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.
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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 -
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
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.
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
Structural Analysis II
1.3.DETERMINATE AND INDETERMINATE STRUCTURAL SYSTEMS
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
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
Structural Analysis II
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
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
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
6 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
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
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
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
Structural Analysis II
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
Structural Analysis II 14 Dept of Civil
Structural Analysis II
Fig 2.5 A Four member Truss
Fig 2.6 Kinematic ally Determinate Structures 15 Dept of Civil
Structural Analysis II
Fig 2.7Unit Displacement along U
16 Dept of Civil
Structural Analysis II 17 Dept of Civil
Structural Analysis II 18 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
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
7.Obtainreactionforce,stress,internalforces,strainenergy. 31 Dept of Civil
Structural Analysis II
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
Structural Analysis II 33 Dept of Civil
Structural Analysis II
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
Structural Analysis II 35 Dept of Civil
Structural Analysis II 36 Dept of Civil
Structural Analysis II
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
Structural Analysis II
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.
elasticstructureitwasvalid.However,an elastic analysisdoesnotgiveinformation about theloadsthatwill actually collapseastructure.An indeterminatestructuremay sustainloads
greaterthantheloadthatfirstcauses ayieldtooccur at anypointinthestructure. Infact,astructurewillstandaslongasitisabletofindredundancies toyield.Itisonly when
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
In thismethod,freeandreactantbendingmomentdiagramsaredrawn.Thesediagramsare overlaidtoidentifythelikelylocations of plastichinges.Thismethodthereforesatisfies the
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.
Using thecollapseMechanism2todeterminereactions,wecan draw thefollowingBMDfor
collapseMechanism2:
FromthisitisapparentthatMechanism2isnottheuniquesolution,andsothedesignplastic moment capacity must be144kNmasimpliedpreviously fromtheUpperboundTheorem.
4.BasicCollapseMechanisms: In frames,the basicmechanismsofcollapseare:
59 Dept of Civil
Structural Analysis II
Beam-typecollapse
SwayCollapse
CombinationCollapse
5.CombinationofMechanisms
Oneofthemostpowerfultoolsinplasticanalysisis CombinationofMechanisms.Thisallows us toworkoutthevirtualworkequationsforthebeamandswaycollapsesseparatelyandthen
Combinationof mechanismsis based onthe ideathat thereareonlya certain number of independentequilibriumequationsforastructure.Anyfurtherequationsareobtainedfroma
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.
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
78 Dept of Civil
Structural Analysis II
5.3. SUSPENSION CABLE
5.3.1. Indroduction
Cablesandarchesareclosely related toeach otherandhencethey aregroupedin thiscoursein
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
Fig 5.5.Cable Subjected to Uniformly Fig 5.6.Free Body Diagram Distributed load
81 Dept of Civil
Structural Analysis II Considera cable which isuniformlyloaded asshown inFig 5.4.
82 Dept of Civil
Structural Analysis II
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.
1. Whatis meantby indeterminatestructures? Structures that do not satisfythe conditions of equilibrium are called indeterminate structure. Thesestructures cannot besolved byordinaryanalysis techniques.
indeterminatestructuremakingit staticallydeterminateis called primarystructure. This is required forsolvingindeterminatestructures byflexibilitymatrixmethod.
17. Give theprimary structures for thefollowing indeterminatestructures. Indeterminatestructure PrimaryStructure
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.
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
St ructural Analysis II
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
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
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
Structural Analysis II
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.
106 Dept of Civil
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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
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
Structural Analysis II
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
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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.
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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
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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
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(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
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.
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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
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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