MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 4 pptx
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 4 pptx
... Z(x) = ax
4
+ bx
3
+ cx
2
Loading: 24 a =
f
0
EI
→ a =
f
0
24 EI
Free end at x = L
Z
′′′
(L) = 24 ax +6b = 0 ⇒ b =−
f
0
L
6EI
Z
′′
(L) =
f
0
L
2
2EI
−
f
0
L
2
EI
+ 2c = 0 → c =
f
0
L
2
4EI
then:
Z(ξ) ... 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...
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 ... by...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 1 ppt
... 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 shear modes 20 5
4 .2. 3. Rayleigh’s quotient 20 7
4 .2. 3.1. Bending of ... attached
concentrated mass 20 7
4 .2. 3 .2. Beam on elastic foundation 20 9
4 .2. 4. Finite element approximation 21 0
4 .2. 4. 1. Longitudinal...
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
; ... 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 =...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 3 ppt
... [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) =
ax
ES
+ b [2. 25]
2. 2.1.3 ... case
of the longitudinal mode of deformation to the shear case.
2. 2 .2. 2 General solution without external loading
Z(x) = a
x
0
dx
GS
+ b [2. 34]
a and b are two arbitrary r...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 6 ppsx
... −6γℓ 0 − 12 −6γℓ
0 −6γℓ 4 ℓ
2
06γℓ 2 ℓ
2
−10 0 1 0 0
0 − 12 6γℓ 0 12 6γℓ
0 −6γℓ 2 ℓ
2
06γℓ 4 ℓ
2
⇒ K
(2)
= K
ℓ
1000
0 12 −6γℓ −6γℓ
0 −6γℓ 12 2 ℓ
2
0 −6γℓ 2 ℓ
2
4 ℓ
2
Then, ... 4 ℓ
2
⇒ K
(1)
= K
ℓ
4 ℓ
2
06γℓ 2 ℓ
2
010 0
6γℓ 0 12 6γℓ
2 ℓ
2
06γℓ 4 ℓ
2
U
2
W
2
ϕ
2
U
3
W
3
ϕ
3
T
⇒
U
2
W...
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 7 pdf
... instance N = 4,
[4. 73] is written as 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...
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...
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 ... symmetries of the problem. The equilibrium equations are:
D...
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. 1 24 ]
The problem is indepen...