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MECHANICAL SYSTEMS |
33 |
T |
T |
T |
θ |
b |
I |
Torsional spring |
Rotational dashpot |
Moment of inertia |
Figure 2.8 Rotational mechanical system components
A rotational spring is similar to a translational spring, but here the spring is twisted. The relationship between the applied torque, T , and the angle θ rotated by the spring is given by
T = kθ , |
(2.30) |
where θ is known as the rotational stiffness constant. In our modelling we are assuming that the mass of the spring is negligible and the spring is linear.
The energy stored in a torsional spring when twisted by an angle θ is given by
E = |
1 |
|
2 kθ 2. |
(2.31) |
A rotary damper element creates damping as it rotates. For example, when a disk rotates in a fluid we get a rotary damping effect. The relationship between the applied torque, T , and the angular velocity of the rotary damper is given by
dθ |
(2.32) |
T = cω = c dt . |
In our modelling the mass of the rotary damper will be neglected, or will be assumed to be negligible. A rotary damper does not store energy.
Moment of inertia refers to a rotating body with a mass. When a torque is applied to a body with a moment of inertia we get an angular acceleration, and this acceleration rotates the body. The relationship between the applied torque, T , angular acceleration, a, and the moment of inertia, I , I is given by
dω
T = I a = I
dt
or, since ω = dθ /dt ,
d2θ T = I dt 2 .
The energy stored in a mass rotating with an angular velocity ω is given by
E = 1 I ω2.
2
Some examples of rotational system models are given below.
(2.33)
(2.34)
(2.35)
Example 2.5
A disk of moment of inertia I is rotated (see Figure 2.9) with an applied torque of T . The disk is fixed at one end through an elastic shaft. Assuming that the shaft can be modelled with a
34 SYSTEM MODELLING
k |
T |
I |
|
b |
θ |
Figure 2.9 Rotational mechanical system
rotational dashpot and a rotational spring, derive an equation for the mathematical model of this system.
Solution
The damper torque and spring torque oppose the applied torque. If θ is the angular displacement from the equilibrium, we can write
T − b |
dθ |
d2θ |
(2.36) |
||||||
− kθ = I |
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dt |
dt 2 |
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or |
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d2θ |
dθ |
(2.37) |
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I |
+ b |
+ kθ = T . |
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dt 2 |
dt |
||||||||
Equation (2.37) is normally written in the form |
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¨ |
˙ |
(2.38) |
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I θ |
+ bθ + kθ = T . |
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Example 2.6
Figure 2.10 shows a rotational mechanical system with two moments of inertia and a torque applied to each one. Derive a mathematical model for the system.
Solution
For the system shown in Figure 2.10 we can write the following equations: for disk 1,
dθ |
dθ |
d2θ |
|||||||||
T1 |
− k(θ1 |
− θ2) − b |
1 |
− |
2 |
= I1 |
1 |
; |
(2.39) |
||
dt |
dt |
dt 2 |
|||||||||
and for disk 2, |
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dθ |
dθ |
= I2 |
d2θ |
||||||||
T2 |
− k(θ2 |
− θ1) − b |
2 |
− |
1 |
2 |
. |
(2.40) |
|||
dt |
dt |
dt 2 |
|||||||||
T1 |
k |
T2 |
|||||||||
I1 |
I2 |
||||||||||
θ1 |
b |
θ2 |
|||||||||
Figure 2.10 Rotational mechanical system
MECHANICAL SYSTEMS |
35 |
Equations (2.39) and (2.40) can be written as |
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¨ |
˙ |
˙ |
+ kθ1 − kθ2 |
= T1 |
(2.41) |
|||||||||||
and |
I1θ1 |
+ bθ1 |
− bθ2 |
|||||||||||||
¨ |
˙ |
˙ |
− kθ1 + kθ2 |
= T2. |
(2.42) |
|||||||||||
I2θ2 |
− bθ1 |
+ bθ2 |
||||||||||||||
Writing the equations in matrix form, we have |
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0 |
I2 |
¨ |
b |
b |
˙ |
+ |
k |
k |
θ2 |
= |
T2 |
|||||
θ¨2 + |
θ˙2 |
(2.43) |
||||||||||||||
I1 |
0 |
θ1 |
b |
−b |
θ1 |
k |
−k |
θ1 |
T1 |
. |
||||||
− |
− |
|||||||||||||||
2.1.2.1 Rotational Mechanical Systems with Gear-Train
Gear-train systems are very important in many mechanical engineering systems. Figure 2.11 shows a simple gear-train, consisting of two gears, each connected to two masses with moments of inertia I1 and I2. Suppose that gear 1 has n1 teeth and radius r1, and that gear 2 has n2 teeth and radius r2. In this analysis we assume that the gears have no backlash, they are rigid bodies, and the moment of inertia of the gears is assumed to be negligible.
The rotational displacement of the two gears depends on their radii and is given by the relationship
r1θ1 = r2θ2
or
θ2 = r1 θ1, r2
where θ1 and θ2 are the rotational displacements of gear 1 and gear 2, respectively. The ratio of the teeth numbers is equal to the ratio of the radii and is given by
r1 = n1 = n, r2 n2
where n is the gear teeth ratio.
