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ME 5720
Mechanics of Composite Materials
Project I
Composite Laminate Design for a Wind Turbine Spar
Poberezhnyy, Mikhail
Miller, Trevor
Latifi, Omar
Aoude, Mohamad
12/10/2015
Department of Mechanical Engineering
Wayne State University
2
1. Introduction
This project was given to analyze the microstructure of a laminate in a wind turbine blade focusing on
the spar. A spar is the main frame of the blade that holds the structure together (see Figure 1 below).
The analysis consist of developing a hybrid laminate with carbon and glass fiber plies at selected fiber
angle orientation that are capable of withstanding wind-loading values found in our research.
Concentrating on high axial and out-of-plane torsional loads, we are able to estimate the mechanical
stresses in each ply. Finally we determined if the proposed hyprid laminate would fail by using Tsi-Hill
Criterion .
A MATLAB sofware code was developed to conduct accurate calculations for the composite laminate
design. The code was developed in such a way to be able to handle any number of plies and fiber
orientation.
Figure 1: Wind turbine blade cross-section.
2. Wind Loading
In order to analyze the stresses present within the laminate, the external loads exerted on the blade
must be determined. In this analysis it is assumed that only the present are the longitudinal axial load
(NX) and the out-of-plane torsional load (MXY).
The axial load can be found from an IEEE Xplore article regarding load analysis of wind turbines [2], table
no. II: For 7 no of plies, the normal load is given as 1958 N/mm. In order to convert into GPa we must
divide by 1000. Therefore:
𝑁 𝑋 = 1.958 𝐺𝑃𝑎 − 𝑚𝑚
Similarly Mxy (Edgewise Bending moment) is found using a report regarding Power and loads of wind
turbines [3], graph (c) on page 21 Fig 2.8, an edgewise bending moment of 1950 N-m can be retrieved
and since we have about 60% fiber glass in our blade, we get the length of the spar box as 912 mm [4]
from the design considerations on wind turbines report. Therefore, after dividing by the spar box
length and 1000 to convert to GPa:
𝑀 𝑋𝑌 = 2.138 𝐺𝑃𝑎 − 𝑚𝑚
3
3. Laminate Composition
A laminate consisting of fiberglass and carbon layers (Figure 2), is used to determine the stresses in the
spar box from the wind loads. The calculations are performed at a section where there is a 7 ply stack.
Figure 2: Example candidate fiberglass-to-carbon spar transition.
Throughout the development of the code and the first set of calculations, it was important to set a
“benchmark” for several steps in the process. Using AS/3501 Carbon/Epoxy and Scotch-ply 1002 E-
Glass Epoxy (see Table 1 for material properties), these benchmark calculations were used to verify
that the code was producing correct results. The ply orientations are [0/45/0/45/0/45/0].
Note that all equations used and also the material properties of from Principles of Composite Material
Mechanics by Gibson[1].
The inputs used in the code for the material properties and ply orientations can be found in Section 8.
Table 1: Elastic Material Properties
Material 𝑬 𝟏 [GPa] 𝑬 𝟐 [GPa] 𝑮 𝟏𝟐 [GPa] 𝝂 𝟏𝟐
AS/3501 Carbon / Epoxy 138 9.0 6.9 0.3
Scotch-ply 1002 E-Glass Epoxy 38.6 8.27 4.14 0.26
Also needed to define the laminate, are the locations of each ply, zk, as shown in Figure 3. These values
can be found from the thickness and number of plies.
Figure 3: Laminated plate geometry and ply numbering system.
