1. Composites Design and Analysis Stress-Strain Relationship Prof Zaffar M. Khan Institute of Space Technology Islamabad

2. Next Generation Aerospace Vehicle Requirements

3. Composite design and analysis

4. MATERIAL SELECTION

6. Composite Analysis
2D vs D State of Stress

7. Required elastic material properties for composites
Metals (isotropic materials)
E, G, ν
2 independent properties – G = __E____
2(1 + ν)
Composite lamina (unidirectional layer, ply)
In plane: E1, E2, G12, ν12
Out of plane : E3, G13, G23, ν13, ν23
But for transverse isotropy (2 = 3):
E2 = E G12 = G13 ν12 = ν G23 = __E23___
2(1 + ν 23)
Therefore 5 independent properties:
E1 E2 G12 ν 12 G23

8. Terminology used in micromechanics
Ef, Em - Young’s modulus of fiber and matrix
Gf, Gm - Shear modulus of fiber and matrix
νf, νm - Poisson’s ratio of fiber and matrix
Vf, Vm - Volume fraction of fiber and matrix
Wf, Wm – Weight fraction of fiber and matrix

9. Elastic properties: Rule of mixtures approach
Parallel model - E1 and ν12 (Constant Strains)
E1 = Ef1Vf + EmVm
ν12 = νf Vf + νmVm
Predictions agree well with experimental data
Series model – E2 and G (Constant Stress)
E2 = __ __ Ef2 Em________
Ef2 Vm + Em Vf
G12 = __ __ Ef2 Em________
Experimental results predicted less accurately

10. Micromechanics example: Volume fraction changes
Knowns : Carbon: E = 34.0 x 106 psi
Epoxy : E = 0.60 x 106 psi
How much does the longitudinal modulus change when the fiber volume fraction is changed from 58% to 65%?
E1 = Ef1 Vf + Em Vm
For Vf = 0.58: E1 = (34.0 x 106 psi)(.58) + (0.60 x 106 psi)(0.42) = 20.0 x 106 psi
For Vf = 0.65: E1 = (34.0 x 106 psi)(.65) + (0.60 x 106 psi)(0.35) = 22.3 x 106 psi
Thus, raising the fiber volume fraction from 58% to 65% increases E1 by 12%

11. Design and analysis of composite laminates: Laminated Plate Theory (LPT)
Used to determine the response of a composite laminate based on properties of a layer (or ply)

12. Laminate Ply Orientation Code
Designate each ply by it’s fiber orientation angle
List plies in sequence starting from top of laminate
Adjacent plies are separated by “/” if their angle is different
Designate groups of plies with same angle using subscripts
Enclose complete laminate in brackets
Use subscript “S” to denote mid plane symmetry, or “T” to denote total laminate
Bar on the top of the ply indicates mid-plane

14. Special types of laminates
Symmetric laminate – for every ply above the laminate mid plane, there is an identical ply(material and orientation) an equal distance below the mid plane
Balanced laminate – for every ply at a +θ orientation, there is another ply at the – θ orientation somewhere in the laminate
Cross-ply laminate – composed of plies of either 0o or 90o (no other ply orientations)
Quasi-isotropic laminate – produced using at least three different ply orientations, all with equal angles between them. Exhibits isotropic extensional stiffness properties (have the same E in all in-plane directions)

15. The response of special laminates
Balanced, unsymmetric, laminate
Tensile loading produces twisting curvature
Ex: [+θ/0/- θ]τ
Symmetric, unbalanced laminate
Tensile loading produces in-plane shearing
Unsymmetric cross-ply laminate
Tensile loading produces bending curvature
Ex: [0/90]τ
Balanced and symmetric laminate
Tensile loading produces extension
Ex: [+θ/- θ]s
Quasi isotropic laminate: [+60/0/- 60]s and [+45/0/+45/90]s
Tensile loading produces extension loading, independent of angle

16. Stress-Strains Relationships in Lamina

20. Elastic Constants-x P σx = P/A εx εy νx = − εy / εx lbs psi in/in 105
105
1872 −
0.140
200
3565 −
0.139
300
5348 −
400
7398 −
0.136
500
8913 −
0.135
600
10695 −
0.134
700
12478 −
0.130
800
14260 −
0.128

21. Elastic Constants-y P σy = P/A εx εy νy = − εx / εy lbs psi in/in 100
100
1751 −
0.139
200
3503 −
0.133
300
5254 −
0.129
400
7005 −
0.128
500
8757 −
0.125
600
10508 −
0.124
700
12259 −
0.119
800
14011 −
0.117

22. Elastic Constants-s P σs = P/2A εμ εν εs = εμ − εν lbs psi in/in 100
100
875 −
200
1748 −
300
2622 −
400
3497 −
500
4371 −
600
5245 −
700
6119 −
800
6993 −

