Lemma, with the r<'sults 35 ( 1. 4 5) with an equivalent set for s 13 and T /3. Equations ( 1 . 45) can now be used to approximate f A (s, p) and f 8 (s, p) in ( 1. 37) with the results f,(s,p) = -b (p)i:: . - Zish[T'(¾)hl (-is)- ¾+o(f)] 0 s ( 1. 46) Similarly , the integrals H Ct and R CY are written as ( 1. 4 7) Approximating these integrals gives (1.48) (cont.) 36 ~3(s,y,p) and '; (s,y,p) 3 in (1.36) are now approximated for !sh! large with the aid of ( 1. 48), yielding = bO(p) e -is(h+y) 4k2 o(.!.) s s ( 1. 49) Also, for Ish I large ( 1. 50) Combining terms from (1. 42) through (1. 50) together gives ~ I (s, Y, p) • -h0 (p) :~ (h-y) e -is(h+y) [r ( ¾ ) h ¼ (-is) ¾+ o(¾)}+ o(¾is(h-yl), 4 s ~ (s, Y, p) = h (p) : : (h-y) e -is(h+y) 2 0 4 s ~ ( ¾)h¼ (-is)¾+ o(¾)] +o(¾ i•(h-yl), (cont . ) 37 fJ.....) = b (p)£ e -is(h+y) O + o(.!.eis(h-y)) 0 k2 \s3 s 4 s (1. 51) ~ 1 (s, y, p) = -h0 (p) ~~: (h-y) e -is (h+y+ ( ¾)h¾ (-isf ¾+ o(¾)] + oC eis(h-y)), s Adding u , u 1 2 and u , v , v and v , it is easily seen that the low2 3 1 3 -is(h+y) est order terms containing e cancel each other. This cancellation appears to occur for all orders in . s. will be exponentially small for and '; 3 y<'h. The terms of o(¾eis(h-y)) Therefore, will not contribute to the solution as :::,. It remains to approximate u u , u , u , v , v , 1 2 1 3 2 x - 0. :::,. 4 and v . 4 Integrating f E (s, y, p) and f G (s, y, p) by parts and approximating yields (1. 52) 38 The other terms can easily be approximated to give ( 1. 53) From ( 1. 51) through ( 1. 5 3), it follows that 1 0n y} - 2 ~(s,y,p) s neven (cont. ) 39 a (p) cos e Y -1+ f-a n 2 2 n S C s neven ~(s,y,p) J co -I n=2 neven The 2 b (p )(k - 1) 8 cos n n s-x Laplace transform can now be easily inverted, giving co ) + n'·--=: 2 .p.even 2 ( 2 ) x en sin 8 y + O(x3 ) a (p) .!s......:._! _ _ n n k2 2 cos -;(x, y, p) 2 ~ b (p)(k -1) 8 cos L n n n= 2 +..!.( l+y/hfi -2-¼J _ 4 neven ( 1. 54) 40 Differentiating with respect to x, (1. 55) yields the same Laplace transformed strains at the base as were assumed in ( 1. 16 ). efficients a (p), a (p), b (p) 0 O n and b (p) n The co- are givenby(l.28)andTableI. Substituting these into (1. 55) and inverting gives u(x,y,t) B sin n 00 +~ ,: _ .) n=2 n even A v(x,y,t) n X cos neven 2 B (k -1)8 cos n n for x small and t large. Substituting the displacements from (1.56) into (1 . 1), it is easily seen that they satisfy the displacement equations of motion to lowest order in x. The near field solution is not singular for x > 0 when y = h. This is clear from (1.37) if it is noted that fE(s,y,p) and hence y = h, ""' :::,. u and v , 4 4 Hy= H, Q' Q' involve only integrable singularities. Ry= R , Q' and f 0 (s,y,p), Q' etc., fE(s,y,p) and fG(s,y,p) Since for can be 41 approximated using (1. 45) and (1. 48), giving ( 1. 57) Also, ( 1. 58) ~ fH(s,h,p) = L (-1) n=2 neven while n/2 [an(p) 2 - 2- - bn(p)(k -1) s f K (s,h,p) is given by the last of (1. 53). From (1. 57), (1. 58) it follows that ( 1. 59) 00 -I 2 (-l)n/lbn(p){k -1)8n-\n =2 s neven Substituting for the coefficients a (p), an(p), b (p) and bn(p) from 0 0 42 ( 1 . 28) and inverting yields u(x, h , t) (1.60) v(x , h,t) = ~ ( - 1t 12 1A x-B (k:-l ) J.-;;z Ln n neven If (1-y/h) gets small at the same rate as x does , t he b e h avi or of ( 1. 60) is in agreement with that of ( 1. 56 ) . (1. 56), of course, get large as absence at y = h. y ~ h which is not consistent with t hei r As Miklowitz shows in parts integrations of fE (s,y,p) The singular te r m s in [ 2], through furthe r and fG(s , y,p), a series of sin g ular n n-~ terms (valid for small x) of the form x /(1-y/h) 4 are obtained for ""' u 4 and ""' v . 4 These terms alternate in sign (cf .(54) in [ 2]) and must cancel the singular term in ( 1. 56 ). It follows then , that for the leadin g singular term in ( 1. 56) to represent u, x must vanish at a f aste r r a te 1 than ( 1- y/ h) 4 does , i.e., the asymptotic solution , ( 1. 56) is limited t o smaller and smaller x values as the corner at y = h is app r oache d. Otherwise the use of additional terms involving needed, which will still have limitations as y X 3 , X 5 , . . . ' will be gets closer and close r to h. b. Inversion by Residue Theory for the Domain h< x<< est The doubly transformed displacements are even functions of a, and /3; ::,. u(s, y, p) ""' and v(s, y, p) thus they have no branch points in the s-plane, and as a result they can be inverted solely by residue theory . 43 From (1. 13), s .(p) is defined as the roots of J L[s.(p),p] = 0 J Then the poles of ~(s,y,p) and ~(s,y , p) are s = 0 (1. 6 1) and s=s.(p) J where, as a result of the boundedness condition, ( 1. 14), Re[ s. (p)] < 0. J This gives u.(x,y,p) = R u.1 ~ /_, (0) + 1 • J= u.1 ( R. 0 J s.(p) J ) (1.62) V. v.(x,y,p)=R 1 CO (0)+ 1 \ L j =0 "" sx and v.(s,y,p) e , 1 V R.i(s.(p)) J J .. respectively, at the pole s.(p). J Now ~ (s,y,p), ~ (s,y,p) and ~ (s,y,p) have no poles in the 4 3 3 s -plane and the only pole of ~ ( s, y, p) comes from the pole of 4 fK(s,y,p) ats=O. Therefore (1.63) 44 On the other hand, the othe r doubly transformed displacements do not have a pole at s = 0. of poles at the modes s = s.(p) , Re[s.(p)] < 0. J J They do, however, have an infinite numbe r Lowest Mode Contribution First consider the contribution of the lowest mode to the solution. From (1. 21), for p small, the lowest mode is given by - yz;-;- y -~ p g Of these, only the roots s (p) = -(l±i) y will have non-zero residues. 0 By definition ~.(s,y,p) = limit S-+s (p) 0 1 ( s - so(p)) \ e sx ~.(s,y,p) 1 Expanding R:i(s (p)) 0 Exp a nd L(s, p) gives rn a Taylor s e ri e s about s = s (p). 0 Then, sinc e ( 1. 64) 45 (1. 65) Along a mode of the Rayleigh- Lamb frequency e quation, L(s, p) = 0. Therefore d s + op oL d P = 0 dL - ~Ls u ( 1. 