3. bhavikatti, s.s, “structural analysis - · pdf filece6602 structural analysis ......

CE6602 STRUCTURAL ANALYSIS – II L T P C 3 1 0 4 OBJECTIVE This course is in continuation of Structural Analysis – Classical Methods. Here in advanced

method of analysis like Matrix method and Plastic Analysis are covered. Advanced topics

such as FE method and Space Structures are covered.

UNIT I FLEXIBILITY METHOD 12 Equilibrium and compatibility – Determinate vs Indeterminate structures – Indeterminacy -

Primary structure – Compatibility conditions – Analysis of indeterminate pin-jointed plane

frames, continuous beams, rigid jointed plane frames (with redundancy restricted to two).

UNIT II STIFFNESS MATRIX METHOD 12 Element and global stiffness matrices – Analysis of continuous beams – Co-ordinate

transformations – Rotation matrix – Transformations of stiffness matrices, load vectors and

displacements vectors – Analysis of pin-jointed plane frames and rigid frames( with redundancy

vertical to two) UNIT III FINITE ELEMENT METHOD 12 Introduction – Discretisation of a structure – Displacement functions – Truss element –

Beam element – Plane stress and plane strain - Triangular elements UNIT IV PLASTIC ANALYSIS OF STRUCTURES 12 Statically indeterminate axial problems – Beams in pure bending – Plastic moment of resistance

– Plastic modulus – Shape factor – Load factor – Plastic hinge and mechanism – Plastic

analysis of indeterminate beams and frames – Upper and lower bound theorems UNIT V SPACE AND CABLE STRUCTURES 12 Analysis of Space trusses using method of tension coefficients – Beams curved in plan

Suspension cables – suspension bridges with two and three hinged stiffening girders TOTAL: 60

PERIODS 53 TEXT BOOKS 1. Vaidyanathan, R. and Perumal, P., “Comprehensive structural Analysis – Vol. I & II”, Laxmi

Publications, New Delhi, 2003

2. 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., New Delhi, 2008 REFERENCES 1. Ghali.A, Nebille,A.M. and Brown,T.G. “Structural Analysis” A unified classical and

Matrix approach” –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, 1990 3. Structural Analysis – A Matrix Approach – G.S. Pandit & S.P. Gupta, Tata McGraw Hill 2004.

4. Matrix Analysis of Framed Structures – Jr. William Weaver & James M. Gere,

CBS Publishers and Distributors, Delhi.

Structural Analysis II

CHAPTER 1

FLEXIBILITY METHOD

Equilibrium 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).

1.1 INTRODUCTION

These are the two basic methods by which an indeterminate skeletal structure is

analyzed. In these methods flexibility and stiffness properties of members are employed.

These methods have been developed in conventional and matrix forms. Here conventional methods are discussed.

Thegivenindeterminatestructureisfirstmadestaticallydeterminatebyintroducing∝ suitable numberof releases. The number of releases required is equal to staticalindeterminacy s. Introductionofreleasesresultsin displacementdiscontinuitiesatthesereleases under the externally applied loads. Pairs ofunknown biactions(forces andmoments)areappliedatthesereleasesinordertorestorethecontinuityorcompatibility of structure.

The computation of these unknown biactions involves solution of∝ linear simultaneousequations.Thenumberoftheseequationsisequaltostaticalindeterminacy s. Aftertheunknownbiactionsarecomputedall theinternalforcescanbecomputedintheentirestructureusingequationsofequilibriumandfreeb odiesofmembers.Therequired displacements can also be computed using methods of displacement computation.

Inflexibilitymethodsinceunknownsareforces atthereleasesthemethodisalsocalled

force method.Since computation of displacement is also required at releases for imposing

conditions of compatibility the method is also called compatibility method. In

computationofdisplacementsuseismadeof flexibilityproperties,hence,themethodis also

called flexibility method.

1.2 EQUILIBRIUM and COMPATABILITY CONDITIONS

Thethreeconditionsofequilibriumarethesumofhorizontalforces,verticalforcesandmom

ents at anyjoint should beequal to zero.

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

i.e.∑FX=0;∑FY=0;∑FZ=0

i.e.∑MX=0;∑MY=0;∑MZ=0 Displacement of a structure should be compatable The compatibility conditions for the supports can be given as 1.Roller Support δV=0

2.Hinged Support δV=0, δH=0

3.Fixed Support δV=0, δH=0, δө=0

1 Dept of Civil


1.3.DETERMINATE AND INDETERMINATE STRUCTURAL SYSTEMS

Ifskeletalstructureissubjectedtograduallyincreasingloads,withoutdistortingthe

initialgeometryofstructure,thatis,causingsmalldisplacements,thestructureissaidto be stable.

Dynamic loads and buckling or instability of structural system are not

consideredhere.Ifforthestablestructureitispossibletofindtheinternalforcesinall the members

constituting the structure and supporting reactions at all the supports providedfrom

staticallyequationsofequilibrium only,thestructureissaidtobe determinate.

Ifitispossibletodetermineallthesupport reactionsfromequationsof equilibrium

alonethestructureissaidtobeexternallydeterminateelseexternally indeterminate.If structureis

externallydeterminatebutitisnotpossible todetermineall

internalforcesthenstructureissaidtobe internallyindeterminate. Thereforeastructural

systemmaybe:

(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 SYSTEM

The 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

–c P = 6 for space frames subjected to general loading P = 3 for plane frames subjected to inplane or normal to plane loading. N = Numberof nodes in structural system. M=Numberofmembersofcompletelystiffstructurewhichincludesfoundationas

singlyconnectedsystem ofmembers. Incompletelystiffstructurethereisnorelease present.Insinglyconnectedsystem

ofrigidfoundationmembersthereisonlyoneroute

betweenanytwopointsinwhichtracksarenotretraced. Thesystemisconsidered comprising of closed 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) 2 Dept of Civil


For plane and space trusses∝sreduces to:

∝s=M- (NDOF)N+ P M= 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+ 6

For 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 determinate

and if∝s<0 Structure is a mechanism.

Itmaybenotedthatstructuremaybemechanismevenif ∝s >0ifthereleasesare

presentinsuchawaysoastocausecollapseasmechanism.Thesituationofmechanism is

unacceptable.

Statically Indeterminacy Itisdifferenceoftheunknownforces(internalforcesplusexternalreactions)andthe

equations of equilibrium.

Kinematic Indeterminacy Itisthenumberofpossiblerelativedisplacementsofthenodesinthedirectionsofstress

resultants.

