THEORY OF STRUCTURES
THEORY OF STRUCTURES
TAKSIAH A. MAJID
CHOONG KOK KEONG
MUSTAFASANIE M. YUSSOF
PENERBIT UNIVERSITI SAINS MALAYSIA
PULAU PINANG
© Penerbit Universiti Sains Malaysia, 2013
EPUB, 2014
Perpustakaan Negara Malaysia
Cataloguing-in-Publication Data
Taksiah A. Majid
Theory of structures / Taksiah A. Majid, Choong Kok Keong, Mustafasanie M. Yussof
Includes index
ISBN 978-983-861-567-9 (Print)
e-ISBN 978-983-861-722-2
1. Structural engineering. 2. Structural design.
Mustafasanie M. Yussof. III. Title.
624.17
I. Choong, Kok Keong.
Copy Editor: Nur Naimah Jaafar
Cover Designer: Intan Suhaila Kassim
Proofreader: Rosni Habib
Typesetter: Norizan Mohammad Noor
Published by Penerbit Universiti Sains Malaysia (Universiti Sains Malaysia Press),
11800 USM Pulau Pinang, Malaysia.
II.
A member of the Malaysian Scholarly Publishing Council (MAPIM).
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CONTENTS
Preface
Chapter 1
Basic Concept of Structural Analysis
1.1 Introduction
1.2 Objectives
1.3 Types of Structures
1.4 Types of Supports
1.5 Reaction Forces
1.6 Equation of Equilibrium
1.7 Free Body Diagram
1.8 Statical Determinacy
1.9 Computation of Reactions Using Equations of Equilibrium
Exercises
Chapter 2
Analysis of Plane Trusses
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Introduction
Objectives
Common Types of Trusses
Classification of Plane Trusses
Determinacy and Stability of Plane Trusses
Sign Convention and Member Forces Representation
Method of Joints
2.7.1 Identification of zero-force members
2.8 Method of Sections
2.8.1 Procedure of analysis
2.9 Introduction to Other Types of Planar Structures
2.9.1 Planar trusses: Compound and complex
2.9.2 Space trusses
Exercises
Chapter 3
3.1
3.2
3.3
3.4
3.5
Analysis of Statically Determinate Beams and Frames
Introduction
Objectives
Internal Forces in Beams
Sign Convention and Computation of Shear Force and Bending Moment
Shear Force and Bending Moment Diagrams for Beams
3.5.1 Characteristics of SFD and BMD
3.5.2 Qualitative deflected shape
3.6 Frame
3.7 Statical Determinacy of Frames
3.8 Analysis of Plane Frames
Exercises
Chapter 4
Analysis of Cables
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Introduction
Objectives
Scope of Analysis
Types of Cables
Symmetrical Cable Subjected to Uniformly Distributed Load
Forces in Cables Subjected to Uniformly Distributed Load
Asymmetrical Cable Subjected to Uniformly Distributed Load
Length of Cables
4.8.1 Cables loaded with point loads
4.8.2 Symmetrical cables
4.8.3 Asymmetrical cables
4.9 Support Conditions
Exercises
Chapter 5
Analysis of Three-Hinged Arch
5.1 Introduction
5.2 Objectives
5.3 Comparison Between a Beam and Arch
5.3.1 Advantage of the arch
5.4 Determinate and Indeterminate Arches
5.5 Equation of a Parabolic Arch
5.6 The Bending Moment in the Arch Rib
5.7 The Thrust and Shear Force in the Arch Rib
Exercises
Chapter 6
Deflection
6.1
6.2
6.3
6.4
6.5
6.6
Introduction
Objectives
Elastic-Beam Theory
Direct Integration Method
Moment-Area Method
Conjugate Beam Method
6.6.1 Conjugate beam support
6.6.2 Procedure for analysis
Exercises
Chapter 7
Influence Lines
7.1 Introduction
7.2 Objectives
7.3 Influence Lines
7.3.1 Quantitative influence lines
7.3.2 Qualitative influence lines
7.4 Maximum Shear Force and Bending Moment at Sections in Beams
7.4.1 Concentrated force
7.4.2 A series of concentrated load
7.4.3 Distributed load
Exercises
Bibliography
The Authors
Index
PREFACE
This book is written with the intention of providing students of civil engineering with basic
skill of structural analysis of statically determine structures. The emphasis of the book is on
development of students’ ability to formulate procedures needed to solve statically determinate
problems. Importance of using appropriate free body diagrams to assist in the process of
analysis is emphasized through the use of diagrams in the examples given in the book. The
students are expected to be able to develop proficiency in the determination of internal forces
and deflections in statically determinate structures through worked examples given in the book.
Chapters of the book are arranged in such a way that it guides students to develop their basic
skill of structural analysis starting from understanding of basic consideration in structural
analysis to development of ability to analyse statically determinate structures. Types of
structures covered are pin-jointed truss, beam, and frame. Three-pinned arch and cable are
also included. This book also covers the topic of evaluation of deflection in beam by means of
moment-area and conjugate beam methods and influence lines.
Each chapter is planned to start with introduction to the topic covered in the chapter and the
objective. The inclusion of objective in each chapter is in line with engineering education
approach nowadays which is outcome-based. Step-by-step description of theory or principle
and analysis procedures is then given. Worked out examples are used whenever necessary in
order to reinforce students’ understanding of application of theory or principle and analysis
procedures in problem solving. Exercises are included at the end of each chapter together with
the answers for students to attempt.
This book is made up of seven chapters which are labelled as Chapter 1 to Chapter 7. Chapter
1 describes basic concepts which are important in the subsequent chapters. Chapters 2, 3, 4,
and 5 cover analysis to determine internal forces in pin-jointed truss, beams and frames, cables
and three-hinged arch, respectively. After chapters on analysis for the determination internal
forces, evaluation of deflection in statically determinate beam is covered in Chapter 6. Finally,
in Chapter 7, analysis procedures for the determination of influence line for statically
determinate beam are described.
Many people have contributed directly or indirectly in the course of writing this book. We
would like to firstly thank USM Press for having confidence in us along the process of
publication of this book. We would also like to express our utmost gratitude to the following
former students and current graduate students for their invaluable help in the compilation,
editing, and checking of worked examples of the book: Kong Siong Yung, Bong Sell Feng, Tan
Chee Ghuan, and Ng Wai Keun. Special thank is due to our colleague Dr. Lau Tze Liang for his
valuable comments and suggestions for the improvement of the book, especially on Chapter 6
and Chapter 7. We would also like to acknowledge the advice and guidance of Professor Meor
Othman Hamzah as member of USM Science Publication Committee (Jawatankuasa Sidang
Pengarang Sains) with regards to the publication of this book. To publication officer of USM
Press, we would like to express our sincere appreciation for the help and patience. Last but not
least, a word of thanks to our family for their continuous and unwavering support from the
beginning until the end. We hope this book will serve as valuable source of learning on the
topic of structural analysis.
Taksiah A. Majid
Choong Kok Keong
Mustafasanie M. Yussof
July 2012
CHAPTER 1
BASIC CONCEPT OF STRUCTURAL
ANALYSIS
1.1 Introduction
A structure can be defined as a form or shape such as buildings, bridges, dams, stadiums, and
many others which are designed to resist any applied loads with sufficient strength and
acceptable degree of deformation. A structure must be able to transmit all applied loads to the
supports and ultimately to the ground underlying the supports. Effects of applied loadings on a
structure must be known. For this purpose, structural analysis is needed.
Structural analysis can be explained as a calculation process carried out with the aim of
understanding the behaviour of the structure under the action of a specified set of loads. Before
learning about the calculation process, types of structures, their supports, and the associated
reactions will be first explained in this chapter. Furthermore, concepts of planar structures,
principle of equilibrium, free body diagram (FBD), and statical determinacy will be also
explained.
1.2 Objectives
At the end of Chapter 1, students should be able to
1. determine statical determinacy of structures,
2. identify and determine the reaction forces for different types of supports, and
3. solve for reaction forces using the principle of equilibrium.
1.3 Types of Structures
Structures can be categorized into framed types and mass types. Framed structures resist the
applied loads by virtue of their geometry, while mass structures are resist the applied loads by
virtue of their weight. Structures are commonly composed of different structural parts or
elements. The most common structural elements are beams, columns, arches, walls, trusses,
and foundations. Figures 1.1(a)–1.1(f) show examples of these structural elements under the
action of loading.
The illustrations shown in Figure 1.1 are diagrams used to represent actual structures. It is
noted that all the diagrams are two dimensional. Such two dimensional diagrams are idealized
representations of actual structures. A structure lying in a plane as shown in Figure 1.1 is
called planar structure. In a planar structure, all structural members and external loading lie in
the same plane. It should be emphasized that planar structure is an idealization of actual
structure which is always three dimensional in nature. However, such idealization simplifies
the analysis process to be considered later in this chapter and further in the book.
Figure 1.1 Some typical structural elements
1.4 Types of Supports
Structures could be understood as an assemblage of structural members. A structure must be
properly supported and joined together in order to be stable and able to carry loads. Structures
are in general made from structural members which are joined using connections. These
connections can be used to join member to member as well as member to support. Three types
of the most commonly used connections are
1. pinned connection,
2. fixed or rigid connection, and
3. roller connection.
Figure 1.2 shows two sketches of pinned connections which include connections at support and
between members.
Figure 1.2 Non-zero relative rotation of pinned connection (a) at support and (b) between
members
An important characteristic of pinned connection is that relative rotation between connecting
members or, between member and supporting point are permitted. It is noted that relative
rotation as depicted in Figure 1.2 has been highly magnified for clarify purpose. In actual
structures, rotation is normally very small.
On the other hand, a fixed connection differs from a pinned connection in the sense that relative
rotation between connecting members or, between member and support are not permitted.
Figure 1.3 shows examples of fixed connection.
Figure 1.3 Zero relative rotation of fixed connections (a) at support and (b) between
members
In the process of analysis, standard symbols are generally used to represent pinned and fixed
connections. They are shown in Figure 1.4, where symbols used to denote pinned and fixed
connections at support are shown in Figure 1.4(a) and those for connections at joints between
members in structures are shown in Figure 1.4(b).
Figure 1.4 Symbols commonly used in analysis model to represent connections (a) at
supports and (b) between members
The third type of commonly used connection is roller connection. Symbols used for roller
connection are as shown in Figure 1.5.
Figure 1.5 Symbols for roller connections
Comparing with pinned connections, rollers are
1. similar to pinned connections in the sense that relative rotation is permitted, and
2. different from pinned connections in the sense that only translation in the direction
perpendicular to the supporting plane is not allowed. Translation in the direction parallel to
supporting plane is allowed.
1.5 Reaction Forces
All applied loadings on a framed structure will be transferred to the supports. Reacting force
(or reactions) will be induced at supports in order to maintain equilibrium. Some structures are
constrained by supports that do not allow any rigid-body movement at the supports. Other
supports resist translational movement but not rotation. Such restraints in translational
movement and rotation have important effect on the behaviour of the structure.
For the purpose of analysis, it is necessary to make some idealization regarding the nature of
supports. As mentioned earlier in section 1.4, common types of supports are fixed, pinned, and
roller supports. Figure 1.6 summarizes the three common supports with the corresponding
standard symbols. The corresponding number of reaction forces for each of support is also
shown.
Using Figure 1.6, the corresponding restraints for each type of support is again explained. At
the point of roller support, the corresponding structure is not free to translate in the direction
perpendicular to the supporting plane. For the case of pinned supports, the translation is not
allowed in directions parallel and perpendicular to the supporting plane. For rigid supports,
apart from no freedom to translate in directions parallel and perpendicular to the supporting
plane, the rotation of structure at the point of support is also restrained.
Due to the above restraints provided by the different types of supports, different types and
numbers of reaction forces will occur at the supports. As shown in Figure 1.6(a), at the point
of rigid support, apart from reaction forces in the directions parallel and perpendicular to the
supporting plane, moment will also occur at the support point since the rotation of structure at
the point of rigid support is restrained. For the case of pinned support in Figure 1.6(b),
reaction forces in the directions parallel and perpendicular to the supporting plane will occur
due to the restraints provided by the support in these two directions. However, no reaction
moment will occur as the rotation is not restrained. As shown in Figure 1.6(c), since
translation in the direction perpendicular to the supporting plane is restrained at a point of
roller support, reaction force will occur in that direction. Since the direction parallel to the
supporting plane is not restrained, no reaction force will occur in the direction parallel to the
supporting plane in a roller support. Similarly, no reaction moment will occur in the case of
roller support since the relative rotation is allowed at a roller support.
From the above explanation, it is clear that a reaction force will occur whenever freedom to
translate or rotate is restrained. This is an important point to remember.
Figure 1.6 Types of support systems and reactions
1.6 Equation of Equilibrium
A body will move when subjected to a force or moment. If many forces and moments are
applied to such a body, the resultant of all forces (Σ F) and all moment (Σ M) will cause the
body to move.
A body is said to be in equilibrium if it is initially at rest, so it will remain at rest when
subjected to a system of forces and moments. If a body is in equilibrium, then all its members
and parts are also in equilibrium. Figure 1.7 shows a general body lying in x-y plane. The
object is acted upon by a series of loading F1, F2, F3, and F4 and moments M1, M2, and M3.
Figure 1.7 A general structure in x-y plane acted upon by a series of loads
For the object or structure shown in Figure 1.7 to be in equilibrium, the following three
equations must be satisfied:
The three equations given in equation (1.1) have the following meanings:
Algebraic sum of the x component of all forces acting on the structure is zero.
Algebraic sum of the y component of all forces acting on the structure is zero.
Algebraic sum of the moments of all forces about any point in the plane of the structure
and the moment of any couple acting on the structure is zero.
1.7 Free Body Diagram
When writing down equilibrium equations as shown in equation (1.1), FBD is very commonly
used. An FBD could be understood as – a diagram representing the structure (members or
parts) where all supports have been removed and replaced with all forces and moment
(including reactions) acting on the structure (members or parts). Figure 1.8 shows a simple
example of FBD for a structure subjected to a load P.
Figure 1.8 Example of FBD for a structure
Ax, Ay, and By in Figure 1.8 represent reaction forces at supports A and B, respectively. These
reaction forces are considered as external forces acting on the structure as shown in the
accompanying FBD. FBD could also be usefully employed to show forces which are internal
to the structure. Figure 1.9 shows another example of FBD of a structure.
In Figure 1.9(a), the structure is imaginary cut at section a-a. The corresponding FBD for this
segment is shown in Figure 1.9(b). In Figure 1.9(b), Ax and Ay are again reactions at support A
while N, S, and M acting on section a-a are called internal forces. FBDs as shown in Figures
1.8 and 1.9 will be extensively used later in the calculation of forces.
Figure 1.9 Example of FBD for a segment of a structure
1.8 Statical Determinacy
One of the purposes of structural analysis is to determine the unknown forces such as reaction
forces and internal forces. In the process of calculating the unknown forces, equilibrium
equations are used. In relation to the use of equilibrium equations in the calculation of unknown
forces, an important concept called statical determinacy will be firstly explained.
Statical determinacy is an important concept in structural analysis. It is a concept which is used
to determine whether equilibrium equations alone are sufficient for the purpose of calculating
all the unknown forces in a given structure. Statical determinacy could be explained as
follows:
If all unknown forces can be determined by using only equilibrium equations, then
that structure is called a statically determinate structure. On the other hand, a
structure where the number of unknown forces is more than the number of
equilibrium equations is called a statically indeterminate structure.
In order to further elaborate the concept of statical determinacy and its determination, the
structure as shown in Figure 1.10 is used.
Figure 1.10 A structure subjected to a point load P
Figure 1.11 shows the corresponding FBD for the structure shown in Figure 1.10.
Figure 1.11 FBD for the structure in Figure 1.10
Ax, Ay, and MA are the three unknown reactions at support A due to load P applied. These
three reactions will be calculated by carrying out structural analysis. In order to calculate Ax,
Ay, and MA, the three equilibrium equations as given in equation (1.1) will be used. Using
these three equilibrium equations, the three unknown reactions could then be calculated. In this
example, it can be seen that the number of equilibrium equations (which is three) is the same as
the number of unknowns (which is also three). Using the three equilibrium equations, the three
unknown forces could be determined. Therefore, the structure in Figure 1.10 is a statically
determinate structure.
Another example as shown in Figure 1.12 is considered.
Figure 1.12 A structure fixed at both ends subjected to point load P
The difference between example in Figure 1.12 and the one in Figure 1.10 is that both ends of
the structure are fixed in Figure 1.12, instead of only one end in Figure 1.10. Figure 1.13
shows the FBD for the structure shown in Figure 1.12.
Figure 1.13 FBD for the structure shown in Figure 1.12
In order to calculate the unknown forces as shown in Figure 1.13, the set of three equilibrium
equations given in equation (1.1) will be again used. However, in the case of structure in
Figure 1.12, in comparison to the number of equilibrium equations available which is three, the
number of unknown forces in this case has become six which are Ax, Ay, MA, Bx, By, and MB.
There is no other equilibrium equation which can be used. As a consequence, not all of the six
unknown forces can be determined. Three extra equations are needed in order to determine all
of the six unknown reaction forces. Therefore, the structure in Figure 1.12 is a statically
indeterminate structure.
From the above two examples, it can be seen that:
1. In example shown in Figure 1.10, the set of three equilibrium equations alone given in
equation (1.1) is sufficient to determine all of the three unknown forces. Thus, this is an
example of statically determinate structure.
2. In example shown in Figure 1.12, the set of three equilibrium equations alone is not
sufficient to determine all of the six unknown forces. Thus, this is an example of statically
indeterminate structure. Three extra equations are required in order that all of the six
unknown forces in the example can be determined completely.
