Table of Contents Chapter No Title Page No STRUCTURAL ANALYSIS – II L T P C 3 1 0 4 OBJECTIVE This...

Table of Contents

Chapter No Title Page No

1 Construction Planning

1.1 Basic Concepts in the Development of ConstructionPlans

1

1.2 Choice of Technology and Construction Method 2

1.3 Defining Work Tasks 3

1.4 Defining Precedence Relationships Among Activities 6

1.5 Estimating Activity Durations 10

1.6 Estimating Resource Requirements for Work Activities 14

1.7 Coding Systems 15

1.8 References 17

2 Fundamental Scheduling Procedures

2.1 Relevance of Construction Schedules 18

2.2 The Critical Path Method 19

2.3 Calculations for Critical Path Scheduling 20

2.4 Activity Float and Schedules 22

2.5 Presenting Project Schedules 25

2.6 Critical Path Scheduling for Activity-on-Node and withLeads, Lags, and Windows

30

2.7 Calculations for Scheduling with Leads, Lags andWindows

33

2.8 Resource Oriented Scheduling 34

2.9 Scheduling with Resource Constraints and Precedence 36

2.10 Use of Advanced Scheduling Techniques 38

2.11 Scheduling with Uncertain Durations 38

2.12 Crashing and Time/Cost Tradeoffs 42

2.13 Improving the Scheduling Process 45

2.14 References 46

3 Cost Control, Monitoring and Accounting

3.1 The Cost Control Problem 47

3.2 The Project Budget 47

3.3 Forecasting for Activity Cost Control 48

3.4 Financial Accounting Systems and Cost Accounts 49

3.5 Control of Project Cash Flows 51

3.6 Schedule Control 52

3.7 Schedule and Budget Updates 54

3.8 Relating Cost and Schedule Information 54

3.9 References 56

4 Quality Control and Safety During Construction

4.1 Quality and Safety Concerns in Construction 57

4.2 Organizing for Quality and Safety 57

4.3 Work and Material Specifications 58

4.4 Total Quality Control 59

4.5 Quality Control by Statistical Methods 61

4.6 Statistical Quality Control with Sampling by Attributes 61

4.7 Statistical Quality Control with Sampling by Variables 66

4.8 Safety 71

4.9 References 71

5 Organization and Use of Project Information

5.1 Types of Project Information73

5.2 Accuracy and Use of Information74

5.3 Computerized Organization and Use of Information 76

5.4 Organizing Information in Databases 78

5.5 Relational Model of Databases 80

5.6 Other Conceptual Models of Databases 81

5.7 Centralized Database Management Systems 84

5.8 Databases and Applications Programs 85

5.9 Information Transfer and Flow 87

5.10 References 88

CE2351 STRUCTURAL ANALYSIS – II L T P C3 1 0 4OBJECTIVEThis course is in continuation of Structural Analysis – Classical Methods. Here in advancedmethod of analysis like Matrix method and Plastic Analysis are covered. Advanced topics suchas FE method and Space Structures are covered.UNIT I FLEXIBILITY METHOD 12Equilibrium and compatibility – Determinate vs Indeterminate structures – Indeterminacy -Primary structure – Compatibility conditions – Analysis of indeterminate pin-jointed planeframes, continuous beams, rigid jointed plane frames (with redundancy restricted to two).UNIT II STIFFNESS MATRIX METHOD 12Element and global stiffness matrices – Analysis of continuous beams – Co-ordinatetransformations – Rotation matrix – Transformations of stiffness matrices, load vectors anddisplacements vectors – Analysis of pin-jointed plane frames and rigid frames( with redundancyvertical to two)UNIT III FINITE ELEMENT METHOD 12Introduction – Discretisation of a structure – Displacement functions – Truss element – Beamelement – Plane stress and plane strain - Triangular elementsUNIT IV PLASTIC ANALYSIS OF STRUCTURES 12Statically indeterminate axial problems – Beams in pure bending – Plastic moment of resistance– Plastic modulus – Shape factor – Load factor – Plastic hinge and mechanism – Plasticanalysis of indeterminate beams and frames – Upper and lower bound theoremsUNIT V SPACE AND CABLE STRUCTURES 12Analysis of Space trusses using method of tension coefficients – Beams curved in planSuspension cables – suspension bridges with two and three hinged stiffening girdersTOTAL: 60 PERIODS53TEXT BOOKS1. Vaidyanathan, R. and Perumal, P., “Comprehensive structural Analysis – Vol. I & II”, LaxmiPublications, New Delhi, 20032. L.S. Negi & R.S. Jangid, “Structural Analysis”, Tata McGraw-Hill Publications, New Delhi,2003.3. BhaviKatti, S.S, “Structural Analysis – Vol. 1 Vol. 2”, Vikas Publishing House Pvt. Ltd., NewDelhi, 2008REFERENCES1. Ghali.A, Nebille,A.M. and Brown,T.G. “Structural Analysis” A unified classical and Matrixapproach” –5th edition. Spon Press, London and New York, 2003.2. Coates R.C, Coutie M.G. and Kong F.K., “Structural Analysis”, ELBS and Nelson, 19903. Structural Analysis – A Matrix Approach – G.S. Pandit & S.P. Gupta, Tata McGraw Hill2004.4. Matrix Analysis of Framed Structures – Jr. William Weaver & James M. Gere, CBSPublishers and Distributors, Delhi.

CE2351 Structural Analysis II

SCE 1 Dept of Civil

CHAPTER 1

FLEXIBILITY METHOD

Equilibrium and compatibility – Determinate vs Indeterminate structures –Indeterminacy -Primary structure – Compatibility conditions – Analysis of indeterminatepin-jointed planeframes, continuous beams, rigid jointed plane frames (with redundancyrestricted to two).

1.1 INTRODUCTION

These are the two basic methods by which an indeterminate skeletal structure isanalyzed. In these methods flexibility and stiffness properties of members are employed.These methods have been developed in conventional and matrix forms. Here conventionalmethods are discussed.

Thegivenindeterminatestructureisfirstmadestaticallydeterminatebyintroducingsuitable numberof releases. The number of releases required is equal tostaticalindeterminacy∝s. Introductionofreleasesresultsindisplacementdiscontinuitiesatthesereleases under the externally applied loads. Pairsofunknown biactions(forcesandmoments)areappliedatthesereleasesinordertorestorethecontinuityorcompatibility ofstructure.

The computation of these unknown biactions involves solution of linearsimultaneousequations.Thenumberoftheseequationsisequaltostaticalindeterminacy∝s.Aftertheunknownbiactionsarecomputedalltheinternalforcescanbecomputedintheentirestructureusingequationsofequilibriumandfreebodiesofmembers.Therequired displacements can also be computed using methods ofdisplacement computation.

Inflexibilitymethodsinceunknownsareforces atthereleasesthemethodisalsocalledforce method.Since computation of displacement is also required at releases forimposing conditions of compatibility the method is also called compatibility method. Incomputationofdisplacementsuseismadeof flexibilityproperties,hence,themethodis alsocalled flexibility method.

1.2 EQUILIBRIUM and COMPATABILITY CONDITIONS

Thethreeconditionsofequilibriumarethesumofhorizontalforces,verticalforcesandmoments at anyjoint should beequal to zero.

i.e.∑H=0;∑V=0;∑M=0Forces should be in equilibrium

i.e.∑FX=0;∑FY=0;∑FZ=0i.e.∑MX=0;∑MY=0;∑MZ=0

Displacement of a structure should be compatableThe compatibility conditions for the supports can be given as1.Roller Support δV=02.Hinged Support δV=0, δH=03.Fixed Support δV=0, δH=0, δө=0


SCE 2 Dept of Civil

1.3.DETERMINATE AND INDETERMINATE STRUCTURAL SYSTEMS

Ifskeletalstructureissubjectedtograduallyincreasingloads,withoutdistortingtheinitialgeometryofstructure,thatis,causingsmalldisplacements,thestructureissaidto be stable.Dynamic loads and buckling or instability of structural system are notconsideredhere.Ifforthestablestructureitispossibletofindtheinternalforcesinall the membersconstituting the structure and supporting reactions at all the supports providedfromstaticallyequationsofequilibrium only,thestructureissaidtobe determinate.

Ifitispossibletodetermineallthesupport reactionsfromequationsof equilibriumalonethestructureissaidtobeexternallydeterminateelseexternally indeterminate.If structureisexternallydeterminatebutitisnotpossible todetermineallinternalforcesthenstructureissaidtobe internallyindeterminate. Thereforeastructuralsystemmaybe:

(1)Externally indeterminate but internally determinate(2)Externally determinate but internally indeterminate(3)Externallyand internallyindeterminate(4)Externally and internallydeterminate

1.3.1.DETERMINATEVs INDETERMINATESTRUCTURES.

Determinatestructurescanbesolvingusingconditionsofequilibriumalone(∑H=0;∑V=0;∑M=0). No otherconditions arerequired.

Indeterminatestructurescannotbesolvedusingconditionsofequilibriumbecause(∑H≠0;∑V≠0;∑M≠0).Additionalconditionsarerequiredforsolvingsuchstructures.Usuallymatrixmethods areadopted.

1.4 INDETERMINACYOF STRUCTURAL SYSTEMThe indeterminacy of a structure is measured as statically (∝s) or kinematical

(∝k)Indeterminacy.∝s= P (M – N + 1) – r = PR– r ∝k= P (N – 1) + r – s+∝k= PM –cP = 6 for space frames subjected to general loadingP = 3 for plane frames subjected to inplane or normal to plane loading.N = Numberof nodes in structural system.M=Numberofmembersofcompletelystiffstructurewhichincludesfoundationassinglyconnectedsystem ofmembers.

Incompletelystiffstructurethereisnorelease present.Insinglyconnectedsystemofrigidfoundationmembersthereisonlyoneroutebetweenanytwopointsinwhichtracksarenotretraced. Thesystemisconsidered comprising ofclosed rings or loops.R = Numberof loops or rings in completely stiff structure.r = Number of releases in the system.c = Number of constraints in the system.R = (M – N + 1)


SCE 3 Dept of Civil

For plane and space trusses∝sreducesto:∝s=M- (NDOF)N+ PM= Number ofmembers in completely stifftruss.P = 6 and 3 for space and plane trussrespectively

N= Number of nodes in truss.NDOF = Degrees of freedomat node which is 2 for plane truss and 3 for space truss.For space truss∝s=M- 3N+ 6For plane truss∝s= M- 2 N+ 3

Test for static indeterminacy of structural system

If ∝s> 0 Structure is statically indeterminate

If ∝s= 0 Structure is statically determinateand if∝s<0 Structure is a mechanism.

Itmaybenotedthatstructuremaybemechanismevenif ∝s >0ifthereleasesarepresentinsuchawaysoastocausecollapseasmechanism.Thesituationofmechanism isunacceptable.

