DET
Structures
Statics: Equilibrium in Frameworks and
Simple Stress
(Intermediate 2)
8083
.
November 2000
HIGHER STILL
DET
Structures
Statics:
Equilibrium in Frameworks and
Simple Stress
(Intermediate 2)
Support Materials


CONTENTS
Part 1 – Lecturer/teacher information and support materials
Outcomes
Delivery
Advice on starting levels
Assessment
Resource materials
Sources of information and learning materials
Acknowledgements
Part 2 – Student materials
Integrative learning approaches
Evidencing superior performance
Casework solutions
Tutorial examples
Resource materials
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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PART 1 – LECTURER/TEACHER INFORMATION AND SUPPORT
MATERIALS
Outcomes
The outcomes and PC’s are listed in sequence for the unit. After each PC a note is
included to indicate the extent of knowledge, understanding and application required.
Outcome 1 – Determine the support reactions for frame structures in equilibrium
Performance Criteria
a) Equilibrium in frame structures is analysed in relation to forces induced in the
members.
A restatement to the candidates of static equilibrium as:
No out of balance forces or components of forces in any direction or moments of
forces about any point occur for equilibrium to exist is a good starting point. These
conditions can be shown to exist in various structures by the simple observation that
neither liner nor angular movements occur.
A small step from this simple principle allows the deduction to be made that when
considering frame structures:
 the externally applied forces and moments must balance
 for any part of the frame, for example, a joint or node, equilibrium must exist
 for each member which makes up the structure equilibrium must exist.
The full understanding of these principles should not be laboured at this point since
continuous reinforcement will occur throughout the delivery of the unit.
b) The general conditions of equilibrium are correctly stated.
There is no real difference in general equilibrium as opposed to static Equilibrium
except that the general conditions allow for uniform movement of the structure to
occur either linearly or angularly provided that no acceleration in the direction of
motion is caused since this would require an out of balance force or moment to act. A
simple consideration of this realisation that equilibrium exists as long as no change in
the state of motion of the structure considered occurs allows the more general
statement known as the general conditions of equilibrium to be made.
No change in the state of a bodies motion will occur unless an out of balance force or
couple is caused to act on the body to cause this change. A body in this steady motion
state is said to be in equilibrium and the converse is also true namely
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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For a body to be in equilibrium no change in state of the bodies motion will occur and
consequently two conditions must apply.
The sum of all forces acting on the body must equal zero. The sum of all moments of
forces acting about any point in the plane of the body must equal zero.
c) Free body diagrams are accurately produced for frame structures in terms of force
Equilibrium.
Free body diagrams should generally be produced to aid with the identification of
possible points of application of reactionary forces induced at supports of frame
structures. Different common types of support can be introduced at this stage and
their characteristic behaviour discussed. Only point supports such as roller supports
and pin supports should be considered. Free body diagrams should be kept as simple
as possible showing all external forces and reactions along with related angles and
dimensions.
d) The reactionary forces are calculated correctly for frame structures in equilibrium.
A range of tutorial examples should be attempted initially where there are only a
total of three external forces and reactions acting on a series of frame structures.
Methods of solution should always start with a free body diagram and approaches
should include graphical approaches employing the principle of concurrency as
well as a moments and analytical force balance should be carried through to a
complete solution. The examples should continue with application containing
four or more forces using a moments and analytical force balance approach.
Outcome 2 – Draw vector diagrams and evaluate force induced in the members of
idealised frame structures
Performance Criteria
a) All assumptions made when considering plane frames are in accordance with
established practice.
A few real examples of frame structures could be looked at in detail and sketches
made showing the position of members, supports and loading. It should then be
explained that the loading transferred to the members of real frames can be very
complex causing bending and torsional effects as well as direct loading. To simplify
the force analysis of such frames then assumptions are generally made to simplify the
analysis in as much as only direct loading is considering to act on each member.
Generally these assumptions are:
 all members are straight
 all Joints are held together by frictionless pins
 forces and reactions only occur at joints
 the whole structure and all its parts are in equilibrium
 force induced in any member can only act along its length.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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b) Frames are analysed correctly to determine all external forces.
This work has already been covered in Outcome 1 since essentially the reactions
require to be determined by analysis of the applied forces. It is now completed as a
common step in the process of establishing the size and nature of the forces induced in
the members and this performance criteria can be overcome either before or during
the drawing of force vector diagrams.
c) Combined vector diagrams for frames are drawn to a scale which allows accurate
determination of forces in members.
This is really the main work of the unit and considerable time should be spent
developing the necessary techniques.
Starting with a simple triangular frame with a load at the apex supported at each
bottom corner the reactions could first be determined by inspection or by moments.
The vector diagram for each joint can then be completed and the following
conclusions drawn:

The vector for each member of the frame always appears in two joint vector
diagrams but the arrows on the two diagrams are opposite in direction. This sense
of direction can be transferred back to the space diagram for the frame clearly
indicating that tension or compression is induced in each member.

The distinction can be made that the vector diagrams indicate force caused by the
members on the joints and generally the requirement is for force caused by the
joints on the members so that arrows pointing towards each other on a frame
member in the space diagram indicates the member is in tension, sometimes called
a tie. Conversely arrows pointing away from each other on a member in the space
diagram indicates the member is in compression, sometimes called a strut.

This method of considering one joint at a time is long and as already observed
duplicates many of the vectors. Clearly some method of drawing the diagrams on
top of each other to produce a combined diagram would be quicker and more
compact but some method other than using arrows to indicate sense of direction of
vectors is required.
