NATIONAL QUALIFICATIONS CURRICULUM SUPPORT
Civil Engineering
Structural Analysis and Design
[ADVANCED HIGHER]
James Dunbar
THERMOCHEMISTRY
ECONOMI CS: MI CROECON
OMI CS ( AH)
III
Acknowledgements
Learning and Teaching Scotland gratefully acknowledge t
his contribution to the National
Qualifications support programme for Civil Engineering. In particular, the assistance of Bill
McKenzie, Mike Scully and Charlie Smith in the preparation of this material is
acknowledged with thanks.
Electronic version 200
2
© Learning and Teaching Scotland 2002
This publication may be reproduced in whole or in part for educational purposes by
educational establishments in Scotland provided that no profit accrues at any stage.
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
i i i
CONTENTS
Overview
1
Tutor Guide
3
S
tudent Guide
7
Study Guide 1:
Analysis of statically determinate pin

jointed
frames
11
Study Guide 2:
Determination of beam deflections by standard formulae
and by Macaulay’s method
21
Study Guide 3:
Design of reinforced concrete elements
41
Study G
uide 4:
Design of structural steelwork elements
75
Study Guide 5:
Design of masonry and timber elements
99
THERMOCHEMISTRY
ECONOMI CS: MI CROECON
OMI CS ( AH)
V
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
1
OVERVIEW
These support materials are provided to assist teachers/lecturers in delivery of
the Advanced Higher Civil Engineering course unit
Structural Analysis and
Design. They will also help to prepare students for assessment.
The Tutor Guide offers brief advice on the entry requirements for the unit, on
the design documents to be issued to candidates with each of the Study
Guides and the
design procedures to be adopted.
The Student Guide provides a brief introduction to the unit, explains the
content of each Study Guide and offers advice on preparation for assessment.
Student support materials are provided in the form of five Study Guide
s, each
covering one or two outcomes of the unit.
The National Assessment Bank support material for this unit contains five
assessment instruments that take the form of ‘end of topic’ tests. These may
be used to provide feedback on candidates’ progress a
s well as being used for
summative unit assessment.
The Study Guides in this pack provide the support notes required for the
outcomes covered by each instrument of assessment. ‘End of Study Guide’
tests are also provided, and these are of a similar sta
ndard to the instruments
of assessment of the National Assessment Bank.
The Study Guides are as follows:
Study Guide 1: Analysis of statically determinate pin

jointed frames
This covers all the performance criteria of Outcome 1.
Outcome 1: Analyse, by
mathematical means, statically determinate
pin

jointed frames.
OVERVIEW
2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Study Guide 2: Determination of beam deflections by standard formulae
and Macaulay’s method
This covers all the performance criteria of Outcome 2.
Outcome 2: Determine the deflections of
statically determinate beams
using standard formulae and Macaulay’s method.
It is recommended that Study Guide 1 be used before Study Guide 2, as the
meaning of the term
statically determinate
is considered in Study Guide 1.
It is also recommended that
Study Guide 2 be used before Study Guides 3, 4
and 5, since knowledge of deflection calculation is required for reinforced
concrete, steelwork and timber design.
Study Guide 3: Design of reinforced concrete elements
This covers all the performance criter
ia of Outcomes 3 and 4.
Outcome 3: Design statically determinate singly reinforced beams and
slabs in reinforced concrete.
Outcome 4: Design short, braced, axially loaded columns in reinforced
concrete.
Study Guide 4: Design of structural steelwork elem
ents
This covers all the performance criteria of Outcomes 5 and 6.
Outcome 5: Design statically determinate structural steel beams.
Outcome 6: Design axially loaded single

storey steel stanchions.
Study Guide 5: Design of masonry and timber elements
This covers all the performance criteria of Outcomes 7 and 8.
Outcome 7: Design vertically loaded single

leaf and cavity walls in
structural masonry.
Outcome 8: Design flooring, simply supported floor joists and axially
loaded columns in structural timbe
r.
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
i i i
TUTOR GUIDE
The Study Guides cover all the performance criteria of each outcome. The
‘End of Study Guide’ tests are extensive and of a standard equivalent to that
of the assessment instruments of the National Assessment Bank. However,
centres migh
t need to develop additional formative assessment material.
General note
It is expected that candidates have previously undertaken the component units
of the Civil Engineering Higher course and are fully conversant with:
•
the conditions of static equil
ibrium
•
mathematical integration techniques
•
the calculation of loads on structural elements
•
the load paths through structural frames
•
the concept of design loads, partial load factors and material safety factors
•
the construction methods for reinfor
ced concrete and masonry elements
•
the fabrication and erection methods for structural steelwork
•
the nature of timber as a building material.
These are not covered to any depth in the Study Guides.
If students are not fully conversant with the proced
ure for determining design
loads, from characteristic (unfactored) loads and partial safety factors,
teachers/lecturers will need to spend some teaching time on this and provide a
number of worked examples.
Study Guide 1: Analysis of statically determinat
e pin

jointed frames
It is recommended that Study Guide 1 be used at the start of the course as it
provides knowledge of statically determinate structures, which is required as
a general concept for all outcomes. This seems to be a difficult concept for
students to grasp and it is expected that individual centres will develop
additional formative assessments.
TUTOR GUIDE
4
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Study Guide 2: Determination of beam deflections by standard formulae
and Macaulay’s method
Prior to issue of the Study Guide, integration tech
niques should be revised.
Candidates should also be issued with a data sheet listing the standard case
deflection formulae for the following cases:
•
a simply supported beam with a uniformly distributed load over the entire
length
•
a simply supported be
am with a concentrated load at mid

span
•
a cantilever beam with a uniformly distributed load over the entire length
•
a cantilever beam with a concentrated load at the end.
In the Study Guide, ‘w’ is used to refer to a uniformly distributed load and
‘W’
to refer to a concentrated load.
Design procedures (Study Guides 3
–
5)
The notes for these guides were developed using PP 7312: 1998 ‘Extracts
from British Standards for students of structural design’ as the design
reference. The use of any other pub
lication may lead to answers that differ to
those given in the examples. Study Guide 2 should be undertaken before the
design Study Guides, as the standard case deflection formulae are widely
used in these design guides.
Study Guide 3: Design of reinfor
ced concrete elements
Each centre should provide candidates with ‘Tables of areas of
reinforcement’ when issuing the Study Guide. The design methods are based
on BS8110 Part 1: 1997 and the notes concentrate on the design equations
rather than the desig
n charts. At the time of publication of the Study Guide
the design charts in PP 7312 were extracted from BS 8110 Part 3: 1985. As
the charts were developed using a materials factor for steel g
m
of 1.15 and
not 1.05 as used in the 1997 version of the code
, there is now an inherent
error in the charts. Areas of reinforcement derived using the charts must
therefore be multiplied by the factor 1.05/1.15, as illustrated on page 51.
TUTOR GUIDE
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
5
Study Guide 4: Design of structural steelwork elements
Each centre should p
rovide candidates with the following documents when
issuing the Study Guide:
•
the Structural Steel section tables for UB and UC sections
•
the safe load tables for UC and UB subject to axial load
•
the safe load tables for web bearing and buckling of UB
sections.
Copies of the most up

