10 ELASTIC CONSTANTS OF UNIDIRECTIONAL COMPOSITES In this chapter we examine a distinct combination of two materials (matrix and fiber),with simple geometry and loading conditions,in order to estimate the elastic properties of the equivalent material,i.e.,of the composite. 10.1 LONGITUDINAL MODULUS E The two materials are shown schematically in Figure 10.1 where m stands for matrix. fstands for fiber. Hypothesis:The two materials are bonded together.More precisely,one makes the following assumptions: Both the matrix m and the fiber f have the same longitudinal strain 8. The interface between the two materials allows the z normal strains in the two materials to be different. ez≠ez m The state of stresses resulting from a force F can therefore be written as: Ot 0 00 Σ→ 0 0 0 Σ→ 0 0 0 m 0 0 ① 0 o① and the corresponding state of strains: e 0 0 Et 0 0 e→ 0 0 e→ 0 0 m 0 0 ① 0 0 2003 by CRC Press LLC
10 ELASTIC CONSTANTS OF UNIDIRECTIONAL COMPOSITES In this chapter we examine a distinct combination of two materials (matrix and fiber), with simple geometry and loading conditions, in order to estimate the elastic properties of the equivalent material, i.e., of the composite. 10.1 LONGITUDINAL MODULUS E The two materials are shown schematically in Figure 10.1 where m stands for matrix. f stands for fiber. Hypothesis: The two materials are bonded together. More precisely, one makes the following assumptions: Both the matrix m and the fiber f have the same longitudinal strain el . The interface between the two materials allows the z normal strains in the two materials to be different. The state of stresses resulting from a force F can therefore be written as: and the corresponding state of strains: e z π e z m f S m s 0 0 0 00 0 00 Æ m S f s 0 0 0 00 0 00 Æ f e m e 0 0 0 et 0 0 0 e z Æ m e f e 0 0 0 et 0 0 0 e z Æ f TX846_Frame_C10 Page 213 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
em m 1 Figure 10.1 Longitudinal Modulus E Each material is assumed to be linear elastic and isotropic,with the following stress-strain relation: e=1+VE-y (10.1) E trace ()I E in whiche represents the strain tensor,E represents the stress tensor,and I the unity tensor.E and v are the elastic constants of the considered material. For the composite (m+A),one uses Equation 9.5 with restriction to the plane I,t.It reduces to: Et E 0 OE 置 0 0 Yer tes 0 0 68 The stress cm can be written as(see Figure 10.1 above): F F em+ot e mter @+① @ which can be written in terms of the volume fractions of the fiber and the matrix as' oVm+o听 @+ @ TSee Section 3.2.2. 2003 by CRC Press LLC
Each material is assumed to be linear elastic and isotropic, with the following stress–strain relation: (10.1) in which e represents the strain tensor, S represents the stress tensor, and I the unity tensor. E and n are the elastic constants of the considered material. For the composite (m + f ), one uses Equation 9.5 with restriction to the plane l,t. It reduces to: The stress s(m+f ) can be written as (see Figure 10.1 above): which can be written in terms of the volume fractions of the fiber and the matrix as1 Figure 10.1 Longitudinal Modulus E 1 See Section 3.2.2. e 1 + n E ------------S n E = – -- trace ( ) S I e et Óg t˛ Ô Ô Ì ˝ Ô Ô Ï ¸ 1 E ---- nt Et –------ 0 nt E –------ 1 Et ---- 0 0 0 1 Gt ------- s st Ót t˛ Ô Ô Ì ˝ Ô Ô Ï ¸ = s F S -- F em + ef ( ) ¥ 1 ------------------------------ s em em + ef --------------- s ef em + ef == = + --------------- m + f m f s = sVm + sVf m f + m f TX846_Frame_C10 Page 214 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