1.5 PRIMARY STRUCTURE Astructure formed bythe removingthe excess orredundant restraints froman

indeterminatestructuremakingit staticallydeterminateis called primarystructure. This is required forsolvingindeterminatestructures byflexibilitymatrixmethod.

Indeterminatestructure PrimaryStructure

3 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 is

statically indeterminate to a degree more than one, then the approach presented in the force method is suitable.

Problem 1.1 Calculate the support reactions in the continuous beam ABC due to loading as shown in

Fig.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 beamis

statically indeterminate to second degree. In this case the primary structure is a

cantilever beam 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 C

can be readily calculated by moment-area method. Thus

4 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 the compatibility 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

5 Dept of Civil

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.Theprimarystructureinthiscaseisacantileverbeamwhichcould

beobtainedbyreleasingtheredundant R1 andR2.

TheR1 isassumedto

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

counterclockwisedirection.Now,calculatedeflectionat B duetoonlyapplied

loading.Let ( L )bethetransversedeflectionat1 B and( L 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

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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

load method. Thus,

(Δ L1) =wL4/8EI – 3wL

4/8EI = −wL

4/2EI (1)

(Δ L2) = wL3/6EI – wL

3 /2EI = − 2wL

3/3EI (2)

7 Dept of Civil

The negativesign indicates that ( L

)isdownwards and rotation( is

1 L2)

clockwise.

Problem 1.3.

A continuous beam ABC is carrying a uniformly distributed loadof 1 kN/m in addition toaconcentratedloadof10kNasshowninFig.7.5a, Draw bending momentandshearforce

diagram.Assume EItobeconstantforallmembers. Fig1.10.Continuous Beam

Fig1.11.Primary Structure

8 Dept of Civil

Fig1.12.Flexibility Coefficients

Fig1.13.Reactions Itisobservedthatthecontinuousbeamisstaticallyindeterminatetofirstdegree.

ChoosethereactionatB, RBy astheredundant.Theprimarystructureisa 9 Dept of Civil

simplysupportedbeamasshowninFig.1.11.Now, compute the deflection at B, in the

releasedstructure due to uniformly distributed load and concentrated load. Thisis accomplished by unit load method.Thus,

−2083.33−1145.84 L

=

EI EI

L =−3229.17

(1)

EI Inthenextstep,applyaunitloadatBinthedirectionof

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

L3

166.67 a

11 =

48EI =

(2)

EI

Now,deflectionatBintheprimary structure due to redundant RB is,

= 166.67 × RB

B (3)

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

L+ B=0(4) Theothertworeactionsarecalculatedbystaticequilibriumequations(videFig. 1.13)

RA =7.8125kN

RB =2.8125kN 10 Dept of Civil


UNIT II STIFFNESS MATRIX METHOD Element and global stiffness matrices – Analysis of continuous beams – Co-ordinate

transformations – Rotation matrix – Transformations of stiffness matrices, load vectors and

displacements vectors – Analysis of pin-jointed plane frames and rigid frames( with redundancy

vertical to two)

2.1 INTRODUCTION

Thegivenindeterminatestructureisfirstmadekinematic allydeterminatebyintroducing

constraints atthenodes.Therequirednumberofconstraintsisequaltodegrees offreedomatthe

nodesthatis kinematicindeterminacy∝k.Thekinematic allydeterminatestructurecomprises

offixedendedmembers,hence,allnodal displacementsarezero.Theseresultsinstress resultant

discontinuitiesatthesenodesundertheactionofappliedloadsorin otherwordstheclamped

jointsarenotinequilibrium. Inordertorestoretheequilibriumofstressresultantsatthenodes

thenodesareimpartedsuitableunknowndisplacements.Thenumberofsimultaneousequationsrepresen

tingjointequilibriumofforcesisequaltokinematicindeterminacy∝k.Solutionof

theseequationsgivesunknownnodaldisplacements.Usingstiffnesspropertiesofmembersthe

memberendforcesarecomputedandhencetheinternalforcesthroughoutthestructure. 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 by

expressingtheunknownjointdisplacementsintermsofloadsby usingload-displacementrelations. Theunknownjointdisplacements(thedegreesoffreedomof thestructure)are calculated by solving

equilibriumequations.Theslope-deflection andmoment-distributionmethodswereextensively used beforethehigh speedcomputingera.Aftertherevolutionincomputerindustry,only directstiffness

methodisused.

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 member

axis 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 is

called system on global coordinates (Common coordinate system dealing with the entire structure) 11 Dept of Civil


Transformationmatrix

The connectivitymatrixwhich relates theinternalforcesQ and theexternal forces R is

known as 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

12 Dept of Civil

Structural Analysis II 13 Dept of Civil


Fig 2.5 A Four member Truss

Fig 2.6 Kinematic ally Determinate Structures 15 Dept of Civil


Fig 2.7Unit Displacement along U

16 Dept of Civil

Structural Analysis II 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.

19 Dept of Civil


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

20 Dept of Civil


Fig 2.10 Frame Member in Local Coordinate System

21 Dept of Civil


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

22 Dept of Civil


Fig 2.12 Rigid Frame 23 Dept of Civil

Structural Analysis II Fig 2.13 Node and Member Numbering

24 Dept of Civil


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

Fig 2.14 Equivalent Joint Load

27 Dept of Civil


Fig 2.15 Support Reactions

28 Dept of Civil


CHAPTER III

FINITE ELEMENT METHOD

Introduction – Discretization of a structure – Displacement functions – Truss element –

Beamelement – Plane stress and plane strain - Triangular elements 3.1.INTRODUCTION TherearetwoversionofFEM:

1. FlexibilityMethodorForceMethod

2. StiffnessMethodorDisplacementMethod.

Thesetofequationsinthestiffnessmethodaretheequilibriumequationsrelatingdisplacement sofpoints.

Rayleigh-Ritzisanapproximatemethodbasedonenergyprincipleby whichwecanobtainequilibriumequationsinmatrixform.

3.1.1 IMPORTANT DEFINITION Nodesarepointsonthestructureatwhichdisplacementsandrotations are tobefoundorprescribed. Element is a small domainonwhichwecan solvethe boundaryvalue

problemintermsofthedisplacementsandforcesofthe nodesonthe element. Thediscrete representationofthe structuregeometrybyelements and nodesis called a mesh. Theprocessofcreating a mesh(discreteentities) is called discretization. Interpolationfunctionisakinematicallyadmissibledisplacementfunctiondefinedonanelement

that canbeusedforinterpolatingdisplacement valuesbetweenthe nodes. Themesh,boundaryconditions,loads,andmaterial propertiesrepresentingthe actual structureis

called a model. Element stiffnessmatrix relate thedisplacementstothe forcesat the elementnodes. Globalstiffnessmatrix is anassemblyofelement stiffnessmatrix that relates

the displacementsofthe nodesonthemeshtoappliedexternal forces. 3.1.2.StepsinFEMprocedure

1.Obtainelementstiffnessandelementloadvector.

