76 Mechanics of Composite Materials,Second Edition FIGURE 2.9 Cartesian coordinates in a three-dimensional body. elastic and has small deformations,stresses and strains at a point are related through six simultaneous linear equations called Hooke's law. Note that 15 unknown parameters are at a point:six stresses,six strains, and three displacements.Combined with six simultaneous linear equa- tions of Hooke's law,six strain-displacement relations-given by Equa- tion (2.8),Equation (2.11),Equation (2.15),and Equation (2.16)-and three equilibrium equations give 15 equations for the solution of 15 unknowns.Because strain-displacement and equilibrium equations are differential equations,they are subject to knowing boundary conditions for complete solutions. For a linear isotropic material in a three-dimensional stress state,the Hooke's law stress-strain relationships at a point in an x-y-z orthogonal system(Figure 2.9)in matrix form are H 0 0 0 -￥ 0 0 0 E E Ox 0 0 0 Ez vE O: (2.17 0 0 ty 0 0 Ta 0 0 0 0 0 0 2006 by Taylor Francis Group,LLC
76 Mechanics of Composite Materials, Second Edition elastic and has small deformations, stresses and strains at a point are related through six simultaneous linear equations called Hooke’s law. Note that 15 unknown parameters are at a point: six stresses, six strains, and three displacements. Combined with six simultaneous linear equations of Hooke’s law, six strain-displacement relations — given by Equation (2.8), Equation (2.11), Equation (2.15), and Equation (2.16) — and three equilibrium equations give 15 equations for the solution of 15 unknowns.1 Because strain-displacement and equilibrium equations are differential equations, they are subject to knowing boundary conditions for complete solutions. For a linear isotropic material in a three-dimensional stress state, the Hooke’s law stress–strain relationships at a point in an x–y–z orthogonal system (Figure 2.9) in matrix form are (2.17) FIGURE 2.9 Cartesian coordinates in a three-dimensional body. z y x x y z yz zx xy E ε ε ε γ γ γ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = − 1 ν ν ν ν ν ν E E EEE EEE G − − − − − 000 1 000 1 000 0 0 01 0 0 0 0 0 01 G G 0 0 0 0 0 01 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ x y z yz zx xy σ σ σ τ τ τ , 1343_book.fm Page 76 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 77 E1-v) VE vE 00 0 Ox (1-2v)1+v) (1-2v)1+v) (1-2v)(1+v) Ex vE E(1-v) vE 0 0 0 Oy Ey (1-2v)(1+v) (1-2v)1+v) (1-2v)(1+v) Oz VE VE E(1-v) 0 0 Ty (1-2v)1+v) (1-2v)1+v) (1-2v)(1+v) 0 0 0 G 0 0 0 0 0 G 0 0 0 0 0 G (2.18) where v is the Poisson's ratio.The shear modulus G is a function of two elastic constants,E and v,as E G- (2.19) 21+v) The 6 x 6 matrix in Equation(2.17)is called the compliance matrix [S]of an isotropic material.The 6x 6 matrix in Equation(2.18),obtained by invert- ing the compliance matrix in Equation(2.17),is called the stiffness matrix [C]of an isotropic material. 2.2.4 Strain Energy Energy is defined as the capacity to do work.In solid,deformable,elastic bodies under loads,the work done by external loads is stored as recoverable strain energy.The strain energy stored in the body per unit volume is then defined as w=1 2xex+Gyey+o:e:+t Yyty+tY) (2.20) Example 2.2 Consider a bar of cross-section A and length L(Figure 2.10).A uniform tensile load P is applied to the two ends of the rod;find the state of stress and strain, and strain energy per unit volume of the body.Assume that the rod is made of a homogeneous isotropic material of Young's modulus,E. 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 77 (2.18) where ν is the Poisson’s ratio. The shear modulus G is a function of two elastic constants, E and ν, as (2.19) The 6 × 6 matrix in Equation (2.17) is called the compliance matrix [S] of an isotropic material. The 6 × 6 matrix in Equation (2.18), obtained by inverting the compliance matrix in Equation (2.17), is called the stiffness matrix [C] of an isotropic material. 