This approach leads to the introduction of Bow’s Notation. The analysis can then be
done again using Bow’s Notation and the combined vector diagram produced.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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d) The magnitude and nature of forces caused to act on specified members are
determined correctly using combined vector diagrams.
The sense of the forces at each end of each member can now be transferred to the
space diagram and finally a table drawn up indicating the letter identification of each
member along with the force induced in the member and the nature of this force.
It should be stressed at this stage that the vector diagram for each joint can only be
completed when not more than two unknown forces occur at that joint.
A set of graded examples can now be completed on prepared sheets to practice the
techniques required and to increase the confidence of the candidates.
e) The conditions of redundancy and collapse are indicated correctly during the
production of appropriate combine vector diagrams.
Although in practical situations the number of members in a plane frame could be 20
or more no real benefit can be derived at this stage by considering frames with more
than 7 members. Once about three of four examples have been completed
successfully with considerable help by the presenter the conditions of insufficient and
overabundance of members can be considered. Insufficient members will result in
collapse indicated during the drawing of the combined vector diagram by the
impossibility of proceeding with the diagram due to lettered points not matching up at
the required angle. An overabundance of members will result in what is often called
redundancy, where the extra members do not carry any of the external load indicated
by one or more points in the combined vector diagram having more than one letter
meaning the member represented by these letter having a vector of zero length. One
or two examples having each of these conditions can then be overtaken.
Outcome 3 – Calculate direct stress and strain and select materials for components
Performance Criteria
a) Direct stress and strain are defined accurately in terms of established practice.
Once the forces acting on a particular member of a frame or any other loaded
component have been established the next logical step would be to deduce what size
of cross section is required by the component to safely support this load. Candidates
will realise that a strong material will require less of a cross section to support a
particular load than a weak material leading to the definition of stress and its function
as an indication of load bearing ability.
The consideration that change in length is also a direct consequence of directly
applied load will lead to the requirement to predict the change in length liable to occur
and the definition of strain and its function as an indicator of resulting change in
length.
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A range of materials which behave in a linear elastic way can then be considered
using their stress/strain curves and a brief explanation given of the material behaviour
under the three stages of elastic, partly elastic and plastic matched between the
diagram and the behaviour of the a component.
Candidates will realise that loading beyond the limit of proportionality is not good
practice.
b) The relationship between stress and strain within the elastic limit is stated in
terms of established theory.
The linear relationship between stress and strain up to the elastic limit will lead to the
conclusion that stress divided by corresponding strain will give a constant value for a
material and that constant is given the name modulus of elasticity.
The range of materials considered can then have numbers attached to them indicating
typical values of ultimate strength, yield (or 0.1% elongation) stress and modulus of
elasticity and the relative merits of the material behaviour considered with some
indication of their common applications.
c) Calculations of stress and strain are accurately performed for loaded components.
A set of graded examples can now be completed possibly as an assisted tutorial
session starting with fairly simple calculations of stress and strain and leading to
calculating the overall extension of a loaded member having two different cross
sections along its length.
d) Component materials are correctly selected to meet simple loading requirements.
This should essentially be presented as an investigative procedure where the
requirements of a loaded component are given and a selection required to be made
from four available materials. Details of ultimate strength, yield stress, modulus of
elasticity and possibly forms of supply should be given a simple tabular form and the
candidate asked to select and justify the selection.
This process should be completed at least three times and the candidate encouraged to
consider all the given parameters and make a selection on a basis of elimination and
optimisation.
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Delivery
It is likely that most candidates attempting this unit will have already achieved a
Standard Grade Mathematics at grade 3 and at least one other Standard Grade 3 in a
technological subject.
Some experience will have been attained in vector quantities and their presentation,
mathematical manipulation and the presentation of problem solutions. It is also likely
that the companion unit in this Intermediate 2 course, Fundamentals of Technology:
Structures, has also been completed recently by the candidate. Thus giving
experience in the general conditions of equilibrium, calculation of moments, vector
diagrams for forces and considerable experience of problem solving involving the
external forces on structures. If candidates have completed this other structures unit
then more time can be spent on application when presenting this unit.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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Advice on starting levels
The requirements of Outcome 1
The conditions common to structures in static equilibrium can be determined by
considering a series of structural analyses derived from commonly used frames, such
as cranes and roof trusses. The general conditions of equilibrium can then be stated
and reactions calculated with the use of free body diagrams. The special case of three
force non-parallel systems should be solved using the principle of concurrency and
also the conditions of equilibrium.
The requirements of Outcome 2
The normal technique of graphically determining the forces acting in the members of
frames caused by external loading can be built up using a simple three member frame
with a single applied load and two reactions. The introduction of Bow’s notation
allows a combined vector diagram to be developed. The various requirements such as
the magnitude and nature of forces in members can then be overtaken. The system
can be extended up to frames with 7 members and the use of roller and pin supports.
Prepared sheets could be used in most instances but a blank sheet approach should be
attempted on one or two occasions to develop the skills of scale selection and layout.
Redundancy and inadequate members could be introduced once frames up to 7
members can be successfully solved. Collapse should only be treated in a very simple
way as one vector diagram is being generated showing that the addition of a member
to the frame allows the vector diagram to be completed. Redundancy, which is more
common should be encountered in two or three examples.
Computer programmes can be used possibly on the same range of examples once the
hand graphical techniques have been mastered and are understood.
The requirements of Outcome 3
Discussions and demonstrations should be used to highlight the concepts of stress and
strain for tension and compression loading only.
The relationship between stress and strain within the elastic limit and the examination
of modulus of elasticity values for a range of materials should be introduced once a
few elementary stress and strain calculations have been mastered. A tutorial approach
would then allow individual experience to be built up of the calculations and
mathematical manipulations concerned with stress, stain and modulus of elasticity up
to examples where two different cross sections must be considered for a simply
loaded component.