to

date tables can be obtained from the Corus Group’s
web site www.corusconstruction.com
Any differences in dimensions or properties of UC or UB sections may be as
a result of different versions of the structural steel s
ection tables being used.
The design methods are based on BS 5950 Part 1: 1990. The use of any other
version of the code may lead to variations in answers to the examples.
Candidates are expected to have prior knowledge of fabrication and erection
metho
ds for simply supported beams and columns and of the definition of
length of a member.
Study Guide 5: Design of masonry and timber elements
The design procedures for masonry and timber are based on BS 5628 Part 1:
1992 and BS 5628 Part 2: 1996 respective
ly. The use of any other versions of
the code may lead to variations in design procedures.
The issue of brick manufacturers’ data sheets may enhance the candidate’s
understanding of the design process for masonry.
6
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
7
STUDENT GUIDE
Introduction
The uni
t Structural Analysis and Design will appeal to you if you are
interested in problem solving. It will broaden your skills in the application of
scientific and technological principles to the area of structural design.
Gaining this award will enable you t
o continue development of the
competences required of the Incorporated Engineer. It will provide a strong
base for further study at HND and Degree level. You will achieve a level of
competence required of a person in a design office who has the responsib
ility
for the design of basic structural elements.
Unit content
The unit stresses the importance of structural engineering in the creative and
safe development of the built environment. It is designed to bring together
the study of structural mechanics,
previously studied and now further
developed, with the processes of structural design. It will introduce you to
the British Standard Codes of Practice used in the design of reinforced
concrete, steelwork, masonry and timber structures
–
all problem

solvi
ng
activities.
The unit has eight outcomes and will be assessed by five ‘end of topic’ tests.
The teaching and learning materials have been prepared as five Study Guides,
which provide the support notes for the outcomes covered by each instrument
of asse
ssment. At the end of each Study Guide you will find an ‘End of
Study Guide’ test that contains questions that are of a standard similar to that
which you can expect in the assessment.
Study Guide 1: Analysis of statically determinate pin

jointed frames
This covers Outcome 1. It will introduce you to the analytical methods used
to determine the forces in pin

jointed frames
STUDENT GUIDE
8
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Study Guide 2: Determination of beam deflections by standard formulae
and Macaulay’s method
This covers Outcome 2. It will enab
le you to determine the deflections of
beams under standard and non

standard loading. Deflection formulae have
been developed for standard loading, which may be used to determine the
maximum deflections of beams. If non

standard load conditions are appli
ed,
Macaulay’s method may be used.
Study Guide 3: Design of reinforced concrete elements
This covers Outcomes 3 and 4. It will introduce you to the design of
reinforced concrete elements: beams; slabs; and columns. You will learn how
to use the design
procedures of BS 8110 to determine the area of tension
reinforcement in beams and slabs, the area of shear reinforcement required in
beams, the area of longitudinal and link steel in axially loaded columns and
how to prepare suitable arrangements of reinfo
rcement.
Study Guide 4: Design of structural steelwork elements
This covers Outcomes 5 and 6. You will learn how to design structural
steelwork elements to BS 5950 Part 1. Simply supported fully restrained
steel beams, and axially loaded columns are co
vered by the Study Guide. In
addition to learning how to use the design code you will learn to use the
structural section tables and safe load tables for UB and UC sections.
Study Guide 5: Design of masonry and timber elements
This covers Outcomes 7 and
8. Two materials will be considered in this
guide: timber and masonry. The design procedures for masonry walls are to
BS 5628 Part 1 and those for timber are to BS 5268 Part 2. In the timber
design section, flooring elements such as boarding, joists, t
rimmer beams and
axially loaded columns will be studied.
Assessment
The assessment of the unit takes the form of five ‘end of topic’ tests, all of
which are closed book. You will not be allowed to use the Study Guides.
However, you will have access to
standard case deflection formulae, relevant
clauses from the design standards and published tables such as Structural
Section tables or areas of reinforcement tables, as applicable. Use the
opportunity during classroom time to develop your skills in the u
se of British
Standards. All the information is there if you know where to look for it!
STUDENT GUIDE
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
9
Part of the classroom time will be allocated to assessment. However, you
will have to spend additional time in preparing yourself for assessment.
Learn how to use
the design codes: what clauses (or page numbers) do you
have to look up for (say) bending moments applied to beams; what tables are
applicable; do the values from the tables have to be modified in some way?
The assessment will be carried out under the su
pervision of an invigilator
(normally your teacher/lecturer), under strict time constraints. These will be
outlined to you prior to undertaking the assessment. You must learn to use
the design codes quickly. Use the ‘End of Study Guide’ tests as a guide
to
your preparedness for final assessment.
Core skills
The assessment tasks of the unit will also be tailored to allow you to develop
a number of core skills, including problem solving. Completion of the unit
may result in automatic certification of c
ertain core skills components.
Successful completion of the Advanced Higher Course in Civil Engineering
will result in automatic certification of other components. You should be
aware of the evidence you must gather to demonstrate attainment of core
skil
ls and your tutor will guide you in this area.
10
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
STUDY GUIDE 1
11
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
STUDY GUIDE 1
Analysis of statically determinate pin

jointed frames
Introduction
This study guide covers Outcome 1 of the unit.
Outcome 1
Analyse, by mathematical means, statically determinate pin

j
ointed
frames.
On completion of the Study Guide you should be able to:
•
distinguish between
statically determinate
and
statically indeterminate
frames
•
calculate the magnitude and nature of forces in pin

joined frames using the
method of joint resolu
tion
•
calculate the magnitude and nature of forces in pin

joined frames using the
method of sections.
STUDY GUIDE 1
12
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
What does the term ‘statically determinate pin

jointed frames’ mean?
What will be considered will be the analysis of trusses where the external
load
s are applied at the node points only (intersection of the individual
elements of the frame), such that no bending effects can be developed in the
members. As only axial compressive and tensile forces are developed in the
frame members the frame is referre
d to as ‘pin

jointed’
–
at a pin only direct
forces can be carried and no bending effects can be developed.
‘Statically determinate’
–
the frame can be solved using the three conditions
of equilibrium only.
The conditions are:
Algebraic sum of moments of
forces must equal zero
M = 0
Algebraic sum of vertical of forces must equal zero
V = 0
Algebraic sum of horizontal of forces must equal zero
H = 0
When considering the frame and its reactions there are three conditions of
equilibrium to solve the reactions, thus there can be
no more than three
unknowns.
In the frame shown:
The support at the left

hand side is a hinge (or pin) which can have both
horizontal and vertical components of force and the support at the right

