Module 1 : Introduction : Review of Basic Concepts in MechanicsLecture 1 : Introduction

Objectives

In this course you will learn the following

Introduction to structural mechanics, with respect to previous courses of engineering.

Solid mechnics; scope of structural mechanics, example of different structure types and load types.

1.1Introduction

Structural Mechanics can be briefly described as the study of the behaviour of structures using the knowledge ofmechanics. Such a description needs some understanding of the terms “structure” and “mechanics”. Structuresinclude a wide variety of systems, such as buildings, bridges, dams, aircrafts, etc., that are built to serve somespecific human needs (for example, habitation, transportation, storage, etc.). Students of Structural Mechanicsshould already have some basic knowledge of mechanics through the prerequisite courses of EngineeringMechanics (or Rigid-body Mechanics or Vector Mechanics) and Solid Mechanics (Mechanics of Deformable Solidsor Mechanics of Materials). In Structural Mechanics, we apply our knowledge of the mechanics of rigid bodies andof deformable solids to the understanding of the behaviour of engineered structures.

In Structural Mechanics, we mostly deal with mechanics of solids (i.e. deformable bodies). However, here wemove on from studying the behaviour of structural members/materials (as in a course of Solid Mechanics) tostudying the behaviour of real structures, or parts thereof. For example, instead of dealing with a beam or acolumn, we study how a building frame (Figure 1.1), composed of several beams and columns, behaves. In asimilar way, we first learn about the loads that are applied to the whole structure, and not to individualmembers. Our knowledge of Structural Mechanics enables us to find the forces that act on individual membersbased on the loads that are acting on the whole structure. Stresses, strains, internal forces and deformations inmembers, then, can be obtained by using what we have already learned about the behaviour of deformablesolids.

Fig. 1.1 Frame in a building

Contents of this course of Structural Mechanics will focus on Civil Engineering structures only. Such structures areclassified into various categories depending on the system/mode of classification:

(a)On the basis of its intended function/usage: Buildings, bridges, dams, industrial sheds, cable ways,chimneys, etc. (Figure 1.2)

(b)On the basis of its form/load transfer mechanism: Beams, columns, floor slabs, arches, shells, trusses,

frames,footings, etc. (Figure 1.3) (c)Considering the analysis perspective: 2-dimensional, 3-dimensional, determinate, indeterminate, etc.

(Figure 1.4)

(a) Building

(b) Bridge

(c) Dams (d) Industrial sheds

(e) Cable ways (f) Chimneys

Figure 1.2 Various types of structures

(a) Building frame

(b) Arched Bridge

(c) Trussed Bridge Figure 1.3 Various structural forms

(a) 2-dimensional structure (b) 3-dimensional structure

Figure 1.4 2-dimensional and 3-dimensional model of structures

Similarly loads are also put into different categories, based on various criteria:

a(a)Based on the source/origin: Wind load, earthquake load, self weight, live load, blast load, temperaturestress, etc.

a(b) Based on the direction of action: Gravity loads, lateral loads, etc. (Figure 1.5)

a(c) Based on time-variation: Static, dynamic, impulse, pseudo-static, etc.

a(d) Based on the mode of action/analysis point-of-view: Concentrated or point load, distributed load, moment,pressure, aaaaetc. (Figure 1.6).

aaaa(a) Lateral loads aaaa(b) Gravity loadsFigure 1.5 Load types based on direction of action

(a) Concentrated or point loads on(b)

Distributed loads on a simply

a simply supported beam supported beam

(c) Pressure acting on inner surface of a cylinder Figure 1.6 Load types based on analysis point of view Recap In this course you have learnt the following

Introduction to structural mechanics, with respect to previous courses of engineering.

Solid mechnics; scope of structural mechanics, example of different structure types and load types.

Module 1 : Introduction : Review of Basic Concepts in MechnicsLecture 2 : Equilibrium

Objectives

In this course you will learn the following

Review of the concepts of equilibrium.

Static equilibrium equations in 3-D and 2-D.

Concept of free body diagrams.

1.2 Equilibrium

The concept of equilibrium is the most central one in the subject of Statics. When the net effect or the resultantof all the forces (and couples) acting on a system is zero, the system is said to be in equilibrium. Thus, based onthe resultant of all the forces R , and the resultant of all the moments (couples) M , the vector equations ofequilibrium are

(1.1)

The two vector equations of equilibrium can be expressed alternatively as scalar equations of equilibrium for asystem of forces in 3 dimensions ( x , y & z ), as

(1.2)

(1.3)

Here, represents the algebraic summation of components of all the forces in x-direction. This summation

is same as the resultant (net effect) of all the forces in x-direction.

This set of six equilibrium equations can be narrowed down to three scalar equations in case of a planer forcesystem (forces acting in two dimensions only)

(1.4)

Figures 1.7 & 1.8 illustrate how resultants are obtained for a two-dimensional (planer) force system.

Figure. 1.7 Obtaining resultants for a truss

Figure 1.8 Obtaining resultants for a cantilever system

These equations provide the necessary and sufficient forces to keep a system in equilibrium. The omission of a forcethat is acting on a system or the inclusion of a force that is not acting on the system produces erroneous results inanalyzing the behaviour of the system. Hence, it is of utmost importance to understand exactly what the mechanicalsystem under consideration is and the forces that are acting on the specific system. A system is a body or acombination of connected bodies. The bodies can be either rigid or deformable (even fluids can be treated as body).For Structural Mechanics, we will restrict ourselves to the study of rigid and deformable solids only. For theimportant task of identifying the forces (and couples) acting on a system, we take the help of Free Body Diagrams.Thus, drawing a free body diagram becomes the first and foremost task in solution of problems in mechanics.

The free body diagram of a body (or its part, or a connected system of bodies) is obtained by isolating it from the allother surrounding bodies. The diagram detaches the system in consideration from all mechanical contacts with otherbodies and sets it free . The other bodies are not shown in the diagram, but they are replaced by the forces (andcouples) that they apply on the system for which we are drawing a free body diagram. The following examples showhow to obtain the free body diagram for a system and also the equilibrium equations for the same system.

Recap In this course you have learnt the following

Review of the concepts of equilibrium.

Static equilibrium equations in 3-D and 2-D.

Concept of free body diagrams.

Module 1 : Introduction : Review of Basic Concepts in MechnicsLecture 3 : Constitutive Relations and Compatibility Conditions

Objectives

In this course you will learn the following

Review of the concept of constitutive relations.

Stress-strain diagrams.

Definitionof prameters related to material properties : - Modulus of elasticity, poison's ratio, shear modulusetc.

Hooke's law.

Stress and strain tensors.

Review of the concepts of compatibility conditions; Interpretation as boundary conditions.

1.3 Constitutive Relations

Equilibrium equations help us obtain the forces that are acting, both internally and externally, at various parts ofa body. However, for deformable solid bodies, understanding their deformation behaviour under the givenstress/loading condition (based on the equilibrium) is of primary importance. The deformation behaviour in sucha system is studied through various parameters, such as strain, displacement, rotation, etc. These deformationparameters are obtained based on the stress-strain relations of the material which the deformable solid is madeof. These are known as Constitutive Relations and are material-specific. The stress-strain diagram for ductilesteel (Figure 1.9) based on a tension test is an example of constitutive relations. It gives us a relation betweenthe engineering (tensile) stress ( ) and engineering (tensile) strain ( ) for ductile steel at different stress (orstrain) values.

Figure 1.9 Stress-strain diagram for ductile steel

Similar stress-strain diagram can be obtained (through experiments) for different materials (aluminium, wood,tool steel, concrete, etc.) and for different types of deformation (uniaxial tensile and compressive, shear,transverse, dilatational, etc.). For the ease of use, these relations are idealized into simple mathematical rules.In Structural Mechanics, we will limit ourselves to linear elastic isotropic homogeneous materials only.

A material is called linear elastic if its stress-strain relation is linear and if when the material is unloaded ittraces back the same stress-strain (loading) path. In other words, stress is a single-valued linear function ofstrain. The behaviour of ductile steel from point “O” to “A” (Figure 1.9) is a linear elastic one. A material will beisotropic if its constitutive relations are non-directional (same for any direction in space, x , y or z ) and it will behomogeneous if it displays the same properties (e.g. a constitutive relation) at any point of the system (sameproperties at [ ] and [ ]). Some basic constitutive relations for a linear elastic isotropichomogeneous material are briefly discussed in the following sections.

1.3.1 Modulus of Elasticity

Hooke's Law provides us the relation for uniaxial stress

(1.5)

The constant of proportionality is called the elastic modulus , modulus of elasticity or Young's modulus . Since isdimensionless the unit of E is same as that of uniaxial stress (e.g. ).

1.3.2 Poison's Ratio

Uniaxial forces case strains not only in its direction, but also in the transverse/lateral directions. For a tensile strainin the axial direction, there will always be compressive strains in the lateral directions, and vice versa. Poisson'sRatio ( ) relates the lateral strains to the axial strain

(1.6)

Note that this ratio is always a dimensionless positive number.

1.3.3 Coefficient of Linear Thermal Expansion

Linear thermal strain ( ) due to change in temperature ( ) is obtained by using this coefficient ( )

(1.7)

has units of per degrees Centigrade (or Fahrenheit)

1.3.4 Shear Modulus

For shear stress ( ) and shear strain ( ), we have a constitutive relation similar to the Hooke's Law for linearstress and strain.

(1.8)

The constant of proportionality ( G ) is known as the shear modulus or modulus of rigidity . It has same units asmodulus of elasticity ( E ). It can be proved that:

(1.9)

1.3.5 Dilatation and Bulk Modulus

Dilatation ( e ) is defined as the change of volume per unit volume

(1.10)

If a three-dimensional body is subjected to uniform hydrostatic pressure p , then the ratio of this (compressive)pressure to the dilatation is known as the bulk modulus ( k )

(1.11)

k is also called the modulus of compression .

1.3.6 Generalized Hooke's Law

This is an extension of the Hooke's Law to three dimensions considering both linear and shears deformations. It isbased on the generalized definitions of strain. As for the Hooke's Law for linear strain/deformation, the equations forGeneralized Hooke's Law are applicable for linear elastic isotropic homogeneous materials only. The 6 equations forlinear and shear strains are:

(1.12a)

(1.12b)

(1.12c)

(1.12d)

(1.12e)

(1.12f)

which can also be expressed alternatively as expressions for stress:

(1.13a)

(1.13b)

(1.13c)

(1.13d)

(1.13e)

(1.13f)

where and are the Lame parameters which are related to the Young's modulus E and Poisson's ratio :

(1.14)

These equations can also be expressed as relation between the stress and strain tensors

Stress tensor Strain tensor

Note that, in a strain tensor, the shear strain (e.g. ) is replaced by the pure or irrotational shear strain (

).

1.4 Compatibility Conditions

Compatibility conditions represent restriction on deformations at specific locations in a system. The location canbe both inside the system and at its boundary. The deformations in a system have to be compatible with thegeometry of the surrounding (both external and internal), and this compatibility is assured through theseconditions. In other words, compatibility conditions specify that deformations in a member/part of a system haveto be compatible with the support conditions (external), as well as with other members/parts of the system(internal). For example, in the case of bar ABC in (Figure 1.10), various compatibility conditions on horizontaldisplacements are:

Figure. 1.10 Axially loaded bar ABC

(1.15)

(1.16)

(1.17)

Where is the deflection of bar AB at point B.

The deformation behaviour of a structural element is usually expressed through differential equations and theassociated compatibility conditions are represented as boundary conditions for those equations.

Recap In this course you have learnt the following

Review of the concept of constitutive relations.

Stress-strain diagrams.

Definitionof prameters related to material properties : - Modulus of elasticity, poison's ratio, shear modulusetc.

Hooke's law.

Stress and strain tensors.

Review of the concepts of compativility conditions; Interpretation as boundary conditions.

Module 1 : Introduction : Review of Basic Concepts in MechnicsLecture 4 : Static Indeterminacy of Structures

Objectives In this course you will learn the following

Review of the concepts of determinate and indeterminate structures.

Unstable systems.

Degree of static indeterminacy : - External and internal.

1.5 Static Indeterminacy of Structures

If the number of independent static equilibrium equations (refer to Section 1.2) is not sufficient for solving for allthe external and internal forces (support reactions and member forces, respectively) in a system, then the systemis said to be statically indeterminate . A statically determinate system, as against an indeterminate one, is that forwhich one can obtain all the support reactions and internal member forces using only the static equilibriumequations. For example, for the system in Figure 1.10, idealized as one-dimensional, the number of independentstatic equilibrium equations is just 1 ( ), while the total number of unknown support reactions are 2 (

), that is more than the number of equilibrium equations available. Therefore, the system is considered

statically indeterminate. The following figures illustrate some example of statically determinate (Figures 1.11a-c)and indeterminate structures (Figures 1.12a-c).

Figure 1.11 Statically determinate structures

Figure 1.12 Statically indeterminate structures

In Section 1.2, the equilibrium equations are described as the necessary and sufficient conditions to maintain theequilibrium of a body. However, these equations are not always able to provide all the information needed to obtainthe unknown support reactions and internal forces. The number of external supports and internal members in asystem may be more than the number that is required to maintain its equilibrium configuration. Such systems areknown as indeterminate systems and one has to use compatibility conditions and constitutive relations in addition toequations of equilibrium to solve for the unknown forces in that system.

For an indeterminate system, some support(s) or internal member(s) can be removed without disturbing itsequilibrium. These additional supports and members are known as redundants . A determinate system has the exactnumber of supports and internal members that it needs to maintain the equilibrium and no redundants. If a systemhas less than required number of supports and internal members to maintain equilibrium, then it is consideredunstable .

For example, the two-dimensional propped cantilever system in (Figure 1.13a) is an indeterminate system because itpossesses one support more than that are necessary to maintain its equilibrium. If we remove the roller support atend B (Figure 1.13b), it still maintains equilibrium. One should note that here it has the same number of unknownsupport reactions as the number of independent static equilibrium equations. The unknown

reactions are , and (Figure 1.13c) and the equilibrium equations are:

(1.18)

(1.19)

(1.20)

An indeterminate system is often described with the number of redundants it posses and this number is known as itsdegree of static indeterminacy . Thus, mathematically:

Degree of static indeterminacy = Total number of unknown (external and internal)forces

- Number of independent equations of equilibrium

(1.21)

It is very important to know exactly the number of unknown forces and the number of independent equilibriumequations. Let us investigate the determinacy/indeterminacy of a few two-dimensional pin-jointed truss systems.

Let m be the number of members in the truss system and n be the number of pin (hinge) joints connecting thesemembers. Therefore, there will be m number of unknown internal forces (each is a two-force member) and 2 nnumbers of independent joint equilibrium equations ( and for each joint, based on its free

body diagram). If the support reactions involve r unknowns, then:

Total number of unknown forces = m + r

Total number of independent equilibrium equations = 2 n

So, degree of static indeterminacy = ( m + r ) - 2 n

For the trusses in Figures 1.14a, b & c, we have:

Figure. 1.14a Determinate truss

1.14a: m = 17, n = 10, and r = 3. So, degree of static indeterminacy = 0, that means it is a statically determinatesystem.

Figure 1.14b (Internally) indeterminate truss

1.14b: m = 18, n = 10, and r = 3. So, degree of static indeterminacy = 1.

Figure 1.14c (Externally) indeterminate truss

1.14c: m = 17, n = 10, and r = 4. So, degree of static indeterminacy = 1.

It should be noted that in case of 1.14b, we have one member more than what is needed for a determinate system(i.e., 1.14a), where as 1.14c has one unknown reaction component more than what is needed for a determinatesystem. Sometimes, these two different types of redundancy are treated differently; as internal indeterminacy andexternal indeterminacy . Note that a structure can be indeterminate either externally or internally or both externallyand internally.

We can group external and internal forces (and equations) separately, which will help us understand easily the casesof external and internal indeterminacy. There are r numbers of external unknown forces, which are the supportreactions components. We can treat 3 system equilibrium equations as external equations. This will lead us to:

Degree of external static indeterminacy = r - 3.

The number of internal unknown forces is m and we are left with (2 n -3) equilibrium equations. The 3 systemequilibrium equations used earlier were not independent of joint equilibrium equations, so we are left with (2 n - 3)equations instead of 2 n numbers of equations. So:

Degree of internal static indeterminacy = m - (2 n - 3).

Please note that the above equations are valid only for two-dimensional pin-jointed truss systems. For example, forthree-dimensional ( “space” ) pin-jointed truss systems, the degree of static indeterminacy is given by ( m + r - 3 n). Similarly, the expression will be different for systems with rigid (fixed) joints, frame members, etc.

Recap

In this course you have learnt the following

Review of the concepts of determinate and indeterminate structures.