2.Transformfromlocalorientationtoglobalorientation.

3.Assembletheglobalstiffnessmatrixandloadvector.

4.Incorporatetheexternalloads

5.Incorporatetheboundaryconditions.

6.Solvethealgebraicequationsfornodaldisplacements.

7.Obtainreactionforce,stress,internalforces,strainenergy. 31 Dept of Civil


8.Interpretandchecktheresults.

9.Refinemeshifnecessary,andrepeattheabovesteps.

3.2.DISCRETISATION OF STRUCTURE Discretizationis the process of separating the length, area or volume we want to analyze into

discrete (or separate) parts or elements. 32 Dept of Civil


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 on

their 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 defines

strain within an element in terms of nodal displacements. These strains with any initial strain, together with elastic properties define

the stress state. 34 Dept of Civil


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

discretized using 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 of

2D elements. Triangular elements are the most used ones. when our 2D domain has curved

boundaries 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

that the visualizations is complex. The size of the stiffness matrix to be handled can become

enormous and unwieldy.

37 Dept of Civil


3.5 PLANE STRESS AND PLANE STRAIN

The plane stress problem is one in which two dimensions ,length and breadth are

comparable and thickness dimension is very small (less than 1/10).Hence normal stress σ2 and

shear 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 stress

problem is very small. In plane strain problem, on the contrary the structure is infinitely long in the 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.

38 Dept of Civil


CHAPTER 4

PLASTIC ANALYSIS OF STRUCTURES

Statically indeterminate axial problems – Beams in pure bending – Plastic moment of resistance

Plastic modulus – Shape factor – Load factor – Plastic hinge and mechanism – Plastic

analysis of indeterminate beams and frames – Upper and lower bound theorems

4.1.Statically indeterminate axial problems

Intheseanalysesweused superposition often,knowing thatforalinearly

elasticstructureitwasvalid.However,an elastic analysisdoesnotgiveinformation about theloadsthatwill actually collapseastructure.An indeterminatestructuremay sustainloads

greaterthantheloadthatfirstcauses ayieldtooccur at anypointinthestructure. Infact,astructurewillstandaslongasitisabletofindredundancies toyield.Itisonly when

astructurehasexhaustedallofitsredundancieswillextraloadcausesit tofail.Plasticanalysisis themethodthroughwhichtheactualfailureloadof astructureis

calculated,andaswillbeseen,thisfailureloadcanbesignificantly greaterthan the elasticload capacity. Tosummarizethis,Prof.SeandeCourcy(UCD)usedtosay:

“astructureonlycollapseswhenithas exhaustedallmeans ofstanding”. Before analyzingcomplete structures, we review material and cross section behaviorbeyondtheelasticlimit. 4.2. Beams in pure bending

4.2.1. MaterialBehavior

Auniaxialtensilestressonaductile materialsuchasmild steeltypicallyprovidesthe

followinggraphofstress versus strain:

Ascanbeseen,thematerialcansustainstrainsfarinexcessofthestrainatwhichyield occurs

beforefailure.This propertyofthematerialis calledits ductility.Thoughcomplex models do exist to

accurately reflect theabovereal behaviourofthe

material,themostcommon,andsimplest,modelistheidealizedstress-straincurve.Thisis

thecurveforanidealelastic-plasticmaterial(whichdoesn‟texist), andthegraphis:

39 Dept of Civil


As canbeseen, oncetheyieldhasbeenreacheditis takenthatanindefiniteamountofstraincan

occur. Sincesomuchpost-yieldstrainismodeled, theactualmaterial(orcross section)mustalso

becapableofallowingsuchstrains.Thatis,itmustbesufficientlyductilefortheidealized stress-

straincurvetobevalid.Nextweconsiderthebehaviourof acrosssectionofanideal elastic-

plasticmaterialsubjecttobending.In doingso,weseektherelationshipbetween applied

momentandtherotation(ormoreaccurately, thecurvature)ofacross section.

4.2.2.Moment-RotationCharacteristics ofGeneralCross Section

Weconsider anarbitrarycross-sectionwithaverticalplaneofsymmetry,whichisalsotheplane ofloading.Weconsiderthecrosssectionsubjecttoanincreasingbendingmoment,andassess thestresses

ateach stage.

Cross sectionandStresses 40 Dept of Civil


Moment-Rotation Curve Stage1– ElasticBehaviour

Theappliedmomentcauses stresses overthecross-sectionthatareallless thantheyieldstress of thematerial. Stage2–YieldMoment

Theappliedmomentisjustsufficientthattheyieldstressof thematerialisreachedatthe outermostfibre(s)ofthecross-section.Allotherstressesinthecrosssectionarelessthanthe yieldstress.Thisislimitofapplicabilityofanelasticanalysisandof elasticdesign.Sinceall fibresareelastic, theratioofthedepthoftheelastictoplasticregions, Stage3–Elasto-PlasticBending

Themomentapplied to thecrosssection hasbeenincreasedbeyond theyieldmoment.Sinceby theidealizedstress-strain curvethematerial cannotsustain astressgreaterthanyieldstress,the

fibresattheyieldstresshaveprogressedinwardstowardsthecentreof thebeam.Thusoverthe cross

sectionthereisanelasticcoreandaplasticregion.Theratioofthedepthoftheelasticcore totheplasticregionis .Sinceextramomentis beingappliedandnostressisbiggerthantheyield

stress,extrarotationof thesectionoccurs:themoment-rotationcurvelossesitslinearityand curves, givingmorerotationperunitmoment(i.e.loosesstiffness). Stage4–PlasticBending

Theappliedmomenttothecrosssectionissuch thatallfibresin thecrosssection areatyield stress.ThisistermedthePlasticMomentCapacityof thesection sincetherearenofibresatan

elasticstress,Alsonotethatthefullplasticmomentrequiresaninfinitestrainattheneutralaxis

41 Dept of Civil

Structural Analysis II andsoisphysicallyimpossibletoachieve.However,itisclosely approximatedin practice.Any attemptatincreasingthemomentat thispointsimply resultsinmorerotation,oncethecross-

sectionhassufficientductility.Thereforeinsteelmembersthecrosssectionclassificationmust beplasticandinconcretemembers thesectionmustbeunder-reinforced.