hand
side is a roller which can have only a verti
cal component of force. There are
three unknowns and there are three conditions of equilibrium with which to
solve them
–
the frame reactions are ‘statically determinate’.
If the frame is provided with two hinges as supports, as shown below,
STUDY GUIDE 1
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
13
ther
e are four unknowns and only three conditions of equilibrium with which
to solve them
–
the frame reactions are ‘statically indeterminate’ and cannot
be solved by using the conditions of equilibrium only.
In a similar manner the elements of the frame mus
t conform to the equation
shown below if the frame is statically determinate:
n =(2j
–
3)
Where n = number of members
j = number of nodes
For the above frame:
n = 9
j = 6
2j
–
3 = 2
6
–
3 = 9
frame is statically determinate
Consider this frame:
n = 11
j = 6
n > (2j
–
3) = 2
6
–
3 = 9
frame is statically indeterminate
to the second degree, since 11
–
9 = 2
At the start of each example ensure the frame (and its reacti
ons) are statically
determinate.
STUDY GUIDE 1
14
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Method of joint resolution
The method for analysis of the forces in frames by joint resolution is best
explained by a worked example and the application of a few simple rules.
Determine the forces in each member for th
e frame shown below.
Step 1: letter each of the nodes (step illustrated on frame)
Step 2: consider the frame as a whole and determine the magnitude and
direction of the forces at the reactions
(a)
Take moments about the hinge and determine roll
er reaction
Take moments about A,
M = 0, clockwise moments positive
(12
3) + (48
3)
–
V
C
6 = 0
V
C
= (36 + 144)/6 = 30 kN
(b)
Apply
V = 0 and
H = 0 to find the magnitude and direction of the
hinge reactions
V = 0
V
A
+ VC
–
48 = 0
upwards
positive
V
A
= 48
–
30 = 18 kN
H = 0
12
–
H
A
= 0
forces to right positive
H
A
= 12 kN
Step 3: select a node with only two unknowns
Note: As no bending effects are present in the frame elements, the condition
of equilibrium
M = 0 cannot be app
lied. As there are only two equilibrium
equations remaining in order to solve them there can be no more than two
unknown forces at any node.
STUDY GUIDE 1
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
15
Redrawing frame
Only nodes A and C are suitable. B has five unknown forces. D, E and F all
have three
unknown forces.
Using only
V = 0 and
H = 0, firstly node A then node C
Node A
V = 0
As the reaction is 18 kN upwards, a balancing force of 18 kN downwards is
required.
This can only occur in a vertical element, thus force AF is 18 kN
H = 0
As the reaction is 12 kN to
the left, a balancing force of 18 kN to the right is
required.
This can only occur in a horizontal element, thus force AB is 12 kN
Node C
V = 0
As the reaction is 30 kN upwards, a balancing force of 30 kN downwards is
required.
This can only occur in
a vertical element, thus force CD is 18 kN
H = 0
As there is only one horizontal element at node C and no external horizontal
forces, the force in the single element must be 0. Force CB = 0.
Step 4: superimpose the known forces on the frame
Remember
the algebraic sum of forces in an element must balance. For
example if the force at one end (node) of an element is 18 kN downwards, for
equilibrium at the other end (node) it must be 18 kN upwards.
STUDY GUIDE 1
16
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Redrawing the frame
Step 5: repeat steps 3 and
4 with the remaining nodes of the frame
There are now only two unknowns at nodes F and D; node E still has three
unknowns.
Considering node F then node D
The inclined forces FB and DB can be split onto horizontal and vertical
components of force,
either by knowing the ratio of the sides or by knowing
the values of the angles.
Node F
V = 0
As the force from member AF is 18 kN upwards, a balancing force of 18 kN
downwards is required.
This can only occur in the vertical component of element
FB, thus the vertical
component of FB is 18 kN. However, FB is an inclined member so the actual
direction of the force along the length of the member must be down and to the
right. The magnitude is
18/cos 45 = 25.45 kN or
18
2 = 25.45 kN
H = 0
At
this node there are three horizontal forces FE, the horizontal
component of FB and the external 12 kN force.
If the force in FB is acting down and to the right, the horizontal
STUDY GUIDE 1
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
17
component is acting to the right. FBh = 25.45 sin45 or 25.45/
2 = 18 kN
FE +
FBh + 12 = 0
forces to right positive
FE +18 +12 = 0
FE =
–
30 kN
30 kN acting to the left
Node D
As member DB is inclined, it can split into its horizontal and vertical
components
DBv = DB sin 45 or DB/
2
DBh = DB cos 45 or DB/
2
Considering the node
, there is a vertical force of 30 kN acting upwards in
element DC.
This must be balanced by a downwards force of 30 KN. This can only occur
in DBv.
DBv = 30 kN
The force in DB must therefore be acting down
and to the left.
DBv = DB sin 45 or DB/
2
DB = 42.4 kN
H = 0
DE is unknown, but must balance DBh as there are no other horizontal
elements at this node. DBh is acting to the left
–
DE must act to the right.
DE = DBh = 42.4 cos 45 or 42.4/
2 = 30 kN
Repeat Step 4: superimpose the known fo
rces on the frame
Now consider node E
The vertical force EA is the only unknown
This must balance the external 48 kN force
EA = 48 kN
Finished frame
STUDY GUIDE 1
18
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Method of sections
The method for analysis of the forces in frames by sections is used when on
ly
the forces in specific elements are required. The three conditions of
equilibrium are available for use, so the section should cut across no more
than three elements in which the forces are unknown.
Consider the frame used in the joint resolution examp
le.
The first step (as before) is to calculate the reactions, giving the result:
The section considered to cut the frame shows that the forces in ED, BD and
BC are to be found.
The external equilibrium of the part of the frame to the left

han
d side of the
section is considered. For each condition of equilibrium equation used there
can be only one unknown. Splitting BD into its horizontal and vertical
components, BD
h
and BD
v
respectively:
H = 0 becomes
Taking forces acting to the right as po
sitive
12
–
12 + BD
h
+ ED + BC = 0
three unknowns
V = 0 becomes
Taking upwards forces as positive
18
–
48 + BD
v
= 0
only one unknown
M = 0
is dependent on where moments are taken.
As a general rule if the section cuts across three
elements, two of th
em will intersect at a node. Take
moments about this node leaving one unknown. Node
D in this example:
STUDY GUIDE 1
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
19
Take moments about D
Clockwise moments positive
(18
6) + (12
3)
–
(48
3) +(BC
3) = 0
108 + 48
–
144 +3.BC = 0
0 + 3BC = 0
BC = 0
V =
0 becomes
Taking upwards forces as positive
18
–
48 + BDv = 0
–
30 + BDv = 0
BDv = 30 kN
towards node D
Force in BD acts along the line of the element, the direction is up and to the
right.
Magnitude of force BD = 30 /cos 45 or 30
2 = 42.4 kN
H =
0
Taking forces acting to the right as positive
12
–
12 + BDh + ED + BC = 0
12
–
12 + 42.4sin 45 +ED = 0
30 + ED = 0
ED =
–
30 kN
From D the force acts to the left.
Superimpose the results on the frame.
Remember the algebraic sum of forces in an elemen
t must equal zero.
Answers agree with those found by method of joint resolution.
s
s
STUDY GUIDE 1
20
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Study Guide 1 End Test
The figure below shows in outline a pin

jointed frame and the loads applied
to it.
(a)
Show that the frame is statically determinate.
(b
)
Calculate the support reactions.
(c)
Using the
method of joint resolution
, determine the magnitude and
nature of the force in each element of the frame. Show the results in an
outline sketch of the frame.
(d)
Using the
method of sections
, check the v
alidity of the results found
using the method of joint resolution, by determining the forces in
elements GF, CF and CD.
Answers:
Roller reaction:
82.5 kN
Hinge reactions:
horizontal
12 kN
vertical
73.5 kN
all in directions shown in diagram
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
21
STUDY GUIDE 2
Determination of beam deflections by standard formulae
and by Macaulay’s method
Introduction
This study guide covers Outcome 2 of the unit.
Outcome 2
Determine the deflections of statically determinate beams using
standard formulae
and Macaulay’s method.
On completion of the Study Guide you should be able to:
•
calculate the maximum deflections of statically determinate beams using
standard formulae
•
calculate critical deflections in beams subject to non

