ME2004 ELASTICITY Homework # Solution Set | University of California, Davis

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University of California, Davis - ME 2004The Linearized Theory of Elasticity ME2004 ELASTICITY Homework # Solution Set Problem 1.1 The axial stress is σx = P A = 15 × 103 N 14 π(0:01 m) ... 2 = 191:0 MPa ; the axial strain is "x = δ L = 2:86 × 10−2 mm 30 mm = 9:53 × 10−4 ; and the lateral strain is " y = δd d = −3:06 × 10−3 mm 10 mm = −3:06 × 10−4 : Thus, the Young’s modulus is E = σx "x = 200 GPa ; the Poisson’s ratio is ν = − " y "x = 0:321 ; the shear modulus is G = E 2(1 + ν) = 75:7 GPa ; and the bulk modulus is K = E 3(1 − 2ν) = 186:2 GPa : Problem 1.2 Since σ = P=A, " = δ=L, and σ = E", it follows that δ = "L = σL E = P L AE : Problem 1.3 The area of the glued joint is A = A0 sin α ; A0 = (0:1 m)(0:06 m) = 6 × 10−3 m2 ; where A0 is the cross sectional area. The components of the axial load P normal and tangential to the joint are F = P sin α ; V = P cos α ; P = 6 kN : Thus, the normal and shear stresses on the joint are σ = F A = P A0 sin2 α = (1 MPa) sin2 α ; τ = V A = P A0 sin α cos α = (1 MPa) sin α cos α ; 1and the factor of safety is F.S. = min σσU ; ττU  : (a) α = 20◦ σ = 0:117 MPa =) σU σ = 1:26 0:117 = 10:8 ; τ = 0:321 MPa =) τU τ = 1:50 0:321 = 4:67 ; ) F.S. = 4:67 (will fail in shear). (b) α = 35◦ σ = 0:329 MPa =) σU σ = 1:26 0:329 = 3:83 ; τ = 0:470 MPa =) τU τ = 1:50 0:470 = 3:19 ; ) F.S. = 3:19 (will fail in shear). (c) α = 45◦ σ = 0:500 MPa =) σU σ = 1:26 0:500 = 2:52 ; τ = 0:500 MPa =) τU τ = 1:50 0:500 = 3:00 ; ) F.S. = 2:52 (will fail in tension). Problem 1.4 For rod AB the axial load is PAB = 175 − 250 = −75 kN, the length is LAB = 0:75 m, the cross sectional area is AAB = 1 4π(0:075 m)2 = 4:42 × 10−3 m2, and the Young’s modulus is EAB = 105 GPa. Thus, the change in length of AB is δAB = PABLAB AABEAB = (−75 × 103 N)(0:75 m) (4:43 × 10−3 m2)(105 × 109 Pa) = −1:209 × 10−4 m : Similarly, for rod BC, PBC = 175 kN, LAB = 1 m, AAB = 14π(0:05 m)2 = 1:963 × 10−3 m2, EAB = 200 GPa, and the change in length of BC is δBC = PBCLBC ABCEBC = (175 × 103 N)(1 m) (1:963 × 10−3 m2)(200 × 109 Pa) = 4:46 × 10−4 m : (a) δABC = δAB + δBC = 3:25 × 10−4 m (b) δB = δAB = −1:209 × 10−4 m 2Problem 1.5 For the wires AE and CF , the cross sectional area is A = π(0:001 m)2 = 3:1416 × 10−6 m2, the Young’s modulus is E = 200 GPa, and the lengths are LAE = 1:5 m and LCF = 0:8 m. The axial loads, PAE and PCF , are yet to be determined. By considering the free-body diagram of bar AD, it follows that a necessary condition for equilibrium is P = 1 2 PCF + 3 2 PAE : (1) Likewise, the kinematic constraints δAE = 3δCF ; δB = 2δCF (2) follow from the geometry of the system, where δAE and δCF are the changes in length of wires AE and CF , and δB is the deflection of point B. (a) It follows from (2) that PAELAE AE = 3 PCF LCF AE =) PAE = 3LLCF AE PCF = 8 5PCF : It then follows from (1) that P = 1 2 PCF + 3 2 (8 5 PCF ) = 29 10 PCF ; so that, when P = 200 N, PCF = 69:0 N ; PAE = 8 5 PCF = 110:3 N : (b) The change in length of wire CF is δCF = PCF LCF AE = (68:97 N)(0:8 m) (3:1416 × 10−6 m2)(200 × 109 Pa) = 8:78 × 10−5 m : Thus the deflection of point B is δB = 2δCF = 1:756 × 10−4 m : Problem 1.6 For the steel spindle AB, JAB = 1 2 π(0:05 m)4 = 9:8175 × 10−6 m4 and the allowable torque is (TAB)all = (τall)ABJAB cAB = (80 × 106 Pa)(9:8175 × 10−6 m4) 0:05 m = 1:571 × 104 N·m : For the aluminum sleeve CD, JCD = 1 2 π[(0:125 m)4 − (0:1 m)4] = 2:2642 × 10−4 m4 3and the allowable torque is (TCD)all = (τall)CDJCD cCD = (70 × 106 Pa)(2:2642 × 10−4 m4) 0:125 m = 1:268 × 105 N·m : Since, from equilibrium, TAB = TCD = T , the allowable applied torque is Tall = min [(TAB)all ; (TCD)all] = 1:571 × 104 N·m : The total twist φ = φAB + φCD when T = Tall is φ = TallLAB JABGAB + TallLCD JCDGCD = (1:571 × 104 N·m)(0:7 m) (9:8175 × 10−6 m4)(77 × 109 Pa) + (1:571 × 104 N·m)(0:5 m) (2:2642 × 10−4 m4)(26 × 109 Pa) = 1:5880 × 10−2 rad : Thus, the largest angle through which A may be rotated is φall = 1:588 × 10−2 rad = 0:910◦ : Problem 1.7 The bending moment distribution in the beam is M(x) = (Pb Pa LL x ; (L − x) ; if 0 if a ≤≤ xx ≤≤ aL;; where x is the distance measured from A. Integrating d2v dx2 = M(x) EI for the deflection v yields v = (6 6EIL EIL Pa Pb x (L3 +− Cx)13x−+CC32(L; − x) + C4 ; if 0 if a ≤≤ xx ≤≤ aL;; The boundary conditions vjx=0 = 0 and vjx=L = 0 give C2 = C4 = 0 and the requirements that v and dv=dx be continuous at x = a give C1 = −P b(L2 − b2) 6EIL ; C3 = C1 + P ab 2EI : Assume that the location xm of the maximum deflection satisfies 0 ≤ xm ≤ a. Then, setting dv dx = P b 2EIL x2 − P b(L2 − b2) 6EIL (0 ≤ x ≤ a) 4equal to zero and solving for x = xm yields xm = rL2 −3 b2 : Note that, since L − b = a and b < L=2, L2 − b2 3 = a L +3 b < a(L − b) = a2 ; thus showing that the assumption 0 ≤ xm ≤ a is correct. It follows that the maximum deflection, δm = −vjx=xm, is δm = P b(L2 − b2)3=2 9p3EIL : Problem 1.8 The bending moment distribution in the beam is M(x) = wLx 2 − wx2 2 ; where x is the distance measured from A. Integrating d2v dx2 = M(x) EI for the deflection v yields EIv = wLx3 12 − wx4 24 + C1x + C2 : The boundary conditions vjx=0 = 0 and vjx=L = 0 give C1 = −wL3 24 ; C2 = 0 : (a) Thus, the equation of the elastic deflection curve of the beam is v = − wx 24EI (x3 − 2Lx2 + L3) : (b) The slope of the beam at A is dv dx x=0 = − wL3 24EI : (c) The deflection of the beam at its midpoint is vjx=L=2 = −−384 5wL EI4 : Problem 2.1 5(a) For the expression ams = bm(cr − dr) [which can be rewritten as ams = bmcr − bmdr], the index m is a free index since it appears once in each term. However, the indices s and r are inconsistent|they are neither free indices nor do they form dummy index pairs. Thus, there is no unambiguous way to expand this expression. This expression is not a valid indicial notation expression. (b) The expression ams = bm(cs − ds) is a valid indicial notation expression, for which m and s are free indices (giving 32 = 9 equations) and there are no dummy index pairs (so there are 3 terms per equation). The corresponding expanded equations are a11 = b1(c1 − d1) ; a12 = b1(c2 − d2) ; a13 = b1(c3 − d3) ; a21 = b2(c1 − d1) ; a22 = b2(c2 − d2) ; a23 = b2(c3 − d3) ; a31 = b3(c1 − d1) ; a32 = b3(c2 − d2) ; a33 = b3(c3 − d3) : (c) The expression ti = σjinj is a valid indicial notation expression, for which i is a free index (giving 31 = 3 equations) and j is a dummy index pair (giving 1 + 31 = 4 terms per equation). The corresponding expanded equations are t1 = σ11n1 + σ21n2 + σ31n3 ; t2 = σ12n1 + σ22n2 + σ32n3 ; t3 = σ13n1 + σ23n2 + σ33n3 : (d) The expression ti = σjini is not a valid indicial notation expression. The index i looks like a free index on the left-hand side and looks like a dummy index pair on the right-hand side and the index j appears in only the term on the right-hand side. (e) The expression σij = 2µ"ij + λ"kkδij is a valid indicial notation expression, for which i and j are free indices (giving 32 = 9 equations) and k is a dummy index pair (giving 1 + 1 + 31 = 5 terms per equation). The corresponding expanded equations are σ11 = 2µ"11 + λ("11 + "22 + "33)δ11 ; σ12 = 2µ"12 + λ("11 + "22 + "33)δ12 ; σ13 = 2µ"13 + λ("11 + "22 + "33)δ13 ; ... σ33 = 2µ"33 + λ("11 + "22 + "33)δ33 : (f) The expression xixi = r2 is a valid indicial notation expression, with no free indices (giving 30 = 1 equation) and for which i is a dummy index pair (giving 31 + 1 = 4 terms). The corresponding expanded equation is x1x1 + x2x2 + x3x3 = r2 ; or, equivalently, x2 1 + x2 2 + x2 3 = r2 : (g) The expression "rs = hr(ds −hskrr) [which can be rewritten as "rs = hrds −hrhskrr] is not a valid indicial notation expression, because the index r is not consistent. (In the second term on the right-hand side, is r a free index of a dummy index pair? It cannot be both.) 6(h) The expression bijcj = 3 is a valid indicial notation expression, for which i is a free index (giving 31 = 3 equations) and j is a dummy index pair (giving 31 + 1 = 4 terms per equation). Note that this is an example of the one exception to the general rule that a free index must appear once and only once in every term of a valid indicial notation expression| the numerical value \3" on the right-hand side is assumed to carry over to each equation. The corresponding expanded equations are b11c1 + b12c2 + b13c3 = 3 ; b21c1 + b22c2 + b23c3 = 3 ; b31c1 + b32c2 + b33c3 = 3 : Problem 2.2 (a) It follows from the substitution property of the Kronecker delta and the result δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3, that δijδjkδkpδpi = δijδjkδki = δijδji = δii = 3 : (b) From the substitution property of the Kronecker delta, δijeijk = eiik = e11k + e22k + e33k : However, from the definition of the permutation symbol, e11k = e22k = e33k = 0 for all three possible values of k. Thus, δijeijk = 0 : (c) Recalling the definition ‘kj ≡ e^0 k · e^j of the direction cosines, the relation e^i = ‘kie^0 k, and the result e^i · e^j = δij relating the orthonormal base vectors and the Kronecker delta, it follows that ‘ki‘kj = ‘ki(e^0 k · e^j) = (‘kie^0 k) · e^j = e^i · e^j = δij : (d) Consider first the brute force approach. The expression eqrsdqds expands into three expressions (one for each value of