eml 4230 introduction to composite materials
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EML 4230 Introduction to Composite Materials. Chapter 4 Macromechanical Analysis of a Laminate Laminate Analysis: Example Dr . Autar Kaw Department of Mechanical Engineering University of South Florida, Tampa, FL 33620 Courtesy of the Textbook Mechanics of Composite Materials by Kaw. - PowerPoint PPT PresentationTRANSCRIPT
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EML 4230 Introduction to Composite Materials
Chapter 4 Macromechanical Analysis of a Laminate
Laminate Analysis: Example
Dr. Autar KawDepartment of Mechanical Engineering
University of South Florida, Tampa, FL 33620
Courtesy of the TextbookMechanics of Composite Materials by Kaw
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Laminate Stacking Sequence
Fiber Direction
x
z
y
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Problem A [0/30/-45] Graphite/Epoxy
laminate is subjected to a load of Nx = Ny = 1000 N/m. Use the unidirectional properties from Table 2.1 of Graphite/Epoxy. Assume each lamina has a thickness of 5 mm. Find
a) the three stiffness matrices [A], [B] and [D] for a three ply [0/30/-45] Graphite/Epoxy laminate.
b) mid-plane strains and curvatures.c) global and local stresses on top
surface of 300 ply.d) percentage of load Nx taken by
each ply.
0o
30o
-45o
5mm
5mm
5mm
z = -2.5mm
z = 2.5mm
z = 7.5mmz
z = -7.5mm
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SolutionA) The reduced stiffness matrix for the Oo Graphite/Epoxy ply is
0
Pa)10(
7.1700
010.352.897
02.897181.8
= [Q] 9
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Pa)10(
7.1700
010.352.897
02.897181.8
= ]Q[ 90
Pa)10(
36.7420.0554.19
20.0523.6532.46
54.1932.46109.4
= ]Q[ 930
Pa)10(
46.5942.87-42.87-
42.87-56.6642.32
42.87-42.3256.66
= ]Q[ 945-
Qbar Matrices for Laminas
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The total thickness of the laminate is h = (0.005)(3) = 0.015 m.
h0=-0.0075 mh1=-0.0025 mh2=0.0025 mh3=0.0075 m
0o
30o
-45o
5mm
5mm
5mm
z = -2.5mm
z = 2.5mm
z = 7.5mm z
z = -7.5mm
Coordinates of top & bottom of plies
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(-0.0075)]-[(-0.0025) )10(
7.1700
010.352.897
02.897181.8
= [A] 9
(-0.0025)]-[0.0025 )10(
36.7420.0554.19
20.0523.6532.46
54.1932.46109.4
+ 9
0.0025]-[0.0075 )10(
46.5942.87-42.87-
42.87-56.6642.32
42.87-42.3256.66
+ 9
)h - h( ]Q[ = A 1 -k kkij
3
1 =k ij Calculating [A] matrix
)(][ 1
3
1h - h Q = A k - kkij
k = ij
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The [A] matrix
m- Pa)4.525(10)1.141(10)5.663(10)1.141(10)4.533(10)3.884(10
)5.663(10)3.884(10)1.739(10 = [A]
887
888
789
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)h - h( ]Q[
21 = B 2
1 -k 2kkij
3
1 =k ij
)] )(-0.0075 - )[(-0.0025 )10( 7.17
00
010.352.897
0
2.897181.2
21 = [B] 229
)(-0.0025 - )(0.0025)10( 36.7420.0554.19
20.0523.6532.46
54.1932.46109.4
21 + 229
])(0.0025 - )[(0.0075 )10( 46.5942.8742.8742.8756.6642.3242.8742.3256.66
21 + 229
Calculating the [B] Matrix
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The [B] Matrix
2
566
665
656
108559100721100721100721101581108559100721108559101293
m Pa.........
[B] =
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)h - h( ]Q[
31 = D 3
1 -k 3kkij
3
1 =k ij
339 )00750()00250()10(17700035108972089728181
31 . .
...
.. [D] =
339 )00250()00250()10(743605201954052065234632195446324109
31 . .
...
...
... +
339 )00250)00750()10(594687428742874266563242874232426656
31 . - (.
.........
+
Calculating the [D] matrix
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The [D] matrix
3
333
333
334
m- Pa107.663105.596105.240105.596109.320106.461105.240106.461103.343
= [D]
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B) Since the applied load is Nx = Ny = 1000 N/m, the mid-plane strains and curvatures can be found by solving the following set of simultaneous linear equations
κ
κ
κ
γ
ε
ε
)(.)(.-)(.-)(.)(.-)(.-
)(.-)(.)(.)(.-)(.)(.
)(.-)(.)(.)(.-)(.)(.-
)(.)(.-)(.-)(.)(.-).
)(.-)(.)(.)(.-)(.).
)(.-)(.)(.-)(.)(.).
=
xy
y
x
xy
y
x
0
0
0
333566
333665
334656
566887
665888
656789
106637105965102405108559100721100721
105965103209104616100721101581108559
102405104616103433100721108559101293
10855910072110072110525410141110(6635
10072110158110855910141110533410(8843
10072110855910129310663510884310(7391
0
0
0
0
1000
1000
Setting up the 6x6 matrix
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/m
.
.
.
m/m
.
.
.
=
κ
κ
κ
γ
ε
ε
xy
y
x
xy
y
x
1
)10(1014
)10(2853
)10(9712
)10(5987
)10(4923
)10(1233
4
4
5
7
6
7
0
0
0
Mid-plane strains and curvatures
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C) The strains and stresses at the top surface of the 300 ply are found as follows. The top surface of the 300 ply is located at z = h1 = -0.0025 m.
