单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,*,力矩分配法的,理论基础：位移法；,计算对象：杆端弯矩；,计算方法：逐渐逼近的方法；,适用范围：,连续梁和无侧移刚架,。,1,、,转动刚度,S,：,表示杆端对转动的抵抗能力。,在数值上,=,仅使杆,端发生单位转动时需在杆端施加的力矩。,1,S,AB,=4,i,1,S,AB,=3,i,S,AB,=,i,1,S,AB,=,0,S,AB,与杆的,i,（材料的性质、横截面,的形状和尺寸、杆长）及远端支承,有关，而与近端支承无关。,S,AB,=4,i,1,8-1,力矩分配法的基本概念,力矩分配法的理论基础：位移法；1、转动刚度S：表示杆端对转,1,因此，在确定杆端转动刚度时：,近端看位移,（是否为单位位移）,远端看支承,（远端支承不同，转动刚度不同）。,M,AB,1,M,AB,M,AB,1,1,1,M,AB,下列哪种情况的杆端弯矩,M,AB,=S,AB,转动刚度,S,AB,=4,i,是（）,i,i,i,i,i,i,A,A,A,A,A,4,iS,AB,3,i,B,B,B,B,B,因此，在确定杆端转动刚度时：近端看位移（是否为单位位移）M,2,转动刚度,S,：,表示杆端对转动的抵抗能力。,在数值上,=,仅使杆端发生单位转动时需在杆端施加的力矩。,1,S,AB,=4,i,1,S,AB,=3,i,S,AB,=,i,1,S,AB,=,0,S,AB,与杆的,i,（材料的性质、横截面的形状和尺寸、杆长）及远端支承有关，,S,AB,=4,i,1,而与近端支承无关。,传递系数,C,：,杆端转动时产生的远端弯矩与近端弯矩的比值。,即：,传递系数仅与远端支承有关,2,i,i,0,C=1/2,C=,1,C=0,远端支承,转动刚度,传递系数,固定,铰支,定向支座,4,i,3,i,i,1/2,1,0,转动刚度S：表示杆端对转动的抵抗能力。,3,2,、单结点结构在结点力偶作用下的力矩分配法,i,B,C,A,M,M,iA,=,4,i=S,iA,M,iB,=,3,i=S,iB,M,iC,=i=S,iC,M,=,M,iA,+M,iB,+M,iC,M=0,a）分配力矩,注,：,1,）,称为力矩分配系数。且,=,1,2,）分配力矩是杆端转动时产生的近端弯矩,。,3,）结点集中力偶荷载顺时针为正。,2,、传递系数,C,：,杆端转动时产生的远端弯矩与近端弯矩的比值。即：,b,）传递弯矩,M,ji,=CM,ij,j=A,，,B,，,C,注,：,1,）传递力矩是杆端转动时产生的远端弯矩,。,2,）只有分配弯矩才能向远端传递。,2、单结点结构在结点力偶作用下的力矩分配法 iBCAM,4,3,、单结点结构在跨中荷载作用下的力矩分配法,200kN,20kN/m,3,m,3,m,6,m,3,i,4,i,A,B,C,200kN,20kN/m,A,B,C,A,B,C,1,）,锁住结点,150,150,90,2,）,放松结点,M,B,=150,90=60,M,B,M,B,m,BA,m,BC,M,B,=,60,S,BA,=43,i=,12,i,S,BC,=34,i=,12,i,BA,=12,i,/24,i,=1/2,BC,=12,i,/24,i,=1/2,30,30,15,3）叠加1）、2）得到最后杆端弯矩,1/2,1/2,150,m,150,90,30,30,15,M,165,120,120,A,B,C,M,图（,kN.m),165,120,90,300,不平衡力矩,=,固端弯矩之和,节点不平衡力,矩要变号分配,.,3、单结点结构在跨中荷载作用下的力矩分配法200kN,5,i,=1,i,=1,i,=2,2,m,2,m,4,m,4,m,A,B,C,D,40kN/m,100kN,15kN,m,用力矩分配法计算,画M图。