4
4. Stress Calculations
The first step is to calculate the stiffness matrices [Q] for the orthotropic lamina of each material using
equation (2.26) and material properties:
[ 𝑄] = [
𝑄11 𝑄12 0
𝑄21 𝑄22 0
0 0 𝑄66
]
Where 𝑄𝑖𝑗 are components of a lamina stiffness matrix that can be found by using material properties
and equation (2.27):
𝑄11 =
𝐸1
1 − 𝜈12 𝜈21
𝑄22 =
𝐸2
1 − 𝜈12 𝜈21
𝑄12 = 𝑄21 =
𝜈12 𝐸2
1 − 𝜈12 𝜈21
𝑄66 = 𝐺12
The resulting stiffness matrices for each ply from the code:
Q(:,:,1) =
39.1673 2.1818 0
2.1818 8.3915 0
0 0 4.1400
Q(:,:,2) =
45.2250 31.4250 32.4404
31.4250 45.2250 32.4404
32.4404 32.4404 35.6090
Q(:,:,3) =
138.8148 2.7159 0
2.7159 9.0531 0
0 0 6.9000
Q(:,:,4) =
45.2250 31.4250 32.4404
31.4250 45.2250 32.4404
32.4404 32.4404 35.6090
Q(:,:,5) =
39.1673 2.1818 0
5
2.1818 8.3915 0
0 0 4.1400
Q(:,:,6) =
17.1206 8.8406 7.6939
8.8406 17.1206 7.6939
7.6939 7.6939 10.7988
Q(:,:,7) =
39.1673 2.1818 0
2.1818 8.3915 0
0 0 4.1400.
Those stiffness matrices are then transformed to the ply angles specified. For the generally orthotropic
lamina stiffness matrix in an off axis with a specific lamina orientation angle from equation (2.35):
[ 𝑄̅] = [
𝑄̅11 𝑄̅12 𝑄̅16
𝑄̅12 𝑄̅22 𝑄̅26
𝑄̅16 𝑄̅26 𝑄̅66
]
Where the 𝑄̅𝑖𝑗 are the components of the transformed lamina stiffness matrix which are defined as
follows from equation (2.36):
𝑄̅11 = 𝑄11 𝑐4
+ 𝑄22 𝑠4
+ 2( 𝑄12 + 2𝑄66 ) 𝑠2
𝑐2
𝑄̅12 = ( 𝑄11 + 𝑄22 − 4𝑄66 ) 𝑠2
𝑐2
+ 𝑄12( 𝑐4
+ 𝑠4)
𝑄̅22 = …
Now that the generally orthotropic lamina stiffness matrices are calculated, the extensional, coupling,
and bending matrices can be found using equations (7.41), (7.42), and (7.43), respectively.
𝐴𝑖𝑗 = ∑(𝑄̅𝑖𝑗)
𝑘
( 𝑧 𝑘 − 𝑧 𝑘−1)
𝑁
𝑘=1
𝐵𝑖𝑗 =
1
2
∑(𝑄̅𝑖𝑗)
𝑘
( 𝑧 𝑘
2
− 𝑧2
𝑘−1)
𝑁
𝑘=1
𝐷𝑖𝑗 =
1
3
∑(𝑄̅𝑖𝑗 )
𝑘
( 𝑧 𝑘
3
− 𝑧3
𝑘−1)
𝑁
𝑘=1
The resulting A, B, and D matrices calculated in the code are as follows:
A = 1.0e+003 *
1.0917 0.2429 0.2177
0.2429 0.4254 0.2177
0.2177 0.2177 0.304
B = 1.0e+003 *
-1.4027 -0.4113 -0.4454
-0.4113 -0.5118 -0.4454
-0.4454 -0.4454 -0.4714
D = 1.0e+004 *
3.1393 0.5723 0.4498
0.5723 1.1602 0.4498
0.4498 0.4498 0.755
The inverse form of the laminate force deformation equation (7.51) below can be used to determine
the mid-plane strains (𝜀°
) and curvature (κ) from the exerted wind loads.
{ 𝜀°
κ
} = [
𝐴 𝐵
𝐵 𝐷
]
−1
{
𝑁
𝑀
}
The loads calculated from Section 2 can be entered in a dialog as shown below in Figure 2.
