PART II MECHANICAL BEHAVIOR OF LAMINATED MATERIALS We have introduced in the previous part the anisotropic properties of composite materials from a qualitative point of view,'and we have indicated the characteristic elastic constants for the behavior of an anisotropic layer in its plane. We have also mentioned the relations that allow one to predict the mechanical behavior of a fiber/matrix combination from the properties of the individual constituents.2In Chapter 5,3 we have also given the elements necessary for the sizing of the laminates made of carbon/epoxy,Kevlar/epoxy,and glass/epoxy, in terms of strength and deformation. This second part is dedicated to the justification and application of these properties and results.It requires a detailed study of the behavior of anisotropic composite layers and of the stacking that makes up the laminate.It is useful to note that the basics of the mechanics of continuous media-namely,the state of stress and strain at a point-already described in great details in many texts on mechanics of materials,are supposed to be known. See Section 3.1. 2 See Section 3.3.1. 3 See Section 5.2/5.3. 2003 by CRC Press LLC
PART II MECHANICAL BEHAVIOR OF LAMINATED MATERIALS We have introduced in the previous part the anisotropic properties of composite materials from a qualitative point of view,1 and we have indicated the characteristic elastic constants for the behavior of an anisotropic layer in its plane. We have also mentioned the relations that allow one to predict the mechanical behavior of a fiber/matrix combination from the properties of the individual constituents.2 In Chapter 5,3 we have also given the elements necessary for the sizing of the laminates made of carbon/epoxy, Kevlar/epoxy, and glass/epoxy, in terms of strength and deformation. This second part is dedicated to the justification and application of these properties and results. It requires a detailed study of the behavior of anisotropic composite layers and of the stacking that makes up the laminate. It is useful to note that the basics of the mechanics of continuous media—namely, the state of stress and strain at a point—already described in great details in many texts on mechanics of materials, are supposed to be known. 1 See Section 3.1. 2 See Section 3.3.1. 3 See Section 5.2/5.3. TX846_Frame_C09 Page 205 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC
ANISOTROPIC ELASTIC MEDIA 9.1 REVIEW OF NOTATIONS 9.1.1 Continuum Mechanics We consider the following classical notions and notations of the mechanics of continuous media: State of stress at a point:This is defined by a second order tensor with the symbol E.The 3 by 3 matrix associated with this tensor is symmetric. In this matrix,there are six distinct terms,which are denoted as oy 011;022;033;023;13;012 State of strain at a point:This is defined as a second order tensor 8.The 3 by 3 matrix for this tensor is symmetric.It consists of six distinct terms denoted as sy C11;C22633;E23;E13i612 Linear elastic medium:The strains are linear and homogeneous functions of the stresses.The corresponding relations are: Ey Pyke x Gut Homogeneous medium:In this case,the matrix terms oy characterizing the elastic behavior of the medium are not point functions.They are the same at all points in the considered medium. 1 For example: E11=p1111011+01112012+p1113013+p1121021+p1122022+p1123023+p1131031+p1152032+p011503分 2003 by CRC Press LLC
