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MECHANICAL SYSTEMS

29

Similarly, when n springs are in series, then the reciprocal of the equivalent stiffness constant keq is equal to the sum of all the reciprocals of the individual spring stiffnesses ki :

1

1

1

1

(2.4)

=

+

+ . . . +

.

keq

k1

k2

kn

As an example, if there are two springs k1 and k2 in series, then the equivalent stiffness constant is given by

k1k2

keq = . (2.5) k1 + k2

A dashpot element is a form of damping and can be considered to be represented by a piston moving in a viscous medium in a cylinder. As the piston moves the liquid passes through the edges of the piston, damping to the movement of the piston. The force F which moves the piston is proportional to the velocity of the piston movement. Thus,

d y

(2.6)

F = c dt .

A dashpot does not store energy.

When a force is applied to a mass, the relationship between the force F and the acceleration a of the mass is given by Newton’s second law as F = ma. Since acceleration is the rate of change of velocity and the velocity is the rate of change of displacement, we can write

d2 y

(2.7)

F = m dt 2 .

The energy stored in a mass when it is moving is the kinetic energy which is dependent on the velocity of the mass and is given by

E =

1

2 mv2.

(2.8)

This energy is released when the mass stops.

Some examples of translational mechanical system models are given below.

Example 2.1

Figure 2.3 shows a simple mechanical translational system with a mass, spring and dashpot. A force F is applied to the system. Derive a mathematical model for the system.

k

m

F

c

y

Figure 2.3 Mechanical system with mass, spring and dashpot

30 SYSTEM MODELLING

Solution

As shown in Figure 2.3, the net force on the mass is the applied force minus the forces exerted by the spring and the dashpot. Applying Newton’s second law, we can write

d y

d2 y

(2.9)

F − k y − c

= m

dt

dt 2

or

d2 y

d y

(2.10)

F = m

+ c

+ k y.

dt 2

dt

Equation (2.10) is usually written in the form

F = m y¨ + c y˙ + k y.

(2.11)

Taking the Laplace transform of (2.11), we can derive the transfer function of the system as

F (s) = ms2Y (s) + csY (s) + kY (s)

or

Y (s)

1

(2.12)

=

.

F (s)

ms2 + cs + k

The transfer function in (2.12) is represented by the block diagram shown in Figure 2.4.

Example 2.2

Figure 2.5 shows a mechanical system with two masses and two springs. Drive an expression for the mathematical model of the system.

Solution

Applying Newton’s second law to the mass m1,

d y

d y

− k1 y1 = m1

d2 y

−k2(y1 − y2) − c

1

−

2

1

,

(2.13)

dt

dt

dt 2

and for the mass m2,

d y

d y

d2 y

F − k2(y2 − y1) − c

2

−

1

= m2

2

,

(2.14)

dt

dt

dt 2

we can write (2.13) and (2.14) as

m1 y¨1 + c y˙1 − c y˙2 + (k1 + k2)y1 − k2 y2 = 0,

(2.15)

m2 y¨2 + c y˙2 − c y˙1 + k2 y2 − k2 y1 = F.

(2.16)

F(s)

1

Y(s)

ms2

+ cs + k

Figure 2.4 Block diagram of the simple mechanical system

MECHANICAL SYSTEMS

31

Figure 2.5 Example mechanical system

Equations (2.15) and (2.16) can be written in matrix form as

y2

=

F

0

m2

y¨

2

+

c c

y˙

2 +

k2

k2

m1

0

y¨

1

c

−c

y˙

1

k1 + k2

−k2

y1

0

.

(2.17)

−

−

Example 2.3

Figure 2.6 shows a mechanical system with two masses, and forces applied to each mass. Drive an expression for the mathematical model of the system.

Solution

Applying Newton’s second law to the mass m1,

d y

d y

d2 y

F1 − k(y1 − y2) − c

1

−

2

= m1

1

,

dt

dt

dt 2

and to the mass m2,

d y

d y

d2 y

F2 − k(y2 − y1) − c

2

−

1

= m2

2

,

dt

dt

dt 2

we can write (2.13) and (2.14) as

m1 y¨1 + c y˙1 − c y˙2 + k y1 − k y2 = F1,

m2 y¨2 + c y˙2 − c y˙1 + k y2 − k y1 = F2.

Equations (2.20) and (2.21) can be written in matrix form as

F2

0

m2

y¨

2

+

c c

y˙

2 +

k k

y2 =

m1

0

y¨

1

c

−c

y˙

1

k

−k y1

F1 .

−

−

(2.18)

(2.19)

(2.20)

(2.21)

(2.22)

k

F1

m1

c

y1

m2 F2

y2

Figure 2.6 Example mechanical system

32 SYSTEM MODELLING

Figure 2.7 Example mechanical system

Example 2.4

Figure 2.7 shows a mechanical system with three masses, two springs and a dashpot. A force is applied to mass m3 and a displacement is applied to spring k1. Drive an expression for the mathematical model of the system.

Solution

Applying Newton’s second law to the mass m1,

k1 y − k1 y1 − k2(y1 − y2) + k2 y2 = m1

d2 y1

(2.23)

dt 2

to the mass m2,

d y

d y

d2 y

−c

2

−

3

− k2(y2 − y1) − k3(y2 − y3) = m2

2

,

(2.24)

dt

dt

dt 2

and to the mass m3,

F − c

d y

d y

d2 y

3

−

2

− k3(y3

− y2) = m3

3

,

(2.25)

dt

dt

dt 2

we can write (2.23)–(2.25) as

m1 y¨1 + (k1 + k2)y1 − k2 y2 m2 y¨2 + c y˙2 − c y˙3 − k2 y1 + (k2 + k3)y2 − k3 y3 m3 y¨3 + c y˙3 − c y˙2 + k3 y3 − k3 y2

= k1 y,

(2.26)

= 0,

(2.27)

= F.

(2.28)

The above equations can be written in matrix form as

k2−

y2

0 .

0

m2

0

y¨

2

0

c

c y˙

2

k2

k3

k3

m1

0

0

y¨

1

0

0

0

y˙

1

k1 + k2

k2

0

y1

k1 y

0

0

m3 y¨3 + 0

−c

−

y˙3 +

−

+

−

y3

= F

c

0

−k3

k3

(2.29)

2.1.2 Rotational Mechanical Systems

The basic building blocks of rotational mechanical systems are the moment of inertia, torsion spring (or rotational spring) and rotary damper (Figure 2.8). The input to a rotational mechanical system may be the torque, T , and the output the rotational displacement, or angle, θ .

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