Examples of useful tutorial questions, integration of applications and possible sources
of useful materials are indicated later in this pack cross referenced to outcome
requirements.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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Assessment
Several assessment packs are available for this unit from National Assessment Bank
and should be used in conjunction with the Structures Arrangement Document and the
Structures Subject Guide already distributed to all centres by the Higher Still
Development Unit.
The first of the assessment packs available suggests how and when assessment tasks
should be administered, the recommended duration of each assessment and conditions
under which assessment should take place.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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Organisation and conditions for assessment: using instrument(s) of assessment
(Part of Section 1 – Organisation and conditions for assessment, this has been
reproduced from the first assessment pack).
There are three assessments in this unit each one can be administered as an end of
topic test, possibly best completed at the end of each outcome.
INSTRUMENT
OF
ASSESSMENT
OUTCOMES
& PC'S
COVERED
1
2
TYPE OF INSTRUMENT
3 structured questions.
1b
1 requires the conditions of equilibrium to be stated.
1 a, c, d
2 and 3 require a free body diagram to be sketched and the
reactions determined.
2 a, b, c, d, e
1 structured question.
Requires a frame with 7 members to be graphically analysed to
determine the force in each member .
3
3 a, b, c, d
1 structured question.
The equations for direct stress, strain and the elastic
relationship between them are correctly stated, selected and
manipulated to calculate stress and extension induced in a
loaded member with two different cross sections along its
length. A material is selected from those available to suit the
loading requirements.
The timing and duration of assessments
The 40 hours could be divided into approximately:
OUTCOME
TEACHING TIME
ASSESSMENT TIME
(INCLUDING REASSESSMENT)
1
12 hours
1 hour
2
16 hours
1 hour
3
9 hours
1 hour
Assessments are best held as end of topic tests.
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Reassessment
Time is allowed within units for assessment and reassessment of outcomes. Where a
candidate has not attained the standard necessary to pass a particular outcome or
outcomes, they should have the opportunity to be reassessed. Reassessment should
focus on the outcome(s) concerned and, as a general rule, should be offered on a
maximum of two occasions following further work on areas of difficulty. Evidence
from the original unit assessment should assist teachers and lecturers to identify why
an individual candidate has failed to achieve a particular outcome and to plan focused
support for learning.
General guidelines for reassessment
 Candidate may be given the opportunity to orally correct minor errors and these
oral responses recorded on the original documentation.
 Where a complete or part of an assessment instrument is required to be repeated
this should be done by the candidate as soon as possible after the original attempt
and again attached to the original documentation.
The conditions under which assessment takes place
Arrangement documents refer to assessment being carried out under controlled
conditions to ensure reliability and credibility. For the purposes of internal
assessment, this means that assessment evidence should be completed under
supervision to ensure that it is the candidates' own work. Supervision may be carried
out by a teacher, invigilator or other responsible person, for example, a workplace
provider.
All three assessments should be 'closed book' and carried out under controlled
conditions. Candidates should be allowed to use data sheets and non programmable
calculators. Those responsible for supervision of the assessments should ensure that
unmarked copies of the data sheets are available for candidate use.
During the presentation of the learning materials for each outcome candidates should
be informed of the form and nature of each assessment also that it is their knowledge
and understanding that is being assessed. Before assessment is attempted candidates
must have experienced the answering of similar types of question via dedicated
learning materials, homework or tutorial sessions. This experience should be acquired
during the delivery of the unit.
Using internal assessment evidence to contribute to course estimates
Unit assessments are designed exclusively for internal assessment. Due to the nature
of this unit it would be difficult for a candidate who is capable of superior
performance to evidencing this when answering each question of each assessment
instrument since solutions involve the statement of knowledge and the following of
fairly predictable techniques of logical and mathematical analysis.
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However the Assessment for Outcome 3 in this unit does allow the candidate to make
a selection for a material and the reasoning for this selection could indicate superior
performance. One other unit in this course – Fundamentals of Manufacture and
Assembly Techniques, Outcome two, when planning a component manufacturing and
assembly procedure, also allows superior performance to be evidenced. When
attempting similar, but perhaps more extended, work during the additional 40 hours
further evidence of superior performance can be produced. The integrated and
broadening nature of the course external assessment would establish the level reached
by the candidate performance.
A ‘prelim’ assessment used to prepare the candidates for the external assessment
would be a more reliable source of estimating candidate performance.
Basically three Assessment Instruments are recommended:


Assessment Instrument 1 – 3 questions
Recommended duration – 1 hour
Presented under controlled, supervised conditions to ensure the completed attempt is
the candidate’s own work.
The first question requires the candidates to state the general conditions of
equilibrium. In the process of completing this question the candidate demonstrates
that they can:
 correctly state the general conditions of equilibrium.
The second question requires the candidate to draw a free body diagram for a frame
structure with one load showing all external forces and reactions, as well as their
relative position to each other. The candidate is then asked to determine the value and
direction of any reactions. In the process of completing this question the candidate
demonstrates that they can:
 analyse the external acting forces
 use the existing conditions of equilibrium
 produce a free body diagram
 calculate the reactionary forces on the frame.
