《纺织复合材料》课程参考文献（Mechanics of Composite Materials with MATLAB）13 Solutions to Problems

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

210 Solutions to Problems >sigma=[0;0;sigma3;0;0;0] sigma 0 0 0.0938 0 0 0 >[S]IsotropicCompliance(72.4,0.3) S= 0.0138 -0.0041 -0.0041 0 0 0 -0.0041 0.0138 -0.0041 0 0 0 -0.0041-0.0041 0.0138 0 0 0 0 0 0 0.0359 0 0 0 0 0 0 0.0359 0 0 0 0 0 0 0.0359 >epsilon S*sigma epsilon -0.0004 -0.0004 0.0013 0 0 0 >format short e >epsilon epsilon -3.8847e-004 -3.8847e-004 1.2949e-003 0 0 0 >>d1=epsi1on(1)*40 d1= -1.5539e-002

210 Solutions to Problems >> sigma = [0;0; sigma3 ; 0 ; 0 ; 0] sigma = 0 0 0.0938 0 0 0 >> [S] = IsotropicCompliance(72.4,0.3) S = 0.0138 -0.0041 -0.0041 0 0 0 -0.0041 0.0138 -0.0041 0 0 0 -0.0041 -0.0041 0.0138 0 0 0 0 0 0 0.0359 0 0 0 0 0 0 0.0359 0 0 0 0 0 0 0.0359 >> epsilon = S*sigma epsilon = -0.0004 -0.0004 0.0013 0 0 0 >> format short e >> epsilon epsilon = -3.8847e-004 -3.8847e-004 1.2949e-003 0 0 0 >> d1 = epsilon(1)*40 d1 = -1.5539e-002

Solutions to Problems 211 >d2 epsilon(2)*40 d2= -1.5539e-002 >>d3=eps11on(3)*40 d3= 5.1796e-002 Problem 2.9 >s1gma2=100/(60*60) sigma2 0.0278 >s1ga=[0;s1gma2;0;0;0;0] sigma 0 0.0278 0 0 0 0 >[S]= 0 rthotropicComp1 iance(155.0,12.10,12.10,0.248,0.458,0.248, 4.40,3.20,4.40) S= 0.0065 -0.0016 -0.0016 0 0 0 -0.0016 0.0826 -0.0379 0 0 0 -0.0016 -0.0379 0.0826 0 0 0 0 0 0 0.3125 0 0 0 0 0 0 0.2273 0 0 0 0 0 0 0.2273

Solutions to Problems 211 >> d2 = epsilon(2)*40 d2 = -1.5539e-002 >> d3 = epsilon(3)*40 d3 = 5.1796e-002 Problem 2.9 >> sigma2 = 100/(60*60) sigma2 = 0.0278 >> sigma = [0 ; sigma2 ; 0 ; 0 ; 0 ; 0] sigma = 0 0.0278 0 0 0 0 >> [S] = OrthotropicCompliance(155.0,12.10,12.10,0.248,0.458,0.248, 4.40,3.20,4.40) S = 0.0065 -0.0016 -0.0016 0 0 0 -0.0016 0.0826 -0.0379 0 0 0 -0.0016 -0.0379 0.0826 0 0 0 0 0 0 0.3125 0 0 0 0 0 0 0.2273 0 0 0 0 0 0 0.2273

212 Solutions to Problems >EpsilonMechanical S+sigma EpsilonMechanical -0.0000 0.0023 -0.0011 0 0 0 >format short e >EpsilonMechanical EpsilonMechanical -4.4444e-005 2.2957e-003 -1.0514e-003 0 0 0 >>Epsi1 onTherma1(1)=-0.01800e-6*30 EpsilonThermal -5.4000e-007 >EpsilonThermal(2)=24.3e-6*30 EpsilonThermal -5.4000e-0077.2900e-004 >EpsilonThermal(3)=24.3e-6*30 EpsilonThermal -5.4000e-0077.2900e-0047.2900e-004 >EpsilonThermal(4)=0 EpsilonThermal -5.4000e-0077.2900e-0047.2900e-004 0 >EpsilonThermal(5)=0