,解：,1,）求,大家算,AB,=,AC,=,AD,=,4/9,2/9,3/9,2,）求,m,大家算,m,AB,=,m,BA,=,m,AD,=,50,50,80,M=15,M,A,m,AB,m,AD,m,AC,M+M,A,=m,AB,+m,AD,+m,AC,M,=50+8015=45,结点,B,A,C,D,杆端,分配系数,BA,AB,AD,AC,CA,DA,4/9,3/9,2/9,分配与传递,20,固端弯矩,50,50,80,10,15,10,10,最后弯矩,40,70,65,10,10,40,70,100,10,80,M,图（,kN.m),i=1i=1i=22m2m4m4mABCD,6,作业：,8-1(a),8-4,作业：,7,8-2,多结点的力矩分配,24kN/m,50kN,128,128,75,M,B,=,128,M,C,=53,M,B,24kN/m,50kN,4,m,4,m,8,m,8,m,A,B,C,D,2EI,2EI,EI,取EI=8,i,=2,i,=2,i,=1,BA,=0.6,BC,=0.4,CB,=0.4,CD,=0.6,76.8 51.2,25.6,M,C,=78.6,M,C,=,78.6,31.4 47.2,15.7,15.7,15.7,9.4 6.3,25.6,分配系数,逐次放松结点进行分配与传递,固端弯矩,最后弯矩,0.6,0.4,0.4,0.6,128,128,75,51.2,76.8,25.6,31.4,47.2,15.7,6.3,9.4,3.2,1.3,1.9,0.7,0.3,0.4,0.2,0.1,0.1,0,86.6,86.6,124.2,124.2,8-2 多结点的力矩分配24kN/m,8,4,m,4,m,8,m,8,m,24kN/m,50kN,A,B,C,D,固端弯矩,最后弯矩,128,128,75,0,86.6,86.6,124.2,124.2,86.6,124.2,192,100,M,图（,kN.m),4m4m8m8m24kN/m50kNA,9,变形条件的校核：,ij,ik,ij,ik,i,i,M,M,=,D,D,ij,ik,ji,ij,ki,ik,i,i,M,M,M,M,=,D,-,D,D,-,D,3,3,2,1,2,1,ij,ji,ij,ki,ki,ik,i,M,M,i,M,M,D,-,D,=,D,-,D,3,2,1,3,2,1,ik,ki,ik,ik,ki,ki,ik,ik,ik,i,M,M,i,m,M,m,M,D,-,D,=,-,-,-,=,3,2,1,3,),(,2,1,),(,j,ik,ik,ik,m,M,M,-,=,D,ki,ik,ik,M,M,M,D,-,D,=,D,2,1,由此可见，在同一结点上各杆杆端转角相等的前提下，两杆,i,端的,ki,ik,ik,M,M,M,D,-,D,=,D,2,1,值之比等于其线刚度之比。,变形条件的校核：ijikijikiiMM=DDijikj,10,注意：,1,）单结点力矩分配法得到精确解；多结点力矩分配法得到,渐近解。,2,）首先从结点不平衡力矩,绝对值较大的结点开始,。,3,）结点不平衡力矩要变号分配。,4,）,结点不平衡力矩,的计算：,结点不平衡力矩,（第一轮第一结点）,固端弯矩之和,（第一轮第二三结点）,固端弯矩之,和加传递弯矩,传递弯矩,（其它轮次各结点）,总等于附加刚臂上的约束力矩,5）不能同时放松相邻结点（因定不出其转动刚度和传递系数）,但可以同时放松所有不相邻的结点，以加快收敛速度。,注意：结点不平衡力矩（第一轮第一结点）固端弯矩之和（第一轮第,11,4I,4I,5I,3I,3I,1,1,1,0.75,0.5,i=,1,1,1,0.75,0.5,A,B,C,D,E,F,5m,4m,4m,4m,2m,20kN/m,例题8-3 求图示刚架,的弯矩图。