Figure 2: Wind Load Input Dialog
The resulting strain – curvature matrix [𝜀 𝑥
°
, 𝜀 𝑦
°
𝜀 𝑥𝑦
°
κ 𝑥,κ 𝑦 , κ 𝑥𝑦] resulted as:
0.0022
-0.0007
-0.0007
0.0001
-0.0002
0.0004
7
From the mid-plane strain, curvature, and stiffness matrices, the lamina stresses (results in Table 2) at
the top and bottom of each ply can be determined using the lamina stress-strain relationship from
equation (7.34):
{
𝜎𝑥
𝜎𝑦
𝜏 𝑥𝑦
} = [
𝑄̅11 𝑄̅12 𝑄̅16
𝑄̅12 𝑄̅22 𝑄̅26
𝑄̅16 𝑄̅26 𝑄̅66
] {
𝜀 𝑥
°
+ zΚ 𝑥
𝜀 𝑦
°
+ zΚ 𝑦
𝛾𝑥𝑦
°
+ zΚ 𝑥𝑦
}
Table 2: Stresses at top and bottom of each ply.
Location 𝝈 𝒙(GPa) 𝝈 𝒚(GPa) 𝝉 𝒙𝒚(GPa)
#1 Top 0.0679 0.0118 -0.02
#1 Bottom 0.0733 0.0082 -0.0151
#2 Top -0.0188 -0.0373 -0.0534
#2 Bottom 0.0124 -0.0148 -0.021
#3 Top 0.2783 0.0058 -0.0169
#3 Bottom 0.2994 0.002 -0.0087
#4 Top 0.0435 0.0077 0.0114
#4 Bottom 0.0591 0.019 0.0276
#5 Top 0.0865 -0.0006 -0.0028
#5 Bottom 0.0892 -0.0024 -0.0003
#6 Top 0.0313 0.0047 0.0102
#6 Bottom 0.0391 0.0073 0.0207
#7 Top 0.0945 -0.006 0.0046
#7 Bottom 0.0998 -0.0095 0.0095
5. Tsai-Hill Failure Criteria
After the stresses are calculated for each ply, it is important to determine if any of the plies will fail.
Tsai-Hill criteria will be used to analyze failure within the laminate. In the Tsai-Hill criteria, equation
(4.14), if the left hand side is greater than 1, then the ply will fail. The strength values used for the
failure calculations are displayed in Table 3.
𝜎1
2
𝑆 𝐿
2
−
𝜎1 𝜎2
𝑆 𝐿
2
+
𝜎2
2
𝑆 𝑇
2
+
𝜏12
2
𝑆 𝐿𝑇
2
= 1
It can be seen that the above equation uses the material coordinate stresses (𝜎1, 𝜎1, and 𝜏12). To get
stresses in material coordinates equation (2.31) is used to transform the stresses:
{
𝜎1
𝜎2
𝜏12
} = [
𝑐2
𝑠2
2𝑐𝑠
𝑠2
𝑐2
−2𝑐𝑠
−𝑐𝑠 𝑐𝑠 𝑐2
− 𝑠2
] {
𝜎𝑥
𝜎𝑦
𝜏 𝑥𝑦
}
8
Table 3: Typical values of lamina strengths for the chosen composites table (4.1)
Material 𝑺 𝑳
+
[MPa] 𝑺 𝑳
−
[MPa] 𝑺 𝑻
+
[MPa] 𝑺 𝑻
−
[MPa] 𝑺 𝑳𝑻(𝑴𝑷𝒂)
AS/3501 Carbon / Epoxy 1448 1172 48.3 248 62.1
Scotchply 1002 E-Glass
Epoxy
1103 621 27.6 138 82.7
The code results for the Tsai-Hill Failure Criteria can be seen below. Since none of the values are
greater than 1, it can be concluded that the proposed laminate structure will not fail under the
expected wind loads.
Location
#1 Top 0.2439
#1 Bottom 0.1262
#2 Top 0.3066
#2 Bottom 0.2183
#3 Top 0.1251
#3 Bottom 0.0641
#4 Top 0.1716
#4 Bottom 0.1634
#5 Top 0.0074
#5 Bottom 0.0072
#6 Top 0.1071
#6 Bottom 0.0471
#7 Top 0.0131
#7 Bottom 0.0275
6. Conclusion
The results of the code show that the originally proposed hybrid laminate is a solution for the spar box
design. To summarize the composite will consist of:
 Fiberglass Material: Scotch-ply 1002 E-Glass Epoxy
 Carbon Layers: AS/3501 Carbon/Epoxy
 Ply Orientation: [0/45/0/45/0/45/0]
Based on “benchmark” calculations done outside of the coding, each step such as the compliance
matrices, A, B, and D stiffness matrices, as well as the stresses as strains are correct for this proposed
laminate structure. The code was developed to be able to calculate failure for any number of plies as
long as the material data and ply orientation of each ply is designated in the input.