The procedures to check for statical determinacy of a given structure can be summarized as
follows:
1. Draw FBD of the structure.
2. Compare the number of unknown forces with the number of equilibrium equations.
Let: α = number of unknown forces
β = number of equilibrium equations
Then if, α = β: Statically determinate structure,
α > β: Statically indeterminate structure.
(1.2)
3. For planar structure, the maximum number of equilibrium equation available is three.
Hence, condition for checking of statical determinacy as given in equation (1.2) could be
further simplified as follows:
If,
α = 3: Statically determinate structure,
α > 3: Statically indeterminate structure.
(1.3)
4. If a structure is statically indeterminate, the difference between α and β is as given below:
(1.4)
i = α − β.
It is called degree of statical indeterminacy. It could be understood as the number of
additional equations required together with equilibrium equations in order that all
unknown forces could be determined.
In order to further understand the concept of statical determinacy and its determination, the
following two more examples are considered.
Example 1.1
Figure 1.14 shows a structure called two-span continuous beam. Support A is a fixed
support, support B is a pinned support, whereas support C is a roller support. Check
statical determinacy of the structure.
Figure 1.14 A two-span continuous beam
Solution
1. Draw FBD.
Figure 1.15
2. Count the number of unknown forces in Figure 1.15.
The unknown forces are: Ax, Ay, MA, By, Bx, and Cy.
the number of unknown forces, α = 6.
3. Compare with the number of equilibrium equation, β.
β = 3, for planar structure.
α > β,
beam is a statically indeterminate structure.
4. Degree of statical indeterminacy.
i = α − β = 6 − 3 = 3.
Example 1.2
Figure 1.16 shows a structure called a simply supported planar truss structure. Support A is
a pinned support and support B is a roller support. Check statical determinacy of the truss.
Figure 1.16 A simply supported truss
Solution
1. Draw the FBD.
Figure 1.17
2. Count the number of unknown forces, α.
The unknown forces are: Ax, Ay, and By.
α = 3.
3. Compare with the number of equilibrium equation, β.
β = 3, for planar structure.
α = β,
truss is a statically determinate structure.
For the structures shown in Examples 1.1 and 1.2, connections are only located at supporting
points. There are cases where connections are found within the structure. In the following
example, a structure with a pinned connection within the structure is considered. Procedures to
check for statical determinacy for such kind of structure are explained.
Example 1.3
Figure 1.18 shows a structure with a pinned joint at B. A is a fixed support, while both C
and D are pinned supports. Check statical determinacy of the structure.
Figure 1.18 A beam structure with pinned joint
Solution
1. Draw FBD.
Figure 1.19
2. Count the number of unknown forces.
The unknown forces are: Ax, Ay, MA, Bx, By, Cx, Cy, Dx, and Dy.
α = 9.
3. Compare with the number of equilibrium equations, β. Since, there are two parts,
β = 2(3) = 6.
α > β,
beam is a statically indeterminate structure.
4. Degree of statical indeterminacy.
i = α − β = 9 − 6 = 3.
The above procedures show that for the case of a structure with pinned joints, the structure is
separated at pinned joints into different parts (called rigid parts). FBD is then drawn for each
of the rigid part. When drawing the corresponding FBD for each part, it is important to
remember that at the point of separation (meaning the point of pinned joint), for example at
point B, two internal forces will occur, that are Bx and By. This is due to the reason that a
pinned joint can only transmit forces and not moment. These internal forces at a pinned joint
must be taken into consideration when counting the number of unknown forces. Another point to
be noted from the above procedures is that the total number of equilibrium equations, β is
equal to 3 × nr (nr = number of rigid parts). In the above example, the rigid parts are AB and
BD. Therefore, nr = 2 and β = 3 × 2 = 6.
It is noted that for the case of structure with roller joint, the number of internal force needed to
be considered in the FBD after separation of the structure at the roller joint will be only one.
This is due to the reason that a roller joint can only transmit force perpendicular to the plane of
the joint.
1.9
Computation of Reactions Using Equations of Equilibrium
Basics about supports such as restraints and corresponding reactions have been explained.
Computation of reactions will be next considered. The set of equilibrium equations to be used
is again restated as follows:
Since ability to formulate correctly the above set of equilibrium equations is essential for the
computation of reaction forces, consideration of forces and moment in equilibrium equations is
briefly reviewed. Examples 1.4 and 1.5 are used for this purpose.
Example 1.4
Compute the reaction forces for the structure called cantilever beam shown in Figure 1.20.
Figure 1.20 Cantilever beam
Solution
1. Draw FBD.
Figure 1.21
2. In Figure 1.21, there are three external loads acting on the beam:
a. Concentrated loads, P1 and P2.
b. Concentrated moment, M1.
3. Since the beam is fixed at support D, there are three reactions at support D, which are
horizontal reaction force Dx, vertical reaction Dy, and reaction moment MD. These three
reactions have to be determined by using equilibrium equations.
4. Formulation of equilibrium for forces in horizontal and vertical directions, which are
equations Σ Fx = 0 and Σ Fy = 0 will be first considered. Referring to Figure 1.21, the
corresponding equations will be as follows:
In the process of formulating the above two equations, any force in which the direction of
action is either in horizontal or vertical direction has to be resolved into components in
these two directions, for example P1 cos θ in x direction and P1 sin θ in y direction, before
the forces can be considered in the corresponding equilibrium equations.
5. For the equilibrium equation of moment, the following two formulations will be considered:
From the two formulations above, it can be seen that:
a. For concentrated loads: if their directions of action pass through the point around which
moment is taken, these forces have no contribution to the equilibrium of moment, for
example Dy, Dx, and P1 cos θ in Σ MD = 0 and Dy, P1 cos θ in Σ MA = 0.
b. For concentrated moments: they will always contribute to equilibrium of moment no
matter which point is taken as reference for the formulation of the equilibrium equation,
for example MD and M1 appear in both formulations of Σ MA = 0 and Σ MD = 0. It is
important to remember this fact as failure to consider couple will result in wrong values
being calculated for the other reactions.
Example 1.5
The simply supported beam shown in Figure 1.22 is subjected to distributed loads. Compute
the reaction for the beam.
Figure 1.22 Simply supported beam
Solution
1. Draw FBD.
Figure 1.23
2. In Figure 1.23, two types of distributed loads are present:
a. Uniformly distributed load: acting uniformly at intensity w1.
b. Linearly distributed load: increasing from load intensity w2 to w3.
It is noted that the intensity of distributed load is expressed in unit of force/length, for
example N/m or kN/m.
3. Since support A is pinned, two reactions will occur at A, that are Ax and Ay, while support
F is a roller support, therefore the only reaction is Fy.
4. Formulation of equilibrium equation of forces in horizontal and vertical directions, Σ Fx = 0
and Σ Fy = 0 again will be considered first:
Since there is no forces with horizontal component, Ax = 0. As for Σ Fy = 0, it can be seen
from the corresponding equation that distributed load is considered as equivalent to a point
load with magnitude equal to the area under the distributed load diagram, for example:
a. For uniformly distributed load: magnitude is equal to area of a rectangle since the
corresponding load diagram is a rectangle.
b. For linearly distributed load: magnitude is equal to the area of a trapezium since the
load diagram is a trapezium.
In the illustration, point load equivalent to area of distributed load diagram has been denoted
using broken arrows.
5. After the formulation of Σ Fx = 0 and Σ Fy = 0, equilibrium of moment Σ M = 0 is next
considered. Taking moment at support A, the following equation can be written:
Moment is force times perpendicular distance to the point where moment is taken. Keeping this
in mind and by studying the formulation as shown above, it can be seen that contribution by
distributed load to moment equilibrium equation is equal to equivalent load corresponding to
the distribution load times distance of action of the corresponding load to the point where
moment is taken. Next figures show the individual contribution from the distributed load to the
moment equilibrium equation.
In the previous illustrations on consideration of moment due to distributed load, the sign of
moment has been omitted. The sign will depend on the positive sense which has been adopted
during the formulation of the moment equilibrium equation. It is again emphasized that moment
due to distributed load is equal to:
Point load equivalent to area of distributed load diagram times distance of the
equivalent point load to the point where moment is to be taken.
The above fundamental definition of moment must be always borne in mind in order that
correct equation of moment is formulated.
After reviewing the basics of consideration of external loads in the formulation of equilibrium
equation, calculation process to determine reaction forces using equilibrium equation as shown
in equation (1.1) will be considered.
Example 1.6
Determine the reaction forces for the structure shown in Figure 1.24.
Figure 1.24
Solution
1. Draw FBD.
2. Reactions
There are three reactions to be determined: Ax, Ay, and By.
Taking moment at A:
It is noted that in the above calculation, the value By is found to be negative. A negative
solution indicates that the direction of reaction assumed at the start of solution process is
incorrect. It is important to note that force is a vector quantity. Therefore, a force has both
magnitude and direction. In the above calculation process, the direction of the reaction has
also been indicated using arrow.
3. The computed reactions.
Example 1.7
Determine the reaction forces for the structure shown in Figure 1.25.
Figure 1.25
Solution
1. Draw FBD.
Statical determinacy,
r = 3,
β = 3.
the beam is statically determinate.
2. Reactions
Checking:
In the above example, checking of statical determinacy has also been included as part of the
solution process. In the subsequent examples, such step of checking of statical determinacy
will be included. Checking of solutions obtained for the reactions is included in order to show
that summation of forces or moments should be zero if the solutions are correct. It is noted that
equation of moment is used for checking since it involves both force and moment. For the
purpose of checking, the reference point for taking of moment should not be the same as that
used in the solution process; for example since moment at point A has already been used in the
solution process, thus a point other than A should be used in the checking of solutions’
correctness.
Example 1.8
Determine the reaction forces for the structure with two-pinned joints shown in Figure 1.26.
Figure 1.26
Solution
1. Draw FBD.
Statical determinacy,
α = 9,
β = 3(nr)
= 3(3)
= 9.
the beam is statically determinate.
2. Support reactions
Checking:
Example 1.9
Determine the reaction forces for the structure with horizontal and vertical elements as
shown in Figure 1.27.
Figure 1.27
Solution
This is an example of a structure called frame. Eventhough it has members which are oriented
horizontally and vertically, the calculation process is similar to that shown in previous
examples.
1. Draw FBD.
Statical determinacy,
α = 3,
β = 3.
the frame is statically determinate.
2. Support reactions.
Example 1.10
Determine the reaction forces for the frame structure shown in Figure 1.28.
Figure 1.28
Solution
1. Draw FBD.
Statical determinacy,
α = 3,
β = 3nr
= 3(1)
= 3.
the frame is statically determinate.
2. Support reactions
Checking:
Example 1.11
Determine the reaction forces for the structure with curved member shown in Figure 1.29.
Solution
Figure 1.29
This is an example of structure called three-pinned arch. It can be seen at the end of the
solution process that the analysis procedure is basically the same as that used in examples with
straight members.
1. Draw FBD.
Statical determinacy,
α = 6,
β = 3nr
= 3(2)
= 6.
the three-pinned arch structure is a statically determinate structure.
2. Support reactions.
Checking:
Example 1.12
Determine the reactions at supports for the beam shown in Figure 1.30.
Figure 1.30
Solution
This example is similar to the structure considered in Example 1.8. The structures in both
examples have two-pinned joints. Such pinned joint is also called hinge or internal hinge.
1. Draw FBD.
Statical determinacy,
α = 9,
β = 3(3)
= 9.
the structure is statically determinate.
2. Support reactions.
Solve equations (1.5) and (1.6) simultaneously, then we have:
Checking:
Example 1.13 Gable Frame
A structure called gable frame as shown in Figure 1.31 is subjected to wind loading.
Determine the reactions at the supports due to the loading.
Solution
The structure in this example is similar in type to the structures considered in Examples 1.9
and 1.10. They are frame structures. However, the frame structure in this example has a
member which is inclined with respect to the horizontal direction. Way to consider example
with inclined member is shown in the solution process.
Figure 1.31
1. Draw FBD.
Statical determinacy,
α = 6,
β = 3(2)
= 6.
the frame is statically determinate.
2. Support reactions.
Checking:
As can be seen from the solution process, attention has to be paid to the consideration of loads
acting on inclined member in the formulation equilibrium equation for forces and moments. In
the case of equilibrium equation for forces, resolving of forces into horizontal and vertical
components must be made before consideration in the respective equilibrium equation for
forces. For the case of equilibrium equation for moment, basic definition of moment should be
remembered as pointed out earlier in section 1.9, where moment is defined as force times
perpendicular distance of the force action to the point where moment is taken. Consideration of
moment due to load on inclined member is further elaborated using Figure 1.32. Figure 1.32
shows two different ways to consider moment at A due to the load 50 kN applied at C. The
required moment can be evaluated by multiplying the load 50 kN (in the original inclined line
of action) with the perpendicular distance (6 + 3 sin 30°) [Figure 1.32(b)] as one possible
expression. Alternatively, the required moment can also be evaluated by summation of moment
due to horizontal and vertical components of the load 50 kN at A [Figure 1.32(c)]. It is
important to note that exactly the same results will be obtained regardless of which formulation
is used. The important point is, correct definition of moment must be used in the formulation.
Figure 1.32
EXERCISES
Exercise 1
Determine the determinacy and the corresponding degree of statical indeterminacy where
applicable for the structures shown in Figure 1.33.
Figure 1.33
Answers: (a) Statically determinate, (b) Statically determinate, (c) Statically determinate
Exercise 2
For the structures shown in Figures 1.34(a), 1.34(b), and 1.34(c), check the statical
determinacy.
Figure 1.34
Answers:
a. Statically indeterminate (i = 2)
b. Statically indeterminate (i = 3)
c. Statically indeterminate (i = 1)
Exercise 3
Determine the support reactions of the beam shown in Figure 1.35.
Figure 1.35
Exercise 4
Calculate the support reactions of the structure shown in Figure 1.36.
Figure 1.36
Exercise 5
Calculate the support reactions of the structure shown in Figure 1.37.
Figure 1.37
CHAPTER 2
ANALYSIS OF PLANE TRUSSES
2.1 Introduction
A truss is an assemblage of straight members connected at their ends by flexible connections to
form a rigid configuration. Because of their light weight and high strength, trusses are widely
used, and their applications are supporting bridges and roof of buildings.
In this chapter, the procedures will be developed for analysing statically determinate trusses.
The objective of analysing the trusses is to determine the reactions and member forces. There
are many ways to do this. Two methods will be used to carry out the analysis where the
equilibrium equations of the whole or parts of the structures will be considered. By
highlighting or considering only parts of the structure (FBD), the FBD will be analysed and the
unknowns will be solved. The two methods that will be used are:
1. Method of joints
2. Method of section
2.2 Objectives
At the end of Chapter 2, students should be able to
1. classify statical determinacy of plane trusses and to check geometric stability, and
2. determine member forces using joints and section methods.
2.3 Common Types of Trusses
Trusses are used to carry loads over relatively long spans compared to ordinary beams. They
are widely used for roofs over large open areas, such as in warehouses, sport centres,
highway, railroad bridges, or industrial buildings as shown is Figures 2.1(a) and 2.1(b).
Figure 2.1 (a) A typical roof truss and (b) truss bridge (pedestrian)
2.4 Classification of Plane Trusses
If all the members of a truss and the applied loads lie in a single plane, the truss is called a
plane truss. For plane trusses, the applied loading acts on each truss in its own plane. The
typical forms of bridge and roof trusses, many of which have been named after their original
designers are shown in Figure 2.2.
Figure 2.2 Common bridge and roof trusses
2.5 Determinacy and Stability of Plane Trusses
Statical determinacy of a plane truss can be checked using equilibrium equations. A truss is
considered to be statically determinate if all reactions of its member forces can be determined
by using the equations of equilibrium. If a plane truss contains m (members), j (joints), and is
supported by r (reactions), and let:
α = number of unknown forces = m + r,
β = number of equilibrium equations = 2j.
Then if,
α = β: Statically determinate truss,
α > β: Statically indeterminate truss,
α < β: Statically unstable truss.
(2.1)
Statically indeterminate trusses have more members and/or external reactions than the
minimum required for stability. The excess members and reactions are called redundants, and
the number of excess members and actions is referred to as the degree of statical
indeterminacy, i which can be expressed as
i = α − β,
i = ( m + r ) − 2j.
(2.2)
Example 2.1
Classify each of the plane trusses shown in Figure 2.3 as statically unstable, determinate,
or indeterminate.
Solution
Figure 2.3(a)
The truss shown in Figure 2.3(a) contains 20 members and 12 joints.
m = 20,
2j = 24,
r = 3.
m + r < 2j,
the truss is statically unstable.
Figure 2.3(b)
The truss shown in Figure 2.3(b) contains 21 members and 12 joints.
m = 21,
2j = 24,
r = 3.
m + r = 2j,
the truss is statically determinate.
Figure 2.3(c)
The truss shown in Figure 2.3(c) contains 25 members and 12 joints.
m = 25,
2j = 24,
r = 3.
m + r > 2j,
2.6
the truss is statically indeterminate.
Sign Convention and Member Forces Representation
The sign convention used in the analysis of plane trusses are described in three equations of
equilibrium similar to section 1.6, as shown in Figure 2.4:
Algebraic sum of the x component of all forces acting on the structure is zero. The
horizontal force is assumed positive if it is to the right.
Algebraic sum of the y component of all forces acting on the structure is zero. The
vertical force is assumed positive if it is upward.
Algebraic sum of the moments of all forces about any point in the plane of the
structure is zero. The clockwise moment is positive.
Figure 2.4 Sign convention
The internal axial forces at any section of the truss member is equal in magnitude but opposite
in direction to the resultant of the components in the direction parallel to the axis of the truss of
all the external loads. A tensile member axial force is always represented by an arrow pulling
away from the joint and compressive member axial force represented by an arrow pushing
towards the joint as shown in Figure 2.5. It is usually convenient to assume the unknown
member forces to be tensile.
