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The shear modulus of a composite material with a transversely isotropic matrix and a fibre
Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
Contents lists available at ScienceDirect
Journal of Applied Mathematics and Mechanics journal homepage: www.elsevier.com/locate/jappmathmech
The shear modulus of a composite material with a transversely isotropic matrix and a fibre夽 S.N. Grebenyuk Zaporizhia, Ukraine
a r t i c l e
i n f o
Article history: Received 27 November 2012
a b s t r a c t A relation is proposed for determining of the shear modulus of a fibrous composite material with a transversely isotropic matrix and a fibre as a function of the elastic constants of the matrix and the fibre as well as the volume fraction of each of them in the composite material. The isotropy planes of the matrix and fibre coincide and are perpendicular to the fibre axis. Two boundary value problems are solved in order to obtain the required relation: the problem of the longitudinal shear of a transversely isotropic solid cylinder that simulates the fibrous composite material and the problem of the combined longitudinal shear of a hollow and solid cylinder that simulate the matrix material and the fibre material respectively. Calculations using the proposed formula are compared with the available experimental data. © 2014 Elsevier Ltd. All rights reserved.
In the majority of studies, a composite material is represented by a homogeneous anisotropic material with mechanical characteristics that depend on the mechanical characteristics of the matrix and the reinforcing fibres, their volume fractions in the composite material, etc. The determination of these mechanical characteristics is a fairly complex mathematical problem and, for the majority of fibrous composite materials, it has therefore been assumed that the frequency of the reinforcement with the fibres is quite large and the reinforced layer can be considered as a transversely isotropic layer. In order to determine the elastic characteristics of the composite material it is necessary to find five independent quantities: the elastic moduli E1 and E2 , the shear moduli G12 and G13 and Poisson’s ratio v12 . Data on just four of the elastic characteristics for a fibrous composite material were obtained in a number of papers,1–4 which restricts the possibility of using them to solve plane problems. More general and widely used relations5,6 enable three-dimensional problems to be solved. In all of the papers listed above it was assumed that the matrix material and the material of the fibre are isotropic. However, in a number of cases it is necessary to take account of the anisotropic properties of the fibres and the matrix. Relations have been proposed7 for solving a plane problem in the mechanics of rubber-cord materials that take account of the transversely isotropic properties of the fibre, and relations for a composite material with a transversely isotropic fibre and an isotropic matrix have been obtained for the three-dimensional case.8 The elastic modulus E11 , the shear modulus G12 and Poisson’s ratio v12 have been determined for a composite material,9,10 taking account of the transversely isotropic properties of the matrix and the fibre. The equality of the corresponding components of the displacement vector was used as the matching condition. The longitudinal modulus of elasticity has been determined taking account of the transversely isotropic properties of the matrix and the fibre11 using an energy matching criterion for a composite material. Unlike the investigations mentioned above, in this paper the shear modulus of the composite material G12 is determined using an energy matching criterion when both the matrix and the fibre possess transversely isotropic properties and the isotropy planes coincide. 1. Basic assumptions and relations Hexagonal packing of the fibres is one of the schemes for arranging the fibres in unidirectional composite materials. We will conceptually cut out an elementary hexagonal cell containing a single fibre and the matrix material surrounding it from the bulk of the composite material and assume that both the matrix and the fibre are made of transversely isotropic materials and that the isotropy planes coincide and are directed perpendicularly to the axis of the fibre.
夽 Prikl. Mat. Mekh., Vol. 78, No. 2, pp. 270-276, 2014. E-mail address: [email protected] http://dx.doi.org/10.1016/j.jappmathmech.2014.07.012 0021-8928/© 2014 Elsevier Ltd. All rights reserved.
188
S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
Fig. 1.
