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SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS
SEMESTER 2 EXAMINATIONS 2011/2012
ENGINEERING MECHANICS
FY016
DR. S.M. BEGG
Time allowed:
TWO hours
Answer:
Any FOUR questions
The total number of questions is SIX
Each question carries 25 marks
Items permitted:
Any approved calculator
Items supplied:
Formulae sheet (pages 7-9)
Marks for whole and part questions are indicated in brackets ( )
May/June 2012
Page 1 of 9
Question 1
(a)
Describe tensile and shear stress and show how they occur in a simply
supported beam subjected to a load at its midpoint.
(5 marks)
(b)
A beam of length 10 m and negligible weight rests on supports at each end. A
concentrated load of 650 N is applied at 3 m from the left-hand end. A second
load of 325 N is applied at 8 m from the same end. Sketch the beam and
calculate the reaction force at each of the supports.
(10 marks)
(c)
Plot the bending moment as a function of distance along the beam.
(7 marks)
(d)
Plot the shear force as a function of distance along the beam.
(3 marks)
Question 2
(a)
Explain Newton’s first law of motion.
(2 marks)
(b)
A mass of 50 kg, travelling at a velocity of 45 ms-1, is struck from behind by a
mass of 60 kg travelling in the same direction at a velocity of 75 ms-1. The
coefficient of restitution for the two materials is 0.9. Using the principle of
conservation of linear momentum and Newton’s law of impact, calculate the
final velocities of the two masses.
(17 marks)
(c)
What is the force required to accelerate the lighter mass from its initial velocity
to its final velocity in 3 s?
(6 marks)
FY016 (2011/2012)
Page 2 of 9
Question 3
A circular, horizontal pipe uniformly tapers from a diameter of 160 mm to a diameter
of 70 mm. Hydraulic oil, of density 1200 kgm-3, flows through the pipe from the
greater diameter cross-section to the smaller diameter cross-section. The pressure
at the greater diameter cross-section is 160 kPa whilst the pressure at the smaller
diameter cross-section is 50 kPa. Using the continuity principle and Bernoulli’s
equation:
(a)
Calculate the velocity of the oil at both cross-sectional areas.
(17 marks)
(b)
Calculate the volume flow rate of oil through the pipe.
(3 marks)
(c)
The oil exits the pipe as a round jet with a velocity of 12 ms-1 where it strikes
perpendicular to a flat, solid plate. What is the reaction force that must be
applied to the plate to resist motion once the jet impacts?
(5 marks)
Question 4 commences on the next page
FY016 (2011/2012)
Page 3 of 9
Question 4
A new high-performance racing bicycle wheel has a diameter of 700 mm and a mass
of 871 g. In a competition, the rider accelerates from a linear velocity of 36 kmhr-1 to
47 kmhr-1 in a time of 12.4 s. Determine the following:
(a)
the final angular velocity of the wheel (in rads-1).
(3 marks)
(b)
the number of revolutions of the wheel.
(4 marks)
(c)
the final centripetal acceleration of the wheel.
(3 marks)
(d)
the torque applied to the wheel (consider the wheel as a hoop).
(9 marks)
(e)
the work done and the average power applied.
(6 marks)
Question 5 commences on the next page
FY016 (2011/2012)
Page 4 of 9
Question 5
A particle of mass 150 g is subjected to a system of four forces as shown below in
Figure Q5:
7N
15 N
30°
6N
39°
5°
18 N
Figure Q5
(a)
Sketch the vector force diagram, approximately to scale, clearly indicating the
resultant force vector.
(4 marks)
(b)
Calculate the resultant force vector.
(15 marks)
(c)
The particle moves with a constant acceleration of 1.4 ms-2, from an initial
velocity of 22 ms-1. After 150 s, the particle stops accelerating and immediately
decelerates at a constant rate of 2.65 ms-1. What is the maximum amount of
kinetic energy that it will have lost by 190 s?
(6 marks)
FY016 (2011/2012)
Page 5 of 9
Question 6
A spring and mass system is suspended vertically. Once the system is displaced, it
undergoes simple harmonic motion. The natural frequency of the system must be
3 Hz to avoid resonance effects. Determine the following:
(a)
The required spring stiffness for a suspended mass of 52 kg.
(6 marks)
(b)
If the maximum amplitude of the vibration is 42 mm, calculate the maximum
force in the spring, taking into account the initial steady-state displacement.
(8 marks)
(c)
Calculate the maximum acceleration of the mass and describe at what point in
the simple harmonic motion that this will occur.
(5 marks)
(d)
Calculate the maximum velocity of the mass and describe at what point in the
simple harmonic motion that this will occur.
(6 marks)
FY016 (2011/2012)
Page 6 of 9
FORMULA SHEET
Data:
Acceleration due to gravity, g = 9.81 ms-2
Density of water, ρ = 1000 kgm-3
Formulae:
Projectiles:
Maximum height, h
v 2 (sin  ) 2
h
2g
Time of flight, t
t
2v sin 
g
Range of flight, R
R
Newton’s second law:
v 2 sin( 2 )
g
Force = rate of change of momentum
Equations of linear motion with constant acceleration:
s  ut 
1 2
at
2
v  u  at
v 2  u 2  2as
s
u  v  t
2
Coefficient of restitution, e:
e 
v1  v2 
u1  u 2 
Fluid flow
Volumetric flow rate,
V  Av
Mass flow Rate,
m  V
Bernoulli’s equation:
P  gh 
FY016 (2011/2012)
1 2
v  constant
2
Page 7 of 9
Venturi meter:
2 gH
A12  A22
Volume flow rate = C d A1 A2
Bending moment equation:
M  E
 
I
y R
Second moments of area:
Rectangle:
I=
Solid cylinder:
bd 3
12
I=
d 4
64
Circular motion
Angle turned through:
  t
Linear velocity:
v  r
Angular acceleration:

 2  1
t
or  
a
r
Equations of motion for constant angular acceleration:
 2  1   t
 1   2 
t
2 

1
  1t  t 2
2
2
2
 2  1  2
 
v2
r
Centripetal acceleration:
ac   2 r 
Centripetal force:
P  m 2 r  m
Angular momentum:
A  I
1 2
I
2
Rotational kinetic energy:
=
Torque:
T = I
FY016 (2011/2012)
v2
r
Page 8 of 9
Moments of inertia:
Solid disc:
Hoop:
I
1
mr 2
2
I  mr 2
Simple harmonic motion
Displacement:
  t
x  r sin( t )
Velocity:
v y  v sin    (r 2  y 2 )
Acceleration:
a y  a cos   2 y
maximum displacement occurs at y = r
Period, T:
ω = 2π/T
T=2πr/v
Frequency, f
f= 1/T Hz
Restoring force:
Fr = ky
ky-ma = 0
Period:
T = 2π√(m/k)
Spring-mass
Simple pendulum
Restoring torque:
Torque = mgx = mglθ for small θ
Torque = Iα = ml2α
Period:
T = 2π√(l/g)
Angular acceleration:
α = gθ/l
Maximum angular velocity: ωp = ωr/l when θ = 0
FY016 (2011/2012)
Page 9 of 9