Figure 2.5 Internal member axial forces
2.7 Method of Joints
The following step-by-step procedure can be used for the analysis of statically determinate
simple plane trusses by the method of joints.
1. Check the trusses for static determinacy, as discussed in preceding chapter. If the truss is
found to be statically determinate and stable, proceed to step 2. Otherwise, end the analysis
at this stage. (The analysis of statically indeterminate trusses is considered in Chapter 4 of
this book.)
2. Identify by inspection any zero-force member of trusses.
3. Determine the slopes of the inclined members (except the zero-force member) of the trusses.
4. Draw the FBD of the whole truss, showing all external loads and reactions. Write zeros at
the members that have been identified as zero-force members.
5. Examine the FBD of the truss to select a joint that has no more than two unknown forces
(which must not be collinear) acting on it. If such a joint is found, then go directly to the next
step. Otherwise, determine reactions by applying the three equations of equilibrium and
equations of condition (if any) to the free body of the whole truss; then select a joint with
two or fewer unknowns, and go to the next step.
a. Draw an FBD of the selected joint, showing tensile force by arrows pulling away
from the joint and compressive force by arrows pushing into the joint. It is usually
convenient to assume the unknown member forces to be tensile.
b. Determine the unknown force by applying the two equilibrium equations ∑ Fx = 0 and
∑ Fy = 0. A positive answer for a member force means that the member is in tension,
as initially assumed, whereas a negative answer indicates that the member is in
compression.
6. If all the desired member forces and reactions have been determine, then go to the next step.
Otherwise, select another joint with no more than two unknowns, and return to step 5.
7. If the reaction were determine in step 5 by using the equilibrium equations and condition of
the whole truss; then apply the remaining joint equilibrium equations that have not been
utilized so far to check the calculations.
2.7.1 Identification of zero-force members
Because trusses are usually designed to support several different loading conditions, it is not
uncommon to find members with zero forces in them when a truss is being analysed for a
particular loading condition. Zero-force members are also added to trusses to brace
compression members against buckling and slender tension members against vibrating. The
analysis of trusses can be expedited if the zero-force members can be identified by inspection.
Two common types of member arrangements that result in zero-force members are the
following:
1. If only two non-collinear members are connected to a joint that has no external loads or
reactions applied to it, then the force in both members is zero [Figure 2.6(a)].
2. If three members, two of which are collinear, they are connected to a joint that has no
external loads or reactions applied to it, then the force in the member that is non-collinear
is zero [Figure 2.6(b)].
Figure 2.6 Identification of zero-force member
Example 2.2
Using method of joints, determine the force in each member of the truss shown in Figure
2.7.
Figure 2.7
Solution
FBD: Assume all members are in tension.
Joint A:
Joint B:
Joint D:
Joint C:
Checking Joint E:
Final Diagram
Example 2.3
Using method of joints, determine the force in each member trusses shown in Figure 2.8.
Figure 2.8
Solution
FBD: Assume all members are in tension.
Joint A:
Joint G:
Joint B:
Joint D:
Joint E:
Joint F:
Checking Joint C:
Final Diagram
2.8 Method of Sections
This method is an effective method when the forces in some members of a truss are to be
determined. Often the force in just one member with the greatest force in it need to be known.
The method of sections will yield the force in the particular member without the labour of
working out the forces in the rest of the truss.
The method of sections involves cutting the truss into two portions by passing an imaginary
section through the members whose forces are desired. The desired member forces are then
determined by considering the equilibrium equations of one of the two portions of the truss.
There are only three equilibrium equations available, so they cannot be used to determine more
than three unknown forces. Thus, in general, section should be chosen that do not pass through
more than three members with unknown force.
2.8.1 Procedure for analysis
The following step-by-step procedure can be used to determine the member forces of statically
determinate plane trusses based on the method of sections.
1. Select the section that passes through as many members as possible whose forces are
desired, but not more than three members with unknown forces. The section should cut the
truss into two parts.
2. Although either of the two portions of the truss can be used to compute the member forces,
select the portion that will require the least amount of computational effort in determining
the unknown forces. To avoid the necessity for the calculation of reactions, if the two
portions of the truss does not have any reactions acting on it, then select this portion for the
analysis of member forces and go to the next step. If both portions of the truss are attached
to external supports, then calculate reaction by applying the equations of equilibrium and
condition (if any) to the free body of the entire truss. Next, select the portion of the truss for
analysis of member forces that has the least number of external loads and reactions applied
to it.
3. Draw the FBD of the portion of the truss selected, showing all external loads and reactions
applied to it and the forces in the members that have been cut by the section. The unknown
member forces are usually assumed to be tensile and therefore they are shown on the FBD
by arrows pulling away from the joints.
4. Determine the unknown forces by applying the equations of equilibrium. This can be
achieved by applying the equilibrium equations (∑ Fx = 0, ∑ Fy = 0, ∑ M = 0) at any joint
location either of the two portions of the truss.
5. Apply an alternative equilibrium equation, which was not used to compute member forces,
to check the calculations. This alternative equation should preferably involves all three
member forces determined by the analysis. If the analysis has been performed correctly,
then this alternative equilibrium equation must be satisfied.
Example 2.4
Using method of sections, determine the force in members AB, AG, HG, BC, CF, and EF for
the truss shown in Figure 2.9.
Figure 2.9
Solution
FBD:
Section a-a:
Section b-b:
2.9 Introduction to Other Types of Planar Structures
Although a great majority of trusses can be analysed as plane trusses, there are some truss
systems such as transmission towers (Figure 2.10) and lattice domes that cannot be treated as
plane trusses because of their shape, arrangement of members, or applied loadings.
2.9.1 Planar trusses: Compound and complex
Compound trusses are constructed by connecting two or more simple plane trusses to form a
single rigid body as shown in Figure 2.11. Analysis of compound trusses can usually be
expedited by using a combination of the method of joints and method of sections described in
the preceding chapters.
Figure 2.10 Transmission tower
Figure 2.11 Compound trusses
Trusses that can be classified neither as simple trusses nor as compound trusses are referred to
as “complex trusses”. Two examples of complex trusses are shown in Figure 2.12. From an
analytical viewpoint, the main difference between simple or compound trusses, and complex
trusses stems from the fact that the methods of joints and method of sections cannot be used for
the analysis of complex trusses.
m = 9, j = 6, r = 3
m + r = 2j
(a)
m = 17, j = 10, r = 3
m + r = 2j
(b)
Figure 2.12 Complex trusses
2.9.2 Space trusses
Because of its shape, arrangement of members, or applied loading, space trusses cannot be
subdivided into plane trusses for the purpose of analysis and must be analysed as threedimensional structures subjected to three-dimensional force system. To simplify the analysis of
space trusses, it is assumed that the truss members are connected at their ends by frictionless
ball-and-socket joints. All external loads and reactions are applied only at the joints, and the
centriodal axis of each member coincides with the line connecting the centres of the adjacent
joints. Because of these simplifying assumptions, the members of space trusses can be treated
as axial force members.
Figure 2.13 Space trusses
EXERCISES
Exercise 1
a. Figure 2.14(a) shows four different types of plane trusses. Check for the statical
determinacy of the trusses.
b. Figure 2.14(b) shows a plane truss with pinned supports at A and G. Find the reactions at
supports A and G. Identify any zero force member.
c. Determine the force in member BC, AC, and BD for the truss shown in Figure 2.14(b) by
using method of sections and classify whether they are in tension or compression.
d. Determine the forces in member GF and FD for the truss shown in Figure 2.14(b) by using
method of joints. Classify whether they are in tension or compression.
Figure 2.14(a)
Figure 2.14(b)
Answers:
a. (i) Statically determinate
(ii) Unstable
(iii) Statically determinate
(iv) Statically indeterminate, i = 1
b. Reactions: Gy = 25 kN ( ), Ay = 25 kN ( ), Gx = 25 kN ( ), Ax = 15 kN ( ); EF = EG =
ED = 0
c. BC = −14.14 kN (C), AC = 15.82 kN (T), BD = −28.28 kN (C)
d. GF = −35.36 kN (C), FD = −35.36 kN (C)
Exercise 2
a. Figure 2.15 shows a pinned-jointed truss with two supports at A and D, where A is roller
and D is pinned. Check for the statical determinacy of the truss.
b. Determine the forces in member BC, BE, and DE for the truss shown in Figure 2.15 by
using method of sections and classify whether they are in tension or compression.
c. Determine the forces in member CF, EF, and CE by using method of joints and classify
whether they are in tension or compression.
Figure 2.15
Answers:
Reaction (optional), Dx = 30 kN ( ), Ay = 50 kN ( ), Dy = 50 kN ( )
a. Statically determinate
b. CB = −40 kN (C), ED = 70 kN (T), EB = −42.43 kN (C)
c. CF = −28.28 kN (C), FE = 20 kN (T), CE = 28.28 kN (T)
Exercise 3
a. Describe briefly three characteristics of truss structure and the two methods of analysing
the truss structure.
b. Determine the forces in member DF, DE, and CE for the truss shown in Figure 2.16 by
using method of sections and classify whether they are in tension or compression.
c. Find the forces in member AB, AC, and BC by using method of joints and classify whether
they are in tension or compression.
Figure 2.16
Answers:
Reactions: Cy = 50 kN ( ), Cx = 10 kN ( ), Ay = 25 kN ( ),
b. CE = −42.43kN (C), DF = 14.14 kN (T), DE = 30 kN (T)
c. AB = 35.35 kN (T), AC = −25 kN (C), BC = −22.98 kN (C)
CHAPTER 3
ANALYSIS OF STATICALLY
DETERMINATE BEAMS AND FRAMES
3.1 Introduction
In Chapter 1, examples of beam and frame structures have been used. In this chapter, analysis
of both structures will be specifically considered. Beams and frames are very commonly used
structures. A beam is a structural element normally oriented horizontally for the purpose of
carrying vertical loads. On the other hand, a frame is a structure composed of horizontal,
vertical, and sometimes inclined members. In addition to vertical loads, frames are also
subjected to horizontal and sometimes inclined loads.
In this chapter, the analysis of beams and frames will be considered. Analysis steps to
calculate shear force (S), axial force, and bending moment (M) in beams and frames will be
explained.
3.2 Objectives
At the end of Chapter 3, students should be able to
1. draw SFD and BMD for statically determinate beams and frames using equilibrium of
section method.
2. draw qualitative deflected shape of beams and frames based on information provided by
BMD.
3.3 Internal Forces in Beams
Analysis of beam will be first considered. Figure 3.1 shows an example of a simply supported
beam.
Figure 3.1 A simply supported beam is subjected to point load P
Under the action of external loading such as P in Figure 3.1, a beam will bend as illustrated in
Figure 3.2.
Figure 3.2 Bending deformation of a simply supported beam when subjected to point load P
Beams carry loads primarily by means of bending action. When subjected to external loads,
forces will develop within a beam for the maintance of equilibrium. Such forces are called
internal forces. Internal forces developed in beam are shear force and bending moment. They
shall be denoted as S and M, respectively. Figure 3.3 shows the internal forces in a beam.
Internal forces S and M on a section of a beam can be defined as follows:
1. S is a force acting parallel to a section and pointing in upward or downward direction.
2. M is a moment acting around an axis perpendicular to the plane of the beams (x-y plane in
Figure 3.3) in anti-clockwise or clockwise direction.
Another illustration of S and M over a beam section is shown in Figure 3.4. Plane x in Figure
3.4 represents the axis of the beam.
Figure 3.3 Internal forces in a beam – S and M
Figure 3.4 Definition of S and M on a section of a beam
3.4
Sign Convention and Computation of Shear Force and Bending
Moment
Before explaining the importance of sign convention, a simple calculation of internal forces in
a beam (Figure 3.5) is first considered. The concept of FBD explained earlier in Chapter 1 is
used in the calculation.
The beam is first cut through section a-a and separated into two parts as shown in Figure
3.5(b). Internal forces S and M are then indicated on both sections belonging to the parts to the
right and left of the cut location. By using equations of equilibrium Σ Fy = 0 and Σ Mz = 0, S
and M could be calculated. There are two choices of FBD which could be used: the one on the
left or right of the cut section [Figures 3.5(c) and 3.5(d)]. Results of calculation using each of
the FBD are shown. Magnitude of S and M are found to be the same for both sets of
calculation.
Figure 3.5 A simple example showing the computation of internal forces S and M in a beam
In the above example, S and M have been indicated on portions to the left [Figure 3.5(c)] and
right [(Figure 3.5(d)] of cut section a-a following a certain convention. For portion on the left
of cut section a-a, S has been indicated as a force acting downwards and M has been indicated
as a moment acting in anticlockwise direction [Figure 3.5(c)]. On the other hand, S and M have
been indicated on portion on the right of cut section in direction opposite to that used in portion
on the left [Figure 3.5(d)]. Such sign convention used when calculating S and M ensure that the
same results are obtained regardless of each portion (left or right of cut section) is used. The
sign convention adopted as shown in Figures 3.5(b), 3.5(c), and 3.5(d) is called the positive
sign convention for S and M.
The following illustrations in Figure 3.6 show the meaning of positive sign convention for S
and M in a beam.
Positive S:
Portion of the member to the left of the cut section tends to be pushed upward relative to the
portion on the right of the cut section.
P ositive M:
Beam tends to be bent concave upwards, causing compression on the upper fibres and
tension in the lower fibres of the beam at the cut section.
Figure 3.6 Meaning of positive sign convention for S and M
A few examples of process involved in the calculation of S and M on section of a beam will be
next considered.
Example 3.1
For the beam shown in Figure 3.7, calculate the internal forces S and M at the sections x-x,
y-y, and z-z.
x-x : just to the right of A
y-y : just to the left of B
z-z : just to the left of D
Figure 3.7
Solution
1. Draw FBD and calculate the reaction forces.
Take moment at C.
2. Evaluation of reactions.
Section x-x:
Section y-y:
Section z-z:
In the above example, the description of section just to the left or just to the right of a certain
point has been used, for example section x-x is a section just to the right of A. Such distinction
of section, whether it is to the right or to the left, it is used whenever calculation of S or M at
point load or couple is involved. S and M are defined on sections. In the calculation process, a
cut is must be made through the section where S and M are required. Whenever there is a point
load (including reaction from a support), S and M on section just to the left or just to the right
of the particular load (or couple) are needed. It is meaningless to talk about S and M at the
location of point load (couple), since it means that calculation of S and M is carried out on
section cut through the load (or couple).
The above example shows that if appropriate portion after the cut is used, then evaluation of
reactions is not necessary. This will result in saving of calculation time. When calculating S
and M on section z-z, portion to the right of section z-z has been used. With the use of portion
to the right, there are no other forces in the FBD except S and M on the cut section. This has
resulted in a rapid calculation of S and M.
Example 3.2
For the beam shown in Figure 3.8, compute the S and M at section x-x which is just to the
left of support B. Note that reactions at supports have been calculated and the values are
indicated in Figure 3.8.
Figure 3.8
Solution
Since the reactions have been calculated, the S and M can be directly computed. Using FBD of
segment AB, S and M at section x-x can be computed as follows:
S + 10 (40) − 100 = 0
S = − 300 kN.
M + 10(40)(20) − 100(40) = 0
M = − 4000 kNm.
Example 3.3
For the beam shown in Figure 3.9, compute the S and M at section D-D which is located at
a distance of 2 m from support B.
Figure 3.9
Solution
1. Reaction forces:
It is noted that calculation of Ay and By has been carried out using proportional method for
the case of simply supported beam under concentrated load.
2. The S and M at section D-D:
By using portion AD:
By using portion BD:
From the Example 3.3, it can be seen that if portion BD is used instead of portion AD,
calculation of S and M is much simpler since it involves less number of forces in the FBD.
Appropriate choice of FBD will result in simpler equation of equilibrium in the calculation of
S and M. This will not only save time but also avoid calculation error. Furthermore, it is noted
that when calculating M, equilibrium of moment has been taken with respect to the cut section.
Although moment can be taken at any point in the FBD, the practice of taking moment at the cut
section will avoid error in M due to wrongly calculated S. When moment is taken with respect
to cut section, there is no need to consider S in the moment equilibrium equation.
Example 3.4
Determine S and M at section A and section B for the beam shown below.
Figure 3.10
Solution
1. Statical determinacy.
nr = 2, r = 4 + 2 = 6.
3nr = 3 × 2 = 6.
3nr = r,
the beam is a statically determinate structure.
2. S and M at B.
To determine Ey, by using portion EC:
From Examples 3.1 to 3.4, it can be seen that general procedures for the computation of S
and M could be summarized as follows:
1. Compute reaction forces.
2. Pass an imaginary cut through the section where S and M are required. Draw the
corresponding FBD to be used. Indicate on the FBD, S and M using the corresponding
position sign convention.
3. Apply equilibrium of force in vertical direction, ∑ Fy = 0, to compute S.
4. Apply equilibrium of moment at the cut section, ∑ Mcut section = 0, to compute M.
Although equilibrium of moment with respect to any other point could also be used,
consideration of moment at cut section will eliminate the necessity to consider moment
caused by S. This will avoid mistake in value of M due to mistake in S.
Step 1 is sometimes not needed if appropriate choice of FBD is made. One such case has been
mentioned in Example 3.1. It is emphasized that when calculating S and M, the section where S
and M were acting is important to be indicated clearly.
3.5
Shear Force and Bending Moment Diagrams for Beams
In previous section, calculation of S and M at particular section in a beam has been explained.
In the design of beams, the information on how S and M change with the change in the location
of section is required. Such information is conveniently provided by SFD and BMD. In this
section, computational procedures to produce these two diagrams for beams will be
considered.
