MPA104
Mechanical Design
Engineering Statics
Duncan Price
IPTME, Loughborough University
© Copyright: D.M.Price@lboro.ac.uk (2006)
1
Mechanics of Materials
Mechanics of Materials can be divided into
three categories:
 Mechanics of Rigid Bodies
 Statics – bodies at rest
 Dynamics – bodies in motion
 Mechanics of Deformable Bodies
 Mechanics of Fluids
2
Statics
Statics is thoroughly used in the analysis of
structures, for instance in architectural and
structural engineering. Strength of materials
is a related field of mechanics that relies
heavily on the application of static
equilibrium.
3
Course Content
12 Lectures:
1-5
7
8-11
12
4
Forces, springs
Free body diagrams
Resolution of forces (tutorial sheet)
Equilibrants and resultants of
forces (tutorial sheet)
Class test (calculators needed)
Levers, moments, reactions and
centre of gravity (tutorial sheet)
Online test
Newton’s Laws
 Sir Isaac Newton (1642-1729)
 Principia 1687
 Formulated three laws on which all
conventional motion is based
5
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 don’t understand
 The only one that doesn’t have an equation
6
Newton II
The resultant force on a particle is equal to the
rate of change of momentum of the particle.
d
F  mv ;
dt
dv
F m ;
dt
F  ma
The form F=ma is only valid if the mass is
constant.
7
Newton III
The forces of action and reaction between
interacting bodies are equal in magnitude and
opposite in direction.
F AB  F BA
 The force of the Earth on the Sun has the same
magnitude 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
8
Newton’s 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)
9
Newton’s Law of Gravitation (2)
On Earth:
F=mg
Where:
m is the mass of the body in question (in kg)
g = 9.81 m/s2
10
Spring Forces (1)
If a spring is stretched from L0 to L it exerts a force on
the object to which it is attached:
LO
F = K (L – L0)
F
F
L
K = spring constant = force required to stretch spring
by unit length (N/m).
K depends on spring material and design.
11
Spring Forces (2)
F = K (L – L0) is also known as Hooke’s Law:
The force exerted by a spring is proportional to
its extension.
Slope = F/ (L – L0) = K
F
(L-Lo)
12
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 the
body - not interested in internal forces
 all other bodies in contact with the one we are
interested in are replaced by vectors
13
Free body diagrams (2)
mg
Sketch of person standing
R1
R2
F=ma
R1+R2-mg=ma, but no acceleration so,
R1+R2=mg
14
Free body diagrams (3)
T
mg
sketch
free body
diagram
15
Free body diagrams (4)
Rules:
 clear sketches
 draw in the correct orientation
 show all forces acting on the body
 don’t show any internal forces between
different parts of the body
 show the forces not the components
16
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.
17
Vector addition
F1
F2
F3
These three forces act together
on the particle. Their effect is
equivalent to a single force
which is the vector sum of the
forces.
F1
FT is the resultant of the
forces F1, F2 and F3
F2
FT
F3
FT=F1+F2+F3
18
Trigonometry (1)
Pythagoras
Theorem:
A2 = B2 + C2
Internal angles
of a triangle
add up to 180°
90 - 
A
B

C
90°
19
Trigonometry (2)
sine() = B/A
cosine() = C/A
A
tangent() = B/C
(hypotenuse)
B
(opposite)

C
(adjacent)
20
Trigonometry (3)
sin(45°) = 1/√2 = 0.707
cos(45°) = 1/√2 = 0.707
45°
√2 m
tan(45°) = 1/1 = 1
1m
sin-1(0.707) = 45°
cos-1(0.707) = 45°
tan-1(1) = 45°
21
45°
1m
Trigonometry (4)
sin(30°) = 1/2 = 0.5
cos(30°) = √3/2 = 0.866
tan(30°) = 1√3 =0.577
60°
sin(60°) = √3/2 = cos(30°)
2m
1m
cos(60°) = 1/2 = sin(30°)
tan(60°) = √3/1 = 1.73
Generally:
cos() = sin(90°- )
tan() = sin()/cos()
sin2() + cos2() = 1
22
30°
√3 m
Sine and Cosine functions
1
sin()
0.8
0.6
0.4
0.2

0
-360
-315
-270
-225
-180
-135
-90
-45
0
-0.2
45
90
135
-0.4
-0.6
-0.8
-1
23
cos()
180
225
270
315
360
Tangent function
10
tan()
8
6
4
2

0
-90
-45
0
-2
-4
-6
-8
-10
24
45
90
Resolving forces (1)
Forces can be broken down into any number of
component forces
It is often convenient to choose two
perpendicular directions for resolution
F = Fx+Fy
Fy
F
FX
25
F = (Fx2+Fy2)1/2
Scalar magnitudes
Resolving forces (2)
Fy
Fx=F cos 
F

Fy=F sin 
=tan-1(Fy/Fx)
Fx
If the components are
perpendicular, they may
be added independently
FT=F1+F2+F3
26
F1
FTx=F1x+F2x+F3x
F2
F3
FTy=F1y+F2y+F3y
Resolving forces (3)
Fy=500 sin 30°
= 250 N
500 N
30°
Fx=500 cos 30°
= 433 N
27
Resolving forces (4)
F2 = 50 N
30°
F1 = 200 N
80°
FTx= F1x+F2x+F3x
= 200 – 50 cos 30° + 80 cos 80°
= 171 N
FTy= F1y+F2y+F3y
= 0 – 50 sin 30° – 80 sin 80°
= - 104 N (i.e. downwards)
28
F3 = 80 N
Resolving forces (5)
Fx = 171 N
FT= √(Fx2+Fy2)