Statically IndeterminacyItisdifferenceoftheunknownforces(internalforcesplusexternalreactions)andthe

equations of equilibrium.Kinematic Indeterminacy

Itisthenumberofpossiblerelativedisplacementsofthenodesinthedirectionsofstressresultants.

1.5 PRIMARY STRUCTUREAstructure formed bythe removingthe excess orredundant restraints froman

indeterminatestructuremakingit staticallydeterminateis called primarystructure. This isrequired forsolvingindeterminatestructures byflexibilitymatrixmethod.

Indeterminatestructure PrimaryStructure

SCE 4 Dept of Civil

1.6.ANALYSIS OF INDETERMINATE STRUCTURES :BEAMS

1.6.1Introduction

Solvestaticallyindeterminate beams of degree more than one. Tosolvetheprobleminmatrixnotation. Tocomputereactionsatallthesupports. To compute internal resisting bending moment at any section of the

continuousbeam.Beamswhicharestaticallyindeterminatetofirstdegree,wereconsidered. If the structure isstatically indeterminate to a degree more than one, then the approach presented in theforce method is suitable.

Problem 1.1Calculate the support reactions in the continuous beam ABC due to loading as shown inFig.1.1 Assume EI to be constant throughout.

Fig 1.1

Fig 1.2

Select two reactions vise, at B(R1 ) and C(R2 ) as redundant, since the given beamisstatically indeterminate to second degree. In this case the primary structure is a cantileverbeam AC.The primary structure with a given loading is shown in Fig. 1.2

In the present case, the deflections (Δ L)1 and (Δ L) 2 of the released structure at B and Ccan be readily calculated by moment-area method. Thus

SCE 5 Dept of Civil

(Δ L) 1 = − 819.16 / EI

(Δ L) 2 = − 2311.875/ EI (1)

Forthepresentproblemthe flexibility matrix is,

a11= 125/3EI ,a21= 625/6EI

a12= 625/6EI , a22 = 1000/3EI (2)

In the actual problem the displacements atBandCare zero. Thus thecompatibility conditions for the problem may be written as,

a11 R1+ a12 R2 + (Δ L) 1 = 0

a21 R1+ a22 R2+ (Δ L) 2 = 0(3)

Substituting the value of E and I in the above equation,

R1 = 10.609 KN and R2 = 3.620 KN

Using equations of static equilibrium, R3 = 0.771 KN m and R4 = −0.755KN m

Problem 1.2

AFixedbeamAB ofconstantflexuralrigidityisshowninFig.1.3Thebeam

issubjectedtoauniformdistributedloadofwmomentM=wL2

kN.m.DrawShearforceandbendingmomentdiagramsbyforcemethod.

Fig 1.3 Fixed Beam

SCE 6 Dept of Civil

1 )

Fig 1.3 Fixed Beam with R1 and R2 as Redundant

Select vertical reaction(R1)and the support moment(R2) at B as the

redundant.Theprimarystructureinthiscaseisacantileverbeamwhichcouldbeobtainedbyreleasingtheredundant R1 andR2.

TheR1 isassumedto

positive in the upward direction andR2 is assumed to be positive in the

counterclockwisedirection.Now,calculatedeflectionat B duetoonlyappliedloading.Let (ΔL )bethetransversedeflectionat B and(ΔL 2 betheslopeatB

duetoexternalloading.Thepositivedirectionsoftheselectedredundantare showninFig.8.3b.

Fig 1.4 Primary Structure with external loading

Fig 1.5 Primary Structure with unit load along R1

SCE 7 Dept of Civil

Fig 1.6 Primary Structure with unit Moment along R2

Fig 1.7 Reaction

Fig1.8.Bending Moment Diagram

Fig1.9.Shear Force Diagram

The deflection(Δ L1)and(Δ L2)of the released structure can be evaluated from unit loadmethod. Thus,

(Δ L1) =wL4/8EI – 3wL4/8EI = −wL4/2EI (1)(Δ L2) = wL3/6EI – wL3 /2EI = − 2wL3/3EI (2)

SCE 8 Dept of Civil

1The negativesign indicates that (ΔL)isdownwards and rotation(ΔL2)is

clockwise.

Problem 1.3.A continuous beam ABC is carrying a uniformly distributed loadof 1 kN/m in addition

toaconcentratedloadof10kNasshowninFig.7.5a, Draw bending momentandshearforcediagram.Assume EItobeconstantforallmembers.

Fig1.10.Continuous Beam

Fig1.11.Primary Structure

SCE 9 Dept of Civil

Fig1.12.Flexibility Coefficients

Fig1.13.Reactions

Itisobservedthatthecontinuousbeamisstaticallyindeterminatetofirstdegree.ChoosethereactionatB, RBy astheredundant.Theprimarystructureisa

SCE 10 Dept of Civil

simplysupportedbeamasshowninFig.1.11.Now, compute the deflection at B, in thereleasedstructure due to uniformly distributed load and concentrated load. Thisisaccomplished by unit load method.Thus,

Δ =−2083.33−1145.84

L EI EI

Δ =−3229.17(1)

L EI

Inthenextstep,applyaunitloadatBinthedirectionofRBy(upwards)and

calculatethedeflectionat B of the following structure.Thus(seeFig.7.5c),

L3

a11 =48EI

=166.67

EI(2)

Now,deflectionatBintheprimary structure due to redundant RB is,

ΔB= 166.67

EI

×RB(3)

In theactual structure, the deflection at B is zero. Hence, thecompatibility equation may bewritten as

ΔL+ΔB=0(4)

Theothertworeactionsarecalculatedbystaticequilibriumequations(videFig.1.13)

RA =7.8125kN

RB =2.8125kN



UNIT II STIFFNESS MATRIX METHOD

Element and global stiffness matrices – Analysis of continuous beams – Co-ordinatetransformations – Rotation matrix – Transformations of stiffness matrices, load vectors anddisplacements vectors – Analysis of pin-jointed plane frames and rigid frames( with redundancyvertical to two)

2.1 INTRODUCTION

Thegivenindeterminatestructureisfirstmadekinematic allydeterminatebyintroducingconstraints atthenodes.Therequirednumberofconstraintsisequaltodegrees offreedomatthenodesthatis kinematicindeterminacy∝k.Thekinematic allydeterminatestructurecomprisesoffixedendedmembers,hence,allnodal displacementsarezero.Theseresultsinstress resultantdiscontinuitiesatthesenodesundertheactionofappliedloadsorin otherwordstheclampedjointsarenotinequilibrium.

Inordertorestoretheequilibriumofstressresultantsatthenodesthenodesareimpartedsuitableunknowndisplacements.Thenumberofsimultaneousequationsrepresentingjointequilibriumofforcesisequaltokinematicindeterminacy∝k.Solutionoftheseequationsgivesunknownnodaldisplacements.Usingstiffnesspropertiesofmembersthememberendforcesarecomputedandhencetheinternalforcesthroughoutthestructure.

Since nodal displacements are unknowns, the method is also called displacement method.Since equilibriumconditionsareappliedatthejointsthemethodisalsocalledequilibriummethod.Sincestiffness properties ofmembers areusedthemethodis alsocalledstiffnessmethod.

In the displacement method of analysis the equilibrium equations are written byexpressingtheunknownjointdisplacementsintermsofloadsby usingload-displacementrelations.Theunknownjointdisplacements(thedegreesoffreedomof thestructure)are calculated by solvingequilibriumequations.Theslope-deflection andmoment-distributionmethodswereextensively usedbeforethehigh speedcomputingera.Aftertherevolutionincomputerindustry,only directstiffnessmethodisused.

2.1.1.PROPERTIES OFTHESTIFFNESS MATRIX

Theproperties ofthestiffness matrixare: It isasymmetricmatrix Thesum of elements in anycolumn must be equal to zero. It is an unstableelementthereforethedeterminantis equal to zero.

2.2.ELEMENT AND GLOBAL STIFFNESS MATRICES

Local co ordinates

In the analysis for convenience we fix the element coordinates coincident with the memberaxis called element (or) local coordinates (coordinates defined along the individual member axis )

Global co ordinates

It is normally necessary to define a coordinate system dealing with the entire structure iscalled system on global coordinates (Common coordinate system dealing with the entire structure)



Transformationmatrix

The connectivitymatrixwhich relates theinternalforcesQ and theexternal forces R is knownas the forcetransformation matrix. Writingit in amatrixform,

{Q} =[b]{R}whereQ=member forcematrix/vector, b=forcetransformationmatrix

R = external force/loadmatrix/ vector

2.3 ANALYSIS OF CONTINUOUS BEAMS

Fig 2.1 Cantilever Beam Fig 2.2 Cantilever Beam with unit load along P1

Fig 2.3 Cantilever Beam with unit Moment along P2

Fig 2.4 Cantilever Beam with unit Displacement along U1



Fig 2.5 A Four member Truss

Fig 2.6 Kinematic ally Determinate Structures



Fig 2.7Unit Displacement along U



2.4.ANALYSIS OF PIN JOINTED PLANE FRAMES

An introduction to thestiffnessmethodwasgivenin thepreviouschapter.Thebasicprinciples

involvedin the analysisof beams,trusseswerediscussed.Theproblemsweresolvedwith hand

computation by thedirectapplicationofthebasicprinciples. Theprocedurediscussedin theprevious

chapterthough enlighteningarenotsuitableforcomputerprogramming.Itisnecessary to keephand

computation to aminimumwhileimplementingthisprocedureon thecomputer.

In thischaptera formalapproachhasbeen discussedwhichmay bereadily programmedon a

computer.In thislesson thedirectstiffnessmethod asapplied toplanar truss structureisdiscussed.



Planetrussesaremadeupofshortthinmembersinterconnectedathingestoformtriangulated

patterns.Ahingeconnectioncanonlytransmitforcesfromonemembertoanothermemberbutnot

themoment. For analysispurpose, thetruss is loaded atthe joints. Hence, atruss member is

subjectedtoonlyaxialforcesandtheforcesremain constant alongthelengthofthemember.The forcesin

thememberatitstwo endsmustbeof thesamemagnitudebutactin theoppositedirections for

equilibriumas shown in Fig.2.8

Fig 2.8 Truss member in Equilibrium

Fig 2.9 Force Displacement Relationship



Fig 2.10 Frame Member in Local Coordinate System



Fig 2.11Plane Frame Member in (a) Local Coordinate System (b) Global coordinate System



Fig 2.12 Rigid Frame



Fig 2.13 Node and Member Numbering



Fig 2.14 Fixed end action due to external loading in element 1 and 2

Fig 2.14 Equivalent Joint Load



Fig 2.15 Support Reactions



CHAPTER IIIFINITE ELEMENT METHOD

Introduction – Discretization of a structure – Displacement functions – Truss element –Beamelement – Plane stress and plane strain - Triangular elements

3.1.INTRODUCTION

TherearetwoversionofFEM:

1. FlexibilityMethodorForceMethod2. StiffnessMethodorDisplacementMethod. Thesetofequationsinthestiffnessmethodaretheequilibriumequationsrelatingdisplacement

sofpoints. Rayleigh-Ritzisanapproximatemethodbasedonenergyprincipleby

whichwecanobtainequilibriumequationsinmatrixform.