66) After a considerable amount of algebra, it can be shown that (1.67) (1.68) Combining (1. 66) through (1. 68) gives ( 1. 69) Write " f A(s,p) = fA(s,p) + r A(s,p) where (cont.) 46 cosh CY h(S -T ) + 2 f3 cosh f3h(SA-T )} CY CY ~ f3 ( 1. 70) Then the following approximations are valid for cosh CY y} { coshf3y s = s (p) 0 and p small -· 2 = 1 + O(p) y (1. 71) sinh CY Y} 3 3 . = is (p) y + O(p~) y 0 { srnh f3 y (1. 72) Equations (1. 69), (1. 70) and (1. 72) are now used to give ( 1. 73) 47 was. First let where (1.74) f, (s, p) = - :~ h 0 (p) {zs 2 2 sinh a, h (Sa, -T) - ( 2s -k!) sinh ~ h (S~-T ~)} , s Approximating gives fc(s 0 ,p) = -n!Z [(-J/'/2bn(p)( + an(p)O(p½)] neven Combining all these yields (1. 75) 48 n even 1 1 + O( 1) + a (p) O(p 2 ) + an (p) O(p1: ) + b (p) O(p) 0 (1. 76) 0 The total contribution of the pole at s (p) to u(x, y, p) 0 can be found by 2 1 adding R: (s (p)) from (1. 73) to R: (s (p)) from (1. 76). 0 0 of the last term in (1. 73) plus the last term in (1. 76) equal to Set the sum ¾~ 0 (p)) , i.e . , ( 1. 77) Approx imatin g R~(s (p)) 0 g ives, after a great deal of algebra, ( 1. 78) Then from (1. 73), (1. 76) and (1. 78), 49 u oo .!. 5 -2 4 7 bo(p) + \ - Ro(so(p)) = { } k 2 3 (p) x (- l)n/2 b (p).3_ o hy e o P n n1r 4 k2 s L n::: 2 n even s ( ) s (1. 79) Now we substitute the coefficients a 0 (p), an(p), b (p) and bn(p) from 0 ( 1. 28) into ( 1. 79). Then, from the boundedness equations (the first and second equations of ( 1. 29)), the following equations hold: ao(p) = 00 -2 -¼ -5 b (p) 7 0 +l n=2 neven AO ,./p n (-1) 2 VO = --2cs y 2b n(p) n1r VO = -cdp 2 Ak ~1 -v) Using these relations, (1. 79) reduces to ( 1. 80) Define the contribution to u(x, y, p) by the lowest mode by ~l(x,y,p). Then uL(x,y,p) = R~[(-1 +i)y]+R~[-(l+i)y]. this expression gives Evaluating 50 -L u (x,y,p) (l.81 ) The lowest mode contribution to v (x, y, p) same way. Starting with the second of ( 1. 64), R is calculated the vi I \ \ s (p) ) 0 0 is expanded using the same approximations as were used to calculate The only new approximations needed are (1. 82) The lowest mode contribution, -L v (x,y,p) = VQ 2 ~L(x,y,p) , is (cosyx+sinyx)e- y x + O(p). (1. 83) p Contribution of the Complex Segments of the Higher Branches s., For the higher modes, s. s.(p) == j 2 1 where is a comJ J J plexconstantsatisfying(l.22). For to be a pole, Re(s.) must be J J s. negative. Let A s. J where that ,.R ,.J s. and s. are real, positive numbers. J J (1. 84) It can easily be shown 51 = K.(y) 1··· p e ±."I ... Rx 1s.x -s. J e J (1. 85) R~'i (s.) J J s.,J K.J (y) is only a function of y. where, for a given mode So the higher modes give edge waves that decay exponentially with x. Hillman and Salzer, [lo], the smallest sf st is = 3. 7488/h. From So the decay of the higher modes contribution is quite rapid and they may be neglected for x/ h greater than about one. Their contribution is quite important, however, for x small where they supply the difference between the near field asymptotic solution and the lowest mode contribution. Since the higher modes will not contribute for x/h > 1, in this region, from (1. 63), (1. 81) and (1. 83) we have that u(x,y,p) 2v 1 0 -yx = - 2- y y sin y x e + O(p-~ ) p ( 1. 86) From the second of ( 1. 2), cr yy ~ - (x,y,p) + k2(x,y,p) = µ (k 2 -2)u v (x,y,p) X y [ Differentiating (1. 86) gives, to lowest order in p, 52 2 u (x,y,p) = 2vo -y y y (cos yx - sin yx)e-yx X p ( 1. 87) Now the O(p) terms in v(x, y, p) are of two types. First there are the terms that come from retaining more terms in the approximations that wereusedfor £A(s ,p), 0 oL I ( )]- l [ os s = so p • f 8 (s ,p), fc(s ,p), 0 0 f 0 (s ,p) 0 and These terms will be _independent of y. Secondly, there are the terms that come from retaining additional terms in the approximations for I)s ,y,p) 0 and J)s ,y,p). 0 Some of these terms will come from keeping the second term in the expansions for cosh o- y and cosh f3y. The first of ( 1. 71) shows that these terms will be propor- . 1 to y 2 tiona The refore v (x, y, p) = y O(p) y Now, from the boundary conditions a is linear in y, yy (x, ±h, p) it must be zero everywhere. = 0. Since o yy (x, y, p) Therefore, using (1. 87), it can be shown that v (x,y,p) y to lowest order in p, i.e., (1. 88) o(½). Keeping the next highest order 53 terms gives, for p small, cr yy (x, y, p) = 0( 1) ( 1. 89) From the first of ( 1, 2) Substituting from (1.87) and (1.88) gives ( 1. 90) Similarly axy (x,y,p) = µ[u y (x,y,p) + V (x,y,p)lj X From (1.86), to lowest order in p, uy (x,y,p) = -v (x,y,p). So the X 3 O(p-~) terms cancel. Keeping the next order terms gives 1. ax/x, y, p) = O(p-r:) As (1. 89) through (1. 91) show, for long-time provided f > 1. o xx (x,y,t) (1.91) will be the dominant stress 54 Inversion of the Time Transform Behind the body wave fronts, but for hX > 1, the time Laplace transformed displacements for the residual problem are given by (1. 86) and the transformed stresses by (1. 89) through (1. 91). To obtain the VQ solution to the original problem, the term 2 must be added to p v(x, y, p) in ( 1. 86). The displacements and stresses are then inverted using the tables in Abramowitz and Stegun, [ 11]. This gives, for t large -· I t U(x Y t) - 4v Y ' ' 0 y2TTC r ~in t, + %'1 (1-2C2 (t>) )} p g o(¾) ( 1. 92) CT Here C ('6) 2 yy (x, y, t) = 0 and S ('6) 2 ( 1. 93) are the Frenel integrals defined by (cont.) 55 D. s2 (<'>) = }z:.Jo ( 1. 94) ";.-z dz and X 2 D. = 4c r t p g ( 1. 