1. Structural Analysis 7 th Edition in SI Units Russell C. Hibbeler Chapter 12: Displacement Method of Analysis: Moment Distribution

2. General Principles & Definition Moment distribution is a method of successive approximations that may be carried out to any desired degree of accuracyMoment distribution is a method of successive approximations that may be carried out to any desired degree of accuracy The method begins by assuming each joint of a structure is fixedThe method begins by assuming each joint of a structure is fixed By unlocking and locking each joint in succession, the internal moments at the joints are “ distributed ” & balanced until the joints have rotated to their final or nearly final positionsBy unlocking and locking each joint in succession, the internal moments at the joints are “ distributed ” & balanced until the joints have rotated to their final or nearly final positions © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

3. General Principles & Definition Member stiffness factorMember stiffness factor Joint stiffness factorJoint stiffness factor The total stiffness factor of joint A isThe total stiffness factor of joint A is © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

4. General Principles & Definition Distribution Factor (DF)Distribution Factor (DF) That fraction of the total resisting moment supplied by the member is called the distribution factor (DF)That fraction of the total resisting moment supplied by the member is called the distribution factor (DF) © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

5. General Principles & Definition Member relative stiffness factorMember relative stiffness factor Quite often a continuous beam or a frame will be made from the same materialQuite often a continuous beam or a frame will be made from the same material E will therefore be constantE will therefore be constant © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

6. General Principles & Definition Carry-over (CO) factorCarry-over (CO) factor Solving for  and equating these eqn,Solving for  and equating these eqn, The moment M at the pin induces a moment of M ’ = 0.5M at the wallThe moment M at the pin induces a moment of M ’ = 0.5M at the wall In the case of a beam with the far end fixed, the CO factor is +0.5In the case of a beam with the far end fixed, the CO factor is +0.5 © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

7. General Principles & Definition Carry-over (CO) factorCarry-over (CO) factor The plus sign indicates both moments act in the same directionThe plus sign indicates both moments act in the same direction Consider the beamConsider the beam © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

8. General Principles & Definition Note that the above results could also have been obtained if the relative stiffness factor is usedNote that the above results could also have been obtained if the relative stiffness factor is used © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

9. General Principles & Definition We begin by assuming joint B is fixed or lockedWe begin by assuming joint B is fixed or locked The fixed end moment at B then holds span BC in this fixed or locked positionThe fixed end moment at B then holds span BC in this fixed or locked position To correct this, we will apply an equal but opposite moment of 8000Nm to the joint and allow the joint to rotate freelyTo correct this, we will apply an equal but opposite moment of 8000Nm to the joint and allow the joint to rotate freely © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

10. General Principles & Definition As a result, portions of this moment are distributed in spans BC and BA in accordance with the DFs of these spans at the jointAs a result, portions of this moment are distributed in spans BC and BA in accordance with the DFs of these spans at the joint Moment in BA is 0.4(8000) = 3200NmMoment in BA is 0.4(8000) = 3200Nm Moment in BC is 0.6(8000) = 4800NmMoment in BC is 0.6(8000) = 4800Nm These moment must be carried over since moments are developed at the far ends of the spanThese moment must be carried over since moments are developed at the far ends of the span © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

11. General Principles & Definition Using the carry-over factor of +0.5, the results are shownUsing the carry-over factor of +0.5, the results are shown The steps are usually presented in tabular formThe steps are usually presented in tabular form CO indicates a line where moments are distributed then carried overCO indicates a line where moments are distributed then carried over In this particular case only one cycle of moment distribution is necessaryIn this particular case only one cycle of moment distribution is necessary The wall supports at A and C “ absorb ” the moments and no further joints have to be balanced to satisfy joint equilibriumThe wall supports at A and C “ absorb ” the moments and no further joints have to be balanced to satisfy joint equilibrium © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

12. General Principles & Definition © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

13. Determine the internal moment at each support of the beam. The moment of inertia of each span is indicated. Example 12.2 © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

14. A moment does not get distributed in the overhanging span AB So the distribution factor (DF) BA =0 Span BC is based on 4EI/L since the pin rocker is not at the far end of the beam Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

15. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

16. The overhanging span requires the internal moment to the left of B to be +4000Nm. Balancing at joint B requires an internal moment of – 4000Nm to the right of B. -2000Nm is added to BC in order to satisfy this condition. The distribution & CO operations proceed in the usual manner. Since the internal moments are known, the moment diagram for the beam can be constructed. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

17. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

18. Stiffness-Factor Modifications The previous e.g. of moment distribution, we have considered each beam span to be constrained by a fixed support at its far end when distributing & carrying over the momentsThe previous e.g. of moment distribution, we have considered each beam span to be constrained by a fixed support at its far end when distributing & carrying over the moments In some cases, it is possible to modify the stiffness factor of a particular beam span & thereby simplify the process of moment distributionIn some cases, it is possible to modify the stiffness factor of a particular beam span & thereby simplify the process of moment distribution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

19. Stiffness-Factor Modifications Member pin supported at far endMember pin supported at far end As shown the applied moment M rotates end A by an amt As shown the applied moment M rotates end A by an amt  To determine , the shear in the conjugate beam at A ’ must be determinedTo determine , the shear in the conjugate beam at A ’ must be determined © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

20. Stiffness-Factor Modifications Member pin supported at far end (cont ’ d)Member pin supported at far end (cont ’ d) The stiffness factor in the beam isThe stiffness factor in the beam is The CO factor is zero, since the pin at B does not support a momentThe CO factor is zero, since the pin at B does not support a moment By comparison, if the far end was fixed supported, the stiffness factor would have to be modified by ¾ to model the case of having the far end pin supportedBy comparison, if the far end was fixed supported, the stiffness factor would have to be modified by ¾ to model the case of having the far end pin supported © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

21. Stiffness-Factor Modifications Symmetric beam & loadingSymmetric beam & loading The bending-moment diagram for the beam will also be symmetricThe bending-moment diagram for the beam will also be symmetric To develop the appropriate stiffness-factor modification consider the beamTo develop the appropriate stiffness-factor modification consider the beam Due to symmetry, the internal moment at B & C are equalDue to symmetry, the internal moment at B & C are equal Assuming this value to be M, the conjugate beam for span BC is shownAssuming this value to be M, the conjugate beam for span BC is shown © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

22. Stiffness-Factor Modifications Symmetric beam & loading (cont ’ d)Symmetric beam & loading (cont ’ d) Moments for only half the beam can be distributed provided the stiffness factor for the center span is computedMoments for only half the beam can be distributed provided the stiffness factor for the center span is computed © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

23. Stiffness-Factor Modifications Symmetric beam with asymmetric loadingSymmetric beam with asymmetric loading Consider the beam as shownConsider the beam as shown The conjugate beam for its center span BC is shownThe conjugate beam for its center span BC is shown Due to its asymmetric loading, the internal moment at B is equal but opposite to that at CDue to its asymmetric loading, the internal moment at B is equal but opposite to that at C © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

24. Stiffness-Factor Modifications Symmetric beam with asymmetric loadingSymmetric beam with asymmetric loading Assuming this value to be M, the slope  at each end is determined as follows:Assuming this value to be M, the slope  at each end is determined as follows: © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

25. Determine the internal moments at the supports of the beam shown below. The moment of inertia of the two spans is shown in the figure. Example 12.4 © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

26. The beam is roller supported at its far end C. The stiffness of span BC will be computed on the basis of K = 3EI/L We have: Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

27. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

28. The forgoing data are entered into table as shown. The moment distribution is carried out. By comparison, the method considerably simplifies the distribution. The beam ’ s end shears & moment diagrams are shown. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

29. Moment distribution for frames: No sidesway Application of the moment-distribution method for frames having no sidesway follows the same procedure as that given for beamApplication of the moment-distribution method for frames having no sidesway follows the same procedure as that given for beam © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

30. Determine the internal moments at the joints of the frame as shown. There is a pin at E and D and a fixed support at A. EI is constant. Example 12.5 © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

31. By inspection, the pin at E will prevent the frame will sidesway. The stiffness factors of CD and CE can be computed using K = 3EI/L since far ends are pinned. The 60kN load does not contribute a FEM since it is applied at joint B. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

32. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

33. The data are shown in table. The distribution of moments successively goes to joints B & C. The final moment are shown on the last line. Using these data, the moment diagram for the frame is constructed as shown. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

