Chapter 4 Macromechanical Analysis of a Laminate Laminate Analysis: Example Dr. Autar Kaw Department of Mechanical Engineering University of South Florida,

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

Solution

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 mh1=-0.0025 mh2=0.0025 mh3=0.0075 m

0o

30o

-45o

5mm

5mm

5mm

z = -2.5mm

z = 2.5mm

z = 7.5mmz

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

1

h - h Q = A k - kkijk =

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[ 2

1 = B 2

1 -k 2kkij

3

1 =k ij

)] )(-0.0075 - )[(-0.0025 )10(

7.17

0

0

0

10.35

2.897

0

2.897

181.2

2

1 = [B] 229

)(-0.0025 - )(0.0025)10(

36.74

20.05

54.19

20.05

23.65

32.46

54.19

32.46

109.4

2

1 + 229

])(0.0025 - )[(0.0075 )10(

46.5942.8742.87

42.8756.6642.32

42.8742.3256.66

2

1 + 229

Calculating the [B] Matrix

The [B] Matrix

2

566

665

656

108559100721100721

100721101581108559

100721108559101293

m Pa

...

...

...

[B] =

)h - h( ]Q[ 3

1 = D 3

1 -k 3kkij

3

1 =k ij

339 )00750()00250()10(

17700

035108972

089728181

3

1. .

.

..

..

[D] =

339 )00250()00250()10(

743605201954

052065234632

195446324109

3

1. .

...

...

...

+

339 )00250)00750()10(

594687428742

874266563242

874232426656

3

1. - (.

...

...

...

+

Calculating the [D] matrix

The [D] matrix

3

333

333

334

m- Pa

107.663105.596105.240

105.596109.320106.461

105.240106.461103.343

= [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

, top101014

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 ply

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 = 7.5mmz

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 =

1001000

223.2

% 53.15 =

100 1000

531.5

% 24.52 =

100 1000

245.2

END