,解：,1,）求分配系数,S,BA,=3 S,BC,=4,S,BE,=43/4=3 S,CB,=4,S,CD,=3 S,CF,=41/2=2,结点,B,：,S=S,BA,+S,BC,+S,BE,3+4+3=10,BA,=3/10=0.3,BC,=4/10=0.4,BE,=3/10=0.3,CB,=4/9=0.445,CD,=3/9=0.333,CF,=2/9=0.222,结点,C,：,S=S,CB,+S,CD,+S,CF,4+3+2=9,2）求固端弯矩,m,kN,ql,m,CB,.,7,.,41,12,2,=,=,m,kN,ql,m,BC,.,7,.,41,12,2,-,=,-,=,m,kN,ql,m,BA,.,40,8,2,=,=,4I4I5I3I3I1110.750.5i=1110.750,12,A,B,C,D,E,F,BA,=3/10=0.3,BC,=4/10=0.4,BE,=3/10=0.3,CB,=4/9=0.445,CD,=3/9=0.333,CF,=2/9=0.222,m,kN,ql,m,Bc,.,7,.,41,12,2,-,=,-,=,m,kN,ql,m,BA,.,40,8,2,=,=,m,CB,=41.,7kN.m,BA,BE,BC,CB,CF,CD,0.3,0.3,0.4,0.445,0.222,0.333,40,41.7,41.7,18.5,9.3,13.9,9.3,3.3 3.3 4.4,40,41.7,9.3=,11,2.2,1.0,0.5,0.7,0.5,0.15 0.15 0.2,43.4 3.5,46.5,24.4,9.8,14.8,EB,1.6,0.1,1.7,FC,4.7,0.2,4.9,ABCD EFBA=3/10=0.3BC=,13,A,B,C,D,E,F,20kN/m,43.4,46.9,24.4,14.6,9.8,4.9,3.5,1.7,M,图（,kN.m),A,B,C,D,E,F,29.1,50.9,54.5,45.5,3.7,1.3,2.5,F,Q,图,(kN),A,B,C,D,E,F,49.2,105.4,2.5,F,N,图,(kN),1.2,ABCDEF20kN/m43.,14,静定伸臂的处理,例题8-4 试作弯矩图。,50kN,1,m,1,m,5,m,50kN,50,S,BA,=15,EI=,常数,5,i=,5,i=,1,S,BC,=,3,BA,=5/6,BC,=1/6,5/6,1/6,20.8,m,M,50,25,4.2,20.8,20.8,50,50kN,20.8,50,M,图（,kN.m),A,B,C,静定伸臂的处理例题8-4 试作弯矩图。50kN1m1m5m,15,4EI,4EI,2EI,2EI,用力矩分配法计算，作M图。,取EI=5,i,=4,i,=4,i,=2.5,i,=2.5,5,10kN/m,20kN,5,m,5,m,1,m,4m,20kN,20,结点,杆端,A,E,B,C,F,AB,EB,BE,BA,BC,CB,CF,FC,m,0.263,0.316,0.421,0.615,0.385,0,0,0,31.25,20.83,20.83,0,0,(,20),2.74,3.29,4.39,1.37,2.20,M,B,=31.25,20.83=10.42,M,C,=20.83,20,2.2=,1.37,0.84 0.53,0.27,0.42,0.10,0.14,0.18,0.05,0.09,A,B,C,E,F,4EI4EI2EI2EI用力矩分配法计算，作M图。取EI=5,16,结点,杆端,A,E,B,C,F,AB,EB,BE,BA,BC,CB,CF,FC,m,0.263,0.316,0.421,0.615,0.385,0,0,0,31.25,20.83,20.83,0,0,(-20),2.74,3.29,4.39,1.37,2.20,0.84 0.53,0.27,0.42,0.10,0.14,0.18,0.05,0.09,0.06 0.03,0.02,0.03,0.01,0.01,0.01,M,0,1.42,2.85,27.80,24.96,19.94,0.56,0.29,计算之前,去掉静定伸臂,将其上荷载向结点作等效平移。,有结点集中力偶时,结点不平衡力矩,=,固