Classical Lamination Theory
NamHuuTran
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Apr 20, 2023
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About This Presentation
Classical Lamination Theory
Slide Content
PendahuluanMaterial Komposit
BAB 4 MacromechanicalAnalysis of a Laminate
Classical Lamination Theory
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
BAB 4 MacromechanicalAnalysis of a Laminate
Classical Lamination Theory
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
Laminate Stacking SequenceFiber Direction
x
z
y
Gambar 4.1
Schematic of a lamina
x
z
y
Gambar 4.1
Schematic of a lamina
Laminate Behavior
•Modulus Elastis
•The Stacking Position
•Thickness
•Angles of Orientation
•Coefficients of Thermal Expansion
•Coefficients of Moisture Expansion
•Modulus Elastis
•The Stacking Position
•Thickness
•Angles of Orientation
•Coefficients of Thermal Expansion
•Coefficients of Moisture Expansion
x
P
PP
P
z
x
(a)
z
(c)
x
z
M M
(b)
x
MM 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
P
PP
P
z
x
(a)
z
(c)
x
z
M M
(b)
x
MM 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
Types of loads allowed in CLTanalysisx
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
N
x= normal force resultant in the x direction (per unit length)
N
y= normal force resultant in the y direction (per unit length)
N
xy= shear force resultant (per unit length)
Gambar 4.3
Resultant forces and moments on a
laminate.
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
N
x= normal force resultant in the x direction (per unit length)
N
y= normal force resultant in the y direction (per unit length)
N
xy= shear force resultant (per unit length)
Gambar 4.3
Resultant forces and moments on a
laminate.
Gambar4.3x
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
M
x= bending moment resultant in the yz plane (per unit length)
M
y= bending moment resultant in the xz plane (per unit length)
M
xy= twisting moment resultant (per unit length)
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
M
x= bending moment resultant in the yz plane (per unit length)
M
y= bending moment resultant in the xz plane (per unit length)
M
xy= twisting moment resultant (per unit length)
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.
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.
Gambar4.4
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.
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.
Global Strains in a Laminate
yx
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
00
0
0
2 .
κ
κ
κ
+ z
γ
ε
ε
xy
y
x
xy
y
x
0
0
0
yx
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
00
0
0
2 .
κ
κ
κ
+ z
γ
ε
ε
xy
y
x
xy
y
x
0
0
0
Gambar4.5z
Mid-Plane
Strain VariationStress VariationLaminate
Gambar 4.5
Strain and stress variation through the thickness of the laminate.
Mid-Plane
Strain VariationStress VariationLaminate
Gambar 4.5
Strain and stress variation through the thickness of the laminate.
END
PendahuluanMaterial Komposit
BAB 4 MacromechanicalAnalysis of a Laminate
Relating Loads to Midplane
Strains/Curvatures
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
BAB 4 MacromechanicalAnalysis of a Laminate
Relating Loads to Midplane
Strains/Curvatures
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
Laminate Stacking SequenceFiber Direction
x
z
y
Gambar 4.1
Schematic of a lamina
x
z
y
Gambar 4.1
Schematic of a lamina
Types of loads allowed in CLT
analysisx
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
N
x= normal force resultant in the x direction (per unit length)
N
y= normal force resultant in the y direction (per unit length)
N
xy= shear force resultant (per unit length)
Gambar 4.3
Resultant forces and moments on a
laminate.
analysisx
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b)
N
x= normal force resultant in the x direction (per unit length)
N
y= normal force resultant in the y direction (per unit length)
N
xy= shear force resultant (per unit length)
Gambar 4.3
Resultant forces and moments on a
laminate.
x
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b) M
x= bending moment resultant in the yz plane (per unit length)
M
y= bending moment resultant in the xz plane (per unit length)
M
xy= twisting moment resultant (per unit length)
Types of loads allowed in CLT
analysis
y
z
Ny
Nx
Nxy
Nyx
(a)
y
x
z
My
Myx
Mxy
Mx
(b) M
x= bending moment resultant in the yz plane (per unit length)
M
y= bending moment resultant in the xz plane (per unit length)
M
xy= twisting moment resultant (per unit length)
Types of loads allowed in CLT
analysis
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.
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.
