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EXAMPLE 14-4: Kelvin State with Unit Loads in Coordinate Directions–Cont’d For the case with the force in the x direction, that is, the state S x (x), we get the following fields: u R ¼ 2C(1 2n) mR sin f cos y u f ¼ C(3 4n) 2mR cos f cos y u y ¼ C(3 4n) 2mR sin y s R ¼ 2C(2 n) R 2 sin f cos y s y ¼ s f ¼ C(1 2n) R 2 sin f cos y t Rf ¼ C(2n 1) R 2 cos f cos y t Ry ¼ C(1 2n) R 2 sin y, t yf ¼ 0 (14:2:11) Notice that for the Kelvin state the displacements are of order O(1/R), while the stresses are O(1=R 2 ), and that ð S T n ds ¼ P, ð S R T n ds ¼ 0 for any closed surface S enclosing the origin. Using the basic Kelvin problem, many related singular states can be generated. For example, define S 0 (x) ¼ S, a ¼ {u , a , s , a , e , a }, where a ¼ 1, 2, 3. Now, if the state S 0 is generated by the Papkovich potentials f(x) and c(x), then S 0 is generated by f 0 (x) ¼ f , a þ c a c 0 (x) ¼ c i, a e i (14:2:12) Further, define the Kelvin state S a (x;j) as that corresponding to a unit load applied in the x a direction at point j , as shown in Figure 14-7. Note that S a (x;j) ¼S a (x j). Also define the set of nine states S ab (x) by the relation S ab (x) ¼S a , b (x) (14:2:13) or equivalently, ^ SS ab (x) ¼ S a (x 1 , x 2 , x 3 ) S a (x 1 d b1 h, x 2 d b2 h, x 3 d b3 h) h ¼ S a (x 1 , x 2 , x 3 ) S a (x 1 , x 2 , x 3 ;d b1 h, d b2 h, d b3 h) h (14:2:14) Continued Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 379 Micromechanics Applications 379 TLFeBOOK EXAMPLE 14-4: Kelvin State with Unit Loads in Coordinate Directions–Cont’d and thus S ab ¼ lim h!0 ^ SS ab (14:2:15) EXAMPLE 14-5: Force Doublet Consider the case of two concentrated forces acting along a common line of action but in opposite directions, as shown in Figure 14-8. The magnitude of each force is specified as 1/h, where h is the spacing distance between the two forces. We then wish to take the limit as h ! 0, and this type of system is called a force doublet. Recall that this problem was first defined in Chapter 13; see Exercise 13-18. From our previousconstructions, the elastic state for this case is givenby S aa (x) with no sum over a. This formmatches the suggested solution scheme presented in Exercise 13-18. x y z P x x FIGURE 14-7 Generalized Kelvin state. h x α - direction 1/h 1/h FIGURE 14-8 Force doublet state. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 380 380 ADVANCED APPLICATIONS TLFeBOOK EXAMPLE 14-6: Force Doublet with a Moment (About g-Axis) Consider again the case of a double-force system with equal and opposite forces but acting along different lines of action, as shown in Figure 14-9. For this situation the two forces produce a moment about an axis perpendicular to the plane of the forces. Again the magnitudes of the forces are taken to be 1/h, where h is the spacing between the lines of action, and the limit is to be taken as h ! 0. The elastic state for this case is specified by S ab (x), where a 6¼ b, and the resulting moment acts along the g-axis defined by the unit vector e g ¼ e a e b . It can be observed from Figure 14-9 that S ab ¼S ba . From the previous equations (14.2.7), (14.2.12), and (14.2.13), the Papkovich potentials for state S ab (x) are given by f ab ¼C d ab R , c ab i ¼ Cd ai x b R 3 , C ¼ 1 8p(1 n) (14:2:16) and this yields the following displacements and stresses: u ab i ¼ C 2mR 3 3x a x b x i R 2 þ (3 4n)d ai x b d ab x i d bi x a (14:2:17) s ab ij ¼ C R 3 15x a x b x i x j R 4 þ 3(1 2n) R 2 (d ai x b x j þ d aj x b x i d ij x a x b ) 3 R 2 (d bi x a x j þ d bj x a x i þ d ab x i x j ) (1 2n)(d ai d bj þ d aj d bi þ d ij d ab ) (14:2:18) Note the properties of state S ab ¼ {u ab , s ab , e ab }: u ab ¼ O(R 2 ), s ab ¼ O(R 3 ), and ð S T ab dS ¼ 0, ð S R T ab dS ¼ e gab e g for any closed surface S enclosing the origin. h x α - direction 1/h 1/h x β - direction FIGURE 14-9 Double-force system with moment. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 381 Micromechanics Applications 381 TLFeBOOK EXAMPLE 14-7: Center of Compression/Dilatation A center of compression (or dilatation) is constructed by the superposition of three mu- tually perpendicular force doublets, as shown in Figure 14-10. The problem was intro- duced previously in Exercise (13.19). The elastic state for this force system is given by S o (x) ¼ 1 2(1 2n)C S aa (x) (14:2:19) with summation over a ¼ 1, 2, 3. This state is then associated with the following potentials: f o ¼ 3 2(1 2n) 1 R , c o i ¼ x i 2(1 2n) 1 R (14:2:20) and these yield the displacements and stresses u o i ¼ x i 2mR 3 s o ij ¼ 1 R 3 3x i x j R 2 d ij (14:2:21) Note that this elastic state has zero dilatation and rotation. In spherical coordinates the displacements and stresses are given by u o R ¼ 1 2mR 2 , u o y ¼ u o f ¼ 0 s o R ¼ 2 R 3 , s o y ¼ s o f ¼ 1 R 3 , t o Ry ¼ t o Rf ¼ t o yf ¼ 0 (14:2:22) A center of dilatation follows directly from the center of compression with a simple sign reversal and thus can be specified by S o (x). x y z FIGURE 14-10 Center of compression. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 382 382 ADVANCED APPLICATIONS TLFeBOOK EXAMPLE 14-8: Center of Rotation Using the cross-product representation, a center of rotation about the a-axis can be expressed by the state a S(x) ¼ 1 2 e abg S bg (x) (14:2:23) where summation over b and g is implied. Thus, centers of rotation about the coordinate axes can be written as 1 S(x) ¼ 1 2 (S 23 S 32 ) 2 S(x) ¼ 1 2 (S 31 S 13 ) 3 S(x) ¼ 1 2 (S 12 S 21 ) (14:2:24) Using the solution (14.2.16), the potentials for this state become a f ¼ 0, a c i ¼ C 2R 3 e aij x j (14:2:25) with the constant C defined in relation (14.2.16). The corresponding displacements and stresses follow as a u i ¼ 1 8pmR 3 e aij x j a s ij ¼ 3 8pR 5 (e aik x k x j þ e ajk x k x i ) (14:2:26) This state has the following properties: Ð S a TdS ¼ 0, Ð S R a TdS ¼ d ai e i , where the integration is taken over any closed surface enclosing the origin. In order to develop additional singular states that might be used to model distributed singular- ities, consider the following property. Definition: Let S(x;l) ¼ {u(x;l), s(x;l), e(x;l)} be a regular elastic state for each par- ameter l 2 [a, b] with zero body forces. Then the state S defined by S (x) ¼ ð b a S(x;l)dl (14:2:27) is also a regular elastic state. This statement is just another form of the superposition principle, and it allows the construction of integrated combinations of singular elastic states as shown in the next example. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 383 Micromechanics Applications 383 TLFeBOOK EXAMPLE 14-9: Half Line of Dilatation A line of dilatation may be created through the superposition relation (14.2.27) by combining centers of dilatation. Consider the case shown in Figure 14-11 that illustrates a line of dilatation over the negative x 3 -axis. Let S o (x;l) be a center of compression located at (0, 0, l) for all l 2 [0, 1). From our previous definitions, it follows that z S o (x) ¼ ð 1 0 S o (x;l)dl where S o (x;l) ¼S o (x 1 , x 2 , x 3 þ l) (14:2:28) will represent the state for a half line of dilatation along the negative x 3 -axis, that is, x 3 2 [0, 1). Using the displacement solution for the center of compression (14.2.21) in (14.2.28) yields the following displacement field for the problem: z u o 1 ¼ x 1 2m ð 1 0 dl ^ RR 3 z u o 2 ¼ x 2 2m ð 1 0 dl ^ RR 3 z u o 3 ¼ 1 2m ð 1 0 (x 3 þ l)dl ^ RR 3 (14:2:29) which can be expressed in vector form as z u o ¼ 1 2m ð 1 0 = 1 ^ RR d l ¼ 1 2m ð 1 0 dl ^ RR ¼ = log (R þx 3 ) (14:2:30) The potentials for this state can be written as z f o ¼ log (R þx 3 ), z c o i ¼ 0 (14:2:31) x 1 x 2 x 3 Line of Dilatation (0,0,−l) R x R ˆ FIGURE 14-11 Line of dilatation along the negative x 3 -axis. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 384 384 ADVANCED APPLICATIONS TLFeBOOK EXAMPLE 14-9: Half Line of Dilatation–Cont’d Notice the singularity at R ¼x 3 , and of course this behavior is expected along the negative x 3 -axis because of the presence of the distributed centers of dilatation. In spherical coordinates the displacement and stress fields become z u o R ¼ 1 2mR , z u o f ¼ sin f 2mR(1 þ cos f) , z u o y ¼ 0 z s o R ¼ 1 R 2 , z s o f ¼ cos f R 2 (1 þcos f) , z s o y ¼ 1 R 2 (1 þ cos f) z t o Rf ¼ sin f R 2 (1 þcos f) , z t o Ry ¼ z t o yf ¼ 0 (14:2:32) 14.3 Elasticity Theory with Distributed Cracks Many brittle solids such as rock, glass, ceramics, and concretes contain microcracks. It is generally accepted that the tensile and compressive strength of these materials is determined by the coalescence of these flaws into macrocracks, thus leading to overall fracture. The need to appropriately model such behaviors has lead to many studies dealing with the elastic response of materials with distributed cracks. Some studies have simply developed moduli for elastic solids containing distributed cracks; see, for example, Budiansky and O’Connell (1976), Hoenig (1979), and Horii and Nemat-Nasser (1983). Other work (Kachanov 1994) has investigated the strength of cracked solids by determining local crack interaction and propa- gation behaviors. The reviews by Kachanov (1994) and Chau, Wong, and Wang (1995) provide good summaries of work in this field. We now wish to present some brief results of studies that have determined the elastic constants of microcracked solids as shown in Figure 14-12. It is assumed that a locally isotropic elastic material contains a distribution of planar elliptical cracks as shown. Some studies have assumed a random crack distribution, thus implying an overall isotropic response; other investigators have considered preferred crack orientations, giving rise to anisotropic behaviors. Initial research assumed that the crack density is dilute so that crack interaction effects can be neglected. Later studies include crack interaction using the well-established self-consistent approach. In general, the effective moduli are found to depend on a crack density parameter, defined by (Cracked Elastic Solid) (Elliptical Shaped Crack) FIGURE 14-12 Elastic solid containing a distribution of cracks. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 385 Micromechanics Applications 385 TLFeBOOK e ¼ 2N p A 2 P () (14:3:1) where N is the number of cracks per unit volume, A is the crack face area, P is the crack perimeter, and the angle brackets indicate the average value. Space limitations prevent going into details of the various analyses, and thus only effective moduli results are given. Three particular examples are presented, and all cases assume no crack closure. EXAMPLE 14-10: Isotropic Dilute Crack Distribution Consider first the special case of a random dilute distribution of circular cracks of radius a. Note for the circular crack case the crack density parameter defined by (14.3.1) reduces to e ¼ Nha 3 i. Results for the effective Young’s modulus EE, shear modulus mm, and Poisson’s ratio nn are given by EE E ¼ 45(2 n) 45(2 n) þ16(1 n 2 )(10 3n)e mm m ¼ 45(2 n) 45(2 n) þ32(1 n)(5 n)e e ¼ 45(n nn)(2 n) 16(1 n 2 )(10 nn 3n nn n) (14:3:2) where E, m, and n are the moduli for the uncracked material. EXAMPLE 14-11: Planar Transverse Isotropic Dilute Crack Distribution Next consider the case of a dilute distribution of cracks arranged randomly but with all crack normals oriented along a common direction, as shown in Figure 14-13. For this case results for the effective moduli are (Transverse Cracked Solid) FIGURE 14-13 Cracked elastic solid with common crack orientation. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 386 386 ADVANCED APPLICATIONS TLFeBOOK EXAMPLE 14-11: Planar Transverse Isotropic Dilute Crack Distribution–Cont’d EE E ¼ 3 3 þ 16(1 n 2 )e mm m ¼ 3(2 n) 3(2 n) þ16(1 n)e (14:3:3) where EE and mm are the effective moduli in the direction normal to the cracks. A plot of this behavior for n ¼ 0:25 is shown in Figure 14-14. It is observed that both effective moduli decrease with crack density, and the decrease is more pronounced for Young’s modulus. EXAMPLE 14-12: Isotropic Crack Distribution Using Self-Consistent Model Using the self-consistent method, effective moduli for the random distribution case can be developed. The results for this case are given by EE E ¼ 1 16(1 nn 2 )(10 3 nn)e 45(2 nn) mm m ¼ 1 32(1 nn)(5 nn)e 45(2 nn) e ¼ 45(n nn)(2 nn) 16(1 nn 2 )(10n 3n nn nn) (14:3:4) Continued 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6 0.8 1 Crack Density, e Effective Moduli EE / mm / FIGURE 14-14 Effective elastic moduli for transversely cracked solid (n ¼ 0:25). Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 387 Micromechanics Applications 387 TLFeBOOK EXAMPLE 14-12: Isotropic Crack Distribution Using Self-Consistent Model–Cont’d It is interesting to note that as e ! 9=16, all effective moduli decrease to zero. This can be interpreted as a critical crack density where the material will lose its coherence. Although it would be expected that such a critical crack density would exist, the accuracy of this particular value is subject to the assumptions of the modeling and is unlikely to match universally with all materials. In the search for appropriate models of brittle microcracking solids, there has been a desire to find a correlation between failure mechanisms (fracture) and effective elastic moduli. However, it has been pointed out (Kachanov 1990, 1994) that such a correlation appears to be unlikely because failure-related properties such as stress intensity factors are correlated to local behavior, while the effective elastic moduli are determined by volume average procedures. External loadings on cracked solids can close some cracks and possibly produce frictional sliding, thereby affecting the overall moduli. This interesting process creates induced anisotropic behavior as a result of the applied loading. In addition to these studies of cracked solids, there also exists a large volume of work on determining effective elastic moduli for heterogeneous materials containing particulate and/or fiber phases; that is, distributed inclusions. A review of these studies has been given by Hashin (1983). Unfortunately, space does not permit a detailed review of this work. 14.4 Micropolar/Couple-Stress Elasticity As previously mentioned, the response of many heterogeneous materials has indicated depend- ency on microscale length parameters and on additional microstructural degrees of freedom. Solids exhibiting such behavior include a large variety of cemented particulate materials such as particulate composites, ceramics, and various concretes. This concept can be qualitatively illustrated by considering a simple lattice model of such materials as shown in Figure 14-15. Using such a scheme, the macro load transfer within the heterogeneous particulate solid is modeled using the microforces and moments between adjacent particles (see Chang and Ma 1991; Sadd, Qui, Boardman, and Shukla 1992; Sadd, Dai, Parmameswaran, and Shukla 2004b). Depending on the microstructural packing geometry (sometimes referred to as fabric), Network of Elastic Elements (Heterogeneous Elastic Material) = ? Inner Degrees of Freedom (Equivalent Lattice Model) FIGURE 14-15 Heterogeneous materials with microstructure. Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 388 388 ADVANCED APPLICATIONS TLFeBOOK [...]... @y @txy @sy þ ¼0 @x @y @mxz @myz þ þ txy À tyx ¼ 0 @x @y (14: 4:7) The in-plane strains can be expressed as @u @v , ey ¼ @x @y @v @u exy ¼ À fz , eyx ¼ þ fz @x @y (14: 4:8) 1 @v @u fz ¼ À 2 @x @y (14: 4:9) ex ¼ while using (14. 4.6) gives Notice that substituting (14. 4.9) into (14: 4:8)2 gives the result exy ¼ eyx The constitutive equations (14. 4.4) yield the following forms for the stress components:... mb r2 f ¼ 0 lþm For this case relation (14. 5.5) for balance of equilibrated forces reduces to a b mb 2 2 _ ¼ r cÀ f ar f À 2 f À !f h 3l þ 2m lþm (14: 5:11) (14: 5:12) The parameter h is defined by a b2 ¼xÀ h2 lþm (14: 5:13) and has units of length, and thus can then be taken as a microstructural length measure for this particular theory Relations (14. 