The third question requires the candidate to determine the value and direction of two
reactions, given their position for a more complex loaded frame than in question 2. In
the process of completing this question the candidate demonstrates that they can:
 analyse the externally acting forces
 use the general conditions of equilibrium
 calculate the reactionary forces on the frame.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
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

Assessment Instrument 2 – 1 question
Recommended duration – 1 hour
The candidate is asked to state three assumptions made when determining the forces
in the members of an idealised frame structure then to graphically solve a problem
involving a frame with 7 members to determine the force in each member of the
frame. A worksheet is provided showing the loaded frame, giving a starting point for
the combined vector diagram and a table to be completed to show the force and its
nature in each member. In the process of completing this question the candidate
demonstrates that they can:
 make the usual assumptions in solving plain frames
 determine reactions where appropriate
 draw a combined vector diagram for the forces on the members to scale
 determine the magnitude and direction of forces
 induced in the members from the combined vector diagram
 indicate the presence of redundancy or collapse if appropriate to the particular
problem being attempted.


Assessment Instrument 3 – 1 question
Recommended duration – 1 hour
The candidate is asked to define stress and strain, give the basic units used for these
properties and relate stress and strain with the modulus of elasticity for a linear elastic
material. The question goes on to set a problem associated with a loaded component
having two different cross sections along its load line. The candidate is asked to
determine the stress induced by the loading, select a suitable material to use
considering the modulus of elasticity and yield stress of a selection of given materials
then continue by calculating the overall change in length of the component made from
the material chosen. In the process of completing this question the candidate
demonstrates they can:
 define stress and strain
 relate stress and strain within the elastic limit
 calculate stress and strain
 select a material to meet simple loading requirements.
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Resource materials
Several visible examples of real frames should be used to provide initial interest, these
can be static or kit models but ideally should be part of the centre structure (the hall
roof trusses) or nearby (the side of a steel bridge). Structures of this type can be used
to point out load and reaction application points and to conclude that force would be
passed to the members. Later in the presentation redundancy and collapse
(mechanism) can be considered for real situations.
Examples of worksheets that can be used to draw combined vector diagrams and
deduce the value and nature of forces induced in frame members are given later in this
pack.
Sources of information and learning materials
This unit, although written to satisfy the requirements of an Intermediate 2 Structures
course, contains many elements that have already been covered by earlier SQA
Modules. Materials already developed for these earlier modules can provide a useful
source of examples and applications.
The Subject Guide Engineering 2: Structures at Intermediate 2 – Appendix 4 maps the
content of this unit with:
 74012 Statics: Components and Structures (1/2)
 64007 Strength of Materials.
The main sources of information used to develop this pack were the following Higher
Still Development Units Publications:
 Arrangements – Structures Intermediate 2 May 1997
 Subject Guide – Structures Intermediate 2 1997
 NABS – Statics: Equilibrium in Frameworks and Simple Stress January 1999.
The learning materials contained in this pack were developed from first principles but
similar material is available from any textbook on Statics published since 1980.
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PART 2 – STUDENT MATERIALS
Integrative learning approaches
Once the candidates have gained some experience in calculating the reactions of
simple frames, can use Bow’s notation and can produce a combined vector diagram
for the forces on the members of a frame, then extended questions will allow all the
requirements of Outcomes 1 and 2 to be overtaken in an integrated way. Following
elementary work on stress, strain, Modulus of Elasticity and the behaviour of
materials under load then extended questions involving these principles plus material
selection can be attempted. Extended questions can now be attempted in small groups
and on an individual basis which sample across the whole unit.
During the additional 40 hours component of the structures course this work could be
further extended using similar techniques spanning across two or more of the three
units making up the course.
Four examples of this type of learning vehicle are given below.
1. Assignment 1 – An overhung frame
The worksheet, Assignment 1, shows the space diagram for an overhung frame.
The frame is pinned to a vertical wall at its upper and lower left hand ends and
supports a load of 30 KN at its lower right hand end. No vertical force is caused
in the wall.
Your task is:
a) to construct a free body diagram on a separate sheet showing the relative
position
of all forces and reactions;
b) go on to determine any reactions
i)
graphically
ii)
analytically.
c) to state three assumptions made when determining the forces caused in the
members of the frame by the external loading;
d) to use worksheet Assignment 1 to graphically determine the forces in each
member of the frame;
e) to state the member supporting the greatest force and the size and nature of
this force.
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Worksheet: Assignment 1
Use to complete parts d) and e)
d)
Vector diagram
Scale 1kN = 2mm
MEMBER
FORCE kN
e) The member supporting the greatest load is:
NATURE
the size of this force is:
The nature of this force is:
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2. Assignment 3 – A stressed component
State the parameter or property which measures the land per unit of cross sectional
area supported by a component.
State the parameter or property which measures the change in length per unit
length of a component under direct loading.
State how these two parameters or properties are related to each other for linear
elastic materials loaded within their elastic limit.
Show the common symbols used for all three of these parameters or properties
along with both basic units and normally used units used to specify them.
A component has an overall length of 300mm and an outside diameter of 40mm.
The middle 100mm of the component is reduced in diameter to 30mm. A direct
tensile load of 60kN is applied to the ends of the component.
Determine:
a) the stress induced in the two different cross sections of the component;
b) a suitable material to use, in order to manufacture the component, selected
from the table below:
MATERIAL
MODULUS OF
ELASTICITY
YIELD
STRESS
GN M-2
MN M-2
Mild Steel
200
120
Aluminium
70
90
Brass
85
100
c) the overall extension of the component under load.
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3. Assignment 4 – The change in length of a frame member under load
The pin jointed frame shown is supported by a pin reaction at its top right hand
end and a cable at its top left hand end. The two bottom joints of the frame both
support external loads.
The frame itself consists of three equilateral triangles, all the same size, with each
member having a length of 2m.
Determine the member supporting the greatest load and the size and nature of this
load.
Given that the members are made of aluminium having a yield stress of 70MN m-2
and a modulus of elasticity of 90 GNm-2. All members have a cross sectional area of
500mm2. Determine the change in length of the member which supports the greatest
load.
