Structural Analysis I

STRUCTURAL
THEORY
I
Shieh-Kung Huang
黃謝恭
1

Shieh-Kung
Huang
Copyright © 2018 by McGraw
-Hill Education. All rights reserved.
STRUCTURAL THEORY I
Position in Civil Engineering
5
Engineering
Mechanics –
Statics
Engineering
Mechanics –
Dynamics
Structural
Theory
I
Structural
Theory
II
Mechanics of
Materials
Engineering
Materials
Reinforced
Concrete
Advanced
Mechanics of
Material
Structural
Theory
III
Dynamics of
Structures
Design of
Reinforced
Concrete
Design of Steel
Structures
I
Design of Steel
Structures
II
Prestressed
Concrete

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STRUCTURAL THEORY I
Basic Information about This Course
6
• Instructor
Shieh-Kung Huang (黃謝恭)
Assistant Professor, Civil Engineering, National Chung Hsing University
Office: 508 Concrete Technology Building (混凝土科技研究中心)
E-mail: skhuang@nchu.edu.tw
Tel: (04) 2287-2221 ext. 508
• Course Hour
Tuesday 14:10 – 17:00
• Classroom
201 Civil & Environmental Engineering Building (土木環工大樓)
• Office Hour
Wednesday 2:00 – 4:00, or by appointment
• Teaching Assistant
TBD (To be determined)
• Prerequisites
Engineering Materials, Engineering Mechanics – Static, and Mechanics of Materials

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STRUCTURAL THEORY I
Course Objectives and Text Book
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• Leet, Kenneth, Chia-Ming Uang, and Anne M. Gilbert. Fundamentals of structural analysis. McGraw-
Hill, 2010.
• Hibbeler, Russell C. Structural analysis. Pearson Prentice Hall, 2018.

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INTRODUCTION TO STRUCTURE SYSTEMS AND LOADS
Chapter Objectives
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Students will be able to
1. get overview of structural analysis
2. understand the need of code and various loads
CHAPTER 1
1.1 Overview of Structural Analysis
1.2 Building and Design Code
1.3 Loads
1.4 Load Combinations
Chapter Outline

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1.1 OVERVIEW OF STRUCTURAL ANALYSIS
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A structure refers to a system of connected parts used to support a load.
When designing a structure to serve a specified function for public use, the engineer must account for its
safety, esthetics, and serviceability, while taking into consideration economic and environmental
constraints.
Once a preliminary design of a structure is proposed, the structure must then be analyzed to ensure that it
has its required stiffness and strength.
• Types of Structures
− Trusses
Trusses consist of slender elements, usually arranged in triangular fashion.
− Cables and Arches
Cables are usually flexible and carry their loads in tension.
The arch achieves its strength in compression, since it has a reverse curvature to that of the cable.
− Frames
Frames are often used in buildings and are composed of beams and columns that are either pin or
fixed connected.
− Surface Structures
A surface structure is made from a material having a very small thickness compared to its other
dimensions.
Chapter 1 Introduction to Structure Systems and Loads

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Trusses

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Cables and Arches

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Frames

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Surface Structures

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1.2 BUILDING AND DESIGN CODE
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A code is a set of technical specifications and standards that control major details of analysis, design, and
construction of buildings, equipment, and bridges.
• Building and Design Code
− Design (Structural) Code
A structural code is written by engineers and other specialists who are concerned with the design
of a particular class of structure (e.g., buildings, highway bridges, or nuclear power plants) or
who are interested in the proper use of a specific material (steel, reinforced concrete, aluminum,
or wood).
− Building Code
A building code is established to cover
construction in a given region (often a city
or a state).
It contains provisions pertaining to architectural,
structural, mechanical, and electrical
requirements.
It should be realized, however, that codes provide only a general guide for design. The ultimate
responsibility for the design lies with the structural engineer.

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1.3 LOADS
Dead Loads
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The load associated with the weight of the structure and its permanent components (floors, ceilings, ducts,
and so forth) is called the dead load.
When designing a structure, the member dead loads must first be
estimated since member sizes are initially unknown, yet must still be
accounted for in the total load carried by the structure.
After members are sized and
architectural details finalized, the
dead load can be computed more
accurately.

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1.3 LOADS
Lives Loads
21
Loads that can be moved on or off a structure are classified as live loads, and they include the weight of
people, furniture, machinery, and other equipment.
Live loads can vary over time especially if the function of the building changes.
The live loads specified by codes for various types of
buildings represent a conservative estimate of the
maximum load likely to be produced by the intended
use and occupancy of the building.

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1.3 LOADS
Other Loads
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• Snow Loads
In some parts of the country, roof loading due to snow can be quite severe, and therefore protection
against possible failure is of primary concern.
• Impact Loads
Moving vehicles may bounce or sidesway as they move over a bridge, and therefore they impart an
impact to the deck.
• Hydrostatic and Soil Pressure
When structures are used to retain water, soil, or granular materials, the pressure developed by these
loadings becomes an important criterion for their design.
Extreme winds, large earthquakes, and tsunamis are among several natural phenomena which generate
lateral loading capable of damaging structures and even result in loss of life.
• Wind Loads
When the speed of the wind is very high, it can cause massive damage to a structure.
• Earthquake Loads
Earthquakes produce lateral loadings on a structure through the structure’s interaction with the ground.
• Tsunami Loads
The basic tsunami loads considered by structural designers are hydrostatic and hydrodynamic
pressures, buoyancy forces, and forces from debris impact.

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1.4 LOAD COMBINATIONS
Other Loads
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The forces (e.g., axial force, moment, shear) produced by various combinations of loads discussed need
to be combined in a proper manner and increased by a factor of safety (load factor) to produce the
desired level of safety.
The combined load effect, sometimes called the required factored strength, represents the minimum
strength for which members need to be designed.
• Allowable-stress design (ASD)
ASD methods include both the material and load uncertainties into a single factor of safety.
Typical load combinations as specified by the ASCE 7-10 Standard include
• Load and resistance factor design (LRFD)
Since uncertainty can be considered using probability theory, there has been an increasing trend to
separate material uncertainty from load uncertainty.