The general procedures can be summarized as follows:
1. Evaluate support reactions.
2. Make necessary cut through section where equations for S and M are to be evaluated
(depending on the structure and loading condition given, such cutting process might need to
be repeated). After choosing suitable FBD to be used, apply
a. equilibrium condition for vertical force in order to obtain equation for S, and
b. equilibrium equation of moment at cut section in order to obtain equation for M.
3. Make use of the equations for S and M obtained in step (2) above, draw the corresponding
SFD and BMD.
From the above description of procedures, it can be seen that the process is similar to the one
used in the calculation of S and M at a particular section. In the calculation of S and M at a
particular section, the location of section is fixed, for example at 2 m from support B.
However, in the case of calculation process to obtain SFD and BMD, the location of the
section is not fixed; it varies within a particular portion of the beam. Such changing location is
represented using distance x from a suitably chosen reference point. Due to the changing
location of section, S and M will be obtained in general as equations in term of x. These two
equations are used in the drawing of SFD and BMD.
The following examples are used to illustrate the drawing process of SFD and BMD.
Example 3.5
Determine S and M at the section which is located at distance x from the fixed end A. Draw
the SFD and BMD.
Figure 3.11 A cantilever beam subjected to point load
Solution
1. Draw the FBD and determine the reaction forces.
2. Computation of S and M.
If the above computational process is examined, it can be seen that S and M at section
located at a distance x from point A is expressed in terms of known forces and the distance
x. Since this distance x can be anywhere between A and B, equations for S and M are
actually equations representing the variation of S and M on sections between A and B. This
variation could be seen by changing the value of x between 0 and 4 m. If a graph of S (y
axis) versus x (x axis) is drawn by using the equation S above, the following SFD will be
obtained:
Similarly, the graph of M versus x (BMD) is plotted based on equation for M calculated. The
resulting BMD is shown below.
Example 3.6
Draw the SFD and BMD for the simply supported beam shown in Figure 3.12.
Figure 3.12
Solution
1. Draw the FBD and calculate the reaction forces.
2. Computation of S and M.
Portion AB:
Portion BC:
Portion BC (alternative):
3. Draw the SFD and BMD by using equations obtained.
Portion AB:
S = 71.429 kN
M = 71.429x kNm
0 ≤x≤ 2m
Portion BC:
S = − 28.571 kN
M = − 28.571x + 200 kNm
2≤x≤7m
Note that if the results of alternate calculation for portion BC as follows:
S = − 28.571 kN
M = 28.571x kNm
0≤x≤5m
have been used, the same diagram will be obtained for portion BC.
SFD and BMD are shown together with the structure analysed. It is important to indicate the
title of SFD and BMD and the corresponding units used.
Example 3.7
For the beam shown in Figure 3.13, draw the corresponding SFD and BMD.
Figure 3.13
Solution
1. Draw the FBD and calculate the reactions forces.
2. Computation of S and M.
Portion AB:
Portion BC:
Portion CD:
3. Draw the SFD and BMD by using equations obtained.
Example 3.8
Draw SFD and BMD for the beam shown.
Figure 3.14
Solution
1. Statical determinacy.
α =3,
β = 3(1) = 3.
α = β,
the beam is statically determinate.
2. Support reactions.
3. Computation of S and M.
Portion AB ( 0 < x < 3 m ):
Portion BC (3 < x < 6 m):
Portion CD (3 < x1 < 6 m):
Portion DE (0 < x1 < 3 m):
4. Draw the SFD and BMD.
BMD drawn in this chapter with positive bending moment that is (1) defined as shown in
Figures 3.3, 3.4, and 3.6, and (2) drawn above the x axis representing the axis of beam, is
referred to as “bending moment drawn on compression side”. Figure 3.15 illustrates the idea
of “drawing of BMD on the compression side”. In the figure, a dotted line superimposed on the
edge of a beam (top or bottom) has been used to indicate state of compression of the fiber (top
or bottom) of the beam section.
Figure 3.15 Drawing of BMD on the compression side
3.5.1 Characteristics of SFD and BMD
There are some characteristics of the shapes of SFD and BMD. These characteristics are
closely related to (1) effect of different types of supports, (2) effect of different types of
loadings, and (3) effect of concentrated load and couple. These three different effects will be
considered one by one.
1. Effect of different types of supports
Figure 3.16 Effect of support condition on shape of SFD and BMD
2. Effect of different types of loadings
Figure 3.17 Effect of loading types on shape of SFD and BMD
3. Effect of concentrated load and couple
Figure 3.18 Effect of concentrated load and concentrated moment (couple) on shape of SFD
and BMD
Figure 3.18 shows there are points where “jump” occurs in SFD and BMD. This is a
characteristic of SFD and BMD whenever concentrated loading such as point load or couple is
encountered. Figure 3.19 illustrates further the occurrence of jump in SFD and BMD.
Figure 3.19 Sudden change in S or M at point of concentrated load and couple
3.5.2 Qualitative deflected shapes
Qualitative deflected shape (QDS) is a rough (usually exaggerated) sketch of the neutral
surface of the structure in the deformed position under the action of a given loading condition.
It is also called the elastic curve of a structure. The use of QDS is as follows:
1. Valuable insights into the behaviour of structures can be obtained.
2. Useful in the computation of the numerical value of deflection.
An illustration of QDS is shown in Figure 3.20. The QDS is represented by the shape of
neutral surface (in deformed position). It is noted that in a QDS, actual magnitude of deflection
is unknown. The deflected shape is normally drawn not to scale and highly exaggerated.
Figure 3.20 QDS of a beam structure
The ability to draw QDS is an important skill to be acquired. It provides important information
about curvature of a member. Such information about curvature of a member provides
important information about state of stress in a section, that is whether it is in compression or
tension.
QDS is drawn based on information provided by BMD and support conditions of the given
problem. To sketch QDS, it is important to know the meaning of positive bending moment and
negative bending moment. This is related to sign convention that has been decided.
Figure 3.21 Effect of positive and negative bending moments
Another aspect that needs to be considered to sketch QDS is support condition. At different
support conditions, different restraint with respect to deflection exists. Figure 3.22 shows how
support conditions influence the QDS. Next, Examples 3.9 to 3.12 illustrate four examples of
QDS sketching.
Figure 3.22 Effect of support condition on QDS
In Example 3.9, SFD and BMD have been drawn. Based on BMD and the understanding in the
meaning of positive and negative bending moment, the corresponding QDS can be drawn. At
supports A and B, no deflection is allowed. However, rotation is allowed. This support
conditions are reflected in the QDS. Included in Figure 3.23 is a sketch showing a portion of
the beam with top fiber of the section in a state of compression. This is in accordance with the
meaning of positive sign convention introduced earlier. A positive bending moments means that
the beam is deflected in such a way that the top fiber is in compression and the bottom fiber is
in tension. This fact will be further stressed in the following examples when QDS are drawn.
Example 3.9
Figure 3.23 QDS sketching of a simply supported beam
BMD in Example 3.10 shows that bending moment of all sections of the cantilever beam is
negative. The beam is supported at the left hand side with a fixed support. As explained
earlier, both translation and rotation at a fixed support are restrained. Based on BMD and the
fixed support condition, the QDS can be drawn. In Example 3.10, the cantilever beam has
deflected in such a way that the bottom fiber of all sections along the beam is in the state of
compression. This is in accordance to the meaning of negative bending moment as mentioned
earlier.
Example 3.10
Figure 3.24 Sketching of QDS of a cantilever beam
In Example 3.11, both fixed support and roller support exist in the beam structure. Keeping in
mind that (1) translation and rotation are restrained at A, and (2) translation is restrained at B,
and remembering the meaning of positive and negative bending moment, the QDS can be
drawn. For the beam structure in Example 3.11, two points where change in sign of curvature
of deflected shape are encountered. This point is the so-called point of contraflexure. Such
point is encountered whenever there is a point of zero bending moment on the BMD, where
change in sign of bending moment occurs.
Example 3.11
Figure 3.25 Sketching of QDS of a propped cantilever with overhang
Example 3.12
Figure 3.26 QDS of a simply supported beam subjected to linearly distributed loading
3.6 Frame
A frame is composed of straight members connected with either rigid connections or hinged
connections to form stable structure. Figure 3.27 shows an example of a frame.
Figure 3.27 A frame subjected to both vertical and lateral loads
Under the action of external loads, the members of a frame may be, in general subjected to
bending moment, shear, and axial forces. Figure 3.28 shows the internal forces of sections a-a
and b-b in the frame shown in Figure 3.27.
Figure 3.28 Internal forces in members of a frame
3.7 Statical Determinacy of Frames
A frame is considered to be statically determinate if the bending moment, shear, and axial
forces in all its members, as well as all the external reactions, can be determined by using the
equations of equilibrium. For better understanding, the following example will be considered.
Example 3.13
Check the statical determinacy of the frame shown in Figure 3.29.
Figure 3.29
Solution
1. Draw the FBD.
2. Number of external reactions: Ax, Ay, Dy = 3.
Number of equilibrium equations: 3.
Statically determinate frame.
Once Ax, Ay, and Dy are determined, bending moment, shears and axial forces in all members
AB, BC, and CD could be determined.
Whenever hinged or roller-type joints are found in members of a frame, additional equilibrium
equations can be formulated. Such equilibrium equation is called equation of condition. Figure
3.30 shows examples of frame structures with hinged and roller joints in a frame.
Figure 3.30 Frames with hinged and roller joints
For frames with only rigid joints, statical determinacy could be determined by using either of
the following approaches:
1. Comparison between number of unknown forces and number of equilibrium equations.
Similar to the case of truss structures, the following equation is used:
3m + r = 3j,
where m = number of members
r = number of external reactions
j = number of rigid joints
For example:
m= 3
j = 4 (including the supports)
r=6
3m + r = 3(3) + 6 = 15.
3j = 3( 4) = 12.
3m + r > 3j,
the frame is statically indeterminate.
Degree of statical indeterminacy, i = 15 − 12 = 3.
2. Alternative approach
In the approach, enough cuts are passed through members of the frame by imaginary sections
and/or enough supports are removed in order to render the structure statically determinate.
The total number of internal (cuts through members) or external restraints (removal of
supports) thus removed equals the degree of statical indeterminacy. An example as shown in
Figure 3.31 is used to illustrate the alternative approach.
Figure 3.31 Removal of internal restraints through passing of imaginary section through a
member of the frame
With one imaginary section, Part ABe and DCe’ become statically determinate. Since one
imaginary section is needed to separate the frame into portions which are all statically
determinate and there are three internal forces associated with any section of a frame, the
frame is statically indeterminate with degree of statical indeterminacy, i = 3 (which is equal to
number of internal restraints removed).
Figure 3.32 Removal of external restraints through removal one of the fixed support from the
frame
In the illustration above, imaginary section has been used to separate the frame into portions
which are statically determinate (Figure 3.31). Alternatively, the frame can also be made
statically determinate by removing one of the fixed supports as shown in Figure 3.32 in the
order to render the frame statically determinate. Removal one of the fixed support from the
frame as shown in Figure 3.32 corresponds to removal three external restraints since two
ranslations and one rotation are not allowed at a fixed support.
External restraints to be removed: Dx, Dy, and MD. With the removal of the three restraints, the
frame ABCD becomes statically determinate. Since degree of statical indeterminacy is equal to
the number of external restraint removed, i = 3.
The following examples are considered to illustrate the use of alternative approach in the
determination of the degree of static indeterminacy of multistorey building frames (without
internal hinges or rollers).
Example 3.14
Check the statical determinacy of the multi-storey frame shown in Figure 3.33.
Figure 3.33 A multi-storey frame with fixed supports
Solution
1. Removal of internal restraints through passing of imaginary sections through members of the
frame.
2. Number of unknown forces and number of equilibrium equations.
Total number of internal restraints (bending moment, shear,
and axial force at each cut)
= 12 × 3 = 36.
Total number of external reactions
Total number of unknown forces, α
Total number of equilibrium equations, β
= 4 × 3 = 12.
=36 + 12 = 48.
= 4 × 3 = 12.
α > β,
the multi-storey frame is statically indeterminate.
Degree of statical indeterminacy, i = 48 − 12 = 36.
Example 3.15
Check the statical determinacy of the double-storey frame shown in Figure 3.34.
Figure 3.34 A double-storey frame with pinned supports
Solution
1. Removal of internal restraints through passing of imaginary sections through members of the
frame.
2. Number of unknown forces and number of equilibrium equations.
Total number of internal restraint
= 2 × 3 = 6.
Total number of unknown forces, α
=6 + 4 = 10.
Total number of equations of equilibrium, β = 2 × 3 = 6.
α > β,
the double-storey frame is statically indeterminate.
Degree of statical indeterminacy, i = 10 − 6 = 4.
Example 3.16
Check the statical determinacy of the box-shaped frame shown in Figure 3.35.
Figure 3.35 A box-shaped frame with pinned and roller supports
Solution
1. Removal of internal restraints through passing of imaginary sections through members of the
frame.
2. Number of unknown forces and number of equilibrium equations:
Total number of internal restraints
= 2 × 3 = 6.
Total number of external reactions
= 1 + 2 = 3.
Total number of unknown forces, α
=6 + 3 = 9.
Total number of equilibrium equations, β = 2 × 3 = 6.
α > β,
the box-shaped frame is statically indeterminate.
Degree of statical indeterminacy, i = 9 − 6 = 3.
Example 3.17
Check the statical determinacy of the frame structure shown in Figure 3.36.
Figure 3.36 A frame structure
Solution
1. Removal of internal restraints through passing of imaginary sections through members of the
frame
2. Number of unknown forces and number of equilibrium equations.
Total number of internal restraints
= 1 x 3 = 3.
Total number of external reactions
= 2 + 1 + 2 + 3 = 8.
Total number of unknown forces, α
=3 + 8 = 11.
Total number of equilibrium equations, β
= 2 × 3 = 6.
α > β,
the frame structure is statically indeterminate.
Degree of statical indeterminacy, i = 11 − 6 = 5.
3.8 Analysis of Plane Frames
Similar to the case of beam, the information on how S and M change with the change in the
location of section is required in the design of frames. Such information is conveniently
provided by SFD and BMD. In this section, determination of shear force and bending moment
in plane statically determinate frames will be considered. The general procedures of
computing the shears and bending moment are the same as those used in beams.
Example 3.18
Draw the SFD, BMD, and the QDS for the frame shown in Figure 3.37.
Figure 3.37
Solution
1. Check the statical determinacy.
Total number of internal restraints
Total number of external reactions
Total number of unknown forces, α
Total number of equilibrium equations, β
α =β,
= 1 × 3= 3.
= 2 + 1= 3.
=3 + 3 = 6.
= 2 × 3 = 6.
the frame is statically determinate.
2. Compute the reactions.
3. Shear force and bending moment in the members.
Shear force and bending moment are defined with respect to sections which are
perpendicular to member axis.
Member AB (S and M at general section a-a):
The corresponding SFD and BMD for this portion:
Member BC (S and M at general section b-b):
The corresponding SFD and BMD for this portion:
Member CD (S and M at general section c-c):
Alternatively:
The corresponding SFD and BMD for this portion:
The SFD and BMD.
When sketching the QDS for frame structures, it should be noted that the relative angle between
rigidly joined members should remain unchanged; for example angle between (1) members AB
and BC, and (2) members BC and CD, should be maintained as 90° in the deflected shape in
Example 3.18.
Example 3.19
Draw the SFD, BMD, and the QDS for the frame shown in Figure 3.38.
Figure 3.38
Solution
1. Check statical determinacy.
Number of reactions, α =3.
Number of equilibrium equations, β = 3.
α =β,
the frame is statically determinate.
2. Compute support reactions.
If calculation is proceeded from C, calculation of reactions Ax, Ay, and MA at A is not
needed. Such approach is adopted in the example.
3. S and M in the members.
Portion CB (S and M at general section a-a):
Portion BA (S and M at general section b-b):
The SFD and BMD.
Example 3.20
Draw the SFD, BMD, and the QDS for the frame shown in Figure 3.39.
Figure 3.39
Solution
1. Check statical determinacy.
Number of external reactions, α =3.
Number of equilibrium equations, β = 3.
α =β,
the frame is statically determinate.
2. Compute reaction forces.
If the calculation of shear force and bending moment is started from point C, there is no need
to compute the reactions Ax, Ay, and MA. This approach is adopted in the example.
3. Shear force and bending moment.
Portion BC:
In this example, a vertical uniformly distributed load with intensity 10 kN/m acts on inclined
member BC. Since shear force and bending moment are defined with respect to section
perpendicular to axis of a member in a frame, components of the loading which are
perpendicular and parallel to the axis of inclined member BC are first computed.
Conversion of distributed load 10 kN/m to component perpendicular and parallel to axis of
member BC:
The load 10 kN/m which is distributed over the length of inclined member BC is resolved into
components perpendicular 8 kN/m and parallel 6 kN/m to the member as follows:
The calculated components are shown acting on the inclined member BC as follows:
For the calculation of shear force and bending moment, only component perpendicular to
member axis, that is 8 kN/m, needs to be considered. The calculation of S and M for portion
BC is as shown below:
Portion BA:
Example 3.21
A gable frame is subjected to a snow loading, as shown in Figure 3.40. Draw the SFD,
BMD, and the QDS for the frame.
Figure 3.40
Solution
1. Check statical determinacy.
Total number of unknown forces, α =2 + 2 + 2 = 6.
Total number of equilibrium equations, β = 2 × 3 = 6.
α =β,
the frame is statically determinate.
2. Compute reaction forces.
From the symmetry of the structure geometry and the loading condition, it can be concluded
that
Joint C is a hinge. Using part ABC (part EDC can also be used):
3. Computation of S and M.
Since this is a symmetrical problem (symmetry with respect to a vertical line passing through
point C), only the SFD and BMD for either part ABC or part EDC needs to be determined.