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Evidencing superior performance
The information reproduced earlier in Part 1 of this pack from the first National
Assessment Bank entry for this unit indicates that the third assessment instrument
would allow candidates to make qualified choices and apply knowledge and
understanding when making material selections at a level above that for a C pass in
the subject. The integrative and extended nature of the work required when
attempting the assignments could also provide a source of evidencing superior
performance. Any evidence produced to support superior performance would have to
be retained for each candidate to back up the lecturer’s/teacher’s predictions of this
performance or to b used as part of an appeals procedure.
Examples of where to look for evidence indicating superior performance are given for
each assignment.
For Assignment 1 – An overhung frame
a) This requires the candidate to construct a free body diagram showing all forces
and reactions in relation to each other.
Superior performance could not easily be demonstrated.
b) This requires the candidate to determine the reactions acting on the frame
graphically and analytically.
The graphical solution involves recognising that the principle of concurrency
applies, using a scale drawing of the free body diagram to establish the point of
concurrency and so to determining the direction of R1. A vector diagram of the
30kN force and the two reactions allows both reactions to be determined by
measuring the vectors.
The analytical solution involves applying the conditions of equilibrium to
calculate the reactions.
Superior performance could not easily be demonstrated.
c) A statement of three assumptions made when determining the forces caused in
the members of a frame by external loading is simply the recall of knowledge.
Superior performance could not easily be demonstrated.
d) Employing the given worksheet to generate the combined vector diagram to a
given scale then using this to establish the forces and their nature in each of the
frame members again is mainly using well practised techniques.
Superior performance could not easily be demonstrated.
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e) Selecting the member supporting the greatest force and stating the size and nature
of the force comes directly from the table generated in the last task.
Superior performance could not easily be demonstrated.
For Assignment 2 – A stressed component
The first part of this assignment where stress, strain and modulus of elasticity are all
considered in detail is the recall of knowledge.
Superior performance could not easily be demonstrated.
During the second part of this assignment after the stress values have been calculated
and a material has to be selected a fairly complex reasoning has to be applied.
Depending on how the material selection is made and how the reasoning for this is
recorded then superior performance could be evidenced.
For Assignment 3 – The change in length of a frame member under load
The first part of this assignment requires the candidate to determine the member
supporting the greatest load and the size and nature of this load. Only a sketch of the
loaded structure is provided. The candidate has to consider how to break down this
task themselves then successfully complete the task involving:
 calculation of at least one reaction
 selecting a scale and starting point for the combined vector diagram
 drawing the diagram without falling off the paper
 using the diagram to measure and nature the force in all members.
The method of attempting the first three items and the setting out of the solution gives
scope for evidencing superior performance.
The second part of this assignment requires the candidate to determine the change in
length of the member bearing the greatest load. The treatment requires the candidate
to use the load determined from the first part of the assignment, the modulus of
elasticity for aluminium and stress/strain theory to calculate the required change in
length.
Superior performance could not easily be demonstrated.
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Assignment solutions
Assignment 1 – An overhung frame
a)
b)
i) The force system is a 3-force non parallel type so the principle of
concurrency applies.
For no vertical force to occur in the wall R2 must act horizontally, R2 and the
30 kN force meet at the point of concurrence, that is the top right hand end of
the frame.
R1 must also pass through this point hence the direction of R1 can be
determined from the free body diagram drawn to scale.
Angle X measures 26.5o.
The vector diagram for the forces R1, R2 and 30 kN can now be completed.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
23
R2 = 60 kN to scale
b)
Scale 1 kN = 2 mm
ii) Take moments about R1
30 x 8 = R2 x 4
R2 = 60 kN acting horizontally to the left
R1Sin x
60 kN
R1Cos x
30 kN
For horizontal force balance
60 = R1 Cos X
----1)
For vertical force balance
30 = R1 Sin X
----2)
divide equation 2) by 1)
Sin X = 30
Cos X 60
henceTan X = 0.5
Hence Angle X = 26.56o
c) Any three statements from:
 all joints or nodes are frictionless pin joints
 all external forces are applied at joints or nodes
 each member can be represented by a straight line
 the whole structure and hence each member is in equilibrium
 all forces induced in the members of the frame act along the length of the
member.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
24
Worksheet: Assessment 1
Use to complete parts d) and e)
d)
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
25
Vector diagram
Scale 1kN = 2mm
MEMBER
FORCE kN
NATURE
CB
60
Tension
BD
41
Compression
DA
30
Compression
DE
30
Tension
EC
30
Tension
EA
41
Compression
e) The member supporting the greatest load is CB, the size of this force is 60kN
and the member is in tension.
Note: One way of considering member deficiency or collapse could be to consider the
frame used in this assignment again the remove the member BD.
An attempt to draw the vector diagram will result points b and c being generated in
the combined vector diagram, these points should be horizontal to each other but are
not when drawn clearly indicating that something is wrong.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
26
The vector diagram for the first joint CAD can be drawn as before. If joint DAB is
now drawn then joint CDB attempted, the vector bc should be horizontal and this is
not possible indicating that equilibrium cannot occur at this joint.
Further consideration will result in the observation that a member to support the
vertical component of R1 does not exist at joint BA, hence a member is missing
between joints BA and CDB.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
27
Assignment 2 – A stressed component
Stress is the measurement of the load per unit cross sectional area supported by a
component.
Strain is the property which measures the change in length per unit length of a
component.
When a linear elastic material is loaded within the elastic limit the resulting stress is
directly proportional to the corresponding strain. The value of stress divided by
corresponding strain is called the modulus of elasticity for the material.
