CONTROL SYSTEMS ENGINEERING D227 S.A.E. SOLUTIONS

TUTORIAL 1 – CREATING MODELS OF ENGINEERING SYSTEMS

SELF ASSESSMENT EXERCISE No.1

1. A mass – spring –system has the following parameters. Stiffness K = 1200 N/m Mass M = 15 kg Damping Coefficient kd = 120 Ns/m

i. Calculate the time constant, critical damping coefficient and the damping ratio. ii. If a constant force of 22 N is applied, what will be the static position of the mass? iii. Calculate the force needed to make the mass move with a constant acceleration of 12 m/s2 at

the point where the velocity is 1.2 m/s. i. T = √(M/k) = √(15/1200) = 0.0112 seconds cc = √4MK = √(4 x 15 x 1200) = 268.3 Ns/m δ = kd/ cc = 120/268.3 = 0.447 ii. Static deflection = F/K = 22/1200 = 0.0183 m iii. For a constant acceleration s2x = a (acceleration) and sx = v (velocity)

( )( )

( )xxx

x

1200 s 6s 0.15F

1 0.0112s x 0.447 x 2 s 0.0112 2001F

1Ts δ 2sTkxF 1Ts δ 2sT

1/k(s)Fx

2

22

2222

++=

++=

++=++

=

F = 0.15 a + 6 v + 1200 x For velocity = 1.2 m/s and a = 12 m/s2

F = 0.15 (12) + 6 (1.2) + 1200 x The deflection x would need to be evaluated from other methods x = v2/2a = 0.06 m F = 81 N

SELF ASSESSMENT EXERCISE No.2 Derive the transfer function for a mass on a torsion bar fitted with a damper and show it is another

example of the second order transfer function. T is torque and J is moment of inertia.

1Ts δ 2sT

1/k 1/k)s(Jk(J/k)s

1/k(s)TθG(s) 22

d2 ++

=++

==

The input is the force F and the output is the movement x, both being functions of time. Spring Torque Ts = kθ Damping Torque Td = kd dθ/dt

Inertia Torque Ti = Id2θ/dt2

The three torques oppose motion so if the total torque on the system is zero then T = Ti + Td + Ts

1/k)s(k(I/k)s1/k(s)

TG(s) kskIsT(s) k

dtdk

dtdIT(t)

d2d

2d2

2

++==++=++=

θθθθθθθ If we

examine the units of (I/k)1/2 we find it is seconds and this is the second order time constant also with the symbol T. The transfer function may be written as

1Ts δ 2sT1/k(s)

TG(s) 22 ++

==θ

SELF ASSESSMENT EXERCISE No.3 A DC Servo motor has a moment of inertia of 12 kg m2. It is coupled to an aerial rotator through a

gear reduction ratio of 4. The driven mass has a moment of inertia of 15 kg m2. The damping on the motor is 0.2 N m s/rad and on the rotator bearings it is 0.4 N m s/rad. Calculate the torque required from the motor to

i. Turn the aerial at a constant rate of 0.5 rad/s. ii. Accelerate the rotator at 0.02 rad/s2 at the start when ω = 0

i. Ie = (Im +Gr2 Io) = (12 + 42 x 15) = 242 kg m2.

Kde = (kdm +Gr2 kdo) = (0.2 + 42 x 0.4) = 6.6 N m s/rad. θ (s)/Tm (s) = (1/Ie)/s{s + Kde/Ie}

( )

ω6.6 242α TωKαIT

/Ikss1/I(s)

Tθ

m

deem

ede

e

m

+=+=

+=

If the rotator is moving at constant speed α (acceleration) is zero. Hence: Tm = 6.6 ω = 6.6 x 0.5 = 3.3 Nm ii. When accelerating at 0.02 rad/s2 the motor acceleration is 4 times larger at 0.08 rad/s2. Tm = 242α + 6.6ω = 19.36 Nm when ω = 0

SELF ASSESSMENT EXERCISE No.4 1. A hydraulic motor has a nominal displacement of 5 cm3/radian. Calculate the torque produced at a

pressure of 120 bar. T =kq p = 5 x 10-6 x 120 x 105 = 60 Nm 2. A hydraulic cylinder has bore of 50 mm and is controlled with a valve with a constant kv = 0.05

m2/s Calculate the time constant T. Given that xi and xo are zero when t = 0, calculate the velocity of the

piston and the output position after 0.2 seconds when the input is changed suddenly to 4 mm. A = π x 0.052/4 = 0.00196 m2

T = A/kv = 0.00196/0.05 = 0.0393 s

m/s 102.0 s 0.0393

m 0.004Tx velocity

dtdx x

dtdxT xTsx

Ts1(s)

xxG(s) io

io

ioi

o ========

Velocity = distance /time distance = xo = v t = 0.102 x 0.2 = 0.204 or 20.4 mm assuming

the velocity is constant.