Elasticity Part 13 potx

of revolution under axisymmetric loading.For this case the displacements and stresses in Cartesian coordinates become EXAMPLE 13-1: Kelvin’s Problem: Concentrated Force Acting in the Int

In Cartesian coordinates, these expressions would give

u¼ 12m

In the previous sections, the displacement vector was represented in terms of first derivatives

of the potential functions f and w Galerkin (1930) showed that it is also useful to represent the

displacement in terms of second derivatives of asingle vector function The proposed sentation is given by

repre-2mu¼ 2(1  n)r2V =(=  V) (13:3:1)where the potential function V is called the Galerkin vector Substituting this form intoNavier’s equation gives the result

=
w ¼2(1 n)

2m r2V

(13:3:3)

Notice that if V is taken to be harmonic, then the curl of w will vanish and the scalar potential

f will also be harmonic This case then reduces to Lame´’s strain potential presented in theprevious section With zero body forces, the stresses corresponding to the Galerkin representa-tion are given by

@z@x= V

(13:3:4)

As previously mentioned, for no body forces the Galerkin vector must be biharmonic InCartesian coordinates, the general biharmonic vector equation would decouple, and thus eachcomponent of the Galerkin vector would satisfy the scalar biharmonic equation However, incurvilinear coordinate systems (such as cylindrical or spherical), the unit vectors are functions

of the coordinates, and this will not in general allow such a simple decoupling Equation(1.9.18) provides the general form for the Laplacian of a vector, and the expression for polarcoordinates is given in Example 1-3 by relation (1:9:21)7 Therefore, in curvilinear coordinatesthe individual components of the Galerkin vector do not necessarily satisfy the biharmonicequation For cylindrical coordinates, only thez component of the Galerkin vector satisfies thebiharmonic equation, while the other components satisfy a more complicated fourth-orderpartial differential equation; see Exercise 13-8 for details

Before moving on to specific applications, we investigate a few useful relationships dealingwith harmonic and biharmonic functions Consider the following identity:

r2(xf )¼ xr2fþ 2@f

@xTaking the Laplacian of this expression gives

r4(xf )¼ r2xr2f

þ 2 @

@x(r2f )and thus iff is harmonic, the product xf is biharmonic Obviously, for this result the coordinate

x could be replaced by y or z Likewise we can also show by standard differentiation that theproductR2f will be biharmonic if f is harmonic, where R2¼ x2þ y2þ z2 Using these results,

we can write the following generalized representation for a biharmonic functiong as

of revolution under axisymmetric loading.

For this case the displacements and stresses in Cartesian coordinates become

EXAMPLE 13-1: Kelvin’s Problem: Concentrated Force Acting

in the Interior of an Infinite Solid

Consider the problem (commonly referred to asKelvin’s problem) of a single trated force acting at a point in the interior of an elastic solid For convenience wechoose a coordinate system such that the force is applied at the origin and acts in thezdirection (see Figure 13-1) The general boundary conditions on this problem wouldrequire that the stress field vanish at infinity, be singular at the origin, and give theresultant force systemPezon any surface enclosing the origin

concen-The symmetry of the problem suggests that we can choose the Love/Galerkinpotential as an axisymmetric formVz(r, z) In the absence of body forces, this function

is biharmonic, and using the last term in representation (13.3.5) withf4¼ 1=R gives thetrial potential

Continued

EXAMPLE 13-1: Kelvin’s Problem: Concentrated Force Acting

in the Interior of an Infinite Solid–Cont’d

Vz¼ AR ¼ A ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

r2þ z2

p

(13:3:9)whereA is an arbitrary constant to be determined We shall now show that this potentialproduces the correct stress field for the concentrated force problem under study

The displacement and stress fields corresponding to the proposed potential followfrom relations (13.3.8)

ð1 02prsz(r, a)dr

ð1 02prsz(r, a)dr þ

ða

a2prtrz(r, z)dzþ P ¼ 0 (13:3:11)

The first two terms in (13.3.11) can be combined, and in the limit asr! 1 the thirdintegral is found to vanish, thus giving

FIGURE 13-1 Kelvin’s problem: concentrated force in an infinite medium.

EXAMPLE 13-1: Kelvin’s Problem: Concentrated Force Acting

in the Interior of an Infinite Solid–Cont’d

P¼ 2

ð1 a2pRsz(r, a)dR

¼ 4pA (1  2n)a

ð1 a

RdR

R3 þ 3a3

ð1 a

EXAMPLE 13-2: Boussinesq’s Problem: Concentrated Force

Acting Normal to the Free Surface of a Semi-Infinite Solid

Several other related concentrated force problems can be solved by this method Forexample, considerBoussinesq’s problem of a concentrated force acting normal to thefree surface of a semi-infinite solid, as shown in Figure 13-2 Recall that the corres-ponding two-dimensional problem was solved in Section 8.4.7 (Flamant’s problem) andlater using complex variables in Example 10-5