Stage5–StrainHardening

Duetostrainhardeningofthematerial,asmallamountofextramomentcanbesustained.

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

With thisidealizedmoment-rotation curve,thecrosssectionlinearlysustainsmomentupto the

plasticmomentcapacityofthesectionandthenyieldsinrotationanindeterminateamount. Again,tousethisidealization,theactual sectionmustbecapableofsustaininglargerotations–

thatisitmustbeductile. Analysis ofRectangularCross Section

Sincewenowknowthatacross

sectioncansustainmoreloadthanjusttheyieldmoment,weareinterestedinhowmuchmore.Inotherword

swewanttofindtheyieldmomentandplasticmoment,andwedosoforarectangularsection.Takingthestre

ssdiagramsfromthoseofthe moment-rotationcurveexaminedpreviously,wehave:

42 Dept of Civil

Structural Analysis II 4.3.ShapeFactor

Thus theratioofelastictoplasticmomentcapacityis:

Thisratioistermedtheshapefactor,f,andisapropertyofacrosssectionalone.Fora rectangularcross-section,wehave:

Andsoarectangularsectioncansustain50%moremomentthantheyieldmoment, beforeaplastichingeisformed.Thereforetheshapefactorisagoodmeasureoftheefficiency ofacross sectioninbending.Shapefactors forsomeothercross sections are 43 Dept of Civil

Structural Analysis II 4.4.PlasticHinge

Notethatoncetheplasticmomentcapacityisreached,thesectioncanrotatefreely–

thatis,itbehaveslikeahinge,exceptwithmomentofMpatthehinge.Thisis termedaplastichinge,and

isthebasisforplasticanalysis.Attheplastichingestressesremainconstant,butstrainsand hencerotations

canincrease.

4.4.1.Methods ofPlasticAnalysis

1. TheIncrementalMethod

Thisisprobably themostobviousapproach:theloadson thestructureareincrementeduntilthe

firstplastichingeforms.Thiscontinuesuntilsufficient hingeshaveformedtocollapsethe

structure.Thisisalabour-intensive,„brute-force‟,approach,butonethatismostreadilysuited

forcomputerimplementation.

2. TheEquilibrium(orStatical) Method

In thismethod,freeandreactantbendingmomentdiagramsaredrawn.Thesediagramsare overlaidtoidentifythelikelylocations of plastichinges.Thismethodthereforesatisfies the

equilibriumcriterionfirstleavingthetwo remainingcriteriontoderivedtherefrom.

3.The Kinematic (or Mechanism) Method

In this method, a collapse mechanism is first postulated. Virtual work equations are then written for this collapse state, allowing the calculations of the collapse bending moment diagram. This method satisfies 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 Method to 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:

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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;

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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:

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ForStep7, wesolvethis equationforthecollapseload: 48 Dept of Civil

Structural Analysis II 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.

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Structural Analysis II 4.5.CollapseMechanism

Soforourpreviousbeam,we know thatwe require twohingesforcollapse(onemore thanits

degreeofredundancy),andwethinkthatthehinges willoccurunderthepoints ofpeakmoment, AandC.Thereforeimposeaunitvirtual displacementatCandrelatethecorrespondingvirtual rotations

ofthehinges using,

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4.5.1 OtherCollapseMechanisms

Forthecollapsemechanismlookedatpreviously,itseemed obviousthat theplastichingein the

spanshouldbebeneaththeload.Butwhy?Usingvirtual workwecan examineanypossible

collapsemechanism.Solet‟sconsiderthefollowingcollapsemechanismandseewhytheplastic hingehas

tobelocatedbeneaththeload.

PlasticHinge betweenAandC: ImposingaunitvirtualdeflectionatB,wegetthefollowingcollapsemechanism:

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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. 52 Dept of Civil

Structural Analysis II WeneedtocheckthatthisisthecorrectsolutionusingtheUniquenessTheorem:

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AndsotheappliedloadisinequilibriumwiththefreeBMDofthecollapseBMD. 2.Mechanism:

Fromtheproposedcollapsemechanismitis apparentthatthebeamis

amechanism. 3.Yield:

FromthecollapseBMDitcanbeseenthatnowhereis exceeded.PM

Thusthesolutionmeetsthethreeconditionsandso,bytheUniquenessTheorem,isthecorrect solution.

Example3–ProppedCantileverwithTwoPointLoads

Forthefollowingbeam,foraloadfactorof2.0,findtherequiredplasticmoment capacity:

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Allowingfortheloadfactor,weneedtodesignthe beamfor thefollowingloads: Once againwe try to picture possible failure mechanisms. Since

maximummomentsoccurunderneathpointloads,thereare tworeal possibilities:

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Mechanism-1

Mechanism-2

Therefore,we analyseboth and apply the UpperboundTheoremto find the

designplasticmomentcapacity. Mechanism1:PlasticHingeatC:

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Mechanism2:PlasticHingeatD: 57 Dept of Civil


1.Equilibrium: UsingtheBMDatcollapse,we‟llcheckthattheheightofthefreeBMDisthatof

theequivalentsimply-supportedbeam.FirstlythecollapseBMDfromMechanism1is:

Hence,thetotalheightsofthefreeBMDare:

Checkingtheseusingasimply-supportedbeamanalysis

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Thus,usingappropriatefreebodydiagrams ofACandDB: AndsotheappliedloadisinequilibriumwiththefreeBMDofthecollapseBMD.

2.Mechanism: Fromtheproposedcollapsemechanismitisapparentthatthebeamisamechanism.Also,since

itisaproppedcantileverand thusonedegreeindeterminate,we

requiretwoplastichingesforcollapse,andthesewehave 3.Yield:

FromthecollapseBMDitcanbeseenthatnowhereis thedesignexceeded.144kNmThus

bytheUniqueness

Theoremwehavethecorrectsolution.Lastly,we‟llexaminewhytheMechanism2collapseisnotthecorrect

solution.Sincethevirtual workmethodprovidesan upperbound,then,by

theUniquenessTheorem,itmustnotbethe correctsolutionbecauseitmustviolatetheyieldcondition.

Using thecollapseMechanism2todeterminereactions,wecan draw thefollowingBMDfor

collapseMechanism2:

FromthisitisapparentthatMechanism2isnottheuniquesolution,andsothedesignplastic moment capacity must be144kNmasimpliedpreviously fromtheUpperboundTheorem.

