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