Stresses in a Lamina in a Laminatek
xy
y
x
k
k
xy
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
xy
y
x
k
k
xy
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
x
h/
-h/
xy
y
x
2
2 dz,
τ
σ
σ
=
xy
y
x
k
h
h
n
k=
k
k-
1
1
τ
σ
σ
=
N
N
N
xy
y
x
h/
-h/
xy
y
x
2
2 dz,
τ
σ
σ
=
xy
y
x
k
h
h
n
k=
k
k-
1
1
Forces & Midplane
Strains/Curvatures dz
γ
ε
ε
QQQ
QQQ
QQQ
=
N
N
N
xy
y
x
k
h
h
n
k =
xy
y
x
k
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
Strains/Curvatures dz
γ
ε
ε
QQQ
QQQ
QQQ
=
N
N
N
xy
y
x
k
h
h
n
k =
xy
y
x
k
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 & Midplane
Strains/Curvatures
0
xy
0
y
0
x
h
h
662616
262212
161211
k
n
1 = k
xy
y
x
dz
QQQ
QQQ
QQQ
=
N
N
N
k
1 - k
xy
y
x
h
h
662616
262212
161211
k
n
1 = k
dz z
QQQ
QQQ
QQQ
+
k
1 k
Strains/Curvatures
0
xy
0
y
0
x
h
h
662616
262212
161211
k
n
1 = k
xy
y
x
dz
QQQ
QQQ
QQQ
=
N
N
N
k
1 - k
xy
y
x
h
h
662616
262212
161211
k
n
1 = k
dz z
QQQ
QQQ
QQQ
+
k
1 k
Integrating terms,hh dz =
k - k
h
h
k
k -
)(
1
1
,h h zdz =
k - k
h
h
k
k -
)(
2
1
2
1
2
1
k - k
h
h
k
k -
)(
1
1
,h h zdz =
k - k
h
h
k
k -
)(
2
1
2
1
2
1
Forces & Midplane
Strains/Curvatures
κ
κ
κ
BBB
BBB
BBB
+
γ
ε
ε
AAA
AAA
AAA
=
N
N
N
xy
y
x
xy
y
x
xy
y
x
662616
262212
161211
0
0
0
662616
262212
161211 ,,,; j = ,,, i = h - h Q = A k - kkij
n
k =
ij 621621)()][(
1
1
621621)()][(
2
1
2
1
2
1
,,; j = ,,, i = h - h Q = B k - kkij
n
k =
ij
Strains/Curvatures
κ
κ
κ
BBB
BBB
BBB
+
γ
ε
ε
AAA
AAA
AAA
=
N
N
N
xy
y
x
xy
y
x
xy
y
x
662616
262212
161211
0
0
0
662616
262212
161211 ,,,; j = ,,, i = h - h Q = A k - kkij
n
k =
ij 621621)()][(
1
1
621621)()][(
2
1
2
1
2
1
,,; j = ,,, i = h - h Q = B k - kkij
n
k =
ij
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.
[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.
Stresses in a Lamina in a Laminatek
xy
y
x
k
k
xy
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
xy
y
x
k
k
xy
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
Moments and Stresses zdz,
τ
σ
σ
=
M
M
M
xy
y
x
h/
-h/
xy
y
x
2
2 zdz,
τ
σ
σ
=
xy
y
x
k
h
h
n
k=
k
k-
1
1
τ
σ
σ
=
M
M
M
xy
y
x
h/
-h/
xy
y
x
2
2 zdz,
τ
σ
σ
=
xy
y
x
k
h
h
n
k=
k
k-
1
1
Moments & Midplane
Strains/Curvatures zdz
γ
ε
ε
QQQ
QQQ
QQQ
=
M
M
M
xy
y
x
k
h
h
n
k =
xy
y
x
k
k-
0
0
0
662616
262212
161211
1
1 dz z
κ
κ
κ
QQQ
QQQ
QQQ
+
xy
y
x
k
h
h
n
k =
k
k
2
662616
262212
161211
1
1
Strains/Curvatures zdz
γ
ε
ε
QQQ
QQQ
QQQ
=
M
M
M
xy
y
x
k
h
h
n
k =
xy
y
x
k
k-
0
0
0
662616
262212
161211
1
1 dz z
κ
κ
κ
QQQ
QQQ
QQQ
+
xy
y
x
k
h
h
n
k =
k
k
2
662616
262212
161211
1
1
0
xy
0
y
0
x
h
h
662616
262212
161211
k
n
1 = k
xy
y
x
zdz
QQQ
QQQ
QQQ
=
M
M
M
k
1 - k
xy
y
x
h
h
662616
262212
161211
k
n
1 = k
dz z
QQQ
QQQ
QQQ
+
k
1 k
2
0
xy
0
y
0
x
h
h
662616
262212
161211
k
n
1 = k
xy
y
x
zdz
QQQ
QQQ
QQQ
=
M
M
M
k
1 - k
xy
y
x
h
h
662616
262212
161211
k
n
1 = k
dz z
QQQ
QQQ
QQQ
+
k
1 k
2
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),()][(
3
1
3
1
3
1
621621)()][(
2
1
2
1
2
1
,,; j = ,,i = , h - h Q = B k - kkij
n
k =
ij
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),()][(
3
1
3
1
3
1
621621)()][(
2
1
2
1
2
1
,,; j = ,,i = , h - h Q = B k - kkij
n
k =
ij
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.
[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.
Forces, Moments, MidplaneStrains,
MidplaneCurvatures
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
MidplaneCurvatures
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
END
PendahuluanMaterial Komposit
BAB 4 MacromechanicalAnalysis of a Laminate
Laminate Analysis Steps
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
BAB 4 MacromechanicalAnalysis of a Laminate
Laminate Analysis Steps
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
Laminate Stacking SequenceFiber Direction
x
z
y
Gambar 4.1
Schematic of a lamina
x
z
y
Gambar 4.1
Schematic of a lamina
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
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
Steps1. 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).
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).
Steps6. 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).
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).
Step 1: Analysis Procedures for Laminate
Step 1: Find the reduced stiffness matrix [Q] for each plyν ν -
E
= Q
1221
1
11
1 νν
E ν
= Q
1221
212
12
1 ν ν
E
= Q
1221
2
22
1 G = Q
1266
Step 1: Find the reduced stiffness matrix [Q] for each plyν ν -
E
= Q
1221
1
11
1 νν
E ν
= Q
1221
212
12
1 ν ν
E
= Q
1221
2
22
1 G = Q
1266
Step 2: Analysis Procedures for Laminate csQ+Q+sQ+cQ = Q
22
6612
4
22
4
1111
)2(2 )()4(
44
12
22
66221112
s+cQ+csQQ+Q = Q csQQQscQQQ = Q
3
661222
3
66121116
)2()2( csQ+Q+ cQ+sQ = Q
22
6612
4
22
4
1122
)2(2 scQQQcsQQQ = Q
3
661222
3
66121126
)2()2( )()22(
44
66
22
6612221166
c + sQ+csQQQ+Q = Q
Step 2: Find the transformed stiffness matrix [Q] using the
reduced stiffness matrix [Q] and the angle of the ply.
22
6612
4
22
4
1111
)2(2 )()4(
44
12
22
66221112
s+cQ+csQQ+Q = Q csQQQscQQQ = Q
3
661222
3
66121116
)2()2( csQ+Q+ cQ+sQ = Q
22
6612
4
22
4
1122
)2(2 scQQQcsQQQ = Q
3
661222
3
66121126
)2()2( )()22(
44
66
22
6612221166
c + sQ+csQQQ+Q = Q
Step 2: Find the transformed stiffness matrix [Q] using the
reduced stiffness matrix [Q] and the angle of the ply.
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.
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.
Step 4: Analysis Procedures for Laminates
Step 4: Find three stiffness matrices [A], [B], and [D]621621)()][(
1
1
,,; j = ,,, i = h - h Q = A k - kkij
n
k =
ij 621621)()][(
2
1
2
1
2
1
,,; j = ,,, i = h - h Q = B k - kkij
n
k =
ij 621621),()][(
3
1
3
1
3
1
, , ; j = , , i = h - h Q = D k - kkij
n
k =
ij
Step 4: Find three stiffness matrices [A], [B], and [D]621621)()][(
1
1
,,; j = ,,, i = h - h Q = A k - kkij
n
k =
ij 621621)()][(
2
1
2
1
2
1
,,; j = ,,, i = h - h Q = B k - kkij
n
k =
ij 621621),()][(
3
1
3
1
3
1
, , ; j = , , i = h - h Q = D k - kkij
n
k =
ij
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
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
Step 5: Substitute the three stiffness matrices [A], [B], and [D]
and the applied forces and moments.