Although the first proposed laminate passed the Tsai-Hill failure criteria, it doesn’t mean that it is the
only solution. Other ply orientations and material combinations result in a laminate that passes the
failure criteria. In order to distinguish between the laminates that passed, analysis on the cost, mass,
safety factor, etc. will need to be completed as well.
9
7. References
[1] Gibson, Ronald F. Principals of Composite Material Mechanics. 3rd ed. Boca Raton: CRC, 2012. Print.
[2] "Loading Analysis and Strength Calculation of Wind Turbine Blade Based on Blade Element
Momentum Theory and Finite Element Method." IEEE Xplore. N.p., n.d. Web. 06 Dec. 2015.
[3] Boorsma, K., November 2012, and Ecn-E--12-04. Power and Loads for Wind Turbines in Yawed
Conditions (n.d.): n. pag. Web.
[4] AIAA-2003-0696 ALTERNATIVE COMPOSITE MATERIALS FOR MEGAWATT-SCALE WIND TURBINE
BLADES: DESIGN CONSIDERATIONS AND RECOMMENDED TESTING (n.d.): n. pag. Web.
8. Appendix A: MATLAB Code Inputs for Ply Material and Orientations
>>
TsaiHill(38.6,8.27,4.14,.26,3,0,1.103,.621,.0276,.138,.0827,138,9,6.9,0.3,3,45,1.448,1.172,.048,.248,.06
2,138,9,6.9,0.3,3,0,1.448,1.172,.048,.248,.062,138,9,6.9,0.3,3,45,1.448,1.172,.048,.248,.062,38.6,8.27,
4.14,.26,3,0,1.103,.621,.0276,.138,.0827,38.6,8.27,4.14,.26,3,45,1.103,.621,.0276,.138,.0827,38.6,8.27,
4.14,.26,3,0,1.103,.621,.0276,.138,.0827)

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ME 5720 Fall 2015 - Wind Turbine Project_FINAL

  • 1. ME 5720 Mechanics of Composite Materials Project I Composite Laminate Design for a Wind Turbine Spar Poberezhnyy, Mikhail Miller, Trevor Latifi, Omar Aoude, Mohamad 12/10/2015 Department of Mechanical Engineering Wayne State University
  • 2. 2 1. Introduction This project was given to analyze the microstructure of a laminate in a wind turbine blade focusing on the spar. A spar is the main frame of the blade that holds the structure together (see Figure 1 below). The analysis consist of developing a hybrid laminate with carbon and glass fiber plies at selected fiber angle orientation that are capable of withstanding wind-loading values found in our research. Concentrating on high axial and out-of-plane torsional loads, we are able to estimate the mechanical stresses in each ply. Finally we determined if the proposed hyprid laminate would fail by using Tsi-Hill Criterion . A MATLAB sofware code was developed to conduct accurate calculations for the composite laminate design. The code was developed in such a way to be able to handle any number of plies and fiber orientation. Figure 1: Wind turbine blade cross-section. 