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SUPPORTS AND REACTION
Chapter Objectives
24
1. familiarize different forces
2. familiarize different supports
3. idealize structures
4. draw free-body diagrams
CHAPTER 2
2.1 Forces
2.2 Supports
2.3 Idealizing Structures
2.4 Free-Body Diagrams
Chapter Outline

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2.1 FORCES
Force System
25
To solve typical structural problems, we use equations involving forces or their components.
Forces may consist of either a linear force that tends to produce translation or a couple that tends to
produce rotation of the body on which it acts.
Although we often represent a moment by a curved arrow to show that it acts in the clockwise or
counterclockwise direction, we can also represent a moment by a vector—usually a double-headed
arrow—using the right-hand rule.
Chapter 2 Supports and Reaction

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2.1 FORCES
Force Multiplication, Addition and Subtraction
26
• Multiplication and Division of a Vector by a Scalar
Product of vector A and scalar a is aA
Magnitude of scaled vector
Law of multiplication applies e.g.
• Vector Addition
Addition of two vectors A and B gives a resultant vector R by the parallelogram law
Result R can be found by triangle construction
Communicative e.g.
Special case: Vectors A and B are collinear
(both have the same line of action)
aA
( )
1 0
a a a
= 
A A
= + = +
R A B B A

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2.1 FORCES
Law of Cosines and Sines
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If a force is to be resolved into components that are not parallel to an x-y coordinate system, the law of
cosines provides a simple relationship between length of sides and interior
2 2
2 cos
c a b ab C
= + −
If a force is to be resolved into components that are not parallel to an x-y coordinate system, the law of
sines provides a simple relationship between length of sides and interior angles opposite the respective
sides
sin sin sin
a b c
A B C
= =

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2.1 FORCES
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2.1 FORCES
Resultant of a Force System
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• Resultant of a Planar Force System
In certain structural problems we will need to determine the magnitude and location of the resultant of
a force system.
• Resultant of a Distributed Load
As an alternative procedure the designer may replace a distributed load that varies in a complex
manner by a statically equivalent set of concentrated loads.
x x y y O
O i i i
R F R F M M
M Rd Fd M
= = =
= = +
  
 

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2.1 FORCES
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2.1 FORCES
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2.1 FORCES
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2.1 FORCES
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2.1 FORCES
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2.1 FORCES
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2.2 SUPPORTS
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To ensure that a structure or a structural element remains in its required position under all loading
conditions, it is attached to a foundation or connected to other structural members by supports.
Although the devices used as supports can vary widely in shape and form, we can classify most supports
in one of four major categories based on the restraints or reactions the supports exert on the structure.

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2.2 SUPPORTS
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2.2 SUPPORTS
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2.2 SUPPORTS
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Hibbeler, Russell C. Structural analysis. Pearson Prentice Hall, 2018

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2.2 SUPPORTS
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Hibbeler, Russell C. Structural analysis. Pearson Prentice Hall,
2018

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2.3 IDEALIZING STRUCTURES
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Before a structure can be analyzed, the designer must develop a simplified physical model of the
structure and its supports as well as the applied loads.

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2.4 FREE-BODY DIAGRAMS
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As a first step in the analysis of a structure, the designer will typically draw a simplified sketch of the
structure or the portion of the structure under consideration.
This sketch, which shows the required dimensions together with all the external and internal forces acting
on the structure, is called a free-body diagram.

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STATICS OF STRUCTURES
Chapter Objectives
43
1. review statics
2. calculate reactions for different structures
3. classify determinate and indeterminate structures
4. determine the degree of indeterminacy
5. differentiate if a structure is stable or unstable
CHAPTER 3
3.1 Equations of Static Equilibrium
3.2 Equations of Condition
3.3 Influence of Reactions on Stability and
Determinacy of Structures
3.4 Classifying Structures
3.5 Comparison between Determinate and
Indeterminate Structures
Chapter Outline

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3.1 EQUATIONS OF STATIC EQUILIBRIUM
Equilibrium Equations
44
As you learned in dynamics, a system of planar forces acting on a rigid structure can always be reduced
to two resultant forces:
where m is the mass of the body, I is the mass moment of inertia of the body with respect to its center of
gravity, a is the linear acceleration of the center of gravity, and a is the angular accelerations of the body
about the center of gravity.
If the body is at rest—termed a state of static equilibrium—both the linear acceleration and the angular
acceleration equal zero,
and, therefore,
Chapter 3 Statics of Structures
R ma
M Ia
=
=
0
0
R
M
=
=
0
0
0
x
y
O
F
F
M
=
=
=




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Alternative Sets of Equilibrium Equations
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As you may remember from your course in statics, either or both of the equilibrium equations can also be
replaced by moment equations.
The first one is
When using these equations it is required that a line passing through points A and B is not parallel to
the y axis.
The second one is
Here it is necessary that points A, B, and C do not lie on the same line.
0 0 0
x A B
F M M
= = =
  
0 0 0
A B C
M M M
= = =
  

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?

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Can we solve
the problem
using this
formulation?

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Statically Determinate and Indeterminate
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• Statically Determinate
If the forces acting on a structure—including both the reactions and the internal forces—can be
computed using any of the foregoing sets of equations of static equilibrium, the structure is said to
be statically determinate or, more simply, determinate.
The equations of static equilibrium can be used to compute the reactions of a determinate structure
that can be treated as a single rigid body.
Equilibrium
Eqs.
Compatibility
Eqs.
Constitutive Law
Force
Stress
Displacement
Strain
Rigid Body
Elastic (Deformed) Body

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Statically Determinate and Indeterminate
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• Statically Indeterminate
If the structure is stable but the equations of equilibrium do not provide sufficient equations to analyze
the structure, the structure is termed indeterminate.
To analyze indeterminate structures, we must derive additional equations from the geometry of the
deformed structure to supplement the equations of equilibrium.
Equilibrium
Eqs.
Compatibility
Eqs.
Constitutive Law
Force
Stress
Displacement
Strain
Rigid Body
Elastic (Deformed) Body

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3.2 EQUATIONS OF CONDITION
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The reactions of many structures can be determined by treating the structure as a single rigid body.
Other stable determinate structures, which consist of several rigid elements connected by a hinge or
which contain other devices or construction conditions that release certain internal restraints, require
that the structure be divided into several rigid bodies in order to evaluate the reactions.
This additional equation is called an equation of condition or an equation of construction.

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3.3 INFLUENCE OF REACTIONS ON STABILITYAND
Unstable
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To produce a stable structure, the designer must supply a set of supports that prevents the structure or
any of its components from moving as a rigid body.
The number and types of supports required to stabilize a structure depend on the geometric arrangement
of members, on any construction conditions built into the structure (hinges, for example), and on the
position of supports.
• Unstable
Supports Supply Less Than Three Restraints: R < 3 (R = number of restraints or reactions)
If the supports supply less than three reactions, then one or more of the equations of equilibrium
cannot be satisfied, and the structure is not in equilibrium.
A structure not in equilibrium is unstable.
Mathematicians would say the set of equations above is inconsistent or incompatible.
DETERMINACY OF STRUCTURES

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Unstable
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• Unstable

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Statically Determinate
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• Statically Determinate
Supports Supply Three Reactions: R = 3
If supports supply three reactions, it will usually be possible to satisfy the three equations of
equilibrium (the number of unknowns equals the number of equations).
Further, if the equations of equilibrium are satisfied, the values of the three reactions are uniquely
determined, and we say that the structure is statically determinate.
If a system of supports supplies three reactions that are configured in such a way that the equations of
equilibrium cannot be satisfied, the structure is called geometrically unstable.
In summary, we conclude that for a single rigid body a minimum of three restraints is necessary to
produce a stable structure (one that is in equilibrium)—subject to the restriction that the restraints
not be equivalent to either a parallel or a concurrent force system.