We will represent the element of the fibrous composite material possessing transversely isotropic properties in the form of a combination of two transversely isotropic infinite coaxial cylinders, a hollow cylinder of external radius b that simulates the matrix and a solid cylinder of radius a that simulates the fibre inserted into it. For this purpose we will approximate the volume of the hexagonal cell with the volume of a cylinder and take the radius of the cylinder such that the volume content of the fibre in the hexagonal cell and the volume content of the fibre in the cylindrical cell are the same. The volume content of the fibres in the composite material is obviously equal to the ratio of the squares of the internal and external radii of the hollow cylinder: (1.1) The elastic constants are found by solving two boundary value problems. Initally, the boundary value problem for the combined deformation of the transversely isotropic matrix and the transversely isotropic fibre is solved and, as a result, we obtain the components of the stress-strain state (SSS) as functions of the elastic constants of the matrix and fibre materials as well as the volume fraction of each of them in the composite material. We then solve a similar boundary value problem for a composite material represented by a homogeneous transversely isotropic material with as yet unknown elastic constants. As a result, we obtain the components of the SSS as functions of the unknown elastic constants of a homogeneous transversely isotropic material that simulates the composite material. We find the elastic constants of the transversely isotropic material from any matching condition as functions of the elastic constants of the matrix and fibre materials, as well as the volume fractions of each of them in the composite material. It is clear that the more accurately the boundary value problem is solved, the more accurate the unknown elastic constants of the composite material will be. Analytical solutions for such a combination are only possible for a restricted number of boundary value problems: uniform longitudinal and transverse stretching and pure transverse or longitudinal shear. A solution of the problem of the longitudinal shear for a transversely isotropic cylindrical body has been given in Ref .8 We will now present the basic relations characterizing pure longitudinal shear in a cylindrical region (Fig. 1) when the SSS is defined by the relations
Suppose the following load is applied to the external cylindrical surface of the region: (1.2) For the axial displacement we have8 (1.3) where C1 and C2 are constants. Using the Cauchy relations, we obtain relations for the strains (1.4)
S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
189
and, when Hooke’s law is taken into account, the relations for the stresses (1.5) 2. The combined longitudinal shear for the matrix and fibre We now consider the problem of the joint longitudinal shear of the hollow cylinder (a ≤ r ≤ b) simulating the matrix and the solid cylinder (0 ≤ r ≤ a) simulating the fibre. Quantities referring to the matrix are denoted by an asterisk and quantities referring to the fibre are denoted by a zero superscript. We represent the basic relations describing the SSS of the matrix (the hollow cylinder) in the form (relabelling the constants) (2.1)
(2.2) Taking account of the finiteness of the displacements when r = 0, we represent the basic relations describing the SSS of the fibre (the solid cylinder) in the form (the constant C2 in relations (1.3)–(1.5) is equal to zero and we relabel C1 as C) (2.3) (2.4) In order to find the unknown constants of relations (2.1)–(2.4) for the problem of the joint longitudinal shear of the matrix and fibre, we use boundary condition (1.2) and the conditions for the continuity of the axial displacement uz and the stress zr at the interface of the materials (2.5) From condition (1.2), we have
whence, when the first equality of (2.5) is taken into account, we obtain
and, finally, from the second equality of (2.5) when account is taken of the last two relations, after introducing the notation
we have (2.6) Then, (2.7) Taking account of expressions (2.6) and (2.7), we obtain, in the final form, the basic relation describing, in the case of the joint longitudinal shear, the SSS of the matrix (2.8)
(2.9) and the SSS of the fibre (2.10) (2.11) 3. Longitudinal shear of the composite material We now solve a similar problem on pure longitudinal shear for the transversely isotropic material simulating the composite material, representing it in the form of a solid infinite cylinder of radius b. The boundary conditions take the form of (1.2). The SSS will be described ◦ , the by relations that are similar to relations (2.3) and (2.4), where now, instead of the constant C, 0 /G12 will appear and, instead of G12 quantity G12 will appear.
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S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
Fig. 2.
