![Forces & Midplane
Strains/Curvatures
κ
κ
κ
B
B
B
B
B
B
B
B
B
+
γ
ε
ε
A
A
A
A
A
A
A
A
A
=
N
N
N
xy
y
x
xy
y
x
xy
y
x
66
26
16
26
22
12
16
12
11
0
0
0
66
26
16
26
22
12
16
12
11
,
,
,
; j =
,
,
, i =
h
-
h
Q
=
A k -
k
k
ij
n
k =
ij 6
2
1
6
2
1
)
(
)]
[( 1
1
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 ](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-22-320.jpg)
![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.](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-23-320.jpg)
![
κ
κ
κ
D
D
D
D
D
D
D
D
D
+
γ
ε
ε
B
B
B
B
B
B
B
B
B
=
M
M
M
xy
y
x
xy
y
x
xy
y
x
66
26
16
26
22
12
16
12
11
0
0
0
66
26
16
26
22
12
16
12
11
.
,
,
; j =
,
,
i =
h
-
h
Q
=
D k -
k
k
ij
n
k =
ij 6
2
1
6
2
1
),
(
)]
[(
3
1 3
1
3
1
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 ](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-28-320.jpg)
![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.](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-29-320.jpg)
![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).](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-35-320.jpg)
![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](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-37-320.jpg)
![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.](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-38-320.jpg)
![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 ](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-40-320.jpg)
![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](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-41-320.jpg)
![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](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-45-320.jpg)
![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](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-46-320.jpg)
![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.](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-50-320.jpg)
![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
](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-51-320.jpg)
![Pa
)
10
(
7.17
0
0
0
10.35
2.897
0
2.897
181.8
=
]
Q
[ 9
0
Pa
)
10
(
36.74
20.05
54.19
20.05
23.65
32.46
54.19
32.46
109.4
=
]
Q
[ 9
30
Pa
)
10
(
46.59
42.87
-
42.87
-
42.87
-
56.66
42.32
42.87
-
42.32
56.66
=
]
Q
[ 9
45
-
Matrices Qbar Untuk Laminas](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-52-320.jpg)
![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 ](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-54-320.jpg)
![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]
8
8
7
8
8
8
7
8
9
](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-55-320.jpg)
![Calculating the [B] Matrix
)
h
-
h
(
]
Q
[
2
1
=
B
2
1
-
k
2
k
k
ij
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] 2
2
9
)
(-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
+ 2
2
9
]
)
(0.0025
-
)
[(0.0075
)
10
(
46.59
42.87
42.87
42.87
56.66
42.32
42.87
42.32
56.66
2
1
+ 2
2
9
](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-56-320.jpg)
![The [B] Matrix
2
5
6
6
6
6
5
6
5
6
10
855
9
10
072
1
10
072
1
10
072
1
10
158
1
10
855
9
10
072
1
10
855
9
10
129
3
m
Pa
.
.
.
.
.
.
.
.
.
[B] =
](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-57-320.jpg)
![Calculating the [D] matrix
)
h
-
h
(
]
Q
[
3
1
=
D
3
1
-
k
3
k
k
ij
3
1
=
k
ij
3
3
9
)
0075
0
(
)
0025
0
(
)
10
(
17
7
0
0
0
35
10
897
2
0
897
2
8
181
3
1
.
.
.
.
.
.
.
[D] =
3
3
9
)
0025
0
(
)
0025
0
(
)
10
(
74
36
05
20
19
54
05
20
65
23
46
32
19
54
46
32
4
109
3
1
.
.
.
.
.
.
.
.
.
.
.
+
3
3
9
)
0025
0
)
0075
0
(
)
10
(
59
46
87
42
87
42
87
42
66
56
32
42
87
42
32
42
66
56
3
1
.
- (
.
.
.
.
.
.
.
.
.
.
+
](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-58-320.jpg)
![The [D] matrix
3
3
3
3
3
3
3
3
3
4
m
-
Pa
10
7.663
10
5.596
10
5.240
10
5.596
10
9.320
10
6.461
10
5.240
10
6.461
10
3.343
=
[D]
](https://image.slidesharecdn.com/classicallaminationtheory-230420160949-9e6d4670/85/Classical-Lamination-Theory-59-320.jpg)
Classical Lamination Theory
- 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
- 2. Laminate Stacking Sequence Fiber Direction x z y Gambar 4.1 Schematic of a lamina
- 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.