standard loading
using
Macaulay’s method.
STUDY GUIDE 2
22
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Macaulay’s method
This is a method of analysis that allows the slope and deflection of a beam to
be determined.
From the equation of simple bending
M E
=
I R
where:
M
=
bending moment at a section
E
=
modulus of e
lasticity
I
=
second moment of area of the section
R
=
radius of curvature
When both E and I are constant for a given section, M and R are the only
variables.
The expression for M is then
EI
M =
R
If the deflection of the member is
y
, and as deflection is a function of the
radius of curvature R, then:
2
2
d y I
dx R
(R is the second derivative of deflection)
then:
2
2
d y
M = EI
dx
where
x
= distance along the length of the beam to position of bending
moment
M.
To obtain deflection:
2
2
d y M
=
dx EI
Bending moment expression
dy M
= + A
dx EI
Slope expression
M
y = + Ax + B
EI
Deflection expression
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
23
Where A and B are constants of integration. They are determined by
conside
ring the boundary conditions relating to the beam (i.e. the known
values of slope and deflection).
It is therefore possible to find the slope and deflection at any point along a
beam by providing a general expression for bending moment at any section in
t
erms of
x
and integrating the equation twice.
The procedure for determining the bending moment expression is as follows:
1.
Assume one end of the beam to be the origin (generally the left

hand
side).
If the beam is statically determinate find the value
of the reactions.
2.
Consider a section
x
–
x
as far from the origin as possible (beyond the
last applied load) and take moments about
x
–
x
considering all loads to
the left

hand side of the section. All the bending moment terms will be
functions of
x
.
3.
Integrate the bending expression with respect to
x
.
Integrate each loading term as a whole
–
don’t break it down into its
components.
4.
Determine the constants of integration A and B for slope and deflection
using the boundary conditions relating to th
e beam, for example:
•
Deflections at supports are assumed zero unless otherwise stated.
•
Slopes at built

in supports are zero.
•
Slope at the centre of a symmetrically loaded beam is zero,
deflection is a maximum.
•
When deflection is a maximum, slo
pe is zero.
•
Bending moments at free ends are zero.
5.
Substitute values of
x
to determine the slope and deflection at any
section along the beam
Note:
When determining quantities, omit any terms inside brackets that are
negative or zero.
STUDY GUIDE 2
24
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Example 1:
Point loads
A beam is simply supported as shown. For the illustrated loading system,
determine:
(a)
the slope and deflection under the 200 kN load
(b)
the magnitude and position of the maximum deflection.
E = 205 kN/mm
2
I = 900
10
6
mm
4
Fi
nd the value of the reactions
Take moments about R
a
M = 0, clockwise moments are positive
(200
2 ) + (350
5 )
–
(R
b
7 ) = 0
R
b
= 307.1 kN
V = 0, upwards forces are positive
R
a
= 200 + 350
–
307.1 = 242.9 kN
Apply Macaulay’s method at a
section x
–
x beyond the last applied load
Take moments about x
–
x
Distance to load from
Moment = Force x distance
section x
–
x
(m)
R
a
x
242.9 x
200 kN
x
–
2
200[x
–
2]
350 kN
x
–
5
350[x
–
5]
X
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
25
The total moment at
x
may be written as:
2
x
2
d y
M = EI = 242.9[x] – 200 [x–2] – 350 [x–5
]
dx
Integrate with respect to ‘x’
2
2
d y
EI = 242.9[x] – 200 [x–2] – 350 [x–5]
dx
moment (kNm)
2 2 2
dy [x] [x–2] [x–5]
EI = 242.9 – 200 – 350 + A
dx 2 2 2
slope equation
(kNm x m = kNm
2
)
3 3 3
[x] [x–2] [x–5]
EI y = 242.9 – 200 – 350 + Ax + B
6 6 6
deflection equation
(kNm
2
x m = kNm
3
)
Now deflections are zero at the sup
ports thus:
When x = 0, y =0 and x = 7, y = 0
Substituting in the deflection eqn. for x = 0, y = 0
3 3 3
[0] [0–2] [0–5]
EI 0 = 242.9 – 200 – 350 + A0 + B
6 6 6
Note:
When determining quantities, omit any term inside a bracket that is
negative or zero.
Thus:
0 =
(0)
–
(ignore)
–
(ignore) + (0) + B
Thus constant of integration B = 0
Substituting in the deflection eqn. for x = 7, y = 0
3 3 3
7 7–5 7–5
EI 0 = 242.9 – 200 – 350 +A [7]
6 6 6
0 = 13886
–
4667
–
467 + 7A
A =
–
1250
STUDY GUIDE 2
26
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Thus genera
l equations for slope and deflection at any point along the length
of the beam are
2 2 2
dy [x] [x–2] [x–5]
EI = 242.9 – 200 – 350 – 1250
dx 2 2 2
slope
3 3 3
[x] [x–2] [x–5]
EI y = 242.9 – 200 – 350 – 1250[x]
6 6 6
deflection
To find the slope and deflection at the 200 kN load substitute for x=2
2 2 2
dy [2] [2–2] [2–5]
EI = 242.9 – 200 – 350 – 1250
dx 2 2 2
=
486
–
(0)
–
(ignore)
–
1250
kNm
2
dy –764
=
dx EI
1 kNm
2
= 10
9
Nmm
1 kN/mm
2
= 10
3
N/mm
2
Units are now consistent
=
9
6
3
764 10
900 10
205 10
=
–
0.0041 radians
3 3 3
[2] [2–2] [2–5]
EI y = 242.9 – 200 – 350 – 1250[2]
6 6 6
= 324
–
0
–
0
–
2500
kNm
3
–2176
y =
EI
=
9
6
3
–2176 10
900 10
205 10
2 4
Nmm
= mm
N/mm mm
=
–
11.8 mm (11.8 mm downwards)
To determine the position of the maximum deflection equate slope equation
to zero.
2 2 2
dy [x] [x–2] [x–5]
EI = 242.9 – 200 – 350 – 1250
dx 2 2 2
0 = 121.5[x]
2
–
100 [x
2
–
4.x + 4]
–
175[x
2
–
10.x + 25]
–
1250
0 =
–
153.5x
2
+ 2150x
–
6025
ax
2
+ bx +c
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
27
Solving for x, x =
2
–2150 (2150 – 4 153.3 6025)
2 153.5
x = (
–
b
b
2
–
4ac)/2a
x = 10.3 m or 3.88 m.
for x = 3.88 m
defl ecti on y =
–
2706/EI =
–
16.7 mm
Exampl e 2: Uni forml y di st ri but ed l oads
A si mpl y support ed beam i s Lm l ong and i s requi red t o carry a uni forml y
di st ri but ed l oad of w kN/m. In general t erms, det ermi
ne t he maxi mum
defl ecti on of t he beam:
Fi nd val ue of react i ons:
As beam i s symmet ri cal ly l oaded, R
a
= R
b
= w
L/2 = wL/2
Appl y Macaul ay’s met hod at a secti on x
–
x beyond t he l ast appl i ed l oad
In t hi s exampl e consi der t he secti on x
–
x i mmedi at el y
t o t he l eft of react i on R
b
Take moment s about x
–
x
Val ue of l oad
Di st ance t o cent re of
Moment = Force x di st ance
l oad from sect i on x
–
x (m)
R
a
wL/2
x
wL.x/2
w kN/m
w.x
x/2
w.x.x/2
STUDY GUIDE 2
28
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
This may be written as:
2 2
x
2
wLx wx d y
M = – = EI
2 2 dx
Integrate
with respect to ‘x’
2 3
3 4
dy wLx wx
EI = – + A slope equation
dx 4 6
wLx wx
EI y = – + Ax + B deflection equation
12 24
In order to find the constants of integration A and B apply the boundary
conditions.
For a simply supported beam with symmetrical loading:
Deflection at supports is zero.
Deflection at mid

