Chapter-2: Slope Deflection Method By Prof. H.P.Sudarshan ...

1 Chapter-2: Slope Deflection Siddhartha Institute Of Tech,TumkurExample:Analyze the propped cantilever shown by using Slope defectionmethod. Then draw Bending moment and shear force :End A is fixed henceA =0 End B is Hinged henceB 0 Assume both ends are fixed and therefore fixed end moments are12wLF,12wLF2BA2AB The Slope Deflection equations for final moment at each end are )2(LEI412wL2 LEI2FM)1(LEI212wL2 LEI2 FMB2 ABBABAB2 BAABAB In the above equations there is only one unknownB .To solve we have boundary condition at B;Since B is simply supported, the BM at B is zeroie. MBA= (2)equationFrom3BB2BA Substituting the value ofBEI in equation (1) and (2) we have end moments048wLL412wLMiseanticlockwismoment indicatessignve-8wL48wLL212wLM32BA232AB MBAhas to be zero, because it is consider the free body diagram of the beam and find reactions usingequations of Problem can be treated asThe bending moment diagram for the givenproblem is as belowThe max BM occurs where SF=0.

2 Consider SF equation at a distance of xfrom right support22 XmaxXwL1289L832wL83wL83 MMBsupportfromL83atoccursBMmaxtheHenceL8 3X0wXwL83S And point of contra flexure occurs where BM=0, Consider BM equation ata distance of x from right For shear force diagram, consider SF equation from BwL85 SLSwL83S0 SwXwL83 SAXBXX Example:Analyze two span continuous beam ABC by Slope Deflection draw Bending moment & Shear force diagram. Take EI constantSolution:Fixed end moments Since A is fixed0A ,,0,0CB Slope Deflection equations are: )2( )1( )4( )3( In all the above four equations there are only two unknownB andC . Andaccordingly the boundary conditions arei-MBA-MBC=0 MBA+MBC=0iiMCB=0since C is end simply support.

3 6( )5( Solving simultaneous equations 5 & 6 we getEIB = = in the Slope definition equationsMAB= + MBA= + + MBC= + MCB= + + Reactions:Consider the free body diagram of the reactions using equations of AB: MA= 0RB 6 = 100 4+ RB= KN V = 0RA+RB= 100KN RA= KNSpan BC: MC= 0RB 5 = 20 5 25+75 RB= 65 KN V=0RB+RC= 20 5 = 100 KNRC= 100-65 = 35 KNUsing these data BM and SF diagram can be BM:Span AB:Max BM in span AB occurs under point load and can be foundgeometricallyMmax= Span BC:Max BM in span BC occurs where shear force is zero orchanges its sign. Hence consider SF equation 35-20x= 02035x = BM occurs at from C Mmax= 35 = KNME xample:Analyze continuous beam ABCD by Slope Deflection Method and thendraw bending moment diagram.)

4 Take EI :0,0,0 CBA Slope Deflection equations: KNM30 MCD In the above equations we have two unknown rotationsCBand , accordinglythe boundary conditions are:0MM0 MMCDCBBCBA ,NowCBCBBBCBA ,AndCBBCCDCB Solving (5) and (6) we Substituting value ofBEI andCEI in Slope Deflection equations we have Reactions:Consider free body diagram of beam AB, BC and CD as shown Maximum Bending Moments:Span AB: Occurs under point load BC: where SF=0, consider SF equation with C as Example:Analyse the continuous beam ABCD shown in figure by slopedeflection Method . The support B sinks by Solution:In this problemA =0, B0, C0, =15 FEM due to yield of support BFor span AB:KNM6100015101201062006 LEI6mm6522baab For span Slope Deflection equation )2( )2( )2( )L32(LEI2 FMCDBCBC2 BCCBCBCBCB2 CBBCBCBB2 ABBABABB2 BAABBAABAB There are only two unknown rotationsB andC.

5 Accordingly the boundaryconditions areNow, Solving theseequations we getockwise Substituting these values in Slope deflections we get the final moments: Consider the free body diagram of continuous beam for finding reactionsReactions:Span AB:RB 6 = 100 x 4 + 100 RB= KNSpan BC:RB 5 = 20 x 5 x25+ 30RB= KNRC= 20 x 5-RB= KNExample:Three span continuous beam ABCD is fixed at Aand continuous overB, C and D. The beam subjected to loads as shown. Analyse the beam by slopedeflection Method and draw bending moment and shear force :Since end A is fixed0,0,0,0 DcBA FEMs:KNM30-84608 WlFAB KNM3084608 WlFBA Slope Deflection equations: BAABAB2 LEI2FM 04EI230-B ABBABA2 LEI2FM 024EI230B 2---------EI30B CBBCBC2 LEI2FM BCCBCB2 LEI2FM DCCDCD2 LEI2FM CDDCDC2 LEI2FM In the above Equations there are three unknowns, EIDCBEI&EI, ,accordingly the boundary conditions are:)hinged(0 Miii0 MMii0 MMiDCCDCBBCBA Now 0 MDC By solving (7), (8) & (9), we By substituting the values ofDcBand, in respective equations we get Reactions.

6 Consider the free body diagram of Beam Beam Example:Analyse the continuous beamshownusing Slope Deflection draw bending momentand shear force :In this problemfixedisAend,0A Slope Deflection equations: BAABAB2 LEI2FM ABBABA2 LEI2FM CBBCBC2 LEI2FM BCCBCB2 LEI2FM In the above equation there are two unknownCBand , accordingly theboundary conditions are:0 Mii024 MMiCBBCBA ,NowCBCBBBCBA (6) Substituting in eqn. (5) from equation (6) in the Slope Deflection equationwe get Final Moments: ) ( Reactions:Consider free body diagram of beams as shownSpan Span Max BMSpan AB:Max BM occurs where SF=0, consider SF equation with A as Span BC:Max BM occurs under point Example:Analyse the beam shown in figure.

7 End support C is subjected to ananticlockwise moment of 12 :In this problemfixedisend,0A Slope Deflection equations: BAABAB2 LEI2FM 04I2E20B 1---------EIB ABBABA2 LEI2FM 024I2E20B 2---------EI2B CBBCBC2 LEI2FM BCCBCB2 LEI2FM In the above equation there are two unknownsCBand , accordingly theboundary conditions are012M0 MMCBBCBA (5) ,NowCBCBBBCBA (6) ,andCBBCCB From (5) and (6) From (6) equationsdeflectionslopeisEIandEIngSubst itutiCB KNM12) (43) ( ) ( ) ( Reaction:Consider free body diagrams of beamSpan Span Example:Analyse the simple frame shown in figure. End A is fixed and ends B &C are hinged. Draw the bending moment :In this problem,0,0,0,0 DCBA Slope deflections are )1( )4( )3( )2( )6(EI21EI1024EI2102 LEI2FM)5(EI21EI1024EI2102 LEI2 FMBDBDBDDBDBDBDBDBBDBD In the above equations we have three unknown rotationsB ,C ,D accordinglywe have three boundary 0 MCB Since C and D are hinged0 MDB Now(9)-----0 EIEI2110M(8) (7) Solving equations 7, 8, & 9 we Substituting these values in Slope equations0) (21) ( ) (21) (10M0) (43) ( ) (43) ( ) ( ) ( Reactions:Consider free body diagram of each membersSpan Span Column BD: Example:Analyse the portal frame shown in figure and also drawn bendingmoment and shear force diagramSolution.

8 Symmetrical problem-Sym frame + Sym loading0,0,0,0 DCBA Slope Deflection equations: 1--------EI2104EI202 LEI2 FMBBBAABAB 2-------EI024EI202 LEI2 FMBBABBABA )2( )2( 5-------EI)02(4EI202 LEI2 FMCCDCCDCD 6-------EI21)0(4EI202 LEI2 FMCCCDDCDC In the above equation there are two unknown rotations. Accordingly the boundaryconditions are0MM0 MMCDCBBCBA Now(7) (8) Multiply by (7) and (8) by Using equation (7)ckwise Here we findCB . It is obvious because the problem is KNM-326421 MKNM64 MKNM646432)64( Consider free body diagram s of beam and columns as shownBy symmetrical we can writeKNM80 RRKNM60 RRCDBA Now consider free body diagram of column ABApplyKN24H32644H0 MAAB Similarly from free body diagram of column CDApplyKN24H32644H0 MDAC Check:0HH0 HDA Hence okayNote:Since symmetrical, only half frame may be analysed.

9 Using first threeequationsand takingCB Example:Analyse the portal frame and then draw the bending moment diagramSolution:This is a symmetrical frame andunsymmetrically loaded, thus it is anunsymmetrical problem and there is a swayAssume sway to ,0,0,0 DBDA Slope Deflection equations 2--------EI83EI430242EI0L32 LEI2FM1--------EI83EI2143042EI0L32 LEI2 FMBBABBABABBBAABAB In the above equation there are three unknowns and,CB, accordingly theboundary conditions are,0 MMMM04MM4MM, ,NowCBCBBBCBA ,AndCBCBCCDCB 90EI23EI23EI23EI`83EI21EI83 EIEI83 EIEI83EI21 MMMM,AndCBCCBBDCCDBAAB (8)&(7)inSubstituteEIEIEI(9)FromCB (7)Eqn )8(Eqn Solving equations (10) & (11) we By Equation (10) , , Substituting these values in Slope Deflection equations, we have Reactions.

10 Consider the free body diagram of beam and columnsColumn Span Column Check: H = 0HA+ HD= = 0 Hence okayExample:Frame ABCD is subjected to a horizontal force of 20 KN at joint C asshown in figure. Analyse and draw bending moment :Frame is Symmetrical and unsymmetrical loaded hence there is a sway towards rightFEMS0 FFFFFFDCCDCBBCBAAB Slope Deflection equations are 2EI32EI343323EI2L32 LEI2FM1---------EI32EI32333EI2L32 LEI2 FMBBABBABABBBAABAB 6---------EI32EI32333EI2L32 LEI2 FMCcCDDCDC The unknown are &,C, , 9060EI38EI2EI260EI32EI32EI32EI34EI32EI34 EI32EI3260 MMMMandCBCCBBDCCDBAAB Solving (7).(8) & (9) we , , Substituting the value of and,CBin Slope Deflection equations Reactions:Consider the free body diagram of the membersMember Member Member Check: H = 0HA+ HD+ P = 0+10 + 10 20 = 0 Hence okayExample:Analyse the portal frame subjected to loads as shown.