Unstable systems.

Degree of static indeterminacy : - External and internal.

Module 1 : Introduction : Review of Basic Concepts in Mechanics

Lecture 5 : Symmetry and Antisymmetry

Objectives

In this course you will learn the following

Concept of symmetry, asymmetry and antisymmetry in structures.

Symmetry and antisymmetry in equilibrium and compatibility conditions.

Use of symmetry and antisymmetry in analyzing a structure.

1.6 Symmetry and Antisymmetry

Symmetry or antisymmetry in a structural system can be effectively exploited for the purpose of analyzingstructural systems. Symmetry and antisymmetry can be found in many real-life structural systems (or, inthe idealized model of a real-life structural system). It is very important to remember that when we saysymmetry in a structural system, it implies the existence of symmetry both in the structure itself includingthe support conditions and also in the loading on that structure. The systems shown in Fig. 1.15 aresymmetric because, for each individual case, the structure is symmetric and the loading is symmetric aswell. However, the systems shown in Fig. 1.16 are not symmetric because either the structure or theloading is not symmetric.

Figure 1.15 Symmetric structural systems

Figure 1.16 Non-symmetric (asymmetric) structural systems

For an antisymmetric system the structure (including support conditions) remains symmetric, however, the

loading is antisymmetric. Fig. 1.17 shows examples of antisymmetric structural systems.

Figure 1.17 Antisymmetric structural systems

It is not difficult to see that the deformation for a symmetric structure will be symmetric about the sameline of symmetry. This fact is illustrated in Fig. 1.18, where we can see that every symmetric structureundergoes symmetric deformation. It can be proved using the rules of structural mechanics (namely,equilibrium conditions, compatibility conditions and constitutive relations), that deformation for a symmetricsystem is always symmetric. Similarly, we always get antisymmetric deformation for antisymmetricstructural systems, as illustrated in Fig. 1.19.

Figure 1.18 Deformation in symmetric systems

Figure 1.19 Deformation in antisymmetric systems

Let us look at beam AB in Fig. 1.20(a), which is symmetric about point C. The deformed shape of thestructure will be symmetric as well (Fig. 1.20(b)). So, if we solve for the forces and deformations in part ACof the beam, we do not need to solve for part CB separately. The symmetry (or antisymmetry) indeformation gives us additional information prior to analyzing the structure and these information can beused to reduce the size of the structure that needs to be considered for analysis.

Figure 1.20 Symmetric beam system AB and its deformation under load

To elaborate on this fact, we need to look at the deformation condition at the point/line of symmetry (orantisymmetry) in a system. The following general rules about deformation can be deduced looking at theexamples in Fig. 1.18 and Fig. 1.19:

1. For a symmetric structure: slope at the point/line of symmetry is zero.

2. For an antisymmetric structure: deflection at the point/line of symmetry is zero.

These information have to be incorporated when we reduce a symmetric (or antisymmetric) structure to asmaller one. If we want to reduce the symmetric beam in Fig. 1.20 to its one symmetric half AC , we have tointegrate the fact the slope at point C for the reduced system AC will have to be zero. This will be a necessaryboundary condition for the reduced system AC . We can achieve this by providing a support at C, which restrictsany rotation, but allows vertical displacement, as shown in Fig. 1.21 (Note: this specific type of support isknown as a “shear-release” or “shear-hinge”). Everything else (loading, other support conditions) remainsunchanged in the reduced system. We can use this system AC for our analysis in stead of the whole beam AB .

Figure 1.21 Reduced system AC is adopted for analysis for beam ABSimilarly, let us consider an antisymmetric system, a simply-supported beam AB which is antisymmetric aboutthe mid-point C (Fig. 1.22(a)). We know that the deformed shape will also be antisymmetric (Fig. 1.22(b)), andthe displacement at point C will be equal to zero. Therefore, for the reduced system, we consider oneantisymmetric half AC , with a support condition at C which allows rotation but does not allow verticaldisplacements there (Fig. 1.22(c)). Everything else remains same as in AB .

Figure 1.22 (a) Antisymmetric simply-supported beam AB ; (b) Antisymmetric deformation patternfor AB ; (c) Reduced system AC is used for analysis

Having a priory knowledge about symmetry/antisymmetry in the structural system and in its deformed shape

helps us know about symmetry/antisymmetry in internal forces in that system. (Symmetry in the system impliessymmetry in equilibrium and constitutive relations, while symmetry in deformed shape implies symmetry ingeometric compatibility.) Internal forces in a symmetric system are also symmetric about the same axis andsimilarly antisymmetric systems have antisymmetric internal forces. Detailed discussion on different types ofinternal forces in various structural systems and on internal force diagrams are provided in the next module(Module 2: Analysis of Statically Determinate Structures). Once we know about these diagrams we can easilysee the following:

1. A symmetric beam-column system has a symmetric bending moment diagram.

2. A symmetric beam-column system has an antisymmetric shear force diagram.

3. An antisymmetric beam-column system has an antisymmetric bending moment diagram.

4. An antisymmetric beam-column system has a symmetric shear force diagram.

Symmetry/antisymmetry for internal forces can be appreciated in a better way after we go through Module 2 .However, two examples are illustrated in Fig. 1.23 for internal forces in symmetric and antisymmetric systems.

Figure. 1.23 Internal force diagrams for a) a symmetric system, and b) an antisymmetric system

One should remember that although the examples shown here are for (primarily) one-dimensional (or linear)systems, the concept and use of symmetry/antisymmetry is not only limited to these systems. It is applicable totwo- and three-dimensional systems as well. Therefore, we will also find “line of symmetry” and “plane ofsymmetry” in addition to “point of symmetry”. However, these concepts are complex and are not explored indetail in this course.

Recap In this course you have learnt the following

Concept of symmetry, asymmetry and antisymmetry in structures.

Symmetry and antisymmetry in equilibrium and compatibility conditions.

Use of symmetry and antisymmetry in analyzing a structure.

Module 1 : Introduction : Review of Basic Concepts in Mechanics

Lecture 6 : Tutorial Problems

Objectives

In this course you will learn the following

Some tutorial problems related to this module.

TUTORIAL PROBLEMS

T1.1 Find the degree of static indeterminacy in the following truss. What are the degree of internaland external indeterminacy?

Figure T1.1

T1.2 Find the degree of static indeterminacy in the following truss. What are the degree of internaland external indeterminacy?

Figure T1.2

T1.3 What is the bending moment at the base of the column CD in Figure T1.3? (Hint: Considersymmetry/antisymmetry)

Figure T1.3

Recap In this course you have learnt the following

You have learned some tutorial problems related to this module.

Answers of tutorial problems

T1.1 Total = 3,

Internal = 2,

External = 1.

T1.2 Total = 0, Internal = – 1, External = 1.

T1.3 Ans: 0

Module 2 : Analysis of Statically Determinate Structures

Lecture 1 : Internal Force on a System

Objectives

In this course you will learn the following

The concept of internal forces in a system.

Understanding internal forces in a solid body.

2.1 Internal Force on a System

Internal forces (or moments) are generated within a solid body (or structural system) when it is acted uponby external forces (including support reactions and other contact forces as well). To illustrate how internalforces are generated or why they exist, let us consider a three-dimensional solid body (Figure 2.1a),supported at points A and B. , and are external loads applied on the system. To study theequilibrium of the whole body, we draw its free body diagram (Figure 2.1b). The supports are replaced byreactions and in the free body diagram. We consider an internal surface by taking an arbitrary cutthrough the system (Figure 2.1c). For equilibrium of the part at the right of the section, there has to beforces acting on the internal surface which balance the external loads and (Figure 2.1d). and are internal forces acting on the surface of the cut.

Figure 2.1 Internal forces at a cross-sectional surface of a solid body

It is important to know the internal forces acting at different sections of a system. The material, of whichthe body is made, should be strong enough to carry these forces. Otherwise the system fails (by crushing,breaking, etc.) under the loading condition.

The general procedure of obtaining internal forces includes these following steps:

Obtain the system configuration (dimension and support conditions) and external loadings applied on it.

1. Draw a free body diagram of the whole system.2. Find the support reactions by using equations of (static) equilibrium.3. Take a cut through the body where internal forces have to be obtained.4. Consider equilibrium of the part of the system at any one side of the cut by drawing a free body

diagram of that part.5. Obtain the unknown internal forces acting on the cut surface by solving these equilibrium equations.

Recap In this course you have learnt the following

The concept of internal forces in a system.

Understanding internal forces in a solid body.

Module 2 : Analysis of Statically Determinate Structures

Lecture 2 : Internal Forces Acting on Typical Structural Members

Objectives

In this course you will learn the following

Internal forces in different types of structural members.

Categorization of internal forces in 2D and 3D systems.

2.2 Internal Forces Acting on Typical Structural Members

It is mentioned in Section 1.1 that one method of classifying structural systems is on the basis of their loadtransfer mechanisms. To elaborate, a system (or a structural member) is identified based on thepredominant types of internal forces carried by it. Thus we have: bars , cables , beams , columns , arches ,etc. Below is a list of such members along with the predominant internal forces that they carry. Cable : Acable or wire can carry axial tension only. Internal forces in cables are not discussed in this chapterbecause cables are very different from all other systems due to their flexible geometry. Internal forces andgeometry of cable systems are discussed in detail in chapter 3.

1. Cable : A cable or wire can carry axial tension only. Internal forces in cables are not discussed in thischapter because cables are very different from all other systems due to their flexible geometry.Internalforces and geometry of cable systems are discussed in detail in chapter 3.

2.Bar : A bar carries only axial forces – tension and compression both. That is why it is also known asaxially loaded bar .

3.Beam : A beam's primary function is to transfer lateral loads applied externally on the beam. These loadsproduce bending moments and shear forces on beam a cross-section.

4.Column : The predominant internal force in a column is axial compression.

5.Beam-Column : A beam-column, as the name suggests, carries all kinds of internal forces that areproduced in a beam or a column, which include: bending moment, shear force and axial force.

6.Arch: An arch is a curved member which carries primarily axial compression under lateral loads appliedexternally.

There is no difference in the shapes of a beam, a column, a beam-column or a bar. All are straightlongitudinal members (one dimension is much larger than the other two) and we will not be able todistinguish one from the other unless we know the load transfer mechanism. Figure 2.2 illustrates thisissue.

Figure 2.2 Different member types having similar shapes

All the members discussed above are primarily one-dimensional geometrically. Two-dimensional membersare also categorized similarly, such as: plates , shells ( thin & thick ), slabs , etc. We also have specificnames for systems formed by combination of members, such as a truss or a frame . A frame is acombination of beams and columns, whereas all the members in a truss are axially loaded bars.

Recap In this course you have learnt the following

Internal forces in different types of structural members.

Categorization of internal forces in 2D and 3D systems.

Module 2 : Analysis of Statically Determinate Structures

Lecture 3 : Axial Force, Shear Force and Bending Moment

Objectives

In this course you will learn the following

Three primary types of internal force, based on their orientation.

Definitions of axial force, shear forces and bending moments.

2.3 Axial Force, Shear Force and Bending Moment

We will restrict our discussions to primarily one-dimensional members (in reality these are three-dimensional structural members, but the other two dimensions are relatively much smaller). When theloading on such a member is on a plane same as the member itself, we call it a two-dimensional (planar)case (see Figure 2.3a for example). In such cases, the internal forces also lie on the same plane. Theinternal forces on any cross-section can be expressed with two orthogonal force components and onemoment in the plane of loading ( , , M in Figure 2.3b). We can align x -axis along the centroidal axis

of the member and we can also align one of the forces, let's say , along this centroidal axis (along theprimary dimension). Then this internal force will be known as the axial force (Figure 2.3c). In general, weconsider an internal surface perpendicular to the centroidal axis (transverse cross-section, also called the yz-plane or x-plane ). Then the other force component acts tangentially to this surface and it is known asthe shear force . The internal moment, which is acting on the transverse cross-section, is known as thebending moment .

Figure 2.3 Axial force, shear force and bending moment on a cross-section of a two-dimensional(planar) system

For a three-dimensional case, that is when the loading is not restricted to one plane, we have threeorthogonal force components and three orthogonal moment components on an internal surface (Figures 2.4a & b).

We align the centroidal axis of the member along, say, x -axis, and consider an internal surfaceperpendicular to it (Figures 2.4c). Then is the axial force and and are the two shear forces .

Moments and are the two bending moments . The moment is known as torsion . This set of

forces is the most generic case of internal forces for such structural members.

Figure 2.4 Axial force, two shear forces and two bending moments for three-dimensionalsystems

Note that these internal forces are defined according to their orientation respective to the structuralmember. The axial force acts along the centroidal axis of the member. The shear forces act in a plane whichis perpendicular to this centroidal axis and the bending moments act along directions perpendicular to thisaxis as well. The torsion acts along the centroidal axis.

Recap In this course you have learnt the following

Three primary types of internal force, based on their orientation.

Definitions of axial force, shear forces and bending moments.

Module 2 : Analysis of Statically Determinate Structures

Lecture 4 : Sign Convention and Notations for Internal Forces

Objectives

In this course you will learn the following

Standard sign conventions and notations for internal forces through illustrations.

2.4 Sign Convention and Notations for Internal Forces

The sign convention for internal forces depends on the internal surface on which these forces are beingconsidered. So, we need to define the internal surface first. Let us assume that the centroidal axis of alongitudinal structural member is aligned along the x -axis, and we consider the internal forces on an x-plane (or x-surface ). If this cross-section is facing positive x -direction, then it is called a positive x -surface, and vice-versa. Figures 2.5a & b show the positive internal forces on positive and negative x -surfaces, respectively.

Figure 2.5 Direction of positive internal forces on a positive x -surface (a) and a negative x -surface (b)

This sign convention is followed throughout this course and relations involving these forces (and otherparameters, such as stresses and deformations) are derived based on this sign convention. It will not beillogical to adopt any other sign convention for internal forces. However, in that case one will have to

develop the equations involving these forces independently following the new sign convention.

Let us restrict our discussion to planer loading (that is, two-dimensional) cases with no torsion. For aninternal segment of a member, we can show the internal forces both on the positive and the negativeinternal surfaces. Positive directions for each internal force (an axial force, a shear force and a bendingmoment) are shown individually in Figure 2.6. This is an easy and standard way of defining sign conventionsfor two-dimensional cases, and students are encouraged to define (for each specific case) their adopted signconvention similarly.

Fig. 2.6 Defining sign convention for internal forces in a planar system

Notations that we follow for these internal forces are: P for axial force, V for shear force, and M for bendingmoment. However, please note that in a three-dimensional case, we need suffixes to distinguish betweenthe two shear forces and also between the two bending moments. General notation for torsion is T .

Recap In this course you have learnt the following

Standard sign conventions and notations for internal forces through illustrations.

Module 2 : Analysis of Statically Determinate Structures

Lecture 5 : Obtaining Internal Forces in a System: General Procedure

Objectives

In this course you will learn the following

How to obtain internal forces using equilibrium conditions.

2.5 Obtaining Internal Forces in a System: General Procedure

The general method of obtaining internal forces at certain cross-section of a system under a given loading(and support) condition is by applying the concepts of equilibrium (Lecture 2). To illustrate, let us considerthe beam-column AB in Figure 2.7 for which we have to find the internal forces at section a - a . As we havelearned earlier, equilibrium conditions are best studied through free body diagrams. We can find thereactions at supports A and B using a free body diagram of the whole beam-column AB (Figure 2.8). Wesolve the three equations for static equilibrium for this free body:

Figure 2.7 Loading and support conditions for planar beam-column system AB

Figure 2.8 Free body diagram of AB

If a system is in static equilibrium condition, then every segment of it is also in equilibrium. So, we canconsider the equilibrium for each of AC or CB independently. Let us consider the equilibrium of part AC , anddraw its free body diagram (Figure 2.9). In addition to externally applied forces and the support reaction

and , this free body is acted upon by forces P , V and M on the surface a - a . These are nothing but the

internal forces (axial force, shear force and bending moment, respectively) acting at the cross-section a - aof AB . Note that these forces are drawn in their respective positive directions in order to avoid signconfusion. Solving the three static equilibrium equations for AC we find these internal forces:

Figure 2.9 Free body diagram of part AC

Thus we obtain the internal forces at section a - a . These could also be obtained by considering theequilibrium of the part at the other side of section a - a , that is of part CB . Figure 2.10 shows the freebody diagram of CB . Again, the internal forces are drawn in their positive directions on surface a - a , whichis a negative x-surface for this free body. Solving the three equations we find the values for these internalforces:

Figure 2.10 Free body diagram of part CB

Note that these values match exactly with the values obtained previously by considering the equilibrium ofsegment AC . This is true for any system because there is always a unique set of internal forces on aninternal surface for a given loading condition.

