Evaluating the analytic solution to a three-dimensional boundary value problem in elasticity using...

Evaluating the analytic solution to a three-dimensional boundary value problem in elasticity using Shanks' transformation

Alexandria Oakes

Department of Physics and Astronomy, Eastern Michigan University, Yœsilanti, Michigan 48197

(Received 1 February 1993; revised 26 May 1993; accepted 16 November 1993)

This paper is the second of a three-part series in which solutions are given to two-dimensional and three-dimensional boundary value problems in elasticity. Using the convergence technique developed in the first part to evaluate the stress in the near zone of a semi-infinite slab subject to static loading, this same technique is now applied in two ways to evaluate the dynamic response of a three-dimensional solid cylinder subject to time-dependent end loading.

PACS numbers: 43.40.Hb, 43.20.Gp

INTRODUCTION I. THEORY

In the first article of this series, 1 various schemes were tried for increasing the rates of convergence of the series solutions on or near the end of the material, called the near zone. The tests were carried out on the solutions to a two-

dimensional, semi-infinite slab problem with static loading, with the solutions evaluated at the end of the slab and

compared to the specified end boundary conditions. It was concluded that the procedure identified in the first article as the Shanks' transformation of order two (Shanks' II) is very successful in accurately summing the series solutions.

In this article, Shanks' transformation II is used in two ways to evaluate the solution to a three-dimensional elas- tadynamic problem. First, the transformation is used to sum the series of complex modes that appears in the solution. In addition, the inverse Laplace transform integral appearing in the solution is evaluated by approximating the integral by a Fourier series, which is summed by use of Shanks' transformation II. A complicating feature of this calculation is the need to solve the Pachhammer-Chree 2'3 equation for the complex wave numbers •, as a function of complex frequencies. The problem considered here is the one solved by Kennedy 4 for a semi-infinite cylinder by the finite difference method. The analytic solution obtained is numerically evaluated by the scheme just described, and the results are compared to those reported by Kennedy. As a further application of this numerical procedure, the strain is evaluated at the point on a cylindrical bar at which G. Fax 5 measured the strain as a function of time when the bar was subjected to a step-function normal stress on its end. The theoretical and experimental results are compared.

The transform method developed by Folk, et aL 6 is used in this chapter to obtain the analytic solution of the problem solved by Kennedy 4 by the finite difference method. Shanks' transformation of order two is combined

with the method given by Crump 7 for numerically inverting Laplace transforms to evaluate the analytic solution in the near zone. These results are compared with those reported by Kennedy. Also, we will evaluate the analytic solution outside the near zone, and compare the answer with the experimental results found by Fax. 5

The problem considered is to find the response of a semi-infinite elastic cylinder for r < a, z > 0, and t > 0 (see Fig. 1 ) when the following boundary and initial conditions are imposed:

Trr: Trz:O at r=a, (1)

Tzz=P(r)X(t) at z=0, (2)

Ur=O at z=0. (3)

The bar is initially undisturbed, and $(t) is the unit step function. We will later limit the analysis to the special case of P(r) = --P0, a constant.

The equations of motion in cylindrical coordinates (r,O,z) for an elastic material with no 0 dependence as given by Lave, 8 are

a2 Ur aA a• p -•t = (X-•'-2•) '•rr -•-- 2• •z-zJi--Fr, (4)

a2 Uz aA 2• a( r• ) p • =(•+2•) •- r • +Fz, (5)

10(rUr) OUz a=-•+•, (6)

r O r O z

l (au, aUz] (7)

Here, Ur and Uz are displacements in the radical and axial directions, respectively, F is a body force that is zero in the present problem, p is the density, 2 and p are the Lame elastic constants, A is the dilatation potential, l• is the rotation potential, t is the time, and T is the stress.

