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MPA104
Mechanical Design
Engineering Statics
Duncan Price
IPTME, Loughborough University
Copyright: D.M.Price@lboro.ac.uk (2006)
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Mechanics of Materials
Mechanics of Materials can be divided into
three categories:
Mechanics of Rigid Bodies Staticsbodies at rest
Dynamicsbodies in motion
Mechanics of Deformable Bodies
Mechanics of Fluids
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Statics
Statics is thoroughly used in the analysis of
structures, for instance in architectural andstructural engineering. Strength of materials
is a related field of mechanics that relies
heavily on the application of static
equilibrium.
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Course Content
12 Lectures:1-5 Forces, springs
Free body diagramsResolution of forces (tutorial sheet)
Equilibrants and resultants offorces (tutorial sheet)
7 Class test (calculators needed)
8-11 Levers, moments, reactions and
centre of gravity (tutorial sheet)12 Online test
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Newtons Laws
Sir Isaac Newton (1642-1729)
Principia 1687
Formulated three laws on which all
conventional motion is based
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Newton I
A particle remains at rest or continues to move
at a constant speed in a straight line unless
there is a constant force acting on it.
The most important law
The one that most people dont understand
The only one that doesnt have an equation
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Newton II
The resultant force on a particle is equal to therate of change of momentum of the particle.
amFdtvd
mFvmdt
d
F ;;
The form F=ma is only valid if the mass is
constant.
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Newton III
The forces of action and reaction between
interacting bodies are equal in magnitude and
opposite in direction.
BAAB FF
The force of the Earth on the Sun has the samemagnitude as the force of the sun on the earth
The force of a tennis ball on a racket has the same
magnitude as the force of the racket on the ball
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Newtons Law of Gravitation (1)
F = G M m / r2
Where:
M & m are particle masses
G is the universal constant of gravitation
(6.673 x 10-11 m3/kg s2)
r is the distance between the particles.
F in Newtons (kg m/s2)
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Newtons Law of Gravitation (2)
On Earth:
F = m g
Where:
m is the mass of the body in question (in kg)g = 9.81 m/s2
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Spring Forces (1)
If a spring is stretched from L0to L it exerts a force on
the object to which it is attached:
F = K (LL0)
K = spring constant = force required to stretch springby unit length (N/m).
K depends on spring material and design.
LO
L
FF
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Spring Forces (2)
F = K (LL0) is also known as Hookes Law:The force exerted by a spring is proportional to
its extension.
Slope = F/ (LL0) = KF
(L-Lo)
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Free body diagrams (1)
FBD is an essential step in the solution of all
problems involving forces on bodies
it is a diagram of the external surface of thebody - not interested in internal forces
all other bodies in contact with the one we are
interested in are replaced by vectors
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Free body diagrams (2)
Sketch of person standing
mg
R1 R2
F=ma
R1+R2-mg=ma, but no acceleration so,
R1+R2=mg
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Free body diagrams (3)
sketch
mg
T
free bodydiagram
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Free body diagrams (4)
Rules:
clear sketches
draw in the correct orientation show all forces acting on the body
dont show any internal forces between
different parts of the body show the forces not the components
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Representation of a force by a vector
A force has the following characteristics:
Magnitude
Direction Point of application
A quantity which has magnitude and direction is
a vector.A quantity which has magnitude only is a scalar.
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Vector addition
F1F2
F3
F1
F2
F3FT
These three forces act together
on the particle. Their effect isequivalent to a single force
which is the vector sum of the
forces.
FTis the resultant of the
forces F1, F2and F3
FT=F1+F2+F3
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Trigonometry (1)
A
B
C
Pythagoras
Theorem:
A2= B2+ C2
Internal anglesof a triangle
add up to 180
90 -
90
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Trigonometry (2)
A
(hypotenuse) B(opposite)
C
(adjacent)
sine() = B/Acosine() = C/Atangent() = B/C
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Trigonometry (3)
sin(45) = 1/2 = 0.707
cos(45) = 1/2 = 0.707
tan(45) = 1/1 = 1
sin-1(0.707) = 45
cos-1(0.707) = 45
tan-1(1) = 45
2 m
1 m
1 m
45
45
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Trigonometry (4)sin(30) = 1/2 = 0.5
cos(30) = 3/2 = 0.866
tan(30) = 13 =0.577
sin(60) = 3/2 = cos(30)
cos(60) = 1/2 = sin(30)tan(60) = 3/1 = 1.73
Generally:
cos() = sin(90- )
tan() = sin()/cos()
sin2() + cos2() = 1
1 m
2 m
3m
30
60
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-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-360 -315 -270 -225 -180 -135 -90 -45 0 45 90 135 180 225 270 315 360
Sine and Cosine functions
sin()
cos()
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-10
-8
-6
-4
-2
0
2
4
6
8
10
-90 -45 0 45 90
Tangent function
tan()
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Resolving forces (1)
Forces can be broken down into any number ofcomponent forces
It is often convenient to choose two
perpendicular directions for resolution
FFy
F = Fx+Fy
F = (Fx2+Fy
2)1/2
Scalar magnitudesFX
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Resolving forces (2)
F
Fx
Fy
Fx=F cos
Fy=F sin
=tan-1(Fy/Fx)
If the components are
perpendicular, they may
be added independently
F1F2
F3
FT=F1+F2+F3 FTx=F1x+F2x+F3x FTy=F1y+F2y+F3y
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Resolving forces (3)
500 N
30Fx=500 cos 30
= 433 N
Fy=500 sin 30
= 250 N
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Resolving forces (4)
FTx= F1x+F2x+F3x
= 20050 cos 30 + 80 cos 80
= 171 N
FTy= F1y+F2y+F3y
= 050 sin 3080 sin 80
= - 104 N (i.e. downwards)
F1= 200 N
F2 = 50 N
F3= 80 N
30
80
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Resolving forces (5)Fx = 171 N
Fy= 104
N
FT= (Fx2+Fy2)
= (1712+ 1042)
= 200 N
= tan-1
(Fy/Fx)
= tan-1(-104/171)
= -31
FT
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Resolving forces (6)
30 75
100 N
A
B C horizontal components:
FACcos75 = FABcos30
FAC= FAB(cos30/cos75)
vertical components:
FACsin75 + FABsin30 = 100 N
FAB
(cos30/cos75)sin75 + FAB
sin30 = 100
FAB= 26.8 N FAC= 89.7 N
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Resultant
If two or more forces at a point they can bereplaced by a single force known as a
resultant.
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Equilibrium
When two or more forces act upon a body and
are so arranged that the body remains at rest
or moves at a constant speed in a straightline*, the forces are said to be in equilibrium.
* i.e. Newtons 1stlaw
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Y-hangs
What happens if we change the angle?
90
750 N
A B
C
Horizontally
ACsin45 = BCsin45
AC = BC
Vertically
ACcos45 + BCcos45=750 N
AC + BC =750 N/cos45
2 AC = 750 N/cos45
AC = 530 N = BC
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Trusses (1)
One of the basic methods to determine
loads in individual truss members is called
the Method of Joints. Each joint is treated as
a separate object and a free-body diagram
is constructed for the joint.
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Trusses (2)
30 45
A
B C
500 N
Horizontal forces
ABcos30 = ACcos45
AC = ABcos30/cos45
Vertical forces
ABsin30 + ACsin45 = 500 N
ABsin30 + AB(cos30/cos45)sin45 = 500 N
AB = 366 N
AC = 448 N
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Trusses (3)
30 45
A
B C
366 N
Horizontal forces
BC = ABcos30 = ACcos45
= 316 N
Vertical forces at B
RB= ABsin30
= 183 N
Vertical forces at C
RC= ACsin45
= 317 N
Check
RB+ RC= 183 N + 317 N = 500 N
448 N
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Moment of a force
The product of the force and the perpendiculardistance from the line of action of the force tothe axis.
Moment about a = F d (units N m)
d F
a
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Principle of Moments (1)
If a body is at rest under the action of several forces, the total
clockwise moment of the forces about any axis is equal to
the anticlockwise moment of the forces about the same axis.
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Principle of Moments (2)
2 m 1.5 m
x N60 N
ab
y N
Moments about a: 60 2= x 1.5
x = 120/1.5 = 80 N
Moments about b: (2 + 1.5) 80 = 2 y
y = 280/2 = 140 N (ignoring mass of beam)
Or more simply: y = 60 + 80 = 140 N
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Principle of Moments (3)
2 m 1.5 m
x N60 N
ab
y N
Now let the beam weigh 5 kg
Additional force of 49 N acting centrally.