95) X Tl = J2rrc r t p g Cd - - v(x, 0, t), and the stress v h 0 The displacement at the centerline, Cd h at Y = h, v Ou o x)x, 2 , t), we re calculated for two times t = 100 th and t = 400th where th= h/cd. The displacements are graphed in Fig. 4 and the stresses in Fig. 5. Observing the stresses in Fig. 5, it is seen that the shortest wavelength (highest frequency) waves lead in the disturbance, progressively becoming longer and longer as x given time. x But since the solution is valid only for low frequency, the large response in these curves is inadmissable. gion x>s 100 For t = 100 th, has arbitrarily been ruled out. are not valid for in the region decreases for a x < h, h < x < !;, Since the re- (1.92) and (1.93) the solutions graphed in Figs. 4 and 5 only hold 100 for' t = 100 th. the solutions hold in the region h < x < x Similarly, for 400 . t = 400 th, It is clear, therefore, that as time increases the solution is valid for larger and larger x. As was mentioned earlier (see discussion following (1. 34)), the problem can be solved approxin1ately using Euler- Bernoulli approximate theory. When this is done, (1. 92) results for the displacements .. "O u l> - 0 ..c > >< ..__.. 0 ----. 100 0 100 200 \ 0 L y Fig. 4 I \ I' \ I I 300r 400 \ \ \/ 75 Centerline Displacement, 50 ~ t = I 00 th - t = 400 th 25 x=h x/h 125 150 v{x, 0, t), for the Semi-Infinite Plate. 100 Ul O" u"IJ- b >< >< >< ..__.. . ..c .......... (\J -+- -I 0 01 Fig. 5 Stress a 25 \ xx (x . y,t) 50 I .I rt::; I 00 th ,rt= 400 th \ \ .I I\\ -. 21\\ 3 x/h 100 I ,\ 125 I .\ 150 I \ at y = h/2 for the Semi-Infinite Plate. 75 \. (J1 --.J 58 and (1. 93) for the stresses except that the order terms are missing. So for long-time in the region x > h, the Euler-Bernoulli approximate theory gives the dominant terms in the solution. 0 ~ x < h, In the region the exact solution differs from the approximate solution by a series of terms that decay exponentially with x. At the base, x = 0, the Euler-Bernoulli approximate theory gives the total moment and the net shear force to lowest order in t. However, it must be emphasized that it gives no information at all about the important singularities at the corners y = ±h, x = 0. 59 II. THE FINITE PLATE 1. Formulation, Formal Solution and Entirety Condition Once the semi-infinite cantilevered plate problem has been solved, the problem of a similar finite cantilevered plate, built-in at the base (x = 0) and stress free at the other end (x = £), the long-time response. can be solved for The plate is depicted in Fig. 6. The problem is formulated in exactly the same way that the semi-infinite plate problem was, i.e., with the displacement equations of motion, (1.1), the stress-strain relations, (1. 2), the initial conditions, (1. 3), and the same boundary conditions at the base (y = ± h), (1. 4). (x = 0) and on the plate faces The only changes are that the radiation conditions, ( 1. 5), are replaced by a xx (£,y,t) = a xy (£,y,t) = 0 for -h ~ y ~ h, t ~ 0 ( 2. 1) and that u(x, y, t) = v(x, y, t) = 0 for x > £, -h < y ~ h, t ~ 0 (2. 2) In the work that follows, it will be assumed that the length of the plate, £, is greater than the width, 2h. As before, the problem is decomposed into a rigid body motion and a residual problem. The residual problem will still satisfy the initial and boundary conditions, (1. 7) and (1. 8), but now the radiation condition, (1.9), is replaced by (2.1). To derive the formal solution, first use the one-sided Laplace transform with respect to t, parameter p. Transforming the Fig. 6 Vt= Vo u =0 u O"yy =a-xy=o · h h (J"yy = (J"x.y = 0 x=f X CTxx=CTxy=O Geometry and Boundary Conditions for the Finite Plate . V y O" 0 61 displacement equations of motion, (1. 1), gives 22 2k u (x, y, p) + (k - 1) v (x, y, p) + u (x, y, p) = ks u(x, y, p) xx xy yy (2. 3) v 2 22VO (x, y, p) + (k - 1) u(x, y, p) + k v (x, y, p) = k v(x, y, p) + XX yy S C2 s VO where the 2 C term comes from the non-zero initial condition in ( 1. 7). s Noting (2. 2), introduce the one-sided finite Laplace transform with respect to x, parameter s, defined by p_ ';;'(s, y, p) = J u(x, y, p) e -sxdx 0 (2. 4) Transforming (2. 3) gives 2 2 d~ ~ 2 2 2)= -d ';;' 2 (s,y,p) + (k -l)sdy(s,y,p) + ks -ks u(s,y,p) = g(s,y,p) dy (2. 5) ~ d 2:::,,. v k2- 1 a';;' s 2 - ks2) """ dyz (s,y,p) + (-;;z-)•dy(s,y,p) + kz v(s,y,p) = h(s,y,p) where 62 ( 2 . 6) 2 - (0 , y ,p ) + VO] +-;X (O,y,p) + (k-l)u h(s, y,p) = k1 [sv(O,y,p) 2 2 y SC s - (£,y ,p) - 21 [sv(£,y,p) + -v (£, y,p) +(k 2 -l)u k y X Equation (2. 5) has exactly the same form as (9) in [ 5]. Solving thi s equation once again yields (1. 10) through (1. 12) for the formal solution where g(s,y,p) and h(s,y,p) are now given by (2.6). In [ 12], Widder proves that the function f(s)= I b e - st da(t) a is entire . Comparing with the first of (2. 4) shows that ~(s, y, p) and ~(s, y, p) can have no poles in the s- plane . For the semi-infinite waveguide it was only necessary to rule out Re[ s /P)] > 0 whe re s. (p) J was defined by (1.13). Now if s.(p) satisfies (1.13) , so do s'.(p), -s .(p) J J J and - s'.(p). Two of these will have real parts greater than zero and two J will have negative real parts. For the finite waveguide it is necessary the poles in the right half s- plane , i.e. for to set the residue of all four of these roots equal to zero. This gener- ates an entirety condition which finally will determine the transformed 63 edge unknowns. their conjugates. So ( 1. 14) now must hold for s .(p) and for - s .(p) and J J This gives four coupled integral equations for the edge unknowns at x = 0 and x = £. These allow four of the edge unknowns, two at each end, to be calculated. The boundary conditions on the ends of the plate can now be used to completely determine the rest of the transformed edge unknowns and hence the formal solutiop. 