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Statically Indeterminate
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• Statically Indeterminate
Restraints Greater Than 3: R > 3
If a system of supports, which is not equivalent to either a parallel or a concurrent force system,
supplies more than three restraints to a single rigid structure, the values of the restraints cannot be
uniquely determined because the number of unknowns exceeds the three equilibrium equations
available for their solution.
Since one or more of the reactions cannot be determined, the structure is termed indeterminate, and
the degree of indeterminacy equals the number of restraints in excess of 3

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Statically Indeterminate
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Structures Composed of Several Rigid Bodies
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• Structures Composed of Several Rigid Bodies
If a structure consists of several rigid bodies interconnected by devices (hinges, for example) that
release C internal restraints, C additional equations of equilibrium (also called condition equations)
can be written to solve for the reactions.

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3.4 CLASSIFYING STRUCTURES
Overall Procedure
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In the majority of cases, we check for stability by verifying that the reactions are not equivalent to a
parallel or a concurrent force system.
Next, to establish if a structure is determinate or indeterminate, we simply compare the number of external
reactions to the equilibrium equations available for the solution—that is, three equations of statics plus
any condition equations.
If any doubt still exists, as a final test, we apply a load to the structure and carry out an analysis using the
equations of static equilibrium. If a solution is possible—indicating that the equations of equilibrium are
satisfied—the structure is stable.
Alternatively, if an inconsistency develops, we recognize that the structure is unstable.

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Examples
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Examples
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INDETERMINATE STRUCTURES
3.5 COMPARISON BETWEEN DETERMINATEAND
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Since determinate and indeterminate structures are used extensively, it is important that designers be
aware of the difference in their behavior in order to anticipate problems that might arise during
construction or later when the structure is in service.
• Determinate Structures
If a determinate structure loses a support, immediate failure occurs
because the structure is no longer stable.
• Indeterminate Structures
On the other hand, in an indeterminate structure alternative paths
exist for load to be transmitted to supports (redundancy).

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INDETERMINATE STRUCTURES
3.5 COMPARISON BETWEEN DETERMINATEAND
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If a determinate structure loses a support, immediate failure occurs
because the structure is no longer stable. An example of the collapse
of a bridge composed of simply supported beams during the 1964
Niigata earthquake is shown in the following photo. As the
earthquake caused the structure to sway, in each span the ends of
the beams that were supported on rollers slipped off the piers and fell
into the water. Had the ends of girders been continuous or connected,
the bridge in all probability would have survived with minimum
damage. In response to the collapse of similar, simply supported
highway bridges in California during earthquakes, design codes have
been modified to ensure that bridge girders are connected at
supports.

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ANALYSIS OF STATICALLY DETERMINATE TRUSSES
Chapter Objectives
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1. understand the characteristic and behavior of trusses
2. analyze statically determinate trusses using the method
of joints
2. analyze statically determinate trusses using the method
of sections
3. differentiate if the structure is stable and determine
4. analyze the compound and complex trusses
5. familiarize the space trusses
CHAPTER 4
4.1 Types of Trusses
4.2 Analysis of Trusses
4.3 Method of Joints
4.4 Zero Bars
4.5 Method of Sections
4.6 Determinacy and Stability
4.7 Analysis of Compound and Complex Trusses
4.8 Space Trusses
Chapter Outline

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4.1 TYPES OF TRUSSES
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Chapter 4 Analysis of Statically Determinate Trusses
A truss is a structural element composed of a stable arrangement of slender interconnected bars.
• Types of Trusses
The members of most modern trusses are arranged in triangular patterns because even when the
joints are pinned, the triangular form is geometrically stable and will not collapse under load.
− Simple Truss
One method to establish a stable truss is to construct a basic
triangular unit and then establish additional joints by extending
bars from the joints of the first triangular element.

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4.1 TYPES OF TRUSSES
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• Types of Trusses
− Compound Truss
If two or more simple trusses are connected by a pin or a pin and a tie, the resulting truss is
termed a compound truss.
− Complex Truss
Finally, if a truss—usually one with an unusual shape—is neither a simple nor a compound truss, it
is termed a complex truss.

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4.2 ANALYSIS OF TRUSSES
80
A truss is completely analyzed when the magnitude and sense (tension or compression) of all bar forces
and reactions are determined.
• Assumptions of Trusses
− Bars are straight and carry only axial load (i.e., bar forces are directed along the longitudinal axis
of truss members). This assumption also implies that we have neglected the deadweight of the bar.
If the weight of the bar is significant, we can approximate its effect by applying one-half of the bar
weight as a concentrated load to the joints at each end of the bar.
− Members are connected to joints by frictionless pins. That is, no moments can be transferred
between the end of a bar and the joint to which it connects. (If joints are rigid and members stiff,
the structure should be analyzed as a rigid frame.)
− Loads are applied only at joints.
As a sign convention (after the sense of a bar force is
established) we label a tensile force positive and
a compression force negative.
Bar forces may be analyzed by considering the equilibrium
of a joint—the method of joints—or by considering the
equilibrium of a section of a truss—the method of
sections.

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4.3 METHOD OF JOINTS
81
To determine bar forces by the method of joints, we analyze free-body
diagrams of joints. The free-body diagram is established by imagining
that we cut the bars by an imaginary section just before the joint.
Because all forces acting at a joint pass through the pin, they constitute
a concurrent force system. For this type of force system, only two
equations of statics (that is, and ) are available to
evaluate unknown bar forces.
Since only two equations of equilibrium are available, we can only
analyze joints that contain a maximum of two unknown bar forces.
To determine bar forces by writing out the equilibrium equations, we
must assume a direction for each unknown bar force (known bar
forces must be shown in their correct sense, in force vector direction
and/or numeric sign).
The analyst is free to assume either tension or compression for any
unknown bar force. For the student, it may be helpful to assume that
all bars are in tension, showing all unknown bar forces acting outward
from the center of the joint
0
x
F =
 0
y
F =


