2
Next Generation Aerospace Vehicle Requirements

Laminate
Theory
Manufacturing
Methods
Materials
Composite
Materials Design /
Analysis Engineer
Design
Guidelines
Design Data
Quality
Assurance

4
MATERIAL SELECTION

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 mid-plane, with both in-plane 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

2D
3D
6

• Metals (isotropic materials)
– E, G, ν
– 2 independent properties – G = __E____
2(1 + ν)
• Composite lamina (unidirectional layer, ply)
–
–
In plane: E
1
, E
2
, G
Out of plane : E
3
12
, G
, ν
13
12
, G
23
, ν
13
, ν
• But for transverse isotropy (2 = 3):
23
E
2
= E
3
G
12
= G
13
ν
12
= ν
13
G
23
= __E
23
___
2(1 + ν
23
)
Therefore 5 independent properties:
E
1
E
2
G
12
ν
12
G
23

•
E f
, E m
•
G f
, G m
- Young’s modulus of fiber and matrix
- Shear modulus of fiber and matrix
• ν f
, ν m
•
V f
, V m
- Poisson’s ratio of fiber and matrix
•
W f
, W m
- Volume fraction of fiber and matrix
– Weight fraction of fiber and matrix

• Parallel model - E
1 and ν
12
(Constant Strains)
E
1
= E f1
V f
+ E m
V m
ν
12
=
f
V f
+ ν m
V m
Predictions agree well with experimental data
• Series model – E
2 and G
12
(Constant Stress)
E
2
=
__ __
E f2
E m
________
E f2
V m
+ E m
V f
G
12
=
__
__
E f2
E m ________
E f2
V m
+
E m
V f
Experimental results predicted less accurately

Knowns : Carbon: E = 34.0 x 10 6 psi
Epoxy : E = 0.60 x 10 6 psi
• How much does the longitudinal modulus change when the fiber volume fraction is changed from 58% to 65%?
E
1
= E f1
V f
+ E m
V m
For
V f
= 0.58:
E
1
= (34.0 x 10 6 psi)(.58) + (0.60 x 10 6 psi)(0.42) = 20.0 x 10 6 psi
For
V f
= 0.65:
E
1
= (34.0 x 10 6 psi)(.65) + (0.60 x 10 6 psi)(0.35) = 22.3 x 10 6 psi
Thus, raising the fiber volume fraction from 58% to 65% increases E 1 by
12%

• Used to determine the response of a composite laminate based on properties of a layer (or ply)

• 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

• 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 0 o or 90 o (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)

• Balanced, unsymmetric, laminate
– Tensile loading produces twisting curvature
– Ex: [+θ/0/- θ]τ
• Symmetric, unbalanced laminate
– Tensile loading produces in-plane shearing
– Ex: [+θ/0/- θ]τ
• 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

(1) The lamina is homogeneous, orthotropic and linear-elastic
(2) The lamina is in a state of plane stress (very thin), therefore, the stresses associated with the z-direction are negligible: σ z
= τ xz
= τ yz
= 0

The normal strain in the x-direction for the lamina in Figure 2a is the sum of the normal strains in the x-direction for the laminae in Figures 2b, 2c and 2d:
From Figure 2b, ε x
= 0
 x
From Figure 2c, E x
=
 x
From Figure 2d, ν y
= −

 x y
or ε x
= or ε

E x x x
= − ν y
ε y
= − ν y
 y
E y
Consequently, the total normal strain in the x-direction is or
ε
ε x
= 0 +
 x
E x x
=
 x
E x
− ν y
− ν y

E y y
 y
E y

The normal strain in the y-direction for the lamina is determined from superposition as follows:
From Figure 2b, ε y
= 0
From Figure 2c, ν x
= −
 y
 x
 y
From Figure 2d, E y
=
 y or ε y
= − ν x
ε x
= − ν x
or ε y
=

E y y
 x
E x
Consequently, the total normal strain in the y-direction is
ε y
= 0 − ν x or
ε y
= − ν x

E x x
 x
E x
+
+
 y
E y
E y
 y
(2)
The shear strain, which is independent of the normal strains, is determined as follows:
From Figure 2b, E s
=

 s s
or ε s
=

E s s
From Figure 2c, ε s
= 0
From Figure 2d, ε s
= 0
 s
Therefore, the shear strain associated with the x-y coordinate system is; ε s
=
E s
(3)