r), each with 9 terms. The expression for r = 1 expands as follows: e q1sdqds = e111d1d1 + e112d1d2 + e113d1d3 + e211d2d1 + e212d2d2 + e213d2d3 + e311d3d1 + e312d3d2 + e313d2d3 = −d2d3 + d3d2 = 0 : Similarly, the expressions for r = 2 and r = 3 give e q2sdqds = d1d3 − d3d1 = 0 ; e q3sdqds = d2d1 − d1d2 = 0 : Thus it is shown that e qrsdqds = 0. However, brute force methods like this should be avoided when possible. Here is a more concise proof that demonstrates the utility of indicial notation: e qrsdqds = eqrsdsdq [dqds = dsdq] = −e srqdsdq [eqrs = −esrq] = −e qrsdqds [q ! s and s ! q (dummy index pairs)] = 0 [x = −x () x = 0] 7Alternatively, this result can be shown as a special case of that seen in Example 2.2 on page 38 by letting Aij = didj, which implies that Aij = Aji. Problem 2.3 (a) Taking a brute force approach, e123 = det 2 4δ δ δ11 21 31 δδ δ12 22 32 δ δ δ13 23 333 5 = det 2 41 0 0 0 1 0 0 0 13 5 = 1 : Similarly, e231 = det 2 40 1 0 0 0 1 1 0 03 5 = 1 ; e312 = det 2 40 0 1 1 0 0 0 1 03 5 = 1 : Likewise, e321 = det 2 40 0 1 0 1 0 1 0 03 5 = −1 ; e213 = det 2 40 1 0 1 0 0 0 0 13 5 = −1 ; e132 = det 2 41 0 0 0 0 1 0 1 03 5 = −1 : Finally, e112 = det 2 41 0 0 1 0 0 0 1 03 5 = 0 ; : : : ; the remaining combinations give a value of zero. Thus, the relation is established. (b) Using the result from (a), eijkepqr = det 24 δi1 δi2 δi3 δj1 δj2 δj3 δk1 δk2 δk3 35 det 2 4δδδpqr111 δδδpqr2 22 δδδpqr3 333 5 = det 2 4δδδkji111 δδδkji222 δδδkji3333 5 det 2 4δ δ δp p p1 2 3 δ δ δq q q1 2 3 δ δ δr r r1 2 33 5 = det 0 @2 4δδδkji111 δδδkji222 δδδkji3333 5 2 4δ δ δp p p1 2 3 δ δ δq q q1 2 3 δ δ δr r r1 2 33 51 A = det 2 4δδδkji111δδδppp111 +++δδδkji222δδδppp222 +++δδδijk333δδδppp333 δδδkji111δδδqqq111 +++δδδkji222δδδqqq222 +++δδδijk333δδδqqq333 δδδkji111δδδrrr111 +++δδδkji222δδδrrr222 +++δδδijk333δδδrrr3333 5 = det 2 4δδδks js isδδδps ps ps δδδks js isδδδqs qs qs δδδks js isδδδrs rs rs3 5 = det 2 4δδδkp jp ip δδδkq jq iq δδδkr jr ir3 5 : 8(c) Setting p ! i in (b), eijkeiqr = det 24 δii δiq δir δji δjq δjr δki δkq δkr 35 = δiiδjqδkr + δiqδjrδki + δirδjiδkq − δirδjqδki − δiqδjiδkr − δiiδjrδkq = 3δjqδkr + δkqδjr + δjrδkq − δkrδjq − δjqδkr − 3δjrδkq = δjqδkr − δjrδkq : (d) Setting q ! j in (c), eijkeijr = δjjδkr − δjrδkj = 3δkr − δkr = 2δkr : (e) Setting r ! k in (d), eijkeijk = 2δkk = 2 · 3 = 6 : Problem 2.4 (a) Given that det[A] = eijkA1iA2jA3k and using the (2.2.25), it follows that elmn det[A] = eijkelmnA1iA2jA3k = (δilδjmδkn + δimδjnδkl + δinδjlδkm − δinδjmδkl − δimδjlδkn − δilδjnδkm)A1iA2jA3k = A1lA2mA3n + A1mA2nA3l + A1nA2lA3m − A1nA2mA3l − A1mA2lA3n − A1lA2nA3m = A1lA2mA3n + A3lA1mA2n + A2lA3mA1n − A3lA2mA1n − A2lA1mA3n − A1lA3mA2n = eijkAilAjmAkn : (b) Multiplying both sides of (a) by elmn and recalling that elmnelmn = 6, if follows that det[A] = 1 6 eijkelmnAilAjmAkn : Problem 2.5 (a) Given that, in an orthonormal basis fe^ig, one has from (2.2.20) that e^i × e^j = eijpe^p, it follows that e^i × e^j · e^k = eijpe^p · e^k = eijpδpk = eijk : (b) Using (a), and being careful not to use the same dummy index pair more than once in a term, u × v · w = (uie^i) × (vje^j) · (wke^k) = uivjwk(e^i × e^j · e^k) = eijkuivjwk : Problem 2.6 9(a) One way to determine the scalar components of v = A · u in the vector basis fe^ig is to use matrix multiplication: [v] = [A][u] = 24 5 0 0 −4 0 0 0 0 2 35 24 −2 03 35 = 24 −10 86 35 : It follows that the dyadic representation of v in the vector basis fe^ig is v = −10e^1 + 8e^2 + 6e^3 : (b) The matrix [L] of direction cosines ‘ij ≡ e^0 i · e^j for this transformation is [L] = 24 cos 30◦ − sin 30◦ 0 sin 30◦ cos 30◦ 0 0 0 1 35 = 264 p3 2 − 12 0 12 p3 2 0 0 0 1 375 : Thus, [A]0 = [L][A][L]T = 264 15 4 + p3 1 + 5p4 3 0 −3 + 5p3 4 54 − p3 0 0 0 2 375 ; [u]0 = [L][u] = 2 4−−p3133 5 ; [v]0 = [L][v] = 2 4− −4 5 + 4 −65pp3 33 5 : (c) Since [A]0[u]0 = 264 15 4 + p3 1 + 5p4 3 0 −3 + 5p3 4 54 − p3 0 0 0 2 375 24 −p3 −1 3 35 = 24 −4 − 5p3 −5 + 4p3 6 35 ; it is seen that [A]0[u]0 = [v]0. (d) The dyadic representations of A, u, and v in the vector basis fe^0 ig are A = 15 4 + p3!e^0 1e^0 1 + 1 + 5p4 3!e^0 1e^0 2 + − 3 + 5p4 3!e^0 2e^0 1 + 5 4 − p3!e^0 2e^0 2 + 2e^0 3e^0 3 ; u = −p3e^0 1 − e^0 2 + 3e^0 3 ; v = (−4 − 5p3)e^0 1 + (−5 + 4p3)e^0 2 + 6e^0 3 : Problem 2.7 10In an orthonormal basis fe^ig, u × (v × w) = uie^i × (vje^j × wke^k) = uivjwke^i × (e^j × e^k) = ejkruivjwke^i × e^r = ejkreirsuivjwke^s = ejkresiruivjwke^s = (δjsδki − δjiδks)uivjwke^s = (uiwi)(vje^j) − (uivi)(wke^k) = (w · u)v − (v · u)w : Problem 2.8 Using the identity u × (v × w) = (w · u)v − (v · u)w ; let u = w = µ^: µ^ × (v × µ^) = (µ^ · µ^)v − (v · µ^)µ^ : Since µ^ · µ^ = 1, it follows that v = (v · µ^)µ^ + µ^ × (v × µ^) : Since v · µ^ is the orthogonal projection of v onto the direction of µ^, it follows that (v · µ^)µ^ and µ^ × (v × µ^) are the vector components of v parallel and orthogonal to µ^, respectively. Problem 2.9 Recalling the definition for the transpose of a second-order tensor, it follows that u · A · u = 1 2 (u · A · u + u · A · u) [trivially] = 1 2 (u · A · u + u · AT · u) [u · A · v = v · AT · u and v ! u] = 1 2 u · (A + AT) · u : Thus, it is seen that u · A · u = 0 for all vectors u if and only if A + AT = 0. Problem 2.10 Using an orthonormal basis fe^ig, (a) (u · A) · v = (uiAije^j) · v = uiAijvj u · (A · v) = u · (Aijvje^i) = uiAijvj (b) (A · B) · u = (AikBkje^ie^j) · u = AikBkjuje^i A · (B · u) = A · (Bkjuje^k) = AikBkjuje^i 11(c) u · (A · B) = u · (AikBkje^ie^j) = uiAikBkje^j (u · A) · B = (uiAike^k) · B = uiAikBkje^j Problem 2.11 (a) Recalling that the dot product of two second-order tensors is another second-order tensor and that the (pre- or post-) dot product of a vector and a second-order tensor is a vector, the following relations hold for all vectors u and v and all second-order tensors A and B: u · (A · B)T · v = v · (A · B) · u [definition of the transpose] = (v · A) · (B · u) [identity] = (B · u) · (v · A) [a · b = b · a for all vectors a and b] = (u · BT) · (AT · v) [property of the transpose] = u · (BT · AT) · v [identity] : Since u and v are arbitrary, it follows that (A · B)T = BT · AT : Alternatively, in terms of scalar components in an orthonormal vector basis fe^ig, where A = Aije^ie^j and B = Bije^ie^j: (A · B)T = (AikBkje^ie^j)T = AjkBkie^ie^j [(Tije^ie^j)T = Tjie^ie^j] = (Bkie^ie^k) · (Ajle^le^j) = BT · AT (b) It follows similarly that (A · u) · (B · v) = (u · AT) · (B · v) [property of the transpose] = u · (AT · B) · v [identity] : (c) Note first that (A · B) · (B−1 · A−1) = A · B · B−1 · A−1 = A · A−1 = I ; (B−1 · A−1) · (A · B) = B−1 · A−1 · A · B = B−1 · B = I : From the definition of the inverse of a second-order tensor, C · D = D · C = I if and only if D = C−1. Thus, it is seen that B−1 · A−1 = (A · B)−1 : 12(d) Note first that, since (A · B)T = BT · AT, (A · A−1)T = (A−1)T · AT : However, A · A−1 = I and IT = I, so it follows that (A−1)T · AT = I : Thus, (A−1)T is the inverse of AT: (A−1)T = (AT)−1 : Problem 2.12 (a) Multiplying both sides of bi = 1 2 eijkAkj by eqpi gives eqpibi = 1 2 eqpieijkAkj = 1 2 eqpiejkiAkj [eijk = ejki] = 1 2 (δqjδpk − δqkδpj)Akj [identity] = 1 2 (Apq − Aqp) [substitution property] = A pq [Aqp = −Apq (skew-symmetry of A)] : (b) Noting from Apq = eqpibi that A11 = e111b1 + e112b2 + e113b3 = 0 ; A12 = e211b1 + e212b2 + e213b3 = −b3 ; A13 = e311b1 + e312b2 + e313b3 = b2 ; ... it follows that [A] = 24 0 −b3 b2 b3 0 −b1 −b2 b1 0 35 : (c) In an orthonormal basis fe^ig, in which A = Aije^ie^j, b = bie^i, and c = cie^i, one has that b × c = eiqpbicqe^p and A · c = A pqcqe^p = eqpibicqe^p [Apq = eqpibi] = eiqpbicqe^p [eqpi = eiqp] : Thus, it is seen that A · c = b × c. 13Problem 2.14 Given that Aij = Aji, NeA(^ µ1; µ^2; µ^3; λ) = Aijµ^iµ^j − λ(^ µiµ^i − 1) ; and @µ^i=@µ^k = δik, it follows that @NeA @µ^k = Aij @@µµ ^^ki µ^j + ^ µi @@µµ^^kj  − λ @@µµ ^^ki µ^i + ^ µi @@µµ ^^ki  = Aij(δikµ^j + ^ µiδjk) − λ(δikµ^i + ^ µiδik) = Akjµ^j + Aikµ^i − 2λµ^k = Akjµ^j + Ajkµ^j − 2λµ^k = (Akj + Ajk)^ µj − 2λµ^k = 2Akjµ^j − 2λµ^k = 2(Akjµ^j − λµ^k) : Thus, @NeA=@µ^k = 0 is satisfied if and only if Akjµ^j = λµ^k : Problem 2.15 Given that C = FT · F, it follows from the properties of the transpose that CT = (FT · F)T = FT · (FT)T = FT · F = C : Thus, C is symmetric. One also has that, for a vector u, u · C · u = u · (FT · F) · u = (u · FT) · (F · u) = (F · u) · (F · u) = jF · uj2 : Since jF · uj2 ≥ 0, where jF · uj2 = 0 if and only if F · u = 0, it follows that u · C · u ≥ 0 8u ; and u · C · u = 0 if and only if F · u = 0. It F is nonsingular, then F · u = 0 if and only if u = 0 and, therefore, C is positive definite. On the other hand, if F is singular, then there exists at least one vector a 6= 0 such that F · a = 0 and, therefore, a · C · a = 0. Thus, in this case, C is positive semidefinite. Problem 2.16 Recall that, for any symmetric second-order tensor A and any positive integer n, one has the spectral representation An = 3X k =1 λ(k)n µ^(k)µ^(k) ; where λ(k) and µ^(k) are the eigenvalues and corresponding (orthonormal) eigenvectors of A. It follows that A3 − IAA2 + IIAA − IIIAI = 3X k =1 λ(k)3 − IA λ(k)2 + IIA λ(k) − IIIA µ^(k)µ^(k) : 14Since, for each value of k, λ(k)3 − IA λ(k)2 + IIA λ(k) − IIIA = 0 ; it then follows that A3 − IAA2 + IIAA − IIIAI = 0 : Problem 2.18 (a) Given that F = aijxixj, @F @xr = aij @xi @xr xj + aijxi @xj @xr = aijδirxj + aijxiδjr = arjxj + airxi = (ari + air)xi : (b) Given the result from part (a), @2F @xr@xs = @ @xs @x @Fr  = (ari + air) @xi @xs = (ari + air)δis = ars + asr : Problem 2.19 The following identities can most easily be established in a Cartesian coordinate system. If they are shown to hold in one coordinate system, then they must be true in every other coordinate system, whether Cartesian or not. (a) r(r2ζ) = (ζ;ii );j e^j = ζ;iij e^j = (ζ;j e^j);ii = r2(rζ) : (b) r(v · x) = (vixi);j e^j = vixi;j e^j + vi;j xie^j = viδije^j + xkvi;j δkie^j = vie^i + xke^k · vi;j e^ie^j [e^k · e^ie^j = (e^k · e^i)e^j = δkie^j] = v + x · (rv) : 15(c) r · (r × v) = (eijkvj;i e^k);p ·e^p = eijkvj;ip e^k · e^p = eijkvj;ip δkp = eijkvj;ik = 0 : [since vj;ik = vj;ki] (d) r · (r2v) = (v;ii );j ·e^j = (v;j ·e^j);ii = r2(r · v) : (e) r · (ζv) = (ζv);i ·e^i = ζv;i ·e^i + ζ;i v · e^i = ζ(v;i ·e^i) + (ζ;i e^i) · v = ζr · v + (rζ) · v : (f) r × (r2v) = e^i × (v;jj );i = e^i × (v;i );jj = (e^i × v;i );jj = r2(r × v) : (g) r2(ζx) = (ζx);ii = (ζx;i +ζ;i x);i = (ζe^i + ζ;i x);i = ζ;i e^i + ζ;i x;i +ζ;ii x = 2ζ;i e^i + xζ;ii = 2rζ + xr2ζ : (h) r2(v · x) = (v · x);ii = (v · x;i +v;i ·x);i = (v · e^i + v;i ·x);i = v;i ·e^i + v;i ·x;i +v;ii ·x = 2v;i ·e^i + x · v;ii = 2r · v + x · r2v : 16(i) r × (r × v) = e^p × (eijkvj;i e^k);p = (e^p × e^k)eijkvj;ip = (epkqe^q)eijkvj;ip = eqpkeijkvj;ip e^q = (δqiδpj − δqjδpi)vj;ip e^q = vj;ij e^i − vj;ii e^j = (vj;j );i e^i − (vje^j);ii = r(r · v) − r2v : Problem 2.20 Note first the following properties of dyads. It follows from the definition (2.3.31) of a dyad that for any three vectors u, v, and w, the dyad (u + v)w formed by the vectors u + v and w is related to the dyads uw and vw by (u + v)w = uw + vw : Similarly, w(u + v) = wu + wv : Likewise, for any two vectors u and v and any scalar α, (αu)v = u(αv) = αuv : It follows that, in a Cartesian coordinate system with a = aie^i and b = bie^i, one has that ab = aibje^ie^j and (ab);i = (apbqe^pe^q);i = a p;i bqe^pe^q + apbq;i e^pe^q = (ape^p);i (bqe^q) + (ape^p)(bqe^q);i = a;i b + ab;i : Thus, r2(ab) = (ab);ii = (a;i b + ab;i );i = a;ii b + ab;ii +2a;i b;i = a;ii b + ab;ii +2a;i b;j δij = a;ii b + ab;ii +2a;i b;j (e^i · e^j) = a;ii b + ab;ii +2(a;i e^i) · (e^jb;j ) = a;ii b + ab;ii +2(a;i e^i) · (b;j e^j)T = (r2a)b + a(r2b) + 2ra · (rb)T : Problem 2.21 Note first that, in Cartesian coordinates, R = (xixi)12 ; x = xie^i ; 17while, in a spherical coordinate system, R is one of the coordinate variables and x = Re^R : (c) In Cartesian coordinates: r · (Rnxx) = [(xixi)n2 (xje^j)(xke^k)];p ·e^p = [(xixi)n2 xjxk];p e^je^k · e^p = hn2 (xixi)n2 −1(xqxq);p xjxk + (xixi)n2 (δjpxk + xjδkp)i e^j(e^k · e^p) = hn2 (xixi)n−2 2 (2δqpxq)xjxk + (xixi)n2 (δjpxk + xjδkp)i e^jδkp = nRn−2xpxjxkδkp + Rn(δjpxkδkp + xjδkpδkp) e^j = nRn−2(xpxp)xj + Rn(xj + 3xj) e^j = nRn−2R2xj + 4Rnxj e^j = (n + 4)Rnxje^j = (n + 4)Rnx In spherical coordinates, if one lets S = Rnxx then, since x = Re^R, it follows that S = Rn+2e^Re^R and the only nonzero component of S = Rnxx is SRR = Rn+2. Then, using equation (2.5.55) for the divergence of a second-order tensor, r · (Rnxx) = r · S = @S @RRR + R2 SRR e^R = (n + 2)Rn+1 + 2Rn+1 e^R = (n + 4)Rn(Re^R) = (n + 4)Rnx Problem 2.22 The spherical coordinate components of h = αR−2e^R are hR = αR−2 ; hθ = 0 ; hφ = 0 : It follows from (2.5.48) that rh = −2αR−3e^Re^R + αR−3e^θe^θ + αR−3e^φe^φ : It follows from (2.5.49) that r · h = −2αR−3 + 2αR−3 = 0 : It follows trivially from (2.5.58) that r × h = 0 : Problem 2.24 Recall the definition (2.4.41) for the curl of a second-order tensor ", which can be written as r × " = e qpr"ip;q e^re^i : 18Let S = r × " so that Sri = eqpr"ip;q. Then, r × (r × ") = r × S = ejikSri;j e^ke^r = ejikeqpr"ip;qj e^ke^r = eijkepqr"ip;jq e^ke^r ; r × (r × ") = eijkepqr"ip;jq e^ke^r : The Cartesian coordinate indicial notation form of r × (r × ") = 0 is therefore eijkepqr"ip;jq = 0 ; which represents 9 equations with 34 + 1 = 82 terms per equation (all but 4 of which are zero). The expanded equations are as follows: (k; r) = (1; 1) : 0 = e231e231"22;33 +e231e321"23;32 +e321e231"32;23 +e321e321"33;22 = "22;33 −"23;32 −"32;23 +"33;22 ; (k; r) = (1; 2) : 0 = e231e312"23;31 +e231e132"21;33 +e321e312"33;21 +e321e132"31;23 = "23;31 −"21;33 −"33;21 +"31;23 ; (k; r) = (1; 3) : 0 = e231e123"21;32 +e231e213"22;31 +e321e123"31;22 +e321e213"32;21 = "21;32 −"22;31 −"31;22 +"32;21 ; (k; r) = (2; 1) : 0 = e312e231"32;13 +e312e321"33;12 +e132e231"12;33 +e132e321"13;32 = "32;13 −"33;12 −"12;33 +"13;32 ; (k; r) = (2; 2) : 0 = e312e312"33;11 +e312e132"31;13 +e132e312"13;31 +e132e132"11;33 = "33;11 −"31;13 −"13;31 +"11;33 ; (k; r) = (2; 3) : 0 = e312e123"31;12 +e312e213"32;11 +e132e123"11;32 +e132e213"12;31 = "31;12 −"32;11 −"11;32 +"12;31 ; (k; r) = (3; 1) : 0 = e123e231"12;23 +e123e321"13;22 +e213e231"22;13 +e213e321"23;12 = "12;23 −"13;22 −"22;13 +"23;12 ; (k; r) = (3; 2) : 0 = e123e312"13;21 +e123e132"11;23 +e213e312"23;11 +e213e132"21;13 = "13;21 −"11;23 −"23;11 +"21;13 ; (k; r) = (3; 3) : 0 = e123e123"11;22 +e123e213"12;21 +e213e123"21;12 +e213e213"22;11 = "11;22 −"12;21 −"21;12 +"22;11 ; If " is symmetric, so that "ij = "ji, then the equations for (k; r) = (1; 2), (2; 3), and (3; 1) are the same as those for (k; r) = (2; 1), (3; 2), and (1; 3), respectively. The six independent equations 19reduce to "11;22 +"22;11 −2"12;12 = 0 ; "22;33 +"33;22 −2"23;23 = 0 ; "33;11 +"11;33 −2"31;31 = 0 ; ("12;3 −"23;1 +"31;2 );1 −"11;23 = 0 ; ("23;1 −"31;2 +"12;3 );2 −"22;31 = 0 ; ("31;2 −"12;3 +"23;1 );3 −"33;12 = 0 : Problem 3.8 The deformed configurations are shown below. Note that, for displacement fields (a) and (b), straight lines remain straight. To see this, consider displacement field (a). Since x = X + u, it follows in this case that x1 = X1 + κX2 ; x2 = X2 + κX1 ; and, hence, that dx1 = dX1 + κ dX2 dx2 = dX2 + κ dX1) =) dx dx21 = κ((dX dX22=dX =dX11) + ) + 1 κ : Thus, dx2=dx1 is constant whenever dX2=dX1 is constant. For displacement (c), one has that x1 = X1 ; x2 = X2 + κX12 ; from which it follows that dx1 = dX1 dx2 = dX2 + 2κx1 dX1) =) dx dx2 1 = dX dX2 1 + 2κx1 : It follows that dx2=dx1 becomes unbounded as dX2=dX1 becomes unbounded, which implies that vertical lines are deformed into vertical lines. Otherwise, dx2=dx1 is not constant when dX2=dX1 is constant. Thus, except for vertical lines, straight lines do not remain straight after deformation. x 1 x1 x1 x 2 x2 x2 (0,0) (0,0) (0,0) (1,κ) (1,κ) (1,κ) (1+ κ,1+ κ) (κ,1) (1−κ,1+ κ) (−κ,1) (1,1+ κ) (0,1) (a) (b) (c) Problem 3.9 20(a) One could determine the Green-St. Venant (Lagrangian) strain tensor E by noting that the deformation x = χ(X) is given by x = X + u, determining the deformation gradient tensor F = (@xi=@XA)e^ie^A, then the right Cauchy-Green deformation tensor C = FT ·F, and finally the strain tensor E = 1 2(C − I). However, it is probably more efficient in this case to use the strain-displacement relation (3.2.48). Noting first that the matrix of scalar components of the displacement gradient is [Grad u] = 264 @u1 @X1 @u1 @X2 @u1 @X3 @u2 @X1 @u2 @X2 @u2 @X3 @u3 @X1 @u3 @X2 @u3 @X3 375 = 24 0 κ 0 κ 0 0 0 0 0 35 ; (3) it follows from (3.2.48) that, in matrix notation, [E] = 1 2 n[Grad u] + [Grad u]T + [Grad u]T [Grad u]o = 1 2 8<: 24 0 κ 0 κ 0 0 0 0 0 35 + 24 0 κ 0 κ 0 0 0 0 0 35 + 24 0 κ 0 κ 0 0 0 0 0 35 24 0 κ 0 κ 0 0 0 0 0 35 9=; = 1 2 8<: 24 0 κ 0 κ 0 0 0 0 0 35 + 24 0 κ 0 κ 0 0 0 0 0 35 + 24 κ2 0 0 0 κ2 0 0 0 0 35 9=; = 24 12 κ2 κ 0 κ 1 2κ2 0 0 0 0 35 ; Thus, E = 1 2 κ2(e^1e^1 + e^2e^2) + κ(e^1e^2 + e^2e^1) : The infinitesimal strain and rotation tensors can similarly be determined using matrix notation (where recall that ru = Grad u): ["] = 1 2 n[ru] + [ru]To = 2 4κ00 0 0 κ0 003 5 ; [!] = 1 2 n[ru] − [ru]To = 2 40 0 0 0 0 0 0 0 03 5 : Thus, " = κ(e^1e^2 + e^2e^1) ; ! = 0 ; θ = 0 : The scalar components of the displacement gradient (3) are all much less than one (i.e., the deformation is infinitesimal) if κ  1, in which case it is seen that E ≈ ". (b) The exact longitudinal strain in the reference material direction e^1 is given by (3.2.13b): H(e^1) = (1 + 2e^1 · E · e^1)1=2 − 1 = (1 + 2E11)1=2 − 1 = (1 + κ2)1=2 − 1 = 1 2 κ2 − 1 8 κ4 + 3 24 κ6 − · · · 21(c) The approximate longitudinal strain in the reference material direction e^1 is given by (3.4.24): H(e^1) ≈ e^1 · " · e^1 = "11 = 0 : Note that this is consistent with the exact result in the case of infinitesimal deformation (i.e., when κ  1). (d) Consider a material line that lies along the X1-axis in the reference configuration, with end points A and B at (X1; X2; X3)A = (0; 0; 0) and (X1; X2; X3)B = (L; 0; 0). The reference length of this material line is L. The corresponding displacements are uA = 0 and uB = κLe^2, so the end points of the material line in the deformed configuration are at (x1; x2; x3)A = (0; 0; 0) and (x1; x2; x3)B = (L; kL; 0). Thus, the deformed length of the material line is ‘ = [L2 + (κL)2]1=2 = L(1 + κ2)1=2 : Thus, the change in length over the original length is ‘ − L L = L(1 + κ2)1=2 − L L = (1 + κ2)1=2 − 1 ; which is just the exact longitudinal strain determined in part (b). Problem 3.10 (a) Since ["] = 24 5 0 0 0 −3 0 0 0 1 35 × 10−3 is a diagonal matrix, it follows immediately that the diagonal elements are the