After the corresponding diagrams have been obtained for either one of the part, diagrams for
the other part could be obtained through suitable inverting or rotation.
Member BC (= member DC):
The distributed load is first converted to components in direction perpendicular and
parallel to member axis.
In this example, the distributed load is not directly distributed over the included member axis.
The load 12 kN/m is defined over the projected length of member BC (or DC). For distributed
loading which is not defined with respect to the inclined length of member, a first step whereby
the distributed load is redefined to be acting along the inclined member axis needs to be
carried out. The calculcation procedure is shown below.
The above figures shows that the distributed load 12 kN/m has been converted to a load
distributed over the length of inclined member BC 9.6 kN/m. Such conversion is calculated
using the following relation:
The total load of 12 kN/m × 4 m = 48 kN is redistributed over the inclined length of 5 m. After
the above conversion is carried out, the components perpendicular and parallel to inclined
member axis are then calculated.
To obtain the perpendicular and parallel components of 7.68 kN/m and 5.76 kN/m,
respectively; the following relation has been used:
The perpendicular component of 7.68 kN/m is applied on member BC for the calculation of S
and M:
Location of S = 0:
M at x = xF = 4.063 m = 3.375 kNm.
In the drawing of complete SFD and BMD, shape of the corresponding diagram under the
condition of symmetrical problem has been used. One example of SFD and BMD for
symmetrical problem is shown in Example 3.9. For SFD, once the diagram for one part of the
structure is obtained, the corresponding diagram for the other part is obtained by (1) inversion
with respect to the axis of symmetry, and (2) further inversion with respect to member axis.
The corresponding sign of shear force is changed after the inversion process. For the case of
BMD, the corresponding process of inversion needs to be carried out only once with respect to
the axis of symmetry.
As mentioned earlier in the section on analysis of beam, all the BMD shown in the examples in
this chapter are BMD drawn on compression side. However, it should be noted that BMD
could also be drawn on tension side. When designing reinforced concrete structures, BMDs
are often drawn on the tension side of the member to facilitate the placement of reinforcement
bars on the tension side of the member. A tension-side BMD can be obtained by simply
inverting the corresponding compression-side BMD about the member axis. Example 3.22
shows several examples on how BMD on tension side is obtained from the corresponding
BMD in compression side for three different problems.
Example 3.22
Figure 3.41
EXERCISES
Exercise 1
Verify that each of the plane frames shown below is statically indeterminate and determine the
corresponding degree of statical indeterminacy.
Figure 3.42
Answers: (a) i = 12, (b) i = 15
Exercise 2
Draw the SFD, BMD, and QDS for the frame shown in Figure 3.43.
Figure 3.43
Answers:
2. (a)
2. (b)
Exercise 3
Figure 3.44 shows a frame with an internal hinge at C. It is supported by two-hinged supports
at A and E. Draw the corresponding SFD and BMD. Also sketch the QDS.
Figure 3.44
Answers:
Exercise 4
Figure 3.45 shows a rigidly jointed frame with pinned and roller support at A and D,
respectively. A horizontal point load of magnitude 40 kN acts at joint B and a uniformly
distributed load with intensity of 3 kN/m acts along portion C to D of the horizontal member
BCD. Draw the corresponding SFD and BMD. Also sketch the QDS.
Figure 3.45
Answers:
Exercise 5
Draw the SFD and BMD for the frame shown in Figure 3.46. Also sketch the QDS.
Figure 3.46
Answers:
CHAPTER 4
ANALYSIS OF CABLES
4.1 Introduction
Cables constructed of high-strength steel wires are completely flexible and have a tensile
strength four to five times greater than structural steel. Because of their strength-to-weight
ratio, engineers use cables to construct long-span structures, including suspension bridges and
roofs over large arenas and convention halls. Figure 4.1 shows Penang Bridge as an example
of cable-stayed bridge, which is the longest bridge in Malaysia.
Figure 4.1 Penang Bridge – a cable-stayed bridge
A cable is a form of structure which is unable to resist bending moments. Since a member in
tension requires lesser cross-sectional area if compared to a member in compression, both
carrying equal amount of load, therefore it is more economical if the girder is suspended on the
cable via hangers as shown in Figure 4.2. In this case the cable and hangers are in tension.
For comparison purpose, a structural parabolic arch that carries a uniformly distributed load is
subjected to a very small bending moment at any point on the arch. Therefore, an economical
bridge can be constructed as shown in Figure 4.3. Girder on the columns is supported by the
structural arch. In this case, the structural arch and the columns are in compression.
Figure 4.2
Figure 4.3
4.2 Objectives
At the end of Chapter 4, students should be able to calculate reactions and tension in cables
using equilibrium equations.
4.3 Scope of Analysis
In this analysis, the cable is considered perfectly flexible in bending so that the bending
moment on any section of the cable must be zero, and the cable can only transmit load to the
supports by means of tension action along its length. It should be noted that:
1. Maximum tension occurs at maximum reaction point.
2. Minimum tension occurs at the lowest point of the cable.
The bending flexibility of a cable will be intuitively understood by a comparison with a cotton
thread from which small weight are hung as shown in Figure 4.4.
Figure 4.4 Suspended cable subjected to concentrated loads
Clearly, the thread has no resistance to bending forces and, since the weight of the thread will
be insignificant in comparison with the loads, it will assume a shape consisting of a series of
straight line and between the applied loads. It is such as an idealization which has been
assumed for the cable in Figure 4.4.
As the applied loads are all vertical, the horizontal reactions at A and B must be equal and
opposite (since they are only horizontal force acting on the structure) and these are denoted by
Ax and Bx. Simple equilibrium will confirm that the horizontal component of the tensile force
at any section will also of value Ax or Bx (Ax = Bx). See Figure 4.5.
Figure 4.5
Using ∑ MB = 0 for the whole structure,
Consider any typical point C (x, y) and in this case positioned between the concentrated loads
w1 and w2. As the bending moment at C is zero, taking moments of all the forces to the left
(refer to Figure 4.6) of C gives:
Note: The sum of the moments of all forces to one side of a point must be zero as the cable
cannot sustain an internal bending moment.
Figure 4.6
4.4 Types of Cables
There are two types of cables that will be considered:
1. Symmetrical cable which is supported at the same level.
2. Asymmetrical cable which is supported at different levels.
4.5 Symmetrical Cable Subjected to Uniformly Distributed Load
In many practical structures, the suspension bridge for example, the dead load of the suspended
superstructure is carried at such numerous positions along the span so that it is sufficiently
accurate to consider the load on the cable as uniformly distributed across the span.
Figure 4.7 illustrates the typical problem. The symmetry of the arrangement demonstrates that:
1. the vertical relative forces at the supports are and
2. the tensile force in the cable at the mid span section is Ax or Bx (Ax = Bx).
Figure 4.7
For the whole structure:
Ay + By = wL.
For symmetrical cable:
and Ax = Bx = H.
Where,
L = horizontal distance (span between two supports)
h = vertical distance from the lowest point to supports
Consider a point at mid span (as shown in Figure 4.8).
Figure 4.8
For equilibrium:
Consider a general point, C (x, y), as shown in Figure 4.9.
Figure 4.9
Substituting the previous value of H and rearranging:
where y = vertical distance from any point on the cable to supports.
This is a parabolic equation and it defines the shape of the cable when subjected to a load
distributed uniformly with respect to the horizontal projection of the cable.
4.6 Forces in Cable Subjected to Uniformly Distributed Load
The tensile force, T in the cable at any position can be obtained from knowledge of the
horizontal and vertical components at the position. For example, at point C (refer to Figure
4.10) in the cable subjected to a uniformly distributed load,
where:
Figure 4.10
Horizontal equilibrium shows that,
and vertical equilibrium for the forces to the left of C gives
and acts in the direction shown. The tensile force in the cable at C is the resultant of H and V.
(4.1)
Maximum tensile force in the cable occurs at x = 0, which is at supports, and it is given by
(4.2)
Minimum tensile force in the cable occurs at
substitute into equation (4.1)
(4.3)
Example 4.1
Figure 4.11 shows a cable subjected to both point load and uniformly distributed load.
Determine
1. the reactions at A and B
2. Tmin and Tmax
3. size of the cable if the allowable stress is 660 kN/m2
Figure 4.11
Solution
1. Draw the FBD and calculate the reactions.
it is symmetrical,
Ay= By= 60 kN.
To prove:
2. To determine Tmin and Tmax.
Tmin (minimum tension).
Consider a point at mid span, x = 10 m.
Tmin occurs at mid span = H.
Tmax (maximum tension) which occurs at support A.
3. To determine the area of cable.
4.7
Asymmetrical Cable Subjected to Uniformly Distributed Load
Figure 4.12 shows an asymmetrical cable (supported at different level) where,
C = lowest point of the cable
d1 = vertical distance from the lowest point to support A
d2 = vertical distance from the lowest point to support B
l1 = horizontal distance from A to C
l2 = horizontal distance from C to B
Figure 4.12 Asymmetrical cable subjected to uniformly distributed load
Consider the whole structure:
Consider left hand side (AC), as shown in Figure 4.13.
Figure 4.13
Consider right hand side (BC), as shown in Figure 4.14.
Figure 4.14
Equate right hand side = left hand side,
(4.4)
Example 4.2
For the cable shown in Figure 4.15, find H.
Figure 4.15
Solution
4.8 Length of Cables
4.8.1 Cables loaded with point loads
For cable loaded with point loads, it is assumed that the self weight is negligible. Hence the
length is the sum of the length of each segment, as shown in Figure 4.16.
Figure 4.16 Cable loaded with point loads
Length of cable, S = s1+ s2 + s3 + s4.
Each segment is assumed to be straight.
4.8.2 Symmetrical cables
For cables loaded with uniformly distributed load, it can be showed that for symmetrical
cables:
Figure 4.17
To proof, refer to Figure 4.17,
Expanding this by the binomial theorem produces the following expression (neglecting higher
order terms).
4.8.3 Asymmetrical cables
For cables supported at different level, it can be showed that:
Figure 4.18
Example 4.3 – Cable subjected to point load
For the Figure 4.19 shown below, find:
1. Force (tension) in segments AC, CD, DE, and EB
2. The length of cable, S
Figure 4.19
Solution
1. Draw the FBD and calculate the reactions.
Consider the whole structure:
a. To calculate H.
Consider segment BE:
b. To calculate forces in segments of the cable.
Consider segment AC:
To calculate yC
To calculate force in T AC
Consider segment AD:
To calculate yD
To calculate force in TCD
Consider segment EB:
To calculate force in EB
Consider segment BED:
2. To find the length of the cable.
S = length of segment AC + length of segment CD + length of segment DE + length of
segment EB
Therefore.
4.9 Support Conditions
There are two types of support conditions for cables:
1. Pinned support, as example support B as shown in Figure 4.20.
Figure 4.20
2. Roller support, say support B as shown in Figure 4.21.
Figure 4.21
There is no horizontal reaction for roller support.
Example 4.4
The cable system as shown in Figure 4.22 is loaded with a uniformly distributed load of 10
kN/m between the roller supports. The horizontal distance between supports is 80 m and the
vertical distance between the lowest point and the left hand support is 3 m. The right hand
support is 6 m higher than the left support. Determine
1. the total length of cable, S
2. the maximum and minimum tension between A and B (TA and TB)
3. the tension in anchor cables (TA’and TB’)
4. vertical and horizontal reactions at supports (pressure on tower)
Figure 4.22
Solution
1. Length of asymmetrical cable is given by
2. To calculate Tmin (which occurs at the lowest point). Consider the whole structure.
Consider left hand side (portion AC, where C is the lowest point).
Check with right hand side.
To calculate Tmax:
3. To calculate tension in anchor cables (depends on a or β and type of support ).
Consider support A:
Tension in anchor cable at support A.
Consider support B:
Tension in anchor cable at support B.
4. To calculate vertical reaction at support. (Refer to figure in section c.)
At support A:
There are no horizontal reactions since these are roller supports.
RAx = RBx = 0.
Example 4.5
The cable system in Figure 4.23 is loaded with a uniformly distributed load of 12 kN/m and
having a self weight of 6 kN/m. Both supports are pinned. The horizontal distance between
supports is 80 m and the vertical distance between the lowest point and the left hand
support is 15 m. The right hand support is 9 m lower than the left support. Determine
1. the lowest point of the cable, x
2. the maximum and minimum tension between A and B (Tmax and Tmin)
3. tension and reactions in anchor cables (TA', TB', RBx, RBy, RAx, RAy )
4. size of cable, if allowable stress is 14400 kN/m2
Figure 4.23
Solution
1. The lowest point of the cable, x. Consider the whole structure.
Note:
cannot be applied becausethere are two point loads and uniformly distributed load.
Figure 4.24
(4.4)
(4.5)
(4.6)
Consider left hand side (portion AC, where C is the lowest point).
(4.7)
Consider right hand side portion.
(4.8)
Equate equations (4.7) and (4.8)
2. To calculate Tmax and Tmin.
Tmin occurs at the lowest point or at x = 48.9 m.
From equation (4.7),
H = 93.93 + 0.6x2,
H = 1528kN = Tmin.
From equation (4.6),
From equation (4.4),
Therefore, Tmax occurs at the highest point or at support,
3. Tension and reactions in anchor cables [(TA, TB, RBx , RBy RAx, RAy) depends on a or β and
type of support].
Consider support A (pinned).
Consider support B (pinned)
4. Size of cable, if allowable stress is 14400 kN/m2
EXERCISES
Exercise 1
A symmetrical cable system shown in Figure 4.25 carries a uniformly distributed load of 3
kN/m and point loads of 20 kN and 40 kN between the support A and B with a span of 40 m.
The lowest point of the cable is located at a distance of 8 m vertically from the supports.
Determine
a. the horizontal distance of the lowest point of the cable from supports
b. the maximum and minimum tension between A and B (Tmin and Tmax)
c. the tension in anchor cable (TA’ and TB’), vertical and horizontal reactions at supports
(RAy , RAx and RBy, RBx). Repeat the calculations if the angle of anchor cable is reduced
to 30°. State the relation between changes in angle and in tension of anchor cable
d. the allowable stress of the cable if the diameter of the circular cable is 0.1 m
Figure 4.25
Answers:
a. x = 21.67 m
b. Tmin = H = 113.02 kN; Tmax = 147.64 kN
c. T ’A = 141.42 kN; T ’B = 147.54 kN; RAy = 175.90 kN; RAx = 4.67 kN;RBy = 189.84 kN;
and RBx = 0
If the anchored angle is reduced to 30°, some values are reduced to: T’A = 141.42 kN (
unchanged); T’B = 130.50 kN; = 155.71 kN; Rax = −9.47 kN; = 160.25 kN; and RBx = 0
(unchanged)
d. Allowable stress = 18798 kN/m2
Exercise 2
The cable system shown in Figure 4.26 loaded with uniformly distributed load of 6 kN/m
between the pinned supports. The horizontal distance between supports is 60 m and the
vertical distance between the lowest point and the left hand support is 3 m. The right hand
support is 6 m higher than the left support. Determine
b. the maximum and minimum tension between A and B
c. the tension in anchor cables
d. size of the cable, if the allowable stress = 14000 kN/m2
Figure 4.26
Answers:
a. S = 61.693 m
b. Tmin = H = 482.33 kN; Tmax = 533.60 kN
c. T’A = 500 kN; T’B= 533.60 kN
d. d = 0.22 m
Exercise 3
The cable system shown in Figure 4.27 carries a uniformly distributed load of 5 kN/m between
the supports and three point loads of 10 kN, 15 kN, and 20 kN at 15 m interval. The horizontal
distance between supports is 60 m and the vertical distance between the lowest point and the
left support is 3 m. The right support is 6 m higher than the left support. Determine
a. the lowest point of the cable (x)
b. the maximum and minimum tension between A and B (Tmin and Tmax)
c. the tension in anchor cable (TA’ and TB ’)
d. vertical and horizontal reactions at supports (RAy, RAx and RBy, RBx)
e. size of the cable, if the allowable stress = 14000 kN/m2
Figure 4.27
Answers:
a. x = 22.54 m
b. Tmin = H = 473.38 kN; Tmax = 522.99 kN
c. T’A = 489.01 kN; T’B = 546.61 kN
d. RAy = 468.44 kN; RAx = 127.60 kN; RBy = 495.64 kN and RBx = 0
e. d = 0.218 m
CHAPTER 5
ANALYSIS OF THREE-HINGED ARCH
5.1 Introduction
Arches have been used for a very long time to span large distance. Bridges, building, and many
others have used the ability of the arch to carry transverse loading efficiently.
The arch carries most of the load axially with bending moment greatly reduced due to the
curvature of the arch. In this chapter, only three-hinged arch will be considered, which is
similar to three-hinged portal frame.
The restraints at both ends reduce horizontal thrusts which are in the opposite direction to the
loading and thus reduce the bending moments.
Bridges are often constructed as three-hinged arches. This type of construction is particularly
suitable when compression-proof building materials are available. Horizontal thrust occurs in
the arch at the supports. It permits much lower bending moments in the arch than in the case of
a beam on two supports with the same span. A significant longitudinal compressive force is
active in the arch to produce this effect. The Permaisuri Tuanku Bainun Bridge, Manjung, Perak
shown in Figure 5.1 is an example of arch bridge.
Figure 5.1 Permaisuri Tuanku Bainun Bridge, Manjung, Perak
5.2 Objectives
At the end of Chapter 5, students should be able to
1. calculate reactions of the three-hinged arch, and
2. draw shear force, thrust, and BMDs for three-hinged arch.
5.3 Comparison Between a Beam and an Arch
For the arch, the loads give their usual downward bending effect but are counteracted by the
upward bending of the thrusts at the ends to reduce the bending moment. Figure 5.2 shows a
sketch of reactions for beam and arch.
