Dynamical Response of a Beam in a Centrifugal Field …
105
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
u
v
w
α
β
γ
m x Oz
m x Oy
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
=
⎡
⎣
N
N
∗
N
∗∗
⎤
⎦
δ 1
δ 2
(16)
The internal energy stored in the beam shall be calculated. The internal energy
due to bending is given by the relation:
E pi =
1
2
L
0
E I y
d
2 w
dx 2
2
+ E I z
d
2 v
dx 2
2
dx
=
1
2
L
0
E I y β
2
+ E I z γ
2
dx
=
1
2
{δ e }
T
⎡
⎣
L
0
E I y
N
(w)
T
N
(w)
+ E I z
N
(v)
T
N
(v)
dx
⎤
⎦ {δ e }
=
1
2
{δ e }
T [k eb ]{δ e }
(17)
where E is Young’s modulus, I y and I z represent the geometrical moment of inertia
around the axis Oy and Oz.
The energy due to the tension/compression is:
E pa =
1
2
L
0
E A
du
dx
2
dx =
1
2
{δ e }
T
L
0
N
u
T
N
u
E Adx
{δ e }
=
1
2
{δ e }
T [k ea ]{δ e }
(18)
where A is the area of the cross section of the beam.
The axial load P in an axial section of the beam gives the energy if in a first
approximation the axial deformations are neglected:
E a =
1
2
L
0
P tot
dv
dx
2
+
dw
dx
2
dx =
1
2
{δ e }
T
k
G
e
{δ e }
(19)