### Problem

```Macromechanics of a
Laminate
Textbook: Mechanics of Composite Materials
Author: Autar Kaw
Figure 4.1
Fiber Direction

x
y
z
CHAPTER OBJECTIVES
 Understand the code for laminate stacking sequence
 Develop relationships of mechanical and hygrothermal
loads applied to a laminate to strains and stresses in each
lamina
 Find the elastic stiffnesses of laminate based on the
elastic moduli of individual laminas and the stacking
sequence
 Find the coefficients of thermal and moisture expansion
of a laminate based on elastic moduli, coefficients of
thermal and moisture expansion of individual laminas,
and stacking sequence
Laminate Behavior
• elastic moduli
• the stacking position
• thickness
• angles of orientation
• coefficients of thermal expansion
• coefficients of moisture expansion
x
P
P
z
(a)
M
M
x
z
(b)
M
M
x
P
x
P
z
(c)
Figure 4.2
Figure 4.3
x
x
z
z
Mxy
Myx
Nx
y
Nyx
Nxy
My
y
Ny
(a)
(b)
Mx
Classical Lamination Theory
 Each lamina is orthotropic.
 Each lamina is homogeneous.
 A line straight and perpendicular to the middle surface remains
straight and perpendicular to the middle surface during




deformation. ( γxz = γyz = 0 ) .
The laminate is thin and is loaded only in its plane (plane stress)
(σz = τxz = τyz = 0 ) .
Displacements are continuous and small throughout the laminate
(| u |, | v |, | w |  | h |) , where h is the laminate thickness.
Each lamina is elastic.
No slip occurs between the lamina interfaces.
Figure 4.4
U0
h/2
Mid-Plane
h/2
x
z

A
z
A 
z
Cross-Section
Cross-Section
wo
Global Strains in a Laminate
 εx
 
 εy
 
γ xy 
2



 w0 
 u0
 
2 



x 
x 






 2 w0 

 v0 
=
.
 + z 
2 
y 
y 






  u0


2
 v0 

w
0
+


 2

x 

 y
xy 
 0
 κx
ε
x
 


 0
 
  ε y  + z  κ y .
 
 
κ xy 
γ0xy 
 
Figure 4.5
Mid-Plane
z
Laminate
Strain Variation
Stress Variation
Figure 4.6
1
2
h0
3
h1
h/2
h2
h3
Mid-Plane
hk-1
hk
tk
k-1
k
k+1
hn-1
hn
n
h/2
z
Stresses in a Lamina in a Laminate
 σ x
 
 σ y =
 
 τ xy  k
Q11 Q12 Q16   ε 

  x
Q Q
  ε y
Q
22
26
 12
  

 γ 
Q16 Q 26 Q 66  k  xy  k
Q11 Q12 Q16


= Q12 Q 22 Q 26




Q
Q
Q
26
66  k
 16
 ε0 
Q11 Q12 Q16
x
 


 0 + z 

Q
Q
Q
ε
y
22
26
 
 12

 0


Q
Q
Q
γ
 xy
26
66  k
 16
 κ x
 
 κ y .
