179306253 lectures in mechanics of materials ion simulescu
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Lectures in Mechanics of Materials Ion SimulescuTRANSCRIPT
- III -
TECHNICAL UNIVERSITY OF CIVIL ENGINEERING
Lectures in
Mechanics of Materials
Eng. Ion S. Simulescu, MPh, PhD, PE
Associate Professor
Bucharest 2004
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PREFACE
Twenty-five years ago, in the late summer of 1979, I left Romania with the vigor and
determination of a thirty year old to conquer the West. Doing so, I left behind family,
friends and especially, what I loved the most, my teaching position at the
Construction Institute of Bucharest, the former name of today’s Technical University
for Civil Engineering. It took the persuasion, the encouragements and the friendship
of my colleague and friend Dr. Eng. Dan Cretu, Chairmen of the Strength of Materials
and Theory of Elasticity Department, to convince me to dramatically alter my life
again and return to Romania. After much hesitation, hours of long distance
conversations and sleepless nights, what appeared for many years to be just an
impossible reverie has become reality. Starting with the fall semester of 2004-2005
academic year, I will be once more part of the faculty in his department.
Since 1980, living in United States, through my own education and professional life, I
have gain a deep understanding and first hand knowledge of the American educational
system and the way structural engineering is practiced in the most advanced country
in the world. I had the opportunity, during my graduate years at Columbia University,
an ivy-league school located in the hart of New York City, to know a number of
renowned Professors including: Masanobu Shinozuka (Probabilistic Theory of
Structural Dynamics and Reliability), Majiek Bienek (Theory of Elasticity,
Viscoelasticity and Plasticity) and Frank Dimagio (Analytical Mechanics). Taking
courses, in preparation for my PhD dissertation, I was completely captivated by the
deeply theoretical character of the lectures and also by the reduced number of
practical applications solved during theses lectures. The successful understanding of
the course theoretical domain rested exclusively with the students required, through
an intensive self-study of the course notes and additional reading, to solve an
extensive number of problems. The students were willing participants in research, and
the best of them spent their summer vacations in the school laboratories or computer
rooms. The exposure to commercial computer programs in solving practical
applications came very early during the instruction years. In contrast, the structural
engineering practice, called generically design, is extremely standardized and in
general limited to a number of simple procedures that every student passing through
an undergraduate program will learn.
It is my intention to adapt and use this accumulated experience into our Romanian
educational system. I strongly believe that the Romanian education system, which
remains almost unchanged since my departure, is in need of improvement and
modernization, especially, by injecting into it the usage of computer applications, as
early as possible in the instruction.
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In the spring of 2004, in preparation for my new academic activity, I began writing a
textbook, entitled Lectures in Mechanics of Materials. This textbook is an
introductory course to the general topic of Mechanics of Materials and is intended to
gradually expose the students, enrolled in the Civil Engineering Section of the
Technical University of Bucharest, to this fundamental subject during the first
semester of their sophomore year. The subject is known in the Romanian technical
literature as Strength of Materials, but the more general title of Mechanics of
Materials is widely employed for undergraduate studies at the engineering colleges in
the United States. Although I followed the departmental approved syllabus, the
lectures are presented in a modern manner by following the basic approach of the
modern theory of continua, namely, the usage of the three basic groups of equations:
the equilibrium, the strain-displacement and the constitutive material equations. I
considered as essential, the early implementation of these three aspects in the
theoretical developments of this extremely important engineering course. This
presentation provides a clear perspective for the students and gives them the
opportunity to compare and follow the methods used in the more sophisticated
theoretical courses of the following study years.
Even though I tried very hard to make the text as perfect and intelligible as possible
and despite the tedious and relentless effort of my young colleague John Creighton, a
structural engineer at Washington Group International, in reading and correcting it, I
am almost certain that many improvements will become necessary as the result of the
interaction with the students.
In this important stage time of my professional carrier a special gratitude is due to
Emeritus Professor Eng. Panait Mazilu, the past Chairman of the Strength of
Materials and Theory of Elasticity Department, for bringing me on in 1975 as an
Assistant Professor and providing support for my professional activity and
development until I left the country. He represents, without a doubt, the brightest
mind I have had the pleasure knowing in my entire professional carrier. A special
thanks to my first teacher in strength of materials, and later colleague and friend, Dr.
Eng. Mihaela Kunst-Germanescu, who through her total dedication made an essential
difference in my technical education. From the American side, I wish to express a
deep appreciation to Professor Masanobu Shinozuka for his support during my
graduate study at Columbia University and his continue guidance during my PhD
preparation.
I also express my appreciation to my wife Catrinel, who sacrificed greatly by
abandoning a life long successful drama carrier to follow me to the United States and
had the audacity to endure hours of technical discussions, many moments of success
as well as difficult times during the past ten years.
Finally, the author thanks Dr. Eng. Dan Cretu without this endeavor would not be
possible.
Dr. Ion S. Simulescu
Princeton, NJ, USA.
May, 2004
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TABLE OF CONTENTS
Lecture 1 Introduction to Mechanics of Materials 1
1.1 What is a Deformable Body?
1.2 Geometrical Classification of the Deformable Body
1.3 Exterior Action Classification
1.4 Stress and Strain
1.5 Engineering Aspects: Analysis, Verification, Optimization
and Design
1.6 Applications of the Mechanics of Materials
Lecture 2 Fundamentals of Mechanics of Materials 10
2.1 Static Equilibrium of Deformable Solid Body
2.2 Support Reaction Forces and Beam Connections
2.3 Generalized Stress Tensor and Components
2.4 Saint Venant’s Principle
2.5 Beam Stresses and Cross-Section Resultants
2.6 Extensional and Shear Strains
2.7 Constitutive Properties of Materials 2.7.1 Tension Tests
2.7.2 Mechanical Properties of Materials
2.7.3 Elasticity, Plasticity and Creep
2.7.4 Linear Elasticity, Hook’s Law and Poisson’s Ratio
2.7.5 Generalized Hook’s Law for Isotropic Materials
2.8 Fundamental Hypotheses and Equations 2.9 Examples
2.9.1 Stress Distribution and Stress Resultants in a
Rectangular Cross-Section
2.9.2 Extensional and Shear Strain
2.9.3 Volumetric Strain
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Lecture 3 Geometrical Characteristics of the Beam
Cross-Section 49
3.1 Definitions
3.2 Parallel-Axis Theorems for Moment of Inertia
3.3 Moment of Inertia about Inclined Axes
3.4 Principal Moments of Inertia
3.5 Maximum Product Moment of Inertia
3.6 Mohr’s Circle Representation of the Moments of Inertia
3.7 Radii of Gyration
3.8 Examples 3.8.1 Rectangle Cross-Section
3.8.2 Composite Cross-Section
Lecture 4 Equilibrium of the Plane Linear Members 68
4.1 Type of Beams, Loads and Reactions
4.2 Equilibrium of Beams Using the Free-Body Diagrams
4.3 Differential Relations between Loads and Cross-Section
Internal Resultants
4.4 Shear Force and Bending Moment Diagrams
4.5 Examples of Shear and Moment Diagrams for Statically
Determined Beams 4.5.1. Simple-Supported Beam Loaded with a Concentrated
Vertical Force
4.5.2 Simple-Supported Beam Loaded with a Concentrated Moment
4.5.3 Beam with Overhang
Lecture 5 Axial Deformation 86
5.1 Basic Theory of the Axial Deformation
5.1.1 Strain-Displacement Equation
5.1.2 Constitutive Equation 5.1.3 Cross-Section Stress Resultants
5.1.4 Equilibrium Equation 5.1.5 Thermal Effects on Axial Deformation
5.2 Uniform-Axial Deformation 5.2.1 Member Subjected to Uniform-Axial Deformation
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5.2.2 Statically Determinate Structure
5.2.3 Statically Indeterminate Structure
5.3 Nonuniform-Axial Deformation 5.3.1 Non-homogeneous Cross-Section Members
5.3.2 Non-homogeneous Cross-Section Members Subjected to Thermal
Changes
5.3.3 Heated Member with a Linear Temperature Variation
5.4 Special Aspects 5.4.1 Normal Stress in the Vicinity of the Load Application
5.4.2 Stress Concentrations
5.4.3 Limits of Poisson’s Ratio
5.5 Design of Members Subjected to Axial Deformation
5.6 Examples 5.6.1 Statically Determinate Structure
5.6.2 Statically Indeterminate Structure
Lecture 6 Pure Shear 112
6.1 Basic Theory
6.2 Discrete Connectors Calculation
6.3 Weld Calculation
Lecture 7 Bending 123
7.1 Definitions
7.2 Pure Bending 7.2.1 Strain-Displacement Equation
7.2.2 Constitutive Equation
7.2.3 Cross-Section Stress Resultants
7.2.4 Normal Stress Distribution
7.3 Nonuniform Bending
7.3.1 Basic Assumptions
7.3.2 Shear Stress Distribution in Rectangular Cross-Section
7.3.3 Limitations in the Usage of the Shear Stress Formula for
Compact Cross-Sections
7.3.3.1 Cross-Section Shape
7.3.3.2 Beam Length
7.3.4 Shear Stress Distribution in Thin-Wall Cross-Sections
7.3.4.1 Assumptions
7.3.4.2 Wide-Flange Cross-Section
7.3.4.3 Closed Thin-Wall Cross-Section
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7.4 Build-Up Beams Connectors 7.4.1 Linear Shear Connectors
7.4.2 Discrete Shear Connectors
7.4.3 Examples
7.4.3.1 Welded Connection
7.4.3.2 Nail Connection
Lecture 8 Beam Deflection 153
8.1 Introduction
8.2 Qualitative Interpretation of the Deflection Curve
8.3 Differential Equations of the Deflection Curve 8.3.1 Integration of the Moment-Curvature Differential Equation
8.3.2 Integration of the Load-Deflection Differential Equation
8.3.3 Boundary and Continuity Integration Conditions
8.4 Examples
8.4.1 Application of the Moment-Curvature Equation
8.4.2 Application of the Load-Deflection Equation
Lecture 9 Torsion 167 9.1 Introduction
9.2 Torsional Deformation of a Member with Circular Cross-
Section 9.2.1 Strain-Displacement Equation
9.2.2 Constitutive Equation
9.2.3 Cross-Section Stress Resultants
9.3 Torsional Deformation of a Member with Closed Thin-Wall
Cross-Section
9.4 Torsional Deformation of a Member with Rectangular Cross-
Section
9.5 Torsional Deformation of a Member with Elliptical Cross-
Section
9.6 Examples 9.6.1 Rectangular Thin-Wall Cross-Section
9.6.2 Capacity Comparison of Different Shaped Solid Cross-Sections
Lecture 10 Plane Stress Transformations 181 10.1 Introduction
10.2 Plane Stress Transformation Equations
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10.3 Principal Stresses
10.4 Maximum Shear Stress
10.5 Mohr’s Circle for Plane Stresses
10.6 Principal Stresses Distribution in Beams
10.7 Example
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LECTURE 1
Introduction to Mechanics of Materials
This textbook is an introductory course to the general topic of Mechanics of Materials
and is intended to gradually expose the students, enrolled in the Civil Engineering
Section of the Technical University of Bucharest, to this fundamental subject during the
first semester of their sophomore year instruction. The subject is known in the Romanian
technical literature as Strength of Materials, but the more general title of Mechanics of
Materials is widely employed for undergraduate studies at the engineering colleges in
the United States of America. The American title has been retained for two main reasons:
(a) English is the language employed for academic instruction of this course and (b)
familiarity with the advanced American educational system is considered essential for the
students of this engineering school.
The theory contained in this textbook represents the base knowledge for the professional
life of any structural engineer. Without a doubt, the theoretical subjects covered in this
textbook represent the foundation for thorough understanding of more advanced subjects
that will confront the student during the undergraduate and graduate study years.
The Classical Mechanics, discipline studied in the previous year, had introduced the
fundamental principals of Statics and Dynamics and applied them in the investigation of
the particle and rigid body behavior, cases representing a drastic idealization of the real
physical systems. The Mechanics of Materials concentrates on the study of the
deformable solid body. The principal objective of this textbook is to familiarize the
student with the basic concepts of deformation, stress and strain induced in a special
category of deformable bodies, namely the beam, by external environmental actions. The
validation of the theoretical derivations is achieved by confronting them with
experimental results obtained during laboratory tests. The understanding of the
mechanical behavior of the beams subjected to exterior actions is the most essential
factor required for successful design of all types of structures such as: bridges, high-rise
buildings, industrial buildings, power plants, aircrafts, spacecrafts, etc.
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Note: Before elaborating about the scope and the main directions of the course,
several conventions used in the text are described. Throughout the entire body
of this textbook, the first instance usage of important terminology necessary to
establish a base technical vocabulary is shown in boldface. Additionally, a
selection of the most important definitions are enclosed in a box and presented
using the italic font.
1.1 What is a Deformable Body?
The entire domain of Mechanics of Materials is concerned with developing the
mathematical methodology necessary to completely define the behavior of the
deformable solid body.
The mathematical description of the physical changes that take place, the modification
of volume and shape, when the three-dimensional body is subjected to exterior
actions, requires that two Cartesian orthogonal coordinate systems be employed:
(1) a global coordinate system OXYZ, being considered as fixed in three-dimensional
space, and (2) a local coordinate system oxyz, rigidly attached to the three-
dimensional body. Consequently, a material point pertinent to the three-dimensional
body can then be defined by two position vectors, each one relative the two
coordinate systems. A schematic representation of the three-dimensional deformable
body subjected to various actions and the Cartesian orthogonal coordinate systems
described above are shown in Figure 1.1. The exterior actions are shown as
concentrated vectors acting in some particular points (points 1 and 2), but in general
the exterior actions have a more complex nature. The symbols attached to some
material points (points 4, 5 and 6) of the three-dimensional body are the schematic
representation of the supports or, more appropriately, constraints.
Figure1.1 Schematic Illustration of a Three-Dimensional Solid Deformable Body
Changes in the volume or shape are generically referred to as deformations.
Consequently, any point of the three-dimensional body moves after the application of
the exterior action from its initial position to a new position in space. For example the
Definition 1.1
A deformable body is a solid three-dimensional body that changes volume and
shape in response to the application of external actions
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material point located in the position marked with the numeral 3 deforms into the
new position *3 . The deformation of the solid body can be small or large, but it plays
a vital role in the analysis of the behavior the three-dimensional solid under the given
exterior actions.
1.2 Geometrical Classification of the Deformable Body
In terms of geometry, any solid body located in three-dimensional space is
represented by its volume and exterior surface. For this study the exterior surface is
considered continuous without any holes or interruptions. The volume is characterized
by three dimensions (length ,l width w and high h ) or by one dimension and two
ratios. For example, if the length l , typically the dimension of greatest magnitude, is
retained, than the ratios width to length lw and height to length lh fully define the
body. The following categories of bodies can be defined as:
(a) Member or Beam is a three-dimensional solid body which has the length
l much larger than the other two dimensions. To be more précised the major
dimension l can be described by a curve called the member longitudinal axis. By
intersecting the three-dimensional volume with a normal plane to the longitudinal
axis a surface is obtained. This surface is the beam cross-section. Usually the beam is
defined:
25.0),max(
l
wh
(1.1)
The cross-section represented by a single surface is called compact, while the surface
which has a central hole is called tubular. A special type of tubular cross-section is
the thin-walls cross-section characterized by very thin wall thicknesses.
The analytical description of the beam longitudinal axis as a curve in space or plane
curve can further differentiate this category into spatial and plane curved members.
The plane members are linear (beam) or curved (arch) members.
Due to the simplicity of manufacturing, the linear member, known in the engineering
practice under the generic name of the beam, is the most commonly employed
structural element. The behavior of the linear beam is the main subject of this course.
Examples of members are shown in Figure 1.2.
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Figure 1.2 Examples of Member Types
(a) Rectangular-Linear, (b) Circular-Linear and (c) Circular-Curved
(b) Plate or Slab is a three-dimensional solid body which has two of its three
dimensions much larger than the third one. The smallest dimension, usually called the
thickness t , is replacing the height in the original definition. The geometry of the plate
median plane or neutral plane, the plane separating the thickness of the plate into
two equal parts, can be represented as a plane or a curved surface. The plate having a
curved median plane is called shell. By definition a plate (slab) has the following
ratios:
1.0l
t and 1.0
w
t
(1.2)
where the dimensions l and w are the length and width of the median plane,
respectively.
If the thickness t is very small comparison with the other two dimensions, the plate is
called membrane.
The plate is also a widely used structural component, but its behavior is the subject of
another course. Examples of plates are depicted in Figure 1.3
Figure 1.3 Plate Types
(a) Slab (Plane), (b) Cylindrical Shell and (c) Parabolic Shell
(c) Block is a three-dimensional solid body which has all of its three dimensions
comparable.
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Figure 1.4 Isolated Foundation
In structural engineering this type of solids are encountered with predilection in all
kind of buildings and machine foundations. An isolated foundation supporting a
structural column is illustrated in Figure 1.4.
Note: It is important to emphasize that the geometrical characterization of the solid
deformable body is also reflected in the mathematical model used in the
description of each body type.
1.3 Exterior Action Classification
The exterior action contributing to the deformation of the three-dimensional solid can
of mechanical or a thermal nature. In the engineering practice, these actions are
labeled under the generic name of loads. The mechanical load is the direct result of
the interaction of the three-dimensional deformable solid under investigation with
other solids, liquids or gases. For example the wind action is mechanical load induced
by the hydrodynamic pressure of the air movement. Similarly, the liquid contained in
the reservoir impinges on the walls and a hydrostatic pressure results.
In general a mechanical load is considered as continuous function of two
variables ),( trp
, where r
the position vector of the material is point where the
function is described and t is the time.
For the case of linear beam type solid, the case of interest for this course, the spatial
variable is described only by the one-dimensional position vector and the function can
be written as ),( txp . The mechanical load is a vectorial function, which is
characterized by direction and intensity.
The classification of the mechanical loads is conducted base on two criteria, both
related with the intensity: (a) the time dependency and (b) the spatial variation.
Definition 1.2
If the intensity of the applied load changes with time the load is called a
dynamically applied load or a dynamic load. As a result the inertial forces,
conforming to Newton’s law, are induced and must be considered in the equations
of equilibrium.
If the intensity of the applied load does not change in time the load is called a
statically applied loads or a static load. In this case there are no inertial forces to
be included in the equations of equilibrium.
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The classification of the mechanical and thermal action as dynamic or static has only
theoretical merit, because, any action is to some extent of dynamic nature. In fact, in
some new textbooks the static loads are called quasi-dynamic loads in order to
incorporate them in the same theory as the dynamic loads.
The spatial variability of the function ),( trp
can be theoretically defined by any
continuous mathematical function. In engineering practice, these functions are
typically limited for simplicity to the following: constant function, linearly varying
function and parabolically varying function. For the rare case of more complicated
spatial variability the concept of piecewise–constant (stepped) or piecewise-linear
functions may be employed.
Examples of mechanical loads commonly used in the analysis of the linear plane
beams are shown in Figure 1.5.
Figure 1.5 Line Mechanical Loads
(a) Concentrated, (b) Uniformly Distributed, (c) Linearly Distributed, and (d)
Parabolically Distributed
In the reality all mechanical loads are applied on the surface of the solid body, which
can be small or large. The application surface degenerates into a line segment in the
theoretical case of the linear solid (beam). The situation when the area of the load
application surface or the length of load application segment is small relative to the
overall area or length of the three-dimensional body suggests the theoretical definition
of the concentrated load.
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Obviously the concentrated load can be static or dynamic in nature. This engineering
simplification can be easily accommodated when dealing with the beam case, but
creates some theoretical difficulties when the other two categories of deformable
bodies (plates and blocks) are investigated.
A special category of loads are the body forces.
A special type of the body force frequently used in structural engineering is the self-
weight of the structural element. Because the gravitational acceleration is considered
constant, this load looses its dynamic nature and can be represented as a static
mechanical load with constant magnitude.
Note: Theoretically, it is very important to differentiate between the pure mechanical
loads and body forces, even from a practical point of view, some types of body
forces are treated as mechanical loads. The self-weight is a typical example.
Thermal loads can be categorized in a manner identical to that of the mechanical
loads, as time dependent or time independent. In engineering practice, the thermal
loads are defined as expansion-contraction thermal loads and gradient thermal
loads.
For the case of the linear member (beam) the expansion-contraction thermal load
manifests when the temperature in the beam varies only along the length of the beam
(mathematically, this means that the intensity of the load depends only on the
variable x ). In contrast, the thermal gradient load manifests only when the
temperature varies only trough the thickness of the beam.
Definition 1.3
The concentrated load is a load which has the intensity obtained by the
following limiting process:
S
trptP S
),(lim)( 0
where S is the application surface
Definition 1.4
A body force is a force representing the existence of the buoyancy, magnetic,
gravitational and inertia forces. It is obtained by multiplying the unit mass of the
material contained in the solid by the acceleration of the solid.
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In general the thermal loading is time dependent (transient), but is reasonably
represented as steady-state where the intensity is presumed to be constant.
All of these categories of loads will be used and studied in-depth in the lectures
concerned with the development of the linearly plane beam theory.
1.4 Stress and Strain
The deformation of the solid body subjected to various types of loads modifies the
internal equilibrium of the body by forcing the three-dimensional solid to move from
its position at rest to a new equilibrium position. The analysis of the local effect of
these deformations conducts to the theoretical concepts of the stress and strain
distribution in the deformable body, the two essential concepts of the Mechanics of
Materials.
This textbook is primarily orientated towards the study of stress and strain
distributions proper for the linearly plane beam type solids. These concepts will be
introduced in the following lectures together with the constitutive law, the functional
relation between stress and strain.
1.5 Engineering Aspects: Analysis, Verification,
Optimization and Design
The problems solved by the theoretical approaches which will be developed during
the instruction time span of this course can be organized in three categories: (a)
analysis, (b) design and (c) optimization.
The engineering activities conducted to determine the deformations, and the stress and
strain distribution of a deformable solid body, when the loads and the geometry of the
body are known, is generically called analysis. This activity will be continuously
emphasized throughout the entire length of the course.
The analysis activities precede the verification activities. The object of the
verification activity is to check the maximum stress distribution and deformations,
calculated during the analysis, against some established limiting values called
allowable stress and deformation, respectively. This aspect is also emphasized in the
following chapters.
If the loads acting on the deformable body and the allowable related to the stress
distribution and deformation have been established, various types of solids can be
analyzed and found proper to the allowable limits. Therefore the described problem is
indeterminate and additional mathematical conditions must be imposed for the
complete solution to be obtained. These additional criteria are called optimization
criteria. The minimum weight criterion is an example. The optimization is a difficult
mathematical problem which is beyond the scope of this text book. However, for a
great many engineering applications, optimization can be achieved without
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mathematical complexity by a simple trial-and-error analysis-verification iterative
process, drawing on the judgment, experience and creativity of the structural engineer.
The three previously described engineering activities encompass the complete
engineering effort from initial concept to final design. In engineering practice, the
loads are established according to the functionality of the structural element, while the
geometrical characteristics are established through an iterative scheme involving
analysis and verification activities. The allowable limits (criteria) are established and
standardized for the material used to manufacture the structural element. Design is the
generic name of the activity for which a structural engineer is trained and tested.
1.6 Applications of the Mechanics of Materials
Applications of the theory of deformable solid bodies, especially beams and plates,
can be found in our daily life. In the past, the more sophisticated structures were
designed by a careful assemblage of simple structural elements which could be
analyzed and verified. Since 1960, with the development and increased usage of the
digital computer, the Mechanics of Materials methods evolved and were transcribed
employing scientific programming languages (FORTRAN, etc.). This way the modern
computer programs or computer codes were born. In today engineering practice a
number of commercial computer codes, such as NASTRAN, ANSYS, ABAQUS
and GT-STRUDL, are extensively employed. These computer codes, sometimes
referred to as Computer-Aided-Engineering (CAE) codes, have the capability of
conducting the analysis as well as the verification activities for the investigated
structure.
Creation of three-dimensional models and drawings of the complex structures are
conducted today utilizing the Computer-Aided-Design (CAD) codes. Many of these
codes have the capability to idealize the constructed model and submit this model
directly or with minimal analyst intervention for analysis and verification.
The analytical methods used in the development of the commercial computer codes
are beyond the scope of this introductory course in Mechanics of Materials, but will
be an integral part of the educational process of the following years.
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LECTURE 2
Fundamentals of Mechanics of Materials
In this lecture the fundamental definitions and concepts used in the Mechanics of
Materials are presented.
2.1 Static Equilibrium of Deformable Solid Body
In this textbook only the time independent loads, the static loads, are considered to act
on the three-dimensional solid. Consequently, the inertial forces are zero and the
necessary conditions for equilibrium are expressed as:
0***1
kRjRiRFR ZYX
n
i
i
(2.1)
kMjMiMrFM ZYX
n
i
iiO
***)(
1
(2.2)
where the notation stands for:
n - is the number of concentrated forces;
R
- is the force resultant vector and its components;
ZYX RRR
,, are the magnitude of the components of R
along the coordinate
axes;
iF
- is a particular force;
iZiYiX FFF
,, - are the magnitude of the components of iF
along the coordinate
axes;
kji
,, - are the Cartesian unit vectors of the OXYZ system;
OM
- is the moment resultant vector and its components calculated at the
arbitrary point O , the origin of the OXYZ coordinate system;
OzOyOx MMM ,, - are the magnitude of the components of OM
along the
coordinate axes;
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ir
- is the position vector extending from the coordinate origin to the point
where the concentrated force iF
is applied.
The vectorial equations (2.1) and (2.2) can be re-written in a scalar form of six
algebraic equations:
01
n
i
iXX FR
(2.3)
n
i
iYY FR1
0
(2.4)
01
n
i
iZZ FR
(2.5)
0OXM (2.6)
0OYM (2.7)
0OZM (2.8)
The equations of equilibrium (2.3) through (2.8) were previously developed and
extensively used in the Classical Mechanics course, where the three-dimensional rigid
solid was studied.
The mathematical statements, (2.1) and (2.2), imply that the three-dimensional solid
body is free of constraints, a situation contradictory to the norm where the constraints
are present. It should be noted that the equilibrium equations do not refer in any way
to the nature of the material present in the solid or to any type of deformation.
Consequently, those equations also apply fully in the general case of a deformable
three-dimensional solid.
To utilize the equations (2.1) and (2.2) under the usual conditions, the three-
dimensional deformable solid is released from its constraints, and these are replaced
with the corresponding reaction forces. As described in section 1.3, the drawing that
illustrates the geometry of the three-dimensional solid, the exterior actions, and the
Definition 2.1
The necessary and sufficient conditions for an unconstrained three-dimensional solid
body to be in static equilibrium (in the absence of the inertia forces) are expressed by
the vectorial equations (2.1) and (2.2)
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unknown reaction forces is called a free-body diagram. A generic example of a free-
body diagram is shown in Figure 2.1.
Figure 2.1 Generic Free-Body Diagrams
(a) Constrained Body, (b) Free Body
The attempt to write the equations of equilibrium, (2.3) through (2.8), using the active
forces and the reaction forces, results in a system of six algebraic equations with
constant coefficients, containing the reaction forces as unknown quantities. This
system of equations may be solved using any linear algebra method of solution for
algebraic equation systems with constant coefficients. If the equations contain only
six unknown reactions it is called a statically determinate system and, consequently,
can be solved. When the number of unknown reactions is higher than six, the equation
system is called a statically indeterminate system and additional equations are
required for the reactions to be calculated. If a number of supports can be removed
without destroying the static equilibrium of the body, those supports may be identified
as redundant supports and the corresponding reaction forces are called redundants.
In the particular case of the plane linear beam when the exterior acting loads and the
reaction forces are contained in the vertical plane OXY only three equations of
equilibrium from the initial six are required for solution as follows:
01
n
i
iXX FR
(2.9)
n
i
iYY FR1
0
(2.10)
0OZM (2.11)
- 51 -
Note: It can be shown that only the number of equilibrium equation is fixed at six,
but the option of writing these equations vary. For the plane case the equations
(2.9) through (2.11) may be replaced, for example, by one force equation and
two moment equations.
2.2 Support Reaction Forces and Beam Connections
The external loads applied on the three-dimensional deformable body are carried to
the supports, which are the points where the interaction with the environment (other
bodies) is considered to take place. At the supporting points the displacements,
represented by a change in the position of the point, are known quantities with
prescribed value. In general, the supported point is constrained against movement in
some direction and thus, the displacement in that direction is zero. There are also
cases where a displacement of known magnitude and/or direction is imposed a
particular support point. Due to the Newton’s first law, reaction forces are developed
at the supporting points. These reaction forces are vector quantities with known
direction, but unknown intensity.
If the studied system is composed of more than one single body, the common points
between each adjoining body are called connections. To write the equilibrium
equations for a multi- body system the connections should also be replaced by
reaction forces. A free-body diagram and a set of equilibrium equations may then be
constructed for each body allowing for solution of all external support reaction as well
as for all internal connection reactions.
Tables 2.1 through 2.3 contain the most common idealized types of supports and their
corresponding reaction forces. The two-dimensional cases are illustrated in Tables 2.1
and 2.2 which are directly related with the behavior of the linear plane beam, the case
of interest for this course. The supports and reactions pertinent to the three-
dimensional case, only occasionally considered in this course, are depicted in Table
2.3 for completeness.
Table 2.1 Plane Supports and Reactions
- 52 -
Table 2.2 Plane Connections and Reactions
Table 2.3 Three-Dimensional Supports and Reactions
2.3 Generalized Stress Tensor and Components
When subjected to external actions, the three-dimensional solid deforms and the
internal original equilibrium is disturbed. To determine the internal forces located in
the volume of the deformed body the method of sections is employed. The
application of the methodology is depicted in Figures 2.2 and 2.3.
Figure 2.2 Three-Dimensional Free-Body Sectioned by a Cutting Plane
- 53 -
Practically, the original body is divided into two parts, each one carrying its portion of
exterior and reaction forces, by an arbitrarily orientated plane. In a manner of
speaking, the cutting plane plays, the role of the member connection. As in the
connection case, the broken continuity of the body is replaced by internal forces
acting at every point of the common surface.