34. Moment distribution for frames: Sidesway To determine sidesway and the internal moments at the joints using moment distribution, we will use the principle of superpositionTo determine sidesway and the internal moments at the joints using moment distribution, we will use the principle of superposition The frame shown is first held from sidesway by applying an artificial joint support at CThe frame shown is first held from sidesway by applying an artificial joint support at C Moment distribution is applied & by statics, the restraining force R is determinedMoment distribution is applied & by statics, the restraining force R is determined © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

35. Moment distribution for frames: Sidesway The equal but opposite restraining force is then applied to the frame The moments in the frame are calculatedThe equal but opposite restraining force is then applied to the frame The moments in the frame are calculated © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

36. Moment distribution for frames: Sidesway Multistory framesMultistory frames Multistory frameworks may have several independent joints dispMultistory frameworks may have several independent joints disp Consequently, the moment distribution analysis using the above techniques will involve more computationConsequently, the moment distribution analysis using the above techniques will involve more computation © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

37. Moment distribution for frames: Sidesway Multistory framesMultistory frames The structure shown can have 2 independent joint disp since the sidesway of the first story is independent of any disp of the second storyThe structure shown can have 2 independent joint disp since the sidesway of the first story is independent of any disp of the second story © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

38. Moment distribution for frames: Sidesway Multistory framesMultistory frames These disp are not known initiallyThese disp are not known initially The analysis must proceed on the basis of superpositionThe analysis must proceed on the basis of superposition 2 restraining forces R 1 and R 2 are applied2 restraining forces R 1 and R 2 are applied The fixed end moments are determined & distributedThe fixed end moments are determined & distributed Using the eqn of eqm, the numerical values of R 1 and R 2 are then determinedUsing the eqn of eqm, the numerical values of R 1 and R 2 are then determined © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

39. Moment distribution for frames: Sidesway Multistory framesMultistory frames The restraint at the floor of the first story is removed & the floor is given a dispThe restraint at the floor of the first story is removed & the floor is given a disp This disp causes fixed end moment (FEMs) in the frame which can be assigned specific numerical valuesThis disp causes fixed end moment (FEMs) in the frame which can be assigned specific numerical values By distributing these moments & using the eqn of eqm, the associated numerical values of R 1 ’ and R 2 ’ can be determinedBy distributing these moments & using the eqn of eqm, the associated numerical values of R 1 ’ and R 2 ’ can be determined © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

40. Moment distribution for frames: Sidesway Multistory framesMultistory frames In a similar manner, the floor of the second story is then given a dispIn a similar manner, the floor of the second story is then given a disp With reference to the restraining forces we require equal but opposite application of R 1 and R 2 to the frame such that:With reference to the restraining forces we require equal but opposite application of R 1 and R 2 to the frame such that: © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

41. Moment distribution for frames: Sidesway Multistory framesMultistory frames Simultaneous solution of these eqn yields the values of C ’ and C ”Simultaneous solution of these eqn yields the values of C ’ and C ” These correction factors are then multiplied by the internal joint moments found from moment distributionThese correction factors are then multiplied by the internal joint moments found from moment distribution The resultant moments are found by adding these corrected moments to those obtained for the frameThe resultant moments are found by adding these corrected moments to those obtained for the frame © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

42. Determine the moments at each joint of the frame shown. EI is constant. Example 12.6 © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

43. First, we consider the frame held from sidesway The stiffness factor of each span is computed on the basis of 4EI/L or using relative stiffness factor I/L Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

44. The DFs and the moment distribution are shown in the table. The eqn of eqm are applied to the free body diagrams of the columns in order to determine A x and D x From the free body diagram of the entire frame, the joint restraint R has a magnitude of Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

45. An equal but opposite value of R = 0.92kN must be applied to the frame at C and the internal moments computed. We assume a force R ’ is applied at C causing the frame to deflect as shown. The joints at B and C are temporarily restrained from rotating. As a result, the FEM at the ends of the columns are determined. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

46. Since both B and C happen to be displaced the same amount and AB and DC have the same E, I and L, the FEM in AB will be the same as that in DC. As shown we will arbitrarily assumed this FEM to be The moment distribution of the FEM is shown below. Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution

47. From the eqm, the horizontal reactions at A and D are calculated. For the entire frame, we require: R ’ =56kN creates the moments tabulated below Corresponding moments caused by R = 0.92kN can be determined by proportion Solution © 2009 Pearson Education South Asia Pte Ltd Structural Analysis 7 th Edition Chapter 12: Displacement Method of Analysis: Moment Distribution