Expressing the stresses in terms of the strains for each material yields Eee EmeeVm+ErEeVr then: Ee=EmVm+E☑ (10.2) Note:Among the real phenomena that are not taken into account in the expression of E is the lack of perfect straightness of the fibers in the matrix.Also, the modulus E,depends on the sign of the stress (tension or compression).In rigorous consideration,the material is "bimodulus." Example:Unidirectional layers with 60%fiber volume fraction (V=0.60)with epoxy matrix: Kevlar “HR"Carbon “HM"Carbon E tension(MPa) 85,000 134,000 180,000 Ec compression(MPa) 80,300 134,000 160,000 10.2 POISSON COEFFICIENT Considering again the loading defined in the previous paragraph,the transverse strain for the matrix m and fiber f can be written as: e,=-o,=-ve E and for the composite (m+f): Vux Gl =-VuEt @+① @+① The strain in the transverse direction can also be written as: E △(ente2=△eVn+9 em+er em er @+ E Vm+eV @+ ① Because e has the same value in m and f -VtrEt =-VmEiVm -VrEeVr ,=vnVm+y☑ (10.3) 2003 by CRC Press LLC
Expressing the stresses in terms of the strains for each material yields then: (10.2) Note: Among the real phenomena that are not taken into account in the expression of El is the lack of perfect straightness of the fibers in the matrix. Also, the modulus El depends on the sign of the stress (tension or compression). In rigorous consideration, the material is “bimodulus.” Example: Unidirectional layers with 60% fiber volume fraction (Vf = 0.60) with epoxy matrix: 10.2 POISSON COEFFICIENT Considering again the loading defined in the previous paragraph, the transverse strain for the matrix m and fiber f can be written as: and for the composite (m + f ): The strain in the transverse direction can also be written as: Because e has the same value in m and f: (10.3) Kevlar “HR” Carbon “HM” Carbon E tension (MPa) 85,000 134,000 180,000 E compression (MPa) 80,300 134,000 160,000 Ee = Eme Vm + Ef e Vf E = EmVm + Ef Vf et n E = = –--s –ne et m f + nt E ------ s m f + = = ¥ –nte et m f + D em + ef ( ) em + ef ------------------------ Dem em ---------Vm Def ef = = + -------Vf et m f + et m Vm et f = + Vf –nt e = –nme Vm -nf e Vf nt = nmVm + nfVf TX846_Frame_C10 Page 215 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
10.3 TRANSVERSE MODULUS E To evaluate the modulus along the transverse direction E,the two materials are shown in the Figure 10.2.In addition,one uses the following simplifications: Hypothesis:At the interface between the two materials,assume the following: Freedom of movement in the direction allows for different strains in the two materials: Et+Et @① Freedom of movement in the z direction allows for different strains in the two materials: Ez+Ez @① Then,the state of stress created by a load F(see Figure 10.2),can be reduced for each material to the following: 00 Σ→ 0 0,0 o 0 0 em 3 Figure 10.2 Transverse Modulus E 2003 by CRC Press LLC
10.3 TRANSVERSE MODULUS Et To evaluate the modulus along the transverse direction E, the two materials are shown in the Figure 10.2. In addition, one uses the following simplifications: Hypothesis: At the interface between the two materials, assume the following: Freedom of movement in the l direction allows for different strains in the two materials: Freedom of movement in the z direction allows for different strains in the two materials: Then, the state of stress created by a load F (see Figure 10.2), can be reduced for each material to the following: Figure 10.2 Transverse Modulus Et e m e f π e z m e z f π S 000 0 st 0 000 Æ TX846_Frame_C10 Page 216 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