The subsequent evaluation of any of the stresses re- quires the stress-strain equations to be given in terms of the displacements Uz and Ur. Note that because of the axial symmetry in this problem, there is no displacement in the 0 direction, i.e., Uo=O. This means there is no 0 dependence in any of the equations of motion, or in the stress- strain equations. All derivatives with respect to 0 are also zero. The stress-strain equations with no 0 dependence are given by

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T•(z,a,t) --T•(z,a,t) --0

FIG. 1. Semi-infinite bar with radius r=a.

Aa(rUr) auz Tzz=• a•+(A+2//) -•z' (8)

Trr=(A+2//) •+•+ • , (9)

r=(OUr OUr] •+•j. (10) The pu•ose of using the transfo• method is to

change our equations from being pa•ial differential equations in several independent variables to ordina• differential equations. In this problem, we will get fid of the z variable by an appropriate choice of sine or cosine transfo•s, and we will eliminate the t variable by use of the Laplace Transform. The notation for these transfo•s are

f(7) = f(z)sin(•pz)dz(sine),

,•(7) = f(z)cos(7z)dz(cosine),

(11)

(12)

and

f(co) = f(t)e iø•t dt(Laplace).

The inverses of these transforms are given by

f (z) = ---- 7)e irz d7,

f (z) =- 7)e dr,

and

(13)

(14)

(15)

= f(w)e-iø•tdw. (16) f(t) • In the calculations of Kennedy 4 and in the experi-

ments of Fox, 5 they determined the sum of the strains e•=OU•/U• and %0 = U•/r. In order to compare with their results, we will do the same. Let

OU z U r E=eøø+ezz=-•-z + r (17)

The Laplace transform in t and the sine transform of z for this quantity is

E=-U--7U

pw2 1 ß + 72Jo(hr) (21•7B1•

X (Jl(kr) a'or •-kJo(kr) + (•[_.l_2//)oh2, where •41 and B1 are given by

A1(7,o) = iPopo•,7( k2 - 72)J1 (ka)

( •, -+- 2 // ) 2// h2 D ( 7,o ) and

(18)

(19)

B1 (7,o) =

where

iPop•,le2o.)J1 (ha) , (20)

and

D(r,o•) = (2h/a) ( k 2 q- 72)J1 (ha)J 1 ( ka) -- ( k 2- 72) 2 XJo(ha)J1(ka)-472hkJ1(ha)Jo(ka), (21)

h 2 = p02/(•l, + 2//) -- 72 (22) and

k2 = pa)2///-- 72. (23) Also,

Jo = Bessel function of zeroth order, (24)

J1 =Bessel function of first order. (25)

In the next step, we take the inverse sine transform of Eq. (18) by using Eq. (14), to obtain

i oo - e i7z = -- E(r,r,,o) d7. (26)

Since the singularities of E are poles, the value of the integral in Eq. (26) is equal to 2rri times the residue at these poles. Folk 9 showed that all the poles of the integrand are due to zeroes in the function D(r,,o) that is in the denom- inator of Eqs. (19) and (20) for ,4 • and B•. The function D(r,,o) is given by Eq. (21). The value of Eq. (26) is therefore

_ 2/APo E ( r'ø'z ) = ( A q- 2// ) //ø r(k-r

h2 J• (ha)

hJ (hr) ) X r +72Jø(hr)

273J1(ha) (Jl ( kr) _ kJo( kr) ) + h

1 X e irz , ( 27 )

aD/a7 7M, L

where the set 7M, L(W) are all the solutions of the 3 10

Pochhammer-Chree dispersion equation' given by

2023 d. Acoust. Soc. Am., Vol. 95, No. 4, April 1994 Alexandria Oakes: Shanks' transformation in boundary valve problems 2023


TABLE I. Sample of roots of the Pochhammer-Chree frequency equation for a complex frequency

For co = ( -- 1.0308, 0.0719)

Y1,1 = (--1.3451, 0.1121) 72,1 = ( 1.2930, 2.1584) ?'2,2 = (- 1.4755, 2.1535) Y3,1 = ( 1.6208, 5.8415) ]/3,2 = (- 1.6847, 5.8446) Y4,1 = ( 1.8172, 9.1259) y4,2 = (- 1.8578, 9.1280) ?'5,1 = ( 1.9584, 12.3380) ]/5,2 = ( -- 1.9885, 12.3394)

TABLE II. Sample calculation of the contributions to E at z/a----10 for the first 9 modes of vibration for a complex frequency.