23. Determination of Elastic Constants
Type
Material Ex Ey νx Es t
Msi
inches
T300/5208
Graphite/Epoxy
26.25
1.49
0.28
1.04
0.005
B(4)/5505
Boron/Epoxy
29.59
2.68
0.23
0.81
AS/3501
20.01
1.30
0.30
1.03
Scotchply 1002
Glass/Epoxy
5.60
1.20
0.26
0.60
Kevlar 49/Epoxy
Aramid/Epoxy
11.02
0.80
0.34
0.33
Type
Material Ex Ey νx Es νf
Specific gravity
Typical thickness
GPa
meters
T300/5208
Graphite/Epoxy
181
1.3
0.28
7.17
0.70
1.6
B(4)/5505
Boron/Epoxy
204
18.5
0.23
5.59
0.50
2.0
AS/3501
138
8.96
0.30
7.1
0.66
Scotchply 1002
Glass/Epoxy
38.6
8.27
0.26
4.14
0.45
1.8
Kevlar 49/Epoxy
Aramid/Epoxy 76
5.5
0.34
2.3
0.60
1.46

24. Transformation of Stress and Strain

25. Area
Stresses
Stresses
Forces

26. These three equilibrium equations may be combined in matrix form as follows:

27. The equations for the transformation of strain are the same as those for the transformation of stress:

28. Stress-Strain Relationships in Global Coordinates
Develop the relationship between the stresses and strains in global coordinates.
[σxys] = [Qxys] [εxys] (1)
[σxys] = [T] [ σ126] (2)
[εxys] = [Ť] [ε126] (3)
Substitution of Equations (2) and (3) into (1) yields 
[T] [ σ126] = [Qxys] [Ť] [ε126] (4)
Pre multiplying both sides of Equation (4) by [T] −1:
[T] −1 [T] [σ126] = [T] −1 [Qxys] [Ť] [ε126] (5)
or [ σ126] = [T] −1 [Qxys] [Ť] [ε126] (6)
or = [T] −1 [Qxys] [Ť] (7)

29. or = [Q126] or [ σ126] = [Q126] [ε126] where [Q126] = [T] −1 [Qxys] [Ť] Note that [T(+θ)] −1 = [T(−θ)] The elements of the matrix [Q126] are as follows: Q11 = Qxx cos4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy sin4θ Q22 = Qxx sin4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy cos4θ Q12 = Q21 = (Qxx + Qyy − 4Qss) sin2θ cos2θ + Qxy (sin4θ +cos4θ) Q66 = (Qxx + Qyy − 2 Qxy − 2Qss) sin2θ cos2θ + Qss (sin4θ +cos4θ) Q16 = Q61 = (Qxx – Qxy − 2Qss) sinθ cos3θ + (Qxy – Qyy + 2Qss) sin3θ co Q26 = Q62 = (Qxx – Qxy − 2Qss) sin3θ cosθ + (Qxy – Qyy + 2Qss) sinθcos3θ

30. [Q126] −1 [ σ126] = [Q126] −1 [Q126] [ε126]
or [Q126] −1 [ σ126] = [ε126]
or [ε126] = [Q126] −1 [ σ126]
or [ε126] = [S126] [ σ126]   or
The elements of the matrix [S126] are as follows: 
S11 = (Q22 Q66 – Q262) / ∆
S12 = S21 = (Q16 Q26 – Q12 Q66) / ∆
S16 = S61 = (Q12 Q26 – Q22 Q16) / ∆
S22 = (Q11 Q66 – Q162) / ∆
S26 = S62 = (Q12 Q16 – Q11 Q26) / ∆
S66 = (Q11 Q22 – Q122) / ∆  
where ∆ = Q11Q22 Q Q12Q26 Q61 − Q22Q162 – Q66Q122 – Q11Q622

31. Summary: Lamina Analysis

32. The Symmetric Laminate with In-Plane Loads A
The Symmetric Laminate with In-Plane Loads A. Stress Resultants and Strains: The “A”

33. Relationship between strains and the stress resultants
Relationship between strains and the stress resultants. Related by constants Aij Deriving expressions for these constants: Expression for is:

34. Substitute the expressions for stress:
Expression for N1:
N1 = (σ1)1 (h1-h2) + (σ1)2 (h2-h3) + (σ1)3 (h3-h4) + …
Substitute the expressions for stress:

35. Above equation can be rewritten as: Similarly for N2 and N6:

36. A11 = (Q11)k tk A12 = (Q12)k tk A16 = (Q16)k tk A22 = (Q22)k tk
Above can be combined into matrix form as: Or
Where
A11 = (Q11)k tk
  A12 = (Q12)k tk 
A16 = (Q16)k tk 
A22 = (Q22)k tk 
A26 = (Q26)k tk 
A66 = (Q66)k tk

