macromechanics of a laminate - usf college of engineeringkaw/class/composites/ppt/chapter4.pdf ·...
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Macromechanics of a Laminate
Textbook: Mechanics of Composite Materials Author: Autar Kaw
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
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. )0 = γ = γ( yzxz . • The laminate is thin and is loaded only in its plane (plane stress)
)0 = τ = τ = σ( yzxzz . • Displacements are continuous and small throughout the laminate
|)h| |w| |,v| |,u(| << , where h is the laminate thickness. • Each lamina is elastic. • No slip occurs between the lamina interfaces.
Figure 4.4
Cross-Section after Loading
x
U0
z
zα
A′
α
α
z A
Mid-Plane
wo
Cross-Section Before Loading
h/2
h/2
Global Strains in a Laminate
.
yxw
yw
xw
+ z
xv +
yu
yv
xu
=
γ
ε
ε
xy
y
x
∂∂∂−
∂∂−
∂∂−
∂∂
∂∂
∂∂
∂∂
02
20
2
20
2
00
0
0
2
.
κ
κ
κ
+ z
γ
ε
ε
xy
y
x
xy
y
x
=
0
0
0
Stresses in a Lamina in a Laminate
kxy
y
x
k kxy
y
x
γ
ε
ε
QQQ
QQQ
QQQ
=
τ
σ
σ
662616
262212
161211
.
κ
κ
κ
QQQ
QQQ
QQQ
+ z
γ
ε
ε
QQQ
QQQ
QQQ
=
xy
y
x
kxy
y
x
k
662616
262212
161211
0
0
0
662616
262212
161211
Forces and Stresses
dz,
τ
σ
σ
=
N
N
N
xy
y
xh/
-h/
xy
y
x
∫2
2
dz,
τ
σ
σ
=
N
N
N
xy
y
x
k
h
h
n
k=
xy
y
xk
k-
∫∑11
Forces and Strains
dz
γ
ε
ε
QQQ
QQQ
QQQ
=
N
N
N
xy
y
x
k
h
h
n
k =
xy
y
xk
k-
∫∑0
0
0
662616
262212
161211
1 1
z dz
κ
κ
κ
QQQ
QQQ
QQQ
+
xy
y
x
k
h
h
n
k =
k
k
∫∑−
662616
262212
161211
1 1
Forces and Strains
∫∑γ
ε
ε
0xy
0y
0x
h
h
662616
262212
161211
k
n
1 = k
xy
y
x
dz
QQQ
QQQ
QQQ
=
N
N
Nk
1 - k
∫∑κ
κ
κ
xy
y
xh
h
662616
262212
161211
k
n
1 = k
dz z
QQQ
QQQ
QQQ
+k
1 k
Forces and Strains
κ
κ
κ
γ
ε
ε
xy
y
x
662616
262212
161211
0xy
0y
0x
662616
262212
161211
xy
y
x
BBB
BBB
BBB
+
AAA
AAA
AAA
=
N
N
N
,,,; j = ,,), i = h - h (])Q [( = A k - kkij
n
k = ij 6216211
1∑
62162121 2
12
1
,,; j = ,,), i = h - h (])Q [( = B k - kkij
n
k = ij ∑
Moments and Strains
κ
κ
κ
DDD
DDD
DDD
+
γ
ε
ε
BBB
BBB
BBB
=
M
M
M
xy
y
x
xy
y
x
xy
y
x
662616
262212
161211
0
0
0
662616
262212
161211
., , ; j = , , ) i = h - h (])Q [( = D k - kkij
n
k = ij 621621
31 3
13
1∑
Forces, Moments, Strains, Curvatures
κ
κ
κ
γ
ε
ε
DDDBBB
DDDBBB
DDDBBB
BBBAAA
BBBAAA
BBBAAA
=
M
M
M
N
N
N
xy
y
x
xy
y
x
xy
y
x
xy
y
x
0
0
0
662616662616
262212262212
161211161211
662616662616
262212262212
161211161211
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).
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
Pa)10(
7.1700
010.352.897
02.897181.8
= [Q] 9
• From Equation
(2.99), the transformed reduced stiffness matrix for each of the three plies are
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-
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
From Equation (4.28a), the
extensional stiffness matrix [A] is
(-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 ∑
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
−−
From Equation (4.28b), the coupling stiffness matrix [B] is
)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
−−−−
The B Matrix
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
m Pa-.........
[B] = 2
566
665
656
108559100721100721100721101581108559100721108559101293
−−−−−
From Equation (4.28c), the bending stiffness matrix [D] is
)h - h( ]Q[
31 = D 3
1 -k 3kkij
3
1 =k ij ∑
[ ]). - (-).(-) (.
....
[D] = 339 00750002501017700035108972089728181
31
[ ]). - (-).() (.........
+ 339 002500025010743605201954052065234632195446324109
31
[ ]). - ().() (.........
+ 339 002500075010594687428742874266563242874232426656
31
−−−−
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 = 1000N/m, the mid-plane strains and curvatures can be found by solving the following set of simultaneous linear equations (Equation 4.29).
κ
κ
κ
γ
ε
ε
)107.663()105.596(-)105.240(-)109.855()101.072(-)101.072(-
)105.596(-)109.320()106.461()101.072(-)101.158()109.855(
)105.240(-)106.461()103.343()101.072(-)109.855()103.129(-
)109.855()101.072(-)101.072(-)104.525()101.141(-)105.663(
)101.072(-)101.158()109.855()101.141(-)104.533()103.884(
)101.072(-)109.855()103.129(-)105.663()103.884()101.739(
=
M
M
M
N
1000
1000
xy
y
x
0xy
0y
0x
333566
333665
334656
566887
665888
656789
xy
y
x
xy
1/m
)104.101(
)103.285(-
)102.971(
m/m
)107.598(-
)103.492(
)103.123(
=
κ
κ
κ
γ
ε
ε
4-
4-
5-
7-
6-
7-
xy
y
x
0xy
0y
0x
Mid-plane strains and curvatures
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),
)104.101(
)103.285(-
)102.971(
(-0.0025) +
)107.598(-
)103.492(
)103.123(
=
γ
ε
ε
4-
4-
-5
7-
6-
-7
xy
y
x
top,300
m/m
)101.785(-
)104.313(
)102.380(
=
6-
6-
-7
Table 4.1 Global strains (m/m) in Example 4.3
γ xyPly # 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)
Using the stress-strain Equations (2.98) for an angle ply,
)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
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
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
Table 4.3 Local strains (m/m) in Example 4.3
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
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 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 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.
σ xx
σ xx
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 ×=