Figure 5.2 (a) Beam and (b) arch
At any point, bending moment in the arch is equal to the bending moment of point on the
equivalent beam minus Hy.
Now, Hy is proportional to y which is a parabola. Therefore, Hy is also parabola since H is
constant. BMD for simply supported beam and arch are shown in Figure 5.3.
Figure 5.3
5.3.1 Advantage of the arch
It has been pointed earlier that the bending moments in the arch rib have been considerably
reduced by the action of the horizontal thrust at the abutments. In conclusion, the arch rib
experiences a high thrust, N with small moments for which a much smaller section than that for
a corresponding beam is required. Therefore, arch spans can be much longer than the span of
beams.
5.4 Determinate and Indeterminate Arches
The method for determining the determinacy of a structure is discussed in Chapter 1. To
determine the statical determinacy of an arch, the following example will give a clearer idea.
Example 5.1
Classify each of the arches shown in Figure 5.4 as statically determinate or statically
indeterminate.
Solution
The arch shown in Figure 5.4(a) contains one member, six reaction forces.
Number of equilibrium equation available = 3.
Number of reaction forces
= 6.
Statically indeterminate arch.
The arch shown in Figure 5.4(b) contains one member, four reaction forces.
Number of equilibrium equation available = 3.
Number of reaction forces
= 4.
Statically indeterminate arch.
The arch shown in Figure 5.4(c) contains two members, six reaction forces.
Figure 5.4
Number of equilibrium equation available = 6.
Number of reaction forces
= 6.
Statically determinate arch.
5.5 Equation of a Parabolic Arch
There are two types of equation for arch, namely:
1. Parabolic arch (as shown in Figure 5.5)
Figure 5.5
2. Semi circle arch (as shown in Figure 5.6)
x2 + y2 = R2.
Figure 5.6
Since there are four reactions, Ax, Ay, Bx, By, and four equations of equilibrium (including one
at the third hinge). Therefore, all unknown reactions can be found.
The most common arch curve is the parabolic shape. An arch shown in Figure 5.7 referred to x
and y axes since the forces and moments in an arch depend upon its shape, and it is necessary
to know the equation.
Figure 5.7
Consider a point A (0, 0), any point P (x, y), C (L/2, h), and B (L, 0). By using parabolic
equation, y = ax2 + bx + c.
(5.1)
(5.2)
Substituting equations (5.2) into (5.1),
Therefore,
equation for a parabolic arch.
It is also important to know the angle, and the arch make with the horizontal.
Where θ = tan -1(slope).
Example 5.2
Calculate the reaction for the three-hinged arch shown in Figure 5.8.
Figure 5.8
Solution
1. Draw the FBD as shown in Figure 5.9.
Figure 5.9
2. To calculate the reactions.
Consider the whole structure:
(5.3)
(5.4)
Substituting equations (5.4) into (5.3),
(5.5)
Consider right hand side of the arch as shown in Figure 5.10.
Figure 5.10
(5.6)
From equation (5.5),
Ax= Bx= 268.75 kN.
Example 5.3
Calculate the reaction for the three-hinged arch shown in Figure 5.11.
Figure 5.11
Solution
1. Draw the FBD as shown in Figure 5.12.
Figure 5.12
2. To calculate the reactions.
Consider the whole structure:
Consider the left hand side of the arch as shown in Figure 5.13.
Figure 5.13
5.6 The Bending Moment in the Arch Rib
Bending moments in an arch, though much lesser than those in a beam of the same span, they
are nevertheless significant and need to be calculated. Consider Figure 5.14:
Figure 5.14
March + Axy – Ayx + (moment of loads) = 0
March + Axy = Ayx – (moment of loads).
But for normal beam moment (refer to Figure 5.15).
Figure 5.15
When March + Axy = beam moment
March = beam moment − Axy.
This shows that the bending moment in an arch is very much less than bending moment in
a beam.
5.7 The Thrust and Shear Force in the Arch Rib
The thrust (N) and shear force (Q) maybe found by simply resolving the forces on the arch in
the correct direction. Care needs to be taken in calculating their values at the position of a
point load. The fact is that there is an abrupt change in value of the thrust and shear on either
side of a point load. Hence, the thrust and shear force need to be calculated on both sides of a
point load.
Example 5.4
Determine the variation in thrust and shear force at every 10 m across the arch from
support A to B as shown in Figure 5.16.
Figure 5.16
Solution
1. Draw the FBD as shown in Figure 5.17.
Figure 5.17
2. Calculate the reactions.
Consider the whole structure:
(5.7)
(5.8)
(5.9)
Substitute By into equation (5.8), Ay = 300 kN.
Consider the right hand side of point C of the arch as shown in Figure 5.18.
Figure 5.18
(5.10)
From equation (5.7), Ax= Bx= 375 kN.
At 10 m from support A, consider the left hand side of point C of the arch as shown in
Figure 5.19.
With loading 15 kN/m:
Figure 5.19
Resolving in N direction (θ = 21.8°).
Resolving in Q direction.
Continue to calculate shear forces and thrusts at 20 m, 30 m, and 40 m from support A.
Example 5.5
Draw the SFD, BMD, and axial thrust diagram for the arch shown in Figure 5.20.
Figure 5.20
Solution
1. Draw the FBD as shown in Figure 5.21.
Figure 5.21
2. To calculate the reactions.
Consider the whole structure:
(5.11)
(5.12)
(5.13)
Subtituting equation (5.13) into equation (5.12),
Ay = 75 kN.
Consider the right hand side of point C of the arch as shown in Figure 5.22.
(5.14)
From equation (5.11), Ax = Bx = 78.125 kN.
3. Calculate N and Q. Draw the assumed direction of N (tangential to point D) and Q
(perpendicular to N) as shown in Figure 5.23.
At x = 25 m, point D:
Figure 5.23
Without loading:
Resolving in N direction.
Resolving in Q direction.
With 100 kN loading:
Draw the assumed direction of N (tangential to point D) and Q (perpendicular to N) as shown
in Figure 5.24.
Figure 5.24
Resolving in N direction.
Resolving in Q direction.
At x = 50 m, point C:
Draw the assumed direction of N (tangential to point C) and Q (perpendicular to N) as
shown in Figure 5.25.
Figure 5.25
Resolving in N direction.
Resolving in Q direction.
At x = 0 m, point A:
Draw the assumed direction of N (tangential to point A) and Q (perpendicular to N) as
shown in Figure 5.26.
Figure 5.26
Resolving in N direction.
Resolving in Q direction.
At x = 100 m, point B:
Draw the assumed direction of N (tangential to point B) and Q (perpendicular to N) as
shown in Figure 5.27.
Figure 5.27
Resolving in N direction.
Resolving in Q direction.
At x = 75 m, point E:
Draw the assumed direction of N (tangential to point B) and Q (perpendicular to N) as
shown in Figure 5.28.
Figure 5.28
Resolving in N direction.
Resolving in Q direction.
4. Calculate bending moment on the arch.
MA = 0; MB = 0; MC = 0.
At x = 25 m (from left hand side of point C) as shown in Figure 5.29:
Figure 5.29
At x = 25 m (from right hand side of point C) as shown in Figure 5.30:
Figure 5.30
At x = 20 m (from left hand side of point C) as shown in Figure 5.31:
Figure 5.31
At x = 40 m (from left hand side of point C) as shown in Figure 5.32:
Figure 5.32
At x = 40 m (from right hand side of point C) as shown in Figure 5.33:
Figure 5.33
At x = 20 m (from right hand side of point C) as shown in Figure 5.34:
Figure 5.34
5. Draw the BMD, thrust diagram, and SFD as shown in Figure 5.35.
Figure 5.35 (a) BMD, (b) Thrust diagram, and (c) SFD
EXERCISES
Exercise 1
A symmetrical three-hinged arch shown in Figure 5.14 is in the form of
where L =
40 m and h = 8 m. Support A is 2.88 m lower than support E. It is designed to carry a uniformly
distributed load of 5 kN/m spanning 26 m on span BCDE. A horizontal uniformly distributed
load of 2 kN/m is applied from A to C. Joint A, C, and E are hinged. Determine
a. support reactions at A and E
b. bending moment at B and D
c. shear force and thrust at point B and D (with loading)
d. sketch the BMD of the arch
Figure 5.14
Answers:
a. Ay = 56.22 kN; Ax = 101.29 kN; Ey = 83.78 kN; Ex = 117.29 kN
b. MB = − 81.58 kNm (hogging); MD = 46.75 kNm (sagging)
c. At B: θB = 21.8°; NB = 126.07 kN; QB = 10.12 kN
At D: θD = 21.8°; ND = 125.16 kN; QD = –2.91 kN
d. BDM
Exercise 2
A three-hinged arch shown in Figure 5.15 is in the form of
. It is designed to carry
a uniformly distributed load of 2 kN/m spanning 30 m on span ABCD and a point load of 8 kN
and 5 kN at point B and D, respectively. Joint A, C, and E are hinged. Determine
a. support reactions at A and E
b. bending moment at B and D
c. shear force and thrust at point B and D (with loading)
Figure 5.15
Answers:
a. Ay = 36.87 kN; Ax = 27.56 kN; Ey = 30.13 kN; Ex = 27.56 kN
b. MB = 21.55 kNm (sagging); MD = 1.09 kNm (sagging)
c. At B: θB = 38.66°; NB = 33.3 kN; QB = 2.48 kN
At D: θD = 43.83°; ND = 37.28 kN; QD = −0.95 kN
Exercise 3
A symmetrical three-hinged arch shown in Figure 5.16 is in the form of
, where L =
40 m and h = 10 m. Support A is 3.6 m higher than support E. It is designed to carry a
uniformly distributed load of 2 kN/m spanning 26 m on span ABCD and a point load of 8 kN
and 5 kN at point B and D, respectively. Joint A, C, and E are hinged. Determine
a. support reactions at A and E
b. bending moment at B and D
c. shear force, Q and thrust, N on the right hand side of point C and D (with loading)
Figure 5.16
Answers:
a. Ay = 37.22 kN; Ax = 40.56 kN; Ey = 27.78 kN; Ex = 40.56 kN
b. MB = 29.14 kNm (sagging); MD = –26.4kNm (hogging)
c. At C: θC = 0°; NC = 40.56 kN; QC = –2.78 kN
At D: θD = 26.57°; NQ = 46.47 kN; QD = –2.23 kN
CHAPTER 6
DEFLECTION
6.1 Introduction
Structures may deform and change shape when being subjected to forces. Other common causes
of structure deformation include temperature changes, construction errors, and support
settlements. If the structure regains its original shape after the deformation-causing actions are
removed, then the deformation is termed “elastic deformation”. In this chapter, the deformation
of the beam is assumed linearly elastic.
Deflection is an important quantity to be calculated in structural analysis and design.
Deflections are determined and used to solve problems in the analysis of statically
indeterminate structures. Deflections are also important in dynamic analyses where the
response of structures subjected to dynamic loadings (earthquake, wind, or vibration) is
determined. In structural design, deflections are calculated and checked in order to make sure
they are within the maximum allowable limits stipulated in the design codes. Large deflections
can cause cracks in non-structural elements (non-load-bearing walls or ceilings), distortion of
door and window frames, and water ponding of flat slab of buildings or undesirable vibrations
of structures. Excessive deflections can cause structure unsightly, unoccupied, and possibly
unsafe.
Deflections can be computed using (1) geometric methods, and (2) work energy methods. In
this chapter, the elastic beam theory and the direct (double) integration method are briefly
reviewed. Next, the geometric methods are discussed particularly to determine the slopes and
vertical displacement of statically determinate beams. Geometric methods are based on a
consideration of the geometry of the deflected shapes of structures. There are two approaches
in geometric method, which are moment-area method and conjugate beam method.
6.2 Objectives
At the end of Chapter 6, students should be able to
1. understand the elastic beam theory, and
2. calculate the deflection of simple beam using the moment-area method and the conjugate
beam method.
6.3 Elastic-Beam Theory
To derive the deflection equation, an initially straight beam that is elastically deformed by
arbitrary loads is considered. The loads are applied perpendicular to the x axis of the beam
and lying in x-y axes as shown in Figure 6.1(a). Due to the loads, the beam deforms into a
curve under shear and bending forces. The longitudinal axis of the beam in the deformed state
is known as the elastic curve of the beam. Consider a differential element dx of the beam, the
angle between the cross-sections and the width of the small element of the deflected beam
shown in Figure 6.1(b) are dθ and dx, respectively.
Figure 6.1
The radius of curvature, R for this arc is measured from the centre of curvature O' to the neutral
axis of the element. The strain in arc, ds that is located at a distance of y from the neutral axis
is given by:
(6.1)
However,
ds = dx = Rdθ,
(6.2)
and
ds′ = (R − y) dθ.
(6.3)
Substitute equations (6.2) and (6.3) into equation (6.1), the following equation is obtained.
which can be further simplified as follows:
(6.4)
The material of the beam is assumed to be homogeneous and it behaves in a linear elastic
manner, thus Hooke’s Law applies:
(6.5)
The flexure formula also applies:
(6.6)
Combine equations (6.5) and (6.6), the following equation is obtained: 1M
(6.7)
where R = the radius of curvature at a specific point on the elastic curve
M = the internal moment in the beam at the point where R is to be determined
E = the material’s modulus of elasticity
I = the beam’s moment of inertia
The product of EI in the equation is flexural rigidity of the beam.
Since
dx = Rdθ,
(6.8)
Substitute equation (6.7) into equation (6.8), thus
(6.9)
Since θ = dy/dx, equation (6.9) can also be expressed as
(6.10)
This differential equation for the deflection of beams is also known as the Bernoulli-Euler
beam equation.
6.4 Direct Integration Method
For simple beams subject to simple loading in which M/EI can be expressed as a single
continuous function of x over the entire length of the beams, the direct (double) integration
method is the most simple and convenient approach for computing slopes and deflections of
beams. The direct integration method essentially involves writing the expression for M/EI in
terms of the distance x along the axis of the beam and integrating this expression successively
to obtain equations for the slope and deflection of the elastic curve. The constants of
integration are determined from the boundary conditions.
Applying equation (6.10) and integrating in x, the slope of beams is obtained:
(6.11)
Integrating equation (6.11) twice, the vertical deflection of beams is obtained:
(6.12)
When a beam is subjected to several loads or a beam consists of different cross sections for
which M/EI function is not continuous, each discontinuity due to a change in loading and/or
flexural rigidity (EI) introduces two additional constants of integration in the analysis and the
problem becomes complicated. Superposition method is applied to determine slope and
deflection caused by the combined effect of loads by superimposing (algebraically adding) the
slopes and deflections (calculated from the direct integration method) due to each of the loads
acting individually on the beam. The difficulty can also be simplified by employing the
singularity functions.
Example 6.1
Determine the equations for slope and deflection of the beam shown in Figure 6.2. Also
determine the slopes and deflections at points B and C of the beam. Given that E = 200 GPa
and I = 360(106) mm4.
Figure 6.2
Solution
1. Draw the FBD of the beam and determine the reaction forces at support A using static
equilibrium equations.
2. Determine the equation for bending moment. Make a cut section at a distance x from support
A. Considering the free body to the left of this section, the equation obtained as follows:
3. The EI of the beam is constant. The equation for M/EI can be written as
4. Integrating the equation for M/EI in x and the equation for slope of the beam as
where C1 is constant.
5. Integrating once more, the equation for deflection of the beam is obtained as
where C2 is constant.
6. The constants of integration, C1 and C2 are evaluated by applying the boundary conditions:
At end A, x = 0, θ = 0; C1 =0.
At end C, x = 0, y = 0; C2 = 0.
7. The equations for slope and deflection of the beam are
8. Substitute numerical data for E, I, and x = 5 m and x = 10 m for points B and C, respectively,
the slopes and deflections at ends B and C are obtained as
The “negative sign” indicates that the angle is measured clockwise from A and the deflection
is downward.
6.5 Moment-Area Method
The moment-area method was developed based on two theorems by Otto Mohr and later stated
formally by Charles E. Greene in 1873. It employs a semi-graphical technique to determine the
slope and vertical displacement of the elastic curve due to bending. Furthermore, it relates the
geometry of the beam’s elastic curve to its M/EI diagram.
The following procedures are used to determine the displacement and slope at a point on the
elastic curve of a beam using the moment-area theorems.
Step 1 : Draw M/EI diagram
The analysis of statically determinate beam is carried out using static equilibrium equations
and from the analysis, the BMD is drawn. To draw the M/EI diagram, the BMD is divided by
the flexural rigidity of the beam. Figure 6.3 shows the steps in drawing the M/EI diagram. Note
that the flexural rigidity throughout the beam “may not be the same” as shown in Figure 6.3(b).
Figure 6.3
Step 2: Draw the elastic curve of the beam under the applied load
Draw an exaggerated view of the elastic curve of the beam. Note that the points of zero slope
occur at fixed supports while zero displacement at fixed, pinned, and roller supports. The
bending moment or M/EI diagram can be referred if the general shape of the elastic curve is
difficult to be drawn. Realize that when the beam is subjected to a positive moment, the beam
bends concave up whereas negative moment bends the beam concave down. The displacement
and slope which are to be determined should be indicated on the elastic curve.
Step 3: Apply Theorem 1 and Theorem 2 of the moment-area method
Theorem 1
Figures 6.4(b) and 6.4(c) show the elastic curve and M/EI diagram of the beam under the
applied load. By using equation (6.9) and Figure 6.4, dθ on either side of the element is equal
to the shaded area under the M/EI diagram. This equation forms the basis for the first momentarea theorem.
Theorem 1
The change in slope between any two points on the elastic curve equals to the area of the M/EI diagram
between the two points.