Moments about a: (60 2) + (49 0.25)= x 1.5
x = 144.5/1.5 = 88.2 N
Reaction of support y = 60 + 49 + 88.2 = 197.2 N
49 N1.75 m
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Principle of Moments (5)
1.5 m 1.5 mab
y N
Now support the unloaded beam at a, b and c:
Moments about b: 49 1.25 = y 1.5 + z 3
Moments about a: 49 0.25 = x 1.5 - z 1.5
Moments about c: 49 1.75 = x 3 + y 1.5
49 N1.75 m
x N
0.5 m c
z N
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Principle of Moments (6)
1.75 m
Now support the unloaded beam at c only
49 N1.75 m
c
z N
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Centre of Gravity (2)If a vertical line through the centre of gravity falls outside the
base upon which the body relies for stability, overturning will
result, unless precautions, such as tying the base down, are
taken.
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Centre of Gravity (3)
For a rectangle or square, the centre ofgravity occurs at the centre of the section.
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Centre of Gravity (4)
For a triangular section the centre ofgravity occurs at a point 1/3 of the height
from the base.
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Centre of Gravity (5)
For a circular section the centre of gravityoccurs at the centre of the circle.
r
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Centre of Gravity (6)To determine the centre of gravity for a compound section:
1. Divide the section into several parts (i.e. rectangles and
triangles, circles) so that the centre of gravity of each
individual part is known.
2. Determine the area and position of centre of gravity of each
part.
3. Take moments about a convenient axis to determine the
centre of gravity of the whole body.
This is based on the principle that, along any one axis (or in any
one direction):
when the distance is measured from the same point in each case.
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Centre of Gravity (7)
Determine the position of the centre ofgravity of the L-section shown below:
0.5 m 0.5 m
1 m
0.75 m
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Centre of Gravity (8)
Divide up section into two rectangles,identify c.o.g of each relative to O and
calculate area.
0.5 m
0.25 m
Area 1 = 0.5 m2
0.25 m
0.625 m Area 2 = 0.125 m2
O
Total area = 0.625 m2
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Centre of Gravity (9)
Let X = location in X axis of c.o.g0.5 m
0.25 m
0.5 m2
0.125 m2
O
Total area = 0.625 m2
0.625 X = 0.50.5 + 0.1250.25X = 0.45 m
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Centre of Gravity (10)
0.5 m2
0.25 m
0.625 m
0.125 m2
O
Total area = 0.625 m2
Let Y = location in Y axis of c.o.g
0.625 Y = 0.125 0.625 + 0.5 0.25
Y = 0.325 m
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Centre of Gravity (11)
Now let us drill a hole in the object, whereis the new centre of gravity?
0.4 m
0.25 m
0.25 m
Area 3 = 0.12564 m2
Total area = 0.49936 m2
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Centre of Gravity (12)
Find new X and Y:
0.25 m 0.12564 m2
Total area = 0.49936 m2
0.49936 X = 0.50.5 + 0.1250.250.125640.25X = 0.5 m
0.49936 Y = 0.50.25 + 0.1250.6250.125640.25
Y = 0.344 m
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Centre of Gravity (14)
Divide the section up into rectangles andtriangles:
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Centre of Gravity (15)
Select the first solution (but either methodwould give the same answer).
Area 1 3x1.5/2 = 2.25m2
Area 2 4x1.5 = 6.00m2
Area 3 3x1.0 = 3.00m2
Total = 11.25m2
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Centre of Gravity (16)
Select two axes A-A and B-B, at the extreme edge
of the figure. The distance to the centre of gravityof each section can then be calculated from these
axes.
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Centre of Gravity (17)
Let X be the horizontal distance to the
centre of gravity of the whole figuremeasured from A-A and Y be the vertical
distance to the centre of gravity of the
whole figure measured from B-B.
Taking moments about A-A:
11.25X = (2.25x2.25)+6.0x2.25)+(3.0x1.5)
= 5.6+13.5+4.5
= 23.6
X = 23.6/11.25 = 2.1 m
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Centre of Gravity (18)
Taking moments about B-B:
11.25Y = (2.25x6.0) + (6.0x3.0) + (3.0x0.5)= 13.5 + 18.0 + 1.5
= 33.0
Y = 33.0/11.25 = 2.9m
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Centre of Gravity (19)
Now let us support the section at the middle
and far rightWhat is the ratio of forces on each support?
3 mX
Y
1.5 m
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Centre of Gravity (20)
Moments about Y.
1.5 RX = 0.9
RX= 0.6
therefore RY= 0.4
3 mX
Y
1.5 m