9 ANISOTROPIC ELASTIC MEDIA 9.1 REVIEW OF NOTATIONS 9.1.1 Continuum Mechanics We consider the following classical notions and notations of the mechanics of continuous media: State of stress at a point: This is defined by a second order tensor with the symbol Â. The 3 by 3 matrix associated with this tensor is symmetric. In this matrix, there are six distinct terms, which are denoted as sij: s11 ; s22; s33 ; s23 ; s13 ; s12 State of strain at a point: This is defined as a second order tensor e. The 3 by 3 matrix for this tensor is symmetric. It consists of six distinct terms denoted as eij: e11; e22; e33; e23; e13; e12 Linear elastic medium: The strains are linear and homogeneous functions of the stresses. The corresponding relations are: 1 Homogeneous medium: In this case, the matrix terms jijkl characterizing the elastic behavior of the medium are not point functions. They are the same at all points in the considered medium. 1 For example: e11 = j1111 s11 + j1112 s12 + j1113 s13 + j1121 s21 + j1122 s22 + j1123 s23 + j1131 s31 + j1132 s32 + j1133 s33. eij = jijk ¥ sk TX846_Frame_C09 Page 207 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC
9.1.2 Number of Distinct ike Terms The above stress-strain relation can be written in matrix form as: 7 11 e22 33 6×6 6×3 039 23 %33 e13 e12 Yigki 03 012 年。e果e车e年e年年e年 e32 032 Eg1 3×6 3×3 0g1 E21 021 Due to the symmetry of the stresses (ot=o,the corresponding coefficients are the same,i.e.,ye=Pye. Due to the symmetry of the strains (E=s,the corresponding coefficients are the same,i.e.,yee=e.In other words,the knowledge of only the coefficients of the 6 x 6 matrix written above is required. In addition,application of the theorem of virtual work on the stresses shows that the coefficients are symmetric,meaning:uke=Pty. Therefore,the 6 x 6 matrix mentioned previously is symmetric.The number of distinct coefficients is: 6(6+1)=21 coefficients 2 Consider two simple stress states: State No.1:One single stress,(o),which causes the strain: ()=Pikt (Oxe) State No.2:One single stress,(2,which causes the strain: (Emn)2=Pmnpq (pg)2 One can write that the work of the stress in state No.I on the strain in state No.2 is equal to the work of the stress in state No.2 on the strain in state No.1,as: (ok)h×(ekd)2=(Cpg)2×(epgh which means:(ceth×Pkipa×(opg2=(opg2×Ppake×(okt)h from which one has: elpg Ppgke 2003 by CRC Press LLC
9.1.2 Number of Distinct jijk Terms The above stress–strain relation can be written in matrix form as: Due to the symmetry of the stresses ( ), the corresponding coefficients are the same, i.e., . Due to the symmetry of the strains (eij = eji), the corresponding coefficients are the same, i.e., . In other words, the knowledge of only the coefficients of the 6 ¥ 6 matrix written above is required. In addition, application of the theorem of virtual work on the stresses shows that the coefficients jijk are symmetric, meaning: .2 Therefore, the 6 ¥ 6 matrix mentioned previously is symmetric. The number of distinct coefficients is: 2 Consider two simple stress states: • State No. 1: One single stress, (sk)1, which causes the strain: (eij)1 = jijk (sk)1 • State No. 2: One single stress, (spq)2, which causes the strain: (emn)2 = jmnpq (spq)2 One can write that the work of the stress in state No. 1 on the strain in state No. 2 is equal to the work of the stress in state No. 2 on the strain in state No. 1, as: which means: from which one has: jkpq = jpqk sk = sk jijk = jijk jijk = jjik jijk = jkij sk ( )1 e k ( ) ¥ 2 spq ( )2 e pq ( ) = ¥ 1 sk ( )1 jkpq spq ( ) ¥ ¥ 2 spq ( )2 j pqk sk ( ) = ¥ ¥ 1 66 1 ( ) + 2 -------------------- = 21 coefficients TX846_Frame_C09 Page 208 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC
■In summary: stress reciprocity:Piee=Pytk strain definition:uee= (9.1) symmetry:ike =ety There remain 21 distinct coefficients The previous stress-strain relation can then be written as: > Sn 01111 0112201133 201123 201113 201112 6 62 p211 02222 02233 202225 202213 20212 62 6 03311 03322 φ3333 203325 203313 203312 033 E25 02311 023220233 202323 202313 2p2312 023 E13 01311 01322 01333 201323 201313 201312 013 E12 012110122201233 2p1223201213 201212 012 The matrix above does not have the general symmetry as in the general form (9x 9)presented previously.(Note the coefficients 2 in this matrix).One can get around this inconvenience by doubling the terms 823,E13,812,introducing the shear strains: Y23=2E23;Y13=2813;Y12=2812 from which the stress-strain behavior can then be written in a symmetric form as: 11 01111 01122 p1133 201123 201113 201112 011 62 0211 02222 02233 20223 202213 202212 02 633 03311 03322 03333 203323 203313 203512 63 (9.2) 2623 Y23 202311 202322 202333 402323 402313 402312 025 2813 Y13 201311 201322 201333 401323 401313 401312 013 2612 Y12 2p121 201222 201233 401223 401213 401212 012 9.2 ORTHOTROPIC MATERIALS Definition:An orthotropic material is a homogeneous linear elastic material having two planes of symmetry at every point in terms of mechanical properties,these two planes being perpendicular to each other. 2003 by CRC Press LLC
In summary: The previous stress–strain relation can then be written as: The matrix above does not have the general symmetry as in the general form (9 ¥ 9) presented previously. (Note the coefficients 2 in this matrix). One can get around this inconvenience by doubling the terms e23, e13, e12, introducing the shear strains: from which the stress–strain behavior can then be written in a symmetric form as: (9.2) 9.2 ORTHOTROPIC MATERIALS Definition: An orthotropic material is a homogeneous linear elastic material having two planes of symmetry at every point in terms of mechanical properties, these two planes being perpendicular to each other. stress reciprocity: strain definition: symmetry: There remain 21 distinct coefficients (9.1) jijk = jijk jijk = jjik jijk = jkij jijk e 11 e 22 e 33 e 23 e 13 e 12 Ó ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ j1111 j1122 j1133 2j1123 2j1113 2j1112 j2211 j2222 j2233 2j2223 2j2213 2j2212 j3311 j3322 j3333 2j3323 2j3313 2j3312 j2311 j2322 j2333 2j2323 2j2313 2j2312 j1311 j1322 j1333 2j1323 2j1313 2j1312 j1211 j1222 j1233 2j1223 2j1213 2j1212 s11 s22 s33 s23 s13 s12 Ó ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ = g 23 2e 23; g 13 2e 13 = = ; g 12 = 2e 12 e 11 e 22 e 33 2e 23 2e 13 2e 12 Ó ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ g 23 g 13 g 12 j1111 j1122 j1133 2j1123 2j1113 2j1112 j2211 j2222 j2233 2j2223 2j2213 2j2212 j3311 j3322 j3333 2j3323 2j3313 2j3312 2j2311 2j2322 2j2333 4j2323 4j2313 4j2312 2j1311 2j1322 2j1333 4j1323 4j1313 4j1312 2j1211 2j1222 2j1233 4j1223 4j1213 4j1212 s11 s22 s33 s23 s13 s12 Ó ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ = TX846_Frame_C09 Page 209 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC
Then one can show that'the number of independent elastic constants is nine. The constitutive relation expressed in the so-called "orthotropic"axes,defined by three axes constructed on the two orthogonal planes and their intersection line,can be written in the following form,called the engineering notation because it utilizes the elastic modulus and Poisson ratios: 811 V 0 0 0 011 E E 22 岩 言 0 0 0 022 E33 E 0 0 0 033 9.3) Y25 0 0 0 023 Y13 0 0 0 0 013 1 Y12 0 0 0 0 0 012 where E,E2,E3 are the longitudinal elastic moduli. G23,Gs,G2 are the shear moduli. Vi2,Vi3,V23,V2,V3,V32 are the Poisson ratios. In addition,the symmetry of the stress-strain matrix above leads to the following relations: (9.4) 9.3 TRANSVERSELY ISOTROPIC MATERIALS Definition:A transversely isotropic material is a homogeneous linear elastic material such that any plane including a preferred axis,is a plane of mechanical symmetry. One can show that'the constitutive relation has five independent elastic constants.For the fiber/matrix composite shown in Figure 9.1 the preferred axis is e.The fibers are distributed uniformly in the direction along e.All directions perpendicular to the fibers characterize the transverse direction t. 3 Proof is shown in Section 13.1. 4Proof is shown in Section 13.2. 2003 by CRC Press LLC
