Composites Design and Analysis. Stress Strain Relationship


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1 Composites Design and Analysis Stress Strain Relationship
2 Composite design and analysis Laminate Theory Manufacturing Methods Materials Composite Materials Design / Analysis Engineer Design Guidelines Design Data Quality Assurance
3 The design/analysis engineer should be able to determine for the following question: How do composite laminates behave under load? Which composite material should be used? What are the guidelines for design and the analysis methods? What design data is required and how is it obtained? Can the composite product be fabricated easily? How can the quality of the composite product be assured? The purpose of the material presented here is to introduce the reader to the principles of the mechanical behavior of thin laminates (how laminates behave under loads). This includes both symmetric and unsymmetric laminates relative to the midplane, with both inplane and flexural loads. Also included are the effects of temperature and moisture, and laminate strength. A liberal use is made of flow charts for aids in analysis
4 Stress Strains Relationships in Lamina (1) The lamina is homogeneous, orthotropic and linearelastic (2) The lamina is in a state of plane stress (very thin), therefore, the stresses associated with the zdirection are negligible: σ z = τ xz = τ yz = 0
5 The normal strain in the xdirection for the lamina in Figure 2a is the sum of the normal strains in the xdirection for the laminae in Figures 2b, 2c and 2d: From Figure 2b, ε x = 0 x x From Figure 2c, E x = or εx = x Ex x y From Figure 2d, ν y = or εx = ν y ε y = ν y y Ey Consequently, the total normal strain in the xdirection is x νy y ε x = 0 + Ex Ey x y or ε x = νy Ex Ey
6 The normal strain in the ydirection for the lamina is determined from superposition as follows: From Figure 2b, ε y = 0 y x From Figure 2c, ν x = or εy = ν x ε x = ν x x Ex y y From Figure 2d, E y = or εy = y Ey Consequently, the total normal strain in the ydirection is or x + y y ε y = 0 ν x Ex E x y ε y = ν x + Ex Ey (2) The shear strain, which is independent of the normal strains, is determined as follows: s s From Figure 2b, E s = or εs = s Es From Figure 2c, ε s = 0 From Figure 2d, ε s = 0 Therefore, the shear strain associated with the xy coordinate system is; ε s = s E s (3)
7 If we express Equations (1), (2) and (3) in matrix form, we have s y x = s y x x y y x E E E E E s y x = ss yy yx xy xx S S S S S s y x (4) Equation (4) may also be expressed as follows: s y x = s y x y y x y x y x x y y x x E E E E E s y x = ss yy yx xy xx Q Q Q Q Q s y x (5) [ε] = [C] [σ] [σ] = [M] [ε]
8 Elastic Constants x P σ x = P/A ε x ε y ν x = ε y / ε x lbs psi in/in in/in ν x = (average)
9 Elastic Constants y σ y vs ε y P σ y = P/A ε x ε y ν y = ε x / ε y lbs psi in/in in/in σ y (psi) 6000 E y = σ y / ε y = 3.05 x 10 6 psi ν y = (average) ε y (in/in)
10 Elastic Constants s P σ s = P/2A ε μ ε ν ε s = ε μ ε ν lbs psi in/in in/in in/in
11 Determination of Elastic Constants Type Material E x E y ν x E s ν f Specific gravity Typical thickness Type Material E x E y ν x E s t GPa GPa GPa meters T300/5208 Graphite/Epoxy T300/5208 Graphite/Epox y Msi Msi Msi inches B(4)/5505 Boron/Epoxy B(4)/5505 Boron/Epoxy AS/3501 Graphite/Epoxy Scotchply 1002 Glass/Epoxy AS/3501 Graphite/Epox y Kevlar 49/Epoxy Aramid/Epoxy Scotchply 1002 Glass/Epoxy Kevlar 49/Epoxy Aramid/Epoxy
12 Transformation of Stress and Strain
13 Area Stresses Stresses Forces
14 These three equilibrium equations may be combined in matrix form as follows:
15 The equations for the transformation of strain are the same as those for the transformation of stress:
16 Stress Strain Relationships in Global Coordinates Develop the relationship between the stresses and strains in global coordinates.
