ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 1 ppt
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 1 ppt
... (GPa)
Specific
gravity
Wood 15 –20 16 0 23 1. 5
Bamboo 15 –30 550 36 0.8
Jute 10 –50 580 22 1. 5
Cotton 15 –40 540 28 1. 5
Wool 75 17 0 5.9 1. 32
Coir 10 –20 250 5.5 1. 5
Bagasse 25 18 0 9 1. 25
Rice 5 15 10 0 6 1. 24
Natural silk 15 ... 12 3
3.6. Composites with High Fiber Fraction 12 7
3.7. Phenomenological Homogeneous Model of a Ply 12 9
3.8. References 13 1
Chapter 4...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 3 ppt
... study
ε
1
=
σ
1
E
1
−ν
12
σ
2
E
2
ε
2
=
σ
2
E
2
−ν
21
σ
1
E
1
(3.58)
γ
12
=
1
G
12
τ
12
The inverse form of these equations is
σ
1
=
E
1
(ε
1
+ν
12
ε
2
)
σ
2
=
E
2
(ε
2
+ν
21
ε
1
) (3.59)
τ
12
= G
12
γ
12
where
E
1, 2
=
E
1, 2
1 −ν
12
ν
21
and the ... Bundle number
12 345
10 .60.70.85 0.90.95
20.80.90.90.85 0.95
31. 01. 21. 11. 00.95...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 4 ppt
... 2 .1 1.55 1. 6 1. 32 2 .1 2.65 1. 75 3.45
Longitudinal modulus,
E
1
(GPa)
60 14 0 14 0 95 210 260 17 0 260
Transverse modulus, E
2
(GPa)
13 11 10 5 .1 19 14 0 19 15 0
Shear modulus, G
12
(GPa)
3.4 5.5 5 .1 ... under constant stress σ
1
= σ
(1)
1
.
As follows from Fig. 3.69, the final strain is
ε
∗
1
=
σ
(1)
1
E
∗
1
110 Advanced mechanics of composite ma...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 6 ppt
... we get
1 +ν
21
E
1
<
1 +ν
12
E
2
≤
1
2G
12
or
1 +ν
21
E
1
<
1
2G
12
Chapter 4. Mechanics of a composite layer 16 3
−300
−250
−200
15 0
10 0
−50
50
10 0
15 0
−2 1. 8 1. 6 1. 4 1. 2 1 −0.8 ... 11 00
σ
(2)
x
225 690 11 25 590 840 400 10 0 520
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
/ε
2
9.7 3...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 13 pptx
... 0
κ
xyT
=
A
14
(A
T
11
A
22
−A
T
22
A
12
) +A
24
A
T
22
A
11
−A
T
11
A
12
−A
T
12
h
A
3
A
44
+
1
4
2A
14
A
24
A
12
−A
2
14
A
22
−A
2
24
A
11
where A = A
11
A
22
−A
2
12
.
Thus, the ... (7.88)–(7. 91) yield
u =
T x
E
1
+E
2
E
1
α
1
+E
2
α
2
+6(E
1
−E
2
)
E
1
E
2
(α
2
−α
1
)
E
2
1
+14 E
1
E
2
−E
2
2
v =
T y
E
1
+E
2
E
1
α
1
+...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 2 ppsx
... i.e.,
cos α =
dx
1
dx
1
+dy
1
dy
1
+dz
1
dz
1
ds
1
ds
1
(2.23)
Here, dx
1
,dy
1
, and dz
1
are specified with Eq. (2 .17 ), ds
1
can be found from Eq. (2 .18 ), and
dx
1
=
1 +
∂u
x
∂x
dx
+
∂u
x
∂y
dy
+
∂u
x
∂z
dz
(x,y,z)
ds
1
= ... form
{
σ
}
=
[
S
]
{
ε
}
(2.46)
where
{
σ
}
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
σ
x
σ
y
σ
z
τ
xy
τ
xz
τ
yz
⎫
⎪
⎪
⎪
⎪...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 5 potx
... (4 .15 ), (4.27)
and (4. 61) –(4.63) yield
ε
1
=a
1
σ
1
+d
1
σ
2
+nσ
n 1
1
R
1
(b
11
σ
1
+c
12
σ
2
)+
1
R
2
2
d
11
σ
2
1
+2e
12
σ
1
σ
2
+e
21
σ
2
2
ε
2
=b
1
σ
2
+d
1
σ
1
+nσ
n 1
1
R
1
(b
22
σ
2
+c
12
σ
1
)+
1
R
2
2
d
22
σ
2
2
+2e
21
σ
2
σ
1
+e
12
σ
2
1
(4.64)
γ
12
=c
1
τ
12
+2nσ
n 1
b
12
R
1
τ
12
where
σ ......
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 7 potx
... C
11
ε
1
+C
12
ε
2
−C
12
ω
2
,σ
2
= C
12
ε
1
+C
22
ε
2
−C
22
ω
2
,
τ
12
= C
44
γ
12
−C
44
ω
12
(4 .15 1)
in which
C
11
=
c
22
c
,C
22
=
c
11
c
,C
44
=
1
c
44
,C
12
=−
c
12
c
,c= c
11
c
22
−c
2
12
Repeating ... +ε
2
)
[ (1 +ε
y
)
2
− (1 + ε
x
)
2
]
1 +ε
2
=
(1 +ε
x
) (1 +ε
y
)
1 +ε
1
sin φ
=
1 +ε
y
1 +ε
1
sin φ, cos φ
=
1 +ε
x
1 +ε
1...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 8 doc
... 4
l
2
x1
l
x2
l
x3
E
1
+
l
2
y1
l
y2
l
y3
E
2
+
l
2
z1
l
z2
l
z3
E
3
−
ν
12
E
2
(l
x3
l
y2
+l
x2
l
y3
)l
x1
l
y1
−
ν
13
E
3
(l
x2
l
z3
+l
z2
l
x3
)l
x1
l
z1
−
ν
23
E
3
(l
y2
l
z3
+l
y3
l
z2
)l
y1
l
z1
+
1
G
12
(l
x1
l
y3
+l
x3
l
y1
)(l
x1
l
y2
+l
x2
l
y1
)
+
1
G
13
(l
x1
l
z3
+l
x3
l
z1
)(l
x1
l
z2
+l
x2
l
z1
)
+
1
G
23
(l
y1
l
z3
+l
y3
l
z1
)(l
y1
l
z2
+l
y2
l
z1
) (1, 2,...
ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 9 potx
... G
12
h, I
(1)
11
=
δ
2
2
E
1
p
j =1
[4j (1 +α) − (3 +α)],
I
(1)
22
=
δ
2
2
E
1
p
j =1
[4j (1 +α) − (3α + 1) ],I
(1)
12
=
1
2
E
1
ν
12
h
2
,
I
(1)
44
=
1
2
G
12
h
2
,
I
(2)
11
=
δ
3
3
E
1
p
j =1
12 j
2
(1 ... (5.44)
B
11
= B
22
=
h
2
E
1
+E
2
,B
12
= E
1
ν
12
h, B
44
= G
12
h,
C
11
=−C
22
=
h
2
8
E
2
−E
1
,C
12
= 0,C
44
=...
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