The strains can be written as: e-→ 0 e 0 0 e ⑩ or ① Then for the composite (m+f),one has 1 E1= On the other hand,using direct calculation leads to (see Figure 10.2) △(em+ep em+er =EVm+EV m ① then: 1 1 =,+耳oy 1 E, V型+ or E=En a-+月 (10.4) Remarks: Due to the above simplifications that allow the possibility for relative sliding along the and z directions at the interface,the transverse modulus E, above may not be accurate. One finds in the technical literature many more complex formulae giving E.However,none can provide guaranteed good result. Taking into consideration the load applied (see Figure 10.2),:the modulus E,that appears in Equation 10.4 is the modulus of elasticity of the fiber in a direction that is perpendicular to the fiber axis.This modulus can be very different from the modulus along the axis of the fiber,due to the anisotropy that exists in fibers.2 This point was discussed in Paragraph 3.3.1. 2003 by CRC Press LLC
The strains can be written as: Then for the composite (m + f ), one has On the other hand, using direct calculation leads to (see Figure 10.2) then: (10.4) Remarks: Due to the above simplifications that allow the possibility for relative sliding along the l and z directions at the interface, the transverse modulus Et above may not be accurate. One finds in the technical literature many more complex formulae giving Et . However, none can provide guaranteed good result. Taking into consideration the load applied (see Figure 10.2),: the modulus Ef that appears in Equation 10.4 is the modulus of elasticity of the fiber in a direction that is perpendicular to the fiber axis. This modulus can be very different from the modulus along the axis of the fiber, due to the anisotropy that exists in fibers.2 2 This point was discussed in Paragraph 3.3.1. e e 0 0 0 et 0 0 0 e z m or f Æ et 1 Et = ----st et D em + ef ( ) em + ef ------------------------ et m Vm et f = = + Vf 1 Et ----st 1 Em ------st Vm 1 Ef = + ---- st Vf 1 Et ---- Vm Em ------ Vf Ef + ---- or Et Em 1 1 – Vf ( ) Em Ef + ------Vf = = ----------------------------------- TX846_Frame_C10 Page 217 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
10.4 SHEAR MODULUS Ga Load application that can be used to evaluate the shear modulus G is shown schematically in the Figure 10.3,both with the angular deformations that are produced.The state of stress,identical for both the matrix and fiber material,can be written as: 0 Tur Σ→ 0 0 10 0 0 The corresponding strains can be written as: 0 Et → Ett 0 0 @ or① 10 0 Using the constitutive equation,one has Et= 1+v E Tu= 2G then: 碧 Ya= Also,from Figure 10.3,one has Ye,(em+e）=Yem+Ye @+① m e → e 2 Figure 10.3 Shear modulus Ga 2003 by CRC Press LLC
10.4 SHEAR MODULUS Gt Load application that can be used to evaluate the shear modulus Gt is shown schematically in the Figure 10.3, both with the angular deformations that are produced. The state of stress, identical for both the matrix and fiber material, can be written as: The corresponding strains can be written as: Using the constitutive equation, one has then: Also, from Figure 10.3, one has Figure 10.3 Shear modulus Gt S 0 t t 0 t t 0 0 0 00 Æ e m or f 0 e t 0 e t 0 0 0 00 Æ e t 1 + n E ------------t t t t 2G = = ------- g t t t G = ----- g t m f + em + ef ( ) g tem m g tef f = + TX846_Frame_C10 Page 218 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