3.1.1 IMPORTANT DEFINITION

Nodesarepointsonthestructureatwhichdisplacementsandrotations are tobefoundorprescribed.

Element is a small domainonwhichwecan solvethe boundaryvalueproblemintermsofthedisplacementsandforcesofthe nodesonthe element.

Thediscrete representationofthe structuregeometrybyelements and nodesis called a mesh.

Theprocessofcreating a mesh(discreteentities) is called discretization.

Interpolationfunctionisakinematicallyadmissibledisplacementfunctiondefinedonanelement thatcanbeusedforinterpolatingdisplacement valuesbetweenthe nodes.

Themesh,boundaryconditions,loads,andmaterial propertiesrepresentingthe actual structureiscalled a model.

Element stiffnessmatrix relate thedisplacementstothe forcesat the elementnodes.

Globalstiffnessmatrix is anassemblyofelement stiffnessmatrix that relates thedisplacementsofthe nodesonthemeshtoappliedexternal forces.

3.1.2.StepsinFEMprocedure

1.Obtainelementstiffnessandelementloadvector.

2.Transformfromlocalorientationtoglobalorientation.

3.Assembletheglobalstiffnessmatrixandloadvector.

4.Incorporatetheexternalloads

5.Incorporatetheboundaryconditions.

6.Solvethealgebraicequationsfornodaldisplacements.

7.Obtainreactionforce,stress,internalforces,strainenergy.



8.Interpretandchecktheresults.

9.Refinemeshifnecessary,andrepeattheabovesteps.

3.2.DISCRETISATION OF STRUCTURE

Discretizationis the process of separating the length, area or volume we want to analyze intodiscrete (or separate) parts or elements.



3.3.DISPLACEMENT FUNCTIONS

The continuum is separated by imaginarylines or surfaces into a number of finite element

The elements are assumed to beconnected at discrete number of nodal points situated ontheir boundaries.

Generalized displacements are the basic unknowns. A function uniquely defines displacement field in terms of nodal displacements.

Compatibility between elements.

2D – 3D elasticity problems, displacement compatibility.

Plates and shells, displacements and their partial derivatives.

All possible rigid body displacements included (if not will not converge).

All uniform strain states included.The displacement function, uniquely definesstrain within an element in terms of nodal displacements.

These strains with any initial strain, together with elastic properties define thestress state.



3.4 TYPES OF ELEMENT

Three are three types of elements are available.

1D Elements 2D Elements 3D Elements

3.4.11D Elements (Beam Element)A beam can be approximated as a one dimensional structure. It can be split into one

dimensional beam elements. So also, a continuous beam or a flexure frame can be discretizedusing 1D beam elements.

A pin jointed truss is readily made up of discrete 1D ties which are duly assembled.3.4.22 D Elements(Triangular Element)

A planewall ,plate, diaphragm, slab, shell etc., can be approximated as an assemblage of2D elements. Triangular elements are the most used ones. when our 2D domain has curvedboundaries it may be advantageous to choose elements that can have curved boundaries.3.4.33 D Elements(Truss Element)

Analysisof solid bodies call for the use of 3 D elements. These have the drawback thatthe visualizations is complex. The size of the stiffness matrix to be handled can becomeenormous and unwieldy.



3.5 PLANE STRESS AND PLANE STRAIN

The plane stress problem is one in which two dimensions ,length and breadth arecomparable and thickness dimension is very small (less than 1/10).Hence normal stress σ2 andshear stresses τxz,τyzare zero.

{σ }= [D]{e }[D]=Stress strain relationship matrix (or) constitutive matrix for plane stress problems.

We have seen that in the Z direction the dimension of the plate in the plane stressproblem is very small. In plane strain problem, on the contrary the structure is infinitely long inthe Z direction. Moreover the boundary and body forces do not vary in the Z directions.

{σ }= [D]{e }[D]=Stress strain relationship matrix (or) constitutive matrix for plane strain problems.



CHAPTER 4

PLASTIC ANALYSIS OF STRUCTURES

Statically indeterminate axial problems – Beams in pure bending – Plastic moment of resistancePlastic modulus – Shape factor – Load factor – Plastic hinge and mechanism – Plasticanalysis of indeterminate beams and frames – Upper and lower bound theorems

4.1.Statically indeterminate axial problems

Intheseanalysesweused superposition often,knowing thatforalinearlyelasticstructureitwasvalid.However,an elastic analysisdoesnotgiveinformation abouttheloadsthatwill actually collapseastructure.An indeterminatestructuremay sustainloadsgreaterthantheloadthatfirstcauses ayieldtooccur at anypointinthestructure.

Infact,astructurewillstandaslongasitisabletofindredundancies toyield.Itisonly whenastructurehasexhaustedallofitsredundancieswillextraloadcausesit tofail.Plasticanalysisisthemethodthroughwhichtheactualfailureloadof astructureiscalculated,andaswillbeseen,thisfailureloadcanbesignificantly greaterthan the elasticload capacity.

Tosummarizethis,Prof.SeandeCourcy(UCD)usedtosay:“astructureonlycollapseswhenithas exhaustedallmeans ofstanding”.Before analyzingcomplete structures, we review material and cross sectionbehaviorbeyondtheelasticlimit.

4.2. Beams in pure bending4.2.1. MaterialBehavior

Auniaxialtensilestressonaductile materialsuchasmild steeltypicallyprovidesthefollowinggraphofstress versus strain:

Ascanbeseen,thematerialcansustainstrainsfarinexcessofthestrainatwhichyield occursbeforefailure.This propertyofthematerialis calledits ductility.Thoughcomplex models do exist toaccurately reflect theabovereal behaviourofthematerial,themostcommon,andsimplest,modelistheidealizedstress-straincurve.Thisisthecurveforanidealelastic-plasticmaterial(whichdoesn’texist), andthegraphis:



As canbeseen, oncetheyieldhasbeenreacheditis takenthatanindefiniteamountofstraincanoccur. Sincesomuchpost-yieldstrainismodeled, theactualmaterial(orcross section)mustalsobecapableofallowingsuchstrains.Thatis,itmustbesufficientlyductilefortheidealized stress-straincurvetobevalid.Nextweconsiderthebehaviourof acrosssectionofanideal elastic-plasticmaterialsubjecttobending.In doingso,weseektherelationshipbetween appliedmomentandtherotation(ormoreaccurately, thecurvature)ofacross section.

4.2.2.Moment-RotationCharacteristics ofGeneralCross Section

Weconsider anarbitrarycross-sectionwithaverticalplaneofsymmetry,whichisalsotheplaneofloading.Weconsiderthecrosssectionsubjecttoanincreasingbendingmoment,andassess thestressesateach stage.

Cross sectionandStresses



Stage1– ElasticBehaviour

Moment-Rotation Curve

Theappliedmomentcauses stresses overthecross-sectionthatareallless thantheyieldstress ofthematerial.Stage2–YieldMoment

Theappliedmomentisjustsufficientthattheyieldstressof thematerialisreachedattheoutermostfibre(s)ofthecross-section.Allotherstressesinthecrosssectionarelessthantheyieldstress.Thisislimitofapplicabilityofanelasticanalysisandof elasticdesign.Sinceallfibresareelastic, theratioofthedepthoftheelastictoplasticregions,Stage3–Elasto-PlasticBending

Themomentapplied to thecrosssection hasbeenincreasedbeyond theyieldmoment.Sincebytheidealizedstress-strain curvethematerial cannotsustain astressgreaterthanyieldstress,thefibresattheyieldstresshaveprogressedinwardstowardsthecentreof thebeam.Thusoverthe crosssectionthereisanelasticcoreandaplasticregion.Theratioofthedepthoftheelasticcoretotheplasticregionis .Sinceextramomentis beingappliedandnostressisbiggerthantheyieldstress,extrarotationof thesectionoccurs:themoment-rotationcurvelossesitslinearityand curves,givingmorerotationperunitmoment(i.e.loosesstiffness).Stage4–PlasticBending

Theappliedmomenttothecrosssectionissuch thatallfibresin thecrosssection areatyieldstress.ThisistermedthePlasticMomentCapacityof thesection sincetherearenofibresatanelasticstress,Alsonotethatthefullplasticmomentrequiresaninfinitestrainattheneutralaxis



andsoisphysicallyimpossibletoachieve.However,itisclosely approximatedin practice.Anyattemptatincreasingthemomentat thispointsimply resultsinmorerotation,oncethecross-sectionhassufficientductility.Thereforeinsteelmembersthecrosssectionclassificationmustbeplasticandinconcretemembers thesectionmustbeunder-reinforced.

Stage5–StrainHardening

Duetostrainhardeningofthematerial,asmallamountofextramomentcanbesustained.

Theabovemoment-rotationcurverepresents thebehaviourofacrosssectionofaregular elastic-plasticmaterial.However,itis usuallyfurthersimplifiedasfollows:

With thisidealizedmoment-rotation curve,thecrosssectionlinearlysustainsmomentupto theplasticmomentcapacityofthesectionandthenyieldsinrotationanindeterminateamount.Again,tousethisidealization,theactual sectionmustbecapableofsustaininglargerotations–thatisitmustbeductile.Analysis ofRectangularCross Section

Sincewenowknowthatacrosssectioncansustainmoreloadthanjusttheyieldmoment,weareinterestedinhowmuchmore.Inotherwordswewanttofindtheyieldmomentandplasticmoment,andwedosoforarectangularsection.Takingthestressdiagramsfromthoseofthe moment-rotationcurveexaminedpreviously,wehave:



4.3.ShapeFactorThus theratioofelastictoplasticmomentcapacityis:

Thisratioistermedtheshapefactor,f,andisapropertyofacrosssectionalone.Forarectangularcross-section,wehave:

Andsoarectangularsectioncansustain50%moremomentthantheyieldmoment,beforeaplastichingeisformed.Thereforetheshapefactorisagoodmeasureoftheefficiencyofacross sectioninbending.Shapefactors forsomeothercross sections are



4.4.PlasticHinge

Notethatoncetheplasticmomentcapacityisreached,thesectioncanrotatefreely–thatis,itbehaveslikeahinge,exceptwithmomentofMpatthehinge.Thisis termedaplastichinge,andisthebasisforplasticanalysis.Attheplastichingestressesremainconstant,butstrainsand hencerotationscanincrease.

4.4.1.Methods ofPlasticAnalysis

1. TheIncrementalMethod

Thisisprobably themostobviousapproach:theloadson thestructureareincrementeduntilthefirstplastichingeforms.Thiscontinuesuntilsufficient hingeshaveformedtocollapsethestructure.Thisisalabour-intensive,‘brute-force’,approach,butonethatismostreadilysuitedforcomputerimplementation.