2. Wind Loading In order to analyze the stresses present within the laminate, the external loads exerted on the blade must be determined. In this analysis it is assumed that only the present are the longitudinal axial load (NX) and the out-of-plane torsional load (MXY). The axial load can be found from an IEEE Xplore article regarding load analysis of wind turbines [2], table no. II: For 7 no of plies, the normal load is given as 1958 N/mm. In order to convert into GPa we must divide by 1000. Therefore: 𝑁 𝑋 = 1.958 𝐺𝑃𝑎 − 𝑚𝑚 Similarly Mxy (Edgewise Bending moment) is found using a report regarding Power and loads of wind turbines [3], graph (c) on page 21 Fig 2.8, an edgewise bending moment of 1950 N-m can be retrieved and since we have about 60% fiber glass in our blade, we get the length of the spar box as 912 mm [4] from the design considerations on wind turbines report. Therefore, after dividing by the spar box length and 1000 to convert to GPa: 𝑀 𝑋𝑌 = 2.138 𝐺𝑃𝑎 − 𝑚𝑚
  • 3. 3 3. Laminate Composition A laminate consisting of fiberglass and carbon layers (Figure 2), is used to determine the stresses in the spar box from the wind loads. The calculations are performed at a section where there is a 7 ply stack. Figure 2: Example candidate fiberglass-to-carbon spar transition. Throughout the development of the code and the first set of calculations, it was important to set a “benchmark” for several steps in the process. Using AS/3501 Carbon/Epoxy and Scotch-ply 1002 E- Glass Epoxy (see Table 1 for material properties), these benchmark calculations were used to verify that the code was producing correct results. The ply orientations are [0/45/0/45/0/45/0]. Note that all equations used and also the material properties of from Principles of Composite Material Mechanics by Gibson[1]. The inputs used in the code for the material properties and ply orientations can be found in Section 8. Table 1: Elastic Material Properties Material 𝑬 𝟏 [GPa] 𝑬 𝟐 [GPa] 𝑮 𝟏𝟐 [GPa] 𝝂 𝟏𝟐 AS/3501 Carbon / Epoxy 138 9.0 6.9 0.3 Scotch-ply 1002 E-Glass Epoxy 38.6 8.27 4.14 0.26 Also needed to define the laminate, are the locations of each ply, zk, as shown in Figure 3. These values can be found from the thickness and number of plies. Figure 3: Laminated plate geometry and ply numbering system.
  • 4. 4 4. Stress Calculations The first step is to calculate the stiffness matrices [Q] for the orthotropic lamina of each material using equation (2.26) and material properties: [ 𝑄] = [ 𝑄11 𝑄12 0 𝑄21 𝑄22 0 0 0 𝑄66 ] Where 𝑄𝑖𝑗 are components of a lamina stiffness matrix that can be found by using material properties and equation (2.27): 𝑄11 = 𝐸1 1 − 𝜈12 𝜈21 𝑄22 = 𝐸2 1 − 𝜈12 𝜈21 𝑄12 = 𝑄21 = 𝜈12 𝐸2 1 − 𝜈12 𝜈21 𝑄66 = 𝐺12 The resulting stiffness matrices for each ply from the code: Q(:,:,1) = 39.1673 2.1818 0 2.1818 8.3915 0 0 0 4.1400 Q(:,:,2) = 45.2250 31.4250 32.4404 31.4250 45.2250 32.4404 32.4404 32.4404 35.6090 Q(:,:,3) = 138.8148 2.7159 0 2.7159 9.0531 0 0 0 6.9000 Q(:,:,4) = 45.2250 31.4250 32.4404 31.4250 45.2250 32.4404 32.4404 32.4404 35.6090 Q(:,:,5) = 39.1673 2.1818 0