span is a
maximum and slope is zero.
Applying the deflection equation at left

hand support,when x = 0 y = 0
3 4
wL[0] w[0]
EI 0 = – + A[0] + B
12 24
hence B = 0
Applying the deflection equation again at right

hand support, when
x = L y = 0
3 4
4 3
wL L wL
EI 0 = – + AL
12 24
wL –wL
EI 0 = + AL A = note negative sign
24 24
Th
us equations become:
2 3 3
3 4 3
dy wLx wx 3wL
EI = – – slope equation
dx 4 6 24
wLx wx wL x
EI y = – – deflection equation
12 24 24
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
29
To determine maximum deflection, substitute in deflection equation for x =
L/2 as the beam is symmetrically loaded.
3 4 3
3 4 3
4 4 4
4 4 4
4
wL(L/2) w(L/2) wL (L/2)
EI y = – –
12 24 24
wL.L wL wL.L
EI y = – –
12 24 48
wL wL wL
EI y = – –
96 384 48
4wL wL 8wL
EI y = – –
384 384 384
–5wL
y =
384EI
8 16
//
Example 3: Cantilever
A cantilever beam is 2m long and is
required to carry a uniformly distributed
load of 20 kN/m and a point load of 64 kN at the tip.
(a)
Using Macaulay’s method, determine the maximum deflection of the
beam in terms of EI.
(b)
Check the answer obtained in (a) by applying the standard equat
ions for
deflection
Additional information
Standard deflection formulae for cantilevers:
Uniformly distributed load
= wL
4
/8EI
Point load at tip
= WL
3
/3EI
STUDY GUIDE 2
30
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Find value of reactions:
Reaction must balance downwards forces, as
V = 0
R
a
= 20
2 + 64 = 104 kN
Taking moments about R
a
,
M = 0, clockwise moments are positive
–
M
a
+ (20
2
1) + 64
2 = 0
M
a
= 168 k
Nm (anticlockwise)
Apply Macaulay’s method, at a section x
–
x as far along the beam as possible.
Take moments about x
–
x
Value of load
Distance to centre of
Moment = Force
distance
load from section x
–
x (m)
104
x
104x
20.x
x/2
20.x.x/2
Consid
ering also the moment at the support, this may be written as:
2 2
x
2
d y 20x
M = EI = – 168 + 104x –
dx 2
Integrate with respect to ‘x’
2
3 4
dy 104x 20x
EI = – 168x + – + A slope equation
dx 2 6
–168x 104x 20x
EI y = + – + Ax + B deflection equatio
n
2 6 24
In order to find the constants of integration A and B apply the boundary
conditions.
For a cantileve
r beam, deflection at the support is zero, and slope is zero at a
built

in support.
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
31
Applying the deflection equation at support, when x = 0 y = 0
2 3 4
–168[0] 104[0] 20[0]
EI 0 = + – + A[0] + B
2 6 24
Therefore B = 0
Applying the slope equation at the built

in support, when x=
0,
dy
= 0
dx
3 4
2
104.[0] 20.[0]
EI 0 = –168.[0] + – + A
2 6
Therefore A = 0
Equations become:
2
3 4
dy 104x 20x
EI = –168x + – slope equation
dx 2 6
–168x 104x 20x
EI y = + – deflection equation
2 6 24
Maximum deflection occurs at the tip of the cantilever, x = 2 m
2 3 4
–168.2 104.2 20.2
EI y = + –
2 6 24
EI y =
–
336 + 138.67
–
13.33
EI y
=
–
210.67
–210.67
y =
EI
Negative sign indicates that deflection is downwards.
STUDY GUIDE 2
32
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
(b)
Compare with standard equations:
Standard deflection formulae for cantilevers:
Uniformly distributed load
= wL
4
/8EI
Point load at tip
= WL
3
/3EI
4
3
20.2 40
For udl = =
8EI EI
64.2 170.67
for a point load = =
3EI EI
210.67
total deflection =
EI
Answers are exactly the same.
Note:
Standard equations assume deflection is downwards and negative sign
is omitted.
Example 4: Simply supported beam with an overhang
Note on deal
ing with variation of uniformly distributed load between spans.
Consider a simply supported beam with an overhang. Three conditions of
uniformly distributed load will be examined and the general expression for
moment derived.
1
.
Constant udl along the
length of the beam
Considering a section x
–
x to towards the end of the beam
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
33
Take moments about x
–
x
M
x
= R
a
.x
–
w.x.x/2
–
R
b
.[x
–
a] = R
a
.x
–
w.x
2
/2
–
R
b
.[x
–
a]
2.
udl on the main span and a larger udl on the overhang
For analysis
purposes this is treated as a constant udl over the entire
beam and an additional load of (z
–
w kN/m) on the overhang. As in (1),
moments are taken about the section x
–
x.
Take moments about x
–
x
(same as condition 1) (additional term)
M
x
= R
a
.x
–
w.x.x/2
–
R
b
.[x
–
a]
–
(z
–
w).[x
–
a].[x
–
a]/2
= R
a
.x
–
w.x
2
/2
–
R
b
.[x
–
a]
–
(z
–
w).[x
–
a]
2
/2
3.
udl on the main span and a smaller udl on the overhang
For analysis purposes this is treated as a constant udl over the entire
beam
les
s
an additional load of (w
–
z kN/m) on the overhang. Load w
–
z
acts
upwards
and gives a positive moment about section x
–
x. As in the
other cases moments are taken about the section x
–
x.
STUDY GUIDE 2
34
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Take moments about x
–
x
(same as condition 1)
(additional t
erm)
M
x
= R
a
.x
–
w.x.x/2
–
R
b
.[x
–
a] + (z
–
w).[x
–
a].[x