 
κ xy
Forces and Stresses
 N x
 
 N y =
 
 N xy 
 N x
 
 N y =
 
 N xy 
h/ 2

-h/ 2
 σ x
 
 σ y  dz,
 
 τ xy 
n
hk
k=1
h k- 1
 
 σ x
 
 σ y  dz,
 
 τ xy k
Forces and Strains
 N x
 
 N y =
 
 N xy
Q11 Q12 Q16  ε0x 

  
 0  dz
Q Q

Q 26  ε y 
12
22



  0
Q16 Q 26 Q66k γ xy
hk
n
 
k = 1 h k-1
n
+
hk

k = 1 h k 1
Q11 Q12 Q16


Q Q Q 
22
26
 12



Q16 Q 26 Q66k
 κ x
 
 κ y  z dz
 
κ xy 
Forces and Strains
   0
 Q11 Q12 Q16 
 N x 
x




 hk
   n 
  0


 N y  =   Q12 Q 22 Q 26  dz   y
 
 h
   k =1 

k -1
 0


 N xy 
  xy
Q16 Q 26 Q66 k

 

 
 Q11 Q12 Q16 
 n 
   x
 hk




+   Q12 Q 22 Q 26  z dz    y 
 h
 k =1 
 
k 1


 xy


Q
Q
Q
26
66 k
 16


Integrating terms
hk

dz = ( hk  hk - 1) ,
hk - 1
hk

hk - 1
1 2
zdz =
( hk  h2k - 1) ,
2
Forces and Strains
 N x   A11
  
 N y  =  A12
  
 N xy  A16
A12
A22
A26
 0
A16   x   B11 B12 B16    x 
  0 
 
A26    y  +  B12 B22 B26    y 
  
 
A66   0   B16 B26 B66   xy
 xy
n
Aij =

[( Qij ) ] k ( hk - hk - 1 ), i = 1,2,6; j = 1,2,6,
k =1
1
Bij =
2
n

k =1
[( Qij ) ] k ( h2k - h2k - 1 ), i = 1,2,6; j = 1,2,6
Moments and Strains
 M x   B11 B12 B16  ε0x   D11 D12 D16  κ x 
 

 
  0 
 
 M y  =  B12 B22 B26 ε y +  D12 D22 D26  κ y 
 

 
  
 
 M xy  B16 B26 B66 γ0

 
 xy  D16 D26 D66 κ xy
1
Dij =
3
n

k =1
[( Qij ) ] k ( h3k - h3k - 1 ) i = 1, 2, 6; j = 1, 2, 6.
Forces, Moments, Strains,
Curvatures
 N x   A11 A12 A16 B11 B12 B16

 

 N y   A12 A22 A26 B12 B22 B26

 

 N xy   A16 A26 A66 B16 B26 B66

=

 M x   B11 B12 B16 D11 D12 D16

 

 M y   B12 B22 B26 D12 D22 D26

 

 M xy   B16 B26 B66 D16 D26 D66
 ε0x 
 
 ε0y 
 
 γ0 
 xy 
 
 κ x
 
 κ y
 
κ xy 
Steps
1.
Find the value of the reduced stiffness matrix [Q] for each ply using its four
elastic moduli, E1, E2, v12, G12 in Equation (2.93).
2.
Find the value of the transformed reduced stiffness matrix [ Q ] for each ply
using the [Q] matrix calculated in Step 1 and the angle of the ply in Equation
(2.104) or Equations (2.137) and (2.138).
3.
Knowing the thickness, tk of each ply, find the coordinate of the top and
bottom surface, hi, i = 1, . . . . . . . , n of each ply using Equation (4.20).
4.
Use the [ Q ] matrices from Step 2 and the location of each ply from Step 3 to
find the three stiffness matrices [A], [B] and [D] from Equation (4.28).
5.
Substitute the stiffness matrix values found in Step 4 and the applied forces
and moments in Equation (4.29).
Steps
6. Solve the six simultaneous Equations (4.29) to find
the midplane strains and curvatures.
7. Knowing the location of each ply, find the global
strains in each ply using Equation (4.16).
8. For finding the global stresses, use the stress-strain
Equation (2.103).
9. For finding the local strains, use the transformation
Equation (2.99).
10. For finding the local stresses, use the
transformation Equation (2.94).
Figure 4.7
z = -7.5mm
0o
5mm
30o
5mm
z = -2.5mm
z = 2.5mm
-45o
z = 7.5mm
z
5mm
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.
Solution
• A) From Example 2.4, the reduced stiffness matrix for
the 00 Graphite/Epoxy ply is
0
 181.8 2.897

 9
[Q] = 2.897 10.35
0 (10 ) Pa


0
0
7.17


0
• From Equation
(2.99), the
transformed
reduced
stiffness matrix
for each of the
three plies are
0
 181.8 2.897


[Q ]0 = 2.897 10.35
0 (109) Pa


0
0
7.17


109.4 32.46 54.19


[Q ]30 = 32.46 23.65 20.05 (109) Pa
54.19 20.05 36.74


 56.66 42.32 - 42.87


[Q]-45 =  42.32 56.66 - 42.87 (109) Pa
- 42.87 - 42.87 46.59


The total thickness of the laminate is
h = (0.005)(3) = 0.015 m.