Recap In this course you have learnt the following

How to obtain internal forces using equilibrium conditions.

Module 2 : Analysis of Statically Determinate Structures

Lecture 6 : Internal Force Diagrams

Objectives

In this course you will learn the following

Why we need internal force diagrams for structural members.

The use of internal force diagrams in structural analysis and design.

2.6 Internal Force Diagrams

Let us consider the beam-column AB of the previous example with the same loading condition, but adifferent cross-section b - b (Figure 2.11). Following the same procedure as in the previous example we canfind the internal forces at b - b . The values of internal forces at b - b are not same as of those at a - a.

Thus, internal forces vary according to the cross-section under consideration.

Figure 2.11 Cross-section b - b of beam column AB

A structural member should be able to carry the internal forces at each section without failure so as toperform its intended function. So, in order to check the integrity or effectiveness of a structural member,one needs to check its capacity against internal forces at its each and every cross-section. This makes thestudy of the variation of internal forces in a member very important to Structural Mechanics. Such a study isbest done through internal force diagrams, which provide, at one glance, several critical information onthese internal forces.

We use individual diagrams for each type of internal force. Thus we have axial force diagram,shear forcediagram and bending moment diagram for a beam. For the beam-column AB of Figure 2.7, we can findinternal forces at each cross section and obtain the internal force diagrams. Figure 2.12 shows three internalforce diagrams for this beam.

Figure 2.12 Axial force (a), shear force (b) and bending moment (c) diagrams for AB

Note that, we have marked +ve and –ve signs in these diagrams and also put our sign convention to definethe direction of the internal force under consideration. In addition to that, it is also important that we labelthese diagrams with values at key points (that means, the maximum positive and negative value points,zero-value points, points where the variation changes, for example from linear to parabolic).

Recap In this course you have learnt the following

Why we need internal force diagrams for structural members.

The use of internal force diagrams in structural analysis and design.

Module 2 : Analysis of Statically Determinate Structures

Lecture 7 : Internal Force as a Function of x

Objectives

In this course you will learn the following

How to express an internal force as a function of the distance measured along the length of the member.

Use of this function in obtaining internal force diagrams.

2.7 Internal Force as a Function of x

Another alternative of studying internal force variations in a structural member is to express the internalforce as a mathematical function of the longitudinal dimension ( x ). Thus, the axial force, shear force andbending moment at a section are expressed as P ( x ), V ( x ) and M ( x ), respectively, where x is thedistance measured along the primary dimension from one end of the member (Figure 2.13). For this course,we will consider the left end of the member as origin unless otherwise specified. Note that equationsinvolving these internal forces change if the direction for positive x or its origin changes.

Figure 2.13. Internal forces at a distance x from the origin

Considering the example of Figure 2.7 again, let us obtain these internal force functions for the wholelength. After obtaining the support reactions, we can investigate internal forces at different sections. Let usfirst consider the portion x = 0 → 6 m . Since no force or moment is acting between these two points, theinternal force functions will be continuous in this section. We draw the free body diagram of the beam upto adistance x from the left end of the beam (Figure 2.14a). Using equilibrium equations, we can find theinternal forces:

Figure 2.14 Free body diagrams upto a distance x from the origin

Similarly, we can find out the internal forces in the portions x = 6 m → 10m (Figure 2.14b) and x = 10 m→ 12 m (Figure 2.14c). For x = 6 m → 10 m :

and for x = 10 m → 12 m :

If we look at these expressions carefully, we see that:

•We measure x always from the same origin and in the same direction. As noted earlier, it is not absolutelynecessary to follow this convention, but it is easier this way.

•The internal force expressions change at points where concentrated forces/moments (including support

reactions) act. We will see later that these forces also change if a distributed force changes its distribution.Using singularity functions , we can combine different expressions for different segments of the beamtogether into a single expression, which we will discuss later.

•We need to obtain mathematical expressions of internal forces first in order to plot the force variationdiagrams. Although these expressions provide adequate information on variation of internal forces, a pictorialrepresentation is always very useful.

Recap In this course you have learnt the following

How to express an internal force as a function of the distance measured along the length of the member.

Use of this function in obtaining internal force diagrams.

Module 2 : Analysis of Statically Determinate Structures

Lecture 8 : Internal Force Diagrams for Various Systems

Objectives

In this course you will learn the following

The different types of internal force diagrams that are needed for different types of structural member.

2.8 Internal Force Diagrams for Various Systems

We have discussed the general procedure for obtaining internal force variations in a planer system. We canapply that procedure for various types of structural system. Here, we discuss the significance of internalforces (and internal force diagrams) for different structural system types.

Truss : A truss members carries only axial force (tension or compression) and no shear force or bendingmoment. The axial force comes from loads applied only at the two ends of a member. Therefore, the axialforce remains constant along the length of a single truss member. So, we do not really need to plotdiagrams or express axial force as function of length ( x ) in case of a truss member.

Cable : A cable is similar to a truss member except for that it carries only axial tension. For further detailon internal forces in cables, see chapter 3.

Axially Loaded Bar : Only axial force exists in these members (such as columns ). However, unlike a trussmember a bar may be acted upon by external forces along its length. Hence, it is important to study thevariation of axial force through diagrams/mathematical expressions.

Beam and Beam-Column : A beam carries shear force and bending moment and if it carries axial force aswell, then we call it a beam-column. It is for these structural members that internal force diagrams are mostimportant, because deformation and failure behaviour of these members can be directly linked to thesediagrams.

Frame : Frames are two/three-dimensional structural systems made of beams and columns. A framemember, in general, carries internal forces similar to a beam-column. Therefore, it is equally important toobtain internal force diagrams for these systems. Note that for a frame, we may need to specify signconvention for each member individually, as these members may have different orientations.

Arch : Arches can be treated as curved beams (or beam-columns). We will discuss later (in Section 2.12)how to deal with a curved centroidal axis, and with orientations of axial and shear forces.

In the next few sections we will discuss specific cases of determining forces in different types of staticallydeterminate systems, such as trusses, beams, arches, etc.

Recap In this course you have learnt the following

The different types of internal force diagrams that are needed for different types of structural member.

Module 2 : Analysis of Statically Determinate StructuresLecture 9 : Example for Trusses,Beams,Frames and Arches

Objectives In this course you will learn the following

Some examples of trusses.

Example 2.1 (a) Find the forces in AB , AD and AC in the following Figure E2.1.

(b) Find the forces in EG , FG and FH in the following Figure E2.1.

Figure E2.1

Solution:

FBD of the whole system:

(a) FBD of the joint B :

FBD of the joint A :

Force in AB = 22.5 kN (Compression).

Force in AD = 37.5 kN (Tension).

Force in AC = 30.0 kN (Compression).

(b) Take a section through EG , FG , FH and consider the equilibrium of part at the right side

Force in EG = 36.0 kN (Compression).

Force in FG = 22.5 kN (Tension).Force in FH = 18.0 kN (Compression).

Example 2.2 Find the forces in all members in the Figure E2.2.

Figure E2.2Solution:

From equilibrium of the whole body

Looking at joint A :

AB is a zero-force member and

Looking at joint B :

Both BC and BD are zero-force member.

Looking at joint D :

Both CD and DF are zero-force member.

Looking at joint C :

CF is a zero-force member and

Equilibrium of joint E :

and

Equilibrium of joint F :

and

Equilibrium of joint H :

Sign convention: Tension +ve, compression –ve.

Note that we have not obtained the support reactions before finding the member forces. It was not necessary for thisspecific problem. Find out these reactions at supports G and H and check if joint equilibrium is satisfied at these twojoints with the member forces that we have found already.

Recap In this course you have learnt the following

Some examples of trusses.

Example of Beams

Objectives

In this course you will learn the following

Some examples of beams.

Example 2.3 Obtain internal force diagrams for the beam in Figure E2.3.

Figure E2.3

Solution:

FBD of the whole system:

FBD for portion BC :

Consider a section at a distance x from A :

( )

( )

The two sets of equation can be written in a combined form by using singular function

where for

for

for

Note that bending moment is zero at the internal hinge location B .

Example 2.4 Obtain internal force diagrams for the beam in Figure E2.4.

Figure E2.4Solution:

FBD of AE :

Taking a section at a distance x from A :

( )

( )

( )

The internal forces remain same for as well, because .

Combining the expressions using singular functions:

Recap In this course you have learnt the following

Some examples of beams.

Example of Frames

Objectives In this course you will learn the following

Some examples of frames.

Example 2.5 Obtain internal force diagrams for the frame in Figure E2.5.

Figure E2.5

Solution:

FBD of the whole system:

Take a section at a distance from A along AB :

These expressions for internal force are valid from point A to B , using which we can obtain the internal forcediagrams for AB . Using these we can find the internal forces at point B ( ) from which we also get the

forces applied on member BD .

At

Internal forces on BD can be obtained from its free body diagram. Let us take a section at a distance from Balong BD .For

For

Example 2.6 Obtain internal force diagrams for the frame in Figure E2.6.

Solution:

FBD of the whole system:

From 6 static equilibrium equations:

For member AB :

For member BC :

For member CD :

For Member DE :

If we draw bending moments on tension side for the whole system

Recap In this course you have learnt the following

Some examples of frames.

Example of Arches

Objectives In this course you will learn the following

Some examples of arches.

Example for Arches

In this section we consider only three-hinged arches . A three-hinged arch has two hinges at the ends and oneinternal hinge along its curved span as shown in Figure 2.15. If we draw the free body diagram of the whole system,we have four unknown support reactions ( , , & ). Using the three static equilibrium conditions and the

internal hinge condition (total moment about that hinge is equal to zero), in addition, we can solve for these fourreactions. Therefore, three-hinged arches are statically determinate structures.

Figure 2.15 A three-hinged arch and its free-body diagram

Unlike other types of structural member we have dealt with so far, the centroidal axis of an arch is curvilinear.Therefore, the direction of axial force, which is aligned along the centroidal axis, does not remain same along thespan with respect to global coordinates. This is true for the direction of shear force as well. However, bendingmoment does not get affected due to the curved nature of the centroidal axis. In case of circular arches, theproblem with axial and shear force directions can be easily handled with circular coordinates as shown in

Example 2.7. The alternative is to consider internal forces in global x and y directions ( and ) in stead of

forces axial and transverse directions ( P and V ), as shown in Example 2.8.

Example 2.7 Obtain the internal force diagram ( AFD , SFD and BMD ) for the arch in Figure E2.7.

Figure E2.7

Solution:FBD of the whole system:

(i) (ii)

(iii)

FBD of part BC :

(iv)

From Equation (iii) and (iv):

Internal forces at an angle in AB :

At point B :

Internal forces in BC :

For curved structures, for example an arch, internal forces are often expressed as x and y components instead ofaxial and shear forces. Note that bending moment remains same.

Following same procedure as before we can obtain the variation of these internal forces as well, Try to find outdiagrams for , for the same arch. The next example illustrates such internal force diagrams.

Example 2.8 Obtain the internal force diagram ( , and bending moment diagram ) for the symmetric arch in

Figure E2.8.

Figure E2.8

Solution:FBD of the ABC :

It is a symmetric structure, so we need to consider only half of the system, either AB or BC .

Recap In this course you have learnt the following

Some examples of arches.

Module 2 : Analysis of Statically Determinate Structures

Lecture 10 : Tutorial Problems

Objectives In this course you will learn the following

Some tutorial problems related to this module.

TUTORIAL PROBLEMS

Find the forces in EC , EF and HF in the following Figure T2.1.

Figure T2.1

Find the forces in all members in the following Figure T2.2.

Figure T2.2

Find the internal force at moment A in Figure T2.3.

Figure T2.3

T2.4 Find Bending Moment Diagram (BMD) and Shear Force Diagram (SFD) of the beams in FigureT2.4.

Figure T2.4

T2.5 Find the shape ( y as a function of x ) of the parabolic three hinges arch for which bendingmoment will be zero at every section.

Figure T2.5

T2.6 Find the bending moment under the load for the circular three hinged arch.

Figure T2.6

Recap In this course you have learnt the following

You have learned some tutorial problems related to this module.

Answers of tutorial problems

T2.1

T2.2

T2.3

Answers of tutorial problems

T2.4

T2.5

T2.6

Module 3 : Cables

Lecture 1 : Introduction

Objectives In this course you will learn the following

Introduction to cables, how cables are different from other structural components.

Use of cables in structural systems.

3.1 Introduction

Cables are flexible wire-like systems having no flexural (bending) stiffness, and they can carry only axialtension and no other type of force. Being fully flexible against bending the shape of a cable is determined bythe external forces that are acting on the cable. Figure 3.1 illustrates how the shape of the cable between twosupports A and B depends on the location and magnitude of the external forces and .

Figure 3.1 Shape of a cable is determined by external loads

A cable is unable to carry bending moment, shear force, torsion or axial compression. Nevertheless, cables canbe very effectively used in achieving long-span light-weight systems, such as bridges or roofs for large arenas.Two kinds of bridge structural systems where cables are used are the suspension-cable systems and cable-stayed systems . Figures 3.2 and 3.3 show examples of suspension-cable bridge and cable-stayed bridge,respectively.

Figure 3.2 A suspension-cable bridge (Golden gate bridge, San Francisco , USA)

Figure 3.3 A cable-stayed bridge (ANZAC bridge, Sydney , Australia )

Cables are usually made of multiple strands of cold-drawn high-strength steel wires twisted together.Generally, they have strength four to five times that of structural steel and practically inextensible underoperating loading conditions. Since cables carry only axial tension, full potential of the cable cross-sectioncan be utilized in transferring forces. Therefore, cables are able to carry the same amount of force with amuch smaller cross-section compared to other structural systems. This high strength-to-weight ratio makescables very useful where light-weight systems are needed. On the other hand, a beam over a very longspan would require a very large (and deep) cross-section, and most of its potential will be used in carryinginternal forces due to its own weight. If we use cables replacing this beam or in combination with a beam instead, a lighter structure will be required, whose self-weight will not add significantly to load effects.

The primary disadvantage with cables is due to their flexible geometry. As the loading on a cable systemchanges (as in the case of moving loads on a bridge) there can also be large change in the cable geometry,and subsequently on forces acting in the cable. Unexpected forces may destabilize a cable system, causingexcessive deformations. A designer should be very careful on this regard while designing a cable system,along with other issues such as, large forces at the anchors, large oscillations, etc.

Recap In this course you have learnt the following

Introduction to cables, how cables are different from other structural components.

Use of cables in structural systems.

Module 3 : Cables

Lecture 2 : The General Cable Theorem

Objectives In this course you will learn the following

Statement and derivation for the general cable theorem.

3.2 The General Cable Theorem

The general cable theorem helps us determine the shape of a cable supported at two ends when it is actedupon by vertical forces. It can be stated as: “ At any point on a cable acted upon by vertical loads, theproduct of the horizontal component of cable tension and the vertical distance from that point to the cablechord equals the moment which would occur at that section if the loads carried by the cable were acting onan simply-supported beam of the same span as that of the cable.”

Figure 3.4 Explanation of the general cable theorem: (a) Cable under vertical loads, and (b) Simply supported beam with equal span under the same set of loads

To explain, let us consider the cable AB in Figure 3.4a, which is acted upon by the vertical loads , ,

and at known locations. The line AB joining the two supports is known as the chord of the cable and thehorizontal distance between the supports is known as its span . The vertical distance between the chord andthe cable at any cross section is known as the dip . This is vertical distance that is mentioned in the general

cable theorem . The cable in Figure 3.4a has a span L and the dip at a distance x from A is y . Thehorizontal reactions at supports A and B have to be equal to satisfy static equilibrium, and let it be H . Thevertical reactions at supports A and B are and , respectively. Figure 3.4b shows a simply-supportedbeam AB of same span ( L ) and acted upon by the same set of forces as the cable AB in Figure 3.4a.

For moment equilibrium about support B for the cable:

(3.1)

where, is the summation of moments due to external forces ( , , and ) about point B

. Since the cable is totally flexible against bending, bending moment at any cross-section is zero. Byequating bending moment at a distance x from A to zero, we get:

(3.2a)

or, (3.2b)

where, is the summation of moments due to external forces ( , , and to the left of x )

about section x . Substituting from Equations 3.1 and 3.2b:

(3.3)

Now, let us consider the simply-supported beam in Figure 3.4b. From moment equilibrium about support B ,we get the vertical reaction at support A :

(3.4)

So, the bending moment at a distance x from A is:

(3.5)

which, is same as the right side of Equation 3.3. Therefore:

Moment at x for the simply-supported beam (3.6)

which is the claim as per the general cable theorem .

