1. No category

Composites Design and Analysis
Stress-Strain Relationship
Prof Zaffar M. Khan
Institute of Space Technology
Islamabad
Next Generation Aerospace Vehicle Requirements
2
Composite design and analysis
Laminate
Theory
Manufacturing
Methods
Materials
Composite
Materials Design /
Analysis Engineer
Design
Guidelines
Design Data
Quality
Assurance
MATERIAL SELECTION
4
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
Composite Analysis
2D
vs.
3D
State of Stress
6
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 = E3
G12 = G13
ν12 = ν13
G23 = __E23___
2(1 + ν 23)
Therefore 5 independent properties:
E1
E2
G12
ν 12
G23
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
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 G12 (Constant Stress)
E2 =
G12
=
Ef2 Em________
Ef2 Vm + Em Vf
__ __
Ef2 Em
Ef2 Vm + Em Vf
__ __
________
Experimental results predicted less accurately
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%?
E = E V +E V
1
-
For Vf = 0.58:
-
For Vf = 0.65:
E1 =
E1 =
f1
f
m
m
(34.0 x 106 psi)(.58) + (0.60 x 106 psi)(0.42) = 20.0 x 106 psi
(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%
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)
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
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)
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
– 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
Stress-Strains Relationships in Lamina
(1)
(2)
The lamina is homogeneous, orthotropic and linear-elastic
The lamina is in a state of plane stress (very thin), therefore, the stresses associated with
the z-direction are negligible: σz = τxz = τyz = 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, εx = 0
x
x
From Figure 2c, Ex= or εx =
x
Ex
x
y
From Figure 2d, νy = − or εx = − νy εy = − νy
y
Ey
Consequently, the total normal strain in the x-direction is
or
x
y
εx = 0 + − νy
Ex
Ey
x
y
εx = − νy
Ex
Ey
The normal strain in the y-direction for the lamina is determined from superposition as follows:
From Figure 2b, εy = 0
y
x
or εy = − νx εx = − νx
x
Ex
y
y
From Figure 2d, Ey=
or εy =
y
Ey
Consequently, the total normal strain in the y-direction is
From Figure 2c, νx = −
 x y
+
E x Ey
 x y
εy = − νx
+
E x Ey
εy = 0 − νx
or
(2)
The shear strain, which is independent of the normal strains, is determined as follows:
s
s
From Figure 2b, Es =
or εs =
s
Es
From Figure 2c, εs = 0
From Figure 2d, εs = 0
Therefore, the shear strain associated with the x-y coordinate system is; εs =
s
Es
(3)
If we express Equations (1), (2) and (3) in matrix form, we have
 1  y

0


x   Ex Ey
x   Sxx Sxy 0  x 

y  =  x 1
0  y  =  Syx Syy 0  y 
   Ex Ey
 
  
s  
1  s   0 0 Sss s 
0
 0

Es 

Equation (4) may also be expressed as follows:
yEx
 Ex

x  1 xy 1  xy
Ey
y  =  xEy
  1 xy 1  xy
s   0
0


[ε] = [C] [σ]
[σ] = [M] [ε]