4. Computational formula We will now find the elastic constants of the composite material using an energy criterion as the matching condition which comprises the following: the elastic strain energy U of a transversely isotropic cylinder (or another body) simulating a composite material is equal to the sum of the elastic strain energies of the hollow cylinder simulating the matrix (U *) and the solid cylinder simulating the fibre (U0 ). In the case of longitudinal shear, we have the following governing relations:
According to the energy matching criterion, the right-hand side of the first of the relations must be equated to the sum of the right-hand ˆ 0 and F± into account, we have sides of the second and the third relations whence, taking the notation for (4.1) It is seen that the shear modulus of the composite material G12 only depends on one independent elastic characteristic of the matrix ∗ and is independent of the remaining four parameters describing the transversely isotropic properties of the matrix. A similar material G12 conclusion can be drawn regarding the relation between the shear modulus of the composite material and the properties of the fibre ◦ . If the matrix material or the fibre material can be assumed to be isotropic, material: the modulus G12 only depends on the quantity G12 ∗ = G∗ or, correspondingly, G◦ = G◦ . The formula for the shear modulus of the composite then, in formula (4.1), we can assume that G12 12 material G12 obtained for a transversely isotropic fibre and an isotropic matrix8 is a special case of formula (4.1). Analysing formula (4.41), a number of conclusions can be drawn: the shear modulus of the composite material G12 increases both when ∗ and shear modulus of the fibre material G◦ increase and, with respect to the volume fraction the shear modulus of the matrix material G12 12 ◦ > G∗ increase and a decreasing function if G∗ > G◦ . of the fibre f, the function G12 is an increasing function if G12 12 12 12 5. Numerical results We will now compare the value of the shear modulus G12 obtained using formula (4.1) with the experimental data given in the literature.8 We shall consider a UD PFRP composite material with an epoxide resin (Ciba-Geigy 913) as the matrix and a fibre made of highdensity polyethylene (VHDPE Tenfor SNIA). The modulus of elasticity of the matrix E* = 5.55 GPa, Poisson’s ratio for the matrix v* = 0.37, the longitudinal modulus of elasticity of the fibre E1◦ = 60.4 GPa, and the transverse modulus of elasticity of the fibre E2◦ = 4.68, Poisson’s ◦ = 1.65 GPa. ratios for the fibre ◦12 = 0.38 and ◦23 = 0.55, and the shear modulus of the fibre G12 Values of the shear modulus G12 for different volume contents of the fibre f are presented below
It should be noted that the error in the calculations using formula (4.1) is less than 2.5% when 0.36 ≤ f ≤ 0.67 and less then 5% when f = 0.75.
S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
191
∗ = G∗ = 2.03 GPa In obtaining the results presented above the matrix material was assumed to be isotropic and the value G12 (G* = E*/[2(1 + *)]) was used. To simulate the transversely isotropic properties of the matrix, we vary the value of the shear modulus of the matrix, which will take ∗ = kG∗ (where G* = 2.03 GPa). The relation between the shear modulus G the values G12 12 and the volume fibre content is shown in Fig. 2 for 0.1 ≤ f ≤ 0.8. The shear modulus of the composite material barely changes when the elastic properties of the fibre are varied in a similar way. ∗ by a factor It can be noted from the graph that the transversely isotropic properties of the matrix (the increase in the modulus G12 of four) has a considerable effect on the elastic characteristics of the composite material: the shear modulus of the composite material G12 decreases from a factor of 3.5 times the initial magnitude, when there is a considerable amount of matrix material in the composite material (f = 0.1), to one and one-half times when the fraction of the matrix material in the composite material is reduced (f = 0.8).
References [1]. [2]. [3]. [4]. [5]. [6]. [7]. [8]. [9].
Lapin AA. Plane deformation of rubber-cord fabrics. In: Stability Calculations in Machine Construction. Moscow: Mashgiz; 1955. p. 87–9. Ashton JE, Helpin JC, Petit PH. Primer on Composite Materials: Analysis. Stamford: Technomic; 1969. p. 124 p. Gough VE. Stiffness of cord and rubber constructions. Rubber Chemistry and Technology 1968;41(4):988–1021. Tangorra G. Fiber-reinforced, oriented rubber sheets. In: Proc Intern Rubber Conf. Moscow: Khimiya; 1971. p. 459–66. Abolinsh DS. Compliance tensor for an elastic material reinforced in one direction. Polymer Mechanics 1965;1(4):28–32. Van Fo Fy GA. Elastic constants and state of stress of a glass-reinforced strip. Polymer Mechanics 1966;2(4):368–72. Rasteryayev YuK, Agaltsov GN. Composite rubber-cord materials and the mechanics of their deformation. Geotekh Mekh 2005;60:200–48. Klastorny M, Konderla P, Piekarski R. An exact stiffness theory for unidirectional xFRP composites. Mechanics Composite Mater 2009;45(1):77–104. Grebenyuk SN. Elastic characteristics of a composite material with a transtropic matrix and fibre. In: Methods of Solving Applied Problems in Solid Mechanics. Dnipropetrovsk: Nauka i Osvita; 2011. Issue12: 62–8. [10]. Grebenyuk SN. Determination of the shear modulus of a composite material with a transtropic matrix and fibre. In: Methods of Solving Applied Problems in Solid Mechanics. Dnipropetrovsk: Nauka i Osvita; 2012. Issue13: 92–8. [11]. Grebenyuk SN. Determination of the longitudinal elastic modulus of a composite material based on an energy matching condition. Vestnik Kherson Nats Tekh Inst 2012;2(45):106–10.