Note that the horizontal component of the axial force at any section of a cable (under vertical externalforces only) is same as the horizontal reaction ( H ) at the end supports. This can be proved considering theequilibrium of horizontal forces on any segment of the cable.

We can solve internal forces in a cable using the general cable theorem, and also we can obtain for theshape of the cable. If the cable length (not the span) is known to us, we can express this length in terms ofthe dip y . Using this information along with the general cable theorem we can solve for both the unknownsH and y . Alternatively, the dip at a certain point, instead of the total length of the cable, may be known tous. This information, along with the general cable theorem helps us solve for both H and y .

Recap In this course you have learnt the following

Statement and derivation for the general cable theorem.

Module 3 : Cables

Lecture 3 : Application of the General Cable Theorem for Distributed Loading

Objectives

In this course you will learn the following

Use of the general cable theorem for cables with distributed loading.

3.3 Application of the General Cable Theorem for Distributed Loading

We have seen that we can apply the general cable theorem to find the cable geometry under vertical loadingcases. The theorem also applies for distributed loading, since bending moment definitions for thecorresponding simply-supported beam ( and ) do not change. For a cable under uniformly

distributed load w , we have:

(3.7)

Let us consider the specific case of a cable AB under uniformly distributed loading w , with the cable'ssupports being at the same horizontal level (Figure 3.5). Note that the system is symmetric about its mid-span where the cable has its maximum dip. Let the span of the cable be L and its dip at the mid-span (pointC ) be . We can find, from the equilibrium of vertical forces and from symmetry, that the vertical supportreactions at both A and B are wL/2. Now, applying the general cable theorem (Equation 3.7) at point C , weget:

(3.8)

Figure 3.5 Free body diagram of a cable under uniformly distributed load

Due to symmetry, we can see that the cable tension (axial force) is horizontal at the mid-span. This can beobserved also if we draw the free body diagram of either the right or the left half of the cable (Figure 3.6).

Figure 3.6 Free body diagram of the right half (CB) of the cable

We can also use Equation 3.8 to define a general shape of the cable in terms of the mid-span dip, . Thus,the dip y at a (horizontal) distance x from the left support now is:

(3.9)

Let T be the axial tension in the cable at a distance x . This axial tension acts along the tangent of the cablegeometry. Let us measure the length of the cable by s , which is measured along the cable curve. Therefore

(3.10)

dy / dx is the slope of the cable and it can be obtained from Equation 3.9 which defines the shape of the cable.Substituting in Equation 3.10, we get:

(3.11)

This equation also shows that the maximum tension occurs at the end supports, that is at x = 0 and x = L ,which is also where the slope of the cable is maximum. The minimum tension occurs at the mid-span and isequal to H .

The shape, as defined in Equation in 3.9, can be used obtain the total length of the cable ( S ) as well.

(3.12)

The expression simplifies if the dip becomes very small compared to the span, that is,

(3.13)

One should remember that Equations 3.8 to 3.12 are valid for only cables with both end supports at the samehorizontal level.

The shape of a flexible cable supported at two ends and hanging only under its self-weight is known as acatenary . It is the shape that a cable attains under uniformly distributed vertical load (self-weight, in thiscase). Therefore, the shape of the cable should be a parabola as per Equation 3.9 and this was what Galileoclaimed. However, Leibniz and other scientists later found the proper equation for a catenary to be differentfrom a parabola. This is because the self-weight of the cable is uniform along its curved length and not alongits span. The distributed loading w that we have considered for obtaining Equation 3.9 is uniform along thespan ( x ) and not along its curved shape ( s ). The equation of a catenary is:

(3.14)

Recap In this course you have learnt the following

Use of the general cable theorem for cables with distributed loading.

Module 3 : CablesLecture 4 : Examples

Objectives

In this course you will learn the following

Some examples of cable systems.

Example 3.1 A 35 m cable is supported at ends A and B which are at the same horizontal level and are 25 m apart.A vertical load of 25 kN is acting at point C which is at a distance of 9 m from A . Find the horizontal reaction at Aand the dip at C .

Figure E3.1

Solution: Free body diagram of the cable:

If the dip at point C is , then applying the general cable theorem, we get:

Where and

Therefore

The other equation based on total length of the cable is

Note: Using the static equilibrium conditions we can find that:

Example 3.2 A light cable (that is, self weight of cable is negligible compared to external loads) is carryinguniformly distributed load of 30 kN / m . The span of the cable is 75 m and its length is 77 m , where thesupports are at same horizontal level. What will be the percentage change in minimum tension if there is a riseof temperature by 35 ° C? Coefficient of thermal expansion of the cable material is (12× 10 -6 / ° C).

Solution: if is the dip at mid point, then using equation 3.13

Change in length due to temperature rise

Differentiating equation 3.13, we get:

Differentiating equation 3.8

decrease

This is the change in horizontal reaction, that is, in minimum tension in the cable.

Recap

In this course you have learnt the following

Some examples of cable systems.

Module 3 : CablesLecture 5 : Tutorial Problems

Objectives

In this course you will learn the following

Some tutorial problems related to this module.

TUTORIAL PROBLEMS

T3.1 A light cable with span 40m is under uniformly distributed load of 1 kN / m . If the support are at the samelevel and the maximum tension allowed in the cable is 30 kN . what is the maximum allowable dip of the cable?

T3.2 Find the tension in the cable at point B for the cable shown in Figure T3.1.

Figure T3.1 Recap In this course you have learnt the following

You have learned some tutorial problems related to this module.

Answers of tutorial problems

T3.1 8.94 m

T3.2 74.56 kN

Module 4 : Deflection of StructuresLecture 1 : Moment Area Method

Objectives

In this course you will learn the following

Importance of computation of deflection.

Computation of deflection using moment area method.

4.1 Introduction

When a structure is subjected to the action of applied loads each member undergoes deformation due towhich the axis of structure is deflected from its original position. The deflections also occur due totemperature variations and lack-of-fit of members. The deflections of structures are important for ensuringthat the designed structure is not excessively flexible. The large deformations in the structures can causedamage or cracking of non-structural elements. The deflection in beams is dependent on the acting bendingmoments and its flexural stiffness. The computation of deflections in structures is also required for solvingthe statically indeterminate structures.

In this chapter, several methods for computing deflection of structures are considered.4.2 Moment Area Method

The moment-area method is one of the most effective methods for obtaining the bending displacement inbeams and frames. In this method, the area of the bending moment diagrams is utilized for computing theslope and or deflections at particular points along the axis of the beam or frame. Two theorems known asthe moment area theorems are utilized for calculation of the deflection. One theorem is used to calculate thechange in the slope between two points on the elastic curve. The other theorem is used to compute thevertical distance (called tangential deviation) between a point on the elastic curve and a line tangent to theelastic curve at a second point.

Consider Figure 4.1 showing the elastic curve of a loaded simple beam. On the elastic curve tangents aredrawn on points A and B . Total angle between the two tangents is denoted as . In order to find out

, consider the incremental change in angle over an infinitesimal segment located at a distanceof from point B . The radius of curvature and bending moment for any section of the beam is given by theusual bending equation.

(4.1)

where R is the radius of curvature; E is the modulus of elasticity; I is the moment of inertia; and M denotesthe bending moment.

The elementary length and the change in angle are related as,

(4.2)

Substituting R from Eq. (4.2) in Eq. (4.1)

(4.3)

The total angle change can be obtained by integrating Eq. (4.3) between points A and B whichis expressed as

(4.4a)

or,

Area of M / EI diagram between A and B (4.4b)

The difference of slope between any two points on a continuous elastic curve of a beam is equal to the areaunder the M / EI curve between these points.

The distance dt along the vertical line through point B is nearly equal to.

(4.5)

Integration of dt between points A and B yield the vertical distance between the point B and the tangentfrom point A on the elastic curve. Thus,

(4.6)

since the quantity M /EI represents an infinitesimal area under the M /EI diagram and distance from thatarea to point B, the integral on right hand side of Eq. (4.6) can be interpreted as moment of the area under theM/EI diagram between points A and B about point B . This is the second moment area theorem.

If A and B are two points on the deflected shape of a beam, the vertical distance of point B from the tangentdrawn to the elastic curve at point A is equal to the moment of bending moment diagram area between thepoints A and B about the vertical line from point B , divided by EI .

Sign convention used here can be remembered keeping the simply supported beam of Figure 4.1 in mind. Asagging moment is the positive bending moment diagram and has positive area. Slopes are positive ifmeasured in the anti-clockwise direction. Positive deviation indicates that the point B lies above the tangentfrom the point A .

Example 4.1 Determine the end slope and deflection of the mid-point C in the beam shown belowusing moment area method .

Solution: The M / EI diagram of the beam is shown in Figure 4.2(a). The slope at A , can be obtained by

computing the using the second moment area theorem i.e.

(clockwise direction)

The slope at B can be obtained by using the first moment area theorem between points A and B i.e.

(anti-clockwise)

(It is to be noted that the . The negative sign is because of the slope being in the clockwise

direction. As per sign convention a positive slope is in the anti-clockwise direction)

The deflection at the centre of the beam can be obtained with the help of the second moment area theorembetween points A and C i.e.

(downward direction)

Example 4.2 Using the moment area method, determine the slope at B and C and deflection at C ofthe cantilever beam as shown in Figure 4.3(a). The beam is subjected to uniformly distributed loadover entire length and point load at the free end.

Solution: The moment curves produced by the concentrated load, W and the uniformly distributed load,w areplotted separately and divided by EI (refer Figures 4.3(b) and (c)). This results in the simple geometric shapesin which the area and locations of their centroids are known.

Since the end A is fixed, therefore, . Applying the first moment-area theorem between points A and C

(negative sign is due to hogging moment)

(clockwise direction)

The slope at B can be obtained by applying the first moment area theorem between points B and C i.e.

–

(clockwise direction)

The deflection at C is equal to the tangential deviation of point C from the tangent to the elastic curve at A (seeFigure 4.3(d)).

= moment of areas under M / EI curves between A and C in Figures 4.3(b) and (c) about C

(downward direction)

Example 4.3 Determine the end-slopes and deflection at the center of a non-prismatic simply supportedbeam. The beam is subjected to a concentrated load at the center.

Solution: The M/EI diagram of the beam is shown in Figure 4.4(b).

Applying second moment-area theorem between points A and B ,

(clockwise direction)

Applying first moment area theorem between A and C .

(anti-clockwise direction)

Applying second moment area theorem between A and C .

(downward direction)

Example 4.4 Determine the slope and deflection at the hinge of the beam shown in the Figure 4.5(a).

Solution: The bending moment diagram is shown in Figure 4.5(b).

Since the end A is fixed, therefore, . Applying the first moment-area theorem between points A and B(refer Figure 4.5(c))

(clockwise direction)

Applying second moment area theorem between points A and B ,

(downward direction)

Applying second moment area theorem between points B and D ,

(anti-clockwise direction)

From the first moment area theorem between points B and D

(clockwise direction)

Example 4.5 Determine the vertical deflection and slope of point C of the rigid-jointed plane frame shown inthe Figure 4.6(a).

Solution: The M/EI and deflected shape of the frame are shown in the Figures 4.6(a) and (b), respectively. Asthe point A is fixed implying that . Applying first moment area theorem between points A and B ,

(looking from the left side)

(anti-clockwise direction)

Applying second moment area theorem between points B and C

The vertical displacement of point C

(downward direction)

Applying first moment area theorem between point B and C

(anti-clockwise direction)

Recap In this course you have learnt the following

Importance of computation of deflection.

Computation of deflection using moment area method.

Module 4 : Deflection of StructuresLecture 2 : Conjugate Beam Method

Objectives

In this course you will learn the following

Computation of deflection using conjugate beam method.

4.3Conjugate Beam Method

The conjugate beam method is an extremely versatile method for computation of deflections in beams. Therelationships between the loading, shear, and bending moments are given by

(4.7)

where M is the bending moment; V is the shear; and w ( x ) is the intensity of distributed laod.

Similarly, we have the following

(4.8)

A comparison of two set of equations indicates that if M / EI is the loading on an imaginary beam, theresulting shear and moment in the beam are the slope and displacement of the real beam, respectively. Theimaginary beam is called as the “ conjugate beam ” and has the same length as the original beam.

There are two major steps in the conjugate beam method. The first step is to set up an additional beam,called "conjugate beam,” and the second step is to determine the “ shearing forces ” and “ bendingmoments ” in the conjugate beam.

The loading diagram showing the elastic loads acting on the conjugate beam is simply the bending-moment diagram of the actual beam divided by the flexural rigidity EI of the actual beam. This elastic load isdownward if the bending moment is sagging.

For each existing support condition of the actual beam, there is a corresponding support condition for theconjugate beam. Table 4.1 shows the corresponding conjugate beam of different types of actual beams. Theactual beam as well as the conjugate beam are always in static equilibrium condition .

The slope of (the centerline of) the actual beam at any cross-section is equal to the “ shearing force ” atthe corresponding cross-section of the conjugate beam. This slope is positive or anti-clockwise if the “shearing force ” is positive — to rotate the beam element anti-clockwise — in beam convention . Thedeflection of (the centerline of) the actual beam at any point is equal to the “ bending moment ” of theconjugate beam at the corresponding point. This deflection is downward if the “ bending moment ” ispositive — to cause top fiber in compression — in beam convention . The positive shearing force and bendingmoment are shown below in Figure 4.7.

Table 4.1 Real and Conjugate beams for different structures

REAL SRUCTURE CONJUGATE STRUCTURE

Example 4.6 Determine the slope and deflection of point A of the of a cantilever beam AB of length L anduniform flexural rigidity EI. A concentrated force P is applied at the free end of beam.

Solution: The conjugate beam of the actual beam is shown in Figure 4.8(b). A linearly varying distributedupward elastic load with intensity equal to zero at A and equal to PL/EI at B. The free-body diagram for theconjugate beam is shown in Figure 4.8(c). The reactions at A of the conjugate beam are given by

The slope at A , and the deflection at the free end A of the actual beam in Figure 4.8(d) are respectively,

equal to the “shearing force” and the “bending moment” at the fixed end A of the conjugate beam inFigure 4.8(d).

Note that points downward because causes tension in bottom fiber of the beam at A (i.e. saggingmoment)Example 4.7 Determine the slope at A and deflection of B of the beam shown in Figure 4.9(a) usingthe conjugate beam method.

Solution: The vertical reaction at A in the real beam is given by

The bending moment at any point X at a distance x from A is given by

The corresponding conjugate beam and loading acting on it are shown in Figure 4.9(b). The loading on the beam

varies parabolically with maximum value as .

The slope at A , in the original beam will be equal to the shear force at A in the conjugate beam, thus,

(clockwise direction)

The deflection of B in the real beam will be equal to the bending moment at B in conjugate beam i.e.

(downward direction)

Example 4.8 Determine the deflection at the free end of the beam shown in Figure 4.10 using conjugate beammethod and verify by moment area method.

Solution:

(a) Conjugate beam method

The corresponding conjugate beam and loading are shown in Figure 4.10(b). The loading is upward linearly

distributed load with maximum value of at B . Taking moment about point B, the vertical reaction at A in

the conjugate beam is given by

The bending moment at C (by taking moment about C ) is given by

(sagging type)

Hence, the deflection of point C will be equal to in the downward direction.

(b) Verification by moment-area method

Applying second moment area theorem between points A and B will give the slope at A i.e.

Further, applying moment area theorem between point A and C

(clockwise direction)

Since

(downward direction)

Recap

In this course you have learnt the following

Computation of deflection using conjugate beam method.

Module 4 : Deflection of Structures

Lecture 3 : Principle of Virtual Work

Objectives In this course you will learn the following

Computation of deflection using principle of virtual work ( PVW ).

Application to pin-jointed structure.

Application of PVW to beams and frames.

Simplified PVW for beams and frames using multiplication of bending moment diagram.

4.4 Principle of Virtual Work

Consider a structural system subjected to a set of forces ( … referred as P force) under stableequilibrium condition as shown in Figure 4.11(a). Further, consider a small element within the structuralsystem and stresses on the surfaces caused by the P forces are shown in Figure 4.11(b) and referred as .

Let the body undergoes to a set of compatible virtual displacement . These displacements are imaginaryand fictitious as shown by dotted line. While the body is displaced, the real forces acting on the body movethrough these displacements. These forces and virtual displacements must satisfy the principle ofconservation of energy i.e.

(4.8)

(4.9)

This is the principle of virtual work

If a system in equilibrium under a system of forces undergoes a deformation, the work done bythe external forces ( P ) equals the work done by the internal stresses due to those forces, ( ).

In order to use the above principle for practical applications, we have to interchange the role of the forcesand displacement. Let the structure acted upon by a virtual force is subjected to real displacements then theEq. (4.9) can be written as

(4.10)

This is the principle of complimentary virtual work and used for computing displacements.