5.11) and (14. 5.12) now form the governing equations... ¼ mxz ¼ @x @y txy ¼ À (14: 4 :14) where F ¼ F(x, y) and C ¼ C(x, y) are the stress functions for this case If C is taken to be zero, the representation reduces to the usual Airy form with no couple stresses Using form (14. 4 .14) in the compatibility equations (14. 4.13) produces r4 F ¼ 0 @ @ (C À l2 r2 C) ¼ À2(1 À n)l2 (r2 F) @x @y @ @ (C À l2 r2 C) ¼ 2(1 À n)l2 (r2 F) @y @x 392 (14: 4:15) ADVANCED APPLICATIONS... 2m l þ 2m sij ¼ (14: 5:9) where all indices are taken over the limited range 1,2 Using a stress formulation, the single nonzero compatibility relation becomes Continued Micromechanics Applications 399 TLFeBOOK Sadd / Elasticity Final Proof 6.7.2004 6:35pm page 400 EXAMPLE 14- 14: Stress Concentration Around a Circular Hole: Elasticity with Voids–Cont’d skk, mm À mb f ¼0 l þ m , mm (14: 5:10) Introducing... antisymmetric part of the stress tensor disappears from the constitutive relations In order to remedy this, we can solve for the antisymmetric stress term from the moment equilibrium equation (14: 4:7)3 to get 1 1 @mxz @myz g ¼ À r2 fz t[xy] ¼ (txy À tyx ) ¼ À þ 2 2 @x 2 @y (14: 4:11) By cross-differentiation we can eliminate the displacements from (14. 4.8) and (14. 4.9) and develop the particular compatibility... circular or spherical particles A pair of such particles represents a doublet, as shown in Figure 14- 20 Corresponding to the doublet (A,B) there exists a doublet or branch vector z a connecting the adjacent particle centers and defining the doublet axis a The magnitude of this vector Za ¼ jz a j is simply the sum of the two radii for particles in contact However, in general the particles need not be... @y sr ¼ sy try tyr mrz (14: 4:21) Using constitutive relations (14. 4.19), the compatibility equations (14. 4.20) can be expressed in terms of stresses, and combining this result with (14. 4.21) will yield the governing equations for the stress functions in polar coordinates @ 1 @ (C À l2 r2 C) ¼ À2(1 À n)l2 (r2 F) 1 2 @r r @y 1 @ @ (C À l2 r2 C) ¼ 2(1 À n)l2 (r2 F) 1 2 r @y @r (14: 4:22) g(m þ k) g , l2... in-plane element are shown in Figure 14- 16 Notice the similarity of this force system to the microstructural system illustrated previously in Figure 14- 15 For the two-dimensional case with no body forces or body couples, the equilibrium equations (14. 4.3) reduce to sy myz tyx txy mxz sx FIGURE 14- 16 390 Couple stresses on planar element ADVANCED APPLICATIONS TLFeBOOK Sadd / Elasticity Final Proof 6.7.2004... problem reduces to classical elasticity It should also be pointed out that only three of the four equations in set (14. 4.13) are independent because the second relation can be established from the other equations Proceeding along similar lines as classical elasticity, we introduce a stress function approach (Carlson 1966) to solve (14. 4.13) A self-equilibrated form satisfying (14. 4.7) identically can be... 6.7.2004 6:35pm page 401 EXAMPLE 14- 14: Stress Concentration Around a Circular Hole: Elasticity with Voids–Cont’d Àx p A3 (l þ m) þ [F(r) À 1] cos 2y Mb!s þ bx(M À 3) mh2 b ! mb 2 A2 pr 2 pr2 h fþ c¼ þ A1 log r þ 2 þ A3 À cos 2y lþm 4 r 4 f¼ (14: 5:15) where F is given by F(r) ¼ 1 þ ! 4mxh2 1 2hK2 (r=h) þ 0 (l þ m)M!s þ 4mxN r2 a3 K2 (a=h) (14: 5:16) and p is the Laplace . form (14. 4 .14) in the compatibility equations (14. 4.13) produces r 4 F ¼ 0 @ @x (C l 2 r 2 C) ¼2(1 n)l 2 @ @y (r 2 F) @ @y (C l 2 r 2 C) ¼ 2(1 n)l 2 @ @x (r 2 F) (14: 4:15) Sadd / Elasticity. vector e g ¼ e a e b . It can be observed from Figure 14- 9 that S ab ¼S ba . From the previous equations (14. 2.7), (14. 2.12), and (14. 2.13), the Papkovich potentials for state S ab (x) are. ¼ 1 2 (S 12 S 21 ) (14: 2:24) Using the solution (14. 2.16), the potentials for this state become a f ¼ 0, a c i ¼ C 2R 3 e aij x j (14: 2:25) with the constant C defined in relation (14. 2.16). The corresponding
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