which can be rewritten as: Yer YeVm+Yevr @+① @ ① Gm tuvj 1 Gu=Gm a-叨+号 (10.5)3 10.5 THERMOELASTIC PROPERTIES 10.5.1 Isotropic Material:Recall When the influence of temperature is taken into consideration,Hooke's law for the case of no temperature influence: e=1z-营ace(21 is replaced by the Hooke-Dubamel law: e=z-首ace(②I+aaTI (10.6 where g=Strain tensor Σ=Stress tensor I=Unity tensor E,v Elastic constants for the considered material a Coefficient of thermal expansion' AT=Change in temperature with respect to a reference temperature at which the stresses and strains are nil 10.5.2 Case of Unidirectional Composite The coefficient of thermal expansion of the matrix is usually much larger (more than ten times)than that of the fiber.'In Figure 10.4,one can imagine that even in the absence of mechanical loading,a change in temperature AT will produce a 3A few values of the shear modulus are shown in Section 3.3.1. See Section 16,"Principal Physical Properties." 2003 by CRC Press LLC
which can be rewritten as: (10.5) 3 10.5 THERMOELASTIC PROPERTIES 10.5.1 Isotropic Material: Recall When the influence of temperature is taken into consideration, Hooke’s law for the case of no temperature influence: is replaced by the Hooke–Duhamel law: (10.6) where e = Strain tensor S = Stress tensor I = Unity tensor E, n = Elastic constants for the considered material a = Coefficient of thermal expansion 4 DT = Change in temperature with respect to a reference temperature at which the stresses and strains are nil 10.5.2 Case of Unidirectional Composite The coefficient of thermal expansion of the matrix is usually much larger (more than ten times) than that of the fiber. 4 In Figure 10.4, one can imagine that even in the absence of mechanical loading, a change in temperature DT will produce a 3 A few values of the shear modulus Gf are shown in Section 3.3.1. 4 See Section 1.6, “Principal Physical Properties.” g t m + f g tVm m g tVf f = + t t Gt ------- t t Gm -------Vm t t Gf = + -----Vf 1 Gt ------- Vm Gm ------- Vf Gf = + ----- Gt Gm 1 1 – Vf ( ) Gm Gf + -------Vf = ------------------------------------ e 1 + n E ------------S n E = – -- trace( ) S I e 1 + n E ------------S n E = – -- trace( ) S I + aDT I TX846_Frame_C10 Page 219 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
em m m Figure 10.4 Unidirectional Composite longitudinal strain in the composite.This longitudinal strain has a value that is intermediate between the strain of the fiber alone and that of the matrix alone. Then,in the composite one finds internal stresses along the direction /(along the direction t,the fiber and matrix can expand differently).One then has: ■For the stresses: Or 00 6 0 0 E 0 00 Σ→ 0 0 m 00 m ① 0 0 ■For the strains: 0 0入 [e0 0入 ￡→ 0 0 e→ 0 E 0 @ 0 0 m 00 10.5.2.1 Coefficient of Thermal Expansion along the Direction I One has for the fiber and the matrix,respectively: Ot Et= m ① The external equilibrium can be written as (see Figure 10.4): Gexem+oixef=0 ① 2003 by CRC Press LLC
longitudinal strain in the composite. This longitudinal strain has a value that is intermediate between the strain of the fiber alone and that of the matrix alone. Then, in the composite one finds internal stresses along the direction l (along the direction t, the fiber and matrix can expand differently). One then has: For the stresses: For the strains: 10.5.2.1 Coefficient of Thermal Expansion along the Direction l One has for the fiber and the matrix, respectively: The external equilibrium can be written as (see Figure 10.4): Figure 10.4 Unidirectional Composite S m s 0 0 0 00 0 00 m S f s 0 0 0 00 0 00 f Æ Æ e m e 0 0 0 et 0 0 0 e z m e f e 0 0 0 et 0 0 0 e z f Æ Æ e m s Em ------ m + amDT e f s Ef = == ----- f + af DT s ¥ em m s ¥ ef f + = 0 TX846_Frame_C10 Page 220 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