Assuming that a torque T is applied to the system, we can write
d2θ1
I1 dt 2 = T − T1
(2.44)
(2.45)
(2.46)
(2.47)
Figure 2.11 A two gear-train system
36 SYSTEM MODELLING
and
I2 |
d2θ2 |
(2.48) |
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= T2. |
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dt 2 |
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Equating the power transmitted by the gear-train, |
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T1 |
dθ1 |
dθ2 |
or |
T1 |
dθ2/dt |
(2.49) |
||||||||||||||
= T2 |
= |
= n. |
||||||||||||||||||
dt |
dt |
T2 |
dθ1/dt |
|||||||||||||||||
Substituting (2.49) into (2.47), we obtain |
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I1 |
d2θ1 |
(2.50) |
||||||||||||||||||
= T |
− nT2 |
|||||||||||||||||||
dt 2 |
||||||||||||||||||||
or |
||||||||||||||||||||
d2θ |
I2 |
d2θ |
; |
|||||||||||||||||
I1 |
1 |
= T − n |
2 |
(2.51) |
||||||||||||||||
dt 2 |
dt 2 |
|||||||||||||||||||
then, since θ2 = nθ1, we obtain |
d2θ1 |
|||||||||||||||||||
(I1 + n2 I2) |
= T . |
(2.52) |
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dt 2 |
||||||||||||||||||||
It is clear from (2.52) that the moment of inertia of the load, I2, is reflected to the other side of the gear-train as n2 I2.
An example of a system coupled with a gear-train is given below.
Example 2.7
Figure 2.12 shows a rotational mechanical system coupled with a gear-train. Derive an expression for the model of the system.
Solution
Assuming that a torque T is applied to the system, we can write
I1 |
d2θ1 |
+ b1 |
dθ1 |
+ k1θ1 |
= T − T1 |
(2.53) |
|
dt 2 |
dt |
||||||
k1 |
T1 |
||||||
T |
I1 |
n1, r1 |
|||||
θ1 |
b1 |
k2 |
|||||
n2, r2 |
I2 |
||||||
T2 |
b2 |
θ2 |
|||||
Gear-train |
|||||||
Figure 2.12 Mechanical system with gear-train
ELECTRICAL SYSTEMS |
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and |
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I2 |
d2θ2 |
dθ2 |
|||||||||||||||||||||||||||||
+ b2 |
+ k2 |
θ2 = T2. |
|||||||||||||||||||||||||||||
dt 2 |
dt |
||||||||||||||||||||||||||||||
Equating the power transmitted by the gear-train, |
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dθ1 |
dθ2 |
or |
T1 |
dθ2/dt |
|||||||||||||||||||||||||||
T1 |
= T2 |
= |
= n. |
||||||||||||||||||||||||||||
dt |
dt |
T2 |
dθ1/dt |
||||||||||||||||||||||||||||
Substituting (2.55) into (2.53), we obtain |
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d2θ1 |
dθ1 |
||||||||||||||||||||||||||||||
I1 |
+ b1 |
+ k1θ1 |
= T − nT2 |
||||||||||||||||||||||||||||
dt 2 |
dt |
||||||||||||||||||||||||||||||
or |
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d2θ |
dθ |
− n I2 |
d2θ |
dθ |
. |
||||||||||||||||||||||||||
I1 |
1 |
+ b1 |
1 |
+ k1θ1 |
= T |
2 |
+ b2 |
2 |
+ k2θ2 |
||||||||||||||||||||||
dt 2 |
dt |
dt 2 |
dt |
||||||||||||||||||||||||||||
Since θ2 = nθ1, this gives |
|||||||||||||||||||||||||||||||
(I1 + n2 I2) |
d2θ1 |
+ n2b2) |
dθ1 |
+ (k1 + n2k2)θ1 = T . |
|||||||||||||||||||||||||||
+ (b1 |
|||||||||||||||||||||||||||||||
dt 2 |
dt |
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37
(2.54)
(2.55)
(2.56)
(2.57)
(2.58)
2.2 ELECTRICAL SYSTEMS
The basic building blocks of electrical systems are the resistor, inductor and capacitor (Figure 2.13). The input to an electrical system may be the voltage, V , and current, i .
The relationship between the voltage across a resistor and the current through it is given by
Vr = Ri, |
(2.59) |
where R is the resistance.
For an inductor, the potential difference across the inductor depends on the rate of change
of current through the inductor, given by |
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vL = L |
di |
(2.60) |
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, |
||||||||||||||||||
dt |
||||||||||||||||||
where L is the inductance. Equation (2.60) can also be written as |
||||||||||||||||||
i = |
1 |
vL dt . |
(2.61) |
|||||||||||||||
L |
||||||||||||||||||
The energy stored in an inductor is given by |
||||||||||||||||||
E = |
1 |
Li 2. |
(2.62) |
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2 |
||||||||||||||||||
R |
L |
C |
||||||||||||||||
i |
i |
i |
||||||||||||||||
Figure 2.13 Electrical system components