PROPERTY
COMMON SYMBOL
Nm-2
Stress
E
Ø 40mm
NORMAL UNIT
MNm-2
Strain
Modulus of Elasticity
BASIC UNIT
-
-
Nm-2
GNm-2
Ø 40mm
Ø 30mm
100mm
a) Longer cross sectional area =
Smaller cross sectional area =
100mm
100mm
x 402 = 1256.6mm2 or 1.2566 x 10-3m2
4
x 302 = 706.9mm2
4
or 0.7069 x 10-3m2
Stress = Load
Area
Hence stress in the longer cross section =
60 x 103
1.2566 x 10-3
Stress = 47.75 MNm-2
Stress in the smaller cross section =
60 x 103
0.7069 x 10-3
Stress = 84.88 MNm-2
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
28
b) The maximum stress induced in the bar is rather near the elastic limit (yield
Stress) for Aluminium and Brass, so the more suitable material to use would
be mild steel.
c) Using mild steel as the component material the modulus of elasticity is
200 GNm-2.
The extension in one 40mm dia section of bar can be calculated from
E = Stress hence Strain = Stress
Strain
E
Hence strain = 47.75 x 106 = 0.2388 x 10-3
200 x 109
Strain = change in length hence extension = strain x original length
Original length
Hence extension = 0.2388 x 10-3 x 100 = 0.02388mm
for both lengths of 40mm dia bar the total extension would be
2 x 0.02388 = 0.04776mm.
In the same way the extension in the 30 mm dia section of bar can be calculated
from
Strain = 84.88 x 106 = 0.4244 x 10-3
200 x 109
extension = 0.4244 x 10-3 x 100 = 0.04244mm
Hence the total extension of the component under load is
0.04776 + 0.04244 = 0.0902mm
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
29
Assignment 3 – The change in length of a frame member under load
Both loaded joints have more than 2 unknown vectors so one reaction must be
determined.
R1
30º
A
6kN
3kN
R2
X
4X
Taking moments about R2
6 x X + 3 x 3X = R1 Sin 30 x 4X
6 + 9 = 2R1
R1 = 15 = 7.5kN
2
This will allow the combined vector diagram for the frame to be drawn staring at the
upper left hand joint.
The space diagram should now be drawn again on a blank sheet of paper and the first
attempt at the combined vector diagram drawn lightly from a convenient starting point
a using a reasonable scale of say 1kN = 10mm.
The scale and starting point can then be modified as required until a satisfactory
vector diagram is achieved (see next page).
The force and nature in all members can now be recorded in a suitable table.
Inspection of the table will determine the member supporting the greatest load is
GB, the load is 9.5kN in compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
30
Combined Vector diagram produced by trial and error.
MEMBER
FORCE kN
NATURE
AE
4.3
Tension
EB
8.5
Compression
EF
1.0
Compression
FD
2.7
Tension
FG
1.0
Tension
GB
9.5
Compression
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
31
The stress induced in member GB is from:
Stress = Load
Area
Stress = 9.5 x 103 = 19 MN m-2
500 x 10-6
This is well within the yield stress of the material.
Corresponding strain can be calculated from:
E = Stress
Strain
hence strain = Stress
E
Strain = 19 x 106 = 0.2lll x 10-3
90 x 109
and strain = Change in length
Original length
Hence change in length = Strain x Original length
= 0.2111 x 10-3 x 2 x 103
= 0.4222 mm
Since the stress is compressive the change in length will be a reduction.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
32
Tutorial examples
The following tutorials are intended to be used directly after the required principles
have been described and discussed. While attempting the tutorial questions the
candidates may require to be given some assistance in the early stages, but candidates
should be encouraged to work through examples and complete prepared worksheets
themselves as far as possible.
Outcome 1 and 2 tutorial
1. A loaded structure made up from straight members pinned together at its joints
is stationary. Indicate which of the following statements are true and which are
false, giving one reason for your answer in each case:
a) no out of balance external force can exist;
b) no out of balance force can exist in the frame members;
c) each member can only be in tension or compression;
d) all members of the frame must support some of the load;
e) the vector diagram for each joint must form a closed figure;
f) the space diagram for the frame shows all external forces, all members
and all important dimensions;
g) the free body diagram only shows all external forces and reactions
including their direction and sense;
h) reactions can exist even if no external forces are applied;
i) a frame will collapse if it has insufficient members to support the load or
reactions acting on each joint.
2. a) In each of the following situations, draw a free body diagram for the loading
on the frame adding the direction and sense of any reactions.
b) i) In each situation making use of the free body diagram, determine the
magnitude and sense of the reactions.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
33
b)
ii)
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
34
3. Each of the worksheets supplied shows, a space diagram for a frame structure in
equilibrium with the spaces lettered according to Bow’s notation, a starting point
to begin the combined vector diagram for the joints and a suitable scale to draw
this diagram.
In each case draw the combined vector diagram and use this to complete the
table at the bottom of each worksheet.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
35
Worksheet for question 3
+
a Starting Point
Scale 1kN = 20mm
MEMBER
FORCE kN
NATURE
DA
DB
DC
CA
BC
For nature enter T for tension and C for compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
36
Worksheet 2 for question 3
+
a Starting point
Scale 1kN = 20mm
MEMBER
FORCE kN
NATURE
DA
4.3
T
DB
2.5
C
DC
1.3
T
CA
5.4
24o
BC
2.2
90o
For nature enter T for tension and C for compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
37
Worksheet 3 for question 3
+ a Starting point
Scale 1kN = 20mm
MEMBER
FORCE kN
NATURE
DA
4.3
T
DB
2.5
C
DC
1.3
T
CA
5.4
24o
BC
2.2
90o
For nature enter T for tension and C for compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
38
Worksheet 4 for question 3
+
a Starting point
Scale 1kN = 10mm
MEMBER
FORCE kN
NATURE
EA
EB
EF
FD
FG
GB
GC
For nature enter T for tension and C for compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
39
Outcome 3 tutorial
1. a) Define stress in terms of load and area supporting the load, state the
symbol and basic units used.
b) Define strain in terms of change in length under direct loading and the
length of the component, state the symbol used.
c) State the relationship between stress and strain within the elastic limit.