2. Forms for the Edge Unknowns from Euler-Bernoulli Approximate Theory In the theme of the earlier work. by Miklowitz and Sinclair, [4] and [ 5], representations will be set down here for the transformed edge unknowns. They will be found from the Euler-Bernoulli theory since then they would be expected to be at least a part of the total representations for the edge unknowns for long-time. To solve the entirety equations, it is necessary to assume forms for the edge unknowns at x = 0 and at x = £. The same representa- tions for the unknown strains at the base will be used as were used for the semi-infinite plate problem. Sinclair and Miklowitz showed in [ 5] that, for an antisymmetric plate with both stresses prescribed on the end, the edge unknowns agreed to lowest order in time with the forms found for them '.Ising Euler-Bernoulli approximate theory. It seems reasonable to expect similar behavior for the edge unknowns at x = .R.. for the present problem. The Euler-Bernoulli approximation to the present problem is forn,ulated in the usual way. The governing equations are 64 4 2 8 v(x, t) + _l_ 8 v(x, t) = 0 n 4 2 2 n 2 ux c r ut p g u(x, y, t) = -y 8v(x, t) Bx 2 (2. 7) M(x, t) = -2h c 2 r 2 8 v(x, t) p g ax 2 Q(x, t) where M(x, t) at x, and Q(x, t) are the net moment and the net shear force respectively (see (1.31) and (1.33)). Initial and boundary condi- tions are v(x, O) = vt(x, 0) = 0 for x > 0 (2. 8) and v(O , t)=O X for t ~ 0 (2. 9) M(i, t) = Q(£, t) = 0 After Laplace transforming with respect to time and using the initial conditions, (2. 8), the first of (2. 7) becomes d 4- 2 v(x, p) + ____e__ -( )= 0 4 2 2 V X, p dx c r p g (2. 10) Next, after finite Laplace transforming with respect to x as in (2. 4), (2 . 10) becomes ~ 4 ""' 23s +-1:__ 2 2 2 v(s,p)=vXXX (0,p)+svXX (0,p)+s v X (0,p)+s v(O,p) C r [ p g (2. 11) -[-;; XXX (£, p) + s-;; XX 2 3 (£, p) ·+ s -;; (£, p) + s -;;(£, p)]e -s£ X Using the boundary conditions and the last two of (2. 7) reduces (2. 11) to """ -; v( s, p) XXX (0,p) + s-; XX (0,p) • (2 . 12) -r )£, 1,2V p) + S 7 3-s£ v(£, p)J e To obtain the doubly transformed solution, the edge unknowns, -;; XXX (0,p),-;; XX (0,p),-;; (£,p) explained earlier, X and -;;(£,p), must be found. Now, as was ~(s, p) must be an entire function of s. Therefore the numerator of (2. 12) must be zero whenever the denominator equals zero. Now the denominator of (2. 12) has zeros at s = ± (l±i)Y, 66 Y= J tp r g Substituting into the nwnerator gives the following four equa2 tions for the edge unknowns: -v XXX (O,p)+(l+i)yv XX (O,p)+ E-2i y 2v (£,p)+2(1-i)y 3v(£,p) ] e -(l+i)y.e = X 3 voy 2(1-i)-2p -v XXX {O , p)+(l-i)yv , (O,p)+ [ 2i y 2v (£,p)+2(l+i)y 3v(£,p) ] e -(l-i)y£ = XX X . 3 Vy . 2(l+i) ~ p , (2. 13) -v XXX (O,p)-(l+i)yv J XX (O,p)+ [ -2iy 2v (£,p)-2(1-i)y 3v(£,p) e (l+i)y£ = X 3 voy -2(1-i) --2p -v XXX (O,p)-(1-i)yv , •(O,p)+ [ 2i y 2v (£,p)-2(l+i)y 3v(£,p) XX X J e {l-i)y£ = 3 voy -2(l+i) - 2 p Solving (2. 13) for the edge unknowns yields ~ D(p) (cont.) 67 -;; xx (O,p)= Vo F1(n )1 ~ c r p D(p) (2. 14) p g -;;(£,p) = X (2. 15) 2v __ 0 H(p) v(f, p) 2 D( ) p p where 2 2 D(p) = cosh y P. + cos y P. E(p) = cosh y P. sinh y P. + cos y P. sin y £ 2 2 F(p) = cosh y £ - cos v £ (2. 16) G(p) = sinh y £ cos y £ - cosh y £ sin y £ H(p) = co sh y £ cos y £ The edge unknowns obtained here are for the original problem where the base of the plate has a constant velocity (see ( 1. 4a)). The residual problem is found by subtracting a rigid body motion v(x, t) = v t from 0 the original problem. problem v /p 0 2 So to obtain the edge unknowns for the residual must be subtracted from v(.£, p). This gives _ ___Q _ ] V [ 2H(e) v(i., p) - p2 D(p) 1 Using the first of (2 . 15) and the second of (2. 7) shows that (2.17) 68 2i.E.l voy (2. 18 ) c r p y D(p) p g 3. Solution of the Entirety Equations The estimates for _ will be called u u(f,y,p) EB _ (£,y,p) and and v (f,y,p) v (e,y,p) where the show how ti,e terms were found. EB superscripts To these estimates will be added a supplementary set of edge unknowns A- in (2.17) and (2.18) EB distinguished by the superscript to account for the difference between the exact and Euler-Bernoulli theories. The exact theory forms for the edge unknowns are _EB _ A (£,y,p) + u (£,y,p) u(£,y,p) = u (2.19) _EB _A of y the p A u (£, y, p) and p. A (£,y,p) + V (£,y,p) v(.R,y,p) == V where _ and v (£, y, p) are two additional unknowns functions Each will be r e pr e sented by a Fourier series in d e pendence incorporated into the s e ri e s coefficients. y with Now from _A the symmetry of the problem u (£, y, p) will be odd in y and hence will _A be represented by a Fourier sine series while and represented by a cosine series . v (£, y, p) will be even Quarter range Fourier series, similar to those used for the semi-infinite plate (see (1. 16)), will be used. Thus co , ( ) . mry u ( £, y, p) = ) c p sin Zh L n n =1 _A (cont.) 69 00 ; \11, y, p) = do(p) + { dn (p) cos ~~y n =1 Later it will be shown that only the n (2. 20) even terms are needed to rep- resent the edge unknowns. Differentiating u(£,y,p) and ;(.R.,y,p) withrespectto y gives (2. 21) 00 ln= 1 ~~ d n (p) sin ~~y v (£, y, p) = - \ y Now use is made of the boundary conditions at x = P.. From ( 1. 2) and ( 2. 1) l -0 XX (£, y, p) = 0 = µ [ k 2u (£, y, p) + (k 2 - 2) -v (£, y, p) . X y ., (2.22) Therefore 00 2 -ux(£, y, p) = (k 2) -2. nTT 2h d (p) sin n2TThY k n -~ 1 n -. 1 ( 2. 23) Similarly, the other boundary condition, crxy (£,y,p) = 0 = u[u (£,y,p) +; (£,y,p)-, _ y X j ( 2. 24) 70 implies that 00 -:;;. (1, y, p) = x vO c r p y P g G((p)) D p _) mr c (p) cos nrry ,__ 2h n 2h n =l The boundary conditions at y = h, (2. 25) from ( 1. 8b) and ( 1. 2), are (2. 26) Letting x .... £ and combining (2. 2 2) and (2. 2 6) shows that v (£,h,p) = ;- (P.,h,p) = 0 y X This corner condition suggests that only the n . (2. 27) even terms be used for the series in (2. 23) and in the second of (2. 20) and (2. 21). boundary condition at y The other = h does not restrict ;-y (.£, h, p) or -:;;.X (£, h, p). Therefore, the series may contribute at y = h and so the n even terms should be used in the Fourier series for these edge unknowns. In summary, the following forms will be assumed for the edge unknowns ;-(0,y,p) = ;- (0,y,p) = -:;;.