For co = -- 1.0308 + tD.0719 Contribution to Contribution to

Y(3t, r) Re E Im E

1,1 --0.8208 (10 -4) --0.1033 (10 -3) 2,1 0.1847 (10 -12) 0.5659 (10 -12) 2,2 0.1069 ( 10 -12) 0.1994 ( 10 -12) 3,1 --0.5045 ( 10 -29) --0.3526 ( 10 -29) 3,2 0.2684 ( 10 -29) --0.5389 ( 10 -29) 4,1 0.1813 (10 -43) --0.1128 (10 -43) 4,2 -0.2009 ( 10 -43) -0.6007 (10 -44) 5,1 0.1162 (10 -57) 0.1337 (10 -57) 5,2 --0.9213 (10 -58) 0.1485 (10 -57)

D(yM, r,(o) =0, (28)

with Im YM, r > 0 or, as shown by Folk 9, those real solutions of Eq. (28) with dy/dw > O.

The labeling of the roots Y3t, z is chosen so that M orders the y's according to their magnitude at (o =0, and L specifies the quadrant of the y plane in which Y3t, r occurs at w=0. This notation makes it easy to identify the solutions Y3t, r that are included, and only the first two quad- rants, L= 1 and L= 2, are needed.

There are two roots that are zero for w=0; we label these separately as y•,• and Yl,2. These roots are real for all real w, with dy•,•/dro > 0. Therefore, y•,• is included in Eq. (27) and Yl,2 is not. Values of YM, r for 1 <M < 0, which includes 18 complex roots were calculated by Goldberg •ø as functions of real w. For the new procedure for inverting the Laplace transform of Eq. (27), which we develop in the next section, we need to know the values of the eigen- values YM, r for a range of complex w. In Table I, we give a sample list of values of Y3t, r for M> 1 for some complex w. All the functions Y3t, r for M> 1 have branch points, and so care is needed in calculating values of each y as ro is varied through or near the location of the branch points.

Goldberg •ø showed that for w=0 the y functions are given approximately by

yM, r(0) =« In [4(M-- 1 )rr] +i(M-- 1 )rr and

YM,2(0) = -- ]/•/,1 (0). Using this equation to determine a first guess, a complex Newton-Raphson method was used to solve the Pochhammer-Chree equation (28) as to was varied in small increments starting at w=0.

II. APPLICATION OF CONVERGENCE TECHNIQUE IN EVALUATING THE SUM OVER MODES AND THE

INVERSE LAPLACE TRANSFORM OF EQ. (27) FOR E(r,z,o•)

The expression for E(r,z,w) in Eq. (27) involves a sum over the complex YM, r. Because the Im YM, r > 0 for large M, the factor exp iyz in each term is small if z and M are large (see Table II). Consequently, the sum converges rapidly for large z. For small z, the convergence of the sum is slow, and as we found in the previous article, • the rate of

convergence is slowest for z=O. In this case, we again use Shanks' transformation of order two to increase the rate of

convergence for small z. In the next step, we evaluate the inverse Laplace trans-

form of E(a,z,t); that is, evaluate

1 foo +ie = •(a,z, to )e -iot dw, E(a,z,t) • where from Eq. (27), we have

E ( a,z, oo )

2Vo E ( 1 + 212)pro ru, L ( (k2+ y2)j•(ha)J•(ka) ha

2y2kJ• ( ha)Jo( ha)

(29)

y2 ( k2-- y2 )jø( ha )J• ( ha ) ) h 2

re'rZ • x aD/ar) . (3o) YM, L

The procedure we used to invert the Laplace transform combines the method proposed by Kenny Crump7 with the Shanks' transformation of order two.

