结构力学12渐近法1

1、第十二章,2020/5/3,1,12-1,渐近法概述,1,、线性代数方程组的解法,:,直接法,2,、结构力学的渐近法,?,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,k,ii,?,k,ij,，,适于渐近解法。,k,ik,k,ij,k,ii,k,ir,k,is,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

2、度的各种刚架。,2020/5/3,它们都属于位移法的渐近解法。,2,12-2,力矩分配法的基本概念,理论基础：位移法；,力矩分配法,一、转动刚度,S,：,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。,S,AB,=4,i,1,S,AB,=3,i,1,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，,2020/5/3,而与近端支承无关。,3,二、分配系数,设,A,点有力矩,M,，求,M,AB,、,M,AC,

3、和,M,AD,D,M,A,B,如用位移法求解：,S,AB,= 4,i,i,AD,?,A,i,AB,M,AB,?,4,i,AB,?,A,?,S,AB,?,A,M,AC,?,i,AC,?,A,?,S,AC,?,A,i,AC,C,M,M,AD,M,AC,M,AB,M,AD,?,3,i,AD,?,A,?,S,AD,?,A,?,1,于是可得,S,AB,= 3,i,S,AB,=,i,1,S,AB,M,AB,M,1,?,?,S,A,?,m,?,0,A,M,?,(,S,AB,?,S,AC,?,S,AD,),?,A,M,AC,S,AC,?,M,?,S,A,M,M,?,A,?,?,S,AB,?,S,AC,?,S,

4、AD,?,S,M,AD,?,Aj,?,2020/5/3,S,Aj,A,S,AD,?,M,?,S,A,?,?,S,A,M,Aj,?,?,Aj,?,M,分配系数,4,?,?,?,1,三、传递系数,M,AB,= 4,i,AB,?,A,近端,A,?,A,l,M,BA,= 2,i,AB,?,A,远端,B,C,AB,M,BA,1,?,?,M,AB,2,M,AB,= 3,i,AB,?,A,A,?,A,M,AB,=,i,AB,?,A,B,C,AB,M,BA,?,?,0,M,AB,M,BA,= -,i,AB,?,A,B,A,?,A,C,AB,M,BA,?,?,?,1,M,AB,在结点上的外力矩按各杆分配系数分配

5、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。,2020/5/3,5,12-3,单结点的力矩分配,基本运算,B,固端弯矩带本身符号,A,M,AB,M,BA,M,B,M,BC,C,M,B,M,BA,M,BC,=,A,M,AB,P,M,BA,P,B,-,M,B,M,BC,P,C,M,B,=,M,BA,+,M,BC,-,M,B,+,0,C,?,M,BA,A,?,M,BC,M,?,AB,?,?,B,M,BC,M,BA,最后杆端弯矩：,?,M,BA,=,M,BA,P,+,M,BA,?,?,?,BA,?,(,?,M,B,),M,BA,?,?,?,BC,?,(,?,M,B,),M,BC,?

6、,M,BC,=,M,BC,P,+,M,BC,M,AB,=,M,AB,P,+,M,?,然后各跨分别叠加简支梁的弯矩图，即得最后弯矩图。,AB,2020/5/3,6,例,1.,用力矩分配法作图示连续梁,（,1,）,B,点加约束,的弯矩图。,167.2,M,图,(kN,m),200,?,6,115.7,200,kN,?,?,150,kN,?,m,M,AB,=,?,20,kN/m,8,90,300,M,BA,=,150,kN,?,m,EI,EI,C,B,A,2,20,?,6,?,?,90,kN,?,m,M,BC,=,?,3m,6m,3m,8,M,B,=,M,BA,+,M,BC,=,60,kN,?,m,

7、200,kN,60,20,kN/m,（,2,）放松结点,B,，即加,-60,进行分配,C,设,i,=,EI/l,B,A,计算转动刚度：,-150,150,-90,S,BA,=4,i,S,BC,=3,i,+,-60,4,i,0.571,0.429,?,BA,?,?,0,.,571,分配系数,：,4,i,?,3,i,C,B,A,-17.2,-34.3,-25.7,0,0.571,A,-150,-17.2,-167.2,2020/5/3,0.429,B,-90,-25.7,-115.7,?,BC,150,-34.3,115.7,分配力矩,:,C,?,?,0,.,571,?,(,?,60,),?,?,