Hence, the area under M/EI diagram between points A and B on the elastic curve represents
the internal angle between the two tangent lines at A and B, respectively, which can be
computed as follows:
(6.13)
Theorem 2
The second moment-area theorem is based on the relative derivation of tangents to the elastic
curve. Figures 6.4(c) and 6.4(d) show a greatly exaggerated view of the vertical deviation, dt
of the tangents on each side of the differential element, dx. Since the slope of elastic curve and
its deflection are assumed to be very small, it is satisfactory to approximate the length of each
tangent line by x and the arc ds’ by dt as shown in Figures 6.4(c) and 6.4(d). From the equation
of calculating the arc length, s = rθ, thus
(6.14)
dt = xdθ.
Substitute equation (6.9) into equation (6.14),
(6.15)
Figure 6.4
Theorem 2
The vertical deviation of the tangent at a point (A) on the elastic curve with respect to the tangent extended from
another point (B) equals to the “moment” of the area under the M/EI diagram between the two points (A and B).
Thus, the vertical deviation of the tangent at A with respect to the tangent at B can be
determined by integration,
(6.16)
Example 6.2
Determine the slope at points B and C of the beam as shown in Figure 6.5. Given that E =
200 GPa and I = 360(106) mm4.
Figure 6.5
Solution
1. Draw the FBD of the beam and determine the reaction forces at support A using static
equilibrium equations.
2. Draw the BMD and then divide it by the EI of the beam to obtain M/EI diagram. It is easier
to solve the problem in terms of EI.
3. The elastic curve of the beam subjected to 10 kN concentrated load is shown in figure
below. Draw tangent line at A, B, and C. The tangent line at A is horizontally straight (no
angle of rotation at A since the beam is fixed at A).
4. The angle between tan A and tan B, θAB is equivalent to θB.
θB = θAB; θC = θAC.
5. Applying Theorem 1, θAB is equal to the area under the M/EI diagram between points A and
B.
6. Substitute the numerical data for E and I.
7. The “negative sign” indicates that the angle is measured clockwise from A.
8. In a similar manner, the area under the M/EI diagram between points A and C (θAC) is
equivalent to θC, hence
9. Substitute numerical data for E and I.
Example 6.3
Determine the vertical displacement at points B and C of the beam shown in Figure 6.6.
Values for the moment of inertia of each segment are indicated as shown below. Given that
E = 200 GPa and I = 4(106) mm4.
Figure 6.6
Solution
1. Draw the FBD of the beam and calculate the reaction forces at A.
2. Draw the BMD and then divide it by the EI of the beam to obtain M/EI diagram as shown in
figure below. It is easier to solve the problem in terms of EI. Note that the EI of span AB
and span BC are not the same.
3. The couple moment at C causes the beam to deflect as shown in the figure below. The
tangent lines at A, B, and C are indicated.
4. Vertical displacement at point A and B can be related directly to the deviations between the
tangent lines. So that from the elastic curve of the beam above,
ΔB
= tBA and ΔC = tCA.
5. Applying Theorem 2, vertical displacement at B is equivalent to the moment of the area
under the M/EI diagram between A and B which is computed about B. Hence
6. Likewise for tCA, the vertical displacement at C is equivalent to the moment of the entire
M/EI diagram from A to C which is computed about C.
6.6 Conjugate Beam Method
The conjugate beam method was developed by H. Müller-Breslau in 1865. Essentially, it
requires the same amount of computation as the moment-area theorems to determine a beam’s
slope or deflection. However, this method is based on the similarity between the relationships
for loading and internal shear, also internal shear and moment as listed below.
The slope and deflection of the elastic curve are related to the internal moment by the
following expressions [equations (6.17) and (6.18)].
(6.17)
or
(6.18)
The following expressions [equations (6.19) and (6.20)] relate the internal shear and moment
to the applied load.
(6.19)
or
(6.20)
Compare the expression for slope, θ and internal shear, V. If the applied load, w is replaced by
the term M/EI, the expression for slope and shear are identical. The same explanation goes to
the expression for displacement, v and internal moment, M. These comparisons can be
considered by having a beam with the same length as the real beam which referred as the
“conjugate beam”. Conjugate beam is loaded with the M/EI diagram derived from the load,
won the real beam. Therefore, the two theorems that are related to the conjugate beam method
can be written as follows.
Theorem 1
The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam.
Theorem 2
The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the
conjugate beam.
Figure 6.7 shows two examples of conjugate beam corresponding to the real beam. Note the
sign of loading, w and the M/EI on the conjugate beam.
Figure 6.7
6.6.1 Conjugate beam support
When drawing the conjugate beam, it is important that the shear force and bending moment
developed at the supports of conjugate beam correspond to the slope and displacement of the
real beam. For example, a cantilever beams as shown in Figure 6.7(a). The real beam is fixed
at A while free end at C which provides zero slope and displacement at A, but the beam has a
non-zero slope and displacement at C. Thus, from Theorem 1 and Theorem 2, the conjugate
beam must be fixed supported at C′ and free end at A′ as the shear force and bending moment at
the support of the conjugate beam correspond to the slope and displacement of the real beam.
Figure 6.8 shows more examples of the real beam and the conjugate beam subjected to their
support conditions.
6.6.2 Procedure for analysis
The following procedure provides a method that could be used to determine the displacement
and slope at a point on the elastic curve of a beam using the conjugate beam method.
1. Draw the BMD of the real beam and consequently draw the M/EI diagram. Note that the EI
of the beam which in some cases, they are not the same throughout the beam.
2. Draw the conjugate beam and it is important to make sure the support condition of the
conjugate beam corresponds to the real beam.
3. The conjugate beam is loaded with M/EI diagram from the real beam. This loading is
assumed to be distributed over the conjugate beam and is directed upward when M/EI is
positive or downward when M/EI is negative. In other words, the loading always acts away
from the beam.
4. Analyse the conjugate-beam using the equations of equilibrium to determine the reaction
forces at the conjugate beam’s supports
5. Section the conjugate beam at the point where the slope θ and displacement Δ of the real
beam are to be determined. At the section, show the unknown shear, V′ and moment, M′
according to their positive convention.
6. Determine the shear and moment using method of section or moment equation method. V′ and
M′ in the conjugate beam is equivalent to θ and Δ in the real beam, respectively.
Figure 6.8
Example 6.4
Determine the slope and vertical displacement at point C of the steel beam shown in Figure
6.9. Take E = 200 GPa and I = 475(106) mm4.
Figure 6.9
Solution
1. Draw the FBD and compute the reaction forces at support A.
2. Draw the BMD and M/EI diagram correspond to the applied load and EI of the beam,
respectively.
3. Draw the conjugate beam as shown in figure below. The end condition at A′ and C′ in the
conjugate beam are corresponding to the support A and the free end C in the real beam.
4. Load the conjugate beam with the M/EI diagram from the real beam. The M/EI diagram is
negative, so it acts downwards on the conjugate beam which is away from the beam as
shown in figure below.
5. Draw the FBD of the conjugate beam and calculate the reaction forces at support C'. As
shown in figure below, the slope and vertical displacement at C in the real beam are
equivalent to the reaction force and moment at support C′ in the conjugate beam,
respectively.
6. Calculate reaction force at C′ in the conjugate beam.
The negative sign indicates that the angle is measured clockwise from A.
7. Calculate moment at C′ in the conjugate beam.
The negative sign indicates that the deflection is downward.
Example 6.5
Determine the maximum vertical displacement of the steel beam shown in Figure 6.10. Take
E = 200 GPa and I = 60(106) mm4.
Figure 6.10
Solution
1. Draw the FBD of the beam and compute the reaction forces at both supports A and C as
shown in figure below.
2. Draw the BMD and M/EI diagram as shown below.
3. Draw the conjugate beam which corresponds to the real beam particularly end/support
conditions. Load the conjugate beam with M/EI diagram from the real beam as shown in
figure below. Since M/EI diagram is positive, it loads the conjugate beam in upward
direction.
a. Draw the FBD of the conjugate beam and calculate the reaction forces at both supports A′
and C′ by using static equilibrium equations. The reaction forces at A′ and C′ in the
conjugate beam are shown in figure below.
Maximum displacement of the real beam occurs at the point where the slope of the beam is
zero. Assume this point acts within span A′B′ (0 ≤ x ≤ 9 m) which the origin of x is taken
from point A′. Note that the peak value of the distributed load is determined from
proportional triangles.
When V′ = 0,
From the above calculation, it shows that the maximum displacement occurs at x = 6.71 m
from point A′ which is within span A′B′. The maximum displacement in the real beam
corresponds to the moment, M′ in the conjugate beam can be calculated as follows:
The negative sign indicates the deflection is downward.
EXERCISES
Exercise 1
Sketch the elastic curve for the beams shown in Figures 6.11(a) and 6.11(b). The
corresponding BMD is shown directly beneath each beam.
Figure 6.11
Answers:
Exercise 2
Using moment-area method and conjugate beam method, calculate the slope and vertical
displacement at point C for each of the structure shown [Figures 6.12(a)–6.12(d)]. The flexural
rigidity of all beam sections is indicated in the figure.
Figure 6.12
Answers:
CHAPTER 7
INFLUENCE LINES
7.1 Introduction
Influence lines have important application in the design of structures that resist large live
loads. In this chapter, steps to draw the influence lines for a statically determinate beam will
be discussed.
7.2 Objectives
At the end of Chapter 7, students should be able to
1. draw influence lines of reaction forces, shear force, and bending moment for beams, and
2. determine location of moving load which produces maximum reaction forces, shear force,
and bending moment for beams using influence lines.
7.3 Influence Lines
The variation of shear and moment for a structure which is subjected to a moving load can be
best described using the influence lines. A moving unit load is placed on the structure to create
the ordinate of shear and moment for a specified point along the beam. The magnitude of the
associated shear, moment at the point can then be calculated based on the ordinate on the
influence lines. Note that the SFD or BMD and influence lines are two different diagrams even
though the latter can also be drawn for shear and moment functions. Influence lines represent
the effect of a moving load only at a specific point while SFD and BMD represent the effect of
fixed loads at all points along the axis of the member. Both quantitative and qualitative
approaches for determining influence lines are discussed in this chapter.
7.3.1 Quantitative influence lines
Either of the following two procedures can be used to construct the influence lines for any
functions (reaction, shear, or moment) at a specific point of the beam.
1. Tabulated ordinate values
a. Draw the FBD of the beam.
b. Place a unit load at the selected points on the beam (i.e. at uniform interval along the
beam). At each point, calculate the ordinate value of the influence line for any functions
(reaction, shear, or moment) using static equilibrium equations. The influence line is
drawn based on the calculated ordinate value.
c. All influence lines for statically determinate beams only consist of straight line segments.
Example 7.1
Construct the influence lines for the following functions (1) vertical reaction at A, (2) shear
force at B, and (3) moment at B for the beam in Figure 7.1.
Figure 7.1
Solution
1. Influence line for vertical reaction at A.
a. Draw the FBD of the beam that indicates a unit load at every 3 m interval on the beam
(including at x = 0 and x = 9 m). The figure below shows a unit load moving from point A to
C.
b. Calculate reaction force at A(Ay) for each of the FBD above using static equilibrium
equation.
c. Tabulate all the reaction forces at A corresponding to each FBD in a table as shown below.
Plot a graph based on the data which then yields the influence line for the reaction force at A
as shown below. The values of Ay at every interval are known as the ordinate of influence
line for reaction force at A.
2. Influence line for shear force at B (x = 6 m)
a. Make use of the reaction forces at A correspond to each FBD in solution (1) of example 7.1.
The ordinate of influence line for shear force at B (x = 6 m) can be solved using method of
section. Consider span AB (span BC can also be used to solve this problem) and draw it
subjected to different unit load position (at every 3 m interval including x = 0 and x = 9 m).
Calculate the ordinate values of influence line for shear force based on each of the span AB
drawn.
i. A unit load at x = 0 and Ay = 1
ii. A unit load at x = 3 m and Ay = 2/3
iii. A unit load at x = 6 m and Ay = 1/3. There are two cases need to be considered for a unit
load at B which are just to the left and just to the right of point B. For the latter case, a unit
load is out of span AB, hence it will not be included in the calculation.
iv. A unit load at x = 9 m and Ay = 0
b. Tabulate all the ordinate values of the influence line for shear force at B in a table as shown
below. Plot a graph based on the tabulated data which yields the influence line diagram for
the shear force B as shown in the figure below.
3. Influence line for bending moment at B.
a. Consider span AB which has been drawn in solution (2) (i)–(iv). Calculate the ordinate
values of influence line for bending moment based on each of the span AB drawn.
i. A unit load at x = 0 and Ay = 1
ii. A unit load at x = 3 m and Ay = 2/3
iii. A unit load at x = 6 m and Ay = 1/3
iv. A unit load at x = 9 m and Ay = 0
b. Tabulate all the ordinate values of the influence line for shear force at B in the table as
shown below. Plot a graph based on the tabulated data which yields the influence line
diagram for the shear force B as shown in the figure below.
2. Influence-line equations
The influence lines can also be constructed by placing a unit load at a variable position x on
the beam and then derive an equation of reaction, shear, or bending moment forces at that point
as a function of x. A graph can easily be plotted based on the derived equation which
represents the influence lines. For statically determinate beam, the influence line equation is a
linear function.
Example 7.2
Construct the influence line for (1) reaction force at A, (2) shear force at B, and (3) bending
moment at B of the beam shown in Figure 7.2 by using influence line equation method.
Figure 7.2
Solution
1. Influence line diagram for reaction force at C.
a. Draw the FBD of the beam as shown in figure below. Place a unit load at variable distance x
from A on the beam.
b. The reaction force at C as a function of x can be determined as follows:
c. Draw a graph based on the equation above which yields the influence line diagram for
reaction force at A. Note that the equation is linear (y = mx + c). The ordinate value of the
influence line for reaction force at A can be calculated from the derived equation.
2. Influence line diagram for shear force at B.
a. Consider span AB as shown in the figure below. Figure (a) illustrates a unit load moves
within span AB while figure (b) shows the load moves beyond point B. Hence, there will be
two different equations of influence line for shear force at B.
(a) A unit load within span AB (0 ≤ x ≤ 5 m)
(b) A unit load within span BC (5 ≤ x ≤ 10 m)
b. A unit load within span AB (0 ≤ x < 5 m)
c. A unit load within span BC (5 < x ≤ 10 m)
d. Draw a graph based on both equations which according to their range of x. The graph drawn
below is the influence line for shear force at B. The ordinate value of influence line can be
calculated from the derived equations.
3. Influence line diagram for bending moment at B.
a. Consider span AB as shown in figure below. Figure (a) illustrates a unit load moves within
span AB while figure (b) shows the load moves beyond point B. Hence, there will be two
different equations of influence line for bending moment at B.
(a) A unit load within span AB (0 ≤ x < 5 m)
(b) A unit load within span BC (5 ≤ x < 10 m)
b. A unit load within span AB (0 ≤ x < 5 m)
c. A unit load within span BC (5 < x ≤ 10 m)
d. Draw a graph based on both equations which according to their range of x. The graph drawn
below is the influence line for shear force at B. The ordinate value of influence line can be
calculated from the derived equations.
7.3.2 Qualitative influence lines
Instead of quantitative method, there is another method that can be used to develop influence
lines for different function. This method is based on the Müller-Breslau Principle which
develops the influence lines qualitatively. The principle states that the ordinate value of an
influence line for any function on any structure is proportional to the ordinates of the deflected
shape that is obtained by removing the restraint corresponding to the function from the structure
and introducing a force that causes a unit displacement in the positive direction.
1. Vertical reaction
For example, to draw a qualitative influence line for the vertical reaction at A of the beam
shown in Figure 7.3(a), the pin support at A need to be removed and it is replaced by a roller
guide as shown in Figure 7.3(b). Apply a force at A in a positive direction that causes a unit
displacement which represents the general shape of the influence line as shown in Figure
7.3(c). Note that the beam rotates as a rigid body without any curvature and it is only
applicable for statically determinate beam.
Figure 7.3
2. Shear force
To draw a qualitative influence line for the shear force at C of the beam shown in Figure
7.4(a), the shear resistance at C need to be removed and it is replaced by a roller guide as
shown in Figure 7.4(b). Apply a positive shear forced, VC on the beam at C and the beam will
deflect as indicated by the dashed line. This line is the influence line for the shear force at C as
shown in Figure 7.4(c).
Figure 7.4
3. Bending moment
To draw a qualitative influence line for the bending moment at C of the beam shown in Figure
7.5(a), the moment resistance at the section need to be removed by placing an internal hinge or
pin. Apply a couple of positive moment MC respectively on the left and right sides of the hinge
that will introduce a unit relative rotation between the two tangents of the deflected shape at
the hinge as shown in Figure 7.5(b). The corresponding deflected beam which is illustrated by
the dashed line is the influence line for bending moment at C as shown in Figure 7.5(c).
The Müller-Breslau Principle can be verified using the principle of virtual work:
Work = a linear displacement x force in the direction of displacement,
or
Work = rotational displacement x moment in the direction of displacement.
Figure 7.5
The sum of all forces and moments on an equilibrium rigid body must equal to zero. If the body
is given an imaginary or virtual displacement, work done by all these forces and couple
moments must equal to zero too. Let consider Figure 7.6(b), if the beam is given a virtual
displacement δY at support A, then the work due to AY can be calculated as follow:
Work = AY δY
While the work done by a unit load can be calculated as follows. Note that the work done by a
unit load is negative.
Work = −1δY′.
Since the beam is in equilibrium, the virtual work sums to zero.
AY δY − 1δY′ = 0.
If δY = 1,
AY = δY′
Hence, the value of AY represents the ordinate of the influence line at the position of the unit
load. Since this value is equivalent to the displacement δY′ at the position of the unit load, it
shows that the shape of the influence line has been established.
