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8/6/2019 Eng Principles
1/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Mechanics
8/6/2019 Eng Principles
2/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
INTRODUCTIONBackground:
Engineers need to understand the effects that forces have on the bodies
which make up most mechanical systems. This field is called mechanics
and it is subdivided into two branches:
Statics
Where the forces acting on a body are balanced and the body is inequilibrium and:
Dynamics
Where the effects of forces on bodies and their subsequent motions are
studied.
Booklist:
Author Title Publisher CommentsHannah J. & Hillier M.J. Applied Mechanics Longman Still the best
introduction.
Johnson A. & Sherwin K. Foundations of Mechanical
Engineering
Nelson
Thornes
Statics, dynamics
and thermofluids.
Dyke P. & Whitworth R. Guide to Mechanics Palgrave
Stroud K.A. Engineering Mathematics Longman For maths
revisionBeer F.P & Johnston E.R. Vector Mechanics for Engineers McGraw Hill In 2 parts,
Good worked
examples
8/6/2019 Eng Principles
3/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
STATICS
Statics is the study of forces in equilibrium. A force in a balanced system
must be opposed by an equal and opposite force:
Multiple forces can also balance:
A force can be resloved into to components at right angles to each other:
To check if a system is in equilibrium, resolve all forces into x and y
components and check that balance:
F1x + F2x = 0 and F1y + F2y = 0
F1F2
F1
F1
F3
F4
F1x
F1
F2x
F2y F1
8/6/2019 Eng Principles
4/22
8/6/2019 Eng Principles
5/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Free Body Diagrams
A diagram showing the particular body or bodies of interest in isolationfrom all other parts of the mechanism, with all the forces acting on that
body shown as vectors.
Rules:1. Draw the body of interest2. Mark the centre of gravity and the force due to gravity Mg3. Add normal forces at all points of contact4. Add friction forces at these points5. Add any other external forces
Example:
8/6/2019 Eng Principles
6/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
STRESS AND STRAIN
Stress
Is intensity of loading stress = load [N/m2]area
It may be tensile (+ve) or compressive (-ve)
Strain
Is distortion of the body caused by stress
Strain = change in length [-]
original length
Hookes law
States that strain is directly proportional to applied stress up to the elastic
limit. The constant slope is known as the modulus of elasticity or
Young's modulus E. The modulus is a measure of the stiffness or rigidityof the material.
(for steel E = 210 GN/m2
)
Example
Find the minimum diameter of a 1m long steel wire required to support a
mass of 50kg without extending by more than 1 mm.
Take Young's modulus for steel to be 210 GN/m2
8/6/2019 Eng Principles
7/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Poisson's ratio
When a bar is loaded axially it stretches longitudinally aut it also
contracts laterally. Poisson's ration expresses the ratio of these twostrains:
Lateral strain [-]
Longitudinal strain
Shear stress
If two equal and opposite forces F, not in the same straight line, act on
parallel faces of a member , then it is said to be loaded in shear.
And the shear stress:
= F/A [N/m2]
Shear strain is the angle of deformation.
Shear stress = G ( the modulus of rigidity ) [N/m2]
Shear strain
(for steel G = 84 GN/m2
)
BENDING OF BEAMS
F
F
A
8/6/2019 Eng Principles
8/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
A beam under load usually experiences bending moments and shearforces which vary along and within the beam material.
There is a maximum bending moment and shear force that can be
sustained (different for each material) and understanding their
distribution can help an engineer to design efficient structures.
Both can be presented in diagrammatic form.
Shear force diagrams
w
w
w
+ve
w/2 w/2
w
w/2
-w/2
+ve
8/6/2019 Eng Principles
9/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Bending moment diagrams
.
w
-wd
d
Fixing moment
w/2 w/2
w
Wd/4
8/6/2019 Eng Principles
10/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Example
Freebody diagram
Must be in equilibrium therefore: RA+RB = 785 + 687
RA 785,2 + 687.5 = RB.4
so RB = 1250N
and RA = 785+687 - 1250
= 222N
2m
3m
4m 1m
George (80 kg) Vinny (70 kg)
BRIDGE
RA RB
80x9.81=785N 70x9.81=687N
8/6/2019 Eng Principles
11/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Shear force diagram
Bending moment diagram
222 N
687 N
222.2 = 444 Nm
222.4-785.2 = -682 Nm
8/6/2019 Eng Principles
12/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
More examples draw shear force and bending moment diagrams
1m
200N
100N
200N
2m10m
3m
100N
226
4030
2000N
2.5m
6m
Cable
1m
Cable
8/6/2019 Eng Principles
13/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Bending of beams
Bending equation for simple bending
( see Hannah + Hillier p 395 for derivation )
M = = EI Y R
Where M = bending moment [Nm]
I = 2nd
moment of area [m4]
= stress [N/m2]
y = distance from neutral axis [m]E = Youngs moduls [N/m
2]
R = radius of curvature [m]
For rectangular cross section
I = bd
3
12
y = d/2
For a circular cross section I = d4
64
y = d/2
So the stress at any point in a beam can be calculated or the radius of
curvature.
b
d
N A
N Ad
8/6/2019 Eng Principles
14/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
eg. For the practice example:
Find the maximum stress due to bending moment Mmax = 682 Nm
Need cross section
I = bd3
= 1.0.023
= 0.67.10-3
m4
12 12y = d/2 = 0.1m
M/I = /y
so = My = 682.0.1I 0.67.10
-3
= 102 kNm-2
1m
0.2m
8/6/2019 Eng Principles
15/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
DYNAMICSNewtons laws of motion
1st Law: The velocity of a body remains constant unless an external
force acts on it.
[the concept of inertia]
2nd Law: When a force acts on a body its momentum changes at a rate
proportional to the magnitude of the force and in the direction
of the force.
[the concept of acceleration]
3rd Law: To every action there is an equal and opposite reaction.
[the concept of equilibrium]
NOTES
Weight and Mass:
Mass is a fixed property of a body (kg)
Weight is the force on a body due to a gravitational field (N)
Momentum = mass x velocity (kgm/s)
Translation and Rotation:
Force (N) = Mass (kg) * Acceleration (m/s2)
Torque (Nm) = Moment of inertia (kgm2) * Angular acceleration (rad/s
2)
Friction
friction force f = N (where N = normal force
= coefficient of friction )
always opposes the motion
Impulse = force x time acting (Ns)
8/6/2019 Eng Principles
16/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Equations governing motion with constant acceleration
V = U + a x t
s =1
2(U+V) x t
s = U x t +1
2a x t
2
V2
= U2
+ 2 x a x s
where V = final velocity
U = initial velocity
a = acceleration
t = time
s = displacement
(for derivations see e.g. Hannah and Hillier page 73)
Example 1: Projectiles
A projectile is a moving body that is acted on only by gravity. The
motion of a projectile can be fully described if its initial speed (Sp) and
angle to the horizontal () are known.
Sp
8/6/2019 Eng Principles
17/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
It is convenient to separate vertical and horizontal motions for all
calculations. e.g.:
1. Time to maximum height
use V = u + a x t on vertical motion:
t1 =
2. Maximum heightuse s = U x t +
1
2a x t
2on vertical motion:
h =
3. Overall time (actually 2.t1 but need to proove)
use s = U x t + 12
a x t2 on vertical motion from top to bottom:
t2 =
4. Distance travelled
use s = U x t +1
2a x t
2on horizontal motion:
R =
(Now calculate the velocity at impact)
8/6/2019 Eng Principles
18/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Motion in a circle
From Newtons first law we can see that a body moving in a circular path
must have a force acting on it. The general term for this force is the
centripetal force and it acts towards the centre of the circle.
For a body of mass M [kg] moving in a circular path of radius r [m] and
with constant velocity v[m/s]
the centripetal force required is F =Mv
r
2
Example 2: A cyclist in a curve
Draw a free body diagram of the cyclist:
8/6/2019 Eng Principles
19/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Calculate:
1. The angle at which the cyclist must lean:
tan =v
rg
2
2. The maximum velocity the cyclist can travel without sliding:
v rg=
8/6/2019 Eng Principles
20/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
ENERGY
Background:
Principle of conservation of energy:
Energy cannot be lost or gained in a closed system but
can be interchanged in form (at specific rates of
conversion).
There are many different type of stored energy (chemical, electrical, heat
etc. ) but in mechanical systems we are mainly interested in the
interchange between energy stored in a mass in a gravitational field
(potential energy) and energy associated with a moving body (kinetic
energy).
Equations governing energy
Potential energy: P.E. = mgh
Kinetic energy: K.E. =1
2
2mv (for translational motion)
K.E. =1
2
2I (for rotational motion)
Where: m = mass [kg]v = velocity [m/s]
I = moment of inertia [kgm2]
= angular velocity [rad/s]
8/6/2019 Eng Principles
21/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
A note on I
If we consider a generalbody made up of n particles
each with mass mn and that
body rotates about a pivot at
0 with an angular velocity
:
The kinetic energy of each particle isK.E. =
1
2
2m vi i
and therefore the kinetic energy of the whole body is
K.E. =1
2
2
1
m vi ii
n
=
K.E. =1
21
2 2
i
n
i i im r
=
K.E. =1
2
2
1
2
i
n
i im r
=
and the boxed part is I.
I can be calculated by integrating mr2 over the total cross sectional area
of the body concerned (perpendicular to the axis of rotation).
Or it can be looked up in a table!........
8/6/2019 Eng Principles
22/22
Manchester Metropolitan University Simon Iwnicki
BSc(Hons) Mechanical Engineering Network Room E354
Unit 64ET4903 Mechanical Engineering Principles [email protected]
C:\teaching\bscep.doc
Some common values of I:
Cylinder I =1
2
2Mr
Sphere I =2
5
2Mr
Thin rod I =1
12
2ML
I =1
3
2ML
Example
A 25 mm dimeter cylinder is allowed to roll from the top of a plane 1.25
m long inclined at an angle of of 20o
to the horizontal. Neglecting friction
and air resistance and assuming that the cylinder does not slip determine
the linear velocity of the cylinder on reaching the bottom of the plane.
[2.3 m/s]