Composite Materials Design and Applications Part 3 doc
Composite Materials Design and Applications Part 3 doc
... Figure 3. 24).
Figure 3. 22 Tensile Test
Figure 3. 23 Short Beam Shear Test
Figure 3. 24 Variation In Stiffness During Curing
TX846_Frame_C 03 Page 52 Monday, November 18, 2002 12:05 PM
© 20 03 by ... proportion of composite materials design.
They appear in almost all applications. Historically they were the first light and
high-performance structures.
1
In the majority...
Composite Materials Design and Applications Part 4 docx
... information on modulus and strength of a basic ply: see Section
3. 3 .3)
0∞:
e
x
e
x
e
y
e
xy
++
0 .3=
90∞:
e
y
e
x
e
y
e
xy
++
0 .34 =
45∞:
e
xy
e
x
e
y
e
xy
++
0 .36 =
±
TX846_Frame_C05 Page ... -45∞.
(More information on modulus and strength of a basic ply: see Section
3. 3 .3)
e
xy
T
xy
t
rupture
=
TX846_Frame_C05 Page 93 Monday, November 18, 2002 12:09 PM
© 20 03 by...
Composite Materials Design and Applications Part 9 docx
... III
j
33 33
=
j
2222
s
4
j
33 33
+ c
4
1–()2s
2
c
2
j
2 233
2
j
232 3
+()+ 0=
j
2222
j
33 33
=
j
2222
c
4
s
4
1–+()2s
2
c
2
j
2 233
2
j
232 3
+()+ 0=
2– s
2
c
2
j
2222
2s
2
c
2
j
2 233
2
j
232 3
+()+ ... II¢
j
–
33 12
j
33 12
0; F
II¢ III¢I¢ III¢
j
–
231 3
j
231 3
0=== ===
e
11
e
22
e
33
g
23
g
13
g
12
Ó ˛
ÔÔ
ÔÔ
ÔÔ
ÔÔ
Ì ˝
ÔÔ
ÔÔ
ÔÔ
ÔÔ
ϸ
j
1111
j
1122
j
1 133
000
j
2211...
Composite Materials Design and Applications Part 12 doc
... Section 5.4.2, Table 5.1:
s
rupture
= 1,059 MPa
and with
r
= 1, 530 kg/m
3
(Table 3. 4 of Section 3. 3 .3, or the calculation in Section
3. 2 .3) :
V
max
= 832 m/s
dm
w
2
r¥ 2
s
eb
d
q
2
=
r
r d
q
... –()+
E
y
➀
H
1
3
H
2
3
–()E
y
➁
H
2
3
+
¥¥=
H
2
2
z
H
1
2
££ ;
t
yz
Q
y
6
E
y
➀
H
1
2
4
z
2
–()
E
y
➀
H
1
3
H
2
3
–()E
y
➁
H
2
3
+
¥¥=
k
x
a
x
8 E
x
➀
H
1
3...
Composite Materials Design and Applications Part 1 pot
... 5 :38 PM
© 20 03 by CRC Press LLC
COMPOSITE
MATERIALS
DESIGN AND APPLICATIONS
© 20 03 by CRC Press LLC
3. 3 Unidirectional Ply
3. 3.1 Elastic Modulus
3. 3.2 Ultimate Strength of a Ply
3. 3 .3 ... Ply
18 .3 Level 3
18 .3. 1 Cylindrical Bonding
18 .3. 2 Double Bonded Joint
18 .3. 3 Composite Beam with Two Layers
18 .3. 4 Buckling of a Sandwich Beam
18 .3. 5 Shear D...
Composite Materials Design and Applications Part 5 potx
... Press LLC
Figure 6 .32 Composite Piece Under Tensile Load
Figure 6 .33 Composite Piece under Compression Load
Figure 6 .34 Composite Piece under Tension-compression Load
Figure 6 .35 Arrangement to ... bolting, and bonding:
Ⅲ
of a composite part to another composite part and
Ⅲ
of a composite part to a metallic part.
6.1 RIVETING AND BOLTING
Ⅲ
In all mec...
Composite Materials Design and Applications Part 7 pot
... Comparison Between Metallic and Composite Springs
5
S.M.C. process: See Section 2.1 .3 and 3. 2.
6
See Section 3. 3.2, comparison of load-elongation diagrams for a metal and a unidirectional.
TX846_Frame_C08 ... 12:22 PM
© 20 03 by CRC Press LLC
8
COMPOSITE MATERIALS
FOR OTHER APPLICATIONS
We have given in Chapter 1 an idea on the diversity of the products which can
be...
Composite Materials Design and Applications Part 8 pot
...
g
23
g
13
g
12
j
1111
j
1122
j
1 133
2
j
11 23
2
j
11 13
2
j
1112
j
2211
j
2222
j
2 233
2
j
22 23
2
j
22 13
2
j
2212
j
33 11
j
33 22
j
33 33
2
j
33 23
2
j
33 13
2
j
33 12
2
j
231 1
2
j
232 2
2
j
233 3
4
j
232 3
4
j
231 3
4
j
231 2
2
j
131 1
2
j
132 2
2
j
133 3
4
j
132 3
4
j
131 3
4
j
131 2
2
j
1211
2
j
1222
2
j
1 233
4
j
12 23
4
j
12 13
4
j
1212
s
11
s
22
s...
Composite Materials Design and Applications Part 10 ppsx
...
s
11
s
22
s
33
++()
2
º
{=
º
3
s
11
s
22
t
12
2
–()
s
22
s
33
t
23
2
–()
s
33
s
11
t
31
2
–()++()– }
dW
dV
˯
ʈ
distorsion
1
12G
s
11
s
22
–()
2
s
22
s
33
–()
2
+
º
{=
º
s
33
s
11
–()
2
6
t
12
2
t
23
2
t
31
2
++()++}
a
s
11
s
22
–()
2
s
22
s
33
–()
2
s
33
s
11
–()
2
º
++{
...
s
I
2
s
II
2
s
III
2
s
I
s
II
s
III
++
3
˯
ʈ
2
–++
Ó ˛
Ì ˝
ϸ
=
dW
dV
˯
ʈ
dis...
Composite Materials Design and Applications Part 13 potx
... titanium:
E
x
= 55 ,33 3 MPa E
x
= 31 ,979 MPa
G
xy
= 16 ,31 5 MPa G
xy
= 28, 430 MPa
800 900 34 0++()
100
2.64 2.17–
2.17
21%=
TX846_Frame_C18a Page 36 8 Monday, November 18, 2002 12:40 PM
© 20 03 by CRC ... complete study of the shear stresses in
a composite beam (Equation 15.16).
s
max
M
f
p
r
2
e
=
F
max
26 10
6
¥
7200
36 00 N.==
M
f
()
F
max M
f
()
a
63 332 9¥
36 00...
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