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4.4 ZERO BARS
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For one or more positions of the load, certain bars may remain unstressed, and these unstressed bars are
termed zero bars (zero-force members).
The designer can often speed the analysis of a truss by identifying bars in which the forces are zero.
• If No External Load Is Applied to a Joint That Consists of Two Bars, the Force in Both Bars
Must Be Zero
• If No External Load Acts at a Joint Composed of Three Bars—Two of Which Are Collinear—the
Force in the Bar That Is Not Collinear Is Zero

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4.4 ZERO BARS
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4.4 ZERO BARS
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4.5 METHOD OF SECTIONS
90
In method of sections, we imagine that the truss is divided into two free bodies by passing an imaginary
cutting plane. The cutting plane must, of course, pass through the bar whose force is to be determined.
Although there is no restriction on the number of bars that can be cut, we often use sections that cut three
bars since three equations of static equilibrium are available to analyze a free body.
If the force in a diagonal bar of a truss with parallel chords is to be computed, we cut a free body by
passing a vertical section through the diagonal bar to be analyzed.
If three bars are cut, the force in a particular bar can be determined by extending the forces in the other
two bars along their line of action until they intersect.

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4.6 DETERMINACY AND STABILITY
Determinacy
106
Thus far the trusses have all been stable determinate structures; however, since indeterminate trusses
are also used in practice, an engineer must be able to recognize a structure of this type because
indeterminate trusses require a special type of analysis.
Since we can write two equilibrium equations for each joint in a truss, the total number of equilibrium
equations available to solve for the unknown bar forces B and reactions R equals 2N (where N
represents the total number of joints).
Therefore, it must follow that if a truss is stable and determinate, the relationship between bars, reactions,
and joints must satisfy the following criteria:
In addition, as we discussed in Chapter 3.3, the restraints exerted by the reactions must not constitute
either a parallel or a concurrent force system.
2
R B N
+ =

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Indeterminacy
107
If , the number of unknown forces exceed the available equations of statics and the truss is
indeterminate and the degree of indeterminacy D equals
Finally, if , there are insufficient bar forces and reactions to satisfy the equations of equilibrium,
and the structure is unstable.
Moreover, as we discussed in Chapter 3.3, you will always find that the analysis of an unstable structure
leads to an inconsistent equilibrium equation, and therefore, if you are uncertain about the stability of a
structure, analyze the structure for any arbitrary loading. If a solution that satisfies statics results, the
structure is stable.
Recall from Chapter 3.4
− verify that the reactions are not equivalent to a parallel or a concurrent force system.
− preliminarily establish if a structure is determinate or indeterminate
− apply a load to the structure and carry out an analysis using the equations of static equilibrium.
− recognize that the structure is unstable if an inconsistency develops
2
R B N
+ 
2
D R B N
= + −
2
R B N
+ 

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Examples
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Examples
109

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4.7 ANALYSIS OF COMPOUND AND COMPLEX TRUSSES
114
• Superposition Principle
If a structure behaves in a linearly elastic manner, the
force or displacement at a particular point produced by a set of
conservative loads acting simultaneously can be evaluated by
adding (superimposing) the forces or displacements at the
particular point produced by each load of the set acting
individually. In other words, the response of a linear, elastic
structure is the same if all loads are applied simultaneously or
if the effects of the individual loads are combined.

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• Compound Trusses
In Chapter 4.1, it was stated that compound trusses are
formed by connecting two or more simple trusses
together either by bars or by joints.
Occasionally this type of truss is best analyzed by applying
both the method of joints and the method of sections.
• Complex Trusses
The member forces in a complex truss can be determined
using the method of joints; however, the solution will
require writing the two equilibrium equations for each
of the j joints of the truss and then solving the complete
set of 2j equations simultaneously.
This approach may be impractical for hand calculations,
especially in the case of large trusses.
Therefore, a more direct method for analyzing a complex
truss, referred to as the method of substitute members,
will be presented here.

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• Method of Substitute Members
1. Choose a member to release the force (generally, the
member which we want to solve)
2. Choose a new member to retain the determinacy
3. Accordingly, separate the truss into two trusses
(a) with the new member and the external force
(b) with the new member and the released force
4. Solve the two trusses differently and make the
summation of the forces of the new members equal
5. The solved released force are the answer

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55
16
5
5
2
2
−
5
2
2
55
16
−
5
73
16
5
73
16
−
0
0
5
2
35
8
35
8
5
11
16
−
1
1
2
2
−
1
2
2
−
11
16
−
1
73
16
1
73
16
5
6
5
6
7
6
−
5 7 15
0
2 6 7
x x
− =  =
5 5 5 55 5 55 5
2 2 0 0 73 73
2 2 16 16 16 16 2
1 1 5 5 1 11 1 11 7
2 2 73 73
2 2 6 6 16 16 16 16 6
15
7
2.02 5.05 1.79 1.79 1.53 4.91 3.81 1.96 0
CB CD FA FE EB ED DA AB AE
F F F F F F F F F
− − −
− − − − −
− − −

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4.8 SPACE TRUSSES
119
A space truss consists of members joined together at their ends to form a stable three-dimensional
structure.
• Determinacy and Stability
Realizing that in three dimensions there are three equations of equilibrium available for each joint
( , , and ), then for a space truss with j number of joints, 3j equations are
available.
If the truss has B number of bars and r number of reactions, then like the case of a planar truss, we
can write
unstable true
statically determinate – check stability
statically indeterminate – check stability
The assumptions and analysis of space trusses are the same as a planar
truss except for the equations of equilibrium are available for three
different axes.
0
x
F =
 0
y
F =
 0
z
F =

3
3
3
R B N
R B N
R B N
+ 
+ =
+ 
0
0
0
0
0
x
x
y
y
z
F
F
F
F
F
=
=
 =
=
=






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4.8 SPACE TRUSSES
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4.8 SPACE TRUSSES
121

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4.8 SPACE TRUSSES
122

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ANALYSISOFSTATICALLYDETERMINATEBEAMS&FRAMES
Chapter Objectives
123
1. determine the internal loadings at specified points
2. draw shear and moment curves
3. use the principle of superposition
4. classify determinate and indeterminate beam or frame
structures
5. determine the degree of indeterminacy
CHAPTER 5
5.1 Internal Loadings at a Specified Point
5.2 Shear and Moment Curves
5.3 Principle of Superposition
5.4 Degree of Indeterminacy
Chapter Outline

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5.1 INTERNAL LOADINGS AT A SPECIFIED POINT
Shear and Moment
124
Chapter 5 Analysis of Statically Determinate Beams and Frames
As you may remember from the study of beams in mechanics of materials or statics courses, shear and
moment are the internal forces in a beam or frame produced by the applied transverse loads.
The shear acts perpendicular to the longitudinal axis, and the moment represents the internal couple
produced by the bending stresses.

