DET
Technological Studies
Structures and Materials
(Advanced Higher)
7135
June 2000
HIGHER STILL
DET
Technological Studies
Structures and Materials
(Advanced Higher)
Support Materials
CONTENTS
Overview
Content
Outcome 1
Apply the general bending equation in solving problems on idealised beams
Outcome 2
Evaluate the distribution of sheer force and bending moment in loaded beams
Outcome 3
Apply the method of sections in solving problems on complex framed structure
systems
DET: Technological Studies: Structures and Materials (Advanced Higher)
1
DET: Technological Studies: Structures and Materials (Advanced Higher)
2
Overview
The support materials for Technological Studies courses in Higher Still have been
created to specifically address the outcomes and PC in each unit at the appropriate
level. These support materials contain a mixture of formal didactic teaching and
practical activities.
The support materials for each unit have been divided into outcomes. This will
facilitate assessment as well as promoting good practice.
The materials are intended to be non-consumable, however it is at the discretion of
each centre how to use these materials.
Each package of Support Materials follows a common format:
 statement of the outcome
 statement of what the student should be able to do on completion of the outcome
 learning and teaching activities
 sequence of structured activities and assignments.
It is important to note that the National Assessments have been designed to allow
assessment either after each outcome has been completed or as an end of unit
assessment when all outcomes have been completed depending on the needs of the
centre.
SQA past paper questions have been used throughout the materials and the further use
of these questions is encouraged.
Using past questions provides an opportunity for students to:
 work at the appropriate level and rigour
 prepare for external assessment
 consolidate learning and teaching
 integrate across units.
Homework is a key factor in effective learning and teaching. The use of resources
such as P & N practice questions in Technological Studies is very useful for homework
activities and also in preparation for assessment.
The use of integrated questions across units is essential in preparation for the
External Assessment.
DET: Technological Studies: Structures and Materials (Advanced Higher)
3
DET: Technological Studies: Structures and Materials (Advanced Higher)
4
CONTENT
Outcome 1 – Apply the general bending equation in solving problems on idealised
beams.
The purpose of this outcome is to introduce students to the concept of beam design and
the use of the general bending equation.
When students have completed the outcome they should be able to:
 apply the general bending equation to simple beam problems and appreciate the
factors involved in beam design
 use tabulated data to calculate maximum deflection and bending moment in
standard beam configurations
 use tabulated materials properties in simple beam designs
 understand how cross-sectional shape and member length influence how efficiently
material is used in buckling situations.
Outcome 2 – Evaluate the distribution of shear force and bending moment in
loaded beams.
The purpose of this outcome is to explain how the tabulated data relating to beams
found in the SQA data booklet has been calculated. The unit demonstrates how to
construct shear force and bending moment diagrams.
When students have completed this outcome they should be able to:
 interpret loading information on a beam from a shear force or bending moment
diagram
 construct a simple shear force and bending moment diagram.
Outcome 3 – Apply the method of sections in solving problems on complex
framed structure systems.
The purpose of this outcome is to introduce students to the method of sections as
applied to solving problems with framed structures.
When students have completed this outcome they should be able to:

apply the method of sections to pin-jointed structures.
DET: Technological Studies: Structures and Materials (Advanced Higher)
5
DET: Technological Studies: Structures and Materials (Advanced Higher)
6
OUTCOME 1 – APPLY THE GENERAL BENDING EQUATION IN SOLVING
PROBLEMS ON IDEALISED BEAMS
The purpose of this outcome is to explain the theory of bending as applied to beams.
When you have completed this unit you should be able to:
 apply the general bending equation to simple beam problems and appreciate the
factors involved in beam design
 use tabulated data to calculate maximum deflection and bending moment in
standard beam configurations
 use tabulated materials properties in simple beam designs
 understand how cross-sectional shape and member length influence how efficiently
material is used in buckling situations.
Before you start on this unit you should have a clear understanding of:
 Young's modulus
 stress
 strain
 equilibrium.
DET: Technological Studies: Structures and Materials (Advanced Higher)
7
Beams and bending
The use of a beam as a structural element has been applied since ancient times. Early
structures to use beams were built in places of worship such as Stonehenge.
SM AH O.1 fig 1
Later cultures such as that of the Greeks and Romans made extensive use of stone
beams in the construction of their buildings.
SM AH O.1 fig 2
The problems faced by these early structural designers in using beams are exactly the
same as those faced by modern engineers. When a beam is placed across a gap
between two columns, the beam will tend to sag due to the effects of its own weight.
Fw
SM AH O.1 fig 3
The bending effect on the beam will increase when a structural load is applied to it.
DET: Technological Studies: Structures and Materials (Advanced Higher)
8
If attempts are made to improve the strength of the beam by increasing its crosssection, a larger proportion of the beam's strength is required to support its own weight.
At the same time, the size and strength of the supporting structure will require to be
increased proportionally to support the beam. A point is reached in designing any
structure where it becomes counter-productive to keep increasing the size of a
structural member.
If the loading on a member is too great or the distance between beam supports is too
large for the applied load, the beam will fracture.
Fw
SM AH O.1 fig 4
Greek and Roman engineers were well aware of these problems and for this reason
standard spans for standard sections of beams for specific loadings, were used in the
design of structures.
SM AH O.1 fig 5
From these restrictions, an elegant style of structural design evolved.
The principal building material of the Romans and Greeks was stone. Whilst stone is
an excellent material if used in compression, it has poor ductile properties which
means that if it is bent a relatively small amount it will break. Modern engineers face
the same problems in using materials such as concrete, which is very brittle. This
problem is overcome in concrete beam design by using pre-tensioned steel ties. These
steel ties perform well in tension and are fixed into the concrete beams when they are
cast.
DET: Technological Studies: Structures and Materials (Advanced Higher)
9
TIE RODS
SM AH O.1 fig 6
When a bending load is applied to the beam, the rods resist the effects of bending by
pulling the bottom of the beam inwards.
PRE-STRESSED
CONCRETE
Fw
RODS PULL THE
FR TIE
BOTTOM OF THE
FR
BEAM IN
SM AH O.1 fig 7
Steel has revolutionised structural design. It has a combination of good strength, high
modulus of elasticity and relatively good ductility. Steel can be shaped into joists by
hot roiling to improve resistance to bending.
For example, an 'I' Beam like the one shown below, is shaped so that the maximum
amount of material is located towards the top and bottom surfaces of the beam.
SM AH O.1 fig 8
DET: Technological Studies: Structures and Materials (Advanced Higher)
10
This type of structural section behaves similarly under load to a solid steel beam of the
same section but, of course, uses less material, weighs less and therefore less of its
strength is required to support its own weight and more is available to support the load.
SM AH O.1 fig 9
Bending stresses in beams
When a bending moment exists at a cross-section of a beam there is an internal force
distribution causing stresses to exist in the material of the beam. The bending moment
is equal to the resultant moment of these stresses if the beam is in equilibrium.
Assumptions involved in the theory
The first assumption to be made is that when a beam is bent, the material on the
convex side is stretched and the material on the concave side is compressed. Tensile
stresses will be set up on the convex side and compressive stresses on the concave side.
The material may be considered to be in simple tension or simple compression.
COMPRESSIVE STRESSES
TENSILE STRESSES
SM AH O.1 fig 10
Other assumptions in the theory are:
 that the material is homogeneous and obeys Hooke’s Law
 the modulus of elasticity has the same value in tension and compression
 the effect of shearing forces on the distribution of stresses can be neglected
 cross-sectional planes originally perpendicular to the longitudinal axis of the beam
remain plane and normal to the axis after bending.
These assumptions are arbitrary since they cannot be predicted from any underlying
theory. They are reasonably accurate however, as the theory of bending stresses is
found to give satisfactory experimental results.
DET: Technological Studies: Structures and Materials (Advanced Higher)
11
Derivation of the stress/curvature relationship
The diagrams below shows part of a beam which is originally straight and which bends
to the form of a circular arc. When it is deformed, part of the beam is in tension and
part in compression. Hence there will be a plane in the material separating the tensile
and compressive regions where no stresses exist. This plane is called the neutral
surface and the line NN which lies in this plane is called the neutral axis (NA).
A
P
B
A
P
y
N
B/
/
P
N
A
/
P
y
/
/
N
N
A/
B
/
B/
R
SM AH O.1 fig 11
Consider the following:
 when the beam is deformed, the material in planes, AA, BB and NN shift to form
planes A’A’, B’B’ and the curved neutral surface N’N’
 since the neutral surface is free of stress and hence there can be no strain, it follows
that there is no alteration in the length of NN i.e. NN = N’N’
 the material originally in layer PP at a distance y from the neutral surface deforms
to form P’P’.
The longitudinal strain in this layer is given by:

L P ' P ' PP

L
PP
In its unbent state PP = NN = N’N’
 
P ' P ' N ' N '
NN
Using arc length = radius x subtended angle (in radians)
NN  R and P' P'  R  y 
R  y   R
 
R
y
y
 

R R
DET: Technological Studies: Structures and Materials (Advanced Higher)
12
Hence: Stress ( in P’P’ = E x Strain ()
  E


y

y
R
E
R
Moment of resistance
Since the beam is in equilibrium the resultant moment of the internal forces must be
equal to the external bending moment due to the applied loads.
A
y
N
N
SM AH O.1 fig 12
In figure 12 above:
Force due to stresses in 
F  A
E
 yA
R
The moment of this force about the neutral axis = Force x distance
 Moment 
E
E 2
yA  y =
y A
R
R
DET: Technological Studies: Structures and Materials (Advanced Higher)
13
The total moment of all internal forces about NN is called the Moment of Resistance,
given by:
E
 R y A
In the limit as A  0
Total Moment of Resistance 
2
E 2
y A
R
The quantity  y 2A is called the 2nd Moment of Area about the neutral axis and is
given the symbol I.
i.e. Moment of Resistance =
E
I
R
For equilibrium this internal moment of resistance equals the applied bending moment
(M) at the cross-section.
E
R
M
E


I
R
M 
Complete, the theory gives:
M  E
 
I
y R
This is known as the general bending equation and is used extensively in structural
design.
DET: Technological Studies: Structures and Materials (Advanced Higher)
14
2nd moment of area
The 2nd moment of area of a section about a particular axis (usually the neutral axis) is
given by the formula:
I   y 2A
For simple geometric shapes the integral is easily calculated.
A rectangular section is shown.
b
y
d
2
y
x
x
d
2
SM AH O.1 fig 13
 A  b  y
I XX 

y
2
dA
d
2
 by
2
dy
d
2
d
b  2
  y3 
 3  d 2
3
3
b  d   d  
      
3  2   2  
b d 3 d 3 
   
3 8
8
b d3
 
3 4
bd 3

8
DET: Technological Studies: Structures and Materials (Advanced Higher)
15
Factors affecting the shape of the beam cross-section
Effect of bending stress
The bending stress is directly proportional to the distance from the neutral axis.
If the material has the same yield strength in tension and compression, then it is logical
to make the beam cross-section symmetrical about the neutral axis. If this is not
possible (e.g. railway lines) the material is distributed between head and base in such a
way that the centroid is still at half the depth of the cross-section.
TENSILE
y
STRESS
NEUTRAL
AXIS
COMPRESSIVE
SM AH O.1 fig 14
Effect of area distribution
Increasing the area does not always increase the load carrying capacity of the beam.
Addition of area certainly increases I but may also increase ymax, and if ymax increases
relatively more than I, then the stress is actually increased by the addition of area.
(from  
My
)
I
Material economy
When designing a beam the aim is usually to obtain maximum strength with the use of
minimum material. The sections illustrated below meet this requirement.
I SECTION
CHANNEL
SECTION
BOX SECTION
TUBULAR
SECTION
SM AH O.1 fig 15
DET: Technological Studies: Structures and Materials (Advanced Higher)
16
Using tabulated ‘I’ values
Although an engineer will be able to calculate an 'I' formula for a given beam section
using integration, there is little to be achieved in doing this on a regular basis.
For this reason, where possible, engineers use tabulated data when 'I' formulae are
required.
'I' formulae for standard common sections are given below and should be used in your
beam design calculations. They are also provided in your Data Booklet for use in
assessments.
D
Common sections
RECTANGLE
x
x
3
Ixx = Bd
12
B
SOLID ROD
TUBE
D
I=
D
t
I=
 D4
64


3
Dt
SM AH O.1 fig 16
DET: Technological Studies: Structures and Materials (Advanced Higher)
17
Worked example
h
H
A box girder of outside dimensions 150 x 300 mm and wall thickness 20 mm is to be
used as a beam.
x
x
b
B
SM AH O.1 fig 17
a) calculate the second moment of area of the beam
b) use the bending equation to calculate the maximum bending moment which can be
applied to the beam if the working stress on the beam is 200N/mm2
c) calculating ‘I’ from the data booklet:
I XX
I XX
 I XX
BH 3  bh 3

12
150  300 3  110  260 3

12
 2116.64  10 9 mm 4

 