)(.
)(.-
)(.
) . + (-
) (.-
) (.
) (.
=
γ
ε
ε
-
-
-
-
-
-
xy
y
x
, top 101014
102853
109712
00250
105987
104923
101233
4
4
5
7
6
7
300
m/m
)(.-
)(.
)(.
=
-
-
-
107851
103134
103802
6
6
7
0o
30o
-45o
5mm
5mm
5mm
z = -2.5mm
z = 2.5mm
z = 7.5mmz
z = -7.5mm
Global Strains/Stresses at top of 30o ply
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Global strains (m/m)
xy
Ply # Position εx εy
1 (00) TopMiddleBottom
8.944 (10-8)1.637 (10-7)2.380 (10-7)
5.955 (10-6)5.134 (10-6)4.313 (10-6)
-3.836 (10-6)-2.811 (10-6)-1.785 (10-6)
2 (300) TopMiddleBottom
2.380 (10-7)3.123 (10-7)3.866 (10-7)
4.313 (10-6)3.492 (10-6)2.670 (10-6)
-1.785 (10-6)-7.598 (10-7) 2.655 (10-7)
3(-450) TopMiddleBottom
3.866 (10-7)4.609 (10-7)5.352 (10-7)
2.670 (10-6)1.849 (10-6)1.028 (10-6)
2.655 (10-7)1.291 (10-6)2.316 (10-6)
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)101.785(-
)104.313(
)102.380(
)10(
36.7420.0554.19
20.0523.6532.46
54.1932.46109.4
=
τ
σ
σ
6-
6-
-7
9
xy
y
x
top,300
Pa
)103.381(
)107.391(
)106.930(
=
4
4
4
Global stresses in 30o ply
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Global stresses (Pa)
Ply # Position σx σy τxy
1(00)
TopMiddleBottom
3.351 (104)4.464 (104)5.577 (104)
6.188 (104)5.359 (104)4.531 (104)
-2.750 (104)-2.015 (104)-1.280 (104)
2(300)
TopMiddleBottom
6.930 (104)1.063 (105)1.434 (105)
7.391 (104)7.747 (104)8.102 (104)
3.381 (104)5.903 (104) 8.426 (104)
3 (-450)
TopMiddleBottom
1.235 (105)4.903 (104)-2.547 (104)
1.563 (105)6.894 (104)-1.840 (104)
-1.187 (105)-3.888 (104)4.091 (104)
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The local strains and local stress as in the 300 ply at the top surface are found using transformation equations as
2)/ 101.785(-
)104.313(
)102.380(
0.50000.43300.4330-
0.8660-0.75000.2500
0.86600.25000.7500
=
/2γ
ε
ε
6-
6-
-7
12
2
1
m/m
.
.
.
=
γ
ε
ε
-
-
-
)10(6362
)10(0674
)10(8374
6
6
7
12
2
1
Local Strains/Stresses at top of 30o ply
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Local strains (m/m)
Ply # Position ε1 ε2 γ12
1 (00) TopMiddleBottom
8.944 (10-8)1.637 (10-7)2.380 (10-7)
5.955(10-6)5.134(10-6)4.313(10-6)
-3.836(10-6)-2.811(10-6)-1.785(10-6)
2 (300) TopMiddleBottom
4.837(10-7)7.781(10-7)1.073(10-6)
4.067(10-6)3.026(10-6)1.985(10-6)
2.636(10-6)2.374(10-6) 2.111(10-6)
3 (-450) TopMiddleBottom
1.396(10-6)5.096(10-7)
-3.766(10-7)
1.661(10-6)1.800(10-6)1.940(10-6)
-2.284(10-6)-1.388(10-6)-4.928(10-7)
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)103.381(
)107.391(
)106.930(
0.50000.43300.4330-
.8660-0.75000.2500
.86600.25000.7500
=
τ
σ
σ
4
4
4
12
2
1
Pa
)101.890(
)104.348(
)109.973(
=
4
4
4
Local stresses in 30o ply
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Local stresses (Pa)Ply # Position σ1 σ2 τ12
1 (00) TopMiddleBottom
3.351 (104)4.464 (104)5.577 (104)
6.188 (104)5.359(104)4.531 (104)
-2.750 (104)-2.015 (104)-1.280 (104)
2 (300) TopMiddleBottom
9.973 (104)1.502 (105)2.007 (105)
4.348 (104)3.356 (104)2.364 (104)
1.890 (104)1.702 (104)1.513 (104)
3 (-450) TopMiddleBottom
2.586 (105)9.786 (104)-6.285 (104)
2.123 (104)2.010 (104)1.898 (104)
-1.638 (104)-9.954 (103)-3.533 (103)
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D) Portion of load taken by each plyPortion of load Nx taken by 00 ply = 4.464(104)(5)(10-3) = 223.2 N/mPortion of load Nx taken by 300 ply = 1.063(105)(5)(10-3) = 531.5 N/mPortion of load Nx taken by -450 ply = 4.903(104)(5)(10-3) = 245.2 N/m
The sum total of the loads shared by each ply is 1000 N/m, (223.2 + 531.5 + 245.2) which is the applied load in the x-direction, Nx.
0o
30o
-45o
5mm
5mm
5mm
z = -2.5mm
z = 2.5mm
z = 7.5mmz
z = -7.5mm
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Percentage of load Nx taken by 00 ply
Percentage of load Nx taken by 300 ply
Percentage of load Nx taken by -450 ply
% 22.32 =
1001000223.2
% 53.15 =
100 1000531.5
% 24.52 =
100 1000245.2
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END