Figure 2.3 Lower Half of the Sectioned Original Body
The internal force )(PR
is called the stress vector or traction and acts on the
elementary surface A located in the vicinity of point P (see Figure 2.3). Using a
local coordinate system Pnts attached to the current point P , where n
represents the
unit vector normal to the cutting plane, the stress vector )(PR
is decomposed into
three components, )(PF
, )(PVs
and )(PVt
. The decomposition process is
illustrated in Figure 2.4. The first component of the stress vector, )(PF , is orientated
parallel to the surface outside normal unit vector n
, while the other two components,
)(PVs
and )(PVs
, are located in the cutting plane.
Note: The normal n
is univocally defined by the choice of cutting plane, while the
other two directions, s
and t
are arbitrarily chosen under the condition that
the three are mutually perpendicular.
Figure 2.4 Stress Vector and Components
The components of the stress tensor located in a three-dimensional solid body are
defined through a limiting process.
- 54 -
The position of the point P is uniquely defined by its position vector r
. The stress
tensor )(rn
is called normal stress and acts normal to the cutting plane. The in-
plane stress tensors )(rns
and )(rnt
are called shear stresses.
Note: The normal and shear stresses belong to a special mathematical category of
algebra named tensors. It is beyond the scope of this course to elaborate on
the mathematical properties of tensors, but it must be emphasized that they
may not be manipulated as vectors. The stress tensor is transformed into a
corresponding stress vector through multiplication with an area and, only then,
manipulated through the familiar vectorial algebra.
Some clarifications regarding the sign convention of the stress tensors are necessary.
The subscript n indicates that the stresses are pertinent to the surface which has the
outward normal vector n
. It is understood that the normal stress )(rn
is positive
when parallel to n
, while the in-plane shear stresses )(rns
and )(rnt
are positive
when they are parallel to the s
and t
axes, respectively.
If a new stress vector )(rpn
pertinent to the surface characterized by the outward
normal n
is defined the following vectorial equation can be written:
tArsArnArArp ntnsnn
**)(**)(**)(*)( (2.15)
After the algebraic simplification the fundamental relation between the surface stress
tensor )(rpn
and the normal and shear tensors can be expressed as:
trsrnrrp ntnsnn
*)(*)(*)()( (2.16)
If a set of three mutually orthogonal planes, defined by a Cartesian general system
and passing through the arbitrarily chosen point P are used, as shown in Figure 2.5,
Definition 2.2
The stress tensor components are defined:
A
rFr
An
)(lim)(
0
(2.12)
A
rVr s
Ans
)(lim)(
0
(2.13)
A
rVr t
Ant
)(lim)(
0
(2.14)
- 55 -
then three corresponding sets of stress tensors are obtained by successively
considering the unit vectors i
, j
and k
as the normal vector of the cutting plane.
Figure 2.5 Set of Three Mutually Orthogonal Planes
Consequently, the equation (2.16) is written for each one of the three orthogonal
planes as follows:
krjrirrp xzxyxx
*)(*)(*)()( (2.17)
krjrirrp yzyyxy
*)(*)(*)()( (2.18)
krjrirrp zzyzxz
*)(*)(*)()( (2.19)
The resulting stress tensors are depicted in Figure 2.6.
The following close examination and discussion of the orientation of the three sets of
tensors shown in Figure 2.6 defines the sign convention used.
Figure 2.6 Three-Dimensional Stress Tensors (Cartesian Axes)
The normal stresses are positive when orientated parallel to the outward normal of the
surface. They tend to pull the plane and consequently are said to be positive in
- 56 -
tension. The shear stress tensor is considered to be positive when orientated parallel
to and in the positive direction of the corresponding in-plane axis. These shear stress
tensors are identified by two subscripts: the first subscript indicates the plane on
which the shear stress acts by the axis defining its normal; the second subscript
indicates the shear stress direction by identifying the coordinate system axis to which
the shear tensor is aligned.
In Figure 2.6, for clarity reasons, only the stress tensors pertinent to the visible faces
are shown. For the parallel planes having negative axes as normal the stresses have
opposite signs. Why? The infinitesimal volume has to be in equilibrium.
Figure 2.7 Three-Dimensional Free-Body Diagram of an Infinitesimal Volume
Employing the notation convention described above, it is observed that the shear
stress tensors converging towards any particular edge have the subscript order
reversed. Consider the situation depicted in Figure 2.7, where for clarity only the
stress tensors which can be used in writing a moment equation around the oz axis are
retained.
Note: The Figure 2.7 is simply a free-body diagram for an elementary volume
located around a point P.
The sign convention previously explained is used. The usage of the superscript )( or
(-) indicates the variation of the same stress tensor in-between two parallel faces. The
following equations relating identical stress tensors located on parallel faces, where
the position vector r
has been drop in the notation, are written as:
xxx (2.20)
yyy (2.21)
xyxyxy
(2.22)
- 57 -
yxyxyx (2.23)
zyzyzy (2.24)
zxzxzx (2.25)
Note: The variation of the stresses between the parallel faces is due to the existence
of the exterior actions.
For the infinitesimal volume to be in equilibrium the total moment induced by the
stress vectors must be zero.
0oxM (2.26)
0oyM (2.27)
0ozM (2.28)
The moment component ozM written about the oz axis considering the right-hand
rule is expressed as:
2***
2***
2***
2***
******2
***
2***
2***
2***
xyx
xyx
yyx
yyx
yzxxyzx
zx
xzx
yyz
yyzM
zy
zyzxzx
yxxyy
yxxoz
(2.29)
Substituting the relations (2.20) through (2.25) and (2.29) into the equilibrium
equation (2.28), equation (2.30) is obtained:
0***)(
xyzM yxxyoz (2.30)
Note: The final expression of the equation (2.30) is obtained by neglecting the
product of the change in stress ( or ) and the geometrical elementary
quantities ( x , y and z ) as being very small quantity as x , y and
z tends to zero.
The equation (2.30) indicates that for the equilibrium to be enforced, the edge
converging shear stress tensors are equal. The equality of the edge converging shear
tensors is called the duality principle of the shear stress tensors and is expressed as:
- 58 -
yxxy (2.31)
It can be concluded that even in the presence of the normal stresses the shear stresses
must satisfy the equation (2.31)
In a similarly manner, the other two equilibrium equations (2.26) and (2.27) involving
moments around ox and oy axes are written and the duality of the other shear stresses
is obtained:
zxxz (2.32)
zyyz (2.33)
The stress tensors can be collected in a tabular form called the generalized stress
tensor:
)()()(
)()()(
)()()(
)(
rrr
rrr
rrr
rT
zzyzx
yzyyx
xzxyx
(2.34)
The generalized stress tensor )(rT
, which plays an important role in the deformable
solid theory, has nine stress tensors components, but due to the duality principle only
six are independent.
Note: It has to be emphasized that the general stress tensor )(rT
is point dependent.
2.4 Saint Venant’s Principle
The Saint Venant’s Principle is not a rigorous law of Mechanics of Materials, but has
a great practical significance in the analysis of beams and other deformable bodies.
Definition 2.3
Saint Venant’s Principle is stated as follows:
The stresses in a deformable solid body at a point sufficiently remote from points of
application of load depend only on the static resultant of the loads and not on the
local distribution of the loads.
- 59 -
The theoretical explanation of the Saint Venant’s Principle is depicted in the Figure
2.8.
Figure 2.8 Saint Venant’s Principle
The original free-body diagram of the three-dimensional body in equilibrium is shown
in Figure 2.8.a. The three-dimensional solid is loaded with a force q distributed over a
small surface of area A . Suppose that the generalized stress tensor )(rT
is known for
any current point )(rP
of the solid body, identified by its position vector r
. Two
concentrated loads of magnitude Q opposing each other and acting collinearly with
the resultant of the distributed force q are superimposed on the three-dimensional
body as shown in Figure 2.8.b. Their magnitude Q is obtained by integrating the
variation of the distributed force q over area A :
A
dAqQ * (2.35)
Proceeding in this manner, the equilibrium of the three-dimensional solid body
remained unchanged and consequently, the generalized stress tensor )(rT
remains
unaffected. The new loading situation shown in Figure 2.28.b may then be divided
into two load cases as indicated in Figures 2.8.c and 2.8.d. The original generalized
tensor )(rT
is written as the sum of two new generalized stress tensors )()1( rT
and )()2( rT
, each one corresponding to the loading situation illustrated in Figure 2.8.c
and 2.8.d, respectively.
)()()( )2()1( rTrTrT
(2.36)
Note: This relation is valid only under the restrictions of the Principle of Linear
Superposition discussed in section 2.8.
Because the loads shown in Figure 2.8.c are self-equilibrated (force and moment
resultants are zero) the generalized stress tensor )()1( rT
calculated in points away
from the loading is expected to tend toward a zero value.
0)()1( rT
(2.37)
- 60 -
Consequently, the original generalized tensor )(rT
is approximated by the
generalized stress tensor )()2( rT
pertinent to the loading situation contained in Figure
2.28.d:
)()( )2( rTrT
(2.38)
Equation (2.34) provides validation for Saint Venant’s Principle.
2.5 Beam Stresses and Cross-Section Resultants
The geometrical description of the linear member or beam was given in Section 1.3.
Here, it is re-emphasized that the length of the member is much larger than the other
two dimensions. Of principal importance in characterizing the behavior of the beam is
a geometrical description of its cross-section. The geometrical properties of the cross-
section and its theoretical role in the development of the beam theory will be
elaborated on during the entire length of these lectures.
Consider the case of a beam with a general, undefined, geometrical cross-section as
shown in Figure 2.9. The volume of the beam was sectioned by a normal cutting plane
with its outward normal parallel to the longitudinal local axis ox . The general
coordinate system Pnts , previously used, is identical now with the local coordinate
system oxyz attached to the beam.
Figure 2.9 Cross-Section Beam Stresses
Definition 2.4
The surface obtained by intersecting the linear member (beam) volume with a cutting
plane having as outward normal n
coincident with the longitudinal local axis of the
member is called the beam cross-section.
- 61 -
In this section the general formulas, expressed in equations (2.12) through (2.14), will
be applied to the case in point of this textbook, the linear beam.
The normal stress )(rn
is renamed as )(rx
, while the shear stresses )(rns
and
)(rnt
became )(rxy
and )(rxz
. The local position vector is written as:
kzjyixr
*** (2.39)
where i
, j
and k
are the unit vectors of the local Cartesian coordinate system.
The stress tensor components )(rx
, )(rxy
and )(rxz
can be recomposed in the
corresponding stress vectors F
, yV
, and zV
, respectively, by multiplying them
with the elementary surface A . By integrating over the entire surface of the cross-
section the components of the resultant force and moment particular to the cross-
section are obtained. They are named cross-sectional resultants and are expressed as:
A
x dArxF *)()(
(2.40)
A
xyy dArxV *)()(
(2.41)
A
xzz dArxV *)()(
(2.42)
A
xy
A
xz dArzdAryxT *)(**)(*)(
(2.43)
A
xy dArzxM *)(*)(
(2.44)
A
xz dAryxM *)(*)(
(2.45)
The cross-section resultants are depicted in Figure 2.10.
Fig 2.10 Cross-Sectional Resultants
- 62 -
The cross-section forces expressed in the equations (2.40) through (2.42) are called
force resultants. Force )(xF is named cross-section normal force or cross-section
axial force, while forces )(xVy and )(xVz are called cross-section shear forces. The
cross-section normal force is positive if orientated parallel to the outward normal of
the cross-section, while the cross-section shear forces are positive when orientated
parallel to the oy and oz positive axes. The normal force is called the tension or
compression force if the action is in the positive or negative direction of the cross-
section normal, respectively.
The equations (2.43) through (2.45) are written considering that the local coordinate
system follows a right-hand rule and they are called the moment resultants. The
moment vector )(xT parallel to ox axis is named cross-sectional torsion moment or
torque. It has a twisting action on the cross-section. The other two cross-sectional
moments, )(xM y and )(xM z , acting parallel with oy and oz positive axes are
named cross-sectional bending moments.
2.6 Extensional and Shear Strains
When subjected to exterior actions the deformable solid body exhibits volume and
shape changes. Geometrically speaking, this means that the geometry of the body
before the action starts differs from the final geometry. During the deformation
process all of the geometrical characteristics (line segments, angles, surface and
volume) of the solid body are altered. A particular point P moves at the end of the
deformation process to a new position *P . Similarly, the point Q , located in the
vicinity of point P , is relocated in the new position *Q . The initial undeformed and
final deformed conditions of the three-dimensional solid body are depicted in Figures
2.11.a and 2.11.b, respectively.
Figure 2.11 Deformation of the Infinitesimal Line Segment
(a) Undeformed Body and (b) Deformed Body
The infinitesimal arc PQ , which has arclength s , is deformed into a new
infinitesimal arc **QP with arclength *s . The definition of the extensional strain is
given below.
- 63 -
The modification of the original right angle defined by the intersection of two line
segments as a result of the deformation process is called shear strain and is
schematically shown in Figure 2.12.
Figure 2.12 Modification of the Right Angle
(a) Undeformed Body and (b) Deformed Body
The original right angle 2
changes after the deformation into a new angle * .
Definition 2.5
The extensional strain in the direction n
at a point )(rP
, identified by its position
vector r
, is defined as:
)*
(lim)(s
ssr
nalong
PQn
(2.46)
Definition 2.6
The shear strain at point )(rP
, identified by its position vector r
, represents the
change of the right angle between two line segments attached to the point and
orientated in directions n
and t
, is defined as:
)2
(lim)( *
talong
PRnalong
PQnt r
(2.47)
- 64 -
The extensional and shear strain definitions are generalized for three mutually
orthogonal planes in a similar manner to the development given for the generalized
stress tensor in Section 2.3. By choosing the Cartesian unit vectors successively as
normal to the cutting planes, the generalized strain tensor )(rT
is obtained:
)()()(
)()()(
)()()(
)(
rrr
rrr
rrr
rT
zzyzx
yzyyx
xzxyx
(2.48)
where:
rx
( ), )(ry
and )(rz
– are the elongation strain measured in the ox , oy
and oz , respectively;
)(rxy
, )(rxz
and )(ryz
- are the shear strains.
Situations involving specific the use of specific components of the generalized strain
tensor )(rT
are commonly encountered and thus, the subject will be revisited in the
following lectures.
The rational used in the development of the Saint Venant’s principal considered only
the generalized stress tensor, but a similar argument can be made for the case of the
generalized strain tensor.
2.7 Constitutive Properties of Materials
The functional expressions relating the generalized stress tensor components to the
generalized strain tensor components are called constitutive equations. They reflect
the nature of the material contained in the deformable body. Mathematically, a
general expression such as equation (2.45) may be written to define the
aforementioned functional relationship:
),...),(),,((),( trdt
dtrftr nmnmijij
(2.49)
where ),( trij
- is the component of the generalized stress tensor;
),( trnm
- is the component of the generalized strain tensor;
ijf - is a given function;
dt
d - is the time derivative
- 65 -
The functions ijf are chosen such that they satisfy the Laws of Thermodynamics and
with parameters established by laboratory testing to give reliable results in physical
usage. The change in the subscripts, from i and j to m and n , indicates that in
general a stress tensor component may be a function of all the strain tensor
components. Only in a few simple cases does a one-to-one relation exist.
2.7.1 Tension Tests
In a laboratory control environment, using a hydraulically actuated testing machine, a
material specimen is subjected to a tension test. The schematics of the theoretical
tension test are shown in Figure 2.13.
The tension test is conducted in the following steps: (a) the original (undeformed)
gage length 0L and cross-sectional area 0A of the specimen are measured before the
test; (b) the ends of the specimen shown in Figure 2.13.a are gripped in the testing
machine; (c) force is slowly applied to the ends of the specimen until the rupture of
the material is obtained; (d) during the load application measurements of the gage
length *L , cross-sectional area *A and the value of the applied forced *P are
tabulated. The steps (c) and (d) are repeated at regular time intervals and the loading
of the specimen is conducting slowly in order to avoid the rise in temperature and
dynamic effects. When the rupture of the specimen occurs the area minA is measured
once again. This kind of laboratory experiments will be carried out in the Department
Laboratory during the course.
Figure 2.13 Tension Test of Structural Steel Specimen
(a) Undeformed Specimen, (b) Deformed Specimen
Figure 2.14 shows the test results obtained from the testing of a structural steel
specimen, characterized by low carbon content. To emphasize the test results obtained
at lower strain values the beginning end of the test was plotted at a magnified scale.
This type of steel is commonly used in structural applications (buildings, bridges,
etc.).
- 66 -
Figure 2.14 Stress-Strain Diagram
The values of the normal stress and elongation strain can be calculated for each
loading step as follows:
0
* *
A
Px (2.50)
0
0* *
L
LLx
(2.51)
where (*) indicates a particular measurement.
The normal stress defined by the equation (2.50) is called engineering stress, while
the extension strain expressed by equation (2.51) is known as the engineering strain.
If the normal stress is calculated using the value of the step measured area *A , the
value obtained is called the true stress and the corresponding strain is called the true
strain. They are calculated as follows:
*
**
A
P
truex (2.52)
)ln()*
ln(0
*
0*
L
L
A
A
truex (2.53)
Note: The definition (2.53) is based on the fact that the volume remained
unchanged **
00 ** LALA .
The engineering strain and the true strain can are related as expressed by equation
(2.54).
- 67 -
)1ln( **
xtruex (2.54)
Note: Analyzing the expressions of the engineering and true stress and strain it has
to be noted that the true values are larger than the engineering values.
2.7.2 Mechanical Properties of Materials
Several important material properties are obtained from the analysis of the stress-
strain diagram. The theoretical behavior of a typical structural steel is illustrated by
the stress-strain diagram shown in Figure 2.15.
The loading and deformation process begins at point A and continues until point B is
reached. This is the elastic region and is characterized by a proportionality of the
ratio between stress and strain. The point B is named proportional limit occurring
when the stress reaches PL . The ratio of stress to strain in the linear range (elastic
region shown) is called the modulus of elasticity or Young’s modulus and is
designated by the letter E .
x
xE
when PLx (2.55)
The units for the modulus of elasticity are stress units (F/L2). Typical units are ksi or
GPa.
Figure 2.15 Theoretical Description of the Stress-Strain Diagram
Continuing beyond the stress and strain characterizing the proportional limit at point
B, the specimen begins to yield. Smaller incremental loading steps are necessary to
induce additional elongation. Two new points, C and D, named the upper yielding
point, uYL )( , and the lower yielding point, lYL )( , are obtained. For practical
- 68 -
purposes only the lower yielding point D is retained and will be called the yielding
point YL . From point D to point E increased elongation of the specimen occurs with
no increase in stress and consequently, the stress at point E has an identical value with
the stress at point D. As shown in Figure 2.15, the diagram is horizontal and the
region is called the perfectly plastic region.
The stress begins again to increase with increased in elongation after passing point E.
The increase continues until point F, where the ultimate stress or the ultimate
strength, is reached. The phenomenon and the zone are called strain hardening and
strain hardening region, respectively.
If the elongation of the specimen continues a decrease in the stress is recorded and the
so-called “necking” appears. The “necking” represents the visual reduction of the
cross-section and continues until the fracture of the specimen is attended at point G.
The stress calculated at point F, is the maximum stress characterizing the entire stress-
strain diagram. This value is called the ultimate stress or ultimate strength and is
designated by U .
If the material stress and strain remain in the elastic region where the stress is
proportional the strain, the material exhibits elastic behavior. If the material is loaded
over the proportionality limit PL the material exhibits plastic behavior.
The material behavior described above is based on the engineering characterization of
the stress and strain. For comparison the true stress and strain behavior is also plotted
in Figure 2.15. The true stress-strain diagram is plotted using a dashed line, while the
engineering stress-strain diagram is plotted with a heavy continuous line.
The area contained under the stress-strain diagram represents the energy necessary to
deform the specimen. It can be noted, from Figures 2.14 and 2.15, that the energy
characterizing the elastic deformation is considerable smaller than the energy
characterizing the plastic deformation. This is an important characteristic typical of
ductile materials.
The behavior of the structural steel, as it was described above, can be explained
without going into the theoretical details by behavior of the metal micro-structure.
Obviously, different metals exhibit different type of stress-strain behavior when
subjected to a tension test. To emphasize this observation, the stress-strain diagrams
for low carbon steel, three high-strength alloy steels and three aluminum alloys are
presented in Figures 2.16.a and 2.16.b, respectively.
- 69 -
Figure 2.16 Examples of Stress-Strain Diagrams
(a) High-Strength Alloy Steels, (b) Aluminum Alloys
Materials, such as structural steel, which undergo large strain before fracture are
called ductile materials. The ductility factor is represented by the ratio of the strain
measured at rupture (point G) and that corresponding to the point B in Figure 2.15.
For metals used in structural engineering the ductility factor can reach values between
10 and 20.
The materials which fail at small strain value are called brittle materials. A
qualitative difference between a ductile and brittle material are shown in Figure 2.17.
Typical examples of brittle behavior are the following materials: ceramics, glass, cast
iron. The welding material, when the weld is improperly made, also exhibits brittle
behavior which in many cases compromises the quality of the entire structural
ensemble. Drastic change in temperature, pressure and load application can very
much influence the material behavior.
Figure 2.17 Ductile and Brittle Stress-Strain Diagrams
The temperature is an important factor in the material behavior. Same materials can
change from ductile to brittle behavior as a function of the temperature.
- 70 -
Figure 2.18 Variation of the Stress-Strain Diagram with Temperature
(a) Low Temperature, (b) High Temperature
In most common applications the materials used in structural engineering are not
subjected to temperature extremes great enough to adversely affect the ductile
behavior. For the special situations when the material is subjected to high or low
temperature corresponding stress-strain diagrams must be constructed. Figures 2.18.a
and 2.18.b illustrate the variation of the stress-strain diagram with temperature for a
type of stainless steel. Note that the strain increases for a given stress with increase in
temperature and vice versa.
The American Society of Mechanical Engineers (ASME) publishes a handbook
containing a large range of stress-strain diagrams for different metal composition and
temperatures.
2.7.3 Elasticity, Plasticity and Creep
The stress-strain diagram described above has been obtained experimentally by
continuous loading of the specimen. If at some time during the loading the load
applied to the specimen is reversed, theoretically following the same increments as
were used during loading, the process is called un-loading.
If the un-loading stress-strain diagram follows the same path as the loading branch the
material is said to be an elastic material. Obviously, the elasticity of the material is
manifested only if the un-loading occurs before the stress and strain reach the elastic
limit, represented by the stress-strain point C shown in Figure 2.19.
As described previously for a typical structural steel, the stress is proportional to
strain (the modulus of elasticity has a constant value) and the behavior is called
linear-elastic. There are materials characterized by a nonlinear-elastic behavior. An
example of non-linear elastic material is shown in Figure 2.19.
- 71 -
Figure 2.19 Non-Linear Elastic Behavior
If the stress-strain values corresponding to point C are exceeded the material exhibits
a plastic behavior and is said to undergo a plastic flow. A typical loading and un-
loading cycle from a plastic range is schematically shown in Figure 2.20. Noted that
during un-loading the material behave as a linear-elastic material.
Figure 2.20 Plastic Behavior
For theoretical applications, the stress-strain diagram shown in Figure 2.15 is replaced
by an artificial but useful stress-strain diagram named Prandtl’s diagram. This type
of diagram is shown in Figure 2.21 and represents a material with a linear-elastic and
perfectly plastic behavior.
- 72 -
Figure 2.21 Prandtl’s Stress-Strain Diagram
The experimental loading and un-loading of the specimen being discussed here are
conducted in a relatively short time interval (a few minutes). An interesting
phenomenon occurs if the specimen is loaded and then left under a constant load for a
long period of time. If the initial strain in the specimen is calculated and the
calculation is repeated at intervals of a few days, an increase in the strain is observed.
This phenomenon is called creep. The result of a creep experiment is shown in Figure
2.22.
Figure 2.22 Creep
Similarly, if the specimen is kept under constant strain for a long period of time a
reduction in the stress value can be observed. This phenomenon is known as
relaxation. A relaxation diagram is depicted in Figure 2.23.
- 73 -
Figure 2.23 Relaxation
2.7.4 Linear Elasticity, Hook’s Law and Poisson’s Ratio
The elastic behavior of materials is important for the structural engineering point of
view. From the explanations regarding the stress-strain diagram it is clear that if the
material is loaded beyond the elastic limit permanent deformations will be present in
the solid body. In general, structural engineers design their structures to behave
elastically.
Analyzing the stress-strain diagram interval between points A and B illustrated in
Figure 2.15 the proportionality between the stress and strain is evident.
Mathematically, the relationship is expressed as:
xx E * for YPx 0 (2.56)
Equation (2.56) is known as Hook’s law. As noted earlier, the constant E is named
modulus of elasticity or Young’s modulus. Structural steels typically have modulus of
elasticity varying around the value of 30000 ksi or 210 GPa.
During the elongation of the tensile specimen, a reduction of the dimensions in the
other two directions normal to the deformation direction is observed. This transversal
contraction, mathematically represented by equation (2.57), is illustrated in Figure
2.24 at an exaggerated scale:
xzy * (2.57)
The constant is called Poisson’s ratio. This constant is non-dimensional and varies
for structural steel around the value of 1/3 value. Theoretically, the Poisson’s ratio is
limited to 0.5. The limits of the Poisson’s ratio are treated in Section 5.4.3.
- 74 -
Figure 2.24 Elongation and Transversal Contraction
2.7.5 Generalized Hook’s Law for Isotropic Materials
The relationship between stress and strain described for the simple case of uni-axial
elongation will be extended to the general case involving generalized stress and strain
generalized tensors. The material constants E and , together with the thermal
coefficient of expansion are the most relevant material characteristics.
Figure 2.25 depicts the general case of the three-axial elongation. Under the influence
of applied load and temperature change, the total extensional deformation can be
decomposed into four independent effect components: (a) the deformation resulting
from the change in temperature, (b), (c) and (d) deformations due to individual
application of the three-directional normal stresses. This approach is called the
Principle of Linear Superposition.
Note: The Principle of Linear Superposition is discussed in Section 2.8. The viability
of this principle is restricted to the case of small deformation and linear elastic
materials.
Definition 2.7
An elastic solid is isotropic if its material characteristics are invariants under any
transformation of the coordinate system.
- 75 -
Figure 2.25 Deformation Induced by a Change in Temperature and
Normal Stress Components
Mathematically, the phenomena illustrated in Figure 2.25 are described with the
following equation:
zyx
xxx
T
xx
(2.57)
zyx
yyy
T
yy
(2.58)
zyx
zzz
T
zz
(2.59)
where the notation stands for:
x - is the total elongation strain in ox direction; T
x - is the elongation strain induced by the thermal expansion;
zyx
yxx
,, - are the elongation strains induced by the action of the normal
strains zyx ,, , respectively, in the ox direction
The total strains y and z are similarly defined by substitution of the appropriate
subscripts.
The existence of the normal stress x acting along the ox direction (see Figure 2.25.c)
introduces, as described in the previous section, the following elongation strains:
E
x
xx
(2.60)
- 76 -
E
x
yx
(2.61)
E
x
zx
(2.62)
Analogously, the formulae (2.60) through (2.62) are written for the case when the
normal stress y is considered (see Figure 2.25.d) as:
E
y
xy
(2.63)
E
y
yy
(2.64)
E
y
zy
(2.65)
Considering the application of the normal stress z depicted in Figure 2.25.e the
following equations are written:
E
zx
z
(2.66)
E
zy
z
(2.67)
E
zz
z
(2.68)
The thermal strains, which are only elongations, are expressed as:
TT
z
T
y
T
x * (2.69)
where:
- is the coefficient of thermal expansion;
T - is the change in temperature.
Introducing the formulae (2.60) through (2.69) into (2.57) through (2.59) equations
(2.70) through (2.72) are obtained as:
TE
zyxx *)]**[*1
(2.70)
- 77 -
TE
zyxy *)]**[*1
(2.71)
TE
zyxz *)]**[*1
(2.72)
For the case of the isotropic linear material the shear strains are related to the shear
stresses by the following relationships:
xyxyG
*1
(2.73)
xzxzG
*1
(2.74)
yzyzG
*1
(2.75)
Figure 2.26 Deformation induced by the Shear Stress Components
Equations (2.73) through (2.75) are illustrated in Figure 2.26. The constant G is the
shear modulus and is obtained from E and using the following relation:
)1(2
EG (2.76)
Equations (2.70) through (2.75) are named the Generalized Hook’s Law and are
widely used in structural engineering applications. By algebraic manipulations
equation (2.70) through (2.75) may be re-written with the stress tensor components
expressed as functions of the strain tensor components. The resulting equations are:
TEE
zyxx
**)*21(
]***)1[()*21(*)1(
(2.77)
- 78 -
TEE
zyxy
**)*21(
]**)1(*[)*21(*)1(
(2.78)
TEE
zyxz
**)*21(
]*)1(**[)*21(*)1(
(2.79)
xyxy G * (2.80)
xzxz G * (2.81)
yzyz G * (2.82)
Equations (2.77) through (2.82) represent the second form of Hook’s law.
2.8 Fundamental Hypotheses and Equations
The methods employed in Mechanics of Materials are based on three basic
hypotheses. They are as follows:
(H1) The solid body is continuous and remains continuous when subjected to exterior
actions or changes in temperature;
(H2) Hook’s Law;
(H3) There exists a unique unstressed state of body to which the body returns
whenever all the exterior actions are removed.
The body satisfying those thee hypotheses is defined as a linear elastic body. Under
the hypotheses (H1) the real microscopic atomic structure of the solids is ignored and
the solid body is idealized as a geometrical copy in the Euclidian space whose points
are identical with the material points of the body. This way the continuity is simulated
and no cracks or holes may open in the interior of the solid during the exterior actions.
A material satisfying the hypotheses (H1) is called a continuum. The hypotheses
(H2) represent the mathematical base of the definition of material elasticity. The
direct implication of the Hook’s Law is the Principle of Linear Superposition, a
principle frequently used in the Mechanics of Materials.
- 79 -
The fundamental concepts presented in the preceding sections are grouped in three
fundamental equations extensively used in Mechanics of Materials: (1) the
equilibrium equations, (2) the geometry of deformation and (3) the material behavior.
