结构力学下多结点力矩分配法.ppt

单结点弯矩分配法,1、附加刚臂，写固端弯矩、求约束力矩RP,2、计算分配系数,3、RP反号分配、传递,4、统计（合计）、作图,多结点力矩分配法源出于单结点，使用时保证放松的结点是单结点即可,多结点的力矩分配,解：把悬臂杆等效如下结构,R1=R2P+R,右结点可分配了,把上述过程写在一起,就是如下的步骤:,右结点约束反号分配左结点,左结点约束右结点反号分配,。如此完成渐近过程,右结点加约束左结点反号分配,锁C，放B,2,2,B,C,锁B，放C,-4/7,-3/7,锁C，放B,1/7,1/7,锁B，放C,-4/98,-3/98,0,合计,多结点的力矩分配,具体演算过程,弯矩图,多结点的力矩分配,-41/14,43/7,-43/7,119/14,-119/14,16,解：,8kN,12kN/m,16kNm,分配系数,固端弯矩,-8.73,0,分配、传递,-1.75,3.49,分配、传递,分配、传递,-1.27,0,分配、传递,-0.25,0.51,分配、传递,-0.19,0,分配,合计,4m,EI,4m,EI,5m,5EI/8,不思，故有惑；不求，故无得；不问，故无知,弯矩图,EI=常数,不思，故有惑；不求，故无得；不问，故无知,解：,0.5,0.5,20,4/11,4/11,3/11,-2.67,-2.67,2.67,-5,-5,1.40,1.40,-0.35,-0.35,你能画出弯矩图吗？,弯矩图,不思，故有惑；不求，故无得；不问，故无知,例题4,解：取半结构计算。,41.67,2/7,1/7,4/7,0,1/3,1/3,1/3,41.67,-41.67,0,0,0,0,0,41.67,-41.67,-11.9,-23.8,-5.95,-11.9,5.95,3.97,3.97,3.97,1.99,1.99,1.99,-0.57,-1.14,-0.28,0.28,-12.47,18.42,-6.23,6.23,-49.98,45.83,-39.6,3.97,1.99,例5,分配与传递,6.4,1.6,分配与传递,0.4,0.4,分配与传递,-0.16,-0.04,分配与传递,0.04,0.04,合计,例5用力矩分配法计算下列结构。杆长都是L。,解：（1）力矩分配法是计算无侧移刚架，但对剪力静定杆也可使用。因为在B结点附加刚臂后：BC杆为一端固定，另端可滑动杆，AB杆成为A端固定铰，另端可滑动杆。,(2)关于分配系数,不思，故有惑；不求，故无得；不问，故无知,1/4,3/4,0,-PL/2,-PL/2,-2PL,-5PL/2,5PL/8,15PL/8,-5PL/8,-9PL/8,PL/8,15PL/8,-2PL,例题6用力矩分配法计算具有剪力静定杆结构,i=常数,不思，故有惑；不求，故无得；不问，故无知,-50,-50,10,3/4,1/4,1/5,1/5,3/5,-20,-60,12,-12,36,12,-12,24,8,-8,1.6,-1.6,4.8,1.6,-1.6,1.2,0.4,-0.4,0.08,0.24,0.08,25.2,-25.2,-14.72,51.04,-36.32,-63.53,0,0,0,小结：多结点力矩分配时，先附加刚臂求出固端弯矩，然后放松若干结点，保证每个结点都是单结点，分配，传递后，再锁紧这些结点，然后放松先前锁紧的结点，进行分配与传递，然后锁紧，两组结点轮换进行。例如下图。,对下列结构，可进行如下的结点分组。,练习题,1,2,EI=常数,答案,7.5,2.5,-2.5,0,3/11,4/11,4/11,1/3,1/3,1/3,1/4,1/4,1/4,1/4,1/4,1/4,1/4,1/4,15,-10,10,-10,10,-1.36,-1.82,-1.82,-0.91,-0.91,0.23,0.23,0.23,0.23,0.3,0.3,0.3,0.12,0.15,0.12,0.12,0.15,-0.07,-0.07,-0.07,0.07,-0.035,-0.035,13.64,-1.82,-11.82,-0.68,9.4,0.3,-9.7,10.15,0.08,0.12,0.23,0.23,0.12,0.23,0.19,-0.07,