Solutions to Problems Problem 2.1 In this case,[S]is symmetric given as follows: TS11 S12 S13 0 0 07 S12 S22 S23 0 0 0 [9= S13S23 S33 0 0 0 0 0 0 S44 0 0 0 0 0 0S55 0 0 0 0 0 S66] S=[S11(S22S33-S23S23)-S12(S12S33-S13S23) +S13(S12S23-S13S22)]S44S55S66 =(S11S22S33-S11S23S23-S33S12S12 -S22S13S13+2S12S23S13)S44S55S66 Next,use the following formula to calculate the inverse of [S]: [C=S-1=ac5L图 ISI Only Cu will be calculated in detail as follows: Cud5a)SuS Su(SaS) I IST where S is given in the book in (2.5).The same procedure can be followed to derive the other elements of [C]given in(2.5)
Solutions to Problems Problem 2.1 In this case, [S] is symmetric given as follows: [S] = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ S11 S12 S13 000 S12 S22 S23 000 S13 S23 S33 000 000 S44 0 0 0000 S55 0 00000 S66 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ |S| = [S11(S22S33 − S23S23) − S12(S12S33 − S13S23) +S13(S12S23 − S13S22)] S44S55S66 = (S11S22S33 − S11S23S23 − S33S12S12 −S22S13S13 + 2S12S23S13) S44S55S66 Next, use the following formula to calculate the inverse of [S]: [C]=[S] −1 = adj[S] |S| Only C11 will be calculated in detail as follows: C11 = (adj[S])11 |S| = (S22S33 − S23S23) S44S55S66 |S| = 1 S (S22S33 − S23S23) where S is given in the book in (2.5). The same procedure can be followed to derive the other elements of [C] given in (2.5)
206 Solutions to Problems Problem 2.2 The reciprocity relations of (2.6)are valid for linear elastic analysis.They can be derived by applying the Maxwell-Betti Reciprocal Theorem.For more details,see [1]. Problem 2.3 1/E -h2/E1-M3/E 0 0 0 -12/E1 1/E2 -23/E2 0 0 0 -y13/E1 -23/E2 1/E2 0 0 0 [9= 0 0 0 2(1+23) 0 0 E2 0 0 0 1/G12 0 0 0 0 0 0 1/G12」 Problem 2.4 S=S11S22S33-S11S23S23-S22S13S13-S33S12S12+2S12S23S13 (爱)（尝）（贸）（安〉 ()()+2(器)（会）() 1-V2-21221-2122321 EE略 1-/ EE where v is given by: /=+21221+2122321 Next,Ci is calculated in detail as follows: C11=5(S253-523523) 居高（会）（】 E (1-2)E1 1-w
206 Solutions to Problems Problem 2.2 The reciprocity relations of (2.6) are valid for linear elastic analysis. They can be derived by applying the Maxwell-Betti Reciprocal Theorem. For more details, see [1]. Problem 2.3 [S] = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1/E1 −ν12/E1 −ν13/E1 0 00 −ν12/E1 1/E2 −ν23/E2 0 00 −ν13/E1 −ν23/E2 1/E2 0 00 000 2(1 + ν23) E2 0 0 0 0 0 01/G12 0 0 0 0 0 01/G12 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Problem 2.4 S = S11S22S33 − S11S23S23 − S22S13S13 − S33S12S12 + 2S12S23S13 = 1 E1 1 E2 1 E2 − 1 E1 −ν23 E2 −ν23 E2 − 1 E2 −ν12 E1 −ν21 E2 − 1 E2 −ν12 E1 −ν21 E2 + 2 −ν12 E1 −ν23 E2 −ν21 E2 = 1 − ν2 23 − 2ν12ν21 − 2ν12ν23ν21 E1E2 2 = 1 − ν E1E2 2 where ν is given by: ν = ν2 23 + 2ν12ν21 + 2ν12ν23ν21 Next, C11 is calculated in detail as follows: C11 = 1 S (S22S33 − S23S23) = E1E2 2 1 − ν 1 E2 1 E2 − −ν23 E2 −ν23 E2 = 1 − ν2 23 E1 1 − ν
