2 Macromechanical Analysis of a Lamina Chapter Objectives Review definitions of stress,strain,elastic moduli,and strain energy. Develop stress-strain relationships for different types of materials. Develop stress-strain relationships for a unidirectional/bidirec- tional lamina. Find the engineering constants of a unidirectional/bidirectional lam- ina in terms of the stiffness and compliance parameters of the lamina. Develop stress-strain relationships,elastic moduli,strengths,and thermal and moisture expansion coefficients of an angle ply based on those of a unidirectional/bidirectional lamina and the angle of the ply. 2.1 Introduction A lamina is a thin layer of a composite material that is generally of a thickness on the order of 0.005 in.(0.125 mm).A laminate is constructed by stacking a number of such laminae in the direction of the lamina thickness(Figure 2.1).Mechanical structures made of these laminates,such as a leaf spring suspension system in an automobile,are subjected to various loads,such as bending and twisting.The design and analysis of such laminated structures demands knowledge of the stresses and strains in the laminate.Also,design tools,such as failure theories,stiffness models,and optimization algorithms, need the values of these laminate stresses and strains. However,the building blocks of a laminate are single lamina,so under- standing the mechanical analysis of a lamina precedes understanding that of a laminate.A lamina is unlike an isotropic homogeneous material.For example,if the lamina is made of isotropic homogeneous fibers and an 61 2006 by Taylor Francis Group,LLC
61 2 Macromechanical Analysis of a Lamina Chapter Objectives • Review definitions of stress, strain, elastic moduli, and strain energy. • Develop stress–strain relationships for different types of materials. • Develop stress–strain relationships for a unidirectional/bidirectional lamina. • Find the engineering constants of a unidirectional/bidirectional lamina in terms of the stiffness and compliance parameters of the lamina. • Develop stress–strain relationships, elastic moduli, strengths, and thermal and moisture expansion coefficients of an angle ply based on those of a unidirectional/bidirectional lamina and the angle of the ply. 2.1 Introduction A lamina is a thin layer of a composite material that is generally of a thickness on the order of 0.005 in. (0.125 mm). A laminate is constructed by stacking a number of such laminae in the direction of the lamina thickness (Figure 2.1). Mechanical structures made of these laminates, such as a leaf spring suspension system in an automobile, are subjected to various loads, such as bending and twisting. The design and analysis of such laminated structures demands knowledge of the stresses and strains in the laminate. Also, design tools, such as failure theories, stiffness models, and optimization algorithms, need the values of these laminate stresses and strains. However, the building blocks of a laminate are single lamina, so understanding the mechanical analysis of a lamina precedes understanding that of a laminate. A lamina is unlike an isotropic homogeneous material. For example, if the lamina is made of isotropic homogeneous fibers and an 1343_book.fm Page 61 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
62 Mechanics of Composite Materials,Second Edition Fiber cross-section Matrix material FIGURE 2.1 Typical laminate made of three laminae. isotropic homogeneous matrix,the stiffness of the lamina varies from point to point depending on whether the point is in the fiber,the matrix,or the fiber-matrix interface.Accounting for these variations will make any kind of mechanical modeling of the lamina very complicated.For this reason,the macromechanical analysis of a lamina is based on average properties and considering the lamina to be homogeneous.Methods to find these average properties based on the individual mechanical properties of the fiber and the matrix,as well as the content,packing geometry,and shape of fibers are discussed in Chapter 3. Even with the homogenization of a lamina,the mechanical behavior is still different from that of a homogeneous isotropic material.For example,take a square plate of length and width w and thickness t out of a large isotropic plate of thickness t(Figure 2.2)and conduct the following experiments. Case A:Subject the square plate to a pure normal load P in direction 1. Measure the normal deformations in directions 1 and 2,and respectively. Case B:Apply the same pure normal load P as in case A,but now in direction 2.Measure the normal deformations in directions 1 and 2, δs andδ2B,respectively. Note that δ1A=δ2B, (2.1a,b) δ2A=δ1B· However,taking a unidirectional square plate (Figure 2.3)of the same dimensions wx w x f out of a large composite lamina of thickness t and conducting the same case A and B experiments,note that the deformations 2006 by Taylor Francis Group,LLC