d) calculating ‘M’ from the bending equation:
M 

I
y
I
M 
y
y = 150mm
200  2116.64  10 9
Nmm
150
 M  2822  106 Nmm
 M  2.82 MNm
M 
SM AH O.1 fig 18
DET: Technological Studies: Structures and Materials (Advanced Higher)
18
Assignment 1.1
Using the formula for the Second moment of area for common sections, solve the
following questions.
1. Calculate the second moment of area of a timber floor joist of section 50 mm x 220
mm.
2. Two identical mild steel beams of 114 mm x 114 mm are used to support a
balcony. Calculate the second moment of area of each beam.
3. Timber support beams are used in the construction of a bridge across a small
stream and each beam is 300 mm x 200 mm in cross-section. Calculate the second
moment of area.
4. Someone suggests using logs of 300 mm in diameter to span the stream instead of
the rectangular beams. Calculate the second moment of area.
5. An exercise machine uses standard section, mild steel hollow square beams to build
up the supporting structure. Calculate ‘I’ if the beams are 30 mm square with a
wall thickness of 3.5 mm.
6. A road sign is supported by two tubular steel posts of outside diameter 150 mm. If
the wall thickness is 8 mm, calculate ‘I’.
7. A steel pipeline has wall thickness of 20 mm and an outside diameter of 500 mm.
Calculate the second moment of area of the section about the neutral axis.
8. A cast iron bracket has a section of 100 mm x 150 mm. The thickness of the web
and flanges is 20 mm. Calculate the second moment of area of the bracket.
9. A concrete box girder of section 3 m x 2 m and wall thickness 250 mm is used as
part of a bridge support. Calculate the second moment of area of the section about
the neutral axis.
10. The stub axle of a car has a diameter of 50 mm. Calculate the ‘I' value for the axle.
DET: Technological Studies: Structures and Materials (Advanced Higher)
19
Using standard section tables
When ‘I’ values have been calculated, either by applying the beam equation or using
the ‘I’ formula, this information can be used to access data on the dimensions and
properties of standard sections.
Your data booklet contains information on standard sections of square and rectangular
tubes. It also gives similar information on standard ‘I’ joists, a copy of which is shown
below.
Normal
size
Mass
per
metre
Depth
of
section
d
Width
of
section
b
Flange
t
mm
mm
cm2
Second moment
of area
1cm4 =
1x104mm4
Axis
Axis
x-x
y-y
cm4
cm4
Thickness
Depth
between
fillets
d
Area
of
section
1cm2 =
1x102mm2
mm
kg
mm
mm
Web
t
mm
127x144
127x44
127X76
127X76
29.76
29.76
16.37
13.36
127.00
127.00
127.0
127.0
114.3
114.3
76.2
76.2
10.2
7.4
5.6
4.5
11.5
11.4
9.6
7.6
79.4
79.5
86.5
94.2
37.3
34.1
21.0
17.0
979.0
994.8
569.4
475.9
241.9
235.4
60.35
50.18
144X144
26.79
114.3
114.3
9.5
10.7
60.8
34.4
735.4
223.1
SM AH O.1 fig 19
The data given in the table details the size and weight of standard joists. The last two
columns of data show the moment of area of the joists with respect to either a x-x axis
(the usual configuration) or a y-y axis.
DET: Technological Studies: Structures and Materials (Advanced Higher)
20
Worked example
A rolled steel ‘I’ joist is required to support a load. The calculated moment of inertia
for the beam is 70 x 106 mm4. The beam is loaded with the flanges in the horizontal
plane.
Use tabulated data to suggest the nominal joist size required.
Consider the beam configuration.
x
x
SM AH O.1 fig 20
The moment of inertia relates to the x-x axis.
The moment of inertia = 70 x 106 mm4
Therefore I = 700 cm4
From the standard tables a value must be located which is equal to or greater than 700
cm4, i.e. 735.4 cm4.
Using this value the nominal joist size is found:
i.e. nominal size = 114 mm x 114 mm.
DET: Technological Studies: Structures and Materials (Advanced Higher)
21
Beam design
In designing a structural element such as a beam, an engineer will be required to work
to four main performance criteria:
 the type of loading on the beam
 the method of support to be applied to the beam
 the acceptable limit of deflection allowable for the beam
 the required span of the beam.
Types of loading
Two types of load can act upon beams:
1. Point load
F
SM AH O.1 fig 21
A point load is a load which acts through a single point on the beam's surface.
2. Uniformly distributed load
SM AH O.1 fig 22
A uniformly distributed load (UDL) is a load which is spread constantly along the
length of a beam. In practice, this is the usual type of load a beam will require to
support.
‘’ is the load acting along the length of the beam and is given in terms of the total
force acting on each metre length of the beam i.e. kN/m.
DET: Technological Studies: Structures and Materials (Advanced Higher)
22
For example, the beam shown below is used to support a water tank.
FR
FR
L
SM AH O.1 fig 23
The UDL acting on the beam will be constant along the length of the beam due to the
weight of the water and the tank.
SM AH O.1 fig 24
The UDL acting on the beam is given by  
W
kN / m
L
Methods of support
Two basic methods can be used to support a beam:
1. Simply supported
F
KNIFE EDGE
ROLLER
SM AH O.1 fig 25
A beam can be supported on a ‘frictionless’ knife support or a ‘frictionless’ roller.
(Supports are assumed ‘frictionless’ to ease calculations.)
In either case the reaction at the support is assumed to be vertical, and the beam is free
to bend.
DET: Technological Studies: Structures and Materials (Advanced Higher)
23
F
FA
FB
SM AH O.1 fig 26
2. Fixed supports
F
A
B
SM AH O.1 fig 27
A beam can be fixed or anchored (‘built-in’) at either or both ends.
The wall reactions in the situation shown will require to resist the effects of bending as
well as supporting the beam load.
F
MA
MB
FA
FB
SM AH O.1 fig 28
DET: Technological Studies: Structures and Materials (Advanced Higher)
24
Beam deflection
As already shown, when a beam is positioned between two supports, it will bend.
When a load is applied it will bend further.
m (BEAM DEFLECTION)
SM AH O.1 fig 29
The bending of a beam will cause a deflection in the beam from the horizontal level.
The maximum deflection on a beam usually occurs in the middle of the beam.
When using beams in the construction of buildings, civil engineers will work to British
Standards standard specifications, which state what the permissible allowable
deflection is for a given beam to suit a given purpose.
For example, the maximum permissible deflection in a domestic floor joist is less than
1 mm.
Beam span
The span of a beam L, is the distance the beam is required to bridge between supports.
F
A
B
L
SM AH O.1 fig 30
Of course not all beams are supported at both ends. A beam that is supported at one
end only, is called a cantilever.
P
L
SM AH O.1 fig 31
DET: Technological Studies: Structures and Materials (Advanced Higher)
25
In working towards a solution to a beam design problem to meet these performance
criteria an engineer will consider the following factors:
 the material to be used
 the size of structural section required
 the shape of structural section required.
To help consider these factors in a mathematical way, the engineer will be required to
use the ‘General Bending Equation’.
M  E
 
I
y R
He/she will also make use of standard tabulated data on common configurations of
beam.
Standard beam configurations
The information provided in your data booklet is typical of that used by engineers in
solving standard beam design problems. An extract has been shown below.
MAXIMUM BENDING
MOMENT
CONFIGURATION
= FL at A
2
=
L at A
2
MAXIMUM DEFLECTION
3
= FL at B
3EI
3
=
L at B
8EI
F
2
A
B
C
at C
= FL
4
3
= FL at C
48EI
SM AH O.1 fig 32
All beams are of length L, cross-section second moment of area I and Young's
Modulus E. Point C is located at mid-span.
DET: Technological Studies: Structures and Materials (Advanced Higher)
26
The diagrams in the first column of the data sheet show the type of loading on the
beam and the method of support.
CONFIGURATION
F
A
B
C
SM AH O.1 fig 33
The diagram shows a beam simply supported at A and B with a central point load C.
The second column gives a formula for calculating the maximum bending moment M
acting on the beam and gives the point at which this bending moment is applied.
MAXIMUM BENDING
MOMENT
CONFIGURATION
MAXIMUM DEFLECTION
F
3
2
A
B
C
= FL at C
4
=
FL at C
48E I
SM AH O.1 fig 34
The maximum bending moment on the beam is:
2
FL
M 
at 'C' the centre of the beam.
4
The third column gives a formula for calculating the maximum deflection  m.
The maximum deflection on the beam above occurs at 'C' and can be found from the
given formula:
 max 
FL3
at ‘C’ the centre of the beam.
48EI
DET: Technological Studies: Structures and Materials (Advanced Higher)
27
Worked example
A timber beam of 4.5 m span is supported at either end by wall fixings. The beam is
required to support a central point load of 120kN.
Calculate:
a) the maximum bending moment acting on the beam
b) the maximum beam deflection.
The data relating to the timber beam is shown below. Information relating to bending
moment and beam deflection, for various common configurations of beam, can be
found in your data booklet.
MAXIMUM BENDING
MOMENT
CONFIGURATION
= FL at C
8
C
=
L
-FL at A,B
8
MAXIMUM DEFLECTION
3
=
FL at C
192E I
SM AH O.1 fig 25
a) The maximum bending moment for the beam occurs at the beam centre, 'C', and at
the ends 'A' and 'B'. Bending moment is found using the given formula.
FL
8
120  10 3  4.5
M 
Nm
8
 M  67.5kNm
M 