a]/2
= R
a
.x
–
w.x
2
/2
–
R
b
.[x
–
a] + (z
–
w).[x
–
a]
2
/2
Example beam with overhang
For the beam loaded as shown below:
(a)
Using Macaulay’s method, in terms of E and I, derive the equations for
slope and deflection along the length of the beam.
(b)
Determine the deflection at the centre of the main span and at the tip of
the cantilever.
Additional information
Beam section
533
210
92 UB
I = 55330 cm
4
E= 205 kN/mm
2
Find value of re
actions:
Taking moments about R
a
,
M = 0, clockwise moments positive
(30
8
4)
–
R
b
8 + (10
2
9) = 0
R
b
= 142.5 kN
Reactions must balance downwards forces
V =0,
R
a
+ R
b
= 30
8 +10
2
R
a
= 260
–
142.5 = 117.5 kN
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
35
Take moments about x
–
x, co
nsidering a uniform load of 30kN/m over the
entire length of beam and a negative (upward) load of 20kN/m on the
overhang.
Value of load
Distance to centre of
Moment = Force x distance
load from section x
–
x (m)
117.5 kN
x
117.5x
30 kN/m.x
x/2
30
.x.x/2
142.5 kN
x
–
8
142.5(x
–
8)
20 kN/m.(x
–
8)
(x
–
8)/2
20.(x
–
8)
2
/2
This may be written as:
2 2 2
x
2
dy 30x 20[x–8]
M = EI = 117.5x – + 142.5[x–8] +
dx 2 2
Integrate with respect to ‘x’
2 3 2 3
3 4 3 4
dy 117.5x 30x 142.5[x–8] 20[x–8]
EI = – + + + A slope equation
dx 2 6 2 6
117.5x 30x 142.5[x–8] 20[x–8]
EI y = – + + + Ax + B deflection equa
tion
6 24 6 24
In order to find the constants of integration A and B apply the standard
con
ditions.
For a simply supported beam deflection is zero at the supports.
When x = 0 , y = 0
–
apply to deflection equation
3 4 3 4
117.5.0 30.0 142.5[0–8] 20[0–8]
EI 0 = – + + + Ax + B
6 24 6 24
0
= (0)
–
(0)
+ (ignore term) + (ignore term) +
(0) + B
B = 0
STUDY GUIDE 2
36
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Also when x = 8, y = 0
–
apply to
deflection equation
3 4 3 4
117.5.8 30.8 142.5[8–8] 20[8–8]
EI 0 = – + + + Ax
6 24 6 24
0
=
10027
–
5120
+
(0)
+
(0)
+ A.8
A =
–
(10027
–
5120)/8 =
–
613.4
Standard equations become:
2 3 2 3
3 4 3 4
dy 117.5x 30x 142.5[x–8] 20[x–8]
EI = – + + – 613.4 slope equation
dx 2 6 2 6
117.5x 30x 142.5[x–8] 20[x–8]
EI y = – + + – 613.4x deflection equa
tion
6 24 6 24
Actual deflections
At centre of main span, x = 4m
3 4 3 4
117.5.4 30.4 142.5[4–8] 20[4–8]
EI y = – + + – 613.4.4
6 24 6 24
EI y = 1253.3
–
320 + (ignore) + (ignore)
–
2453.6
In terms of EI, y =
–
1520.3/EI
Actual deflection,
y =
–
1520.3
10
12
/(205
10
3
55330
10
4
)
y =
–
13.4 mm (downwards deflection)
Deflection at tip of beam, x = 10m
3 4 3 4
117.5.10 30.10 142.5[10–8] 20[10–8]
EI y = – + + – 613.4 10
6 24 6 24
EI y
=
19583.3
–
12500
+
190
+
13.3
–
6134
In terms of EI, y = 1152.6 /EI
Actual deflection, y = 1152.6
10
12
/(205
10
3
55330
10
4
)
y = 10.2 mm (upwards deflection)
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
37
Study Guide 2 End Test
Standard load case formulae
1.
A series of timber
beams form part of a balcony of a building. The
beams are cantilevered over a 2.4m length as shown in Figure 1. The
beams are at 1.2m centres and are required to support a uniformly
distributed load over the entire length and a point load at the tip. Usin
g
the design formulae and the additional data, determine the deflection at
the tip of the beam.
Figure 1
Additional data:
Uniformly distributed load on floor being carried
by beams
2.4 kN/m
Point load at cantilever tip
1 kN
Modulus of el
asticity of timber section (E)
8800 N/mm
2
Deflection formulae:
Due to udl
=wL
4
/8EI
Due to point load at tip
=WL
3
/3EI
2.
For the 457
191
82 UB beam loaded as shown below, use the
standard case deflection formulae given in the design data to deter
mine
the mid

span deflection.
Figure 2
STUDY GUIDE 2
38
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Design data:
Modulus of elasticity (E)
205 kN/mm
2
Second moment of area
37050 cm
4
Deflection formulae:
Due to udl
=5wL
4
/384EI
Due to point load at mid

span
=WL
3
/48EI
Derivation of formulae
3.
Us
ing Macaulay’s method, prove that the standard formula for a simply
supported beam carrying a point load at mid

span is:
=WL
3
/48EI
Macaulay’s method
4.
For the beam loaded as shown below:
(a)
Calculate the value of the reactions R
a
and R
b
.
(b)
Derive
an equation for the bending moment at any section along
the length of the beam in terms of length ‘x’ from R
a
.
(c)
Derive the equations for slope and deflection.
(d)
Determine the actual deflection of the beam when x = 3m.
Figure 4
E = 10800 N/
mm
2
I = 357
10
6
mm
4
STUDY GUIDE 2
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
39
Answers:
1.
udl
= 5.03 mm
point
= 2.33 mm
tot al
= 7.36 mm
2.
udl
= 5.33 mm
point
= 3.56 mm
tot al
= 8.89 mm
4.
Ra =9.8 kN
Rb = 7.3 kN
Mx = 9.8x
–
2x
2
/2 +7.3(x
–
6)
–
1(x
–
6)
2
/2
Slope equation: 9.8x
2
/2
–
2x
3
/6 +7.3(x
–
6)
2
/2
–
(x
–
6)
3
/6
–
40.
8
Deflection equation: 9.8x
3
/6
–
2x
4
/24 +7.3(x
–
6)
3
/6
–
(x
–
6)
4
/24
–
40.8x
Deflection: 22mm
40
ST
RUCTURAL ANALYSIS AN
D DESIGN ( AH)
STUDY GUIDE 3
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
41
STUDY GUIDE 3
Design of reinforced concrete elements
Introduction
This study guide covers Outcomes 3 and 4 of the unit.
Outcome 3
Design statically det
erminate singly reinforced beams and slabs in
reinforced concrete.
Outcome 4
Design short, braced, axially loaded columns in reinforced concrete.
On completion of the study guide you should be able to:
•
Design singly reinforced beams in reinforced con
crete.
This will involve: determining the design loads on beams; calculating the
areas of reinforcement to resist ultimate bending moments; determining
suitable arrangements of link reinforcement to resist the shear forces in
beams; and assessing the suit
ability of beams in deflection.
•
Design singly reinforced slabs in reinforced concrete.
This will involve: determining the design loads on slabs; calculating the
areas of reinforcement to resist the ultimate bending moments; determining
suitable arrange
ments of secondary (transverse) reinforcement; and
assessing the suitability of slabs in deflection.
•
Design axially loaded reinforced concrete columns.
The design process is from the British Standard:
BS 8110
–
1: 1997 Structural use of concrete
Part 1:
Code of practice for design and construction
STUDY GUIDE 3
42
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
In the design process the following material strengths will be used
throughout:
Characteristic strength of concrete, f
cu
40 N/mm
2
Characteristic strength of main reinforcement, f
y
460 N/mm
2
Characteristic s
trength of shear reinforcement, f
yv
250 N/mm
2
Unit weight of concrete
24 kN/m
3
In addition to the study guide you will require a copy of
Reinforced Concrete
Design