Consider a structure shown in Figure 4.12(a) and subjected to P force and it is required to find thedisplacement of point C in the direction specified. First apply a virtual force at C in the required

direction. Next apply the external (real) loads acting on the structures as shown in Figure 4.12(a) with thevirtual force remain in the position. The displacement of C in the required ditection be and the internalelements deform by an amount . Using Eq. (4.10)

(4.11)

The left hand side of Eq. (4.11) denotes the external work done by the virtual force moving through thereal dispolacement . On the other hand, the right hand side of Eq. (4.11) represents the internal workdone by the virtual internal element forces d f moving through the displacement .

Since is arbitrary and for convenience let =1 (i.e. unit load). The Eq. (4.11) can be re-written as

(4.12)

where f denotes the internal force in the members due to virtual unit load.

The right hand side of Eq. (4.12) will directly provide the displacement of point C due to applied externalforces. This method is also known as unit load method.

Similarly for finding out a rotation, at any point of a loaded structure, the corresponding Eq. (4.12) willtake place as

(4.13)

where denotes the internal force in the members due to virtual unit moment applied in the direction of

interested .

4.4.1 Application to Pin-Jointed Structures

Consider a pin-jointed structure as shown in Figure 4.13 and subjected to external force P 1 , P 2 and P 3 .Let the vertical displacement of point C , is required. Under the action of the real external load, let

the axial force in typical member be and therefore, the deformation of the member ( Land AE are the length and axial rigidity of typical member).

Apply a unit vertical load at C and substituting in Eq. (4.12) leads to

(4.14)

The basic steps to be followed for finding the displacements of the pin-jointed structure are

1. Compute the axial force in various members (i.e. ) due to applied external forces.

2. Compute the axial force in various members (i.e. ) due to unit load applied in the direction ofrequired displacement of the point.

3. Compute the product for all members.

4. The summation will provide the desired displacement.

5. The axial force shall be taken as positive if tensile and negative if compressive.

6. The positive implies that the desired displacement is in the direction of applied unit load

and negative quantity will indicate that the desired displacement is in the opposite direction of theapplied unit laod.

Example 4.9 Find the horizontal and vertical deflection at joint C of the pin-jointed frame shown in Figure4.14. AE is constant for all members.

Solution: Calculate forces i.e. force in various members of the truss due to the applied loading. These

can be obtained by considering the equilibrium of various joints as marked in Figure 4.14(b).

Table 4.2

Member LengthFor For

L L

AB L -P 0 0 0 0BC L 0 0 0 0 0CD L -P -1 PL -1 PLDA L 0 0 0 0 0

AC 0 0

The computation of for two desired displacements of pin-jointed frame are shown in Table 4.2.

Horizontal displacement of joint C ,

Vertical displacement of joint C ,

Example 4.10 For the pin-jointed structure shown in the Figure 4.15, find the horizontal and verticaldisplacement of the joint D . The area of cross-section, A =500 and E =200,000 for all themembers.

Solution: The axial rigidity of the members, kN. The computation of thedesired displacements is presented in Table 4.3

Table 4.3

Member Length L(m)For For

L L

AB 2 1920 0 0

BC 2 480 0 0

CD 2 480 0 0

DE 2 120 1 -

EF 2 - 1080 1 -

CE 2 - - 480 0 0

BF 2 480 0 0

The horizontal deflection of = 55.2 mm

The vertical deflection of = - 8.31 mm = 8.31 mm

4.4.2Application to beams and frames

In order to find out the vertical displacement of C of the beam shown in Figure 4.16(a), apply a unit loadas shown in Figure 4.16(b).

The internal virtual work is considered mainly due to bending and caused due to internal moments

under going the rotation due to the applied loading. (internal virtual work done by shearing forces and

axial forces is small in comparison to the bending moments and hence ignored). Since the

where is the moment due to applied loading, the Eq. (4.12) for the displacement of C will take a shape

of

(4.15)

The basic steps to be followed for finding the displacement or slope of a beams and frames are summarizedas

1. Compute the bending moment (i.e. ) due to applied external forces.

2. Compute the bending moment (i.e. ) due to unit load applied in the direction of requireddisplacement or slope.

3. Compute the integral over the entire members of the beam or frame which will

provide the desired displacement.4. The bending moment shall be taken as positive if sagging and negative if hogging (in case of

beams).

5. The positive implies that the desired displacement is in the direction of applied unit

load and negative quantity will indicate that the desired displacement is in the opposite direction ofthe applied unit load.

Example 4.11 Determine the slope and deflection of point A of the cantilever beam AB with length L andconstant flexural rigidity EI.

Solution: Deflection under the Load - Apply a vertical unit load at point A of the beam as shown in Figure4.17(b). Consider any point X at a distance of x from A ,

The vertical deflection of point A is given by

Slope at the free end: Apply a unit couple at point A of the beam as shown in Figure 4.17(c). Consider anypoint X at a distance of x from A .

The slope at A is given by

Example 4.12. Determine mid-span deflection and end slopes of a simply supported beam of span Lcarrying a udl w per unit length.

Solution: Mid-span deflection : Apply a unit load at mid span as shown in Figure 4.18(b). Consider any pointX at a distance of x from A

(0< x < L)

(0< x < L/2 )

( L /2< x < L )

The vertical deflection of point C is given by

End slopes : Applying a unit couple at A as shown in Figure 4.18(c). Consider any point X at a distance of xfrom A

(0< x < L )

(0< x < L/2 )

The slope at A is given by

Due to symmetry (anti-clockwise direction)

Example 4.13 Determine vertical deflection and rotation of point B of the beam shown in Figure 4.19(a).The beam is subjected to a couple at C .

Solution: Vertical deflection of B : Apply a unit load at B as shown in Figure 4.19(b). Consider any point X ata distance of x from C

(0< x < a + b )

(0< x < b )

= (b < x < a + b)

The vertical deflection of point B is given by

(i.e. in the upward direction)

Rotation of B : Apply a unit couple at B as shown in Figure 4.19(c). Consider any point X at a distance of xfrom C

(0< x < a + b )

(0< x < a + b )

The rotation of point B is given by

Example 4.14. Determine horizontal deflection of C and slope at A of a rigid-jointed plane frame as shownin Figure 4.20(a). Both members of the frame have same flexural rigidity, EI .

Solution: Horizontal deflection of C : Apply a unit load C as shown in Figure 4.20(b).

Consider AB : ( x measured A )

Consider BC : ( x measured C )

The horizontal deflection of point C is given by

Rotation at A : Applying a couple at A as shown in Figure 4.20(d).

Consider AB : ( x measured A )

Consider BC : ( x measured C )

The slope at A is given by

4.4.3Moment diagrams multiplication method for beams and frames

Recall the Eq. (4.15) in which the bending deflection of the beams and frames are obtained by theintegration of the two bending moments variations (i.e. and ) over a length of the members.However, for a uniform beam section (i.e. EI is constant) such integrals can be readily derived depending

upon the various shapes of the bending moment diagrams. The computation of integral is

given in the Table 4.A1. The various steps for this method for finding deflections of the beams and frameare:

1. Draw the bending moment diagram of given beam or frame due to applied external loading (i.e. diagram).

2. Draw the corresponding bending moment diagram due to unit load applied in the direction ofinterested deflection (i.e. diagram).

3. Compute the desired deflection by computing the with the help of results shown in

Table 4.A1.

Example 4.15 Determine the deflection under the load and point D of a simply supported beam withoverhang as shown in Figure 4.21

Figure 4.21

Solution: Bending moment diagram (i.e. diagram) due to concentrated load W is shown in Figure4.21(b).

Deflection under the Load : Apply a vertical unit load in place of W . The bending moment diagram due tothis load is shown in Figure 4.21(c). The vertical deflection under the load is obtained by multiplying thebending moment diagrams of Figure 4.21(b) and (c) and is given by

(refer Table 4.A1)

Deflection of the free end : Apply a unit vertical load acting upward at point D of the beam. The bendingmoment diagram due to this load is shown in Figure 4.21(d). The vertical deflection under the load isobtained by multiplying the bending moments diagrams of Figure 4.21(b) and (d) and is given by

(refer Table 4.A1)

Example 4.16 Using the diagram multiplication method, determine the deflection under the load and endslopes of a non-prismatic simply supported beam.

Solution: Bending moment (B.M.) diagram (i.e. diagram) due to concentrated load W on the beam isshown in Figure 4.22(b).

Mid-span deflection : Apply a unit load in the downward direction at C . Deflection at C is given by multiplyingthe diagrams of Figure 4.22 (b) and (c) as follows

Slope at A : Apply a unit couple at A acting in the clockwise direction and plot the bending moment diagramof the beam as shown in Figure 4.22(d). The slope at A is given by multiplying the diagrams of Figure 4.22(b) and (d) as follows

(clockwise direction)

Slope at B : Apply a unit couple at B acting in the anti-clockwise direction and plot the bending momentdiagram of the beam as shown in Figure 4.22(e). The slope at B is given by multiplying the diagrams ofFigure 4.22 (b) and (e) as follows

Example 4.17 Using the diagram multiplication method, determine the horizontal displacement and rotaionof point C of the rigid-jointed plane frame shown in Figure 4.23. Both the members of the frame have sameEI value.

Solution: The free-body and bending moment diagram (B.M.D.) of the frame due to applied loading areshown in Figures 4.23(b) and (c), respectively.

Horizontal deflection of C : Apply a horizontal force at C as shown in Figure 4.23(d) and plot the bendingmoment diagram as shown in Figure 4.23(e). The horizontal deflection at C is given by multiplying thediagrams of Figure 4.22 (c) and (e) as follows

Rotation of C : Apply a unit couple at C as shown in Figure 4.23(f) and plot the bending moment diagram asshown in Figure 4.23(g). The slope at C is given by multiplying the diagrams of Figure 4.22 (c) and (g) as follows

(anti-clockwise direction)

Recap In this course you have learnt the following

Computation of deflection using principle of virtual work ( PVW ).

Application to pin-jointed structure.

Application of PVW to beams and frames.

Simplified PVW for beams and frames using multiplication of bending moment diagram.

Table 4.A1 Evaluation of the integral

KiL 1/2 KiL 1/2 K L 2/3 KiL 2/3 KiL 1/3 KiL 1/2 KiL

1/2 KiL 1/3 KiL 1/6 K L 1/3 KiL 5/12 KiL 1/4 KiL 1/6(1 + a ) KiL

1/2 KiL 1/6 KiL 1/6 K L 1/3 KiL 1/4 KiL 1/12 KiL 1/6(1 + b ) KiL

1/2 iL

1/6 iL

1/6

L

1/3 iL

1/12 iL

1/12 iL

1/6((1 + b ) +

(1 + a ) )iL

2/3 KiL 1/3 KiL 1/3 K L 8/15 KiL 7/15 KiL 1/5 KiL 1/3(1 + ab ) KiL

2/3 KiL 5/12 KiL 1/12 K L 7/15 KiL 8/15 KiL 3/10 KiL 1/12(5 - b - ) KiL

2/3 KiL 1/4 KiL 1/12 K L 7/15 KiL 11/30 KiL 2/15 KiL 1/12(5 - a - ) KiL

1/3 KiL 1/14 KiL 1/12 K L 1/5 KiL 3/10 KiL 1/5 KiL 1/12(1 + a + )KiL

1/3 KiL 1/12 KiL 1/12 K L 1/5 KiL 2/15 KiL 1/30 KiL 1/12(1 + b + )KiL

1/2 KiL 1/6(1+ a ) KiL1/6 KL ((1 + b )

+ (1 + a ) )

1/3(1 + ab )KiL

1/12(5 - b - ) KiL

1/12(1 + a + ) KiL 1/3 KiL

Module 4 : Deflection of StructuresLecture 4 : Strain Energy Method Objectives In this course you will learn the following

Deflection by strain energy method.

Evaluation of strain energy in member under different loading.

Application of strain energy method for different types of structure.

4.5 Deflection by Strain Energy Method

The concepts of strain, strain-displacement relationships are very useful in computing energy-relatedquantities such as work and strain energy. These can then be used in the computation of deflections. In thespecial case, when the structure is linear elastic and the deformations are caused by external forces only,(the complementary energy U * is equal to the strain energy U ) the displacement of structure in thedirection of force is expressed by

(4.16)

This equation is known as Castigliano's theorem. It must be remembered that its use is limited to thecalculation of displacement in linear elastic structures caused by applied loads. The use of this theorem isequivalent to the virtual work transformation by the unit-load theorem.

4.5.1 Calculation of Strain Energy

When external loads are applied on an elastic body they deform. The work done is transformed into elasticstrain energy U that is stored in the body. We will develop expressions for the strain energy for differenttypes of loads.

Axial Force : Consider a member of length L and axial rigidity AE subjected to an axial force P appliedgradually as shown in the Figure 4.24. The strain energy stored in the member will be equal to the externalwork done by the axial force i.e

(4.17)

Figure 4.24 Member subjected to axial force

Bending Moment: Consider a beam of length L and flexural rigidity EI subjected to a general loading asshown in Figure 4.25. Consider a small differential element of length, dx . The energy stored in the smallelement is given by

(4.18)

The total strain energy in the entire beam will be

(4.19)

Figure 4.25 Member under bending

Shear Force: The strain energy stored in the member due to shearing force is expressed by

(4.20)

where V is the shearing force; and is the shearing rigidity of the member.

Twisting Moment: The strain energy stored in the member due to twisting moment is expressed by

(4.21)

where T is the twisting moment; and GJ is the torsional rigidity of the member.

Example 4.18 Find the horizontal deflection at joint C of the pin-jointed frame as shown in Figure 4.26(a). AEis constant for all members.

Solution: The force in various members of the frame is shown in Figure 4.26(b). Calculation of strain energyof the frame is shown in Table 4.4.

Table 4.4

Member Length ( L ) Force ( P )

AB L P

BC L P

BD CD L 0 0

Horizontal displacement of joint C ,

Example 4.19 A bar of uniform cross-section is bent into a quadrant of circle of radius R . One end of thebent is fixed and other is free. At the free end it carries a vertical load W . Determine the vertical andhorizontal deflection at A .

Solution:

Vertical displacement of A : The vertical displacement of A is given by

For evaluation of the total strain energy in the system, consider a small element as shown in the Figure.

The bending moment at this element, . Thus,

Since there is no horizontal force acting at point A , apply a horizontal force, F at A as shown in Figure4.27(b). From the Castigliano's theorem, the horizontal displacement of A due to applied external load W isgiven by

The bending moment at the small element is . Thus, the horizontaldisplacement of A

(i.e. deflection is in direction)

Example 4.20 Determine the deflection of the end A of the beam as shown in Figure 4.28. The flexibility of

the spring is .

Solution: Reactions at support B and C are (upward) and (downward)

Force in the spring = Reaction,

Deflection under the load is given by

where is the total strain energy stored in the system; is the energy stored in the

member AB ; is the energy stored in the member BC ; and = strain energy in the spring.

Strain energy in the spring is given by

Consider member AB : ( x measured from A )

Consider member BC : ( x measured from C )

Thus,

The deflection of point A,

Recap In this course you have learnt the following

Deflection by strain energy method.

Evaluation of strain energy in member under different loading.

Application of strain energy method for different types of structure.

Module 4 : Deflection of StructuresLecture 5 : Bending Deflection Due to Temperature Variation

Objectives

In this course you will learn the following

Bending deflection of beams due to temperature variation.

4.6 Bending Deflection due to Temperature Variation

Consider a beam member (refer Figure 4.29) subjected to temperature gradient over the depth of beamsuch that

(4.22)

where = temperature at the top of the beam; and = temperature at the bottom of the beam.

The deflection of the beam due to temperature variation is shown in Figure 4.29(b). It is assumed thattemperature varies linearly through the depth, d and is the coefficient of thermal expansion of thematerial.

Consider a small element of length dx . The strain at top and bottom of the small elements are

(4.23a)

(4.23b)

The curvature of the beam is given by

(4.24)

The equation (4.24) can be used for finding out the bending deflection in beams due to temperaturevariation. If the beam is restrained from rotation, the moment induced in the beam will be given by

(4.25)

The equation (4.25) is obtained by equating the right hand side of equation (4.24) to from the simple

bending theory.

Temperature deflections of a cantilever beam:

Consider a cantilever beam as shown in Figure 4.30 subjected to temperature gradient over

the depth. Integrating the equation (4.24)

(4.26)

(4.27)

Boundary conditions: At x = 0, and v = 0 will give the values of arbitrary constants as .

The slope and deflection of the free end of the cantilever beam are

(4.28a)

(4.28a)

Recap In this course you have learnt the following

Bending deflection of beams due to temperature variation.