b) The maximum deflection (  m) of the beam occurs at 'C' and is calculated using the
formula.
FL3
m 
192 EI
Values for F and L have been given in the question, but before a value for  m can be
calculated, values for E and I need to be found.
E is the value of Young's Modulus for the material. 'E' values for common materials
are listed in your data booklet. There are two entries relating to wood.
DET: Technological Studies: Structures and Materials (Advanced Higher)
28
Young’s
Modulus
E
2
kN/mm
Material
Wood parallel to grain,
compression
Wood perpendicular
to grain
Ultimate
Tensile
Stress
2
N/mm
Ultimate
Compressive
Stress
2
N/mm
19 -16
--
55 - 100
6 - 16
0.6 - 1.0
--
--
2-6
SM AH O.1 fig 36
In this example, although the force acts perpendicularly to the grain, the bending
moment produces stress which is parallel to the grain in the beam.
Although any value for E between 19-16 kN/mm2 would be acceptable, in a design
situation it is safer to assume the lower value.
Take E = l6 kN/mm2.
D
I is the second moment of area of the beam. Again 'I' is found using standard
information given in the data booklet.
x
x
3
IX = Bd
12
B
SM AH O.1 fig 37
For the rectangular timber beam:
D
200
100
B
SM AH O.1 fig 38
DET: Technological Studies: Structures and Materials (Advanced Higher)
29
BD 3
12
100  200 3
I
12
 I  66.7  10 6 mm 4
I
Having found E and I for the beam apply these in the deflection formula.
 max
  max
FL3

192 EI
120  10  4.5  10 

192  16  10  66.67  10 
3 3
3
3
6
  max  53.25 mm
DET: Technological Studies: Structures and Materials (Advanced Higher)
30
Using the bending equation
M  E
is used in conjunction with tabulated data to assist
 
I
y R
in the design of beams.
The bending equation
Normally only two elements of the equation require to be used. For example:
M 

I
y
Where:
M = maximum bending moment.
I = second moment of area.
 = maximum value of stress produced in the beam due to bending.
y = the distance this stress will act from the neutral axis.
In the 'I' beam shown below, the maximum bending moment will occur at the centre of
the beam due to the action of the point load P.
F
FLANGE
Cmax
Cmax
y
WEB
NEUTRAL
AXIS
max
max
SM AH O.1 fig 39
The bending moment can normally be calculated using the standard information found
in the data booklet.
If required, the second moment of area can also be found using the formula given in
the data booklet.
The maximum stress will occur in the flanges at the top and bottom of the beam. As
the beam bends, compressive stress is produced in the top flange and tensile stress in
the bottom flange. ‘y’ is the distance from the N/A to the top or bottom edge of the
beam.
DET: Technological Studies: Structures and Materials (Advanced Higher)
31
Worked example
A beam is 2.5 m long and is simply supported at its ends. It carries a UDL of 100
kN/m run over its entire length.
a) calculate the maximum bending moment for the beam
b) if the beam has a cross-section of 100 mm x 300 mm, calculate the maximum value
of bending stress in the beam.

Consider the beam configuration = 100 kN/m.
100
W = L
A
N/A
00
3
N/A
B
SM AH O.1 fig 40
SM AH O.1 fig 41
From the tabulated data for the beam, the maximum bending moment occurs at the
centre of the beam and can be calculated from:
M 
L2
8
100  10 3  2.5 2
M 
8
 M  78.125 kNm



The maximum stress will occur in the top and bottom edge of the beam. We must
use the beam equation.
M 

I
y
My
 
I
DET: Technological Studies: Structures and Materials (Advanced Higher)
32
Before the maximum tensile stress can be found, the value of ‘I’ for the beam is
required. From the data booklet:
BD 3
12
100  300 3
I
12
 I  225  10 6 mm 4
I
My
it is essential to ensure that the units used
I
are compatible i.e. M, y and I share the same basic units.
Before applying the beam equation  

M = 78 x 103 Nm = 78 x 103 x 103 Nmm = 78 x 106 Nmm
y = 150 mm
I
= 225 x 104 mm4
Applying these units in the equation,  will be given in N/mm2.
My
I
78  10 6  150
 