2.9 Examples
2.9.1 Stress Distribution and Stress Resultants in a Rectangular Cross-Section
The cross-section characteristics and the stresses acting on it are depicted in Figure
2.27. The origin of the coordinate system oxyz passes though the center of the
rectangular cross-section.
Figure 2.27 Example 2.9.1
The stresses existing in the cross-section are expressed by the following functions:
zCyCCx ** 210 (2.83)
2
3 * yCxy (2.84)
Definition 2.8
Principle of Linear Superposition is stated as follows
The displacement in a material point of the solid body resulting from the consecutive
action of exterior loads is equal to the sum of the displacements in the same material
point if the loads were to act in independently.
- 80 -
2
54 * zCCxz (2.85)
Substituting the relations (2.83) through (2.85) into equations (2.40) through (2.45)
the stress resultant forces and moment are calculated as:
ACdAzCdAyCdAC
dAzCyCCdAzyxxF
AAA
AA
x
******
*]**[*),,()(
0210
210
(2.86)
12
****
*]*[*),,()(
3
3
2
3
2
3
hbCdAyC
dAyCdAzyxxV
A
AA
xyy
(2.87)
A A
AA
xzz
hbCACdAzCdAC
dAzCCdAzyxxV
12
******
*]*[*),,()(
3
54
2
54
2
54
(2.88)
0********
*]*[**]*[*
*),,(**),,(*)(
2
3
2
54
2
3
2
54
AAA
AA
A
xy
A
xz
dAyzCdAzyCdAyC
dAyCzdAzCCy
dAzyxzdAzyxyxT
(2.89)
12
**
*******
*]**[**),,(*)(
3
2
2
210
210
hbC
dAzCdAyzCdAzC
dAzCyCCzdAzyxzxM
AAA
AA
xy
(2.90)
12
**
*******
*]**[**),,(*)(
3
1
2
2
10
210
hbC
dAzyCdAyCdAyC
dAzCyCCydAzyxyxM
AAA
AA
xz
(2.91)
For numeric application the following cross-section dimensions and constant
coefficients are assigned:
- 81 -
10.0b m 20.0h m
6
0 10*120C Pa 6
1 10*80C3m
N 6
2 10*40C3m
N
6
3 10*100C 4m
N
6
4 10*270C Pa 6
5 10*30C4m
N
The area and the stress resultants obtained by replacing the above values into the
equations (2.86) through (2.91) are:
02.0* hbA 2m
6
0 10*4.2*)( ACxF N
33
3 10*67.612
**)(
hbCxVy N
63
54 10*4.512
***)(
hbCACxVz N
0)( xT mN *
23
2 10*67.612
**)(
hbCxM y mN *
33
1 10*33.512
**)(
hbCxM z mN *
2.9.2 Extensional and Shear Strain
The deformation of a thin square plate with an edge length a is illustrated in Figure
2.28. The corner B moved horizontally by the length b into a new position *B , while
the other three corners kept their original location. It is supposed that the edge **BA
remains linear through the deformation.
The extensional strain in the ox direction, accordingly to the definition (2.46) is:
x
xxyx xx
*
0lim),( (2.92)
- 82 -
To determine the lengths x and *x two points P and Q as shown in Figure 2.28
are selected. After the deformation of the plate they are located in the new positions *P and *Q .
The original area of the plate is theoretically divided in both directions in n equally
spaced divisions and the mesh shown in the Figure 2.28 is obtained. For clarity, in
Figure 2.28 the number of division was limited only to five ( 5n ). The point P is
located at node ),( ji of the mesh, while the point Q is located at node ),1( ji . The
original length x is obtained as:
n
ai
n
ai
n
axxPQx PQ *)1(* (2.93)
Figure 2.28 Elongation Strain in Square Plate
(a) Undeformed and (b) Deformed Shapes
After the deformation the segment **QP is shorter and is calculated as:
yna
b
n
a
n
ya
ba
in
ya
ba
ya
b
in
ya
ba
ya
bxxQPx
PQ
**
*
)](
*
*[-
)]1(
*
*[**
***
(2.94)
The extensional strain is obtained as:
- 83 -
ya
b
n
an
ay
na
b
n
a
x
xxyx nxx *
**limlim),(
2
*
0
(2.95)
It can be concluded that the elongation strain ),( yxx is a function only of variable
y and varies between values of zero and a
b .
The shear strain ),( yxxy which reflects the right angle change is calculated following
a similar rationale. The original right angle is defined by the segments PM and PQ
as shown in Figure 2.29.a. After the deformation, as illustrated in Figure 2.29.b, the
angle changes to angle * defined between the segments **MP and **QP .
Figure 2.29 Elongation Strain in Square Plate
(a) Undeformed and (b) Deformed Shapes
From definition 2.6 the shear strain is given by:
]2
[lim),( *
00
yxxy yx (2.96)
The following geometric relations can be written:
)arctan(2 **
***
PM
PM
yy
xx
(2.97)
in
ya
ba
ya
bx
P*
*
**
(2.98)
- 84 -
in
b
n
b
in
ya
ba
ya
bi
n
n
ay
a
ba
n
ay
a
bx
M
*
*
*
**
)(*
)(*
2
*
(2.99)
jn
ay
P** (2.100)
)1(** jn
ay
M (2.101)
The angle * is calculated using the above formulae:
)]1(*arctan[2
]
)1(*
arctan[2
*
a
x
a
b
n
an
i
n
b
(2.102)
The shear strain is obtained as:
)]1(*arctan[]2
[lim),( *
00
a
x
a
byx
yxxy
(2.103)
The shear strain ),( yxxy is dependent only on variable x and varies from )arctan(a
b
to zero value.
2.9.3 Volumetric Strain
The rectangular parallelepiped of isotropic linear elastic material is subjected to three
normal stresses x , y and z as shown in Figure 2.30.
The absence of the shear stress indicates that only volumetric changes result without
shape change. Consequently, the undeformed volume V is modified and after the
deformation the solid body has a new volume *V . The volumetric deformation is
called dilatation. The volumetric strain is defined by the following ratio:
V
VV
V
VV
*
(2.104)
Using the notation shown in Figure 2.30 the following relations are written:
zyx LLLV ** (2.105)
- 85 -
**** ** zyx LLLV (2.106)
Figure 2.30 Volumetric Strain
The change in the dimensions is expressed as:
xxx LL *)1(*
(2.107)
yyy LL *)1(*
(2.108)
zzz LL *)1(*
(2.109)
Substituting the above expressions into the expression of the volumetric strain the
following equation is obtained:
zyxzyzx
yxzyxzyx
zyx
zyxzzyyxx
VLLL
LLLLLL
1)1(*)1(*)1(
**
***)1(**)1(**)1(
(2.110)
If the strains are small the products can be neglected and the volumetric strain
becomes:
zyxV (2.111)
Substituting the strain equations (2.70) through (2.72) into equation (2.111) and for
the condition 0T the volumetric strain is written as a function of stresses:
)(*21
zyxVE
(2.112)
- 86 -
Equation (2.112) indicates that the volumetric strain is always a positive value in the
presence of three dimensional tensions, which geometrically represents a volume
increase.
- 87 -
LECTURE 3
Geometrical Characteristics of the Beam
Cross-Section
The definitions of the beam and cross-section were specified in the previous lectures.
Some geometrical characteristics of the cross-section, such as the area and moments
of inertia, have a central role in the theoretical development of Mechanics of
Materials and are the main subject of this lecture.
3.1 Definitions
The beam cross-section is a plane area bounded by a closed curve B . For
mathematical convenience the Cartesian plane coordinate system oyz , as illustrated in
Figure 3.1, is defined.
Figure 3.1 Cross-Section Area
The total area of the cross-section is calculated as:
A
dAA (3.1)
The integral contained in equation (3.1) defines the summation of the differential
areas dA over the two defining variables y and z . The area is characterized by units of
length squared [L2], with the symbol [L] representing length.
- 68 -
The first moments of the area A about the coordinate system axes oy and
oz are called the static moments. These are defined as:
A
y dAzS * (3.2)
A
z dAyS * (3.3)
The units of the static moments are [L3].
The geometric center of the cross-section is called the centroid. Equations (3.4) and
(3.5) are used to calculate the plane position ),( zy of the centroid C . The notation is
shown in Figure 3.2.
A
dAyAy ** (3.4)
A
dAzAz ** (3.5)
Figure 3.2 Centroid Location
Note: Analyzing the integrals contained in the equations (3.4) and (3.5), the
following important conclusions regarding the position of the centroid C may
be drawn:
(a) if the cross-section area possesses one axis of symmetry, the centroid C
lies on that axis;
(b) if the cross-section area possesses two axes of symmetry, the centroid C is
located at their intersection;
(c) if the cross-section area is symmetric about a point, the centroid C is
located at the location of that point.
These cases are illustrated in Figure 3.3.
- 69 -
Figure 3.3 Types of Symmetry for Plane Area
(a) One Symmetry Axis, (b) Two Symmetry Axes and (c) Point Symmetry
The second moments of the cross-section area A about the coordinate system
axes oy and oz are called the moments of inertia and are defined as (see
Figure 3.4 for notation) :
dAzIA
y *2
(3.6)
dAyIA
z *2
(3.7)
Figure 3.4 Second Moments of Inertia Notation
The summation of the moment of inertia yI and zI is called the polar
moment of inertia and represents the second moment of inertia about the axis
ox normal to the cross-section. The polar moment of inertia is defined as:
zy
A
po IIdAzydAI *)(* 222 (3.8)
The unit of the second moments of inertia is [L4].
Note: The second moments of inertia are always positive values.
Another important geometric property is the product of inertia of the area A
also called in the Romanian technical literature the centrifugal moment of
inertia. The definition of the centrifugal moment of inertia is given in
equation (3.9):
- 70 -
A
yz dAzyI ** (3.9)
The unit of the product of inertia is [L4].
Note: Contrary to the moments of inertia which are always positive values, the
product of inertia moment of inertia may have either a positive or negative
value. If the area A has an axis of symmetry the product moment of inertia
yzI .calculated for a coordinate system including that axis is zero.
3.2 Parallel-Axis Theorems for Moment of Inertia
The above described moments of inertia are usually calculated relative to a coordinate
system CC zCy anchored at the cross-section centroid C . A new translated coordinate
system Oyz , with axes oy and oz parallel to the centroidal axes CCy and CCz ,
respectively, is defined in Figure 3.5. The correspondence between the moments of
inertia relative to this new coordinate system Oyz and those calculated with respect to
the centroidal coordinate system CC zCy is studied in this section.
Figure 3.5 Parallel-Axis Theorems Notation
The position ),( zy of an arbitrary point located on the cross-section area A
expressed relative to the Oyz coordinate system is written as:
Cyyy (3.10)
Czzz (3.11)
where the distances y and z are the horizontal and vertical distances, respectively,
between the two coordinate systems considered.
The moments of inertia about axes Oy and Oz , yI and zI , are expressed in the
equations (3.6) and (3.7), respectively. Substitution of equations (3.10) and (3.11) into
- 71 -
equations (3.6) and (3.7) yields the following new expressions for the moments of
inertia:
A A
CC
A
C
A
y dAzdAzzAzdAzzdAzI ****2**)(*2222 (3.12)
A A
CC
A
C
A
z dAydAyyAydAyydAyI ****2**)(*2222 (3.13)
Note that the integrals A
C dAz * and A
C dAy * become zero when the static moments
are calculated relatively to the centroidal axes.
The moments of inertia calculated about centroid axes are expressed as:
dAzIA
CyC*
2
(3.14)
dAyIA
CzC*
2
(3.15)
Substituting equations (3.14) and (3.15) into equations (3.12) and (3.13) final
expressions of the moments of inertial calculated about the axes of the translate
system Oyz are obtained as:
Cyy IAzI *2 (3.16)
czz IAyI *2 (3.17)
Equations (3.16) and (3.17) are called the parallel-axis theorem for moments of
inertia.
Note: Examination of equations (3.16) and (3.17) shows that the minimum values
for the moments of inertia are obtained when the axes Oy and Oz are
coincident with the centroidal axes Cy and Cz , respectively.
A similar approach can be used for the case of the polar moment of inertia poI defined
by equation (3.8). Substituting equations (3.10) and (3.11) into equation (3.8), the
polar moment of inertia about point O is obtained as:
CC zyzypo IAyIAzIII ** 22 (3.18)
Grouping the terms in the equation (3.18), the final expression may be written as:
CO pp IAI *2 (3.19)
- 72 -
where 222 zy and dAzyI C
A
CpC*)(
22
Equation (3.19) represents the parallel-axis theorem for the polar moment of
inertia.
The parallel-axis theorem for the product of inertia is derived in a similar manner to
that for the moments of inertia. By substitution of equations (3.10) and (3.11) into
equation (3.9), the following expression is obtained:
A
zyc
A
C
A
CC
A
yz
CCIdAyzdAzyAzy
dAzzyydAzyI
******
*)(*)(**
(3.20)
Since the coordinate system CC yCz passes through the centroid of the cross-section
the integrals representing the static moments are zero and consequently, the equation
(3.20) reduces to:
CC zyyz IAzyI ** (3.21)
Equation (3.21) represents the parallel-axis theorem for the product of inertia.
3.3 Moment of Inertia about Inclined Axes
Consider Figure 3.6 where a new coordinate system '' zOxy is shown rotated by an
angle from the position of the original coordinate system oyz . The rotation angle
is positive when increasing in the trigonometric positive sense (counter-clockwise).
This convention corresponds to the right-hand rule previously used.
Figure 3.6 Axes Rotation Notation
- 73 -
The position ),( '' zy of a current point located in the cross-section A relative to the
coordinate system '' zoxy can be written as:
sin*cos*' zyy (3.22)
cos*sin*' zyz (3.23)
Accordingly, with the definition equations (3.6) and (3.7) the moments of inertia in
the rotated coordinate system '' zOxy follow as:
dAzydAzIAA
y*)cos*sin*(*)( 22'
' (3.24)
dAzydAyIAA
z*)sin*cos*(*)( 22'
' (3.25)
After algebraic manipulations the moments of inertia 'yI and
'zI are obtained as:
22 sin*cos*sin**2cos*' zyzyyIIII (3.26)
22 cos*cos*sin**2sin*' zyzyzIIII (3.27)
Further, the equations (3.26) and (3.27) are expressed in an alternative form by
substitution of the double angle )*2( formulae:
)*2sin(*)*2cos(*22
' yz
zyzy
yI
IIIII
(3.28)
)*2sin(*)*2cos(*22
' yz
zyzy
zI
IIIII
(3.29)
The product of inertia ''zyI is defined as:
A
A
zy
dAzyzy
dAzyI
*)cos*sin*(*)sin*cos*(
*)(*)( ''''
(3.30)
After the algebraic manipulations, the equation (3.30) becomes:
)sin(cos*cos*sin*)( 22'' yzzyzy
IIII (3.31)
Further more, the equation (3.31) is re-written using the double angle )*2( as:
- 74 -
)*2cos(*)*2sin(*2
'' yz
zy
zyI
III
(3.32)
If the derivatives of the moments of inertia shown in equations (3.28) and (3.29) are
taken relative to the double angle )*2( an interesting result is obtained:
''
'
)*2cos(*)*2sin(*2)2( zyyz
zyyII
III
(3.33)
''
'
)*2cos(*)*2sin(*2)2( zyyz
zyz IIIII
(3.34)
Note: The first derivative of the moment of inertia relative to the double
angle )*2( is the product of inertia.
The sum of equations (3.28) and (3.29) reveal the following important relationship:
zyzyIIII '' (3.35)
Note: Equation (3.35) indicates the invariance of the sum of the moments of inertia
with the rotation of the axes.
3.4 Principal Moments of Inertia
The moments of inertia 'yI and
'zI , expressed by equations (3.28) and (3.29), are
functions of the angle of the rotated coordinate system '' zOxy . The extreme values
(the maximum and minimum) of the moments of inertia 'yI and
'zI are called
principal moments of inertia. The corresponding values of the rotation angle
describe the principal axes of inertia. The principal axes of inertia passing through
the centroid of the cross-section area are called centroidal principal axes of inertia.
As known from Calculus, the condition for a real function to have an extreme point (a
maximum or minimum) is that the first derivative of the function be equal to zero at
that point. For the case of the moments of inertia 'yI and 'z
I , using equation (3.33) and
(3.34), the condition of extreme, is written as:
0)*2cos(*)*2sin(*2
)( 000''
yz
zy
zyI
III (3.36)
From equation (3.36) the value of the angle corresponding to the principal
directions is obtained as:
- 75 -
zy
yz
II
I
*2)*2tan( 0 (3.37)
From the trigonometry, it is known that equation (3.37) has two solutions, )*2( 01
and )*2( 02 , related as shown in the equation (3.38):
0102 *2*2 (3.38)
Consequently, it is concluded that the principle directions are perpendicular to each
other:
20102
(3.39)
Using the following trigonometric identities
)*2(tan1
)*2tan()*2sin(
0
2
00
(3.40)
)*2(tan1
1)*2cos(
0
20
(3.41
and substituting them into equations (3.33) and (3.34) the final expressions for the
principle moments of inertia are obtained as:
2
2
max14
)(
2yz
zyzy
p IIIII
II
(3.42)
2
2
min24
)(
2yz
zyzy
p IIIII
II
(3.43)
Note: The invariance of the sum of the moments of inertia is also preserved for the
case of the principal moments of inertia. By addition of equations (3.42) and
(3.43) the invariance is proven:
zyppIIII '' 21
(3.44)
To identify which of the two angles, 01 or 02 , corresponds to the maximum moment
of inertia max1 II p the second derivative of the function 'yI , shown in equation
(3.28), is used. The condition for the point to be a maximum is:
- 76 -
0)]*2sin(*)*2cos(*2
[)()2(
0002
2'
yz
zyyI
III (3.45)
The expression (3.45) is re-written as:
0)*2sin(*])*2tan(
1*
2[ 0
0
yz
zyI
II (3.46)
After the trigonometric manipulations and the usage of the equation (3.37) the
inequality (3.46) became:
0tan
*cos*]4
)([*2 0
0
22
2
yz
yz
zy
II
II (3.47)
The condition for the inequality (3.47) to hold is:
0tan 0
yzI
(3.48)
Note: Here, in order for inequality (3.48) to hold true, the sign of product of inertia
yzI must be opposite to that of the tangent of the angle 0 .
Practically, the angle corresponding to the direction of the maximum moment of
inertia is obtained by successively assigning to angle 0 the values 01 and 02 and
identifying which angle verifies the inequality (3.48).
3.5 Maximum Product Moment of Inertia
Consider the angle of the principal directions 0 established and the original
coordinate system oyz shown in Figure 3.6 rotated such that the y and z axes align
with the principal directions. Then, the following expressions hold:
1py II (3.49)
2pz II (3.50)
0yzI (3.51)
Substituting equations (3.49) through (3.51) into the equations (3.28), (3.29) and
(3.32) the expression for the moments of inertia as functions of the principal moments
of inertia are obtained:
- 77 -
)*2cos(*22
2121'
pppp
y
IIIII
(3.52)
)*2cos(*22
2121'
pppp
z
IIIII
(3.53)
)*2sin(*2
21''
pp
zy
III
(3.54)
From equation (3.54), it is easy to see that the maximum value for the product of
inertia is obtained when:
1)*2sin( (3.55)
Then,
4
(3.56)
The maximum value of the product of inertia is obtained for an angle of rotation
4
measured in the counter-clockwise direction from the position of the principal
axes is expressed in equation (3.57).
2
21
max''
pp
zy
III
(3.57)
Substituting the principal moments of inertia given by equations (3.43) and (3.44) and
(3.42) into equation (3.57) a new expression for the max''zy
I is obtained:
2
2
max 4
)('' yz
zy
zyI
III
(3.58)
The corresponding moments of inertia are obtained by replacing 4
in equations
(3.52) and (3.53):
22
21''
zypp
zy
IIIIII
(3.59)
- 78 -
3.6 Mohr’s Circle Representation of the Moments of
Inertia
A very interesting and useful relationship, shown in equation (3.60), is obtained by
manipulating the equations (3.28) and (3.32) in the following manner: (a) the equation
(3.28) is rearranged by moving in the left hand side the term 2
zy II and then
squaring both sided of the equation, (b) the equation (3.332) is squared and (c) adding
together the previous obtained equations
2
222
4
)(]
2[ ''' yz
zy
zy
zy
yI
III
III
(3.60)
The following notation is employed in the implementation of the equation (3.60):
'yI (3.61)
''zyI (3.62)
2
2
4
)(yz
zyI
IIR
(3.63)
2
zy
C
II (3.64)
Substitution of equations (3.61) through (3.64) into the equation (3.60) yields a new
form for equation (3.60)
222)( RC (3.65)
Geometrically, equation (3.65) represents the equation of a circle located in the
Oyz plane. The circle has center C located at )0,( C and radius R .
The coordinates of the intersection points, 1P and 2P , between the circle and the
horizontal axis O , are obtained by solving the algebraic system composed of
equation (3.65) and the equation of the axis )0( :
RCP 1 (3.66)
RCP 2 (3.67)
- 79 -
Substituting equations (3.63) and (3.64) into equations (3.66) and (3.67) the position
of the intersection points 1P and 2P are expressed as shown in equations (3.68) and
(3.69) and are identified as the principal moments of inertia.
1
2
2
14
)(
2pyz
zyzy
P IIIIII
(3.68)
2
2
2
24
)(
2pyz
zyzy
P IIIIII
(3.69)
The graphical representation of the Mohr’s circle is depicted in Figure 3.7.
Figure 3.7 Morh’s Circle Representation
Note: Practically the Mohr’s circle is constructed using the following steps:
(a) The coordinates system O is drawn as shown in Figure 2.7. The horizontal
axis O represents the moments of inertia, while the vertical axis O
represents the product of inertia (note that the positive axis is drawn
upwards). The drawing should be done roughly to scale. The representation
considers that the following conditions are met: zy II , 0yI , 0zI and
0yzI ;
(b) Using the calculated values of the moments of inertia yI and zI and the product
of inertia yzI two points noted as Y and Z are placed on the drawing. The line
YZ intersects the horizontal axis in point C which represents the center of the
Mohr’s circle;
- 80 -
(c) The distance CY is the radius of the circle. Using the radius CY and the
position of the center C the Mohr’s circle is constructed. The intersection
points, 1P and 2P , between the circle and the horizontal axis represent the
maximum and the minimum moments of inertia;
(d) The absolute value of the )*2tan( 0 can be calculated from the graph;
(e) The angle of the principal direction 1 is the angle measured in the counter-
clockwise direction between lines CY and CP1. To obtain the position of the
two principal directions corresponding to the cross-section system Oyz an
additional point Z’, the reflection of the point Z in reference to axis , has to
be constructed. The lines Z’P1 and Z’P2 represent the principal direction 1
(associated with the maximum moment of inertia) and 2 (associated with the
minimum moment of inertia), respectively. The two directions can then be
transcribed on the cross-section sketch.
3.7 Radii of Gyration
The square root of the ratio of the moment of inertia to the area is called the radius of
gyration and has the unit of [L].
The radii of gyration relative to the original coordinate system oyz are calculated as:
A
Ir
y
y (3.70)
A
Ir z
z (3.71)
For the case of the principal moments of inertia, the corresponding radii of gyration
are:
A
Ir
p
p
1
1 (3.72)
A
Ir
p
p
2
2 (3.73)
3.8 Examples
To clarify the theoretical aspects and the formulae derived in this lecture, two
examples are presented.
- 81 -
3.8.1 Rectangle Cross-Section
A rectangular cross-section is shown in Figure 3.8. The rectangle is characterized by
two symmetry axes and consequently, the centroid C is located at their intersection.
The coordinate system used is the centroidal coordinate system shown in the Figure
3.8.
Figure 3.8 Rectangular Cross-Section
The following cross-sectional characteristics are calculated using the formulae
previously developed:
hbA *
A
y dAzSC
* =0
A
z dAySC
* =0
12
*)*(**
32/
2/
22 hbdzbzdAzI
h
hA
yC
12
*)*(**
32/
2/
22 hbdyhydAyI
b
bA
zC
0CC zyI
32
h
A
Ir C
C
y
y
32
b
A
Ir C
C
z
z
- 82 -
It is shown thus, that for a rectangular cross-section the centroidal coordinate system
represents the principal axes of inertia )0( CC zyI .
If the moment of inertia about the axis coinciding with the lower edge of the rectangle
is required, using the notation shown in Figure 3.8(b), the parallel-axis theorem for
moments of inertia is employed:
3
*
12
*)*(*)
4(*)(
3322'
'
hbhbhb
hIAzI
Cyy
3.8.2 Composite Cross-Section
The L-shaped cross-section illustrated in Figure 3.9(a) is proposed for investigation.
The vertical and horizontal legs have a height of t6 and t9 , respectively, while
thickness t is uniform for the entire figure. The L-shaped cross-section can be
decomposed into two rectangular areas, 1A and 2A , representing the areasof the
individual legs of the cross-section. The distances of the centroids, 1C and 2C , of the
two rectangular areas are positioned relative to the coordinates system Oyz without
any difficulty as depicted in Figure 3.9(b).
Figure 3.9 L-Shaped Cross-Section
The individual area of each leg and total area of the L-shaped cross-section are
calculated as:
2
1 *6)(*)*6( tttA
2
2 *8)(*)*8( tttA
2
21 *14 tAAA
The position of the cross-section centroid C is obtained:
322
2211 *43)*8(*)*5()*6(*)2
(*** ttttt
AyAyAy
- 83 -
tt
ty *07.3
*14
*432
3
322
2211 *22)*8(*)2
()*6(*)*3(*** ttt
ttAzAzAz
tt
tz *57.1
*14
*222
3
A new coordinate system aligned with the system Oyz and with the origin at the
centroid C of the entire cross-section is established as CC zCy . The following
calculations are performed with reference to this centroidal coordinate system. The
moments of inertia about the centroidal coordinate axes are calculated as:
44
4444223
223
2
222
2
111
*10.40*588
23576
*196
1800*
12
8*
196
2400*
12
216])*
14
22
2(*)*8(
12
)(*)*8([
])*14
22*3(*)*6(
12
)*6(*)([]*[]*[(
tt
tttttt
ttt
ttttt
zAIzAII yyyC
44
4444223
223
2
222
2
111
*60.112*588
66208
*196
5832*
12
512*
196
7776*
12
6])*
14
43*5(*)*8(
12
)(*)*8([
])*14
43
2(*)*6(
12
)*6(*)([]*[]*[(
tt
ttttttttt
tt
ttt
yAIyAII zzzC
4
442
22
2222211111
*57.38196
7560
*196
3240*
196
4320)]*
14
22
2(*)*
14
43*5(*)*8(0[
)]*14
22*3(*)*
14
43
2(*)*6(0[]**[]**[(
t
tttt
ttt
tttt
tzyAIzyAII zyzyzy CC
The principal moments of inertia are obtained as:
424244
442
2
1
*28.129)*57.38(4
)*60.112*10.40(
2
*60.112*10.40
4
)(
2
tttt
ttI
IIIII yz
zyzy
p
- 84 -
424244
442
2
2
*42.23)*57.38(4
)*60.112*10.40(
2
*60.112*10.40
4
)(
2
tttt
ttI
IIIII yz
zyzy
p
The angle of the principal direction of inertia is calculated as:
064.1
2
)*60.112*10.40(
)*57.38(
2
)()*2tan(
44
4
0
tt
t
II
I
zy
yz
78.46*2 0
39.230
Using the test contained in the equation (3.48) to determine if the rotation
angle 39.230 is the angle of the principal direction results in the following:
0)*57.38(
)4325.0(
)*57.38(
)39.23tan()tan(44
0
ttI yz
Consequently, the angle 39.230 is the angle of the direction of the minimum
moment of inertia and 39.2302 , while the complementary angle 61.669039.2301 represents the direction of the maximum moment of
inertia.
The angles 01 and 02 are illustrated in Figure 3.10.
Figure 3.10 L-Shaped Cross-Section Principal Directions
The radii of gyrations are obtained as:
tt
t
A
Ir C
C
y
y *69.1*14
*10.402
4
- 85 -
tt
t
A
Ir C
C
z
z *84.2*14
*6.1122
4
tt
t
A
Ir
p
p *04.3*14
*28.1292
41
1
tt
t
A
Ir
p
p *29.1*14
*42.232
42
2
The construction of the Mohr’s circle is conducted as explained in Section 3.6. Using
the moments and the product of inertia calculated above the following values are
determined:
4
24244
2
2
*93.52
)*57.38(4
)*60.112*10.40(
4
)(
t
ttt
III
R yz
zy
444
*35.762
*60.112*10.40
2t
ttIIy
zy
C
4
1 *3.129 tRyI Cp
4
2 *4.23 tRyI Cp
Figure 3.11 L-Shaped Cross-Section Mohr’s Circle
- 86 -
LECTURE 4
Equilibrium of the Plane Linear Members
In this lecture the variations of the internal resultants acting on the cross-section of a
linear member, are examined.
The definition of the plane linear member and its cross-section geometrical
characteristics has been discussed in the previous lectures.
For this discussion, the local coordinate system attached to the plane linear beam is a
right-hand coordinate system having the ox axis aligned along the length of the beam
with origin at the leftmost located point.
4.1 Type of Beams, Loads and Reactions
For better understanding of the following theoretical derivations, a number of
schematic diagrams representing loaded beams commonly found in structural
engineering are depicted in Figure 4.1.
The movement of a point located on the left axis of the beam resulting from the
application of forces in the oxy plane is completely determined by three generalized
components, explicitly, by two in-plane displacements and a rotation. The two
displacements are chosen, for convenience, in the directions ox and oy of the
coordinates system attached to the beam, while the rotation is described by the
angular motion about an axis parallel to the oz axis located at the point of interest
(normal to the plane of the definition plane). The forces and moments resulting from
the constraints induced by the existence of supports, accordingly with the Newton’s
First Law, are called reactions. Replacing the supports by their corresponding
reaction forces, the free-body diagram of the beam can be drawn. For the plane
linear beam the reactions at supports and connections were illustrated in Tables 2.1
and 2.2 contained in Lecture 2. In the figures, a reaction force is distinguished from an
applied force (load) by a slash mark in the shaft of the arrow symbolically describing
the force.