4.BasicCollapseMechanisms: In frames,the basicmechanismsofcollapseare:

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Beam-typecollapse

SwayCollapse

CombinationCollapse

5.CombinationofMechanisms

Oneofthemostpowerfultoolsinplasticanalysisis CombinationofMechanisms.Thisallows us toworkoutthevirtualworkequationsforthebeamandswaycollapsesseparatelyandthen

combinethemtofindthecollapseloadfactorforacombinationcollapsemechanism.

Combinationof mechanismsis based onthe ideathat thereareonlya certain number of independentequilibriumequationsforastructure.Anyfurtherequationsareobtainedfroma

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combinationoftheseindependentequations.

Sinceequilibriumequationscanbeobtainedusing

virtualworkappliedtoapossiblecollapsemechanism,itfollowsthatthereareindependent collapsemechanisms,andothercollapsemechanismsthatmaybeobtainedformacombination

oftheindependentcollapsemechanisms.

6.SimplePortalFrame

Inthisexamplewewillconsiderabasicprismatic(soallmembershavethesame plasticmomentcapacity)rectangularportalframewithpinnedfeet:

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Wewillconsiderthisgeneralcasesothatwecaninferthepropertiesandbehaviourofallsuch frames.Wewillconsidereachofthepossiblemechanisms outlinedabove.

7.Beam collapse:

Thepossiblebeamcollapselooks as follows:

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Structural Analysis II 4.7.CollapseMode

Sincewedon‟tknowtherelativevaluesofHandV,wecannotdeterminethe correctcollapsemode.However,wecanidentifythesecollapsemodesifweplot the

threeloadfactorequationsderivedaboveonthefollowinginteractionchart:

Noticethateachmechanismdefinesaboundary andthatitisonlytheregion

insidealloftheseboundariesthatissafe.Now,foragivenrationofVtoH,we

willbeabletodeterminethecriticalcollapsemechanism.Notealsothatthebeamcollapsemechanismison

lycriticalforthisframeatpointPonthechart–

thispointisalsoincludedintheCombinedmechanism.Thebendingmomentdiagramscorresponding

toeachofthemechanismsare approximately: Aninterestingphenomenon isobserved atpointQonthechart,wheretheSway

andCombinedmechanismsgivethesameresult.Lookingatthebendingmoment diagrams,we cansee 63

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thatthis occursasthe momentatthe topofthe leftcolumn becomesequaltothe mid-spanmomentof the beam:

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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.

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CHAPTER 5

SPACE AND CABLE STRUCTURES Analysis of Space trusses using method of tension coefficients – Beams curved in plan

Suspension 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 Method The 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

member T= Pull in the member

l = 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

support lengths lij of each member using the equation lij =√Xij2+Yij

2

3. Resolve the the applied the forces at the joint in the X and Y directions. Determine the support reactions 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 of equilibrium 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

the support lengths lij of each member using the equation lij =√Xij2+Yij

2 +Zij

2

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

Tetrahedron: simplestelementofstablespacetruss

(sixmembers,fourjoints)expandbyadding 3members and1jointeachtime DeterminacyandStability b+r<3junstable b+r=3jstaticallydeterminate(checkstability)

b+r>3jstaticallyindeterminate(checkstability)

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InternalForces In orderto obtain theinternalforcesataspecifiedpoint,weshouldmakesection cut

perpendiculartotheaxis ofthememberatthis point.This sectioncutdivides thestructureintwo

parts.Theportionofthestructureremovedfromthepartintoconsiderationshouldbereplaced

bytheinternalforces.Theinternalforcesensuretheequilibriumoftheisolatedpartsubjectedto

theactionofexternalloadsandsupportreactions.Afreebody diagramofeithersegmentofthe

cutmemberisisolatedandtheinternalloads couldbederivedbythesixequations ofequilibrium

appliedtothesegmentintoconsideration. 5.1.Example

Inthefollowingexampleweshallconstructtheinternalforces diagramsforthegiveninFig.spaceframestructure. Theintroducedglobalcoordinatesystemis showninthesamefigure.

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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.

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Structural Analysis II 5.2.Example

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5.2 BEAMS CURVED IN PLAN 5.2.1Introduction

Arches 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 beams supporting water tanks,Silos etc.,beams supporting corner lintels and curved balconies etc.,Ramps in traffic interchanges invariably have curved in plan beams.

Curved beams in addition to the bending moments and shears would also develop torsional moments. 5.2.2.Moment,Shear and Torsion

The three diverse force components have one thing in common – the strain energy stored

in a beam due to each type of force. Among the 3 we normally ignore the strain energy due to shear forces as negligible.

U = ∫M2ds/2EI+∫T

2ds/2GJ

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5.3. SUSPENSION CABLE

5.3.1. Indroduction

Cablesandarchesareclosely related toeach otherandhencethey aregroupedin thiscoursein

thesamemodule.Forlongspanstructures(fore.g.incasebridges)engineerscommonlyuse

cableorarchconstructionduetotheirefficiency.Inthefirst lessonofthismodule,cables

subjectedtouniformandconcentratedloadsarediscussed.Inthesecond lesson,arches in

generalandthreehingedarchesinparticularalongwithillustrativeexamplesareexplained. In thelasttwolessons ofthismodule, twohingedarchandhingeless arches

areconsidered.Structure may be classified into rigid and deformable structures depending on

change in geometry ofthestructurewhilesupportingtheload.Rigidstructuressupportexternallyapplied

loadswithout appreciable change intheir shape(geometry). Beamstrussesand framesare examplesof

rigidstructures. Unlikerigidstructures,deformablestructuresundergochangesin

theirshapeaccordingtoexternallyappliedloads.However,itshouldbenotedthatdeformationsarestillsma

ll.Cablesandfabricstructuresaredeformablestructures.Cablesaremainly used to supportsuspension

roofs,bridgesand cablecarsystem. They arealsousedin electrical

transmissionlinesandforstructuressupportingradioantennas.In thefollowingsections,cables

subjectedtoconcentratedloadandcables subjectedtouniformloads areconsidered.

