NAT IONAL QUALIFICAT IONS CURRICULUM SUPPORT
Civil Engineering
Structural Analysis and Design
[ADVANCED HIGHER]
James Dunbar
abc
Acknowledgements
Learning and Teaching Scotland gratefully acknowledge this 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 2002
© 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.
CONTENTS
Overview
1
Tutor Guide
3
Student 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 Guide 4:
Design of structural steelwork elements
75
Study Guide 5:
Design of masonry and timber elements
99
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ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
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 Guides, 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 as 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 standard 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.
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O V ER V I E W
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 criteria 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 elements
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 timber.
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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 might 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 equilibrium
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 reinforced 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 procedure 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 determinate 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.
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TU T O R G U ID E
Study Guide 2: Determination of beam deflections by standard formulae
and Macaulay’s method
Prior to issue of the Study Guide, integration techniques 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 beam 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 publication 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 reinforced 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 design 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.
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TU T O R G U ID E
Study Guide 4: Design of structural steelwork elements
Each centre should provide 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 section 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
methods 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 respectively. 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.
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STUDENT GUIDE
Introduction
The unit 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 to 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 responsibility
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-solving
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 assessment. 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
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S TU D EN T G U ID E
Study Guide 2: Determination of beam deflections by standard formulae
and Macaulay’s method
This covers Outcome 2. It will enable 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 applied,
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 reinforcement.
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 covered 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, trimmer 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 use of British
Standards. All the information is there if you know where to look for it!
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S TU D EN T G U ID E
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 supervision 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 certain 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
skills and your tutor will guide you in this area.
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S TU DY GU ID E 1
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-jointed
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 resolution
• calculate the magnitude and nature of forces in pin-joined frames using the
method of sections.
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S TU DY GU ID E 1
What does the term ‘statically determinate pin-jointed frames’ mean?
What will be considered will be the analysis of trusses where the external
loads 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 referred 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
Algebraic sum of vertical of forces must equal zero
Algebraic sum of horizontal of forces must equal zero
ΣM = 0
ΣV = 0
ΣH = 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 vertical 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,
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S TU DY GU ID E 1
there 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 must 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 reactions) are statically
determinate.
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S TU DY GU ID E 1
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 the 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 roller reaction
Take moments about A, ΣM = 0, clockwise moments positive
(12 × 3) + (48 × 3) – V C × 6 = 0
V C = (36 + 144)/6 = 30 kN ↑
(b)
Apply ΣV = 0 and ΣH = 0 to find the magnitude and direction of the
hinge reactions
ΣV = 0
ΣH = 0
V A + VC – 48 = 0
V A = 48 – 30 = 18 kN ↑
12 – H A = 0
H A = 12 kN ←
upwards positive
forces to right positive
Step 3: select a node with only two unknowns
Note: As no bending effects are present in the frame elements, the condition
of equilibrium ΣM = 0 cannot be applied. 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.
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S TU DY GU ID E 1
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 ΣV = 0 and ΣH = 0, firstly node A then node C
Node A
ΣV = 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 ↓
ΣH = 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
ΣV = 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 ↓
ΣH = 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.
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S TU DY GU ID E 1
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
ΣV = 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
ΣH = 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
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S TU DY GU ID E 1
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
ΣH = 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 forces 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
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S TU DY GU ID E 1
Method of sections
The method for analysis of the forces in frames by sections is used when only
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 example.
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-hand 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:
ΣH = 0 becomes
Taking forces acting to the right as positive
12 – 12 + BD h + ED + BC = 0
three unknowns
ΣV = 0 becomes
Taking upwards forces as positive
18 – 48 + BD v = 0
only one unknown
ΣM = 0
18
is dependent on where moments are taken.
As a general rule if the section cuts across three
elements, two of them will intersect at a node. Take
moments about this node leaving one unknown. Node
D in this example:
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S TU DY GU ID E 1
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
ΣV = 0 becomes
18 – 48 + BDv = 0
–30 + BDv = 0
BDv = 30 kN ↑
Taking upwards forces as positive
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
ΣH = 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 element must equal zero.
!
Answers agree with those found by method of joint resolution.
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S TU DY GU ID E 1
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 validity of the results found
using the method of joint resolution, by determining the forces in
elements GF, CF and CD.
Answers:
Roller reaction:
Hinge reactions:
horizontal
vertical
20
82.5 kN
12 kN
73.5 kN
all in directions shown in diagram
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S TU DY GU ID E 2
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.
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S TU DY GU ID E 2
Macaulay’s method
This is a method of analysis that allows the slope and deflection of a beam to
be determined.
M
E
=
From the equation of simple bending
I
R
where:
M
E
I
R
=
=
=
=
bending moment at a section
modulus of elasticity
second moment of area of the section
radius of curvature
When both E and I are constant for a given section, M and R are the only
variables.
EI
The expression for M is then M =
R
If the deflection of the member is y, and as deflection is a function of the
radius of curvature R, then:
d2 y
I
∝
(R is the second derivative of deflection)
2
dx
R
then:
M = EI
d2 y
dx 2
where x = distance along the length of the beam to position of bending
moment M.
To obtain deflection:
d2 y
M
=
2
dx
EI
Bending moment expression
dy
M
=∫
+A
dx
EI
Slope expression
y=
22
M
∫ ∫ EI + Ax + B
Deflection expression
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S TU DY GU ID E 2
Where A and B are constants of integration. They are determined by
considering 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
terms 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 the 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, slope 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.
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S TU DY GU ID E 2
Example 1: Point loads
A beam is simply supported as shown. For the illustrated loading system,
determine:
(a)
(b)
the slope and deflection under the 200 kN load
the magnitude and position of the maximum deflection.
E = 205 kN/mm2
I = 900 × 10 6 mm 4
Find the value of the reactions
Take moments about R a
ΣM = 0, clockwise moments are positive
(200 × 2 ) + (350 × 5 ) – (R b × 7 ) = 0
R b = 307.1 kN
ΣV = 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
X
Take moments about x–x
Ra
200 kN
350 kN
24
Distance to load from
section x–x
(m)
x
x–2
x–5
Moment = Force x distance
242.9 x
200[x–2]
350[x–5]
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S TU DY GU ID E 2
The total moment at x may be written as:
M x = EI
d2 y
= 242.9[x] – 200 [x–2] – 350 [x–5]
dx 2
Integrate with respect to ‘x’
EI
d2 y
= 242.9[x] – 200 [x–2] – 350 [x–5]
dx 2
EI
moment (kNm)
dy
[x]2
[x–2]2
[x–5]2
= 242.9
– 200
– 350
+A
dx
2
2
2
EI y = 242.9
[x]3
[x–2]3
[x–5]3
– 200
– 350
+ Ax + B
6
6
6
slope equation
(kNm x m = kNm2 )
deflection equation
(kNm 2 x m = kNm3 )
Now deflections are zero at the supports thus:
When x = 0, y =0
and x = 7, y = 0
Substituting in the deflection eqn. for x = 0, y = 0
EI 0 = 242.9
[0]3
[0–2]3
[0–5]3
– 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
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
25
S TU DY GU ID E 2
Thus general equations for slope and deflection at any point along the length
of the beam are
EI
dy
[x]2
[x–2]2
[x–5]2
= 242.9
– 200
– 350
– 1250
dx
2
2
2
EI y = 242.9
slope
[x]3
[x–2]3
[x–5]3
– 200
– 350
– 1250[x]
6
6
6
deflection
To find the slope and deflection at the 200 kN load substitute for x=2
EI
dy
[2]2
[2–2]2
[2–5]2
= 242.9
– 200
– 350
– 1250
dx
2
2
2
=
486
–
(0) –
(ignore)
dy
–764
=
dx
EI
kNm 2
– 1250
1 kNm 2 = 10 9 Nmm
1 kN/mm 2 = 10 3 N/mm 2
Units are now consistent
764 × 109
× 900 × 106
205 × 103
= –0.0041 radians
=
EI y = 242.9
[2]3
[2–2]3
[2–5]3
– 200
– 350
– 1250[2]
6
6
6
= 324
– 0
– 0
–2176
y =
EI
–2176 × 109
× 900 × 106
=
3
205 × 10
kNm 3
– 2500
Nmm
= mm
N/mm 2 × mm 4
= –11.8 mm (11.8 mm downwards)
To determine the position of the maximum deflection equate slope equation
to zero.
EI
dy
[x]2
[x–2]2
[x–5]2
= 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
26
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
ax 2 + bx +c
S TU DY GU ID E 2
Solving for x, x =
–2150 ±
(21502 – 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
deflection y = – 2706/EI = – 16.7 mm
Example 2: Uniformly distributed loads
A simply supported beam is Lm long and is required to carry a uniformly
distributed load of w kN/m. In general terms, determine the maximum
deflection of the beam:
Find value of reactions:
As beam is symmetrically loaded, R a = R b = w × L/2 = wL/2
Apply Macaulay’s method at a section x–x beyond the last applied load
In this example consider the section x–x immediately to the left of reaction R b
Take moments about x–x
Value of load
Distance to centre of
load from section x–x (m)
Ra
w kN/m
wL/2
w.x
x
x/2
Moment = Force x distance
wL.x/2
w.x.x/2
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S TU DY GU ID E 2
This may be written as:
Mx =
wLx wx 2
d2 y
–
= EI 2
2
2
dx
Integrate with respect to ‘x’
dy
wLx 2
wx 3
=
–
+A
dx
4
6
wLx 3
wx 4
EI y =
–
+ Ax + B
12
24
EI
slope equation
deflection equation
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
EI 0 =
wL[0]3
w[0]4
–
+ A[0] + B
12
24
hence B = 0
Applying the deflection equation again at right-hand support, when
x=L y=0
wL L3
wL4
–
+ AL
12
24
wL4
EI 0 =
+ AL
24
EI 0 =
A=
–wL3
24
note negative sign
Thus equations become:
dy
wLx 2
wx 3
3wL3
=
–
–
dx
4
6
24
3
4
3
wLx
wx
wL x
EI y =
–
–
12
24
24
EI
28
slope equation
deflection equation
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 2
To determine maximum deflection, substitute in deflection equation for x =
L/2 as the beam is symmetrically loaded.
EI y =
EI y =
EI y =
EI y =
y=
wL(L/2)3
w(L/2) 4
wL3 (L/2)
–
–
12
24
24
3
4
3
wL.L / 8 wL / 16 wL .L
–
–
12
24
48
4
4
4
wL
wL
wL
–
–
96
384
48
4
4
4wL
wL
8wL4
–
–
384
384
384
4
–5wL
384EI
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 equations for
deflection
Additional information
Standard deflection formulae for cantilevers:
Uniformly distributed load
∆ = wL 4 /8EI
Point load at tip
∆ = WL 3 /3EI
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
29
S TU DY GU ID E 2
Find value of reactions:
Reaction must balance downwards forces, as ΣV = 0
R a = 20 × 2 + 64 = 104 kN
Taking moments about R a ,
ΣM = 0, clockwise moments are positive
–M a + (20 × 2 × 1) + 64 × 2 = 0
M a = 168 kNm (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
104
20.x
Distance to centre of
load from section x–x (m)
x
x/2
Moment = Force × distance
104x
20.x.x/2
Considering also the moment at the support, this may be written as:
M x = EI
d2 y
20x 2
=
–
168
+
104x
–
dx 2
2
Integrate with respect to ‘x’
dy
104x 2
20x
= – 168x +
–
+A
dx
2
6
–168x 104x 3
20x 4
EI y =
+
–
+ Ax + B
2
6
24
EI
slope equation
deflection equation
In order to find the constants of integration A and B apply the boundary
conditions.
For a cantilever beam, deflection at the support is zero, and slope is zero at a
built-in support.
30
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 2
Applying the deflection equation at support, when x = 0 y = 0
EI 0 =
–168[0]2
104[0]3
20[0]4
+
–
+ A[0] + B
2
6
24
Therefore B = 0
Applying the slope equation at the built-in support, when x= 0,
EI 0 = –168.[0]2 +
dy
=0
dx
104.[0]3
20.[0]4
–
+A
2
6
Therefore A = 0
Equations become:
dy
104x 2
20x
= –168x +
–
dx
2
6
3
–168x 104x
20x 4
EI y =
+
–
2
6
24
EI
slope equation
deflection equation
Maximum deflection occurs at the tip of the cantilever, x = 2 m
EI y =
–168.22
104.23
20.24
+
–
2
6
24
EI y = –336 + 138.67 – 13.33
EI y = –210.67
y =
–210.67
EI
Negative sign indicates that deflection is downwards.
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
31
S TU DY GU ID E 2
(b)
Compare with standard equations:
Standard deflection formulae for cantilevers:
Uniformly distributed load
Point load at tip
For udl
for a point load
total deflection
∆ = wL 4 /8EI
∆ = WL 3 /3EI
20.24
40
=
8EI
EI
3
64.2
170.67
=
∆=
3EI
EI
210.67
∆=
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 dealing 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
32
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 2
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]
3.
– (z–w).[x–a] 2 /2
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 less 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.
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
33
S TU DY GU ID E 2
Take moments about x–x
Mx
(same as condition 1)
(additional term)
= 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 cm4
E= 205 kN/mm2
Find value of reactions:
Taking moments about R a ,
ΣM = 0, clockwise moments positive
(30 × 8 × 4) – R b × 8 + (10 × 2 × 9) = 0
R b = 142.5 kN
Reactions must balance downwards forces
ΣV =0,
R a + R b = 30 × 8 +10 × 2
R a = 260 – 142.5 = 117.5 kN
34
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 2
Take moments about x–x, considering 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
117.5 kN
30 kN/m.x
142.5 kN
20 kN/m.(x–8)
Distance to centre of
load from section x–x (m)
x
x/2
x–8
(x–8)/2
Moment = Force x distance
117.5x
30.x.x/2
142.5(x–8)
20.(x–8) 2 /2
This may be written as:
M x = EI
dy 2
30x 2
20[x–8]2
=
117.5x
–
+
142.5[x–8]
+
dx 2
2
2
Integrate with respect to ‘x’
dy 117.5x 2
30x 3 142.5[x–8]2
20[x–8]3
=
–
+
+
+A
slope equation
dx
2
6
2
6
117.5x 3
30x 4
142.5[x–8]3
20[x–8]4
EI y =
–
+
+
+ Ax + B deflection equation
6
24
6
24
EI
In order to find the constants of integration A and B apply the standard
conditions.
For a simply supported beam deflection is zero at the supports.