8、34,.,3,M,BA,0,0,?,?,0,.,429,?,(,?,60,),?,?,25,.,7,M,BC,3,i,?,?,0,.,429,7,i,=,(3),最后结果。合并前面两个过程,7,12-4,多结点的力矩分配,A,B,M,AB,M,BA,M,B,m,BA,M,BC,?,B,C,?,C,M,CD,渐近运算,D,M,CB,M,C,m,BC,-,M,B,m,CB,M,C,固定,放松，平衡了,固定,-,M,C,放松，平衡了,放松，平衡了,固定,2020/5/3,8,例,1.,用力矩分配法列表计算图示连续梁。,100kN,20kN/m,A,EI,=1,B,0.4,0.6,60,-100,EI

9、,=2,C,EI,=1,D,i,AB,i,BC,i,CD,6m,4m,4m,0.667,0.333,100,-66.7,-33.3,22,-14.7,-7.3,2.2,6m,?,m,-60,分,14.7,配,与,传,1.5,递,0.2,M,ij,-43.6,43.6,A,2020/5/3,-33.4,29.4,44,-7.3,2.9,0.3,4.4,1,2,?,S,BA,?,4,?,?,?,?,6,3,?,?,S,?,4,?,1,?,1,BC,?,4,?,1,?,6,2,1,?,?,8,4,1,?,6,B,-0.7,0.4,-1.5,-0.7,92.6,-92.6,92.6,21.9,41.3

10、,-41.3,0,2,?,?,3,?,0,.,4,?,?,?,BA,2,?,1,?,?,3,?,?,?,0,.,6,?,BC,1,?,S,CB,?,4,?,?,1,?,?,4,?,?,S,?,3,?,1,?,1,CD,?,6,2,?,C,41.3,B,133.1,C,D,M,图（,kN,m,）,1,?,?,CB,?,?,0,.,667,?,1,?,1,?,?,2,9,?,?,?,?,CD,?,0,.,333,20kN/m,A,EI,=1,6m,B,EI,=2,4m,100kN,C,4m,41.3,EI,=1,6m,D,92.6,43.6,A,21.9,56.4,B,133.1,C,D,M,图（

11、,kN,m,）,51.8,A,求支座反力,6.9,B,68.2,B,56.4,68.2,124.6,C,D,Q,图（,kN,）,43.6,2020/5/3,10,上题若列位移法方程式，用逐次渐近解法：,10,?,B,?,3,?,C,?,240,?,0,?,?,?,(1),?,B,?,3,?,C,?,200,?,0,?,?,将上式改写成,第一次,近似值,?,B,24,20,2.4,2,0.24,?,C,-66.67,-8,-6.67,-0.8,-0.67,?,B,?,24,?,0,.,3,?,C,?,?,(2),?,C,?,?,66,.,67,?,0,.,334,?,B,?,余数,?,B,?,?

12、,0,.,3,?,C,?,?,?,C,?,?,0,.,334,?,B,?,0.2,(3),结,果,-0.08,?,B,=48.84,48.88,?,C,=-82.89,-82.06,精确值,1,1,M,BC,= 4,i,BC,?,B,+2,i,BC,?,C,-100 =,4,?,?,48,.,84,?,2,?,?,(,?,82,.,96,),?,100,?,?,92,.,6,4,4,2020/5/3,11,力矩分配法小结：,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

13、算：,固端弯矩之和,（第一轮第一结点）,（第一轮第二、三,结点）,固端弯矩之和,结点不平,衡力矩,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。,2020/5/3,12,例,2.,q=,20kN/m,A,1,B,E,4m,D,4,F,5m,1,2,m,2,4m,4,m,3,1,C,1,?,?,BA,?,0,.,3,?,?,?,BC,?,0,.,4,?,?,?,0,.,3,?,BE,B,0.4,?,?,CB,?,0,.,445,?,?,?,CD,?,0,.,333