The mid plane is 0.0075 m from the top and bottom of
the laminate. Hence using Equation (4.20), the
location of the ply surfaces are
h0 = -0.0075 m
h1 = -0.0025 m
h2 = 0.0025 m
h3 = 0.0075 m
3
Aij =  [Qij]k ( hk - hk - 1)
k =1
0
 181.8 2.897


[A] = 2.897 10.35
0 (109) [(-0.0025)- (-0.0075)]

0
0 7.17

From Equation
(4.28a), the
extensional stiffness
matrix [A] is
109.4 32.46 54.19


+ 32.46 23.65 20.05 (109) [0.0025- (-0.0025)]
54.19 20.05 36.74


 56.66 42.32 - 42.87


+  42.32 56.66 - 42.87 (109) [0.0075- 0.0025]
- 42.87 - 42.87 46.59


The [A] matrix
1.739(109 ) 3.884(108 )
5.663(107 ) 

8
8
8 
[A] = 3.884(10) 4.533(10)  1.141(10) Pa - m
5.663(107 )  1.141(108 ) 4.525(108 ) 


From
Equation
(4.28b), the
coupling
stiffness
matrix [B]
is
1 3
2
2
=
[
]
(
Qij k hk hk - 1)
Bij

2 k =1
181.2 2.897 0 
1
[B] = 2.897 10.35 0  (109 ) [(-0.0025)2 - (-0.0075)2 )]

2
0 7.17
 0
109.432.4654.19
1
+ 32.4623.6520.05 (109 ) (0.0025)2 - (-0.0025)2

2
54.1920.0536.74


42.32  42.87
 56.66
1
+  42.32 56.66  42.87 (109 ) [(0.0075)2 - (0.0025)2 ]

2
 42.87  42.87 46.59 
The B Matrix
 3.129106  9.855105   1.072106 

5
6
6 
[B] =  9.85510 
1.15810   1.07210  Pa-m2
  1.072106   1.072106  9.855105  


3
From
Equation
(4.28c), the
bending
stiffness
matrix [D] is
1
3
3
=
[
(
Qij ]k hk hk - 1)
Dij

3 k =1
181.8
1 
[D] =
2.897

3
 0
109.4
1
+ 32.46
3 
 54.19
0 
10.35 0  ( 109 ) (-0.0025 )3 - (-0.0075 )3

0
7.17
2.897


32.46 54.19
23.65 20.05 ( 109 ) ( 0.0025 )3 - (-0.0025 )3

20.05 36.74

42.32  42.87
 56.66
1
+  42.32
56.66  42.87 ( 109 ) ( 0.0075 )3 - ( 0.0025 )3

3 
 42.87  42.87 46.59 



The [D] matrix
 3.343104 
6.461103   5.240103 

3
3
3 
[D] =  6.46110 
9.32010   5.59610  Pa - m 3
 5.240103   5.596103  7.663103  


B) Since the applied load is Nx = Ny = 1000N/m, the midplane strains and curvatures can be found by solving the
following set of simultaneous linear equations (Equation
4.29).