Translated by E. L. S
Contents lists available at ScienceDirect
Journal of Applied Mathematics and Mechanics journal homepage: www.elsevier.com/locate/jappmathmech
The shear modulus of a composite material with a transversely isotropic matrix and a fibre夽 S.N. Grebenyuk Zaporizhia, Ukraine
a r t i c l e
i n f o
Article history: Received 27 November 2012
a b s t r a c t A relation is proposed for determining of the shear modulus of a fibrous composite material with a transversely isotropic matrix and a fibre as a function of the elastic constants of the matrix and the fibre as well as the volume fraction of each of them in the composite material. The isotropy planes of the matrix and fibre coincide and are perpendicular to the fibre axis. Two boundary value problems are solved in order to obtain the required relation: the problem of the longitudinal shear of a transversely isotropic solid cylinder that simulates the fibrous composite material and the problem of the combined longitudinal shear of a hollow and solid cylinder that simulate the matrix material and the fibre material respectively. Calculations using the proposed formula are compared with the available experimental data. © 2014 Elsevier Ltd. All rights reserved.
In the majority of studies, a composite material is represented by a homogeneous anisotropic material with mechanical characteristics that depend on the mechanical characteristics of the matrix and the reinforcing fibres, their volume fractions in the composite material, etc. The determination of these mechanical characteristics is a fairly complex mathematical problem and, for the majority of fibrous composite materials, it has therefore been assumed that the frequency of the reinforcement with the fibres is quite large and the reinforced layer can be considered as a transversely isotropic layer. In order to determine the elastic characteristics of the composite material it is necessary to find five independent quantities: the elastic moduli E1 and E2 , the shear moduli G12 and G13 and Poisson’s ratio v12 . Data on just four of the elastic characteristics for a fibrous composite material were obtained in a number of papers,1–4 which restricts the possibility of using them to solve plane problems. More general and widely used relations5,6 enable three-dimensional problems to be solved. In all of the papers listed above it was assumed that the matrix material and the material of the fibre are isotropic. However, in a number of cases it is necessary to take account of the anisotropic properties of the fibres and the matrix. Relations have been proposed7 for solving a plane problem in the mechanics of rubber-cord materials that take account of the transversely isotropic properties of the fibre, and relations for a composite material with a transversely isotropic fibre and an isotropic matrix have been obtained for the three-dimensional case.8 The elastic modulus E11 , the shear modulus G12 and Poisson’s ratio v12 have been determined for a composite material,9,10 taking account of the transversely isotropic properties of the matrix and the fibre. The equality of the corresponding components of the displacement vector was used as the matching condition. The longitudinal modulus of elasticity has been determined taking account of the transversely isotropic properties of the matrix and the fibre11 using an energy matching criterion for a composite material. Unlike the investigations mentioned above, in this paper the shear modulus of the composite material G12 is determined using an energy matching criterion when both the matrix and the fibre possess transversely isotropic properties and the isotropy planes coincide. 1. Basic assumptions and relations Hexagonal packing of the fibres is one of the schemes for arranging the fibres in unidirectional composite materials. We will conceptually cut out an elementary hexagonal cell containing a single fibre and the matrix material surrounding it from the bulk of the composite material and assume that both the matrix and the fibre are made of transversely isotropic materials and that the isotropy planes coincide and are directed perpendicularly to the axis of the fibre.
夽 Prikl. Mat. Mekh., Vol. 78, No. 2, pp. 270-276, 2014. E-mail address: [email protected] http://dx.doi.org/10.1016/j.jappmathmech.2014.07.012 0021-8928/© 2014 Elsevier Ltd. All rights reserved.
188
S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
Fig. 1.