62 Mechanics of Composite Materials, Second Edition isotropic homogeneous matrix, the stiffness of the lamina varies from point to point depending on whether the point is in the fiber, the matrix, or the fiber–matrix interface. Accounting for these variations will make any kind of mechanical modeling of the lamina very complicated. For this reason, the macromechanical analysis of a lamina is based on average properties and considering the lamina to be homogeneous. Methods to find these average properties based on the individual mechanical properties of the fiber and the matrix, as well as the content, packing geometry, and shape of fibers are discussed in Chapter 3. Even with the homogenization of a lamina, the mechanical behavior is still different from that of a homogeneous isotropic material. For example, take a square plate of length and width w and thickness t out of a large isotropic plate of thickness t (Figure 2.2) and conduct the following experiments. Case A: Subject the square plate to a pure normal load P in direction 1. Measure the normal deformations in directions 1 and 2, δ1A and δ2A, respectively. Case B: Apply the same pure normal load P as in case A, but now in direction 2. Measure the normal deformations in directions 1 and 2, δ1B and δ2B, respectively. Note that (2.1a,b) However, taking a unidirectional square plate (Figure 2.3) of the same dimensions w × w × t out of a large composite lamina of thickness t and conducting the same case A and B experiments, note that the deformations FIGURE 2.1 Typical laminate made of three laminae. Fiber cross-section Matrix material 1A 2B 2A 1B = = δ δ δ δ , . 1343_book.fm Page 62 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 63 Case A Case B Undeformed state Undeformed state w+δ2B w+δ2A w+81A w+δ1B Deformed state p Deformed state FIGURE 2.2 Deformation of square plate taken from an isotropic plate under normal loads. 81A≠δ2B, (2.2a,b) δ2A≠δ1B· because the stiffness of the unidirectional lamina in the direction of fibers is much larger than the stiffness in the direction perpendicular to the fibers. Thus,the mechanical characterization of a unidirectional lamina will require more parameters than it will for an isotropic lamina. Also,note that if the square plate(Figure 2.4)taken out of the lamina has fibers at an angle to the sides of the square plate,the deformations will be different for different angles.In fact,the square plate would not only have 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 63 (2.2a,b) because the stiffness of the unidirectional lamina in the direction of fibers is much larger than the stiffness in the direction perpendicular to the fibers. Thus, the mechanical characterization of a unidirectional lamina will require more parameters than it will for an isotropic lamina. Also, note that if the square plate (Figure 2.4) taken out of the lamina has fibers at an angle to the sides of the square plate, the deformations will be different for different angles. In fact, the square plate would not only have FIGURE 2.2 Deformation of square plate taken from an isotropic plate under normal loads. w 2 w 1 t t w w w Undeformed state Deformed state Undeformed state w + δ2A w + δ1A p p w + δ2B w + δ1B w Case A Case B p p Deformed state 1A 2B 2A 1B δ δ δ δ ≠ ≠ , . 1343_book.fm Page 63 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
64 Mechanics of Composite Materials,Second Edition Fiber cross section Case】 Case B Undeformed state Undeformed state w+δ2A w+δ2B D w+δ1B Deformed state ↓p Deformed state FIGURE 2.3 Deformation of a square plate taken from a unidirectional lamina with fibers at zero angle under normal loads. deformations in the normal directions but would also distort.This suggests that the mechanical characterization of an angle lamina is further complicated Mechanical characterization of materials generally requires costly and time-consuming experimentation and/or theoretical modeling.Therefore, the goal is to find the minimum number of parameters required for the mechanical characterization of a lamina. Also,a composite laminate may be subjected to a temperature change and may absorb moisture during processing and operation.These changes in temperature and moisture result in residual stresses and strains in the lam- inate.The calculation of these stresses and strains in a laminate depends on the response of each lamina to these two environmental parameters.In this chapter,the stress-strain relationships based on temperature change and moisture content will also be developed for a single lamina.The effects of temperature and moisture on a laminate are discussed later in Chapter 4. 2006 by Taylor Francis Group,LLC
