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4.4 Undamped natural frequency of an actuator
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Considering the electrical analogy where pressure is analogous to voltage P→V and flow rate is analogous to current Q→ i then fluid compressibility is sometimes referred to as fluid capacitance:
Capacitance ȕ
& 9
4.3 Force and torque equations for actuators with moving mass or
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This is an undamped 2nd order differential equation, because there is no 1st order differential term dU/dt, and is often written in the following standard form:
GW 8 G Ȧ 8
Q
ȦQ
is called the undamped natural frequency.Cylinder 1P
9 ȕ
$ 9
ȕ N $ VWLIIQHVV VSULQJ
K\GUDXOLF 0 UDGV
Q N
Ȧ
Motor
9 1P ȕ ' 9
ȕ W ' VWLIIQHVV WRUTXH
K\GUDXOLF - UDGV
Q W Ȧ
P
P
(4.8)
(4.9)
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© Deloitte & Touche LLP and affiliated entities.
360° thinking .
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84
Some interesting points arise by considering the undamped natural frequency equation:
• for a double-rod linear actuator $ $
= $. If the volumes are dominated by the actuator $only, that is line volumes are negligible, then it is a simple matter to show that the minimum undamped natural frequency occurs with the actuator in its mid-position with 9 = 9 = 9. This frequency is given by:
ȕ
& 9
$ / 0 /& UDGV
09
ȕ Q $
Ȧ
• for a single rod actuator where $≠ $
, and again if line volumes are negligible, it can also be shown that the minimum undamped natural frequency occurs very close to the central position, even for area ratios up to Ȗ $$• a high undamped natural frequency requires a low mass and/or a high stiffness. The latter may be achieved using a large cross-sectional-area and small volumes
It is possible that steel pipes may contribute to the effective bulk modulus. The pipe material bulk modulus ȕS may be determined from the theory of elasticity assuming a pipe of internal diameter
GL and external diameter GR:
»»
¼ º
««
¬
ª
ȝ
G G
G G ( ȕ
L R
L R S
E is the modulus of elasticity for the material and m is Poisson’s ratio for the material, typically m = 0.3 and E = [1P for steel. For a pipe with a wall thickness t << GLthen GL ≈ GRand equation (4.11) can be approximated by:
W(
G ȕ
S
|
The effective bulk modulus for a pipe ȕH is then obtained by combining the pipe effect ȕS with the fluid volume effect ȕHI
HI S
H ȕ
ȕ
ȕ
(4.10)
(4.11)
(4.12)
(4.13)
85 Example 4.1
SLSHSLSH Consider the single-rod actuator shown.
Pipes 1 and 2 are flexible hose 7mm diameter 2m long Hose bulk modulus ȕK
[1POil bulk modulus ȕ [1P
Bore diameter 50mm, rod diameter 28mm, actuator stroke 150mm Load and actuator mass 550kg.
Assume the actuator is in its mid-position Side 1
Pipe volume = [ P
[ [
ʌ
Actuator area $
P [
ʌ[
Actuator volume = [ [P Total volume 9 [P
Effective bulk modulus:
H
ȕH
[
[
ȕ
¸
¹
¨ ·
©
¸ §
¹
¨ ·
©
§
H [1P
ȕ
Side 2
Pipe volume = [P
Actuator area $
[ P[
ʌ
Actuator volume [ [P
Total volume 9
[P86 Effective bulk modulus:
H
H
H
1P [ ȕ
ȕ
[
[
ȕ
¸
¹
¨ ·
©
¸ §
¹
¨ ·
©
§
[
[ [
[
[ [
9 ȕ
$ 9
ȕ N $ VWLIIQHVV
H
H
[ [
N
1P [
N
undamped natural frequency UDGV +]
[ 0
Q N
Ȧ
• note in this example that the much smaller-volume hose, compared with actuator volumes, significantly influences the effective bulk modulus on each side due to the hose flexibility when compared with a rigid steel pipe
• also note that the hydraulic stiffness at side 1 is almost double that on side 2 due to the area ratio effect.
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