Elasticity Part 13 potx

In Cartesian coordinates, these expressions would give u ¼ 1 2m @f @x , v ¼ 1 2m @f @y , w ¼ 1 2m @f @z e x ¼ 1 2m @ 2 f @x 2 , e y ¼ 1 2m @ 2 f @y 2 , ÁÁÁ s x ¼ @ 2 f @x 2 , s y ¼ @ 2 f @y 2 , t xy ¼ @ 2 f @x@y (13:2:4) Thus, for this case any harmonic function can be used for Lame ´ ’s potential. Typical forms of harmonic functions are easily determined, and some examples include x 2 À y 2 , xy, r n cos ny, log r, 1 R , log (R þz) with r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 p , y ¼ tan À1 y x , R ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z 2 p (13:2:5) 13.3 Galerkin Vector Representation 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 a single vector function. The proposed repre- sentation is given by 2mu ¼ 2(1 À n)r 2 V À = (= Á V) (13:3:1) where the potential function V is called the Galerkin vector. Substituting this form into Navier’s equation gives the result r 4 V ¼À F 1 À n (13:3:2) Note that for the case of zero body forces, the Galerkin vector is biharmonic. Thus, we have reduced Navier’s equation to a simpler fourth-order vector equation. By comparing the representations given by (13.1.1) with that of (13.3.1), the Helmholtz potentials can be related to the Galerkin vector by f ¼À 1 2m = Á V =  w ¼ 2(1 Àn) 2m r 2 V (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 the previous section. With zero body forces, the stresses corresponding to the Galerkin representa- tion are given by Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 349 Displacement Potentials and Stress Functions 349 TLFeBOOK s x ¼ 2(1 Àn) @ @x r 2 V x þ nr 2 À @ 2 @x 2  = Á V s y ¼ 2(1 Àn) @ @y r 2 V y þ nr 2 À @ 2 @y 2  = Á V s z ¼ 2(1 Àn) @ @z r 2 V z þ nr 2 À @ 2 @z 2  = Á V t xy ¼ (1 Àn) @ @y r 2 V x þ @ @x r 2 V y  À @ 2 @x@y = Á V t yz ¼ (1 Àn) @ @z r 2 V y þ @ @y r 2 V z  À @ 2 @y@z = Á V t zx ¼ (1 Àn) @ @x r 2 V z þ @ @z r 2 V x  À @ 2 @z@x = Á V (13:3:4) As previously mentioned, for no body forces the Galerkin vector must be biharmonic. In Cartesian coordinates, the general biharmonic vector equation would decouple, and thus each component of the Galerkin vector would satisfy the scalar biharmonic equation. However, in curvilinear 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 polar coordinates is given in Example 1-3 by relation (1:9:21) 7 . Therefore, in curvilinear coordinates the individual components of the Galerkin vector do not necessarily satisfy the biharmonic equation. For cylindrical coordinates, only the z component of the Galerkin vector satisfies the biharmonic equation, while the other components satisfy a more complicated fourth-order partial differential equation; see Exercise 13-8 for details. Before moving on to specific applications, we investigate a few useful relationships dealing with harmonic and biharmonic functions. Consider the following identity: r 2 (xf ) ¼ xr 2 f þ 2 @f @x Taking the Laplacian of this expression gives r 4 (xf ) ¼r 2 xr 2 f ÀÁ þ 2 @ @x (r 2 f ) and thus if f 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 the product R 2 f will be biharmonic if f is harmonic, where R 2 ¼ x 2 þ y 2 þ z 2 . Using these results, we can write the following generalized representation for a biharmonic function g as g ¼ f o þ xf 1 þ yf 2 þ zf 3 þ 1 2 R 2 f 4 (13:3:5) where f i are arbitrary harmonic functions. It should be pointed out that not all of the last four terms of (13.3.5) are independent. Consider now the special Galerkin vector representation where only the z component