N / mm 2
6
225  10
   52 N / mm 2



DET: Technological Studies: Structures and Materials (Advanced Higher)
33
Assignment 1.2
Using standard tabulated data, solve the following problems.
1. A simply supported timber beam of 2 m span is required to carry out a central point
load of 120 kN. The section of the beam is 500 mm x 120 mm. Calculate:
a) the maximum bending moment acting on the beam
b) the maximum deflection which will result.
2. An aluminium alloy round bar of diameter 125 mm is to be used to span a 5 m gap.
The bar is simply supported at its ends and carries a uniformly distributed load of
20 kN/m. Calculate:
a) the maximum bending moment acting on the bar
b) the maximum deflection which will result.
3. A rolled mild steel girder of 'I' section, measuring 200 mm x 500 mm with an
18 mm thick web and flanges, spans 6 m. The girder is fixed at its ends and
supports a uniformly distributed load of 200 kN/m. Calculate:
a) the maximum bending moment on the girder
b) the maximum deflection for this loading.
4. A simply supported bridge structure is made up from four 150 mm diameter timber
beams and spans a gap of 5 m. The maximum deflection caused in the structure
due to the action of a central point load should be no more than l0 mm.
Calculate the maximum load which can be applied to each beam.
5. A beam of rectangular section, 150 mm x 250 mm and length 2.4 m, is used as a
cantilever. If the maximum bending moment in the beam should not exceed
240 kNm, calculate the maximum bending load on the beam.
If the maximum permissible deflection in the beam is 10 mm, calculate the
modulus of elasticity for the beam and suggest a suitable material from which it
could be made.
DET: Technological Studies: Structures and Materials (Advanced Higher)
34
Assignment 1.3
Using the standard tabulated data provided in your data booklet in conjunction with the
bending equation, solve the following questions.
1. A beam is 2.5 m long and is simply supported at its ends. It carries a UDL of 100
kN/m over its entire length.
a) calculate the maximum bending moment for the beam
b) if the beam has a cross section of 100 mm x 300 mm, calculate the maximum
value of bending stress in the beam.
2. A cantilever of rectangular section is uniformly loaded along its 3 m length at
1.5 kN/m. If the cantilever is to be made from 120 mm thick wood, calculate the
depth of timber required.
The maximum stress due to bending should not exceed 5 N/mm2.
3. A box girder, of outside dimensions 150 mm x 300 mm and wall thickness 20 mm,
is used as a beam. The maximum allowable stress in the girder is 100 N/mm2.
a) if the beam is fixed at 2.5 m centres, calculate the maximum UDL which can be
applied to the beam
b) if the maximum allowable deflection for the beam is 1mm, calculate Young’s
Modulus of the material from which the beam should be made.
4. A mild steel stub axle, 50 mm in diameter and 175 mm long, is subjected to a
140 kN load at its end.
a) calculate the maximum bending moment in the axle
b) calculate the maximum stress in the axle
c) calculate the maximum deflection.
5. A steel pipeline, outside diameter 200 mm and wall thickness 10 mm, is filled with
oil.
If the oil has a specific density of 800 kg/m3 calculate the UDL acting on the
pipeline.
DET: Technological Studies: Structures and Materials (Advanced Higher)
35
Assignment 1.4
1. The maximum SWL of a lift rope must not exceed 20 kN. To avoid the lift being
overloaded, a cut-out switch is used to prevent the lift operating if the force in the
lift rope exceeds 20 kN. The main rope pulley is mounted on a beam, as shown,
and can be assumed to be a point load. The limit switch is placed below this beam
and will be operated by the deflection of the beam if an overload occurs.
3m
3m
LIMIT SWITCH
LIFT
WINCH
SM AH O.1 fig 42
a) If the beam is a mild steel joist, of size 127 mm x 114 mm and mass 26.79 kg/m,
what will be the central deflection of the joist at the maximum operating load?
b) Why should this calculation not include the self weight of the beam?
c) An engineer inspecting the system decided that the limit switch should be
mounted above the beam. Suggest reasons for this decision and describe what
changes should be made in the wiring of the limit switch.
2. A water channel, used to irrigate fields in a Third World country, has to cross a
small ravine. The channel is supported on three round logs, 150 mm in diameter,
spanning a 5 m gap. The logs can be considered to be simply supported and the
trough, when full, can be considered as a distributed load of 0.8 kN/m.
SM AH O.1 fig 43
DET: Technological Studies: Structures and Materials (Advanced Higher)
36
a) calculate the maximum bending stress in one log and show where it occurs
b) refer to the data booklet and discuss the suitability of this size of log with
particular reference to its mechanical strength.
3. A cantilever style balcony is supported by two ‘I’ shaped mild steel beams, 2 m in
length. The weight of the balcony is evenly supported by the two beams and can
be assumed to produce a uniformly distributed load on each beam of 3 kN/m.
a) calculate the total weight of the balcony
b) the nominal size of each beam is 114 mm x 114 mm, calculate the maximum
bending moment for one beam and state where it occurs
c) calculate the maximum deflection.
4. A mild steel beam that is used to support scenery on a stage is shown below.
5m
470 kN
SM AH O.1 fig 44
The beam is 5 m in length and is rectangular in cross-section. The depth is
400 mm and the width is 200 mm.
a) calculate the maximum bending moment produced in the beam
b) calculate the maximum bending stress which results in the beam
c) explain why in practice a solid section beam would not be used
d) draw a diagram to represent a more suitable beam section.
5. In a health club, the exercise equipment is built using mild steel hollow square
beams. On one particular machine, the frame assembly is built into two side
supports. The cross beam is designed to support a maximum central point load of
2.5 kN and is 1.2 m in length. For safety reasons, the beam must not deflect more
than 3 mm when in use.
a) calculate the bending moment acting on the beam
b) the required second moment of area about the x-axis of the beam
c) select an appropriate standard section of the hollow square beam from your data
booklet
d) calculate the maximum bending stress in the beam which you selected in part (c).
DET: Technological Studies: Structures and Materials (Advanced Higher)
37
DET: Technological Studies: Structures and Materials (Advanced Higher)
38
OUTCOME 2 – EVALUATE THE DISTRIBUTION OF SHEAR FORCE AND
BENDING MOMENT IN LOADED BEAMS
The purpose of this outcome is to explain how the tabulated data relating to beams
found in the SQA data booklet has been calculated. The unit demonstrates how to
construct shear force and bending moment diagrams.
When you have completed this outcome you should be able to:
 interpret loading information on a beam from a shear force or bending moment
diagram
 construct a simple shear force and bending moment diagram.
Before you start this outcome you should have a clear understanding of:
M  E
 the bending equation
 
I
y R
 the conditions of equilibrium as applied to a beam
 the principle of moments.
DET: Technological Studies: Structures and Materials (Advanced Higher)
39
Bending moment diagrams
As already seen from work in Bending of Beams the maximum bending moment
produced when a beam is loaded can be calculated either by using the bending
equation or from consulting tabulated data for beams, loaded in a standard
configuration such as is given in the SQA data booklet.
A third method by which bending moments can be ascertained is through the use of
bending moment diagrams. Bending moment diagrams are a graphical method of
representing the effect of load on a beam.
F
L
BEAM LOADING CONFIGURATION
FL
4
L
BENDING MOMENT DIAGRAM
SM AH O.2 fig 1
Bending moment diagrams have two advantages over using the bending equation or
tabulated data when applied to calculating bending moment:
 bending moment diagrams can be used to analyse the effect of loading on any
configuration of beam
 bending moment diagrams show the effect of loading across the whole length of
the beam.
The bending moment at a cross-section ‘x-x’ of a beam is the moment tending to bend
the beam and produces bending stresses in the material of the beam. It is calculated by
determining the algebraic sum of the moments of all the external forces acting to the
right or left of the section.
DET: Technological Studies: Structures and Materials (Advanced Higher)
40
Sign convention
SAGGING
HOGGING
x
x
x
x
POSITIVE BENDING MOMENT
NEGATIVE BENDING MOMENT
SM AH O.2 fig 2
Consider a simply supported beam loaded as shown below:
4N
3m
6N
2m
1m
x
A
B
C
D
x
7N
3N
SM AH O.2 fig 3
Bending moment can be calculated for any position ‘x’ of ‘x-x’ along the length by
considering only the effects of all forces to the left (or right).
Measuring from left:
1. at A, x =0
BM = 3 x 0 = 0 Nm
2. at B, x = 3 m
BM = +(3 x 3) = +9 Nm (beam tending to sag)
3. at C, x = 5 m
BM = +(3 x 5) – (4 x 2) = (15 – 8) = +7 Nm
4. at D, x = 6 m
BM = +(3 x 6) – (4 x 3) – (6 x 1) = 0 Nm
Since moments can be taken to left or right of the section, step 3 bending moment
could be more easily calculated by measuring from the right hand end, thus:
5. at C, x = 1m
BM = +(7 x 1) = +7 Nm (beam tending to sag)
DET: Technological Studies: Structures and Materials (Advanced Higher)
41
We can construct a diagram to show this information.
4N
3m
6N
2m
1m
x
A
B
C
D
x
7N
3N
+9
+7
0
0
BENDING MOMENT DIAGRAM
SM AH O.2 fig 4
It should be noted that:
 in general bending moments are calculated only at points where loads are applied
 the bending moments at the ends of simply supported beams are zero
 bending moments to left or right of a section are chosen as a matter of convenience.
DET: Technological Studies: Structures and Materials (Advanced Higher)
42
Shear force diagrams
The shearing force at any cross-section of the beam is the force tending to shear the
beam at that point. It causes shear stresses to act within the material.
Shearing forces may be positive or negative according to the following sign
convention.
POSITIVE SHEARING FORCE
NEGATIVE SHEARING FORCE
SM AH O.2 fig 5
Consider a simply supported beam loaded as shown below.
5N
A
x
7N
C
B
D
8N
4N
x
SM AH O.2 fig 6
As with bending moment diagrams, the shearing effect at any position of ‘x-x’ is
calculated as the algebraic sum of only those shearing forces to the left (or right) of the
section.
1. From A to B, there is only one force acting to the left of any point in this section
thus, shearing force = +4 N for all points between A and B.
2. From B to C, there are two forces acting to the left of any point in this section thus,
shearing force = (4 – 5) = -1 N for all points between B and C.
3. From C to D, there are three forces acting to the left of any point in this section,
thus shearing force = (4 – 5 – 7) = -8 N for all points between C and D.
DET: Technological Studies: Structures and Materials (Advanced Higher)
43
Since shearing forces can be taken to left or right of the section, step 3 shearing
force could be more easily calculated to the right of the section, thus:
4. From D to C, there is only one force acting to the right of any point in this section,
thus shearing force = -8 N for all points in this section.
Again we can construct a diagram to show this information.
5N
A
x
7N
C
B
D
8N
4N
x
+4N
-1N
-8N
SHEAR FORCE DIAGRAM
SM AH O.2 fig 7
DET: Technological Studies: Structures and Materials (Advanced Higher)
44
Uniformly distributed load
When uniformly distributed loads (UDLs) are applied to beams a similar method can
be applied to the construction of shear force and bending moment diagrams.
Consider the beam shown below.
 = 30kN/m
4m
RA
RB
SM AH O.2 fig 8
Shear force diagram
To find the reactions RA and RB the UDL can be regarded as concentrated at its own
centre of gravity i.e. at the mid-point of its length. This gives a concentrated load of
120 kN (from .L) acting 2 m from A (L/2).
Applying Equilibrium
M
B
0
 R A  4  120  2
 R A  60 kN
2m
 FORCES Vert  0
120kN
 R A  RB  120  0
 RB  60 kN
RA
4m
RB
SM AH O.2 fig 9
For any position of the ‘x-x’ axis the distributed load to the left, can be replaced by a
x
point load of value   x acting at from ‘x-x’.
2
x .x kN
2
RA
x
x
RB
x
DET: Technological Studies: Structures and Materials (Advanced Higher)
45
For all values of x from A to B, the shearing force is given by:
RA  .x