17 Stress Strain Relationships in Global Coordinates [σ xys ] = [Q xys ] [ε xys ] (1) [σ xys ] = [T] [ σ 126 ] (2) and [ε xys ] = [Ť] [ε 126 ] (3) Substitution of Equations (2) and (3) into (1) yields [T] [ σ 126 ] = [Q xys ] [Ť] [ε 126 ] (4) Pre multiplying both sides of Equation (4) by [T] 1 : [T] 1 [T] [σ 126 ] = [T] 1 [Q xys ] [Ť] [ε 126 ] (5) or [ σ 126 ] = [T] 1 [Q xys ] [Ť] [ε 126 ] (6) or = [T] 1 [Q xys ] [Ť] (7)
18 or = [Q 126 ] or [ σ 126 ] = [Q 126 ] [ε 126 ] where [Q 126 ] = [T] 1 [Q xys ] [Ť] Note that [T (+θ) ] 1 = [T ( θ) ] The elements of the matrix [Q 126 ] are as follows: Q 11 = Q xx cos 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy sin 4 θ Q 22 = Q xx sin 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy cos 4 θ Q 12 = Q 21 = (Q xx + Q yy 4Q ss ) sin 2 θ cos 2 θ + Q xy (sin 4 θ +cos 4 θ) Q 66 = (Q xx + Q yy 2 Q xy 2Q ss ) sin 2 θ cos 2 θ + Q ss (sin 4 θ +cos 4 θ) Q 16 = Q 61 = (Q xx Q xy 2Q ss ) sinθ cos 3 θ + (Q xy Q yy + 2Q ss ) sin 3 θ cosθ Q 26 = Q 62 = (Q xx Q xy 2Q ss ) sin 3 θ cosθ + (Q xy Q yy + 2Q ss ) sinθcos 3 θ
19 [Q 126 ] 1 [ σ 126 ] = [Q 126 ] 1 [Q 126 ] [ε 126 ] or [Q 126 ] 1 [ σ 126 ] = [ε 126 ] or [ε 126 ] = [Q 126 ] 1 [ σ 126 ] or [ε 126 ] = [S 126 ] [ σ 126 ] or The elements of the matrix [S 126 ] are as follows: S 11 = (Q 22 Q 66 Q 262 ) / S 12 = S 21 = (Q 16 Q 26 Q 12 Q 66 ) / S 16 = S 61 = (Q 12 Q 26 Q 22 Q 16 ) / S 22 = (Q 11 Q 66 Q 162 ) / S 26 = S 62 = (Q 12 Q 16 Q 11 Q 26 ) / S 66 = (Q 11 Q 22 Q 122 ) / where = Q 11 Q 22 Q Q 12 Q 26 Q 61 Q 22 Q 16 2 Q 66 Q 12 2 Q 11 Q 62 2
20 Summary: Lamina Analysis
21 The Symmetric Laminate with In Plane Loads A. Stress Resultants and Strains: The A
22 Relationship between strains and the stress resultants. Related by constants Aij Deriving expressions for these constants: Expression for is:
23 Expression for N 1: N 1 = (σ 1 ) 1 (h 1 h 2 ) + (σ 1 ) 2 (h 2 h 3 ) + (σ 1 ) 3 (h 3 h 4 ) + Substitute the expressions for stress:
24 Above equation can be rewritten as: Similarly for N2 and N6:
25 Above can be combined into matrix form as: Or Where A 11 = (Q 11 ) k t k A 12 = (Q 12 ) k t k A 16 = (Q 16 ) k t k A 22 = (Q 22 ) k t k A 26 = (Q 26 ) k t k A 66 = (Q 66 ) k t k
26 Example Determine the elements of [A] for a symmetric laminate of 6 layers: [0/45/90] s. The material of each lamina is Gr/Ep T300/5208 and t=0.005 = m 0 Lamina 90 Lamina Q 11 = Q xx = Q 11 = Q yy = Q 22 = Q yy = Q 22 = Q xx = Q 12 = Q xy = Q 12 = Q xy = Q 66 = Q ss = 7.17 Q 66 = Q ss = 7.17 Q 16 = 0 Q 16 = 0 Q 26 = 0 Q 26 = 0 Q 11 = Q xx cos 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy sin 4 θ Q 22 = Q xx sin 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy cos 4 θ Q 12 = Q 21 = (Q xx + Q yy 4 Q ss ) sin 2 θ cos 2 θ + Q xy (sin 4 θ + cos 4 θ) Q 66 = (Q xx + Q yy 2Q xy 2 Q ss ) sin 2 θ cos 2 θ + Q ss (sin 4 θ + cos 4 θ) Q 16 = Q 61 = (Q xx Q xy 2 Q ss ) sinθ cos 3 θ + (Q xy Q yy + 2 Q ss ) sin 3 θ cosθ Q 26 = Q 62 = (Q xx Q xy 2 Q ss ) sin 3 θ cosθ + (Q xy Q yy + 2 Q ss ) sinθ cos 3 θ Q 11 = 0.25 Q xx + 2 (Q xy + 2 Q ss ) (0.25) Q yy = 56.65
27 Q 22 = 0.25 Q xx + 2 (Q xy + 2 Q ss ) (0.25) Q yy = Q 12 = (Q xx + Q yy 4 Q ss ) (0.25) + Q xy (0.5) = Q 66 = (Q xx + Q yy 2 Q xy 2Q ss ) (0.25) + Q ss (0.25) = Q 16 = (Q xx Q xy 2 Q ss ) (0.25) + (Q xy Q yy +2 Q ss ) (0.25) = Q 26 = (Q xx Q xy 2 Q ss ) (0.25) + (Q xy Q yy +2 Q ss ) (0.25) = The Laminate A 11 = ( ) 2 = A 22 = ( ) 2 = A 12 = ( ) 2 = A 66 = ( ) 2 = A 16 = ( ) 2 = A 26 = ( ) 2 = GPa m