where,taking into account the equality of the strains: @+a4r=-0×号×+4A7 er @ = (ca-am)△T_(c-am)△T @ 左+×号 V represents the volume fraction.The longitudinal strain can then be written as: Et=Et= (OI EV+am EmVm)(△T) ErVr+EmVm @① It is also the longitudinal strain that is created only by the effect of temperature: Et =ae△T @+① where a,is the longitudinal coefficient of thermal expansion.One can then equate the above expressions to obtain: ar ErVr+am Em Vm (10.7) ErVr+EmVm 10.5.2.2 Coefficient of Thermal Expansion along the Transverse Direction t The global thermal strain can be written as (see Figure 10.4): △(emte )=e-em+e er em+ef em+er em+er @+① (m then: Er =e,XVm+e,×V @+① 四① Using the Hooke and Duhamel law (Equation 10.6): )+① 5 For the Poisson coefficients of common fibers,see Section 3.3.1. 2003 by CRC Press LLC
where, taking into account the equality of the strains: V represents the volume fraction. The longitudinal strain can then be written as: It is also the longitudinal strain that is created only by the effect of temperature: where a is the longitudinal coefficient of thermal expansion. One can then equate the above expressions to obtain: (10.7) 10.5.2.2 Coefficient of Thermal Expansion along the Transverse Direction t The global thermal strain can be written as (see Figure 10.4): then: Using the Hooke and Duhamel law (Equation 10.6)5 : 5 For the Poisson coefficients of common fibers, see Section 3.3.1. s Em ------ m + amDT –s m em ef ----- 1 Ef = ¥ ¥ ---- + af DT s m ( )D af – am T 1 Em ------ em ef ----- 1 Ef + ¥ ---- ------------------------------ ( )D af – am T 1 Em ------ Vm Vf ------ 1 Ef + ¥ ---- = = ------------------------------ e m e f ( )D af EfVf + am EmVm ( ) T EfVf + EmVm = = ------------------------------------------------------------------- e m + f = aDT a af EfVf + am EmVm EfVf + EmVm = ------------------------------------------------ et m + f D em + ef ( ) em + ef ------------------------ et m em em + ef --------------- et f ef em + ef = = + --------------- et m + f et ¥ Vm m et ¥ Vf f = + et m + f Vm Em ------s m – + amDT Ë ¯ Ê ˆ Vm nf Ef ----s f – + af DT Ë ¯ Ê ˆ = + Vf TX846_Frame_C10 Page 221 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC
Using the expressions for stresses obtained before,one obtains: a++.以4-a EmVm+EV m+f The quantity between the brackets represent the coefficient of thermal expan- sion along the transverse direction t,a,,which can be written as: a,=aa V+avEa-VE 2×(r-0m) (10.8) +子 10.5.3 Thermomechanical Behavior of a Unidirectional Layer Under the combined effect of the stresses and temperature,the global thermo- mechanical strains of a unidirectional layer can be obtained using the following relation: 0 Et ΓE de E E 0 6 +△T C (10.9) Yer 0 0 0 in which the coefficients E,E ve Ger,a and a,have the values given by the Equations 10.2 to 10.8,respectively. 2003 by CRC Press LLC
Using the expressions for stresses obtained before, one obtains: The quantity between the brackets represent the coefficient of thermal expansion along the transverse direction t, at , which can be written as: (10.8) 10.5.3 Thermomechanical Behavior of a Unidirectional Layer Under the combined effect of the stresses and temperature, the global thermomechanical strains of a unidirectional layer can be obtained using the following relation: (10.9) in which the coefficients E, Et , vt , Gt , a and at have the values given by the Equations 10.2 to 10.8, respectively. et m + f amVm + afVf ( ) nfEm – nmEf ( ) EmVm + EfVf + ---------------------------------VmVf( ) af – am Ó ˛ Ì ˝ Ï ¸ = DT at amVm afVf nf Em – nmEf ( ) Em Vf ------ Ef Vm + ------ = + + ---------------------------------- ¥ ( ) af – am e et Óg t˛ Ô Ô Ì ˝ Ô Ô Ï ¸ 1 E ---- nt Et –------ 0 vt E –----- 1 Et ---- 0 0 0 1 Gt ------- s st Ót t˛ Ô Ô Ì ˝ Ô Ô Ï ¸ DT a at 0 Ó ˛ Ô Ô Ì ˝ Ô Ô Ï ¸ = + TX846_Frame_C10 Page 222 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC