This document contains 4 self-assessment exercises involving modeling of engineering systems: 1) A mass-spring-damper system is analyzed to calculate time constants and force required for constant acceleration. 2) A mass on a torsion bar with damping is modeled as a second-order system. 3) Forces required to turn and accelerate a geared DC servo motor are calculated. 4) Torque from a hydraulic motor and velocity/position of a hydraulic cylinder are determined.

1. 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) () () ()xxxx1200 s 6s 0.15F 1 0.0112s x 0.447 x 2 s 0.0112 2001F1Ts δ 2sTkxF 1Ts δ 2sT1/k(s) Fx2222222++= ++= ++= ++ =
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 2sT1/k 1/k)s(Jk(J/k)s1/k(s) TθG(s)22d2++ = ++ ==
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) kdtdkdtdIT(t) d2d2d22++ ==++=++= 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

2. 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) Tθmdeemedeem+= += + =
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 0.0393m 0.004Tx velocity dtdx xdtdxT xTsx Ts1(s) xxG(s)ioioioio========
Velocity = distance /time distance = xo = v t = 0.102 x 0.2 = 0.204 or 20.4 mm assuming the velocity is constant.