In the technical language employed by structural engineers, beams are identified by
the manner in which they are supported. The beam illustrated in Figure 4.1.a is called
- 86 -
a simply supported beam. This beam has one end, point A, constrained only in the
vertical direction by a roller support, while the other end, point B, is constrained in
both vertical and horizontal directions by a pined support or clamped support. The
reactions corresponding to the particular constraints are shown in the figure.
Figure 4.1 Examples of Beams
(a) Simple Supported Beam, (b) Cantilevered Beam, (c) Propped Cantilevered
Beam and (d) Two-span Continuous Beam
The beam depicted in Figure 4.1.b is called a cantilevered beam and has only one
end, point C, constrained against movement in both, horizontal and vertical, directions
and against the rotation around an axis parallel to the oz axis. This type of support is
called a fixed support and the corresponding reactions are is composed of two forces
and a moment.
- 87 -
More complicated beam configurations are depicted in Figures 4.1.c and 4.1.d. The
beam shown in Figure 4.1.c is called a propped cantilever beam. The end point A
has a fixed support, while the point B has a roller support attached. In Figure 4.1.d a
simply supported two-span continuous beam is represented. At point A, a pinned
support is located, while at points B and D, roller supports are attached. The
corresponding reactions are shown.
From the above description it can be concluded that in the analysis of the single-body
plane beams, three types of supports are commonly used:
1. the roller support prevents the displacement in the direction normal to the
rolling plane and is replaced in the free-body diagram by a corresponding
reaction represented by a concentrated force as shown for case1 of Table 2.1.
A similar case is that of a supporting cable illustrated in case 2 of Table 2.1;
2. the pinned support prevents all translational displacement in the beam
description plane and is replaced in the free-body diagram by a reaction,
commonly represented by two orthogonal concentrated force components as
depicted in case 3 of Table 2.1;
3. the fixed support completely prevents all translational displacement and
rotation in the beam definition plane and is replaced in the free-body diagram
by a set of reactions commonly represented by two orthogonal concentrated
force components and a concentrated moment as depicted in case 4 of Table
2.1.
If the beam is composed of a multi-body system, the decomposition of the system into
multiple single-body systems is required. At the connections between the individual
single-body systems the reactions, as shown in Table 2.2, are introduced in the free-
body diagram of each system .
The external forces (loads) applied to the plane beams are classified as:
1. distributed forces;
2. concentrated forces;
3. concentrated moments (couples).
The most commonly used distributed transverse forces are uniform and linearly
distributed. Examples of distributed transversal forces are shown in Figures 4.1.a,
4.1.c and 4.1.d. Concentrated force and moment are illustrated in Figures 4.1.b and
4.1.b.
- 88 -
4.2 Equilibrium of Beams Using the Free-Body Diagrams
The beam depicted in Figure 4.1.a is used to explain the application of the free-body
diagram concept in the case of the plane linear member.
The beam is first conceptually freed of constraints and then, the corresponding
reactions, as described in the previous section, are introduced. As indicated before, the
roller support at point A of Figure 4.1.a is replaced by a vertical concentrated force,
while the pinned support at point B is replaced by horizontal and vertical forces. At
this point, the schematic diagram representing contains only the externally applied
forces and reaction forces and is identified as the free-body diagram. Using this
diagram, the next step is to write the equations of equilibrium and solve them for the
unknown reaction forces. From the six equilibrium equations, (2.3) though (2.8), only
three can be used, the other three are identically zero. If the beam definition plane
oxy also represents the plane of action for exterior loads, and thus, the equilibrium
equations are represented by equations (2.9) through (2.11).
If the number of unknown reactions is equal to the number of independent
equilibrium equations the beam is a statically determinate beam. If the number of
unknown reactions is larger than three the beam is called statically indeterminate
beam. Only the statically determinate beams are treated in this textbook. The beams
shown in Figures 4.1.a and 4.1.b are statically determinate beams, while Figures 4.1.c
and 4.1.d illustrate statically indeterminate beams.
After the reactions have been calculated, the beam is conceptually sectioned, as
illustrated in Figure 4.2, by a cutting-plane coincidental with a cross-section located at
distance Cx from the origin point A. In Lecture 2, this separation operation was called
the method of sections. To maintain intact the equilibrium of the two separated parts,
the corresponding internal resultants are introduced and two free-body diagrams, AC
and CB, are obtained as illustrated in Figure 4.2. In the absence of the externally
applied axial forces, only two cross-sectional internal resultants, the transverse shear
force CV and the bending moment CM , are necessary to maintain the equilibrium of
the two parts. The indices y and z , used in the previous lectures to indicate the oy
and oz axes along which the shear force and moment are acting have been omitted
for clarity.
Figure 4.2 Application of the Method of Sections
- 89 -
If an elementary element of the beam is separated around a particular cross-section, as
shown in Figure 4.3, the sign convention established has a physical meaning, directly
related with the deformation of the element.
Note: Although it would be more consistent to define the positive shear force V in
the positive direction of the axis oy , however, the sign convention adopted
above is universally employed and is directly connected with the physical
significance explained.
Figure 4.3 Sign Convention for Internal Resultants
(a) Shear Force, (b) Bending Moment and (c) Shear Force and Bending Moment
4.3 Differential Relations between Loads and Cross-
Section Internal Resultants
The free-body diagram concept will be extended to write the equilibrium of the
elementary volume around a point of the beam. The free-body diagram of the
elementary volume is shown in Figure 4.4.
Definition 4.1
The sign convention for the cross-section internal resultants is stated as:
1. the positive shear force,V , acts in the negative direction of the oy axis at
the face x of the cross-section;
2. the positive bending moment, M , makes the face y of the beam
concave.
- 90 -
Figure 4.4 Equilibrium of Elementary Beam Volume
The external forces acting on the elementary volume are considered to be uniformly
distributed, their variation being small enough to be neglected over the short length
x quantity. These are represented by two distributed forces, )(xpn and )(xpt , and a
distributed moment )(xm , all acting in the positive sense relative to the respective
axes.
The equilibrium equations are:
0 xF 0* xpFFF t (4.1)
0 yF 0*)( xpVVV n (4.2)
0C
zM
0)*()(
)2
**()*(
xmMM
xxpMxV n (4.3)
Definition 4.2
The sign convention for the external loads is:
1. the positive tangential distributed and concentrated loads on the beam
longitudinal axis ox act in the positive direction of ox axis;
2. the positive normal distributed and concentrated loads on the beam
longitudinal axis ox act in the positive direction of oy axis;
3. the positive distributed and concentrated bending moments act in the
positive direction of the oz axis according to the right-hand rule.
- 91 -
After algebraic manipulations and neglecting the second-order terms equations (4.1)
through (4.3) yield a new format as follows:
)(xpx
Ft
(4.4)
)(xpx
Vn
(4.5)
)()( xmxVx
M
(4.6)
where x is the position of the cross-section.
If the length of the elementary volume x tends to zero, the volume becomes
infinitesimal and the following differential equations are obtained:
)(lim 0 xpdx
dF
x
Ftx
(4.7)
)(lim 0 xpdx
dV
x
Vnx
(4.8)
)()(lim 0 xmxVdx
dM
x
Mx
(4.9)
Equations (4.7) through (4.9) are called the differential relation between the loads
and the cross-sectional internal resultants. If the differential equations (4.7), (4.8)
and (4.9) are integrated the following expressions are obtained:
1*)()( CdxxpxF t (4.10)
2*)()( CdxxpxV n (4.11)
3*)(*)()( CdxxmdxxVxM (4.12)
The integration constants 1C , 2C and 3C are identified using the presumed known
values, boundary values of the cross-section internal resultants at the beginning of the
integration interval located at 0x :
01 FC (4.13)
- 92 -
(4.14)
03 MC (4.15)
Introducing equations (4.13) through (4.15) into equations (4.10) through (4.1) the
following expressions are obtained:
dxxpFxF t *)()( 0 (4.16)
dxxpVxV n *)()( 0 (4.17)
dxxmdxxVMxM *)(*)()( 0 (4.18)
These equations are important and are extensively used in determination of the
variation of the cross-sectional internal resultants for beams subjected to various
loading conditions.
Note: From equations (4.16) through (4.18) important practical conclusions are
derived:
(a) the loading functions )(xpn , )(xpt and )(xm must be continuous functions on
finite-intervals for the integration to be possible;
(b) the cross-sections where concentrated forces and moments are acting represent
discontinuity points. Those cases are treated as limiting cases of the
expressions obtained above;
(c) the integrals in equations (4.16) and (4.17) represent the area contained under
the loading curve over the length of the integration interval. The shear force
related integral in equation (4.18) represents the area contained under the shear
force curve over the length of the integration interval;
(d) values representing the boundary conditions of the cross-section internal
resultants at the beginning of the integration interval must be known in order
to evaluate the cross-sectional internal resultants on that interval.
4.4 Shear Force and Bending Moment Diagrams
The plot of the shear force )(xV and bending moment )(xM variation along the
longitudinal axis ox of the entire beam are called cross-section resultants diagrams
- 93 -
or shear force and bending moment diagrams. They play a very important role in
design and verification of beams. The shear and bending moment diagrams can be
constructed and plotted using a combination of the two methods described in the
previous section: (a) the equilibrium method (method of sections) and (b) the
differential relations.
The application of the two methods is preceded by creation of the free-body diagram
of the beam and calculation of the reaction forces.
The first method, the equilibrium method, requires a relatively large number of
sections to be created and the variation of the resultants between them to be
considered a priori linear. This is an approximation, but if the sections are relatively
closely spaced the induced error is negligible for practical purposes.
Employing the second method, the integration of the differential equations, requires
the beam loading to be decomposed into several intervals of continuity. The variation
of the shear force and bending moment obtained are exact.
Both methods have advantages and disadvantages and for complicated systems of
beams or loading, they are used together in a complementary manner. To become
proficient in the usage of the methods requires practice. A few examples are shown in
the following section.
4.5 Examples of Shear and Moment Diagrams for
Statically Determined Beams
The following examples illustrate the practical application of methods, the method of
sections and the differential relations method, for constructing shear and moment
diagrams.
4.5.1. Simply-Supported Beam Loaded with a Concentrated Vertical Force
The notation, geometry and loading are shown in Figure 4.5.a, while the
corresponding free-body diagram is depicted in Figure 4.5.b.
The roller support located at point A is replaced by a vertical reaction, while the
pinned support attached at point C is replaced by vertical and horizontal reactions. In
the absence of any horizontal load, the horizontal reaction at point B is null and, for
clarity, is not shown. Consequently, only two unknown vertical reactions remain.
- 94 -
Figure 4.5 Simply-Supported Beam Loaded with a Concentrated Vertical Force
(a) Geometry and Load and (b) Free-Body Diagram
Because there is no horizontal load, only two equations of equilibrium can be written.
0 yF 0 CA VPV (4.19)
0C
zM 0)(** aLPLVA (4.20)
Solving equations (4.19) and (4.20) the reaction forces are obtained:
L
aLPVA
)(* (4.21)
L
aP
L
aLPPVPV AC
*)(*
(4.22)
From examination of the free-body diagram represented in Figure 4.5(b) it is seen that
the loading has two intervals of continuity, AB and BC, where the load is zero
(mathematically it may be considered as constant with a zero value). Point B
represents the point where the existence of the concentrated load P creates a
discontinuity. The free-body diagram, with the calculated reactions, is used to
calculate the shear-force and bending moment diagrams.
The shear diagram starts at point A, where the concentrated reaction AV acts. Being a
concentrated force, the point A is a discontinuity point. A similar argument is made
for point C where the other concentrated reaction exists. The concentrated force is
treated using the sections at the right and at the left. Taking sections in different
positions of x and writing the equilibrium on the vertical axis oy the shear diagram is
calculated as follows:
at point A
for 0x 0)( xV (4.23)
- 95 -
for 0x L
aLPVxV A
)(*)(
(4.24)
at point B
for ax L
aLPVxV A
)(*)(
(4.25)
for ax
L
aP
PL
aLPPVxV A
*
)(*)(
(4.26)
at point C
for Lx
L
aP
PL
aLPPVxV A
*
)(*)(
(4.27)
for Lx 0
*)(*
)(
L
aPP
L
aLP
VPVxV CA
(4.28)
The differential relations previously obtained are used to define the variation of the
shear force in-between the points describing the continuity intervals A+B
- and B
+C
- of
the loading. From equation (4.17) it is observed that the shear force, in the absence of
transverse load, has a constant value equal to that calculated at the beginning of the
interval.
Note: Concentrated forces may be treated as “continuity” interval under the limiting
condition where the interval length tends to zero.
The moment diagram is calculated in a similar manner to the shear diagram only
replacing the vertical equation of equilibrium with the moment equation of
equilibrium. The section method is applied as follows:
at point A
for 0x 0)( xM (4.29)
at point B
for ax aL
aLPaVxM A *
)(**)(
(4.30)
- 96 -
at point C
for Lx 0)(**
)(*
)(**)(
aLPLL
aLP
aLPLVxM A
(4.31)
Using the differential relation (4.9) leads to the conclusion that the variation of the
moment is linear in both intervals. The shear and moment diagrams are plotted in
Figure 4.6.
Figure 4.6 Shear and Moment Diagrams
(a) Free-body diagram, (b) Shear diagram and (c) Moment diagram
4.5.2 Simply-Supported Beam Loaded with a Concentrated Moment
The geometry and the notation employed are depicted in Figure 4.7. The free-body
diagram is depicted in Figure 4.7.b.
- 97 -
Figure 4.7 Simply-Supported Beam Loaded with a Concentrated Moment
(a) Geometry and (b) Free-Body Diagram
The roller support located at point A is replaced by a vertical reaction, while the
pinned support attached at point C is replaced by vertical and horizontal reactions. In
the absence of any horizontal load the horizontal reaction at point B is null and, for
clarity, is not shown. Consequently, only two unknown vertical reactions remained
As before, with no horizontal load, the following equations of equilibrium are written
using the free-body diagram as a guide:
0 yF 0 CA VV (4.32)
0C
zM - 0* 0 MLVA (4.33)
Solving equations (4.32) and (4.33) the reaction forces are obtained:
L
MVA
0 (4.34)
L
MVV AC
0 (4.35)
Analyzing the free-body diagram represented in Figure 4.7.b it becomes obvious that
the loading has two intervals of continuity, AB and BC, where the load is nil. The
point B represents the point where the existence of the concentrated moment
0M creates a point of discontinuity. The free-body diagram shown in Figure 4.7.b,
along with the calculated reactions, is used to construct the shear-force and bending
moment diagrams.
The shear diagram starts at point A, where the concentrated reaction AV acts. Because
the existence of the concentrated reaction force acting at point A, the point becomes a
discontinuity point. A similar argument can be made for the point C where the other
concentrated force reaction acts. The concentrated reaction force is treated using
sections to the left and the right of the point of application. Taking sections at
different positions of the ox axis and writing the vertical equilibrium equations, the
shear diagram is calculated as follows:
- 98 -
at point A
for 0x 0)( xV (4.36)
for 0x L
MVxV A
0)( (4.37)
at point B
for ax L
MVxV A
0)( (4.38)
for ax L
MVxV A
0)( (4.39)
at point C
for Lx L
MVxV A
0)( (4.40)
for Lx 0)( CA VVxV (4.41)
The differential relations previously obtained are used to define the variation of the
shear force in-between the points describing the continuity intervals A+B
- and B
+C
- of
the loading. From equation (4.17) it is evident that the shear force, in the absence of
distributed vertical load )(xpn , has a constant value equal with that calculated at the
beginning of the interval.
Note: The concentrated moment may be treated as a “continuity” interval where in
the limit the interval length tends to zero. The concentrated moment does not
create a discontinuity in the shear diagram.
The moment diagram is calculated in a similar manner as the shear diagram except
that the vertical equation of equilibrium is replaced with the moment equation of
equilibrium. The method of section is applied as follows:
at point A
for 0x 0)( xM (4.42)
at point B
for ax aL
MaVxM A **)( 0 (4.43)
- 99 -
for ax
L
aLMMa
L
M
MaVxM A
)(*
*)(
0
0
0
0
(4.44)
at point C
for Lx 0**)( 0
0
0 MLL
MMLVxM A (4.45)
From examining of the differential relation (4.9) it is seen that the variation of the
moment is linear in both intervals. The resulting shear and moment diagrams are
plotted in Figure 4.8.
Figure 4.8 Shear Force and Moment Diagrams
(a) Free-body diagram, (b) Shear diagram and (c) Moment diagram
4.5.3 Beam with Overhang
The overhang beam geometry is shown in Figure 4.9.a. A uniformly distributed load
of 8kN/m is acting between the supports A and B and a concentrated force of 16 kN
acts at the tip of the cantilever as shown.
The pinned support at point A and a roller support at point B are replaced by the
corresponding reaction forces. In the absence of any horizontal force, the horizontal
- 100 -
reaction at the pinned support A is zero. The free-body diagram is shown in Figure
4.9.b.
Figure 4.9 Beam with Overhang
(a) Geometry and (b) Free-Body Diagram
As previously explained the first step in solving the overhang beam is the
determination of the reactions. The two remaining equilibrium equations are:
0 yF 0164*8 CA VV (4.46)
0B
zM 0)2*16()2*4*8()4*( AV (4.47)
Solving the algebraic system of equations (4.44) and (4.45) the reactions are found as:
8AV kN (4.48)
4048 AB VV kN (4.49)
Examination of the loading function reveals the following two continuity intervals: (a)
from support A to support B where the uniformly distributed load acts and (b) from
support B to the tip of the cantilever C where the load is null. The discontinuity points
are easily identified as the points A, B and C. The calculation follows the general
methodology employed in the previous two examples. The free-body diagram and the
shear and moment diagrams are contained in Figure 4.10. Using the method of
sections, the values of the shear force are calculated as:
at point A
for 0x 0)( xV (4.50)
for 0x 8)( AVxV kN (4.51)
at point B
for 4x 24)4*8()( AVxV kN (4.52)
- 101 -
for 4x 1640)4*8()( AVxV kN (4.53)
at point C
for 6x 1640)4*8()( AVxV kN (4.54)
for 6x 01640)4*8()( AVxV kN (4.55)
Using the differential relations (4.8) for the continuity interval AB it is seen that the
shear diagram has a linear variation, while in interval BC the shear is constant.
The moment diagram is calculated in a similar manner by applying the method of
sections as follows:
at point A
for 0x 0)( xM (4.56)
at point B
for 4x 32)2*4*8()4*()( AVxM mkN * (4.57)
at point C
for 6x mkN
VxM A
* 0)2*40(
)4*4*8()6*()(
(4.58)
From the differential relation (4.9) and the previous observations concerning the
variation of shear, it became obvious that the variation of the moment is parabolic in
the interval AB and linear in the interval BC. The maximum local moment in the
interval AB is located where the corresponding shear force is zero. An equation
similar to (4.52) is written for a general location on the interval A to B and then, used
to calculate the position of the local maximum moment:
0)*8( max xVA (4.59)
18
max AVx m (4.60)
The maximum local moment is:
4)2
**8(* max
maxmaxmax xxxVM A mkN * (4.61)
- 102 -
Using an equation similar to (4.57) a calculation is performed to find the location
where the moment on the interval AB becomes zero:
0)2
**8(*)( x
xxVxM A (4.62)
28
*2 AV
x m (4.63)
Figure 4.10 Shear and Moment Diagrams for Beam with Overhang
(a) Free-Body Diagram, (b) Shear Force and (c) Moment Diagram
- 103 -
LECTURE 5
Axial Deformation
Axial deformation has been experimentally introduced in Lecture 2, when the tension
test of a structural steel specimen was described. The case in point of this lecture is a
systematic description of all aspects pertinent to axial deformation.
5.1 Basic Theory of Axial Deformation
The three aspects discussed during Lecture 2, equilibrium equations, strain-
displacement equations and the constitutive relation, are applied to the case of axial
deformation. A local coordinate system oxyz , having ox axis coincident with the
longitudinal axis of the member, is employed.
5.1.1 Strain-Displacement Equation
A direct consequence of part (b) of the definition 5.1 is the fact that axial deformation
depends only on the variable x , which represents the position of a particular cross-
section on the longitudinal axis of the member.
The physical phenomenon described by definition 5.1 is illustrated in Figure 5.1,
where the undeformed and deformed conditions of the linear member are presented.
Definition 5.1
A plane linear member, when subjected to exterior loads and/or change of
temperature, undergoes an axial deformation if after the deformation:
(a) the axis of the member remains straight;
(b) the cross-sections remain plane, perpendicular to the longitudinal axis
of the beam and do not rotate about the same longitudinal axis after the
deformation.
- 112 -
Adjacent cross-sections A and B originally located at distance x from each other, as
shown in Figure 5.1.a, are found after the deformation to be located at distance *x .
The change in the position of the two cross-sections are described by their
displacements )(xu and )( xxu , respectively.
Figure 5.1 Geometrical Aspects of the Axial Deformation
(a) Undeformed Member and (b) Deformed Member
The extensional strain )(xx is expressed as:
dx
du
x
xuxxu
x
xxx xxx
]
)()([lim)(lim)( 0
*
0 (5.1)
Equation (5.1) is called the strain-displacement equation. The cross-section
distribution of the elongation strain )(xx is shown in Figure 5.2. The elongation
strain is a function only of the cross-section position described by the variable x .
Figure 5.2 Extensional Strain Distribution
The rest of the generalized strain tensor components are:
)(*)( xx xy (5.2)
)(*)( xx xz (5.3)
0 yzxzxy (5.4)
- 113 -
From equations (5.2) through (5.4) is evident that a transversal reduction of the cross-
section takes place concomitant with the axial deformation.
The displacement )(xu pertinent to a particular cross-section is obtained by
integration from the equation (5.1):
dxxuxux
x *)()(0
0 (5.5)
where )0(0 xuu is the displacement at the beginning at the integration interval.
Consequently, the total elongation of the member is calculated from equation (5.5) as:
L
x dxxuLue0
*)()0()( (5.6)
where L is the total length of the bar.
5.1.2 Constitutive Equation
The constitutive equation reflects, as was describe in Lecture 2, the relation between
the stress and the strain. If the linear elastic material behavior is considered, the
relation between the normal stress and extension strain )(xx for the case of the axial
deformation is written:
)(*),,(),,( xzyxEzyx xx (5.7)
Equation (5.7) represents the application of Hook’s Law for the case of axial
deformation. The material constant E , the modulus of elasticity, has a value unique to
each specific material and is obtained from tensile tests. The cross-section of the bar
is a small surface and the variation of the modulus of elasticity is negligible on this
surface. In this case the constitutive equation (5.7) is expressed as:
)(*)()( xxEx xx (5.8)
Equation (5.8) implies that the normal stress )(xx varies only along the length of the
member, but has a constant value on the entire cross-section. The representation of the
normal stress )(xx is shown in Figure 5.3.
The rest of the stress tensor components are zero:
0)()( xx zy (5.9)
0 xyxyxy (5.10)
- 114 -
Figure 5.3 Cross-Section Normal Stress Distribution
5.1.3 Cross-Section Stress Resultants
Considering the stress distribution represented by equation (5.8) through (5.10) the
cross-section stress resultants are obtained as:
)(*)( *)( *)()( xAxdAxdAxxF x
A
x
A
x (5.11)
yx
A
x
A
xy SxdAzxdAxzxM *)( **)( *)(*)( (5.12)
zx
A
x
A
xz SxdAyxdAxyxM *)( **)(*)(*)( (5.13)
The relation between the normal stress )(xx and the cross-section resultants )(xF ,
)(xM y and )(xM z is derived using the notation shown in Figure 5.4.
Figure 5.4 Normal Stress and Stress Resultants
(a) Stress Resultants and (b) Normal Stress
- 115 -
If the axes oy and oz of the coordinate system intersect such that the x axis passes
through the cross-section centroid, the static moments yS and zS are zero and the
axial force )(xF remains the only non-zero stress resultant. Then, from equation
(5.11) the normal stress x is calculated as:
)(
)()(
xA
xFxx (5.14)
Consequently, it is to be concluded that a beam made from a linear elastic material
undergoes an axial deformation if the axial force passes through the cross-section
centroid.
5.1.4 Equilibrium Equation
The equilibrium equation pertinent to the case of axial deformation was derived in
Section 4.3 of Lecture 4. The detailed derivation is not repeated and only the final
result expressed as the differential relation (4.7) is employed. Considering the exterior
loading and stress resultants shown in Figure 5.5, acting on an infinitesimal volume of
length x separated from the body of the beam, the following differential equation is
written:
)()(
xpdx
xdFt (5.13)
where )(xpt is the distributed loading parallel to the beam longitudinal axis.
Figure 5.5 Infinitesimal Volume Equilibrium
If the axial stress resultant )(xF is orientated as shown in Figure 5.5 and tends to
lengthen the segment the internal force is called tension. The corresponding normal
stress x is called tension stress. In contrast, if the stress resultant force )(xF is
orientated to shorten or compress the segment the corresponding internal force and
normal stress are called compression and compression stress, respectively.
Integrating equation (5.15) the stress resultant force )(xF is calculated:
- 116 -
x
t dpFxF0
0 *)()( (5.16)
where )0(0 xFF is the value of the axial force at the origin of the integration
interval.
5.1.5 Thermal Effects on Axial Deformation
The equations obtained in the previous sections are derived considering only the
exterior load action and neglecting the change in temperature. In this section the
effects of the thermal change are introduced.
During Lecture 2, the thermal strain along the axis Ox was introduced as:
TT
x * (5.17)
where is the thermal expansion coefficient and T is the change in the member
temperature.
The total elongation strain is the sum of the elongation strain induced by the exterior
load action and thermal effects and is expressed as:
)(*)()(
)()( xTx
xE
xx xT
xxxx
(5.18)
By substitution of equation (5.14) into (5.18) the elongation strain is obtained as:
)(*)()(*)(
)()( xTx
xAxE
xFxx (5.19)
Then, using equation (5.19) in equation (5.6) the total elongation of the member is
written as:
LLL
x dxxTxdxxAxE
xFdxxe
000*)(*)(*
)(*)(
)(*)( (5.20)
5.2 Uniform-Axial Deformation
A special case of axial deformation frequently encountered in structural engineering is
the case of uniform-axial deformation shown in Figure 5.6. Systems composed of
many members subjected to uniform-axial deformation are also commonly used in
structural practice. A typical example is the plane truss, where each individual
member is subjected to uniform-axial deformation under the action of to tension or
compression forces.
- 117 -
The formulation related with the definition of the uniform-axial member and its
application in the investigation of the statically determinate and indeterminate
structures is presented in this section.
5.2.1 Members Subjected to Uniform-Axial Deformation
Definition 5.2 is mathematically transcribed as:
0)( AxA (5.21)
0)( ExE (5.22)
0)( FxF (5.23)
.
Figure 5.6 Member Exhibiting Uniform Axial-Deformation
Note: Equation (5.23) implies the absence of the distributed load )(xpt in equation
(5.15)
Rewriting equations (5.6), (5.8) and (5.14) obtained in the previous section for the
case in point of a member with uniform axial-deformation the following equations are
obtained:
Definition 5.2
The uniform axial-deformation element is a linear member characterized by:
(a) a constant area along the entire length of the member;
(b) is made of a homogeneous elastic material;
(c) is subjected to a constant axial force .F
- 118 -
0
0
0
)(
)()(
A
F
xA
xFxx (5.24)
0
00
0
*)(
)()(
AE
F
xE
xx x
x (5.25)
00
0
00 *
***)(
AE
LFLdxxe
L
x (5.26)
In the computer applications equation (5.26) takes the following form:
ekF * (5.27)
where L
AEk 00 * is called the axial stiffness coefficient.
The axial stiffness coefficient represents the force applied to the member ends when
the elongation is equal with the unit length. The unit is [F/L]. The reciprocal value of
the axial stiffness is called the axial flexibility coefficient. The flexibility coefficient
is calculated:
00 *
1
AE
L
kf (5.28)
Using the flexibility coefficient expression equation (5.28) is cast in a new format:
Ffe * (5.29)
If the change in temperature is also constant along the entire length of the member the
formula (5.20) are amended as follows:
LTAE
LFe **
*
* (5.30)
Using the flexibility coefficient f the elongation is calculated as:
LTFfe *** (5.31)
Rewriting the equation (5.31) the force F is obtained:
]**[* LTekF (5.32)
5.2.2 Statically Determinate Structure
An example of a statically determinate structure is shown in Figure 5.7. A rigid,
weightless beam AC is supported at end A by a column and at end C by a vertical rod
- 119 -
CD. The rod is attached at point D to a “leveling jack”, a component which permits a
limited vertical movement and has the mission to keep the beam AC leveled. The
beam AC is loaded with a vertical force P located at distance aL from the left end A.
To calculate the axial forces acting on the column and rod the system is decomposed
into three components: the beam, the column and the rod.
Figure 5.7 Statically Determinate Structure
The free-body diagram of the beam is depicted in Figure 5.8 and is used to calculate
the reaction forces 1F and 2F . The connection between the column and the beam
requires in general, two reaction forces. In the absence of any horizontal force the
horizontal component is null and is not shown
Figure 5.8 Free-Body Diagram
The reaction forces can be calculated using the following equilibrium equations:
0 yF 021 FPF (5.33)
0C
zM 0)1(***1 aLPLF (5.34)
Solving equations (5.33) and (5.34) the reaction forces are obtained:
)1(*)1(**
1 aPL
aLPF
Compression in column (5.35)
- 120 -
PaPaPPFF *)1(*12 Tension in rod (5.36)
Note: The column and the rod are loaded with forces having opposite directions than
the reaction forces calculated in equations (5.35) and (5.36). This became
obvious if one attempt to draw the free-body diagram for the rod and column.
From equations (5.35) and (5.36) it is evident that the column and the rod are loaded
with compression force 1F and tension force 2F , respectively. These two elements are
characterized, using definition 5.2, as members with uniform axial-deformation.
Using the formulae (5.24) through (5.26) the following related column values are
calculated:
col
col
xA
Fx 1)( (5.37)
colcolcol
col
xcol
xAE
F
E
xx
*
)()( 1
(5.38)
colcol
colAE
LFe
*
* 11 (5.39)
Note: The force 1F is compression and the column is shortened by the loading
action.