2.2.4 Strain Energy Energy is defined as the capacity to do work. In solid, deformable, elastic bodies under loads, the work done by external loads is stored as recoverable strain energy. The strain energy stored in the body per unit volume is then defined as (2.20) Example 2.2 Consider a bar of cross-section A and length L (Figure 2.10). A uniform tensile load P is applied to the two ends of the rod; find the state of stress and strain, and strain energy per unit volume of the body. Assume that the rod is made of a homogeneous isotropic material of Young’s modulus, E. x y z yz zx xy E σ σ σ τ τ τ ν ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = (1− ) ( )( ) ( )( ) ( )( ) 12 1 12 1 12 1 000 −+ −+ −+ ν ν ν ν ν ν ν ν ν E E EE E ( )( ) ( ) 12 1 ( )( ) ( )( ) 1 − + 12 1 12 1 − ν ν −+ −+ ν ν ν ν ν ν 000 12 1 12 1 1 1 2 ν ν ν ν ν ν ν ν E EE ( )( ) ( )( ) ( ) −+ −+ ( )( − − 1 000 0 0 0 00 0 0 00 0 0 0 000 + ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ν) G G G ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ x y z yz zx xy ε ε ε γ γ γ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ , G = E 2(1 + ) . ν W x x y y z z xy xy = + ++ + 1 2 (σε σε σε τ γ τyz yz zx zx γ + τ γ ). 1343_book.fm Page 77 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
78 Mechanics of Composite Materials,Second Edition Cross-section'A' FIGURE 2.10 Cylindrical rod under uniform uniaxial load,P. Solution The stress state at any point is given by P 0x=A0y=0,0:=0,tg=0,tax=0,ty=0. (2.21)） If the circular rod is made of an isotropic,homogeneous,and linearly elastic material,then the stress-strain at any point is related as 0 0 0 0 0 0 E E Ey 1 0 0 0 Ez E E 00 r 0 (2.22) 0 0 0 0 0 0 0 0 0 0 0 P ex=A正y-AE=P AE (2.23) Ye=0,Ya=0,Yw=0. The strain energy stored per unit volume in the rod,per Equation(2.20),is 2006 by Taylor Francis Group,LLC
78 Mechanics of Composite Materials, Second Edition Solution The stress state at any point is given by (2.21) If the circular rod is made of an isotropic, homogeneous, and linearly elastic material, then the stress–strain at any point is related as (2.22) (2.23) The strain energy stored per unit volume in the rod, per Equation (2.20), is FIGURE 2.10 Cylindrical rod under uniform uniaxial load, P. P L Cross-section ‘A’ z y P x x y z yz zx P A σ σστ τ = == == ,,, ,, 00 0 0 τxy = 0. x y z yz zx xy E ε ε ε γ γ γ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = − 1 ν ν ν ν ν ν E E EEE EEE G − − − − − 000 1 000 1 000 0 0 01 0 0 0 0 0 01 G G 0 0 0 0 0 01 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ P A 0 0 0 0 0 , xy z yz P AE P AE P AE ε ε ν ε ν γ = =− =− = ,,, 0, zx xy γ γ = = 0 0 , . 1343_book.fm Page 78 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 79 w=是o(是)}o-器)roo+o0+oo 1p2 2A2E 11 2E (2.24) 2.3 Hooke's Law for Different Types of Materials The stress-strain relationship for a general material that is not linearly elastic and isotropic is more complicated than Equation(2.17)and Equation(2.18). Assuming linear and elastic behavior for a composite is acceptable;however, assuming it to be isotropic is generally unacceptable.Thus,the stress-strain relationships follow Hooke's law,but the constants relating stress and strain are more in number than seen in Equation(2.17)and Equation(2.18).The most general stress-strain relationship is given as follows for a three-dimen- sional body in a 1-2-3 orthogonal Cartesian coordinate system: 01 Cn C12 C13 C14 C15 C16 E1 02 C21 C22 C23 C24 C25 C26 03 C31 C32 C33 C34 C35 C36 E3 (2.25) T23 C42 C43 C4 C45 C46 Ys T31 C51 C52 Cx3 Cs Css C56 Ya T12 C61 C62 C63 C64 Cos C66] Y] where the 6 x 6 [C]matrix is called the stiffness matrix.The stiffness matrix has 36 constants. What happens if one changes the system of coordinates from an orthogonal system 1-2-3 to some other orthogonal system,1'-2'-3'?Then,new stiffness and compliance constants will be required to relate stresses and strains in the new coordinate system 1'-2'-3'.However,the new stiffness and compli- ance matrices in the 1'-2'-3'system will be a function of the stiffness and compliance matrices in the 1-2-3 system and the angle between the axes of the 1'-2'-3'system and the 1-2-3 system. 