NOISE, VIBRATION AND HARSHNESS

Noise, Vibration and Harshness



Noise, Vibration and Harshness

COVENTRY UNIVERSITY

Faculty of Engineering and Computing

NOISE, VIBRATION & HARSHNESS MODULE (313MED)

First Assignment - For hand-in on or before Monday 26th November 2012

Question 1

A wheel and suspension of a car are modelled as seen in Figure Q1 below. You can consider the suspension arm to be a long slender, rigid rod of mass m and the wheel and tyre to be a point-mass with mass M. This gives a representation of the un sprung mass mode of the car as a one degree of freedom system.

Find:

the equation of motion of the system

(8 marks)

the natural frequency of the system

(4 marks)

the value of C that would give a non-dimensional damping value ?????????

(8 marks)

State all assumptions and show all your analysis work.

X(s)

X(w)

k = 1.5 kN/m

kt= 2 kN/m

L = 1 ms

a = 0.6 m

m = 25 kg

M= 40 kg

Car body



r

Figure Q1

Solution:

Using Newton's Second Law of Motion, F=ma, the single degree of freedom equation is represented by M x¨(t) + C x?(t) + K x(t) = f (t) (Equation 1)

In order to use the given data, we will transform the above equation into algebraic equation using Laplace Domain Theory.

L { M x¨ + C x? + K x } = M ( s2 X(s) - s x(0) - x?(0) ) + C ( s X(s) - x(0) ) + K X(s)

L { M x¨ + C x? + K x } = ( M s2 + C s + K ) X(s) - Ms x(0) - M x?(0) - C x(0)

L { f (t) } = F(s)

Therefore the equation 1 becomes,

[M s2 + C s + K] X(s) = F(s) + (M s + C) a + M L)

Consider the two masses m and M are interconnected and Xs, Xw and r are the respective displacements.

k(Xs-Xw)+C(Xs-Xw)+MXw+Kt(Xw-r)+u=0

Xs and Xw are two forces which are acting at the spring with coefficient of elasticity k so k*(difference of forces) and C*(difference of forces) C is the damping coefficient. The force Xw acts on mass M while the difference of two forces times Kt is there.

Let u is the control input. This is the equation of motion of the system.

Replacing the values in the question we get:

1.5(0.4)+C(0.4)+40(0.6)+2(0.6-r)+u = 0

(B) NATURAL FREQUENCY

Natural frequency is the frequency at which vibrations takes place in the absence of damping and is given by

W0 =v

W0=v

W0=0.707 Hz

(C) ???????

???

Which is the required value of damping factor.

Question 2

The two degree of freedom system shown in Figure Q2 represents the suspension of a vehicle. The system parameters are given by m1 = 2000 kg, m2 = 50 kg, k1 = 103 N/m, and k2 = 104 N/m. Damping has been neglected.

(a) Calculate the natural frequencies of the system and the corresponding mode shapes. Normalise the mode shape vectors so the larger value is 1.

(12 marks)

(b) Also calculate the transmissibility for the system at 30 Hz. This is either the force transmitted to the ground through the tire contact patch when a unit force ...