0

0

Es 


x 
y  =
 
s 
Qxx Qxy 0  x 
Qyx Qyy 0  y 

 
 0
0 Qss s 
(4)
(5)
Elastic Constants-x
P
σx = P/A
εx
εy
lbs
psi
in/in
in/in
0
0
0
0
105
1872
0.000520
− 0.000073
0.140
200
3565
0.001005
− 0.000140
0.139
300
5348
0.001495
− 0.000210
0.140
400
7398
0.002165
− 0.000245
0.136
500
8913
0.002515
− 0.000340
0.135
600
10695
0.003022
− 0.000405
0.134
700
12478
0.003545
− 0.000460
0.130
800
14260
0.004050
− 0.000520
0.128
νx = 0.135 (average)
νx = − εy / εx
Elastic Constants-y
σy vs εy
14000
12000
P
σy = P/A
εx
εy
lbs
psi
in/in
in/in
νy = − εx / εy
10000
8000
0
100
0
1751
0
− 0.000075
0
0.000540
0.139
200
300
400
500
600
700
800
3503
5254
7005
8757
10508
12259
14011
− 0.000145
− 0.000210
− 0.000285
− 0.000350
− 0.000415
− 0.000470
− 0.000540
0.001090
0.001628
0.002218
0.002793
0.003351
0.003943
0.004620
0.133
0.129
0.128
0.125
0.124
0.119
0.117
νy = 0.127 (average)
σy (psi)
Ey = σy / εy = 3.05 x 106 psi
6000
4000
2000
0
0
0.001
0.002
0.003
εy (in/in)
0.004
0.005
Elastic Constants-s
P
σs = P/2A
εμ
εν
εs = ε μ − ε ν
lbs
psi
in/in
in/in
in/in
0
0
0
0
0
100
875
0.001075
− 0.000561
0.001632
200
1748
0.002212
− 0.001229
0.003441
300
2622
0.003527
− 0.002065
0.005592
400
3497
0.005175
− 0.003164
0.008339
500
4371
0.007219
− 0.004615
0.011834
600
5245
0.010547
− 0.007250
0.017797
700
6119
0.013412
− 0.009630
0.023042
800
6993
0.019082
− 0.014710
0.033792
Determination of Elastic Constants
Type
Type
Material
Ex
Ey
GPa
GPa
νx
Es
νf
Specific
gravity
GPa
Material
Ex
Ey
Msi
Msi
νx
Es
t
Msi
inches
Typical
thickness
meters
T300/5208
Graphite/Epoxy
181
1.3
0.28
7.17
0.70
1.6
0.000125
B(4)/5505
Boron/Epoxy
204
18.5
0.23
5.59
0.50
2.0
0.000125
AS/3501
Graphite/Epoxy
138
8.96
0.30
7.1
0.66
1.6
0.000125
Scotchply 1002
Glass/Epoxy
38.6
8.27
0.26
4.14
0.45
1.8
0.000125
Kevlar 49/Epoxy
Aramid/Epoxy
76
5.5
0.34
2.3
0.60
1.46
0.000125
T300/5208
Graphite/Epox
y
26.25
1.49
0.28
1.04
0.005
B(4)/5505
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
0.005
Scotchply 1002
Glass/Epoxy
5.60
1.20
0.26
0.60
0.005
Kevlar 49/Epoxy
Aramid/Epoxy
11.02
0.80
0.34
0.33
0.005
Transformation of Stress and Strain
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.
[σxys] = [Qxys] [εxys]
[σxys] = [T] [ σ126]
[εxys] = [Ť] [ε126]
(1)
(2)
(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]
or
[ σ126] = [T] −1 [Qxys] [Ť] [ε126]
or
= [T] −1 [Qxys] [Ť]
(5)
(6)
(7)
or
or
= [Q126]
[ 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
[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 Q66 + 2 Q12Q26 Q61 − Q22Q162 – Q66Q122 – Q11Q622
Summary: Lamina Analysis
The Symmetric Laminate with In-Plane Loads
A.
Stress Resultants and Strains: The “A”
Relationship between strains and the stress resultants. Related by
constants Aij
Deriving expressions for these constants:
Expression for
is:
Expression for N1:
N1 = (σ1)1 (h1-h2) + (σ1)2 (h2-h3) + (σ1)3 (h3-h4) + …
Substitute the expressions for stress:
Above equation can be rewritten as:
Similarly for N2 and N6:
Above can be combined into matrix form as:
Or
Where
A11 =
A12 =
A16 =
A22 =
A26 =
A66 =
(Q11)k tk
(Q12)k tk
(Q16)k tk
(Q22)k tk
(Q26)k tk
(Q66)k tk
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” = 0.000125 m
0° Lamina
Q11 = Qxx = 181.8
Q22 = Qyy = 10.34
Q12 = Qxy = 2.897
Q66 = Qss = 7.17
Q16 = 0
Q26 = 0
90° Lamina
Q11 = Qyy = 10.34
Q22 = Qxx = 181.8
Q12 = Qxy = 2.897
Q66 = Qss = 7.17
Q16 = 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) + 0.25 Qyy = 56.65
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 = 0.000125 (181.8 + 10.34 + 56.25) 2 = 0.0622
A22 = 0.000125 (10.34 + 181.8 + 56.65) 2 = 0.0622
A12 = 0.000125 (2.897 + 2.897 + 42.31) 2 = 0.0120
A66 = 0.000125 (7.17 + 7.17 + 46.59) 2 = 0.0152
A16 = 0.000125 (0 + 0 + 42.87) 2 = 0.0107
A26 = 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:
[A] = [M]
E1
1 A =
11
or
1  1 2
h
1
h
where h = laminate thickness
1 A =  2 E1
1 A =E
12
66
6
A22 =
1  1 2
h
E2
1  1 2
1
h
A21 =
h
 1E 2
1  1 2
Solving last equation, we have the following elastic constants for the
single “equivalent” orthotropic layer:
A12
A12
A11
E1 = (1 –
)
E2 = A22 (1 –
)
2
A11 A22
h
E6 =
1
h
2
A66
A11 A22
h
ν1 =
A12
A22
ν2 =
A12
A11
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]
The properties of the single “equivalent” orthotropic layer is determined by
[a] h = [C]
1
a11h = E 1
or
a21h =
 1
E1
 2
a12h = E 2
a66h =
1
E6
1
E2
a22h =
elastic constants for the single “equivalent” orthotropic layer:
E1 =
1
a11h
ν1 = –
a12
a11
E2 =
1
a 22h
ν2 = –
a12
a 22
E66 =
1
a 66h
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
Figure 1 In-Plane Modulus and Compliance of T300/5208 CrossPly 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 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 unsymmetric.
Example:
[0, 90]
[0, 90, 0]
[+30, −30, −30, +30]
[+30, −30, +30, −30]
[+30, −30, +30]
un-symmetric cross-ply, A16 = A26 = 0
symmetric cross-ply, A16 = A26 = 0
symmetric balanced angle-ply, A16 = A26 = 0
un-symmetric balanced angle-ply, A16 = A26 = 0
symmetric un-balanced angle-ply, A16, A26 ≠ 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:
A11 = A22
A16 = A26 = 0
1
A66 = (A11 – A22) which is equivalent to G = E for an isotropic material.
2 1 
Summary: Laminate Analysis
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.
N1 
Answer:  N 2  =
 
 N 6 
0.402  108

8
0.222  10

0

0.222  108
0.402  10 8
0

0

0

8 
0.254  10 
 1 
 2 
 
 6 
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