We will represent the element of the fibrous composite material possessing transversely isotropic properties in the form of a combination of two transversely isotropic infinite coaxial cylinders, a hollow cylinder of external radius b that simulates the matrix and a solid cylinder of radius a that simulates the fibre inserted into it. For this purpose we will approximate the volume of the hexagonal cell with the volume of a cylinder and take the radius of the cylinder such that the volume content of the fibre in the hexagonal cell and the volume content of the fibre in the cylindrical cell are the same. The volume content of the fibres in the composite material is obviously equal to the ratio of the squares of the internal and external radii of the hollow cylinder: (1.1) The elastic constants are found by solving two boundary value problems. Initally, the boundary value problem for the combined deformation of the transversely isotropic matrix and the transversely isotropic fibre is solved and, as a result, we obtain the components of the stress-strain state (SSS) as functions of the elastic constants of the matrix and fibre materials as well as the volume fraction of each of them in the composite material. We then solve a similar boundary value problem for a composite material represented by a homogeneous transversely isotropic material with as yet unknown elastic constants. As a result, we obtain the components of the SSS as functions of the unknown elastic constants of a homogeneous transversely isotropic material that simulates the composite material. We find the elastic constants of the transversely isotropic material from any matching condition as functions of the elastic constants of the matrix and fibre materials, as well as the volume fractions of each of them in the composite material. It is clear that the more accurately the boundary value problem is solved, the more accurate the unknown elastic constants of the composite material will be. Analytical solutions for such a combination are only possible for a restricted number of boundary value problems: uniform longitudinal and transverse stretching and pure transverse or longitudinal shear. A solution of the problem of the longitudinal shear for a transversely isotropic cylindrical body has been given in Ref .8 We will now present the basic relations characterizing pure longitudinal shear in a cylindrical region (Fig. 1) when the SSS is defined by the relations
Suppose the following load is applied to the external cylindrical surface of the region: (1.2) For the axial displacement we have8 (1.3) where C1 and C2 are constants. Using the Cauchy relations, we obtain relations for the strains (1.4)
S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
189
and, when Hooke’s law is taken into account, the relations for the stresses (1.5) 2. The combined longitudinal shear for the matrix and fibre We now consider the problem of the joint longitudinal shear of the hollow cylinder (a ≤ r ≤ b) simulating the matrix and the solid cylinder (0 ≤ r ≤ a) simulating the fibre. Quantities referring to the matrix are denoted by an asterisk and quantities referring to the fibre are denoted by a zero superscript. We represent the basic relations describing the SSS of the matrix (the hollow cylinder) in the form (relabelling the constants) (2.1)
(2.2) Taking account of the finiteness of the displacements when r = 0, we represent the basic relations describing the SSS of the fibre (the solid cylinder) in the form (the constant C2 in relations (1.3)–(1.5) is equal to zero and we relabel C1 as C) (2.3) (2.4) In order to find the unknown constants of relations (2.1)–(2.4) for the problem of the joint longitudinal shear of the matrix and fibre, we use boundary condition (1.2) and the conditions for the continuity of the axial displacement uz and the stress zr at the interface of the materials (2.5) From condition (1.2), we have
whence, when the first equality of (2.5) is taken into account, we obtain
and, finally, from the second equality of (2.5) when account is taken of the last two relations, after introducing the notation
we have (2.6) Then, (2.7) Taking account of expressions (2.6) and (2.7), we obtain, in the final form, the basic relation describing, in the case of the joint longitudinal shear, the SSS of the matrix (2.8)
(2.9) and the SSS of the fibre (2.10) (2.11) 3. Longitudinal shear of the composite material We now solve a similar problem on pure longitudinal shear for the transversely isotropic material simulating the composite material, representing it in the form of a solid infinite cylinder of radius b. The boundary conditions take the form of (1.2). The SSS will be described ◦ , the by relations that are similar to relations (2.3) and (2.4), where now, instead of the constant C, 0 /G12 will appear and, instead of G12 quantity G12 will appear.
190
S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
Fig. 2.