Module 4 : Deflection of StructuresLecture 6 : Maxwell-Betti Law of Reciprocal Deflections

Objectives

In this course you will learn the following

Maxwell-Betti Law of reciprocal deflection.

Illustrative examples for proving law of reciprocal deflection.

4.7Maxwell-Betti Law of Reciprocal Deflections

Maxwell-Betti Law of real work is a basic theorem in the structural analysis. Using this theorem, it will beestablished that the flexibility coefficients in compatibility equations, formulated to solve indeterminatestructures by the flexibility method, form a symmetric matrix and this will reduce the number of deflectioncomputations. The Maxwell-Betti law also has applications in the construction of influence lines diagrams forstatically indeterminate structures. The Maxwell-Betti law, which applies to any stable elastic structure (abeam, truss, or frame, for example) on unyielding supports and at constant temperature, states:

The deflection of point A in direction 1 due to unit load at point B in direction 2 is equal in themagnitude to the deflection of point B in direction 2 produced by a unit load applied at A indirection 1.

The Figure 4.31 explains the Maxwell-Betti Law of reciprocal displacements in which, the displacement is

equal to the displacement .

In order to prove the reciprocal theorem, consider the simple beams shown in Figure 4.32.

Let a vertical force at point B produces a vertical deflection at point A and at point B as shown

in Figure 4.32(a). Similarly, in Figure 4.32(b) the application of a vertical force at point A produces avertical deflections and at points A and B , respectively. Let us evaluate the total work done by the

two forces and when they are applied in different order to the zero to their final value.

Case 1: applied and followed by

(a) Work done when is gradually applied

s

(b) Work done when is gradually applied with in place

Total work done by the two forces for case 1 is

(4.29)

Case2: applied and followed by

(c) Work done when is gradually applied

(d) Work done when is gradually applied with in place

Total work done by the two forces for case 2 is

(4.30)

Since the final deflected position of the beam produced by the two cases of loads is the same regardless of

the order in which the loads are applied. The total work done by the forces is also the same regardless of theorder in which the loads are applied. Thus, equating the total work of Cases 1 and 2 give

(4.31)

If , the equation (4.31) depicts the statement of the Maxwell-Betti law i.e.

The Maxwell-Betti theorem also holds for rotations as well as rotation and linear displacement in beams andframes.

Example 4.21 Verify Maxwell-Betti law of reciprocal displacement for the direction 1 and 2 of the pin-jointedstructure shown in Figure 4.33(a).

Solution: Apply the forces and in the direction 1 and 2, respectively. The calculation of total strainenergy in the system is given in Table 4.5.

Table 4.5

Member Length Force PAB L -( )

AC L P 1

Since , hence the Maxwell-Betti law of reciprocal displacement is proved.

Example 4.22 Verify Maxwell-Betti law of reciprocal displacement for the cantilever beam shown in Figure4.34(a).

Solution: Apply the forces and in the directions 1 and 2, respectively. The total strain energy storedis calculated below.

Consider any point X at a distance x from B .

Since , the Maxwekk-Betti law of reciprical displacement is proved.

Example 4.23 Verify Maxwell-Betti law of reciprocal displacement for the rigid-jointed plane frame withreference to marked direction as shown in Figure 4.35(a). EI is same for both members.

Solution: Apply the forces and in the directions 1 and 2, respectively as shown in Figure 4.35(b).

Consider AB : ( x measured from A )

Consider BC : ( x measured from B )

Thus

The displacement in the direction 1 due to unit load applied in 2 is

The displacement in the direction 2 due to unit load applied in 1 is

Since , proves the Maxwell-Betti law of reciprocal displacements.

Recap In this course you have learnt the following

Maxwell-Betti Law of reciprocal deflection.

Illustrative examples for proving law of reciprocal deflection.

Module 4 : Deflection of StructuresLecture 7 : Tutorial Problems Objectives In this course you will learn the following

Some tutorial problems related to this module.

TUTORIAL PROBLEMS

T4.1 Using moment area method, determine the end slope and deflection of the mid-span point C inthe beam shown in Figure T4.1.

T4.2 Determine the slope and deflection at the internal hinge of the beam shown in the Figure T4.2.

T4.3 Determine the slope at A and deflection of B of the beam shown in Figure T4.3 using themoment area method.

T4.4 Find the maximum slope and deflection of the simply supported beam shown in Figure T4.4using moment area method.

T4.5 Using conjugate beam method determine the ratio P / Q for the beam shown in Figure T4.5 if(i) slope at C is zero and (ii) deflection at C is zero.

T4.6 Determine the deflection at B for the beam using conjugate beam method. Take EI=7×104 kNm2.

T4.7 Determine the mid-span deflection for the beam using conjugate beam method.

T4.8 Determine the expression for the slope and deflection of the free end of the cantilever beam shown in the Figure T4.8

T4.9 Determine the horizontal and vertical displacement of joint C of the pin-jointed frame asshown in Figure T4.9. All the members of the frame have uniform axial rigidity (AE) . Use the unitload method and verify by the strain energy method.

T4.10 Determine the vertical displacements of joint D and E of the pin-jointed frame as shown inFigure 4.10. All the members of the frame have uniform axial rigidity (AE) . Use the unit loadmethod and verify by the strain energy method.

T4.11 Determine the vertical displacement of joint C and horizontal displacement E of the pin-jointed frame as shown in Figure T4.11. All the members of the frame have uniform axial rigidity(AE) . Use the unit load method and verify by the strain energy method.

T4.12 Determine the horizontal and vertical displacements of joints C and D of the pin-jointedframe as shown in Figure T4.12. All the members of the frame have uniform axial rigidity (AE) . Usethe unit load method and verify by the strain energy method.

T4.13 Determine the vertical displacement of joint C and horizontal displacement of joint D of thepin-jointed frame as shown in Figure T4.13. All the members of the frame have uniform axialrigidity, AE=15×103 kN. Use the unit load method and verify by the strain energy method.

T4.14 Determine the vertical displacement of joint D if member BC of the pin-jointed frame asshown in Figure T4.14 is long by an amount ∆ from the original length L . All the members of theframe have same AE value.

T4.15 Using unit load method and strain energy method, determine the deflection and slope ofpoint C of the uniform beam shown in Figure T4.15.

T4.16 Using unit load method and strain energy method, determine the deflection at the center ofthe beam shown in Figure T4.16 under the distributed load w.

T4.17 Using unit load method and strain energy method, determine the deflection and rotation ofthe point B of the beam shown in Figure T4.17. The beam is carrying a uniformly distributed load,w over the entire length.

T4.18 Using unit load method and strain energy method, determine the deflection under the loadW and horizontal displacement of roller at D of the rigid-jointed plane frame shown in Figure T4.18.

T4.19 Using unit load method and strain energy method, determine the deflection at the center ofAB and horizontal displacement of roller at C of the rigid-jointed plane frame shown in FigureT4.19.

T4.20 Verify the Maxwell-Betti Law of reciprocal displacements for the structures in Figure T4.20.

The required direction 1 and 2 are marked on the structures.

Figure T4.20

Recap In this course you have learnt the following

You have learned some tutorial problems related to this module.

Answers of tutorial problems

T4.1 and

T4.2

T4.3 and

T4.4 and

T4.5 (i) and (ii)

T4.6

T4.7

T4.8 and

T4.9

T4.10

T4.11

T4.12

T4.13

T4.14

T4.15

T4.16

T4.17

T4.18

T4.19

T4.20(a) (b)

(c) (d)

Module 5 : Force Method - Introduction and applicationsLecture 1 : Analysis of Statically Indeterminate Beams Objectives In this course you will learn the following

Introduction to statically indeterminate structure.

Analysis of statically indeterminates beam using moment area and conjugate beam method.

To demonstrate the application of moment area and conjugate beam method through illustrativeexamples.

5.1 Introduction

A strucure in which the laws of statics are not sufficient to determine all the unknown forces or moments issaid to be statically indeterminate. Such structures are analyzed by writing the appropriate equations ofstatic equilibrium and additional equations pertaining to the deformation and constraints known ascompatibility condition.

The statically indeterminate structures are frequently used for several advantages. They are relatively moreeconomical in the requirement of material as the maximum bending moments in the structure are reduced.The statically indeterminate are more rigid leading to smaller deflections. The disadvantage of theindeterminate structure is that they are subjected to stresses when subjected to temperature changes andsettlements of the support. The construction of indeterminate structure is more difficult if there aredimensional errors in the length of members or location of the supports.

This chapter deals with analysis of statically indeterminate structures using various force methods.

5.2 Analysis of Statically Indeterminate Beams

The moment area method and the conjugate beam method can be easily applied for the analysis of staticallyindeterminate beams using the principle of superposition. Depending upon the degree of indeterminacy ofthe beam, designate the excessive reactions as redundant and modify the support. The redundant reactionsare then treated as unknown forces. The redundant reactions should be such that they produce thecompatible deformation at the original support along with the applied loads. For example consider a proppedcantilever beam as shown in Figure 5.1(a). Let the reaction at B be R as shown in Figure 5.1(b) which can beobtained with the compability condition that the downward vertical deflection of B due to applied loading(i.e. shown in Figure 5.1(c)) should be equal to the upward vertical deflection of B due to R (i.e. shown in Figure 5.1(d)).

Example 5.1 Determine the support reactions of the propped cantilever beam as shown in Figure 5.2(a).

Solution: The static indeterminacy of the beam is = 3 – 2 = 1. Let reaction at B is R acting in the upwarddirection as shown in Figure 5.2(b). The condition available is that the .

(a) Moment area method

The bending moment diagrams divided by EI of the beam are shown due to P and R in Figures 5.2(c) and (d),respectively.

Since in the actual beam the deflection of the point B is zero which implies that the deviation of point B fromthe tangent at A is zero. Thus,

or

Taking moment about A , the moment at A is given by

The vertical rection at A is

The bending moment diagram of the beam is shown in Figure 5.2(e).

(b) Conjugate beam method

The corresponding conjugate beam of the propped cantilever beam and loading acting on it are shown inFigure 5.2(f).

The unknown R can be obtained by taking moment about B i.e.

Example 5.2 Determine the support reactions of the fixed beam with internal hinge as shown in Figure 5.3(a).

Solution: The static indeterminacy of the beam is = 4-2-1 =1 Let the shear in the internal hinge be R . The freebody diagrams of the two separated portions of the beam are shown in Figure 5.3(b) along with their M/EIdiagrams. The unknown R can be obtained with the condition that the vertical deflection of the free ends of thetwo separated cantilever beams is identical.

Consider AC : The vertical displacement of C is given by

or

Consider CB : The vertical displacement of C is given by

Equating the from Eqs. (i) and (ii)

Solving for R will give

The reactions at the supports are given by

Example 5.3 Determine the support reactions of the fixed beam with one end fixed and other supported onspring as shown in Figure 5.4(a). The stiffness of spring is .

Solution: The static indeterminacy of the beam is = 3–2 =1. Let the force in the spring be R . The free bodydiagram of the beam along with the M/EI diagram and spring are shown in Figure 5.4(b) and (c), respectively.The unknown R can be obtained with the condition that the vertical deflection of the free end of the beam andspring is identical.

Using moment area theorem, the deflection of free end A of the beam is

The downward deflection of spring is

Equating and

The bending moment at B

The vertical reaction at B

The force in the spring (compressive)

The bending moment diagram of the beam is shown in Figure 5.4(d).

Example 5.4 Determine the support reactions of the fixed beam as shown in Figure 5.5(a). The beam carries

a uniformly distributed load, w over the left half span.

Solution: The static indeterminacy of the beam is = 4-2 =2. Let the reactions at B be the unknown as shownin Figure 5.5(b).

(a) Moment Area Method

The free body diagram of the beam is shown below along with their M/EI diagrams. The unknowns and

can be obtained with the condition that the vertical deflection and slope at B are zero.

Since the change of slope between points A and B is zero (due to fixed supports at A and B ), therefore,according to the first moment area theorem,

or (i)

or (ii)

Solving equations (i) and (ii)

The bending moment diagram of the beam is shown in Figure 5.5(f)

(b) Conjugate Beam Method

The corresponding conjugate beam (i.e. free-free beam) and loading on it are shown in Figure 5.5(g).

Considering vertical equilibrium of all forces acting on Conjugate beam

or (iii)

Taking moment about A

or (iv)

Solving eqs. (iii) and (iv)

Example 5.5 The end B of a uniform fixed beam sinks by an amount D . Determine the end reactions usingmoment area method.

Solution: The degree of indeterminacy is 2. Let end reactions due to settlement at B be and asshown in Figure 5.6(b). The M/EI diagram of the beam is shown in Figure 5.6(c).

Applying first moment area theorem between A and B

or (i)

Applying second moment area theorem between point A and B

or

Solving eqs. (i) and (ii)

By equilibrium conditions, the reactions at support A are

and

Example 5.6 Determine the support reactions of the continuous beam as shown in Figure 5.7(a).

Solution: The static indeterminacy of the beam is = 3-2 =1. Let the vertical reaction at B be the unknown Ras shown in Figure 5.7(b). The M/EI diagrams of the beam are shown in Figure 5.7(c).

Because of symmetry of two spans the slope at B , As a result

or

or

The vertical reaction at A and C are

The bending moment diagram of the beam is shown in Figure 5.7(d).

Recap In this course you have learnt the following

Introduction to statically indeterminate structure.

Analysis of statically indeterminates beam using moment area and conjugate beam method.

To demonstrate the application of moment area and conjugate beam method through illustrativeexamples.

Module 5 : Force Method - Introduction and applicationsLecture 2 : The Force Method Objectives In this course you will learn the following

Concept of force method for analysis of statically indeterminate structure.

Selection of the basic determinate structure.

Illustration of force method by numerical examples.

5.3 The Force Method

The force method is used to calculate the response of statically indeterminate structures to loads and/orimposed deformations. The method is based on transforming a given structure into a statically determinateprimary system and calculating the magnitude of statically redundant forces required to restore thegeometric boundary conditions of the original structure. The force method (also called the flexibilitymethod or method of consistent deformation ) is used to calculate reactions and internal forces in staticallyindeterminate structures due to loads and imposed deformations.

The basic steps in the force method are as follows:

(a) Determine the degree of static indeterminacy, n of the structure.

(b) Transform the structure into a statically determinate system by releasing a number of static constraintsequal to the degree of static indeterminacy, n. This is accomplished by releasing external supportconditions or by creating internal hinges. The system thus formed is called the basic determinate structure.

(c) For a given released constraint j, introduce an unknown redundant force corresponding to the typeand direction of the released constraint.

(d) Apply the given loading or imposed deformation to the basic determinate structure . Use suitablemethod (given in Chapter 4) to calculate displacements at each of the released constraints in the basicdeterminate structure .

(e) Solve for redundant forces ( j =1 to n ) by imposing the compatibility conditions of the originalstructure. These conditions transform the basic determinate structure back to the original structure byfinding the combination of redundant forces that make displacement at each of the released constraintsequal to zero.

It can thus be seen that the name force method was given to this method because its primarycomputational task is to calculate unknown forces , i.e. the redundant forces through .

5.3.1Selection of the basic determinate structure

There is no limit to the number of different basic determinate structure that can be generated for a givenstructure. The choice of structure, however, must ensure that the primary system is stable. In addition, itis recommended that the basic determinate structure be chosen to minimize computational effort andmaximize computational accuracy.

(a) Stability of Basic determinate structure

It is not sufficient merely to release the correct number of statical constraints in generating a basicdeterminate structure. Care must be taken to ensure that the basic determinate structure is stable. Thisfact is explained in the Table 5.1 where any arbitrary release of constraint can result into the unstablebasic determinate structure.

Table 5.1 Selection of basic determinate structure

Given indeterminate structure Unstable basic determinatestructure

Stable basic determinatestructure

(b) Choice of Basic determinate Structure for minimum Computation work

The computational effort required in calculating the response of a given structure using the force methodcan vary significantly depending on the choice of basic determinate structure. In this regard, there are twoissues to consider:

1. Select the basic determinate structure such as the displacements can be easily computed (i.e. convertingit into simple structure).

2. Select the basic determinate structure to maximize the number of flexibility coefficients equal to zero.

These issues are illustrated in the following examples.

Consider a fixed beam as shown in Figure 5.8(a). The beam is non-prismatic with degree of indeterminacy2. Three basic determinate structures are shown in Figures 5.8(b), (c) and (d). Among the three structuresthe computation effort will be minimum for the beam as in Figure 5.8(b) as the resulting basic determinatestructure consists of two uniform cantilever beams.

