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
http://autarkaw.com/books/composite/index.htmlhttp://autarkaw.com/books/composite/index.htmlhttp://autarkaw.com/books/composite/index.html
Fiber Direction
θ
x
z
y
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 = 7.5mm z
z = -7.5mm
A) 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 m h1=-0.0025 m h2=0.0025 m h3=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 kkij3
1 =k ij ∑
)(][ 13
1h - h Q = A k - kkij
k = ij ∑
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
−−−−
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
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 . - (.
.........
+
−−−−
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
3
333
333
334
m- Pa107.663105.596105.240105.596109.320106.461105.240106.461103.343
= [D]
−−−−
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
−
−
−
−
−
−
−
−
)(.
)(.-
)(.
) . + (-
) (.-
) (.
) (.
=
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 = 7.5mm z
z = -7.5mm
γ xy
Ply # Position εx εy
1 (00) Top Middle Bottom
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) Top Middle Bottom
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) Top Middle Bottom
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
Ply # Position σx σy τxy
1 (00) Top Middle Bottom
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) Top Middle Bottom
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) Top Middle Bottom
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)
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
Ply # Position ε1 ε2 γ12
1 (00) Top Middle Bottom
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) Top Middle Bottom
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) Top Middle Bottom
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
Ply # Position σ1 σ2 τ12
1 (00) Top Middle Bottom
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) Top Middle Bottom
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) Top Middle Bottom
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)
Portion of load Nx taken by 00 ply = 4.464(104)(5)(10-3) = 223.2 N/m Portion of load Nx taken by 300 ply = 1.063(105)(5)(10-3) = 531.5 N/m Portion 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 = 7.5mm z
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 ×=
EML 4230 Introduction to Composite MaterialsLaminate Stacking SequenceProblemSolution ������ ����Calculating [A] matrixThe [A] matrixCalculating the [B] MatrixThe [B] MatrixCalculating the [D] matrixThe [D] matrixB) 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 equationsMid-plane strains and curvaturesC) 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. Global strains (m/m)Slide Number 17Global stresses (Pa)The local strains and local stress as in the 300 ply at the top surface are found using transformation equations asLocal strains (m/m)Slide Number 21Local stresses (Pa)D) Portion of load taken by each plySlide Number 24END