= √(1712 + 1042)
= 200 N
 = tan-1(Fy/Fx)
FT
= tan-1(-104/171)
= -31°
29
Fy = 104
N
Resolving forces (6)
B
C
30°
75°
horizontal components:
FACcos75° = FABcos30°
A
FAC = FAB(cos30°/cos75°)
100 N
vertical components:
FACsin75° + FABsin30° = 100 N
FAB(cos30°/cos75°) sin75° + FABsin30° = 100
FAB = 26.8 N FAC = 89.7 N
30
Resultant
If two or more forces at a point they can be
replaced by a single force known as a
resultant.
31
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 straight
line*, the forces are said to be in equilibrium.
* i.e. Newton’s 1st law
32
Equilibrant
If two or more forces act at a point and are not
in equilibrium a force equal in magnitude and
opposite in direction to their resultant must be
applied to restore equilibrium. Such a force is
called the equilibrant.
33
Equilbrant and resultant
50 N
10 N
30 N
20 N
e
a
ae = resultant
ea = equilibrant
34
“Y-hangs”
A
B Horizontally
ACsin45° = BCsin45°
90°
C
AC = BC
Vertically
ACcos45° + BCcos45°=750 N
750 N
AC + BC =750 N/cos45°
2 AC = 750 N/cos45°
AC = 530 N = BC
What happens if we change the angle?
35
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.
36
Trusses (2)
Horizontal forces
ABcos30° = ACcos45°
500 N
AC = ABcos30°/cos45°
A
Vertical forces
ABsin30° + ACsin45° = 500 N
30°
B
ABsin30° + AB(cos30°/cos45°)sin45° = 500 N
45°
C
AB = 366 N
AC = 448 N
37
Trusses (3)
Horizontal forces
BC = ABcos30° = ACcos45°
= 316 N
A
366 N
Vertical forces at B
448 N
RB = ABsin30°
= 183 N
30°
B
45°
Vertical forces at C
C
RC = ACsin45°
= 317 N
Check
RB + RC = 183 N + 317 N = 500 N
38
Moment of a force
The product of the force and the perpendicular
distance from the line of action of the force to
the axis.
d
a
Moment about a = F d (units N m)
39
F
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.
40
Principle of Moments (2)
60 N
xN
b
2m
a
1.5 m
yN
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
41
Principle of Moments (3)
60 N
1.75 m
b
2m
49 N
a
xN
1.5 m
yN
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
42
Principle of Moments (4)
1.75 m
0.5 m b
1.5 m
xN
49 N
a
1.5 m
yN
Now support the unloaded beam at a and b:
Moments about b: 49  1.25 = y  1.5
y = 61.25/1.5 = 40.83 N
Therefore x = 49 – 40.83 = 8.17 N
43
Principle of Moments (5)
1.75 m
0.5 m b
xN
1.5 m
49 N
a
1.5 m
yN
c
zN
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
44
Principle of Moments (6)
1.75 m
49 N
1.75 m
c
zN
Now support the unloaded beam at c only…
45
Centre of Gravity (1)
The centre of gravity of an object is a point
at or near the object through which the
resultant weight of the object acts.
46
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.
47
Centre of Gravity (3)
For a rectangle or square, the centre of
gravity occurs at the centre of the section.
48
Centre of Gravity (4)
For a triangular section the centre of
gravity occurs at a point 1/3 of the height
from the base.
49
Centre of Gravity (5)
For a circular section the centre of gravity
occurs at the centre of the circle.
r
50
Centre of Gravity (6)
To determine the centre of gravity for a compound section:
1.
2.
3.
Divide the section into several parts (i.e. rectangles and
triangles, circles) so that the centre of gravity of each
individual part is known.
Determine the area and position of centre of gravity of each
part.
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.
51
Centre of Gravity (7)
Determine the position of the centre of
gravity of the L-section shown below:
0.75 m
0.5 m
1m
52
0.5 m
Centre of Gravity (8)
Divide up section into two rectangles,
identify c.o.g of each relative to O and
calculate area.
0.25 m
0.625 m
Area 2 = 0.125 m2
0.5 m
Area 1 = 0.5 m2
0.25 m
O
Total area = 0.625 m2
53
Centre of Gravity (9)
0.25 m
0.125 m2
Let X = location in X axis of c.o.g
0.5 m
0.5
m2
0.625 X = 0.50.5 + 0.1250.25
X = 0.45 m
O
Total area = 0.625 m2
54
Centre of Gravity (10)
0.125 m2
0.625 m
0.5 m2
0.25 m
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
O
Total area = 0.625 m2
55
Centre of Gravity (11)
Now let us drill a hole in the object, where
is the new centre of gravity?
0.4 m
Total area = 0.49936 m2
0.25 m
0.25 m
56
Area 3 = 0.12564 m2
Centre of Gravity (12)
Find new X and Y:
0.49936 X = 0.50.5 + 0.1250.25 –
0.125640.25
0.25 m
0.12564 m2
Total area = 0.49936 m2
X = 0.5 m
0.49936 Y = 0.50.25 + 0.1250.625
– 0.125640.25
Y = 0.344 m
57
Centre of Gravity (13)
Another example:
58
Centre of Gravity (14)
Divide the section up into rectangles and
triangles:
59
Centre of Gravity (15)
Select the first solution (but either method
would give the same answer).
60
Area 1
3x1.5/2
=
2.25m2
Area 2
4x1.5
=
6.00m2
Area 3
3x1.0
=
3.00m2
Total =
11.25m2
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 gravity
of each section can then be calculated from these
axes.
61
Centre of Gravity (17)
Let X be the horizontal distance to the
centre of gravity of the whole figure
measured 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
62
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
63
Centre of Gravity (19)
X
Y
3m
1.5 m
Now let us support the section at the middle
and far right…
What is the ratio of forces on each support?
64
Centre of Gravity (20)
Moments about Y.
1.5 RX = 0.9
RX = 0.6
X
3m
1.5 m
65
Y
therefore RY = 0.4