(O,y,p) =-:;;. (0,y,p) = 0 y y 0) ;-x(0,y,p) = b (p{(l-y/hf¼-(l+y/hf¼+ 2-¼y/h] +J= 0 bn(p) sin n rrhy 2 2 neven (cont..) 71 I CX) ;)O,y,p) = ao(p) + an(p) cos ~rr: n=2 n even co VoY .QJEJ_ l ( ) . nrry D( ) + en p srn 2h cpr gp y p n =2 n even ~(1!, y, p) = CX) ~ (£, y' p) = y VO _gJpj_ c r p y D(p) p g ;:;y, y • p) = ~k:22) I ~~ 2 \ nrr ( ) nrry 2h en p cos 2h + l n=2 n even (2. 28) dn (p) sin ~nhy n even VQ ;(1,y,p)=z p l2H(p) _ LD(p) JJ + I dn (p) cos ~rrhy n=2 n even CX) ; y nrr d ( ) s1'n · mry (£, y, p) = - \' )_ 2h n p 2h n=2 n even co .Qi.el _ \' nrr ( ) nrry l 2h c n p cos 2h D(p) n=2 n even The unknown coefficients a (p), an(p), b (p) 0 0 and bn(p) here will differ from the values found for them for the semi-infinite plate case. The ecige unknowns from (2. 28) are substituted into (2. 6) to give 72 g(s,y,p) and h(s,y,p). Substituting these into the entirety equations, (1. 14), and integrating using (1. 17) yields the following infinite set of algebraic equations: 00 a (p)M?(s.,p) + )' a (p)M1:(s . ,p) + b (p)N?(s.,p) 0 0 J J n~2 n J J J J n even (2. 29) -s.£ 00 + \ b (p) N1:1(s., p) + d (p) P?(s., p)e J + \ [c (p) 01:(s., p) 0 n~2 n J J J J n~2 n J J n even n even oo + d (p)P.n (s.,p) ] e n J J -s.£ -s.£ J + Q.(s.,p) + R.(s.,p)e J JJ JJ plus a similar set with = 0 replaced by -s.(p). J n O n M. (s.,p), N. (s . ,p), N. (s.,p) J J J J J J ( 1. 1 9) while and Q.(s.,p) JJ are exactly the same as in n O.(s.,p)-J J In (2. 29), 0 M. (s., p), J J sJ.(p) p/2 (-lJ 28 s.Y.(s.,p) n J J J k2 0 s P. (s.,p) = - 2 Y.(s.,p) J J r3. J J J (2.30) ( cont. ) 73 2 _g_{_p_)[2(k -l)s. R.(s.,p) J J /3~ D(p) Y.(s . , p) J J J Y.(s.,p)H(p) J J D(p) where Y.(s.,p) is as in (1.14). J J Equations (2. 30) are approximated for done for the semi-infinite plate. by ( 1. 2 1), i. e. , s = s ( p) small, just as was For1;,he lowest mode, s 0 ( p) = ± ( 1 ±i) y, ( 1. 1 9) and ( 2 . 3 0) for p y = /zc Pr ' p g and p . s (p) 0 is given Approximating s m all y i e 1 d s ( 1. 2 3 ) for O 0 n O n M (s , p), M (s , p), N (s , p), N (s , p) 0 0 0 0 0 0 0 0 and Q (s , p) while for the 0 0 other terms the following results: so3h3 7 5 + O(p~) (cont.) 74 5 + O(p2) (2.31) For the higher modes ,.. s. (p) = s . ' J J j ~ 1, Approximating now gives (1. 24) for N~(s., p) J J " where s. satisfies (1. 22). J M?(s., p), M~(s ., p), N?(s ., p), J J J J J J and Q.(§., p) while for the other terms the following results: J J 2 " " ( k - l) 8 s. tan s. h n 2 4( 2 ,.. 2) k fl -s. n J + 0 ( p2 ) p/2 2 ,..2 2 (-1 J (k - 1) 4s. 8 tan ;,h n 2 4( 2 "2) k 9 - s. n J + O(p ) ,.. ( - l f1 01:" ( s . , P) = J J /2 0 ,.. P. (s. , p) J J n " P.(s ., p)= J J 2 (cont.) 75 2 tan s.h A R.(s.,p) J J A s. J (2.32) 2v tan ;_h 0 + 2 C .!iLE2. + 0 ( 2) D(p) p s To solve this infinite set of algebraic equations, assume that the unknown coefficients are of th e form b B - _Q_ !1el O(p) - p D(p) B b ( ) _ _E_EJ.El n p - p D(p) C n (p) = en !1el p D(p) D d (p) = _Q_ B..e.l 0 p D(p) D d () =-E.B..e.l n p p D(p) (2.33) 76 where A , An, B 0 , Bn, Cn, D and Dn 0 0 are independent of p D(p), E(p) and F(p) are defined in (2. 16). equations obtained from s and Substituting into the fo u r = s 0 (p) = ±(l±i) y and neglecting all but the lowest order terms in p yields (2 . 34) Substituting the four roots s (p) into (2. 34) and multiplying both sides 0 2 2 2 of the resulting equations by cd /hc r gives the following equations: p g + [-2·1 2 ~2yv0 G(p)~ Y p 2 voy D(p) 3 = 2(1-i)-2P (_ 6 AO E(p)\ \ Jp D(p)/ + ( 1-i) ~ (Mk2 6 !:..(£)_) y \ p D(p) + [2i 2 (2yv O y p2 9iP.2.) D(p) ( cont. ) 77 Qil?.l) 2 2 1• 2( Yvo y 2 D(p) [ p (2. 3 5) 3 . voy = -2(1-1)-2P 2 + [ 2 • 2( yv o ~ ) 2 i y D(p) ' p -2( l+i) y 3 c:o ~~~:)] 3 vo y exp [( 1-i) y i] = -2(1+i)-2- p where 00 M = -2 _ -1_ 4 5 7 n \ (-l)2 BO+ nf 2 2B mr n , and n even C 2 s 6 = 22 C r p g Comparing (2. 3 5) with (2. 13) shows tha t they are identical provided A - o _Q ED((p)) =v (O,p) .jp p XXX , (cont.) 78 (2. 36) 2v O H(p) - 2 D(p) = v(£, p) p Substituting the solution of (2. 13) for v (£, p) and v(.£, p) from (2. 15) X shows that the last two of (2.36) are satisfied identically. This shows that, for long-time, the dominant terms in the edge unknowns at x = £ will be in agreement with those given by the Euler-Bernoulli approximate theory. The first two of (2. 36) are not satisfied identically be- cause of the singularities at the corners . Substituting (2. 14) into the first two of (2. 36) gives A 0 = n \' (-l) 2 -2 4 _7 B + ) 0 L n=2 neven _ .1. 5 00 2B n = mr v0 / _,,,_ 2 _ 2 cd 'v3k (1-v) So the four equations coming from the lowest mode have been reduced to two, which it should be noted are identical to the first two boundedness equations for the semi-infinite plate, (1.29). For the higher mode s, s., j ~ 1, where s.J is a s = s.(p) = J J 79 s.J is a root, then I., -s. J J and -I. will also be roots where I. is the complex conjugate of s .. J J constant which satisfies (1. 22). Now, if J Since we are setting both the real and imaginary parts of (2. 29) equal to zero, it is only necessary to use two of these roots, Consider a root s. in the first quadrant, i.e. J . ,. ,.+ s. = s. J J w -h ere ,.R s. J an d s.J and -s J. . ,.I s. J ,. R + . ,.I = S. 1 S. J (2. 38) J are real, positive numbers. Then, for this root, using (2. 33) for the unknown coefficients and retaining only the lowest order terms in p, the entirety equations become the following set of algebraic equations: 00 ,. +, p) + \ (s. ~ A M1: ( /~ ,_ n J J n= 2 J s: , p) + Bn N1:J ( s:, p) ] J n even (2. 39) oo r • ,.I ,.Rn} + + -1s.£-s. x + \ · C 01: ( p) + D ef.1 ( p)] e J e J =0 L L n J J n J J n=2 n even s. , where M~(s':,p), N?(s":,p) J J JJ and s. , and N~(s':,p) J J are as rn (1.24) and d:(s':,p) J J P1:(s":, p) are given by (2. 