Elasticity Part 9 ppt

EXAMPLE 9-7: Conical Shaft–Cont’d C ¼ C z ffiffiffiffiffiffiffiffiffiffiffiffiffiffi r 2 þ z 2 p  1 3 z 3 (r 2 þ z 2 ) 3=2  (9:7:9) where the constant C has been determined to satisfy the load-carrying relation (9.7.8) C ¼ T 2pm( 2 3  cos j þ 1 3 cos 3 j) (9:7:10) The stresses follow from relations (9.7.4) t ry ¼ Cmr 2 (r 2 þ z 2 ) 5=2 t yz ¼ Cmrz (r 2 þ z 2 ) 5=2 (9:7:11) and the displacement u y can be determined by integrating equations (9.7.2) to get u y ¼ Cr 3(r 2 þ z 2 ) 3=2 þ !r (9:7:12) where !r is the rigid-body rotation term about the z-axis and ! can be determined by specifying the shaft rotation at a specific z location. Additional examples of such problems are discussed in Timoshenko and Goodier (1970) and Sokolnikoff (1956). Before leaving the torsion problem, it should be mentioned that this problem can also be easily formulated and solved using the numerical finite element method. Chapter 15 discusses this important numerical scheme and provides a series of such solutions in Example 15-4 and Figure 15-7. These examples illustrate the power and usefulness of this numerical method to solve problems with complicated geometry that could not be easily solved using analytical means. 9.8 Flexure Formulation We now investigate a final case of deformation of elastic cylinders under end loadings by considering the flexure of elastic beams subject to transverse end forces, as shown in Figure 9-19. The problem geometry is formulated as a cantilever beam of arbitrary section with a fixed end at z ¼ 0 and transverse end loadings P x and P y at z ¼ l. Following our usual procedure, the problem is to be solved in the Saint-Venant sense, in that only the resultant end loadings P x and P y are used to formulate the boundary conditions at z ¼ l. From our general formulation in Section 9.1, s x ¼ s y ¼ t xy ¼ 0. The other three nonzero stresses will be determined to satisfy the equilibrium and compatibility relations and all associated boundary conditions. From our earlier work, the equilibrium and compatibility relations resulted in equations (9.1.2) from which it was argued that t xz and t yz were independ- ent of z, and s z was a bilinear form in x, y, z (see Exercise 9-1). Motivated from strength of materials theory, we choose the arbitrary form for s z as Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 229 Extension, Torsion, and Flexure of Elastic Cylinders 229 TLFeBOOK s z ¼ (Bx þ Cy)(l  z)(9:8:1) where B and C are constants. Using this result in the remaining equilibrium equation in the z direction gives @t xz @x þ @t yz @y  (Bx þ Cy) ¼ 0 which can be written in the form @ @x [t xz  1 2 Bx 2 ] þ @ @y [t yz  1 2 Cy 2 ] ¼ 0(9:8:2) This equilibrium statement motivates the introduction of another stress function F(x, y), such that t xz ¼ @F @y þ 1 2 Bx 2 t yz ¼ @F @x þ 1 2 Cy 2 (9:8:3) This form then satisfies equilibrium identically, and using it in the remaining compatibility relations gives the results @ @y (r 2 F) þ nB 1 þ n ¼ 0  @ @x (r 2 F) þ nC 1 þ n ¼ 0 (9:8:4) x y z l S R P x P y (x o ,y o ) FIGURE 9-19 Flexure problem geometry. Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 230 230 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK This system can be integrated to get r 2 F ¼ n 1 þ n (Cx  By) þ k (9:8:5) where k is a constant of integration. In order to determine this constant, consider the rotation about the z-axis. From the general relation (2.1.9), ! z ¼ [(@v=@x)  (@u=@y)]=2, differentiat- ing with respect to z and using Hooke’s law and our previous results gives @! z @z ¼ 1 2 @ 2 v @x@z  @ 2 u @y@z  ¼ 1 2m @t yz @x  @t xz @y  ¼ 1 2m r 2 F ¼ 1 2m n 1 þ n (Cx  By) þ k ! (9:8:6) From the torsion formulation, the angle of twist per unit length was specified by the parameter a, and selecting the section origin (x ¼ y ¼ 0) at the center of twist, relation (9.8.6) then implies that k ¼2ma. Thus, the governing equation (9.8.5) can be written as r 2 F ¼ n 1 þ n (Cx  By)  2ma (9:8:7) The zero loading boundary condition on the lateral surface S is expressed by t xz n x þ t yz n y ¼ 0 and using the stress function definition, this can be written as @F @x dx ds þ @F @y dy ds þ 1 2 Bx 2 dy ds  Cy 2 dx ds  ¼ 0or dF ds ¼ 1 2 Bx 2 dy ds  Cy 2 dx ds  (9:8:8) It is convenient to separate the stress function F into a torsional part f and a flexural part c, such that F(x, y) ¼ f(x, y) þ c (x, y)(9:8:9) where the torsional part is formulated by r 2 f ¼2ma in R df ds ¼ 0onS (9:8:10) while the flexural portion satisfies r 2 c ¼ n 1 þ n (Cx  By)inR d c ds ¼ 1 2 (Bx 2 dy ds  Cy 2 dx ds )onS (9:8:11) Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 231 Extension, Torsion, and Flexure of Elastic Cylinders 231 TLFeBOOK Because we have already investigated the torsional part of this problem in the preceding sections, we now pursue only the flexural portion. The general solution to (9:8:11) 1 may be expressed as the sum of a particular solution plus a harmonic function c(x, y) ¼ f (x, y) þ 1 6 n 1 þ n (Cx 3  By 3 )(9:8:12) where f is a harmonic function satisfying r 2 f ¼ 0. The boundary conditions on end z ¼ l can be stated as ðð R t xz dxdy ¼ P x ðð R t yz dxdy ¼ P y ðð R [xt yz  yt xz ]dxdy ¼ x o P y  y o P x (9:8:13) Using the first relation of this set gives ðð R [ @ @y (f þ c) þ 1 2 Bx 2 ]dxdy ¼ P x (9:8:14) but from the torsion formulation ðð R @f @y dxdy ¼ 0, and so (9.8.14) can be written as ðð R [ @ @x (x @c @y )  @ @y (x @c @x )]dxdy þ ðð R 1 2 Bx 2 dxdy ¼ P x (9:8:15) Using Green’s theorem and the boundary relation (9:8:11) 2 , the first integral can be expressed as ðð R [ @ @x (x @c @y )  @ @y (x @c @x )]dxdy ¼ ðð R [ 3 2 Bx 2 þ Cxy]dxdy and thus equation (9.8.15) reduces to BI y þ CI xy ¼P x (9:8:16) In a similar manner, boundary condition (9:8:13) 2 gives BI xy þ CI x ¼P y (9:8:17) The expressions I x , I y , and I xy are the area moments of inertia of section R I x ¼ ðð R y 2 dxdy, I y ¼ ðð R x 2 dxdy, I xy ¼ ðð R xydxdy (9:8:18) Relations (9.8.16) and (9.8.17) can be solved for the constants B and C Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 232 232 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK B ¼ P x I x  P y I xy I x I y  I 2 xy C ¼ P y I y  P x I xy I x I y  I 2 xy (9:8:19) The final boundary condition (9:8:13) 3 can be expressed as  ðð R [x @f @x þ y @f @y ]dxdy  ðð R [x @c @x þ y @c @y ]dxdy þ ðð R 1 2 (Cxy 2  Bx 2 y)dxdy ¼ x o P y  y o P x (9:8:20) From the torsion formulation,  ðð R x @f @x þ y @f @y ! dxdy ¼ 2 ðð R fdxdy ¼ T ¼ aJ, and so (9.8.20) becomes aJ þ ðð R 1 2 (Cxy 2  Bx 2 y)  (x @c @x þ y @c @y )  dxdy ¼ x o P y  y o P x (9:8:21) Once the flexural stress function c is known, (9.8.21) will provide a relation to determine the angle of twist a. Relation (9.8.21) can also be used to determine the location (x o , y o ) for no induced torsional rotation, a point commonly called the shear center or center of flexure. Choosing a ¼ 0, this equation can be independently used for the two cases of (P x ¼ 0, P y 6¼ 0) and (P x 6¼ 0, P y ¼ 0) to generate two equations for locations x o and y o . If the x-axis is an axis of symmetry, then y o ¼ 0; and likewise, if the y-axis is one of symmetry, then x o ¼ 0. For a section with two perpendicular axes of symmetry, the location (x o , y o ) lies at the intersection of these two axes, which is at the centroid of the section. However, in general the shear center does not coincide with the section’s centroid and need not even lie within the section. 