State the symbol and basic units used for modulus of elasticity of a
material.
2. A composition rubber pad is to be used as a mounting between a concrete floor
and a heavy machine tool. The mounting compresses 3mm under a load of 6kN.
The stress caused in the rubber must not be more than 250 kNm-2. Determine:
a) the diameter of pad required;
b) the thickness of pad required.
Take the modulus of elasticity for composition rubber as 950kN m-2.
3. A steel strut is 600mm long, 40mm dia for half its length and 20mm dia for the
other half. Under a load of 30kN determine the overall reduction in its length.
Take the modulus of elasticity for steel as 200MNm-2.
4. A alloy steel bar increases in length by 0.05% when subjected to a tensile stress of
108 MN m-2. Calculate the modulus of elasticity for the material. State any
assumptions made.
5. A tie used in a structure is 800mm long. The first 200mm requires to have a hole
drilled along its length causing its cross section to be 36mm outside dia and 20mm
inside diameter. The remaining 600mm length has be have its outside diameter
reduced to 25mm dia.
a) determine the stress induced in each cross section of the bar under an
axial load of 30kN;
b) On the basis of the tale of properties given and the maximum stress
induced in the tie, justify the selection of a material to use to manufacture
the tie;
MATERIAL
YIELD STRESS
MN M-2
ULTIMATE STRENGTH
MN M-2
MODULUS OF
ELASTICITY
Timber
20
70
15
Brass
65
170
84
Aluminium
90
190
70
c) For the material chosen, calculate the overall extension of the tie under load.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
40
Solutions to tutorial examples
Outcome 1 and 2 tutorial
a) True
Since the frame is stationary it can have no out of balance force or couple acting
on it, that is to say, the frame is in equilibrium.
b) True
Since the frame is in equilibrium then each part of it is in equilibrium, hence each
frame member must not have any out of balance effect on it.
c) True
In a pin jointed structure the only direction that forces can have on members is
along the member and since the member itself is in equilibrium only equal and
opposite forces will exist along the member causing tension or compression.
d) False
It is possible for some members of a frame to have no force caused to act in them
by the external forces or passes on by other members.
e) True
Since each joint is in equilibrium then the vector diagram for that joint must close.
f) True
The space diagram is a simplified sketch of the loaded frame showing all the
things stated.
g) False
The free body diagram must show enough information to apply the candidates of
equilibrium to externally applied forces and reactions. Hence the relative position
of these forces must also be shown, that is all the dimensions required to calculate
moment arms.
h) False
Reactions are caused to act when forces are applied, hence no forces mean no
reactions.
i) True
Equilibrium cannot exist at a joint where the members are not sufficient to support
the loading and collapse would result.
6 kN
2 i) a)
1m
R1
1m
R2
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
41
b)
Take moments about R1
6
x 1 = R2 x 2
R2 = 3 kN
From vertical force balance
R1 + 3 = 6
hence R1 = 3 kN
Note: since the reactions are symmetrical to the load it would also be
acceptable to determine their value by inspection.
2. ii) a)
5kN
O
2.5m
A
6m
R2
R1
b) Take moments about R1
5 x 2.5 = R2 x 6
R2 = 0.833 kN
0.833kN
5kN
R1 cos A
R1 sin A
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
42
From Vertical force balance:
R1 Sin A = 0.833
----1)
From Horizontal force balance:
R1 Cos A = 5
----2)
By dividing equation 1) by 2)
Sin A = 0.833
Cos A
5
Tan A = 0.1667
Hence angle A = 9.462o
From R1 Sin A = 0.833
R1 =
0.833
= 5.07 kN
Sin 9.462o
Note: Since this is a 3 force, non-parallel force system the free body diagram can be
drawn to scale, the point of concurrency determined at 0, angle A determined by
joining 0 to the lower left hand joint and a vector diagram drawn to a suitable scale to
determine the size of R1 and R2.