principal strains and the base vectors are the corresponding principal directions of strain. Following the labelling convention that "(1) ≥ "(2) ≥ "(3), the principal strains are "(1) = 5 × 10−3 ; "(2) = 1 × 10−3 ; "(3) = −3 × 10−3 ; and the corresponding principal directions of strain are υ^(1) = e^1 ; υ^(2) = e^3 ; υ^(3) = −e^2 : Note that υ^(3) = −e^2, not +e^2, so that the principal directions form a right-handed system, that is, υ^(1) × υ^(2) = υ^(3). (b) Recall that the orthogonal shear strain is given approximately by γ(υ^0; υ^00) ≈ 2υ^0 · " · υ^00 : Thus, with υ^0 = (e^1 + e^2 + e^3)=p3 and υ^00 = (−e^1 + e^3)=p2, it follows that γ = 2p13p1210−3 1 1 1 2 45 0 0 0 0 0 1 −3 03 5 2 4−0 113 5 = p26 × 10−3 1 1 1 2 4−0 153 5 = − 8 p 6 × 10−3 : 22Problem 3.11 The time-dependent displacement field u = f(X · p^ − ct)d^ describes plane elastic waves that propagate in the direction of p^ with motion of material points in the direction of d^. The function f characterizes the wave shape. A harmonic wave, for instance, would have f(x) = sin x. Using indicial notation in a Cartesian coordinate system, the displacement field is given by ui = f(Xjp^j − ct)d^i : It follows from the chain rule that ui;r = f0(Xjp^j − ct)(Xkp^k − ct);r d^i = f0(X · p^ − ct)Xk;r p^kd^i = f0(X · p^ − ct)δkrp^kd^i = f0(X · p^ − ct)^ prd^i : We can use this result below. (a) Noting that e = ui;i = f0(X · p^ − ct)^ pid^i = f0(X · p^ − ct)p^ · d^ ; one can see that the motion is isochoric if p^ and d^ are orthogonal, so that p^ · d^ = 0. This is known as a shear wave. (b) Noting that θ = 1 2 eijkuj;i e^k = 1 2 f0(X · p^ − ct)eijkp^id^je^k = 1 2 f0(X · p^ − ct)p^ × d^ ; one can see that the motion is irrotational if p^ and d^ are parallel, so that p^ × d^ = 0. This is known as a pressure wave. Problem 3.12 The scalar components of the displacement field u = are^r + brze^θ + c sin θe^z are ur = ar ; uθ = brz ; uz = c sin θ : Thus, using the expressions (3.5.5) for the scalar components of the infinitesimal strain field in cylindrical coordinates, "rr = @ur @r = a "θθ = 1 r @u @θθ + ur = a "zz = @uz @z = 0 "rθ = 12 1r @u @θr + @u @rθ − urθ  = 0 "θz = 12 @u @zθ + 1r @u @θz  = 12 br + rc cos θ "zr = 12 @u @zr + @u @rz  = 0 23In order for the approximate physical interpretations of this strain field to be valid, the scalar components of the displacement gradient must all be much less than one. Using (2.5.19), the displacement gradient is ru = ae^re^r − bze^re^θ + bze^θe^r + ae^θe^θ + bre^θe^z + c r cos θe^ze^θ : Thus, noting that j cos θj ≤ 1, the requirement is that jaj  1 ; jbzj  1 ; jbrj  1 ; jc=rj  1 : Problem 3.13 Using (2.5.49), it follows that the dilatation is given in a spherical coordinate system by e = r · u = @uR @R + 1 R @u @θθ + 2uR + uθ cot θ + R sin 1 θ @u @φφ : Since u = a3R−2 ln(R=a) sin θ cos φ e^R, uR = a3R−2 ln(R=a) sin θ cos φ ; uθ = uφ = 0 ; and e = @uR @R + 2 R uR = −2a3R−3 ln(R=a) sin θ cos φ + a3R−3 sin θ cos φ + 2a3R−3 ln(R=a) sin θ cos φ = a3R−3 sin θ cos φ : Alternatively, you could determine "RR, "θθ, and "φφ and, hence, e = "RR + "θθ + "φφ, but this is a bit more work. Problem 3.14 r × ! = eijk!rj;i e^ke^r = 1 2 eijk(ur;ji −uj;ri )e^ke^r [!rj = 1 2 (ur;j −uj;r )] = − 1 2 eijkuj;ri e^ke^r [eijkur;ji = 0] = −θk;r e^ke^r [θk = 1 2 eijkuj;i ] = −rθ Problem 3.15 The Cartesian scalar components of strain in a simply connected body are given by "11 = AX22 ; "22 = AX12 ; "12 = BX1X2 ; "31 = "32 = "33 = 0 ; where A and B are constants. 24(a) Of the six compatibility equations (3.4.45), only the first is not trivially satisfied: "11;22 +"22;11 −2"12;12 = 2A + 2A − 2B = 0 : Thus, compatibility requires that B = 2A, in which case the components of strain are "11 = AX22 ; "22 = AX12 ; "12 = 2AX1X2 ; "31 = "32 = "33 = 0 : (b) Determining the corresponding displacement field requires solving the six independent scalar partial differential equations given by "ij = 1 2(ui;j +uj;i ). Since "11 = u1;1 = AX22, it follows that u1 = AX1X22 + f(X2; X3) ; (4) where f(X2; X3) is an as-yet-to-be-determined function. Similarly, since "22 = u2;2 = AX12, u2 = AX12X2 + g(X1; X3) ; (5) and, since "33 = u3;3 = 0, u3 = h(X1; X2) : (6) Three of the partial differential equations have now been used. The remaining three will be used to determine the functions f(X2; X3), g(X1; X3), and h(X1; X2) to within an arbitrary rigid motion. Since "12 = 12(u1;2 +u2;1 ) = 2AX1X2, one has that u1;2 +u2;1 = 4AX1X2 : By substituting (4) and (5) for u1 and u2, respectively, one finds that f;2 (X2; X3) + g;1 (X1; X3) = 0 =) f;2 (X2; X3) = −g;1 (X1; X3) = ξ(X3) ; where ξ(X3) is a function that remains to be determined. It follows that f(X2; X3) = X2ξ(X3) + ζ(X3) ; (7) g(X1; X3) = −X1ξ(X3) + η(X3) ; (8) where ζ(X3) and η(X3) also remain to be determined. Combining (4){(8), the results so far are u1 = AX1X22 + X2ξ(X3) + ζ(X3) ; (9) u2 = AX12X2 − X1ξ(X3) + η(X3) ; (10) u3 = h(X1; X2) : (11) There are two equations left to consider. It follows by substituting (10) and (11) into "23 = 12(u2;3 +u3;2 ) = 0 that − X1ξ0(X3) + η0(X3) + h;2 (X1; X2) = 0 =) X1ξ0(X3) − η0(X3) = h;2 (X1; X2) = φ(X1) ; 25and, consequently, that ξ(X3) = aX3 + b ; η(X3) = cX3 + d ; h(X1; X2) = aX1X2 − cX2 + (X1) ; where (X1) is function a that remains to be determined. The results so far are u1 = AX1X22 + aX2X3 + bX2 + ζ(X3) ; (12) u2 = AX12X2 − aX1X3 − bX1 + cX3 + d ; (13) u3 = aX1X2 − cX2 + (X1) : (14) Finally, by substituting (12) and (14) into "31 = 1 2(u3;1 +u1;3 ) = 0, it follows that 0(X1) + ζ0(X3) = −2aX2 =) a = 0 ; 0(X1) = −ζ0(X3) = e ; and, hence, that (X1) = eX1 + f ; ζ(X3) = −eX3 + g : Thus, the most general form of the displacement field is u1 = AX1X22 + bX2 − eX3 + g ; u2 = AX12X2 − bX1 + cX3 + d ; u3 = eX1 − cX2 + f : Since they do not appear in the expressions for the scalar components of strain, the constants of integration b, c, d, e, f, and g correspond to an arbitrary rigid-body motion. As seen below, b, c, and e correspond to a rigid rotation, while d, f, and g characterize a rigid translation. (c) Clearly, ujX=0 = 0 if and only if d = f = g = 0. Noting that !12 = 1 2 (u1;2 −u2;1 ) = b ; !23 = 1 2 (u2;3 −u3;2 ) = c ; !31 = 1 2 (u3;1 −u1;3 ) = e ; it follows that !jX=0 = 0 if and only if b = c = e = 0. Thus, the displacement field in this case is given by u1 = AX1X22 ; u2 = AX12X2 ; u3 = 0 : Problem 4.2 (a) The scalar components of the traction t = n^ · σ acting on the surface with unit outward normal n^ = (e^2 + e^3)=p2 are t1 t2 t3 = p12 0 1 1 2 43 1 1 1 0 2 1 2 03 5 = p12 2 2 2 (MPa) : 26Thus, the traction vector acting on this surface is t = p2e^1 + p2e^2 + p2e^3 (MPa) : The normal traction is ν = t · n^ = p20 + p2p12 + p2p12 = 0 + 1 + 1 = 2 MPa : The normal traction vector is ν = νn^ = p2e^2 + p2e^3 (MPa) : The projected shear traction is τ(P) = pjtj2 − ν2 = p2 + 2 + 2 − 4 = p2 MPa : The projected shear traction vector is τ(P) = t − ν = p2 − 0
e^1 + p2 − p2
e^2 + p2 − p2
e^3 = p2e^1 (MPa) : (b) Typically, you will want to solve for eigenvalues and eigenvectors numerically. However, in this case it turns out that you can find them manually pretty easily. The principal invariants of σ are Iσ = σ11 + σ22 + σ33 = 3 MPa ; IIσ = σ11σ22 + σ22σ33 + σ33σ11 − σ12σ21 − σ23σ32 − σ31σ13 = −6 MPa2 IIIσ = det [σ] = −8 MPa3 ; and the corresponding characteristic equation of σ is σ3 − Iσσ2 + IIσσ − IIIσ = σ3 − 3σ2 − 6σ + 8 = (σ − 1)(σ − 4)(σ + 2) = 0 : Thus, the principal stresses are σ(1) = 4 MPa ; σ(2) = 1 MPa ; σ(3) = −2 MPa : The direction n^(1) corresponding to the first principal stress is a solution to the following: 24 3 − 4 1 1 1 0 − 4 2 1 2 0 − 4 35 264 n^(1) 1 n^(1) 2 n^(1) 3 375 = 24 −1 1 1 1 −4 2 1 2 −4 35 264 n^(1) 1 n^(1) 2 n^(1) 3 375 = 24 000 35 =) n^(1) 1 = 2^ n(1) 2 = 2^ n(1) 3 : Thus, the first principal direction of stress is n^(1) = p26e^1 + p16e^2 + p16e^3 : 27The direction n^(2) corresponding to the second principal stress is a solution to 24 3 − 1 1 1 1 0 − 1 2 1 2 0 − 1 35 264 n^(2) 1 n^(2) 2 n^(2) 3 375 = 24 2 1 1 1 −1 2 1 2 −1 35 264 n^(2) 1 n^(2) 2 n^(2) 3 375 = 24 000 35 =) n^(1) 1 = −n^(1) 2 = −n^(1) 3 : Thus, the second principal direction of stress is n^(2) = p13e^1 − p13e^2 − p13e^3 : Finally, the third principal direction of stress is n^(3) = n^(1) × n^(2) = p12e^2 − p12e^3 : As a check, it is also seen that this is a solution of 24 3 − (−2) 1 1 1 0 − (−2) 2 1 2 0 − (−2) 35 264 n^(3) 1 n^(3) 2 n^(3) 3 375 = 24 5 1 1 1 2 2 1 2 2 35 264 n^(3) 1 n^(3) 2 n^(3) 3 375 = 24 000 35 : Problem 4.4 (a) The scalar components of stress, in units of MPa, are σ11 = 6X1X32 ; σ22 = 1 ; σ33 = 3X12 ; σ12 = 0 ; σ23 = 2 ; σ31 = −2X32 : Substituting these into the Cartesian coordinate equilibrium equations (f = 0), given in indicial notation by σji;j = 0, σ11;1 +σ21;2 +σ31;3 = 6X32 + 0 − 6X32 = 0 ; σ12;1 +σ22;2 +σ32;3 = 0 + 0 + 0 = 0 ; σ13;1 +σ23;2 +σ33;3 = 0 + 0 + 0 = 0 ; it is seen that this stress field does satisfy equilibrium. (b) At the point X = 2e^1 + 3e^2 + 2e^3 (m), the stress field has the value [σ] = 24 48 0 −16 0 1 2 −16 2 12 35 (MPa) ; and the unit normal to the plane 2X1 + X2 − X3 = 5 (m) is n^ = (2e^1 + e^2 − e^3)=p6. Thus, the traction t = n^ · σ is [t] = p16 2 1 −1 2 4−48 0 0 1 2 16 2 12 −163 5 = p16 112 −1 −42 (MPa) ; or t = 112 p6 e^1 − p16e^2 − p426e^3 (MPa) : 28(c) The normal traction is ν = t · n^ = 112 p6 p26 + p−16p16 + −p42 6 p−16 = 224 6 − 16 + 42 6 = 265 6 MPa : The normal traction vector is ν = νn^ = 265 6p6(2e^1 + e^2 − e^3) = 36:06e^1 + 18:03e^2 − 18:03e^3 (MPa) : The projected shear traction is τ(P) = pjtj2 − ν2 = s112 p6 2 + −p162 + −p4262 − 265 6 2 = 20:84 MPa : The projected shear traction vector is τ(P) = t − ν = 112 p6 − 3265 p6e^1 +  − p16 − 6265 p6e^2 +  − p426 + 6265 p6e^3 = 71p6 18 e^1 − 271p6 36 e^2 + 13p6 36 e^3 (MPa) = 9:66e^1 − 18:44e^2 + 0:88e^3 (MPa) : (d) Using Mathematica to find the eigenvalues and eigenvectors of [σ] = 24 48 0 −16 0 1 2 −16 2 12 35 (MPa) ; the principal stresses are σ(1) = 54:09 (MPa) ; σ(2) = 6:55 (MPa) ; σ(3) = 0:36 (MPa) ; and the corresponding principal directions of stress are n^(1) = −0:9345e^1 + 0:0134e^2 + 0:3558e^3 ; n^(2) = −0:3413e^1 − 0:3189e^2 − 0:8842e^3 ; n^(3) = 0:1016e^1 − 0:9477e^2 + 0:3026e^3 : Problem 4.5 (a) The scalar components of stress, in units of ksi (= 103 psi), are σ11 = σ22 = σ33 = 1 ; σ12 = 0 ; σ23 = 4X1 ; σ31 = 2X2 : Substituting these into the Cartesian coordinate equilibrium equations (f = 0), given in indicial notation by σji;j = 0, σ11;1 +σ21;2 +σ31;3 = 0 + 0 + 0 = 0 ; σ12;1 +σ22;2 +σ32;3 = 0 + 0 + 0 = 0 ; σ13;1 +σ23;2 +σ33;3 = 0 + 0 + 0 = 0 ; it is easily seen that this stress field does satisfy equilibrium. 