32). J N e xt use s.(p) = -s": in the entirety equations. This will generJ J a te another set of algebraic e quations. Usin g (2. 33) for th e unknown co- J efficients, retaining only the lowest order terms in p, and multiplying 80 ,.+ n the resulting equations by - s . ,t e J gives the following set of algebraic equations: e O oo + + + {R - BON . ( s . , p) + ) ~A M1: ( s . , p) - B N1: ( s . , p J J ... • n J J n J J I m }{~ n= 2 n even ~ 1- ,.Inx 1• s. J e - s,.R . P. e J ( 2. 40) 00 \ r n ,.+ +L LC 0 . (s. , p) n~2 n J J n even As was mentioned earlier, the smallest s:lJ is s,.R1 = 3. 7488 h. -s~ £ Since P./h :::=>: 2, e J will be small. that are multiplied by e -s~J £ Neglecting the terms in (2 . 39) shows that (2.39) reduces to the last of ( 1. 29). Therefore, the coefficients Table I. Equations (2. 40) can now be reduced to An, B 0 and Bn are given by (2. 41) oo -) -~ • ,.I ,.R P.} n + -1s.P. -s. 2AM.(s . ,p)e J e J n J J n=2 n even wh e re the right hand side is known. be d ete rmin e d from (2 .4 1). The coefficients C n and D n can Since the right hand side b e comes small as 81 AR -S . e n X so will J C and D . n n So the Fourier series parts of the edge unknowns at x = £ are small corrections to the Euler-Bernoulli terms. Once the edge unknowns have been found, can be calculated from (2. 6). operator, e - s P. , g(s, y, p) and h{s, y, p) Any term that is multiplied by the shift will not contribute to the solution for x< £. So, for 0 ~ x< £, the doubly transformed displacements are once again given by ( 1. 3 5) through ( 1. 39) where the coefficients are now given by (2. 33). Since the terms multiplied by e-s £ will not contribute for 0 $'. x < £, for x the doubly transformed displacements will still have poles in this region. So the verted by residue theory. changed from ( 1. 63). s-x Laplace transform can still be in- The contribution of the pole at s = 0 is un- The contribution of the lowest mode is calculated in exactly the same way as it was in Chapter I giving ( 1. 81) for R~(s (p)). 0 Then using (2.33) and (2.35) gives, for a root s (p), 0 t u( ) vo s~(p) lliEl s;(p) F(p)l so(p)x -i RO sO(p) = - 2 -sO(p) + - - D(p) - - - 3 D(p) ye + O(p ) 4p y 2y I All four of the lowest mode roots will contribute to the solution. ming the residues for each of the ~ {[(l+ ;,t (x, y, p) = : : s (p) 0 gives ZEg,;;r(p)) sin y x + (1 - +ii(l - 2E(p)- F(p)) sin y X ~ D(p) Sum- (1- !J.E.l) D(p) ~rnn cos y cos y+-y x j (2. 43) x] ey + O(p-i). 82 Similarly, the lowest mode contribution to v(x, y, p) and to the stresses i~ given by ?(x y p) = v O {1/1 + ~ ) cos y x + /E(p)+F(p)) sin y x]e -y x ' ' Zp2 ~ D(p) \ D(p) (2. 44) - [(E(15(:,(p)) cos y x - ~ ~~~:) - sin y x] e Y x}+ 0( I) (2. 45) 0yy (x, y, p) = Q( 1) L The higher mode roots with Re(s.) < 0 will once again give J edge waves of the same form as (1. 85). On the other hand, for the roots with Re(§.)> 0, exponentially increasing waves will result. CalJ culating the residues for these roots shows that ,.. u K. (y) J - F(p) - pD(p) "'V K. (y) J 00 { ,.. + , p) +• \/ [A Mn. ( s., "+ p ) B N.0 (s. 0 J J -n J J n=2 n even s~x s1:'x + B N1: (s :, p) ]} e J n J J e J + 0( 1) 83 Au where K. (y) J "V and K. (y) J are, for a given J, a function only of y. Using (2. 39) we have "u K. (y) oo F(p) = J pD(p) \ [ c o.n (s.+ , p) 1 n=2 n J J n even (2.46) + s. , + D P1:1 ( n J J 1 -i s. (£-x) -s:Z(.e-x) p )] e J e J + 0 ( 1) So the contribution of these terms will decay exponentially away from x = f. At x = f, summing up all of the residues should just give the Fourier series terms in and D u(£, y, p) and ~(£, y, p) in (2. 28). Since C n are small (see discussion after (2. 41)) these terms will be ne- n glected. So, for X h > 1, the transformed displacements and stresses are given by (2. 43) through (2. 45). In the next two sections, these will be inverted in two different ways. The first method will show the solution as traveling waves while the second will bring out the vibrational form. 4. Inversion of the Time Transform-Traveling Wave Form For the region away from the b2se but behind the body wave- fronts the transformed displacements and stresses are given by (2. 43) through (2. 45). Of these, only ~(x, y, p) and the dominant stress axx (x, y, p) will be inverted here. ~(x, y, p) can be inverted in the same way that ~(x, y, p) will be while the other two stresses vanish for 84 long-time. The traveling wave form of the solution is found by using the bi nomial series expansion 00 = ) (-l)m 6 m (2. 47) ~ ·=o to expand the denominator of the transformed solutia.n until a form i s obtained that can be inverted directly. see that ~(x, y, p) has has pD(p). Observing (2. 44) and (2. 45) we 2 p D(p) in its denominator while So it is necessary to expand D~p) axx (x, y, p) which, from the first of (2. 15), is 1 1 D(p) = --2.,-------2-=---cosh y£+cos y£ Now by Lerch's Theorem, Then p can be required to be real and positive. y £ will also be real and positive; h ence 1 (2. 4 7) to expand D(p) with t::. = cos 2 cos y £ < 1. 2 cosh y £ Using 2 y£ 2 cosh y £ yields (2. 48) Using the binomial series expansion, (2. 47), two more times gives 85 1 2 cosh y P. This series can be reduced to (2 . 49 ) Substituting this into (2. 48) yields 1 D(p) (2 . 50) . . -2k(m+l)v£ . The terms rn (2. 50) decay as e so for any given value of y £ only a finite number of terms will be needed to calculate the solution to a given level of accuracy. Note that as y P. gets smaller and smaller, corresponding to longer and longer time, more terms will be needed. Evaluating the first few terms in (2. 50) gives 2 1 = 4{e- Y 1-(4+2cos 2yf)e- 4 Y 1+(17+ 16cos 2 y £+2 cos 4 v f)e-b y £ D(p) (2 . 51) -(80+96 cos 2vl+24 cos4yl+2 cos 6yl)e-Svl }+O(e-!O y l) Using (2. 51) m (2. 