1. Pendahuluan Material Komposit
BAB 4 Macromechanical Analysis of a Laminate
Classical Lamination Theory
Qomarul Hadi, ST,MT
Teknik Mesin
Universitas Sriwijaya
Sumber Bacaan
Mechanics of Composite Materials by Kaw
3. Laminate Behavior
• Modulus Elastis
• The Stacking Position
• Thickness
• Angles of Orientation
• Coefficients of Thermal Expansion
• Coefficients of Moisture Expansion
4. x
P
P
P
P
z
x
(a)
z
(c)
x
z
M M
(b)
x
M
M
A
P
=
xx
(4.1)
Strains in a
Gambar 4.2
A beam under (a) axial load, (b) bending moment,
and (c) combined axial and bending moment.
AE
P
=
xx
I
Mz
=
xx
z
=
xx
M
EI
z
+
P
AE
1
=
xx
1
z
+
= 0
z
+
= 0
5. Types of loads allowed in CLT analysis
x
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
Nx = normal force resultant in the x direction (per unit length)
Ny = normal force resultant in the y direction (per unit length)
Nxy = shear force resultant (per unit length)
Gambar 4.3
Resultant forces and moments on a
laminate.
7. 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
( yz
xz .
The laminate is thin and is loaded only in its plane (plane stress)
)
0
=
τ
=
τ
=
σ
( yz
xz
z .
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.
12. Pendahuluan Material Komposit
BAB 4 Macromechanical Analysis of a Laminate
Relating Loads to Midplane
Strains/Curvatures
Qomarul Hadi, ST,MT
Teknik Mesin
Universitas Sriwijaya
Sumber Bacaan
Mechanics of Composite Materials by Kaw
14. Types of loads allowed in CLT
analysis
x
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
Nx = normal force resultant in the x direction (per unit length)
Ny = normal force resultant in the y direction (per unit length)
Nxy = shear force resultant (per unit length)
Gambar 4.3
Resultant forces and moments on a
laminate.
15. x
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
Mx = bending moment resultant in the yz plane (per unit length)
My = bending moment resultant in the xz plane (per unit length)
Mxy = twisting moment resultant (per unit length)
Types of loads allowed in CLT
analysis
23. Stiffness Matrices
[A] – Extensional stiffness matrix relating the resultant in-
plane forces to the in-plane strains.
[B] – Coupling stiffness matrix coupling the force and
moment terms to the midplane strains and midplane
curvatures.
29. Stiffness Matrices
[A] – Extensional stiffness matrix relating the resultant in-
plane forces to the in-plane strains.
[B] – Coupling stiffness matrix coupling the force and
moment terms to the midplane strains and midplane
curvatures.
[D] – Bending stiffness matrix relating the resultant
bending moments to the plate curvatures.
30. Forces, Moments, Midplane Strains,
Midplane Curvatures
κ
κ
κ
γ
ε
ε
D
D
D
B
B
B
D
D
D
B
B
B
D
D
D
B
B
B
B
B
B
A
A
A
B
B
B
A
A
A
B
B
B
A
A
A
=
M
M
M
N
N
N
xy
y
x
xy
y
x
xy
y
x
xy
y
x
0
0
0
66
26
16
66
26
16
26
22
12
26
22
12
16
12
11
16
12
11
66
26
16
66
26
16
26
22
12
26
22
12
16
12
11
16
12
11
32. Pendahuluan Material Komposit
BAB 4 Macromechanical Analysis of a Laminate
Laminate Analysis Steps
Qomarul Hadi, ST,MT
Teknik Mesin
Universitas Sriwijaya
Sumber Bacaan
Mechanics of Composite Materials by Kaw
35. 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).
36. Steps
6. Solve the six simultaneous Equations (4.29) to find the mid-plane 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).
37. Step 1: Analysis Procedures for Laminate
Step 1: Find the reduced stiffness matrix [Q] for each ply
ν
ν
-
E
=
Q
12
21
1
11
1 ν
ν
E
ν
=
Q
12
21
2
12
12
1
ν
ν
E
=
Q
12
21
2
22
1 G
=
Q 12
66
38. Step 2: Analysis Procedures for Laminate
c
s
Q
+
Q
+
s
Q
+
c
Q
=
Q
2
2
66
12
4
22
4
11
11
)
2
(
2
)
(
)
4
( 4
4
12
2
2
66
22
11
12 s
+
c
Q
+
c
s
Q
Q
+
Q
=
Q
c
s
Q
Q
Q
s
c
Q
Q
Q
=
Q
3
66
12
22
3
66
12
11
16
)
2
(
)
2
(
c
s
Q
+
Q
+
c
Q
+
s
Q
=
Q
2
2
66
12
4
22
4
11
22
)
2
(
2
s
c
Q
Q
Q
cs
Q
Q
Q
=
Q
3
66
12
22
3
66
12
11
26
)
2
(
)
2
(
)
(
)
2
2
( 4
4
66
2
2
66
12
22
11
66 c
+
s
Q
+
c
s
Q
Q
Q
+
Q
=
Q
Step 2: Find the transformed stiffness matrix [Q] using the
reduced stiffness matrix [Q] and the angle of the ply.