When x = 0 , y = 0 – apply to deflection equation
EI 0 =
0 =
117.5.03
30.0 4 142.5[0–8]3
20[0–8]4
–
+
+
+ Ax + B
6
24
6
24
(0)
– (0) + (ignore term) + (ignore term) + (0) + B
B=0
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
35
S TU DY GU ID E 2
Also when x = 8, y = 0 – apply to deflection equation
EI 0 =
117.5.83 30.84
142.5[8–8]3
20[8–8]4
–
+
+
+ Ax
6
24
6
24
0 = 10027
– 5120 +
(0)
+
(0)
+ A.8
A = –(10027–5120)/8 = –613.4
Standard equations become:
dy 117.5x 2
30x 3 142.5[x–8]2
20[x–8]3
=
–
+
+
– 613.4
dx
2
6
2
6
117.5x 3
30x 4
142.5[x–8]3
20[x–8]4
EI y =
–
+
+
– 613.4x
6
24
6
24
EI
slope equation
deflection equation
Actual deflections
At centre of main span, x = 4m
EI y =
117.5.43
30.44
142.5[4–8]3
20[4–8]4
–
+
+
– 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
EI y =
117.5.103 30.104
142.5[10–8]3
20[10–8]4
–
+
+
– 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)
36
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 2
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. Using
the design formulae and the additional data, determine the deflection at
the tip of the beam.
Figure 1
Additional data:
2.
Uniformly distributed load on floor being carried
by beams
Point load at cantilever tip
Modulus of elasticity of timber section (E)
2.4 kN/m
1 kN
8800 N/mm2
Deflection formulae:
Due to udl
Due to point load at tip
∆=wL 4 /8EI
∆=WL 3 /3EI
For the 457 × 191 × 82 UB beam loaded as shown below, use the
standard case deflection formulae given in the design data to determine
the mid-span deflection.
Figure 2
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
37
S TU DY GU ID E 2
Design data:
Modulus of elasticity (E)
Second moment of area
205 kN/mm 2
37050 cm 4
Deflection formulae:
Due to udl
Due to point load at mid-span
∆=5wL 4 /384EI
∆=WL 3 /48EI
Derivation of formulae
3.
Using 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)
(d)
Derive the equations for slope and deflection.
Determine the actual deflection of the beam when x = 3m.
Figure 4
E = 10800 N/mm2
I = 357 × 10 6 mm 4
38
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 2
Answers:
1.
∆ udl = 5.03 mm
∆ po int = 2.33 mm
∆ to t al = 7.36 mm
2.
∆ udl = 5.33 mm
∆ po int = 3.56 mm
∆ to t 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
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
39
40
ST RU CT U R AL AN ALYS I S AND DE SI GN ( AH )
S TU DY GU ID E 3
STUDY GUIDE 3
Design of reinforced concrete elements
Introduction
This study guide covers Outcomes 3 and 4 of the unit.
Outcome 3
Design statically determinate 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 concrete.
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 suitability 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 arrangements 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
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
41
S TU DY GU ID E 3
In the design process the following material strengths will be used
throughout:
Characteristic strength of concrete, f cu
Characteristic strength of main reinforcement, f y
Characteristic strength of shear reinforcement, f yv
40 N/mm2
460 N/mm2
250 N/mm2
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 tension reinforcement)
A s – area of main tension reinforcement
A sv – area of link (shear) reinforcement
42
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 3
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
cover 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 should 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 conditions 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
protection, 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.
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
43
S TU DY GU ID E 3
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 distance 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.
Extract from table 3.28
Spacing
fy
N/mm 2
250
460
mm
280
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 checked 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
44
ST RU CT U R AL AN ALYS I S AN D DE SI GN ( AH )
S TU DY GU ID E 3
(b)
where 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 lesser 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 any other design considerations. From Table 3.25:
Situation
Definition of
percentage
Minimum percentage
f y = 250N/mm 2 f y = 460N/mm 2
Tension reinforcement
0.24
0.13
(c) Rectangular sections
100 A s /A c
(in solid slabs this minimum
should be provided in both
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 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
The distribution steel is always placed inside the main steel thus giving the
tension reinforcement the greatest effective depth.
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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
Columns
Neither the area of tension reinforcement nor area of
compression should exceed 4% of the gross crosssectional area of the concrete
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%
Effective span for calculations
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.
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Reinforced 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,
b – breadth of section
d – effective depth of section
Notes:
(1) for a slab always consider a typical 1 m width
(2)
f cu = 40 N/mm 2
b =1000 mm
d – effective depth
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 = overall 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.
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The equation becomes:
K = M × 10 6 /bd 2 f cu
2.
Check that K ≤ K′ where K′ = 0.156
Only singly reinforced 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 always assume f y = 460 N/mm 2
Singly reinforced sections only.
Area of reinforcement, A s = M × 10 6 /0.95 f y z mm2
5.
Calculate minimum and maximum areas of reinforcement and compare
with calculated area
Minimum area
Maximum area
0.13%bh
4%bh
3.12.5.3
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 always 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.
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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 moderate 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 =
=
=
z =
Find A s
Area of tension steel required
d(0.5 + √(0.25–K/0.9))
d(0.5+√(0.25–0.13/0.9))
0.825 d
< 0.95 d
128 mm
A s = M × 10 6 /(0.95 f y z)
= 125 × 10 6 /(0.95 × 460 × 128)
= 2335 mm2 /m
Provide T20@ 125 mm crs (2510 mm2 /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.
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As this falls within Code limits of 0.13% and 4%, this is satisfactory. 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 right 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 mm2 /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.
2.
The concrete grade shown as a curve on each chart.
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/mm2 )
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/mm2
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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
Hence 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
reinforcement.
Solution
Cover
From Table 3.3 of BS 8110, minimum cover to all steel for f cu = 40 N/mm2 an
exposure condition severe is 40 mm.
From Table 3.4 of BS 8110, minimum cover needed to all steel for simplysupported 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.
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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 – 40 – 12 – 25/2= 485.5 mm
485 mm(say)
Longitudinal reinforcement
Applied-moment factor K =
=
=
Since applied-moment factor K
lever arm distance
z =
=
=
M ×10 6 /(bd 2 f cu )
335 × 10 6 /(300 × 485 2 × 40)
0.119 < 0.156
< K′ beam is suitable for design
d(0.5+ √(0.25–K/0.9))
d(0.5+√0.25–0.119/0.9)) = d(0.84) ≤ 0.95d
0.84 × 485 = 407 mm,
Area of tension steel needed
Provide as tension steel 4T25 mm bars
A s = M/0.95 f y z
A s = 335 × 10 6 /0.95 × 460 × 407
= 1883 mm2
(A s = 1963 mm2 )
An alternative arrangement may be to provide 2T32 and 1T20
(1608 + 314 = 1922 mm 2 )
Percentage of tension steel provided
= 100A s /(bh)
= 100 × 1963/(300 × 550)
= 1.18 % of gross section
As this is within Code limits of 0.13% and 4%, this is satisfactory.
Design of shear reinforcement
Procedure for determining shear reinforcement for beams and slabs
A.
Shear resistance of solid slabs
3.5.5
1. Calculate shear stress at the point of highest shear force
The design shear stress, v c , at any section should not exceed the shear stress
at calculated at any section using equation 21
v=
V
bd
b =1000 mm
The form and area of shear reinforcement are found using the
recommendation Table 3.16
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Extract from Table 3.16
Value of v
v < vc
v c < v < (v c + 0.4)
(v c + 0.4) < v < 5 N/mm 2
Form of shear
reinforcement to be
provided
None
Minimum links for the
whole length of the beam
Designed links
Area of shear
reinforcement to be
provided
None
A s v ≥ 0.4b v s v /0.95f yv
A s v ≥ (v–v c )b v s v /0.95f yv
In no case should v exceed 0.8√f cu or 5 N/mm2
As f cu is taken as 40 N/mm 2 then 0.8√f cu = 0.8√40 = 5.06 N/mm 2
thus
v < 5 N/mm2
In slabs the general condition is that no shear reinforcement is provided.
Always ensure that:
v < vc
2.
Determine concrete shear stress v c
Using Table 3.8 find 100A s /b v d
b v = b = 1000 mm
In previous slab example T20@ 125 mm crs (2510 mm 2 /m) was provided
Percentage area of reinforcement 100As/b v
d = 100 × 2510/(1000 × 150)
= 1.67 %
d = 150 mm
From Table 3.8 for d =150 and 100A s /b v d = 1.67
Shear resistance = 0.95 N/mm2
However the note at the foot of Table 3.8 must be applied
For characteristic concrete strengths greater than 25 N/mm 2 , the values in this
table may be multiplied by (f cu /25) 1/3 . The value of f cu should not be taken as
greater than 40.
For (f cu /25) 1/3 when f cu = 40 N/mm 2
(40/25) 1/3 = 1.17
Shear resistance v c = 1.17 × Table 3.8 value
In this example, v c = 1.17 × 0.95 = 1.11 N/mm2 .
The requirement is that no shear reinforcement is used in slabs.
Check v < v c .
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B.
Shear resistance of beams
1.
Calculate shear stress
3.4.5
3.4.5.2
The design shear stress, v c , at any section is determined and compared with
the shear stress calculated at any section using equation 3. The difference
between the values indicates the amount of shear reinforcement required.
v=
V
bvd
b v = b = breadth of section
In no case should c exceed 0.8√f cu or 5 N/mm2
As f cu is taken as 40 N/mm 2 then 0.8√f cu = 0.8√40 = 5.06 N/mm 2
thus
v < 5 N/mm2
The form and area of shear reinforcement are found using the
recommendation Table 3.7
Extract from Table 3.7:
Value of v
Less than 0.5v c
0.5v c < v < (v c + 0.4)
(v c + 0.4) < v < 5 N/mm 2
2.
Form of shear
reinforcement to be
provided
None
Minimum links for the
whole length of the beam
Designed links
Area of shear
reinforcement to be
provided
None
A s v ≥ 0.4b v s v /0.95f yv
A s v ≥ (v–v c )b v s v /0.95f yv
Determine concrete shear stress v c
As for slabs the percentage area of reinforcement 100A s /b v d and the effective
depth of the beam are used to determine v c .
However care should be taken when calculation 100A s /b v d – see
Clause 3.4.5.4:
‘The term As is that area of longitudinal tension reinforcement which
continues for a distance at least equal to d beyond the section being
considered. At supports the full area of tension reinforcement at the section
may be applied in the table provided the requirements for curtailment and
anchorage are met (see 3.12.9).’
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For simply supported beams carrying a uniformly distributed load – maximum
moment occurs at mid-span – hence this is where the maximum amount of
tension reinforcement is needed. At the supports the bending moment is zero,
and in theory no reinforcement needs to be provided. However in practice, in
accordance with clause 3.12.8 and Figure 3.24, at least 50% of the
reinforcement must continue over to the supports.
From beam example, provide as tension steel or
4T25 bars (A s =1963 mm2 )
or
2T32 & 1T20 (1608 + 314 = 1922 mm2 )
At the support this would be 2T25 (i.e. at least 50% of reinforcement) or the
2T32’s.
Thus at the support, where shear is highest, A s is based on 2T25 or 2T32’s,
not the full area. Conversely at mid-span A s is based on the full area.
For the previous beam example
A s (2T25) = 981 mm 2
b = 300 mm
d =480 mm
100A s /b v d = 100 × 981/(300 × 480) = 0.68%
v c = 0.54 N/mm 2
From table 3.8 for 100A s /b v d = 0.68 and d ≥ 400 mm
Design concrete shear stress v c = (f cu /25) 1/3 × 0.63 = 0.74 N/mm 2
At mid-span – 4T25
100A s /b v d=100 × 1963/(300 × 480)= 1.36% d = 480 mm
From table 3.8
v c = 0.69 N/mm2
Design concrete shear stress v c = 1.17 × 0.69 = 0.81 N/mm2
This procedure tends to result in a variation of shear reinforcement along the
length of the beam.
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3.
Determining the size and spacing of shear reinforcement
Shear reinforcement in the form of links or stirrups are always provided in
beams. The form and area are determined from Table 3.7
f yv =strength of shear reinforcement – in this course only mild steel links are
considered
f yv = 250 N/mm2
A sv = total cross-section of links at neutral axis.
The links are designed to go round the outside of the main reinforcement
Area A sv refers to (in its simplest form) two legs of reinforcement
s v = spacing of links along the member
To begin with A sv and s v are unknowns as they refer to the links. Generally a
size of bar is chosen for the shear reinforcement and the spacing varied along
the length of the beam.
Typically 8, 10, 12 or 16mm diameter bars are used as links. If a bar size is
chosen then this leaves the spacing s v as the only unknown
The equations may be written as:
sv =
sv =
0.95 f yv A sv
0.4 b v
0.95 f yv A sv
(v – v c ) b v
minimum link spacing
close space links
Clause 3.4.5.5 states that, regardless of the above calculation, the spacing of
links should not exceed 0.75d.
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Design for deflection
Procedure for determining deflection limits for beams and slabs
The actual span/depth ratio is compared with the appropriate ratio obtained
from Table 3.9, which is modified by the value, obtained from Table 3.10
1.
Calculate the actual span/depth ratio:
Effective span
Effective depth
2.
Find the basic l/d ratio from Table 3.9.
Support conditions
Cantilever
Simply supported
Continuous
Rectangular sections
7
20
26
Thus for a simply supported beam or slab
3.
basic l/d = 20
Determine the modification factor for tension reinforcement from Table
3.10
See also Clause 3.4.6.5
The value K was previously calculated using K = M × 10 6 /bd 2 f cu
Table 3.10 requires M/bd 2 to be calculated
This may be done directly or by rearranging the above equation so:
M/bd 2 =K × f cu
The value of service stress f s is calculated using
f s = 2f y A s req /3 A s pro v – Eqn 8
Where the Area of reinforcement provided is approximately the same as
that required, then:
f s = 2/3 f y = 307 N/mm 2
Thus the modification factor may be determined from Table 3.10.
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4.
Determine permissible deflection ratio and compare with actual value
Permissible span/depth ratio = Table 3.9 value × Table 3.10 value
This is now compared with the actual span/depth ratio
Beam or slab complies with code when actual < permissible.
Full beam design example
BS 8110: Part 1:1997
A series of simply supported beams carry an imposed load of 5 kN/m 2 , in
addition to the dead loading calculated from the cross-section shown below
and the application of finishes.
Using the additional data, design a suitable arrangement of reinforcement for
the beam and check the beam’s suitability in deflection.