of V is nonvanishing; that is, V ¼ V z e z . For this case, the displacements are given by Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 350 350 ADVANCED APPLICATIONS TLFeBOOK 2mu ¼ 2(1 À n)r 2 V z e z À = @V z @z  (13:3:6) With zero body forces, V z will be biharmonic, and this case is commonly referred to as Love’s strain potential. A special case of this form was introduced by Love (1944) in studying solids of revolution under axisymmetric loading. For this case the displacements and stresses in Cartesian coordinates become 2mu ¼À @ 2 V z @x@z ,2mv ¼À @ 2 V z @y@z ,2mw ¼ 2(1 À n)r 2 V z À @ 2 V z @z 2 s x ¼ @ @z nr 2 À @ 2 @x 2  V z , t xy ¼À @ 3 V z @x@y@z s y ¼ @ @z nr 2 À @ 2 @y 2  V z , t yz ¼ @ @y (1 À n)r 2 À @ 2 @z 2  V z s z ¼ @ @z (2 À n)r 2 À @ 2 @z 2  V z , t zx ¼ @ @x (1 À n)r 2 À @ 2 @z 2  V z (13:3:7) The corresponding relations in cylindrical coordinates are given by 2mu r ¼À @ 2 V z @r@z ,2mu y ¼À 1 r @ 2 V z @y@z ,2mu z ¼ 2(1 Àn)r 2 V z À @ 2 V z @z 2 s r ¼ @ @z nr 2 À @ 2 @r 2  V z , t ry ¼À @ 3 @r@y@z V z r  s y ¼ @ @z nr 2 À 1 r @ @r À 1 r 2 @ 2 @y 2  V z , t yz ¼ 1 r @ @y (1 À n)r 2 À @ 2 @z 2  V z s z ¼ @ @z (2 À n)r 2 À @ 2 @z 2  V z , t zr ¼ @ @r (1 À n)r 2 À @ 2 @z 2  V z (13:3:8) We now consider some example applications for axisymmetric problems where the field variables are independent of y. EXAMPLE 13-1: Kelvin’s Problem: Concentrated Force Acting in the Interior of an Infinite Solid Consider the problem (commonly referred to as Kelvin’s problem) of a single concen- trated force acting at a point in the interior of an elastic solid. For convenience we choose a coordinate system such that the force is applied at the origin and acts in the z direction (see Figure 13-1). The general boundary conditions on this problem would require that the stress field vanish at infinity, be singular at the origin, and give the resultant force system Pe z on any surface enclosing the origin. The symmetry of the problem suggests that we can choose the Love/Galerkin potential as an axisymmetric form V z (r, z). In the absence of body forces, this function is biharmonic, and using the last term in representation (13.3.5) with f 4 ¼ 1=R gives the trial potential Continued Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 351 Displacement Potentials and Stress Functions 351 TLFeBOOK EXAMPLE 13-1: Kelvin’s Problem: Concentrated Force Acting in the Interior of an Infinite Solid–Cont’d V z ¼ AR ¼ A ffiffiffiffiffiffiffiffiffiffiffiffiffiffi r 2 þ z 2 p (13:3:9) where A is an arbitrary constant to be determined. We shall now show that this potential produces the correct stress field for the concentrated force problem under study. The displacement and stress fields corresponding to the proposed potential follow from relations (13.3.8) 2mu r ¼ Arz R 3 ,2mu y ¼ 0, 2mu z ¼ A 2(1 À2n) R þ 1 R þ z 2 R 3  s r ¼ A 2(1 À 2n)z R 3 À 3r 2 z R 5  , t ry ¼ 0 s y ¼ A (1 À 2n)z R 3 , t yz ¼ 0 s z ¼ÀA (1 À2n)z R 3 þ 3z 3 R 5  , t zr ¼ÀA (1 À 2n)r R 3 þ 3rz 2 R 5  (13:3:10) Clearly, these stresses (and displacements) are singular at the origin and vanish at infinity. To analyze the resultant force condition, consider an arbitrary cylindrical surface enclosing the origin as shown in Figure 13-1. For convenience, we choose the cylinder to be bounded at z ¼Æa and will let the radius tend to infinity. Invoking vertical equilibrium, we can write ð 1 0 2prs z (r, a)dr À ð 1 0 2prs z (r,Àa)dr þ ð a Àa 2prt rz (r, z)dz þP ¼ 0 (13:3:11) The first two terms in (13.3.11) can be combined, and in the limit as r !1the third integral is found to vanish, thus giving x y z P x y z P Resultant boundary condition evaluation FIGURE 13-1 Kelvin’s problem: concentrated force in an infinite medium. Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 352 352 ADVANCED APPLICATIONS TLFeBOOK EXAMPLE 13-1: Kelvin’s Problem: Concentrated Force Acting in the Interior of an Infinite Solid–Cont’d P ¼À2 ð 1 a 2pRs z (r, a)dR ¼ 4pA (1 À 2n)a ð 1 a RdR R 3 þ 3a 3 ð 1 a RdR R 5 ! ¼ 8p(1 Àn)A (13:3:12) The constant is now determined and the problem is solved. Of course, the stress field is linearly related to the applied loading, and typically for such three-dimensional prob- lems the field also depends on Poisson’s ratio. 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. For example, consider Boussinesq’s problem of a concentrated force acting normal to the free 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) and later using complex variables in Example 10-5. This problem can be solved by combining a Galerkin vector and Lame ´ ’s strain potential of the forms V x ¼ V y ¼ 0, V z ¼ AR f ¼ B log (R þz) (13:3:13) Using similar methods as in the previous example, it is found that the arbitrary constants become A ¼ P 2p , B ¼À (1 À 2n)P 2p (13:3:14) The displacements and stresses are easily calculated using (13.2.4) and (13.3.7); see Exercise 13-9. x y z P FIGURE 13-2 Boussinesq’s problem: normal force on the surface of a half space. Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 353 Displacement Potentials and Stress Functions 353 TLFeBOOK EXAMPLE 13-3: Cerruti’s Problem: Concentrated Force Acting Parallel to the Free Surface of a Semi-Infinite Solid Another related example is Cerruti’s problem of a concentrated force acting parallel to the free surface of an elastic half space (see Figure 13-3). For convenience, the force is chosen to be directed along the x-axis as shown. Although this problem is not axisym- metric, it can be solved by combining a particular Galerkin vector and Lame ´ ’s strain potential of the following forms: V x ¼ AR, V y ¼ 0, V z ¼ Bx log (R þz) f ¼ Cx R þz (13:3:15) Again, using methods from the previous examples, the constants are found to be A ¼ P 4p(1 À n) , B ¼ (1 À 2n)P 4p(1 À n) , C ¼ (1 À2n)P 2p (13:3:16) The displacements and stresses follow from relations (13.2.4) and (13.3.7); see Exercise 13-10. 13.4 Papkovich-Neuber Representation Using scalar and vector potential functions, another general solution to Navier’s equations was developed by Papkovich (1932) and later independently by Neuber (1934). The completeness of this representation was shown by Eubanks and Sternberg (1956), and thus all elasticity solutions are representable by this scheme. We outline the development of this solution by first writing Navier’s equation in the form r 2 u þ 1 1 À 2n = (= Á u) ¼À F m (13:4:1) x y z P FIGURE 13-3 Cerruti’s problem: tangential force on the surface of a half space. Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 354 354 ADVANCED APPLICATIONS TLFeBOOK Using the Helmholtz representation (13.1.1) and relation (13.1.3), this previous equation can be written as r 2 [u þ 1 (1 À 2n) =f] ¼À F m (13:4:2) Define the vector term in the brackets as h ¼ u þ 1 (1 À 2n) =f (13:4:3) We note that r 2 h ¼ÀF=m, = Áh ¼ 2(1 À n) 1 À2n r 2 f (13:4:4) Using the identity r 2 (R Á h) ¼ R Ár 2 h þ 2(= Á h), it can be shown that = Á h ¼ 1 2 r 2 (R Á h) þ R Á F m  (13:4:5) Combining results (13.4.5) with (13.4.4) gives r 2 2(1 À n) 1 