28 Equivalent Engineering Constants for the Laminate Equation for the composite laminate is Or [N] = [A] [ε] Matrix equation for a single orthotropic lamina or layer is Or [σ] = [M] [ε]
29 The properties of the single equivalent orthotropic layer can be determined from the following equation: [A] = [M] where h = laminate thickness or 1 A 11 = E1 1 A 12 = 2E1 1 A 66 = E 6 h h h 1 h A 22 = E 2 1 A 21 = h 1E Solving last equation, we have the following elastic constants for the single equivalent orthotropic layer: A11 A22 E 1 = (1 ) E 2 = (1 ) h A12 2 A11A 22 h A A A 22 1 h 12 E 6 = A 66 ν 1 = ν 2 = A A 22 A A 12 11
30 The equation for the composite laminate is or [ε] = [a] [N] where [a] = [A] 1 The equation for a single orthotropic lamina is or [ε] = [C] [σ]
31 The properties of the single equivalent orthotropic layer is determined by [a] h = [C] 1 2 or a 11 h = a 12 h = a 66 h = E E1 2 1 E 6 1 a 21 h = a 22 h = E 1 1 E 2 elastic constants for the single equivalent orthotropic layer: E 1 = 1 E 2 = E 66 = a11h 1 a22h 1 a66h a a 12 ν 1 = ν 2 = 11 a a 12 22
32 1) Cross Ply Laminates Special Cases:
33
34 Balanced Angle Ply Laminates:
35 Example: [0, 90] un symmetric cross ply, A 16 = A 26 = 0 [0, 90, 0] symmetric cross ply, A 16 = A 26 = 0 [+30, 30, 30, +30] symmetric balanced angle ply, A 16 = A 26 = 0 [+30, 30, +30, 30] un symmetric balanced angle ply, A 16 = A 26 = 0 [+30, 30, +30] symmetric un balanced angle ply, A 16, A 26 0
36 3) Quasi Isotropic Laminates: Has m ply groups spaced at ply orientations of 180/m degrees. Examples: [0/60/ 60]s m = 3 180/m = 60 [0/90/45/ 45]s m = 4 180/m = 45 [0/60/ 60/60/0/ 60]s m=3 180/m = 60 The modulus of the quasi isotropic laminate has the following properties: A 11 = A 22 A 16 = A 26 = 0 A 66 = (A 11 A 22 ) which is equivalent to G = 1 2 E 1 for an isotropic material.
37 Summary: Laminate Analysis
38 The Symmetric Laminate with Bending and Twisting Loads: The D Matrix M 1 and M 2 are bending moments per unit length M 6 is a twisting couple per unit length. The basic difference between in plane loading and flexural loading is that the strains are not the same in all layers of the laminate that has flexural loading.
39 The Symmetric Laminate with Bending and Twisting Loads: The D Matrix Stress Resultants and Curvatures: The D Matrix 1 2 d w dx ε x = (z ) = (z 2 ) = z K where ρ is the radius of curvature, and K is defined as the curvature. Note (from Figure 2) that z is positive and is negative. Thus ε x is positive, which is evident from inspection of the figure.
40 ε 1 = z = z K 1 ε 2 = z = z K 2 ε 6 = 2 z = z K 6 = z [ε 126 ] = z [K 126 ] where K 1, K 2 and K 6 are referred to as the curvatures [σ 126 ] = [Q 126 ] [ε 126 ] [σ 126 ] = z[q 126 ] [K 126 ] X w X w X X w K K K
41 The objective of this section is to determine the elements of the matrix which relates the flexural loads M 1, M 2, M 6 to the curvatures K 1, K 2, K 6 for a symmetric laminate. This matrix is defined as the D matrix: Furthermore, we wish to determine the relationship between the loads M 1, M 2, M 6 and the stresses and strains in a particular layer a distance z from the laminate mid plane.