Related quantities for the rod CD are similarly calculated:
rod
rod
xA
Fx 2)( (5.40)
rodrodrod
rod
xrod
xAE
F
E
xx
*
)()( 2
(5.41)
rodrod
rodAE
LFe
*
* 22 (5.42)
Note: The force 2F is tension and the rod is elongated by the loading action.
For the beam to stay in perfect horizontal balance the displacement at point A and C
should be equal and manifest in the same direction:
CA uu (5.41)
- 121 -
The expressions of the displacements at points A and C are calculated using the
formula (5.5).The displacement at point A is equal with the column elongation,
because the initial displacement at ground connecting point is considered zero:
colcolcolcol
colgroundAAE
LF
AE
LFeuu
*
*
*
*0 1111 (5.42)
The displacement at point C is calculated as:
rodrod
DrodDCAE
LFueuu
*
* 22 (5.43)
where Du is the displacement allowed by the “leveling jack”.
The necessary displacement allowed by the leveling device is calculated from
equation (5.41):
0*
*
*
* 2211 rodrodcolcol
rodcolDAE
LF
AE
LFeeu (5.44)
Note: The displacement Du has to be positive for the device to work properly (the
device works only in tension). This means that the deformation of the column
has to be larger than the rod deformation.
The numerical application for this example is presented in section 5.6.1.
5.2.3 Statically Indeterminate Structure
A typical example of a statically indeterminate structure is shown in Figure 5.9. This
type of structures requires a more involved methodology for in order to calculate the
stress and strain distributions.
Figure 5.9 Statically Indeterminate Structure
The system shown in Figure 5.9 is composed of a rigid member AD, pinned into the
wall at point A, and two unequal linear elastic rods, BE and CF. The rods, BE and CF,
- 122 -
are attached to the ceiling at points E and F, respectively. The system is loaded with a
concentrated vertical force P at point D. The free-body diagram used to write the
equilibrium equations is shown in Figure 5.10.
Figure 5.10 Free-Body Diagram
The following equilibrium equations are written:
0 xF 0AH (5.47)
0 yF 021 PFFVA (5.48)
0A
zM 0*** 21 cPbFaF (5.49)
The equilibrium equations (5.48) and (5.49) contain three unknown reaction forces 1F ,
2F and AV . In order to solve these unknown quantities one additional equation is
necessary. This equation is obtained from the deformation compatibility condition
schematically described in Figure 5.11. Because the beam AD is rigid, purely
geometric relations between the rod elongations, 2e and 2e , and the rotation angle
are written as:
*1 ae (5.50)
*2 be (5.51)
Figure 5.11 Deformation Notation
Using equation (5.29) and the relations (5.50) and (5.51) the forces in the rods are
expressed as:
****1
1
1
1 akaf
F (5.52)
- 123 -
****1
2
2
2 bkbf
F (5.53)
The stiffness coefficients, 1k and 2k , are calculated from the geometrical and material
properties characteristics of the rods:
1
111
*
L
AEk (5.54)
2
222
*
L
AEk (5.55)
Substituting equations (5.52) and (5.53) into the equilibrium equations (5.48) and
(5.49), 1F and 2F are eliminated from the system leaving only two unknowns, AV
and :
0**** 21 PbkakVA (5.56)
0***** 2
2
2
1 cPbkak (5.57)
Solving the algebraic system, the two unknowns are found as:
PPbkak
c**
**12
2
2
1
(5.58)
where 1 is the rotation angle corresponding to 1P
Pbkak
cbbkcaak
bkakPVA
***
)(**)(**
*)**(
2
2
2
1
21
21
(5.59)
Introducing the rotation angle into equations (5.52) and (5.53) the rod forces are
calculated:
PFPbkak
cakF **
**
**1_12
2
2
1
11
(5.60)
PFPbkak
cbkF **
**
**1_22
2
2
1
22
(5.61)
where 1_1F and 1_2F are the forces in the rods in 1P
The rod stresses are also calculated:
- 124 -
1
11
A
Fx (5.62)
2
22
A
Fx (5.63)
The rod elongations are obtained using equations (5.50) and (5.51) as:
Pbkak
cae *
**
*2
2
2
1
1
(5.64)
Pbkak
cbe *
**
*2
2
2
1
2
(5.65)
If the allowable vertical force P is required, the stress in the rods must be compared
against the allowable stress all :
allxA
F
1
11 (5.66)
allxA
F
2
22 (5.67)
At limit, the relations (5.66) and (5.67) are rewritten as:
allAPFF ** 11_11 (5.68)
allAPFF ** 21_22 (5.69)
Introducing the rod forces, equations (5.60) and (5.61), into equations (5.68) and
(5.69) the allowable vertical force admitted by the system is calculated as:
]**
)**(**,
,**
)**(**min[
]*
,*
min[
2
2
2
2
12
1
2
2
2
11
1_2
2
1_1
1
cbk
bkakA
cak
bkakA
F
A
F
AP
all
all
allallall
(5.70)
The numerical application for this example is presented in section 5.6.2.
- 125 -
5.3 Nonuniform-Axial Deformation
The definition of a member with uniform-axial deformation is specified in Section
5.3. If any one of the assumptions contained in definition 5.2 is violated the axial
deformation is called nonuniform-axial deformation. The most common cases of
nonuniform-axial deformation are treated in the following subsections.
5.3.1 Non-homogeneous Cross-Section Members
The theory developed in the previous sections assumed that the cross-section is made
from a homogeneous material described by its modulus of elasticity. In structural
engineering practice it is not uncommon to have a case in which a member
constructed from two different materials bounded together at their interface is forced
to undergo an axial deformation. The members made from different materials but
behaving together as a single member are called composite sections.
For the purpose of analysis it is assumed that the member is made from two materials,
each being characterized by a specific modulus of elasticity ( 1E and 2E ) and area ( 1A
and 2A ). The strain distribution is, as before, assumed to be a function of only the
variable x , the position of the particular cross-section of interest:
)(),,( xzyx xx (5.71)
Considering that both materials are homogeneous linear elastic materials the Hook’s
law may be written for each material as:
)(*)( 1
1 xEx xx (5.72)
)(*)( 2
2 xEx xx (5.73)
The total axial force )(xF , as expressed in equations (5.11), is divided into two axial
forces, each acting at the centroid of the corresponding bar cross-section.
)()(
*)(*)(*)()(
21
2
2
1
1
21
xFxF
dAxdAxdAxxFA
x
A
x
A
x
(5.74)
where
)(*)(*)()( 1
1
1
1
1
1
xAxdAxxF x
A
x (5.75)
)(*)(*)()( 2
2
2
2
2
2
xAxdAxxF x
A
x (5.76)
- 126 -
Substituting the axial stresses expressed in equations (5.72) and (5.73) into equations
(5.75) and (5.76) the equilibrium equation (5.74) is written as:
)(*)](*)(*[
)(*)(*)(*)(*
)()()(
2211
2211
21
xxAExAE
xAxExAxE
xFxFxF
x
xx
(5.77)
Consequently, the elongation strain is obtained:
)(*)(*
)()(
2211 xAExAE
xFxx
(5.78)
Accordingly, the normal stresses are calculated employing the equations (5.72) and
(5.73) as:
)(*)(*)(*
)(2211
11 xFxAExAE
Exx
(5.79)
)(*)(*)(*
)(2222
22 xFxAExAE
Exx
(5.80)
In general, the stress resultant moments, as expressed in equations (5.12) and (5.13),
do not vanish and additional restrictions regarding the geometrical characteristics of
the cross-section must be imposed. Nullification of the stress resultant moments is
obtained by using symmetric cross-section about one or both centroidal axes.
The frequently encountered practical case when the areas )(1 xA and )(2 xA are
constants along the entire length and the member is subjected to a constant force P is
considered below. It is additionally assumed that the cross-section is symmetrical only
about the vertical axis oy . Then, the normal stresses developed in each material zone
are calculated, following the equations (5.79) and (5.80), as:
constPAEAE
Exx
*
**)(
2211
11 (5.81)
constPAEAE
Exx
*
**)(
2222
22 (5.82)
The assumption regarding the symmetrical aspect of the cross-section about the
vertical axis nullify the stress resultant moment yM when the bar local coordinate
system is centroidal. The second stress resultant moment zM can be made zero by
manipulating the position of the application of the force P . It was previously argued
that the individual cross-section resultants )(1 xF and )(2 xF must be located at the
- 127 -
centroids of the cross-sections 1A and 2A . This raises the question about the location of
the applied force P in order that only axial forces are induced in each of the
individual bars composing the cross-section. This situation is solved by replacing the
axial forces )(1 xF and )(2 xF with a resultant force and a null moment written about a
horizontal axis. The stress resultant moment about the vertical axis has a zero value
due to the imposed symmetry of the cross-section about that axis. The moment
equation, written about an horizontal axis passing through the application point of the
force P , is:
0*)(* 21 PP dFddF (5.83)
where d is the distance between the centroids of cross-sections 1A and 2A , while Pd
is the distance between the application point of the force P and the centroid of the
area 2A .
The distance Pd is then calculated:
dFF
FdP *
21
1
(5.84)
Note: In general the application point of the force P does not correspond with the
centroid of the composite cross-section.
5.3.2 Non-homogeneous Cross-Section Members Subjected to Thermal Changes
Consider the composite section described in Section 5.3.1 subjected to change in
temperature. For generality, in this discussion, let be assumed that each individual bar
undergoes a different change in temperature 1T and 2T , respectively. The notation
employed in the Section 5.3.1 is maintained. In the absence of the exterior forces, the
equation of equilibrium is:
0)()( 21 xFxF (5.85)
The axial forces pertinent to each one of the bars is constant along their respective
lengths and thus, the equation (5.85) becomes:
021 FF (5.86)
The equilibrium equation contains two unknown cross-sectional forces, )(1 xF and
)(2 xF , and consequently the system is statically indeterminate. An additional
equation is necessary to obtain these forces. This equation is derived from the
condition of equality of the elongations for the two individual bars imposed by the
existence of the rigid members attached at their ends. This equation is written as:
- 128 -
21 ee (5.87)
The elongations may be express using the formulation of equation (5.31):
LTFfe *** 11111 (5.88)
LTFfe *** 22222 (5.89)
Substituting equations (5.88) and (5.89) into equation (5.87) the second necessary
equation is obtained:
0****** 22221111 LTFfLTFf (5.90)
Solving the algebraic equation system (5.86) and (5.90) the cross-section forces are
calculated:
21
112221
*)**(
ff
LTTFF
(5.91)
The axial stress in the bars is:
1
11
A
F (5.92)
2
22
A
F (5.93)
5.3.3 Heated Member with a Linear Temperature Variation
The slender beam shown in Figure 5.12.a is heated by a heating coil capable of
producing a linearly varying temperature as shown in Figure 5.12.b. The beam has a
constant cross-section area A and is made from a linear elastic material. The beam is
attached to rigid supports at ends A and B.
Figure 5.12 Heated Uniform Member
(a) Geometry and (b) Temperature Variation
- 129 -
In the absence of any applied forces between A and B, the equilibrium equation is
written as:
0 BA FF (5.94)
The system is statically indeterminate containing two unknown reaction forces
AF and BF . An additional equation is necessary. This equation is obtained by
observing that the total elongation of the beam is null:
0e (5.95)
Following the equations (5.20) and (5.31) the total elongation is calculated:
L
dxxTFfe0
*)(** (5.96)
where AE
Lf
*
Accordingly, with the temperature variation shown in Figure 5.12.b the thermal
variation in a particular cross-section is:
L
xTTTxT ABA *)()( (5.97)
The integral contained in the equation (5.96) is calculated using the expression (5.97):
2*)(
2*)(*
*]*)([*)(00
LTT
LTTLT
dxL
xTTTdxxT
AB
ABA
L
ABA
L
(5.98)
Substituting (5.98) into (5.96) the total elongation is expressed as:
2*)(**
LTTFfe AB (5.99)
Imposing the condition (5.95) the force F is found:
2
)(***2*)(*
ABAB TTEA
f
LTT
F
(5.100)
Using the equation (5.94) results that:
- 130 -
FFF BA (5.101)
The normal stress is then calculated as:
2
)(**
)(
)()( AB
x
TTE
A
F
xA
xFx
(5.102)
5.4 Special Aspects
5.4.1 Normal Stress in the Vicinity of the Load Application
A vertical prismatic bar characterized by a constant cross-section along its entire
length L is loaded at one end by a concentrated force Q and supported at the other end
as illustrated in Figure 5.13. Based on Newton’s Law the constraint at the support
generates a uniform distributed reaction q opposed to the action. The free-body
diagram is shown in Figure 5.13.a. The example considered falls in the category of
members with uniform axial-deformation studied in Section 5.2. The exact
determination of the distribution of normal stress along the length of the beam
requires advanced methodologies employed in the Theory of Elasticity and for this
reason only the results are presented. Analyzing the results illustrated in Figure 5.13 it
is found that the formulae obtained in Section 5.2 are valid in the majority of the
cross-sections except of those located in the vicinity of the ends. The perturbation
zone has a length b roughly equal to the width of the cross-section.
Figure 5.13 Normal Stress Distribution
Note: It can be concluded that for practical purposes the formulae obtained in
Section 5.2 based on the assumptions contained in definition 5.1 are valid
especially if the ratio 25.0L
b. Special attention has to be given to the areas
located near the point of load application or near abrupt changes in the cross-
section. The application of Saint Venant’s Principle is valid for the case of the
beam.
- 131 -
5.4.2 Stress Concentrations
In the theoretical development pertinent to the axial deformation of linear members,
the area )(xA of the cross-section is considered as a smoothly varying function of the
position x . If discontinuities appear in the definition of the cross-sectional area, the
formulae obtained in the preceding sections are invalid and the concept of stress
concentration must be introduced.
Figure 5.14 Concentration of Stress
A typical case is shown in Figure 5.14 where a prismatic type linear member having a
circular hole of diameter d is subjected to a constant tension forceQ . The member
cross-section is described by the height b and thickness t .
The normal stress max around the hole can be significantly greater than tdb
Q
*)(
and varies as a function of the ratiob
d. For practical applications the coefficient K ,
called the stress-concentration factor is introduced. This factor is defined as the
ratio of the maximum normal stress max around the hole to the normal stress
calculated in the absence of the hole nom , called nominal stress.
nom
K
max (5.103)
tc
Q
tdb
Qnom
**)(
. (5.104)
The average normal stress calculated with the formula (5.14) is:
tb
Pavg
* . (5.105)
The variation of the stress-concentration factor K , calculated using the Theory of
Elasticity methods is illustrated in Figure 5.15. The stress concentration factor for the
configuration under consideration varies from 2.3 to 3.0.
- 132 -
Figure 5.15 Variation of the Stress-Concentration Factor
Analyzing the variation, it is concluded that if the diameter d decreases the
concentration factor K also increases. This is somewhat misleading and is due to the
way the chart in Figure 5.15 is constructed. Using the average normal stress avg in the
definition of the stress concentration factor K , equation (5.103) the following formula
is obtained:
)1(** maxmax
b
d
b
cK
avgavg
(5.106)
Considering the extreme cases shown in Figure 5.15 and applying these within
equation (5.106) it can be concluded that the maximum normal stress max increases
with the increase in the diameter of the hole and varies as:
avgavg *6.4*0.3 max (5.107)
For small diameter holes the stress concentration disappears at relatively small
distance from the hole. This is an example of the application of Saint Venant’s
Principle.
5.4.3 Limits of Poisson’s Ratio
The volumetric strain V was calculated in Section 2.9.3 for the general case when the
normal stresses x , y and z are applied simultaneously. For the case of axial
deformation only the normal stress x has a non-zero value and, consequently, the
equation (2.111) is written as:
xxVEV
V
*)*21()(
*21
(5.108)
- 133 -
Because the volume can not decrease during the tensioning of the axially deformed
member the volumetric strain is a positive value. Mathematically this condition is
enforced as:
0
V
VV (5.109)
Consequently, from physical reality and equation (5.108) it is seen that:
5.00 (5.110)
The limits established for Poisson’s ratio by the expression (5.110) are generally
valid for all materials used in structural engineering. The cases representing the
limiting values, 0 and 5.0 , are pertinent to cork and water, respectively.
Structural steel has a Poisson’s ratio of 0.33. There are some cases when the material
has a negative Poisson’s ratio. These materials are called swollen solids and this
unusual behavior is characteristic of certain materials subjected to radiation.
The transversal contraction of the cross-section is similarly obtained as:
xxAEA
A
**2)(
*2
(5.111)
5.5 Design of Members Subjected to Axial Deformation
In the design of the members subjected to axial deformation two important factors, the
load L and the resistance R , are considered. The load L represents the maximum
axial force that occurs in the specific member when subjected to the action of exterior
forces or change in temperature. The most common definition for the resistance R is
the force which is developed in the member when the normal stress reaches the
yielding value Y .
The design of the member subjected to axial deformation is conducted under the
condition that the capacity of the member, represented by its resistance force R , must
always be greater or equal to the demand force L . Mathematically, this assumption is
expressed as:
RL (5.112)
In the calculation of the load L and resistance R forces there are typically a number
of factors (load magnitudes and directions, material characterization, manufacturing
tolerance, etc.) which are not known with absolute certainty. The degree of
uncertainty may be treated using the concepts proper of Probability Theory. In order
to circumvent the difficulties inherent in the rigorous application of probabilistic
methods, a global factor SF , called the safety factor, encompassing all possible
uncertainties is introduced as:
- 134 -
allR
RSF (5.113)
The allowable resistance allR is then used in place of the resistance R in equation
(5.112).
allRL (5.114)
The safety factor is always greater than or equal to unity:
1SF (5.115)
This is the approach used by the method called ultimate strength design method and
was adopted by American Concrete Institute (ACI) and American Institute of Steel
Construction (AISC).
If the relation between the stress and strain is linear, than a similar safety factor may
be defined by limiting the value of the normal stress in the axially deformed member.
all
YSF
(5.116)
The design formula (5.114) is modified using the relationship between maximum
normal stress max and the allowable normal stress all :
all max (5.117)
The formula (5.117) was used for a long period of time in a procedure known as the
allowable-stress design. Due to the simplicity of application, this method is still
commonly used in United States for the design of steel structures.
5.6 Examples
The application of the theoretical formulae developed in this lecture is illustrated in
the following examples.
5.6.1 Statically Determinate Structure
The structure illustrated in Figure 5.7 was investigated in Section 5.5.2. The following
numerical values are considered in the numerical application of the generic
formulation:
10P kN
- 135 -
0.5L m 0.31 L m 5.12 L m 4.0a
0.12colA 2cm 0.5rodA 2cm
1110*2colE Pa 1110*2rodE Pa
The reaction forces are:
61 F kN Compression in column
42 F kN Tension in rod
The column related values are:
610*5col
x Pa
510*5.2 col
x
510*5.7 cole m
The rod related values are:
610*8rod
x Pa
510*4 rod
x
510*6 rode m
The displacement Du of the leveling device is calculated as:
555 10*5.110*610*5.7 rodcolD eeu m
5.6.2 Statically Indeterminate Structure
The statically indeterminate structure illustrated in Figure 5.9 is solved in Section
5.2.3. For the numerical application of the formulation developed the following data
are employed:
5.1a m 5.2b m 0.4c m
0.21 L m 3.12 L m
- 136 -
11
21 10*2 EE Pa
0.61 A 2cm 0.32 A 2cm
810*1.2all Pa
The stiffness coefficients are:
7
1 10*0.6km
N
7
2 10*615.4km
N
The rotation angle 1 is:
9
1 10*446.9 radians
The rod forces 1_1F and 1_2F are calculated as:
85.01_1 F N
09.11_2 F N
The allowable vertical force allP for the system is:
410*780.5allP N
- 137 -
LECTURE 6
Pure Shear
In the previous lecture normal stress and elongation strain concepts were presented
and discussed. The concepts of shear stress and strain are the subject of this lecture.
6.1 Basic Theory
In accordance with the shear stress definitions (2.13) and (2.14), if on a particular
cross-section there exists only a stress resultant V , orientated parallel to any axis
describing the cross-section plane, the shear stress is defined as:
A
VA
0lim (6.1)
where V is the tangential stress vector acting on the infinitesimal area A .
The resultant shear forceV is obtained by integrating the shear stress vector over the
entire area of the cross-section A :
A
dAV * (6.2)
The shear force V and the shear stress are illustrated in Figure 6.1.
Figure 6.1 Pure Shear
(a) Shear Force and (b) shear stress
- 123 -
The average shear stress avg is calculated by dividing the shear force by the total area
of the cross-section:
A
Vavg (6.3)
The distribution of shear stress is not uniform on the cross-section and in general, can
not be determined from the integral equation (6.2). In some cases the average shear
stress avg may be used in place of the true shear stress without introducing
significant an error. This is commonly done for the analysis of parts used for
connection of axially deformed members to other members or supports. These parts,
called connectors, are typically bolts, nails, rivets pins, welds and glue. The bolts,
rivets, pins and nails are discrete connectors, while the welds and glue are
continuous connectors.
When the exterior force acts parallel to the particular surface of the part the
phenomenon is called direct shearing. A typical example of direct shearing is the
punching shear of a metal sheet illustrated in Figure 6.2.a. A metal sheet is placed
between two rigid blocks. Through holes existing in the rigid blocks a punching rod is
pushed with a force P . The resulting small circular metal disk “sheared” from the
sheet is called a slug. The free-body diagram is shown in Figure 6.2.b.
Figure 6.3 Sheet-Metal Punch
(a) Punching Machinery and (b) Free-Body Diagram
Using the notation shown in the free-body diagram, where the exterior force and the
shear stress acting around the metal slug are depicted, the average shear stress is
calculated as:
td
Pavg
** (6.4)
- 124 -
Another example of direct shearing is the hinge pin in an ordinary pair of pliers
shown in Figure 6.3. When force P is exerted on the arms of the pliers, the pin cross-
section is subjected to shear stress as depicted in Figure 6.3.b.
Figure 6.3 Pair of Pliers
The direct shear is characterized as single shear or multiple shear as a function of
the number of planes (cross-sections) on which the shear stress acts. Examples of
single shear and double shear are illustrated in Figures 6.5 and 6.7, respectively,
where two steel parts in tension are connected by a bolt.
The existence of shear stress implies the existence of shear strain, which represents
the change of the right angle. The original right angle shown in Figure 6.4.a is
modified into an acute angle * as depicted in Figure 6.4.b. The angular
change representing the shear strain is calculated as:
*
2
(6.5)
Figure 6.4 Shear Strain
(a) Undeformed angle and (b) Deformed angle
- 125 -
The shear strain , measured in radians, is a small angle and for the sake of simplicity,
is approximated by its tangent:
s
s
L
)
2tan( * (6.6)
If the material is homogeneous and linear elastic, the constitutive equation is
described by Hook’s law:
*G (6.7)
where G is the shear modulus characterizing the material.
6.2 Discrete Connectors Calculation
In Figure 6.5.a, under the action of force P the two metal parts press against the bolt
in bearing and consequently, contact stresses called bearing stresses, will develop.
The tendency of the two connected pieces to pull apart induce induce a direct shearing
in the bolt. To clarify the phenomenon which takes place the connecting bolt is
isolated and shown in Figure 6.5.c.
Figure 6.5 One-Bolt Single Shear Connection
The bearing stress b exerted on the bolt is shown on the bolt free-body diagram and
is calculated as:
bear
bear
bearA
F (6.8)
where bearF and bearA are the bearing force and the bearing area, respectively.
- 126 -
In the case of single shear the bearing stress bearF is equal the axial applied force P .
The actual distribution of the bearing stress around the contact area is a complex
distribution, which, for practical purposes, is simplified by considering that the stress
is uniformly distributed over the area obtained by multiplying the bolt diameter by the
thickness of the part bearing on the bolt. For the case shown in Figure 6.5 the
maximum bearing stress is obtained as:
brb Ftd
P
min
max_*
(6.9)
where d and mint are the bolt diameter and the minimum thickness of the two parts
connected by the bolt.
The limiting bearing stress brF is a value obtained from laboratory tests and is
uniquely determined for each type of connector.
The shearing force is transferred through the bolt section mn and by approximating
the shear stress as uniformly distributed over the bolt cross-section, the average shear
stress avgb _ is obtained as:
vavgb Fd
P
2_*
*4
(6.10)
where vF is the limiting value for the connector shear stress obtained also from
laboratory tests. The failure of a bolt in a single shear connection is illustrated in
Figure 6.6.
Figure 6.6 Failure of Bolt in Single Shear
Equations (6.9) and (6.10) may be combined into one giving the allowable force in the
connector as:
PFd
FtdP vbrallb )*4
*,**min(
2
min_
(6.11)
- 127 -
Equation (6.11) represents the design formula for one-connector single shear
connections.
The double shear type connection is shown in Figure 6.7.a. The bearing action is
transferred in this case over three surfaces. Considering the same simplified
distribution explained above the bearing stress is calculated by application of equation
(6.9) to each of the three contact surfaces. The difference appears when analyzing the
transfer of the shearing force through the body of the connector. For double shear the
bearing force is equal to P for the middle layer contact and 2/P for each of the outer
two layer contact surfaces. Also, as illustrated in Figure 6.7.b, a resultant shear force
of 2/P occurs on each of two cross-sections of the bolt mn and pq .
Considering that the shear stress is uniformly distributed over the shearing areas the
average shear stress is calculated as:
vavgb Fd
P
2_*
*2
(6.12)
Figure 6.7 One-Bolt Double Shear Connection
The design formula is written as:
PFd
FtdP vbrallb )*2
*,**min(
2
min_
(6.13)
where ),min( 231min tttt .
Multiple shearing surfaces may occur if there are more than two parts held together by
one connector. Considering that there are 1n contact parts pulling in one direction and
- 128 -
2n contact parts pulling in the opposite direction, the maximum bearing stress
max_b is calculated as:
bb Ftd
n
P
td
n
P
]*
,*
max[min_2
2
min_1
1max_ (6.14)
where min_1t and min_2t are the minimum thicknesses of the parts acting in opposite
directions.
The average shear stress in the bolt or pin of the case considered above is:
v
t
avgb Fdn
P
2
sec
_**
*4
(6.15)
where tnsec is the number of shearing sections.
In structural engineering practice, many connections require more than one connector
to transfer the force between the attached parts. Examples of multiple connector
connections are illustrated in Figures 6.8 and 6.9. These figures illustrate connections
with only one line of connectors.
Figure 6.8 Two-Connector Single-Shear Connection
- 129 -
Figure 6.9 Three-Connector Single-Shear Connection
It is commonly assumed when the geometry of the connection is symmetrical and the
load passes through the connection centroid, that the load is equally sheared by all
connectors.
The most frequently encountered type of axially loaded connection is the single lap
connection. An example of two plate single lap with multiple fasteners connection is
shown in Figure 6.10. To calculate the axial load pertinent to the cross-sections 1-1
through 4 -4 the free-body diagrams are constructed and shown in Figures 6.10.b
through 6.10.e. The verification of the reduced cross-sections of the plate is required
and consequently, the evaluation of the reduced area in all perforated cross-sections is
necessary. The effective normal stress is then calculated in the usual manner by
dividing the corresponding axial force acting on the cross-section by the reduced area
of the cross-section.
Figure 6.10 Axial Load Distributions in the Single Lap Connection
- 130 -
Suppose that plate A has a thickness t and width b and the connector diameter is d .
Accordingly, with the assumption made above, each one of the ten connectors carries
an equal load of 10
P. The reduced areas are calculated as:
dttbA **3*11 (6.16)
dttbA **2*22 (6.17)
dttbA **3*33 (6.18)
dttbA **2*44 (6.19)
Consequently, the maximum normal stress is obtained as:
]*7.0
,max[2211
A
P
A
Pplate (6.20)
Additional details pertinent to the design and verification of connections using
discrete connectors, such as the verification of the zigzagging cross-section going
through connectors 1-2-3-2-1, are discussed in specialty courses.
6.3 Weld Calculation
Welds are categorized in four basic groups: groove, fillet, slot and plug welds. These
categories are illustrated in Figure 6.11.
Figure 6.11 Basic Weld Types
- 131 -
In the widely used Manual of Steel Construction Allowable Stress Design published
by the American Institute of Steel Construction (AISC) groove welds are further
classified in many subcategories. In engineering practice, the fillet weld is the most
commonly used type of weld, because of it is the easiest to fabricate.
Figure 6.12 Effective Throat Dimension for Fillet Welds
(a) Equal Legs and (b) Unequal Legs
The allowable stress on various types of welds is dependent upon the effective area
of the weld. The effective area of the groove or fillet weld is calculated as the product
of the effective throat dimension et and the length of the weld.
For fillet welds the effective throat dimension is defined as the shortest dimension
measured from the weld root to the face of the weld. Assuming that the weld has
equal legs of length a the effective throat is equal to 2
a. The calculation of the
effective throat accordingly to AISC is shown in Figure 6.12. For full penetration
groove welds, the AISC defines the effective throat as a function of the thickness of
the parts joined in the connection as pictured in Figures 6.13.a and 6.13.b. In the case
of partial penetration welds, the effective throat is a function of the depth of the
preparation as shown in Figures 6.13.c and 6.13.d.
Figure 6.13 Groove Weld Effective Throat
- 132 -
The design formula for the fillet weld is:
PFtl weldalleffi
n
weldi )**( ___ (6.21)
where: n is the number of welds parallel to the axial force P ;
weldil _ , effit _ are the effective length and throat of the weld i ;
weldallF _ is the allowable shear stress in the weld.
The weld allowable shear stress weldallF _ is calculated as:
electrodeyweldall FF __ *3.0 (6.22)
where electrodeyF _ is the ultimate tensile strength of the electrode material.
For groove weld the design formula is:
PFtl weldalleffweld _** (6.23)
The allowable stress weldallF _ in the weld is:
yweldall FF *6.0_ (6.24)
where yF is the base metal yielding stress.
- 133 -
LECTURE 7
Bending
7.1 Definitions
A linear beam subjected to transversal loading (forces and moments) deflects
laterally, a phenomenon known in structural practice as bending. In order to simplify
the study of plane linear beam bending, it has to be assumed that the beam has a
geometric longitudinal plane of symmetry as illustrated in Figure 7.1. The local
coordinate system oxyz , similar with those used in the previous lectures, has the ox
axis aligned along the length of the beam. The supports and the loading are also
considered to be symmetric in relation to the longitudinal symmetry plane of the
beam. As a direct consequence of the existence of the vertical plane of symmetry
oxy , the beam deformation manifests in this plane, and thus, it is also called the
plane of bending.