64 Mechanics of Composite Materials, Second Edition deformations in the normal directions but would also distort. This suggests that the mechanical characterization of an angle lamina is further complicated. Mechanical characterization of materials generally requires costly and time-consuming experimentation and/or theoretical modeling. Therefore, the goal is to find the minimum number of parameters required for the mechanical characterization of a lamina. Also, a composite laminate may be subjected to a temperature change and may absorb moisture during processing and operation. These changes in temperature and moisture result in residual stresses and strains in the laminate. The calculation of these stresses and strains in a laminate depends on the response of each lamina to these two environmental parameters. In this chapter, the stress–strain relationships based on temperature change and moisture content will also be developed for a single lamina. The effects of temperature and moisture on a laminate are discussed later in Chapter 4. FIGURE 2.3 Deformation of a square plate taken from a unidirectional lamina with fibers at zero angle under normal loads. w w p p p p w + δ2B w + δ1B w + δ1A w + δ2A w w w w 2 1 Fiber cross section Case A Case B Undeformed state Deformed state Deformed state Undeformed state t t 1343_book.fm Page 64 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 65 ● Fiber cross section Undeformed state Deformed state FIGURE 2.4 Deformation of a square plate taken from a unidirectional lamina with fibers at an angle under normal loads. 2.2 Review of Definitions 2.2.1 Stress A mechanical structure takes external forces,which act upon a body as surface forces (for example,bending a stick)and body forces (for example, the weight of a standing vertical telephone pole on itself).These forces result in internal forces inside the body.Knowledge of the internal forces at all points in the body is essential because these forces need to be less than the strength of the material used in the structure.Stress,which is defined as the intensity of the load per unit area,determines this knowledge because the strengths of a material are intrinsically known in terms of stress. Imagine a body(Figure 2.5)in equilibrium under various loads.If the body is cut at a cross-section,forces will need to be applied on the cross-sectional area so that it maintains equilibrium as in the original body.At any cross- 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 65 2.2 Review of Definitions 2.2.1 Stress A mechanical structure takes external forces, which act upon a body as surface forces (for example, bending a stick) and body forces (for example, the weight of a standing vertical telephone pole on itself). These forces result in internal forces inside the body. Knowledge of the internal forces at all points in the body is essential because these forces need to be less than the strength of the material used in the structure. Stress, which is defined as the intensity of the load per unit area, determines this knowledge because the strengths of a material are intrinsically known in terms of stress. Imagine a body (Figure 2.5) in equilibrium under various loads. If the body is cut at a cross-section, forces will need to be applied on the cross-sectional area so that it maintains equilibrium as in the original body. At any crossFIGURE 2.4 Deformation of a square plate taken from a unidirectional lamina with fibers at an angle under normal loads. w t t w w p p 2 1 Fiber cross section Undeformed state Deformed state 1343_book.fm Page 65 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
66 Mechanics of Composite Materials,Second Edition Arbitrary plane AP FIGURE 2.5 Stresses on an infinitesimal area on an arbitrary plane. section,a force AP is acting on an area of AA.This force vector has a com- ponent normal to the surface,AP,and one parallel to the surface,AP.The definition of stress then gives △P =lim aM-0△A' t;lim △P (2.3a,b) AM4-→0△A The component of the stress normal to the surface,o is called the normal stress and the stress parallel to the surface,t,is called the shear stress.If one takes a different cross-section through the same point,the stress remains unchanged but the two components of stress,normal stress,o,and shear stress,t,will change.However,it has been proved that a complete definition of stress at a point only needs use of any three mutually orthogonal coordi- nate systems,such as a Cartesian coordinate system. Take the right-hand coordinate system x-y-z.Take a cross-section parallel to the yz-plane in the body as shown in Figure 2.6.The force vector AP acts 2006 by Taylor Francis Group,LLC