44) shows that ~(x, y, p) has the form 86 d::, -;(x, y, p) = v O ~ T. (x, y, p) . LO J J = where 1 [ ] - y (2i-x) T 1 (X, y , p) = 2 2 CO s y X + CO s y ( 2.£- X) - sin y ( 2.£- X) e p (2. 52) T (x,y,p) = - ~ [2cos v x+4sin y x-2sin y (2.£-x)+cosy(21'+x) 2 p + sin y (2£+x)] e- Y ( 2 f+x) T (x,y,p) = - ~ [8cosyx+2sinyx+6cos y (2.£-x)-4siny(2i-x) 3 p + 2 cos y (21'+x) +cos y (4£-x) - sin y (4£-x) ] e -T (x,y,p) = o(e- y (41'+x)) 4 etc. Inspection of (2. 52) shows that T (x, y, p) O second of (1 . 89), i.e., T O T O - y (41'-x) is exactly the same as the is just -; for the semi-infinite plate. So will be a wave traveling in the positive x-direction. T (x, y, p) 1 is proportional to e - y (2£-x) . So T 1 represents a 87 wave traveling in the negative x-direction. of T T 1 O from the boundary at x == £. from the boundary x == 0 x-direction. The other This wave is the reflection Similarly T 2 is the reflection of and is a wave traveling in the positive T. 1 s represent further reflections from the J ends of the plate. At the base of the plate we have that T {0,y,p):: 0 and 0 T (0,y,p):: -T (0,y,p), 1 2 T (0,y,p):: -T (0,y,p), etc. 3 4 So the dis- placement changes sign when it reflects from the built-in end which is, of course, required by the boundary condition. The transformed normal stress, ten as a series of traveling waves. crxx (x, y, p), can also be writ- Using (2. 51) in the first of (2.45) yields (X) crxx (x,y,p) :: Vol U.(x,y,p) (2. 53) J• :: 0 J where == Zuy (1-v)c r p p g (l 2 µ{ [cosyx-sinyx]e-yx -V Cr p [-2sinvx+cosy(2£-x) p g + sin y (2£-x) ] e-y (2£-x) = ( 2 ) Y [ -4 cosy x + 2 sin y x - 2 cos v (2£-x) 1 -v cprgp . ] -y (2£+x) - cos v (2£+x) + srn y (2£+x) e (cont.) 88 -U3 (x, Y, p) = o(e-y (4f-x)) etc. Note that U 0 (x, y, p) is, once again, just the normal stress for the Now we have that at the end x = P., semi-infinite plate, (1. 90). normal stress at the end as required by the boundary condition. Equations (2. 52) and (2. 53) can be inverted by using the tables in Abramowitz and Stegun, [ 11] . This gives (X) v(x,y,t) =Vo) T.(x,y,t) . - ·o J J= (2. 54) (X) CT xx {x,y,t) = Vo j U.(x,y,t) ·- o J J= where the outgoing wave is given by (2.55) 6 2 = __x_ _ 0 4C r t p g Here C (6 ) and S (6 ) are the Frenel integrals which were defined 2 0 2 0 in (1.94). Inverting T (x,y,p) 1 flected wave is given by and U (x,y,p) 1 shows that the first re- 89 = - 2t[C2(L~l)+ (2£-x) J2rrc r t p g 2 2 + 2 R e{ 4 c~r g [(2l- x) - x - 2i x (21-xJerfc z 1 2t - z Ji U (x,y,t) = 1 I 2uy (2. 56) (1- v ) c r p g where 6 = (2£-x) 2 1 4c r t p g = (2£-x)-i X Z 1 v 8c r t p g and erfc 1s the complementary error function. T/x,y,p) and and D/x,y , p) yields terms that are similar to U (x, y, t). 1 v{x, 0, t) where shown. Inverting the other is shown in Fig. 7 for the case f= 20 T (x,y,t) 1 and t = 700 th The outgoing wave and the first four reflections are As can be seen, the fourth reflection is quite small so the fifth - , > u'01.c0 > ......... >< 0 !'- 0 0 .-..... I I I 0 -200 -100 0 100 200 300 t t. - Fig. 7 CD ® 4 • x/h 12 l.TOTAL ! 16 = 20. /~WAVE OUTGOING Displacement v(x, 0 , 700) for the Finite Plate with 8 ~ NO. OF REFLECTION 20 0 '° 91 and succeeding reflections will not contribute very much to the solution. The total displacement of the plate, which was obtained by adding up the outgoing wave and the four reflections is also shown. CJ xx h (x, - , t), 2 is graphed in Fig. 8. The stress, The outgoing wave, the first three reflections and the total stress are shown. 5. Inversion of the Time Transform-Vibrational Form The Laplace transformed displacements and stresses, given by (2. 43) through (2. 45) for the region h < x < P. , can also be inverted by means of a contour integration and residue theory. As shown by Miklowitz in [ 9], for example, the long-time behavior of the solution is determined by the singularities of the transformed solution closest to the Bromwich contour. Observing (2. 43) through (2. 4 5), we see that the transformed solution has a branch point at p = 0 and poles wherever D(p) = cosh 2 y P. + cos 2 y P. (2. 57) =0 Using l 'Hopital' s rule shows that p = 0 is not a pole of (2. 43) through (2. 45). A branch cut is made along the negative real axis and the branch 1s chosen so that jp will be real and positive when p is. It can be shown that (2. 57) does not have any roots rn the half plane Re p > 0. Singularities in the half plane with time and can be neglected for long-time. Re p < 0 will decay Integrating over a small circle around the branch point at p = 0 shows that it also does not contribute to the solution. The contributions of the integrals along the (.) 0 > -ol ::L b )( )( >< ..__. ..c "... C\I . 0 0 I'- - - 2 0 -,b \ 01 2 3 Fig. 8 TOTAL ~ @ 8 I x/h I2 CD ~ Stress axx(x,l 700) for the Finite Plate with 4 I WAVE ! = 20 . I6 CD I®,·.. NO. OF REFLECT I ON 20 I '°N 93 branch cut will also decay with time. So it is only necessary to con- sider the roots of (2 . 57) on the imaginary axis. Setting p = iw and substituting into (2 . 57) gives D(iw) = co sh P. j c Wr cos P. p g J w c r p g + 1 =O (2. 58) Equation (2. 58) has an infinite number of roots corresponding to the natural frequencies of vibration of the plate . (2 . 58), w is a root of Note that if -w will be also. The first ten roots of (2. 58) were calculated numerically . TTC frequency was nondimensionalized by dividing by ws = h 2 ing values for O =J&... ws s The The result - are given in Table 2. The solution given by (2. 43) through (2. 45) is only valid for small. p Comparing the exact Rayleigh-Lamb frequency spectrum with the small p approximations, (1. 21) and (1. 22), that we are using, shows that the latter are valid for at least O ~ O ~ 0. 10. Selecting O = 0 . 