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
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
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
Step 6: Solve the six simultaneous equations to find the
midplane strains and 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
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 + =
Step 7: Find the global strains in each ply.
xy
y
x
0
xy
0
y
0
x
xy
y
x
z + =
Step 8: Analysis Procedure for Laminates
Step 8: Find the global stresses using the stress-strain
equation.
xy
y
x
662616
262212
161211
xy
y
x
QQQ
QQQ
QQQ
=
Step 8: Find the global stresses using the stress-strain
equation.
xy
y
x
662616
262212
161211
xy
y
x
QQQ
QQQ
QQQ
=
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
][][][
200
010
001
][ = R
s-csc-sc
sc-cs
scsc
= T
22
22
22
2
2
][ )cos( = c )sin( = s
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
][][][
200
010
001
][ = R
s-csc-sc
sc-cs
scsc
= T
22
22
22
2
2
][ )cos( = c )sin( = s
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
][
scscsc
sccs
scsc
=T
22
22
22
1
2
2
][ )cos(= c )sin(s
Step 10: Find the local stresses using the transformation
equation.
τ
σ
σ
T =
xy
y
x
12
2
1
1
][
scscsc
sccs
scsc
=T
22
22
22
1
2
2
][ )cos(= c )sin(s
END
PendahuluanMaterial Komposit
BAB 4 MacromechanicalAnalysis of a Laminate
Laminate Analysis: Example
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
BAB 4 MacromechanicalAnalysis of a Laminate
Laminate Analysis: Example
QomarulHadi, ST,MT
TeknikMesin
UniversitasSriwijaya
SumberBacaan
Mechanics of Composite Materials by Kaw
Laminate Stacking SequenceFiber Direction
x
z
y
Gambar 4.1
Schematic of a lamina
x
z
y
Gambar 4.1
Schematic of a lamina
Problem
A [0/30/-45] Graphite/Epoxy
laminate is subjected to a load of
N
x= N
y= 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 30
0
ply.
d)percentage of load N
xtaken by
each ply.0
o
30
o
-45
o
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.
A [0/30/-45] Graphite/Epoxy
laminate is subjected to a load of
N
x= N
y= 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 30
0
ply.
d)percentage of load N
xtaken by
each ply.0
o
30
o
-45
o
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.
Solution
A)ThereducedstiffnessmatrixfortheO
o
Graphite/Epoxyply
is0 Pa)10(
7.1700
010.352.897
02.897181.8
= [Q]
9
A)ThereducedstiffnessmatrixfortheO
o
Graphite/Epoxyply
is0 Pa)10(
7.1700
010.352.897
02.897181.8
= [Q]
9
Pa)10(
7.1700
010.352.897
02.897181.8
= ]Q[
9
0
Pa)10(
36.7420.0554.19
20.0523.6532.46
54.1932.46109.4
= ]Q[
9
30
Pa)10(
46.5942.87-42.87-
42.87-56.6642.32
42.87-42.3256.66
= ]Q[
9
45-
Matrices Qbar Untuk Laminas
7.1700
010.352.897
02.897181.8
= ]Q[
9
0
Pa)10(
36.7420.0554.19
20.0523.6532.46
54.1932.46109.4
= ]Q[
9
30
Pa)10(
46.5942.87-42.87-
42.87-56.6642.32
42.87-42.3256.66
= ]Q[
9
45-
Matrices Qbar Untuk Laminas
Thetotalthicknessofthelaminateis
h=(0.005)(3)=0.015m.
h
0=-0.0075m
h
1=-0.0025m
h
2=0.0025m
h
3=0.0075m0
o
30
o
-45
o
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.
h=(0.005)(3)=0.015m.
h
0=-0.0075m
h
1=-0.0025m
h
2=0.0025m
h
3=0.0075m0
o
30
o
-45
o
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.
Calculating [A] matrix(-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 )(][
1
3
1
h - h Q = A k - kkij
k =
ij
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 )(][
1
3
1
h - h Q = A k - kkij
k =
ij
The [A] matrixm- 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
)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
Calculating the [B] Matrix)h - h( ]Q[
2
1
= B
2
1 -k
2
kkij
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
2
1
= B
2
1 -k
2
kkij
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
The [B] Matrix
2
566
665
656
108559100721100721
100721101581108559
100721108559101293
m Pa
...
...
...
[B] =
2
566
665
656
108559100721100721
100721101581108559
100721108559101293
m Pa
...
...
...
[B] =
Calculating the [D] matrix)h - h( ]Q[
3
1
= D
3
1 -k
3
kkij
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
. - (.
...
...
...
+
3
1
= D
3
1 -k
3
kkij
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
. - (.
...
...
...
+
The [D] matrix
3
333
333
334
m- Pa
107.663105.596105.240
105.596109.320106.461
105.240106.461103.343
= [D]
3
333
333
334
m- Pa
107.663105.596105.240
105.596109.320106.461
105.240106.461103.343
= [D]
B)SincetheappliedloadisN
x=N
y=1000N/m,themid-
planestrainsandcurvaturescanbefoundbysolvingthe
followingsetofsimultaneouslinearequations
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
x=N
y=1000N/m,themid-
planestrainsandcurvaturescanbefoundbysolvingthe
followingsetofsimultaneouslinearequations
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
Mid-plane strains and
curvatures/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
curvatures/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
C)Thestrainsandstressesatthetopsurfaceofthe30
0
plyarefound
asfollows.Thetopsurfaceofthe30
0
plyislocatedatz=h
1=-
0.0025m.
)(.
)(.-
)(.
) . + (-
) (.-
) (.
) (.
=
γ
ε
ε
-
-
-
-
-
-
xy
y
x
, top
101014
102853
109712
00250
105987
104923
101233
4
4
5
7
6
7
30
0 m/m
)(.-
)(.
)(.
=
-
-
-
107851
103134
103802
6
6
7 0
o
30
o
-45
o
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.
0
plyarefound
asfollows.Thetopsurfaceofthe30
0
plyislocatedatz=h
1=-
0.0025m.
)(.
)(.-
)(.
) . + (-
) (.-
) (.
) (.
=
γ
ε
ε
-
-
-
-
-
-
xy
y
x
, top
101014
102853
109712
00250
105987
104923
101233
4
4
5
7
6
7
30
0 m/m
)(.-
)(.
)(.
=
-
-
-
107851
103134
103802
6
6
7 0
o
30
o
-45
o
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.
Global strains (m/m)
xy
Ply # Position ε
x
ε
y
1(0
0
) 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(30
0
) 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(-45
0
) 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
Ply # Position ε
x
ε
y
1(0
0
) 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(30
0
) 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(-45
0
) 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
)
)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,30
0 Pa
)103.381(
)107.391(
)106.930(
=
4
4
4
Global stresses in 30
o
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,30
0 Pa
)103.381(
)107.391(
)106.930(
=
4
4
4
Global stresses in 30
o
ply
Global stresses (Pa)
Ply # Position σ
x
σ
y
τ
xy
1(0
0
) Top
Middle
Bottom
3.351 (10
4
)
4.464 (10
4
)
5.577 (10
4
)
6.188 (10
4
)
5.359 (10
4
)
4.531 (10
4
)
-2.750 (10
4
)
-2.015 (10
4
)
-1.280 (10
4
)
2(30
0
) Top
Middle
Bottom
6.930 (10
4
)
1.063 (10
5
)
1.434 (10
5
)
7.391 (10
4
)
7.747 (10
4
)
8.102 (10
4
)
3.381 (10
4
)
5.903 (10
4
)
8.426 (10
4
)
3(-45
0
) Top
Middle
Bottom
1.235 (10
5
)
4.903 (10
4
)
-2.547 (10
4
)
1.563 (10
5
)
6.894 (10
4
)
-1.840 (10
4
)
-1.187 (10
5
)
-3.888 (10
4
)
4.091 (10
4
)
Ply # Position σ
x
σ
y
τ
xy
1(0
0
) Top
Middle
Bottom
3.351 (10
4
)
4.464 (10
4
)
5.577 (10
4
)
6.188 (10
4
)
5.359 (10
4
)
4.531 (10
4
)
-2.750 (10
4
)
-2.015 (10
4
)
-1.280 (10
4
)
2(30
0
) Top
Middle
Bottom
6.930 (10
4
)
1.063 (10
5
)
1.434 (10
5
)
7.391 (10
4
)
7.747 (10
4
)
8.102 (10
4
)
3.381 (10
4
)
5.903 (10
4
)
8.426 (10
4
)
3(-45
0
) Top
Middle
Bottom
1.235 (10
5
)
4.903 (10
4
)
-2.547 (10
4
)
1.563 (10
5
)
6.894 (10
4
)
-1.840 (10
4
)
-1.187 (10
5
)
-3.888 (10
4
)
4.091 (10
4
)
Thelocalstrainsandlocalstressasinthe30
0
plyatthe
topsurfacearefoundusingtransformationequationsas
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
0
plyatthe
topsurfacearefoundusingtransformationequationsas
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 (m/m)
Ply # Position ε
1
ε
2
γ
12
1(0
0
) 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(30
0
) 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(-45
0
)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
)
Ply # Position ε
1
ε
2
γ
12
1(0
0
) 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(30
0
) 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(-45
0
)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
Local stresses in 30
o
ply
)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 30
o
ply
Local stresses (Pa)
Ply # Position σ
1
σ
2
τ
12
1(0
0
) Top
Middle
Bottom
3.351 (10
4
)
4.464 (10
4
)
5.577 (10
4
)
6.188 (10
4
)
5.359(10
4
)
4.531 (10
4
)
-2.750 (10
4
)
-2.015 (10
4
)
-1.280 (10
4
)
2(30
0
) Top
Middle
Bottom
9.973 (10
4
)
1.502 (10
5
)
2.007 (10
5
)
4.348 (10
4
)
3.356 (10
4
)
2.364 (10
4
)
1.890 (10
4
)
1.702 (10
4
)
1.513 (10
4
)
3(-45
0
) Top
Middle
Bottom
2.586 (10
5
)
9.786 (10
4
)
-6.285 (10
4
)
2.123 (10
4
)
2.010 (10
4
)
1.898 (10
4
)
-1.638 (10
4
)
-9.954 (10
3
)
-3.533 (10
3
)
Ply # Position σ
1
σ
2
τ
12
1(0
0
) Top
Middle
Bottom
3.351 (10
4
)
4.464 (10
4
)
5.577 (10
4
)
6.188 (10
4
)
5.359(10
4
)
4.531 (10
4
)
-2.750 (10
4
)
-2.015 (10
4
)
-1.280 (10
4
)
2(30
0
) Top
Middle
Bottom
9.973 (10
4
)
1.502 (10
5
)
2.007 (10
5
)
4.348 (10
4
)
3.356 (10
4
)
2.364 (10
4
)
1.890 (10
4
)
1.702 (10
4
)
1.513 (10
4
)
3(-45
0
) Top
Middle
Bottom
2.586 (10
5
)
9.786 (10
4
)
-6.285 (10
4
)
2.123 (10
4
)
2.010 (10
4
)
1.898 (10
4
)
-1.638 (10
4
)
-9.954 (10
3
)
-3.533 (10
3
)
D)Portionofloadtakenbyeachply
PortionofloadN
xtakenby0
0
ply=4.464(10
4
)(5)(10
-3
)=223.2N/m
PortionofloadN
xtakenby30
0
ply=1.063(10
5)
(5)(10
-3
)=531.5N/m
PortionofloadN
xtakenby-45
0
ply=4.903(10
4
)(5)(10
-3
)=245.2N/m
Thesumtotaloftheloadssharedbyeachplyis1000N/m,(223.2+
531.5+245.2)whichistheappliedloadinthex-direction,N
x.0
o
30
o
-45
o
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.
PortionofloadN
xtakenby0
0
ply=4.464(10
4
)(5)(10
-3
)=223.2N/m
PortionofloadN
xtakenby30
0
ply=1.063(10
5)
(5)(10
-3
)=531.5N/m
PortionofloadN
xtakenby-45
0
ply=4.903(10
4
)(5)(10
-3
)=245.2N/m
Thesumtotaloftheloadssharedbyeachplyis1000N/m,(223.2+
531.5+245.2)whichistheappliedloadinthex-direction,N
x.0
o
30
o
-45
o
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.
Percentage of load N
xtaken by 0
0
ply
Percentage of load N
xtaken by 30
0
ply
Percentage of load N
xtaken by -45
0
ply % 22.32 =
100
1000
223.2
% 53.15 =
100
1000
531.5
% 24.52 =
100
1000
245.2
xtaken by 0
0
ply
Percentage of load N
xtaken by 30
0
ply
Percentage of load N
xtaken by -45
0
ply % 22.32 =
100
1000
223.2
% 53.15 =
100
1000
531.5
% 24.52 =
100
1000
245.2
END
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