- 6. Gambar 4.3 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)
- 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.
- 8. Gambar 4.4 Cross-Section after Loading x u0 z z A z A Mid-Plane wo Cross-Section Before Loading h/2 h/2 Gambar 4.4 Gambar showing the relationship between displacements through the thickness of a plate to midplane displacements and curvatures.
- 9. Global Strains in a Laminate y x w y w x w + z x v + y u y v x u = γ ε ε xy y x 0 2 2 0 2 2 0 2 0 0 0 0 2 . κ κ κ + z γ ε ε xy y x xy y x 0 0 0
- 10. Gambar 4.5 z Mid-Plane Strain Variation Stress Variation Laminate Gambar 4.5 Strain and stress variation through the thickness of the laminate.
- 11. END
- 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
- 13. Laminate Stacking Sequence Fiber Direction x z y Gambar 4.1 Schematic of a lamina
- 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
- 16. Stacking Sequence 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.
- 17. Stresses in a Lamina in a Laminate k xy y x k k xy y x γ ε ε Q Q Q Q Q Q Q Q Q = τ σ σ 66 26 16 26 22 12 16 12 11 . κ κ κ Q Q Q Q Q Q Q Q Q + z γ ε ε Q Q Q Q Q Q Q Q Q = xy y x k xy y x k 66 26 16 26 22 12 16 12 11 0 0 0 66 26 16 26 22 12 16 12 11
- 19. Forces & Midplane Strains/Curvatures dz γ ε ε Q Q Q Q Q Q Q Q Q = N N N xy y x k h h n k = xy y x k k- 0 0 0 66 26 16 26 22 12 16 12 11 1 1 z dz κ κ κ Q Q Q Q Q Q Q Q Q + xy y x k h h n k = k k 66 26 16 26 22 12 16 12 11 1 1
- 21. Integrating terms , h h dz = k - k h h k k - ) ( 1 1 , h h zdz = k - k h h k k - ) ( 2 1 2 1 2 1
- 22. Forces & Midplane Strains/Curvatures κ κ κ B B B B B B B B B + γ ε ε A A A A A A A A A = N N N xy y x xy y x xy y x 66 26 16 26 22 12 16 12 11 0 0 0 66 26 16 26 22 12 16 12 11 , , , ; j = , , , i = h - h Q = A k - k k ij n k = ij 6 2 1 6 2 1 ) ( )] [( 1 1 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
- 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.
- 24. Stresses in a Lamina in a Laminate k xy y x k k xy y x γ ε ε Q Q Q Q Q Q Q Q Q = τ σ σ 66 26 16 26 22 12 16 12 11 . κ κ κ Q Q Q Q Q Q Q Q Q + z γ ε ε Q Q Q Q Q Q Q Q Q = xy y x k xy y x k 66 26 16 26 22 12 16 12 11 0 0 0 66 26 16 26 22 12 16 12 11
- 26. Moments & Midplane Strains/Curvatures zdz γ ε ε Q Q Q Q Q Q Q Q Q = M M M xy y x k h h n k = xy y x k k- 0 0 0 66 26 16 26 22 12 16 12 11 1 1 dz z κ κ κ Q Q Q Q Q Q Q Q Q + xy y x k h h n k = k k 2 66 26 16 26 22 12 16 12 11 1 1
- 28. κ κ κ D D D D D D D D D + γ ε ε B B B B B B B B B = M M M xy y x xy y x xy y x 66 26 16 26 22 12 16 12 11 0 0 0 66 26 16 26 22 12 16 12 11 . , , ; j = , , i = h - h Q = D k - k k ij n k = ij 6 2 1 6 2 1 ), ( )] [( 3 1 3 1 3 1 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
- 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
- 31. END
- 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
- 33. Laminate Stacking Sequence Fiber Direction x z y Gambar 4.1 Schematic of a lamina
- 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
- 47. END