2. TheEquilibrium(orStatical) Method

In thismethod,freeandreactantbendingmomentdiagramsaredrawn.Thesediagramsareoverlaidtoidentifythelikelylocations of plastichinges.Thismethodthereforesatisfies theequilibriumcriterionfirstleavingthetwo remainingcriteriontoderivedtherefrom.

3.The Kinematic (or Mechanism) Method

In this method, a collapse mechanism is first postulated. Virtual work equations are then written forthis collapse state, allowing the calculations of the collapse bending moment diagram. This methodsatisfies the mechanism condition first, leaving the remaining two criteria to be derived there from.

We will concentrate mainly on the Kinematic Method, but introduce now the Incremental Methodto illustrate the main concepts.

4.4.1.1. IncrementalMethod

Example1– ProppedCantilever

We now assess the behaviorof a simple statically indeterminate structure under

increasingload.Weconsideraproppedcantilever withmid-spanpointload:



Sincethepeakmomentsarelessthan theyieldmoments,weknow

thatyieldstresshasnotbeen reachedatany pointin

thebeam.Also,themaximummomentoccursatAandsothispointwill firstreachtheyieldmoment.

4.4.1.2.EquilibriumMethod

Introduction

Toperformthis analysis wegenerallyfollowthefollowingsteps:

1.Findaprimary structurebyremoving redundantuntilthestructureis staticallydeterminate;

2.Drawtheprimary(orfree)bendingmomentdiagram;

3.DrawthereactantBMDforeachredundant,as appliedtotheprimary structure;

4.ConstructacompositeBMDbycombingtheprimaryand reactantBMDs;



5.DeterminetheequilibriumequationsfromthecompositeBMD;

6.Choosethepointswhereplastichingesarelikelytoformandintroduceintotheequilibrium equations;

7.Calculatethecollapseloadfactor,orplasticmomentcapacityas required.

Fordifferentpossiblecollapsemechanisms,repeatsteps 6and7,varyingthehingelocations.

WenowapplythismethodtotheIllustrativeExamplepreviouslyanalyzed.

Steps 1to3oftheEquilibriumMethodareillustratedinthefollowingdiagram:

ForStep4,inconstructingtheCompositeBMD,wearbitrarilychoosetensionontheunderside of

thebeamaspositive.ByconventionintheEquilibrium Method,insteadof drawingthetwo BMDson

oppositesides(asisactually thecase),thereactantBMDisdrawn‘flipped’overthe

lineandsubtractedfromtheprimaryBMD:thenetremainingareais thefinal BMD.Thisisbest

explainedbyillustrationbelow:



ForStep7, wesolvethis equationforthecollapseload:



4.4.1.3 KinematicMethodUsingVirtualWork

Introduction

Probably theeasiestway tocarry outaplasticanalysisisthrough theKinematicMethodusing

virtual work.Todothisweallowthepresumedshapeatcollapsetobethecompatible

displacementset,andtheexternalloadingandinternalbendingmomentstobetheequilibrium set.Wecan

thenequate externalandinternalvirtualwork,andsolveforthecollapseloadfactor

forthatsupposedmechanism.

Remember:

Equilibrium set:theinternalbendingmoments atcollapse;

Compatibleset:thevirtualcollapsedconfiguration (seebelow).

Notethatin theactual collapseconfiguration thememberswillhaveelasticdeformationin

betweentheplastichinges.However,sinceavirtual displacementdoesnothavetobereal,only

compatible,wewillchoosetoignoretheelasticdeformationsbetweenplastichinges,andtake

themembers tobestraightbetweenthem.



4.5.CollapseMechanism

Soforourpreviousbeam,we know thatwe require twohingesforcollapse(onemore thanits

degreeofredundancy),andwethinkthatthehinges willoccurunderthepoints ofpeakmoment,

AandC.Thereforeimposeaunitvirtual displacementatCandrelatethecorrespondingvirtual rotations

ofthehinges using,



4.5.1 OtherCollapseMechanisms

Forthecollapsemechanismlookedatpreviously,itseemed obviousthat theplastichingein the

spanshouldbebeneaththeload.Butwhy?Usingvirtual workwecan examineanypossible

collapsemechanism.Solet’sconsiderthefollowingcollapsemechanismandseewhytheplastic hingehas

tobelocatedbeneaththeload.

PlasticHinge betweenAandC:ImposingaunitvirtualdeflectionatB,wegetthefollowingcollapsemechanism:



Andsoweseethatthecollapseloadfactorforthismechanismdependsonthepositionofthe

plastichingeinthespan.

4.6.PlasticAnalysisofBeams

Example2–Fixed-FixedBeamwithPointLoad

Tostarttheproblem, weexaminetheusual elasticBMDtoseewheretheplastic hingesare

likelytoform:

Wealsoneedtoknowhowmanyhinges arerequired.Thisstructureis 3˚staticallyindeterminate

andsowemightexpectthenumberofplastichingesrequiredtobe4.However,sinceoneofthe

indeterminaciesishorizontalrestraint,removingitwouldnotchangethebendingbehaviourof thebeam.

Thusforabendingcollapseonly2indeterminaciesapplyandsoitwill only take3 plastichinges

tocausecollapse.SolookingattheelasticBMD,we’llassumeacollapsemechanismwiththe3plastichinges

at thepeakmomentlocations:A,B,andC.



WeneedtocheckthatthisisthecorrectsolutionusingtheUniquenessTheorem:



AndsotheappliedloadisinequilibriumwiththefreeBMDofthecollapseBMD.

2.Mechanism:

Fromtheproposedcollapsemechanismitis apparentthatthebeamis amechanism.

3.Yield:

FromthecollapseBMDitcanbeseenthatnowhereis exceeded.PM

Thusthesolutionmeetsthethreeconditionsandso,bytheUniquenessTheorem,isthecorrect solution.

Example3–ProppedCantileverwithTwoPointLoads

Forthefollowingbeam,foraloadfactorof2.0,findtherequiredplasticmoment capacity:



Allowingfortheloadfactor,weneedtodesignthe beamfor thefollowingloads:

Once againwe try to picture possible failure mechanisms. Since

maximummomentsoccurunderneathpointloads,thereare tworeal possibilities:



Mechanism-1

Mechanism-2

Therefore,we analyseboth and apply the UpperboundTheoremto find the

designplasticmomentcapacity.

Mechanism1:PlasticHingeatC:



Mechanism2:PlasticHingeatD:



1.Equilibrium:UsingtheBMDatcollapse,we’llcheckthattheheightofthefreeBMDisthatof

theequivalentsimply-supportedbeam.FirstlythecollapseBMDfromMechanism1is:

Hence,thetotalheightsofthefreeBMDare:

Checkingtheseusingasimply-supportedbeamanalysis



Thus,usingappropriatefreebodydiagrams ofACandDB:

AndsotheappliedloadisinequilibriumwiththefreeBMDofthecollapseBMD.2.Mechanism:

Fromtheproposedcollapsemechanismitisapparentthatthebeamisamechanism.Also,sinceitisaproppedcantileverand thusonedegreeindeterminate,werequiretwoplastichingesforcollapse,andthesewehave3.Yield:

FromthecollapseBMDitcanbeseenthatnowhereis thedesignexceeded.144kNmThusbytheUniquenessTheoremwehavethecorrectsolution.Lastly,we’llexaminewhytheMechanism2collapseisnotthecorrectsolution.Sincethevirtual workmethodprovidesan upperbound,then,bytheUniquenessTheorem,itmustnotbethe correctsolutionbecauseitmustviolatetheyieldcondition.Using thecollapseMechanism2todeterminereactions,wecan draw thefollowingBMDforcollapseMechanism2:

FromthisitisapparentthatMechanism2isnottheuniquesolution,andsothedesignplastic moment

capacity must be144kNmasimpliedpreviously fromtheUpperboundTheorem.

4.BasicCollapseMechanisms:In frames,the basicmechanismsofcollapseare:



Beam-typecollapse

SwayCollapse

CombinationCollapse

5.CombinationofMechanismsOneofthemostpowerfultoolsinplasticanalysisis CombinationofMechanisms.Thisallows us

toworkoutthevirtualworkequationsforthebeamandswaycollapsesseparatelyandthen

combinethemtofindthecollapseloadfactorforacombinationcollapsemechanism.

Combinationof mechanismsis based onthe ideathat thereareonlya certain number of

independentequilibriumequationsforastructure.Anyfurtherequationsareobtainedfroma



combinationoftheseindependentequations.

Sinceequilibriumequationscanbeobtainedusing

virtualworkappliedtoapossiblecollapsemechanism,itfollowsthatthereareindependent

collapsemechanisms,andothercollapsemechanismsthatmaybeobtainedformacombination

oftheindependentcollapsemechanisms.

6.SimplePortalFrame

Inthisexamplewewillconsiderabasicprismatic(soallmembershavethesame

plasticmomentcapacity)rectangularportalframewithpinnedfeet:



Wewillconsiderthisgeneralcasesothatwecaninferthepropertiesandbehaviourofallsuch

frames.Wewillconsidereachofthepossiblemechanisms outlinedabove.

7.Beam collapse:

Thepossiblebeamcollapselooks as follows:



4.7.CollapseMode

Sincewedon’tknowtherelativevaluesofHandV,wecannotdeterminethe

correctcollapsemode.However,wecanidentifythesecollapsemodesifweplot the

threeloadfactorequationsderivedaboveonthefollowinginteractionchart:

Noticethateachmechanismdefinesaboundary andthatitisonlytheregioninsidealloftheseboundariesthatissafe.Now,foragivenrationofVtoH,wewillbeabletodeterminethecriticalcollapsemechanism.NotealsothatthebeamcollapsemechanismisonlycriticalforthisframeatpointPonthechart–thispointisalsoincludedintheCombinedmechanism.Thebendingmomentdiagramscorrespondingtoeachofthemechanismsare approximately:

Aninterestingphenomenon isobserved atpointQonthechart,wheretheSwayandCombinedmechanismsgivethesameresult.Lookingatthebendingmoment diagrams,we cansee



thatthis occursasthe momentatthe topofthe leftcolumn becomesequaltothe mid-spanmomentof thebeam:



4.8. Upper bound(Unsafe)Theorem:

This canbestatedas:

If abendingmomentdiagramisfoundwhichsatisfiesthe conditionsof equilibrium

andmechanism(butnotnecessarilyyield),thenthecorresponding loadfactoris eithergreaterthanor

equalto the trueloadfactorat collapse.

Thisis calledtheunsafetheorembecauseforanarbitrarilyassumedmechanismtheloadfactoris

eitherexactly right(when theyieldcriterionismet)oriswrongandistoolarge,leadinga designer

tothinkthattheframecancarrymoreloadthanis actuallypossible.

4.9. Lowerbound(Safe)Theorem:

Ifabendingmomentdiagramisfoundwhichsatisfiestheconditionsofequilibriumandyield

(butnotnecessarilythatofmechanism),then thecorrespondingload factoriseitherlessthanor

equaltothetrueloadfactor atcollapse.



CHAPTER 5SPACE AND CABLE STRUCTURES

Analysis of Space trusses using method of tension coefficients – Beams curved in planSuspension cables – suspension bridges with two and three hinged stiffening girders

5.1ANALYSIS OF SPACE TRUSSES USING METHOD OF TENSION COEFFICIENTS

5.1.1.Tension Co-efficient MethodThe tension co efficient for a member of a frame is defined as the pull or tension in that

member is divided by its length.t = T/l

Where t = tension co efficient for the memberT= Pull in the memberl = Length

5.1.2.Analysis Procedure Using Tension Co-efficient – 2D Frames

1.List the coordinates of each joint (node)of the truss.2.Determine the projected lengths Xij and Yij of each member of the truss. Determine the supportlengths lij of each member using the equation lij =√Xij

2+Yij2

3. Resolve the the applied the forces at the joint in the X and Y directions. Determine the supportreactions and their X and Y components.4.Identify a node with only two unknown member forces and apply the equations of equilibrium.The solution yields the tension co efficient for the members at the node.5.Select the next joint with only two unknown member forces and apply the equations ofequilibrium and apply the tension co efficient.6.Repeat step 5 till the tension co efficient of all the members are obtained.7.Compute the member forces from the tension co efficient obtained as above using

Tij= tijx lij

5.1.3.Analysis Procedure Using Tension Co-efficient – Space Frames

1.In step 2 above the projected lengths Zij in the directions are also computed.Determine thesupport lengths lij of each member using the equation lij =√Xij

2+Yij2 +Zij

2

2.In step 3 above the components of forces and reactions in the Z directions are also to bedetermined.3.In step 4 and 5,each time, nodes with not more than three unknown member forces are to beconsidered.

Tetrahedron: simplestelementofstablespacetruss (sixmembers,fourjoints)expandbyadding3members and1jointeachtimeDeterminacyandStabilityb+r<3junstableb+r=3jstaticallydeterminate(checkstability)b+r>3jstaticallyindeterminate(checkstability)



InternalForcesIn orderto obtain theinternalforcesataspecifiedpoint,weshouldmakesection cut

perpendiculartotheaxis ofthememberatthis point.This sectioncutdivides thestructureintwoparts.Theportionofthestructureremovedfromthepartintoconsiderationshouldbereplacedbytheinternalforces.Theinternalforcesensuretheequilibriumoftheisolatedpartsubjectedtotheactionofexternalloadsandsupportreactions.Afreebody diagramofeithersegmentofthecutmemberisisolatedandtheinternalloads couldbederivedbythesixequations ofequilibriumappliedtothesegmentintoconsideration.

5.1.Example

InthefollowingexampleweshallconstructtheinternalforcesdiagramsforthegiveninFig.spaceframestructure. Theintroducedglobalcoordinatesystemisshowninthesamefigure.



The introduced local coordinate systems of the different elements of the space frame are

presentedinFig.Thetypical sectionswheretheinternalforcesmustbecalculated,inorderto

constructtherelevant diagrams,arenumbered from1to8inthesamefigure.Thetypical

sectionsareplacedatleastatthebeginningandattheendofeachelement(segment)ofthe

frame.Theinternalforces diagrams,inthelimitsof eachelement,couldbederivedbyusingthe

corresponding referenceandbasediagrams.



5.2.Example



5.2 BEAMS CURVED IN PLAN

5.2.1IntroductionArches are in fact beams with an initial curvature. The curvature is visible only in

elevation.In plan they they would appear in straight.the other cases of curved beams are ring beamssupporting water tanks,Silos etc.,beams supporting corner lintels and curved balconies etc.,Rampsin traffic interchanges invariably have curved in plan beams.

Curved beams in addition to the bending moments and shears would also develop torsionalmoments.

5.2.2.Moment,Shear and Torsion

The three diverse force components have one thing in common – the strain energy storedin a beam due to each type of force. Among the 3 we normally ignore the strain energy due toshear forces as negligible.

U = ∫M2ds/2EI+∫T2ds/2GJ



5.3. SUSPENSION CABLE5.3.1. Indroduction

Cablesandarchesareclosely related toeach otherandhencethey aregroupedin thiscourseinthesamemodule.Forlongspanstructures(fore.g.incasebridges)engineerscommonlyusecableorarchconstructionduetotheirefficiency.Inthefirst lessonofthismodule,cablessubjectedtouniformandconcentratedloadsarediscussed.Inthesecond lesson,arches ingeneralandthreehingedarchesinparticularalongwithillustrativeexamplesareexplained.

In thelasttwolessons ofthismodule, twohingedarchandhingeless archesareconsidered.Structure may be classified into rigid and deformable structures depending onchange in geometry ofthestructurewhilesupportingtheload.Rigidstructuressupportexternallyappliedloadswithout appreciable change intheir shape(geometry). Beamstrussesand framesare examplesofrigidstructures.

Unlikerigidstructures,deformablestructuresundergochangesintheirshapeaccordingtoexternallyappliedloads.However,itshouldbenotedthatdeformationsarestillsmall.Cablesandfabricstructuresaredeformablestructures.Cablesaremainly used to supportsuspensionroofs,bridgesand cablecarsystem. They arealsousedin electricaltransmissionlinesandforstructuressupportingradioantennas.In thefollowingsections,cablessubjectedtoconcentratedloadandcables subjectedtouniformloads areconsidered.

Theshapeassumedby aropeorachain(withnostiffness)undertheactionofexternalloadswhenhungfromtwosupportsisknownasafunicularshape. Cableisafunicular structure.Itiseasytovisualizethatacablehungfromtwosupportssubjectedtoexternal loadmustbeintenscable.Acablemaybedefinedasthestructureinpuretensionhavingthefunicularshapeof theload.(videFig.5.1and5.2).

As stated earlier, the cables are considered to be perfectly flexible (no flexuralstiffness)and inextensible.Astheyareflexibletheydonotresistshearforceandbendingmoment.Itissubjected toaxial tension only anditisalwaysacting tangentialtothecable at anypoint along thelength.If theweightof thecableisnegligibleascomparedwith theexternally appliedloadsthenitsself weightisneglectedintheanalysis.In thepresent analysisself weightisnotconsidered.

Consideracableasloadedin Fig.5.3.Letusassume thatthecablelengthsandsagat()areknown.Thefour reactioncomponentsatACDEBandB, cable tensionsineach ofthefour segmentsand threesagvalues:a totalof eleven unknown quantitiesaretobedetermined.Fromthegeometry,onecouldwritetwoforceequilibriumequations(0,0==ΣΣyxFF)ateachofthepointandDCBA,,,Ei.e.atotal of tenequationsandtherequiredonemoreequationmay bewrittenfromthegeometryof thecable.Forexample,ifoneof thesagisgiven then theproblemcan besolvedeasily.Otherwiseif the total lengthofthecableisgiventhentherequired equationmaybewritten as



Fig 5.1 Deformable Structure

Fig 5.2Unloaded Cable

Fig 5.3Cable in Tension

Cablesubjectedtouniformload.

Cablesareusedtosupportthedeadweightandliveloads ofthebridgedeckshavinglongspans.

Thebridgedecksaresuspendedfromthecableusingthehangers.Thestiffeneddeckprevents

thesupportingcablefrom changingitsshapebydistributingtheliveloadmovingoverit,fora

longerlengthofcable.In suchcases cableis assumedtobeuniformlyloaded.



Fig 5.4.Cable subjected to concentrated load

Fig 5.5.Cable Subjected to Uniformly Fig 5.6.Free Body Diagram

Distributed load



Considera cable which isuniformlyloaded asshown inFig 5.4.



Duetouniformlydistributedload,thecabletakesaparabolicshape.Howeverduetoits owndeadweightittakesashapeof acatenary. Howeverdeadweight of thecableis neglected in the presentanalysis.

5.3.ExampleDeterminereactioncomponentsatA andB,tensioninthecable andthesag ofthecable showninFig.5.7.Neglectthe selfweightofthe cable in the analysis.



Fig 5.7.

Sincetherearenohorizontalloads,horizontalreactionsatAandBshouldbethesame. TakingmomentaboutE,yields



Fig 5.8

Fig 5.9



UNIT-I FLEXIBILITYMATRIXMETHOD FORINDETERMINATE STRUCTURES

1. Whatis meantby indeterminatestructures?Structures that do not satisfythe conditions of equilibrium are called indeterminatestructure. Thesestructures cannot besolved byordinaryanalysis techniques.

2. Whataretheconditions ofequilibrium?Thethreeconditionsofequilibriumarethesumofhorizontalforces,verticalforcesandmomentsat anyjoint should beequal to zero.

i.e.∑H=0;∑V=0;∑M=0

3. Differentiatebetween determinateand indeterminatestructures.Determinatestructurescanbesolvingusingconditionsofequilibriumalone(∑H=0;∑V=0;∑M=0). No otherconditions arerequired.Indeterminatestructurescannotbesolvedusingconditionsofequilibriumbecause(∑H≠0;∑V≠0;∑M≠0).Additionalconditionsarerequiredforsolvingsuchstructures.Usuallymatrixmethods areadopted.

4. Definedegreeofindeterminacy (i).Theexcessnumberofreactionsthatmakeastructureindeterminateiscalleddegreeofindeterminacy,andisdenotedby(i).Indeterminacyisalsocalleddegreeofredundancy.Indeterminacyconsists ofinternal andexternal indeterminacies.

i =II+EIwhereII=internal indeterminacyand EI=external indeterminacy.

5. Defineinternal and external indeterminacies.Internalindeterminacy(II)istheexcessnoofinternalforcespresentinamemberthatmakeastructureindeterminate.Externalindeterminacy(EI)isexcessnoofexternalreactionsinthememberthatmakethestructureindeterminate.

i =II+EI;EI=r– e;wherer=no ofsupport reactions and e=equilibrium conditions

II=i –EIe=3 (planeframes) ande=6 (spaceframes)

6. Write theformulaefordegreeofindeterminacy for:(a)Two dimensional pinjointed truss(2D Truss)

i =(m+r)– 2j wherem=no ofmembersr=no ofreactions j=no ofjoints

(b)Two dimensional rigid frames/plane rigid frames (2DFrames)i =(3m+r)– 3j wherem=no ofmembers

r=no ofreactions j =no ofjoints



(c)Threedimensional spacetruss (3D Truss)i =(m+r)– 3j wherem=no ofmembers

r=no ofreactions j=no ofjoints

(d)Threedimensional spaceframes (3DFrame)i =(6m+r)– 6j wherem=no ofmembers

r=no ofreactions j=no ofjoints

7. Determine thedegreeofindeterminacy for the following2D truss.i =(m+r)-2jwherem=19

r=4j =10e=3

∑i =(19+4)–2x10=3External indeterminacyEI=r–e=4–3=1∑Internal indeterminacy II=i–EI=3-1=2

8. Determine the total, internal and external degreeofindeterminacy for theplanerigid framebelow.

i =(3m + r)– 3jwherem=7

r=4 j=6e=3

∑i =(3x7+ 4)– (3x6) =7External indeterminacyEI=r–e=4–3=1∑Internal indeterminacy II=i–EI=7-1=6

9. Determinei, EI, II for thegiven plane truss.i =(m + r)– 2jwherem=3

r=4 j=3e=3

∑i =(3+ 4)–(2x3) =1External indeterminacyEI=r–e=4–3=1∑Internal indeterminacyII=i–EI=1-1=0



10. Find theindeterminacyfor thebeams given below.Forbeams degreeofindeterminacyisgiven byi =r–e

(a)

i =r–e wherer=no of reactions, e=no of equilibrium conditions r=4ande=3∑i =4–3=1

(b)

i =r–ewherer=5 ande=3

∑i =5–3=2

11. Find theindeterminacyfor thegiven rigid planeframe.i =(3m + r)– 3jwherem=3

r=4j =4

∑i =(3x3+ 4)– (3x4) =1External indeterminacyEI=r–e=4–3=1∑Internal indeterminacy II=i–EI=1-1=0

12. Find theindeterminacyofthespace rigid frame.i =(6m + r)– 6jwherem=8

r=24 (i. e. 6persupportx4)j =8 e=6

∑i =(6x8+24)– (6x 8) =24External indeterminacyEI=r–e=24–6=18

∑Internal indeterminacy II=i–EI=24-18=6



13. Find theindeterminacyfor thegiven space truss.i =m +r-3jwherem=3

r=18 (i. e. 6reactions persupport x3)j =4

∑i =(3+18)– (3x4) =9External indeterminacyEI=r–e=18–6=12∑Internal indeterminacy II=i–EI=9-12=-3

14. Whatare thedifferent methods ofanalysis of indeterminatestructures.Thevarious methods adopted forthe analysis ofindeterminatestructures include:(a)Flexibilitymatrixmethod.(b) Stiffness matrixmethod(c)Finite Element method

15. Briefly mention the two types ofmatrixmethods ofanalysis ofindeterminatestructures.

Thetwo matrixmethodsof analysis ofindeterminatestructuresare:(a)Flexibilitymatrixmethod– This method is also called the forcemethod inwhich the

forces in thestructurearetreated as unknowns. Theno of equations involved is equal tothedegreeofstaticindeterminacyofthestructure.

(b)Stiffness matrixmethod– This is also called thedisplacement method in which thedisplacements thatoccurin thestructurearetreated as unknowns. Theno ofdisplacements involved is equal to theno ofdegrees of freedom ofthestructure.

16. Definea primary structure.Astructure formed bythe removingthe excess orredundant restraints fromanindeterminatestructuremakingit staticallydeterminateis called primarystructure. This isrequired forsolvingindeterminatestructures byflexibilitymatrixmethod.

17. Give theprimary structures for thefollowing indeterminatestructures.Indeterminatestructure PrimaryStructure

18. Definekinematicindeterminacy(Dk)orDegreeofFreedom (DOF)Degrees offreedom is defined as theleast no ofindependent displacementsrequired todefinethedeformed shapeofastructure. Therearetwo types of DOF: (a)Nodal typeDOFand(b)Joint typeDOF.



19. Briefly explain the twotypes ofDOF.(a)Nodal type DOF– This includes the DOFat thepoint of application of concentrated

load ormoment, at asection wheremoment ofinertia changes, hingesupport, rollersupport and junction oftwo ormoremembers.

(b)Joint type DOF– This includes the DOFat thepoint wheremoment ofinertiachanges, hingeand rollersupport, and junction oftwo ormoremembers.

20. Forthevarious support conditions shown belowgive the DOFs.

(a) No DOF

(b) 1– DOF(c) 2– DOF(d) 1– DOF

21. Forthe truss shown below,whatis the DOF?Pin jointed planeframe/truss

DOF/ Dk = 2j–rwherer=no of reactions

j = no ofjoints

22. Define compatibility in forcemethod ofanalysis.Compatibilityis definedas the continuitycondition on thedisplacements ofthestructureafterexternal loads are applied to thestructure.

23. Define theForceTransformation Matrix.The connectivitymatrixwhich relates theinternalforcesQ and theexternal forces R isknown as the forcetransformation matrix. Writingit in amatrixform,

{Q} =[b]{R}whereQ=member forcematrix/vector

b= forcetransformationmatrixR = external force/loadmatrix/ vector

24. Whatare therequirements to besatisfied whileanalyzing a structure?Thethreeconditions to besatisfied are:(a)Equilibrium condition(b)Compatibilitycondition(c)Forcedisplacement condition



25. Defineflexibility influence coefficient(fij)Flexibilityinfluence coefficient(fij)is defined as thedisplacement at joint ‘i’dueto aunitload at joint ‘j’, while all otherjoints arenot load.

26. Write theelementflexibility matrix(f)fora truss member.The element flexibilitymatrix(f) foratruss memberis given by

=





=





=



UNIT –II STIFFNESS MATRIXMETHOD

1. Whatarethebasicunknowns in stiffness matrix method?In thestiffness matrixmethod nodal displacementsaretreatedas thebasicunknowns forthesolution ofindeterminatestructures.

2. Definestiffness coefficientkij.Stiffness coefficient ‘kij’is defined as the forcedeveloped at joint ‘i’duetounitdisplacementat joint ‘j’while all otherjoints arefixed.

3. Whatis thebasicaimofthestiffness method?The aim ofthestiffnessmethod is to evaluatethevalues ofgeneralized coordinates ‘r’knowingthestructurestiffness matrix‘k’ and nodal loads ‘R’through thestructureequilibrium equation.

{R} =[K]{r}

4. Whatis thedisplacement transformationmatrix?The connectivitymatrixwhich relates theinternaldisplacement ‘q’and theexternaldisplacement‘r’is known as thedisplacement transformation matrix ‘a’.

{q} =[a]{r}

5. Howarethebasicequations ofstiffness matrixobtained?Thebasicequations ofstiffness matrixareobtained as:

EquilibriumforcesCompatibilityofdisplacementsForcedisplacement relationships

6. Whatis theequilibriumcondition usedin thestiffness method?Theexternal loads and theinternal memberforcesmust bein equilibrium atthenodalpoints.

7. Whatis meantby generalized coordinates?Forspecifyingaconfiguration ofasystem, acertain minimum no ofindependentcoordinatesarenecessary. Theleast no ofindependent coordinates thatareneeded tospecifytheconfigurationis known as generalized coordinates.

8. Whatis thecompatibility condition used in theflexibility method?Thedeformed elements fit togetherat nodal points.

9. Writeabout theforcedisplacement relationship.Therelationship ofeachelement must satisfythestress-strain relationship oftheelementmaterial.

10. Writetheelementstiffness fora truss element.Theelement stiffness matrixforatruss element isgiven by







{R} =[K]{r}


{q} =[a]{r}














{R} =[K]{r}


{q} =[a]{r}










11. Writetheelementstiffnessmatrix fora beamelement.Theelement stiffness matrixforabeam element is given by

12. Compareflexibility method and stiffness method.Flexibilitymatrixmethod

Theredundant forces aretreated as basicunknowns.Thenumberofequationsinvolved is equal to thedegreeofstaticindeterminacy

ofthestructure.Themethod is thegeneralization of consistent deformation method.Different proceduresareused fordeterminateandindeterminatestructures

Stiffness matrixmethodThejoint displacements aretreatedas basicunknownsThenumberofdisplacements involved is equal to theno ofdegrees offreedom of

thestructureThemethod is thegeneralization oftheslopedeflection method.Thesameprocedureis used forboth determinateand indeterminatestructures.

13. Is itpossibleto developtheflexibility matrix foran unstablestructure?Inorderto develop theflexibilitymatrixforastructure, it has to bestableanddeterminate.

14. Whatis therelation between flexibility and stiffness matrix?The element stiffness matrix‘k’is theinverseofthe element flexibilitymatrix‘f’ and isgivenbyf=1/k ork =1/f.

15. Whatarethetypeofstructures that can besolved using stiffness matrix method?Structures such as simplysupported, fixed beams and portal frames can besolved usingstiffness matrixmethod.

16. Givetheformula forthesizeoftheGlobal stiffness matrix.Thesizeoftheglobal stiffness matrix(GSM) =No: ofnodes xDegrees offreedom pernode.

17. List theproperties ofthestiffness matrixTheproperties ofthestiffness matrixare:

It isasymmetricmatrix Thesum of elements in anycolumn must be equal to zero. It is an unstableelementthereforethedeterminantis equal to zero.

18. Whyis thestiffness matrixmethodalso called equilibrium method ordisplacementmethod?

































Stiffness method is based on thesuperposition ofdisplacements and henceis also known asthedispalcement method. And sinceit leads to the equilibrium equations themethod is alsoknown as equilibrium method.

19. Ifthe flexibilitymatrixis givenas

Writethe correspondingstiffness matrix.

Stiffness matrix= 1/(Flexibilitymatrix)i.e. [K]=[F]-1

20. Writethen stiffness matrixfora 2Dbeam element.Thestiffness matrixfora2Dbeam element is given by

















UNIT III FINITE ELEMENT METHOD

1. Whatis meantbyFinite elementmethod?Finite element method (FEM)is anumerical technique forsolvingboundaryvalue problemsin which alargedomain is divided into smallerpieces or elements. Thesolution isdetermined byasuumingcertianploynomials.Thesmall pieces arecalled finiteelementand thepolynomials arecalled shapefunctions.

2. Listout theadvantages ofFEM. Sincetheproperties of each element are evaluatedseparatelydiffernt material

properties can beincorporated foreach element. Thereis no restriction in theshapeofthemedium. Anytypeofboundarycondition can be adopted.

3. Listout thedisadvantages ofFEM.The computational cost is high.

Thesolution is approximate and severalchecksare required.

4. Mention thevarious coordinates in FEM. Local or element coordinates

Natural coodinates Simplenatural coodinates

Areacoordiantesor Triangularcoordiantes Generalisedcoordinates

5. Whatare thebasicsteps in FEM? Discretization ofthestructure Selection ofsuitabledisplacement fuction

Findingtheelement properties Assemblingthe elementproperties

Applyingtheboundaryconditions Solvingthesystem of equations Computingadditional results

6. Whatis meantby discretization?Discretization is theprocess ofsubdividingthegiven bodyinto anumberof elementswhich results in asystem of equivalent finite elements.

7. Whatare thefactors governing theselection of finite elements?



ThegeometryofthebodyThenumberofindependent spacecoordinates

Thenatureofstress variation expected

8. Definedispalcementfunction.Displcementfunction is defined as simple functions which areassumed toapproximatethedisplacements foreach element. Theymayassumed in the form ofpoynomials, ortrignometricalfunctions.