Figure 7.1 Beam with Vertical Plane of Symmetry
- 153 -
After application of the load, the beam deforms as schematically depicted in Figure
7.2. One imagines that the beam consists of a number of parallel longitudinal fibers as
pictured in Figure 7.2.a. Then, after the deformation it is evident from the figure that
some of the fibers are stretched and some are shortened. The horizontal plane
containing the fibers which remained unchanged is called the neutral surface. The
curve resulting from the intersection between the neutral surface and the longitudinal
plane of symmetry is called the deflection curve. The above discussed terminology is
depicted in Figure 7.2.b.
Figure 7.2 Beam Deflected Shape from Bending and Descriptive Terminology
7.2 Pure Bending
Two examples of beams subjected to pure bending are shown in Figures 7.3 and 7.4.
The corresponding bending moment diagrams are also included.
The simple supported beam pictured in Figure 7.3 is loaded with two equal
concentrated moments acting at the ends A and B. Consequently, the shear force is
Definition 7.1
The segment of a plane linear beam is in pure bending if along its entire length:
(a) the cross-section is constant and symmetrical about a longitudinal plane ;
(b) the material properties are constant;
(c) it is subjected only to a constant bending moment;
(d) after the load application cross-sections remain plane and perpendicular to
the deflection curve.
- 154 -
zero for the entire length of the beam, while the bending moment is a constant
function along the length of the member.
Figure 7.3 Simply-Supported Beam in Pure Bending
The cantilever shown in Figure 7.4 is also in pure bending, because the shear force is
zero and the bending moment, as indicated by the moment diagram, is constant for the
entire length of the beam.
Figure 7.4 Cantilever Beam in Pure Bending
An interesting situation is depicted in Figure 7.5. The beam is simply-supported and
loaded symmetrically by two concentrated forces P located at equal distance a from
the supports A and B .
Figure 7.5 Simply-Supported Beam with Central Region in Pure Bending and the End
Regions in Non-uniform Bending
- 155 -
The shear diagram is constant in the segments AC and DB, and zero only on the
central segment CD. The bending moment, as shown in the moment diagram, has a
linear variation within the end segments AC and DB, and is constant within the
central segment CD. In accordance with definition 7.1, only the central segment CD is
in pure bending. In contrast to the situation of pure bending when only the constant
moment exists, the presence of the shear force results in a variation of the bending
moment and is referred to as non-uniform bending. The beam pictured in Figure 7.5
has its end regions in non-uniform bending and only the central region in pure
bending.
7.2.1 Strain-Displacement Equation
Assumption (d) listed in Definition 7.1 is known as the Bernoulli-Euler hypothesis.
This name was later extended to encompass the entire theory of the beam in bending
known today as the Bernoulli-Euler beam theory. This assumption, which
practically is seldom realized, introduces an important kinematic assumption very
similar in nature with that postulated in the axial deformation of the beam. The main
implication of assumptions (a) to (d) is that the fibers contained in any vertical plane
behave identically to those in the longitudinal plane of symmetry oxy .
Mathematically, this implies the independence of the deformation on the variable z
and, consequently, the beam deformation can be represented by the deformation of the
fibers located in the longitudinal plane of symmetry.
Figure 7.6 illustrates a beam segment in undeformed and deformed conditions. The
radius and the center C are called radius and center of curvature, respectively.
The radius of curvature is a function only of the variable x as a direct consequence
of assumption (d).
Figure 7.6 Beam Segment Deformation
(a) Undeformed and (b) Deformed
The undeformed segment PQ transforms during the loading application into the
curved segment P*Q*. Similarly, the undeformed segment AB located on the
deflection curve becames A*B* after the deformation, with the difference being that
- 156 -
its original length remains unchanged. Using the general definition (2.42) the
elongation strain x is expressed as:
][lim
][lim),(),,(
*
**
x
xx
PQ
PQQPyxzyx
QP
QPxx
(7.1)
By studying Figure 7.6, the following geometrical relations can be written:
xABPQ (7.2)
*** *)( xABBA (7.3)
**** *])([ yxxQP (7.4)
Introducing the geometric relations (7.2) through (7.4) into the elongation strain
equation (7.1):
)(
]*)(
*)(*))(([lim),(
*
**
x
y
x
xyxyx QPx
(7.5)
For the case of small displacements the radius of curvature )(x is related to the
vertical displacement )(xv by the following relation:
2
2 )(
)(
1
dx
xvd
x
(7.6)
Note: Detailed explanation pertinent to equation (7.6) is provided in Section 8.3.
Substituting equation (7.6) into equation (7.5), the following ordinary differential
equation is obtained:
ydx
xvdyxx *
)(),(
2
2
(7.7)
Equation (7.7) represents the strain-displacement equation of the linear plane
member in pure bending.
Examination of equation (7.6) reveals that the elongation strain ),( yxx is inversely
proportional to the radius of curvature )(x . The deformation of the beam as a
- 157 -
function of the curvature is illustrated in Figure 7.7. The inverse of the radius of
curvature )(x is called curvature and is notated by the letter )(/1 xk . For a
positive radius of curvature the center C is located above the beam deflection curve
and the deformation is concave as depicted in Figure 7.7.a. In contrast, the
deformation is convex if the curvature is negative and the center C is located below
the beam deflection curve as shown in Figure 7.7.b.
Figure 7.7 Beam Deformation and Radius of Curvature
(a) Positive Curvature and (b) Negative Curvature
Variation of the elongation strain ),( yxx , illustrated in Figure 7.8.a, for a particular
cross-section corresponding to a positive radius of curvature is a linear function. The
fibers located above the deflection curve are shortened, while these below the
deflection curve are extended. Consequently, the areas above and below the deflection
curve are in compression and tension, respectively. For the case of the negative
curvature, shown in Figure 7.8.b, the distribution is similar, but the compression and
tension zones are reversed. Considering the relations (7.5) and (7.8) for a particular
cross-section (x=constant) the strain ),( yxz is positive above the neutral plane and
negative below the neutral plane.
Figure 7.8 Elongation Strain Distribution
The transversal strains are expressed as:
- 158 -
),(*),(),( yxyxyx xzy (7.8)
Note: In the classical theory of the beam transversal deformation of the cross-section
is neglected.
7.2.2 Constitutive Equation
The constitutive equation represents, as described in Lecture 2, the relation between
stress and strain. For the case of isotropic elastic material behavior under axial
deformation, the relation between the stress and strain is written as follows for the
beam in bending:
)(*)(),(*)(),(),,(
x
yxEyxxEyxzyx xxx
(7.9)
where )(xE is the modulus of elasticity, which has a value proper to each specific
material.
The normal stress ),( 0 yxxx pertinent to a particular cross-section is a function of
the variable y and varies linearly as it is directly proportional to the elongation
strain ),( yxx .
The rest of the stress tensor components are zero:
0 zy (7.10)
0 xyxyxy (7.11)
7.2.3 Cross-Section Stress Resultants
Considering the normal stress ),( yxx distribution represented by the equation (7.9)
the cross-section stress resultants )(xF , )(xM y and )(xM z illustrated in Figure 7.9
are calculated as:
z
A
AA
x
Sx
xEdAy
x
xE
dAx
yxEdAyxxF
*)(
)(*
)(
)(
*)(
*)(*),()(
(7.12)
- 159 -
yz
A
AA
xy
Ix
xEdAzy
x
xE
dAx
zyxEdAyxzxM
*)(
)(**
)(
)(
*)(
**)(*),(*)(
(7.13)
z
A
AA
xz
Ix
xEdAy
x
xE
dAx
yxEdAyxyxM
*)(
)(*
)(
)(
*)(
*)(*),(*)(
2
2
(7.14)
where zS , yzI and zI are the static moment, the product of inertia and the
moment of inertia, respectively.
Figure 7.9 Normal Stress and Stress Resultants
(b) Stress Resultants and (b) Normal Stress
If the coordinate system oxyz is considered passing through the cross-section centroid
and the cross-section is symmetric about y axis, the cross-section stress
resultants )(xFx , )(xM z and )(xM z became:
0)( xF (7.15)
0)( xM y (7.16)
'*)(
)(zz I
x
ExM
(7.17)
where 'z is the axis passing through the cross-section centroid and parallel to z .
7.2.4 Normal Stress Distribution
Combining equation (7.17) with equation (7.9), the normal stress ),( yxx is
expressed as:
- 160 -
yI
xMyx
z
zx *
)(),(
'
(7.18)
Equation (7.18), called Navier’s formula, expresses the linear distribution of the
normal stress ),( 0 yxxx in the beam cross-section. The variation of the normal
stress for positive and negative bending moments is illustrated in Figure 7.10.
Figure 7.10 Normal Stress Distribution
By definition the neutral axis of the cross-section is the axis where the normal stress
),( 0 yxxx is zero.
Note: In the case of pure bending, the neutral axis passes through the centroid of the
cross-section and is horizontal.
The maximum normal stresses, tension or compression, are obtained at extreme
located fibers. The ratio obtained by dividing the moment of inertia by the distance
measured from the centroid to the extreme fiber is called the-section modulus.
Obviously if the cross-section has only one axis of symmetry two section moduli can
be defined:
)(
)()(
'
xy
xIxW
bottom
zbottom (7.19)
top
ztop
y
xIxW
)()(
'
(7.20)
Consequently, if the bending moment is assumed positive, 0)( xM z , as illustrated in
Figure 7.10.a, the maximum tensile and compressive stresses are obtained as:
)(
)()(max__
xW
xMx
bottom
ztensionx (7.21)
)(
)()(max__
xW
xMx
top
zncompressiox (7.22)
- 161 -
If the bending moment assumes a negative value 0)( xM z , the maximum stresses are
as shown in Figure 7.10.b:
)(
)()(max__
xW
xMx
top
ztensionx (7.23)
)(
)()(max__
xW
xMx
bottom
zncompressiox (7.24)
Note: If the cross-section has two axes of symmetry the section moduli are equal
and, consequently, the magnitudes of the maximum normal stresses are equal.
7.3 Nonuniform Bending
The main limitation of the pure bending theory is the requirement that only the
bending moment )(xM z is allowed to be present in the cross-section of the beam. In
general, the beams are acted on by transversal loads and, consequently, the existence
in the cross-section of the shear force )(xVy is expected. The bending of a beam which
manifests in the presence of the bending moment and shear force is called
nonuniform bending. The problem of the nonuniform bending was first studied by
D. J. Jurawski, a Russian engineer, in connection with the behavior of rectangular
timber beams. In the case of nonuniform bending, both, the normal stress and shear
stress, are present in the beam cross-section. Because the analysis of the shear stress
distribution on the cross-section is a complicated task requiring mathematical and
theoretical knowledge beyond the level of this textbook, the complexity of the
theoretical development presented herein will be gradually increased. First, the
simplest case of a rectangular cross-section will be investigated. Later, the
formulations obtained for the case of the rectangular cross-section are extended to
other geometrical type of cross-sections.
7.3.1 Basic Assumptions
As illustrated in Figure 7.11 the concurrent existence of bending moment )(xM z and
shear force )(xQy induces normal stress ),( yxx and shear stress ),,( zyxxy ,
respectively.
According to the duality principle the shear stress ),,( zyxyx acting in a longitudinal
plane exists and has a non-zero value.
),,( zyxxy = ),,( zyxyx (7.25)
- 162 -
Figure 7.11 Nonuniform Bending of Rectangular Cross-Section
The duality of the shear stresses expressed in equation (7.25) is illustrated in Figure
7.12. In the absence of tangential forces from the exterior surface of the beam it is
expected that ),,( zyxxy to be zero at the upper and lower fibers, marked in the Figure
7.12 by points a and c .
Figure 7.12 Shear Stresses Duality
The shear stress ),,( zyxxy is related with the angular strain and ultimately involves
change in shape of the deformable body. Figure 7.13 depicts a qualitative comparison
considering the same cantilever beam in pure and nonunniform bending.
Figure 7.13 Cantilever in Pure and Nonuniform Bending
(a) Pure Bending and (b) Nonuniform Bending
- 163 -
In accordance with the theoretical developments derived in Section 7.1, every plane
cross-section of the cantilever beam in pure bending remains planar through the
deformation as shown in Figure 7.13.a. In contrast, the cantilever shown in Figure
7.13.b, which is in nonuniform bending due to the existence of the shear force along
its entire length, results in warping of the planar cross-sections after deformation.
Note: The finding that in nonuniform bending the cross-section does not remain
plane after the deformation contradicts the basic kinematic assumption of pure
bending described by definition 7.1. However, detailed investigation using
advanced methods of the Theory of Elasticity indicate that the warping of the
cross-section due to the shear stress is insignificant and does not affect the
longitudinal strain when the ratio of cross-section height to length of beam is
small. This is the case for most ordinary structural members.
To use the distribution of ),( yxx derived for the case of pure bending in the case of
the beam subjected to nonuniform bending a supplementary assumption must be
made: the distribution of the normal stress in a given cross-section is independent of
the shear induced deformation.
Note: As a consequence of imposing the above stated assumption, the pure bending
strain-displacement relation remains valid and the influence of the angular
deformations on the longitudinal strain is neglected.
7.3.2 Shear Stress Distribution in a Rectangular Cross-Section
In the absence of a strain-displacement relation for the shear stress, the distribution of
the shear stress is obtained from equilibrium considerations instead of the approach
used for normal stress in the previous section. In the American technical literature the
shear distribution is called the shear flow, by analogy to the flow of the liquid.
A thin slice isolated from the rectangular cross-section of a linear elastic beam is
illustrated in Figure 7.14.a. To simplify the figure only the cross-section stress
resultants, the shear force and the bending moment are shown. The corresponding
stresses are depicted in Figure 7.14.b.
By further subdividing the thin element with a horizontal plane ab the free- diagram
of the remaining upper portion is obtained. Using the notation shown in Figure 7.14.c
the horizontal equilibrium of this remaining upper portion can be written as:
12 FFH (7.26)
The horizontal force H represents the resultant force of the longitudinal shear
stresses ),,(),,( zyxzyx xyyx acting on the plane ab , while the horizontal forces 1F
and 2F are the resultant forces of the normal stresses ),( yxx acting on the vertical
areas. To determine the shear stress distribution in the cross-section the duality
principle is applied.
- 164 -
Figure 7.14 Stress Resultants and Stresses
(a) Cross-section Resultants, (b) Stresses and (c) Free-body Diagram
Using the notation shown in Figure 7.15 and the Navier’s normal stress formula
(7.18) the forces 1F and 2F are expressed as:
)(*)(
**)(
*),(
'
''
''
1
ySI
xM
dAI
xMdAxF
A
z
z
z
Az
z
A
x
(7.27)
)(*)(
**)(
*),(
'
''
''
2
ySI
xxM
dAI
xxMdAxxF
A
z
z
z
Az
z
A
x
(7.28)
where )('
yS A
z and zI is the static moment of the section 'A and the moment of inertia
with respect to the neutral axis, respectively.
Note: The Figure 7.15 illustrates the rectangular cross-section after the deformation
emphasizing the lateral deformation of the rectangular cross-section.
Figure 7.15 Notation
The static moment )('
yS A
z of the section 'A is expressed as the product of the area 'A and the distance from the centroid of 'A centroid to the neutral axis.
- 165 -
'
0
' *)('
yAyS A
z (7.29)
where '
0y is the distance from the centroid of area 'A to the neutral axis.
Substituting equations (7.27) and (7.28) into equation (7.26) yields the following
equation:
)(*])()(
[
)(*)(
)(*)(
'
''
ySI
xM
I
xxM
ySI
xMyS
I
xxMH
A
z
z
z
z
z
A
z
z
zA
z
z
z
(7.30)
The shear flow q which is the shear force per unit length is found by taking the limit
of the expression (7.30):
z
A
zy
z
A
zzzxx
I
ySxV
I
yS
x
xMxxM
x
Hyxq
)(*)(
)(*]
)()([limlim),(
'
'
00
(7.31)
Note: The relationship between bending moment and shear force was established in
Lecture 4 (equation (4.6)).
The average shear stress is calculated in a similar manner with the calculation of the
shear flow if instead of the length x the elementary area tx * is used:
tI
ySxV
tI
yS
x
xMxxM
tx
Hyx
z
A
zy
z
A
zzzx
xavryx
*
)(*)(
*
)(*]
)()([lim
*lim),(
'
'
0
0_
(7.32)
Equation (7.32) is called Jurawski’s formula for shear stress. In Figure 7.16 a typical
rectangular cross-section is shown. The distribution of the average shear stress
),(_ yxavryx calculated according to Jurawski’s formula (7.32) is illustrated in Figure
7.16.b.
- 166 -
Figure 7.16 Shear Stress Distribution for a Rectangular Cross-Section
Using the notation shown in Figure 7.16.a, the static moment )('
yS A
z of the area above
the cutting plane cd is calculated:
]4
[*2
]2
)2
(
[*)2
(*)( 22
'
yhb
y
yh
yh
byS A
z
(7.33)
The moment of inertia about the horizontal axis passing through the cross-section
centroid is:
12
* 3hbI z (7.34)
Introducing equations (7.33) and (7.34) into equation (7.32) the variation of the
average shear stress is obtained:
]*4
1[**2
)(*3
*12
*
]4
[*2
*)(
),(),(
2
2
3
22
__
h
y
A
xV
bhb
yhb
xV
yxyx
y
y
avryxavryx
(7.35)
The distribution of the average shear stress in the rectangular cross-section is
parabolic and has the maximum value at the neutral axis location, where the normal
stress is zero. The maximum average shear stress is:
A
xVyx
y
avrxy
)(*5.1)0,(max
_ (7.36)
- 167 -
Note: The formulae (7.31) and (7.32) derived for the case of the rectangular cross-
section can be extended without any theoretical difficulty to the general case
as follows:
)(
),(*)(),(
'
xI
yxSxVyxq
z
A
zy (7.37)
),(*)(
),(*)(),(
'
_yxtxI
yxSxVyx
z
A
zy
avryx (7.38)
The general formulae (7.37) and (7.38) are mainly restricted by the assumption that
the shear stress ),( yxxy is parallel to the shear force acting on the cross-section, an
assumption which is not always valid.
7.3.3 Limitations in the Usage of the Shear Stress Formula for Compact Cross-
Sections
Cross-sections without holes such as solid rectangular or circular cross-sections are
called compact cross-sections in contrast to the “thin-wall” or “built-up” cross-
sections. Jurawski’s formula for shear stress expressed by equation (7.32) is derived
under the assumption that the cross-section is a rectangular shape, but in engineering
practice this formula is applied to a large variety of compact cross-section shapes.
Additional limitations in the usage of Jurawski’s formula for shear stress are
discussed below.
7.3.3.1 Cross-Section Shape
The correct distribution of shear stress can be calculated using advanced methods of
the Theory of Elasticity. Some of these results are presented and compared with the
average shear stress calculated using equation (7.36).
The exact shear stress distribution at the neutral axis (NA) obtained from analyzing
the case of the cantilever beam subjected to a concentrated load applied at the tip
point and having different cross-sectional ratios bh is illustrated in Figure 7.17. The
cross-sections of the cantilevers shown in Figure 7.17.a and 7.17.b have an aspect
ratio of 25.0/ bh and 2/ bh , respectively.
The distribution of the shear stress calculated at the neutral axis of the wider cross-
section ( 25.0/ bh ) indicates a substantial departure from the average shear stress
calculated from equation (7.36). Contrary, the maximum shear stress calculated for
the narrow rectangular cross-section ( 2/ bh ) shows a very good agreement with the
value obtained from the application of the same formula (7.36). The above findings
limit the application of the formula (7.36) to the case of narrow rectangular cross-
sections. The following limit in the cross-section aspect ratio is imposed:
- 168 -
0.2b
h (7.39)
Figure 7.17 Rectangular Cross-Section Beams
(a) Wide and (b) Narrow
The case of the cantilever with a circular cross-section is shown in Figure 7.18.a. The
difference between the value of the shear stress from the exact solution and the
average shear stress from Jurawski’s formula at the neutral axis location is negligible.
Figure 7.18 Circular Cross-Section
7.3.3.2 Beam Length
The verification of the average shear stress value calculated with equation (7.38) by
comparison with the exact solution from the Theory of Elasticity it can be shown that
- 169 -
the error in using Jurawski’s formula for shear stress is negligible when the ratio of
cross-section height h to beam length l is limited as follows:
0.4h
l (7.40)
7.3.4 Shear Stress Distribution in Thin-Wall Cross-Sections
The compact sections are seldom used in the structural engineering. The majority of
beams used in steel structures are classified as beams with “thin-wall” cross-sections.
The thin-wall cross-section is categorized in (a) open and (b) closed cross-sections as
illustrated in Figures 7.19.a and 7.19.b, respectively.
Figure 7.19 Thin-Wall Cross-Sections
The two most common types of thin-wall cross-sections used in structural
engineering, tubular and wide-flange cross-sections, are illustrated in Figures 7.21.
The thin-wall cross-sections are characterized by a reduced aspect ratio between the
wall thickness and the overall dimensions of the cross-section. The characteristic ratio
0/ dt and ht f / of the tubular and wide-flange cross-section, respectively, are
extremely small.
Figure 7.20 Thin-Wall Beams
(a) Tubular and (b) Wide-Flange
7.3.4.1 Assumptions
The reduced thickness of the wall of the cross-section facilitates the introduction of
two important assumptions:
(a) the shear flow is always tangent to the local centerline of the cross-section;
- 170 -
(b) the shear stress is constant in the thickness of the cross-section.
These two assumptions mimics the situation described in the case of the rectangular
cross-section.
The shear flow ),( sxq and the shear stress ),( sx characteristic to a thin-wall cross-
section are shown in Figure 7.21.
Figure 7.21 Shear Flow and Shear Stress in Thin-Wall Cross-Section
(a) Shear Flow and (b) Shear Stress
They are related through the following formula:
),(
),(),(
sxt
sxqsx (7.41)
7.3.4.2 Wide-Flange Cross-Section
The calculation of shear stress for wide-flange cross-section will closely follow the
methodology employed for the shear stress determination in the case of the
rectangular cross-section. A thin slice with length x is isolated from the body of the
beam. The normal stress and the shear flow are illustrated in Figure 7.22.a and 7.22.b,
respectively.
Figure 7.22 Isolated Slice of a Wide-Flange Beam
- 171 -
The shear flow in the web is calculated using the equilibrium of the upper remaining
portion of the slice after further subdividing it with a horizontal plane cutting at
y above the NA as shown in Figure 7.23. The free-body diagram of the upper portion
is pictured in Figure 7.23.b
Figure 7.23 Free-Body Diagram and Notation
(a) Free-Body Diagram Perspective, (b) Free-Body Diagram Longitudinal View and
(c) Cross-Section View and Notation
The horizontal equilibrium equation of the isolated upper body is written as:
021 wFFF (7.42)
Using a rational similar to that used for the rectangular cross-section, equation (7.42)
is solved to obtain the distribution of the shear flow in the web:
z
A
zy
webxyI
SxVyxq
w'
*)(),(_ (7.43)
Using the notation from in Figure 7.23.c, the static moment 'wA
zS and the moment of
inertia zI calculated about the neutral axis are:
)4
(*24
)(*
2
]2
[*2
1*]
2[*]
22[*
2
1*]
22[*
2
222
'
yhthhb
yh
yh
thhhh
bS
www
www
wwA
zw
(7.44)
)***(*12
1 333
wwwz hthbhbI (7.45)
Incorporating equations (7.44) and (7.45) in equation (7.43) the following expression
for the web shear flow is obtained:
- 172 -
]****
*4
***
***[*
2
)(*3),(
2
333
333
222
_
yhthbhb
t
hthbhb
hthbhbxVyxq
www
w
www
wwwy
webxy
(7.46)
Consequently, the shear stress in the web is obtained:
]****
*4
***
***[*
*2
)(*3),(
2
333
333
222
_
yhthbhb
t
hthbhb
hthbhb
t
xVyx
www
w
www
www
w
y
webxy
(7.47)
Equations (7.47) can be written as:
]*[**
)(*5.1),( 2
21_ yCCht
xVyx ww
w
y
webxy (7.48)
where the coefficients 1wC and 2wC are constants and depend only on the cross-section
dimensions.
Formula (7.47) indicates a parabolic variation of the shear stress xy valid in the web
for the following range of y values.
22
ww hy
h (7.49)
If the horizontal cutting plane passes into the flange the static moment 'fA
zS is
calculated as:
]4
[**2
1 22
'
yh
bS fA
z (7.50)
The resulting shear flow and shear stress equations follow:
]****
*4
***
*[*
2
)(*3),(
2
333
333
2
_
yhthbhb
b
hthbhb
hbxVyxq
www
www
y
flangexy
(7.51)
]****
*4
***
*[*
*2
)(*3),(
2
333
333
2
_
yhthbhb
b
hthbhb
hb
b
xVyx
www
www
y
flangexy
(7.52)
- 173 -
Equation (7.52) shows that the shear stress flangexy _ is zero as expected for 2
hy
and reaches its maximum value at2
why . The shear stress calculated with equations
(7.47) and (7.52) exhibit a discontinuity at 2
why , the border between web and
flange. Equation (7.52) has been applied to a wide rectangular shape with ratio of
height to width in violation of the limits established in section 7.3.3.1. Consequently,
an erroneous linear variation between the zero value and the value calculated from the
equation (7.47) is suggested for the shear stress ),(_ yxflangexy in the flange.
The shear flow in the flange is obtained by sectioning the thin slice of beam with a
vertical plane as shown in Figures 7.22.a and 7.22.b. The shear horizontal force
fF necessary to equilibrate the forces 1F and 2F induced by the normal stress
variation is shown in Figure 7.24.a.
Figure 7.24 Free-Body Diagram and Notation
(a) Free-Body Diagram Perspective, (b) Free-Body Diagram Longitudinal View and
(c) Cross-Section View and Notation
The horizontal equilibrium equation is written:
021 fFFF (7.53)
The shear flow ),( yxq f in the flange calculated using the notation shown in Figure
7.24c is:
z
A
zy
fI
SxVyxq
f'
*)(),( (7.54)
The geometrical quantities used above are expressed functions of the cross-section
dimensions:
)***(*12
1 333
wwwz hthbhbI (7.55)
- 174 -
)(*8
)22
(*** 22'''
w
f
fff
A
z hhsth
tsyAS f (7.56)
Inserting equations (7.55) and (7.56) into equation (7.54) the shear flow in the flange
is obtained:
sxVhthbhb
hhsxq y
www
w
flangezx *)(*]***
[*2
3),(
333
22
_
(7.57)
The shear stress calculated in accordance with the relation (7.41) is:
f
y
www
w
flangezxt
sxV
hthbhb
hhsx
*)(*]
***[*
2
3),(
333
22
_
(7.58)
The shear stress calculated in equation (7.58) acts on the vertical plane parallel to the
longitudinal axis of the beam. The duality of the shear stresses indicates that an equal
shear stress acts horizontally on the flange but in the plane of the cross-section. This
is a flangexz _ shear stress.
The variation of the shear stress xy in a particular cross-section is shown in Figure
7.25 and is obtained by giving particular values to y and s in equations (7.47) and
(7.58), respectively.
Figure 7.25 Shear Stress Distribution in Wide-Flange Cross-Section
From equations (7.47) and (7.58) it is evident that the shear stress varies linearly on
the flange and parabolicly on the web.
The resultant of the shear stress acting on the web is equal to:
)](*3
2[** minmaxmin wwwwwweb thT (7.59)
- 175 -
Note: In the case of the structural steel shapes (W, S, etc.) the shear force taken by
the web accounts for more than 90% of the total shear force acting on the
entire cross-section. To obtain conservative results steel design standards
require the entire shear force to be sustained by the web alone.
7.3.4.3 Closed Thin-Wall Cross-Section
The most commonly used closed thin-wall cross-sections are the box and tubular
cross-sections depicted in Figure 7.19.b. The analysis of the distribution of shear
stress for a closed thin-wall cross-section is conducted in a similar manner with the
theoretical development employed for open cross-sections. A thin slice of length x is
isolated from the body of the closed thin-wall beam. As illustrated in Figure 7.26.b
the beam has the longitudinal vertical plane as a plane of symmetry. The shear flow
shown in Figure 7.26.b must to be directed in such a manner that the resultant is equal
to the vertical shear force acting on the cross-section.
Figure 7.26 Closed Thin-Wall Cross-Section
The main problem arising in evaluation of shear stress in closed thin-wall members is
how to position the cutting plane. If the vertical cutting plane passing through the axis
of symmetry oy is used the beam is separated in two symmetric halves. According to
assumption (a) stated in Section 7.3.4.1 the shear flow is tangent to the centerline of
the cross-section. Thus, for example, in Figure 7.26.c it is obviously that the shear
flow is horizontal at ends a andb . The shear flow must be zero in the plane of
symmetry and as a consequence of the duality principle the shear flow in the cross-
section at these locations is also zero. Since the shear flow is zero at the cross-section
points intersecting the axis of symmetry oy , the general distribution is obtained by
using cutting planes symmetrical to this axis and writing the horizontal equilibrium
for the beam segment contained between these planes.
The tubular circular cross-section illustrated in Figure 7.27 is used to exemplify the
methodology. Two symmetric cutting planes located by the angle are shown in
Figure 7.27.a. The free-body diagram obtained is pictured in Figure 7.27.b. The
equilibrium equation is written as:
12*2 FFH (7.60)
- 176 -
Figure 7.27 Circular Thin-Wall Cross-Section
The unbalanced force H is expressed as:
xqH * (7.61)
Following the same steps as in the previous sections the shear flow is obtained as:
)(*2
),(*)(),(
xI
xSxVxq
z
Ay (7.62)
Consequently the shear stress is:
txI
xSxVx
z
Ay
yz*)(*2
),(*)(),(
(7.63)
Using the notation shown in Figure 7.27.c, the static moment ),( xS A and the
moment of inertia )(xI z relative to the neutral axis oz are calculated as:
sin***2),( 2 trxSA (7.64)
trxI z **)( 3 (7.65)
Substituting equations (7.64) and (7.65) into equations (7.62) and (7.63) the shear
flow and stress are calculated as:
r
xVxq
y
*
sin*)(),(
(7.66)
tr
xVx
y
yz**
sin*)(),(
(7.67)
- 177 -
The angle varies in the interval from zero to 2
and, consequently, the shear flow
and stress attain their maximum values at the neutral axis. The maximum shear stress
is:
tr
xVx
y
yzyz**
)()
2,(max_
(7.68)
If the expression of the area is employed equation (7.68) becomes:
)(
)(*2)
2,(max_
xA
xVx
y
yzyz
(7.69)
where trxA ***2)( is the area of the cross-section.