at A, shearing force = (+60–0) = +60 kN
from A to B the shearing force due to the UDL is directly proportional to x and by
convention, negative
at B, shearing force =  60  30  4 = -60 kN
alternatively, considering x to the right of x-x, the shearing force at
B = (-60+0) = -60 kN.
30kN/m
60N
60N
+60N
-60N
SHEAR FORCE DIAGRAM
SM AH O.2 fig 11
Bending moment diagram
From A to B, the bending moment at any point x to the left of B, including the effect of
the UDL, is given by:
x

M X   60 x    .x  
2

30 x 2
 M X  60 x 
kNm
2
This equation, applying only between A and B, forms a curve and values of bending
moment can be tabulated and the curve plotted.
BENDING MOMENTX FROM A TO B
x (m)
0
1.0
2.0
3.0
4.0
BMx(kNm)
0
45
60
45
0
SM AH O.2 fig 12
DET: Technological Studies: Structures and Materials (Advanced Higher)
46
30kN/m
60N
60N
+60
+45
+45
0
0
BENDING MOMENT DIAGRAM
SM AH O.2 fig 13
By inspection of the BMD the maximum bending moment is 60 kNm acting at midspan (i.e. 2 m from A). In this case the symmetry of the diagram gives an easy
solution. More generally the value is found mathematically by differentiating the
equation with respect to x, thus:
M X  60 x 

30 x 2
2
dM
 60  30 x
dx
The maximum turning value and hence the maximum bending moment occurs when:
dM
60
 0 i.e. when x 
2m
dx
30
and:
 30  2 2
M max  60  2   
 2
 120  60



 60 kNm
DET: Technological Studies: Structures and Materials (Advanced Higher)
47
Worked example
A simply supported beam supports a point load 1m from the left hand end and a UDL
of 30 kN/m over a 2m run as shown below.
120kN
A
 = 30kN/m
C
1m
B
2m
SM AH O.2 fig 14
Draw the shear force and bending moment diagrams and determine the maximum
shear force and bending moment.
a) calculating reactions at A and B:
i)
replace UDL by point load at mid-point of UDL.
SM AH O.2 fig 15
M
0
 R A  3  120  2   30  2  1  0
240  60
 RA 
3
 R A  100 kN
B
 Forces
Vert
0
 R A  RB  120  60  0
 RB  180  100
 RB  80 kN
DET: Technological Studies: Structures and Materials (Advanced Higher)
48
b) calculating shear forces:
i)
from A to C, shear force at any point to left of C = +RA = +100 kN
ii)
from C to B, shear force at any point to left of B includes the effect of the
UDL, thus shearing forcex = +100 – 120 - .x = -(20 + x) kN
at x = 0, shearing force = -20 kN
at x =2m, shearing force = -[20 + (30 x 2)] = -80kN
as x increases, the shearing effect is directly proportional to x and varies
linearly between the two values calculated.
.x kN
120kN
X
A
C
B
100kN
80N
+100kN
-20kN
SHEAR FORCE DIAGRAM
-80kN
SM AH O.2 fig 16
The maximum shear force is+100 kN.
c) bending moments
i)
from A to C, the bending moment at any point,x, to the left of C is given by:
SM AH O.2 fig 17
DET: Technological Studies: Structures and Materials (Advanced Higher)
49
Mx = 100x kNm
at x = 0, Mx = 0 kNm
at x = 1m, Mx = +100 kNm
as x increases, the bending moment is directly proportional to x and varies
linearly between the two values calculated.
From C to B, the bending moment at any point,x, to the left of C, including the effect
of the UDL, is given by:
SM AH O.2 fig 18

M X  1001  x   120 x    x 

30 x 2
 M X  100  100 x  120 x 
2
30 x 

 M X  100  x 20 
 kNm
2 

x

2
This equation, applying only between C and B, forms a curve and values of bending
moment can be tabulated and the curve plotted.
BENDING MOMENTX FROM B TO C
x (m)
0
0.5
1.0
1.5
2.0
BMx(kNm)
100
86.25
65.0
36.25
0
SM AH O.2 fig 19
120kN
A
30kN/m
C
100
B
86.25
65
36.25
0
0
BENDING MOMENT DIAGRAM (kNm)
SM AH O.2 fig 20
The maximum bending moment is +100kNm acting 1m from left hand end.
DET: Technological Studies: Structures and Materials (Advanced Higher)
50
Worked example
A horizontal beam of uniform cross-section is simply supported at A and B and is
loaded as shown.
20kN
20kN/m
A
B
4m
C
3m
RA
RB
Construct the shear force and bending moment diagrams and determine the value and
position of the maximum bending moment(s).
a) reactions A and B:
i) point load due to UDL = (20 x 4) kN acting at 2m from RA
2m
20kN
80kN
4m
RA
RB
7m
SM AH O.2 fig 22
M
RA
 RB 
0
20  7   20  4  2
4
 RB  75 kN
 Forces
Vert
0
 R A  RB  20  80  0
 R A  25 kN
b) shear force and bending moment diagrams:
i)
the beam must be dealt with in two parts, A to B and B to C.
From A to B, the shear force acting at any point to the left of B comprises a constant
element due to the point load (RA) and an variable element due to the UDL. This is
given by the equation:
Shear Force x  25    x 
= +25-20x kN …(i)
DET: Technological Studies: Structures and Materials (Advanced Higher)
51
From A to B, the bending moment acting at any point to the left of ‘X-X’ comprises a
positive element due to RA and a negative element due to the UDL.
SM AH O.2 fig 23
This is given by the equation:
x