- 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
- 49. Laminate Stacking Sequence Fiber Direction x z y Gambar 4.1 Schematic of a lamina
- 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
- 52. Pa ) 10 ( 7.17 0 0 0 10.35 2.897 0 2.897 181.8 = ] Q [ 9 0 Pa ) 10 ( 36.74 20.05 54.19 20.05 23.65 32.46 54.19 32.46 109.4 = ] Q [ 9 30 Pa ) 10 ( 46.59 42.87 - 42.87 - 42.87 - 56.66 42.32 42.87 - 42.32 56.66 = ] Q [ 9 45 - Matrices Qbar Untuk Laminas
- 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
- 56. Calculating the [B] Matrix ) h - h ( ] Q [ 2 1 = B 2 1 - k 2 k k ij 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] 2 2 9 ) (-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 + 2 2 9 ] ) (0.0025 - ) [(0.0075 ) 10 ( 46.59 42.87 42.87 42.87 56.66 42.32 42.87 42.32 56.66 2 1 + 2 2 9
- 57. The [B] Matrix 2 5 6 6 6 6 5 6 5 6 10 855 9 10 072 1 10 072 1 10 072 1 10 158 1 10 855 9 10 072 1 10 855 9 10 129 3 m Pa . . . . . . . . . [B] =
- 58. Calculating the [D] matrix ) h - h ( ] Q [ 3 1 = D 3 1 - k 3 k k ij 3 1 = k ij 3 3 9 ) 0075 0 ( ) 0025 0 ( ) 10 ( 17 7 0 0 0 35 10 897 2 0 897 2 8 181 3 1 . . . . . . . [D] = 3 3 9 ) 0025 0 ( ) 0025 0 ( ) 10 ( 74 36 05 20 19 54 05 20 65 23 46 32 19 54 46 32 4 109 3 1 . . . . . . . . . . . + 3 3 9 ) 0025 0 ) 0075 0 ( ) 10 ( 59 46 87 42 87 42 87 42 66 56 32 42 87 42 32 42 66 56 3 1 . - ( . . . . . . . . . . +
- 59. The [D] matrix 3 3 3 3 3 3 3 3 3 4 m - Pa 10 7.663 10 5.596 10 5.240 10 5.596 10 9.320 10 6.461 10 5.240 10 6.461 10 3.343 = [D]
- 60. 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 3 3 3 5 6 6 3 3 3 6 6 5 3 3 4 6 5 6 5 6 6 8 8 7 6 6 5 8 8 8 6 5 6 7 8 9 10 663 7 10 596 5 10 240 5 10 855 9 10 072 1 10 072 1 10 596 5 10 320 9 10 461 6 10 072 1 10 158 1 10 855 9 10 240 5 10 461 6 10 343 3 10 072 1 10 855 9 10 129 3 10 855 9 10 072 1 10 072 1 10 525 4 10 141 1 10 ( 663 5 10 072 1 10 158 1 10 855 9 10 141 1 10 533 4 10 ( 884 3 10 072 1 10 855 9 10 129 3 10 663 5 10 884 3 10 ( 739 1 0 0 0 0 1000 1000 Setting up the 6x6 matrix
- 62. 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 , top 10 101 4 10 285 3 10 971 2 0025 0 10 598 7 10 492 3 10 123 3 4 4 5 7 6 7 300 m/m ) ( . - ) ( . ) ( . = - - - 10 785 1 10 313 4 10 380 2 6 6 7 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.
- 63. Global strains (m/m) 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)
- 65. Global stresses (Pa) 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)
- 66. The local strains and local stress as in the 300 ply at the top surface are found using transformation equations as 2 )/ 10 1.785( - ) 10 4.313( ) 10 2.380( 0.5000 0.4330 0.4330 - 0.8660 - 0.7500 0.2500 0.8660 0.2500 0.7500 = /2 γ ε ε 6 - 6 - -7 12 2 1 m/m . . . = γ ε ε - - - ) 10 ( 636 2 ) 10 ( 067 4 ) 10 ( 837 4 6 6 7 12 2 1
- 67. Local strains (m/m) 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)
- 69. Local stresses (Pa) 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)
- 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
- 72. END