Details of Reinforcing Steel
.
Symbols and terms used in reinforced concrete design
For
a simply supported beam with tension on the bottom surface due to
bending.
b
–
breadth of the section
h
–
overall depth of the section
d
–
effective depth of section (this is the depth from the compression
surface to the centre of the t
ension reinforcement)
A
s
–
area of main tension reinforcement
A
sv
–
area of link (shear) reinforcement
STUDY GUIDE 3
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
43
Reinforced concrete design
BS 8110
–
1:1997
Clause reference
Design considerations
Notes for design and detailing concrete elements
Concrete cover to
reinforcement
3.3
Cover to the steel reinforcement is necessary to ensure that the bond of the
steel with the concrete is fully developed, so that both the steel and the
concrete are effective in resisting the applied forces. In addition the nominal
co
ver specified should be such that the concrete protects the steel against
corrosion and fire. To this effect the nominal cover, that is the minimum
cover to all the reinforcement, should at least:
•
be the size of the main longitudinal reinforcement
•
be
the size of the nominal maximum aggregate
•
satisfy the durability requirements (i.e. exposure).
When casting concrete against uneven surfaces, such as against earth, the
value should be not less than 75mm; when cast against a blinding layer the
cover sho
uld be specified as not less than 40mm.
The cover to protect the steel from corrosion is given in Table 3.3 of
BS 8110: Part 1 and depends on the exposure conditions that may be expected
and the quality of the concrete.
Definitions for exposure conditio
ns are given in Table 3.2 and quality is
defined in terms of the concrete grade i.e.
C30, C35
,
etc.
Table 3.4 gives the nominal cover required to protect the steel from the
effects of fire, with the values being dependent on time periods of fire
protectio
n, e.g. 1 hour, 2 hours, etc.
Spacing of reinforcement
(a)
Minimum
distance between bars
3.11.12.1
During the concreting operation the aggregate must be allowed to move freely
between the bars to obtain the maximum compaction and bond. For this
reason
the bar spacing should be greater than the nominal maximum size of
the aggregate.
STUDY GUIDE 3
44
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Minimum distance between bars = h
agg
+ 5 mm
where h
agg
is the nominal maximum aggregate size
For normal concrete work a 20 mm aggregate
is specified, thus minimum dista
nce between
bars = h
agg
+ 5 mm = 20 + 5 = 25 mm
(b)
Maximum distance between bars in tension (beams)
3.12.11.2.3
This clause is used to ensure a limit on the crack widths on the tension face of
the concrete. The clear distance between adjacent bars
should be not greater
than the value given in table 3.28 of the code. The value of spacing indicated
is for the condition zero redistribution of steel
–
redistribution will not be
considered in this course and may be considered as being equal to zero.
Ext
ract from table 3.28
f
y
Spacing
N/mm
2
mm
250
280
460
155
Spacing of bars
–
slabs
3.12.11.2.7
‘In no case should the clear spacing between bars exceed the lesser of three
times the effective depth or 750 mm
In addition, unless the crack widths are chec
ked by direct calculation, the
following rules will ensure adequate control of cracking for slabs subject to
normal internal and external environments:
(a)
no further check is required if either:
(1)
grade 250 steel is used and the slab depth does not
exceed 250
mm
(2)
grade 460 steel is used and the slab depth does not exceed 200
mm
(3)
the reinforcement percentage (100 A
s
/bd) is less than 0.3%
where
A
s
–
area of tension reinforcement
b
–
breadth of section
d
–
effective depth
STUDY GUIDE 3
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
45
(b)
whe
re none of the conditions (1), (2) or (3) apply, the bar spacing
should be limited to the values given in table 3.28 for slabs where the
percentage of reinforcement exceeds 1% or the values given in table
3.28 divided by the reinforcement percentage for le
sser amounts.’
Example
If a slab is 300 mm deep and from design calculations 0.45% of high yield
reinforcement is required, then the maximum distance between bars can be
determined as follows:
From Table 3.28
max. spacing = 155 mm
However this figure is
based on 1% or more of reinforcement being provided
From Cl 3.12.11.2.7
maximum spacing = 155/0.45 = 344 mm
Minimum area of reinforcement
3.12.5
Enough reinforcement should be provided to control the crack widths in the
tension face regardless of a
ny other design considerations. From Table 3.25:
Situation
Definition of
Minimum percentage
percentage
f
y
= 250N/mm
2
f
y
= 460N/mm
2
Tension reinforcement
(c) Rectangular sections
100 A
s
/A
c
0.24
0.13
(in solid slabs this minimum
should be provided in b
oth
directions)
For high yield reinforcement minimum permissible area is 0.13% of gross
section,
therefore minimum A
s
= 0.0013bh
Distribution or secondary steel is required in slabs. This reinforcement runs at
right angles to the main tension reinforcem
ent and is tied to it. The purpose
of the secondary steel is to tie the slab together and to assist in distributing
the loading through the slab. The area of this steel must be
at least
equal to
the minimum area of steel found from Table 3.25
The dis
tribution steel is always placed inside the main steel thus giving the
tension reinforcement the greatest effective depth.
STUDY GUIDE 3
46
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
Maximum area of reinforcement
3.12.6
Physically in order to compact the concrete properly and ensure adequate
bond develops between
the concrete and the steel reinforcing bars a maximum
must be put on the amount of reinforcement allowed in elements
Beams and slabs
Neither the area of tension reinforcement nor area of
compression should exceed 4% of the gross cross

sectional area of
the concrete
Columns
The longitudinal reinforcement should not exceed the
following amounts, calculated as percentages of the
gross cross

sectional area:
(a)
vertically cast columns
6%
(b)
horizontally cast columns
8%
(c)
laps in columns
10%
Eff
ective span for calculations
3.4.1.2
For a simply supported beam it may be taken as the smaller distance:
(a)
centres of bearings, or
(b)
clear distance between supports plus the effective depth d
An example is provided in the design notes.
STUDY GUIDE 3
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
47
Reinforc
ed concrete design
BS 8110: Part 1:1997
Procedure
Design of tension reinforcement
Procedure for determining the main area reinforcement in a beam/slab
using equations of BS 8110: Part 1: 1997 Clause 3.4.4.4
1.
Determine the value of K
K = M/bd
2
f
cu
where M
–
applied bending moment
f
cu
–
characteristic strength of concrete,
f
cu
= 40 N/mm
2
b
–
breadth of section
d
–
effective depth of section
Notes:
(1)
for a slab always consider a typical 1 m width
b =1000 mm
(2)
d
–
effective dept
h
This is the depth from the compression surface to the centre of the tension
reinforcement. The size of the reinforcing bars is not known nor is the
size of the stirrups (beam links) so an initial estimate must be made.
Beams
Typically for a beam
the main bar size is of the order of 25 mm
and the links are generally 8, 10 or 12 mm diameter.
Effective depth, d = overall depth (h)
–
cover
–
link diameter
–
main bar
dia/2
Assuming 30 mm cover, a link size of 10 mm and main bars of 20 mm
d = h
–
30
–
10
–
20/2 = h
–
50 mm
Slabs
The bar size in a slab is generally smaller than would be required for a beam,
say 16 mm.
Slabs are designed so that links are not required and the cover is generally for
mild exposure conditions.
Effective depth, d = o
verall depth (h)
–
cover
–
main bar dia/2
d = h
–
20
–
16/2 = h