14、,?,?,?,0,.,222,?,CF,C,m,BA,= 40kN,m,m,BC,= -,41.7kN,m,m,CB,= 41.7kN,m,0.3,B,0.445,-41.7,-18.5,2.2,-1.0,-0.5,-0.7,24.4,-9.8,-14.6,-4.65,-0.25,-4.90,C,0.333,0.222,-9.3,-13.9,A,43.5,46.9,40,3.3,0.3,24.5,14.7,9.8,-41.7,-9.3,3.3,4.4,D,3.45,1.7,4.89,-0.5,0.15,0.15,0.2,43.45,3.45,-46.9,1.65,0.07,1.72,M,图,(

15、,kN,?,m,),2020/5/3,F,E,13,例,3.,带悬臂杆件的结构的力矩分配法。,A,B,1m,A,B,1m,5/6,1/6,25,-20.8,-4.2,-20.8,+20.8,5m,50,+50,5m,EI,=,常数,EI,=,常数,C,1m,50kN,m,C,1m,50kN,D,50kN,D,M,2020/5/3,A,B,14,M,/2,用力矩分配法计算，作,M,图。,取,EI,=5,M,B,=31.25,20.83=10.42,M,C,=20.83,20,2.2=,1.37,A,2kN/m,20,20kN,20kN,4EI,i,=4,B,4EI,i,=4,2EI,i,=2.5

16、,C,2EI,i,=2.5,F,E,5,m,结点,杆端,A,AB,E,EB,B,BA,0.316,5,m,1,m,BE,0.263,BC,0.421,C,(,20),F,CB,CF,FC,0.615,0.385,m,0,0,0,31.25,20.83,20.83,0,1.37,2.74,3.29,4.39,2.20,0.84 0.53,0.42,0.05,0.10,0.14,0.18,0.09,0.27,15,2020/5/3,4,m,0,结点,杆端,A,AB,0,E,EB,BE,0.263,B,BA,0.316,BC,0.421,C,(-20),CB,CF,0.615,0.385,F,FC,

17、0,0.27,0.02,0.29,m,0,0,31.25,20.83,20.83,0,1.37,2.74,3.29,4.39,2.20,0.84 0.53,0.42,0.05,0.10,0.14,0.18,0.09,0.03,0.01,0.01,0.01,0.06 0.03,0.56,M,0,1.42,2.85,27.80,24.96,19.94,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,),2020/5/3,16,20kN/m,20kN/m,A,2,i,3,m,i,i,i,3,m,结点,杆端,AG,

18、0.5,15,2020/5/3,4,i,G,S,AG,=4,i,S,AC,=4,i,S,CA,=4,i,S,CH,=2,i,S,CE,=4,i,AG,=0.5,AC,=0.5,CH,=0.2,2,i,i,C,2i,i,H,CA,=0.4,CE,=0.4,i,3,m,E,1.5,m,20,?,1,.,5,m,AG,?,?,?,?,15,kN,.,m,3,C,E,CE,0.4,CH,CH,0.2,A,AC,0.5,CA,0.4,m,17,7.11,结点,杆端,A,20kN/m,7.11,C,m,0.78,AG,AC,0.5,0.5,2.63,2.63,15,7.5 7.5,1.58,1.58,0.75,CA,0.4,3.75,CH,0.2,CE,0.4,E,CH,1.50,0.75,1.50,0.75,0.04,0.79,0.37 0.38,0.19,M,图（,kN.m),0.04,0.79,0.08,0.03,0.08,0.02 0.02,M,7.11,7.11,2.36,0.78,1.58,0.79,2020/5/3,18,例、,求矩形衬砌在上部土压力作用下的弯矩图。,q,A,I,2,I,1,E,B,解：取等代结构如图。,q,l,1,q,设梁柱的线刚度为,i,1,，,i,2,S,BE,?,2,i,1,S,BF,?,2,i,2,i,2,i,1,?,BF,?,?,BE,?,i,