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Sign Convention
125
Generally, we can define shear as positive if it tends to produce clockwise rotation of the free body on
which it acts.
Moment will be considered positive if it produces compression stresses in the top fibers of the cross
section and tension in the bottom fibers; negative moment, on the other hand, bends a member concave
down.
If a flexural member is vertical, the engineer is free to define the positive and negative sense of both the
shear and moment. For the case of a single vertical member, one possible approach for establishing the
positive direction for shear and moment is to rotate the computation sheet containing the sketch 90°
clockwise so that the member is horizontal, and then apply the same conventions.

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Sign Convention
126
For single-bay frames many analysts define moment as positive when it produces compression stresses
on the outside surface of the member, where inside is defined as the region within the frame.
The positive direction for shear is then arbitrarily defined, as shown by the arrows on the following figure.
Perhaps an easy way to remember this sign convention is to isolate a small segment of the member and
note that positive normal force tends to elongate the segment; positive shear tends to rotate the
segment clockwise; and positive bending moment tends to bend the segment concave upward, so as to
“hold water.”

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x3

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5.2 SHEAR AND MOMENT CURVES
Overview of Shear and Moment Curves
137
To design a beam, we must establish the magnitude of the shear and moment (and axial load if it is
significant) at all sections along the axis of the member.
To provide this information graphically, we construct shear and moment curves.
Although we can construct shear and moment curves by cutting free bodies at intervals along the axis of a
beam and writing equations of equilibrium to establish the values of shear and moment at particular
sections, it is much simpler to construct these curves from the basic relationships between load, shear,
and moment.

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Relationship Between Load, Shear, and Moment
138
• Relationship between Load and Shear
Considering equilibrium of forces acting in the y direction on the element, we can write
This equation states that the slope of the shear curve at a particular point along the axis of a member
equals the ordinate of the load curve at that point.
• Relationship between Shear and Moment
To establish the relationship between shear and moment, we sum moments of the forces acting on the
element about an axis normal to the plane of the beam and passing through point o
This equation states that the slope of the moment curve at any point along the axis of a member is the
shear at that point.
dV
V dV V wdx dV wdx w
dx
+ = +  =  =
dM
M dM M Vdx dM Vdx V
dx
+ = +  =  =

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Relationship Between Load, Shear, and Moment
139
• Relationship between Shear and Moment under Concentrated Force
A concentrated force produces a sharp change in the ordinate of a shear and moment curve.
To construct the shear and moment curves for a beam supporting distributed and concentrated loads, we
first compute the shear and moment at the left end of the member. We then proceed to the right,
locating the next point on the shear curve by adding algebraically, to the shear at the left, the force
represented by (1) the area under the load curve between the two points or (2) a concentrated load. To
establish a third point, load is added to or subtracted from the value of shear at the second point.
Typically, we evaluate the ordinates of the shear curve at each point where a concentrated load acts or
where a distributed load begins or ends.
In a similar manner, points on the moment curve are established by adding algebraically to the moment, at
a particular point, the increment of moment represented by the area under the shear curve between a
second point.

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Some Basic Examples
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Some Basic Examples
141

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Shear and Moment Curves for Frames
142
A frame is composed of several connected members that are either fixed or pin connected at their ends.
Generally, we will always draw the moment diagram positive on the compression side of the member for a
frame.

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X
X

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119.76

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-30

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shear is negative since it tends
to produce counterclockwise
rotation
moment draws on the right-
hand side of the column
since the compression is
on the right

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5.3 PRINCIPLE OF SUPERPOSITION
Overview of Superposition Principle
165
The principle of superposition states:
If a structure behaves in a linearly elastic manner, the force or displacement at a particular point
produced by a set of conservative loads acting simultaneously can be evaluated by adding
(superimposing) the forces or displacements at the particular point produced by each load of the set
acting individually. In other words, the response of a linear, elastic structure is the same if all loads are
applied simultaneously or if the effects of the individual loads are combined.

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Some Basic Examples
166
Most loadings on beams in structural analysis will be a combination of the loadings shown in the following
figure.

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Some Basic Examples
167

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Invalid Examples
168
The principle of superposition does not apply to beam-columns
or to structures that undergo large changes in geometry when
loaded.
A second case in which superposition is invalid is that a
flexible cable supports two loads at the third points of
the span.

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Appendix Table A.1
Page 759

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5.4 DEGREE OF INDETERMINACY
173
The basic approaches we discussed in Chapter 3 still apply when we want to extend our discussion to
include indeterminate frames—structures composed of members that carry shear, axial load, and
moment at a given section.

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5.4 DEGREE OF INDETERMINACY
174

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CABLES AND ARCHES
Chapter Objectives
175
1. learn the characteristics, types, and behavior of cable and
arch structures
2. analyze the cable supporting concentrated loads
3. analyze the cable uniform distributed loads
4. analyze the three-hinged arches
CHAPTER 6
6.1 Cables
6.2 Analysis of a Cable Supporting Concentrated
Gravity Loads
6.3 General Cable Theorem
6.4 Analysis of a Cable Supporting Uniform
Distributed Loads
6.5 Arches
6.6 Three-Hinged Arches
Chapter Outline

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6.1 CABLES
176
Chapter 6 Cables and Arches
Cables are often used in engineering structures for support and to transmit loads from one member to
another.
In the force analysis of such systems, the weight of the cable itself may be neglected; however, when
cables are used as guys for radio antennas, electrical transmission lines, and derricks, the cable weight
may become important and must be included in the structural analysis.
When deriving the necessary relations between the force in the cable and its slope, we will make the
assumption that the cable is perfectly flexible and inextensible.
• Due to its flexibility, the cable offers no resistance to shear or bending and, therefore, the force
acting in the cable is always tangent to the cable at points along its length.
• Being inextensible, the cable has a constant length both before and after the load is applied.
As a result, once the load is applied, the geometry of the cable remains fixed, and the cable or a segment
of it can be treated as a rigid body.

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GRAVITY LOADS
6.2ANALYSIS OFACABLE SUPPORTING CONCENTRATED
177
When a set of concentrated loads is applied to a cable of negligible weight, the cable deflects into a series
of linear segments. The resulting shape is called the funicular polygon.