9.9 Flexure Problems Without Twist Because we have previously studied examples of the torsion problem, we shall now develop flexure solutions to problems that do not include twist. The two examples to be investigated include simple symmetric cross-sections with single end loadings along an axis of symmetry. EXAMPLE 9-8: Circular Section Consider the flexure of an elastic beam of circular section, as shown in Figure 9-20. The end loading (P x ¼ 0, P y ¼ P) passes through the center of the section, which coincides with the centroid and center of twist. Thus, for this problem there will be no torsion (a ¼ 0), and so f ¼ 0 and F ¼ c. It is convenient to use polar coordinates for this problem, and the governing equation (9:8:11) 1 can then be written as r 2 c ¼ n 1 þ n P I x r cos y (9:9:1) Continued Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 233 Extension, Torsion, and Flexure of Elastic Cylinders 233 TLFeBOOK EXAMPLE 9-8: Circular Section–Cont’d while the boundary condition (9:8:11) 2 becomes 1 a @c @y ¼ 1 2 P I x a 2 sin 3 y on r ¼ a (9:9:2) The solution to (9.9.1) can then be taken as c ¼ P I x [ f  1 6 n 1 þ n r 3 cos 3 y](9:9:3) Using trigonometric identities, relations (9.9.3) and (9.9.2) can be rewritten c ¼ P I x [ f  1 24 n 1 þ n r 3 ( cos 3y þ 3 cos y)] @c @y ¼ 1 8 P I x a 3 (  sin 3y þ 3 sin y)onr ¼ a (9:9:4) Based on the previous relations, we look for solutions for the harmonic function in the form f ¼ P n A n r n cos ny and consider the two terms f ¼ A 1 r cos y þ A 3 r 3 cos 3y (9:9:5) Combining (9.9.4) and (9.9.5) yields c ¼ P I x [A 1 r  nr 3 8(1 þ n) ] cos y þ [A 3  n 24(1 þ n) ]r 3 cos 3y ! (9:9:6) Boundary condition (9:9:4) 2 yields two relations to determine the constants A 1 and A 3 A 1 ¼ 3 þ 2n 8(1 þ n) a 2 A 3 ¼ 1 þ 2n 24(1 þ n) (9:9:7) x y z a P l FIGURE 9-20 Flexure of a beam of circular section. Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 234 234 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK EXAMPLE 9-8: Circular Section–Cont’d and back-substituting this result into (9.9.6) gives the final form of the stress function c ¼ P I x  3 þ 2n 8(1 þ n) a 2 x  1 þ 2n 8(1 þ n) xy 2 þ 1  2n 24(1 þ n) x 3 ! (9:9:8) The stresses corresponding to this solution become t xz ¼ P 4I x 1 þ 2n 1 þ n xy t yz ¼ P I x 3 þ 2n 8(1 þ n) a 2  y 2  1  2n 3 þ 2n x 2 !  z ¼ P I x y(l  z) (9:9:9) Note for this section I x ¼ pa 4 =4. The maximum stress occurs at the origin and is given by t max ¼ t yz (0, 0) ¼ P pa 2 3 þ 2n 2(1 þ n) (9:9:10) This can be compared to the value developed from strength of materials theory t max ¼ 4P=3pa 2 . Differences in the maximum shear stress between the two theories are small, and for the special case n ¼ 1=2, the elasticity solution is the same as the elementary result. Comparison of the shear stress distribution with strength of materials theory for n ¼ 0:1 has been given by Sadd (1979), and again differences were found to be small. Displacements for this problem can be determined through the usual integra- tion process (see Exercise 9-21). EXAMPLE 9-9: Rectangular Section Our second flexure example involves a beam of rectangular section with end loading (P x ¼ 0, P y ¼ P) passing through the shear center, as shown in Figure 9-21. The section dimensions are the same as those given in Figure 9-12. As in the previous example, there is no torsion (a ¼ 0), and so for this case f ¼ 0 and F ¼ c. Formulation equations (9.8.11) then give r 2 c ¼ v 1 þ v P I x x in R d c ds ¼ 1 2 P I x y 2 dx ds on S (9:9:11) For the rectangular section, dc ds ¼  dc dy ¼ 0, x ¼a  dc dx ¼ 1 2 P I x b 2 , y ¼b 8 > > < > > : (9:9:12) Continued Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 235 Extension, Torsion, and Flexure of Elastic Cylinders 235 TLFeBOOK EXAMPLE 9-9: Rectangular Section–Cont’d Based on these boundary relations we are motivated to select a solution of the form c ¼ P I x [ f  1 6 v 1 þ v (x 3  a 2 x)  b 2 x 2 ](9:9:13) with the harmonic function f satisfying f (x, y) ¼ 0, x ¼a v 6(1 þ v) (x 3  a 2 x), y ¼b ( (9:9:14) Because we expect t yz to be an even function in x and y, and t xz to be odd in y, we look for a harmonic solution for f in the form f (x, y) ¼ X 1 n¼1 A n sin npx a cosh npy a (9:9:15) This form satisfies (9:9:14) 1 identically, while (9:9:14) 2 implies that X 1 n¼1 b n sin npx a ¼ v 6(1 þ v) (x 3  a 2 x)(9:9:16) where b n ¼ A n cosh(npb=a). Relation (9.9.16) is recognized as a Fourier sine series, and thus the coefficients follow from standard theory (8.2.28) and are given by b n ¼ 2va 3 ( 1) n =(1 þ v)n 3 p 3 . Putting these results back together gives the final form of the stress function c ¼ P I x  1 6 v 1 þ v (x 3 a 2 x) b 2 x 2 þ 2va 3 (1 þ v)p 3 X 1 n¼1 (  1) n n 3 sin npx a cosh npy a cosh npb a 2 6 4 3 7 5 (9:9:17) x y z 2 a P l 2b Figure 9-21 Flexure of a beam of rectangular section. Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 236 236 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK EXAMPLE 9-9: Rectangular Section–Cont’d The stresses then follow to be t xz ¼ 2va 2 P (1 þ v)p 2 I x X 1 n¼1 (1) n n 2 sin npx a sinh npy a cosh npb a t yz ¼ P 2I x (b 2 y 2 )þ vP 6(1 þ v)I x 3x 2  a 2  12a 2 p 2 X 1 n¼1 (1) n n 2 cos npx a cosh npy a cosh npb a 2 6 4 3 7 5 s z ¼ P I x y(l  z) (9:9:18) The corresponding results from strength of materials gives t yz ¼ P(b 2  y 2 )=2I x , and thus the second term of (9.9.18) 2 represents the correction to the elementary theory. Note that if n ¼ 0, this correction term vanishes, and the two theories predict identical stresses. For the case of a thin rectangular section with b  a, cosh (np b=a) !1, and it can be shown that the elasticity solution reduces to the strength of materials prediction. A similar result is also found for the case of a thin section with a >> b. Comparison of the shear stress distribution t yz with strength of materials theory for n ¼ 1=2 has been presented by Sadd (1979), and differences between the two theories were found to be sizable. As in the previous example, the maximum stress occurs at x ¼ y ¼ 0 t max ¼ t yz (0, 0) ¼ P 2I x b 2  vPa 2 6(1 þ v)I x 1 þ 12 p 2 X 1 n¼1 (  1) n n 2 sech npb a "# (9:9:19) Again, the strength of materials result is given by the first term in relation (9.9.19). This concludes our brief presentation of flexure examples. Solutions to additional flexure problems are given by Sokolnikoff (1956) and Timoshenko and Goodier (1970). References Higgins TJ: A comprehensive review of Saint-Venant torsion problem, Amer. Jour. of Physics, vol 10, pp. 248-259, 1942. Higgins TJ: The approximate mathematical methods of applied physics as exemplified by application to Saint-Venant torsion problem, Jour. of Appl. Physics, vol 14, pp. 469-480, 1943. Higgins TJ: Analogic experimental methods in stress analysis as exemplified by Saint-Venant’s torsion problem, Proc. Soc. of Exp. Stress Analysis, vol 2, pp. 17-27, 1945. Kellogg OD: Foundations of Potential Theory, Dover, New York, 1969. Sadd MH: A comparison of mechanics of materials and theory of elasticity stress analysis, Mech. Eng. News, ASEE, vol 16, pp 34-39, 1979. Sokolnikoff IS: Mathematical Theory of Elasticity, McGraw-Hill, New York, 1956. Timoshenko SP, and Goodier JN: Theory of Elasticity, McGraw-Hill, New York, 1970. Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 237 Extension, Torsion, and Flexure of Elastic Cylinders 237 TLFeBOOK Exercises 9-1. Under the assumption that s x ¼ s y ¼ t xy ¼ 0, show that equilibrium and compatibility equations with zero body forces reduce to relations (9.1.2). Next integrate relations @ 2 s z @x 2 ¼ @ 2 s z @y 2 ¼ @ 2 s z @z 2 ¼ @ 2 s z @x@y ¼ 0 to justify that s z ¼ C 1 x þ C 2 y þ C 3 z þ C 4 xz þ C 5 yz þ C 6 , where C i are arbitrary constants. 9-2. During early development of the torsion formulation, Navier attempted to extend Coulomb’s theory for bars of circular section and assume that there is no warping displacement for general cross-sections. Show that although such an assumed displacement field will satisfy all elasticity field equations, it will not satisfy the boundary conditions and thus is not an acceptable solution. 9-3. Referring to Figure 9-2, if we choose a different reference origin that is located at point (a,b) with respect to the given axes, the displacement field would now be given by u ¼az(y b), v ¼ az(x  a), w ¼ w(x, y) where x and y now represent the new coordinates. Show that this new representation leads to an identical torsion formulation as originally developed. 9-4. In terms of a conjugate function c(x, y) defined by @c @x ¼ 1 a @w @y , @c @y ¼ 1 a @w @x show that the torsion problem may be formulated as r 2 c ¼ 0inR c ¼ 1 2 (x 2 þ y 2 ) þ constant on S 9-5. A function f (x,y) is defined as subharmonic in a region R if r 2 f  0 at all points in R.It can be proved that the maximum value of a subharmonic function occurs only on the boundary S of region R. For the torsion problem, show that the square of the resultant shear stress t 2 ¼ t 2 xz þ t 2 yz is a subharmonic function, and thus the maximum shear stress will always occur on the section boundary. 9-6. Employing the membrane analogy, develop an approximate solution to the torsion problem of a thin rectangular section as shown. Neglecting end effects at y ¼b, the membrane deflection will then depend only on x, and the governing equation can be integrated to give z ¼ f ¼ ma(a 2  x 2 ), thus verifying that the membrane shape is parabolic. Formally compute the maximum membrane slope and volume enclosed to justify relations (9.5.15). Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 238 238 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK [...]... Example 9- 9 For each theory, calculate and plot the dimensionless shear stress tyz (0, y)b2 =P versus y=b for an aspect ratio b=a ¼ 1 9- 23 Solve the flexure problem without twist of an elastic beam of elliptical section as shown in Figure 9- 7 with Py ¼ P Show that the stress results reduce to (9. 9 .9) for the circular case with a ¼ b 242 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK Sadd / Elasticity Part. .. ellipse 9- 9 Develop relation (9. 4.14) for the load-carrying torque of an equilateral triangular section 9- 10 For the torsion of a bar of elliptical section, express the torque equation (9. 4.6) in terms of the polar moment of inertia of the section, and compare this result with the corresponding relation for the equilateral triangular section 9- 11* For the triangular section shown in Figure 9- 9, calculate... Comprehensive texts on this solution method include Muskhelishvili ( 195 3, 196 3), Milne-Thomson ( 196 0), Green and Zerna ( 196 8), and England ( 197 1) Additional briefer sources of information can also be found in Sokolnikoff ( 195 6) and Little ( 197 3) The purpose of this chapter is to introduce the basics of the method and to investigate its application to particular problems of engineering interest We shall first briefly... Starting with the stress solution (9. 