9. Briefly explain afewterminology used in FEM.Thevarious terms used in FEM are explained below.

Finite element–Small elements used forsubdividingthegiven domain tobeanalysedarecalled finiteelements. Theseelements maybe1D, 2Dor 3Delements dependin on thetypeofstructure.

Nodes and nodal points– Theintersection ofthediffernt sides ofelementsarecalled nodes. Nodes areoftwo types – external nodes and internal nodes.

oExternal nodes – Thenodal point connectingadjacent elements.oInternal nodes– The extranodes used to increasethe accuracyofsolution.

Nodal lines – Theinterfacebetween elements arecalled nodal lines.

Continuum– Thedomain in which matter existsat everypoint is calledacontinuum.It can be assumed as havinginfinitenumberof connected particles.

Primaryunknowns– Themain unknowns involved in the formulation oftheelement properties areknown as primaryunknowns.

Secondaryunknowns– Theseunknowns arederived from primaryunknowns areknown as secondaryunknowns.In displacement formulations, displacementsaretreatedas primaryunknowns and stress, strain, moments and shear force aretreated as secondaryunknowns.

10. Whatarediffernt types ofelements used in FEM?Thevarious elements used in FEM are classifiedas:

Onedimensional elements(1D elements) Two dimensional elements(2D elements)

Threedimensional elements(3D elements)

11. Whatare1-D elements?Give examples.Elements havingaminimum oftwo nodes arecalled1Delements. Beamsareusuallyapproximated with 1Delements. Thesemaybestraight orcurved. There can beadditional nodes within the element.



Basic1-D element 1-D element with3nodes

Curved element with 3nodes

12. Whatare2-D elements?Give examples.Aplane wall, plate, diaphragm, slab, shell etc. can be approximated as anassemblageof2-D elements.Most commonlyused elements aretriangular, rectangularand quadrilateralelements.

Triangular elements Curved triangularelement

Rectangularand Quadrilateral elements

13. Whatare3-D elements?Give examples.3-D elements areusedformodelingsolid bodies andthevarious 3-Delements aretetrahedron, hexahedron,and curvedrectangularsolid.



14. Whatareaxisymmetricelements?Axisymmetricelements areobtained byrotatinga1-Dlineabout an axis. Axisymmetricelements areshown in thefigurebelow.

15. DefineShapefunction.Shapefunction is also called an approximatefunction or an interpolation function whosevalueis equal to unityat thenodeconsidered andzeros at all othernodes.Shapefunction isrepresented byNi wherei =nodeno.

16. Whataretheproperties ofshapefunctions?Theproperties ofshape functions are:

Theno ofshapefunctions will beequal to theno ofnodes present in theelement.

Shapefunction will haveaunit valueat thenode considered and zero valueatothernodes.

Thesum ofall theshapefunction is equal to 1. i. e.Ni =1

17. Defineaspect ratio.Element aspect ratio is defined as theratio ofthelargest dimension oftheelement to itssmallest dimension.

18. Whatarepossiblelocations fornodes?Thepossiblelocations fornodes are:

Point of application of concentrated load.Location wherethereis achangein intensityofloads

Locations wheretherearediscontinuities in thegeometryofthestructureInterfaces between materials ofdifferent properties.

19. Whatarethecharacteristics ofdisplacementfunctions?Displacement functions should havethefollowing characteristics:

Thedisplacement field should becontinuous.



Thedisplacement function should becompatiblebetweenadjacent elements Thedisplacement field must representconstant strain states of elements

Thedisplacement function must represent rigid bodydisplacements of anelement.

20. Whatis meantby planestrain condition?Planestrain is astateofstrain in which normal strain and shearstrain directedperpendicularto theplaneofbodyis assumed to bezero.



UNIT –IVPLASTICANALYSIS OFSTRUCTURES

1. Whatis a plastichinge?When asection attains full plasticmoment Mp, itacts as hingewhich is called aplastichinge.It is defined as theyielded zonedueto bendingat which largerotations can occurwith aconstant valueofplasticmoment Mp.

2. Whatis a mechanism?When an-degreeindeterminatestructuredevelops n plastichinges, it becomesdeterminateand theformation of an additional hingewill reducethestructureto amechanism. Onceastructurebecomesamechanism, it will collapse.

3. Whatis differencebetween plastichingeandmechanical hinge?Plastichinges modifythebehaviourofstructures in thesamewayas mechanical hinges.Theonlydifferenceis that plastichinges permit rotation with aconstant resistingmomentequal to theplasticmoment Mp. At mechanical hinges, theresistingmoment is equal tozero.

4. Definecollapseload.Theload that causes the(n +1)th hingeto form amechanism is called collapseloadwheren is thedegreeofstaticallyindeterminacy.Oncethestructurebecomes amechanism

5. Listout theassumptions madeforplasticanalysis.Theassumptions forplasticanalysisare:

Planetransversesections remain planeand normal to thelongitudinal axis beforeand afterbending.

Effect ofshearis neglected. Thematerial is homogeneous and isotropicboth in theelasticand plasticstate.

Modulus of elasticityhasthesamevalueboth in tension and compression. Thereis no resultant axial forcein thebeam. Thecross-section ofthebeam is symmetrical about an axis through its centroid

and parallel to theplaneofbending.

6. Defineshapefactor.Shapefactor(S) is defined as theratio ofplasticmoment ofthesection to theyieldmoment ofthesection.





























WhereMp =PlasticmomentM=Yield momentZp =Plasticsection modulusZ=Elasticsection modulus

7. Listout theshapefactors forthefollowing sections.(a)Rectangularsection S =1.5(b)Triangularsection S =2.346(c)Circularsection S =1.697(d)Diamond section S =2

8. Mention thesection having maximumshapefactor.Thesection havingmaximum shapefactorisatriangularsection, S =2.345.

9. Defineloadfactor.Loadfactoris definedastheratio of collapseloadto workingloadand is given by

10. Stateupperbound theory.Upperbound theorystates that of all theassumedmechanisms theexact collapsemechanism is that whichrequires aminimum load.

11. Statelowerbound theory.Lowerbound theorystates that thecollapseloadis determined byassumingsuitablemoment distribution diagram. Themoment distribution diagram is drawn in such awaythat theconditions of equilibrium aresatisfied.

12. Whatarethedifferent types ofmechanisms?Thedifferent types ofmechanisms are:

Beam mechanism

Column mechanism










Beam mechanism

Column mechanism










Beam mechanism

Column mechanism



Panel orswaymechanismCablemechanism

Combined or compositemechanism

13. Depending on thesupportand load conditionsindicatethepossiblelocations ofplastichinges.

14. Mention thetypes of frames.Framesarebroadlyoftwo types:(a)Symmetricframes(b)Un-symmetricframes

15. Whataresymmetricframes and howthey analyzed?Symmetricframes areframes havingthesamesupport conditions, lengths and loadingconditions on thecolumns and beams oftheframe. Symmetricframes can beanalyzedby:(a)Beam mechanism(b)Column mechanism

16. Whatareunsymmetrical frames and howarethey analyzed?Un-symmetricframes havedifferentsupport conditions, lengths and loadingconditionson its columns and beams. Theseframes can beanalyzed by:(a)Beam mechanism(b)Column mechanism



(c)Panel orswaymechanism(d)Combined mechanism

17. Defineplasticmodulusofa section Zp.

Theplasticmodulus ofasection is thefirst moment oftheareaaboveandbelowthe equalareaaxis.It is theresistingmodulus ofafullyplasticized section.

Zp =A/2 (Z1+Z2)

18. Howis theshapefactorofa hollowcircularsection related to theshapefactorofaordinary circularsection?

Theshapefactorofahollow circularsection =AfactorKxshapefactorofordinarycircularsection.

SFofhollow circularsection =SFof circularsection x{(1– c3)/(1– c4)}

19. Givethegoverning equation forbending.

Thegoverningequation forbendingis given by

M/I=/y

WhereM=Bendingmoment

I=Moment ofinertia

=Stress

y=c.g. distance

20. Givethetheorems fordetermining thecollapseload.

Thetwo theorems forthedetermination of collapseload are:

(a)StaticMethod [Lowerbound Theorem]

(b)KinematicMethod [Upperbound Theorem]



UNIT-V CABLE ANDSPACE STRUCTURES

1. Whatarecablestructures?Longspan structuressubjected to tension and usessuspension cablesforsupports.Examples of cablestructures aresuspension bridges, cablestayed roof.

Suspension bridge– cablestructure

2. Whatis thetrueshapeofcablestructures?Cablestructures especiallythecableofasuspension bridgeis in theform ofacatenary.Catenaryis theshapeassumed byastring/cablefreelysuspended betweentwo points.

3. Whatis thenatureof forcein thecables?Cables of cablestructures haveonlytension andno compression orbending.

4. Whatis a catenary?Catenaryis theshapetaken up byacableor ropefreelysuspended betweentwo supports andunderits own selfweight.

5. Mention thedifferenttypes ofcablestructures.Cablestructures aremainlyoftwo types:(a)Cableoveraguidepulley(b)Cableoverasaddle

6. Briefly explain cableovera guidepulley.Cableoveraguidepulleyhas thefollowingproperties:

Tension in thesuspension cable=Tension in theanchorcable

Thesupportingtower will besubjected to vertical pressureand bendingdueto nethorizontal cabletension.



7. Briefly explain cableoversaddle.Cableoversaddlehas thefollowingproperties:

Horizontal component oftension in thesuspension cable=Horizontal componentoftension in theanchor cable

Thesupportingtower will besubjected to onlyvertical pressuredueto cabletension.

8. Whatis thedegreeofindeterminacy ofa suspension bridgewith two hingedstiffening girder?Thetwo hinged stiffeninggirderhas onedegreeofindeterminacy.

9. Whatarethemain functions ofstiffening girders in suspension bridges?Stiffeninggirders havethefollowing functions.

Theyhelp in keepingthecables in shape

Theyresist part ofshearforceand bendingmoment dueto liveloads.

10. Differentiatebetween planetruss and spacetruss.Planetruss

All members liein oneplane All joints areassumed to behinged.

Spacetruss

This is athreedimensional truss

All joints areassumed to beball and socketed.

11. Definetension coefficientofa truss member.Thetension coefficient foramemberofatruss isdefined as thepull ortension in thememberdivided byits length, i. e. theforcein thememberperunit length.

12. Givesomeexamples ofbeamscurved in plan.Curved beams arefoundin thefollowingstructures.

Beams in abridgenegotiatingacurve



Ringbeams supportingawatertank Beams supportingcornerlintels

Beams in ramps13. Whataretheforces developed in beams curvedin plan?

Beamscurved in plan will havethefollowingforces developed in them:

Bendingmoments Shear forces

Torsional moments

14. Whatarethesignificantfeatures ofcircularbeams on equally spaced supports? Slopeon eithersideof anysupport will bezero.