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6.3 GENERAL CABLE THEOREM
178
In the analysis of the cable, engineers have observed that certain parts are similar to the analysis of a
simply supported beam with a equal span and carrying the same loads applied to the cable.
• For cable,
• For simply supported beam
( tan )
0 ( tan ) where
0 ( tan ) 0 tan ( tan )
B
B y B y B i i
z y z z B z z z B z
m H L
M A L m H L A m Pd
L
x x
M A x m H x h m xH m H x h Hh m m
L L
a
a
a a a
 −
= = −  +  =  = 
= = −  + −  =  − −  + −  =  − 


0 where
compare with cable
B
B A B A B i i
z A z z B z z z
m
M R L m R m Pd
L
x
M R x m M m m Hh M
L

= = −   =  = 
= −   =  −   =


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179
A comparison of the computations between those for a cable and those for a simply supported beam that
supports the cable loads leads to the following statement of the general cable theorem:
At any point on a cable supporting vertical loads, the product of the cable sag h and the horizontal
component H of the cable tension equals the bending moment at the same point in a simply supported
beam that carries the same loads in the same position as those on the cable. The span of the beam is
equal to that of the cable.
The relationship above can be stated by the following equation:
where H is horizontal component of cable tension
hz is cable sag at point z where Mz is evaluated
Mz is moment at point z in a simply supported beam carrying the loads applied to the cable
z z
Hh M
=

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DISTRIBUTED LOADS
6.4ANALYSIS OFACABLE SUPPORTING UNIFORM
Distributed Loads
182
In order to analyze the problem of suspension bridge, we will first determine
the shape of a cable subjected to a uniform horizontally distributed vertical
load w0.
Applying the equations of equilibrium yields
Divide the equations by , we have
Neglect the square term, we also have
x

0 tan
o
dy
M
dx

=  =

0 cos ( )cos( ) 0
0 ( ) sin ( )sin( ) 0
0 ( ) cos sin 0
2
x
y
o
F T T T
F w x x T T T
x
M w x x T y T x
  
  
 
=  − + +  +  =
=  −  − + +  +  =

=   −  +  =



( )
( )
cos
0 0
sin
0 ( )
x
y
d T
F
dx
d T
F w x
dx


=  =
=  =



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DISTRIBUTED LOADS
Distributed Loads
183
The equilibrium yields
Since the slope is zero at , then the horizontal force at this point is T,
and we have
which indicates the horizontal component of force at any point along the
cable remains constant.
Realizing that at , gives
Then, we can obtain the slope at any point,
This is the equation of a parabola and, by using the boundary condition, the
finial equation is
Realize that we have neglected the weight of the cable, which is uniform
along the length of the cable, and not along its horizontal projection.
( ) ( )
cos sin
0 ( ) tan
d T d T dy
w x
dx dx dx
 

= = =
0
x =
cos H
T F
 =
sin 0
T  = 0
x =
0
sin
T w x
 =
2
2
0 0 0
2
tan or
2
H H H
dy w x w d y w
y x
dx F F dx F
 = =  = =
2
2
h
y x
L
=
0
( )
w x w
 =

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We can start from the result of a parabola cable
Considering the density of cable is m, the distributed load can be
Substituting back to the function,
We can integrate it to get z
Applying the boundary condition,
We can further integrate it to get x
Applying the boundary condition,
Finally, we have a catenary cable
1
1 1
2
sinh sinh
1 H H H
dz dy
dx z x C x C
F F dx F
z
m m m
−  
=  = +  = +
 
+  
DISTRIBUTED LOADS
Its Own Weight
186
2
2 2
( ) ( ) 1
dy
ds dx dy ds w x dx w x
dx
m m
 
= +  =  = +  
 
( )
1 1
0
0 sinh 0 0
x
dy
C C
dx =
= = +  =
2
2
( )
H
d y w x
dx F
=
2
2
2
2
1 1 where
H H
d y dy dz dy
z z
dx F dx dx F dx
m m
 
= +  = + =
 
 
2
sinh cosh
H
H H
dy F
x y x C
dx F F
m m
m
   
=  = +
   
   
2 2
0
0 H H
x
F F
y C C
m m
=
= = +  = −
cosh cosh 1
H H H
H H
F F F x
y x
F F
m m
m m m
   
= − = −
   
   

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Comparison between a parabola cable and a catenary cable
DISTRIBUTED LOADS
187
Parabola Cable Catenary Cable
Assumption Distributed Loads Its Own Weight
Function
Slop
Tension
Max. Tension
cosh 1
H
H
F x
y
F
m
m
 
= −
 
 
2
0
2 H
w
y x
F
=
sinh
H
dy x
dx F
m
=
0
H
dy w
x
dx F
=
2 2 2
0
( ) H
T x F w x
= +
( )
cosh
H
H
H
T x F y
x
F
F
m
m
= +
=
2 2 2
max 0
H
T F w L
= + max cosh
H
H
L
T F
F
m
=

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6.5 ARCHES
188
Like cables, arches can be used to reduce the bending moments in long-span structures.
It uses material efficiently because applied loads create mostly axial compression on all cross sections.
Eventually, for a particular set of loads, the designer can establish one shape of arch—the funicular
shape—in which all sections are in direct compression (moments are zero).
• Type of Arches
Arches are often classified by the number of hinges they
contain or by the manner in which their bases are constructed.
− Three-hinge Arches
The three-hinged arch is the easiest to analyze and
construct since it is stable and determinate.
− Two-hinge Arches
In long-span bridges, two main arch ribs are used to
support the roadway beams. The possibility of buckling
most also consider since the arch rib is mostly in
compression. It is indeterminate to the first degree.
− Fixed-ended Arches
Fixed-ended arches are often constructed of masonry or
concrete when the base of an arch bears on rock,
massive blocks of masonry, or heavy reinforced concrete
foundations. It is indeterminate to the third degree.

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6.6 THREE-HINGED ARCHES
189
Because of symmetry, the vertical components of the reactions at supports A and C are identical in
magnitude.
To express the required area of the bars in terms of load P, we divide the bar forces by the allowable
compressive stress
We will also express the bar length in terms of  and the span length L as
Finally, we calculate the volume V of bar material required to support the load in terms of the geometry of
the structure and the compressive (note: )
2
sin sin
2 sin
AB CB AB CB
P P
F F F F
 

= =  = =
allow
allow
2 2
sin sin
P P
A A

  
=  =
bar
2
cos
L
L

=
bar
allow
2
2 sin2
PL
V AL
 
= =
sin2 2sin cos
  
=

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INFLUENCELINESFORSTATICALLYDETERMINATESTRUCTURES
Chapter Objectives
190
1. understand the concept of and the need for influence lines
2. draw the influence line for a statically determinate beams
3. draw the influence line for a statically determinate floor
girders
4. draw the influence line for a statically determinate trusses
5. use the influence lines for maximum loads estimation
CHAPTER 7
7.1 Influence Lines
7.2 Influence Line for Determinate Beams
7.3 Müller–Breslau Principle for Determinate Beams
7.4 Use of Influence Lines
7.5 Influence Lines for Determinate Floor Girders
7.6 Influence Lines for Determinate Trusses
7.7 Moving Live Loads
7.8 Absolute Maximum Shear and Moment
Chapter Outline