Let us repeat the definitions of the transforms used so far with a change in notation for the sine and cosine transforms,

P(ro) = f(t)eiotdt, (31)

•(s) = g( t)cos(st)dt, (32)

and

•(s) = g( t)sin(st)dt. (33)

Choose

g(t) =f(t)e -at, or f(t) =g( t)e at,

and write,

P(oo) = g(t)e i(ø-ia)t dt.



Let

(34)

or

o=s+ia, (35)

and now write,

P(s=ia)= fo g( t)eiStdt=g(s) + i•(s). (36)

That is,

g(s)=Re[F(s+ia)] (37)

and

•(s)=Im[F(s+ia)]. (38)

as

The inverse cosine and sine transforms can be written

g(t) --- g(s)cos(st)ds, for t>O,

and

2; •(s)sin(st)ds, g( t) =•r for t>0.

We, of course, may write g = (g-3-g)/2, and use

g(t)--- [g(s)cos(st)+•(s)sin(st) ]ds

1 {Re[F(s+ia) ]cos(st)

+ Im[F(s+ ia) ]sin(st) }ds, (39)

where we used Eqs. (37) and (38). If we only needed to represent g(t) in the interval

2n T < t < 2 (n q- 1 ) T, we could use a Fourier series rather than Fourier integrals; that is,

gnt = • A nO -[- Z A nk COS -[- Bnk sin k=l

=g(t), for 2nT<t<2(n+l)T, (40)

if

Ank=• d 2nT g( t)cos dt (41 ) and

Bnk=• a2nT g(t) sin dt. (42) Note, with reference to Eqs. (32), (33), (37), and (38), that

n=0 k=l

and that

, (43)

gn(t)=g(t) for 2nT<t<2(n+ l)T. (44)

Also note that if the trapezoidal rule is used to numerically evaluate Eq. (39) for g(t), Eq. (43) is the result.. With these points in mind and using Eq. (44), we conclude that [recall that g=exp( --at)f (t)]

eat

k=l

+A,

for 0<t<2T, (45)

where the error A (t) is given by

A (t) =e at • g(t) n=l

=ea' • e-a(2nr+t) f(2nr+t) n=l

= • e-2narf(2nT+t). (46)

Since the function E(t) in Eq. (29) that corresponds to f(t) in the above derivation is bounded, we will use

If(t) I <Me et,

in Eq. (46) and obtain

Me et IA(t) l <meet • e -(a-e)2nc-- . (47) n --(e 2t(a-e)-- 1 )

Set A0= ]A (t) l max acceptable error, and then write Me et

=Meete -:T(a-e) (48) AO-- (e2t(a-e)_ 1 ) ' for a big enough. Define the relative error by

R = Ao/Me et , ( 49 )

and take the logarithm of Eq. (48)-

In R =--2T(a--e),

or

a=e--ln R/2T.

For our case, e----O, and therefore the procedure fol- lowed was to accept a relative error of R = 10 -4 and cal- culate

a=e--ln R/2T=4.605/T. (5O)



0.00045

0.00030 m

0.0•15m

29 58 87 116 145

ß me (microseconds)

o

lO 70 80 40 50 60 90 ,

'rime (microseconds)

FIG. 3. Kennedy graph of strain E vs time t at z?a = 10.

FIG. 2. Graph of strain E vs time t at z?a = 10.

In order to use Eq. (45) to evaluate the inverse La- place transform of F(w), we need to choose a value for T as well as a value for a. Equation (50) is used to determine a after a value for T is chosen. In the derivation, we as- sumed that 0 < t < 2T, which means that if we want to evaluate an inverse transform for times in the range 0 < t < tma x , we must choose T > 0.5tma x . After trying several values of T greater than 0.5tma x, Crump 7 decided on the following rule of thumb'

T=0.8tmax. (51)

With T and a chosen, Eq. (45) can be used to evaluate f(t) in that equation for 0 < t < tmax. However, in these applications, the sum in Eq. (45) does not converge rapidly. We solved this problem by use of Shanks' transformation of order two to evaluate the sum. The accuracy of the summing procedure by Shanks' transformation can be determined by comparing the closeness of the five terms remaining in the second to last iteration of the transformation. For any value of time t for which the mean square deviation of these five terms differed by more than 0.0001% of the final answer, this value for œ(t) was dis- carded from the list of values for the function period.