10 as the highest admissible frequency, we see from Table 2 that for a given ! ratio only a limited number of roots can be used. ber of allowable roots increases as ~ Note that the num - gets larger. The contribution of a pole at p = i w to v(x, y, t) and a xx (x, y, t) is given by v(x,y,p) limit (p- i w) p-+ i w e axx (x,y,p) pt (2.59) 94 TABLE 2 Natural Frequencies of Vibration for the Plate w TT C o =ws ws = 2hs =5 11 = 10 P. h =20 01 0.084 0.021 0.005 02 0.526 0. 131 0. 032 03 1. 474 0.368 0.092 04 2.889 0.722 o. 180 05 4.777 1. 194 0.298 06 7. 136 1. 784 0.446 0. 7 9.966 2.491 0.622 08 13.269 3.317 0.829 °' 9 17.043 4.260 1. 065 O10 21. 290 5. 3 22 1. 330 P. h P. 95 where v(x, y, p) tively. 0xx (x, y, p) and Substituting p = i1v are given by (2. 44) and (2 . 45) respec- into (2. 16) gives E(i(J)) = (l;i) rinh 2r cos 2r + cosh 2r srn 2r] F(iw) = i sinh 2r sin 2r oD(p) op (2.60) i r = -..- [ sinh 2r cos 2r - cosh 2r sin 2r] w p=iw where p_~ r Using =z.J~ p g (2.59) and (2.60) to calculate the residue yields V a I . 4w [ P p=1 cu v.,.. 0 aD R (1 w) = - 2 A -1 . . ,.. 2rx 2rx ll F(1 w) (cosh-p_- - cos -p_-) J - ( 1 -1") (2.61) E("1 w ") ( sin • •h 2rx • 2rx)} - - sin - e iwt p_ p_ w shows that the part of the displacement associated with the frequency w can be written Combining this with the residue from the pole at p = -i as A vw(x, y, t) = V (x, y, w) sin wt He re V(x, y, w) is the mode shape which is given by 96 where (2.62) d(r) -- sinh 2r cos 2r - cosh 2r sin 2r e(r) = sinh 2r cos 2r + cosh 2r sin 2r f(r) = sinh 2r sin 2r Similarly a w (x,y,t)=L(x,y,w)sinwt (2.63) xx where 4vd1h L(x, y, ,~) = - 2 {. . 2r x 2r x) f(r) \cosh-P.- +cos-£- (1-v) P. wrd(r) [ • •h2rx • 2rx)] -e ( r )(sin -£- + sin-£- The lowest mode shape for the displacement, shown in Fig. 9 for the the case Fig. 10. ~ = 5 while I:(x, ' V(x, 0, w ), 1 is h 2 , w1 ) is shown in These mode shapes agree with those found by Den Hartog in [ 13] using Euler- Bernoulli approximate theory. The near-field asymptotic solution, valid as x-+ 0, is obtained exactly as it was in Chapter I with the transformed displacements being given by ( 1. 5 5), i. e . , _J 0 w 3: (/) I- 2 0 0 w (/) I <( w a.. 0 \... - > X - 0 --. I 0 -20 -15 -10 -5 Fig. 9 x/h Lowest Mode Shape for 2 f. v(x, 0, t) for h = 5. 3 4 "' 5 0f-===:---+------t-------t-------t------'---t- '°--.J _j 0 5 w (f) f- ~ 0 0 w U) :r: <! Q_ w 4- 0 ..... b X >< >< .__., ~ .!: '- (\J ---. -3 0 -1L 0 3 5 I Fig. 10 x/h 3 h Lowest Mode Shape for axx(x,"z, t) for 2 £ h = 5. 4 5 00 '° 99 +n!2 an (p) ~k: ;/) x:0 n sin 0nY + O(x3) n even v(x, y, p) O'.) -I 2 bn(p)(k -l) e ncos n =2 n even The coefficients a (p), an(p), b (p) 0 and 0 bn(p) are now given by (2. 32). Calculating the strains at the base of the plate shows that they are the base strains for the semi-infinite plate multiplied by a reflection function, i.e., - uX (0 , y, p ) = US! ( X y ) F(p) p D( p) (2. 6 4) - v X (O,y,p) := v SI X (y) F(p) () p Dp 100 where u 31 X (y) and v 51 X (y) represent the y dependence of the strains for the semi-infinite plate and are shown in Figs. 2 and 3 respectively. The transformed strains, (2. 64), can be inverted by using the s a.me methods as in the last two sections. USI (y) X 1 2TTi = X (0, y, t) vs' (y) X L Formally, we have that cosh2£ J2cp r - cos2£ J2c pr p g p g p eptdp p cosh2£ J2cp r +cos2.eJ2; r p g p g (2. 65) As (2. 6 5) shows, the strains at the base, including the singular term, will be time dependent for the finite plate instead of constant as was the case for the semi-infinite plate. 101 REFERENCES [ 1] J. Miklowitz, "Analysis of Elastic Wavegui des Involving an Edge 11 , in Wave Propagation in Solids , ASME Publications , New York, 1969. [ 2] J. Miklowitz, 11 On Solving Elastic Waveguide Problems Involving Non-Mixed Edge Conditions", ONR Technical Report No . 5 , Contract Nonr-220 (57), NR-064-487, California Institute of Technology, Pasadena , California, August, 1967 . [ 3] J. Miklowitz, "Pulse Propagation in an Elastic Plate with a Built-in Edge", ONR Technical Report No . 7 , Contract Nonr- 220 (57), NR-064-487, California Institute of Technology, Pasadena , California, May, 1968. [4] G. B. Sinclair and J. Miklowitz, -"Two Nonmixed Symmetric End Loadings of an Elastic Waveguide", International Journal of Solids and Structures,v. 11, 1975, pp. 275-294 . [ 5] G. B. Sinclair and J. Miklowitz, "On the Semi-infinite Elastic Waveguide with Nonmixed Antisymmetric End- Loads 11 , Proceedings, Fourth Canadian Congress of Applied Mechanics, CANCAM 73, c/o Division de Me'canique Applique', Ecole Polytechnique de Montre'al, C. P . 501, Snowdon, Montre'al, 1973. [ 6] M. Knein, "Der Spannungszustand bei Ebener Formanderung und Vollkommen Verhinderter Querdehnung", Abh. Aerodyn . Inst. Techn. Hochschule Aachen 7, 1927, pp . 43-62. [7] M. L. Williams, "Stress Singularities Resulting From Various Boundary Conditions in Angular Corners of Plates in Extension" , Journal of Applied Mechanics , 19 , 1952, 526-528 . [ 8] Ya S. Uflyand, "Survey of Articles on the Applications of Integral Transforms in the Theory of Elasticity", translated by W.J.A. Whyte, edited by I.N. Sneddon, North Carolina State University at Raleigh, Department of Mathematics, Applied Mathematics Research Group, Technical Report No . AF OSR-65-1556 , October 1, 1965 , pp. 126-156. [ 9] J . Miklowitz, The Theory of Elastic Waves a:rid Waveguides, North Holland Publishing Co., Amsterdam, 1977. [ 10] A . P. Hillman and H. E. Salzer, 11 Roots of sin z = z 1 ' , Philosophical Magazine,(?), v. 34, 1943, p. 575 . 102 [ 11] M. Abramowitz and I. A. Ste gun , Editors, Handbook of Mathematical Functions , National Bureau of Standards, Applied Mathematics Series, No. 55, 1964. [ 12) D. V. Widder, The Laplace Transform, Princeton University Press, Princeton, 1941, p. 57. [ 13] J.P. Den Hartog, Mechanical Vibrations, McGraw-Hill Book Company, Inc., New York, 1940.