À2n f À 1 2 R Á h ! ¼ R Á F m (13:4:6) Defining the term in brackets by scalar h, we get r 2 h ¼ R Á F m (13:4:7) Using the definition of h, we can eliminate f from relation (13.4.3) and obtain an expression for the displacement vector. Redefining new scalar and vector potentials in terms of h and h, we can write 2mu ¼ A À = B þ A Á R 4(1 À n) ! (13:4:8) where r 2 A ¼À2F, r 2 B ¼ R Á F (1 À n) (13:4:9) This general displacement representation is the Papkovich-Neuber solution of Navier’s equa- tions. For the case with zero body forces, the two potential functions A and B are harmonic. The four individual functions A x , A y , A z , and B, however, are not all independent, and it can be shown that for arbitrary three-dimensional convex regions, only three of these functions are independent. Note that a convex region is one in which any two points in the domain may be connected by a line that remains totally within the region. Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 355 Displacement Potentials and Stress Functions 355 TLFeBOOK 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 it can be easily shown that A ¼ 2(1 À n)r 2 V, B ¼ = ÁV À A Á R 4(1 À n) (13:4:10) As with the Galerkin vector solution, it is convenient to consider the special case of axisym- metry where A r ¼ A y ¼ 0, A z ¼ A z (r, z), B ¼ B(r, z) with r 2 B ¼ 0 and r 2 A z ¼ 0 (13:4:11) For this axisymmetric case, B and A z are commonly called the Boussinesq potentials, and as before with zero body forces they are harmonic functions. EXAMPLE 13-4: Boussinesq’s Problem Revisited We consider again the problem shown previously in Figure 13-2 of a concentrated force acting normal to the stress-free surface of a semi-infinite solid. Because the problem is axisymmetric, we use the Boussinesq potentials defined by (13.4.11). These potentials must be harmonic functions of r and z, and using (13.2.5), we try the forms A z ¼ C 1 R , B ¼ C 2 log (R þ z) (13:4:12) where C 1 and C 2 are constants to be determined. The boundary conditions on the free surface require that s z ¼ t rz ¼ 0 everywhere except at the origin, and that the summation of the total vertical force be equal to P. Calculation of these stresses follows using the displacements from (13.4.8) in Hooke’s law, and the result is s z ¼À 3C 1 z 3 4(1 À n)R 5 t rz ¼ r R 3 C 2 À (1 À 2n) 4(1 À n) C 1 À 3C 1 z 2 4(1 À n)R 2  (13:4:13) Note that the expression for s z vanishes on z ¼ 0, but is indeterminate at the origin, and thus this relation will not directly provide a means to determine the constant C 1 . Rather than trying to evaluate this singularity at the origin, we pursue the integrated condition on any typical plane z ¼ constant P ¼À ð 1 0 s z (r, z)2prdr (13:4:14) Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 356 356 ADVANCED APPLICATIONS TLFeBOOK EXAMPLE 13-4: Boussinesq’s Problem Revisited–Cont’d Invoking these boundary conditions determines the two constants C 1 ¼ 2(1 Àn) p P, C 2 ¼ (1 À 2n) 2p P (13:4:15) The results for the displacements and stresses are given by u r ¼ P 4pmR rz R 2 À (1 À 2n)r R þz ! u z ¼ P 4pmR 2(1 Àn) þ z 2 R 2 ! u y ¼ 0 (13:4:16) s r ¼ P 2pR 2 À 3r 2 z R 3 þ (1 À 2n)R R þ z ! s y ¼ (1 À 2n)P 2pR 2 z R À R R þz ! s z ¼À 3Pz 3 2pR 5 , t rz ¼À 3Prz 2 2pR 5 (13:4:17) Many additional problems can be solved using the Papkovich method, and some of these are given in the exercises. This technique also is used in the next chapter to generate solutions for many singular stress states employed in micromechanics