The distribution of the shear flow is illustrated in Figure 7.28.
Figure 7.28 Shear Flow Distribution
7.4 Built-Up Beam Connectors
The cross-sections of the beams discussed in the previous sections are considered to
be made from continuous material without any interruptions. They are called
homogeneous cross-sections. In structural engineering practice the compact cross-
sections, which are homogeneous, are seldom used in steel structures. For shorter
span beams or lighter loads the wide-flange homogeneous beams are commonly used.
In general, the industrial steel buildings are subjected to heavy loads and the most
commonly used cross-sections, such as the wide-flange and closed thin-wall, are
fabricated from parts joined together by connectors. These types of beams are called
build-up members. The connectors may be glue, welds, rivets, bolts or nails. Glue
(adhesives) and nails are connectors proper to be used in the wood built-up members,
- 178 -
while welds, rivets and bolts are commonly used for the steel built-up members.
Examples of build-up members are illustrated in Figure 7.29.
Figure 7.29 Built-Up Beams
(a) Glued Wood Beam, (b) Box Wood Beam, (c) Welded Steel Beam and
(d) Reinforced Steel Beam
The first two cross-sections shown in Figures 7.29.a and 7.29.b are wood fabricated
beams constructed using glue or nails, respectively. Figures 7.29.c and 7.29.d are
illustrating steel beams constructed using wells or bolts.
The shear flow induced by nonuniform bending must be transferred in-between the
adjoining parts comprising the beam cross-section. This shear transfer is realized in
three ways: (a) by shear distributed over the interface areas, (b) by shear distributed
along a connector line and (c) by discrete shear connectors. The shear transfer type (a)
is common for the laminated wood beams composed of multiple pieces as shown in
Figure 7.29.a. The linear transfer, type (b), is proper to the welded beams as pictured
in Figure 7.29.c. The discrete type shear transfer, shown in Figures 7.29 .b and 7.29.d,
are pertinent to bolt or nail connectors. There are special cases where welds are
discontinuously applied, referred to as stitch welds.
The shear flow determined using equation (7.31) is employed to calculate the required
shear flow recq , which represents the shear force per unit length that must be
transferred from one part of the cross-section to the adjacent part under the given
loading condition. The required shear is calculated as:
z
y
z
A
zy
recI
yAxV
I
SxVq
**)(*)( (7.70)
where A is the area on which the unbalanced flexural stresses are exerted.
7.4.1 Linear Shear Connectors
The required shear flow calculated with equation (7.69) determines the actual
necessary force per unit length which must be transmitted by the weld. The shear
capacity of the weld weldq must be larger than the required shear flow:
- 179 -
weldreq qq (7.71)
The capacity of the weld is determined from laboratory test data.
7.4.2 Discrete Shear Connectors
The discrete connectors have a shear capacity connectorQ , also established by laboratory
testing, and are assumed spaced at a equal distance s . For the discrete connector to
transfer the shear the following relation must be maintained:
connectorreq Qsq * (7.72)
Equation (7.72) is applied in different zones of the beams, considering in the
calculation the maximum value of the vertical shear force in the particular zone. This
way, the distance between the connectors may be increased or decreased within the
various zones of the beam in accordance with the total shear flow requirement for the
zone.
7.4.3 Examples
Two examples of shear connectors are investigated in the following sections. The
first, contained in section 7.4.3.1 exemplifies the requirements for welds within a
built-up steel beam. The second, demonstrates the calculation of discrete connectors,
as applied in the design of a nail connected wooden box beam.
7.4.3.1 Welded Connection
The build-up beam is assembled from three rectangular steel plates connected by
welds. The dimensions of the steel plates and the notation are shown in Figure 7.30.
The cross-section is subjected to a vertical shear force 45yV .kN
Figure 7.30 Steel Build-Up Beam
- 180 -
The required shear flow is calculated using equation (7.64):
z
y
z
A
zy
reqI
yAxV
I
SxVq
**)(*)(
Note: The area considered in the calculation is the flange which during the bending
tries to slip relatively to the web.
Using the dimensions shown in Figure 7.30 the static moment and the moment of
inertia are calculated:
3
1
5.377437)2
15
2
290(*)165*15(
)22
(*)*(*
mm
thbtyAS
f
fflange
flange
z
483
23
3
121
3
10*3045.112
5.7*290])
2
15
2
290(*)165*15(
12
165*15[*2
12
*])
22(*)*(
12
*[*2
mm
ththbt
btI
f
f
f
z
The required shear flow is:
130.010*3045.1
5.377437*458reqq mmkN /
Two welds are located at the interface between the flange and the web and,
consequently, they share the required shear flow. The shear flow in the weld is:
065.02
req
weld
qq mmkN /
7.4.3.2 Nail Connection
The cross-section of a built-up wooden box type beam is illustrated in Figure 7.31.
The beam is assembled from four planks of wood having the dimensions indicated in
the figure. The connection between planks is realized by employing nails along the
four interfaces. The capacity of each nail is NQnail 800 . A vertical shear force
kNVy 15 acts on the cross-section.
- 181 -
Figure 7.31 Wood Box Beam
The shear flow is transferred between the upper flange and the two webs and is
calculated as:
z
y
z
A
zy
reqI
yAxV
I
SxVq
**)(*)(
The static moment and the moment of inertia are obtained as:
3 864000)2
40
2
280(*)40*180( mmS A
z
4833
10*642.212
180*200
12
210*280mmI z
The required shear flow is calculated as:
mmkNqreq / 0491.010*642.2
10*64.8*158
5
There are two rows of nails at this location of the cross-section. The nails are spaced
at the unknown distance s along the length of the beam.
mmkNq rownailreq / 0246.02
0491.0__
Using equation (7.72) the distance between the nails is calculated as:
mmq
Qs
rownailreq
nail 52.326.24
800
__
- 182 -
LECTURE 8 Beam Deflection
8.1 Introduction
A beam with a straight longitudinal axis subjected to transversal loads acting in its
longitudinal plane of symmetry deforms and the resulting curve is called the
deflection curve. In the previous lecture, the curvature of the deflection curve was
used to determine the stress and strain distribution in the cross-section of a beam
under the restrictions established for pure and nonuniform bending. In this lecture, the
equation of the defection curve will be derived and consequently, the displacements at
any point of the beam may be calculated. The calculation of beam deflections is an
important part of structural analysis and design. The deflections are limited to
prescribed tolerances imposed by the functionality of the particular structural element.
The cantilever beam pictured in Figure 8.1 deforms in its vertical plane of symmetry
under the action of the exterior transversal loading.
Figure 8.1 Example of Beam Deflection
The deflection curve and its slope are mathematically represented by the real
functions )(xv and )(x , respectively. The vertical distance measured from a given
point located on the undeformed axis of the beam to the corresponding point located
on the deflection curve is called the transverse displacement.
- 167 -
8.2 Qualitative Interpretation of the Deflection Curve
For the general case of nonuniform bending the equation relating the bending moment
)(xM z with the radius of curvature )(x is:
)(*)(
)()( xI
x
xExM zz
(8.1)
where )(xE - modulus of elasticity;
)(xI z - moment of inertia about the horizontal centroidal axis of the cross-
section;
)(xM z - cross-section bending moment;
)(x - radius of curvature of the deflection curve.
Note: If the material is homogeneous linear elastic and the beam is of uniform cross-
section the product zIE * is called the flexural rigidity and is constant.
From the relation given by equation (8.1), the deflection )(xv can be anticipated
using the sign convention established in Lecture 7 and re-plotted in Figure 8.2. The
moment diagram is always plotted at the beam side where the fibers are subjected to
tension, while the curvature center is placed on the opposite side. When the bending
moment is positive the deflection curvature is concave, while for the negative bending
moment the deflection curvature is convex. At the supports the deflection must
correspond to the prescribed support constraint condition. Locations where the
deflection curve changes from concave to convex, or vice versa, are called inflection
points.
Figure 8.2 Sign Convention for Beams in Bending
An example of qualitative interpretation of the deflection curve is shown in Figure
8.3. The moment diagram changes from positive to negative in the interval AB at
- 168 -
point D. In between points A and D the moment is positive and the curvature center is
located above the deflection curve. Between points D and B the moment diagram is
negative and, thus, the curvature center is located below the deflection curve. The
change in curvature takes place at point D, which is an inflection point. On the
overhanging end BC, the moment is negative and the curvature is convex.
Considering that at the supporting points the beam is constrained not to move
vertically, the qualitative deflection diagram, as shown in Figure 8.3, can be sketched.
Figure 8.3 Example of Qualitative Interpretation of Deflection Curve
8.3 Differential Equations of the Deflection Curve
From Calculus it is known that the slope of a real function at any particular point of
its defined continuity interval is the first derivative of the function. For the case in
point, the real function is represented by the beam displacement curve )(xv . Using the
notation shown in Figure 8.1 the following relation is written:
dx
xdvx
)()(tan (8.2)
Under the assumption of small displacements, the first derivative is a very small
value:
1)(
dx
xdv (8.3)
consequently, the angle can approximate by its tangent:
dx
xdvxx
)()(tan)( (8.4)
In Calculus the relation between the radius of curvature )(x and the deflection )(xv is
established as:
- 169 -
2
3
2
2
2
]))(
(1[
)(
)(
1
dx
xdv
dx
xvd
x
(8.5)
Under the small displacement assumption (8.3), equation (8.5) can be simplified and
written as:
2
2 )(
)(
1
dx
xvd
x
(8.6)
Substituting equation (8.6) into equation (8.1) yields the following equation:
)(*)(
)()(2
2
xIxE
xM
dx
xvd
z
(8.7)
The moment-curvature equation, a second order differential equation, is obtained
from equation (8.7) and expressed in the following standard form:
"*)(*)()( vxIxExM z (8.8)
where the notation 2
2" )(
dx
xvdv is employed.
Consider the differential relation between the transverse loading and the cross-section
stress resultants, )(xV and )(xM , previously obtained from the equilibrium of the
beam infinitesimal volume element:
)()(
xpdx
xdVn (8.9)
)()(
xVdx
xdM (8.10)
where )(xV and )(xM are the vertical shear force and the bending moment,
respectively.
Differentiating the moment-curvature equation (8.8) and using equations (8.9) and
(8.10) the shear-deflection and load-deflection equations are obtained:
]'"*)(*)([)( vxIxExV z (8.11)
]""*)(*)([)( vxIxExp zn (8.12)
- 170 -
The load-deflection equation (8.12) is a forth-order differential equation.
If the beam is made from a homogeneous linear elastic material and has a constant
cross-section in the continuity interval of the bending moment or transversal loading
equations (8.8) and (8.12) became:
"**)( vIExM z (8.13)
IV
zn vIExp **)( (8.14)
The general theory of differential equations with constant coefficients can be
employed to obtain the solution of the second-order differential equation (8.8) and
forth-order differential equation (8.12). The continuity intervals pertinent to the
functions involved in these differential equations, the bending moment and transverse
load, must be recognized in order for the integration process to be properly
conducted.
8.3.1 Integration of the Moment- Curvature Differential Equation
Integration of the second-order differential equation (8.8) on an interval of continuity
for the bending moment )(xM z yields the following relations:
1
' *)(*)(
)()()( Cdx
xIxE
xMxvx
z
z
(8.13)
21 ***)(*)(
)()( CxCdxdx
xIxE
xMxv
z
z (8.14)
For the case of a beam with constant bending stiffness constIE z )*( , the integrals
expressed in equations (8.13) and (8.14) may be simplified as follows:
1
' *)(*
1)()( CdxxM
IExvx z
z
(8.15)
21 **)(*
1)( CxCdxxM
IExv z
z
(8.16)
The integration constants 1C and 2C are calculated by imposing the boundary
conditions of the specific problem at hand.
- 171 -
8.3.2 Integration of the Load-Deflection Differential Equation
Successive integration of the forth-order differential equation (8.12) in a continuity
interval of the transverse loading function yields the following relations:
1
'" *)(])(*)(*)([)( CdxxpxvxIxExV nz (8.17)
21
" ***)()(*)(*)()( CxCdxdxxpxvxIxExM nz (8.18)
321 *)(*)(
1**
)(*)(*
*]**)([)(*)(
1)()(
CdxxIxE
CdxxIxE
xC
dxdxdxxpxIxEdx
xdvx n
z
(8.19)
43
21
***)(*)(
1
**)(*)(
1***
)(*)(*
**]**)([)(*)(
1)(
CxCdxdxxIxE
dxdxxIxE
CdxdxxIxE
xC
dxdxdxdxxpxIxE
xv n
z
(8.20)
The integration constants 1C , 2C , 3C and 4C are calculated by imposing the boundary
conditions applicable to the specific problem.
Note: The moment-curvature equation (8.8) can be used only if the bending moment
variation is known, the case for statically determinate structures. The load-deflection
curve equation (8.14) requires only that the variation of transverse load be known
and, consequently, can be used for either statically determinate or indeterminate
beams.
For the case of a beam with constant bending stiffness constIE z )*( , the integrals
expressed in equations (8.17) through (8.20) may be considerably simplified as
follows:
1
''' *)()(**)( CdxxpxvIExT nz (8.21)
21
" ***)()(**)( CxCdxdxxpxvIExM nz (8.22)
32
2
1 ]*2
*
***)([*
1)()(
CxCx
C
dxdxdxxpIEdx
xdvx n
z
(8.23)
- 172 -
43
2
2
3
1 *]2
*6
*
****)([*
1)(
CxCx
Cx
C
dxdxdxdxxpIE
xv n
z
(8.24)
8.3.3 Boundary and Continuity Integration Conditions
As previously stated, the functions (8.15) though (8.21) obtained above, represent the
general solutions. Only after the boundary conditions are imposed and the integration
constants determined the solutions became representative for a specific case study.
The boundary conditions commonly encountered in the application of the equations
(8.15) and (8.16) or equations (8.21) through (8.24) are presented in Table 8.1.
In general, the transverse load )(xpn and the bending moment )(xM functions are
described for a particular case by a number of continuity intervals. Therefore, the
continuity conditions at the common ends of the intervals must be described in order
for the constants to be calculated. Each continuity interval is treated as an independent
interval and the boundary and continuity conditions are applied. The most common
continuity conditions are summarized in Table 8.2.
Table 8.1 Boundary Conditions
- 173 -
Table 8.2 Continuity Conditions
8.4 Examples
The methodology used in the application and integration of the moment-curvature
(8.8) and load-deflection (8.14) differential equations is illustrated in examples 8.4.1
and 8.4.2, respectively.
8.4.1 Application of the Moment-Curvature Equation
The deflection curve is required for a cantilever beam subjected to a concentrated
force BP and a concentrated bending moment BM both acting at the tip of the beam,
point B. The beam is characterized by a constant cross-section along its entire length,
with geometry and loading as shown in Figure 8.4.
zIE * constant (8.25)
The corresponding reactions, AP and AM , are found using the equilibrium equations:
BA PP (8.26)
LPMM BBA * (8.27)
- 174 -
Figure 8.4 Cantilever Beam
The moment-curvature equation (8.8) requires knowledge of the bending moment
function and identification of the continuity intervals. For the case in point, the
bending moment )(xM diagram is continuous on the entire length of the beam and is
expressed as:
)(**)( xLPMxPMxM BBAA (8.28)
Using equation (8.28) in the moment-curvature equation (8.13), the problem specific
differential equation is obtained:
xIE
P
IE
Mx
IE
P
IE
M
IE
xM
dx
xvdv
z
B
z
A
z
A
z
A
z
***
****
)()("
2
2
(8.29)
Integrating the differential equation (8.29) twice yields the following expressions for
deflection and slope:
1
2***2
**
)()( Cx
IE
Px
IE
M
dx
xdvx
z
B
z
A (8.30)
21
32 ****6
***2
)( CxCxIE
Px
IE
Mxv
z
B
z
A (8.31)
The integration constants 1C and 2C are identified using the boundary conditions at
point A:
for 0x 0)0( x (8.32)
0)0( xv (8.33)
- 175 -
Solving the algebraic equations (8.32) and (8.33) by substitution of equations (8.30)
and (8.31) the integration constants 1C and 2C are found as:
01 C (8.34)
02 C (8.35)
Substituting equations (8.34) and (8.35) into equations (8.30) and (8.31), the final
deflection and slope expressions are obtained:
)2
(***
**
)(x
LxIE
Px
IE
Mx
z
B
z
B (8.36)
)3
(****2
***2
)( 22 xLx
IE
Px
IE
Mxv
z
B
z
B (8.37)
The maximum deflection value is obtained at the tip of the cantilever:
]3
*2*[*
**2
2 LPM
IE
Lv BB
z
B (8.38)
and the corresponding rotation
]2
*[**
LPM
IE
LBB
z
B (8.39)
The deflection curve is a cubic (third-order) polynomial and is schematically plotted
in Figure 8.5
Figure 8.5 Deflection curve
Equations (8.38) and (8.39) may be written for the case when only the concentrated
force BP is considered:
B
z
B PIE
Lv *
**3
3
(8.40)
- 176 -
B
z
B PIE
L*
**2
2
(8.41)
In the absence of the concentrated force BP equations (8.38) and (8.39) become:
B
z
B MIE
Lv *
**2
2
(8.42)
B
z
B MIE
L*
* (8.43)
8.4.2 Application of the Load-Deflection Equation
The simply supported beam shown in Figure 8.6 is subjected to a concentrated
vertical force BP acting at distance a from the fixed support A. The beam has a
constant flexural rigidity along its entire length.
zIE * constant (8.44)
The reaction force and moment at point A are obtained by solution of the equilibrium
equations:
BA PP (8.45)
aPM BA * (8.46)
Figure 8.6 Cantilever Beam
The transverse load has two continuity intervals, AB and BC, and, consequently, the
forth-order differential equation must be integrated for each one of them as:
- 177 -
For interval AB (the interval origin is point A) the transversal load is zero:
0)( xp AB (8.47)
Using equation (8.47) in equations (8.21) through (8.24) yields the following:
AB
ABzAB CxvIExV 1
''')(**)( (8.48)
ABAB
ABzAB CxCxvIExM 21
" *)(**)( (8.49)
ABABAB
z
AB CxCx
CIE
x 32
2
1 ]*2
*[**
1)( (8.50)
ABABABAB
z
AB CxCx
Cx
CIE
xv 43
2
2
3
1 *]2
*6
*[**
1)( (8.51)
For interval BC (the interval origin is point B) the transversal load is also zero:
0)( xpBC (8.52)
Equations (8.21) through (8.24) are written considering equation (8.52) as:
BC
BCzBC CxvIExV 1
"' )(**)( (8.53)
BCBC
BCzBC CxCxvIExM 21
" *)(**)( (8.54)
BCBCBC
z
BC CxCx
CIE
x 32
2
1 ]*2
*[**
1)( (8.55)
BCBCBCBC
z
BC CxCx
Cx
CIE
xv 43
2
2
3
1 *]2
*6
*[**
1)( (8.56)
Eight (8) integration constants must be calculated. Four (4) boundary conditions (at
points A and C) and four (4) continuity conditions (at point B) are employed. The
boundary conditions are expressed as:
Point A at x=0 (interval AB)
0)0( xAB 03 ABC (8.57)
0)0( xvAB 04 ABC (8.58)
- 178 -
Point C at x=b (interval BC)
0)( bxVBC 01 BCC (8.59)
0)( bxM BC 02 BCC (8.60)
The continuity conditions at point B (both intervals) are:
)0()( xvaxv BCAB
BCABAB
z
Ca
Ca
CIE
4
2
2
3
1 ]2
*6
*[**
1 (8.61)
)0()( xax BCAB
BCABAB
z
CaCa
CIE
32
2
1 ]*2
*[**
1 (8.62)
)0()( xMaxM BCAB 0* 221 BCABAB CCaC (8.63)
)0()( xVPaxV BCBAB 011 BC
B
AB CPC (8.64)
Solving the system of algebraic equations (8.57) through (8.64) the integration
constants are calculated:
B
AB PC 1 aPC B
AB *2 03 ABC 04 ABC (8.65)
01 BCC 02 BCC z
BBC
IE
aPC
**2
* 2
3 z
BBC
IE
aPC
**3
* 3
4
(8.66)
Substituting the integration constants (8.65) and (8.66) into equations (8.48) through
(8.51) and (8.53) through (8.56), respectively, the variation of the shear force, bending
moment, slope and deflection for each interval are obtained:
For the interval AB (the interval origin is point A)
BAB PxV )( (8.67)
)(*)( xaPxM BAB (8.68)
)2
(**
*)(
xa
IE
xPx
z
BAB (8.69)
- 179 -
)3
(***2
*)(
2x
aIE
xPxv
z
BAB (8.70)
For the interval BC (the interval origin is point B)
0)( xVBC (8.71)
0)( xM BC (8.72)
z
BBC
IE
aPx
**2
*)(
2
(8.73)
)3
*2(*
**2
*)(
2a
xIE
aPxv
z
BBC (8.74)
The deflection and slope at points B and C are:
z
BB
IE
aPv
**3
* 3
(8.75)
z
BB
IE
aP
**2
* 2
(8.76)
)3
(**2
* 2a
LIE
aPv
z
BC (8.77)
z
BC
IE
aP
**2
* 2
(8.78)
- 180 -
LECTURE 9
Torsion
9.1 Introduction
In the physical sense, torsion of a linear member refers to the action of twisting
(rotation) of a structural member about its longitudinal axis when subjected to a
twisting moment called torque. The twisting moment is a vector collinear with the
member longitudinal axis. Examples of torque are illustrated in Figure 9.1.
Figure 9.1 Torque Examples
In engineering practice, a linear member subjected to torsion is named torsional
member or shaft. The geometrical shape of the member subjected to torsion plays an
important role in the resulting deformation. To substantiate this assertion, two
examples of shafts in torsion, one with circular cross-section and the other with
- 181 -
rectangular cross-section, are illustrated in Figures 9.2 and 9.3. To better identify the
deformation, a two-directional mesh is marked on the lateral surfaces of both
members. The mesh is photographed before and after the application of the torque at
the left end of the member, while the right end is kept fixed.
Figure 9.2 Torsional Deformation of a Bar with Circular Cross-Section
Figure 9.3 Torsional Deformation of a Bar with Rectangular Cross-Section
Examination of Figure 9.2 reveals that adjacent cross-sections rotate about the
longitudinal axis relative to one another without undergoing other type of visible
deformation. In contrast, the member with rectangular cross-section, illustrated in
Figure 9.3, suffers a visible warping of the cross-section in addition to the relative
rotation of adjacent cross-section.
This is an indication that only the torsional-deformation of the members with circular
cross-sections can be treated exactly using simple kinematic assumptions. Members
having more complicated cross-sections solved using the advanced methods of the
Theory of Elasticity.
9.2 Torsional Deformation of a Member with
Circular Cross-Section
9.2.1 Strain-Displacement Equation
Following the observations derived from Figure 9.2 the torsional-deformation of a
member with circular cross-section is schematically illustrated in Figure 9.4.
- 182 -
Figure 9.4 Torsional-Deformation of a Member with Circular Cross-Section
The radial plane ABEF is located in the body of the circular cross-section beam
before the application of torque. After the application of a torsional load T and the
associated deformation, at the fixed end the radius AB remains in its original position,
while the radius EF at the free end rotates to a new position *EF . Similarly, in a
particular cross-section identified by the coordinate x the radius CD rotates, after the
deformation, to the position *CD .The rotation angle measured between the original
position and the deformed position is notated )(x . Consequently, the torsional
deformation of a member with circular cross-section is described by three
fundamental kinematic assumptions.
It is necessary to establish a consistent sign convention for which will be used
throughout the theoretical development. The torque )(xT and rotation angle )(x are
positive on the cross-section if they rotate in the right-hand rule sense of the outer
normal to the cross-section. The sign convention established for the torque )(xT and
angle of rotation )(x is shown in Figure 9.5.
Definition 9.1
A plane linear member when subjected to torsional moment undergoes a torsional
deformation if after the deformation:
(c) the axis of the member remains straight and without longitudinal
extension;
(d) the cross-sections remain plan and perpendicular to the longitudinal
axis of the beam;
(e) radial lines remain straight and radial as the cross-section rotates about
the longitudinal axis of the member.
- 183 -
Figure 9.5 Torque and Rotation Angle Sign Convention
To study the relation between the rotation angle and the shear strain the notation
shown in Figure 9.6.a is employed. Consider the elementary volume of length x and
radius isolated from the body of the beam as shown in Figure 9.6.b.
Figure 9.6 Torsional-Deformation of the Elementary Element
After the rotation of the cross-sections, the original right angle QRS deforms into
angle Q*R*S*, which is no longer a right angle. The shear strain is expressed as:
**'***
2),( SRSSRQx
(9.1)
where x is the position of the cross-section and is varies between zero and the
cross-section radius.
Since ),( x is a small angle its value can be approximated by its tangent:
dx
d
xSR
SSx xx
*
*limlim),( 0'*
'*
0
(9.2)
Equation (9.2) is called the strain-displacement equation for torsional deformation
of a circular cross-section. The shear strain ),( x is a function of the cross-section
position x and the distance measured radially from the center of the circular cross-
section. In a particular cross-section, the shear strain ),( constx varies linearly
with and reaches its maximum value on the cross-section periphery.
- 184 -
Figure 9.7 Shear Strain Distribution for circular Cross-Sections
(a) Solid Cross-Section and (b) Tubular Cross-Sections
The shear strain distribution pertinent to circular and tubular circular cross-sections is
depicted in Figure 9.7.
9.2.2 Constitutive Equation
The constitutive equation reflects, as was describe in Lecture 2, the functional
relationship between stress and strain. If the homogeneous linear elastic material
behavior is considered, the relationship between shear stress and strain is written
according to Hook’s Law as:
),(*),( xGx (9.3)
The material constant G is called shear modulus. Equation (9.3) indicates that the
shear stress follows the linear distribution of the shear strain when the material is
elastic and homogeneous. The distribution of the shear stress in solid and tubular
cross-sections is shown in Figure 9.8.
Figure 9.8 Shear Stress Distribution for circular Cross-Sections
(b) Solid Cross-Section and (b) Tubular Cross-Section
Substituting equation (9.2) into equation (9.3), the shear stress is expressed as a
function of the angle of rotation as:
- 185 -
dx
dGx
**),( (9.4)
Accordingly with the consideration of the duality of shear stress principle, the
torsional deformation induced cross-section shear stress demands the existence of
shear stress acting in the radial-longitudinal planes as illustrated in Figure 9.9. The
shear stress and strain shown in the cross-sectional and longitudinal planes are
pictured in Figures 9.9.a and 9.9.b, respectively.
Note: The orientation of the in-plane cross-section shear stress results from the
positive direction of the torque acting on the cross-section. The orientation of the
longitudinal shear stress on radial planes shown in Figure 9.9.b is a direct result of the
duality principle.
Figure 9.9 Shear Stress and Strain Distribution in Radial Planes
9.2.3 Cross-Section Stress Resultants
The torque )(xT is related with the shear stress ),( x by the following integral
equation:
A
dAxxT *),(*)( (9.5)
Introducing equation (9.4) into equation (9.5):
p
AA
Idx
dGdA
dx
dGdA
dx
dGxT ******]**[*)( 2
(9.6)
where AdIA
p *2 is the polar moment of inertia of the cross-section.
Equation (9.6) can be rewritten as:
- 186 -
pIG
xT
dx
d
*
)(
(9.7)
Equation (9.7) is called the torque-twist equation. Considering the torque )(xT as a
continuous real function on a given interval of continuity, equation (9.7) can be
integrated and the angle of rotation at a particular cross-section is calculated as:
)0(**
)()( xdx
IG
xTx
p
(9.8)
The total angle of twist between the ends of the continuity interval is obtained as the
difference between the angles of rotation at both ends:
L
p
total dxIG
xT
0
**
)( (9.9)
where L is the length of the member.
By combining equations (9.4) and (9.7), a new expression for the shear stress is
obtained:
*)(
),(pI
xTx (9.10)
Equation (9.10) is widely used in the analyses of beams with circular cross-sections.
The maximum value for the shear stress is obtained as:
RI
xTRxx
p
*)(
),()(max (9.11)
where R is the exterior radius of the cross-section.
9.3 Torsional Deformation of a Member with
Closed Thin-Wall Cross-Section
In the previous section, the primary focus of the theoretical development was on
members with solid or thick walled hollow cross-sections. In structural engineering
and especially in the aerospace engineering the usage of members with a closed thin-
wall cross-section is prevalent. The torsional deformation of thin-wall members can
be solved if some simplifying assumptions are imposed. These simplifying
assumptions are:
(a) the member is cylindrical and the cross-section does not vary along the
length;
- 187 -
(b) the cross-section is closed and is comprised of a single-cell with a
small thickness relative to the general dimensions of the cross-section;
(c) the shear stress is constant through the thickness and parallel to the
median curve defining the cross-section;
(d) the member is subjected to end torque only;
(e) the warping of the cross-section is unrestrained at both ends;
The first three assumptions, (a), (b) and (c) are common for the definition of the thin-
wall cross-section and were used before, while the other two have a specific
application for torsion investigation. An example of a single-cell thin-wall torsional
member is illustrated in Figure 9.10.
Note: An important aspect of this discussion is that the cross-section is not required
to be circular.
Figure 9.10 Single Cell Thin-Wall Cross-Section
(a) Geometry and Notation, (b) Shear Flow Distribution and
(c) Shear Stress Distribution
The shear flow q is defined as:
tq * (9.12)
where is the shear stress.