212 Solutions to Problems >> EpsilonMechanical = S*sigma EpsilonMechanical = -0.0000 0.0023 -0.0011 0 0 0 >> format short e >> EpsilonMechanical EpsilonMechanical = -4.4444e-005 2.2957e-003 -1.0514e-003 0 0 0 >> EpsilonThermal(1) = -0.01800e-6*30 EpsilonThermal = -5.4000e-007 >> EpsilonThermal(2) = 24.3e-6*30 EpsilonThermal = -5.4000e-007 7.2900e-004 >> EpsilonThermal(3) = 24.3e-6*30 EpsilonThermal = -5.4000e-007 7.2900e-004 7.2900e-004 >> EpsilonThermal(4) = 0 EpsilonThermal = -5.4000e-007 7.2900e-004 7.2900e-004 0 >> EpsilonThermal(5) = 0

Solutions to Problems 213 EpsilonThermal -5.4000e-0077.2900e-0047.2900e-004 0 0 >EpsilonThermal(6)=0 EpsilonThermal -5.4000e-0077.2900e-0047.2900e-004 0 >EpsilonThermal EpsilonThermal' EpsilonThermal -5.4000e-007 7.2900e-004 7.2900e-004 0 0 0 >Epsilon EpsilonMechanical EpsilonThermal Epsilon -4.4984e-005 3.0247e-003 -3.2242e-004 0 0 0 >d1=Eps11on(1)*60 d1= -2.6991e-003 >>d2=Ep3i1on(2)*60 d2= 1.8148e-001 >>d3=Epsi1on(3)*60

Solutions to Problems 213 EpsilonThermal = -5.4000e-007 7.2900e-004 7.2900e-004 0 0 >> EpsilonThermal(6) = 0 EpsilonThermal = -5.4000e-007 7.2900e-004 7.2900e-004 0 0 0 >> EpsilonThermal = EpsilonThermal’ EpsilonThermal = -5.4000e-007 7.2900e-004 7.2900e-004 0 0 0 >> Epsilon = EpsilonMechanical + EpsilonThermal Epsilon = -4.4984e-005 3.0247e-003 -3.2242e-004 0 0 0 >> d1 = Epsilon(1)*60 d1 = -2.6991e-003 >> d2 = Epsilon(2)*60 d2 = 1.8148e-001 >> d3 = Epsilon(3)*60

214 Solutions to Problems d3= -1.9345e-002 >> Problem 2.10 e1-a1△T-61△T S11 S12 S13 0 0 0 01 e2-a2△T-2△T S12 S22 52g 0 0 0 e3-ag△T-3△T 513 S23 S33 0 0 0 03 Y23 0 0 0 S44 0 0 T23 Y13 0 0 0 0 S55 0 T13 Y12 0 0 0 0 0 S66] T12 01 C11 C12 C13 0 0 0 E1-a1△T-61△T 02 C12 C22 C23 0 0 E2-a2△T-32△T 03 C13 C23 C33 0 0 0 E3-ag△T-3△T T23 0 0 0 CAA 0 0 23 T13 0 0 0 0 C65 0 Y13 T12 0 0 0 0 C66] Y12 Problem 3.1 Let A be the total cross-sectional area of the unit cell and let A and Am be the cross-sectional areas of the fiber and matrix,respectively.Then,we have the following relations based on the geometry of the problem: A+Amm=A Divide both sides of the above equation by A to obtain: A!Am A+A=1 Substituting A/A=V/and A"/A=Vm,we obtain(3.1)as follows: v5+vm=1

214 Solutions to Problems d3 = -1.9345e-002 >> Problem 2.10 ⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎩ ε1 − α1∆T − β1∆T ε2 − α2∆T − β2∆T ε3 − α3∆T − β3∆T γ23 γ13 γ12 ⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎪⎭ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ S11 S12 S13 000 S12 S22 S23 000 S13 S23 S33 000 000 S44 0 0 0000 S55 0 00000 S66 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎩ σ1 σ2 σ3 τ23 τ13 τ12 ⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎪⎭ ⎧ ⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎩ σ1 σ2 σ3 τ23 τ13 τ12 ⎫ ⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎭ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ C11 C12 C13 000 C12 C22 C23 000 C13 C23 C33 000 000 C44 0 0 0000 C55 0 00000 C66 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎧ ⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎩ ε1 − α1∆T − β1∆T ε2 − α2∆T − β2∆T ε3 − α3∆T − β3∆T γ23 γ13 γ12 ⎫ ⎪⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎪⎭ Problem 3.1 Let A be the total cross-sectional area of the unit cell and let Af and Am be the cross-sectional areas of the fiber and matrix, respectively. Then, we have the following relations based on the geometry of the problem: Af + Am = A Divide both sides of the above equation by A to obtain: Af A + Am A = 1 Substituting Af /A = V f and Am/A = V m, we obtain (3.1) as follows: V f + V m = 1