4. Computational formula We will now find the elastic constants of the composite material using an energy criterion as the matching condition which comprises the following: the elastic strain energy U of a transversely isotropic cylinder (or another body) simulating a composite material is equal to the sum of the elastic strain energies of the hollow cylinder simulating the matrix (U *) and the solid cylinder simulating the fibre (U0 ). In the case of longitudinal shear, we have the following governing relations:
According to the energy matching criterion, the right-hand side of the first of the relations must be equated to the sum of the right-hand ˆ 0 and F± into account, we have sides of the second and the third relations whence, taking the notation for (4.1) It is seen that the shear modulus of the composite material G12 only depends on one independent elastic characteristic of the matrix ∗ and is independent of the remaining four parameters describing the transversely isotropic properties of the matrix. A similar material G12 conclusion can be drawn regarding the relation between the shear modulus of the composite material and the properties of the fibre ◦ . If the matrix material or the fibre material can be assumed to be isotropic, material: the modulus G12 only depends on the quantity G12 ∗ = G∗ or, correspondingly, G◦ = G◦ . The formula for the shear modulus of the composite then, in formula (4.1), we can assume that G12 12 material G12 obtained for a transversely isotropic fibre and an isotropic matrix8 is a special case of formula (4.1). Analysing formula (4.41), a number of conclusions can be drawn: the shear modulus of the composite material G12 increases both when ∗ and shear modulus of the fibre material G◦ increase and, with respect to the volume fraction the shear modulus of the matrix material G12 12 ◦ > G∗ increase and a decreasing function if G∗ > G◦ . of the fibre f, the function G12 is an increasing function if G12 12 12 12 5. Numerical results We will now compare the value of the shear modulus G12 obtained using formula (4.1) with the experimental data given in the literature.8 We shall consider a UD PFRP composite material with an epoxide resin (Ciba-Geigy 913) as the matrix and a fibre made of highdensity polyethylene (VHDPE Tenfor SNIA). The modulus of elasticity of the matrix E* = 5.55 GPa, Poisson’s ratio for the matrix v* = 0.37, the longitudinal modulus of elasticity of the fibre E1◦ = 60.4 GPa, and the transverse modulus of elasticity of the fibre E2◦ = 4.68, Poisson’s ◦ = 1.65 GPa. ratios for the fibre ◦12 = 0.38 and ◦23 = 0.55, and the shear modulus of the fibre G12 Values of the shear modulus G12 for different volume contents of the fibre f are presented below
It should be noted that the error in the calculations using formula (4.1) is less than 2.5% when 0.36 ≤ f ≤ 0.67 and less then 5% when f = 0.75.
S.N. Grebenyuk / Journal of Applied Mathematics and Mechanics 78 (2014) 187–191
191
∗ = G∗ = 2.03 GPa In obtaining the results presented above the matrix material was assumed to be isotropic and the value G12 (G* = E*/[2(1 + *)]) was used. To simulate the transversely isotropic properties of the matrix, we vary the value of the shear modulus of the matrix, which will take ∗ = kG∗ (where G* = 2.03 GPa). The relation between the shear modulus G the values G12 12 and the volume fibre content is shown in Fig. 2 for 0.1 ≤ f ≤ 0.8. The shear modulus of the composite material barely changes when the elastic properties of the fibre are varied in a similar way. ∗ by a factor It can be noted from the graph that the transversely isotropic properties of the matrix (the increase in the modulus G12 of four) has a considerable effect on the elastic characteristics of the composite material: the shear modulus of the composite material G12 decreases from a factor of 3.5 times the initial magnitude, when there is a considerable amount of matrix material in the composite material (f = 0.1), to one and one-half times when the fraction of the matrix material in the composite material is reduced (f = 0.8).
References [1]. [2]. [3]. [4]. [5]. [6]. [7]. [8]. [9].
Lapin AA. Plane deformation of rubber-cord fabrics. In: Stability Calculations in Machine Construction. Moscow: Mashgiz; 1955. p. 87–9. Ashton JE, Helpin JC, Petit PH. Primer on Composite Materials: Analysis. Stamford: Technomic; 1969. p. 124 p. Gough VE. Stiffness of cord and rubber constructions. Rubber Chemistry and Technology 1968;41(4):988–1021. Tangorra G. Fiber-reinforced, oriented rubber sheets. In: Proc Intern Rubber Conf. Moscow: Khimiya; 1971. p. 459–66. Abolinsh DS. Compliance tensor for an elastic material reinforced in one direction. Polymer Mechanics 1965;1(4):28–32. Van Fo Fy GA. Elastic constants and state of stress of a glass-reinforced strip. Polymer Mechanics 1966;2(4):368–72. Rasteryayev YuK, Agaltsov GN. Composite rubber-cord materials and the mechanics of their deformation. Geotekh Mekh 2005;60:200–48. Klastorny M, Konderla P, Piekarski R. An exact stiffness theory for unidirectional xFRP composites. Mechanics Composite Mater 2009;45(1):77–104. Grebenyuk SN. Elastic characteristics of a composite material with a transtropic matrix and fibre. In: Methods of Solving Applied Problems in Solid Mechanics. Dnipropetrovsk: Nauka i Osvita; 2011. Issue12: 62–8. [10]. Grebenyuk SN. Determination of the shear modulus of a composite material with a transtropic matrix and fibre. In: Methods of Solving Applied Problems in Solid Mechanics. Dnipropetrovsk: Nauka i Osvita; 2012. Issue13: 92–8. [11]. Grebenyuk SN. Determination of the longitudinal elastic modulus of a composite material based on an energy matching condition. Vestnik Kherson Nats Tekh Inst 2012;2(45):106–10.
Translated by E. L. S