29(b) At the point X = e^1 + 2e^2 + 3e^3 (in.), the stress field has the value [σ] = 24 1 0 4 0 1 4 4 4 1 35 (ksi) ; and the unit normal to the plane X1 + X2 + X3 = 6 (in.) is n^ = (e^1 + e^2 + e^3)=p3. Thus, the traction t = n^ · σ is [t] = p13 1 1 1 2 41 0 4 0 1 4 4 4 13 5 = p13 5 5 9 (ksi) ; or t = 5 p 3 e^1 + p53e^2 + p93e^3 (ksi) : (c) The normal traction is ν = t · n^ = p53p13 + p53p13 + p93p13 = 53 + 5 3 + 9 3 = 19 3 ksi : The normal traction vector is ν = νn^ = 19 3p3(e^1 + e^2 + e^3) (ksi) : The projected shear traction is τ(P) = pjtj2 − ν2 = r25 3 + 25 3 + 81 3 − 361 9 = 4p3 2 ksi : The projected shear traction vector is τ(P) = t − τ = p53 − 319 p3e^1 + p53 − 319 p3e^2 + p93 − 319 p3e^3 = − 4p3 9 e^1 − 4p3 9 e^2 + 8p3 9 e^3 (ksi) : (d) Using Mathematica to find the eigenvalues and eigenvectors of [σ] = 24 1 0 4 0 1 4 4 4 1 35 (ksi) ; the principal stresses are σ(1) = 1 + 4p2 = 6:66 (ksi) ; σ(2) = 1 (ksi) ; σ(3) = 1 − 4p2 = −4:66 (ksi) ; 30and the corresponding principal directions of stress are n^(1) = 1 2 e^1 + 1 2 e^2 + p12e^3 ; n^(2) = −p12e^1 + p12e^2 ; n^(3) = −1 2 e^1 − 1 2 e^2 + p12e^3 : Problem 4.6 The equation for the unit outward normal n^ to a traction-free plane is t = n^ · σ = σT · n^ = 0 ; [t] = 2 4σ2 0 2 1 2 0 11 2 13 5 2 4n n n^ ^ ^1 2 33 5 = 2 40 0 03 5 : In order for this equation to have a nontrivial solution (i.e., n^ 6= 0), the determinant of the stress tensor must be zero; det 2 4σ2 0 2 1 2 0 11 2 13 5 = 0 =) σ11 = 2 MPa The corresponding equation, 24 2 2 1 2 0 2 1 2 0 35 24 n^1 n^2 n^3 35 = 24 000 35 : has the following solution: n^ = 2 3 e^1 − 1 3 e^2 − 2 3 e^3 : Problem 4.7 Given the principal directions of stress, n^(1), n^(2), and n^(3), the unit outward normals to the eight octahedral planes are given by n^oct = p13(an^(1) + bn^(2) + cn^(3)) ; where a, b, and c are each equal to either +1 or −1 (giving 23 = 8 possible combinations). Using the spectral representation of the stress, σ = σ(1)n^(1)n^(1) + σ(2)n^(2)n^(2) + σ(3)n^(3)n^(3) ; it can then be seen that the traction vectors on the octahedral planes are given by toct = n^oct · σ = p13(aσ(1)n^(1) + bσ(2)n^(2) + cσ(3)n^(3)) and it follows that the normal tractions on the octahedral planes are νoct = ν(n^oct) = toct · n^oct = 1 3 (a2σ(1) + b2σ(2) + c2σ(3)) = 1 3 (σ(1) + σ(2) + σ(3)) 31Thus, the normal traction on each of the octahedral planes is the same and, using the definition (2.3.105c) for the principal invariants, it can be expressed as νoct = 13 (σ(1) + σ(2) + σ(3)) = 1 3 Iσ The projected shear tractions on the octahedral planes are τ (P ) oct = qjtoctj2 − νoct 2 = r1 3 a2(σ(1))2 + b2(σ(2))2 + c2(σ(3))2 − 1 9(σ(1) + σ(2) + σ(3))2 = 13 q3 (σ(1))2 + (σ(2))2 + (σ(3))2 − (σ(1) + σ(2) + σ(3))2 = 13 q3 (σ(1))2 + (σ(2))2 + (σ(3))2 − (σ(1))2 − (σ(2))2 − (σ(3))2 − 2σ(1)σ(2) − 2σ(2)σ(3) − 2σ(3)σ(1) = 13 q2(σ(1))2 + 2(σ(2))2 + 2(σ(3))2 − 2σ(1)σ(2) − 2σ(2)σ(3) − 2σ(3)σ(1) = 13 q(σ(1) − σ(2))2 + (σ(2) − σ(3))2 + (σ(3) − σ(1))2 Thus, the projected shear traction on each of the octahedral planes is the same and, using the definition (2.3.105c) for the principal invariants, it can be expressed as τ (P ) oct = 13 q(σ(1) − σ(2))2 + (σ(2) − σ(3))2 + (σ(3) − σ(1))2 = 1 3p2Iσ2 − 6IIσ Problem 4.9 Solving the equilibrium equation r·σT+f = 0 for the force field necessary to maintain equilibrium gives f = −r · σT or, in terms of Cartesian scalar components, fi = −σji;j. With the scalar components of the stress field given by σ11 = BX12X2, σ22 = 1 3B(X23 − 3b2X2), σ33 = 2bBX32, σ12 = B(b2 − X22)X1, and σ23 = σ31 = 0, it follows that the scalar components of the necessary body force field are f1 = −σ11;1 −σ21;2 −σ31;3 = −2BX1X2 + 2BX1X2 + 0 = 0 ; f2 = −σ12;1 −σ22;2 −σ32;3 = −B(b2 − X22) − B(X22 − b2) + 0 = 0 ; f3 = −σ13;1 −σ23;2 −σ33;3 = 0 + 0 − 4bBX3 = −4bBX3 : Thus, the necessary body force field is f = −4bBX3e^3 : Problem 5.1 The scalar components of the strain tensor obey the tensor transformation rule (2.3.68), so, for a 90◦ rotation about an axis parallel to e^1, 24 "0 11 "0 12 "0 13 "0 21 "0 22 "0 23 "0 31 "0 32 "0 33 35 = 24 1 0 0 0 0 1 0 −1 0 35 24 "11 "12 "13 "21 "22 "23 "31 "32 "33 35 24 1 0 0 0 0 −1 0 1 0 35 = 24 "11 "13 −"12 "31 "33 −"32 −"21 −"23 "22 35 32Therefore, in engineering notation, "¯0 1 = ¯ "1 ; "¯0 2 = ¯ "3 ; "¯0 3 = ¯ "2 ; "¯0 4 = −"¯4 ; "¯0 5 = −"¯6 ; "¯0 6 = ¯ "5 : Substituting these results into (5.3.17) and simplifying, one obtains the symmetry requirement that 2(C¯12 − C¯13)(¯ "1"¯2 − "¯1"¯3) + 4C¯14"¯1"¯4 + 2C¯15(¯ "1"¯5 + ¯ "1"¯6) + 2C¯16(¯ "1"¯6 − "¯1"¯5) +(C¯22 − C¯33)(¯ "2 2 − "¯2 3) + 2(C¯24 + C¯34)(¯ "2"¯4 + ¯ "3"¯4) + 2C¯25(¯ "2"¯5 + ¯ "3"¯6) + 2C¯26(¯ "2"¯6 − "¯3"¯5) +2C¯35(¯ "3"¯5 + ¯ "2"¯6) + 2C¯36(¯ "3"¯6 − "¯2"¯5) + 2C¯45(¯ "4"¯5 − "¯4"¯6) + 2C¯46(¯ "4"¯6 + ¯ "4"¯5) +(C¯55 − C¯66)(¯ "2 5 − "¯2 6) + 4C¯56"¯5"¯6 = 0 : This equation is satisfied for all possible values of the components of strain if and only if C¯13 = C¯12 ; C¯33 = C¯22 ; C¯34 = −C¯24 ; C¯66 = C¯55 ; C¯14 = C¯15 = C¯16 = C¯25 = C¯26 = C¯35 = C¯36 = C¯45 = C¯46 = C¯56 = 0 : These are the necessary and sufficient conditions for symmetry with respect to a 90◦ rotation about an axis parallel to e^1. They can be used to infer the conditions for symmetry with respect to a 90◦ rotation about an axis parallel to e^2 via the following changes in the engineering notation indices: 1 ! 2 ; 2 ! 3 ; 3 ! 1 ; 4 ! 5 ; 5 ! 6 ; 6 ! 4 : These changes give C¯21 = C¯23 ; C¯11 = C¯33 ; C¯15 = −C¯35 ; C¯44 = C¯66 ; C¯25 = C¯26 = C¯24 = C¯36 = C¯34 = C¯16 = C¯14 = C¯56 = C¯54 = C¯64 = 0 : A similar consideration for symmetry with respect to a 90◦ rotation about an axis parallel to e^3 yields no new conditions. So, combining the results for symmetry with respect to 90◦ rotations about an axis parallel to e^1 and an axis parallel to e^2, the necessary and sufficient conditions for a material to have cubic symmetry are C¯11 = C¯22 = C¯33 ; C¯12 = C¯23 = C¯31 ; C¯44 = C¯55 = C¯66 ; C¯14 = C¯15 = C¯16 = C¯24 = C¯25 = C¯26 = C¯34 = C¯35 = C¯36 = C¯45 = C¯46 = C¯56 = 0 : The engineering notation stiffness matrix, in a vector basis with base vectors aligned with the axes of symmetry, reduces to the following: 26666664 σ¯1 σ¯2 σ¯3 σ¯4 σ¯5 σ¯6 37777775 = 26666664 C¯11 C¯12 C¯12 0 0 0 C¯12 C¯11 C¯12 0 0 0 C¯12 C¯12 C¯11 0 0 0 0 0 0 C¯44 0 0 0 0 0 0 C¯44 0 0 0 0 0 0 C¯44 37777775 26666664 "¯1 "¯2 "¯3 "¯4 "¯5 "¯6 37777775 : Problem 5.2 33Since σij = 2µ"ij + λ"kkδij and using the definitions (5.3.2) for engineering notation, it follows that σ¯1 = σ11 = (2µ + λ)"11 + λ"22 + λ"33 = (2µ + λ)¯ "1 + λ"¯2 + λ"¯3 ; σ¯2 = σ22 = λ"11 + (2µ + λ)"22 + λ"33 = λ"¯1 + (2µ + λ)¯ "2 + λ"¯3 ; σ¯3 = σ33 = λ"11 + λ"22 + (2µ + λ)"33 = λ"¯1 + λ"¯2 + (2µ + λ)¯ "3 ; σ¯4 = σ23 = 2µ"23 = µ"¯4 ; σ¯5 = σ31 = 2µ"31 = µ"¯5 ; σ¯6 = σ12 = 2µ"12 = µ"¯6 ; Thus, in matrix form, 26666664 σ¯1 σ¯2 σ¯3 σ¯4 σ¯5 σ¯6 37777775 = 26666664 2µ + λ λ λ 0 0 0 λ 2µ + λ λ 0 0 0 λ λ 2µ + λ 0 0 0 0 0 0 µ 0 0 0 0 0 0 µ 0 0 0 0 0 0 µ 37777775 26666664 "¯1 "¯2 "¯3 "¯4 "¯5 "¯6 37777775 : Problem 5.3 Since "ij = [(1 + ν)σij − νσkkδij]=E and using the definitions (5.3.2) for engineering notation, it follows that "¯1 = "11 = 1 E σ11 − ν E σ22 − ν E σ33 = 1 E σ¯1 − ν E σ¯2 − ν E σ¯3 ; "¯2 = "22 = − ν E σ11 + 1 E σ22 − ν E σ33 = − ν E σ¯1 + 1 E σ¯2 − ν E σ¯3 ; "¯3 = "33 = − ν E σ11 − ν E σ22 + 1 E σ33 = − ν E σ¯1 − ν E σ¯2 + 1 E σ¯3 ; "¯4 = 2"23 = 2(1 + ν) E σ23 = 2(1 + ν) E σ¯4 ; "¯5 = 2"31 = 2(1 + ν) E σ31 = 2(1 + ν) E σ¯5 ; "¯6 = 2"12 = 2(1 + ν) E σ12 = 2(1 + ν) E σ¯6 ; Thus, in matrix form, 26666664 "¯1 "¯2 "¯3 "¯4 "¯5 "¯6 37777775 = 266666664 1E − νE − νE 0 0 0 − νE 1E − νE 0 0 0 − νE − νE 1E 0 0 0 0 0 0 2(1+ν) E 0 0 0 0 0 0 2(1+ν) E 0 0 0 0 0 0 2(1+ν) E 377777775 26666664 σ¯1 σ¯2 σ¯3 σ¯4 σ¯5 σ¯6 37777775 : Problem 5.5 The given relation E = µ(3λ + 2µ) λ + µ 34can be solved for µ in terms of E and λ. Multiplying both sides by λ + µ gives (λ + µ)E = µ(3λ + 2µ) =) 2µ2 + (3λ − E)µ − Eλ = 0 : It then follows from the quadratic formula that µ = E − 3λ + r 4 ; where r = p(3λ − E)2 − 4(2)(−Eλ) = p9λ2 − 6Eλ + E2 + 8Eλ = pE2 + 9λ2 + 2Eλ : Substituting this expression for µ into the given relation K = λ + 2 3 µ then gives K = λ + E − 3λ + r 6 = E + 3λ + r 6 : Problem 5.6 Recall that the dilatation can be determined from e = ~ σ=K, where ~ σ = 13σii is the mean stress and K is the bulk modulus. In this case, the mean stress is σ~ = 1 3 (200 MPa) = 66:67 MPa and the bulk modulus is K = E 3(1 − 2ν) = (207 GPa) 3[1 − 2(0:292)] = 165:87 GPa Thus, the dilatation is e = (66:67 × 106 Pa) (165:87 × 109 Pa) = 4:02 × 10−4 : Alternatively, one could determine "11, "22, and "33 from "ij = [(1 + ν)σij − νσkkδij]=E and then determine the dilatation from e = "ii. Problem 5.7 The components of stress are given by σij = 2µ"ij + λ"kkδij where µ = G = 26:6 GPa and λ = µ(E − 2µ) 3µ − E = 53:8 GPa : With "11 = 200×10−6, "22 = 300×10−6, "33 = 0, "12 = 100×10−6, "23 = 400×10−6, and "31 = 0, it follows that "kk = "11 + "22 + "33 = 500 × 10−6 35and σ11 = 2µ"11 + λ"kk = 37:5 MPa ; σ22 = 2µ"22 + λ"kk = 42:9 MPa ; σ33 = 2µ"33 + λ"kk = 26:9 MPa ; σ12 = 2µ"12 = 5:32 MPa ; σ23 = 2µ"23 = 21:3 MPa ; σ31 = 2µ"31 = 0 : In matrix notation, [σ] = 24 37:5 5:32 0 5:32 42:9 21:3 0 21:3 26:9 35 (MPa) : The strain energy density is given by U = 1 2 σij"ij = 1 2 σ11"11 + 1 2 σ22"22 + 1 2 σ33"33 + σ12"12 + σ23"23 + σ31"31 = 19:24 kPa : Problem 5.8 The components of strain are given by "ij = [(1 + ν)σij − νσkkδij]=E where E = 71:0 GPa and ν = E − 2G 2G = 0:335 : With σ11 = 20 MPa, σ22 = 0, σ33 = 15 MPa, σ12 = −4:0 MPa, σ23 = 10 MPa, and σ31 = 5:0 Mpa, it follows that "11 = 1 E [σ11 − ν(σ22 + σ33)] = 2:11 × 10−4 ; "22 = 1 E [σ22 − ν(σ11 + σ33)] = −1:65 × 10−4 ; "33 = 1 E [σ33 − ν(σ11 + σ22)] = 1:17 × 10−4 ; "12 = 1 + ν E σ12 = 1 2Gσ12 = −7:52 × 10−5 ; "23 = 1 2Gσ23 = 1:88 × 10−4 ; "31 = 1 2Gσ31 = 9:40 × 10−5 : In matrix notation, ["] = 24 211 −75:2 94:0 −75:2 −165 188 94:0 188 117 35 × 10−6 : 36The strain energy density is given by U = 1 2 σij"ij = 12 σ11"11 + 12 σ22"22 + 12 σ33"33 + σ12"12 + σ23"23 + σ31"31 = 5640 Pa : Problem 6.1 A cylindrical coordinate system can be defined with the z-axis along the centroidal axis of the sample and the top and bottom planes at z = h and z = 0, respectively. The boundary of the sample is composed of three surfaces: the bottom surface, S1 = fX j z = 0; r < ag ; the top surface, S2 = fX j z = h; r < ag ; and the lateral surface, S3 = fX j r = a; 0 ≤ z ≤ hg : Other means of specifying the boundary surfaces are also acceptable. The lateral surface S3 is traction-free (~t = 0 and n^ = e^r) for both cases below: σrr = σrθ = σrz = 0 8X 2 S3 : Now consider the two cases. (a) When the sample is ideally bonded to the platens at A and B, the material points at the interface are constrained to move with the platens. Thus, at the bottom surface (platen B and surface S1), u~ = 0, and at the top surface (platen A and surface S2), u~ = −δe^z: ur = uθ = uz = 0 8X 2 S1 ; ur = uθ = 0 ; uz = −δ 8X 2 S2 : (b) If the platens are frictionless, that means that the traction components tangent to the interfaces are zero. That is, there are mixed boundary conditions at S1 and S2, with t~r = t~θ = 0, u~z = 0 at S1, and ~ uz = −δ at S2: σzr = σzθ = 0 ; uz = 0 8X 2 S1 ; σzr = σzθ = 0 ; uz = −δ 8X 2 S2 : Problem 6.2 (a) In Cartesian coordinates, the fixed boundary condition at the base gives u1 = u2 = u3 = 0 on − a ≤ X1 ≤ a ; X2 = 0 ; −1 < X3 < 1 : 37For the semicircular surface, ~t = −qe^2 and n^ = cos θe^1 + sin θe^2. It follows from ^ njσji = t~i that σ11 cos θ + σ21 sin θ = 0 σ12 cos θ + σ22 sin θ = −q) on X12 + X22 = a2 ; X2 > 0 ; −1 < X3 < 1 : (b) In cylindrical coordinates, the fixed boundary condition at the base gives ur = uθ = uz = 0 on r ≤ a ; θ = 0; π ; −1 < z < 1 : For the semicircular surface, ~t = −q sin θe^r − q cos θe^θ and n^ = e^r. It follows from ^ njσji = t~i that σrr = −q sin θ σrθ = −q cos θ) on r = a ; 0 < θ < π ; −1 < z < 1 : Problem 6.3 The boundary conditions are σRR = σRθ = σRφ = 0 at R = a : Assuming uR = U(R) and uθ = uφ = 0, it follows from (6.4.5) that the components of stress are given by σRR = (2µ + λ)U0(R) + 2λ R U(R) ; σθθ = σφφ = λU0R + 2 R (µ + λ)U(R) ; σRθ = σθφ = σφR = 0 : The equilibrium equations (4.4.8) reduce in this case to @σRR @R + 1 R (2σRR − σθθ − σφφ) + fR = 0 : Since fR = −ρgR=a, it follows after simplifying that (2µ + λ)U00(R) + 2(2µ + λ) R U0(R) − 2(2µ + λ) R2 U(R) = ρgR a or R2U00(R) + 2RU0(R) − 2U(R) = ρgR3 (2µ + λ)a : The complete solution to this equation is U(R) = AR + B R2 + ρgR3 10(2µ + λ)a The assumed displacement field must be of this form in order to be in equilibrium with the prescribed body force field. However, it is also necessary that the displacement field be bounded as 38R ! 0 (or, equivalently, the displacement should be zero at R = 0), which is true if and only if B = 0, which leaves U(R) = AR + ρgR3 10(2µ + λ)a : The corresponding expression for σRR is given, after simplifying, by σRR = (2µ + 3λ)A + (6µ + 5λ)ρgR2 10(2µ + λ)a : Imposing the boundary condition σRRjR=a = 0 and solving for A, A = − (6µ + 5λ)ρga 10(2µ + 3λ)(2µ + λ) Thus, the displacement field in the sphere is uR = ρgR 10(2µ + λ)a R2 − a2 6 2µ µ + 5 + 3λ λ ; uθ = uφ = 0 : Problem 6.5 The boundary conditions are uR = uθ = uφ = 0 at R = a ; σRR = −p ; σRθ = σRφ = 0 at R = b : Assuming uR = U(R) and uθ = uφ = 0, it follows from (6.4.5) that the components of stress are given by σRR = (2µ + λ)U0(R) + 2λ R U(R) ; σθθ = σφφ = λU0R + 2 R (µ + λ)U(R) ; σRθ = σθφ = σφR = 0 ; and it follows from (6.4.12) that equilibrium is satisfied if and only if U(R) = AR + B R2 =) σRR = (2µ + 3λ)A − 4RµB3 : From the boundary condition uRjR=a = 0 it follows that B = −Aa3 =) σRR = A2µ + 3λ + 4µRa33 : From the boundary condition σRRjR=b = −p, and using the relation 2µ + 3λ = 3K, it follows that A = −b3p (2µ + 3λ)b3 + 4µa3 = −b3p 3Kb3 + 4µa3 : 39(a) Consolidating these results, the displacement field is uR = −b3p 3Kb3 + 4µa3R − Ra32 ; uθ = uφ = 0 ; and the stress field is σRR = − 1 + (4 1 + (4µa µa33))==(3 (3KR Kb33)) p ; σθθ = σφφ = − 11 + (4 − (2µa µa33))==(3 (3KR Kb33)) p ; σRθ = σθφ = σφR = 0 : (b) Letting b ! 1, σRR = −1 + 34KR µa33p ; σθθ = σφφ = −1 − 32KR µa33p : Far from the inclusion (R  a), σRR = σθθ = σφφ = −p =) σ(k) max = p : At the inclusion boundary (R = a), σRR = −1 + 34Kµ p ; σθθ = σφφ = −1 − 32Kµ p =) σ(k) max = 1 + 34Kµ p : Thus, the stress concentration factor is s.c.f. = 1 + 4µ 3K : Problem 6.7 There are different ways of approaching this problem. One way is to view the axial displacement δ of the rigid inner core as being prescribed, which is then related to the force F = Fe^z. Then the boundary conditions are ur = uθ = 0 ; uz = δ at r = a ; ur = uθ = uz = 0 at r = b ; σzr = σzθ = σzz = 0 at z = 0; L : Assuming ur = uθ = 0 and uz = U(r), it follows from (3.5.5) that "rr = "θθ = "zz = "rθ = "θz = 0 ; "zr = 1 2 U0(r) ; and from σij = 2µ"ij + λ"kkδij that σrr = σθθ = σzz = σrθ = σθz = 0 ; σzr = µU0(r) : 40From (4.4.4), equilibrium is satisfied if and only if r2U00(r) + rU0(r) = 0 =) U(r) = A ln r + B : Enforcing the boundary condition uzjr=b = 0 gives B = −A ln b =) U(r) = A(ln r − ln b) = A ln r b : Then, enforcing the boundary condition uzjr=a = δ gives A = δ ln(a=b) =) U(r) = ln( ln(a=b r=b))δ ; σzr = r ln( µδ a=b) : (a) Now to relate δ and F . Since n^ = −e^r on r = a, one has that tz = −σzr and F = Z0L Z02π tz a dθdz = Z0L Z02π −σzrjr=a a dθdz = Z0L Z02π −ln(µδ a=b) dθdz = − 2πLµδ ln(a=b) : Thus, δ = F 2πLµ ln b a ; and the work done is W = 1 2 F δ = F 2 4πLµ ln b a : (b) The boundary condition that σzr = 0 on z = 0; L is not satisfied, so the above solution is an approximate solution valid for values of z sufficiently far from the ends at z = 0 and z = L. Problem 7.1 A Cartesian coordinate for the problem can be defined as shown below (other coordinate systems could be defined that would work equally well). L/2 L/2 a h/2 h/2 X1 X2 With respect to this coordinate system, a complete statement of the boundary conditions is as follows. 