This procedure is accurate to the specified relative error except at values of t where f(t) is changing rapidly. Troubles at these special values of t can be overcome.

Before proceeding, it is worth pointing out the reason we do not use the computer program available through IMSL for inverting Laplace transforms. The IMSL program evaluates F(o) at a different set of over a hundred values of o for each value of t. In this calculation, we want to evaluate the inverse transform for several thousand values

of t. Therefore, if we use the IMSL program, F (o) Would be evaluated at about a half a million values of o. However, evaluating Eq. (30) or F (o) in our problem is even more difficult than it looks. To evaluate this expression for any value of o, we must first solve the Pochhammer-Chree equation (28).

2h -- ( k2 + 72 )J1( ha )J1( ka ) _ ( k2- 72 ) 2Jo( ha )J1( ka ) a

-- 472hkJ1 ( ha )Jo(ka) =0 with

h 2 = po2/(/• + 21a ) - 72 and

k2= pro2/la- T2 ,

for the set of complex roots t'•t,L (co) for each value of co used. In iterating to find the correct values of the t% from the above equation, Bessel functions of complex arguments must be evaluated over and over again.

By combining Shanks' transformation of order two (Ref. 11 ) with the scheme given by Crump, 7 we obtain the same accuracy as the IMSL program, and yet we only evaluate F(o) at about a hundred values of o. This is an enormous saving in computing effort.

III. RESULTS AND CONCLUSIONS

We now return to the problem of inverting the Laplace transform of Eq. (41 ) for the conditions used by Kennedy 4 in her finite difference calculation. That is, set P0 = 1 megabar, z/a=10, Young's modulus Y=2.0 megabars, and Poisson's ratio a= 1/3 and p=7.8 g/cm 3. The Lam6 constants are related to Y and a by

( 1 -2a) ( 1 + 2a)

and

!•= Y/2( l +a).

To make the answers as general as possible, we convert Eq. (30) into dimensionless variables before we take the inverse Laplace transform. That is,

Z=z/a, (52)

T= ( Vz>/a)t, (53)

F=ya, (54)

v=(a/VD)O, (55)

where

2026 J. Acoust. Soc. Am., Vol. 95, No. 4, April 1994 Alexandria Oakes: Shanks' transformation in boundary valve problems 2026


1.25

0.75

0.50

0.25

! i i i i i i

110 220 330 440 550 660

'nme (microseconds)

FIG. 4. Graph of strain E vs time t at z/a= 79.3.

is the dilatation wave velocity. However, in reporting these results for comparison with those of Kennedy, we convert to the units she used. Recall that the end conditions of the

problem are

Tzz= --PoS( t),

and

at z--O

Ur=O , at z=0.

Figure 2 shows the results of the evaluation of the inverse Laplace transform for 0 < t < 100/•s. The results are similar but not identical to the results shown in Fig. 3 that were obtained by Kennedy 4 by the finite difference method. It is difficult to judge the accuracy of Kennedy's finite difference calculation, since the results of a finite difference calculation cannot evaluate oscillations with peri- ods less than the At time increment used in the calculation.

Even so, the rise time of the first pulse and its amplitude match extremely well. The eventual fall off to a steady value of 0.333( 10 -3 ) are also in excellent agreement.