39. Step 3: Analysis Procedures for Laminates
Step 3: Find the coordinate of the top and bottom surface of
each ply.
hk-1
hk
hn
h2
h1
h0
Mid-Plane
1
2
3
n
k-1
k
k+1
h3
z
h/2
tk
hn-1
h/2
Gambar 4.6
Coordinate locations of plies in the laminate.
40. Step 4: Analysis Procedures for Laminates
Step 4: Find three stiffness matrices [A], [B], and [D]
6
2
1
6
2
1
)
(
)]
[( 1
1
,
,
; j =
,
,
, i =
h
-
h
Q
=
A k -
k
k
ij
n
k =
ij
6
2
1
6
2
1
)
(
)]
[(
2
1 2
1
2
1
,
,
; j =
,
,
, i =
h
-
h
Q
=
B k -
k
k
ij
n
k =
ij
6
2
1
6
2
1
),
(
)]
[(
3
1 3
1
3
1
,
,
; j =
,
,
i =
h
-
h
Q
=
D k -
k
k
ij
n
k =
ij
41. Step 5: Analysis Procedure for Laminates
Step 5: Substitute the three stiffness matrices [A], [B], and [D]
and the applied forces and moments.
κ
κ
κ
γ
ε
ε
D
D
D
B
B
B
D
D
D
B
B
B
D
D
D
B
B
B
B
B
B
A
A
A
B
B
B
A
A
A
B
B
B
A
A
A
=
M
M
M
N
N
N
xy
y
x
xy
y
x
xy
y
x
xy
y
x
0
0
0
66
26
16
66
26
16
26
22
12
26
22
12
16
12
11
16
12
11
66
26
16
66
26
16
26
22
12
26
22
12
16
12
11
16
12
11
42. Step 6: Analysis Procedures for Laminates
Step 6: Solve the six simultaneous equations to find the
midplane strains and curvatures.
κ
κ
κ
γ
ε
ε
D
D
D
B
B
B
D
D
D
B
B
B
D
D
D
B
B
B
B
B
B
A
A
A
B
B
B
A
A
A
B
B
B
A
A
A
=
M
M
M
N
N
N
xy
y
x
xy
y
x
xy
y
x
xy
y
x
0
0
0
66
26
16
66
26
16
26
22
12
26
22
12
16
12
11
16
12
11
66
26
16
66
26
16
26
22
12
26
22
12
16
12
11
16
12
11
43. Step 7: Analysis Procedures for Laminates
Step 7: Find the global strains in each ply.
xy
y
x
0
xy
0
y
0
x
xy
y
x
z
+
=
44. Step 8: Analysis Procedure for Laminates
Step 8: Find the global stresses using the stress-strain
equation.
xy
y
x
66
26
16
26
22
12
16
12
11
xy
y
x
Q
Q
Q
Q
Q
Q
Q
Q
Q
=
45. Analysis Procedures for Laminated Composites
Step 9: Find the local strains using the transformation equation.
γ
ε
ε
R
T
R
=
γ
ε
ε
xy
y
x
1
12
2
1
]
[
]
[
]
[
2
0
0
0
1
0
0
0
1
]
[ =
R
s
-
c
sc
-sc
sc
-
c
s
sc
s
c
=
T
2
2
2
2
2
2
2
2
]
[
)
cos(
=
c
)
sin(
=
s
46. Step 10: Analysis Procedures for Laminates
Step 10: Find the local stresses using the transformation
equation.
τ
σ
σ
T
=
xy
y
x
12
2
1
1
]
[
s
c
sc
sc
sc
c
s
sc
s
c
=
T
2
2
2
2
2
2
1
2
2
]
[
)
cos(
=
c
)
sin(
s
48. Pendahuluan Material Komposit
BAB 4 Macromechanical Analysis of a Laminate
Laminate Analysis: Example
Qomarul Hadi, ST,MT
Teknik Mesin
Universitas Sriwijaya
Sumber Bacaan
Mechanics of Composite Materials by Kaw
50. 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 = 7.5mm
z
z = -7.5mm
Gambar 4.7
Thickness and coordinate locations
of the three-ply laminate.
51. Solution
A) The reduced stiffness matrix for the Oo Graphite/Epoxy ply
is
0
Pa
)
10
(
7.17
0
0
0
10.35
2.897
0
2.897
181.8
=
[Q] 9
53. 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
Gambar 4.7
Thickness and coordinate locations
of the three-ply laminate.
54. Calculating [A] matrix
(-0.0075)]
-
[(-0.0025)
)
10
(
7.17
0
0
0
10.35
2.897
0
2.897
181.8
=
[A] 9
(-0.0025)]
-
[0.0025
)
10
(
36.74
20.05
54.19
20.05
23.65
32.46
54.19
32.46
109.4
+ 9
0.0025]
-
[0.0075
)
10
(
46.59
42.87
-
42.87
-
42.87
-
56.66
42.32
42.87
-
42.32
56.66
+ 9
)
h
-
h
(
]
Q
[
=
A 1
-
k
k
k
ij
3
1
=
k
ij
)
(
]
[ 1
3
1
h
-
h
Q
=
A k -
k
k
ij
k =
ij
70. 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 = 7.5mm
z
z = -7.5mm
Gambar 4.7
Thickness and coordinate locations of the three-ply laminate.
71. 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
=
100
1000
223.2
%
53.15
=
100
1000
531.5
%
24.52
=
100
1000
245.2