42 The stress resultant M 1, which has units of pounds per inch of length in the 2 direction, is determined as follows: If we use the expression for σ 1 from Equation 5, M 1 becomes
43 After integration, M 1 becomes where k represents the k th layer and n is the number of layers. Similarly
44 If we combine above three equations, we get The elements of the D matrix are
45 B. Equivalent Engineering Constants for the Laminate with Flexural Loading A composite laminate may sometimes be modeled as an equivalent orthotropic single layer, as discussed in Section III B, with the D matrix as well as the A matrix. The objective here is to determine the elastic constants for this equivalent orthotropic layer, and to explain when it is appropriate to use them. The equation for the composite laminate is The equation for a single orthotropic layer is
46 The properties of the single equivalent orthotropic layer can be determined by equating the modulus of equation (1) with that of Equation (2): If we equate the individual elements of the left hand side of equation (3) with the corresponding elements on the right hand side, then solve for the elastic constants, we have
47 We model the structure with orthotropic plate elements to be used in a finite element program. The elastic constants of these equivalent orthotropic plate elements are determined from Equations (4). We take the results of the finite element program and determine M 1, M 2, and M 6 at the point we are interested in analyzing. Then we use the equations developed in Section II and IV to determine the stresses and strains in the various layers. This method is appropriate for laminates with D 16 = D 26 = 0 and flexural loading. For the more general cases in which D 16 and D 26 are not zero, this method is approximate and must be used with discretion. If the laminate loads are in plane rather than flexural, we should use the equivalent elastic constants derived in Section III B rather than those presented in this section. For laminates with many layers, the equivalent laminate elastic constants of Section III B and those presented here approach each other in value.
48 Cross Ply Laminates: C. Special Cases: All layers are either 0 or 90, which results in D 16 = D 26 = 0. This is sometimes called special orthotropic w. r. t. flexure. Balanced Angle Ply Laminates: There are only two orientations of the plies (same magnitude but opposite in sign) with an equal number of plies with positive and negative orientations. If for every layer at θ above the mid plane, there is an identical layer at θ below the mid plane, then D 16 = D 26 = 0. However, this does not produce a symmetric laminate and leads to coupling between mid plane strains and curvatures. If the laminate is symmetric then, D 16 and D 26 are not zero. However, as the number of layers approaches infinity, the terms D 16 and D 26 approach zero.
49 Example: [0, 90] un symmetric cross ply, A 16 = A 26 = D 16 = D 26 = 0 [0, 90, 0] symmetric cross ply, A 16 = A 26 = D 16 = D 26 = 0 [+30, 30, +30, 30] un symmetric balanced angle ply, A 16 = A 26 = D 16 = D 26 = 0 [+30, 30, 30, +30] symmetric balanced angle ply, A 16 = A 26 = 0, and D 16, D 26 0 [+30, 30, +30] symmetric un balanced angle ply, A 16, A 26 0, and D 16, D 26 0 [+30, 30] 4s or [+30, 30, +30, 30, +30, 30, +30, 30] s or 16 ply symmetric balanced angle ply, A 16 = A 26 = 0, and D 16, D 26 0
50 D. Summary: Laminate Analysis
51 Problems (Chapter IV) The stress in the upper 45 layer for the lamina shown are ρ x = 400 MPa, ρ y = 50 MPa, ρ s = 70 MPa. The thickness of each lamina is meters. Determine the loads (stress resultants M 1, M 2, M 6 ) that would cause these stresses. Use either metric or English units. The [D] matrix for this laminate is (metric): Verify the value for D 12.
52 V: The Symmetric Laminate with both In Plane and Flexural Loads We merely combine these two equations for the laminate shown in Figure 1; The strains of Equation 8, Section III A, must be interpreted as mid plane strains: ε 1o, ε 2o, ε 6o are the mid plane strains of the symmetric laminate K 1, K 2, K 6 are the laminate curvatures as discussed in Section IV A. The elements of the A matrix are defined by Equations (9), Section III A and he elements of the D matrix are defined by Equations 12, Section IV A.