Moment x  25 x    .x  
2

20 x 2
 25 x 
2
= 5(5x-2x2) kNm ….(ii)
Note. These equations apply only to sections between A and B.
Using equations (i) and (ii) above values for shear force and bending moment between
A and B can be tabulated as follows:
x (m)
Shear
Forcex (kN)
Mx (kNm)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
+25
+15
+5
-5
-15
-25
-35
-45
-55
0
+10
+15
+15
+10
0
-15
-35
-60
SM AH O.2 fig 24
From B to C, the shearing force acting at any point to the right of ‘X-X’ is dependent
on the single point load and is constant, thus:
Shearing Force from B to C = +20 kN
From B to C, the bending moment acting at any point to the right of ‘X-X’ is
dependent on the single point load and varies linearly with x, thus:
M x  20  x
 20 x kNm
at C, x = 0  M x  0
at B, x = 3m  M x  20  3  60 kNm
DET: Technological Studies: Structures and Materials (Advanced Higher)
52
A
20kN
20kN/m
B
4m
C
3m
25kN
75kN
+25
+20
-55
SHEAR FORCE DIAGRAM (kN)
+15
+15
0
0
0
-60
BENDING MOMENT DIAGRAM (kNm)
SM AH O.2 fig 25
c) maximum bending moments
i)
from inspection of the diagram, the maximum positive bending moment
occurs between 1.0m and 1.5m from A. Mathematically the exact position
is found by differentiating equation (ii) with respect to x.


M x  5 5 x  2 x 2  25 x  10 x 2

dM
 25  20 x
dx
The maximum turning value and hence maximum positive bending moment occurs
when
dM
25
 0 i.e. when x 
 1.25m
dx
20
and
M max  25  1.25  10  1.25 2


 31.25  15.625
 15.625 kNm
DET: Technological Studies: Structures and Materials (Advanced Higher)
53
The maximum negative bending moment occurs at B with a value of –60 kNm.
Assignment 2.1
1. For the beam shown below:
a) draw the shear force and bending moment diagrams
b) calculate the maximum tensile bending moment (233 kNm)
c) given E = 200GN/m2, calculate the radius of curvature at section B (24 x 103
mm).
B
N
2m
A
200mm
100kN
2m
8
4
INA = 120 x 10 mm
SM AH O.2 fig 26
2. A horizontal beam of uniform cross-section is simply supported at A and B and
carries the loads shown.
a) determine the magnitude of the support forces at A and B
b) sketch the shear force and bending moment diagrams for the beam, stating all
important values
c) calculate the value and state the name of the maximum direct stress due to
bending in the beam material.
600mm
3kN
4kN
A
400mm
N
3m
2m
2m
B
-6
A
4
INA = 12 x 10 m
ENLARGED CROSS-SECTION
SM AH O.2 fig 27
DET: Technological Studies: Structures and Materials (Advanced Higher)
54
3. A horizontal beam of uniform cross-section is simply supported at A and B and
carries the loads shown.
a) determine the magnitude of the support forces at A and B
b) sketch the shear force and bending moment diagrams for the beam and indicate
the significant values
c) determine the radius of curvature at A, given that the modulus of elasticity of
the material is 208 GN/m2.
4m
8kN
UDL = 5kN/m
600 mm
4kN
A
A
400 mm
N
4m
2m
2m
B
-6
INA = 4 x 10m
4
ENLARGED CROSS-SECTION
SM AH O.2 fig 28
4. A cantilever of rectangular cross-section is uniformly loaded along its length of 3
m at 1500 N/m run. If the material is wood, 75 mm wide, find the depth at the
position of maximum bending moment for a maximum stress due to bending of
5 N/mm2.
Draw the shear force and bending moment diagrams for the cantilever.
5. A steel tube of 8 mm bore and 0.8 mm thick walls, is fully charged with mercury
(density 13600 kg/m3) and forms part of some physical apparatus. In use the tube,
which is 500 mm long, is supported at its ends.
Draw the shear force and bending moment diagrams for the tube and calculate the
maximum bending moment applied to the tube.
DET: Technological Studies: Structures and Materials (Advanced Higher)
55
DET: Technological Studies: Structures and Materials (Advanced Higher)
56
OUTCOME 3 – APPLY THE METHOD OF SECTIONS IN SOLVING
PROBLEMS ON COMPLEX FRAMED STRUCTURE SYSTEMS
The purpose of this outcome is to apply the method of sections to framed structures.
When you have completed this outcome you should be able to:
 apply the method of sections to pin-jointed structures.
Before you start on this outcome you should have a clear understanding of:
 nodal analysis
 conditions of equilibrium
 principle of moments.
DET: Technological Studies: Structures and Materials (Advanced Higher)
57
Pin-jointed structures
Pin-jointed structures consist of individual members carrying only axial forces, i.e. no
moments being transmitted at the joints (nodes). Thus it is possible to determine the
equilibrium of the various parts of such structures by considering each node
individually as a free body. This is known as ‘nodal analysis’ or ‘the method of
joints’. (You should already be familiar with this type of analysis).
Each node may be solved by calculation or by applying a graphical method
(triangle/polygon of forces). This type of solution is normally employed when the
forces in all the members of a structure are required.
When the forces in only a few members are required, the above is rather too clumsy
and time consuming a method. Consequently an alternative method, called ‘the
method of sections’, is preferred.
Consider the framed structure below, where only the two reactions and the forces in
members A, B and C are required to be found.
SM AH O.3 fig 1
1. Reactions
In order to calculate the reactions it is convenient to consider the frame as a solid
body acted upon by external forces alone.
SM AH O.3 fig 2
DET: Technological Studies: Structures and Materials (Advanced Higher)
58
Calculating roller reaction:
Taking moments about the hinge (H):
Moments H = 0
= + (500x3)-(100x2 sin 600)-(RRx6)
= + 1500 – 173.2 – 6RR
RR= + 1326.8 = + 221.13kN
6
The reaction at the rollers = 221.13 kN acting vertically upwards.
SM AH O.3 fig 3
Calculating hinge reaction:
SM AH O.3 fig 4
 Forces Horizontal   100 kN
 ForcesVertical  221  13  500
  278  87 kN
The reaction at the hinge (FH) is the equilibrant of the two component forces above
(i.e. 100 kN and –278.87 kN).
FH 

