STRUCTURES
There are three main types of structure - mass, framed and shells.
Mass structures perform due to their own weight.
An example would be a dam.
Frame structures resist loads due to the arrangement of its members. A house roof
truss can support a load many times its own weight. A two dimensional frame such
as this is known as a plane frame. The Eiffel Tower is an example of a three
dimensional frame structure, known as a space frame. Electricity pylons are good
examples of frame structures.
Shells are structures where its strength comes from the
formation of sheets to give strength. A car body is an
example of a shell structure.
Adding Vectors
In the structure below, three forces are acting on it.
The diagram shown below can represent the forces
in the above diagram.
FL1
FL2
FR
FL1 and FL2 represent the force exerted due to the
mass of the people. FR is the reaction force. This type of diagram is known as a free-body
diagram.
The reaction load FR is found by adding the two downward forces together.
In any force system the sum of the vertical forces must be equal to zero.
 Fv = 0
FR = FL1 + FL2
FR = 810N + 740N
There are two other conditions of static
equilibrium  FH = 0
- this states that all
forces acting horizontally must be equal to
zero.
FR = 1550N
 Fv = 0
- this is one of the
conditions of static equilibrium.
We shall meet the other condition of static
equilibrium later in the course.
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Resolution of a force
90N
30º
In some cases the loads may not be acting in the
same direction, and can not therefore be added
together directly.
BEAM
In the situation shown below the force is acting down at an angle.
FV
F
This force can be split into two separate components. A
vertical component FV and a horizontal component FH.
30º
FH
To resolve a force into its components you will have know two things, its magnitude and
direction.
Trigonometry is used to resolve forces.
sin a 
opp
hyp
cosa 
adj
hyp
Where - hyp = force (F)
opp = vertical component (FV)
adj = horizontal component (FH)
FV
The diagram above can be redrawn as below.
To find the horizontal force (FH)
cos 30  FH
90
FH  90  cos 30
 77.94 N
90N
30º
FH
Horizontal force = 77.94 N
To find the vertical force (FV)
sin 30  FV
90
FV  90  sin 30
 45 N
Vertical force = 45 N
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MOMENT OF A FORCE
Principles of Moments - Revision
The Principle of Moments states that if a body is in Equilibrium the sum of the clockwise
moments (given positive sign) is equal to the sum of anti-clockwise moments (negative sign).
This can be written:
MO=0
This is the third condition of static equilibrium.
We can use this principle to find an unknown force or unknown distance.
6N
2m
F
4m
M  0
0
CWM  ACWM
F  4  6  2
4 F  12
F
12
4
F  3N
BEAM REACTIONS
We are now going to study beams with external forces acting on them. We shall resolve
forces into their components and use moments to find the support reactions.
Definitions of some of the terms you have met already:
RESULTANT
The resultant is that single force that replaces a system of
forces and produces the same effect as the system it replaces.
EQUILIBRIUM
Equilibrium is the word used to mean balanced forces. A body
several forces acting on it which all balance each other is to be
in a state of equilibrium. A body in equilibrium can be at rest or
travelling with uniform motion in a straight line.
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Conditions of Equilibrium
In the force system in this section you shall apply the three condition of equilibrium that you
have used before. To solve the force systems the conditions of equilibrium are applied in a
certain order, the correct order is shown below.
1. The sum of all the moments equals zero.
 Mo = 0
2. The sum of the forces in the y direction equals zero.
 Fy = 0
3. The sum of the forces in the x direction equals zero.
 Fx =0
Beam Reactions
A beam is usually supported at two points. There are two main ways of supporting a beam -
1. Simple supports (knife edge)
15KN
20KN
2. Hinge and roller
180KN
200KN
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Worked Examples
1. SIMPLE SUPPORTS
Simple supports are used when there is no sideways tendency to move the beam.
10KN
8KN
0.5M
0.8M
0.8M
RA
RB
1. The forces at the supports called reactions, always act vertically.
2. The beam is in equilibrium; therefore the conditions of equilibrium apply.
The value of Reactions RA and RB are found as follows.
Take moments about RA
M RA  0
CWM  ACWM
(8  0.8)  ( RB  1.6)  (10  0.5)
6.4  1.6 RB  5
1.6 RB  6.4  5
RB 
1. 4
1. 6
RB  0.88 KN
There are two methods available to find
RA
Equating vertical forces
1. Take moments again, this time about
RB, or
2. Equate the vertical forces (since the
beam is in equilibrium)
FV  0
RA  RB  10  8
RA  18  RB
RA  18  0.88
RA  17.12 KN
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2. HINGE AND ROLLER SUPPORTS
Hinge and roller supports are used when there is a possibility that the beam may move
sideways.
20N
4m
2m
2m
30º
RA
40N
RB
Note: The reaction at a roller support is always at right angles to the surface. The
direction of RA is assumed. If any of the components work out as negative values then
the direction will be opposite the assumed direction.
The reaction at the hinge support can be any direction.
(Find the two components of the hinge reaction, then the resultantant)
There are three unknown quantities above: 1. The magnitude of Reaction RB.
2. The magnitude of Reaction RA.
3. The direction of Reaction RA.
Redraw as a free-body diagram showing vertical and horizontal components of the forces.
20N
4m
2m
HRA
2m
H(40)
V(40)
VRA
RB
The vertical and horizontal components of the 40N force are found first V(40) and H(40) V40  sin 30 0  40
H 40  cos 30 0  40
V40  20 N
H 40  34.64 N
To find RB -take moments about VRA, this eliminates one of the unknown vertical forces.
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M A  0
CWM  ACWM
(20  4)  (20  6)  ( R B  8)
8R B  200
RB 
200
8
R B  25 N
Vertical forces
Horizontal forces
FY  0
V RA  R B  20  20
V RA  40  R B
Fx  0
V RA  40  25
H RA  34.64  0
V RA  15 N
H RA  34.64 N
Use VRA and HRA to find RA
RA
15N
34.64N
RA  15 2  34.64 2
 225  1199.93
 37.75 N
Find direction of RA
A
tan a  VR
HRA
 15
34.64
 23.4
20N
37.75N
23.4º
40N
Higher Technological Studies Revision: Moments Outcome 1
25N
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