 1.739(109 ) 3.884(108) 5.663(107 ) - 3.129(106) 9.855(105)
1000 

  3.884(108) 4.533(108) - 1.141(108) 9.855(105) 1.158(106)
1000

 
 0  5.663(107 ) - 1.141(108) 4.525(108) - 1.072(106) - 1.072(106)

=
 0 - 3.129(106) 9.855(105) - 1.072(106) 3.343(104) 6.461(103)

 
0

  9.855(105) 1.158(106) - 1.072(106) 6.461(103) 9.320(103)

 
 0 
6
6
5
3
3
 - 1.072(10 ) - 1.072(10 ) 9.855(10 ) - 5.240(10 ) - 5.596(10 )
- 1.072(106)  ε0x 
 
0
6
- 1.072(10 )  εy 
 
5  0 
9.855(10 ) γxy
 
- 5.240(103)  κ x 
 
- 5.596(103)  κ y 
 
 
7.663(103) κ xy 
Mid-plane strains and curvatures
 ε0x   3.123(10-7)
  

0
 εy   3.492(10-6) m/m
  

 γ0  - 7.598( -7)
10 
 xy  = 
   2.971(10-5)
 κx 

  - 3.285( -4) 1/m
10
 κ y 

  
-4 
4.101(
)
10
κ xy  
C) The strains and stresses at the top surface of
the 300 ply are found as follows. First, the top
surface of the 300 ply is located at z = h1 = -0.0025
m. From Equation (4.16),
 εx 
 3.123(10-7) 
 2.971(10-5)
 




-6
-4
 εy 
=  3.492(10 )  + (-0.0025)- 3.285(10 )
 




 γxy  0
- 7.598(10-7) 
 4.101(10-4)



 30 , top 
 2.380(10-7)


-6
=  4.313(10 ) m/m


- 1.785(10-6)


Table 4.1 Global strains (m/m) in Example 4.3
εx
εy
Ply #
Position
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)
 xy
Using the stress-strain Equations (2.98) for an angle
ply,
-7
2.380(
)

 σx 
10
109.4 32.46 54.19

 

 9 
 σy 
= 32.46 23.65 20.05 (10 )  4.313(10-6)


 
54.19 20.05 36.74
- 1.785(10-6)
τxy  0, top 



30
6.930(104)


= 7.391(104) Pa


3.381(104)


Table 4.2 Global stresses (Pa) in Example 4.3
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)
The local strains and local stress as in the 300 ply at the
top surface are found using transformation Equation
(2.94) as
-7
 ε1  0.7500 0.2500 0.8660  2.380(10 )


 

-6
 ε2 =  0.2500 0.7500 - 0.8660  4.313(10 )



 

 γ12 /2 - 0.4330 0.4330 0.5000 - 1.785(10-6)/ 2
 ε1
 
 ε2
 
γ12 
 4.837(10- 7 )


= 4.067(10- 6 ) m /m


 2.636(10- 6)


Table 4.3 Local strains (m/m) in Example 4.3
ε1
ε2
γ12
Ply #
Position
1 (00)
Top
Middle
Bottom
8.944 (10-8) 5.955(10-6) -3.836(10-6)
1.637 (10-7) 5.134(10-6) -2.811(10-6)
2.380 (10-7) 4.313(10-6) -1.785(10-6)
2 (300)
Top
Middle
Bottom
4.837(10-7) 4.067(10-6) 2.636(10-6)
7.781(10-7) 3.026(10-6) 2.374(10-6)
1.073(10-6) 1.985(10-6) 2.111(10-6)
3 (-450)
Top
Middle
Bottom
1.396(10-6) 1.661(10-6) -2.284(10-6)
5.096(10-7) 1.800(10-6) -1.388(10-6)
-3.766(10-7) 1.940(10-6) -4.928(10-7)
4
6.930(

10 )
 σ1  0.7500 0.2500 .8660


  

4

=
0.2500
0.7500
.8660
7.391(
10 )
 σ2  



  

τ12  - 0.4330 0.4330 0.5000 3.381(104)
9.973(104)


= 4.348(104) Pa


1.890(104)


Table 4.4 Local stresses (Pa) in Example 4.3
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)
D) The portion of the load Nx taken by each ply can be
calculated by integrating the stress  xx through the
thickness of each ply. However, since the stress varies
linearly through each ply, the portion of the load Nx taken
is simply the product of the stress  xx at the middle of
each ply (See Table 4.2) and the thickness of the 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.
Percentage of load Nx taken by
00
223.2
 100
ply 
1000
= 22.32%
Percentage of load Nx taken by 300 ply
Percentage of load Nx taken by -450 ply
531.5
 100
1000
= 53.15%

245.2
 100
1000
= 24.52%

Figure 4.8
Strip 1, E1, 1
Strip 2, E2, 2
```