Additional design data:
Unit weight of concrete
Finishes to concrete
Beam centres
Exposure conditions
Fire period
Characteristic strength of concrete
Characteristic strength of main bars
24 kN/m 3
1 kN/m 2
3m
mild
1 hour
f cu = 40 N/mm2
f y = 460 N/mm 2
Bar is deformed round (Type 2)
Characteristic strength of links
f yv = 250 N/mm2
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Solution
Overall depth of concrete section
h = 400 mm
Breadth of concrete section
b = 275 mm
Cover: ≥ bar diameter
≥ Table 3.3 value = 20 mm
≥ Table 3.4 value = 20 mm
Also Figure 3.2. Minimum beam width = 200 mm in 1-hour fire
If a 25 mm diameter bar is assumed for the design, cover = 25 mm
Link size is not known, assume 10 mm link diameter
Effective depth of concrete section d = h – cover – link – half main bar
= 400 – 25 – 10 – 25/2 = 350 mm (say)
Span
Beam effective span lesser of:
(a) centre to centre of supports
(b) clear span + d
Effective span, L =4.7 m
5000 – 300/2 – 300/2 = 4700 mm
5000 – 2 × 300 + 350 = 4750 mm
Characteristic dead load, g k
slab weight
beam self weight
finishes
24 × 0.2 × 3
24 × 0.275 × 0.4
1×3
gk
Characteristic imposed load, q k
3×5
qk
Design load
Maximum moment
Maximum shear force
14.4
2.64
3
20
kN/m
kN/m
kN/m
kN/m
15 kN/m
= 1.4 g k + 1.6 q k
= 1.4 × 20 + 1.6 × 15
= 52 kN/m
M = (1.4 g k + 1.6 q k ) L 2 /8 = 52 × 4.7 2 /8
= 143. 6 kNm
V = (1.4 g k + 1.6 q k ) L/2 = 52 × 4.7/2 = 122.2 kN
Main reinforcement at position of maximum moment
Limiting factor
K′ = 0.156
Compute
K = M × 10 6 /(bd 2 f cu )
= 143.6 × 10 6 /(275 × 350 2 × 40)
= 0.106
As K <K′,.
lever arm
z = d(0.5 + √(0.25 – K/0.9))
= 355(0.5+ √(0.25 – 0.106/0.9))
= 302 mm < 0.95 d
(0.95d = 337 mm)
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A s = M × 10 6 /(0.95f y z)
= 143.6 × 10 6 /(0.95 × 460 × 302)
= 1088 mm2
Percentage area of reinforcement 100A s /bd = 100 × 1088/(275 × 355)
= 1.11%
As this is within the code limits of 0.3% and 4% this is satisfactory
Selected reinforcement
4T20 (A s = 1256 mm2 )
Area of tension reinforcement
Detailing
Clear distance between bars b – [(no of bars × dia) – 2(cover + link dia.)]/(No
of bars – 1)
=
[275 – (4 × 20) – 2(20 + 10]/(4 – 1)
=
45 mm
Permissible clear distance
minimum 25 mm
(clause 3.12.11)
maximum 155 mm
(table 3.28)
Shear reinforcement at (or near) support
b = 275 mm
Breadth of section for shear
bv =
Design shear stress
v =
V × 10 3 / (b v d)
= 122.2 × 1000/(275 × 355)
= 1.25 N/mm2
Area of reinforcing bars in accordance with Clause 3.4.5.4. (i.e. 50%)
At the support the longitudinal bars effective for shear are 2T20
Area of bars for shear
A s pro v = 628 mm2
Percentage provided
100A s pro v /(b v d)
= 100 × 628/(275 × 355)
= 0.64%
From Table 3.8 for 100A s prov /(b v d) = 0.64% and d = 355 mm
v c = 0.57 N/mm2
Characteristic concrete strength greater than 25 N/mm 2 therefore increase v c
according to footnote in Table 3.8.
Increased shear strength
v c = v c × (f cu /25) 1/3 = 0.67 N/mm2
Provide links to take shear force in accordance with Table 3.7
(v c + 0.4) = (0.67 + 0.4) = 0.97 N/mm 2
v > (v c + 0.4) [ 1.25 > 0.97 ]
Provide close spaced links near the supports
Assume the use of R10 links A sv = 157 mm2
Spacing of links at ends member s v = A sv 0.95f yv /(b v (v–v c ))
= 157 × 0.95 × 250/(275(1.25 – 0.67))
= 233 mm
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Provide R10 links at 225 mm centres near the supports
At (or near mid-span) – minimum links only required
Spacing of links
s v = A sv 0.95f yv /[(b v (0.4)]
= 157 × 0.95 × 250/(275 × 0.4) = 338 mm
but limited to 0.75 d = 0.75 × 355 = 266 mm
Provide R10 links at 250 mm centres near the centre of the beam
X = 2.35 × 0.4/ 1.25 = 0.75 m
Provide minimum links in middle 1.5 m (2 × 0.75 m) of beam, R10’s at 250
mm centres
Provide close spaced links at the ends of the beam, R10 at 225mm centres
Deflection criteria: Full calculation
Actual l/d
4700/355 = 13.2
Basic l/d (Table 3.9)
20
Service stress
f s = 2f yA s req /(3A s pro v )
= 2 × 460 × 1088/(3 × 1256)
= 260 N/mm2
M/bd 2 = 143.6 × 10 6 /(275 × 355 2 ) = 4.1 N/mm 2
Or alternatively
M/bd 2 = K × f cu = 0.1 × 40 = 4 N/mm 2
From Table 3.10 for f s =260 N/mm2 , and M/bd 2 = 4.1
Modification factor for tension steel = 0.92
Permissible simply-supported l/d =20 × 0.92 = 18.4
As actual < permissible beam is serviceable in deflection (13.2 < 18.4)
Deflection criteria: Abridged version of calculation
Actual l/d
4700/355 = 13.2
Basic l/d ( Table 3.9)
20
Service stress
f s ≈ 307 N/mm 2
M/bd 2 = K × f cu = 0.1 × 40 = 4 N/mm2
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From Table 3.10
Modification factor for tension steel = 0.84
Permissible simply-supported l/d =20 × 0.84 = 16.8
As actual < permissible beam is serviceable in deflection
By taking f s = 307 N/mm 2 (f s =2/3 f y) a conservative value is found – always
erring on the safe side.
Simply supported beams (or slabs) with an overhang
BS 8110–1 states in 3.2.1.2.2:
‘It will normally be sufficient to consider the following arrangements of
vertical load:
(a) All spans loaded with the maximum design ultimate load
(1.4G k + 1.6Q k )
(b) Alternate spans loaded with the maximum design ultimate load (1.4G k +
1.6Q k ) and all other spans loaded with the minimum design ultimate
load (1.0 G k ).’
In the design to consider a simply supported beam with an overhang then
three possible loading arrangements must be considered:
1.
2.
3.
maximum load on main span and overhang
maximum load on main span, minimum load on the overhang
minimum load on main span, maximum load on the overhang
The requirements for reinforced concrete design is to determine the shear
force and bending moments in the member and design reinforcement
accordingly. Thus three shear diagrams to be drawn together with their
significant values. These are superimposed on one another and the shear force
envelope created. Similar for the bending moments a bending moment
envelope must be developed.
Two maximum values of bending moment will be found:
• A sagging moment along the length of the main span (which means the
main reinforcement will be positioned near the bottom surface)
• A hogging bending moment over the overhang support (the reinforcement
will be positioned near the top surface).
Thereafter the design procedure in the notes for either a beam or a slab can be
used.
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Example
For the beam shown below and using the loading given in the design data,
determine the maximum sagging and hogging bending moment.
Design data:
Characteristic dead load inclusive of self-weight
Characteristic imposed load
24kN/m
18kN/m
Solution
Maximum load =1.4G k + 1.6Q k = 1.4 × 24 + 1.6 × 18 = 62.4 kN/m
Minimum load = 1.0G k =24 kN/m
Maximum load on all spans
Take moments about A
62.4 × 10 × 10/2 = B × 8
A = 62.4 × 10 – 390 = 234 kN
B = 390 kN
Drawing shear force diagram
Maximum sagging moment = 0.5 × 234 × 3.75 = 438.75kNm
Maximum hogging moment = 0.5 × 124.8 × 2 = 124.8 kNm
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Maximum load on main span, minimum load on the overhang
Take moments about A
62.4 × 8 × 8/2 + 24 × 2 × 9= B × 8 B = 303.6 kN
A = 62.4 × 8 + 24 × 2 – 303.6 = 243.6 kN
Drawing shear force diagram
Maximum sagging moment = 0.5 × 243.6 × 3.9 = 475kNm
Maximum hogging moment = 0.5 × 48 × 2 = 48 kNm
Minimum load on main span, maximum load on the overhang
Take moments about A
24 × 8 × 8/2 + 62.4 × 2 × 9= B × 8 B = 236.4 kN
A = 24 × 8 + 62.4 × 2 – 236.4 = 80.4 kN
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Drawing shear force diagram
Maximum sagging moment = 0.5 × 80.4 × 3.35 = 134.67kNm
Maximum hogging moment = 0.5 × 124.8 × 2 = 124.8 kNm
Design values are:
Sagging bending moment 475 kNm
Hogging bending moment 124.8 kNm
Use design procedures to determine main and shear reinforcement values.
Axially loaded column design
BS 8110: Part 1:1997
Calculations for reinforcement are in accordance with BS 8110:Part 1: 1997
Clause 3.8.4
The first step in the design of a column is to determine whether the proposed
arrangement of column dimensions and height will make it short or slender.
If the column is slender in addition to the axial load it will also be designed
for moments due to deflection, thus for axial design the column must be
short.
Example of design information
Design ultimate axial load on column
Height of column FFL to FFL
Depth of the cross section
Width of column
Characteristic strength of concrete
Characteristic strength of reinforcement
N = 2400 kN
L= 4.5 m
h = 350 mm
b = 350 mm
f cu = 40 N/mm2
f y = 460 N/mm 2
Flooring arrangement supported by column continuous beam and slab floor
construction with beams 600 mm deep by 350 mm wide in both directions.
All spans equal.
The column is connected at the bottom to a base designed to carry a moment
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Effective height of column
Clause 3.8.1.6
The effective height is found using equation 30
le = β lo
The value of β is found using Table 3.19 for braced columns and Table 3.20
for unbraced columns where β is a function of the end restraint conditions of
the column.
The definitions for braced and unbraced columns are given in clause 3.8.1.5.
‘A column may be considered braced in a given plane if lateral stability to
the structure as a whole is provided by walls or bracing or buttressing design
to resist all lateral forces in that plane. It should otherwise be considered as
unbraced.’
The column may also be defined as short or slender as defined in clause
3.8.1.3.
‘A column may be considered as short when both the ratios l ex /h and l ey /b are
less than 15 (braced) and 10 (unbraced). It should otherwise be considered
as being slender.’
For this section of the course the columns will be restricted to short braced
systems.
To use Table 3.19 the end restraint conditions must be known
End condition at top
1
2
3
1
0.75
0.80
0.90
End condition at bottom
2
3
0.80
0.90
0.85
0.95
0.95
1.00
The end conditions are defined in clause 3.8.1.6.2.
Condition 1
The end of the column is connected monolithically to beams on either side
which are at least as deep as the overall dimension of the column in the plane
considered.
The column in the example supports beams 600 mm deep.
If the column is the lowest length of a structure and is connected to a
substantial base then condition 1 may also apply at the base.
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Condition 2
The end of the column is connected monolithically to beams or slabs on either
side which are shallower than the column, e.g. the column in the example is
connected to a floor which is 300 mm deep
Condition 3
The column is connected to shallow members that will provide some nominal
restraint, e.g. a shallow floor
In the example consider the column to be connected at the top to beams 600
mm deep and to a substantial base at the bottom:
For top restraint – condition 1
bottom restraint – condition 1
From Table 3.19
β = 0.75
The clear height l o – this is defined as the clear height between end
restraints.
For this example the height of the column is given as 4.5 m, with 600 mm
deep beams framing in at the top. Then the clear height
l o = 4.5 – 0.6 = 3.9 m
le = βl o =0.75 × 3.9 = 2.925 m
Effective height
Note: The column can have two different effective heights, one based on the
x–x axis the other on the y–y for simplicity in this course l ex = l ey = l e
Slenderness of column
3.8.1.3
In order for the column to be defined as short braced, both the ratios l ex /h and
l ey/b must be less than 15.
As b is defined as the smaller dimension, only one check needs to be applied.
l ey/b < 15
Slenderness of column
2925/350 = 8.35
Slenderness of column within allowable for a short column.
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Axial load considerations
3.8.4
Two equations are used:
Equation 38 – where a column cannot be subject to significant moments due
to the nature of the loading.
N = 0.4 f cu A c + 0.7 A sc f y
Equation 39 – where a column is carrying an approximately equal
arrangement of beams the spans of which do not vary by more than 15% and
the beams are designed to carry uniformly distributed loads.
N = 0.35 f cu A c + 0.7 A sc f y
In both equations it is generally the area of reinforcement A sc that is required,
thus the equations must be rearranged with A sc as the subject.
Note:
Area of concrete A c
= Cross-sectional area less area of reinforcement
= (b × h) – A sc
Consider the example for which design data is given.
Assuming the short braced column is carrying an approximately equal
arrangement of beams, the spans of which do not vary by more than 15%, and
the beams are designed to carry uniformly distributed loads.
Use Equation 39
Main bars
N = 0.35f cu (b.h – A sc ) + 0.7 A sc f y
Area of reinforcement A sc =
N × 103 – 0.35 f cu b h
0.7 f y – 0.35 f cu
2400 × 103 – 0.35 × 40 × 350 × 350
0.7 × 460 – 0.35 × 40
= 2224 mm 2
=
Note: At least four bars must be provided, i.e. one in each corner.
An even number of bars must be provided.
From the Areas of Reinforcing Steel Chart, select
6T25 (A sc = 2944 mm 2 ) or 4T32 (A sc = 3215 mm 2 )
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Minimum area of reinforcement – Table 3.25
General rule
100A sc /A cc = 0.4
Minimum Area
A sc = 0.4 bh/100 = 0.4 × 350 × 350 / 100
A sc = 490 mm 2
Maximum area of reinforcement Cl.3.12.6.2
Vertical cast columns 6% of gross area
A sc = 6/100 bh = 6/100 × 350 × 350
A sc = 7350 mm 2
Area of steel chosen is suitable.
Links
Clause 3.12.7.1
‘When part or all of the main reinforcement is required to resist compression,
links or ties at least one-quarter the size of the largest compression bar or 6
mm, whichever is greater, should be provided at a maximum spacing of 12
times the size of the smallest compression bar.’
Assuming that 6T25’s are used
Minimum diameter of links
Quarter the diameter of largest bar
= 25/4 = 6.25 mm
Make diameter of links
R8
Mild steel bars are used as the links are only required for containment
purposes and not required to carry load.
Maximum spacing of links
12 times diameter of smallest main bar
2 × 25 = 300 mm
Provide R8 links @ 300 mm centres
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Study Guide 3 End Test
For all questions:
Characteristic strength of concrete, f cu
Characteristic strength of main reinforcement, f y
Characteristic strength of shear reinforcement, f yv
40 N/mm2
460 N/mm2
250 N/mm2
Unit weight of concrete
24 kN/m 3
Slab design
A simply supported slab spans between two brick walls as shown. For the
given design information determine:
(a)
the maximum bending moment and shear force in the slab per metre
width
(b)
design a suitable arrangement of main and secondary reinforcement
(c)
check the suitability of the slab to resist the shear forces
(d)
check the slab for crack control and defection requirements
Design information:
Characteristic imposed load
Characteristic dead load due to finishes
Exposure conditions
Fire resistance
4.5 kN/m2
1.5 kN/m2
mild
1 hour
Beam design
A series of beams at 4m centres support pre-cast concrete flooring units. For
analysis purposes the beams may be assumed to have the arrangement of a
simply supported beam with an overhang as shown below:
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For one beam
(a)
Determine:
(i) The maximum sagging bending moment
(ii) The maximum hogging bending moment
(iii) The maximum shear force at support A
(b)
Design a suitable arrangement of main reinforcement at the critical
sections
(c)
Design suitable shear reinforcement at support A
(d)
Check the cantilevered portion of the beam for deflection
Design information:
Characteristic imposed load
Characteristic dead load due to finishes
Characteristic dead load of flooring units
Exposure conditions
Fire resistance
Beam cross-section
5 kN/m 2
2 kN/m 2
3.8 kN/m2
moderate
2 hours
300 mm × 650 mm
Column design
The diagram below illustrates the layout braced multi-storey in-situ
reinforced concrete building.
Using the design data, for the central column ‘X’:
(a)
(b)
(c)
Evaluate the design axial load
If the height of the columns between floor levels is 4.8m and the beams
framing into the columns are 600mm deep, show that the column is
short.
Determine a suitable arrangement of main and link reinforcement.
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Design data:
Characteristic imposed load on floor
Characteristic dead load due to finishes
Axial loads on column ‘X’ for upper floors
Characteristic imposed load
Characteristic dead load
Exposure conditions
Fire resistance
Column cross-section
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5 kN/m 2
2 kN/m 2
1200kN
820kN
moderate
3 hours
350mm × 350 mm
S TU DY GU ID E 3
Answers
Slab Design
Answers where appropriate – per metre width of slab:
Design loading 16.86 kN/m
K = 0.031
A S = 575 mm 2 /m
M = 46.55 kNm
V = 39.6 kN
v =0.2 N/mm2
Actual span/depth ratio = 24.1
vc= 0.47 N/mm 2
permissible span/depth ratio = 24.4
Beam design
Results from analysis of all three load cases:
Reaction
A(kN)
Load case 1
Load case 2
Load case 3
233.2
255.4
70.2
Reaction B
(kN)
moment
505.2
362.2
341.3
Maximum
sagging
moment (kNm)
379.8
459
88.4
Maximum
hogging
moment (kNm)
278.3
109
278.3
Maximum sagging moment = 459 kNm
Maximum hogging moment = 278.3 kNm
Maximum shear at A = 255.4 kN
(b)
Sagging reinforcement
K = 0.108
z = 0.86 d
A s = 2053 mm 2
Hogging reinforcement
K = 0.066
z = 0.92 d
A s = 1164 mm 2
(c)
v =0.58 N/mm2
vc= 0.61 N/mm 2
R12’s @ 200 mm centres or equivalent
(d)
Actual span/depth ratio = 4.7
Permissible span/depth ratio
= 6.2
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Column design
Dead load from floor = 373.4 kN
Imposed load from floor = 240 kN
Design load, N = 3698 kN
A sc = 6438 mm 2
Main reinforcement 8T32 (6430 mm2 )
Links R8’s @350 mm centres, or alternatives
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STUDY GUIDE 4
Design of structural steelwork elements
Introduction
This study guide covers Outcomes 5 and 6 of the unit.
Outcome 5
Design statically determinate structural steel beams.
Outcome 6
Design axially loaded, single-storey steel stanchions.
On completion of the study guide you should be able to:
• Use structural section tables to find the properties of universal beams and
universal columns.
• Design structural steel beams with full lateral restraint.
This will involve determining the suitability of beams for bending
moment, shear force, deflection, and web buckling and bearing at the
supports.
• Design axially loaded steel columns.
The design process is from the British Standard: BS 5950–1:1990 Structural
use of steelwork in building, Part 1: Code of practice for design in simple and
continuous construction.
In the design process Grade 43 steel will be used throughout.
In addition to the study guide you will require a copy of each of the following
documents:
Universal beams – dimensions and properties
Universal columns – dimensions and properties
Safe load tables – bearing/buckling/shear values
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Structural steel struts subject to axial load
BS 5950–1:1990
Compression members
clause 4.7
Compression members are classed as either:
(1)
Plastic
Compact
Semi-compact
in each case the section can resist the full load
(squash load)
It is only necessary for strut design that the section be classified as not
slender.
This is not related to the Euler theory definition of ‘slender’ but relates to the
ability of the cross-section of the member to carry the load without distortion
of the section.
(2)
Slender
the section fails at a load less than the squash load
due to local buckling of the section
The amount of load that the members can carry is dependent on the
slenderness, λ, of the gross section, design strength, p y , and the section
classification.
For plastic, compact and semi-compact sections:
The compressive resistance P c = A g × p c
[load = area × stress ]
The design strength is found using Table 6 of BS 5950. In this course only
Grade 43 steel will be considered:
Design grade
43
76
Thickness less than or
equal to (mm)
16
40
63
80
100
Design strength
p y (N/mm 2 )
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275
265
255
245
235
S TU DY GU ID E 4
Thus for a 457 × 152 × 74 UB in Grade 43 steel:
web thickness
t = 9.9 mm
flange thickness
T= 17.0 mm
Design strength p y = 265 N/mm 2
Note: The design strength depends on the greater material thickness.
For slender sections p y is modified by a stress reduction factor as given in
Table 8 (not applicable to UC sections).
The classification of the section is found using Table 7 of BS5950 and is a
measure of the ease with which the cross-section can distort (or buckle) under
load. This can occur in one of two ways:
(1)
(2)
The outstand of the flange
The web
b/T – ratio
d/t – ratio
Values of b/T and d/t are obtained from the section tables and compared with
the Table 7 limits.
Extract from Table 7
Type of element
Outstand element
of a compression
flange
Web, where whole
section is subject
to compression
Type of
section
Rolled section
Rolled section
(1) Plastic
b/T ≤ 8.5ε
(2)
Compact
b/T ≤ 9.5ε
(3) Semicompact
b/T ≤ 15ε
d/t ≤ 39ε
d/t ≤ 39ε
d/t ≤ 39ε
ε – this is a factor which depends on the material grade
ε = (275/p y) 1/2
For a 203 × 203 × 52 UC section in Grade 43 steel
Flange thickness T = 12.5 mm
Design strength p y = 275 N/mm 2
ε = (275/p y) 1/2 = (275/275) 1/2 = 1.0
From the section tables, the ratios for local buckling are:
b/T = 8.18 < 8.5ε
d/t = 20.1 < 39ε
Flanges are plastic
Web is not slender
b/T = 8.17
d/t = 20.3
Section is not slender
For axially loaded columns all that is required from the classification of the
section is that it is not slender.
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The slenderness is a measure of the ease with which the strut will buckle
over its length and is found using:
λ=
Slenderness
L
Effective length
= E
Radius of gyration
r
Effective length is defined as the length between points of effective restraint
of a member multiplied by a factor to take account of the end conditions and
loading
L E = KL
where L = actual length
K = effective length ratio
K is determined from the conditions of end restraint.
L E = 1.0L or 0.85 L
In this course
The value you should use will be given in each question.
Note: Universal column sections have two major axes x–x and y–y, thus there
are two values of slenderness:
λ =
L Ey
L Ex
and λ =
rx
ry
Both values have to be calculated in order to determine the compressive
strength p c.
To decide which of the Tables 27(a) – (d) are to be used in determining p c
reference must first be made to Table 25.
For UC sections, extracting from Table 25 gives:
Type of section
Rolled H-section
Thickness
up to 40 mm
over 40 mm
x–x axis
27(b)
27(c)
y–y axis
27(c)
27(d)
An H-section is one for which the overall depth of the section divided by the
overall breadth (D/B) is less than 1.2. An I-section is one for which D/B >
1.2.
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All UC sections are H-sections, and nearly all of the UB sections are
I-sections.
Thus for a 203 × 203 × 52 UC section in Grade 43 steel:
Flange thickness T = 12.5 mm
Design strength p y =275 N/mm 2
Look up Table 27(b) for the x–x axis
and Table 27(c) for the y–y axis
The critical value of compressive strength is the lesser of the two values from
the tables.
The compressive resistance is calculated using:
Pc = Ag × pc
A g = gross sectional area (from Section Tables).
Struts
The process for design or checking the adequacy of a column is as follows:
(1)
From the Section Tables, the material (flange) thickness is found.
(2)
The design strength is taken from Table 6 [either 275 or 265 N/mm 2 ].
(3)
The effective length, L E , is found using the appropriate end restraint
conditions.
L E = 1.0 L or 0.85 L
(4)
The slenderness, λ, is calculated for both the x–x and y–y axes.
λ =
L Ey
L Ex
and λ =
rx
ry
The radius of gyration, r, is found using the Section Tables.
(5)
The strut curve tables for both axes are selected from Table 25.
(6)
The compressive strength, p c , is read from the appropriate part of
Tables 27 (a)–(d).
The critical value of compressive strength is selected, i.e. the lower of
the two values.
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(7)
The compressive resistance P c is calculated. P c = A g × p c.
The area of the section A g is found from the Section Tables.
For safe economic design, the compressive resistance P c should just exceed
the design load F c , i.e. P c ≥ F c .
Terms used in the code:
F – the factored design load
P – resistance or capacity (maximum load which can be applied to the
section)
p – strength of the section (permissible stress)
subscripts:
c – compression
t – tension
b – bearing
Example
Check the suitability of a 203 × 203 × 52 UC section to carry the applied
axial loads given, if the actual length between restraints is 4.2 m. It can be
assumed that the effective length of the column L E = 0.85L for both axes.
The applied axial loads are:
Dead
450 kN
Imposed
350 kN
Solution
Design load, F c
(1)
=
=
=
=
1.4 × Dead + 1.6 × Imposed
1.4 × 450 + 1.6 × 350
630 + 560
1190 kN.
From Section Tables
flange thickness, T
ratios for local buckling
radius of gyration r x
radius of gyration r y
area of section A g
80
12.5
Flange
Web
8.90
5.16
66.3
mm
b/T 8.17
d/t 20.3
cm
cm
cm2
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(2)
p y = 275 N/mm2
From Table 6 [as T < 16 mm ]
Section classification ε = 1
b/T = 8.17 < 8.5ε
Plastic
d/t = 20.3 < 39ε
Not slender
thus section is not slender.
(3)
From Table 24
L E = 0.85L
= 0.85 × 4.2m = 3.57 m for both axes
(4)
λx =
L Ey
L Ex
and λ y =
rx
ry
= 3.57 × 100/8.9 = 3.57 × 100/5.16
= 40.1
= 69
Slenderness is a ratio. The height of the stanchion was given in metres,
and from the section tables the radius of gyration is in cm. The units
must cancel out.
In this example the calculation is carried out using cm units.
(5)
From Table 25
For buckling about x–x axis, use strut curve Table 27(b)
For buckling about y–y axis, use strut curve Table 27(c)
with p y = 275 N/mm 2
(6)
Slenderness x–x axis
Slenderness y–y axis
(7)
Pc = Ag × pc
p cx = 250 N/mm 2
p cy = 183 N/mm 2
= (66.3 × 100) × 183
= 1213000 N = 1213 kN
Note: Be careful with the units. The area is given, in the dimensions and
properties table, in cm2 units, and p cy is in N/mm 2 .
cm2 × 100 = mm2
mm 2 × N/mm 2 = N
divide by 1000 to determine load in kN
1213 kN > 1190 kN
therefore P c > F c and section is suitable.
Compare the value of P c with the value given in the Safe Load Tables.
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Stanchions restrained at an intermediate position
Although they are not as efficient as UC sections, Universal Beam sections
are frequently used to carry axial loads. The y-y axis of a UB section is
considerably weaker than a comparable UC section, and to improve its loadcarrying capacity intermediate side rails may be inserted to restrain the y–y
axis. This principle can also be applied to any other form of column section.
For the arrangement shown, the column can buckle about the x–x axis over
the length L x , whereas for the y–y axis the column can buckle over either L y1
or L y2 or L y3 , whichever has the greatest effective length.
Values of p cx and p c y are then calculated and the critical value used to find the
compressive resistance P c .
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Example – Stanchion with intermediate restraints
The diagram below shows the elevation of a column that is to be provided
with intermediate ties providing restraint about the y–y axis only. The
column is to carry axial unfactored loads of 580kN dead and 760kN imposed.
The proposed section for the column is 533 × 210 × 101 UB. Check the
suitability of the proposed section, using the safe load tables for UB
sections, L E = 1.0L
Design load = 1.4 dead + 1.6 imposed = 1.4 × 580 + 1.6 × 760 = 2028kN
Effective length y–y axis
Effective length x–x axis
From the safe load tables:
L EY
For L EY = 3.5m
For L EY = 4.0m
For L EY = 3.6m
L EX
For L EX = 6.5m
For L EX = 7.0m
For L EX = 6.8m
L EY = 3.6m
L EX = 6.8m
P CY = 2060 kN
P CY = 1850 kN
P CY = 2060 –1/5 × (2060–1850) = 2018kN
P CX = 2760 kN
P CX = 2750 kN
P CX = 2760 –3/5 × 10 = 2754 kN
Compressive resistance of stanchion, P C =2018 kN
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Design of members subject to bending
BS 5950–1:1990
The rules for the design of members in bending are given in sections 4.2 and
4.3 of BS 5950: Part 1. The design of such elements is primarily concerned
with bending strength, but since bending moment is subject to variation along
the length of the member and is accompanied by shear action, the
combination of bending and shear must be taken into account. In addition, the
degree of restraint applied to the beam ends and along its length greatly
influence the bending capacity. If the beam is fully restrained along its
compression flange as defined in clause 4.2.2, there is no need to make any
allowance for lateral torsional buckling.
All beams must have an adequate resistance to bending and shear, and beams
that are not fully restrained laterally must be checked for a reduced bending
capacity due to lateral torsional instability, in accordance with section 4.3.
Fully restrained beams
A practical situation where the beam may be considered as having its
compression flange fully restrained is where it is required to carry a concrete
floor (either in-situ or formed from pre-cast concrete units) on the top flange.
A bond can form at the steel/concrete interface which will resist any sideways
movement of the beam compression flange.
‘At critical sections the combination of maximum moment and coexistent
shear, and the combination of maximum shear and coexistent moment should
be checked.’
The beams considered in this course will all be simply supported but will
carry a variety of loading. In this type of example the maximum shear force
will occur at the reactions, where moment is equal to zero. Conversely the
maximum moment will occur at or near the centre of the beam where shear is
zero or has a low value. For design purposes the maximum shear will be
considered together with the maximum moment and the shear will meet the
criteria for low shear load given in clause 4.2.5.