53 Equation 1, Section IV A relates the curvatures K 1, K 2, K 6 and the strains ε 1, ε 2, ε 6 in a particular lamina for a symmetric laminate with flexural loading. For the laminate subjected to both flexural loading and inplane loading, the strains and stresses in a particular lamina must be written in terms of mid plane strains as well as the curvatures:
54 The quantities on the left hand side of Equation (2) are the strains in a particular lamina a distance z from the mid plane and are composed of two parts as shown. The strains [ε o 126] are produced by the in plane loads and the strains z [K 126 ] are produced by the flexural loads. This is shown in detail in Figure 2 for the strain component ε 1. The displacement of point a in the 1 direction is The strain in the 1 direction of point a is, by definition,
55 Summary: Laminate Analysis
56 We are given a set of allowable stresses in local coordinates (σ x, σ y, σ z ) for a particular lamina in a laminate and our objective is to calculate the allowable stress resultants. First, we must realize that we cannot solve for 6 unknowns by prescribing just 3 knowns. If we are given (σ x, σ y, σ z ) k, we can determine (ε x, ε y, ε z ) k, (ε 1, ε 2, ε 6 ) k, and (σ 1, σ 2, σ 6 ) k as shown by Figure 1. However, we can not determine (ε 1o, ε 2o, ε 6o,K 1, K 2, K 6 ) because we do not have 6 independent equations; Note that (ε 1, ε 2, ε 6 ) k and (σ 1, σ 2, σ 6 ) k are not independent. In fact, they are related by Equations (9) or (5). Thus we need to prescribe any 3 of the following 12 quantities: N 1, N 2, N 6, M 1, M 2, M 6, ε 1o, ε 2o, ε 6o, K 1, K 2, K 6. Now we may continue the solution and solve for the remaining unknowns. In summary: If we are given (N 1, N 2, N 6, M 1, M 2, M 6 ), we can solve for all other quantities. If we are given (ε 1o, ε 2o, ε 6o,K 1, K 2, K 6 ), we can solve for all other quantities. If we are given (ε x, ε y, ε z ) k, we must also be given the values for any 3 of the quantities in (1) or (2). Then we can proceed to solve for all other quantities.
57 Problems (Chapter V) A laminate consists of Graphite/Epoxy, T300/5208, layers as follows: [0, 45, 45, 90] s. The applied stress resultants are M 1 = 25 in lb/in, M 2 = 0, M 6 = 0, N 1 = 1500 lb/in, N 2 = 0, N 6 = 0. Determine the stresses (σ x, σ y, σ z ) midway between the upper and lower surface of the upper 45 lamina. Verify the values for elements A 22 and D 22.
58 VI: The Unsymmetrical Laminate If the laminate is symmetric, there is no coupling between mid plane strains and laminate curvatures, as illustrated by Equation (1) of Section V A. However, if the laminate is unsymmetric, coupling does exist as shown in Equation (1) below: The objective here is to derive the elements of the matrix that relates the in plane and bending stress resultants to the mid plane strains and laminate curvatures. We are especially interested in the B matrix of Equation (1).
59 Before we proceed with this derivation, recall the equations for the strains in the k th lamina, Equation (2) of Section V A: [ε 126 ] k = [ε o 126] + z [K 126 ] We obtain the expression for the strains in the k th lamina by substituting Equation (3) into Equation (9) of Section II D: [σ 126 ] k = [Q 126 ] k [ε 126 ] k [σ 126 ] k = [Q 126 ] k [ε o 126] + z [Q 126 ] k [K 126 ]
60 The expression for N 1 is determined as follows: Substitution of the expression for σ 1 from Equation (4) into equation (5) yields:
61 After integration, Equation (6) becomes: where n is the number of layers in the laminate and k refers to a particular layer. Similarly:
62 The expression for M 1 is: If we obtain the expression for σ 1 from Equation (4) and follow the procedure presented in Section IV A, we have Similarly,
63 If we combine Equations (8), (9) and (10), we have If we combine Equations (12), (13) and (14), we have
64 We now combine Equation (16) and (18) and the derivation is complete: Where A ij = (Q ij ) k (b k a k ) B ij = (Q ij ) k b 2 2 k a k 2 D ij = (Q ij ) k b 3 3 k a k 3 If the laminate is symmetric w. r. t. the mid plane, we see that B ij = 0, and the in plane quantities become uncoupled from the flexural quantities. Otherwise, there is coupling because of the existence of the B matrix. That is, in plane loads cause bending and twisting as well as in plane strains. Also, flexural loads cause in plane strains as well as bending and twisting.
65 Summary: Laminate Analysis
66
67
68 Problems (Chapter VI) A laminate consists of Graphite/Epoxy, T300/5208, layers as follows: [0, 45, 90]. The applied stress resultants are N 1 = 100 lb/in, N 2 = N 6 = M 1 = M 2 = M 6 = 0. Determine the stresses (σ x, σ y, σ z ) in the middle of the 90 lamina.
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