This problem can be solved by combining a Galerkin vector and Lame´’s strainpotential of the forms

EXAMPLE 13-3: Cerruti’s Problem: Concentrated Force Acting Parallel to the Free Surface of a Semi-Infinite Solid

Another related example isCerruti’s problem of a concentrated force acting parallel tothe free surface of an elastic half space (see Figure 13-3) For convenience, the force ischosen to be directed along thex-axis as shown Although this problem is not axisym-metric, it can be solved by combining a particular Galerkin vector and Lame´’s strainpotential of the following forms:

Using the Helmholtz representation (13.1.1) and relation (13.1.3), this previous equation can

be written as

r2[uþ 1(1 2n)=f]¼ 

equa-be shown that for arbitrary three-dimensionalconvex regions, only three of these functions areindependent Note that a convex region is one in which any two points in the domain may beconnected by a line that remains totally within the region

Comparing the Galerkin vector representation (13.3.1) with the Papkovich solution(13.4.8), it is expected that a relationship between the two solution types should exist, and itcan be easily shown that

EXAMPLE 13-4: Boussinesq’s Problem Revisited

We consider again the problem shown previously in Figure 13-2 of a concentratedforce acting normal to the stress-free surface of a semi-infinite solid Because theproblem is axisymmetric, we use the Boussinesq potentials defined by (13.4.11).These potentials must be harmonic functions ofr and z, and using (13.2.5), we try theforms

Az¼C1

R, B¼ C2log (Rþ z) (13:4:12)whereC1andC2are constants to be determined

The boundary conditions on the free surface require that sz¼ trz¼ 0 everywhereexcept at the origin, and that the summation of the total vertical force be equal toP.Calculation of these stresses follows using the displacements from (13.4.8) in Hooke’slaw, and the result is

Note that the expression for szvanishes onz¼ 0, but is indeterminate at the origin, andthus this relation will not directly provide a means to determine the constantC1 Ratherthan trying to evaluate this singularity at the origin, we pursue the integrated condition

on any typical planez¼ constant

P¼ 

ð1 0

EXAMPLE 13-4: Boussinesq’s Problem Revisited–Cont’d

Invoking these boundary conditions determines the two constants

22pR5

(13:4:17)

Many additional problems can be solved using the Papkovich method, and some of theseare given in the exercises This technique also is used in the next chapter to generate solutionsfor many singular stress states employed in micromechanics modeling

An interesting connection can be made for the two-dimensional case between the vich-Neuber scheme and the complex variable method discussed in Chapter 10 For the case ofplane deformation in thex, y-plane, we choose

Papko-Ax¼ Ax(x, y), Ay¼ Ay(x, y), Az¼ 0, B ¼ B(x, y) (13:4:18)Using the general representation (13.4.8), it can be shown that for the plane strain case

2m(uþ iv) ¼ (3  4n)g(z)  zg0(z) c(z) (13:4:19)with appropriate selection of g(z) and c(z) in terms of Ax, Ay, andB It is noted that this form is identical to(10.2.9) found using the complex variable formulation

A convenient summary flow chart of the various displacement functions discussed in thischapter is shown in Figure 13-4 The governing equations in terms of the particular potentialfunctions are for the zero body force case Chou and Pagano (1967) provide additional tablesfor displacement potentials and stress functions.

The previous solution examples employing displacement potentials simply used preselectedforms of harmonic and biharmonic potentials We now investigate a more general scheme todetermine appropriate potentials for axisymmetric problems described inspherical coordin-ates Referring to Figures 1-4 and 1-5, cylindrical coordinates (r, y, z) are related to sphericalcoordinates (R, f, y) through relations

Restricting attention to axisymmetric problems, all quantities are independent of y, and thus

we choose the axisymmetric Galerkin vector representation Recall that this lead to Love’sstrain potential Vz, and the displacements and stresses were given by relations (13.3.6) to(13.3.8) Because this potential function was biharmonic, consider first solutions to Laplace’sequation In spherical coordinates the Laplacian operator becomes

r2¼ @2

@R2þ2R

1sin f

d

df sin f

dFndf

These terms are commonly referred to asspherical harmonics

Our goal, however, is to determine the elasticity solution that requires biharmonic functionsfor the Love/Galerkin potential In order to construct a set of biharmonic functions, we employ thelast term in relation (13.3.5) and thus argue that ifRnFnare harmonic,Rnþ2Fnwill be biharmonic.Thus, a representation for the Love strain potential may be written as the linear combination

in governing equation (13.5.3) will be the same if we were to replacen by ( n  1) This thenimplies that solution forms Rn1Fn1¼ Rn1Fn will also be harmonic functions.Following our previous construction scheme, another set of biharmonic functions for thepotential function can be expressed as

Vz¼B0( 2þ z2)1=2þ B1z(r2þ z2)1=2þ   

þ A0( 2þ z2)1=2þ A1z(r2þ z2)3=2þ    (13:5:8)and this form will be useful for infinite domain problems For example, the solution to theKelvin problem in Example 13-1 can be found by choosing only the first term in relation(13.5.8) This scheme can also be employed to construct a set of harmonic functions for thePapkovich potentials; see Little (1973)