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7.1 INFLUENCE LINES
191
Chapter 7 Influence Lines for Statically Determinate Structures
If a structure is to be safely designed, we must proportion its members and joints so that the maximum
force at each section produced by live and dead load is less than or equal to the available capacity.
To establish maximum design forces at critical sections produced by moving loads, we frequently
construct influence lines.
An influence line is a diagram whose ordinates, which are plotted as a function of distance along the span,
give the value of an internal force, a reaction, or a displacement at a particular point in a structure as a
unit load of 1 kip or 1 kN moves across the structure.
Once the influence line is constructed, we can use it
− to determine where to place live load on a structure to maximize the force (shear, moment, etc.) for
which the influence line is drawn, and
− to evaluate the magnitude of the force (represented by the influence line) produced by the live load.
Although an influence line represents the action of a single moving load, it can also be used to establish
the force at a point produced by several concentrated loads or by a uniformly distributed load.

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7.2 INFLUENCE LINE FOR DETERMINATE BEAMS
192
As you become familiar with the construction of influence lines, you will only have to place the unit load at
two or three positions along the axis of the beam to establish the correct shape of the influence line.

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7.3MÜ LLER–BRESLAUPRINCIPLEFORDETERMINATEBEAMS
203
The Müller–Breslau principle provides a simple procedure for establishing the shape of influence lines for
the reactions or the internal forces (shear and moment) in beams.
The principle states:
The influence line for any reaction or internal force (shear, moment) corresponds to the deflected shape
of the structure produced by removing the capacity of the structure to carry that force and then
introducing into the modified (or released) structure a unit deformation that corresponds to the restraint
removed.

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7.3MÜ LLER–BRESLAUPRINCIPLEFORDETERMINATEBEAMS
204
?

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7.4 USE OF INFLUENCE LINES
Concentrated Load
205
Here, we describe how to use an influence line to compute the maximum value of a function when the live
load, which can act anywhere on the structure, is either a single concentrated load or a uniformly
distributed load of variable length.
• Concentrated Load
Since the ordinate of an influence line represents the value of a certain function produced by a unit
load, the value produced by a concentrated load can be established by multiplying the influence line
ordinate by the magnitude of the concentrated load.
• Uniformly Distributed Load
To establish the maximum value of a function produced by a uniform
load of variable length, we must distribute the load over the
member in the region or regions in which the ordinates of the
influence line are either positive or negative.
The force dP produced by the uniform load w acting on an
infinitesimal beam segment of length dx equals the product of
the distributed load and the length of the segment, that is,
To establish the increment of the function dF produced by the force
dP, we multiply dP by the ordinate y of the influence line at the
same point, to give
dP wdx
=
( )
dF dP y
=

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Uniformly Distributed Load
206
• Uniformly Distributed Load (cont.)
To evaluate the magnitude of the function F between any two points A and B, we integrate both sides
between those limits to give
Recognizing that ydx represents an infinitesimal area dA under the influence line, we can interpret the
integral on the right side as the area under the influence line between points A and B. Thus,
where AAB is the area under the influence line between A and B.
• Example
Maximum Ay is
( ) ( )
B B B B
A A A A
F dF dP y wdx y w ydx
= = = =
   
AB
F wA
=
0
2
w L

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7.5INFLUENCELINESFORDETERMINATEFLOORGIRDERS
209
Occasionally, floor systems are constructed as shown in left figure, where it can be seen that floor loads
are transmitted from slabs to floor beams, then to side girders, and finally supporting columns. An
idealized model of this system is shown in plane view, such as right figure.
Since the girders are main load-carrying members in this system, it is sometimes necessary to construct
their shear and moment influence lines. In this regard, notice that a unit load on the floor slab is
transferred to the girder only at points where it is in contact with the floor beams, i.e., points A, B, C, and
D. These points are called panel points, and the region between these points is called a panel, such as
BC.

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?

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7.6 INFLUENCE LINES FOR DETERMINATE TRUSSES
Construction of Influence Lines for a Truss
217
Since truss members are typically designed for axial force, their cross sections are relatively small
because of the efficient use of material in direct stress.
• Construction of Influence Lines for a Truss
To illustrate the procedure for constructing influence
lines for a truss, we will compute the ordinates
of the influence lines for the reaction at A and
for bars BK, CK, and CD of the truss.

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Construction of Influence Lines for a Truss
218

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Influence Lines for a Trussed Arch
219
• Influence Lines for a Trussed Arch
As another example, we will construct the influence lines for the
reactions at A and for the forces in bars AI, BI, and CD of the
three-hinged trussed arch.

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7.7 MOVING LIVE LOADS
220
In Section 7.4 we discussed how to use an influence line to evaluate the maximum value of a function
when the live load is represented by either a single concentrated load or a uniformly distributed load.
We now want to extend the discussion to include maximizing a function when the live load consists of a
set of concentrated loads whose relative position is fixed. Such a set of loads might represent the forces
exerted by the wheels of a truck or a train.

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• Increase–Decrease Method
Changes for F1 between position 1 and 2 is
Changes for F5 between position 1 and 2 is
For general, changes for all forces between position 1 and 2 is
1 1 1
f Fm x
 =
5 2 5
f F m x
 =
1 2
i i j j
f Fm x F m x
 = +
 

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7.8 ABSOLUTE MAXIMUM SHEAR AND MOMENT
225
So far we have learned how to use the influence line; however, as a designer, it is necessary to determine
which section along the span of the beam is the most critical one.
A single concentrated load acting on a beam produces a triangular moment curve whose maximum
ordinate occurs directly at the load.
The dashed line, termed the moment envelope, represents the maximum value of live load moment
produced by the concentrated load that can develop at each section of the simply supported beam.
The absolute maximum live load moment due to a single load on a simple beam occurs at midspan.