  • 5. 5 2.1818 8.3915 0 0 0 4.1400 Q(:,:,6) = 17.1206 8.8406 7.6939 8.8406 17.1206 7.6939 7.6939 7.6939 10.7988 Q(:,:,7) = 39.1673 2.1818 0 2.1818 8.3915 0 0 0 4.1400. Those stiffness matrices are then transformed to the ply angles specified. For the generally orthotropic lamina stiffness matrix in an off axis with a specific lamina orientation angle from equation (2.35): [ 𝑄̅] = [ 𝑄̅11 𝑄̅12 𝑄̅16 𝑄̅12 𝑄̅22 𝑄̅26 𝑄̅16 𝑄̅26 𝑄̅66 ] Where the 𝑄̅𝑖𝑗 are the components of the transformed lamina stiffness matrix which are defined as follows from equation (2.36): 𝑄̅11 = 𝑄11 𝑐4 + 𝑄22 𝑠4 + 2( 𝑄12 + 2𝑄66 ) 𝑠2 𝑐2 𝑄̅12 = ( 𝑄11 + 𝑄22 − 4𝑄66 ) 𝑠2 𝑐2 + 𝑄12( 𝑐4 + 𝑠4) 𝑄̅22 = … Now that the generally orthotropic lamina stiffness matrices are calculated, the extensional, coupling, and bending matrices can be found using equations (7.41), (7.42), and (7.43), respectively. 𝐴𝑖𝑗 = ∑(𝑄̅𝑖𝑗) 𝑘 ( 𝑧 𝑘 − 𝑧 𝑘−1) 𝑁 𝑘=1 𝐵𝑖𝑗 = 1 2 ∑(𝑄̅𝑖𝑗) 𝑘 ( 𝑧 𝑘 2 − 𝑧2 𝑘−1) 𝑁 𝑘=1 𝐷𝑖𝑗 = 1 3 ∑(𝑄̅𝑖𝑗 ) 𝑘 ( 𝑧 𝑘 3 − 𝑧3 𝑘−1) 𝑁 𝑘=1
  • 6. The resulting A, B, and D matrices calculated in the code are as follows: A = 1.0e+003 * 1.0917 0.2429 0.2177 0.2429 0.4254 0.2177 0.2177 0.2177 0.304 B = 1.0e+003 * -1.4027 -0.4113 -0.4454 -0.4113 -0.5118 -0.4454 -0.4454 -0.4454 -0.4714 D = 1.0e+004 * 3.1393 0.5723 0.4498 0.5723 1.1602 0.4498 0.4498 0.4498 0.755 The inverse form of the laminate force deformation equation (7.51) below can be used to determine the mid-plane strains (𝜀° ) and curvature (κ) from the exerted wind loads. { 𝜀° κ } = [ 𝐴 𝐵 𝐵 𝐷 ] −1 { 𝑁 𝑀 } The loads calculated from Section 2 can be entered in a dialog as shown below in Figure 2. Figure 2: Wind Load Input Dialog The resulting strain – curvature matrix [𝜀 𝑥 ° , 𝜀 𝑦 ° 𝜀 𝑥𝑦 ° κ 𝑥,κ 𝑦 , κ 𝑥𝑦] resulted as: 0.0022 -0.0007 -0.0007 0.0001 -0.0002 0.0004
  • 7. 7 From the mid-plane strain, curvature, and stiffness matrices, the lamina stresses (results in Table 2) at the top and bottom of each ply can be determined using the lamina stress-strain relationship from equation (7.34): { 𝜎𝑥 𝜎𝑦 𝜏 𝑥𝑦 } = [ 𝑄̅11 𝑄̅12 𝑄̅16 𝑄̅12 𝑄̅22 𝑄̅26 𝑄̅16 𝑄̅26 𝑄̅66 ] { 𝜀 𝑥 ° + zΚ 𝑥 𝜀 𝑦 ° + zΚ 𝑦 𝛾𝑥𝑦 ° + zΚ 𝑥𝑦 } Table 2: Stresses at top and bottom of each ply. Location 𝝈 𝒙(GPa) 𝝈 𝒚(GPa) 𝝉 𝒙𝒚(GPa) #1 Top 0.0679 0.0118 -0.02 #1 Bottom 0.0733 0.0082 -0.0151 #2 Top -0.0188 -0.0373 -0.0534 #2 Bottom 0.0124 -0.0148 -0.021 #3 Top 0.2783 0.0058 -0.0169 #3 Bottom 0.2994 0.002 -0.0087 #4 Top 0.0435 0.0077 0.0114 #4 Bottom 0.0591 0.019 0.0276 #5 Top 0.0865 -0.0006 -0.0028 #5 Bottom 0.0892 -0.0024 -0.0003 #6 Top 0.0313 0.0047 0.0102 #6 Bottom 0.0391 0.0073 0.0207 #7 Top 0.0945 -0.006 0.0046 #7 Bottom 0.0998 -0.0095 0.0095 5. Tsai-Hill Failure Criteria After the stresses are calculated for each ply, it is important to determine if any of the plies will fail. Tsai-Hill criteria will be used to analyze failure within the laminate. In the Tsai-Hill criteria, equation (4.14), if the left hand side is greater than 1, then the ply will fail. The strength values used for the failure calculations are displayed in Table 3. 𝜎1 2 𝑆 𝐿 2 − 𝜎1 𝜎2 𝑆 𝐿 2 + 𝜎2 2 𝑆 𝑇 2 + 𝜏12 2 𝑆 𝐿𝑇 2 = 1 It can be seen that the above equation uses the material coordinate stresses (𝜎1, 𝜎1, and 𝜏12). To get stresses in material coordinates equation (2.31) is used to transform the stresses: { 𝜎1 𝜎2 𝜏12 } = [ 𝑐2 𝑠2 2𝑐𝑠 𝑠2 𝑐2 −2𝑐𝑠 −𝑐𝑠 𝑐𝑠 𝑐2 − 𝑠2 ] { 𝜎𝑥 𝜎𝑦 𝜏 𝑥𝑦 }