Theshapeassumedby aropeorachain(withnostiffness)undertheactionofexternal loadswhenhungfromtwosupportsisknownasafunicularshape. Cableisafunicular structure.Itiseasy

tovisualizethatacablehungfromtwosupportssubjectedtoexternal loadmustbeintens cable.Acablemaybedefinedasthestructureinpuretensionhavingthefunicularshapeof the

load.(videFig.5.1and5.2). As stated earlier, the cables are considered to be perfectly flexible (no flexuralstiffness)

and inextensible.Astheyareflexibletheydonotresistshearforceandbendingmoment.Itissubjected to

axial tension only anditisalwaysacting tangentialtothecable at anypoint along thelength.If the

weightof thecableisnegligibleascomparedwith theexternally appliedloadsthenitsself weightis

neglectedintheanalysis.In thepresent analysisself weightisnotconsidered. Consideracableasloadedin Fig.5.3.Letusassume thatthecablelengthsandsagat()areknown.

Thefour reactioncomponentsatACDEBandB, cable tensionsineach ofthefour segmentsand three

sagvalues:a totalof eleven unknown quantitiesaretobedetermined.Fromthegeometry,onecould

writetwoforceequilibriumequations(0,0==ΣΣyxFF)ateachofthepointandDCBA,,,Ei.e.atotal of ten

equationsandtherequiredonemoreequationmay bewrittenfromthegeometryof thecable.

Forexample,ifoneof thesagisgiven then theproblemcan besolvedeasily.Otherwiseif the total length

ofthecableisgiventhentherequired equationmaybewritten as

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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.

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Fig 5.4.Cable subjected to concentrated load

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

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Structural Analysis II Considera cable which isuniformlyloaded asshown inFig 5.4.

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

5.3.Example DeterminereactioncomponentsatA andB,tensioninthecable andthesag ofthecable shown inFig.5.7.Neglectthe selfweightofthe cable in the analysis.

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Structural Analysis II Fig 5.7.

Sincetherearenohorizontalloads,horizontalreactionsatAandBshouldbethesame.

Taking momentaboutE,yields

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Fig 5.8

Fig 5.9

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UNIT-I FLEXIBILITYMATRIXMETHOD FORINDETERMINATE STRUCTURES

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

2. Whataretheconditions ofequilibrium? Thethreeconditionsofequilibriumarethesumofhorizontalforces,verticalforcesand

momentsat 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). Theexcessnumberofreactionsthatmakeastructureindeterminateiscalleddegreeof

indeterminacy,andisdenotedby(i).Indeterminacyisalsocalleddegreeofredundancy.

Indeterminacyconsists ofinternal andexternal indeterminacies.

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

5. Defineinternal and external indeterminacies. Internalindeterminacy(II)istheexcessnoofinternalforcespresentinamemberthat

makeastructureindeterminate.

Externalindeterminacy(EI)isexcessnoofexternalreactionsinthememberthatmake

thestructureindeterminate. i =II+EI;

EI=r– e;wherer=no ofsupport reactions and e=equilibrium

conditions II=i –EI

e=3 (planeframes) ande=6 (spaceframes)

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

i =(m+r)– 2j wherem=no ofmembers r=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 95 Dept of Civil


(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)-2j

wherem=19

r=4 j =10 e=3

∑i =(19+4)–2x10=3 External indeterminacyEI=r–e=4–3=1

∑Internal indeterminacy II=i–EI=3-1=2

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

i =(3m + r)–

3j wherem=7 r=4 j =6

e=3 ∑i =(3x7+ 4)– (3x6) =7 External indeterminacyEI=r–e=4–3=1


9. Determinei, EI, II for thegiven plane

truss. i =(m + r)– 2j wherem=3

r=4 j =3 e=3

∑i =(3+ 4)–(2x3) =1 External indeterminacyEI=r–e=4–3=1

∑Internal indeterminacyII=i–EI=1-1=0

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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=4 ande=3

∑i =4–3=1

(b)

i =r–e

wherer=5 ande=3 ∑i =5–3=2

11. Find theindeterminacyfor thegiven rigid

planeframe. i =(3m + r)– 3j

wherem=3 r=4

j =4 ∑i =(3x3+ 4)– (3x4) =1 External indeterminacyEI=r–e=4–3=1


12. Find theindeterminacyofthespace rigid

frame. i =(6m + r)– 6j

wherem=8 r=24 (i. e.

6persupportx4) j =8 e=6 ∑i =(6x8+24)– (6x 8) =24 External indeterminacyEI=r–e=24–6=18


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13. Find theindeterminacyfor thegiven space

truss. i =m +r-3j

wherem=3 r=18 (i. e. 6reactions persupport

x3) j =4 ∑i =(3+18)– (3x4) =9 External 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

ofindeterminate structures. 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 to thedegreeofstaticindeterminacyofthestructure.

(b)Stiffness matrixmethod– This is also called thedisplacement method in which

the displacements thatoccurin thestructurearetreated as unknowns. Theno of

displacements involved is equal to theno ofdegrees of freedom ofthestructure.

16. Definea primary structure. Astructure formed bythe removingthe excess orredundant restraints froman

indeterminatestructuremakingit staticallydeterminateis called primarystructure. This is required 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

to definethedeformed shapeofastructure. Therearetwo types of DOF: (a)Nodal type

DOFand(b)Joint typeDOF.

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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, roller

support and junction oftwo ormoremembers. (b)Joint type DOF– This includes the DOFat thepoint wheremoment ofinertia

changes, 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–

r wherer=no of reactions

j = no ofjoints

22. Define compatibility in forcemethod ofanalysis. Compatibilityis definedas the continuitycondition on thedisplacements

ofthestructure afterexternal loads are applied to thestructure.

23. Define theForceTransformation Matrix. The connectivitymatrixwhich relates theinternalforcesQ and theexternal forces R

is known as the forcetransformation matrix. Writingit in amatrixform, {Q} =[b]{R}

whereQ=member forcematrix/vector

b= forcetransformationmatrix R = external force/loadmatrix/ vector

24. Whatare therequirements to besatisfied whileanalyzing a structure?

Thethreeconditions to besatisfied

are: (a)Equilibrium condition

(b)Compatibilitycondition

(c)Forcedisplacement condition

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St ructural Analysis II

25. Defineflexibility influence coefficient(fij)

Flexibilityinfluence coeff icient(fij)is defined as thedisplacement at joint „i‟dueto

aunit load at joint „j‟, while all otherjoints arenot load.

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

by =

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UNIT –II STIFFNESS MATRIXMETHOD

1. Whatarethebasicunknow ns in stiffness matrix method? In thestiffness matrixmet hod nodal displacementsaretreatedas thebasicunknowns for thesolution ofindeterminatestructures.