41• On the surfaces at X1 = ±L=2, the unit outward normal is n^ = ±e^1 and the prescribed traction is ~t = ±σ0e^1. Thus σ11 = σ0 ; σ12 = 0 on X1 = ±L=2 : • On the surfaces at X2 = ±h=2, the unit outward normal is n^ = ±e^2 and the prescribed traction is ~t = 0. Thus σ21 = σ22 = 0 on X2 = ±h=2 : • On the surface at r = a, the unit outward normal is n^ = − cos θe^1 −sin θe^2 and the prescribed traction is ~t = 0, where r = pX12 + X22, cos θ = X1=r, and sin θ = X2=r. Thus, σ11 cos θ + σ21 sin θ = 0 σ12 cos θ + σ22 sin θ = 0) on r = a : Problem 7.2 The plane strain boundary conditions are ur = uθ = 0 at r = a ; σrr = 0 ; σrθ = τ0 at r = b : Assuming ur = 0 and uθ = U(r), it follows from (7.2.17) that "rr = "θθ = 0 ; "rθ = 12 hU0(r) − 1r U(r)i and from (7.2.3) that σrr = σθθ = 0 ; σrθ = µhU0(r) − 1r U(r)i Then, the equilibrium equations (7.2.18) are satisfied if and only if r2U00(r) + rU0(r) − U(r) = 0 =) U(r) = Ar + B r : Enforcing the boundary condition uθjr=a = 0 gives B = −Aa2 =) U(r) = Ar − ar2 ; σrθ = 2Aµa r2 2 ; and enforcing the boundary condition σrθjr=b = τ0 then gives A = b2τ0 2µa2 : Thus, the displacement field is ur = 0 ; uθ = τ0b2 2µa ar − ar  and the stress field is σrr = σθθ = 0 ; σrθ = b2τ0 r2 : 42One should also note that σzz = ν(σrr + σθθ) = 0 : Problem 7.3 The three-dimensional boundary conditions are ur = uθ = uz = 0 at r = a ; σzr = σzθ = σzz = 0 at z = 0; L ; and the prescribed body force field is fr = fθ = 0 ; fz = −γ : (a) Assuming antiplane strain with uz = U(r), the equilibrium equation (7.1.15) is satisfied if and only if r2U00(r) + rU0(r) = γr2 µ =) U(r) = A ln r + B + γ 4µ r2 : Note first that one must have A = 0 in order for uz to bounded at r = 0: U(r) = γ 4µ r2 + B : Then, enforcing the boundary condition uzjr=a = 0 gives B = −γa2 4µ =) uz = − γ 4µ (a2 − r2) : The stress field is given by (7.1.13): σθz = 0 ; σrz = 12 γr : (b) Since the boundary conditions σzr = 0 and z = 0; L are not satisfied, this is an approximate solution valid for values of z sufficiently far from the ends at z = 0; L. Problem 7.4 Given the Airy stress function Φ = C 1 2X22 ln(X12 + X22) + X1X2 tan−1 X X2 1 − X22 ; 43note first the following: Φ;1 = CX2 tan−1 X X2 1 ; Φ;11 = −CX22 X2 1 + X22 ; Φ;111 = 2CX1X22 (X12 + X22)2 ; Φ;1111 = 2C(X24 − 3X12X22) (X12 + X22)3 ; Φ;2 = C X2 ln(X12 + X22) + X1 tan−1 X X2 1 − X2 ; Φ;22 = C ln(X12 + X22) + X12X+22X22 ; Φ;222 = C X124+X2X22 − (X122+XX23 22)2 ; Φ;2222 = 2C(2X14 − 3X12X22 − X24) (X12 + X22)3 ; Φ;112 = −2CX12X2 (X12 + X22)2 ; Φ;1122 = 2C(3X12X22 − X14) (X12 + X22)3 : It follows that r4Φ = Φ;1111 +2Φ;1122 +Φ;2222 = 2C(X24 − 3X12X22 + 6X12X22 − 2X14 + 2X14 − 3X12X22 − X24) (X12 + X22)3 = 0 Thus, the Airy stress function satisfies the required compatibility condition, independent of the value of C. (a) The components of stress from this Airy stress function are σ11 = Φ;22 = C ln(X12 + X22) + X12X+22X22 ; σ22 = Φ;11 = −CX22 X2 1 + X22 ; σ12 = −Φ;12 = −C tan−1 X X2 1 + XX12 1+XX2 22 : The boundary condition on the positive X1-axis (t = 0, n^ = −e^2) gives σ21 = σ22 = 0 when X2 = 0 and X1 > 0 : 44Noting that tan−1(X2=X1) = θ = 0 on the positive X1-axis, it is seen that these conditions are satisfied. The boundary condition on the negative X1-axis (t = τ0e^1, n^ = −e^2) gives σ21 = −τ0 ; σ22 = 0 when X2 = 0 and X1 < 0 : Noting that tan−1(X2=X1) = θ = π on the negative X1-axis, it is seen that these conditions are satisfied if and only if C = τ0 π : (b) Noting that X1 = r cos θ, X2 = r sin θ, X12+X22 = r2, and tan−1(X2=X1) = θ, the (Cartesian) components of stress can be reexpressed in terms of the cylindrical coordinates r and θ: σ11 = C(ln r2 + sin2 θ) ; σ22 = −C sin2 θ ; σ12 = −C(θ + sin θ cos θ) : Thus, it is seen that σ22 and σ12 are independent of r, while σ11 becomes unbounded as r ! 0. Alternate Approach One could use cylindrical coordiantes from the outset: Φ = C 1 2r2 sin2 θ ln r2 + r2θ sin θ cos θ − r2 sin2 θ = C2 [r2(ln r − 1)(1 − cos 2θ) + r2θ sin 2θ] : Then, @Φ @r = C2 [r(2 ln r − 1)(1 − cos 2θ) + 2rθ sin θ] ; @2Φ @r2 = C2 [(2 ln r + 1)(1 − cos 2θ) + 2θ sin θ] ; @Φ @θ = C2 [r2(2 ln r − 1) sin 2θ + 2r2θ cos 2θ] ; @2Φ @θ2 = C2 [4r2 ln r cos 2θ − 4r2θ sin 2θ] ; and it follows that r2Φ = @2Φ @2Φ @θ2 = 2C ln r ; r4Φ = r2 (2C ln r) = 2C −r12 + r12 = 0 : The cylindrical coordinate components of stress are σrr = − cos 2θ) + 2θ sin 2θ] ; σrθ = − @ @r 1r @@θ Φ = −C2 [(2 ln r + 1) sin 2θ + 2θ cos 2θ] : 45The boundary conditions, σθr = σθθ = 0 when θ = 0 ; σθr = −τ0 ; σθθ = 0 when θ = π ; are satisfied if and only if C = τ0=π. Problem 7.5 The plane strain/stress boundary conditions for the problem are: • The surfaces at r = a (n^ = −e^r) and r = b (n^ = e^r) are traction-free (t = 0), which gives σrr = σrθ = 0 on r = a and r = b : • The surface at θ = π=2 is fixed: ur = uθ = 0 on θ = π 2 : • The traction distribution on the surface θ = 0 (n^ = −e^θ =) tr = −σθr ; tθ = −σθθ) is statically requivalent to P = −Pe^1 = −Pe^r: F1 = Zab tr dr = −P =) Zab σθrjθ=0 dr = P ; F2 = Zab tθ dr = 0 =) Zab σθθjθ=0 dr = 0 ; M0 = Zab rtθ dr = 0 =) Zab rσθθjθ=0 dr = 0 : For the proposed Airy stress function of the form Φ = Ar3 + Br + Cr ln r sin θ ; check first that the compatibility condition r4Φ = 0 is satisfied: @θ2 = 8Ar + 2rC  sin θ ; r4Φ = r2(r2Φ) = 4rC3 + 8rA − 2rC3 − 8rA − 2rC3  sin θ = 0 : Next, the components of stress are @r 1r @@θ Φ = − 2Ar − 2rB3 + Cr  cos θ : 46The boundary conditions σrr = σrθ = 0 on r = a and r = b are satisfied if and only if 2Aa − 2B a3 + C a = 0 2Ab − 2B b3 + C b = 0 9>=>; =) B = −Aa2b2 ; C = −2A(a2 + b2) : Now the components of stress are σrr = 2A r + ar23b2 − a2 +r b2 sin θ ; σθθ = 2A 3r − ar23b2 − a2 +r b2 sin θ ; σrθ = −2A r + ar23b2 − a2 +r b2 cos θ : For the average boundary conditions at θ = 0, since σθθjθ=0 = 0, Zab σθθjθ=0 dr = 0 ; Zab rσθθjθ=0 dr = 0 ; are identically satisfied. Finally, since Zab σrθjθ=0 dr = −2A Zab r + ar23b2 − a2 +r b2 dr = −2A b2 − a2 − (a2 + b2) ln ab  = P ; it follows that A = − P 2N ; B = P a2b2 2N ; C = P(a2 + b2) N ; where N ≡ b2 − a2 − (a2 + b2) ln b a : Note that the fixed-displacement boundary condition at θ = π=2 is not satisfied|the solution is approximate in the sense of Saint-Venant’s principle. Problem 7.6 The plane strain/stress boundary conditions for the problem are: • The surfaces at θ = 0 (n^ = e^θ) has prescribed traction ~t = −pe^θ, which gives σθr = 0 ; σθθ = −p on θ = 0 : • The surfaces at θ = −α (n^ = −e^θ) is traction-free, ~t = 0, which gives σθr = σθθ = 0 on θ = −α : • The surface at X1 = r cos θ = L is fixed: ur = uθ = 0 on X1 = L : 47For the proposed Airy stress function of the form Φ = Cr2(α + θ − sin θ cos θ − cos2 θ tan α) ; note first that @Φ @r = 2Cr(α + θ − sin θ cos θ − cos2 θ tan α) ; @2Φ @r2 = 2C(α + θ − sin θ cos θ − cos2 θ tan α) ; @Φ @θ = Cr2(1 − cos2 θ + sin2 θ + 2 sin θ cos θ tan α) ; @2Φ @θ2 = 2Cr2(2 sin θ cos θ + cos2 θ tan α − sin2 θ tan α) ; and check that the compatibility condition r4Φ = 0 is satisfied: @@θ Φ = −2C(sin2 θ + sin θ cos θ tan α) : These stress components satisfy the traction-free boundary condition on θ = −α: σrθjθ=−α = −2C sin2 α − sin α cos αcos sin αα = 0 ; σθθjθ=−α = 2C α − α + sin α cos α − cos2 αcos sin αα = 0 : For θ = 0, σrθjθ=0 = 0 ; σθθjθ=0 = 2C(α − tan α) : Thus the boundary conditions on θ = 0 are satisfied if and only if C = p 2(tan α − α) Note that the fixed-displacement boundary condition at X1 = L is not satisfied|the solution is approximate in the sense of Saint-Venant’s principle. Problem 7.7 The plane strain/stress boundary conditions for the problem are: 48• The surfaces at X2 = h=2 (n^ = e^2) has prescribed traction ~t = −τ0e^1, which gives σ21 = −τ0 ; σ22 = 0 on X2 = h=2 : • The surfaces at X2 = −h=2 (n^ = −e^2) is traction-free, ~t = 0, which gives σ21 = σ22 = 0 on X2 = −h=2 : • The surface at X1 = L is fixed: u1 = u2 = 0 on X1 = L : • The surfaces at X1 = 0 (n^ = −e^1) is traction-free, ~t = 0, which gives σ11 = σ12 = 0 on X1 = 0 : Using the semi-inverse method, and taking a cue from the boundary conditions on σ21 at the top and bottom of the beam, assume that σ21 is independent of X1; σ21 = η0(X2) : where η(X2) is an arbitrary function. Then, since σ12 = −Φ;12, it follows that the most general corresponding form of the Airy stress function is Φ = −X1η(X2) + ξ(X1) + ζ(X2) and the most general forms of the remaining stresses consistent with the equilibrium equation are σ11 = Φ;22 = −X1η00(X2) + ζ00(X2) ; σ22 = Φ;11 = ξ00(X1) ; where ξ(X1) and ζ(X2) are arbitrary functions. In order to satisfy the boundary conditions that σ22 = 0 on X2 = ±h=2, it follows that ξ00(X1) = 0 and hence that ξ(X1) = αX1 +β. However, the values of α and β have no effect on the components of stress, so they can be set equal to zero with no loss of generality; Φ = −X1η(X2) + ζ(X2) ; σ11 = −X1η00(X2) + ζ00(X2) ; σ22 = 0 ; σ21 = η0(X2) : In order to satisfy the boundary condition that σ11 = 0 on X1 = 0, it follows that ζ00(X2) = 0 and hence that ζ(X2) = γX2 + δ. However, the values of γ and δ have no effect on the components of stress, so they can be set equal to zero with no loss of generality; Φ = −X1η(X2) ; σ11 = −X1η00(X2) ; σ22 = 0 ; σ21 = η0(X2) : 49Looking next at the compatibility condition r4Φ = −X1η(IV )(X2) = 0, it follows that η(X2) = AX3 2 + BX22 + CX2 + D. However, the value of D has no effect on the components of stress, so it can be set equal to zero with no loss of generality; Φ = −AX1X23 − BX1X22 − CX1X2 ; σ11 = −6AX1X2 − 2BX1 ; σ22 = 0 ; σ21 = 3AX22 + 2BX2 + C : In order to satisfy the boundary conditions that : Finally, one needs to consider the boundary condition that σ12 = 0 on X1 = 0. Clearly, this boundary condition can not be satisfied exactly, so instead one requires that statically equivalent condition from which it follows that A = τ0=(2h2). The the solution stress field (approximate in the sense of Saint-Venant’s Principle) is σ11 = − 3τ0 h2 X1X2 + τh0 X1 ; σ22 = 0 ; σ21 = 3τ0 2h2 X22 − τh0 X2 − 78τ0 : Note that the fixed displacement boundary condition at X1 = L is also only approximately satisfied in the sense of Saint-Venant’s Principle. Problem 7.8 The plane strain/stress boundary conditions for the problem are: 50• The edges of the wedge, θ = ±β, have unit outward normal n^ = ±e^θ and are traction-free, so σθr = σθθ = 0 on θ = ±β : • The distribution of tractions on a surface r = a must be in equilibrium with the concentrated moment; X F1 = Z−ββ[σrr(a; θ) cos θ − σrθ(a; θ) sin θ]a dθ = 0 X F2 = P + Z−ββ[σrr(a; θ) sin θ + σrθ(a; θ) cos θ]a dθ = 0 X M3 = Z−ββ[aσrθ(a; θ)]a dθ = 0 for arbitrary a. Using the semi-inverse method, make the same assumption as was made in Section 