MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 5 pptx
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 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
+ ... exceeds about 25 N, to be compared with the threshold of
20 N prod...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 1 ppt
... 28 0
5. 3 .5. Examples of application in thermoelasticity 28 3
5. 3 .5. 1. Thermoelastic law 28 3
5. 3 .5 .2. Thermal stresses 28 4
5. 3 .5. 3. Expansion joints 28 5
5.3 .5. 4. Uniaxial plate expansion 28 6
5. 3.6. In-plane, ... 196
4 .2. 2.1. Longitudinal modes 196
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 shea...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 2 pdf
... 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 transformed into:
γ cos 2 −sin 2
sin 2 γcos 2
R
C
=
−γ ... 2
R
C
=
−γ cos 2
sin 2
[1.113]
which has the non-trivial solution:
R =
sin 2 sin 2 −(γ cos 2 )
2
sin 2 sin 2 +(γ cos 2 )
2
; C =
2 sin 2 cos...
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 ... [2. 23]
The solution is:
X(x) = a
dx
ES
+ b [2. 24]
where a and b are two arbitrary constants.
If the beam is homogeneous and of constant cross-section, [2. 24] gives:
X(x...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 6 ppsx
... −6γℓ 0 − 12 −6γℓ
0 −6γℓ 4γℓ
2
06γℓ 2 ℓ
2
−10 0 1 0 0
0 − 12 6γℓ 0 12 6γℓ
0 −6γℓ 2 ℓ
2
06γℓ 4γℓ
2
⇒ K
(2)
= K
ℓ
1000
0 12 −6γℓ −6γℓ
0 −6γℓ 12 2 ℓ
2
0 −6γℓ 2 ℓ
2
4γℓ
2
Then, ... 4γℓ
2
⇒ K
(1)
= K
ℓ
4γℓ
2
06γℓ 2 ℓ
2
010 0
6γℓ 0 12 6γℓ
2 ℓ
2
06γℓ 4γℓ
2
U
2
W
2
ϕ
2
U
3
W
3
ϕ
3
T
⇒
U
2
W
2
ϕ...
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 ... homogeneous system to
be solved is:
2( γ
L
+ ̟
2
) 2
L
2
L
2
L
2
L
2
L
+ π
2
− ̟
2
2γ
L
2
L
2
L
2
L
2
L
+ 4π
2
− ̟
2
2γ
L...
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
2L...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 10 pdf
... ςdn ⇒ (d
P)
2
= (d r)
2
+ (dς )
2
+ ς
2
(d n)
2
+ 2 dr ·d n
[7 .28 ]
dr is in the tangent plane so
(dr)
2
= g
2
α
(dα)
2
+ g
2
β
(dβ)
2
[7 .29 ]
The determination of (d n)
2
is obtained ... [6.68],
D[[Z]]=p
0
⇒
∂
2
∂r
2
+
1
r
∂
∂r
+
1
r
2
∂
2
∂θ
2
∂
2
Z
∂r
2
+
1
r
∂Z
∂r
+
1
r
2
∂
2
Z
∂θ
2
=
p
0
D
[6. 124 ]
The problem is independ...