Torsional moment on everysupport will bezero

15. Givetheexpression forcalculating equivalentUDL on a girder.Equivalent UDLon agirderis given byWe:

16. Givetherangeofcentral dip ofa cable.Thecentral dip ofacableranges from 1/10 to 1/12 ofthespan.

17. Givethehorizontal and vertical components ofa cablestructuresubjected to UDL.

Thehorizontal and vertical reactions aregiven by:

and respectively







Torsional moments







and respectively







Torsional moments







and respectively



18. Givetheexpression fordetermining thetension T in thecable.

Thetension developed inthecableisgiven by whereH=horizontalcomponent and V=vertical component.

19. Givethetypes ofsignificant cablestructuresLinear structures

Suspension bridges

Drapedcables Cable-stayed beams ortrusses

Cabletrusses Straight tensioned cables

Three-dimensional structures

Bicyclewheelroof

3D cabletrusses Tensegritystructures

Tensairitystructures

20. Whatarecablesmadeof?Cablescanbeofmildsteel,highstrengthsteel,stainlesssteel,orpolyesterfibres.

Structuralcablesare made of a seriesof smallstrandstwistedor boundtogether toformamuchlarger cable.Steelcablesare eitherspiralstrand,wherecircularrodsaretwisted togetherorlockedcoilstrand,whereindividualinterlockingsteelstrandsformthecable (often withaspiral strand core).Spiralstrand isslightly weaker thanlockedcoilstrand. Steelspiralstrand cableshaveaYoung'smodulus,Eof150±10kN/mm²andcomeinsizesfrom3to90mmdiameter.Spiralstrandsuffers fromconstructionstretch,where thestrandscompactwhenthecable isloaded.






Suspension bridges




Bicyclewheelroof










Suspension bridges




Bicyclewheelroof





Question Paper Code: 10230

B.E.IB.Tech. DEGREE EXAMINATION, MAY/JUNE 2012.

Seventh Semester

Civil Engineering

CE 2351/CE 61/CE 1352/10111 CE 602/080100036 - STRUCTURAL ANALYSIS - II

(Regulation 2008)

Time: Three hours Maximum: 100 marks

Answer ALL. questions.

PART A - (10 x 2 = 20 marks)

1. What are the conditions to be satisfied for determinate structures and how are indeterminate structures identified?

2. Write down the equation for the degree of static indeterminacy of the pinjointed plane frames, explaining the notations used.

3. Define degree of freedom of the structure with an example.

4. Write a short note on global stiffness matrices.

5. What are the needs to satisfy the shape function?

6. What is constant strain triangle?

7. Define plastic hinge with an example.

8. What is collapse load and define load factor?

9. Define a space frame and what is the nature of joint provided in the space trusses?

10. What are the types of stiffening girders?

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PART B - (5 x 16 = 80 marks)

11. (a) Analyze the pin-jointed plane frame shown in Figure Q. 11 (a) by flexibility matrix method. The flexibility for each member. is 0.0025 mrn/kN.

to ~N

4 WV Figure Q. 11 (a)

Or

(b) Analyze the continuous beam ABC shown in Figure Q. 11 (b) by flexibility matrix method and draw the bending moment diagram.

A

EI = Constant

Figure Q. 11 (b)

12. (a) Analyze the continuous beam ABC shown in Fig. Q. 12 (a) by stiffness method and also sketch the bending moment diagram.

EI = Constant

Figure Q. 12 (a)

Or

2 10230





(b) Analyze. the portal frame ABeD shown in Figure Q. 12 (b) by stiffness method and also sketch the bending moment diagram .

. 4-h\..

A D

Figure Q. 12 (b)

13. (a) With a two dimensional triangular element model, derive for the displacement in the matrix form

Or

(b) For the two dimensional truss structure shown in Figure Q. 13 (b), formulate the global stiffness matrix [K]. The geometry and loading are symmetrical about centre line. Assume the area of cross section of all members is the same. Take E = 2 X 108 kN / m 2 •

Figure Q. 13 (b)

3 10230





14. (a) Determine the shape factor of a T-section beam of flange dimension 100 x 12 mm and web dimension 138 x 12 mm thick.

Or

(b) Determine the collapse load 'W', for a three span continues beam of constant plastic moment 'Mp', loaded as shown in Fig. Q. 14(b).

f L. l'

Figure Q. 14 (b)

15. (a) A quarter circular beam of radius 'R' curved in plan is fixed at A and free at B as shown in Fig. Q. 15(a). It carries a vertical load- P at its free end. Determine the deflection at free end and draw the bending moment and torsional moment diagrams. Assume flexural rigidity (El) = torsional rigidity (GJ).

A_

o

Figure Q. 15 (a)

Or

(b) A three hindged stiffening girder of a suspension bridge of 100 m span subjected to two point loads 10 kN each placed at 20 m and 40 m, respectively from the left hand hinge. Determine the bending moment and shear force in the girder at section 30 m from each end. Also determine the maximum tension in the cable which has a central dip of 10m.

4 10230





Reg. No. :1,---,-_,---,- [I In

B.E./B.Tech. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2011.

Sixth Semester

Civil Engineering

CE 2351 - STRUCTURAL ANALYSIS - II

(Regulation 2008)

Time : Three hours Maximum: 100 marks

Answer ALL questions.

PART A - (10 x 2"" 20 marks)

1. Give the mathematical expression for the degree of static indeterminacy of rigid jointed plane frames.

2. What are the properties which characterize the structure response by

means afforce-displacement relationship?

3. What is degree of kinematic indetenninacy and give an example?

4. Write down the equation of element stiffness matrix as applied to 2D plane

element.

5. What is the basic idea of mesh generation scheme?

6. State the stress-stain relationship in Cartesian co-ordinates.

7. Define plastic modulus and shape factor.

8. What are meant by load factor and collapse load?

9. Define tension coefficient For what type of structures tension coefficient

method is employed?

10. What are the components of forces a.cting on the beams curved in plan

and show the sign conventions of these forces?



PART B - (5 x 16 == 80 marks)

11. (a) Analyze the continuous beam ABC shown in Fig, Q 11 (a) by flexibility matrix method and sketch the bending moment diagram.

h~~.fic ..... 4- '/'Vt-- I Z,.M.- I

e I ~ Cc"~sbA"f-

Fig. Q 11 (a)

Or

(b) Analyze the portal frame ABeD shown in Fig. Q 11 (b) by flexibility matrix method and sketch the bendi.."lg moment diagram.

Fig. Q 11 (b)

12. (a) Analyze the continuous beam ABC shown in Fig. Q12(a) by stiffness method and also draw the shear force diagram.

IJft.N .£.5' /J..N/M..-A~ . I'~'M I.~ fP:'Cla~c. ~L_ 3~ .-}--- .1' .

'3M.-E:':r. ::. ~bt '11 t-

Fig. Q12 (a)

Or

2 66144



(b) Analyze the portal frame ABeD shown in Fig. Q 12(b) by stiffness method and also draw the bending moment diagram.

:3pft ...

_k f :----

,. .. ~ h1 f..2.. 5'~--

F;y.~. p A

~:t::: c.c....sh:tt1\"" "

Fig. Q12 (b)

13. (aj Draw the typical finite elements. Explain with a triangular element model for displacement formulation.

Or

(b) Write a note on constant strain triangle. Explain in detail about the 4-nodded rectangular element to arrive the stiffness matrix.

14. (a) A simply supported beam of span 5m is to be designed for a UDL of 25 kN / m. Design a suitable I section using plastic theory, assuming yield stress in steel as fy = 250 N/mm2 •

Or

(b) Analyze a propped cantilever of length 'L and subjected to UDL of w/m length for the entire span and find the collapse load.

15. (a) A suspension bridge has a span 50 m with a 15 m wide runway. It is subjected to a load of 30 kN/m including self weight. The bridge is supported by a pair of cables having a central dip of 4m. Find the cross sectional area of the cable necessary if the maximum permissible stress in the cable material is not to exceed 600 MPa.

Or

3 66144



(b) A semi circular beam of radius 'R' in plan is subjected to UDL and simply supported by three columns spaced equally_ Derive the expression for bending moment and torsional moment at x-be a point on the beam making an angle a' with axis passing through the base of the circle,

4 66144



Reg. No.: 1 \ 1212101'6 1\ b 1:,:1; ls,,!:;T'''",",,'', .... t':'i'!1" \.t r.~~:~

'I Question Paper Code

B.E.IB.Tech. DEGREE EXAMINATION, APRILIMA¥Q<1l1

Sixth Semester

Civil Engineering

CE 2351- STRUCTURAL ANALYSIS - II

(Regulation 2008)

-~

~ -"'"4!:-

.~

~-

Time : Three hours Maximum: 100 marks

1.

2.

3. Define static indeterminacy.

4. Define flexibility of a structure.

5. State any two advantages of FEM. . :~~~.~ ~ ,:ft,'t. ':,:-. _.

D,efi:e plan;-~tress. 6.

7. :~ : ~_ ~." W· :i'. ;0>1"'" '" Whtlhs shape faCtor.:;,::

I·Sta"J;e upperbound."theorem. 'f'. (,' ~"

9. c,Qive aiily:two"~xainples of beams curved in plan.

10. WEite stress$s in suspended wires due to self weight. '!i.e; .J.

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PART B - (5 x 16 = 80 marks)

11. (a) Analyse the continuous beam show in Fig .1 using force method.

100 kN

EI const

Fig. 1

Or

(b) Analyse the portal frame ABeD shown in Fig .2 using force method.

EI canst

Fig.2

12. (a) Analyse the continuous beam shown in Fig.3 using di;,;placement method.

240 kN

EI const

Fig.3

Or

2 11217



(b) Analyse the truss shown in FigA using displacement method.

FigA

13. (a) Develop the shape functions for an 8 noded brick element.

Or

(b) Construct the shape functions of a 2D beam element.

14. (a) Derive the shape factor for I section and circular section.

(b) Find the fully plastic momerit req!rtrt\Q. forCtb,~ frame shown in Fig.5 if all the members have same vahle o{,;,Mp • '\

T~~

2k~.] ................. , 2m .. . ,··4m

··:,'···.c .. 7'

6m

. "ii;·;'!·\'4,,,;;,,. .:'

'''Fr~.5 15. (a) A suspension cable is supported at two point "A" and "B" , "A" being one

metre above "B". The distance AB being 20m. The cable is subjected to 4 loads of 2kN, 4kN, 5kN and 3kN at distances of 4m, 8m, 12m and 16m

,respectively from "A". Find the maximum tension in the cable, if the dip of the cable at point of application of first load is 1m with respect to level at A. Find. also the length of the cable.

Or

(b) Derive the expressions for BM, SF and TM in a semicircular beam simply supPdt-ted oni,'three supports equally spaced.

,,5:."_ ",,=-

3 11217

/



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