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Shear capacity
Clause 4.2.3
The applied shear force along the span, F v , should nowhere exceed the shear
capacity, P v
i.e. F v < P v
where P v = 0.6p y A v
From Clause 4.2.3(a)
for rolled I, H and channel sections, loaded parallel to web
A v = tD
Where t = thickness of beam web
D = overall depth of beam
These values are found from Section Tables
p y = design strength found from Table 6 and based on the beam
flange thickness.
Moment capacity with low shear load
Clause 4.2.5
The moment capacity M c of any section will depend on its section
classification from Table 7.
When F v < 0.6 P v the influence of co-existent shear on the moment capacity
may be ignored and the following capacities applied. All beams on this course
will be subject to the low shear load condition
Classification of the section
Extract from Table 7
Type of element
Type of section
(1) Plastic
Outstand element
of compression
flange
Web, with neutral
axis at mid-depth
Rolled sections
b/T ≤ 8.5ε
Rolled sections
d/t ≤ 79e
Class of section
(2) Compact
(3) Semicompact
d/t ≤ 79ε
b/T ≤ 15ε
d/t ≤ 98ε
d/t ≤ 120ε
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For plastic or compact sections
M c = p yS x or 1.2 p y Z x whichever is the lesser.
S x = plastic modulus of beam about major axis
Z x = elastic modulus of beam about major axis
These values are found from Section Tables.
The section reaches its full plastic moment of resistance.
For semi-compact sections
Mc = p y Zx
Only the extreme fibres of the section reach the design strength.
For slender sections
M c = p y′ Z x
The capacity of the section is further reduced by local buckling
p y′ = p y × stress reduction factor of Table 8 (not generally applicable to UB
sections).
Deflection
The beam must be checked for the serviceability limit state of deflection,
such that the actual deflection due to unfactored imposed load only is less
than the deflection limits given in Table 5, i.e.
actual deflection < span/360 or span/200
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Web buckling and bearing
Buckling
4.5.2.1
At points of concentrated load such as the condition shown below, where the
beam is supported, it is possible that the beam web can buckle. The
requirements of clause 4.5.2.1 must be met.
The buckling resistance P w = (b 1 + n 1 ) t p c
Typical values of bearing length are:
for a beam resting on brick supports b 1 =100 mm
for a beam resting on seating cleats b 1 = 21 mm
n 1 = load dispersal length – for supports this is equal to half the depth
of the beam
t = beam web thickness
p c = compressive strength of the web – found using Table 27(c) of the
code for a slenderness, λ = 2.5d/t
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Bearing
It is also possible that the junction between the beam flange and web may be
subject to a bearing failure.
The local capacity of the interface between the web and the beam is given by:
P crip = (b 1 + n 2 ) t p y w
where n 2 =
the length obtained by dispersion through the flange to the
flange-to-web connection at a slope of 1:2.5 to the plane of the
flange.
n 2 = 2.5( T + r )
T = thickness of beam flange
r = root radius of beam
p yw = design strength of the web found using Table 6 and web
thickness
Example – fully restrained steel beam
A simply supported beam spans over a 6m length and is required to carry the
unfactored loading shown in the diagram below.
DEAD 24kN/m
IMPOSED 30kN/m
Check the suitability of a 610 × 229 × 113 universal beam for:
(a)
(b)
(c)
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bending
shear
bearing over a 75 mm wide support
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Solution
Section properties
Section size
Depth of steel section
Width of steel section
Thickness of flange
Thickness of web
Root radius
Inertia about major axis
Plastic modulus about major axis
Elastic modulus
Area of section
610 × 229 × 113 Universal Beam
D = 607.6 mm
B = 228.2 mm
T = 17.3 mm
t = 11.1 mm
r = 12.7 mm
I x = 87320 cm 4
S x = 3281 cm 3
Z = 2874 cm 3
A = 144 cm2
Strength of steel
For material thickness of 17.3 mm, from Table 6
Design strength (Grade 43)
p y = 265 N/mm2
Young’s modulus
E = 205 kN/mm2
Factored loading
Distributed load
w = 1.4 dead + 1.6 imposed
= 1.4 × 24 + 1.6 × 30
= 81.6 kN/m
W = 1.4 × 124 + 1.6 × 84
= 308 kN
wL/2 + W/2
81.6 × 6/2 + 308/2
= 398.8 kN
This is also the maximum shear
wL 2 /8 + WL/4
81.6 × 6 2 /8 + 308 × 6/4 = 829.2 kNm
Point load
Reactions
Maximum moment
Section classification
Ratios for local buckling
Shear
Shear area
mm 2
Shear capacity
Clause 3.1.1
b/T = 6.60 < 8.5e
d/t = 49.3 < 79e
Therefore section is plastic
Av
4.2.3
= tD = 11.1 × 607.6 = 6744.4
P v = 0.6 p y A v
= 0.6 × 265 × 6744.4/10 3
= 1072.4 kN
Design shear force
F v = 398.8 kN
Clause 4.2.5
Since F v < 0.6 P v low shear load condition applies
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Note: this is the worst arrangement of shear force.
The beam should be checked for two conditions:
(1)
(2)
maximum shear force and coexistent moment
maximum moment and coexistent shear force
If the low shear load condition applies at the reactions, where the shear force
is the highest, it applies along the length of the beam. Thus the beam needs
only to be checked for maximum moment.
Moment capacity
Moment capacity for plastic section
or
4.2.5
M c = p yS x ≤ p y Z
M c = p yS x
= 265 × 3281/10 3
= 869.5 kNm
M c = 1.2p y Z
=1.2 × 265 × 2874/10 3
= 914 kNm
Critical value is the lesser of the two
Since M < M c (829.2 kNm < 869.5 kNm) applied moment is within moment
capacity.
Deflection
Apply imposed loads only
Uniform load
30 kN/m
Maximum deflection for a uniformly distributed load
∆ = 5wL 4 /384EI
= 5 × 30 × 6000 4 /384 × 205 × 10 3 ×
= 2.83 mm
Maximum deflection for a point load at mid-span
Imposed point load
W = 84 kN
∆ = WL 3 /48EI
= 84 × 10 3 × 6000 3 /48 × 205 × 10 3 ×
Total deflection
87320 × 10 4
87320 × 10 4
= 2.11 mm
∆ = 2.83 + 2.11 = 4.94 mm
From Table 5 the limiting defection for a beam carrying brittle finishes
span/360 =6000/360 = 16.7 mm
Since actual deflection < limiting deflection (4.94 mm < 16.7 mm), the beam
is serviceable in deflection.
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Web bearing and buckling
Bearing on the web
Local capacity of web
where:
Stiff bearing length
Spread to fillet
4.5.3
P cr it = (b 1 +n 2 )t.p yw
b 1 = 75 mm
n 2 = 2.5 (r+T)
= 2.5 (12.7+17.3) = 75 mm
For beam web thickness of 11.1 mm
Design strength of web
p yw = 275 N/mm2
Local capacity of web
P c = (b 1 +n 2 )t.p yw
= (75+75)11.1 × 275/10 3
= 458 kN
Force applied through flange
F v = 398.8 kN
Since value of reaction is less than the capacity of the web, no bearing
stiffener is required.
Buckling of the web
4.5.2
Buckling capacity of web
P w = (b 1 + n 1 )t.p c
where:
Stiff bearing length
b 1 = 75 mm
Spread to centre of section
n 1 = D/2 = 607.6/2 = 303.8 mm
p c is the compressive strength based on p y for the web (t = 11.1 mm from
Table 6: 275 N/mm 2 )
and slenderness ratio λ
λ = 2.5 d/t
d is the depth between the fillets obtained from the Section Tables.
Alternatively d/t is the ratio for local buckling obtained when checking the
beam classification (d/t = 49.3).
λ = 2.5 × 547.6/11.1 =123.3
or
λ = 2.5 × 49.3 = 123.3
From Table 27(c)
Compressive strength
Buckling capacity of web
p c = 94 N/mm2
P w = (b 1 + n 1 )t.p c
= (75+ 303.8)11.1 × 94/10 3
= 395.2 kN
Since the reaction force is greater than the capacity (398.8 kN > 395.2 kN) a
load carrying stiffener will be required to prevent the web from buckling.
This is not a failure condition and will not change the suitability of the beam
to carry the loads. However stiffeners will have to be designed to help
distribute the end reaction from the beam web to the support.
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Example 2 – Use of the moment capacity table and ‘bearing and buckling
values for unstiffened webs’ tables.
A series of beams at 3m centres are required to carry pre-cast concrete
flooring units. The beams are simply supported over a span of 7m.
Using the design data:
(a) Select a suitable UB section.
(b) Check the suitability of the chosen beam in shear and deflection.
(c) Check the suitability of the section, using the appropriate safe load
table, for web bearing and buckling, given that the stiff bearing length
at the supports is 40mm.
Design data:
Dead load due to pre-cast concrete units
Dead load allowance for finishes and self weight
Imposed load
2.5 kN/m2
1.5 kN/m2
6 kN/m 2
Solution
Design load = 1.4 × dead + 1.6 × imposed
(1.4 × [2.5 + 1.5]) + 1.6 × 6) × 3 = 45.6 kN/m
Maximum moment, M = wL 2 /8 = 45.6 × 7 2 /8 == 279.3 kNm
Maximum shear, F v =wL/2 = 45.6 × 7/2 = 159.6 kN
(a)
From the moment capacity table select a UB section with a moment
capacity of at least 279.3 kNm.
406 × 178 × 54 UB ( M c = 289 kNm)
No other checks have to be done – don’t classify the section; don’t
apply clause 4.2.5, the safe load table has already carried out the
necessary checks.
(b)
Before shear can be checked using Cl. 4.2.3, the design strength, p y,
must be found.
For material thickness of 10.9
Design strength (Grade 43)
Shear area
Shear capacity
Design shear force
mm, from Table 6
p y = 275 N/mm2
A v = tD = 7.7 × 402.6 = 3100 mm 2
P v = 0.6 p y A v
= 0.6 × 275 × 3100/10 3
= 511.5 kN
F v =159.6 kN
The beam is suitable in shear
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Deflection
Apply unfactored imposed loads only
Uniform load
6 × 3 = 18 kN/m
Maximum deflection for a uniformly distributed load
∆ = 5wL 4 /384EI
= 5 × 18 × 7000 4 /384 × 205 × 10 3 ×
18720 × 10 4
= 14.7 mm
From Table 5, the limiting defection for a beam carrying brittle finishes
L/360
7000/360 = 19.4 mm
Since actual deflection < limiting deflection (14.7 mm < 19.4 mm).
Beam is serviceable in deflection
Using the safe load tables –bearing and buckling values for unstiffened webs
At the bottom of the table ‘buckling and bearing values for unstiffened webs’
the following formula is given:
web capacity = C 1 +b 1 C 2 +t p C 3
The third term only applies if additional plates have been welded to the
flange of the beam, thus for a universal beam only
web capacity = C 1 +b 1 C 2
Bearing
For the 406 × 178 × 54 UB end bearing
C 1 = 111
C 2 = 2.09
Stiff bearing length, b 1 = 40 mm
Web bearing capacity = 111 + 40 × 2.09
= 111 + 83.6
= 194.6 kN > 159.6 kN
Beam web does not require a bearing stiffener
Buckling
For end buckling
C 1 = 152
C 2 = 0.753
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Web buckling capacity = 152 + 40 × 0.753
= 152 + 30.1
= 182.1 kN > 159.6 kN
Beam web does not require a buckling stiffener
The end column of the table also gives the shear capacity of the section found
using
From table P v = 505 kN
Shear capacity P v = 0.6 py A v
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Universal beams subject to bending moment capacity for grade 43 steel
Serial
Size
mm
914 x 419
Designation
Mass/
metre
kg
388
343
Moment
Capacity
kNm
4690
4110
289
253
224
201
3340
2890
2520
2220
838 x 292
226
194
176
2430
2030
1800
762 x 267
197
173
147
1900
1640
1370
686 x 254
170
152
140
125
1490
1320
1210
1060
610 x 305
238
179
149
1980
1460
1210
610 x 229
140
125
113
101
1100
975
872
792
122
109
101
92
82
848
747
694
652
566
98
89
82
74
67
82
74
67
60
52
914 x 305
533 x 210
457 x 191
457 x 152
Serial
Size
mm
406 x 178
Designation
Mass/
metre
kg
74
67
60
54
Moment
Capacity
kNm
412
371
327
289
406 x 140
46
39
244
198
356 x 171
67
57
51
45
333
278
246
213
356 x 127
39
33
180
148
305 x 165
54
46
40
232
199
172
305 x 127
48
42
37
194
168
148
305 x 102
33
28
25
132
112
92
254 x 146
43
37
31
156
133
109
254 x 102
28
25
22
97
84
72
591
533
503
456
404
203 x 133
30
25
86
71
203 x 102
23
63
477
429
396
352
300
178 x 102
152 x 89
19
16
47
34
127 x 76
13
23
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Study Guide 4 End Test
Figure 1 shows the part layout of multi-storey steelwork structure with precast concrete floor units. The construction is such that all beams may be
assumed to have full lateral restraint to their compression flanges and all
connections can be assumed to be simple.
Using the design information provided, complete the following tasks:
Beams
Type A
(a) Determine the design uniformly distributed load on a typical internal
beam.
(b)
Determine the maximum bending moment and shear force on beam type
A.
(c)
Using the appropriate clauses of BS 8110–1, check the suitability of a
406 × 178 × 74 UB section for:
(i)
(ii)
(ii)
bending
shear
deflection
Type B
(a) Determine the design concentrated load on a typical internal beam.
(b)
Determine the maximum bending moment and shear force on beam type
B.
(c)
Using the moment capacity table for Universal Beams select a suitable
UB section.
(d)
Check the suitability of the selected beam for shear and deflection.
(e)
If the end reactions of beam B bear on to a stiff bearing length of 50mm
at the stanchions, check the beam for web bearing and web buckling.
Stanchion ‘X’
(a) Determine the design axial load on the column due to the flooring.
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(b)
In addition to the floor loading the stanchion is required to carry a
design axial load of 3160kN from upper floors. Using the appropriate
clauses of BS 8110–1, check the suitability of a 305 × 305 × 158 UC
section if the length between restraints is 4m and the effective length L E
=0.85L.
(c)
If the UC were to be replaced by a UB section, from the safe load tables
select a suitable section if side rails were used to restrain the y–y axis at
mid-height. L E =1.0L for both axes.
Design information
Characteristic dead load due to floor units
Characteristic dead load due to self weights
and finishes
Characteristic imposed load
Deflection formulae:
Uniformly distributed load
Concentrated load at third points
2.8 kN/m2
2.2 kN/m2
5 kN/m 2
∆=5wL 4 /384EI
∆=23WL 3 /684EI
Figure 1
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Answers
Type A beam
(a)
Design load 49.8 kN/m
(b)
Maximum moment M = 398.4 kNm
Maximum shear force F v = 199.2 kN
(c )
M c = 412 kNm, P v = 647 kN
Deflection = 17.1 < span/360
Type B beam
(a)
Concentrated load 398.4 kN
(b)
Maximum moment M = 1195 kNm
(c)
Section 610 × 305 × 149 UB or 686 × 254 × 140 UB for both
M c = 1210 kNm
(d)
For 610 × 305 × 149 UC Pv = 1149 kN
deflection = 14.4 < span/360
(e)
Web bearing capacity 456 kN
Web buckling capacity 431 kN
Column
Axial design load
4255 kN
P c = 4526 kN
UB section: 686 × 254 × 170 or alternative P cy = 4700 kN, P cx = 5040 kN
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STUDY GUIDE 5
Design of masonry and timber elements
Introduction
This study guide covers Outcomes 7 and 8 of the unit.
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 timber.
On completion of the study guide you should be able to:
• Design single-leaf and cavity walls formed from bricks and/or blocks.
• Design timber floor boards, floor joists and trimmer beams.
This will involve determining the suitability of beams for bending
moment, shear force, deflection and bearing at the supports.
• Design axially loaded timber columns.
The design processes are from the following British Standards:
Brickwork and blockwork
BS 5628–1:1992 Code of practice for masonry
Part 1: Structural use of unreinforced masonry
Timber
BS 5268–2:1996 Structural use of timber
Part 2: Code of practice for permissible stress design, materials and
workmanship
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Design of masonry to BS 5628–1: 1992
Materials
Masonry is the name given to the construction consisting of brickwork (or
blockwork) and mortar. The final strength of the structural elements formed
is dependent on:
• The strength of the brick (obtained from the manufacturer’s data sheets)
• The strength of the mortar (dependent on mortar constituents and
proportions)
Bricks may be divided into the categories:
• Clay – which may be further subdivided into common, facing and
engineering
• Calcium silicate
• Concrete bricks
Standard brick format is 215 mm long × 102.5 mm wide × 65 mm high
Blocks may be defined as:
• Clay blocks
• Dense concrete blocks
• Aerated (or lightweight) concrete blocks
Sizes can vary from 100 to 200 mm wide, up to 300 mm high and 440 mm
long.
Mortar varieties are:
• Cement: lime: sand
• Masonry cement and sand
• Cement: sand with plasticiser
Mortars are as specified by their mix proportions as a ratio of cement to lime
to sand, e.g. 1:2:8 (cement: lime: sand). Cement sand mortars have strength
but accommodate movement poorly. Introducing varying amounts of lime
allows the mortar to accommodate settlement, temperature and moisture
changes to which the final structure may be subject, but has the disadvantage
of reducing the final strength of the mortar. Table 1 of BS 5628 designates
four ranges of mortars assigned (i), (ii), (iii) and (iv).
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(i) indicates a mortar with high strength but poor at accommodating
movement.
(iv) indicates a mortar good at accommodating movement but having low
strength.
Table 1 further specifies the conditions of use between the three main mortar
types:
Cement: lime: sand mortars have good bonding properties and hence high
resistance to rain penetration but low resistance to frost attack. Conversely
cement: sand with plasticiser mortars have a high resistance to frost attack
but do not bond well with the bricks or blocks and may be subject to rain
penetration. Masonry mortars attempt to provide a medium between the other
two mortar types by having better bonding properties than cements with
plasticisers and an increased resistance to frost attack over cement: lime: sand
mortars.
Design considerations
BS 5628–1
Loads
In determining the loads on elements of the structure the following
characteristic loads and their combinations are considered in this course.
Dead:
The characteristic dead load (G k ) is the weight of the structure complete with
finishes, fixtures and partitions.
Imposed:
The characteristic imposed load (Q k ) is calculated in accordance with BS
6399: Part 1, based on the activity/occupancy for which the floor area will be
used, or in accordance with BS 6399: Part 3 for roof loads.
Combinations of the above loads form the basis of the design loads used in
the analysis of the structural elements. Clause 22 of the code outlines, for
combinations of load, the design loads taken as the sum of the products of the
component loads multiplied by the appropriate partial safety factor.
For example, the combination of dead and imposed load used for the design
load is taken as the most severe condition of:
Design dead load = 0.9 G k or 1.4 G k
Design imposed load = 1.6 Q k
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Characteristic strength of masonry
Clause 23
The characteristic strength f k of any masonry wall or column must take into
consideration the following factors:
1.
Mortar type and proportions – Table 1.
2.
The brick type and strength – Table 2 (a) to (d) dependent on the brick
or block unit used and ratio of block dimensions – see clause 23.1.
3.
The horizontal cross-sectional area.
If the plan area is less than 0.2 m 2 the characteristic compressive
strength f k is multiplied by the factor:
(0.70 + 1.5A)
23.1.1
A = horizontal cross-sectional area (m2 ).
4.
The thickness of the wall
For example 23.1.2 states that the thickness of the wall (single leaf) is
equal to the width of a standard brick format. The value of f k obtained
from table 2(a) may be multiplied by 1.15
Example
Question
A brick wall is to be constructed of standard bricks having a compressive
strength of 50 N/mm 2 as specified in the manufacturer’s literature. The mortar
is required to have high strength and a good resistance to frost during
construction. The wall thickness is not of standard brick thickness. Select a
suitable mortar and hence determine the characteristic strength of the
masonry f k .
Solution
Using Table 1
Good resistance to frost and high strength is required. Therefore use a
cement: sand mortar with plasticiser – highest strength designation (ii).
Therefore use mortar designation (ii).
Using Table 2(a) masonry constructed from standard brick format
For mortar designation (ii) and brick compressive strength 50 N/mm 2
f k = 12.2 N/mm2
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It is a narrow brick wall so clause 23.1.2 does apply.
Thus the characteristic strength of the masonry f k =12.2 × 1.15 =
14 N/mm2 .
Partial safety factor for material strength
27
Clause 27 of BS 5628-1 allows the partial safety factor g m to be applied to
design procedures.
Values of g m can be found using Table 4 and are dependent on:
• Category of manufacturing control of structural units.
• Category of construction control.
In both cases control may be expressed as normal or special.
In manufacturing control, normal is taken as the standard assumed to be
supplied if not specifically stated otherwise. Special indicates that the
manufacturer has high quality control limits such that the bricks or blocks
provided consistently meet a high standard.
In construction control normal is assumed to be that obtained on site without
any rigorous supervision and testing requirements. Special indicates that
supervision and control on site are to a high level of quality control.
It is therefore possible to specify from the manufacture of the brick or blocks
special control, but on site only have normal construction control. In this case
from Table 4 this would give a value γ m of 3.1.
Detailed design consideration
Section four BS 5628-1
As with the other materials used in the design of structural elements, when
dealing with compressive members (columns and walls in the case of
masonry) slenderness is a key factor.
The slenderness ratio is calculated using:
slenderness ratio =
effective height or length
effective thickness
For normal construction this should not exceed 27.
The effective height or length is found using clause 28.3 which lists for walls,
columns and piers the effective height as a function of the clear distance
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between lateral supports and the resistance to lateral movement. This is
similar to the methods given in the structural steelwork and reinforced
concrete codes (for example L E = 1.0L in steel and l e = βl o in concrete
design).
For masonry design the explanation of the terms lateral support and enhanced
resistance are given in clause 28.2.
Lateral supports
The lateral support can be based on the horizontal or vertical dimension,
depending on whether the support is provided on a horizontal or vertical line.
Examples of the type of support given are listed below.
For this course only dead and imposed load on walls will be considered, thus
only the effective height of the structure need be calculated, as walls required
to carry wind loading are considered to span vertically.
Horizontal lateral supports
Simple resistance
For houses not exceeding three storeys, with timber floors and joists spaced
not more than 1.2 m apart, and connected by suitable joist hangers.
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Enhanced resistance
(a)
Floors or roofs of any form of construction spanning from both sides at
the same level, i.e. interior load bearing walls
(b)
In-situ concrete floors which bear on to at least half the thickness of the
wall or inner leaf of a cavity wall, but not less than 90 mm
(c)
In the case of houses of not more than three storeys, a timber floor
spanning onto a wall from one side and has a bearing not less than 90
mm
similar arrangement to above.
Effective height
28.3
Effective height of a wall with:
(1)
Simple resistance is taken as the clear distance between supports.
(2)
Enhanced resistance is taken as 0.75 times the clear distance between
supports.
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Effective thickness
28.4
The effective thickness of a wall is illustrated in Figure 3 of the code with
reference to Table 5 and is explained in further detail in clauses 28.4.1 and
28.4.2
Example
A load bearing internal single leaf wall having a clear height of 3.4 m is to be
formed using standard format bricks. It may be assumed that the floors are
formed using timber joints at 450mm centres. Determine the slenderness of
the wall.
Solution
The floor will provide enhanced resistance to the wall.
Effective height 0.75 × clear height = 0.75 × 3400 = 2550 mm
Effective thickness Table 3
Effective thickness = width of single leaf wall = 102.5 mm
Slenderness ratio =
Effective height
2550
=
= 24.9 < 27 OK
Effective thickness 102.5
Eccentricity of loading
The design of the wall must take into account any eccentricity of loading that
may occur. Clause 31 gives guidance on the application of dead and imposed
loads in walls.
For external walls with the floor bearing directly onto the wall, the load is
assumed to act at 1 / 3 the bearing depth from the loaded face.
For interior walls with continuous flooring each side of the floor, the load
may be taken as bearing on to 1 / 2 the width of the wall and each portion is
assumed to act at 1 / 3 the bearing depth from the loaded face.
For joist hangers the load is assumed to act at the face of the wall.
Loads from upper storeys are assumed to act axially on the wall in all cases.
The resultant eccentricity, e x , is calculated by equating moments due to the
applied loading with the moment due to the total load on the wall, and is
expressed in terms of the wall thickness t.
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Design procedure for loading
32
The actual load on the wall is compared with the design vertical resistance of
the wall. Clause 32.2.1 gives the formula for calculating the design vertical
resistance of a wall per unit length and the definitions of the terms.
β.t.f k
γm
β – capacity reduction factor allowing for the effects of slenderness and
eccentricity, and is obtained from table 7
t – thickness of the wall
f k – characteristic strength of the masonry
γ m – partial safety factor for the material
Example
Single leaf wall
A single leaf internal brick wall of a traditionally built house is subject to the
loading given in the design data. It may be assumed that the floor consists of
timber joists at 450mm centres spanning across the wall.
Check the suitability of the wall in compression.
Design data
Axial design load from upper floor
Characteristic imposed loading on floors
Characteristic dead load inclusive of self-weight of floors
Span of floor on left hand side of wall
Span of floor on right hand side of wall
Clear height of wall
Compressive strength of standard brick units
Mortar designation
Category of manufacturing control of structural units
Category of construction control
Solution
Calculate design loads
Upper floors
Design load on floor
Load from l.h.s floor
Load from r.h.s floor
48 kN/m
1.5 kN/m2
0.6 kN/m2
3m
4m
2.7 m
10 N/mm 2
(iv)
normal
normal
48 kN/m
=
=
=
=
1.4 × 0.6 + 1.6 × 1.5
3.24 kN/m2
3.24 × 3/2
R 1 = 4.9 kN/m
3.24 × 4/2
R 2 = 6.5 kN/m
R T = 59.4 kN/m
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Find eccentricity at top wall
The eccentricity from the timber floors is assumed to be applied at t/3
Applying
R T .e x =
59.4 × e x =
=
SR.t/3
6.5 × t/3 – 4.9 × t/3
0.009t
Slenderness
28
Timber floor spanning across the wall would be assumed to give enhanced
resistance (see 28.2.2.2(a)).
Effective height = 0.75 × clear height = 0.75 × 2.7 = 2025 mm
From Figure 3 effective thickness of columns and walls effective thickness, t
= 102.5 mm
Slenderness ratio =
effective height
2025
=
= 19.8 < 27 suitable
effective thickness 102.5
Design vertical resistance of a wall per unit length
32.2
β – from Table 7 for slenderness =19.8 and e x = 0.009 t
As e x < 0.05 t, eccentricity is taken as 0.05 t
Interpolating for
Slenderness
18
20
19.8
β
0.77
0.70
0.706
Material safety factor from Table 4 for normal conditions of manufacture and
construction control γ m is 3.5.
f k – characteristic strength of masonry
23
The compressive strength from the manufacturer’s data 10 N/mm 2 and mortar
designation is (iv) from Table 2.
f k = 3.5 N/mm2
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This can be multiplied by 1.15, as it is a single leaf brick wall that is equal in
width to the width of a standard format brick (see cl.23.1.2)
f k = 3.5 × 1.15 = 4.0 N/mm2
Resistance of wall
β.t.f k
γm
0.706 × 102.5 × 4.0
= 82.7N/mm = 82.7kN/m
3.5
As this value exceeds the total design load on wall (59.4 kN/m), wall is
suitable.
Cavity walls under compression
The design process is the same as that for the single leaf wall but additional
points listed below should be considered:
• Floor loadings are normally only carried on the inner leaf.
• The tabulated values of brick and block arrangements are acceptable:
Outer leaf
brick
brick
block
Inner leaf
brick
block
block
• The two leaves of masonry are connected by ties in accordance with clause
29.1 and Table 6 of BS 5628.
• As the walls are connected by ties, the effective height of the outer leaf is
taken as being the same as that of the inner leaf.
• The rules of Figure 3 are used to obtain the effective thickness:
the greatest of 2/3(t 1 + t 2 )
or t 1
or t 2
• In design, each leaf is considered separately
The slenderness ratio calculations are based on both leaves.
The strength calculations are based on a single leaf.
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Example
Cavity wall
The external wall of a traditionally built low-rise structure is shown below.
Using the design data, check the suitability of the prescribed concrete blocks
and select a suitable compressive strength of brick.
Design data:
Compressive strength of units:
Inner leaf – block 150 mm thick × 225 mm
high × 440 mm long
Characteristic loads from upper floor on inner leaf:
Dead
Imposed
Imposed loading on floors
Dead load inclusive of self weight of floors
Span of pre-cast concrete units forming floor
(units bear in inner leaf only)
Clear height of wall
Mortar designation
Category of manufacturing control of structural units
Category of construction control
Solution
Calculate design loads
Outer leaf – axial
1.4 × 13 + 1.6 × 6
Inner leaf
Axial
1.4 × 40 + 1.6 × 18
Floor
(1.4 × 3.5 + 1.6 × 3) × 4.6/2
Total load
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5 N/mm2
40 kN/m
18 kN/m
3 kN/m 2
3.5 kN/m 2
4.6 m
3.15 m
(iii)
normal
normal
27.8 kN/m
84.8 kN/m
22.3 kN/m
R1 =
R T = 107.1 kN/m
S TU DY GU ID E 5
Find eccentricity at top wall
The eccentricity from the floor may be taken as t / 3
Applying
R T e x = R 1 t/3
107.1 × e x = 22.3 × t/3
22.3t
ex =
= 0.07t
107.1 × 3
Slenderness
28
Pre-cast floor bearing on to the wall would be assumed to give enhanced
resistance, see 28.2.2.2(a).