The another structure under consideration is a four-span continuous beam as shown in Figure 5.9. Thedegree of static indeterminacy of the beam is 3. Two basic determinate structures are illustrated. On the left-hand side of the figure, the basic determinate structure is formed by releasing moment in the beam at thethree interior supports. On the right-hand side of the figure, the basic determinate structure is formed byreleasing the vertical reaction at the three interior supports.

For each basic determinate structure, bending moments (bending moment in the basic determinate

structure due to applied loading) (due to ), (due to ), and (due to ) are plotted. Forthe basic determinate structure on the left-hand side of the diagram, all integrations required for calculating

the deflections can be easily performed using integration tables. In addition, for the left-hand side of thediagram, several deflection coefficients are zero. On the right-hand side, however, all coefficients arenonzero. The choice of basic determinate structure on the left allowed the influence of a given redundantforce to be restricted to a relatively small portion of the structure (two spans in this particular case). For

the structure on the right-hand side, the influence of a given redundant force R j is felt throughout thestructure. It can be concluded that the basic determinate structure on the left-hand side is preferablebecause it reduces the computational effort.

Given staticallyindeterminate

beam

Basic determinate structure

Figure 5.9

Example 5.7 Analyze the continuous beam shown in Figure 5.10(a) using the force method. Also, draw thebending moment diagram. EI is constant for entire beam.

Solution: The degree of static indeterminacy = 3–2 =1. The moment at B is taken as redundant R and thebasic determinate structure will be then two simply supported beams as shown in Figure 5.10(b).

Rotation of point B due to applied loads

Rotation of point B due to R

Equating the rotation of point B due to applied loads and R i.e.

or

The reaction at A is given by

The vertical reaction at C is given by

The vertical reaction at B is

The bending moment diagram of the beam is shown in Figure 5.10(e).

Example 5.8 Analyze the uniform continuous beam shown in Figure 5.11(a) using the force method. Also,draw the bending moment diagram.

Solution: The degree of static indeterminacy of the beam = 4 – 2 = 2. The moment at A and B are taken asunknown and , respectively.

Equating the rotation at A due to applied loading and unknown and equal to zero i.e.

or (i)

Equating the rotation at B due to applied loading and and equal to zero i.e.

or (ii)

Solving equations (i) and (ii)

The bending moment diagram of the beam is shown in Figure 5.11(d)

Example 5.9 Find the force in various members of the pin-jointed frame shown in Figure 5.12(a). AE isconstant for all members.

Solution: The static indeterminacy of the pin-jointed frame = 1. The vertical reaction at C is taken asunknown force R . The computation of deflection of point C due to applied loading and R are shown in Tables5.2 and 5.3, respectively.

Table 5.2

Member Length

AB

AC L 0 1 0AD

The vertical displacement of joint C due to applied loading = ( )

Table 5.3

Member Length

AB

AC L R 1 RAD

The vertical displacement of joint C due to R = ( )

Adding the displacement of point C due to applied loading and R and equating it to zero i.e.

kN

The force in various members of the frame are as follows

kN (Tensile)

kN (Compressive)

kN (Compressive)

Example 5.10 Find the force in various members of the pin-jointed frame shown in Figure 5.13(a). AE isconstant for all members.

Solution: The static indeterminacy of the pin-jointed frame = 1. The horizontal reaction at B is taken asunknown force R . The computation of horizontal deflection of point B due to applied loading and R are shownin Tables 5.4 and 5.5, respectively.

Table 5.4

Member Length(m) L

AC 2 -100.44 1 -200.88

AD 3.46 161.912 -970.32

CD 2 -96.967 2 -387.869

BC 3.46 53.97 -323.83

BD 2 -93.48 1 -186.96

-2069.86

The horizontal displacement of joint B due to applied loading = ( )

Table 5.5Member Length(m) L

AC 2 R 1 2R

AD 3.46 R 10.392 R

CD 2 2R 2 8 R

BC 3.46 R 10.392 R

BD 2 R 1 2R

32.784 R

The horizontal displacement of joint B due to R = ( )

Adding the horizontal displacemet of point B due to applied loading and R and equating it to zero i.e.

R = 63.136 kN

The force in various members are as follows

= –37.305 kN

= 52.558 kN

= 29.328 kN

= –55.383 kN

= –30.345 kN

( –ve indicates compressive force and +ve indicates tensile force)

Example 5.11 Analyze the non-prismatic fixed beam shown in Figure 5.14(a) using force method.

Solution: Degree of indeterminacy of the system = 2. We choose shear force and moment at section C asredundant and , respectively.

Total displacement in the direction of

or (i)

Total rotation in the direction of

or (ii)

The reactions at support are given by

The bending moment diagram of the beam is shown in Figure 5.14(d)

Example 5.12 Determine the horizontal thrust in a two-hinged trapezoidal arch. EI is constant.

Figure 5.15(c) BMD due toapplied loading

Solution: The degree of static indeterminacy = 4–3 = 1. Let H be the unknown horizontal reaction at A andB . Let and be the vertical reactions at support A and B respectively

Due to symmetry,

Displacement in the direction of H due to applied loading is calculated by multiplication of two bendingmoment diagrams of Figure 5.15(c) and (e).

Displacement in the direction of H due to H obtained by multiplying diagram of Figure 5.10(d) and (e).

Since the net deflection in the direction of H is zero, therefore

or

Example 5.13 Determine the reaction of the propped cantilever beam if the beam is assumed to besubjected to a linear temperature gradient such that the top surface of the beam is at temperature and

lower at . The beam is uniform having flexural rigidity as EI and depth d . The coefficient of thermal

expansion for beam material is .

Solution: The degree of static indeterminacy =1. Remove the support at B and allow the beam deflect freelyunder the temperature variation. The deflection of the free end of the beam due to temperature variation(from eqs 4.28 of Chapter 4).

(i)

Apply the force R at point B such that the deflection in the direction of R is equal to . Since deflection of

a cantilever beam due to force R is equal , therefore

(ii)

Equating the from two expressions of Eqs. (i) and (ii)

The vertical reaction and bending moment at A will be

Recap In this course you have learnt the following

Concept of force method for analysis of statically indeterminate structure.

Selection of the basic determinate structure.

Ellustration of force method by numerical examples.

Module 5 : Force Method - Introduction and applicationsLecture 3 : Analysis of Statically Indeterminate Structure Objectives In this course you will learn the following

Energy method for analysis of statically indeterminate structures.

Illustrative examples for analysis of statically indetrminate structures using every method.

5.4 Analysis of Statically Indeterminate Structures by Energy Method

Let a statically indeterminate structure has degree of indeterminacy as n . On the selected basicdeterminate structure apply the unknown forces , ..... and . Using the Eq. (4.16) the displacement

in the direction of is expressed by

( j = 1, 2, .. ……n) (5.1)

The equations (5.1) will provide the n linear simultaneous equations with n unknowns , ..... and .

Since the is known, therefore, the solution of simultaneous equations will provide the desired ( j =1,

2,…., n ).

For structures with members subjected to the axial forces only (i.e. pin-jointed structures), the equation(5.1) is re-written as

(5.2)

where P is the force in the member due to applied loading and unknown ( j =1, 2,…., n ); and L and AE

are length and axial rigidity of the member, respectively.

For structures with members subjected to the bending moments (i.e. beams and rigid-jointed frames), theequation (5.1) is re-written as

(5.3)

where M is the bending moment due to applied loading and unknown ( j =1, 2,…., n ) at a small

element of length dx ; and EI is the flexural rigidity.

Example 5.14 A beam is suspended by three springs as shown in Figure 5.17(a). The flexibility of thesprings AD , BE and CF are , and respectively. The beam carries a load W at the middle of DE.Determine the force in the spring BE assuming (i) the beam to be stiff in comparison to the springs and (ii)flexible with flexural rigidity EI .

Solution: The degree of static indeterminacy = 3–2 = 1. Let the force in the spring BE be R as shown inFigure 5.17(b). Taking moment about point F , we have

Similarly, taking moment about point D , we have

(i) When beam is rigid

Total energy stored in the system is due to springs only as the beam is rigid. Thus,

Since the displacement of point E is zero in the vertical direction implying that

or

(ii) When beam is flexible

The total energy stored in the beam

Span DG : ( x measured from D )

Span GE : ( x measured from D )

Span EF : ( x measured from F )

Thus,

Total strain energy in the system

Hence,

Since

Example 5.15 The free ends of two cantilever beams each of length L and flexural rigidity EI are joinedtogether with a spring as shown in Figure 5.18(a). The stiffness of the spring is . Determine theforce in the spring due to a concentrated load W acting at center of the lower cantilever.

Solution: Let the force in the spring be R as shown in Figure 5.18(b). According to the Castigliano'stheorem

where U is the total strain energy stored in the system.

Consider beam DE : ( x measured from D ),

and

Consider the spring,

Consider the beam AC ,

Since

Example 5.16 Find the expression for the prop reaction in the propped cantilever beam shown in theFigure 5.19(a).

Solution: Let reaction at support A be R . According to the Castigliano's theorem

The bending moment at any point X at a distance x from A is given by

Since

Example 5.17 Determine the force in various members of the pin-jointed frame shown in Figure 5.20(a).Length and AE is constant for all members.

Solution: The static indeterminacy of the pin-jointed frame is =12+3–7×2 = 1. Let the force in the memberBG be R as shown in Figure 5.20(b). According to the Castigliano's theorem

The computation of is made in Table 5.6.

Table 5.6

Member Length, L ( m ) P Final force (kN)

AB 2 -120 +R 1 2 R- 240 -40AG 2 120 -R -1 2 R- 240 40AF 2 -60 +R 1 2 R- 120 20BC 2 -120 +R 1 2 R- 240 -40BG 2 -R -1 2 R -80CD 2 -120 +R 1 2 R- 240 -40CG 2 120 -R -1 2 R- 240 40DE 2 -60 +R -1 2 R- 120 20DG 2 60 -R -1 2 R- 120 -20EF 2 -60 +R 1 2 R- 120 20EG 2 60 -R -1 2 R- 120 -20FG 2 60 -R -1 2 R- 120 -20

or

R = 80 kN

The final force in various members of the frame is shown in Table 5.6.

Example 5.18 Determine the force in various members of the pin-jointed frame as shown in Figure 5.21(a),if the member BC is short by an amount of . All members of the frame have same axial rigidity as AE.

Solution: The static indeterminacy of the pin-jointed frame is =5 + 4 - 2×4=1. Since the member BC isshort by an amount of , therefore, apply a force R in the member BC such that displacement in thedirection of R is . Thus, according to the Castigliano's theorem.

The computation of is made in Table 5.7.

Table 5.7

Member Length,L ( m ) F Final force

AB L R 1 RL 1AC

BC L R 1 RL 1BD L

CD R 1 RL 1

or

The final force in various members of the frame is shown in Table 5.7.

Example 5.19 Determine the horizontal reaction of the portal frame shown in Figure 5.22(a) by energymethod. Also, calculate the horizontal reaction when the member BC is subjected to distributed load, w overentire length.

Solution: Static indeterminacy of the frame = 1.

Let the horizontal reaction, H at D be the redundant. The reaction at A and D are

For the span AB ( x measured from A ),

For the span BE ( x measured from B ),

For the span CD ( x measured from D ),

For the span CE ( x measured from C ),

Since

Horizontal reaction due to udl, w over BC :

The horizontal reaction due to small incremental load wdx is given by

(using the expression derived earlier for concentrated force and putting P = wdx , a = x and b = L – x ).

The horizontal reaction due to entire distributed load

Example 5.20 Analyze the portal frame shown in Figure 5.23 by strain energy method.

Solution: Static indeterminacy of the frame = 2. Horizontal and vertical reactions at A are taken asredundant.

For the span AB ( x measured from A ),

Since

or 6 V + 4 H = 3 wL (i)

and

or 32 V + 12 H = 15 wL (ii)

Solving Eqs (i) and (ii) for H and V ,

Example 5.21 Determine the support reactions of the continuous beam as shown in Figure 5.24(a) if thebeam is assumed to be subjected to a linear temperature gradient such that the top surface of the beam is attemperature and lower at . The beam is uniform having flexural rigidity as EI and depth d . The

coefficient of thermal expansion for beam material is .

Solution: The degree of static indeterminacy =2. Remove the supports at B and C and allow the beam todeflect freely under the temperature variation. The deflection of the points B and C of the beam due totemperature variation

(i)

(ii)

Apply the forces and at point B and C , respectively. According to Castigliano's theorem

(iii)

(iv)

Consider BC : ( x measured from C )

Consider AB : ( x measured from B )

;

Thus,

or (v)

Similarly,

or (vi)

Solving eqs. (v) and (vi)

and

The reactions of the beam are shown in Figure 5.24(d).

Recap In this course you have learnt the following

Energy method for analysis of statically indeterminate structures.

Illustrative examples for analysis of statically indetrminate structures using every method.

Module 5 : Force Method - Introduction and applicationsLecture 4 : Three Moment Equation Objectives In this course you will learn the following

Derivation of three moment equation for analysis of continous beams.

Demonstration of three moment equation using numerical examples.

5.5 Three Moment Equation

The continuous beams are very common in the structural design and it is necessary to develop simplifiedforce method known as three moment equation for their analysis. This equation is a relationship that existsbetween the moments at three points in continuous beam. The points are considered as three supports ofthe indeterminate beams. Consider three points on the beam marked as 1, 2 and 3 as shown in Figure5.25(a). Let the bending moment at these points is , and and the corresponding vertical

displacement of these points are , and , respectively. Let and be the distance betweenpoints 1 – 2 and 2 – 3, respectively.

The continuity of deflected shape of the beam at point 2 gives

(5.4)

From the Figure 5.25(d)

and (5.5)

where

and (5.6)

Using the bending moment diagrams shown in Figure 5.25(c) and the second moment area theorem,

(5.7)

(5.8)

where and are the areas of the bending moment diagram of span 1-2 and 2-3, respectivelyconsidering the applied loading acting as simply supported beams.

Substituting from Eqs. (5.7) and Eqs. (5.8) in Eqs. (5.4) and Eqs. (5.5).

(5.9)

The above is known as three moment equation . Sign Conventions

The and are positive for sagging moment and negative for hogging moment. Similarly, areas

and are positive if it is sagging moment and negative for hogging moment. The displacements

and are positive if measured downward from the reference axis.

Example 5.22 Analyze the continuous beam shown in Figure 5.26(a) by the three moment equation. Drawthe shear force and bending moment diagram.

Solution: The simply supported bending moment diagram on AB and AC are shown in Fig 5.26 (b). Sincesupports A and C are simply supported

Applying the three moment equation to span AB and BC ( = = = 0)

or =-56.25 kN.m

The reactions at support A , B and C are given as

= 41.25 kN

= 41.25 kN

= 120 + 40 3 – 41.25 – 41.25 = 157.5 kN

The bending moment and shear force diagram are shown in Figures 5.26(c) and (d), respectively

Example 5.23 Analyze the continuous beam shown in Figure 5.27(a) by the three moment equation. Drawthe shear force and bending moment diagram.

Solution: The effect of a fixed support is reproduced by adding an imaginary span as shown in Figure

5.27 (b). The moment of inertia, of the imaginary span is infinity so that it will never deform and thecompatibility condition at the end A , that slope should be is zero, is satisfied.

Applying three moment equation to the span and AB :

or + = – 135 (i)

Span AB and BC :

or + = – 225 (ii)

Solving Eqs. (i) and (ii), = – 45 kNm and = – 45 kNm

The shear force and bending moment diagram are shown in Figures 5.27(d) and (e), respectively.

Example 5.24 Analyze the continuous beam shown in Figure 5.28(a) by the three moment equation. Drawthe shear force and bending moment diagram.

Solution: The simply supported moment diagram on AB , BC and CD are shown in Figure 5.28(b). Since thesupport A is simply supported, The moment at D is .

Applying three moment equation to the span AB and BC :

or (i)

Span BC and CD : ( )

or (ii)

Solving Eqs. (i) and (ii) will give and .

The bending moment and shear force diagram are shown in Figures 5.28(d) and (c), respectively.

Example 5.25 Analyze the continuous beam show in Fig. 5.29(a) by the three moment equation method ifsupport B sinks by an amount of 10 mm. Draw the shear force and bending moment diagram. Take flexuralrigidity .

Solution: Since support A and D are simply supported, .

Applying the three moment equation for span AB and BC : ( )

or (i)

Span BC and CD :

or (ii)

Solving Eqs. (i) and (ii), and .

The bending moment diagram is shown in Figure 5.29(b).

Recap In this course you have learnt the following

Derivation of three moment equation for analysis of continous beams.

Demonstration of three moment equation using numerical examples.

Module 5 : Force Method - Introduction and applicationsLecture 5 : Tutotial Problems Objectives In this course you will learn the following

Some tutorial problems related to this module.