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 79 . (2.24) 2.3 Hooke’s Law for Different Types of Materials The stress–strain relationship for a general material that is not linearly elastic and isotropic is more complicated than Equation (2.17) and Equation (2.18). Assuming linear and elastic behavior for a composite is acceptable; however, assuming it to be isotropic is generally unacceptable. Thus, the stress–strain relationships follow Hooke’s law, but the constants relating stress and strain are more in number than seen in Equation (2.17) and Equation (2.18). The most general stress–strain relationship is given as follows for a three-dimensional body in a 1–2–3 orthogonal Cartesian coordinate system: (2.25) where the 6 × 6 [C] matrix is called the stiffness matrix. The stiffness matrix has 36 constants. What happens if one changes the system of coordinates from an orthogonal system 1–2–3 to some other orthogonal system, 1′–2′–3′? Then, new stiffness and compliance constants will be required to relate stresses and strains in the new coordinate system 1′–2′–3′. However, the new stiffness and compliance matrices in the 1′–2′–3′ system will be a function of the stiffness and compliance matrices in the 1–2–3 system and the angle between the axes of the 1′–2′–3′system and the 1–2–3 system. W P A P AE P AE = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + −⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 1 2 ( ) 0 ν P AE ( ) ( ) ( ) ( )( ) ( )( 0 0 0 00 0 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ++ ν 0) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 1 2 2 2 P A E = 1 2 2 σx E 1 2 3 23 31 12 σ 11 12 σ σ τ τ τ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = C CCCCC CCCCCC CCC 13 14 15 16 21 22 23 24 25 26 31 32 33 34 CCC CCCCCC CCCCC 35 36 41 42 43 44 45 46 51 52 53 54 55 56 C CCCCCC 61 62 63 64 65 66 ⎡ 1 ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ε 2 3 23 31 12 ε ε γ γ γ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ , 1343_book.fm Page 79 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
80 Mechanics of Composite Materials,Second Edition Inverting Equation(2.25),the general strain-stress relationship for a three- dimensional body in a 1-2-3 orthogonal Cartesian coordinate system is 分 S11 S12 S13 S14 S15 S16 61 S21 S22 S23 S24 S25 S26 02 E3 S31 S32 S38 S34 S35 S36 03 (2.26) Yz Sa S42 S43 S44 S4 S46 T23 Y31 S51 S52 S53 S54 55 S56 T31 S61 S62 S63 S64 S65 S66JLt12】 In the case of an isotropic material,relating the preceding strain-stress equation to Equation(2.17),one finds that the compliance matrix is related directly to engineering constants as 1 S11=E=5n=5a8 5B=-岂5B=5a=58=5m=5, (2.27) 1 =S55=S66, and S other than in the preceding,are zero. It can be shown that the 36 constants in Equation(2.25)actually reduce to 21 constants due to the symmetry of the stiffness matrix [C]as follows.The stress-strain relationship (2.25)can also be written as Ci=1.6, (2.28) where,in a contracted notation, 04=T23,05=T31,06=T12, e4=Y23ve5=Y31e6=Y12 (2.29a-f) 2006 by Taylor Francis Group,LLC
80 Mechanics of Composite Materials, Second Edition Inverting Equation (2.25), the general strain–stress relationship for a threedimensional body in a 1–2–3 orthogonal Cartesian coordinate system is . (2.26) In the case of an isotropic material, relating the preceding strain–stress equation to Equation (2.17), one finds that the compliance matrix is related directly to engineering constants as (2.27) and Sij, other than in the preceding, are zero. It can be shown that the 36 constants in Equation (2.25) actually reduce to 21 constants due to the symmetry of the stiffness matrix [C] as follows. The stress–strain relationship (2.25) can also be written as , (2.28) where, in a contracted notation, (2.29a–f) 1 2 3 23 31 12 11 12 ε ε ε γ γ γ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = S SSSSS SSSSSS SSS 13 14 15 16 21 22 23 24 25 26 31 32 33 34 SSS SSSSSS SSSSS 35 36 41 42 43 44 45 46 51 52 53 54 55 56 S SSSSSS 61 62 63 64 65 66 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ σ 2 3 23 31 12 σ σ τ τ τ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 11 22 33 1 S E == = S S S12 13 21 23 31 32 E =− = = = = = SSSSS ν , 44 55 66 1 S G== = S S , σ ε i ij j j = =… C i = ∑ , 16 1 6 στστ στ 4 23 5 31 6 12 === ,,, ε4 23 5 31 6 12 === γ ,,. ε γ ε γ 1343_book.fm Page 80 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC