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Structural Analysis – II10CV53Subject Code : 10CV53No. of Lecture Hours/Week : 04Total No. of Lecture Hours : 52IA Marks : 25Exam Hours : 03Exam Marks : 100STRUCTURAL ANALYSIS – IIPART - AUNIT - 1ROLLING LOAD AND INFLUENCE LINES:Rolling load analysis for simply supported beams for several point loads and UDL. Influenceline diagram for reaction, SF and BM at a given section for the cases mentioned in aboveuinit 16 HoursUNIT - 2SLOPE DEFLECTION METHOD:Introduction, Sign convention, Development of slope-deflection equations and Analysis ofBeams and Orthogonal Rigid jointed plane frames (non-sway) with kinematic redundancyless than/equal to three. (Members to be axially rigid)8 HoursUNIT - 3MOMENT DISTRIBUTION METHOD:Introduction, Definition of terms- Distribution factor, Carry over factor, Development ofmethod and Analysis of beams and orthogonal rigid jointed plane frames (nonsway) withkinematic redundancy less than/equal to three. (Members to be axially rigid)8 HoursUNIT - 4SWAY ANALYSIS:Analysis of rigid jointed plane frames (sway, members assumed to be axially rigid andkinematic redundancy 3) by slope deflection and moment distribution methods. 4 HoursPART - BUNIT – 5KANIS METHODS:Introduction, Basic Concept, Analysis of Continuous beams and Analysis of rigid jointednon-sway plane frames.6 HoursUNIT - 6FLEXIBILITY MATRIX METHOD OF ANALYSIS:Introduction, Development of flexibility matrix for plane truss element and axially rigid planeframed structural elements and Analysis of plane truss and axially rigid plane frames byflexibility method with static indeterminacy 3.7 HoursDept. of Civil Engg., SJBITPage 1

Structural Analysis – II10CV53UNIT - 7STIFFNESS MATRIX METHOD OF ANALYSIS:Introduction, Development of stiffness matrix for plane truss element and axially rigid planeframed structural elements. And Analysis of plane truss and axially rigid plane frames bystiffness method with kinematic indeterminacy 3.7 HoursUNIT - 8BASIC PRINCIPLES OF DYNAMICS:Basic principles of Vibrations and causes, periodic and aperiodic motion, harmonic and nonharmonic motion. Period and frequency, Forced and Free Vibration, Damping and Equationsof Single Degree of Freedom System with and without damping6 HoursREFERENCE BOOKS:1. Basic Structural Analysis- Reddy C.S. - Second Edition, Tata McGraw Hill PublicationCompany Ltd.2. Theory of Structures Vol. 2 - S.P. Gupta, G.S. Pandit and R. Gupta, Tata McGraw HillPublication Company Ltd.3. Structural Dynamics-by M.Mukhopadhyay,4. Structural Analysis-II -S. S. Bhavikatti – Vikas Publishers, New Delhi.5. Basics of Structural Dynamics and Aseismic Design By Damodhar Swamy and KavitaPHI Learning Private Limited6. Structural Analysis- D.S. Prakash Rao,, A Unified Approach, University Press7. Structural Analysis, 4th SI Edition by Amit Prasanth & Aslam Kassimali, ThomsonLearning.Dept. of Civil Engg., SJBITPage 2

Structural Analysis – II10CV53TABLE OF CONTENTUNITTOPICUnit – 1Unit – 2Unit – 3Unit – 4Unit – 5Unit – 6Unit – 7Unit – 8Rolling load and influence linesSlope deflection methodMoment distribution methodSway analysisKanis methodsFlexibility matrix method of analysisStiffness matrix method of analysisBasic principles of dynamicsDept. of Civil Engg., SJBITPAGENO414314357677786Page 3

Structural Analysis – IIUNIT - 1ROLLING LOAD AND INFLUENCE LINES10CV531 Introduction: Variable LoadingsSo far in this course we have been dealing with structural systems subjected to aspecific set of loads. However, it is not necessary that a structure is subjected to a single setof loads all of the time. For example, the single-lane bridge deck in Figure1 may be subjectedto one set of a loading at one point of time (Figure1a) and the same structure may besubjected to another set of loading at a different point of time. It depends on the number ofvehicles, position of vehicles and weight of vehicles. The variation of load in a structureresults in variation in the response of the structure. For example, the internal forces changecausing a variation in stresses that are generated in the structure. This becomes a criticalconsideration from design perspective, because a structure is designed primarily on the basisof the intensity and location of maximum stresses in the structure. Similarly, the location andmagnitude of maximum deflection (which are also critical parameters for design) alsobecome variables in case of variable loading. Thus, multiple sets of loading require multiplesets of analysis in order to obtain the critical response parameters.Figure 1 Loading condition on a bridge deck at different points of timeInfluence lines offer a quick and easy way of performing multiple analyses for a singlestructure. Response parameters such as shear force or bending moment at a point or reactionat a support for several load sets can be easily computed using influence lines.Dept. of Civil Engg., SJBITPage 4

Structural Analysis – II10CV53For example, we can construct influence lines for (shear force at B ) or (bendingmoment at) or (vertical reaction at support D ) and each one will help us calculate thecorresponding response parameter for different sets of loading on the beam AD (Figure 2).Figure 2 Different response parameters for beam ADAn influence line is a diagram which presents the variation of a certain responseparameter due to the variation of the position of a unit concentrated load along the length ofthe structural member. Let us consider that a unit downward concentrated force is movingfrom point A to point B of the beam shown in Figure 3a. We can assume it to be a wheel ofunit weight moving along the length of the beam. The magnitude of the vertical supportreaction at A will change depending on the location of this unit downward force. Theinfluence line for (Figure3b) gives us the value of for different locations of the moving unitload. From the ordinate of the influence line at C, we can say that when the unit load is atpoint C .Figure 3b Influence line of for beam ABDept. of Civil Engg., SJBITPage 5

Structural Analysis – II10CV53Thus, an influence line can be defined as a curve, the ordinate to which at anyabscissa gives the value of a particular response function due to a unit downward load actingat the point in the structure corresponding to the abscissa. The next section discusses how toconstruct influence lines using methods of equilibrium.2 Construction of Influence Lines using Equilibrium MethodsThe most basic method of obtaining influence line for a specific response parameter isto solve the static equilibrium equations for various locations of the unit load. The generalprocedure for constructing an influence line is described below.1. Define the positive direction of the response parameter under consideration througha free body diagram of the whole system.2.For a particular location of the unit load, solve for the equilibrium of the wholesystem and if required, as in the case of an internal force, also for a part of the member toobtain the response parameter for that location of the unit load.This gives the ordinate of theinfluence line at that particular location of the load.3. Repeat this process for as many locations of the unit load as required to determinethe shape of the influence line for the whole length of the member. It is often helpful if wecan consider a generic location (or several locations) x of the unit load.4. Joining ordinates for different locations of the unit load throughout the length of themember, we get the influence line for that particular response parameter. The following threeexamples show how to construct influence lines for a support reaction, a shear force and abending moment for the simply supported beam AB .Example 1 Draw the influence line for (vertical reaction at A ) of beam AB in Fig.1Solution:Free body diagram of AB :Dept. of Civil Engg., SJBITPage 6

Structural Analysis – II10CV53So the influence line of :Example 2 Draw the influence line for (shear force at mid point) of beam AB in Fig.2.Solution:Dept. of Civil Engg., SJBITPage 7

Structural Analysis – II10CV53Example 3 Draw the influence line for (bending moment at ) for beam AB in Fig.3.Solution:Dept. of Civil Engg., SJBITPage 8

Structural Analysis – IIDept. of Civil Engg., SJBIT10CV53Page 9

Structural Analysis – II10CV53Similarly, influence lines can be constructed for any other support reaction or internalforce in the beam. However, one should note that equilibrium equations will not be sufficientto obtain influence lines in indeterminate structures, because we cannot solve for the internalforces/support reactions using only equilibrium conditions for such structures.3 Use of Influence LinesIn this section, we will illustrate the use of influence lines through the influence linesthat we have obtained in Section 2. Let us consider a general case of loading on the simplysupported beam (Figure 4a) and use the influence lines to find out the response parametersfor their loading. We can consider this loading as the sum of three different loadingconditions, (A), (B) and (C) (Figure 4b), each containing only one externally applied force.Figure4: Application of influence lines for a general loading: (a) all the loads, and (b)the general loading is divided into single force systemsDept. of Civil Engg., SJBITPage 10

Structural Analysis – II10CV53For loading case (A), we can find out the response parameters using the threeinfluence lines. Ordinate of an influence line gives the response for a unit load acting at acertain point.Therefore, we can multiply this ordinate by the magnitude of the force to get the response dueto the real force at that point. ThusSimilarly, for loading case (B):And for case (C),By the theory of superposition, we can add forces for each individual case to find theresponse parameters for the original loading case (Figure4a). Thus, the response parametersin the beam AB are:One should remember that the method of superposition is valid only for linear elasticcases with small displacements only. So, prior to using influence lines in this way it isnecessary to check that these conditions are satisfied.It may seem that we can solve for these forces under the specified load case usingequilibrium equations directly, and influence lines are not necessary. However, there may berequirement for obtaining these responses for multiple and more complex loading cases. Forexample, if we need to analyse for ten loading cases, it will be quicker to find only threeinfluence lines and not solve for ten equilibrium cases.The most important use of influence line is finding out the location of a load forwhich certain response will have a maximum value. For example, we may need to find thelocation of a moving load (say a gantry) on a beam (say a gantry girder) for which we get themaximum bending moment at a certain point. We can consider bending moment at point D ofDept. of Civil Engg., SJBITPage 11

Structural Analysis – II10CV53Example3, where the beam AB becomes our gantry girder. Looking at the influence line ofone can say that will reach its maximum value when the load is at point D . Influence linescan be used not only for concentrated forces, but for distributed forces as well, which isdiscussed in the next section.4 Using Influence Lines for Uniformly Distributed LoadConsider the simply-supported beam AB in Figure 6.5, of which the portion CD isacted upon by a uniformly distributed load of intensity w/unit length . We want to find thevalue of a certain response function R under this loading and let us assume that we havealready constructed the influence line of this response function. Let the ordinate of theinfluence line at a distance x from support A be . If we consider an elemental length dx of thebeam at a distance x from A , the total force acting on this elemental length is wdx . Since dxis infinitesimal, we can consider this force to be a concentrated force acting at a distance x.The contribution of this concentrated force wdx to R is:Therefore, the total effect of the distributed force from point C to D is:Figure 5 Using influence line for a uniformly distributed loadingDept. of Civil Engg., SJBITPage 12

Structural Analysis – II10CV53Thus, we can obtain the response parameter by multiplying the intensity of theuniformly distributed load with the area under the influence line for the distance for whichthe load is acting. To illustrate, let us consider the uniformly distributed load on a simplysupported beam (Figure 6). To find the vertical reaction at the left support, we can use theinfluence line for that we have obtained in Example 1. So we can calculate the reaction as:Figure 6.6 Uniformly distributed load acting on a beamSimilarly, we can find any other response function for a uniformly distributed loadingusing their influence lines as well. For non-uniformly distributed loading, the intensity w isnot constant through the length of the distributed load. We can still use the integrationformulation:However, we cannot take the intensity w outside the integral, as it is a function of x .Dept. of Civil Engg., SJBITPage 13

Structural Analysis – II10CV53UNIT – 2SLOPE DEFLECTION METHODIn this method the end moments or support moments expressed in terms of slopes,deflections, stiffness and length of the members. The unknown slope values (slopes) aredetermined from the condition of equilibrium of joints for moments that isMBA MBC 0MBAAMBCBCAssumption1. All the joints of the frame are rigid that is angle between the members do not changeat a joint even after deformation.2. The joints are assumed to rotate as a whole3. Directions due to axial and shear stress are neglected because they are negligible orsmallSign Conventions1. Moments:Clockwise moment veAnticlockwise moment –ve2. Rotation:Clockwise Rotation veAnticlockwise Rotation –ve3. Sinking of supportIf right support sinks down Δ is veIf left support sinks down Δ is –ve4. Bending MomentsSagging BM is ve and Hogging BM is –ve5. Shear ForceLeft side upward the SF is veLeft side downward the SF is -veRight side upward the SF is -veRight side downwardward the SF is veDept. of Civil Engg., SJBITPage 14

Structural Analysis – II1. Draw BMD, Elastic curve and SFD by slope deflection Method40kN10kN/mAB3m10CV532mC6m2I1.5IFixed End Moment:MFAB MFBA MFBC MFCB -19.2 kN-m 28.8 kN-m -30 kN-m 30 kN-mSlope Deflection EquationMAB MFAB -19.2 (2θA θB)0.(2θA θB)(θA 0 due to fixity at support A)MAB -19.2 0.6EI θBMBA MFBA 28.8 (2θA θB)0(2θB θA).MBA 28.8 1.2EI θBMBC MFBC -30 (2θB θC)(2θB θC)MBC -30 1.33EI θB 0.667EI θCMCB MFCB 30 (2θC θB)(2θC θB)MCB 30 1.33EI θC 0.667EI θBDept. of Civil Engg., SJBITPage 15

Structural Analysis – II10CV53Apply the condition of EquilibriumMBA MBC 0 and MCB 0At ‘B’MBA MBC 028.8 1.2EI θB -30 1.33EI θB 0.667EI θC 02.533EI θB 0.667EI θC 1.21At ‘c’MCB 030 0.667EI θC 0.667EI θB 00.667EI θB 1.33EI θC -302Solving Eq 1 and 2θB .θC .Substitute the above values in the S-D EquationMAB -19.2 0.6EI θBMAB -14.76 kN-mMBA 28.8 1.2EI θBMBA 37.68 kN-mMBC -30 1.33EI θB 0.667EI θCMBC -37.68 kN-mMCB 30 1.33EI θC 0.667EI θBMCB 0Support Reaction10kN/m14.840kNVADept. of Civil Engg., SJBIT37.68VB137.68VB20VCPage 16

Structural Analysis – IIVA 11.424kN, VB1 28.576kN,VB VB1 VB2 58.734kNVB2 36.28kN andVC 23.72kN10CV5336.2811.42411.424 ABC 23.7228.57628.576SFD48kN-m40kN-m37.68kN-m 14.8kN-mBMDElastic Curve2. Draw BMD, Elastic curve and SFD by slope deflection Method40kN20kN/mAB1m2mC4m1.5I2I10kN2mIFixed End Moment:MFAB MFBA Dept. of Civil Engg., SJBIT -17.78 kN-m 8.89 kN-mPage 17

Structural Analysis – IIMFBC MFCB 10CV53 -26.67 kN-m 26.67 kN-mMFCD MCD -10 X 2 -20kN-mSlope Deflection EquationMAB MFAB -17.78 (2θA θB)0(2θA θB)(θA 0 due to fixity at support A)MAB -17.78 1.33EI θBMBA MFBA (2θA θB)0(2θB θA) 8.89 MBA 8.89 2.67EI θBMBC MFBC (2θB θC). -26.67 (2θB θC)MBC -26.67 1.5EI θB 0.75EI θCMCB MFCB (2θC θB). 26.67 (2θC θB)MCB 26.67 1.5EI θC 0.75EI θBApply the condition of EquilibriumMBA MBC 0 and MCB MCD 0At ‘B’MBA MBC 08.89 2.67EI θB-26.67 1.5EI θB 0.75EI θC 04.17EI θB 0.75EI θC 17.78Dept. of Civil Engg., SJBIT1Page 18

Structural Analysis – II10CV53At ‘c’MCB MCD 026.67 1.5EI θC 0.75EI θB -20 00.75EI θB 1.5EI θC -6.672Solving Eq 1 and 2θB .θC .Substitute the above values in the S-D EquationMAB -17.78 1.33EI θBMAB -10.3426 kN-mMBA 8.89 2.67EI θBMBA 23.703 kN-mMBC -26.67 1.5EI θB 0.75EI θCMBC -23.703 kN-mMCB 26.67 1.5EI θC 0.75EI θBMCB 20kN-mMCD -20kN-mSupport Reaction20kN/m10.3440kN23.74VAVA 22.21kN,VB1VB1 17.79kN,23.7420VB2VB2 40.935kN and20VCVC 49.065kNVB VB1 VB2 58.734kNDept. of Civil Engg., SJBITPage 19

Structural Analysis – II10CV5340.9422.2122.2110 10 ABC D17.7917.79SFD39.05626.67kN-m40kN-m23.7kN-m 10.38kN-m20kN-mBMDElastic CurveDept. of Civil Engg., SJBITPage 20

Structural Analysis – II10CV53SINKING OF SUPPORTAδMAB BMBA BδMBA AMAB Slope Deflection equationsMAB MFAB (2θA θB -)δ is ve when right side support sinksδ is -ve when left side support sinks1. Analyse the continuous beam by slope deflection method the support Bsinks by 5mmDraw BMD, EC and SFD. Take EI 2 X 104 kN-mm260kN80kNA20kN/mB3m2mC2m2mD2mFixed End Moment:MFAB MFBA Dept. of Civil Engg., SJBIT -28.8 kN-m 43.2 kN-mPage 21

Structural Analysis – IIMFBC MFCB 10CV53 -40 kN-m 40 kN-mMFCD MCD -20 X 2 X 1 -40kN-mSlope Deflection EquationMAB MFAB (2θA θB –0 -28.8 ())(2θA θB -( .)( .).)))(θA 0 due to fixity at support A)MAB -52.8 0.8 x 104θBMBA MFBA (2θA θB -)0 43.2 (2θB θA -)MBA 19.2 1.6 x 104θBMBC MFBC (2θB θC - -40 )(2θB θC -()MBC -2.5 2 x 104θB 104 θCMCB MFCB 40 (2θC θB .)(2θC θB -(.))MCB 77.5 2 x 104θC 104θBApply the condition of EquilibriumMBA MBC 0 and MCB MCD 0At ‘B’MBA MBC 019.2 1.6 x 104θB-2.5 2 x 104θB 104 θC 03.6 x 104θB 104 θC -16.7Dept. of Civil Engg., SJBIT1Page 22

Structural Analysis – II10CV53At ‘c’MCB MCD 077.5 2 x 104θC 104θB -40 0104θB 2 x 104θC -37.52Solving Eq 1 and 2θB 0.66 x 10-4θC -19.1 x 10-4Substitute the above values in the S-D EquationMAB -52.8 0.8 x 104θBMAB -52.275 kN-mMBA 19.2 1.6 x 104θBMBA 20.256 kN-mMBC -2.5 2 x 104θB 104 θCMBC -20.256 kN-mMCB 77.5 2 x 104θC 104θBMCB 40kN-mMCD -40kN-mSupport Reaction80kN52.27560kNVAVA 30.4kN,20.256VB1VB1 29.6kN,20.25640VB2VB2 35.064kN and20kN/m40VCVC 84.94kNVB VB1 VB2 64.664kNDept. of Civil Engg., SJBITPage 23

Structural Analysis – II30.44035.06435.06410CV5330.4 ABC 29.6 D29.6SFD44.972kN-m44.980kN-m52.275kN-m 40kN-m20.256kN-m-BMDElastic CurveDept. of Civil Engg., SJBITPage 24

Structural Analysis – II10CV53Analysis of NON-SWAY Portal FramesA frame is a structure having both horizontal and vertical members, such as beams andcolumns. The joint between any two members is assumed to rotate has a whole when loadsare applied (rigid) hence they are called rigid jointed frames.The frames is which the beams and columns are perpendicular to each other are calledorthogonal frames. The moment of the joints in frames in the lateral direction is called LateralSway or Sway.ΔΔThe frames which do not sway in lateral direction are called non sway portal frames.Examples:Dept. of Civil Engg., SJBITPage 25

Structural Analysis – II1. Analyse the frame shown in the figure and draw BMD10CV5340kN/mB6m, 2I4m ICIADFixed End Moment:MFAB 0MFBA 0MFBC MFCB -120 kN-m 120 kN-mMFCD MFDC 0Slope Deflection EquationMAB MFAB 0 (2θA θB)0(2θA θB)(θA 0 due to fixity at support A)MAB 0.5EI θBMBA MFBA 0 (2θA θB)0(2θB θA)MBA EI θBMBC MFBC -120 (2θB θC)(2θB θC)MBC -120 1.33EI θB

Basic principles of Vibrations and causes, periodic and aperiodic motion, harmonic and non-harmonic motion. Period and frequency, Forced and Free Vibration, Damping and Equations of Single Degree of Freedom System with and without damping 6 Hours REFERENCE BOOKS: 1. Basic Structural Analysis-Reddy C.S. -Second Edition, Tata McGraw Hill Publication