9 .9) , integrate the straindisplacement relations and use boundary conditions that require the displacements and rotations to vanish at z ¼ 0 Compare the elasticity results with strength of materials theory Also investigate whether the elasticity displacements indicate that plane sections remain plane 9- 22* Make a comparison of theory of elasticity and strength of materials... needed to develop the elasticity solutions Further and more detailed information on complex variables may be found in the mathematical texts by Churchill ( 196 0) or Kreyszig ( 199 9) 10.1 Review of Complex Variable Theory A complex variable z is defined by two real variables x and y in the form z ¼ x þ iy (10:1:1) pffiffiffiffiffiffiffi where i ¼ À1 is called the imaginary unit, x is know as the real part of z, that is,... and can be determined from Example 9- 1 or strength of materials theory Show that the stress concentration plot gives (tmax )keyway ! 2, as b=a ! 0, thus indicating that a small notch will result in a (tmax )solid shaft doubling of the stress in circular section under torsion 9- 19 Example 9- 6 provides the torsion solution of a closed thin-walled section shown in Figure 9- 16 Investigate the solution of...Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 2 39 y parabolic membrane b z a x a a x 9- 7 Using the stress results for the torsion of the elliptical section, formally integrate the strain-displacement relations and develop the displacement solution (9. 4.11) 9- 8 For the torsion of an elliptical section, show that the resultant shear... Cylinders 2 39 TLFeBOOK Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 240 y y = m1x x=a x y = −m2x 9- 13 For the torsion problem in Example 9- 3, explicitly justify that the required values for pffiffiffi the constants appearing in the stress function are given by c ¼ 3 À 8 and pffiffiffi K ¼ Àma=[4a2 (1 À 2)] Also calculate the resulting shear stresses and determine the location and value of the maximum stress 9- 14... maximum stress 9- 14 Attempt to solve the torsion of a rectangular section shown in Figure 9- 12 by using the boundary equation method Show that trying a stress function created from the four products of the boundary lines x ¼ Æ a and y ¼ Æ b will not satisfy the governing equation (9. 3 .9) 9- 15* Using the torque relation (9. 5.12) for the rectangular section, compute the nondimensional load-carrying parameter... that the load-carrying behavior can be given by the approximate relation (9. 5.15) 9- 16 Using the relation (9. 5.16), develop an approximate solution for the load-carrying torque of the channel section shown b t a t t b 240 FOUNDATIONS AND ELEMENTARY APPLICATIONS TLFeBOOK Sadd / Elasticity Final Proof 3.7.2004 2:57pm page 241 9- 17 A circular shaft with a keyway can be approximated by the section shown . ¼ X 1 n¼1 A n sin npx a cosh npy a (9: 9:15) This form satisfies (9: 9:14) 1 identically, while (9: 9:14) 2 implies that X 1 n¼1 b n sin npx a ¼ v 6(1 þ v) (x 3  a 2 x) (9: 9:16) where b n ¼ A n cosh(npb=a). Relation (9. 9.16). A 3 r 3 cos 3y (9: 9:5) Combining (9. 9.4) and (9. 9.5) yields c ¼ P I x [A 1 r  nr 3 8(1 þ n) ] cos y þ [A 3  n 24(1 þ n) ]r 3 cos 3y ! (9: 9:6) Boundary condition (9: 9:4) 2 yields two relations to. York, 196 9. Sadd MH: A comparison of mechanics of materials and theory of elasticity stress analysis, Mech. Eng. News, ASEE, vol 16, pp 34- 39, 197 9. Sokolnikoff IS: Mathematical Theory of Elasticity,