If we express Equations (1), (2) and (3) in matrix form, we have







 s x y 




=








1

E
 x x
E x
0
  y
E y
1
E y
0
0
0
1
E s















 s x y 




= 




S xx
S yx
0
S xy
S yy
0 S
0
0 ss












 s x y 




Equation (4) may also be expressed as follows:







 s x y 




=





1



1


 x
 x
E y y

E x
 x

0 y
 y
E x
1
  x
 y
E y
1
  x
 y
0
0
0
E s















 s x y 




= 




Q xx
Q yx
0
Q xy
Q yy
0
0
0
Q ss












 s x y 




[ε] = [C] [σ]
[σ] = [M] [ε]
(4)
(5)

P lbs
σ x
= P/A psi
ε x in/in
ε y in/in
ν x
= − ε y
/ ε x
0
105
0
1872
0 0
0.000520
− 0.000073
200
300
400
500
3565
5348
7398
8913
0.001005
0.001495
0.002165
0.002515
− 0.000140
− 0.000210
− 0.000245
− 0.000340
600 10695 0.003022
− 0.000405
700 12478 0.003545
− 0.000460
800 14260 0.004050
− 0.000520
ν x
= 0.135 (average)
0.140
0.139
0.140
0.136
0.135
0.134
0.130
0.128

P lbs
σ y
= P/A psi
0
100
0
1751
ε
0 x in/in
− 0.000075
ε y in/in
0
0.000540
ν y
= − ε x
/ ε y
0.139
200
300
400
500
3503 − 0.000145
5254 − 0.000210
7005 − 0.000285
8757 − 0.000350
600 10508 − 0.000415
700 12259 − 0.000470
800 14011 − 0.000540
0.001090
0.001628
0.002218
0.002793
0.003351
0.003943
0.004620
ν y
= 0.127 (average)
0.133
0.129
0.128
0.125
0.124
0.119
0.117
σ y
vs ε y
σ y
(psi)
8000
6000
4000
2000
0
0
14000
12000
10000
E y
= σ y
/ ε y
= 3.05 x 10
6
psi
0.001
0.002
ε y
(in/in)
0.003
0.004
0.005

P σ lbs s
= P/2A psi
ε
μ in/in
ε
ν in/in
400
500
600
700
800
0
100
200
300
0
875
1748
2622
3497
4371
5245
6119
6993
0
0.001075
0.002212
0.003527
0.005175
0.007219
0.010547
0.013412
0.019082
0
− 0.000561
− 0.001229
− 0.002065
− 0.003164
− 0.004615
− 0.007250
− 0.009630
− 0.014710
ε s
= ε
μ
− ε
ν in/in
0
0.001632
0.003441
0.005592
0.008339
0.011834
0.017797
0.023042
0.033792

Type Material E x
E y
ν x
E s t
Type
T300/5208
Material E x
E y
ν x
E s
ν f
Specific gravity
Typical thickness
Graphite/Epoxy
GPa
181
GPa
1.3
0.28
GPa
7.17
0.70
1.6
meters
0.000125
B(4)/5505
AS/3501
Boron/Epoxy
Graphite/Epoxy
204
138
18.5
8.96
0.23
0.30
5.59
7.1
0.50
0.66
2.0
1.6
0.000125
0.000125
Scotchply 1002
Kevlar 49/Epoxy
Glass/Epoxy
Aramid/Epoxy
38.6
76
8.27
5.5
0.26
0.34
4.14
2.3
0.45
0.60
1.8
1.46
0.000125
0.000125
T300/5208
B(4)/5505
Graphite/Epox y
Msi Msi Msi
26.25
1.49
0.28
1.04
inches
0.005
Boron/Epoxy 29.59
2.68
0.23
0.81
0.005
AS/3501 Graphite/Epox y
20.01
1.30
0.30
1.03
Scotchply 1002 Glass/Epoxy 5.60
1.20
0.26
0.60
0.005
0.005
Kevlar 49/Epoxy Aramid/Epoxy 11.02
0.80
0.34
0.33
0.005

Area
Stresses
Stresses
Forces

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

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

Stress-Strain Relationships in Global Coordinates
Develop the relationship between the stresses and strains in global coordinates.
[σ
[σ xys xys
] = [Q xys
] [ε
] = [T] [ σ
[ε xys
] = [Ť] [ε
126
] xys
126
]
]
Substitution of Equations (2) and (3) into (1) yields
[T] [ σ
126
] = [Q xys
] [Ť] [ε
126
]
(3)
(1)
(2)
(4)
Pre multiplying both sides of Equation (4) by [T] −1 :
[T] −1 [T] [σ
126
] = [T] −1 [Q or [ σ
126 xys
] [Ť] [ε
126
] = [T] −1 [Q xys
] [Ť] [ε
126
]
] (5)
(6) or = [T] −1 [Q xys
] [Ť] (7)