TUTORIAL PROBLEMS

T5.1 Determine the support reactions of the propped cantilever beam as shown in Figure T5.1. Usemoment area method and verify by conjugate beam method.

T5.2 Determine the support reactions of the propped cantilever beam as shown in Figure T5.2. Use moment area method or conjugate beam method.

T5.3 Determine the shear in the internal hinge and support reactions of the fixed beam shown inFigure T5.3

T5.4 Determine the force in the spring of the beam shown in Figure T5.4.

T5.5 Determine the reaction at the prop end of the cantilever beam shown in the Figure T5.5.

T5.6 Determine the support reactions of the propped cantilever beam (Figure T5.6) if support Asettle downward by an amount of . Take flexural rigidity of member AB as EI . Member BC isrigid.

T5.7 Determine the support reactions of the uniform continuous beam as shown in Figure T5.7. AtB there is an internal hinge.

T5.8 Determine the force in the spring of the beam shown in Figure T5.8. The beam ABC is uniform with flexural rigidity EI . The stiffness of the spring is .

T5.9 Analyze the fixed beam shown in Figure T5.9.

T5.10 Determine the support reactions of the fixed beam shown in Figure T5.10.

T5.11 Determine the support reactions of continuous beam shown in Figure T5.11.

T5.12 Determine the force in various members of the pin-jointed structure as shown in FigureT5.12. All the members of the frame have the same axial rigidity, AE .

T5.13 Determine the force in various members of the pin-jointed structure as shown in FigureT5.13, if the temperature in the member BC rises by an amount . All the members of theframe have the same length, L and axial rigidity, AE . Take coefficient of thermal expansion as .

T5.14 Determine the force in various members of the pin-jointed frame as shown in Figure T5.14.All members of the frame have same axial rigidity as AE .

Figure T5.14

T5.15 Determine the force in various members of the pin-jointed frame as shown in Figure T5.15. Take kN for all members.

T5.16 Determine the support reactions of the rigid-jointed plane frame as shown in Figure T5.16.Both members of the frame have the same flexural rigidity as EI .

T5.17 Determine the support reactions of the rigid-jointed plane frame as shown in Figure T5.17 ifthe member BC is too long by an amount from the original length. Both the members of theframe have the same length and flexural rigidity.

T5.18 Determine the support reactions in the rigid-jointed plane frame as shown in Figure T5.18.Note that the member AB has an and for member BC the EI = .

T5.19 Determine the support reactions of rigid-jointed plane frame as shown in Figure T5.19. All members of frame have same flexural rigidity.

T5.20 Determine the support reactions of the rigid-jointed plane frame as shown in Figure T5.20,if the member BC is assumed to be subjected to a linear temperature gradient such that the topsurface of the beam is at temperature and lower at . The beam is uniform having flexuralrigidity as EI and depth d . The coefficient of thermal expansion for beam material is a .

T5.21 Determine the support reactions of the rigid-jointed plane frame as shown in Figure T5.21.

T5.22 Using theorem of three moment find the reactions of the uniform beam shown in FigureT5.22.

T5.23 Using theorem of three moments, determine the reactions of the uniform continuous beamshown in Figure T5.23.

T5.24 Using the theorem of three moments analyze the uniform continuous beam shown in Figure T5.24.

T5.25 Using the theorem of three moments, determine the support reaction, if support B settlesdown by an amount . Take the flexural rigidity of the entire beam as EI .

Recap In this course you have learnt the following

You have learned some tutorial problems related to this module.

Answers of tutorial problems

T5.1

and

T5.2

and

T5.3

and

T5.4

T5.5

T5.6

and

T5.7

T5.8

T5.9 and

T5.10

T5.11 and

T5.12

Member Force

AD

BD

DE 0

BE 0

CE 0

T5.13 Member Force

AB

BC

BE

CD

CE

T5.14 Member Force

AB 0

BC 50 (T)

CD – 50 (T)

AC 50 (T)

BD –50 (T)

T5.15 Member Force

AD 13.585 (T)

BD 58.248 (C)

CD 104.6 (C)

T5.16

and

T5.17

T5.18

and

T5.19

T5.20

and

T5.21

T5.22 and

T5.23

T5.24

T5.25

Module 6 : Influence LinesLecture 1 : Introduction: Variable LoadingsObjectives In this course you will learn the following

Introduction to variable loading on a structure.

The problems of analyzing a structure for multiple loading cases.

Introduction to the concept of infulence line as a solution to this problem.

6.1 Introduction: Variable Loadings

So far in this course we have been dealing with structural systems subjected to a specific set of loads.However, it is not necessary that a structure is subjected to a single set of loads all of the time. Forexample, the single-lane bridge deck in Figure 6.1 may be subjected to one set of a loading at one point oftime (Figure 6.1a) and the same structure may be subjected to another set of loading at a different point oftime. It depends on the number of vehicles, position of vehicles and weight of vehicles. The variation ofload in a structure results in variation in the response of the structure. For example, the internal forceschange causing a variation in stresses that are generated in the structure. This becomes a criticalconsideration from design perspective, because a structure is designed primarily on the basis of theintensity and location of maximum stresses in the structure. Similarly, the location and magnitude ofmaximum deflection (which are also critical parameters for design) also become variables in case ofvariable loading. Thus, multiple sets of loading require multiple sets of analysis in order to obtain thecritical response parameters.

Figure 6.1 Loading condition on a bridge deck at different points of time

Influence lines offer a quick and easy way of performing multiple analyses for a single structure. Responseparameters such as shear force or bending moment at a point or reaction at a support for several load setscan be easily computed using influence lines.

For example, we can construct influence lines for (shear force at B ) or (bending moment at C ) or

(vertical reaction at support D ) and each one will help us calculate the corresponding responseparameter for different sets of loading on the beam AD (Figure 6.2).

Figure 6.2 Different response parameters for beam AD

An influence line is a diagram which presents the variation of a certain response parameter due to thevariation of the position of a unit concentrated load along the length of the structural member. Let usconsider that a unit downward concentrated force is moving from point A to point B of the beam shown inFigure 6.3a. We can assume it to be a wheel of unit weight moving along the length of the beam. Themagnitude of the vertical support reaction at A ( ) will change depending on the location of this unit

downward force. The influence line for (Figure 6.3b) gives us the value of for different locations of

the moving unit load. From the ordinate of the influence line at C, we can say that when the unitload is at point C .

Figure 6.3b Influence line of for beam AB

Thus, an influence line can be defined as a curve, the ordinate to which at any abscissa gives the value ofa particular response function due to a unit downward load acting at the point in the structurecorresponding to the abscissa. The next section discusses how to construct influence lines using methods ofequilibrium.

Recap In this course you have learnt the following

Introduction to variable loading on a structure.

The problems of analyzing a structure for multiple loading cases.

Introduction to the concept of infulence line as a solution to this problem.

Module 6 : Influence LinesLecture 2 :Construction of Influence Lines using Equilibrium Methods

Objectives

In this course you will learn the following

Construction of influence lines using equilibrium conditions.

Some examples following this method.

6.2 Construction of Influence Lines using Equilibrium Methods

The most basic method of obtaining influence line for a specific response parameter is to solve the staticequilibrium equations for various locations of the unit load. The general procedure for constructing aninfluence line is described below.

1. Define the positive direction of the response parameter under consideration through a free body diagramof the whole system.

2..For a particular location of the unit load, solve for the equilibrium of the whole system and if required,as in the case of an internal force, also for a part of the member to obtain the response parameter for thatlocation of the unit load.This gives the ordinate of the influence line at that particular location of the load.

3.Repeat this process for as many locations of the unit load as required to determine the shape of theinfluence line for the whole length of the member. It is often helpful if we can consider a generic location(or several locations) x of the unit load.

4.Joining ordinates for different locations of the unit load throughout the length of the member,we get theinfluence line for that particular response parameter.

The following three examples show how to construct influence lines for a support reaction, a shear forceand a bending moment for the simply supported beam AB .

Example 6.1 Draw the influence line for (vertical reaction at A ) of beam AB in Fig. E6.1.

Solution:

Free body diagram of AB :

So the influence line of :

Example 6.2 Draw the influence line for (shear force at mid point) of beam AB in Fig. E6.2.

Solution:

For

For

So the influence line for :

Example 6.3 Draw the influence line for (bending moment at ) for beam AB in Fig. E6.3.

Solution:

For

For

So, the influence of :

Similarly, influence lines can be constructed for any other support reaction or internal force in the beam.However, one should note that equilibrium equations will not be sufficient to obtain influence lines inindeterminate structures, because we cannot solve for the internal forces/support reactions using onlyequilibrium conditions for such structures.

Recap In this course you have learnt the following

Construction of influence lines using equilibrium conditions.

Some examples following this method.

Module 6 : Influence LinesLecture 3 : Use of Influence Lines

Objectives

In this course you will learn the following

Use of influence line through an example

6.3 Use of Influence Lines

In this section, we will illustrate the use of influence lines through the influence lines that we have obtainedin Section 6.2. Let us consider a general case of loading on the simply supported beam (Figure 6.4a) anduse the influence lines to find out the response parameters ( , and ) for their loading. We canconsider this loading as the sum of three different loading conditions, (A), (B) and (C) (Figure 6.4b), eachcontaining only one externally applied force.

Figure 6.4 Application of influence lines for a general loading: (a) all the loads, and (b) thegeneral loading is divided into single force systems

For loading case (A), we can find out the response parameters using the three influence lines. Ordinate ofan influence line gives the response for a unit load acting at a certain point.

Therefore, we can multiply this ordinate by the magnitude of the force to get the response due to the realforce at that point. Thus

Similarly, for loading case (B):

And for case (C),

By the theory of superposition, we can add forces for each individual case to find the response parametersfor the original loading case (Figure 6.4a). Thus, the response parameters in the beam AB are:

One should remember that the method of superposition is valid only for linear elastic cases with smalldisplacements only. So, prior to using influence lines in this way it is necessary to check that theseconditions are satisfied.

It may seem that we can solve for these forces under the specified load case using equilibrium equationsdirectly, and influence lines are not necessary. However, there may be requirement for obtaining theseresponses for multiple and more complex loading cases. For example, if we need to analyse for ten loadingcases, it will be quicker to find only three influence lines and not solve for ten equilibrium cases.

The most important use of influence line is finding out the location of a load for which certain response willhave a maximum value. For example, we may need to find the location of a moving load (say a gantry) ona beam (say a gantry girder) for which we get the maximum bending moment at a certain point. We canconsider bending moment at point D of Example 6.3, where the beam AB becomes our gantry girder.Looking at the influence line of , one can say that will reach its maximum value when the load isat point D .

Influence lines can be used not only for concentrated forces, but for distributed forces as well, which isdiscussed in the next section.

Recap In this course you have learnt the following

Use of influence line through an example

Module 6 : Influence LinesLecture 4 : Using Influence Lines for Uniformly Distributed Load

Objectives

In this course you will learn the following

How to use influence lines for distributed loading cases.

6.4 Using Influence Lines for Uniformly Distributed Load

Consider the simply-supported beam AB in Figure 6.5, of which the portion CD is acted upon by a uniformlydistributed load of intensity w/unit length . We want to find the value of a certain response function Runder this loading and let us assume that we have already constructed the influence line of this responsefunction. Let the ordinate of the influence line at a distance x from support A be . If we consider anelemental length dx of the beam at a distance x from A , the total force acting on this elemental length iswdx . Since dx is infinitesimal, we can consider this force to be a concentrated force acting at a distance x .The contribution of this concentrated force wdx to R is:

Therefore, the total effect of the distributed force from point C to D is:

(area under the influence line from C to D )

Figure 6.5 Using influence line for a uniformly distributed loading

Thus, we can obtain the response parameter by multiplying the intensity of the uniformly distributed loadwith the area under the influence line for the distance for which the load is acting. To illustrate, let usconsider the uniformly distributed load on a simply supported beam (Figure 6.6). To find the verticalreaction at the left support, we can use the influence line for that we have obtained in Example 6.1. So

we can calculate the reaction as:

Figure 6.6 Uniformly distributed load acting on a beam

Similarly, we can find any other response function for a uniformly distributed loading using their influencelines as well.

For non-uniformly distributed loading, the intensity w is not constant through the length of the distributedload. We can still use the integration formulation:

However, we cannot take the intensity w outside the integral, as it is a function of x .

Recap In this course you have learnt the following

How to use influence lines for distributed loading cases.

Module 6 : Influence LinesLecture 5 : Müller-Breslau Principle

Objectives

In this course you will learn the following

The Müller-Breslau principle for influence lines.

Derivation of the principle for different types of internal forces.

Example of application of this principle.

6.5 Müller-Breslau Principle

The Müller-Breslau principle uses Betti's law of virtual work to construct influence lines. To illustrate themethod let us consider a structure AB (Figure 6.7a). Let us apply a unit downward force at a distance xfrom A , at point C . Let us assume that it creates the vertical reactions and at supports A and B ,respectively (Figure 6.7b). Let us call this condition “System 1.” In “System 2” (figure 6.7c), we have thesame structure with a unit deflection applied in the direction of . Here is the deflection at point C .

Figure 6.7 (a) Given system AB , (b) System 1, structure under a unit load, (c) System 2, structure with aunit deflection corresponding to

According to Betti's law, the virtual work done by the forces in System 1 going through the correspondingdisplacements in System 2 should be equal to the virtual work done by the forces in System 2 goingthrough the corresponding displacements in System 1. For these two systems, we can write:

The right side of this equation is zero, because in System 2 forces can exist only at the supports,corresponding to which the displacements in System 1 (at supports A and B ) are zero. The negative signbefore accounts for the fact that it acts against the unit load in System 1. Solving this equation we get:

In other words, the reaction at support A due to a unit load at point C is equal to the displacement at pointC when the structure is subjected to a unit displacement corresponding to the positive direction of supportreaction at A . Similarly, we can place the unit load at any other point and obtain the support reaction dueto that from System 2. Thus the deflection pattern in System 2 represents the influence line for .

Following the same general procedure, we can obtain the influence line for any other response parameteras well. Let us consider the shear force at point C of a simply-supported beam AB (Figure 6.8a). We applya unit downward force at some point D as shown in System 1 (Figure 6.8b). In system 2 (Figure 6.8c), weapply a unit deflection corresponding to the shear force, . Note that the displacement at point C isapplied in a way such that there is no relative rotation between AC and CB . This will avoid any virtual workdone by the bending moment at C ( ) going through the rotation in System 2. Now, according toBetti's law:

Figure 6.8 (a) Given system AB , (b) System 1, structure under a unit load, (c) System 2, structure with a

unit deflection corresponding to , (d) System 2, structure with a unit deflection corresponding to

Thus, the deflected shape in System 2 represents the influence line for shear force . Similarly, if we

want to find the influence line for bending moment , we obtain System 2 (Figure 6.8d) by applying aunit rotation at point C (that is, a unit relative rotation between AC and CB ). However, we do not want anyrelative displacement (between AC and CB ) at point C in order to avoid any virtual work done by goingthrough the displacements in System 2. Betti's law provides the virtual work equation:

So, as we have seen earlier, the displaced shape in System 2 represents the influence line for the responseparameter .

Construction of System 2 for a given response function is the most important part in applying the Müller-Breslau principle. One must take care that other than the concerned response function no other force (ormoment) in System 1 should do any virtual work going through the corresponding displacements in System2. So we make all displacements in System 2 corresponding to other response functions equal to zero. Forexample, in Figure 6.8c, displacements corresponding to , and are equal to zero. Example 6.4illustrates the construction of influence lines using Müller-Breslau principle.

Example 6.4 Construct influence lines for , , and for the beam AB in Fig. E6.4.

Solution:

System 2 for : (Note that there is no bending moment at D , i.e. )

System 2 for :

System 2 for : (Note that only contributes to virtual work because even though there is rotation

at point D , )

The deflected shape in each system 2 provides the influence line for the corresponding response function.

Recap

In this course you have learnt the following

The Müller-Breslau principle for influence lines.

Derivation of the principle for different types of internal forces.

Example of application of this principle.

Module 6 : Influence LinesLecture 6 : Tutorial Problems

Objectives

In this course you will learn the following

Some tutorial problems related to this module.

TUTORIAL PROBLEMS

T6.1 Draw the Influence Line Diagram (ILD) for RE . Consider B & B’ and C & C’ to be at aninfinitesimal distance to each other.

Figure T6.1

T6.2 Draw the Influence Line Diagram (ILD) for MG for the following Figure T6.2.

Figure T6.2

T6.3 Find the maximum shear force at C for the moving load combination in Figure T6.3.

Figure T6.3

Recap In this course you have learnt the following

You have learned some tutorial problems related to this module.

Answers of tutorial problems

T6.1

T6.2

T6.3 58.75 kN