7.4.5, namely that σrθ = 0 everywhere. The boundary condition then reduces to σθθ = 0 on θ = ±β : and the equilibrium conditions reduce to Z−ββ aσrr(a; θ) cos θ dθ = 0 ; Z−ββ aσrr(a; θ) sin θ dθ = −P : It follows from the same arguments as were used in Section 7.4.5 that σrr = 1 r (2C cos θ − 2D sin θ) ; σθθ = 0 : Substituting this result into the first equilibrium condition, 0 = Z−ββ(2C cos θ − 2D sin θ) cos θ dθ = 2C Z−ββ cos2 θ dθ − 2D Z−ββ sin θ cos θ dθ = C(2β + sin 2β) : Since β is arbitrary, this condition is only satisfied if C = 0. Using this result, the final condition becomes P = 2D Z−ββ sin2 θ dθ = D(2β − sin 2β) : Thus, D = P 2β − sin 2β 51and the stress field in the wedge is σrr = − 2P sin θ r(2β − sin 2β) ; σθθ = σrθ = 0 : Problem 7.9 The plane strain/stress boundary conditions for the problem are: • The edges of the wedge, θ = ±β, have unit outward normal n^ = ±e^θ and are traction-free, so σθr = σθθ = 0 on θ = ±β : • The distribution of tractions on a surface r = a must be in equilibrium with the concentrated moment; X F1 = Z−ββ[σrr(a; θ) cos θ − σrθ(a; θ) sin θ]a dθ = 0 X F2 = Z−ββ[σrr(a; θ) sin θ + σrθ(a; θ) cos θ]a dθ = 0 X M3 = M + Z−ββ[aσrθ(a; θ)]a dθ = 0 for arbitrary a. One needs to determine the values of the constants A and B such that an Airy stress function of the form Φ = Aθ + B sin 2θ is biharmonic (satisfies compatibility) and the corresponding stress components satisfy these boundary conditions. Consider first the compatibility condition. The Laplacian of this Airy stress function is r2 sin 2θ and the biharmonic is r4Φ = r2(r2Φ) = −24B r4 sin 2θ + 8rB4 sin 2θ + 16 r4B sin 2θ = 0 : Thus, compatibility is satisfied. The corresponding components of stress are σrr = @θ Φ = r12 (A + 2B cos 2θ) One can now consider the boundary conditions. 52• The boundary condition σθθ = 0 on θ = ±β is trivially satisfied. • The boundary condition σθr = 0 on θ = ±β is satisfied if and only if A = −2B cos 2β : The expression for σrθ then reduces to σrθ = 2B r2 (cos 2θ − cos 2β) : • For equilibrium of forces in the X1 direction, X F1 = Z−ββ −4aB2 sin 2θ cos θ − 2aB2 (cos 2θ − cos 2β) sin θ a dθ : Noting that the integrand is an odd function of θ, it follows immediately that P F1 = 0. • For equilibrium of forces in the X2 direction, = − 4B a sin3 β + 4B a sin2 β sin β + 4B a sin3(−β) − 4B a sin2 β sin(−β) = 0 • Finally, for moment equilibrium, X M3 = M + Z−ββ 2B(cos 2θ − cos 2β) dθ = M + 2B 1 2 sin 2θ − θ cos 2β β −β = M + 4B 12 sin 2β − β cos 2β : Thus, P M3 = 0 if and only if B = M 4β cos 2β − 2 sin 2β 53It is thus shown that an Airy stress function of the form Φ = Aθ+B sin 2θ does provide the solution to the problem and the corresponding stress field in the wedge is σrr = 2M sin 2θ (2β cos 2β − sin 2β)r2 σθθ = 0 σrθ = M(cos 2θ − cos 2β) (2β cos 2β − sin 2β)r2 Problem 8.1 The outer and inner boundaries, @S(1) and @S(2), respectively, of the hollow circular cross section are given by @S(1) = (X1; X2) j X12 + X22 = b2 ; @S(2) = (X1; X2) j X12 + X22 = a2 : This is the inspiration for trying a Prandtl stress function of the form φ = m(X12 + X22 − b2) ; since it clearly satisfies the boundary conditions: φ = 0 8(X1; X2) 2 @S(1) ; φ = m(a2 − b2) 8(X1; X2) 2 @S(2) : Fortuitously, the Laplacian of this function happens to be a non-zero constant, r2φ = φ11 + φ22 = 4m ; so it also satisfies the governing compatibility equation and is therefore the Prandtl stress function for the hollow cylinder. Noting that b2) : Problem 8.2 Boundary Conditions • The outer lateral surface of the composite cylinder (r = b, n^ = e^r) is traction free (~t = 0), so, in cylindrical coordinates, σrr = σrθ = σrz = 0 on r = b : (15) • The distribution of tractions on the end of the shaft (and on each cross section) is statically equivalent to the applied torque T . 54Interface Continuity Conditions • The displacement u and the stress components σrr, σrθ, and σrz are continuous across the interface at r = a. Solution - using the semi-inverse method, assume that plane cross sections remain plane and undistorted. For small angles of rotation (Θ = αz  1), this implies that ur = uz = 0 ; uθ = αrz ; where α is the twist per unit length. It follows from the strain-displacement relations (3.5.5) that "rr = "θθ = "zz = "rθ = "zr = 0 ; "θz = 1 2 αr ; and, thus, from the constitutive equation σij = 2µ"ij + λ"kkδij, that σrr = σθθ = σzz = σrθ = σzr = 0 ; σθz = (µ µ1 2αr ; αr ; if if r < a a < r < b ; . It can been seen then that, based on the assumption that plane cross sections remain plane and undistorted, • the equilibrium conditions (4.4.4) are satisfied, • the traction-free lateral surface boundary condition (1) is satisfied, and • the interface continuity conditions are satisfied. Thus, provided that the distribution of tractions on the end of the shaft is statically equivalent to the applied torque T (as will be shown below), the assumption is correct and this is the solution to the problem. (a) The stress distribution is given in terms of the twist per unit length α by σrr = σθθ = σzz = σrθ = σzr = 0 ; σθz = (µ µ1 2αr ; αr ; if if r < a a < r < b ; . (b) For the boundary condition that the distribution of tractions on the end of the shaft be statically equivalent to the applied torque T , the unit outward normal is n^ = e^z and tr = σzr = 0 ; tz = σzz = 0 ; tθ = σzθ : Also, t1 = tr cos θ − tθ sin θ = −σzθ sin θ ; t2 = tr sin θ + tθ cos θ = σzθ cos θ ; t3 = tz = 0 : 55Therefore, X F1 = X F2 = 0 ; since σzθ is independent of θ, and X F3 = X M1 = X M2 = 0 ; since t3 = 0. Finally, X M3 = Z0b Z02π(rtθ) rdθdr = 2π Z0b r2σzθ dr = 2π Z0a r2σzθ dr + 2π Zab r2σzθ dr = 2πµ1α Z0a r3 dr + 2πµ2α Zab r3 dr = 12 πµ1αa4 + 1 2 πµ2α(b4 − a4) = 12 πα[µ1a4 + µ2(b4 − a4)] ; so that, by setting P M3 = T, one gets α = 2T π[µ1a4 + µ2(b4 − a4)] : (c) The torque carried by the annulus is the resultant moment about the z-axis of the distribution of traction on the cross section of the annulus; X M3annulus = Zab Z02π(rtθ) rdθdr = 1 2πµ2α(b4 − a4) : Problem 8.3 Even though φ = 0 on the boundary of the rectangular cross section, since r2φ = 2m a2 Xb222 − 1 + 2bm2 Xa212 − 1 is not a non-zero constant, the function φ will not satisfy compatibility and, therefore, cannot be used as a Prandtl stress function for this cross section. Problem 8.4 The given function is φ = m(r2 − b2) 2a cos r θ − 1 : (a) It must be established that this function satisfies the boundary conditions and the compatibility condition. Boundary Conditions 56• Every point on the portion of the boundary @S1 satisfies r = 2a cos θ and it follows that φ = 0 on @S1. • Every point on the portion of the boundary @S2 satisfies r = b and it follows that φ = 0 on @S2. Thus, it is seen that φ = 0 at every point on the boundary, thus satisfying the traction-free lateral surface boundary condition. Compatibility = −4ma cos θ rb23 − 2m + 2ma cos θ 1r + rb23 − 2m − 2ma cos θ 1r − rb23 = −4m Thus, since r2φ is a non-zero constant, compatibility is satisfied. The twist per unit length α is related to r2φ by α = −r2φ=2µ. It follows that m = 1 2 µα : (b) The project shear traction is given by τ(P)(e^3) = jrφj, where rφ = @φ @r e^r + 1r @φ @θ e^θ = 2ma cos θ 1 + rb22 − 2mr e^r − 2ma sin θ 1 − rb22 e^θ : So, assuming that τmax = τ(P)(e^3) (b;0) = jrφj(b;0) and noting that rφj(b;0) = (4ma − 2mb)e^r ; it follows that τmax = 2m(2a − b) = µα(2a − b) : In the limit as b ! 0, lim b!0 τmax = 2µαa : On the other hand, it was seen in (8.1.28) and (8.1.31) that the maximum shear traction for a circular cylinder of radius a is (τmax)circle = 2T πa3 = µαa : 57Comparing these results shows that limb!0 τmax (τmax)circle = 2 : Problem 8.5 φ = m(a2 − X12 + b2X22)(a2 + b2X12 − X22) = m[a4 − b2(X14 + X24) + (1 + b4)X12X22 − a2(1 − b2)(X12 + X22)] φ;11 = m[−12b2X12 + 2(1 + b4)X22 − 2a2(1 − b2)] φ;22 = m[−12b2X22 + 2(1 + b4)X12 − 2a2(1 − b2)] r2φ = mf[2(1 + b4) − 12b2](X12 + X22) − 4a2(1 − b2)g (a) Recall that r2φ must be a nonzero constant. A necessary and sufficient condition for this is 2(1 + b4) − 12b2 = 0 =) b4 − 6b2 + 1 = 0 =) b2 = 3 ± p8 : The additional constraint jbj < 1 thus gives b2 = 3 − p8 = 0:1716 : (16) There are no constraints on either a or m [other than how they are related in part (b) below]. (b) Given the constraint (??), one has that b2 ; ±p1a− b2 ; it is seen that they will not intersect if jbj > 1. Problem 8.6 The given function is φ = m Xa212 + Xb222 − 1 : (a) It must be established that this function satisfies the boundary conditions and the compatibility condition. Boundary Conditions • At every point on the portion of the boundary @S1 = (X1; X2) j Xa212 + Xb222 = k2 ; so it follows that φ = m(k2 − 1) on @S1. • At every point on the portion of the boundary @S2 = (X1; X2) j Xa212 + Xb222 = 1 ; so it follows that φ = 0 on @S2. Thus, it is seen that dφ=ds = 0 at every point on the boundary, thus satisfying the tractionfree lateral surface boundary condition. Compatibility r2φ = φ;11 +φ;22 = 2m a12 + b12 = 2m(aa22b+2 b2) : 59Thus, since r2φ is a non-zero constant, compatibility is satisfied. The twist per unit length a2 πa43b − π(ka)43(kb) − 2bm2 πab 4 3 − π(ka)( 4 kb)3 = −m(1 − k4)πab = (1 − k4)µαπa3b3 a2 + b2 =) J~ = µα T = (1 − k4)aπa 2 +3bb32 60http://www.springer.com/978-0-8176-4117-7 [Show More]

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