Figure 9.11 Free-Body Diagram of the Thin-Wall Element
Consider the elementary volume element ABCD, shown in Figure 9.10.a and
magnified for clarity in Figure 9.11, isolated from the body of a cylindrical thin-wall
- 188 -
member subjected to torques at both ends. As a result of the duality of shear stress
principle, the existence of shear flow in the transverse cross-sections induces shear
flows in the longitudinal direction. The equilibrium of the shear forces in the
longitudinal direction, in the absence of any forces acting on the exterior surface
planes, is written as:
024 VV (9.13)
Using the notation shown in Figure 9.11, the equilibrium equation (9.13) shows that:
BA qq (9.14)
Equality (9.14) implies that the shear flow q is constant on the entire cross-section
and, consequently, the shear stress is expressed as:
constt
q (9.15)
Equation (9.15) reflects the underlying assumption c that the shear stress is constant
across the thickness of the thin-wall.
Using the notation shown in Figure 9.12, the integral relation between the torque
T and the shear flow q is written as:
mmm CC
sC
dsqdsqFdT ***** (9.16)
It is evident from Figure 9.12.b that the following geometrical relation holds:
dsdA **2
1 (9.17)
Figure 9.12 Thin-Wall Cross-Section Notation
Substituting equation (9.17) into the equilibrium equation (9.17) the torque T is
expressed as:
- 189 -
mAqT **2 (9.18)
where mA is the area enclosed by the median curve.
From equation (9.18) the shear flow is obtained as:
mA
Tq
*2 (9.19)
The shear stress is calculated by substitution of equation (9.19) into equation
(9.15):
tA
T
m **2 (9.20)
9.4 Torsional Deformation of a Member with Rectangular
Cross-Section
As illustrated in Figure 9.3 the prismatic member suffers not only relative the rotation
of adjacent cross-sections, but also warping of the cross-sections when subjected to
torque application. This type of torsional problem is exactly solved using methods of
the Theory of Elasticity. Without going into theoretical details, only some practical
formulae are presented in this section. The distribution of shear stress for a
rectangular cross-section is illustrated in Figure 9.13.
Figure 9.13 Shear Stress Distribution In a Rectangular Cross-Section
It is notable that the shear stress attains a the maximum value at the middle of the
longer edge, while becoming zero at the corners. The maximum shear stress
corresponding to a rectangular cross-section is expressed as:
2max
** td
T
(9.21)
- 190 -
where is the ratio of the dimension d to t satisfying the relation 1td .
The rotation angle is calculated:
JG
LT
*
* (9.22)
where the torsional moment of inertia J is obtained as:
3** tdJ (9.23)
The constants and are listed in Table 9.1.
Table 9.1 Torsional Constants for Rectangular Cross-Section
9.5 Torsional Deformation of a Member With Elliptical
Cross-Section
The shear stress distribution in an elliptical cross-section is illustrated in Figure 9.14.
The maximum value occurs on the periphery of the cross-section at the two ends of
the ellipse minor axis and is calculated using the following formula:
2max
**
*2
ba
T
(9.24)
Figure 9.14 Shear Stress Distribution in the Elliptical Cross-Section
The rotation angle for a member with elliptical cross-section is calculated as:
JG
LT
*
* (9.25)
- 191 -
where the torsional moment of inertia J is obtained as:
22
33 **
ba
baJ
(9.26)
9.6 Examples
9.6.1 Rectangular Thin-Wall Cross-Section
An unknown torque T is applied to an aluminum member with a rectangular thin-
wall cross-section. The geometry of the cross-section is illustrated in Figure 9.15. The
allowable shear stress for the aluminum is 110all MPa .
Figure 9.15 Thin-Wall Rectangular Cross-Section
The unknown torque T is calculated using equation (9.20):
allm tAT ***2
The median curve encloses the area mA as shown in Figure 9.15.b:
232 10*016.2201636*56)2
4
2
432(*)
2
4
2
452( mmmAm
The thickness t of the tube is:
mmmt 004.04
Consequently, the allowable torque for the cross-section is calculated as:
mNT *08.1774)10*110(*)004.0(*)10*016.2(*2 63
If the tube was manufactured such that the thicknesses of the walls varies as depicted
in Figure 9.16, the torque T is calculated as:
- 192 -
qAT m **2
Figure 9.16 Imperfect Thin-Wall Cross-Section
The shear flow q , which is accordingly to equation (9.15) constant, is determined
using the minimum thickness as follows:
mNtq allall /10*85.3)10*110(*0035.0* 56
min
The allowable torque is then calculated as:
mNqAT allm *32.1552)10*110(*)0035.0(*)10*016.2(*2**2 63
9.6.2 Capacity Comparison of Different Shaped Solid Cross-Sections
Three types of solid cross-section shapes are illustrated in Figure 9.17. The allowable
shear stress is the same for all the cross-sections considered.
Figure 9.17 Solid Cross-Sections
(a) Circular, (b) Square and (c) Rectangular
The torsional capacity allT of the circular cross-section is calculated as:
c
IT
allp
all
*
The polar moment of inertia is obtained as:
- 193 -
2
*
32
)2(* 44 ccI p
Consequently, the torsional capacity is:
allall
allp
all cc
c
IT
**57.1*
2
**3
3
The torsional capacity bT of the squared cross-section is calculated using the formula
(9.19) and Table 9.1 as:
allallb aaaT **208.0***208.0 32
The torsional capacity cT of the rectangular cross-section is:
allallc aaaT **141.0*)*5.0(*)*2(*282.0 32
If the condition of equal areas is imposed, the radius of the circular cross-section is
found as:
ac *5642.0
and the capacity aT of the circular cross-section becomes:
allalla acT **282.0**57.1 33
To compare these three cross-sections, the following ratios are calculated considering
the torsional capacity of the circular cross-section as the base for comparison:
7376.0**282.0
**208.01
3
3
all
all
a
b
a
a
T
Tratio
500.0**282.0
**141.02
3
3
all
all
a
b
a
a
T
Tratio
Thus, it is shown that the most effective cross-section for torsional-deformation is the
circular cross-section.
- 194 -
LECTURE 10
Plane Stress Transformation
10.1 Introduction
As discussed in Lecture 2, the general three-dimensional state of stress is obtained, as
was discussed in Lecture 2, by passing a set of three orthogonal planes though a point
of the solid body and isolating the infinitesimal volume located around the point. The
three-dimensional state of stress is illustrated in Figure 10.1 where, for clarity, only
the stresses drawn on the faces with positive normal are represented.
Figure 10.1 Three-Dimensional State of Stress
Mathematically the three-dimensional state of stress is represented by a generalized
stress tensor ),,( zyxT defined by nine distinct stress components:
zzyzx
yzyyx
xzxyx
zyxT
),,( (10.1)
Applying the shear stress duality principle the number of independent tensorial
components is reduced from nine to six and, consequently, the symmetric shear stress
components are related as:
yxxy (10.2)
195
zxxz (10.3)
zyyz (10.4)
There are instances when some of the stress tensor components vanish and the general
three-dimensional stress tensor ),,( zyxT degenerates into a tensor characterized by
only three independent components. This condition is called plane state of stress. For
the present theoretical development, it is assumed that all stress components pertinent
to the planes having normal vector parallel to the z axis are zero:
0z (10.5)
0 xzzx (10.6)
0 yzzy (10.7)
This type of plane stress tensor is illustrated in Figure 10.2.
Figure 10.2 Plane Stress Tensor
Additionally, the remaining non-zero stress tensor components contained in equation
(10.1) are assumed to be independent of the variable z .
),(),,( yxTzyxT (10.8)
Consequently, the plane stress tensor ),( yxT can be represented in plane oxy as
depicted in Figure 10.3:
Figure 10.3 Plane State of Stress
196
The plane stress tensor ),( yxT is defined by three non-zero components:
yyx
xyxyxT
),( (10.9)
Several of the conditions encountered in previous lectures, including the study of the
axial and torsional deformation and pure and non-uniform bending, are characterized
by different states of plane stress. In reality, the plane stress tensor is a direct result of
the assumptions imposed on the deformation.
Examples of plane stress tensors, such as uni-axial, pure shear and bi-axial, are
illustrated in Figure 10.4.
Figure 10.4 Examples of Plane State of Stress
(a) Uni-Axial, (b) Pure Shear and (c) Bi-Axial
10.2 Plane Stress Transformation Equations
Suppose that the components of the plane stress tensor ),( yxT , as expressed by
equation (10.9), are defined at any point ),( yxP of the vertical plane oxy . The
variation of stress components when the reference system attached to point ),( yxP is
rotated with a counterclockwise angle , as illustrated in Figure 10.5, is the subject of
this section.
197
Figure 10.5 Representation of the Plane State of Stress
(a) Normal Planes and (b) Rotated Planes
The transformation relations between stresses n , t and nt pertinent to a rotated
plane and the stresses x , y and xy are obtained by writing the equilibrium
equations for the infinitesimal triangular element depicted in Figure 10.6. The
inclined plane is defined by its positive normal n which is rotated counterclockwise
with an angle from the horizontal direction x .
Figure 10.6 Equilibrium of the Infinitesimal Triangular Element
(a) Stresses and (b) Forces
Using the notation shown in Figure 10.6.b the equilibrium equations are written as:
0 nF
0cos*)*(sin*)*(
sin*)*(cos*)*()*(
yyxxxy
yyxxn
AA
AAA (10.10)
0 tF
0sin*)*(cos*)*(
cos*)*(sin*)*()*(
yyxxxy
yyxxnt
AA
AAA (10.11)
198
From Figure 10.6 the following geometrical relations can be derived:
cos*AAx (10.12)
sin*AAy (10.13)
Substituting equations (10.12) and (10.13) into equations (10.10) and (10.11) and
using the shear stress duality principle the following expressions are obtained for
normal n and shear nt stresses:
cos*sin**2sin*cos* 22
xyyxn (10.14)
)sin(cos*cos*sin*)( 22 xyyxnt (10.15)
Equations (10.14) and (10.15) can be re-written using the trigonometric relations
between the angle and the double angle )*2( :
)*2sin(*)*2cos(*22
xy
yxyx
n
(10.16)
)*2cos(*)*2sin(*2
xy
yx
nt
(10.17)
Equation (10.16) and (10.17) are called the plane stress transformation equations.
In general, two faces are needed to express the plane stress tensor around a point
),( yxP . Consequently, the formulae (10.16) and (10.17) are applied twice: first,
considering the rotation angle 'xx and, secondly, for the complementary
angle 'xy . The notation is illustrated in Figure 10.7. The rotation angles 'xx
and
'xy are related as:
2
''
xxxy (10.18)
The relation between the double angles necessary in equations (10.16) and (10.17) is
obtained as:
'' *2*2xxxy
(10.19)
Consequently, the following trigonometric relations can be established:
)*2sin()*2sin( '' xyxx (10.20)
199
)*2cos()*2cos( '' xyxx (10.21)
Figure 10.7 Stresses on Orthogonal Rotated Faces
The stresses on two orthogonal rotated faces are expressed as:
)*2sin(*
)*2cos(*22
)(
'
'''
xxxy
xx
yxyx
xxnx
(10.22)
)*2cos(*
)*2sin(*2
)(
'
''''
xxxy
xx
yx
xxntyx
(10.23)
)*2sin(*
)*2cos(*22
)90(
'
'''
xxxy
xx
yxyx
xxny
(10.24)
)*2cos(*
)*2sin(*2
)90(
'
''''
xxxy
xx
yx
xxntxy
(10.25)
If equations (10.22) and (10.24) are summed, the invariance of the summation of
normal stresses is established:
yxyx '' (10.26)
200
10.3 Principal Stresses
The maximum and minimum normal stresses are called principal stresses and
mathematically represent the extreme values of the normal stress function )( n . The
extreme values are obtained by imposing the condition that the first derivative of the
normal stress )( n relative to the rotation angle is zero:
0)(
d
d n (10.27)
The explicit expression for equation (10.27) is obtained by differentiating equation
(10.16):
0)*2cos(**2)*2sin(*)( xyyx (10.28)
Dividing by 2* )*2cos( the trigonometric equation (10.28) is transformed into
equation (10.29) relating the tangent of twice the principal directions angle, p , to the
stresses ion the orthogonal planes x and y :
2
)*2tan(yx
xy
p
(10.29)
The angle p represents the angle for which the normal stress )( n reaches its
extreme value. The geometrical illustration of the equation (10.29) is presented in
Figure 10.8.a where the distance R is calculated as:
2
2
4
)(xy
yxR
(10.30)
Figure 10.8 Geometrical Representation of Equation (10.29)
Solution of the trigonometric equation (10.29) yields two solutions, )*2( 1p
and )*2( 2p , where the two angles are related as:
201
12 *2*2 pp (10.31)
From equation (10.31), the orthogonality of the two principal directions is established:
212
pp (10.32)
To calculate the values of the normal stress )( n corresponding to the angles 1p and
2p it is necessary to evaluate the trigonometric functions )*2sin( p and )*2cos( p
contained in equation (10.16). Using the notation shown in Figure 10.8.a and the
trigonometric relations (10.20) and (10.21) these functions can be expressed as:
R
xy
p
)*2sin( 1 (10.33)
R
yx
p2)*2cos( 1
(10.34)
R
xy
pp
)*2sin()*2sin( 12 (10.35)
R
yx
pp2)*2cos()*2cos( 12
(10.36)
Substituting first the trigonometric expressions (10.33) and (10.34) into equation
(10.16) and then (10.35) and (10.36) the principal stresses 1 and 2 are obtained:
Ravxy
yxyx
pn
2
2
114
)(
2)( (10.37)
Ravxy
yxyx
pn
2
2
224
)(
2)( (10.38)
The average normal stress is calculated as:
2
yx
av
(10.39)
The principal stresses are schematically depicted in Figure 10.9.
202
Figure 10.9 Principal Stresses and Directions
The right-hand expression of equation (10.28) being identical with the expression
(10.17), representing the shear stress nt , allows equation (10.28) to be re-written as:
0)(
nt
n
d
d
(10.40)
Equation (10.40) indicates that the principal normal stresses are obtained for a
rotated plane where the shear stress is zero.
The invariance of the sum of the normal stresses is again shown to be valid for the
case of the principal stresses. Summation of equations (10.37) and (10.38) yields:
yxav *221 (10.41)
To identify which of the two angles, 1p or 2p , corresponds to the maximum
principal stress 1 the second derivative of the function )( n relative to the rotation
angle is employed. The condition for the point to be a maximum is:
0)()(
2
2
p
n
d
d
(10.42)
The condition (10.42) is explicitly written as:
0)*2sin(*)*2cos(*2
pxyp
yx
(10.43)
The inequality (10.43) can be manipulated and cast in a new form:
0)*2sin(*])*2tan(
1*
2[
pxy
p
yx
(10.44)
If the )*2sin( p is expanded the following trigonometric expression is established:
203
2)(cos*tan*2cos*sin*2)*2sin( ppppp (10.45)
Substituting equations (10.29) and (10.45), representing the )*2tan( p and
)*2sin( p , into the inequality (10.44) the following expression is obtained:
0tan
*cos*]4
)([*2 22
2
xy
p
pxy
yx
(10.46)
The condition for the inequality (10.46) to hold true is:
0tan
xy
p
(10.47)
Note: It is important to note that for the inequality (10.47) to hold true, the signs of
the tangent of the angle p and shear stress xy must be identical.
The angle corresponding to the direction of the maximum normal stress can also be
obtained by successively assigning to the angle in equation (10.16) the values 1p
and 2p and observing which angle produces the maximum principal stress.
10.4 Maximum Shear Stresses
The maximum shear stresses are determined in a similar manner as the principal
stresses. The extreme condition for the shear stress function )( nt contained in
equation (10.17) is written as:
0)(
)(
d
d nt (10.48)
The explicit format of equation (10.48) is obtained by differentiating the expression
(10.17):
0)*2sin(**2)*2cos(*)( xyyx (10.49)
Dividing by )*2cos(*2 the trigonometric equation (10.49), the tangent of the
principal directions angle s is obtained:
)*2tan(
12)*2tan(pxy
yx
s
(10.50)
204
The angle s represents the angle for which the shear stress )( nt reaches its extreme
value. Equation (10.50) is illustrated in Figure 10.10.
Figure 10.10 Angular Relation between )*2( s and )*2( p
Solving the trigonometric equation (10.50), two solutions )*2( 1s and )*2( 2s are
obtained. They are related as:
12 *2*2 ss (10.51)
Dividing equation (10.50) by two (2), the orthogonality of the two angles 1s and 2s is
obtained:
212
ss (10.52)
Equation (10.50) indicates that a relation between the angles )*2( p and )*2( s can
be established. With this intent, equation (10.50) is recast into a new format as
follows:
0)*2sin(
)*2cos(
)*2cos(
)*2sin(
p
p
s
s
(10.53)
First, multiplying by the terms in the denominator, equation (10.53) becomes:
0)*2cos(*)*2cos()*2sin(*)*2sin( spps (10.54)
Then, equation (10.54) is simplified as:
0)]*2()*2cos[( ps (10.55)
Therefore,
2
)]*2()*2(
ps (10.56)
205
The relationship between angles s and p is calculated from equation (10.56):
4
ps (10.57)
Still, equation (10.57) does not indicate how to identify the direction of the maximum
shear stress. By examination of Figure 10.10, the following angular relations can be
established:
2
*22
*3*2
2*2*2 1111
ppps (10.58)
The relationship between the angles of the maxim principal and shear stress
directions, s and p , is obtained from (10.58) as:
4
11
ps (10.59)
Again using the notation shown in Figure 10.10, the following trigonometric relations
are obtained:
R
yx
s2)*2sin( 1
(10.60)
R
xy
s
)*2cos( 1 (10.61)
R
yx
s2)*2sin( 2
(10.62)
R
xy
s
)*2cos( 2 (10.63)
Successively substituting the two groups of expressions, (10.60) and (10.61), and,
(10.62) and (10.63), into equation (10.17) the maximum and minimum shear stresses
are calculated as:
max
2
2
114
)()(
Rxy
yx
snts (10.64)
min
2
2
224
)()(
Rxy
yx
snts (10.65)
206
The normal stresses corresponding to the maximum and minimum shear stresses are
calculated by substituting the expressions (10.60) through (10.63) into equation
(10.16):
2)( 11
yx
sns
(10.66)
2)( 22
yx
sns
(10.67)
The maximum and minimum shear stresses and the corresponding normal stresses are
illustrated in Figure 10.11.
Figure 10.11 Relationships between Principal Planes and Maximum Shear Stress
Planes
Note: From Figure 10.12.b it can be concluded that, in contrast, to the principal
planes which are free of shear stress, the planes on which the shear stress
achieves extreme values are not necessarily free of normal stresses.
10.5 Mohr’s Circle for Plane Stresses
Mohr’s circle is a graphical construction reflecting the variation of the plane state of
stress around a particular point, including information pertinent to the principal and
maximum shear stresses.
From equations (10.16) and (10.17), first squared and then summed, the following
relation is obtained:
222)( Rntavn (10.68)
207
Equation (10.68) represents the equation of a circle of radius R defined in the ntn ,
plane. The center of the circle is located at point )0,( avC . The values of circle
radius R and average normal stress av are calculated employing equations (10.30) and
(10.39), respectively.
Intersecting the equation of the circle (10.68) with the horizontal axis 0nt the
intersection points )0,( 11 nP and )0,( 22 nP are obtained:
11 Ravn (10.69)
22 Ravn (10.70)
It can be concluded that these intersection points represent the principal stresses
1 and 2 .
The Mohr’s circle for plane stress condition is drawn relative to a Cartesian system
with the abscissa and the ordinate axis representing the normal stresses and the
shear stress , respectively. The following sign convention is employed as illustrated
in Figure 10.12:
(a) the positive shear stress axis is downward;
(b) the positive angle is measured counterclockwise;
(c) the shear stress on a face plots as positive shear if tends to rotate the face
counterclockwise.
Figure 10.12 Mohr’s Circle Notation
208
Note: The positive direction of the vertical axis, representing the shear stress ,
pointing downward (sign convention (a)) is elected in order to be able to
enforce the positive measurement of the angle (sign convention (b)).
Examining the notations shown in Figure 10.12 clarifies that all the angles are
measured from the line XY ( 0 ) in the anticlockwise direction.
Morh’s circle for plane stress is constructed in the following steps:
(f) The coordinate system is drawn as shown in Figure 10.12. The horizontal axis
represents the normal stress , while the vertical axis represents the shear
stress . To gain full advantage of the graphical benefits of the method it is
necessary that the drawing to be made on scale. However, the method is also
helpful as a conceptual tool in combination with the governing equations
wherein it may be drawn more roughly. The representation considers that the
following conditions are met: yx and 0xy ;
(g) Using the calculated values of the normal stresses x and y and the shear
stress xy two points noted as ),( xyxX and ),( xyyY are placed on the
drawing. The line XY intersects the horizontal axis at point C which
represents the center of the Mohr’s circle;
(h) The distance CX represents the radius of the circle. Using the radius CX and
the position of the center )0,( avgC the Mohr’s circle is constructed. The
intersection points, 1P and 2P , between the circle and the horizontal axis
represent the maximum and the minimum principal stresses;
(i) The value of the )*2tan( p can be calculated from the graph. The angle
)*2( p is identified on the graph by the 1XCP angle and is measured from
the 0 to the principal directions line in the counterclockwise direction;
(j) The lines 1XP and 2XP represent the principal direction1 (associated with the
maximum principal stress) and 2 (associated with the minimum principal
stress), respectively.
Every point on the Mohr’s circle corresponds to a pair of stresses and on a
particular face. To emphasize the face involved the point is labeled identically with
the face where it belongs. For example, the face x , y and n are represented on the
Mohr’s circle by the points X , Y and N . To reinforce the shear stress sign convention
(c) two icons indicating the rotation sense induced by the shear stress are shown in
Figure 10.12. The angle measured from 0 to the radius line CN in the
counterclockwise direction is equal to twice the angle of the plane rotation )*2( .
Note: The points X andY represent the case of orthogonal planes having normals
parallel to axes x and y , respectively. The line XY corresponds to angle
209
0 . The double angle )*2( 1p of the maximum principal direction is
measured from the line CX to the line 1CP . By these conventions, the double
angle sense is established as being positive in the counterclockwise direction.
The angle 1p is the angle 12 PXP and has the same direction as the double
angle )*2( 1p . The angle associated with the minimum direction 2p is
perpendicular to the angle 1p . The stresses corresponding to a plane rotated
with an angle are obtained by placing on the circle the radius CN located by
measuring in the counterclockwise direction an angle of )*2( from the
lineCX . The corresponding stresses n and nt are a function of the location
of the point ),( ntnN position in the coordinate system and may be
obtained by scaling them from the figure or by the use of the analytical
equations (10.30), (10.39) and (10.68). The opposite point
),( ntnT represents the stresses on the orthogonal rotated face.
In comparison with the technique used to show the principal axes in the oxy
representation (Figures 10.7 through 10.11) the plot obtained from the Mohr’s circle
appears to be misleading. The cause is that the Mohr’s circle is drawn in the
coordinate system. In the oxy representation the principal directions are correctly
plotted by artificially rotating the principal directions obtained from the Mohr’s
circle around the point C with an angle )*2( 1p measured in the counterclockwise
direction.
10.6 Principal Stresses Distribution in Beams
One of the most important applications of the plane state of stress theory described
above is found in the study of variation of the stresses in beams under non-uniform
bending. Recall from Lecture 7 that under some imposed kinematic assumptions, a
beam subjected to transversal loading is in a state of plane stress. With the exception
of some areas (around the supports or the application points of concentrated loads) the
beam theory characterizes the existence of only two types of stresses: normal stress
x and shear stress xy . The normal stress x is calculated using Navier’s formula
expressed by equation (10.71), while the shear stress xy is obtained employing
Jurawski’s formula contained in equation (10.72):
yI
xMyx
z
zx *
)(),( (10.71)
tI
ySxVyx
z
A
zy
xy*
)(*)(),(
'
(10.72)
210
The notation used in the formulae (10.71) and (10.72) is explained in Lecture 7 and is
not repeated herein.
The plane stress tensor previously expressed in equation (10.9) is written for the case
of the beam in nonuniform bending as:
0
),(yx
xyxyxT
(10.73)
The entire theoretical development described in the previous sections can be without
restriction applied to the study of the particular plane stress tensor (10.73).
Consequently, the variation of the stresses around any point in a beam subjected to
nonuniform bending can be calculated. Figure 10.13 represents an example of the
application of plane stress theory for the case of a simply supported beam.
In the example, the beam has a rectangular cross-section and is subjected to a single
concentrated force P*2 located at the mid-span. It is evident that the ratios of the
beam dimensions and the loading do not violate any of the assumptions related to the
applications of the formulae (10.71) and (10.72). The shear force diagram )(xV and
the bending diagram )(xM , where the axis identification indices were dropped for
clarity, are plotted.
Figure 10.13 Simple Supported Beam
The geometrical characteristics of the rectangular cross-section involved in the
evaluation of the formulae (10.71) and (10.72) are:
33
**3
2
12
)*2(*cb
cbI z (10.74)
211
bt (10.75)
)1(***2
1
2*)(*)(
2
22
c
ycb
cyycbyS z
(10.76)
For the left half of the beam cx 50 the shear force )(xV and the bending moment
)(xM are expressed as:
PxV )( (10.77)
xPxM *)( (10.78)
Substituting equations (10.74) through (10.78) into equations (10.71) and (10.72), the
expressions for normal and shear stresses are obtained as:
yxcb
Py
cb
xPyxx **
**
2
3*
**3
2
*),(
33
(10.79)
)1(**
*4
3
***3
2
)1(***2
1*
*
)(*)(),(
2
2
3
2
22'
c
y
cb
P
bcb
c
ycbP
tI
ySxVyx
z
A
zy
xy
(10.80)
Note: The minus (-) sign appearing in formula (10.81) has been inserted in order to
comply with the shear sign convention (c).
To obtain an illustrative variation of the principal stresses, the rectangular domain of
the beam is divided by superimposing a rectangular mesh. For the case under study,
the mesh has five spaces in the longitudinal x direction and eight spaces in the
vertical y direction. Using Mathcad programming capabilities the principal stresses
and corresponding angular directions can be easily calculated for every point of the
mesh. The principal stresses calculated for two cross-sections cx and cx *4 and
all nine points vertically describing the cross-sections are contained in Table 10.1.
The ratios 0
max
and
0
min
are tabulated instead of the max and min ,
wherecb
P
**20 .
A review of the results presented in Table 10.1 shows that at the extreme fibers the
principal stresses correspond with the normal stresses and reach the maximum values.
At the extreme fiber locations the shear stress xy is zero. The situation is different for
212
the case of wide-flange beams where both x and xy have significant values at the
junction between the web and the flange.
Table 10.1
Today, with the help of modern computer codes, the formulae involved in the
calculation of the principal stresses and directions can be computed using a very
refined mesh. The graph containing the curves tangent to the principal directions in
every point of the mesh is called the stress trajectory. Two sets of curves are drawn
and they are orthogonal at every point. The stress trajectory graph pertinent to the
simply supported beam investigated above is pictured in Figure 10.14. A typical
example of practical usage of the stress trajectory curves is the placement of the
reinforcement in reinforced concrete beam. Because the stress trajectory graph does
not give any indication about the magnitude of the principal stresses another type of
graph is also used. This is called a stress contour plot and contains curves of equal
principal stress magnitudes. The commercial codes employed today can provide these
plots.
213
Figure 10.14 Stress Trajectory Plot
10.7 Example
The theoretical formulation derived above is used to investigate the following
practical case:
20x MPa
10y MPa
10xy MPa
The corresponding stress tensor is written as:
MPayxT *1010
1020),(
The state of stress for the case above is shown in Figure 10.13.
Figure 10.13 Example Plane State of Stress
The following values illustrated in Figure 10.13 are calculated as:
MPayx
avg 52
)10(20
2
214
MPayx
152
)10(20
2
MPaR xy
yx028.18)10(15
4
)(222
2
667.015
10
2
)*2tan(
yx
xy
p
69.33*2 p
Figure 10.14 Geometrical Relations
From equation (10.47) it is established that the angle related to the maximum
principal direction must have a negative tangent. Consequently, the angles of the
principal direction are:
84.161 p
16.739084.162 p
The principal stresses, shown in Figure 10.15, are obtained as:
MPaRavg 028.23028.1851
MPaRavg 028.13028.1852
215
Figure 10.15 Principal Stresses
The angle of the maximum shear stresses is calculated as:
84.614584.164511 ps
The maximum shear stresses are calculated as:
MPaxy
yx
snts 028.184
)()(
2
2
11
MPaxy
yx
snts 028.184
)()(
2
2
22
The normal stresses acting on the maximum shear planes are calculated as:
MPayx
sns 52
)( 11
MPayx
sns 52
)( 22
Figure 10.16 Maximum Shear Stresses
216
The maximum shear stresses and the corresponding normal stresses are illustrated
Figure 10.16
Morh’s circle pertinent to the problem is illustrated in Figure 10.17.
Figure 10.17 Mohr’s Circle
Note: The points 'X and 'Y represent the case of orthogonal planes having the
normal directions rotated with angles of 30 and 120 , respectively, from the
x axis. Successively substituting the above angular values in equations
(10.16) and (10.18) the following stresses pertinent to points 'X and 'Y are
obtained:
for 30
MPaX
84.3'
MPaYX
99.17''
for 120
MPaY
16.6'
MPaXY
99.17'''
217
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Bucuresti, Romania, 1974.
2. Craig, Roy R., Mechanics of Materials, Second Edition, John Willey &
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3. Gere, James M., Mechanics of Materials, Fifth Edition, Brooks/Cole,
Pacific Grove, CA, 2001.
4. Higdon, A., Ohlsen, E.H., and Stiles, W.B., Mechanics of Materials, Third
Edition, John Willey & Sons, New York, 1962.
5. Popov, E.P., Introduction to Mechanics of Solids, Prentice-Hall Inc.,
Englewood Cliffs, New Jersy, 1968.
6. Timoshenko, A., and Goodier, J.N., Theory of Elasticity, Third Edition,
McGraw-Hill Inc., New York, 1970.
7. Timoshenko, S.P., Strength of Materials, Part 1, Elementary Theory and
Problems, Third Edition, CBS Publishers & Distributors, New Deli, India,
2002.
8. Hartog, J.P.D., Strength of Materials, Dover Publications Inc., New York,
1961.