modeling. An interesting connection can be made for the two-dimensional case between the Papko- vich-Neuber scheme and the complex variable method discussed in Chapter 10. For the case of plane deformation in the x, y-plane, we choose A x ¼ A x (x, y), A y ¼ A y (x, y), A z ¼ 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(u þ iv) ¼ (3 À4n)g(z) À z g 0 (z) À c(z) (13:4:19) with appropriate selection of g(z) and c(z) in terms of A x , A y , and B. It is noted that this form is identical to (10.2.9) found using the complex variable formulation. Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 357 Displacement Potentials and Stress Functions 357 TLFeBOOK A convenient summary flow chart of the various displacement functions discussed in this chapter is shown in Figure 13-4. The governing equations in terms of the particular potential functions are for the zero body force case. Chou and Pagano (1967) provide additional tables for displacement potentials and stress functions. 13.5 Spherical Coordinate Formulations The previous solution examples employing displacement potentials simply used preselected forms of harmonic and biharmonic potentials. We now investigate a more general scheme to determine appropriate potentials for axisymmetric problems described in spherical coordin- ates. Referring to Figures 1-4 and 1-5, cylindrical coordinates (r, y, z) are related to spherical coordinates (R, f, y) through relations R ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi r 2 þ z 2 p , sin f ¼ r R , cos f ¼ z R (13:5:1) 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’s strain potential V z , 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’s equation. In spherical coordinates the Laplacian operator becomes r 2 ¼ @ 2 @R 2 þ 2 R @ @R þ 1 R 2 cotf @ @f þ 1 R 2 @ 2 @f 2 (13:5:2) We first look for separable solutions of the form R n F n (f), and substituting this into Laplace’s equation gives Displacement Formulation: Navier’s Equation Helmholtz Representation: f, j Lamé’s Strain Potential: f Galerkin Vector: V Love’s Strain Function: V z V = V z e z ∂z ∂V z Papkovich-Neuber: A, B 4(1−n) 2mu = A −∇ ∇ A . R B + Boussinesq’s Potentials: A z ,B Axisymmetric Problem A=A z (r,z)e z , B = B (r,z) m∇ 2 u + (l + m)∇(∇ . u) + F = 0 u = 2(1 − n)∇ 2 V −∇ ∇ (∇ ∇ . V ) u = 2(1 − n)∇ 2 V z e z −∇ ∇ u = ∇f + ∇ ϫ j (f + 2m)∇∇ 2 f + m∇ ϫ ∇ 2 j + F = 0 ⇒∇ 4 f = 0, ∇ 4 j = 0 ⇒∇ 4 V = 0 ⇒∇ 2 A = 0, ∇ 2 B = 0 ⇒∇ 2 A z = 0, ∇ 2 B z = 0 ∇ 4 V z = 0 ∇ 2 f = 0, j = 0 u = f ∆ FIGURE 13-4 Displacement potential solutions. Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 358 358 ADVANCED APPLICATIONS TLFeBOOK [...]... biharmonic ´ 13- 4 For the case of Lame’s potential, show that strains and stresses are given by (13. 2.3) Displacement Potentials and Stress Functions 365 TLFeBOOK Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 366 13- 5 Justify that the Galerkin vector satisfies the governing equation (13. 3.2) 13- 6 Show that the Helmholtz potentials are related to the Galerkin vector by relations (13. 3.3) 13- 7 Justify... 2pR3 R Rþz (R þ z)2 13- 11 Explicitly justify that the Papkovich functions A and B satisfy relations (13. 4.9) 13- 12 For the axisymmetric case, the Papkovich functions reduced to the Boussinesq potentials B and Az defined by relations (13. 4.11) Determine the general form of the displacements and stresses in cylindrical coordinates in terms of B and Az 13- 13 Using the results of Exercise 13- 12, verify that... ) 13. 