结构力学课件12渐近法

2020/5/7,1,渐近法,第十二章,2020/5/7,2,1、线性代数方程组的解法:,直接法,渐近法,2、结构力学的渐近法,力学建立方程，数学渐近解,不建立方程式，直接逼近真实受力状态。其突出的优点是每一步都有明确的物理意义。,3、位移法方程的两个特点:,(1)每个方程最多是五项式；,(2)主系数大于副系数的总和，即kiikij，适于渐近解法。,4、不建立方程组的渐近解法有：,(1)力矩分配法：适于连续梁与无侧移刚架。,(2)无剪力分配法：适于规则的有侧移刚架。,(3)迭代法：适于梁的刚度大于柱刚度的各种刚架。,它们都属于位移法的渐近解法。,12-1渐近法概述,2020/5/7,3,12-2力矩分配法的基本概念,力矩分配法,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。,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。,SAB=4i,SAB=3i,SAB=i,SAB=0,SAB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。,一、转动刚度S：,2020/5/7,4,二、分配系数设A点有力矩M，求MAB、MAC和MAD,如用位移法求解：,于是可得,2020/5/7,5,三、传递系数,MAB=4iABA,MBA=2iABA,MAB=3iABA,MAB=iABA,MBA=-iABA,在结点上的外力矩按各杆分配系数分配给各杆近端截面，各杆远端弯矩分别等于各杆近端弯矩乘以传递系数。,2020/5/7,6,基本运算,MAB,MBA,MBC,MABP,MBAP,MBCP,MB,0,+,=,最后杆端弯矩：,MBA=,MBAP+,MBC=,MBCP+,MAB=,MABP+,然后各跨分别叠加简支梁的弯矩图，即得最后弯矩图。,12-3单结点的力矩分配,2020/5/7,7,例1.用力矩分配法作图示连续梁的弯矩图。,（1）B点加约束,MAB=,MBA=,MBC=,MB=,MBA+MBC=,-150,150,-90,（2）放松结点B，即加-60进行分配,60,设i=EI/l,计算转动刚度：,SBA=4i,SBC=3i,分配系数：,0.571,0.429,分配力矩:,-34.3,-25.7,-17.2,0,+,(3)最后结果。合并前面两个过程,0.571,0.429,-150,150,-90,-34.3,-25.7,-17.2,0,-167.2,115.7,-115.7,0,167.2,115.7,300,90,M图(kNm),=,2020/5/7,8,12-4多结点的力矩分配,MBA,MBC,MCB,MCD,MAB,mBA,mBC,mCB,放松，平衡了,固定,放松，平衡了,固定,固定,放松，平衡了,渐近运算,2020/5/7,9,C,B,例1.用力矩分配法列表计算图示连续梁。,0.4,0.6,0.667,0.333,m,-60,60,-100,100,分配与传递,-33.3,-66.7,-33.4,29.4,44,22,14.7,-14.7,-7.3,-7.3,4.4,2.9,2.2,-1.5,-0.7,-0.7,0.3,0.4,1.5,0.2,-43.6,92.6,-92.6,41.3,-41.3,Mij,0,43.6,92.6,133.1,41.3,21.9,M图（kNm）,2020/5/7,10,51.8,68.2,56.4,43.6,6.9,Q图（kN）,求支座反力,2020/5/7,11,上题若列位移法方程式，用逐次渐近解法：,(1),将上式改写成,(2),余数,(3),B,C,第一次近似值,24,-66.67,-8,20,2.4,-6.67,2,-0.8,0.24,-0.67,0.2,-0.08,结果,B=48.84,C=-82.89,精确值,48.88,-82.06,MBC=4iBCB+2iBCC-100=,2020/5/7,12,1）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。,力矩分配法小结：,2020/5/7,13,0.222,1,1,1,B,C,mBA=40kNm,mBC=-41.7kNm,mCB=41.7kNm,0.3,0.4,0.3,0.445,0.333,40,-41.7,-41.7,-18.5,-9.3,-13.9,-9.3,3.3,3.3,4.4,2.2,-1.0,-0.5,-0.7,-0.5,0.15,0.15,0.2,-4.65,1.65,-0.25,0.07,43.45,3.45,-46.9,24.4,-9.8,-14.6,1.72,-4.90,M图,例2.,2020/5/7,14,5/6,1/6,50,25,-20.8,-4.2,-20.8,+20.8,+50,例3.带悬臂杆件的结构的力矩分配法。,M/2,2020/5/7,15,用力矩分配法计算，作M图。,取EI=5,20,0.263,0.316,0.421,0.615,0.385,0,0,0,31.25,20.83,20.83,0,0,(20),2.743.294.39,1.37,2.20,MB=31.2520.83=10.42,MC=20.83202.2=1.37,0.840.53,0.27,0.42,0.100.140.18,0.05,0.09,A,B,C,E,F,2020/5/7,16,2.85,0.060.03,0.02,0.03,0.010.010.01,M,0,1.42,27.80,24.96,19.94,0.56,0.29,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=固端弯矩之和结点集中力偶(顺时针为正),2020/5/7,17,4i,2i,SAG=4i,SAC=4i,SCA=4i,SCH=2i,SCE=4i,AG=0.5,AC=0.5,CA=0.4,CH=0.2,CE=0.4,0.5,0.5,0.4,0.2,0.4,15,2020/5/7,18,7.57.5,3.75,1.500.751.50,0.75,0.75,0.370.38,0.19,0.080.030.08,0.04,0.04,0.020.02,M,7.11,7.11,2.36,0.78,1.58,0.79,0.79,M图（kN.m),2020/5/7,19,例、求矩形衬砌在上部土压力作用下的弯矩图。,E,I1,I2,解：取等代结构如图。设梁柱的线刚度为i1，i2,2i1,2i2,2020/5/7,20,M,M图,当竖柱比横梁的刚度大很多时(如i220i1),梁端弯矩接近于固端弯矩ql2/12。此时竖柱对横梁起固定支座的作用。,当横梁比竖柱的刚度大很