66 Mechanics of Composite Materials, Second Edition section, a force ΔP is acting on an area of ΔA. This force vector has a component normal to the surface, ΔPn, and one parallel to the surface, ΔPs. The definition of stress then gives , . (2.3a,b) The component of the stress normal to the surface, σn, is called the normal stress and the stress parallel to the surface, τs, is called the shear stress. If one takes a different cross-section through the same point, the stress remains unchanged but the two components of stress, normal stress, σn, and shear stress, τs, will change. However, it has been proved that a complete definition of stress at a point only needs use of any three mutually orthogonal coordinate systems, such as a Cartesian coordinate system. Take the right-hand coordinate system x–y–z. Take a cross-section parallel to the yz-plane in the body as shown in Figure 2.6. The force vector ΔP acts FIGURE 2.5 Stresses on an infinitesimal area on an arbitrary plane. Arbitrary plane ΔPs ΔP ΔA ΔPn σn A Pn A = →limΔ Δ 0 Δ τs A Ps A = → limΔ Δ 0 Δ 1343_book.fm Page 66 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 67 △P △P AA Cross-section FIGURE 2.6 Forces on an infinitesimal area on the y-z plane. on an area AA.The component AP:is normal to the surface.The force vector AP,is parallel to the surface and can be further resolved into components along the y and z axes:AP and AP.The definition of the various stresses then is x=lim △2 △M-0△A △P. Twy=lim 44-0△A' Ux=lim △2 (2.4a-c） 44-→0△A Similarly,stresses can be defined for cross-sections parallel to the xy and xz planes.For defining all these stresses,the stress at a point is defined generally by taking an infinitesimal cuboid in a right-hand coordinate system 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 67 on an area ΔA. The component ΔPx is normal to the surface. The force vector ΔPs is parallel to the surface and can be further resolved into components along the y and z axes: ΔPy and ΔPz. The definition of the various stresses then is , . (2.4a–c) Similarly, stresses can be defined for cross-sections parallel to the xy and xz planes. For defining all these stresses, the stress at a point is defined generally by taking an infinitesimal cuboid in a right-hand coordinate system FIGURE 2.6 Forces on an infinitesimal area on the y–z plane. z ΔPz ΔA Cross-section ΔPy ΔP ΔPx y x σx A Px A = →limΔ Δ 0 Δ τxy A Py A = → limΔ Δ 0 Δ τxz A Pz A = → limΔ Δ 0 Δ 1343_book.fm Page 67 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
68 Mechanics of Composite Materials,Second Edition G H FIGURE 2.7 Stresses on an infinitesimal cuboid. and finding the stresses on each of its faces.Nine different stresses act at a point in the body as shown in Figure 2.7.The six shear stresses are related as Txy=Tyx Tye=Tay! Tzx=Tz· (2.5a-c） The preceding three relations are found by equilibrium of moments of the infinitesimal cube.There are thus six independent stresses.The stresses o,, and o are normal to the surfaces of the cuboid and the stressest and tw are along the surfaces of the cuboid. A tensile normal stress is positive,and a compressive normal stress is negative.A shear stress is positive,if its direction and the direction of the normal to the face on which it is acting are both in positive or negative direction;otherwise,the shear stress is negative. 2.2.2 Strain Similar to the need for knowledge of forces inside a body,knowing the deformations because of the external forces is also important.For example, a piston in an internal combustion engine may not develop larger stresses than the failure strengths,but its excessive deformation may seize the engine. Also,finding stresses in a body generally requires finding deformations.This is because a stress state at a point has six components,but there are only three force-equilibrium equations (one in each direction). 2006 by Taylor Francis Group,LLC
68 Mechanics of Composite Materials, Second Edition and finding the stresses on each of its faces. Nine different stresses act at a point in the body as shown in Figure 2.7. The six shear stresses are related as , , . (2.5a–c) The preceding three relations are found by equilibrium of moments of the infinitesimal cube. There are thus six independent stresses. The stresses σx, σy, and σz are normal to the surfaces of the cuboid and the stresses τyz, τzx, and τxy are along the surfaces of the cuboid. A tensile normal stress is positive, and a compressive normal stress is negative. A shear stress is positive, if its direction and the direction of the normal to the face on which it is acting are both in positive or negative direction; otherwise, the shear stress is negative. 