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Copyright 2024 University of Southampton Page 1 of 10
UNIVERSITY OF SOUTHAMPTON FEEG2002W1
______________________________________________________
SEMESTER 2 EXAMINATIONS 2023-24
TITLE: MECHANICS, MACHINES AND VIBRATION
DURATION: 4 Hours (Online Open-Book)
We recommend that you spend 180 minutes on this paper.
______________________________________________________
This paper contains 3 questions.
Answer ALL questions.
An outline marking scheme is shown in brackets to the right of each
question.
If a software package is used to plot the graphs for the solution to
question one, then all steps must be clearly explained and all angles
and directions must be clearly defined.
Note that a formula sheet is provided at the end of this paper.
FEEG2002W1
Copyright 2024 University of Southampton Page 2 of 10
Q1
(i) In the 5-bar linkage mechanism with a suspension link that is
shown in Figure 1.1, the crank AD that rotates about fixed axis
A, has a pin D which slides in the straight slot of link CE. Link BC
turns about joint B.
Figure 1.1 Schematic of a hydraulic excavator mechanism
Determine the mobility of this mechanism and state any assumptions
you make in calculating the mobility.
[5 marks]
(ii) In the crank and slotted-lever, quick-return mechanism shown in
Figure 1.2, the link OA with 40 mm length rotates at a constant
angular speed of 100 rad/s. A sliding link which is pin joined to
OA at A, slides along the link BC and hence makes BC rotate
about B, as shown in the figure:
Page 3 of 10
Figure 1.2. Crank and slotted-lever, quick-return mechanism
The ground link length is 70 mm. For the position shown, i.e., when
the angle of OA is 60°, calculate the angular velocity of the link BC.
[16 marks]
(iii) For the same four-bar linkage mechanism in Figure 1.2:
a) Locate all the possible instantaneous centres of velocities.
[7 marks]
b) Calculate the angular velocity of the slider, using the “angular
velocity ratio” theorem.
[6 marks]
TOTAL [34 MARKS]
TURN OVER
FEEG2002W1
Copyright 2024 University of Southampton Page 4 of 10
Q2
A technician is given two identical elastic springs of stiffness k equal
to 2500 Nm-1 and a single viscous damper c. The technician is asked
to support a motor of mass m equal to 5 kg and cannot decide which
of the two configurations shown in Figure 2.1 to use. For any
calculations take g the acceleration due to gravity to be 10 ms-2
.
(a) (b)
Figure 2.1. Mounting configurations using two identical springs, a
single viscous damper and a supported mass.
(i) Mounting options and comparisons.
(a) What is the frequency above which any motor vibration to
the ground will begin to be isolated for the two
configurations
[6 marks]
Page 5 of 10
(b) What will be the static deflection of the mount which gives
the widest frequency range of isolation
[2 marks]
(c) What is the value for the viscous damping constant in the
two cases such that the combined loaded mount (mass
supported by the springs plus damper) has critical
damping
[4 marks]
(d) The technician uses both mount options and measures
equal vibration on the motor mass when the motor is
operating at 3000 rpm. Why is the vibration level the same
irrespective of the mount option chosen Estimate the motor
out of balance ( ) when the motor mass acceleration is
10g at this speed.
[4 marks]
(ii) A novel mount installation
An academic proposes to use a massless bar that is pivoted at its
midpoint with one bar end supporting the mass via one spring and
the other bar end restrained by an identical elastic spring k as shown
in Figure 2.2. One can assume that the bar is free to rotate by an
angle and displacements are small from the equilibrium position.
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Copyright 2024 University of Southampton Page 6 of 10
Figure 2.2. Isolation comprising a massless bar and the two
identical springs and the attached mass.
(a) What are the displacements of points A and B at the bar ends
in terms of the bar rotation Hence draw the free body
diagram for the supported mass m and the massless bar
from the static equilibrium position. [6 marks]
(b) Considering the equation of motion for the bar, show that the
‘novel’ system is identical in its dynamic behaviour to one of
the systems presented in part (i) when the damper is not
present. [7 marks]

(c) A single viscous damper is to be added either in parallel with
the upper or lower spring. Explain why there is no preferable
position to add it. [4 marks]
TOTAL [33 MARKS]
Page 7 of 10
Q3
A uniform horizontal sign of length 8 is represented by a rigid bar
supported by two vertical elastic springs each of stiffness Kv as shown
in Figure 3.1 below. The bar has a mass M equal to 8 , where
is the mass per unit length. The bar has a moment of inertia J about
its centre equal to 6
2
.
Figure 3.1. A horizontal sign in its equilibrium position.
(i)
(a) Derive expressions for the potential and kinetic energies of
the system in terms of the small vertical displacement y of its
centre and small rotation in radians about its centre.
[4 marks]
(b) Using Lagrange’s equations, or otherwise, obtain the
equations of motion for the system in matrix form.
[8 marks]
(c) Show that the system has no rigid body modes in this plane.
[3 marks]
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Copyright 2024 University of Southampton Page 8 of 10
(ii) The bar is observed to be bending when the wind is gusting and
it is necessary to estimate the fundamental natural frequency
treating the bar as having a bending stiffness EI.
(a) Show that the function
( ) ( )
2
x W x Lx = +
is suitable to use in
Rayleigh’s method.
[2 marks]
(b) Using Rayleigh’s method, show that the estimated fundamental
natural frequency
0
, in rad/s, is given by the expression:
0 = √
15(16
65

,792
+

20

5

4)
[11 marks]
(c) The elastic springs have stiffness Kv equal to 40 kNm-1
. The
sign has a total length 8L= 8 m, mass per unit length
equal to 10 kgm-1 and a bending stiffness EI about its neutral
axis of 2 MNm-2
. Evaluate the estimated fundamental natural
frequency in Hz to two decimal places.
[3 marks]
(d) To avoid a resonance due to the wind loading both springs
are repositioned to be at the left-hand end of the beam.
Why is this a wrong decision
[2 marks]
TOTAL [33 MARKS]
END OF PAPER (Formula sheet follows)
Page 9 of 10
USEFUL FORMULAE
SDOF system m, k, c :
Viscous damping ratio =

= 2√

= 2

Half power bandwidth ≈ 2
Energy dissipated per cycle for a viscously damped system
excited harmonically =
2
Undamped natural frequency = √

in rad/s.
Damped natural frequency = √1
2
Receptance

=
1
2 +
=
1
1
1

2

2+ 2

Transmissibility = |
1
1

+

2
2

+
2
2

|,
Response to unbalance
Log decrement = (

+

1
) ≈ 2 for light damping, xi is the i
th
peak in the free response.
Free vibration solution (damped)
( ) =

( + ) =

( ( ) + ( ))
Impulse response function: ( ) =
1

( )
Convolution Integral ( ) = ∫

( ) ( )
2 2
1 2 2 2
n
n n
me X
M i

=
+
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Copyright 2024 University of Southampton Page 10 of 10
Lagrange’s equations of motion:

(

)

+

=
T, U and qj are the kinetic energy, potential energy and generalised
coordinate or degree of freedom, respectively. When the kinetic
energy is not a function of the displacements then

= 0 and this
term is removed.
Rayleigh’s Quotient: 0
2 =

Differential Equations of motion for beams under a
distributed loading:

4
4 +
2

2 = ( , )
For a beam
=

2

2
and =

3

3
using the convention shown in the diagram.
Expressions for the kinetic T and strain (potential) energies U for
a non-uniform beam:
=
1
2
∫ ( ) ( )
2 ; =
1
2
∫ ( ) ( ) (

2
2
)
2