Effective height = 0.75 × clear height = 0.75 × 3.15 m = 2362.5 mm
From figure 3 effective thickness greater
(a) 2/3 (t 1 + t 2 ) = 2/3(102.5 + 150) =
(b) t 1
(c) t 2
of:
168.33 mm
102.5 mm
150 mm
Effective thickness =168.33 mm
Slenderness ratio =
effective height
2362.5
=
= 14 < 27 suitable
effective thickness 168.33
Calculations for inner leaf
Design vertical resistance of a wall per unit length
32.2
b from Table 7 for slenderness ratio =14 and e x = 0.07t
Eccentricity
Slenderness of 14
0.05 t
0.89
0.1 t
0.83
Interpolating for 0.07t
0.89 – 0.02/0.05 × 0.06 = 0.866
Material safety factor from Table 4 for normal conditions of manufacture and
construction control γ m = 3.5
f k – characteristic strength of masonry
23
The compressive strength from the manufacturer’s data 5 N/mm 2 and mortar
designation is (iii).
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Which Table 2 is to be used?
As solid concrete blocks are used with a height to least horizontal dimension
ratio = 225/150 = 1.5
Clause 23.1.6 states that the value of f k should be obtained by interpolation
between the values given in tables 2 (b) and 2(d)
Table 2(b)
f k = 2.5 N/mm2
Table 2(d)
f k = 5.0 N/mm2
f k = 3.75 N/mm2
Resistance of wall
β.t.f k
γm
0.866 × 150 × 3.75
= 139.2N/mm = 139.2kN/m
3.5
As this value exceeds the total design load on wall (107.1 kN/m), wall is
suitable.
Design of masonry walls with piers
28.1
A pier is a thickened section forming an integral part of the masonry
construction. Piers are placed at regular intervals along the wall and have the
advantage of reducing the slenderness ratio of the wall and hence increasing
its load carrying capacity. Piers may be introduced into single leaf or cavity
walls.
Plan view of cavity wall with piers at constant centres
The design procedure is the same as that for a single leaf or cavity wall, with
an additional consideration given in Table 5 when determining the effective
thickness of the wall.
Example
The diagram below shows the outline of a cavity wall with piers at 3m
centres. The wall is to be formed from blockwork 100 mm thick for both
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leaves with the piers being an additional 100 mm thick and 300 mm wide.
The cavity between the inner and outer leaves may be assumed to be 50 mm.
Check the suitability of the inner leaf to carry the loads given in the design
data. It may be assumed that the floor units bear over the entire width of the
inner leaf. The height of the wall may be considered as being 4.2m.
Design data:
Structural units: concrete block
100 mm thick × 200 mm high × 300 mm long
Loads from upper floor:
Inner leaf – dead
imposed
Imposed loading on floor
Dead load inclusive of self weight of floors
Span of pre-cast concrete units forming floor
Mortar designation
Category of manufacturing control of structural units
Category of construction control
7 N/mm2
30 kN/m
20 kN/m
5 kN/m 2
4 kN/m 2
3.5 m
(ii)
normal
normal
Solution
Note: for simplicity the outer leaf calculations have been omitted – they
would be carried out using the procedure adopted in the previous example.
Calculate design loads
Inner leaf
Axial
1.4 × 30 + 1.6 × 20
Floor
(1.4 × 4 + 1.6 × 5) × 3.5/2
Total load
74 kN/m
R 1 = 23.8 kN/m
R T = 97.8 kN/m
Find eccentricity at top wall
The eccentricity from the floor may be taken as t/3
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Applying
R T .e x = SR.t/3
97.8 × e x = 23.8 × t/3
23.8t
ex =
= 0.08t
99.8 × 3
Slenderness
28
Pre-cast floor bearing on to the wall would be assumed to give enhanced
resistance, see 28.2.2.2(a).
Effective height = 0.75 × clear height = 0.75 × 4.2 m = 3150 mm
Coefficient K from Table 5
Ratio pier spacing to
pier width = 3000/300
= 10
Ratio t p /t 2 = 200/100 = 2
Value of K from Table 5
K=1.2 (this value may require to
be interpolated in some
examples)
From Figure 3 effective thickness greater of:
(a)
(b)
(c)
2/3 (t 1 + Kt 2 ) = 2/3(100 + 1.2 × 100)
t1
Kt 2 = 1.2 × 100
= 146.7 mm
100 mm
120 mm
Effective thickness =146.7 mm
Slenderness ratio =
effective height
3150
=
= 21.5 < 27 suitable
effective thickness 146.7
Design vertical resistance of a wall per unit length
β from Table 7 for slenderness ratio = 21.5 and e x = 0.08 t
Eccentricity
Slenderness of 20
Slenderness of 22
0.05 t
0.70
0.62
Interpolating for 0.08t
β = 0.60
114
0.1 t
0.64
0.56
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Material safety factor from Table 4 for normal conditions of manufacture and
construction control γ m = 3.5
f k – characteristic strength of masonry
23
The compressive strength from the manufacturer’s data 7 N/mm 2 and mortar
designation is (ii).
Which Table 2 is to be used?
As solid concrete blocks are used with a height to least horizontal dimension
ratio = 200/100 = 2
Use Table 2(d) for unit strength 7 N/mm 2 and mortar designation (ii)
f k = 6.4 N/mm2
Resistance of wall
β.t.f k
γm
0.60 × 100 × 6.4
= 109.7N/mm = 109.7kN/m
3.5
As this value exceeds the total design load on wall (97.8 kN/m), wall is
suitable.
Structural design of timber to BS 5268–2:1996
The structural design of timber elements is based on permissible stresses and
deflections derived from elastic theory.
Flexural members – members subject to bending
Members subject to bending (i.e. beams) are assumed to behave in accordance
with elastic bending theory provided that the permissible material stresses are
not exceeded. The bending expression can be applied to timber design:
M/I =f/y =E/R
At any point across a section of a beam which is located a distance y from the
neutral axis of a section, a stress f will be developed as a consequence of
applying a bending moment M to the section. The magnitude of the stress
developed will vary with the second moment of area of the section I.
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In the timber design code, σ is the designation for stress, hence the above
equation may be written as:
M/I = σ/y
In timber design generally rectangular sections are used, therefore the
maximum compressive and tensile bending stress will occur at the extreme
fibres. Thus y is equal to half the depth of the section.
As both I and y are geometric properties of the section it is convenient to
combine the two terms in a single property which is referred to as the elastic
modulus and denoted by the symbol Z.
Z = I/y
Further, as rectangular sections are being considered, if b is the width of the
section and h the depth then I and y may be expressed as:
I = bh 3 /12 and y =h/2
Hence elastic modulus, Z = (bh 3 /12)/(h/2) = bh 2 /6
Considering, from the bending expression, M/I = σ/y and combining with the
definition of elastic modulus then:
M = σZ
This can be rearranged to determine the maximum bending stress in the beam
and then compared with the permissible stress that the beam may carry.
Other design requirements
The other checks required to timber beams are shear, bearing, deflection and
the maximum depth to breadth ratio.
As the beams are simply supported, and generally carry uniformly distributed
loads, shear will be a maximum at the supports.
Maximum shear stress = –
3
Load
×
2
Cross-sectional area
The end bearing area is dependent on the contact area with the beam support.
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Maximum bearing stress =
Load
Load
=
Contact area
width of beam × bearing length
Both of these values can then be compared with the permissible values
obtained from the design code.
Deflection, as with other structural materials, is a serviceability requirement.
Maximum deflections are determined using the standard deflection formulae
and compared with the deflection limits given in the design code.
Timber as a structural material
Unlike other construction materials, timber cannot be mixed to a predetermined formula. The cut wood has to be inspected and graded by visual
or mechanical means. The design code allows for a number of strength
classes based on the inspection of the timber, or alternatively, if the species
of timber is known it may be classified as given in Table 8 of the code
according to its standard name.
Appropriate grade stresses are assigned to the graded timber. For flexure the
appropriate grade stresses are:
•
•
•
Bending parallel to the grain
Compression perpendicular to the grain
Shear parallel to the grain
Account must also be taken of the loading and exposure conditions that the
timber will be subject to. The design code lists almost thirty factors that can
be applied to the grade stresses. Only a few will be of concern in this course.
Modification factors
Moisture content of timber related to service class.
It is difficult to artificially dry solid timber more than 100 mm thick, unless it
is specially dried. BS 5268 recognises three services classes that are related
to the conditions of end use. Service classes 1 and 2 generally require the
timber to be artificially dried and the dimensions and properties of the dried
timber can be taken as the grade values. Service class 3 timber is used when
the finished structure is fully exposed or if the timber is more than 100 mm
thick. In this case the grade values must be modified by a factor K 2 found in
Table 13 which allows for the differing load carrying mechanisms of wet and
dry timber.
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Service
class
Examples of end use of
timber
3
2
2
1
External & fully exposed
Covered and unheated
Covered and heated
Internal use and
continuously heated
Average
moisture
content
> 20 %
18 %
15 %
18%
12%
Moisture content in
each piece at time
each piece at time
24 %
20 %
24%
20%
Service classes 1 and 2 use unmodified stresses and moduli.
Service class 3 timber uses modified stresses and moduli.
i.e. K 2 =1
i.e. K 2 <1
Duration of loading K 3
The grade stresses based on the strength classes of the timber apply to longterm loading on the structural element. Table 14 gives a modification factor
for various load durations and list values of K 3 varying from 1.0 to 1.75. K 3
is applied to the grade stresses only and does not apply to the modulus of
elasticity.
Load-sharing systems K 8
A load-sharing system may be considered as being, for example, a series of
four or more floor joists connected by flooring in such a way that act together
– a standard timber floor. Provided that the joists are no farther apart than
610 mm centres then the grade stresses should be modified by the
modification factor K 8 =1.1.
For all other systems K 8 may be taken as being equal to 1.0.
For load-sharing systems the mean modulus of elasticity should be used to
calculate any deflections except in circumstances where dynamic loads may
occur, e.g. gymnasia, where the minimum value should be used.
Depth factor K 7
The grade stresses based on the strength classes of the timber apply to
materials having a depth (h) of 300 mm. A modification factor K 7 is applied
to the grade bending stress of beams having a depth other than 300 mm.
For solid timber beams:
Depth of beam (h) mm
72 mm or less
72 > h < 300
h > 300
118
K 7 value
1.17
(300/h) 0.11
0.81(h 2 + 92300)
(h 2 + 56200)
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Notching of beams K 5
Figure 2
Notching the end of a beam for construction purposes causes stress
concentrations that must be allowed for in the shear calculation. The shear
stress should be calculated by using the effective depth (h e ) shown in Figure
2. The grade shear stress should be multiplied by a modification factor K 5 to
obtain the permissible stress.
For beams notched on the underside K 5 = h e /h
Note: Beams with notches on the top edge are not considered in this unit.
Deflection
The deflection is acceptable if the deflection of the fully loaded beam does
not exceed 0.003 times the span of the member or 14mm whichever is the
lesser.
Timber flexural members design examples
Boarding
Check the suitability of 20mm tongued and grooved floor boarding spanning
between 50mm × 250mm timber joists at 600mm centres. The boards are of
strength class C14.
Note that boarding is normally provided in lengths up to 3m long. Each
board spans over a number of joists and for analysis purposes may be treated
as a continuous beam.
The maximum moment occurs at an internal support and may be found using
M = wL 2 /10. The maximum shear force (reaction) occurs at the outside
support and may be taken as V= 0.4wL.
Additional data:
Dead load inclusive of self-weight of boards
Imposed load
0.15 kN/m2
1.5 kN/m2
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Solution
Consider a width of boarding (b) = 1m
b=1000 mm
(Actual width of floor is immaterial if width of one metre is assumed)
Length between supports
L = 600mm = 0.6m
Load on boarding
w =dead + imposed = 0.15 + 1.5 = 1.65 kN/m2
Considering a typical 1m width of board b=1.65 × 1 = 1.65 kN/m
Bending
Maximum moment M = wL 2 /10 = 1.65 × 0.6 2 /10 = 0.06 kNm
Elastic modulus of board Z = bh 2 /6 = 1000 × 20 2 /6 = 66667 mm 3
Actual bending stress, s = M/Z = 0.06 × 10 6 /66667 = 0.9 N/mm2
Permissible stress =
grade bending stress parallel to the grain × K 2 × K 3 × K 7 × K 8
Grade stress from Table 7 – C14 = 4.1 N/mm 2
K 2 – wet stresses modification factor – material 20 mm
thick – service class 1
K 3 – duration of loading – on floor this may be taken
as long term
K 7 – depth factor – less than 72 mm
K 8 – load-sharing – boards are load-sharing
K2 = 1
K3 = 1
K 7 = 1.17
K 8 = 1.1
Permissible stress = 4.1 × 1.0 × 1.0 × 1.17 × 1.1
= 5.28 N/mm2 > 0.9 N/mm2 boards suitable in bending
Shear
Maximum shear force V = 0.4wL = 0.4 × 1.65 × 0.6 = 0.4 kN
Maximum shear stress v =
120
3V
3 × 0.4 × 103
=
= 0.03 N/mm2
2bh
2 × 1000 × 20
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Permissible stress =
grade shear stress parallel to the grain × K 2 × K 3 × K 8
Grade stress from Table 7 – C14 = 0.6 N/mm 2
The modification factors used for bending are still applicable – except K 7 that
is applied to bending only.
Permissible stress = 0.6 × 1.0 × 1.0 × 1.1
= 0.66 N/mm2 > 0.03 N/mm 2 boards suitable in shear
Deflection
Considering the beam as continuous, ∆ = wL 4 /384EI
E mean from Table 7 – C14 (one board cannot act on its own) E = 6800 N/mm 2
I =bh 3 /12 = 1000 × 20 3 /12 = 666667 mm4
∆ = wL 4 /384EI = 1.65 × 600 4 /(384 × 6800 × 666667) = 0.13 mm
Permissible deflection (clause 2.10.7) = 0.003 × span = 0.003 × 600 = 1.8 mm
Actual deflection less than permissible – beam is suitable.
Floor joists
The floor joists for the boarding example above also require to be checked. It
may be assumed that the joists are simply supported over a span of 3.6 m and
bear on to blockwork supports 100 mm wide. The revised dead load to
include for the self-weight of the beam may be taken as 0.34 kN/m2 . The
joists are strength class C16.