2.2.2 Strain Similar to the need for knowledge of forces inside a body, knowing the deformations because of the external forces is also important. For example, a piston in an internal combustion engine may not develop larger stresses than the failure strengths, but its excessive deformation may seize the engine. Also, finding stresses in a body generally requires finding deformations. This is because a stress state at a point has six components, but there are only three force-equilibrium equations (one in each direction). FIGURE 2.7 Stresses on an infinitesimal cuboid. G F σzz σxx σyy τzy τzx τyx τyz τxz τxy H E A B z y x C τ τ xy = yx τ τ yz = zy τ τ zx = xz 1343_book.fm Page 68 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 69 Q D Ax C Ay (x,y) A 444444= FIGURE 2.8 Normal and shearing strains on an infinitesimal area in the x-y plane. The knowledge of deformations is specified in terms of strains-that is, the relative change in the size and shape of the body.The strain at a point is also defined generally on an infinitesimal cuboid in a right-hand coordi- nate system.Under loads,the lengths of the sides of the infinitesimal cuboid change.The faces of the cube also get distorted.The change in length cor- responds to a normal strain and the distortion corresponds to the shearing strain.Figure 2.8 shows the strains on one of the faces,ABCD,of the cuboid. The strains and displacements are related to each other.Take the two perpendicular lines AB and AD.When the body is loaded,the two lines become A'B'and A'D'.Define the displacements of a point (x,y,z)as u=u(x,y,z)=displacement in x-direction at point (x,y,z) v=v(x,y,z)=displacement in y-direction at point (x,y,z) ww(x,y,z)=displacement in z-direction at point (x,y,z) The normal strain in the x-direction,E,is defined as the change of length of line AB per unit length of AB as A'B'-AB Ex lim AB-0 AB (2.6) where 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 69 The knowledge of deformations is specified in terms of strains — that is, the relative change in the size and shape of the body. The strain at a point is also defined generally on an infinitesimal cuboid in a right-hand coordinate system. Under loads, the lengths of the sides of the infinitesimal cuboid change. The faces of the cube also get distorted. The change in length corresponds to a normal strain and the distortion corresponds to the shearing strain. Figure 2.8 shows the strains on one of the faces, ABCD, of the cuboid. The strains and displacements are related to each other. Take the two perpendicular lines AB and AD. When the body is loaded, the two lines become A′B′ and A′D′. Define the displacements of a point (x,y,z) as u = u(x,y,z) = displacement in x-direction at point (x,y,z) v = v(x,y,z) = displacement in y-direction at point (x,y,z) w = w(x,y,z) = displacement in z-direction at point (x,y,z) The normal strain in the x-direction, εx, is defined as the change of length of line AB per unit length of AB as , (2.6) where FIGURE 2.8 Normal and shearing strains on an infinitesimal area in the x–y plane. Q′ D′ C′ B′ B x C A D y Δy Δx (x,y) θ1 θ2 P′ A′ εx AB A B AB AB = ′ ′ − → lim0 1343_book.fm Page 69 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
70 Mechanics of Composite Materials,Second Edition AB=(AP)+(BP)2, =V△x+(x+△x,y)）-x,y+[o(x+△x,)-o(x,y, AB=△x. (2.7a,b) Substituting the preceding expressions of Equation(2.7)in Equation(2.6), Ex=lim △-→0 Ar Using definitions of partial derivatives 割食 ou Ex=ax' (2.8) because au1, Ox k1, x for small displacements. The normal strain in the y-direction,e is defined as the change in the length of line AD per unit length of AD as Ey=lim. A'D-AD ΓAD0AD (2.9) where 2006 by Taylor Francis Group,LLC
70 Mechanics of Composite Materials, Second Edition (2.7a,b) Substituting the preceding expressions of Equation (2.7) in Equation (2.6), . Using definitions of partial derivatives (2.8) because , , for small displacements. The normal strain in the y-direction, εy is defined as the change in the length of line AD per unit length of AD as , (2.9) where A B′ ′ = (A P′ ′) + (B′ ′ P ) , 2 2 = + [ ( Δ Δ x u x + x y, ) − u(x y, )] +[ ( v x + Δx y, ) − 2 v x( , y)] , 2 AB = Δx. u x x y u x y x x ε = + + − → lim ( , ) ( , / Δ Δ 0 1 2 2 1 ) ( , ) ( , ) Δ Δ x Δ v x x y v x y x ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ + ⎡ + − ⎣ ⎢ ⎤ ⎦ ⎥ ⎧ ⎨ ⎪ 2 ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ − 1 x u x v x ε = + ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ 1 2 2 2 1 / ⎦ ⎥ ⎥ − 1 x = u x ε , ∂ ∂ ∂ ∂ << u x 1 ∂ ∂ << v x 1 y AD A D AD AD ε = ′ ′ − → lim0 1343_book.fm Page 70 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC