Chapter 5 Plane-Stress Stress-Strain Relations in a Global Coordinate System

Chapter 5 Plane-Stress Stress-Strain Relations in a Global Coordinate System One of the most important characteristics of structures made of fiber-reinforced materials, and one which dictates the manner in which they are analyzed, is the use of multiple fiber orientations. Generally structural laminates are made of multiple layers of fiber-reinforced material, and each layer has its own specific fiber orientation. To this point we have studied the response of fiber-reinforced materials in the principal material system, whether it is fully three-dimensional stress-strain behavior, as in equations (2.45) and (2.47), or plane-stress behavior, as in equations (4.4) and (4.14). If we are to accommodate multiple layers of fiber-reinforced materials, each with its own fiber orientation, then we will be confronted with using multiple 1-2-3 coordinate systems, each with its own orientation with respect to some global or structural coordinate system. If we are dealing with an x-y-z Cartesian coordinate system to describe the geometry of the structure, then the orientation of each principal material system must be defined with respect to the x-y-z system. If we are dealing with an x-θ-r cylindrical coordinate system to describe the structure, then the orientation of each principal material system must be defined with respect to the x-θ-r system, and so forth, for a spherical coordinate system. This leads to a large number of coordinates systems and orientations for describing the response of the fiber-reinforced structure. As an alternative approach, we can refer the response of each layer of material to the same global system. We accomplish this by transforming the stress-strain relations from the 1-2-3 coordinate system into the global coordinate system. This will be our approach here, in particular; it will be done for a state of plane stress using the standard transformation relations for stresses and strains learned in introductory strength-of-materials courses. Transformation can also be done for a general state of stress. However, transformation here will be limited to the plane-stress state because it will be useful for the develop- 161

162CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY ment of classical lamination theory, which begins in the next chapter. Equally important, though, is the fact that the transformation of the description of the stress-strain response of fiber-reinforced material from the principal material coordinate system to a global coordinate system results in concepts so different from what one encounters with isotropic materials that it is best to start with a simpler plane-stress state and progress to the more complicated general stress state. When the concepts for the plane-stress stress state response described in a coordinate system other than the principal material coordinate system are fully understood, progression to a three-dimensional stress state is easier. 5.1 Transformation Relations Consider Figure 5.1(a), the familiar picture of an isolated element in the principal material coordinate system. Figure 5.1(b) shows a similar element but one that is isolated in an x-y-z global coordinate system. The fibers are oriented at an angle θ with respect to the +x axis of the global system. The fibers are parallel to the x-y plane and the 3 and z axes coincide. The fibers assumed their orientation by a simple rotation of the principal material system about the 3 axis. The orientation angle θ will be considered positive when the fibers rotate counterclockwise from the +x axis toward the +y axis. Often the fibers not being aligned with the edges of the element are referred to as an off-axis condition, generally meaning the fibers are not aligned with the analysis coordinate system (i.e., off the +x axis). Though we will use the notation for a rectangular Cartesian coordinate system as the global system (i.e., x-y-z), the global coordinate system can be considered to be any orthogonal coordinate system. The use of a Cartesian system is for convenience only and the development is actually valid for any orthogonal coordinate system. The stress-strain relations are a description of the relations between stress and strain at a point within the material. The functional form of these relations does not depend on whether the point is a point in a rectangular Cartesian coordinate system, in a cylindrical coordinate system, or in a spherical, elliptical, or parabolic coordinate system. The stresses on the small volume of element are now identified in accordance with the x-y-z notation. The six components of stress and strain are denoted as σ x σ y σ z τ yz τ xz τ xy ε x ε y ε z γ yz γ xz γ xy (5.1) and the six components of stress are illustrated in Figure 5.2. Although we seem interested in describing the stress-strain relation in another coordinate system now that we have developed it and fully understand it in the 1-2-3 system, we should emphasize that Figure 5.2 should be interpreted quite literally. We should

5.1. TRANSFORMATION RELATIONS 163 interpret the figure as asking, What is the relation between the stresses and deformations denoted in equation (5.1) for a small volume of material whose fibers are oriented at some angle relative to the boundaries of the element rather than parallel to them? This is the real issue! Loads will not be always applied parallel to the fibers; our intuition indicates that unusual deformations are likely to occur. The skewed orientation of the fibers must certainly cause unusual distortions of the originally cubic volume element. What are these deformations? How do they depend on fiber-orientation? Are they detrimental? Are they beneficial? Fortunately these and other questions can be answered by transforming the stress-strain relations from the 1-2-3 system to the x-y-z system. If we consider the special case shown in Figure 5.1(b), where the two coordinate systems are related to each other through a simple rotation θ about the z axis, then the stresses in the 1-2-3 system are related to the stresses in the x-y-z system by σ 1 =cos 2 θσ x +sin 2 θσ y +2sinθ cos θτ xy σ 2 =sin 2 θσ x +cos 2 θσ y 2sinθ cos θτ xy σ 3 = σ z τ 23 =cosθτ yz sin θτ xz (5.2) τ 13 =sinθτ yz +cosθτ xz τ 12 = sin θ cos θσ x +sinθ cos θσ y + ( cos 2 θ sin 2 θ ) τ xy For a state of plane stress, σ 3, τ 23,andτ 13 are zero, and upon rearranging, the third, fourth, and fifth components of equation (5.2) give σ z =0 cos θτ yz sin θτ xz =0 sin θτ yz +cosθτ xz =0 (5.3) sin 2 θ +cos 2 θ =1 (5.4) Because the only solution to the last two equations of equation (5.3) is τ yz = τ xz =0 (5.5) leading to the conclusion that for a plane-stress state in the 1-2-3 principal material coordinate system, the out-of-plane stress components in the x-y-z global coordinate system are also zero. This may have been intuitive but here we have shown it directly. The first, second, and sixth component of equation (5.2) may look more

164CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY Δ 2 3 Δ 1 2 Δ y z Δ x 2 y Δ 3 1 Δ z (a) Element in 1-2-3 system (b) Element in x-y-z system FIGURE 5.1 Elements of fiber-reinforced material in 1-2-3 and x-y-z coordinate systems x 1 θ familiar in the form ( σx + σ y σ 1 = 2 ( σx + σ y σ 2 = 2 ( σx σ y τ 12 = 2 ) + ) ( σx σ y 2 ( σx σ y 2 ) sin 2θ + τ xy cos 2θ ) cos 2θ + τ xy sin 2θ ) cos 2θ τ xy sin 2θ (5.6) This form is derivable from equation (5.2) by using trigonometric identities and is the form usually found in introductory strength-of-materials courses. The form of equation (5.2) will be most often used as it can be put in matrix form as σ 1 σ 2 τ 12 cos 2 θ sin 2 θ 2sinθcos θ = sin 2 θ cos 2 θ 2sinθcos θ sin θ cos θ sin θ cos θ cos 2 θ sin 2 θ σ x σ y τ xy (5.7) z σ z τ yz τ xz y τ xy τ xz τ xy σ y τ yz σ x x FIGURE 5.2 Stress components in x-y-z coordinate system

5.1. TRANSFORMATION RELATIONS 165 This transformation matrix of trigonometric functions will be used frequently in the plane-stress analysis of fiber-reinforced composite materials and it will be denoted by [T ], [T ] being written as m 2 n 2 2mn [T ]= n 2 m 2 2mn (5.8) mn mn m 2 n 2 where m = cosθ, n = sinθ. With the above notation equation (5.7) can be written σ 1 σ x σ 2 =[T ] σ y (5.9) τ 12 τ xy or σ 1 σ 2 τ 12 = m 2 n 2 2mn n 2 m 2 2mn mn mn m 2 n 2 σ x σ y τ xy (5.10) The inverse of equation (5.10) is σ x m 2 n 2 2mn σ y = n 2 m 2 2mn mn mn m 2 n 2 τ xy σ 1 σ 2 τ 12 (5.11) which implies m 2 n 2 2mn [T ] 1 = n 2 m 2 2mn (5.12) mn mn m 2 n 2 In a similar manner, the strains transform according to the specialized relations equation (5.2) as ε 1 =cos 2 θε x +sin 2 θε y +2sinθ cos θε xy ε 2 =sin 2 θε x +cos 2 θε y 2sinθ cos θε xy ε 3 = ε z ε 23 =cosθε yz sin θε xz (5.13) ε 13 =sinθε yz +cosθε xz ε 12 = sin θ cos θε x +sinθ cos θε y +(cos 2 θ sin 2 θ)ε xy

166CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY Note very well that the tensor shear strains, ε, not the engineering shear strains are being used in the above. These two measures of strain are different by a factor of two; that is, ε 23 = 1 2 γ 23 ε 13 = 1 2 γ 13 (5.14) ε 12 = 1 2 γ 12 If engineering shear strain is used instead, then the transformation relations become ε 1 =cos 2 θε x +sin 2 θε y +2sinθ cos θ 1 2 γ xy ε 2 =sin 2 θε x +cos 2 θε y 2sinθ cos θ 1 2 γ xy ε 3 = ε z γ 23 =cosθγ yz sin θγ xz (5.15) γ 13 =sinθγ yz +cosθγ xz 1 2 γ 12 = sin θ cos θε x +sinθ cos θε y +(cos 2 θ sin 2 θ) 1 2 γ xy As a result of the plane-stress assumption, specifically by equation (4.2), γ 23 =0 γ 13 =0 (5.16) and by analogy to equation (5.3), it is concluded from the fourth and fifth equation of equation (5.15) that γ yz =0 γ xz =0 (5.17) Also, due to the third equation of equation (5.15) and equation (4.3), ε z = S 13 σ 1 + S 23 σ 2 (5.18) More importantly, if equation (5.7) is used to transform the stresses, then ε z =(S 13 cos 2 θ + S 23 sin 2 θ)σ x +(S 13 sin 2 θ + S 23 cos 2 θ)σ y +2(S 13 S 23 )sinθ cos θτ xy (5.19) This equation is very important because it indicates that a shear stress in the x-y plane, τ xy, produces an extensional strain, ε z, perpendicular to that plane! For an isotropic material, S 13 = S 23, and this simply will not happen. Shear

5.2. TRANSFORMED REDUCED COMPLIANCE 167 stresses do not cause extensional strains in isotropic materials! This generation of extensional strains by shear stresses is an important characteristic of fiberreinforced composite materials. Returning to equation (5.15) to focus on the strains involved in the planestress assumption, we can write the first, second, and sixth equations as ε 1 ε 2 1 2 γ 12 =[T ] ε x ε y 1 2 γ xy (5.20) or ε 1 ε 2 1 2 γ 12 = m 2 n 2 2mn n 2 m 2 2mn mn mn m 2 n 2 ε x ε y 1 2 γ xy (5.21) It is very important to note that the transformation retains the factor of 1 2 with the engineering shear strain. 5.2 Transformed Reduced Compliance Continuing with the transformation of the stress-strain relations for plane stress to the x-y-z global coordinate system, the stress-strain relations in the 1-2-3 principal material coordinate system, equation (4.4), can be written in a slightly modified form to account for the use of the tensor shear strain rather than the engineering shear strain as ε 1 ε 2 1 2 γ 12 = S 11 S 12 0 S 12 S 22 0 0 0 1 2 S 66 σ 1 σ 2 τ 12 (5.22) Using the transformations given by equations (5.9) and (5.20) in equation (5.22) leads to ε x S 11 S 12 0 σ x ε y S 12 S 22 0 [T ] = 1 2 γ 1 xy 0 0 2 S [T ] σ y (5.23) 66 τ xy

168CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY and multiplying both sides of equation (5.23) by [T ] 1 results in ε x S 11 S 12 0 σ x ε y =[T ] 1 S 12 S 22 0 1 2 γ 1 xy 0 0 2 S [T ] σ y 66 τ xy (5.24) Substituting for [T ]and[t] 1 from equations (5.8) and (5.12), we find that multiplying these three matrices together, and multiplying the third equation through by a factor of 2, yields ε x S 11 S12 S16 σ x ε y = S 12 S22 S26 σ y (5.25) S 16 S 26 S 66 γ xy τ xy The S ij are called the transformed reduced compliances. Note that the factor of 1/2 has been removed and the engineering shear strain reintroduced. Equation (5.25) is a fundamental equation for studying the plane-stress response of fiberreinforced composite materials. The transformed reduced compliances are defined by S 11 = S 11 m 4 +(2S 12 + S 66 )n 2 m 2 + S 22 n 4 S 12 =(S 11 + S 22 S 66 )n 2 m 2 + S 12 (n 4 + m 4 ) S 16 =(2S 11 2S 12 S 66 )nm 3 (2S 22 2S 12 S 66 )n 3 m S 22 = S 11 n 4 +(2S 12 + S 66 )n 2 m 2 + S 22 m 4 (5.26) S 26 =(2S 11 2S 12 S 66 )n 3 m (2S 22 2S 12 S 66 )nm 3 S 66 =2(2S 11 +2S 22 4S 12 S 66 )n 2 m 2 + S 66 (n 4 + m 4 ) Equation (5.25) and the definitions equation (5.26) relate the strains of an element of fiber-reinforced material as measured in the x-y-z global coordinate system to the applied stresses measured in that coordinate system. We can look upon these equations as strictly the end result of simple steps in linear algebra, that is, transformations, inversions, and so forth. Alternatively, we can view them as what they actually are, namely, relations that describe what we shall see to be the complex response of an element of fiber-reinforced material in a state of plane stress that is subjected to stresses not aligned with the fibers, nor perpendicular to the fibers. The most profound results of equation (5.25) are that anormalstressσ x will cause a shearing deformation γ xy through the S 16 term, and similarly a normal stress σ y will cause a shearing deformation through the S 26 term. Equally important, because of the existence of these same S 16 and S 26 terms at other locations in the compliance matrix, a shear stress τ xy will cause strains ε x and ε y. Such responses are totally different from those in metals. In

5.2. TRANSFORMED REDUCED COMPLIANCE 169 metals, normal stresses do not cause shear strains, and shear stresses do not cause extensional strains. This coupling found in fiber-reinforced composites is termed shear-extension coupling. Shear-extension coupling is an important characteristic and is responsible for interesting and important responses of fiber-reinforced composite materials. Recall, equation (5.19) provided another example of shearextension coupling. Through a series of examples we will examine the response of an element of fiber-reinforced material to simple stress states (i.e., σ x only, and then τ xy only) and compare the responses with the response of a similar element of metal. After we work through these specific examples, the implications and meaning of shear-extension coupling will be clear. Figure 5.3 shows the variations with θ of the various elements of the transformed reduced compliance matrix for a graphite-reinforced material. Note that S 12 and all on-diagonal terms are even functions of θ, while the off-diagonal terms S 16 and S 26 are odd functions of θ. The importance of this will be illustrated shortly. Also note the rapid variation of some of the compliances as θ increases or decreases from 0.Atθ =30, S 11 has increased by a factor of 8 relative to its value at θ =0 and S 16 has changed from 0 at θ =0 to nearly its maximum magnitude. Before proceeding with the examples, we should discuss two special cases of equation (5.26). For the first case, consider the situation when the fibers are aligned with the x axis, namely, θ =0.Withθ =0, m =1andn =0and equation (5.26) reduces to S 11 (0 )=S 11 S22 (0 )=S 22 S 12 (0 )=S 12 S26 (0 )=0 (5.27) S 16 (0 )=0 S66 (0 )=S 66 where the argument of 0 is used as a reminder that the S ij s are functions of θ. The results of equation (5.27) simply state that at θ =0 the transformed reduced compliance degenerates to the reduced compliance, that is, the compliance in the principal material coordinate system. In the principal material system there is no S 16 or S 26. The quantities S 11, S 12, S 22,andS 66 are often referred to as the on-axis compliances. The barred quantities of equation (5.26) are frequently called the off-axis compliances. For the second special case, consider isotropic materials. The compliances of

170CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY 300 S ij, (TPa) 1 S ij, (TPa) 1 S 66 200 100 S 22 S 11 0 90 45 0 45 90 θ, degrees (a) On-diagonal terms 100 50 S 26 S 16 0 50 S 12 100 90 45 0 45 90 θ, degrees (b) Off-diagonal terms FIGURE 5.3 Variation of transformed reduced compliance with fiber angle θ for graphite-reinforced composite equation (5.26) reduce to S 11 = 1 E S 12 = ν E S 16 =0 (5.28) S 22 = 1 E S 26 =0 S 66 = 1 2(1 + ν) = G E which can be demonstrated by using the definitions of the compliances for an isotropic material, equation (4.6), in equation (5.26). For example: S 11 = 1 ( E m4 + 2ν ) 2(1 + ν) + n 2 m 2 + 1 E E E n4 (5.29a)

5.2. TRANSFORMED REDUCED COMPLIANCE 171 S 11 = 1 E ( m 4 +2m 2 n 2 + n 4) = 1 E ( m 2 + n 2) 2 (5.29b) But m 2 + n 2 =1 (5.30) so S 11 = 1 E (5.31) The proof that S 22 = 1 E (5.32) is identical. From S 12,wesee ( 1 S 12 = E + 1 ) 2(1 + ν) n 2 m 2 ν ( n 4 + m 4) (5.33a) E E E S 12 = ν ( 2n 2 m 2 + n 4 + m 4) = ν ( n 2 + m 2) 2 E E (5.33b) so by equation (5.30) S 12 = ν E (5.34) From S 16,wesee ( 2 S 16 = E 2ν ) ( 2(1 + ν) 2 nm 3 E E E 2ν ) 2(1 + ν) n 3 m E E (5.35a) Similarly it can be shown Finally, S 66 =2 S 16 =(0)nm 3 (0)n 3 m =0 (5.35b) S 26 =0 (5.36) ( 2 E + 2 E + 4ν ) 2(1 + ν) n 2 m 2 2(1 + ν) ( + n 4 + m 4) (5.37a) E E E 2(1 + ν) ( S 66 = 2n 2 m 2 + n 4 + m 4) = E Again, by equation (5.30), S 66 = E 2(1 + ν) 2(1 + ν) E ( n 2 + m 2) 2 (5.37b) (5.38) Thus, independent of the direction of the coordinate system, for an isotropic material equation (5.28) is true. We now turn to a series of examples that illustrate the shear-extension coupling predicted by the stress-strain relations of equation (5.25), specifically the deformations caused by a tensile normal stress. Consider a thin element of aluminum subjected to a tensile stress σ x = 155 MPa. As in Figure 5.4(a), the

172CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY aluminum element has dimensions 50 mm 50 mm. Thickness is not important at the moment but consider the element to be thin. As σ y and τ xy are zero, the stress-strain relations of equation (5.25) reduce to or, by equation (5.28), ε x = S 11 σ x ε y = S 12 σ x γ xy = S 16 σ x (5.39) ε x = 1 E σ x ε y = ν E σ x (5.40) γ xy =0 σ x Referring to Table 2.1, we note that the strains in the aluminum are 1 ε x = 72.4 10 9 155 106 = 2140 μ mm/mm 0.3 ε y = 72.4 10 9 155 106 = 642 μ mm/mm γ xy =0 250 10 6 =0 (5.41) The dimensional changes of the square element of aluminum are δδ x = ε x Δ x = (2140 10 6 )(50) = 0.1070 mm δδ y = ε y Δ y =( 642 10 6 )(50) = 0.0321 mm (5.42) so the deformed dimensions of the aluminum element, in Figure 5.4(b), are Δ x + δδ x =50.107 mm Δ y + δδ y =49.968 mm (5.43) This behavior is well known; the material stretches in the direction of the applied stress and contracts perpendicular to that direction (both in the y direction and the z direction, though the latter is not shown), and all right corner angles remaining right. Turning to Figure 5.4(c), we now consider a similar-sized square element of graphite-reinforced material with the fibers aligned with the x direction and also subjected to the 155 MPa stress in the x direction. The stress-strain relations of equation (5.25) reduce to ε x = S 11 (0 )σ x ε y = S 12 (0 )σ x γ xy = S 16 (0 )σ x (5.44)

5.2. TRANSFORMED REDUCED COMPLIANCE 173 or, if we use the compliances of equation (5.27), ε x = S 11 σ x ε y = S 12 σ x γ xy =0 σ x (5.45) From equation (4.5) and Table 2.1, or, alternatively, directly from equation (2.56), S 11 =6.45 (TPa) 1 S 12 = 1.600 S 22 =82.6 S 66 = 227 (5.46) y Δ x = 50 mm 50.107 mm Deformed Δ y = 50 mm x σ x = 155.0 MPa 49.968 mm (a) (b) 50.050 mm σ x = 155.0 MPa 49.988 mm (c) (d) 50.394 mm 30 (e) σ x = 155.0 MPa Increase in right angle 0.553 (f) 49.791 mm 50.394 mm 30 σ x = 155.0 MPa 49.791 mm (g) Decrease in (h) right angle 0.553 FIGURE 5.4 Response of aluminum and graphite-reinforced composite to tensile normal stress σ x

174CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY resulting in ε x =(6.45 10 12 )(155 10 6 ) = 1000 μ mm/mm ε y =( 1.600 10 12 )(155 10 6 )= 248 μ mm/mm γ xy = 227 10 6 0=0 (5.47) The dimensional changes of the graphite-reinforced element are δδ x = ε x Δ x = (1000 10 6 )(50) = 0.0500 mm δδ y = ε y Δ y =( 248 10 6 )(50) = 0.0124 mm (5.48) so the deformed dimensions are Δ x + δδ x =50.0500 mm Δ y + δδ y =49.988 mm (5.49) Figure 5.4(d) shows the deformed shape of the graphite-reinforced element, and the deformation is similar to that of the aluminum; the element stretches in the x direction and contracts in the y direction(andinthez direction), and all right corner angles remain right. For the same applied stress level of 155 MPa, the elongation of the fiber-reinforced material in the fiber direction, δδ x in equation (5.48), is about one-half the elongation of the aluminum, δδ x in equation (5.42); the difference is due to the difference between E 1 for the graphite-reinforced material and E for aluminum. Please note that a tensile stress level of 155 MPa in the fiber direction of graphite-reinforced material, which results in the 1000 μ mm/mm elongation strain, is considerably below the ultimate capacity of that material. Now consider a square element of graphite-reinforced material with the fibers oriented at θ =30 with respect to the x axis and also subjected to a stress, as shown in Figure 5.4(e), σ x = 155 MPa. The strains are determined by equation (5.25) as ε x = S 11 (30 )σ x ε y = S 12 (30 )σ x γ xy = S 16 (30 )σ x (5.50) With θ =30, m = 3/2 andn =1/2, and equations (5.26) and (5.46) give S 11 (30 )=50.8 (TPa) 1 S22 (30 )=88.9 S 12 (30 )= 26.9 S26 (30 )= 3.77 (5.51) As a result, from equation (5.50), S 16 (30 )= 62.2 S66 (30 ) = 126.0 ε x =(50.8 10 12 )(155 10 6 ) = 7880 μ mm/mm ε y =( 26.9 10 12 )(155 10 6 )= 4170 μ mm/mm γ xy =( 62.2 10 12 )(155 10 6 )= 9640 μ rad( 0.553 ) (5.52)

5.2. TRANSFORMED REDUCED COMPLIANCE 175 where the notation μ rad has been introduced. The shear strain represents an angle change, in radians, and the prefix μ has been added to represent the factor 10 6,thatis, 1μ rad = 10 6 rad = 57.3 10 6 degrees (5.53) This is the counterpart of μ mm/mm for the strains ε 1 and ε 2. It is interesting to note that at θ =30, the angle change of the shear-extension coupling effect of S 16 results in larger deformations than the Poisson contraction effects of S 12 (i.e., 9640 μ rad versus 4170 μ mm/mm). The relative magnitude of these two effects at all values of θ can be determined from the character of S 12 and S 16 as a function of θ, as in Figure 5.3(b). With the above numbers, dimensional changes of the graphite- reinforced element become δδ x = ε x Δ x =(0.00787)(50) = 0.394 mm δδ y = ε y Δ y =( 0.00417)(50) = 0.209 mm (5.54) Unlike the previous two cases, however, the original right corner angles do not remain right. The change in right angle is given by the value of γ xy in equation (5.52), namely, 9640 μ rad, or 0.553. The deformed dimensions of the element are Δ x + δδ x =50.394 mm Δ y + δδ y =49.791 mm (5.55) and Figure 5.4(f) illustrates the deformed shape of the element. It is important to properly interpret the sign of γ xy. A positive γ xy means that the right angle between two line segments emanating from the origin, one line segment starting from the origin and extending in the +x direction, the other line segment starting from the origin and extending in the +y direction, decreases. Because γ xy in the above example is negative, the angle in the lower left hand corner of the element increases. As a final example of the effects of tension normal stress in the x direction, consider an element of graphite-reinforced composite with the fibers oriented at θ = 30 relative to the +x axis, as in Figure 5.4(g). This example illustrates one of the important characteristics of the S ij as regards their dependence on θ. In this situation the stress-strain relations of equation (5.25) become ε x = S 11 ( 30 )σ x ε y = S 12 ( 30 )σ x (5.56) γ xy = S 16 ( 30 )σ x As noted earlier, inspection of the definitions of the off-axis compliance in equation (5.26) reveals that S 16 and S 26 are odd functions of n, and hence of θ, while

176CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY the remaining S ij are even functions of θ. Therefore S 11 ( 30 )=+ S 11 (+30 )=50.8 (TPa) 1 S 12 ( 30 )=+ S 12 (+30 )= 26.9 S 16 ( 30 )= S 16 (+30 )=62.2 S 22 ( 30 )=+ S 22 (+30 )=88.9 (5.57) S 26 ( 30 )= S 22 ( 30 )=3.77 S 66 ( 30 )=+ S 66 (+30 ) = 126.0 and substituting into equation (5.56) results in ε x =(50.8 10 12 )(155 10 6 ) = 7880 μ mm/mm ε y =( 26.9 10 12 )(155 10 6 )= 4170 μ mm/mm (5.58) γ xy =(62.2 10 12 )(155 10 6 ) = 9640 μ rad (0.553 ) With these numbers, dimensional changes of the 30 graphite-reinforced element become δδ x = ε x Δ x =(0.00787)(50) = 0.394 mm δδ y = ε y Δ y =( 0.00417)(50) = 0.209 mm (5.59) Like the +30 case, the original right corner angles do not remain right and the change in right angle is given by the value of γ xy in equation (5.58), namely, 9640 μ rad, or 0.553. The deformed dimensions of the element are the same as the +30 case, namely, Δ x + δδ x =50.394 mm Δ y + δδ y =49.791 mm (5.60) Figure 5.4(h) illustrates the deformed shape of the element. It is important to note that the change in the right corner angle for the 30 case is opposite the change for the +30 case. This ability to change the sign of the deformation by changing the fiber angle is a very important characteristic of fiber-reinforced composite materials. Here the sign change of S 16 was responsible for the sign of the change in right angle. Because it is also an odd function of θ, S 26 changes sign with θ and in other situations it can be responsible for controlling the change in sign of a deformation. In more complicated loadings, specifically with stress components σ x and σ y both present, both S 16 and S 26 control the sign of the deformation. The potential for using this characteristic is enormous. It is important to note that simply rotating the fiber angles by 30 increases the strain in the direction of the applied stress by a factor of 8. Equation (5.47) indicates that for θ =0, ε x = 1000 μ mm/mm, while equation (5.52) shows that for θ =30, ε x = 7880 μ mm/mm. The loss of stiffness when the fibers rotate away from the loading direction is quite significant. As mentioned, S 16 and S 26 serve double duty in that they couple normal stresses to shear deformation, and they couple the shear stress to extensional

5.2. TRANSFORMED REDUCED COMPLIANCE 177 deformations. Another series of examples will illustrate this latter coupling and further illustrate the influence of the sign dependence of S 16 and S 26 on θ. The series will again start with an element of aluminum and progress through an element of graphite-reinforced material. This progression, though adding nothing to what we already know about the behavior of aluminum, is taken specifically to show the contrasts, and in some cases the similarities, in the response of fiber-reinforced composites and isotropic materials. Consider, as shown in Figure 5.5(a), a 50 50 mm square of aluminum loaded by a 4.40 MPa shear stress τ xy. Of interest are the deformations caused by the application of this shear stress. Because σ x and σ y are zero, the stress-strain relations of equation (5.25) reduce to ε x = S 16 τ xy ε y = S 26 τ xy γ xy = S 66 τ xy (5.61) For aluminum, S 16 and S 26 were shown to be zero, in equation (5.28), and as a result ε x =0 τ xy ε y =0 τ xy (5.62) γ xy = S 66 τ xy = 1 G τ xy Using the value of shear modulus for aluminum from Table 2.1 yields ε x =0 4.40 10 6 =0 ε y =0 4.40 10 6 =0 γ xy = 1 27.8 10 9 4.40 106 = 158.0 μ rad (0.00905 ) (5.63) confirming our experience with aluminum that only a shear deformation results; the angle in the lower left corner decreases by 158.0 μ rad, or 0.00905.The lengths of the sides of the deformed element are still exactly 50 mm; Figure 5.5(b) shows the deformed shape. Applying a shear stress τ xy to an element of graphite-reinforced composite with the fibers aligned with the x axis, as in Figure 5.5(c), leads to ε x = S 16 (0 )τ xy ε y = S 26 (0 )τ xy (5.64) γ xy = S 66 (0 )τ xy Using the compliances of equation (5.27) and the numerical values from equation (5.46) leads to ε x =0 τ xy =0 4.40 10 6 =0 ε y =0 τ xy =0 4.40 10 6 =0 γ xy = S 66 τ xy = (227 10 12 )(4.40 10 6 ) = 1000 μ rad (0.0573 ) (5.65)

178CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY y Δ x = 50 mm 50 mm Δ y = 50 mm τ xy = 4.40 MPa x 50 mm (a) Decrease in right angle 0.00905 (b) 50 mm τ xy = 4.40 MPa 50 mm (c) Decrease in right angle 0.0573 (d) 49.986 mm τ xy = 4.40 MPa 49.999 mm 30 (e) Decrease in right angle 0.0318 (f) 50.01368 mm τ xy = 4.40 MPa 50.000829 mm 30 (g) Decrease in (h) right angle 0.0318 FIGURE 5.5 Response of aluminum and graphite-reinforced composite to a positive shear stress τ xy Again, as in Figure 5.5(d), the only deformation is the shear strain; the 4.40 MPa shear stress τ xy causes a much larger shear strain in the graphite-reinforced material with the fibers aligned with the x axis than in the aluminum. This is because the value of G 12 for a graphite-reinforced compositeismuchlessthan the value of G for aluminum. Attention now turns to the case of Figure 5.5(e), applying the 4.40 MPa shear stress τ xy to an element of graphite-reinforced material with its fibers oriented at

5.2. TRANSFORMED REDUCED COMPLIANCE 179 30 relative to the +x axis. This situation results in an unexpected and unusual response. As with the past cases, the stress-strain relations of equation (5.25) result in ε x = S 16 (30 )τ xy ε y = S 26 (30 )τ xy γ xy = S 66 (30 )τ xy (5.66) and using the appropriate numerical values for the off-axis compliances from equation (5.51) yields ε x =( 62.2 10 12 )(4.40 10 6 )= 274 μ mm/mm ε y =( 3.77 10 12 )(4.40 10 6 )= 16.58 μ mm/mm γ xy = (126.0 10 12 )(4.40 10 6 ) = 555 μ rad (0.0318 ) (5.67) The above numbers indicate that due to the shear stress, both the x and y dimensions decrease! This behavior is unlike anything that happens with an isotropic material and is totally unexpected. This coupling of shear stress and extensional deformation again provides unlimited potential for using fiber-reinforced composite materials to achieve results not possible or even conceivable with metals. The dimensional changes associated with the above strains are δδ x = ε x Δ x =( 0.000274)(50) = 0.01368 mm δδ y = ε y Δ y =( 0.00001658)(50) = 0.000829 mm (5.68) The shear strain is positive so the right corner angle in the lower-left-hand corner of the element decreases by 555 μ rad, or 0.0318. Figure 5.5(f) illustrates the deformed shape of the element; the lengths of the sides being given by Δ x + δδ x =49.986 mm Δ y + δδ y =49.999 mm (5.69) Finally, consider the element of graphite-reinforced composite with the fibers oriented at 30 relative to the +x axis, as in Figure 5.5(g). With an applied stress of τ xy =4.4 MPa, the stress-strain relations of equation (5.25) become ε x = S 16 ( 30 )τ xy ε y = S 26 ( 30 )τ xy γ xy = S 66 ( 30 )τ xy (5.70) If we use numerical values from equation (5.57), equation (5.70) becomes ε x =(62.2 10 12 )(4.40 10 6 ) = 274 μ mm/mm ε y =(3.77 10 12 )(4.40 10 6 )=16.58 μ mm/mm γ xy = (126.0 10 12 )(4.40 10 6 ) = 555 μ rad (0.0318 ) (5.71) These numbers indicate that with the fibers at θ = 30, the sides of the element increase in length! This is exactly opposite the case with θ =+30. However,

180CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY the right angle in the lower left corner decreases the same as for the θ =+30 orientation. The simple switching of the fiber angle has a significant influence on the response. Figure 5.5(h) illustrates the deformed element; the dimensional changes are given by δδ x = ε x Δ x =(0.000274)(50) = 0.01368 mm δδ y = ε y Δ y =(0.00001658)(50) = 0.000829 mm (5.72) and the new dimensions being Δ x + δδ x =50.01368 mm Δ y + δδ y =50.000829 mm (5.73) The examples of Figures 5.4 and 5.5 illustrate one of the most important characteristics of the response of fiber-reinforced materials, namely, the coupling effects when the fibers are oriented at some angle relative to the direction of the applied load. Here because the stress levels considered were small, the deformations were small. For higher stress levels, larger deformations result. The important point is, couplings are present in fiber-reinforced materials, and they can be used to advantage. With some experience, intuition allows one to qualitatively predict these coupling effects for simple cases. With all three components of stress applied, however, relying on intuition can be dangerous and the stressstrain relations of equation (5.25) should be used. Using the stress-strain relations of equation (5.25) is recommended in all cases, even if, in your mind, it is only being used to confirm intuition. Many times your intuition can be fooled, but use of equation (5.25) always leads to the correct answer. It is important to keep in mind the discussion of Section 2.5 regarding specification of either the stress components or the strain components. In the context of plane stress, either σ 1 or ε 1, and either σ 2 or ε 2, and either τ 12 or γ 12 can be stipulated, but not both the stress and the strain component from any one of the pairs. In the context of the fibers being oriented at some angle with respect to the x axis, either σ x or ε x, and either σ y or ε y, and either γ xy or τ xy can be specified. In the above series of examples, the stress component from each of the pairs was stipulated, two of the three stress components being zero in all the cases. In all cases the strains in each of the stress-strain pairs were being sought. By contrast, consider again the 50 mm 50 mm square of graphite-reinforced composite, loaded in tension and with its fibers oriented at 30 with respect to the +x axis. Assume that, instead of being completely free to deform, as in the last examples, the off-axis element is constrained from any deformation in the y direction, as in Figure 5.6(a). Of interest here are the deformations that result from this loading. For this situation σ x is known to be 155 MPa, ε x is unknown, σ y is unknown, ε y is known to be zero, τ xy is known to be zero, and γ xy is unknown. The unknowns involve both stresses and strains, and the knowns involve both stresses and strains. For this particular situation the stress-strain relations

5.2. TRANSFORMED REDUCED COMPLIANCE 181 of equation (5.25) become ε x = S 11 σ x + S 12 σ y + S 16 0 0= S 12 σ x + S 22 σ y + S 26 0 γ xy = S 16 σ x + S 26 σ y + S 66 0 (5.74) where it is understood that the S ij are evaluated at θ = 30. From the second equation σ y = S 12 σ x (5.75) S 22 and using this in the first and third equations results in ( ε x = S 11 S 12 2 ) σ x S 22 ( γ xy = S 16 S ) (5.76) 12 S26 σ x S 22 The terms in parenthesis are identified as before as reduced compliances. Using numerical values for graphite-reinforced composite, from equation (5.57), we find ε x = 6610 μ mm/mm γ xy = 9820 μ rad (0.563 ) σ y =46.9 MPa (5.77) According to these calculations, the applied stress in the x direction causes the x dimension to increase by δδ x = ε x Δ x =(0.00661)(50) = 0.331 mm (5.78) and the lower left hand right corner angle to decrease by 9820 μ rad, or 0.563. The stress in the y direction required to maintain the state of zero deformation in that direction is a tensile value of 46.9 MPa. The deformed shape of the element is shown in Figure 5.6(b), and compared to the case of Figure 5.4(g), the addition y 50 mm 30 50 mm (a) σ x x = 155.0 MPa Decrease in right angle 0.563 50.331 mm FIGURE 5.6 Response of a partially constrained off-axis element of graphitereinforced material to a normal stress σ x (b)

182CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY of the restraint in the y direction decreases the change in length and increases slightly the change in right angle. Though the deformations in the x and y directions are very important,itis equally important to remember that accompanying these dimensional changes are changes in the z direction. Equation (5.19), or its more fundamental form, equation (5.18), is the expression for the through-the-thickness strain. If we consider the situation in Figure 5.5(e) as an example, equation (5.19) becomes ε z =2(S 13 S 23 )sinθ cos θτ xy (5.79) Using numerical values of S 13 and S 23 from equation (2.56) yields ε z =2( 1.600 + 37.9) 10 12 sin(30 )cos(30 ) 4.40 10 6 ε z = 138.3 μ mm/mm (5.80) The element of material becomes thicker due to the application of the shear stress. The change in thickness of the element of material, Δh, is given by Δh = ε z h (5.81) Exercises for Section 5.2 1. Verify equation (5.26). This can be done by going through the steps discussed for arriving at the equation. The factors of 1/2 and 2 in the derivation of that equation can be confusing and lead to errors. Be sure you understand the origins of equation (5.26). 2. Suppose a 50 mm 50 mm square of graphite-reinforced material with its fibers oriented at +30 with respect to the +x axis is somehow restrained from any shear deformation but is free to deform in extension in the x and y directions. The square is compressed by a stress of 25 MPa in the y direction. (a) What is the deformation of the element in the x and y directions? (b) What shear stress is required to maintain this zero shear deformation condition? 3. As we mentioned earlier, it is often easy to forget that ε z is not zero for the plane-stress condition. Consider a 50 mm 50 mm element of graphite-reinforced material subjected to a tensile stress of 25 MPa in the x direction and assume the element thickness h is 0.150 mm. Plot the change in thickness of this element, Δh, as a function of fiber orientation angle θ and note the orientation of maximum and minimum thickness change. Use the range π/2 θ +π/2 and comment on the oddness or evenness of Δh versus θ. 4. Repeat Exercise 3 but consider the thickness change due to an applied shear stress of 25 MPa.

5.3. TRANSFORMED REDUCED STIFFNESSES 183 Computer Exercise It would be useful to write a short computer program to compute the elements of the S matrix as given by equation (5.26). Have the program read in and print out the values of E 1, E 2, ν 12,andG 12 ;alsoprintthevaluesofs 11, S 12, S 22, and S 66. Then have the program compute and print the values of the S ij as a function of θ, π/2 θ π/2. For glass-reinforced composite, plot these six quantities as a function of θ. Dothe S ij for glass-reinforced material vary as rapidly with θ as they do for graphite-reinforced materials? What is the reason for this? 5.3 Transformed Reduced Stiffnesses The inverse of the stress-strain relations of equation (5.25) can be derived by slightly rewriting the stress-strain relation of equation (4.14) to account for the factor of 1/2 in the shear strain as σ 1 σ 2 τ 12 = Q 11 Q 12 0 Q 12 Q 22 0 0 0 2Q 66 ε 1 ε 2 1 2 γ 12 (5.82) substituting the transformations given by equations (5.9) and (5.20) into equation (5.82), premultiplying both sides of the resulting equation by [T ] 1,and multiplying the three matrices together. This is all in analogy to equation (5.24), and the result is σ x Q 11 Q12 Q16 ε x σ y = Q 12 Q22 Q26 ε y (5.83) τ xy Q 16 Q26 Q66 γ xy The factors of 1/2 and 2 have been eliminated and the relation written in terms of the engineering strain. The Q ij are called the transformed reduced stiffnesses, and sometimes the off-axis reduced stiffnesses, and they are defined by Q 11 = Q 11 m 4 +2(Q 12 +2Q 66 )n 2 m 2 + Q 22 n 4 Q 12 =(Q 11 + Q 22 4Q 66 )n 2 m 2 + Q 12 (n 4 + m 4 ) Q 16 =(Q 11 Q 12 2Q 66 )nm 3 +(Q 12 Q 22 +2Q 66 )n 3 m Q 22 = Q 11 n 4 +2(Q 12 +2Q 66 )n 2 m 2 + Q 22 m 4 (5.84) Q 26 =(Q 11 Q 12 2Q 66 )n 3 m +(Q 12 Q 22 +2Q 66 )nm 3 Q 66 =(Q 11 + Q 22 2Q 12 2Q 66 )n 2 m 2 + Q 66 (n 4 + m 4 )

184CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY Figure 5.7 illustrates the variations of the various components of the transformed reduced stiffness matrix with θ. Like the transformed reduced compliances, the S ij, the transformed reduced stiffnesses vary significantly with θ. For example, compared to its value at θ =0,thevalueof Q 11 at θ =30 decreases 50 percent, while the value of Q66 increases by about a factor of 8. As with the S ij, Q12 and the on-diagonal terms are all even functions of θ, while the off-diagonal terms Q 16 and Q 26 are odd functions of θ. Obviously the S ij of equation (5.26) are related to the Q ij of equation (5.84) by [ Q] =[ S] 1 (5.85) However, the form of equation (5.84) is more convenient computationally. In analogy to equation (5.27), Q 11 (0 )=Q 11 Q22 (0 )=Q 22 Q 12 (0 )=Q 12 Q26 (0 )=0 (5.86) Q 16 (0 )=0 Q66 (0 )=Q 66 200 150 Q 22 Q 11 Q ij, GPa 100 50 Q 66 0 90 45 0 45 90 θ, degrees (a) On-diagonal terms 50 25 Q 12 Q 26 Q ij, GPa 0 25 50 90 45 0 45 90 θ, degrees (b) Off-diagonal terms FIGURE 5.7 Variation of transformed reduced stiffness with fiber angle θ for graphite-reinforced composite Q 16

5.3. TRANSFORMED REDUCED STIFFNESSES 185 where the Q 11, Q 12, Q 22,andQ 66 are often referred to as the on-axis reduced stiffnesses and are given by equation (4.17). For an isotropic material, in analogy to equation (5.28), from equation (4.18), Q 11 = E 1 ν 2 Q 12 = νe 1 ν 2 Q 16 =0 Q 22 = E 1 ν 2 Q 26 =0 Q 66 = G = E 2(1 + ν) (5.87) The transformed reduced stiffness of equation (5.84) and the stress-strain relations that use them, equation (5.83), are very important equations in the analysis of fiber-reinforced composite materials. The transformed stiffnesses will be used more frequently than the transformed reduced compliances. Like the transformed compliances, the transformed stiffnesses relate the strains as defined in the x-y-z global coordinate system to the stresses defined in that system, and the existence of the Q 16 and Q 26 terms, like the existence of the S 16 and S 26 terms, represent shear-extension coupling effects. Whereas the reduced compliances represent the deformations that result from a prescribed stress, the stiffnesses represent the stresses that must be applied to produce a prescribed deformation. Although one can use either the transformed reduced stiffnesses or the transformed reduced compliances to solve a given problem, the physical interpretation of the two quantities is so different that in a given problem it is generally more convenient to use one rather than the other. For example, consider a 50 mm 50 mm square of material that has been stretched in the x direction by 0.050 mm. To determine the stresses required to achieve this deformation, it is convenient to use the stress-strain relations of equation (5.83) directly, as opposed to the stress-strain relations of equation (5.25). As a parallel to the series of examples presented in the discussion of the reduced compliance matrix, a similar series of examples will next be discussed to illustrate the physical implications of the terms in the Q matrix. Shear-extension coupling and sign sensitivity of the Q 16 and Q 26 terms will again be evident. The particular examples can be considered the complement of the examples presented previously. They are termed complementary examples because the strain variable in each stress-strain pair is specified, whereas before, the stress variable of the pair was specified. These examples follow. A50mm 50 mm element of aluminum, in Figure 5.8(a), is stretched 0.050 mm in the x direction. The y dimension does not change and the right corner angles remain right. Interest focuses on the stresses required to effect this deformation. Assume the element is in a state of plane stress. Note the comple-

186CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY mentary nature of this problem relative to the problem discussed in Figure 5.4(a). In the present situation the strains of the stress-strain pairs are known and it is the stresses that are unknown and to be solved for. In the situation of Figure 5.4(a) the stresses of the stress-strain pairs were known and it was the strains that were unknown and to be solved for. Specifically, in Figure 5.4(a) σ x was the only nonzero stress, whereas in Figure 5.7(a) ε x is the only nonzero strain. For y 50.0500 mm Δ x = 50 mm Δ y = 50 mm (a) x σ (b) y = 23.9 MPa σ x = 79.6 MPa 50.0500 mm σ y = 3.02 MPa (c) (d) σ x = 155.7 MPa 50.0500 mm σ y = 30.1 MPa τ xy = 46.7 MPa 30 (e) (f) σ x = 92.8 MPa 50.0500 mm σ y = 30.1 MPa τ xy = 46.7 MPa 30 σx = 92.8 MPa (g) (h) FIGURE 5.8 Stress required in aluminum and graphite-reinforced material to produce extensional strain

5.3. TRANSFORMED REDUCED STIFFNESSES 187 the deformation described in Figure 5.8(a) ε x = δδ x = 0.050 = 1000 10 6 Δ x 50 ε y = δδ y = 0 (5.88a) Δ y 50 =0 and because the right corner angles remain right, γ xy =0 (5.88b) To compute the stresses required to produce this deformation, the stress-strain relations of equation (5.83) can be used, resulting in σ x = Q 11 ε x σ y = Q 12 ε x τ xy = Q 16 ε x (5.89) Because aluminum is isotropic, the Q ij of equation (5.87) indicate equation (5.89) becomes σ x = E 1 ν 2 ε x σ y = νe 1 ν 2 ε x τ xy =0 ε x Substituting in numerical values for aluminum from Table 2.1 yields σ x = 72.4 109 1 (0.3) 2 1000 10 6 =79.6 MPa σ y = (0.3)(72.4 109 ) 1 (0.3) 2 1000 10 6 =23.9 MPa τ xy =0 (5.90) (5.91) Figure 5.8(b) illustrates this stress state, and the results correlate with our past experience in that a tensile stress in the y direction is required to overcome the tendency of the material to contract in the y direction. Because ε y = 0, a tensile stress is required to enforce this. Also, the specimen changes thickness in the z direction. Consider next, as in Figure 5.8(c), an element of graphite-reinforced composite in a state of plane stress with its fibers aligned with the x axis and stretched in the x direction by 0.050 mm, with no deformation in the y direction. Equation (5.88) just applied to aluminum defines the strain state. For this case the stress-strain relations, equation (5.83), reduce to σ x = Q 11 (0 )ε x σ y = Q 12 (0 )ε x τ xy = Q 16 (0 )ε x (5.92)

188CHAPTER 5. PLANE-STRESS STRESS-STRAIN RELATIONS IN A GLOBAL COORDINATE SY and by equation (5.86) to σ x = Q 11 ε x σ y = Q 12 ε x τ xy =0 ε x =0 (5.93) For the graphite-reinforced material, we find using equation (4.17) and the numerical values of the engineering properties from Table 2.1, or alternatively, equations (2.58) and (4.15), or equations (2.56) and (4.16), that Q 11 = 155.7 GPa Q 22 =12.16 (5.94) Q 12 =3.02 Q 66 =4.40 Using the applicable stiffness from equation (5.94), the stresses given by equation (5.93) are σ x = (155.7 10 9 )(1000 10 6 ) = 155.7 MPa σ y =(3.02 10 9 )(1000 10 6 )=3.02 MPa (5.95) τ xy =0 Because of the relative values of E for aluminum and E 1 for the graphitereinforced material, stretching the graphite-reinforced material by 0.050 mm takes about twice as much stress as stretching the aluminum the same amount. However, restraining the deformation in the y direction in the graphite-reinforced material, 3.02 MPa, equation (5.95), takes far less than restraining the deformation in the aluminum, 23.9 MPa, equation (5.91), both in terms of absolute stress level, and in terms of stress level relative to σ x. The smaller value of σ y required for the graphite-reinforced material is a direct result of the small value of ν 21. Consider now the situation in Figure 5.8(e), a 50 mm 50 mm element of graphite-reinforced material with the fibers oriented at +30 with respect to the +x axis and stretched 0.050 mm in the x direction. The strains are again given by equation (5.88), and the stresses required to produce these strains are σ x = Q 11 (30 )ε x σ y = Q 12 (30 )ε x (5.96) τ xy = Q 16 (30 )ε x With θ =30, m = 3/2 andn =1/2, and equations (5.94) and (5.84) give Q 11 (30 )=92.8 GPa Q22 (30 )=21.0 Q 12 (30 )=30.1 Q26 (30 )=15.5 Q 16 (30 )=46.7 Q66 (30 )=31.5 (5.97) Using these numerical values, we find that the stresses required to produce the prescribed deformations are σ x =(92.8 10 9 )(1000 10 6 )=92.8 MPa σ y =(30.1 10 9 )(1000 10 6 )=30.1 MPa (5.98) τ xy =(46.7 10 9 )(1000 10 6 )=46.7 MPa

5.3. TRANSFORMED REDUCED STIFFNESSES 189 Remarkably, a shear stress must be applied to the element of material, in addition to the other two stresses, in order to have the element simply elongate in the x direction. Figure 5.8(f) illustrates the stresses for this example. If this shear stress was not applied, then shearing deformations would result and the deformation would not be just a simple elongation in the x direction. Equally remarkable is the fact that the shear stress required to stop the right corner angles from changing is larger than the stress σ y required to restrain the material against deformation in the y direction (i.e., resistance to Poisson effects). As a final example of the stresses required to effect a simple elongation in the x direction, consider an element of graphite-reinforced material with the fibers oriented in the 30 direction, as in Figure 5.8(g). For this situation σ x = Q 11 ( 30 )ε x σ y = Q 12 ( 30 )ε x τ xy = Q 16 ( 30 )ε x (5.99) and from equation (5.97) and the evenness and oddness properties of the Q ij, Q 11 ( 30 )=+ Q 11 (+30 )=92.8 GPa Q 12 ( 30 )=+ Q 12 (+30 )=30.1 Q 16 ( 30 )= Q 16 (+30 )= 46.7 Q 22 ( 30 )=+ Q 22 (+30 )=21.0 (5.100) Q 26 ( 30 )= Q 26 (+30 )= 15.5 Q 66 ( 30 )=+ Q 66 (+30 )=31.5 Substituting these into equation (5.99) yields the stresses required, namely, σ x =(92.8 10 9 )(1000 10 6 )=92.8 MPa σ y =(30.1 10 9 )(1000 10 6 )=30.1 MPa τ xy =( 46.7 10 9 )(1000 10 6 )= 46.7 MPa (5.101) Again, shear stresses must be applied to the element, in addition to the other two stresses, to obtain the simple state of elongation in the x direction. However, reversing the fiber orientation is responsible for the change in the sign of the shear stress required to effect the deformation. The stress to restrain the deformation in the y direction is insensitive to the sign of the off-axis fiber orientation, as is the stress in the x direction. Figure 5.8(h) illustrates the stresses of equation (5.101). This just-completed series of examples illustrates the physical interpretation of the reduced stiffness matrix, and it also illustrates, in a different fashion, the existence, level, and character of shear-extension coupling, a coupling inherent in fiber-reinforced composite materials. As expected, the Q 16 and Q 26 serve double duty in regard to shear-extension coupling. We will illustrate this type of coupling