2. iii) a)
2kM
2m
3kM
4m
2m
R2
R1
b) Take moments about R1
2 x 2 + 3 x 6 = R2 x 8
R2 = 2.75 kN
From Vertical force balance
R1 + 2.75 = 2 + 3
R1 = 2.25 kN
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
43
2. iv)
a)
2kN
4kN
R1
45
1m
1m
2m
A°
b) Take moments about R2
2 x 1 + 4 + 3 = R1 Sin 45 x 4
2 + 12 = 2.8284 R1
R1 = 4.95 kN
4.955 sin 45
R1 sin A
R1 cos A
4.95 cos A
4kN
2kN
From the vertical force balance
R1 Sin A + 3.5 = 4 + 2
R1 Sin A = 2.5 ----1)
From the horizontal force balance
R1 Cos A = 3.5 ------2)
Divide equation 1) by 2)
Sin A = 2.5
Cos A 3.5
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
44
Tan A = 0.7143
Angle A = 35.54o
From R1 Sin 35.54 = 2.5
R1 =
2.5
Sin 35.54
R1 = 4.3 kN
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
45
Worksheet for question 3
6kN
A
B
D
C
R1
R2
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
46
Starting Point
a
+
drawn to a scale of 1kN = 20mm
d
c
b
Scale 1kN = 20mm
MEMBER
FORCE kN
NATURE
DA
3.45
C
DB
3.45
C
DC
1.75
T
CA
3.0
90o
BC
3.0
90o
For nature enter T for tension and C for compression
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
47
Worksheet 2 for question 3
A
B
D
R1
C
R2
d
c
drawn to a scale of 1kN = 20mm
+
Starting point
a
b
Scale 1kN = 20mm
MEMBER
FORCE kN
NATURE
DA
4.3
T
DB
2.5
C
DC
1.3
T
CA
5.4
240
BC
2.2
900
For nature enter T for tension and C for compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
48
Worksheet 3 for question 3
3kN
2kN
B
A
F
E
C
D
+ Starting point a
e
b
f
d
Drawn to a Scale of 1kN = 20mm
MEMBER
FORCE kN
NATURE
EA
2.3
C
EB
1.2
C
EF
0.5
C
FC
3.3
C
FD
1.6
T
c
For nature enter T for tension and C for compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
49
Worksheet 4 for question 3
Cable support
B
A
45°
A
F
A
E
A
D
A
4kN
G
A
C
A
2kN
c
d
f
e
b
g
+ a starting point
Drawn to a Scale of 1kN = 10mm
MEMBER
FORCE kN
NATURE
EA
4.0
T
EB
5.6
C
EF
0.5
C
FD
1.8
T
FG
0.5
C
GB
5.0
C
GC
3.0
T
For nature enter T for tension and C for compression.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
50
Outcome 3 tutorial
1.
a)
The stress induced in a component under direct load is calculated from the
load divided by the cross sectional area of the material supporting the load.
The symbol used to denote stress is the Greek letter (sigma) and the basic unit
is Newton per square metre.
b)
The strain induced in a component under direct load is calculated from the
change in length under the load divided by the original unloaded length of the
component. The symbol used to denote strain is the Greek letter (epsilon).
c)
Stress is directly proportional to the corresponding strain when a material is
loaded within the elastic limit.
The value calculated from stress divided by strain with the elastic limit is
called the modulus of elasticity for the material.
The symbol used for the modulus of elasticity for a material is E and its basic
units are Newtons per square metre.
2.
a)
From Stress = Load
Area
Area = Load = 6 x 103
= 0.024 m2
3
Stress 250 x 10
The pad is circular in cross section hence
A = d2 = 0.024
4
where d is the diameter of the pad
hence d2 = 4 x 0.024 = 0.03056 m2

2
d = 0.1748 m or 174.8 mm
c)
Since E = Stress
Strain
Strain = Stress = 250 x 103 = 0.2632
E
950 x 103
and Strain = change in length
original length
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
51
Original length = change in length
Strain
Original length = 3
0.2632
Original length = 11.4mm
3.
Stress in the 40 mm dia section = Load
Area
Stress (40 mm dia) = 30 x 103 = 23.873 n mm-2
 x 402
4
E = Stress hence Strain = Stress
Strain
E
Strain (40 dia) = 23.873 x 106 = 0.1194 x 10-3
200 x 109
Strain = change in length
Original length
hence Change in length = Strain x Original length
reduction in length (40mm dia) = 0.1194 x 10-3 x 300 = 0.0358mm
Stress (20mm dia) = 30 x 103 = 95.493 N mm-2
 x 202
4
Strain (20mm dia) = 95.493 x 106 = 0.4775 x 10-3
200 x 109
reduction in length (20mm dia) = 0.4775 x 10-3 x 300 = 0.1432mm
Hence the overall reduction in length under load = 0.0358 + 0.1432 =
0.179mm
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
52
4.
Assuming the material is loaded within its elastic limit
Stress = 108 x 106 N M-2
Strain = 0.05 = 0.5 x 10-3
100
Since E = Stress
Strain
E = 108 x 106 = 215 GN m-2
0.5 x 10-3
5. a) Cross section of hollow section =
 (362 - 202) = 703.72 mm2
4
Cross section of solid section =  x 252 = 490.87 mm2
4
Stress = Load
Area
Hence stress in the hollow section = 30 x 103
703.72 x 10-6
and stress in the solid section =
30 x 103
490.87 x 10-6
= 42.63 MN m-2
= 61.12 MN m-2
b) The maximum stress induced in the tie must be less than the yield stress of
the material chosen. The choice is between brass and aluminium, but the
yield stress of brass is rather near to the maximum stress in the tie, so on this
basis Aluminium would be the best material to use.
c) From E = Stress
Strain
hence Strain = Stress
E
If Aluminium is used
Strain in the hollow section = 42.63 x 106 = 0.609 x 10-3
70 x 109
Strain = change in length
original length
hence change in length = Strain x o/length
change in length in the hollow section = 0.609 x 10-3 x 200 = 0.122 mm
Strain in the solid section = 61.12 x 106 = 0.873 x 10-3
70 x 109
change in length of the solid section = 0.873 x 10-3 x 600 = 0.524 mm
Hence the overall extension of the aluminium tie
= 0.122 + 0.524 = 0.646 mm
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
53
Resource materials
Presenting centres who in the past have delivered modules 74012 – Statics:
Components and Structures (1/2) and 64067 Strength of Materials or any other similar
modules, will have materials already available although some reworking may be
necessary. The examples suggested for the tutorials and assignments in this pack
indicate the standard of complexity required and care should be taken to keep the
level similar when extending the range, type or quantity of materials to be used by
candidates.
DET: Structures: Statics: Equilibrium in Frameworks and Simple Stress (Intermediate 2)
54