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
Laminate Stacking Sequence
Fiber Direction
x
z
y
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
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
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
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
(-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
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
)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
The [B] Matrix
2
566
665
656
108559100721100721100721101581108559100721108559101293
m Pa.........
[B] =
)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
The [D] matrix
3
333
333
334
m- Pa107.663105.596105.240105.596109.320106.461105.240106.461103.343
= [D]
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
/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
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
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)
)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
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)
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
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)
)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
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)
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
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
END