28 mm
To calculate K, ensure that units are in N and mm, as moment is quoted in
kNm.
STUDY GUIDE 3
48
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
The equation becomes:
K = M
10
6
/bd
2
f
cu
2.
Check that K
K
where K
= 0.156
Only singly reinforc
ed sections will be considered, therefore you
must
always
check that K
K
always.
3.
Determine the lever arm distance, z
lever arm distance, z = d (0.5 +
(0.25
–
K/0.9))
but
z
0.95 d
4.
Calculate the area of reinforcement
In this course alway
s assume f
y
= 460 N/mm
2
Singly reinforced sections only.
Area of reinforcement, A
s
= M
10
6
/0.95 f
y
z mm
2
5.
Calculate minimum and maximum areas of reinforcement and compare
with calculated area
Minimum area
0.13%bh
3.12.5.3
Maximum area
4%bh
3
.12.6
The calculated value must lie between these limits
If area is less then reinforcement at least equal in area to 0.13%bh must
be provided
If area is greater than 4%bh then section size must be increased
Nominal maximum aggregate size will al
ways be assumed to be 20 mm
in this course
Example 1:
Typical slab reinforced with high

yield steel
A
simply supported slab is required to carry an ultimate moment 125 kNm per
metre width. The slab designated exposure condition is moderate with a
chosen
fire resistance period of two hours. If the slab has an overall depth of
200 mm, determine a suitable arrangement of reinforcement.
STUDY GUIDE 3
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
49
Solution
Cover
From Table 3.3 of BS 8110, minimum cover to all steel required for f
cu
= 40
N/mm
2
and exposure condition mod
erate is 30 mm.
From Table 3.4 of BS 8110, minimum cover needed to all steel for a slab with
a fire period of two hours is 35 mm.
Minimum nominal cover to all steel is 35 mm.
From Figure 3.2 of BS 8110, minimum possible slab thickness complying
with a fire
period of two hours is 125 mm. Thus the thickness provided
complies with fire regulations.
Find K
If the bar diameter is assumed to be 20 mm
Effective depth of section
d
=
h
–
cover
–
bar diameter/2
=
200
–
35
–
20 /2
=
155 mm
Resistance

moment factor
K
=
M
10
6
/(bd
2
f
cu
)
=
125
10
6
/(1000
150
2
40)
=
0.13
K<K
(ie 0.156)
Find z
lever arm distance
z
=
d(0.5 +
(0.25
–
K/0.9))
=
d(0.5+
(0.25
–
0.13/0.9))
=
0.825 d < 0.95 d
z
=
128 mm
Find A
s
Area of tension steel required
A
s
=
M
10
6
/(0.
95 f
y
z)
=
125
10
6
/(0.95
460
128)
=
2335 mm
2
/m
Provide T20@ 125 mm crs
(2510 mm
2
/m)
Applying detailing rules
3.12.11.2.7
Since 3d < 750
mm, maximum clear distance between bars for tension steel
3d= 3
160 = 480 mm,
maximum spacing (centre
to centre) of bars = 3d + dia = 500 mm.
Actual spacing used 125 mm, spacing suitable
Proportion of tension steel provided
100A
s
/(bh)
=
100
2510/(1000
200)
=
1.25 % of gross section.
STUDY GUIDE 3
50
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
As
this falls within Code limits of 0.13% and 4%, this is satisfact
ory. Refer to
Table 3.25 and clause 3.12.6.
As the slab thickness of 200 mm does not exceed 200 mm, no check on the
bar spacing is required with high

yield steel. See Cl.3.12.11.2.7(a)(2).
Distribution steel
The distribution or secondary steel runs at r
ight angles to the main tension
reinforcement and is tied to it. The purpose of the secondary steel is to tie
the slab together and to assist in distributing the loading through the slab.
The area of this steel must be at least equal to the minimum area
of steel
found from Table 3.25, i.e.
0.13%bh
The distribution steel is always placed inside the main steel thus giving the
tension reinforcement the greater effective depth.
Minimum distribution steel required
Minimum steel area (Table 3.25) A
s
=
0.13
%bh
=
0.0013
1000
200
=
260 mm
2
/m width
Using the design charts
The percentage area of steel,
100A
s
/bd
, may be found using the
Design
Charts of
BS 8110
–
3.
This percentage value is dependent on:
1.
The concrete grade shown as a curve on each chart.
2.
The value
of the bending stress,
M/bd
2
, on the vertical axis of the chart.
The course will use only
Design Chart 2
from BS 8110
–
3. This deals with
singly reinforced beams and slabs using high yield (f
y
= 460 N/mm
2
)
reinforcing steel.
From the previous example
M
= 125 kNm, b=1000 mm, d=155 mm, hence:
M/bd
2
=
125
10
6
/1000
155
2
=
5.2 N/mm
2
STUDY GUIDE 3
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
51
Refer to Design Chart 2
(a)
Find this value on the vertical axis
(b)
Project horizontally to the line f
cu
= 40 N/mm
2
(c)
Project vertically and read value from the horizontal axis
(d)
Read value of 100A
s
/bd, 1.
58
(e)
As this chart was intended for a different version of the code, a
multiplication factor based on differing material partial safety factors,
g
m
, must be introduced.
Factor = 1.05/1.15
(f)
Amended value of 100A
s
/bd = 1.58
1.05/1.15 = 1.44
Hen
ce A
s
= 1.44bd/100 =1.44
1000
155/100 = 2332 mm
2
/m
By calculation A
s
= 2335 mm
2
/m
The remainder of the design procedure is carried out as before.
Example 2
:
Typical singly reinforced beam with high

yield steel
A
simply supported beam is required to
carry an ultimate moment 335 kNm.
The beam designated exposure condition is severe with a chosen fire
resistance period of two hours. If the beam has an overall depth of 550 mm
and breadth of 300 mm, determine a suitable arrangement of longitudinal
reinfo
rcement.
Solution
Cover
From Table 3.3 of BS 8110, minimum cover to all steel for f
cu
= 40
N/mm
2
an
exposure condition severe is 40 mm.
From Table 3.4 of BS 8110, minimum cover needed to all steel for simply

supported beam with a fire period of two hours
is 40 mm.
Thus minimum permissible cover to all steel is 40 mm.
The breadth of section is 300 mm from Figure 3.2 of BS 8110, minimum
possible beam width complying with a fire period of two hours is
200 mm. Breadth provided complies with fire regulations
.
STUDY GUIDE 3
52
STRUCTURAL ANALYSIS
AND DESIGN ( AH)
The effective depth of section, d = overall depth
–
cover
–
link
–
main bar
dia/2
assume a main bar and link size say 25 mm and 12 mm
d = 550
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