2. Definestiffness coefficien tkij.

Stiffness coefficient „kij‟ is defined as the forcedeveloped at joint „i‟dueto

unit displacementat joint „j‟w hile all otherjoints arefixed. 3. Whatis thebasicaimofthestiffness method?

The aim ofthestiffnessmethod is to evaluatethevalues ofgeneralized coor dinates

„r‟ knowingthestructurestiffn ess matrix„k‟ and nodal loads „R‟through thestr

ucture equilibrium equation.

{R} =[K]{r}

4. Whatis thedisplacement transformationmatrix? The connectivitymatrixwhich relates theinternaldisplacement „q‟and theexternal

displacement„r‟is known as thedisplacement transformation matrix „a‟.

{q} =[a]{r}

5. Howarethebasicequations ofstiffness matrixobtained? Thebasicequations ofstiffness matrixareobtained

as: Equilibriumforces Compatibilityofdisplacements

Forcedisplacement relationships

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

7. Whatis meantby generali zed coordinates? Forspecifyingaconfiguration ofasystem, acertain minimum no ofindepen dent coordinatesarenecessary. Theleast no ofindependent coordinates thatareneeded to specifytheconfigurationis known as generalized coordinates.

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

9. Writeabout theforcedisp lacement relationship. Therelationship ofeachel ement must satisfythestress-strain relationship oftheelement material.

10. Writetheelementstiffnes s fora truss element. Theelement stiffness matrixforatruss element isgiven by

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11. Writetheelementstiffnes smatrix fora beamelement. Theelement stiffness matrixforabeam element is given by

12. Compareflexibility met hod and stiffness method. Flexibilitymatrixmethod

Theredundant forces aretreated as basicunknowns. Thenumberofequationsinvolved is equal to thedegreeofstaticindete

rminacy ofthestructure. Themethod is thegeneralization of consistent deformation method.

Different proceduresareused fordeterminateandindeterminatestructures

Stiffness matrixmethod

Thejoint displacem ents aretreatedas basicunknowns

Thenumberofdisplacements involved is equal to theno ofdegrees o ffreedom of

thestructure Themethod is thegeneralization oftheslopedeflection method.

Thesameprocedureis used forboth determinateand indeterminatest ructures.

13. Is itpossibleto developth eflexibility matrix foran unstablestructure? Inorderto develop theflexibilitymatrixforastructure, it has to bestableand determinate.

14. Whatis therelation betw een flexibility and stiffness matrix? The element stiffness matrix„k‟is theinverseofthe element flexibilitymatrix„f‟ and is givenbyf=1/k ork =1/f.

15. Whatarethetypeofstructtures that can besolved using stiffness matrix method? Structures such as simply supported, fixed beams and portal frames can besolved using stiffness matrixmethod.

16. Givetheformula forthes izeoftheGlobal stiffness matrix. Thesizeoftheglobal stiffness matrix(GSM) =No: ofnodes xDegrees offree dom per node.

17. List theproperties ofthestiffness matrix Theproperties ofthestiffn ess matrixare:

It isasymmetricm atrix Thesum of eleme nts in anycolumn must be equal to zero.

It is an unstableel ementthereforethedeterminantis equal to zero.

18. Whyis thestiffness matrix methodalso called equilibrium method

ordisplacement method?

102 Dept of Civil


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

also known as equilibrium me thod.

19. Ifthe flexibilitymatrixis givenas

Writethe correspondingstiffness matrix.

Stiffness matrix= 1/(Flexibilitymatrix)

i.e. [K]=[F]-1

20. Writethen stiffness matri xfora 2Dbeam element.

Thestiffness matrixfora2 Dbeam element is given by 103 Dept of Civil


UNIT III FINITE ELEMENT METHOD

1. Whatis meantbyFinite elementmethod? Finite element method (FEM)is anumerical technique forsolvingboundaryvalue

problems in which alargedomain is divided into smallerpieces or elements. Thesolution

is determined byasuumingcertianploynomials.Thesmall pieces arecalled finite elementand 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

elements which results in asystem of equivalent finite elements.

7. Whatare thefactors governing theselection of finite elements? 104 Dept of Civil


Thegeometryofthebody Thenumberofindependent spacecoordinates Thenatureofstress variation expected

8. Definedispalcementfunction. Displcementfunction is defined as simple functions which areassumed

toapproximate thedisplacements foreach element. Theymayassumed in the form

ofpoynomials, or trignometricalfunctions.

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

Finite element–Small elements used forsubdividingthegiven domain

tobe analysedarecalled finiteelements. Theseelements maybe1D, 2Dor

3D elements dependin on thetypeofstructure. Nodes and nodal points– Theintersection ofthediffernt sides ofelementsare

called 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 calleda

continuum.It can be assumed as havinginfinitenumberof connected particles.

Primaryunknowns– Themain unknowns involved in the formulation

ofthe element properties areknown as primaryunknowns. Secondaryunknowns– Theseunknowns arederived from primaryunknowns are

known as secondaryunknowns.In displacement formulations, displacements

aretreatedas primaryunknowns and stress, strain, moments and shear force

are treated 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.

Beamsareusually approximated with 1Delements. Thesemaybestraight

orcurved. There can be additional nodes within the element. 105 Dept of Civil


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 anassemblageof

2-D elements.Most commonlyused elements aretriangular, rectangularand

quadrilateral elements.

Triangular elements Curved triangular element

Rectangularand Quadrilateral elements

13. Whatare3-D elements?Give examples. 3-D elements areusedformodelingsolid bodies andthevarious 3-Delements

are tetrahedron, hexahedron,and curvedrectangularsolid.

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14. Whatareaxisymmetricelements? Axisymmetricelements areobtained byrotatinga1-Dlineabout an axis.

Axisymmetric elements areshown in thefigurebelow.

15. DefineShapefunction. Shapefunction is also called an approximatefunction or an interpolation function whose

valueis equal to unityat thenodeconsidered andzeros at all othernodes.Shapefunction is

represented 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

valueat othernodes. 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

its smallest dimension.

18. Whatarepossiblelocations fornodes? Thepossiblelocations fornodes are:

Point of application of concentrated load. Location wherethereis achangein intensityofloads Locations wheretherearediscontinuities in thegeometryofthestructure Interfaces between materials ofdifferent properties.

19. Whatarethecharacteristics ofdisplacementfunctions? Displacement functions should havethefollowing characteristics:

Thedisplacement field should becontinuous. 107 Dept of Civil


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

Thedisplacement function must represent rigid bodydisplacements of

an element.

20. Whatis meantby planestrain condition? Planestrain is astateofstrain in which normal strain and shearstrain

directed perpendicularto theplaneofbodyis assumed to bezero.

