1、Chapter 8 Moment Distribution Method,8-1 Introduction8-2 Fundamental Concepts 8-3 Single-Joint Moment Distribution8-4 Multi-Joint Moment Distribution8-5 No-shear Moment Distribution8-6 Combined Application of DM & MDM8-7 Other approximation methods8-8 Influence lines of statically indeterminate forc
2、es,8-1 Introduction,1. Structures Applicable,2.Sign Conventions,Successive approximation methods,Moment Distribution,No-shear Moment Distribution Specific Frames,Same as in the Disp. Method: (Slope-deflection method),Continuous BeamsNo-sway frames,End moments End rotations Member rotations,clockwise
3、: +,8-2 Fundamental Concepts,1. Rotational stiffness S,Moment applied to the end of a member to cause a unit rotation of the end alone.,2. Carry-over factor C,Ratio of moment produced at far end to moment applied at near end.,Carry-over factor is related only to the support condition of far end.,Mem
4、ber-end moments:,3. Distribution factor , : Distribution factor,4. Moment Distribution method,Steps: (1) Calculate distribution factors Aj; (2) Calculate Near-end moments (=AjM); (3) Calculate far-end moments (= CAj AjM).,1. Analysis procedure,(1) Add constraint to prevent rotation.,8-3 Single-Joint
5、 Moment Distribution,(2) Calculate fixed-end moments and unbalanced moment.,(3) Unlock constraint, distribute balancing moment & carry over.,(4) Add fixed-end moment to distributed/carry-over moment.,=+,unbalanced moment,balancing moment,Example：,(1) Add constraint,Fixed-end moments:,Unbalanced mome
6、nt:,(2) Unlock constraint,Check: BA+ BC =1,SBA=4i, SBC=3i,(3) Calculate final moments,(4) Draw bending moment diagram,Moment diagram (kNm),Discussion：,(1) How to deal with the following situation,EI,EI,80kNm,8-4 Multi-Joint Moment Distribution,1. Steps:,Example：,R.S.,D.F.,F.E.M.,Take up the half-str
7、ucture:,2. Use of Symmetry,8-5 No-shear Moment Distribution,Structures applicable to: Frames consisting of two specific members:,without relative sidesway at two ends;with shear force statically determinate.,2. Analysis procedure,Steps: Same as the moment distribution method. (1) Lock the joints, ca
8、lculate F.E.M (3) Add up to find F.M.,3. Differences Members with shear force statically determinate (1) No end rotation; (2) free horizontal movement at near end.Fixed end moments (F.E.M.) of the member: (1) Take member as that with one end fixed, and the other sliding. (2) Take shear force at slid
9、ing end as an external load.,F.E.M.SAB , CAB,Properties,Fixed end moments (F.E.M.) of members with shear forces statically determinate:,2) Rotational stiffness of members with shear forces statically determinate:,SBA= iBA , CBA= -1,SBC= iBC , CBC= -1,Example 1：,R.S.,D.F.,F.E.M.,SBA=iAB=3, SBC=3iBC=1
10、2,Distribution and carry over process:,Moment diagram (kNm),Bending moment diagram:,Example 2：,Example 2：,Left-hand half frame,Distribution & carry over process,Example 2：,Moment diagram (kNm),8-6 Combined Application of DM & MDM,Applicable to frames with sidesway of general case.2. Primary unknowns
11、: joint translations; Primary structure: only joint translations restrained.,= +,3. Conical equation: k11Z1 +F1P=0,4. Determination of k11 and F1P by moment Distribution:,Moments due to Z1=1,k11=QBA + QCD=2.17i,Moments due to external loading,F1P=QBAQCD=23.79kN,4. Solution of equation:5. Final momen
12、t:,8-7 Other approximation methods,Neglecting shear and axial deformations.2. Neglecting lateral displacements of frames under vertical loading Layered analysis method.3. Neglecting joint rotations of frames under horizontal loading Shear distribution method (inflection point method).,Shear distribu
13、tion method (1) Only applicable to a frame with linear flexural rigidities of beams much larger (over 3 times) than those of columns. (2) The frame is subjected to horizontal loading only.,Shear distribution factors:,8-8 Influence lines of statically indeterminate forces,Construction methods: the st
14、atic method the kinematic method2. Theoretical foundation of kinematic method:,Primary system,Procedure of kinematic method: (1) Remove the constraint corresponding to Z1; (2) Give a displacement to the structure along Z1 direction; (3) Influence line: (4) Influence line is positive when displacement is upward.,Example： Influence lines of a continuous beam.,Example： Influence lines of a 5-bay continuous beam.,Deflection formula of a simply supported beam:,