Solutions to Problems 207 Similarly,the other elements of [C]are obtained as follows: C2= (1+23)h2E2 1-w C3= 1+3n22=C2 1- C22= (1-221)E2 1-w C23= (23+41221)E2 1-w C33= 1-h22)2=C2 1-w E2 C44=2(1+23】 C65=G12 C66=G12=C65 Problem 2.5 1EEE E 0 0 0 -V E E 0 0 0 E 0 0 0 [S= 0 0 2(1+w) 0 0 0 E 0 0 0 2(1+） E 0 0 0 0 0 0 2(1+) E [1 -V -V 0 0 0 1 -V 0 0 0 [S]=E -V -V 1 0 0 0 0 0 0 2(1+v) 0 0 0 0 0 0 2(1+v) 0 0 0 0 0 2(1+w)」
Solutions to Problems 207 Similarly, the other elements of [C] are obtained as follows: C12 = (1 + ν23)ν12E2 1 − ν C13 = (1 + ν23)ν12E2 1 − ν = C12 C22 = (1 − ν12ν21)E2 1 − ν C23 = (ν23 + ν12ν21)E2 1 − ν C33 = (1 − ν12ν21)E2 1 − ν = C22 C44 = E2 2(1 + ν23) C55 = G12 C66 = G12 = C55 Problem 2.5 [S] = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 E −ν E −ν E 000 −ν E 1 E −ν E 000 −ν E −ν E 1 E 000 000 2(1 + ν) E 0 0 000 0 2(1 + ν) E 0 000 0 0 2(1 + ν) E ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ [S] = 1 E ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 −ν −ν 000 −ν 1 −ν 000 −ν −ν 10 0 0 0 0 0 2(1 + ν)0 0 0 0 0 0 2(1 + ν) 0 0 0 0 0 0 2(1 + ν) ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
208 Solutions to Problems Problem 2.6 「11-v1-w 0 0 0 1-v1 1-w 0 0 0 1-y1-v 1 0 0 0 E (C]= 1+2w (1+v)(1+2w) 0 0 0 0 0 0 0 0 1+2w 0 2 0 1+2w 0 0 0 2 Problem 2.7 >s1gma3=150/(40*40) sigma3 0.0938 >siga=[0;0;s1gma3;0;0;0] sigma 0 0 0.0938 0 0 0 >[S]= 0 rthotrop1 cComp1 iance(50.0,15.2,15.2,0.254,0.428,0.254,4.70, 3.28,4.70) S= 0.0200-0.0051 -0.0051 0 0 0 -0.0051 0.0658 -0.0282 0 0 0 -0.0051-0.0282 0.0658 0 0 0 0 0 0 0.3049 0 0 0 0 0 0 0.2128 0 0 0 0 0 0.2128 >epsilon S*sigma
208 Solutions to Problems Problem 2.6 [C] = E (1 + ν)(1 + 2ν) ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 1 − ν 1 − ν 000 1 − ν 1 1 − ν 000 1 − ν 1 − ν 10 0 0 000 1+2ν 2 0 0 000 0 1+2ν 2 0 000 0 0 1+2ν 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Problem 2.7 >> sigma3 = 150/(40*40) sigma3 = 0.0938 >> sigma = [0;0; sigma3 ; 0 ; 0 ; 0] sigma = 0 0 0.0938 0 0 0 >> [S] = OrthotropicCompliance(50.0,15.2,15.2,0.254,0.428,0.254,4.70, 3.28,4.70) S = 0.0200 -0.0051 -0.0051 0 0 0 -0.0051 0.0658 -0.0282 0 0 0 -0.0051 -0.0282 0.0658 0 0 0 0 0 0 0.3049 0 0 0 0 0 0 0.2128 0 0 0 0 0 0 0.2128 >> epsilon = S*sigma
Solutions to Problems 209 epsilon -0.0005 -0.0026 0.0062 0 0 0 >format short e >epsilon epsilon -4.7625e-004 -2.6398e-003 6.1678e-003 0 0 0 >d1 epsilon(1)*40 d1= -1.9050e-002 >>d2=eps11on(2)*40 d2= -1.0559e-001 >d3=eps11on(3)*40 d3- 2.4671e-001 Problem 2.8 >s1gma3=150/(40*40) sigma3 0.0938
Solutions to Problems 209 epsilon = -0.0005 -0.0026 0.0062 0 0 0 >> format short e >> epsilon epsilon = -4.7625e-004 -2.6398e-003 6.1678e-003 0 0 0 >> d1 = epsilon(1)*40 d1 = -1.9050e-002 >> d2 = epsilon(2)*40 d2 = -1.0559e-001 >> d3 = epsilon(3)*40 d3 = 2.4671e-001 Problem 2.8 >> sigma3 = 150/(40*40) sigma3 = 0.0938