108 Dept of Civil


UNIT –IVP LASTICANALYSIS OFSTRUCTURES

1. Whatis a plastichinge? When asection attains full plasticmoment Mp, itacts as hingewhich is called aplastic

hinge.It is defined as they ielded zonedueto bendingat which largerotation s can

occur with aconstant valueofpla sticmoment Mp.

2. Whatis a mechanism? When an-degreeindeterminatestructuredevelops n plastichinges, it becom es

determinateand theformation of an additional hingewill reducethestructur eto

a mechanism. Onceastructu rebecomesamechanism, it will collapse.

3. Whatis differencebetwe en plastichingeandmechanical hinge? Plastichinges modifythebehaviourofstructures in thesamewayas mechani cal hinges.

Theonlydifferenceis that plastichinges permit rotation with aconstant

resistingmoment equal to theplasticmomen t Mp. At mechanical hinges,

theresistingmome nt is equal to zero.

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

pseload wheren is thedegreeofstat

icallyindeterminacy.Oncethestructurebecomes a mechanism

5. Listout theassumptions m adeforplasticanalysis. Theassumptions forplasticanalysisare:

Planetransversesections remain planeand normal to thelongitudinal axis before

and afterbending. Effect ofshearis n eglected. Thematerial is ho mogeneous and isotropicboth in theelasticand plasticstate. Modulus of elasti cityhasthesamevalueboth in tension and compression. Thereis no resultant axial forcein thebeam.

Thecross-section ofthebeam is symmetrical about an axis through its

centroid and parallel to the planeofbending.

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

theyield moment ofthesection.

109 Dept of Civil


WhereMp =Plasticmoment M=Yield moment Zp =Plasticsectio n modulus Z=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 havin g maximumshapefactor. Thesection havingmaximum shapefactorisatriangularsection, S =2.345.

9. Defineloadfactor. Loadfactoris definedasth eratio of collapseloadto workingloadand is given by

10. Stateupperbound theor y.

Upperbound theorystates that of all theassumedmechanisms theexact col

lapse mechanism is that whichr equires aminimum load.

11. Statelowerbound theory . Lowerbound theorystates that thecollapseloadis determined byassumingsuitable

moment distribution diagram. Themoment distribution diagram is drawn in such

away that theconditions of equi librium aresatisfied.

12. Whatarethedifferent ty pes ofmechanisms? Thedifferent types ofmechanisms are:

Beam mechanism Column mechanism

110 Dept of Civil


Panel orswaymechanism Cablemechanism Combined or compositemechanism

13. Depending on thesupportand load conditionsindicatethepossiblelocations

of plastichinges.

14. Mention thetypes of frames. Framesarebroadlyoftwo

types: (a)Symmetricframes

(b)Un-symmetricframes

15. Whataresymmetricframes and howthey analyzed? Symmetricframes areframes havingthesamesupport conditions, lengths and loading

conditions on thecolumns and beams oftheframe. Symmetricframes can beanalyzed

by: (a)Beam mechanism

(b)Column mechanism

16. Whatareunsymmetrical frames and howarethey analyzed? Un-symmetricframes havedifferentsupport conditions, lengths and

loadingconditions on its columns and beams. Theseframes can beanalyzed by: (a)Beam mechanism

(b)Column mechanism

112 Dept of Civil


(c)Panel orswaymechanism

(d)Combined mechanism

17. Defineplasticmodulusofa section Zp.

Theplasticmodulus ofasection is thefirst moment oftheareaaboveandbelowthe

equal areaaxis.It is theresistingmodulus ofafullyplasticized section.

Zp =A/2 (Z1+Z2)

18. Howis theshapefactorofa hollowcircularsection related to

theshapefactorofa ordinary circularsection?

Theshapefactorofahollow circularsection

=AfactorKxshapefactorofordinary circularsection.

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

4)}

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]

113 Dept of Civil


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

and underits 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

net horizontal cabletension.

114 Dept of Civil


7. Briefly explain cableoversaddle. Cableoversaddlehas thefollowingproperties:

Horizontal component oftension in thesuspension cable=Horizontal

component oftension in theanchor cable

Thesupportingtower will besubjected to onlyvertical pressuredueto

cable tension.

8. Whatis thedegreeofindeterminacy ofa suspension bridgewith two hinged stiffening 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

the memberdivided byits length, i. e. theforcein thememberperunit length.

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

Beams in abridgenegotiatingacurve 115 Dept of Civil


Ringbeams supportingawatertank Beams supporting cornerlintels Beams in ramps

13. Whataretheforces developed in beams curvedin plan? Beamscurved in plan will havethefollowingforces developed in them:

Bendingmoments Shear forces Torsional moments

14. Whatarethesignificantf eatures ofcircularbeams on equally spaced supports? Slopeon eithersid eof anysupport will bezero. Torsional moment on everysupport will bezero

15. Givetheexpression forca lculating equivalentUDL on a girder. Equivalent UDLon agird eris given byWe:

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

17. Givethehorizontal and v ertical components ofa cablestructuresubjec ted to UDL.

Thehorizontal and vertica l reactions aregiven by:

and respectively

116 Dept of Civil


18. Givetheexpression ford etermining thetension T in thecable.

Thetension developed inthecableisgiven by whereH=h orizontal

component and V=vertic al component.

19. Givethetypes ofsignificant cablestructures Linear structures

Suspension bridge s Drapedcables Cable-stayed bea ms ortrusses Cabletrusses Straight tensioned cables

Three-dimensional structures

Bicyclewheelroof 3D cabletrusses Tensegritystructu res Tensairitystructures

20. Whatarecablesmadeof? Cablescanbeofmil dsteel,highstrengthsteel,stainlesssteel,orpolyest erfibres.

Structuralcablesare mad e of a seriesof smallstrandstwistedor boundtogether toforma

muchlarger cable.Steelc ablesare eitherspiralstrand,wherecircularrodsar etwisted together

orlockedcoilstrand,whereindividualinterlockingsteelstrandsformthecable (often with

aspiral strand core).

Spiralstrand isslightly weaker thanlockedcoilstrand. Steelspiralstra nd cableshavea

Young'smodulus,Eof150 ± 10kN/mm²andcomeinsizesfrom3to90mmdiame ter.

Spiralstrandsuffers from constructionstretch,where thestrandscompact whenthecable is

loaded. 117

3. bhavikatti, s.s, “structural analysis - · pdf filece6602 structural analysis ......

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