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
22
Q
12
Q
66
Q
16
Q
26
= Q
= Q
= Q
= Q
= Q xx xx
21
= (Q xx
61
62 cos 4 θ + 2 (Q xy sin 4 θ + 2 (Q xy
= (Q xx
+ Q yy
+ 2 Q ss
) sin 2 θ cos 2 θ + Q
+ 2 Q ss
− 4Q ss
) sin 2 θ cos 2 θ + Q yy sin 4 θ
) sin 2 θ cos 2 θ + Q yy cos 4 θ
(sin 4 θ +cos 4 θ)
+ Q
= (Q
= (Q yy xx xx
− 2 Q xy
– Q xy
− 2Q ss xy
) sin 2 θ cos 2 θ + Q ss
− 2Q ss
) sinθ cos 3 θ + (Q xy
(sin
– Q yy
4 θ +cos 4 θ)
– Q xy
− 2Q ss
) sin 3 θ cosθ + (Q xy
– Q yy
+ 2Q ss
+ 2Q ss
) sin 3 θ co
) sinθcos 3 θ

[Q
126 or
] −1 [ σ
[Q
126
] = [Q
126
] −1 [ σ or
126
] −1 [Q
126
] [ε
126
]
126
] = [ε
126
]
[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
26
2 ) / ∆
S
12
S
S
16
S
22
26
S
66
= S
= S
21
61
= (Q
= S
62
= (Q
11
11
= (Q
= (Q
12
Q
= (Q
12
Q
66
22
16 where ∆ = Q
11
Q
Q
26
Q
– Q
26
16
2
Q
– Q
16
12
2
22
Q
– Q
66
12
) / ∆
) / ∆
Q
66
– Q
22
Q
16
– Q
11
Q
26
+ 2 Q
) / ∆
) / ∆
) / ∆
12
Q
26
Q
61
− Q
22
Q
16
2 – Q
66
Q
12
2 – Q
11
Q
62
2

A.
Stress Resultants and Strains: The “A”

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

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:

Above equation can be rewritten as:
Similarly for N
2 and N
6:

Above can be combined into matrix form as:
Or
Where
A
11
A
12
A
16
A
22
A
26
A
66
= (Q
11
) k
= (Q
12
) k
= (Q
16
) k
= (Q
22
) k
= (Q
26
) k
= (Q
66
) k t k t k t k t k t k t k

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” = 0.000125 m
.
0° Lamina
Q
11
Q
22
Q
12
Q
66
Q
16
Q
26
= 0
= Q xx
= Q yy
= Q xy
= Q ss
= 0
= 181.8
= 10.34
= 2.897
= 7.17
Q
11
Q
22
Q
12
Q
66
90° Lamina
= Q yy
= Q xx
= Q xy
= Q ss
= 10.34
= 181.8
= 2.897
= 7.17
Q
16
Q
26
= 0
= 0
Q
11
Q
22
Q
12
Q
66
Q
16
Q
26
Q
11
= Q xx
= Q xx
= Q
21
= (Q
= Q
= Q xx
61
62 cos 4 θ + 2 (Q xy sin 4 θ + 2 (Q xy
= (Q
+ Q
= (Q
= (Q xx yy xx xx
+ Q yy
– 2Q xy
– Q
– Q xy xy
+ 2 Q
– 4 Q ss ss
) sin
) sin 2 θ cos 2 θ + Q xy
– 2 Q ss
) sin 2
2 θ cos 2 θ + Q
+ 2 Q ss
) sin 2 θ cos 2 θ + Q yy cos 4 θ
(sin 4 θ + cos 4 θ)
θ cos 2 θ + Q ss
– 2 Q ss
) sinθ cos 3 θ + (Q xy
– 2 Q ss
) sin 3 θ cosθ + (Q xy yy sin
(sin
– Q yy
– Q yy
4
4 θ
θ + cos 4 θ)
+ 2 Q ss
) sin 3 θ cosθ
+ 2 Q ss
) sinθ cos 3 θ
= 0.25 Q xx
+ 2 (Q xy
+ 2 Q ss
) (0.25) + 0.25 Q yy
= 56.65