Figure 7.6
If the shear restraint is removed from the beam at B and is replaced with a roller guide as
shown in Figure 7.7(a), the beam will undergo a virtual displacement δY at that section. Hence,
the virtual work done by the shear force at B and a unit load can be calculated as follow:
VCδY − 1δY′ = 0.
If δY = 1,
VC = δY′.
The shape of the influence line for shear force at B is proven based on the virtual work
calculated above. Let consider Figure 7.7(b) which the moment restraint is removed from the
beam at B and replaced by a hinge or pin. If a virtual rotation δΦ is introduced at the hinge, the
virtual work done by the internal moment and a unit load can be calculated as follows:
MCδΦ − 1δY′ = 0.
If δΦ = 1,
MC = δY′.
This indicates that the deflected beam has the same shape as the influence line for the internal
moment at point B.
Figure 7.7
7.4
Maximum Shear Force and Bending Moment at Sections in Beams
Once the influence line for a function has been constructed, it will be possible to position live
loads on the beam which will produce the maximum value of the function. Two types of
loadings, namely concentrated and distributed load will be considered in the analysis.
7.4.1 Concentrated force
For any concentrated force, F acting on the beam, the value of the function can be determined
by multiplying the ordinate of the influence line at position x by the magnitude of F. For the
beam shown in Figure 7.8(a), the influence line for the reaction force at support A is shown in
Figure 7.8(b). The ordinate of influence line for the reaction force A when the unit load at point
B is 0.5. Thus, the magnitude of reaction force A is AY = 0.5F.
Figure 7.8 (a) The beam and (b) influence line for AY
7.4.2 A series of concentrated load
Once the influence line of a function at specified point in the structure has been established, the
maximum effect caused by a live concentrated force is determined by multiplying the peak
ordinate of the influence line by the magnitude of the force. In some cases, several concentrated
loadings such as wheel loads must be placed on the beam. Trial and error procedure can be
used to calculate the magnitude of the required function. Since the load is linear, the principle
of superposition can be utilized to determine the combined effects of the loads.
Example 7.3
Determine the maximum positive shear force created at point C in the beam shown in Figure
7.9 due to the series of point loads.
Figure 7.9
Solution
Shear force
Consider a simply supported beam with associated influence line for shear force at point B as
shown in Figure 7.10(a). Trial and error method can be used to calculate the maximum positive
shear at B due to the series of concentrated loads moving from the right to the left of the beam.
There are three cases to be considered as shown in Figure 7.10(b). The magnitude of shear
force for each case can be calculated as follows:
Case 1: (VC )1 = 4.5(0.75) +18(0.625) +18(0.5) = 23.63 kN.
Case 2: (VC )2= 4.5(−0.125) +18(0.75) +18(0.625) = 24.19 kN.
Case 3: (VC)3 = 4.5(0) +18(−0.125) +18(0.75) = 11.25 kN.
The critical loading condition occurs when the middle load is placed just to the right of B.
Case 2 yields the largest value for VB and therefore represents the critical loading.
Figure 7.10
Example 7.4
Determine the critical position of maximum internal moment at point B in the beam shown
in Figure 7.11 due to the series of point loads.
Figure 7.11
Solution
Moment
As the loads are positioned as shown in Figure 7.12(b), the magnitude of bending moment for
each case can be calculated as follows.
Case 1: (MC)1 = 9(2.25) +18(1.95) +13.5(1.5) = 75.6kNm.
Case 2: (MC)2 = 9(1.35) +18(2.25) +13.5(1.8) = 76.95 kNm.
Case 3: (MC)3 = 9(0) +18(0.9) +13.5(2.25) = 46.58 kNm.
The critical loading condition occurs when the middle load is placed just to the right of B.
Case 2 yields the largest value for MB and therefore represents the critical loading.
Figure 7.12
7.4.3 Distributed load
Each dx segment of this load creates a concentrated force of dF = wo dx. If dF is located at x,
where the influence line ordinate is y, the value of the function is (dF)(y) = (wo dx)y. The effect
of all concentrated force is determined by integrating over the entire length of the beam.
∫ woydx = wo ∫ ydx.
Since ∫ y dx is equivalent to the area under the influence line, generally the value of the function
caused by a uniform load is equal to the area under the influence line multiplied by the intensity
of the distributed load. However, this is only true for cases where the distributed load is longer
than the span.
Figure 7.13 Influence line diagram for any function
On the other hand, for the distributed load that is shorter than the span, the position of the
distributed load should be placed accordingly to achieve the maximum value of the function.
Let consider Figure 7.14(a). It shows that the influence line for shear force at specific point A.
If the distributed load is moving from left to right of the beam, the maximum shear force can be
achieved when the head of the load is placed just to the left of point A [Figure 7.14(b)]. While
for the maximum positive shear force, the tail of load should be placed just to the right of point
A [Figure 7.14(c)].
Figure 7.14
The magnitude of the function can be calculated by multiplying the intensity of the distributed
load with the area under the influence line diagram covered by the load. The maximum bending
moment can be achieved when the unshaded area under the influence line diagram for bending
moment is maximum as shown in Figure 7.15.
Figure 7.15
The area under the influence line for bending moment is maximum when the distributed load is
placed in such a way that Qq = Rr. The following expression can be used to obtain the ordinate
of influence line Qq and Rr:
(7.1)
Rr can be determined using the same manner as equation (7.1)
(7.2)
Since Qq = Rr,
(7.3)
Note: l = AQ + AR.
Example 7.5
Calculate the maximum positive shear force that can be developed at point B in the beam
shown in Figure 7.16(a) due to a moving concentrated load of 4 kN and a moving uniformly
distributed load of 2 kN/m. The influence line for shear at C is given and shown in Figure
7.16(b).
Figure 7.16
Solution
1. Concentrated force
The maximum positive shear force at C will occur when the 4 kN force is located at x = 2.5 m.
The ordinate of influence line at this peak is +0.75, hence
VC = 0.75(4 kN)
= 3kN.
2. Distributed load
The moving uniformly distributed load creates the maximum positive shear for VC when the
load acts on the beam between x = 2.5 m and x = 10 m. The magnitude of VC due to this
loading is
3. Total maximum shear force at C:
(VC)max = 3kN + 5.625kN
= 8.625 kN.
Example 7.6
Calculate the maximum bending moment that can be developed at point B in the beam
shown in Figure 7.17(a) due to a moving load of length 3 m and intensity of 4 kN/m. Figure
7.17(b) shows the influence line for bending moment at B.
Figure 7.17
Solution
The distributed load should be placed in a way that the largest area under the influence line
diagram can be obtained. By using the expression in equation (7.3) and refer to Figure 7.14.
7.4
But,
l = AQ + AR
AR = l − AQ.
(7.5)
Hence, substitute equations (7.5) into (7.4)
Thus, the tail of the distributed load is placed 0.75 m to the left of point B while the head of the
distributed load is placed 2.25 m to the right of point B. The ordinate value of influence line at
both locations can be determined using equation (7.1).
The maximum bending moment at B can be calculated as follows:
EXERCISES
Exercise 1
A beam shown in Figure 7.18(a) is supported by pinned at B and roller at D. Draw the
quantitative influence lines diagram for the vertical reaction at B and D, and bending moment
at C for the beam. Determine the maximum positive bending moment at C if the beam is
subjected to a series of concentrated moving load as shown in Figure 7.18(b).
Figure 7.18
Answers:
a. Influence line for reaction force at B.
b. Influence line for reaction force at D.
c. Influence line for bending moment at C.
d. Maximum positive bending moment at C, MC,MAX = 9 kNm.
Exercise 2
Figure 7.19 shows the beam ABCDE with roller supports at B and D, while pinned support at
E. The beam is composed of two rigid parts, AC and CE which are connected by a hinge joint
at C. A moving load of length 2 m and intensity of 2 kN/m crosses the beam. Calculate the
maximum positive and negative value of shear force and the maximum value of bending
moment at the middle of span DE.
Figure 7.19
Answers:
Exercise 3
Figure 7.20 shows a single overhanging beam.
a. Draw the qualitative influence lines for:
i. The vertical reactions at support A and C.
ii. The shear forces just to the left and the right of support C.
iii. The bending moment at point B.
b. Calculate the maximum value of bending moment at point B due to a moving load of length 3
m and intensity of 5 kN/m.
Figure 7.20
Answers:
a. (i) Influence line for reaction force at A.
Influence line for reaction force at C.
(ii) Influence line for shear force just to the left of support C.
Influence line for shear force just to the right of support C.
(iii) Influence line for bending moment at B.
b. Maximum positive bending moment at B, MB, MAX = 15 kNm.
BIBLIOGRAPHY
Beer, F. P. & Johnston, E. R. Mechanics of Materials (2nd ed.). London: McGraw-Hill, 1992.
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Schodek, D. L. Structures (5th ed.). New Jersey: Pearson Prentice-Hall, 2004.
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THE AUTHORS
Taksiah A. Majid obtained her BSc (Hons) in Civil Engineering with first Class Honours from
Middlesex Polytechnic, London in 1990. She pursued her MSc (Eng) in Structural Engineering
and PhD at the University of Liverpool in the research area of structural dynamics. In 1996, she
started her career at Universiti Sains Malaysia (USM) as a lecturer in the Structural
Engineering Group, School of Civil Engineering. She was appointed as Program Chairman for
the Structural Engineering from 2003–2005, and Deputy Dean-Graduate Studies and Research
from 2005–2007. She has been appointed as external examiner for various IPTAs in Malaysia
and supervised postgraduate students on Numerical Analysis of Seismic Engineering, Wind
Engineering and Industrialized Building Construction (IBS). She is currently the Coordinator of
the Disaster Research Nexus, USM.
Choong Kok Keong holds a doctorate degree in architectural engineering from The University
of Tokyo, Japan. He is currently an Associate Professor at the School of Civil Engineering,
USM. He has 13 years of experience in teaching undergraduate and postgraduate courses in the
area of structural mechanics and structural analysis. His research interest is in the area of
computational analysis of shell and spatial structures. He has served as Program Chairman for
the Structural and Physical at the School of Civil Engineering, USM from 2009 until present.
Mustafasanie M. Yussof is a lecturer at the School of Civil Engineering, USM. He earned his
Bachelor degree in Civil Engineering from the Universiti Teknologi Malaysia and has a MSc
in Structural Engineering from the USM. He is currently pursuing research study towards his
PhD at the University of Surrey, United Kingdom. He has six years of experience in teaching
undergraduate courses in the area of Structural Engineering. His research interest includes the
computational analysis of structural behaviour of Space Structures, particularly the Structural
Glass Facade Systems.
INDEX
A
applied loads
relation between internal shear to
relation between moment to
arch
advantage of
angle between horizontal plane and
comparison between beams to
equation of
parabolic
semi circle
statical determinacy of, Example 5.1
three-hinged
arch rib
bending moment in
thrust and shear force in
asymmetrical cable
example calculation of, Example 4.4
subjected to uniformly distributed load
axial force
axial thrust diagram
example of, Figure 5.35.
B
ball-and-socket joints
beam
bending moment (M)
diagram. See BMD
drawn on compression side
example of, Example 3.22
drawn on the tension side
example of, Example 3.22
in the arch rib
maximum
example of, Example 7.4, Example 7.6
negative
positive
procedure for computation of
qualitative influence line for
Bernoulli-Euler beam equation
BMD
calculation process to obtain
characteristics of
example for drawing process of, Example 3.5, Example 3.6, Example 3.7, Example 3.8
for simply supported beam
for arch
procedures to produce for beams
representation of
boundary conditions
C
cable
analysis of
area/size of, Example 4.1, Example 4.5
asymmetrical
bending flexibility of
construction of
forces in
horizontal equilibrium for
vertical equilibrium for
subjected to point load
example of, Example 4.3
symmetrical
types of
cantilever beam
example of, Example 1.4, Example 3.5, Example 3.10
centriodal axis
Charles E. Greene
collinear members
columns
complex trusses
compound trusses
compression
state of
concentrated force
example of, Example 7.5
from distributed load
concentrated loads
a series of
example of
Example 7.3
Example 7.4
and couple
concentrated moments
conjugate beam
analysis procedure using
method
theorem
theorem
support
connections
hinged
fixed or rigid
pinned
characteristic of
sketches of
roller
comparing with pinned
cross-sectional area
D
deflection
computation of
equation
example of, Example 6.1
excessive
of elastic curve
degree of statical determinacy
determinacy. See also statical determinacy
diagram
bending moment. See BMD
shear force. See SFD
thrust, Figure 5.35.
differential element
direct integration method
constant of
displacement
to determine
distributed load
concentrated force from
diagram
example of, Example 7.5
unit of
dynamic analyses
dynamic loadings
E
elastic beam theory
elastic curve
to draw
elastic deformation
equation of
condition
for frame
for truss
equilibrium
for forces
for moment
of plane structures
equilibrium
of forces
of moment
external
force
loads
restraint
supports
F
fixed support (rigid)
flexibility
flexure formula
flexural rigidity
force
axial
compressive member
equilibrium of
external
horizontal
internal
internal axial
member axial
reaction
tensile
vertical
frame
equation of condition for
example of, Example 1.9, Example 1.10
gable
example of
Example 1.13
SFD, BMD and QDS for, Example 3.20, Example 3.21
statical determinacy of
example of, Example 3.13
structure of
with hinged joints
with only rigid joints
with roller joint
framed structures
free body diagram (FBD)
G
geometric method
approaches in
girder
H
H. Müller-Breslau
Principle
hinge
Hooke’s Law
horizontal equilibrium
horizontal thrust
I
idealization
inclined loads
inclined members
influence line
application of
area under the
equation
for statically determinate beam
examples of construction of
Example 7.1,
Example 7.2
for statically determinate beams
Müller-Breslau Principle
qualitative
for bending moment
for shear force
for vertical reaction
quantitative
procedure of influence-line equations
procedure of tabulated ordinate values
representation of
use of
internal
angle
axial force
tensile
force
example of, Example 3.1
J
joint
ball-and-socket
hinged
pinned
rigid
roller
L
length of
cable
asymmetrical, Example 4.4
example of, Example 4.3 loaded with point loads
symmetrical
linearly
distributed load
elastic
loadings
effect of different types of
longitudinal compressive force
M
M/EI diagram
to draw
mass structures
member axial force
method
conjugate beam
analysis procedure using
theorem
theorem
geometric
influence line equation
integration
direct (double)
moment-area
proportional
superposition
tabulated ordinate values
of joints
of sections
work energy
method of joints
procedure of
method of sections
procedure of
moment
definition of
equilibrium of
fundamental definition of
sign of
moment-area method
theorem
theorem
moment equilibrium equation
Müller-Breslau Principle
N
negative bending moment
neutral axis
neutral surface
non-collinear members
non-zero
slope
relative rotation
O
Otto Mohr
P
parabolic equation
pinned
joint
example of, Example 1.3
support
for cable
plane frame
analysis of
example of, Example 3.18, Example 3.19, Example 3.20
planar structure
plane trusses
classification of
determinacy and stability of
example of, Example 2.1
point
load (couple)
of contraflexure
positive bending moment
proportional method
Q
qualitative deflected shape (QDS), example for sketching of
Example 3.9, Example 3.10, Example 3.11 & Example 3.12
the use of
to sketch
R
radius of curvature
reaction forces
computation of
example of, Example 1.6, Example 1.7
redundants
rigid support. See fixed support
roller
joint
support
for cable
S
section
method of
semi circle arch
equation of
SFD
calculation process to obtain
characteristics of
example for drawing process of,
Example 3.5,
Example 3.6,
Example 3.7,
Example 3.8
procedures to produce
representation of
shear force (S)
diagram. See SFD
in the arch rib (Q)
maximum
example of, Example 7.3, Example 7.5
procedure for computation of
qualitative influence line for
the calculation of
sign convention
positive
simple plane trusses
analysis of statically determinate
simply supported beam
example of, Example 1.5
simply supported planar truss
example of, Example 1.2
singularity functions
slope
equation for
example of, Example 6.1
example determination of
Example 6.2
Example 6.4
relation to internal moment
to determine
space trusses
statical determinacy
procedure to check for
degree of
of an arch, Example 5.1
example to determine, Example 1.1, Example 1.2, Example 1.3
statically determinate
analysis of
beam
plane trusses member forces
simple plane trusses
beam
influence line for
frame
structure
truss
statically indeterminate
truss
strength-to-weight ratio
structural
analysis
elements
members
parabolic arch
equation of
structure
definition of
deformation causes of
elements of
framed
mass
statically determinate
types of
with curved member
example of, Example 1.11
superposition method
supports
conjugate beam
fixed
types of
effect of different
support condition
types of
symmetrical
cable
problem
T
tabulated ordinate values
example of influence line
construction using, Example 7.1
tangents lines
tensile
force
maximum, Example 4.1, Example 4.4, Example 4.5
minimum, Example 4.1, Example 4.4, Example 4.5
tension
maximum, Example 4.1,
Example 4.4
minimum
Example 4.1
Example 4.4
three-dimensional
force system
structures
three-hinged arch
example of
Example 5.2
Example 5.3
three-pinned arch
thrust diagram, Example 5.5
translational movement
truss, Figure 2.2.
applications of
complex
compound
methods of analysing the
stability of plane
statically determinate
statically indeterminate
two-span continuous beam
compound
example of, Example 1.1
U
uniformly distributed load
asymmetrical cable subjected to
forces in cable subjected to
V
vector quantity
vertical
equilibrium
deflection of beams, Example 6.1
displacement, Example 6.3, Example 6.4
maximum, Example 6.5
loads
reaction
qualitative influence line for
virtual
displacement
work
W
work
virtual
work energy method
Z
zero-force members
identification of
member arrangements that result in