For the next application, we evaluate the inverse La- place transform of Eq. (41 ) for the case that z/a = 79.3, or=0.3, the bar velocity V0 = Yx/-•-p = 500 ms, and VD= 5.8( 103) ms. These are the conditions for which Fox 7 measured the strain E. The results of this calculation and

Fox's measurements are shown, respectively, in Figs. 4 and 5. Although not shown in his graphs, the arrival time of the head pulse was reported to be at about 330.1/•s, while ours arrived at 333.3/•s. The static limit reached by the strain was normalized to unity in Fox's calculation. This is in excellent agreement with our value of unity.

Since results shown in Fig. 4 are similar but not identical to those of Fox shown in Fig. 5, one possible expla- nation of this could be due to the sizes of the gauges that were used in Fox's experiments, making rapid oscillations

STRAIN

, A

I!il• I I/I v

20 •

s

TIME -

FIG. 5. Fox's graph of strain E vs time t at z/a= 79.3.

difficult to see. For example, for a gauge length of 1 cm, the time for a signal to travel the length of the gauge is approximately At=O.Ol(m/Vo)=2/•s. Fox reported rapid oscillations that are not seen in Fig. 5 when he tried a gauge of length 3.7 mm. Another concern is that the experimental end condition was somewhat different from the theoretical one. Although the experimental normal stress on the end rose rapidly (less than 1/•s) to the constant P0, this is only approximately a step function. More impor- tantly, the other experimental end condition was Try=0 at z=0, while the condition used in the calculations was Ur=O at z=0. Folk 9 had shown that the response far from the end (and z/a = 79.3 is far from the end) is the same for these two conditions.

The accuracy of this method of inverting the Laplace transform for times less than z/Vo and for large time were checked. For short times, the calculations show, as expected, that the strain E is zero. For large times, the calculations yield the expected static value of 0.333 ( 10- 3). From these, it is concluded that the Shanks' transformation of order two has also been successful in evaluating the analytic solution to a three-dimensional, semi-infinite cylinder subject to time-dependent end loadings. It appears to be a viable way of handling three-dimensional boundary value problems in elasticity.

ACKNOWLEDGMENTS

The author wishes to express her appreciation to Dr. Robert T. Folk at Lehigh University for his encourage- ment, guidance, and assistance during all phases of this study. The author also wishes to thank Mrs. Mary Jane Callison for typing this manuscript; along with a thank you to Mr. Dennis Oakes for editing the subsequent revisions of this article. The author also wishes to thank Mrs. Rose-

marie Williams for typing and editing the original version of this paper.

•A. Oakes, "Testing convergence schemes on the solution of two- dimensional boundary value problems in elasticity," J. Acoust. Soc. Am. 87, 894-900 (1990).

2L. Pochhammer, J. Math. (Crelle) 81, 324 (1876). C. Chree, Quant. J. Math. 21, 287 (1886). L. W. Kennedy, "Elastic strain pulse produced by a sudden application of a radially distributed stress to the end of a cylindrical bar," Ph.D. dissertation, Lehigh University, Bethlehem, PA (1968). G. Fox and C. W. Curtis, "Elastic strain produced by sudden application of pressure to one end of a cylindrical bar--II. Experimental ob- servations," J. Acoust. Soc. Am. 30, 559 (1958).



6R. Folk, G. Fox, C. A. Shook, and C. W. Curtis, "Elastic strain produced by sudden application of pressure to one end of a cylindrical bar--I. Theory, J. Acoust. Soc. Am. 30, 552 (1958).

?K. Crump, "Numerical inversion of Laplace transform using Fourier series approximation," J. Assoc. Comput. Mach. 23 ( 1 ), 89-96 (1976). A. H. Love, The Mathematical Theory of Elasticity (Dover, New York, 1944), 4th ed.

9R. T. Folk, "Time dependent boundary value problems in elasticity," Ph.D. dissertation, Lehigh University, Bethlehem, PA (1958).

1øI. S. Goldberg, "Solutions to dynamic pure end condition problems of elasticity," Ph.D. dissertation, Lehigh University, Bethlehem, PA (1970).

11D. Shanks, "Non-linear transformations of divergent and slowly con- vergent sequences," J. Math. Phys. 34, 1 (1955).



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