Q
22
= 0.25 Q xx
+ 2 (Q xy
+ 2 Q ss
) (0.25) + 0.25 Q yy
= 56.65
Q
12
= (Q xx
+ Q yy
– 4 Q ss
) (0.25) + Q xy
(0.5) = 42.31
Q
66
= (Q xx
+ Q yy
– 2 Q xy
–2Q ss
) (0.25) + Q ss
(0.25) = 46.59
Q
16
= (Q xx
– Q xy
– 2 Q ss
) (0.25) + (Q xy
– Q yy
+2 Q ss
) (0.25) = 42.87
Q
26
= (Q xx
– Q xy
– 2 Q ss
) (0.25) + (Q xy
– Q yy
+2 Q ss
) (0.25) = 42.87
The Laminate
A
11
= 0.000125 (181.8 + 10.34 + 56.25) 2 = 0.0622
A
22
= 0.000125 (10.34 + 181.8 + 56.65) 2 = 0.0622
A
12
= 0.000125 (2.897 + 2.897 + 42.31) 2 = 0.0120
A
66
= 0.000125 (7.17 + 7.17 + 46.59) 2 = 0.0152
A
16
= 0.000125 (0 + 0 + 42.87) 2 = 0.0107
A
26
= 0.000125 (0 + 0 + 42.87) 2 = 0.0107
GPa-m

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]

The properties of the single “equivalent” orthotropic layer can be determined from the following equation: or 1 h
[A] = [M]
A
11
= E
1
1
 
1

2
1 h
A where h = laminate thickness
12
=

2
E
1
1
 
1

2
1 h
A
66
= E
6
1 h
A
22
= E
2
1
 
1

2
1 A
21
= h 1

1
E
2
 
1

2
Solving last equation, we have the following elastic constants for the single “equivalent” orthotropic layer:
E
1
A
11
= (1 – h
A
12
2
A
11
A
22
) E
2
=
A
22 h
(1 –
A
12
2
A
11
A
22
)
E
6
1
= A
66 h
ν
1
=
A
12
A
22
ν
2
=
A
12
A
11

The equation for the composite laminate is or where
[ε] = [a] [N]
[a] = [A] -1
The equation for a single orthotropic lamina is or [ε] = [C] [σ]

The properties of the single “equivalent” orthotropic layer is determined by
[a] h = [C]
1 or a
11 h =
E
1 a
21 h =
 
1
E
1 a
12 h =
 
E
2
2
1 a
22 h =
E
2
1 a
66 h =
E
6 elastic constants for the single “equivalent” orthotropic layer:
E
1
= 1 a
11 h
E
2
= 1 a
22 h
E
66
= 1 a
66 h
ν
1
= – a
12 a
11
ν
2
= – a
12 a
22

•
• Cross-Ply Laminates: All layers are either 0° or 90°, which results in A
16 and Table 1.
= A
26
= 0. This is sometimes called specially orthotropic w. r. t. in-plane force and strains. Refer to Figure 1
• h = total laminate thickness

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

• 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.

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

The balanced angle-ply laminate is another case of “special orthotropy” w.r.t. inplane force and strains; that is A
16
= A
26
= 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 A
16 are not zero.
and A
26
Also remember that A
16 symmetric.
= A
26 for all cross-plies, whether they are symmetric or un-
Example:
[0, 90]
[0, 90, 0]
[+30, −30, −30, +30]
[+30, −30, +30, −30]
[+30, −30, +30] un-symmetric cross-ply, A symmetric cross-ply, A
16
16
= A
= A
26
26
= 0
= 0 symmetric balanced angle-ply, A symmetric un-balanced angle-ply, A
16
, A
26
16
= A
26 un-symmetric balanced angle-ply, A
16
≠ 0
= 0
= A
26
= 0

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
[0/90/45/−45]s
[0/60/−60/60/0/−60]s m = 3 m = 4 m=3
180/m = 60°
180/m = 45°
180/m = 60°
The modulus of the quasi-isotropic laminate has the following properties:
A
11
A
16
A
66
= A
22
= A
26
= (A
11
= 0
– A
22
1
) which is equivalent to G =
2 1
E
  for an isotropic material.

1.
Determine the modulus for the laminate [0/+45/−45/90] s
where the 0° and 90° layers are
Scotch-ply 1002, and the +45° and −45° layers are T300/5208. The thickness of each layer is 0.00125m. Obtain the modulus by (1) hand calculations, and (2) computer program.
Answer: 



N
N
N
6
1
2




= 



0
0
.
402
.
222
0


10

8
10

8
0
0
.
222
.
402
0

10

10

8

8
0 .
254
0
0

10

8











1
6
2 



2.
A stress analysis of a structure results in the stresses shown at a particular point in a composite laminate. The laminate is the same as in problem 5. a.
Determine the strains in global coordinates.
Answers: 0.00358, -0.00198, 0.00158 b.
Determine the stresses in the 0° lamina in local coordinates. c.
Determine the stresses in the 45° lamina in local coordinates.
Answers: 2.887 MPa, 4.68 MPa, -39.84 MPa