(5)--Chapter 8 -1结构力学结构力学.ppt

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1、Chapter 8 Methodof successive approximations and influence lines for indeterminate structure第八章第八章 渐进法渐进法及超静定结构的影响线及超静定结构的影响线Moment distribution method力矩分配法 It can be employed only for the analysis of continuous beams and rigid frames without sidesway.力矩分配法的应用范围是连续梁和没有侧向线位移的刚架连续梁和没有侧向线位移的刚架 The inte

2、rnal forces of structure can be calculated accurately by using the force-method or displacement-method.However the accuracy of the successive approximation may be carried out to any desired degree.The moment distributed method,was initially developed by Hardy Cross in1924,was the most widely used fo

3、r analysis of structure from 1930 through 1960s.力矩分配法的计算精度可以满足一定的工程需要General remarks on methods of successive approximationsNo-shear distribution method无剪力分配法8.1 Concepts in moment distribution method力矩分配法的基本概念Sign conventionAB Clockwise moments that act on the member are considered positive,whereas

4、 counterclockwise moments are negative.对杆端而言，顺时针弯矩为正，而逆时针弯配为负Counterclockwise moments on joints are considered positive.对节点而言，逆时针弯配为正ABLMABMBAFQABFQBA4i2iABAB3i0ABEnd forces in common useABqMABMBAFQABFQBAEnd forces in common useABPABq0AB0P转动刚度1.Bend stiffness 1MAB=4i The moment that must be applied

5、at end of member to cause a unit rotation of the end is defined the bending stiffness.SABABSAB=4inear end far end The far end is fixedAB1MAB=3iSAB=3iThe far end is hinged1MAB=iABSAB=iThe far end is a directional supportABMAB=0SAB=0The far end is freeTerms in common useDistribution and carryover of j

6、oint moment节点弯矩的分配和传递Consider the moment equilibrium condition of the free body of the joint A2.Distribution factor分配系数 The fraction of the total resisting moment supplied by the member is called the distribution factor AlABMAB=4 iAB AMBA=2 iAB AABAMAB=3iABAAABMAB=iABAMBA=-iAB A传递系数3.Carryover facto

7、r The carryover factor represents the ratio of the far end moment to the near end moment when the near end rotates.Reexplain the performance of the member-end moments for rigid frame 结点上有不平衡的力矩M，在节点连接的各杆近端按一定比例、反方向地分配给各个杆端，并按传递系数的比例传递到远端，此后，结点实现力矩平衡。When a moment M is applied at a joint,the moment w

8、ill be distributed among the near ends of various members connected to the joint and carryovered to the far ends of the members.The magnitude of distributed moment is equal to the multiplication of distribution factor and M(but in opposite direction).Whereas the carryover bending moment is equal the

9、 product of carryover factor and the corresponding near end moment.ABCMoment distribution method=+Joint B is locked by a clamp,the continuous beam become two independent single beams.Only span AB is deformed and the fixed end moments at the end of AB are yielded.Rotate joint B,AB and BC are all defo

10、rmed and the end moments at the end of AB and BC are yielded.Superimpose resultsExample:Determine the member end moments for the continuous beam 3631.Lock joint B.The fixed end moments,due to the external loads can be evaluated as follows.附加刚臂锁住B，写出固端弯矩：2.Release join BABCThe lopsided moment at join

11、t B Moment distribution at a single jointThe linear stiffnessBend stiffness363Distribution factorCarryover factorABCMember ends AB BA BC CBDistribution factor 0.571 0.429Fixed end moment -150 150 -90 0distributed andcarryover -34.3 -25.7 -17.20End moments -167.2 115.7 -115.7 0167.2115.73.The procedu

12、re of calculationSolution:1.Bend stiffness2.Distribution factor3.Carryover factor42242EI1.5EI2EIExample:Draw the moment diagram for the frame with single rigid joint42244.Fixed end moments 5.Process of calculationMember ends:AB BA BC BD CB DBDistribute -3 -3 -4 carryover ：0-1.5-2End moments：0 57 -53

13、 -4 48.5 -25753448.5260100 0.3 0.3 0.4Distribution factor:0 60 -50 0 50 0Fixed end moments8.2 Moment distribution at multi joints 4466EI=1EI=2EI=1ABCD All the rigid joins that are free to rotate are temporarily restrained against rotation by imaginary clamps applied to them.The fixed end moments of

14、these joins generally are not in balance.Released one of the joints,allow it to rotate under the negative of unbalanced moment until the equilibrium state is reached.The joint,which is now in equilibrium is reclamped.Next,another joint with unbalanced moment is released,balanced,reclamped in the sam

15、e manner.The procedure is repeated until the unbalanced moments at each joint are negligibly samll.多节点的力矩分配4466EI=1EI=2EI=1ABCD1.Fixed end moments 4.Carryover factor2.Bend stiffness3.Distributed factor4466EI=1EI=2EI=1ABCD5.process of calculation Distributed factor 0.4 0.6 0.667 0.333MF -60 60 -100 1

16、00 0 0Release joint C -66.7 -33.3-33.40Release joint B 29.4 44.014.722.0Release joint C -14.7 -7.30-7.3Release joint B：2.9 4.41.52.2Release joint C：-1.5 -0.7-0.70Release joint B：0.3 0.40.2end-43.692.6-92.641.3-41.304466EI=1EI=2EI=1ABCDABCD43.692.641.390200 力矩分配的过程是逐步放松约束，使刚结点产生转角的过程，多次分配后，转角基本接近实际情况

17、，停止,计算结果是不精确的。计算的精度按工程要求，前后两次相差5%时，结束。通常要求2-3轮。-43.692.6-92.641.3-41.3045464Example:Determine the member end moments by using distribution method and draw the internal force diagrams for the rigid frameSolution1.bending stiffness2.Distribution factor3.Carryover factor4.Fixed end moments 5.Process of

18、 calculationRelease C：-18.5 -9.3 -13.9-9.3-4.7 0Release B：3.3 3.3 4.4 2.201.7Release C：-1.0 -0.5 -0.7-0.5-0.2 0 Release B：0.15 0.15 0.2 endMember end moments：043.43.5-46.924.4-9.8-14.6-4.9 0 1.7AB BA BE BC CB CF CD FC DC EB Member ends 0.3 0.3 0.4 0.445 0.222 0.333Distribution factor0 40 0 -41.7 41.

19、7Fixed end moments6.The bending moment diagram 43.43.51.79.84.914.646.94062.543.43.51.79.84.914.646.9AB43.429.150.954.545.53.749.21.3+-7.Draw the shear diagram B54.550.91.3FNBA=0FNBCFNBEC3.745.549.2FNCBFNCDFNCF105.4-1.3-+1.2-49.28.Draw the axial force diagram Analyze the continuous beam with a hango

20、verScd=0Distributed factor 5/6 1/6 1 0Fixed end moments 0 0 0 -50 0Release C：50 0250Release B:-20.8 -4.2endMember-end moments：-20.8 20.8 50 -50 0ABCD50kN1m1m5mThe moments are distributed at two jointsABSAB=3i ：0 0 25 50 Release B -20.8 -4.2 endMember end moments 0 -20.8 20.8 50The moments are distributed at single jointABCD50kN50kN1m1m5m50kN.m20.850Fixed end moment