EXAMPLE 13-5: Spherical Cavity in an Infinite Medium

Subjected to Uniform Far-Field Tension

Consider the problem of a stress-free spherical cavity in an infinite elastic solid that issubjected to a uniform tensile stress at infinity The problem is shown in Figure 13-5,and for convenience we have oriented thez-axes along the direction of the uniform far-field stressS

We first investigate the nature of the stress distribution on the spherical cavity causedsolely by the far-field stress For the axisymmetric problem, the spherical stresses arerelated to the cylindrical components (see Appendix B) by the equations

sR¼ srsin2fþ szcos2fþ 2trzsin f cos f

sf¼ szsin2fþ srcos2f 2trzsin f cos f

tRf¼ (sr sz) sin f cos f trz( sin2f cos2f)

It is found that the superposition of the following three fields satisfies the problemrequirements:

1 Force doublet in z direction: This state corresponds to a pair of equal and oppositeforces in thez direction acting at the origin The solution is formally determined fromthe combination of two equal but opposite Kelvin solutions from Example 13-1 Thetwo forces are separated by a distance d, and the limit is taken as d! 0 This

x

y z

EXAMPLE 13-5: Spherical Cavity in an Infinite Medium

Subjected to Uniform Far-Field Tension–Cont’d

summation and limiting process yields a state that is actually the derivative (@=@z) ofthe original Kelvin field with a new coefficient ofAd (see Exercise 13-18) Thiscoefficient is denoted asK1

2 Center of dilatation: This field is the result of three mutually orthogonal force pairs from the previous state (1) (see Exercise 13-19) The coefficient of thisstate is denoted byK2

double-3 Particular biharmonic term: A state corresponding to the A1 term from equation(13.5.8)

Combining these three terms with the uniform far-field stress and using the condition ofzero stress on the spherical cavity provide sufficient equations to determine the threeunknown constants Details of this process can be found in Timoshenko and Goodier(1970), and the results determine the coefficients of the three superimposed fields

(13:5:11)

Using these constants, the stress and displacement fields can be determined

The normal stress on thex,y-plane (z¼ 0) is given by

sz(r, 0)¼ S 1 þ 4 5n

2(7 5n)

a3

r3þ 92(7 5n)

Continued

EXAMPLE 13-5: Spherical Cavity in an Infinite Medium

Subjected to Uniform Far-Field Tension–Cont’d

two-dimensional predictions This is to be expected because the three-dimensional fieldhas an additional dimension to decrease the concentration caused by the cavity Bothstress concentrations rapidly decay away from the hole and essentially vanish atr > 5a.Additional information on this problem is given by Timoshenko and Goodier (1970)

1.9 1.95

2.05

2

2.1 2.15 2.2

Poisson's Ratio

FIGURE 13-6 Stress concentration factor behavior for the spherical cavity problem.

0 0.5 1 1.5 2 2.5 3 3.5

Dimensionless Distance,r/a

Two Dimensional Case: s q( ,p/2)/S

Three Dimensional Case:s z( ,0)/S, ν = 0.3

FIGURE 13-7 Comparison of two- and three-dimensional stress concentrations around cavities.

whereFijis some differential operator and F is a tensor-valued variable Normally, the searchlooks for forms that automatically satisfy the equilibrium equations (13.6.1), and these arecalledself-equilibrated forms.

It is apparent that the equilibrium equations will be satisfied if sijis expressed as thecurl ofsome vector function, because the divergence of a curl vanishes identically It can be shown thatone such equilibrated form that provides acomplete solution to the elasticity problem is given by

sij¼ "imp"jklFmk,pl (13:6:4)where F is a symmetric second-order tensor Relation (13.6.4) is sometimes referred to as theBeltrami representation, and F is called the Beltrami stress function It was shown by Carlson(1966) that all elasticity solutions admit this representation It is easily demonstrated that(13.6.4) is an equilibrated form, since

sij,j¼ ("imp"jklFmk,pl),j¼ "imp"jklFmk,plj¼ 0because of the product of symmetric and antisymmetric forms in indicesjl

Property (1.3.5) allows expansion of the alternating symbol product, and thus relation(13.6.4) can be expressed as

sij¼ dijFkk,ll dijFkl,kl Fij,kkþ Fli,ljþ Flj,li Fkk,ij (13:6:5)or

The first invariant of the stress tensor then becomes

13.6.1 Maxwell Stress Function Representation

The Maxwell stress function representation considers the reduced form whereby all diagonal elements of Fijare set to zero; that is,

Notice that theAiry stress function that is used for two-dimensional problems is a special case

of this scheme with F11¼ F22¼ 0 and F33¼ f(x1,x2)

13.6.2 Morera Stress Function Representation

The Morera stress function method uses the general form with diagonal terms set to zero;that is,

3

This approach yields the representation