Solution
Centres of joists
Load/joist
600mm = 0.6m
w = (dead + imposed) × centres
= (0.34 + 1.5) × 0.6 = 1.1 kN/m
Bending
For a simply supported beam
Maximum moment M = wL 2 /8 = 1.1 × 3.6 2 /8 = 1.78 kNm
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Elastic modulus of board Z = bh 2 /6 = 50 × 250 2 /6 = 520833 mm 3
Actual bending stress, σ = M/Z = 1.78 × 10 6 /520833 = 3.42 N/mm 2
Permissible stress = grade bending stress parallel to the grain × K 2 × K 3 ×
K7 × K8
Grade stress from Table 7 – C16 = 5.3 N/mm 2
K 2 – wet stresses modification factor – material 20 mm
thick – service class 1
K 3 – duration of loading – on domestic floor this may be
taken as long term
K 7 – depth factor (clause 2.10.5)
K 7 = (300/h) 0.11 = 300/250) 0.1 = 1.02
K 8 – load-sharing – boards are load-sharing
K 2 =1
K 3 =1
K 7 =1.02
K 8 =1.1
The assumption is that the floor boards are of sufficient length to distribute
the load over at least four joists.
Permissible bending stress
= 5.3 × 1.0 × 1.0 × 1.02 × 1.1
= 5.95 N/mm2 < 3.42 N/mm 2
Beam satisfactory in bending.
Shear
Maximum shear force V = wL/2 = 1.1 × 3.6/2 = 1.98 kN
3V
3 × 1.98 × 103
Maximum shear stress v =
=
= 0.24 N/mm 2
2bh
2 × 50 × 250
Permissible stress = grade shear stress parallel to the grain × K 2 × K 3 × K 8
Grade stress from Table 7 – C16 = 0.67 N/mm 2
The modification factors used for bending are still applicable
(K 7 is only applicable to bending)
Permissible stress = 0.67 × 1.0 × 1.0 × 1.1
= 0.74 N/mm2 > 0.24 N/mm 2
122
joist suitable in shear
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Bearing
Value of reaction = wL/2 = 1.1 × 3.6/2 = 1.98 kN
Joist bears on to a 100mm wide support and width of joist is 50mm
Reaction
1.98 × 103
=
= 0.39 N/mm 2
Actual bearing stress =
Bearing length × width
100 × 50
Permissible stress = compression perpendicular to the grain × K 2 × K 3 ×
K8
Grade stress from Table 7 – compression perpendicular to the grain – 2.2
N/mm 2
Two values of compression perpendicular to the grain are given in Table 7.
Which value should be used? Reference should be made to Note 1 of the
table.
The modification factors used for shear are still applicable
Permissible stress = 2.2 × 1.0 × 1.0 × 1.1
= 2.42 N/mm2 > 0.39 N/mm 2 bearing length is suitable
Deflection
As the beam is simply supported, ∆ = 5wL 4 /384EI
E mean from Table 7 – C14 (one board cannot act on its own) E = 8800 N/mm 2
I =bh 3 /12 = 50 × 250 3 /12 = 65.1 × 10 6 mm4
∆ = 5wL 4 /384EI = 5 × 1.1 × 3600 4 /(384 × 8800 × 65.1 × 10 6 ) = 4.2 mm
Permissible deflection (clause 2.10.7) = 0.003 × span £ 14 mm = 0.003 ×
3600 = 10.8 mm
Actual deflection less than permissible – beam suitable.
Notches
If the beam is notched at the support, then the shear cross-sectional area is
reduced and the modification factor K 5 applies (see clause 2.10.4).
Consider the above beam with a 75mm notch on the underside.
Dimension h e = h – 75 = 250 –75 = 175 mm
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Maximum shear stress v =
3V
3 × 1.98 × 103
=
= 0.34 N/mm 2
2bh
2 × 50 × 174
K5 = he/h = 175/250 = 0.7
Permissible stress = grade shear stress parallel to the grain × K 2 × K 3 × K 8
× K5
Permissible stress = 0.67 × 1.0 × 1.0 × 1.1 × 0.7
= 0.52 N/mm2 > 0.34 N/mm 2
Joist is still suitable in shear.
Timber compression members
As with all structural materials, the design of compression members is
dependent on the slenderness ratio.
Where the slenderness ratio, λ = L e /i
L e = effective length is found using Table 18, which lists for conditions of
end restraint, the ratio of L e /L, where L is the actual length.
Values given for L e are 0.7L, 0.85L, 1.0L, 1.5L and 2 L.
i is the radius of gyration of the section. As only solid rectangular sections
will be dealt with, there are two possible axes of buckling, x–x and the y–y.
Hence there are two values of slenderness ratio:
λ x = L ex /i x
The radius of gyration i x = √I x /A
λ y = L ey /i y
i y = √I y /A
Where I (for a rectangular section) = bh 3 /12 Ix =bh 3 /12 and Iy =hb 3 /12
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Area A = bh
considering x–x axis
considering y–y axis
The critical slenderness ratio is the larger of the two
In no case should the slenderness ratio exceed 180 (see clause 2.11.4)
The permissible stress is based on the comments of clause 2.11.5 which gives
two design procedures:
1.
2.
Compression members with slenderness ratios less than 5 (short
columns)
Compression members with slenderness ratios greater than 5 (slender
columns)
In both cases the permissible stress is taken as the grade compression stress
parallel to the grain multiplied by the modification factors for moisture
content, duration of loading and load sharing.
Permissible stress = grade stress parallel to the grain × K 2 × K 3 × K 8
In addition for members with a slenderness greater than 5, the above formula
is multiplied by K 12 given in Table 19.
Factor K 12 varies with slenderness ratio as calculated above and with
E/σ c ,″
where E = minimum modulus of elasticity of the material, and
σ c ,″ = compression parallel to the grain.
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Example
Single column
A timber column 200mm × 200mm is required to carry a load of 210 kN. The
load has been transferred to the column by timber joists such that the end
restraint conditions top and bottom may be taken as restrained in position but
not in direction. The height of column is 2.8 m and the timber may be taken
as strength class C27. The load may be considered as short term.
Solution
As timber is greater than 100mm thick it would be difficult to dry the section,
so use wet stresses. Values found in Table 7 are modified by factor K 2 found
in Table 13
From Table 7
σ c ,″ = compression parallel to the grain = 8.2 N/mm2
E min = 8200 N/mm 2
L e =1.0L = 2800 mm
K 2 = 0.6
K 2 = 0.8
I = bh 3 /12 = 200 × 200 3 /12 = 1.333 × 10 8 mm4
A = bh = 200 ´ 200 = 40000 mm 2
i= √I/A = 57.7 mm
l = L e /i = 2800/57.7 = 48.5 (for both axes) < 180 suitable
Ratio E/σc ,″ = (8200 × 0.8)/(8.2 × 0.6) = 1333.3 {modified by factor K 2 }
From Table 19:
40
0.809
0.811
1300
1400
50
0.757
0.760
Modification factor K 12 for λ = 48.5 and E/ σc ,″ =1333.3
K 12 = 0.767
Alternatively for Table 19 an equivalent slenderness L e /b may be used for
rectangular sections, in this example 2800/200 = 14
From Table 19:
1300
1400
as before.
126
11.6
0.809
0.811
14.5
0.757
0.760
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S TU DY GU ID E 5
K 3 for short term loading = 1.5
K 8 for non load-sharing member = 1.0
Permissible stress = grade stress parallel to the grain × K 2 × K 3 × K 8 × K 12
= 8.2 × 0.6 × 1.5 × 1 × 0.767
= 5.66 N/mm2
Actual compressive stress = Load/Area = 210 × 10 3 /40000 = 5.25 N/mm 2
As this is less than 5.66 N/mm2 the column is suitable.
Example
Column forming part of a partition wall
A timber column of 72mm × 168mm cross-section supports a medium term
axial load of 24 kN. The column forms part of a partition wall that is 3.9 m
high and the columns are arranged such that there is no load sharing. The
column is restrained in position only top and bottom and is provided with
restraining side rails at the third points about the weaker axis. Check the
suitability of strength class C22 to carry the load.
Solution
As there are two differing effective lengths and hence two different
slenderness ratios, the critical axis must be identified
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L e = 3.9m
I = bh 3 /12
i= √I/A
l = L e /I
Ix =
L e =1.3m
72 × 168
12
3
Iy =
168 × 723
12
28.45 × 10 6 mm 4
5.23 × 10 6 mm 4
√28.45 × 10 6 /(168 × 72)
√5.23 × 10 6 /(168 × 72)
48.5 mm
20.8 mm
3900
= 80.4
48.5
1300
= 62.5
20.8
Critical axis for buckling is the x–x axis
Section is less than 100mm thick so service class 1 or 2 applies (K 2 = 1.0)
From Table 7
σ c ,″ = compression parallel to the grain = 7.5 N/mm2
E min = 6500 N/mm 2
Ratio E/ σc ,″ = 6500/7.5 = 867
From Table 19 for the ratio value of 867 and l = 80.4
K 12 = 0.51
K 3 for medium term loading = 1.25
K 8 for non-load sharing member = 1.0
Permissible stress = grade stress parallel to the grain × K 2 × K 3 × K 8 × K 12
= 7.5 × 1.0 × 1.25 × 1 × 0.51
= 4.78 N/mm2
Actual compressive stress
= Load/Area = 24 × 10 3 /(72 × 168)
= 1.98 N/mm2
As this is less than 4.78 N/mm2 the column is suitable.
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Study Guide 5 End Test
Timber design
In order to gain more space for selling goods a retailer has decided to have a
mezzanine floor erected in High Street premises. The retailer has requested
that the construction should be timber, be left exposed, and be a feature of the
premises.
Figure 1 shows the proposed mezzanine plan and part section. A trimmer
beam will support the floor joists internally, and externally the joists will be
supported by the existing brick walls.
All timber members are shown as dressed sizes.
Floor joists
(a) If the floor joists are at 600mm centres, determine the design loading on
a typical member.
(b)
Check the suitability in bending and shear of an 87mm × 216mm
section.
Trimmer beam
(a) Treating the loading from joists to be uniformly distributed, determine
the maximum bending moment and shear force in the trimmer beam.
(b)
Given that the floor joists provide full lateral restraint to the trimmer,
check the suitability of a 121mm × 321mm section in bending and
deflection.
(c)
If the beam bears on to 100mm wide brick at the external supports
check its suitability in bearing.
(d)
In order to reduce the overall depth of the floor construction a 70mm
notch is to be cut out of the trimmer beam as shown in the part section.
Check the suitability of the beam in shear at the column support
Column
(a) Determine the design loading on the column.
(b)
Check the suitability of the 121mm × 121mm section if the effective
length of the column can be taken as 3m.
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Design data
Duration of loading
Timber grade
Dead loading including self-weight of timber
Imposed loading
Figure 1
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long term
C27
1.5 kN/m2
4 kN/m 2
S TU DY GU ID E 5
Masonry design
Figure 2 shows the layout of a derelict city centre building, the existing
external walls of which are incapable of resisting any additional loads. Your
client has recently purchased the building with a view to renovating it. The
local authority has insisted that the existing facade must remain in place if
possible. After a structural survey had been carried out on the building it was
discovered that one wall was in such a bad state of repair that it must be
demolished.
In consultation with the client and his architect it has been decided that an
inner carcass of brick work or block work will be used to transmit the loads
from the structure to the foundations and a new cavity wall would replace the
wall that has to be demolished. The new masonry will be tied to the existing
facade to provide it with a degree of stability but there will be no load
interaction between the existing and new work.
The client has a supply of bricks that he wishes to use on the contract.
However the architect feels that 150mm blocks for the cavity wall and
200mm blocks for the inner carcass wall would be a better arrangement.
Two designs are thus required for the new masonry adjacent to the facade and
for the wall replacing the demolished side.
Considering the walls that are to be designed to have a clear height of 3.6m
and that the floors framing into the walls provide enhanced resistance and
using relevant design information given:
New masonry adjacent to the facade (single skin masonry design)
Brick
(a) Determine the design axial load per metre on a single skin brick wall
with piers every 3m.
(b)
Check the suitability of this arrangement to carry the design load.
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Block
(c) Determine the design axial load per metre on a 200 mm block wall.
(d)
Choose a suitable compressive strength of block and a suitable mortar
designation to carry the design load.
Replacement wall (cavity masonry construction)
Brick
(a) Determine the design axial load per metre on a cavity brick wall with
piers every 1.5m.
(b)
Check this arrangement of the brick wall to carry the design load.
Block
(c) Determine the design axial load per metre on the 150 mm block cavity
wall
(d)
Choose a suitable compressive strength of block and a suitable mortar
designation to carry the design load. The cavity width between the
skins may be taken as 60 mm.
Design data:
Characteristic dead due to flooring
Characteristic imposed load
Characteristic loads from upper floors:
Single skin walls
For brick wall
For block wall
Cavity wall
Brick
Inner leaf
132
2.6kN/m 2
3kN/m 2
dead
imposed
dead
imposed
40kN/m
50kN/m
44kN/m
50kN/m
dead
imposed
80kN/m
66kN/m
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Block
Inner leaf
dead
imposed
84kN/m
66kN/m
Manufacturing control and construction control can be assumed to be normal
throughout
Bricks supplied by client – standard format – compressive strength of unit
50 N/mm2
Mortar designation (iii)
Proposed blocks – solid concrete 440mm long × 215mm high × 150 or 200
thick.
Figure 2
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Answers
Timber
Floor joists
Design loading
3.3 kN/m
Max. moment
6.6 kNm
Max. shear force
6.6 kN
Permissible bending stress
11.4 N/mm 2 actual bending stress 9.76 N/mm2
Permissible shear stress
1.21N/mm2 actual shear stress 0.53 N/mm2
Trimmer beam
Design loading
11 kN/m
Max. moment
17.8 kNm
Max. shear force
19.8 kN
Permissible bending stress
9.9 N/mm 2 actual bending stress 8.6 N/mm2
Permissible shear stress
0.93N/mm2 actual shear stress 0.9 N/mm2
Permissible bearing stress
2.75N/mm2 actual bearing stress 1.63 N/mm2
Column
Design load 39.6 kN
Permissible compressive stress
4.13 N/mm2
actual compressive stress 2.7 N/mm2
Masonry
Floor loading 27.37 kN/m
Single leaf brick
Total load
163.34 kN/m
minimum eccentricity
Slenderness
23.9
Resistance
164.5 kN/m
Single leaf block
Total load
169 kN/m
minimum eccentricity
Slenderness
13.5
Compressive strength of unit 35 N/mm 2
Minimum f k = 6.42 N/mm2
Cavity wall brick
Inner leaf
Total load
245 kN/m
minimum eccentricity
Slenderness
17
Resistance
245.2 kN/m
Cavity wall block
Inner leaf
Total load
250.5 kN/m
minimum eccentricity
Slenderness
13.5
Compressive strength of unit 10 N/mm 2
Minimum f k = 3.23 N/mm2
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