  • 8. 8 Table 3: Typical values of lamina strengths for the chosen composites table (4.1) Material 𝑺 𝑳 + [MPa] 𝑺 𝑳 − [MPa] 𝑺 𝑻 + [MPa] 𝑺 𝑻 − [MPa] 𝑺 𝑳𝑻(𝑴𝑷𝒂) AS/3501 Carbon / Epoxy 1448 1172 48.3 248 62.1 Scotchply 1002 E-Glass Epoxy 1103 621 27.6 138 82.7 The code results for the Tsai-Hill Failure Criteria can be seen below. Since none of the values are greater than 1, it can be concluded that the proposed laminate structure will not fail under the expected wind loads. Location #1 Top 0.2439 #1 Bottom 0.1262 #2 Top 0.3066 #2 Bottom 0.2183 #3 Top 0.1251 #3 Bottom 0.0641 #4 Top 0.1716 #4 Bottom 0.1634 #5 Top 0.0074 #5 Bottom 0.0072 #6 Top 0.1071 #6 Bottom 0.0471 #7 Top 0.0131 #7 Bottom 0.0275 6. Conclusion The results of the code show that the originally proposed hybrid laminate is a solution for the spar box design. To summarize the composite will consist of:  Fiberglass Material: Scotch-ply 1002 E-Glass Epoxy  Carbon Layers: AS/3501 Carbon/Epoxy  Ply Orientation: [0/45/0/45/0/45/0] Based on “benchmark” calculations done outside of the coding, each step such as the compliance matrices, A, B, and D stiffness matrices, as well as the stresses as strains are correct for this proposed laminate structure. The code was developed to be able to calculate failure for any number of plies as long as the material data and ply orientation of each ply is designated in the input. Although the first proposed laminate passed the Tsai-Hill failure criteria, it doesn’t mean that it is the only solution. Other ply orientations and material combinations result in a laminate that passes the failure criteria. In order to distinguish between the laminates that passed, analysis on the cost, mass, safety factor, etc. will need to be completed as well.
  • 9. 9 7. References [1] Gibson, Ronald F. Principals of Composite Material Mechanics. 3rd ed. Boca Raton: CRC, 2012. Print. [2] "Loading Analysis and Strength Calculation of Wind Turbine Blade Based on Blade Element Momentum Theory and Finite Element Method." IEEE Xplore. N.p., n.d. Web. 06 Dec. 2015. [3] Boorsma, K., November 2012, and Ecn-E--12-04. Power and Loads for Wind Turbines in Yawed Conditions (n.d.): n. pag. Web. [4] AIAA-2003-0696 ALTERNATIVE COMPOSITE MATERIALS FOR MEGAWATT-SCALE WIND TURBINE BLADES: DESIGN CONSIDERATIONS AND RECOMMENDED TESTING (n.d.): n. pag. Web. 8. Appendix A: MATLAB Code Inputs for Ply Material and Orientations >> TsaiHill(38.6,8.27,4.14,.26,3,0,1.103,.621,.0276,.138,.0827,138,9,6.9,0.3,3,45,1.448,1.172,.048,.248,.06 2,138,9,6.9,0.3,3,0,1.448,1.172,.048,.248,.062,138,9,6.9,0.3,3,45,1.448,1.172,.048,.248,.062,38.6,8.27, 4.14,.26,3,0,1.103,.621,.0276,.138,.0827,38.6,8.27,4.14,.26,3,45,1.103,.621,.0276,.138,.0827,38.6,8.27, 4.14,.26,3,0,1.103,.621,.0276,.138,.0827)