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DEFLECTIONS OF BEAMS AND FRAMES
Chapter Objectives
226
1. determine the elastic deflections using the method of
double integration
2. determine the elastic deflections using the moment-area
method
3. determine the elastic deflections using the elastic load
method
4. determine the elastic deflections using the conjugate-beam
method
5. familiarize the integration and differentiation relationship
between forces and deflections
CHAPTER 8
8.1 Double Integration Method
8.2 Moment-Area Method
8.3 Elastic Load Method
8.4 Conjugate Beam Method
Chapter Outline

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8.1 DOUBLE INTEGRATION METHOD
Geometry of Shallow Curves
227
Chapter 8 Deflections of Beams and Frames
The double integration method is a procedure to establish the equations for slope
and deflection at points along the longitudinal axis (elastic curve) of a loaded beam.
The equations are derived by integrating the differential equation of the elastic
curve twice, hence the name double integration.
• Geometry of Shallow Curves
The deflected shape is represented in figures by the displaced position of the
longitudinal axis (also called the elastic curve).
Since the tangents to the curve are perpendicular to the radii at points A and B,
it follows that the angle between the radii is also d. The slope of the curve
at point A equals
From the geometry of the triangular segment ABO, we can write
where y, representing the change in slope per unit length of distance along the
curve, is called the curvature. Since slopes are small in actual beams, ,
and we can express the curvature as
tan
dy
d d
dx
 
= 
1
d
d ds
ds

  y

=  = =
ds dx

2
2
1
d d d y
dx dx dx
 
y y

= =  = =

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Differential Equation of the Elastic Curve
228
• Differential Equation of the Elastic Curve
As load is applied, the element deforms into a trapezoid as the sides of the segment, which remain
straight, rotate about a horizontal axis (the neutral axis) passing through the centroid of the section.
By definition, the strain e at the top surface can be expressed as
If behavior is elastic, the flexural stress can be related to the strain by Hooke’s law, which states that
For elastic behavior the relationship between the flexural stress at the top fiber and the moment acting
on the cross section is given by 2
2
Mc d y M
I dx EI
 =  =
dl cd
=
2
2
d y
E
dx Ec

 e
=  =
2
2
dl d d y
c
dx dx c dx
 e
e e y
=  =  = =

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Elastic Beam Theory
229
2
2
d y
dV dM dy
w V
M EI
dx dx dx dx
M
V wdx M Vdx y dx y dx
EI
EI
EI
w V M y
w V M y
y 
y
y 
y
y 
y 
=
= = =
=
= = = =
=
⎯⎯⎯
→ ⎯⎯⎯
→ ⎯⎯⎯
→ ⎯⎯⎯
→ ⎯⎯⎯
→
   
⎯⎯⎯
→ ⎯⎯⎯
→ ⎯⎯
→ ⎯⎯⎯
→ ⎯⎯⎯
→
⎯⎯
⎯ ⎯⎯
⎯ ⎯
⎯ ⎯⎯
⎯ ⎯⎯
⎯
積分積分積分積分
微分微分微分微分

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8.2 MOMENT-AREA METHOD
Overview of Moment-Area Method
235
As we observed in the double integration method, we will establish a procedure that utilizes the area of
the moment diagrams [actually, the M/EI diagrams] to evaluate the slope or deflection at selected points
along the axis of a beam or frame, call moment-area method.
This method, which requires an accurate sketch of the deflected shape, employs two theorems.
One theorem is used to calculate a change in slope between two points on the elastic curve.
The other theorem is used to compute the vertical distance (called a tangential deviation) between a point
on the elastic curve and a line tangent to the elastic curve at a second point.

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Derivation of the Moment-Area Theorems
236
• Derivation of the Moment-Area Theorems
To establish the total angle change (change in slope), we must sum up the d increments for all
segments of length ds between points A and B by integration.
This relationship constitutes the first moment-area principle, which can be stated as:
The change in slope between any two points on a smooth continuous elastic curve is equal to the
area under the M∕EI curve between these points.
B B
AB AB
A A
M
d dx A
EI
 
 = = =
 

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Derivation of the Moment-Area Theorems
237
• Derivation of the Moment-Area Theorems
To evaluate tBA, we must sum all increments of dt by integrating the contribution of all the infinitesimal
segments between points A and B.
where is the distance from the vertical axis through A to the centroid of the area between A and B.
This result constitutes the second moment-area theorem, which can be stated as follows:
The tangential deviation at point B on a smooth continuous elastic curve
from the tangent line drawn to the elastic curve at a second
point A is equal to the moment about B of the area under the
M/EI curve between the two points.
B B B
BA AB
A A A
M
t dt xd x dx x A
EI

= = = =
  
x
Courteous to: CALC RESOURCE
https://calcresource.com/statics-deflections-moment-area.html

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positive?
what’s the true direction?

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positive?

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positive?
positive?

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Can we find tAC via two triangles (not one)?!

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?
?

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8.3 ELASTIC LOAD METHOD
260
The elastic load method is a procedure for computing slopes and deflections in simply supported beams.
In the elastic load method, we imagine that the M/EI diagram, whose ordinates represent angle change
per unit length, is applied to the beam as a load (the elastic load), and then compute the shear and
moment curves.
We recognize the relationship between slopes, deflections, and moment as
Moreover, we also learned the relationship between uniform loads, shears, and moment as
So, the slopes and deflections can be simply solved by assuming the
M/EI is a new uniform loads and calculating the shears and moments
2
2
and
dy d y d M
dx dx dx EI


= = =
2
and
dM d M dV
V w
dx dx dx
= = =

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8.4 CONJUGATE BEAM METHOD
Overview of Conjugate Beam Method
265
The conjugate beam method permits us to extend the elastic load method
to beams with other types of supports and boundary conditions by
replacing the actual supports with conjugate supports to produce a
conjugate beam.
The effect of these fictitious supports is to impose boundary conditions
which ensure that the shear and moment, produced in a beam loaded by
the M/EI diagram, are equal to the slope and the deflection, respectively,
in the real beam.
In the figure, we can reason that if the slope and deflection at A must be zero,
the values of elastic shear and elastic moment at A must also equal zero.
On the other hand, since both slope and deflection can exist at the free end
of the actual cantilever, a support that has a capacity for shear and
moment must be provided at B.
Therefore, in the conjugate beam we must introduce an imaginary fixed
support, or conjugate fixed support, at B.

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Conjugate Supports
266
Conjugate Beam
Real Beam
V
y M
  
 
y
 =
=
y
 =
=
y
 =
=
y
 =
=
y
 =
=
y
 =
=
V
M
=
=
V
M
=
=
V
M
=
=
V
M
=
=
V
M
=
=
V
M
=
=

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List of Conjugate Supports
267

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Procedure of Conjugate Beam Method
268
In summary, to compute deflections in any type of beam by the conjugate beam method, we proceed as
follows:
• Establish the moment curve for the real structure.
• Produce the M/EI curve by dividing all ordinates by EI. Variation of E or I may be taken into account in
this step.
• Establish the conjugate beam by replacing actual supports or hinges with the corresponding conjugate
supports.
• Apply the M/EI diagram to the conjugate structure as the load, and compute the shear and moment at
those points where either slope or deflection is required.
y
Conjugate Beam
Real Beam

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?

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Thanks for your attention!
See you next semester!
Of course, don’t forget the final exam!
— Date: 2022/6/13 —

Structural Analysis I

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Structural Analysis I