 278  87 2   100 2
77768  5  10000
296  26
100
   tan 1 2.9626
tan  
 296  26 kN
 71.3o
The reaction at the hinge = 296.26 kN acting at 71.30 as shown.
DET: Technological Studies: Structures and Materials (Advanced Higher)
59
2. Members A, B and C
The structure is considered to be cut into two pieces by a section through the
members under consideration. Both parts of the structure can then be treated as
separate structures in equilibrium and free-body diagrams drawn.
Note the relationship of the forces in each of the ‘cut’ members.
Force direction is arbitrary at this point and will be confirmed later by calculation.
SM AH O.3 fig 5
The forces in the sectioned members can be found by resolution of the forces or by
taking moments about a suitable point.
Either the left or right-hand section of the cut structure may be used to calculate the
required member values …. whichever is the more convenient. In this case the
right-hand section is used to ease calculations (avoiding the 71.30 angle).
DET: Technological Studies: Structures and Materials (Advanced Higher)
60
a) Taking moments at 1
SM AH O.3 fig 6
Ignoring all forces acting thro’ fulcrum:
F
C
Moments  0

 2 sin 60 o  221  13  3
 221  13  3
2 sin 60 o
  383 kN
 FC 
The force in member C is 383 kN acting as shown (calculation positive), i.e. the
member is a tie.
b) Taking moments at 2
SM AH O.3 fig 7
Ignoring all forces acting thro’ fulcrum:
Moments  0
500 1  221  13 4  100  2 sin 60 0  FA  2 sin 60 0 
500  884  52 173  2  557  72

1  732
1  732
  322 kN
 FA 
The force in member A is 322 kN acting opposite to direction shown (calculation
negative), i.e. the member is a strut.
DET: Technological Studies: Structures and Materials (Advanced Higher)
61
c) Taking moments at 3
SM AH O.3 fig 8
Ignoring all forces acting thro’ fulcrum:
Moments  0


 221  13 2  100  2 sin 60 0  500 1
0 

 322  2 sin 60 0  FB  2 sin 60 0

 442  26 173  2  500  577.7
 557  75
 FB 

1  732
1  732
  322 kN

 

The force in member B is 322 kN acting opposite to direction shown (calculation
negative), i.e. the member is a strut.
d) Alternative method to calculate force in B.
SM AH O.3 fig 9
DET: Technological Studies: Structures and Materials (Advanced Higher)
62
FBHorizontal  322  383  100
  161kN 
FBVertical  221  3  500
  278  7kN 
SM AH O.3 fig 10
FB 
 1612  278  7 2

25921  77673  7
 321  9 kN
The force in member B is 321.9 kN acting as shown i.e. the member is a strut.
Note: Discrepancy in answers due to rounding errors in calculations.
DET: Technological Studies: Structures and Materials (Advanced Higher)
63
Worked example
For the structure shown below, determine the magnitude and nature of the force in
members A, B and C.
Reactions
By inspection Reaction R1 = R2 = 20kN
Members A, B and C
Take section thro’ members A,
B and C. Disregard left hand
section to simplify calculations.
The direction of forces is
arbitrary at this stage and will
be confirmed by calculation.
a) Taking Moments at 1
Ignoring all forces acting
thro’ fulcrum:
D  3 sin 30 0  cos 60 0
 3 0  5 0  5
 0  75m
DET: Technological Studies: Structures and Materials (Advanced Higher)
64
Moment  0
 10  0  75  15  3  FA  3 sin 300 
 7  5  45  1  5FA
 FA 
 37  5
1 5
  25kN
The force in member A is 25 kN acting opposite to assumed direction (calculation
negative), i.e. the member is a strut.
b) Taking Moments at 2
Ignoring all forces acting thro’
fulcrum:
Moments  0
FB  3 cos 30 0  10  3 cos 30 0
 FB  10 kN
The force in member B is 10 kN acting
in assumed direction (calculation
positive), i.e. the member is a tie.
c) Taking Moments at 3
Ignoring all forces acting
thro’ fulcrum:
Moments  0


  FC  3 sin 60 0  10  1  5  0  75  15  4  5
 22  5  67  5
 45

0
2.598
3 sin 60
  17.32 kN
 FC 
The force in member C is 17.32 kN acting in assumed direction (calculation positive),
i.e. the member is a tie.
DET: Technological Studies: Structures and Materials (Advanced Higher)
65
Worked example
For the structure shown below, determine the magnitude and direction of the reactions
R1 and R2. Apply the ‘method of sections’ to find the magnitude and nature of member
A.
Reactions
D = 12sin300 = 6 m
d = Dcos600 = 3 m
Taking moments about hinge (R1),
Moments  0
R2  12  20  3  15  6
 R2 

150
12
 12  5 kN 
Forces Horiz  0

 2 15 cos 30  R1H 
o
 R1H  2  12  99
 25.98 kN 

ForcesVert  0

 2 15 sin 30 o


 0
 20  12.5  R1V 
 R1V   15  20  12  5
  22.5 kN 
Note: Vertical component acts opposite to assumed direction (calculation negative).
DET: Technological Studies: Structures and Materials (Advanced Higher)
66
R1 
R 1 H  R 1V
2
 25  982  22.52
2
 34  37kN
  tan 1
22  5
25  98
 40  9 o
Reaction at hinge (R2) = 34.37 kN acting at 40.9o to negative X axis as shown.
Members A, B and C
Taking a section thro’ required members and ignoring left hand section.
a) Considering Moments at 1, both FA and
FC have zero moment (acting thro’
fulcrum) thus, for equilibrium to be
maintained, FB must also have zero
moment i.e. FB = 0 N
Member B is redundant.
b) Taking moments at 2
Moments  0


 FC  6 sin 60 0  FA  0

 0
 12  5  12 cos 30 0

 FC   2.83 kN


The force in member C is 2.83 kN acting
in assumed direction (calculation positive), i.e. the member is a tie.
c) Taking moments at 3
Moments  0
 12  5  12  FC  0

 0
0

F

12
sin
30
A


 FA   25 kN


The force in member A is 25 kN acting
in assumed direction (calculation positive), i.e. the member is a tie.
DET: Technological Studies: Structures and Materials (Advanced Higher)
67
Assignment 3.1
1. Consider each of the four frame structures shown. Two of the frames contain nonload bearing (redundant) members.
Identify these frames and indicate the on-load bearing members. Explain your
answers.
2. The frame structure shown below is supported by a hinge at ‘H’ and a ‘frictionless
roller’ at R.
a) calculate the reactions at R and H
b) use the method of sections to calculate the magnitude and nature of the forces in
the members intersected by ‘Z – Z’
c) apply nodal equlibrium at R to calculate the magnitude and nature of the force in
member ‘A’.
DET: Technological Studies: Structures and Materials (Advanced Higher)
68
3. The pylon structure shown is used to support
floodlighting. The floodlighting canopy weighs
50kN, which can be assumed to act vertically
through the top node of the pylon. The maximum
horizontal wind force which the canopy will
transmit to the top of the pylon is 20kN.
By applying the method of sections, calculate the
forces acting in the members intersected by ‘X –
X’.
4. The roof truss shown is simply supported at its ends.
3. calculate the support reactions
4. by applying the method of sections, calculate the forces acting in the members
intersected by ‘Z – Z’.
DET: Technological Studies: Structures and Materials (Advanced Higher)
69
5. For the structure loaded as shown below, calculate the hinge reaction and
magnitude and nature of the forces acting in the members intersected by ‘X – X’.
6. The figure shows a wall mounted frame under a single load. Use the method of
sections to determine the magnitude and nature of the forces acting in members A,
B and C.
DET: Technological Studies: Structures and Materials (Advanced Higher)
70