6.2 Morera Stress Function Representation The Morera stress function method uses the general form with diagonal terms set to zero; that is, 2 0 Fij ¼ 4 F12 F13 F12 0 F23 3 F13 F23 5 0 (13: 6:11) This approach yields the representation 364 ADVANCED APPLICATIONS TLFeBOOK Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 365 s11 ¼ À2F23, 23 s22 ¼ À2F31, 31 s33 ¼ À2F12, 12 s12 ¼ ÀF12, 33 þ F23, 13. .. Theory of Elasticity, McGraw-Hill, New York, 1970 Exercises 13- 1 Using the Helmholtz representation, determine the displacement field that corresponds to the potentials f ¼ x2 þ 4y2 , w ¼ R2 e3 Next show that this displacement field satisfies Navier’s equation with no body forces 13- 2 Explicitly show that the dilatation and rotation are related to the Helmholtz potentials through relations (13. 1.3) 13- 3 For... a complete solution to the elasticity problem is given by sij ¼ "imp "jkl Fmk, pl (13: 6:4) where F is a symmetric second-order tensor Relation (13. 6.4) is sometimes referred to as the Beltrami 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,... given by Timoshenko and Goodier (1970) 362 ADVANCED APPLICATIONS TLFeBOOK Sadd / Elasticity 13. 6 Final Proof 3.7.2004 2:55pm page 363 Stress Functions In the absence of body forces, the stress formulation of elasticity theory includes the equilibrium and Beltrami-Michell equations: sij, j ¼ 0 (13: 6:1) 1 skk, ij ¼ 0 1þn (13: 6:2) sij, kk þ In order to develop a general solution to this system, stress... problem of Example 13- 4 are given by (13. 4.16) and (13. 4.17) Note the interesting behavior of the radial displacement, that ur > 0 only for points where z=R > (1 À 2n)R=(R þ z) Show that points satisfying this inequality lie inside a cone f fo , with fo determined by the relation cos2 fo þ cos fo À (1 À 2n) ¼ 0 13- 14* The displacement field for the Boussinesq problem was given by (13. 4.16) For this case,...Sadd / Elasticity Final Proof 3.7.2004 2:55pm page 359   1 d dFn sin f þ n(n þ 1)Fn ¼ 0 sin f df df (13: 5:3) Next, making the change of variable x ¼ cos f, relation (13. 5.3) becomes (1 À x2 ) d 2 Fn dFn þ n(n þ 1)Fn ¼ 0 À 2x dx2 dx (13: 5:4) and this is the well-known Legendre differential equation The two fundamental solutions... )À3=2 þ Á Á Á (13: 5:8) and this form will be useful for infinite domain problems For example, the solution to the Kelvin 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 the Papkovich potentials; see Little (1973) Displacement Potentials and Stress Functions 359 TLFeBOOK Sadd / Elasticity. .. product, and thus relation (13. 6.4) can be expressed as sij ¼ dij Fkk, ll À dij Fkl, kl À Fij, kk þ Fli, lj þ Flj, li À Fkk, ij (13: 6:5) or s11 ¼ F33, 22 þ F22, 33 À 2F23, 23 s22 ¼ F11, 33 þ F33, 11 À 2F31, 31 s33 ¼ F22, 11 þ F11, 22 À 2F12, 12 s12 ¼ ÀF12, 33 À F33, 12 þ F23, 13 þ F31, 23 s23 ¼ ÀF23, 11 À F11, 23 þ F31, 21 þ F12, 31 s31 ¼ ÀF31, 22 À F22, 31 þ F12, 32 þ F23, 12 (13: 6:6) Displacement Potentials . are easily calculated using (13. 2.4) and (13. 3.7); see Exercise 13- 9. x y z P FIGURE 13- 2 Boussinesq’s problem: normal force on the surface of a half space. Sadd / Elasticity Final Proof 3.7.2004. 365 TLFeBOOK 13- 5. Justify that the Galerkin vector satisfies the governing equation (13. 3.2). 13- 6. Show that the Helmholtz potentials are related to the Galerkin vector by relations (13. 3.3). 13- 7 representation (13. 1.1) and relation (13. 1.3), this previous equation can be written as r 2 [u þ 1 (1 À 2n) =f] ¼À F m (13: 4:2) Define the vector term in the brackets as h ¼ u þ 1 (1 À 2n) =f (13: 4:3) We