Then one can show that3 the number of independent elastic constants is nine. The constitutive relation expressed in the so-called “orthotropic” axes, defined by three axes constructed on the two orthogonal planes and their intersection line, can be written in the following form, called the engineering notation because it utilizes the elastic modulus and Poisson ratios: (9.3) where E1, E2, E3 are the longitudinal elastic moduli. G23, G13, G12 are the shear moduli. n12, n13, n23, n21, n31, n32 are the Poisson ratios. In addition, the symmetry of the stress–strain matrix above leads to the following relations: (9.4) 9.3 TRANSVERSELY ISOTROPIC MATERIALS Definition: A transversely isotropic material is a homogeneous linear elastic material such that any plane including a preferred axis, is a plane of mechanical symmetry. One can show that 4 the constitutive relation has five independent elastic constants. For the fiber/matrix composite shown in Figure 9.1 the preferred axis is . The fibers are distributed uniformly in the direction along . All directions perpendicular to the fibers characterize the transverse direction t. 3 Proof is shown in Section 13.1. 4 Proof is shown in Section 13.2. e 11 e 22 e 33 g 23 g 13 g 12 Ó ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ 1 E1 ---- n21 E2 –------ n31 E3 –------ 000 n12 E1 –------ 1 E2 ---- n32 E3 –------ 000 n13 E1 –------ n23 E2 –------ 1 E3 ---- 000 000 1 G23 ------- 0 0 0 0 00 1 G13 ------- 0 0 0 0 00 1 G12 ------- s11 s22 s33 s23 s13 Ó s12 ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ n21 E2 ------ n12 E1 ------; n31 E3 ------ n13 E1 ------; n32 E3 ------ n23 E2 = = = ------ TX846_Frame_C09 Page 210 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC
Figure 9.1 Transversely Isotropic Unidirectional The engineering stress-strain relation has the form: Ett 1 V Ot 创 E E 0 0 0 Ver 1 0 0 0 Gu E E Eer Vir E 0 0 0 Orr Ee (9.5) Yu 0 0 0 2(1+v) 0 0 Tw E Yer 0 0 0 0 1 0 Gu Tir 1 Yer 0 0 0 0 Gu Remarks:The independent elastic constants are Young modulus along the direction:Ec. Young modulus along any transverse direction t:E. Shear modulus in the plane t:Ga Poisson coefficients:ve and v The symmetry of the coefficients of the constitutive relation leads to E One may also note that the shear modulus in the plane t,t'can be written as: E 2(1+v) This equation is classic for isotropic materials 2003 by CRC Press LLC
The engineering stress–strain relation has the form: (9.5) Remarks: The independent elastic constants are Young modulus along the direction: E. Young modulus along any transverse direction t: Et . Shear modulus in the plane , t: Gt . Poisson coefficients: nt and nt . The symmetry of the coefficients of the constitutive relation leads to One may also note that the shear modulus in the plane t, t¢ can be written as: This equation is classic for isotropic materials. Figure 9.1 Transversely Isotropic Unidirectional e ett et¢t¢ g tt¢ g t¢ g t Ó ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ 1 E ---- nt Et –------ nt Et –------ 0 00 nt E –------ 1 Et ---- nt Et –---- 0 00 nt E –------ nt Et –---- 1 Et ---- 0 00 000 2 1 + nt ( ) Et --------------------- 0 0 000 0 1 Gt ------- 0 000 0 0 1 Gt ------- s stt st¢t¢ ttt¢ t t¢ Ó t t ˛ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ì ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ï ¸ = nt E ------ nt Et = ------ Et 2 1 + nt ( ) --------------------- TX846_Frame_C09 Page 211 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC
