MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 6 ppsx
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 6 ppsx
... 6 ℓ 0 − 12 6 ℓ
0 6 ℓ 4γℓ
2
06 ℓ 2 ℓ
2
−10 0 1 0 0
0 − 12 6 ℓ 0 12 6 ℓ
0 6 ℓ 2 ℓ
2
06 ℓ 4γℓ
2
⇒ K
(2)
= K
ℓ
1000
0 12 6 ℓ 6 ℓ
0 6 ℓ 12 2 ℓ
2
0 6 ℓ 2 ℓ
2
4γℓ
2
Then, ... found to be:
K
(1)
α
= K
ℓ
4γℓ
2
6 ℓS 6 ℓC 2 ℓ
2
6 ℓS C
2
+ 12 S
2
CS( 12 −1) 6 ℓS
6 ℓC CS( 12 −1)S
2
+ 12 C
2
6 ℓC
2 ℓ
2
6 ℓS 6 ℓC 4γℓ
2
...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 13 ppsx
... coefficient 46
law 44, 46
wave 40
relation of constraint 2
resonant range 22 5, 22 6
response 22 6
response spectrum 22 7
resultant stress 72
rigid body 24 , 26 2
mode 23 6
motion 8
rigid connection
mode 25 6
spring ... 26 7
quasi-inertial range 22 5, 22 8
quasi-static
correction 22 8, 23 5
mode 22 8
range 22 5, 22 8
Rayleigh minimum principle 29 4, 29 9
Rayle...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 1 ppt
... 322
6 .2. 2.3. External loading of the edges and inhomogeneous
boundary conditions 322
6 .2. 3. Surface and concentrated loadings 324
6 .2. 3.1. Loading distributed over the midplane
surface 324
6 .2. 3 .2. ... 1 96
4 .2. 2.1. Longitudinal modes 1 96
4 .2. 2 .2. Torsion modes 20 0
4 .2. 2.3. Flexure (or bending) modes 20 0
4 .2. 2.4. Bending coupled with shear modes 20 5
4 .2....
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 2 pdf
... get:
X
2
= U
2
sin θ
2
e
iψ
2
; Z
2
= U
2
cos θ
2
e
iψ
2
;
ψ
2
= ω
t −
x sin θ
2
− z cos θ
2
c
L
(σ
zz
)
2
=
−iωU
2
c
L
λ +2G(cos θ
2
)
2
e
iψ
2
;
(σ
zx
)
2
=
+iωU
2
c
L
G sin 2
2
e
iψ
2
; ... that:
λ +2G(cos θ)
2
G
=
λ +2G(1 −(sin θ)
2
)
G
=
λ +2G(1 −(γ sin β)
2
)
G
= γ
2
(1 2( sin β)
2
) = γ
2
cos 2
The system [1.1 12] is thus...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 3 ppt
... as:
x+dL /2
x−dL /2
(S)
r ×
ρ
∂
2
ξ
∂t
2
−
∂
t
1
∂x
dS dx
=
x+dL /2
x−dL /2
(S)
(r ×
f
(e)
)dS dx +
M
(e)
(x; t)dL
[2. 20]
Using the relations [2. 1], [2. 8], and [2. 9], the following ... In terms of local quantities it
reduces to:
ρ
∂
2
ξ
∂t
2
−
∂
t
1
∂x
=
f
(e)
(x, y, z; t)
ρ
∂
2
ξ
j
∂t
2
−
∂σ
xj
∂x
= f
(e)
j
(x, y, z; t); (j = 1, 2, 3)
[2....
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 4 pptx
... GS
∂Y
s
∂x
2
+
∂Z
s
∂x
2
dx
+
1
2
L
0
GJ
T
∂ψ
x
∂x
2
+
EI
z
∂
2
Y
b
∂x
2
2
+
EI
y
∂
2
Z
b
∂x
2
2
dx
[3 .2]
The kinetic energy E
κ
is:
E
κ
=
1
2
L
0
ρS{
˙
X
2
+ (
˙
Y
s
+
˙
Y
b
)
2
+ (
˙
Z
s
+
˙
Z
b
)
2
}dx
+
1
2
L
0
ρ(J
˙
ψ
2
x
+ I
y
˙
ψ
2
y
+ ... as:
E
es
=
1
2
L
0
GSκ
∂Z
s
∂x
2
dx [3 .21 ]
The bending str...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 5 pptx
... +E
n
I
n
ℓ
n
0
∂
2
Z
n
∂x
2
2
dx
=[X
I
Z
I
Z
′
I
X
J
Z
k
Z
′
J
]
K
11
K
12
K
13
K
14
K
15
K
16
K
21
K
22
K
23
K
24
K
25
K
26
K
31
K
32
K
33
K
34
K
35
K
36
K
41
K
42
K
43
K
44
K
45
K
46
K
51
K
52
K
53
K
54
K
55
K
56
K
61
K
62
K
63
K
64
K
65
K
66
X
I
Z
I
Z
′
I
X
J
Z
J
Z
′
J
where ... is...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 7 pdf
... follows:
2
L
+ ̟
2
1
− ̟
2
2
L
2
L
2
L
2
L
2
L
+ ̟
2
2
− ̟
2
2
L
2
L
2
L
2
L
2
L
+ ̟
2
3
− ̟
2
2
L
2
L
2
L
2
L
2
L
+ ̟
2
4
− ̟
2
q
1
q
2
q
3
q
4
=
0
0
0
0
by ... 1, 2
If four modes are retained in the projection, the following homogeneous system to
be solved is:
2( γ
L
+ ̟
2
) 2
L
2
L
2
L
2
L
2...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 8 potx
... or odd
Ehπ
2
4
n
2
η
+
m
2
η
2
α
2
n,m
+
n
2
η +
m
2
2η
β
2
n,m
otherwise
Ehπ
2
4
n
2
η
+
m
2
η
2
α
2
n,m
+
n
2
η +
m
2
2η
β
2
n,m
+
8n
2
m
2
π
2
(n
2
− m
2
)
2
α
n,m
β
n,m
where ... system:
4Eh
M
π
2
4
n
2
+
m
2
η
2
2
−
ω
′
n,m
2
2n
2
m
2
η
2
(n
2
− m
2
)
2
2n
2
m
2
η
2
(n
2
−...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 9 pps
... Z
ε
xx
=−z
∂
2
Z
∂x
2
+
1
2
∂Z
∂x
2
+
z
2
2
∂
2
Z
∂x
2
2
+
∂
2
Z
∂x∂y
2
ε
xy
=−z
∂
2
Z
∂x∂y
+
z
2
2
∂
2
Z
∂x∂y
∂
2
Z
∂x
2
+
∂
2
Z
∂y
2
+
1
2
∂Z
∂x
∂Z
∂y
ε
yy
=−z
∂
2
Z
∂y
2
+
1
2
∂Z
∂y
2
+
z
2
2
∂
2
Z
∂y
2
2
+
∂
2
Z
∂x∂y
2
[6. 89]
In ... m
2
)
2
; α
n,m
= β
n,m
out -of- phase modes:
f
n,m
=
c...