ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 5 potx
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 5 potx
... Failure of the aluminum–boron plastic in static and cyclic tensile
loading. Mechanics of Composite Materials, 20(1), 44 52 .
Mileiko, S.T. (1982). Mechanics of metal-matrix fibrous composites. In Mechanics ... on Composite Materials (ICCM-10), Vol. 4, Characterization and Ceramic Matrix
Composites
, Canada, 19 95, pp. 171–178.
Chapter 4. Mechanics of a composite layer...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 7 potx
... angle.
210 Advanced mechanics of composite materials
0
0.03
3.6
0.19
0 .58
0.68
0.81
0. 95
0.03
0.014
0. 05
0.1
0.1 75
0.3
0.49
0.74
0.92
0.99
0.06
0.1
0.16
0.26
0.4
0 .59
0.79
0.034
0.94
0.9 95
0
0.1
0.2
0.3
0.4
0 .5
0.6
0.7
0.8
0.9
1
0123 456 78910
F
l
c
1.8
3. 15
0.68 ... c
2
= 5. 2 × 10
−8
(MPa)
−3
.
Chapter 4. Mechanics of a composite layer 2 05
Fig. 4 .52 . Mic...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 9 potx
... D
22
=
h
3
24
E
1
+E
2
,D
12
=
h
3
12
E
1
ν
12
,D
44
=
h
3
12
G
12
x
y
z
h/2
h/2
Fig. 5. 24. An antisymmetric cross-ply laminate.
298 Advanced mechanics of composite materials
p, MPa
1.0
1 .5
− 150
.
10
5
e
x
.
10
5
e
x
.
10
5
e
x
.
10
5
e
y
.
10
5
e
y
.
10
5
e
y
−100 50
0
50
... Fig. 5. 18. From Eqs. (5. 20), we have
two possible constitutive equations, i....
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 1 ppt
... (GPa)
Specific
gravity
Wood 15 20 160 23 1 .5
Bamboo 15 30 55 0 36 0.8
Jute 10 50 58 0 22 1 .5
Cotton 15 40 54 0 28 1 .5
Wool 75 170 5. 9 1.32
Coir 10–20 250 5. 5 1 .5
Bagasse 25 180 9 1. 25
Rice 5 15 100 6 1.24
Natural silk 15 ... Layers and Bulk Materials 243
4.9. References 253
Chapter 5. Mechanics of Laminates 255
5. 1. Stiffness Coefficients of a Generalized Aniso...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 2 ppsx
... form
{
σ
}
=
[
S
]
{
ε
}
(2.46)
where
{
σ
}
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
σ
x
σ
y
σ
z
τ
xy
τ
xz
τ
yz
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎭
{
ε
}
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
ε
x
ε
y
ε
z
γ
xy
γ
xz
γ
yz
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎭
[
S
]
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
S
11
S
12
S
13
S
14
S
15
S
16
S
21
S
22
S
23
S
24
S
25
S
26
S
31
S
32
S
33
S
34
S
35
S
36
S
41
S
42
S
43
S
44
S
45
S
46
S
51
S
52
S...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 3 ppt
... under in-plane loading.
80 Advanced mechanics of composite materials
0
0. 25
0 .5
0. 75
1
1. 25
1 .5
0 5 10 15 20 25 30 35 40 45 50
s
n
x
1
0
2
l
i
3,4
s
Fig. 3.19. Distribution of normal stresses along ... S
n
from
70 Advanced mechanics of composite materials
Table 3.2
Strength of bundles consisting of fibers of different strengths.
Fiber number Bundle n...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 4 ppt
... frame.
Chapter 3. Mechanics of a unidirectional ply 111
0.3
0.4
0 .5
0.6
0.7
0.8
0.9
1
0.3 0. 35 0.4 0. 45 0 .5
m
m
u
m
Fig. 3 .51 . Dependence of parameter µ
m
on the matrix Poisson’s ratio.
Fig. 3 .52 . Test ... Glass–
epoxy
Carbon–
epoxy
Carbon–
PEEK
Aramid–
epoxy
Boron–
epoxy
Boron–
Al
Carbon–
Carbon
Al
2
O
3
–
Al
Fiber volume fraction, v
f
0. 65 0.62 0.61 0.6 0 .5 0 .5 0.6 0...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 6 ppt
... for 45
◦
off-axis tension of a
two-matrix unidirectional composite.
174 Advanced mechanics of composite materials
Consider the first set of inequalities in Eqs. (4. 85) and assume that the first of ... 2160 2 250 2630 1420 2000 890 1100
σ
(2)
x
2 25 690 11 25 590 840 400 100 52 0
ε
1
3 1.43 1 .5 2.63 0.62 0 .50 0.47 0.27
ε
2
0.31 0. 45 0. 75 0.2 0.37 0.1 0. 05 0.13
ε
1...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 8 doc
... xz-plane, Eqs. (5. 15) yield
V
x
= S
55
γ
x
or V
x
= S
55
γ
x
(5. 20)
in which, in accordance with Eq. (5. 17)
S
55
=
s
−e
G
xz
dz (5. 21)
whereas Eq. (5. 19) yields
S
55
=
h
2
s
−e
dz
G
xz
(5. 22)
If ... Eqs. (5. 16) are not the inverse form of Eqs. (5. 15) . Indeed,
solving Eqs. (5. 16), using Eqs. (5. 18), and taking into account that
a
55
= A
66
,a
56
=−A
56
,a
66
=...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 10 ppsx
... τ
12
= τ
45
12
,weget
σ
1
=
1
2
σ
45
1
+σ
45
2
−τ
45
12
σ
2
=
1
2
σ
45
1
+σ
45
2
+τ
45
12
τ
12
=
1
2
σ
45
1
−σ
45
2
(6.24)
Substitution in Eq. (6.23) yields
F
σ
45
1
,σ
45
2
,τ
45
12
=
1
4
2
σ
2
0
+
1
τ
2
0
σ
45
1
2
+
σ
45
2
2
+
1
2
2
σ
2
0
−
1
τ
2
0
σ
45
1
σ
45
2
+
2
σ
2
0
τ
45
12
2
= ... yields
F
σ
45
1
,σ
45
2...