37. 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
Q11 = Qxx = Q11 = Qyy = 10.34
Q22 = Qyy = Q22 = Qxx = 181.8
Q12 = Qxy = Q12 = Qxy = 2.897
Q66 = Qss = Q66 = Qss = 7.17
Q16 = 0 Q16 = 0
Q26 = 0 Q26 = 0
Q11 = Qxx cos4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy sin4θ
Q22 = Qxx sin4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy cos4θ
Q12 = Q21 = (Qxx + Qyy – 4 Qss) sin2θ cos2θ + Qxy (sin4θ + cos4θ)
Q66 = (Qxx + Qyy – 2Qxy – 2 Qss) sin2θ cos2θ + Qss (sin4θ + cos4θ)
Q16 = Q61 = (Qxx – Qxy – 2 Qss) sinθ cos3θ + (Qxy – Qyy + 2 Qss) sin3θ cosθ
Q26 = Q62 = (Qxx – Qxy – 2 Qss) sin3θ cosθ + (Qxy – Qyy + 2 Qss) sinθ cos3θ
Q11 = 0.25 Qxx + 2 (Qxy + 2 Qss) (0.25) Qyy = 56.65

38. Q22 = 0.25 Qxx + 2 (Qxy + 2 Qss) (0.25) + 0.25 Qyy = 56.65
Q12 = (Qxx + Qyy – 4 Qss) (0.25) + Qxy (0.5) = 42.31
Q66 = (Qxx + Qyy – 2 Qxy –2Qss) (0.25) + Qss (0.25) = 46.59
Q16 = (Qxx – Qxy – 2 Qss) (0.25) + (Qxy – Qyy +2 Qss) (0.25) = 42.87
Q26 = (Qxx – Qxy – 2 Qss) (0.25) + (Qxy – Qyy +2 Qss) (0.25) = 42.87
The Laminate
A11 = ( ) 2 = GPa-m
A22 = ( ) 2 =
A12 = ( ) 2 =
A66 = ( ) 2 =
A16 = ( ) 2 =
A26 = ( ) 2 =

39. 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]

40. [A] = [M] where h = laminate thickness
The properties of the single “equivalent” orthotropic layer can be determined from the following equation:
[A] = [M] where h = laminate thickness
or A11 = A12 = A66 = E6
A22 = A21 =
Solving last equation, we have the following elastic constants for the single “equivalent” orthotropic layer:
E1 = (1 – ) E2 = (1 – )
E6 = A66 ν1 = ν2 =

41. 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]

42. [a] h = [C] or a11h = a12h = a66h = a21h = a22h = E1 = E2 = E66 =
The properties of the single “equivalent” orthotropic layer is determined by
[a] h = [C]
or a11h = a12h = a66h =
a21h = a22h =
elastic constants for the single “equivalent” orthotropic layer:
E1 = E2 = E66 =
ν1 = – ν2 = –

43. Special Cases:
Cross-Ply Laminates: All layers are either 0° or 90°, which results in A16 = A26 = 0. This is sometimes called specially orthotropic w. r. t. in-plane force and strains. Refer to Figure 1 and Table 1.
h = total laminate thickness

44. Figure 1 In-Plane Modulus and Compliance of T300/5208 Cross-Ply Laminates.

45. Balanced Angle-Ply Laminates: There are only two orientations of the laminae; same magnitude but opposite in sign. With an equal number of plies with positive and negative orientations. Refer to Figure 2 and table 2.
h = total laminate thickness.

46. Figure 2 In-Plane Modulus and Compliance of T300/5208 Angle-Ply Laminates.

47. [0, 90] un-symmetric cross-ply, A16 = A26 = 0
The balanced angle-ply laminate is another case of “special orthotropy” w.r.t. in-plane force and strains; that is A16 = A26 = 0.
The values of Table 2 apply for un-symmetric laminate as well as symmetric laminates. For example, the values for [+30, −30, +30, −30] or [+30, +30, −30, −30] are the same as those for [+30, -30, -30, +30].
If the laminate is not balanced, such as [+30, −30, +30], then the terms A16 and A26 are not zero.
Also remember that A16 = A26 for all cross-plies, whether they are symmetric or un-symmetric.
Example:
[0, 90] un-symmetric cross-ply, A16 = A26 = 0
[0, 90, 0] symmetric cross-ply, A16 = A26 = 0
[+30, −30, −30, +30] symmetric balanced angle-ply, A16 = A26 = 0
[+30, −30, +30, −30] un-symmetric balanced angle-ply, A16 = A26 = 0
[+30, −30, +30] symmetric un-balanced angle-ply, A16, A26 ≠ 0

48. 3) Quasi-Isotropic Laminates:
Has ‘m’ ply groups spaced at ply orientations of 180/m degrees.
Note: This does not apply for m = 1 and m = 2. The laminate need not be symmetric to be quasi-isotropic. For example, the laminate [0, 45, −45, 90] is quasi-isotropic.
Examples:
[0/60/−60]s m = /m = 60°
[0/90/45/−45]s m = /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:
A11 = A22
A16 = A26 = 0
A66 = (A11 – A22) which is equivalent to G = for an isotropic material.

49. Summary: Laminate Analysis