力法知识讲解双语

1、一一. .超静定结构的静力特征和几何特征超静定结构的静力特征和几何特征The static and geometrical features of statically independent structures (SIS)静力特征静力特征: :仅由静力平衡方程不能求出所有内力和反力仅由静力平衡方程不能求出所有内力和反力. . Static features: Internal forces and reactions can not be determined only by Static equilibrium conditions 超静定问题的求解要同时考虑结构的超静定问题的求解要同时

2、考虑结构的“变形、本构、平变形、本构、平衡衡”. .Deformation and equilibrium must be considered simultaneously 几何特征几何特征: :有多余约束的几何不变体系。有多余约束的几何不变体系。Geometrical features: stable systems with redundant restraints. 与静定结构相比与静定结构相比, , 超静定结构的超静定结构的优点优点为为The advantages of SIS : 1.内力分布均匀内力分布均匀the distribution of internal forces a

3、re distributed more uniformly; 2.抵抗破坏的能力强抵抗破坏的能力强More resistance to fracture1.内力与材料的物理性质内力与材料的物理性质、截面的几何形状和尺寸有关截面的几何形状和尺寸有关。The internal forces are related with the properties of materials, the Geometrical form and sizes of the cross-section of the members二二. .超静定结构的性质超静定结构的性质The properties of SIS2.

4、2.温度变化、支座移动一般会产生内力温度变化、支座移动一般会产生内力。Temperature change, movements of supports usually cause internal forces. 1.1.力法力法force method-以多余约束力作为基本未知量以多余约束力作为基本未知量Take redundant forces as primary unknowns。2.2.位移法位移法displacement method-以结点位移作为基本未以结点位移作为基本未知量知量. . Take joint displacements as primary unknowns。

5、三三. .超静定结构的计算方法超静定结构的计算方法Computation methods3.3.混合法混合法combined method- -以结点位移和多余约束以结点位移和多余约束力作为基本未知量力作为基本未知量. . 4.4.力矩分配法力矩分配法moment distribution method-近似计算方法近似计算方法approximation method.力法等方法的基本思想力法等方法的基本思想the idea of force method: 1.找出未知问题不能求解的原因找出未知问题不能求解的原因find the reasons whywe can not solve the

6、 problem, 2.将其化成会求解的问题将其化成会求解的问题 transform it into solvable problem, 3.找出改造后的问题与原问题的差别找出改造后的问题与原问题的差别find the differencebetween the transformed and original problems, 4.消除差别后消除差别后,改造后的问题的解即为原问题的解改造后的问题的解即为原问题的解removing the difference, the transformed problem becomessolvable.5.5.矩阵位移法矩阵位移法matrix disp

7、lacement method-结构矩阵分析法之一结构矩阵分析法之一.7-2 超静定次数的确定超静定次数的确定Determination of indeterminacy超静定次数超静定次数= =多余约束个数多余约束个数= =去除的多余约束数去除的多余约束数degrees of indeterminacy number of redundant restraintsnumber of released restraints. . 几次超静定结构几次超静定结构?X X比较法比较法: :与相近的静定结构与相近的静定结构 相比相比, , 比静定结构比静定结构 多几个约束即为几多几个约束即为几 次超静

8、定结构次超静定结构. .X X1 12 2X X1 1X X2 2力法基本体系不惟一力法基本体系不惟一Primary system of force method is not unique. .若一个结构有若一个结构有N个多余约束个多余约束,则称其为则称其为N次次超静定结构超静定结构 If a structure has n redundant restraints, then it is called SIS of n-th degrees of indeterminacy.X X1 1X X1 1X X2 2X X2 2X X3 3X X3 3X X1 1X X2 2X X3 32.2.

9、去除一个铰支座或者去除一去除一个铰支座或者去除一个单铰相当于去掉个单铰相当于去掉2个约束。个约束。Removing a hinged support or a single hinge is equivalent to releasing 2 restraints1.1.去掉一个链杆或切断一个链去掉一个链杆或切断一个链杆相当于去掉一个约束杆相当于去掉一个约束 Removing a roller support or cutting a link or two-force member is equivalent to releasing one restraint3.3.去掉一个固定端支座或去

10、掉一个固定端支座或切断一根弯曲杆相当于去切断一根弯曲杆相当于去掉三个约束掉三个约束. .Removing a fixed support or cutting out a flexural member is equivalent to release 3 restraints 1X2X3X1X2X3X1X2X3X4.4.将刚结点变成铰结点或将刚结点变成铰结点或将固定端支座变成固定铰将固定端支座变成固定铰支座相当于去掉一个约支座相当于去掉一个约束束. .Changing a rigid joint of 2 member ends into a hinged joint or changing

11、 fixed support into a hinged support is equivalent to release 2 restraints 2X3X1X2X3X1X几何可变体系不能几何可变体系不能作为基本体系作为基本体系一个无铰封闭框有一个无铰封闭框有三个多余约束三个多余约束. .Closed frame:3 redundant restraints1X2X3X基本结构基本结构指去掉多余约束后的结构指去掉多余约束后的结构Primary structure: structure after Removing the redundant restraints（14 次）次）(1 次）(6

12、 次)(4 次)(6 次)1X2X3X4X5X6X7X8X9X10X(10 次) Basic concept of Force method一一.力法的基本概念力法的基本概念Basic concept of Force method1基本体系基本体系=基本结构基基本结构基本未知力外力本未知力外力Primary system=primary structure + unknown forces + external forces待解的未知问题待解的未知问题problem to be solved01变形条件变形条件 deformation condition1X力法基本未知量力法基本未知量 pr

13、imary unknown of force method10101111P11111X01111PX22/qlMPlM1EIl3311/EIqlP841/)(/831qlXPMXMM1182/qlM1. 确定基本体系确定基本体系 determine primary system2. 写出位移条件写出位移条件, ,力法方程力法方程 write out deformation Condition , develop canonical equations of force method 3. 作单位弯矩图作单位弯矩图, ,荷载弯矩图荷载弯矩图; construct Moment Diagrams

14、 caused by unit force and external forces4. 求出系数和自由项求出系数和自由项 find the coefficients and free terms5. 解力法方程解力法方程solve canonical equations 6. 叠加法作弯矩图叠加法作弯矩图 construct moment diagram by Superposition principle. qllEI2EIqllEI2EIX1X212变形条件变形条件:0021力法的典型方程力法的典型方程qllEI2EIqX1X212变形条件变形条件:0021qX1=11X1121X2=12

15、X2212P1P2012121111PXX022221212PXX-力法的典型方程力法的典型方程Canonical equations of force method)(jiij主系数主系数main coefficients0)(jiij付系数付系数secondary coefficientsiP自由项自由项free termsjiij位移互等位移互等law of reciprocal displacements柔度系数柔度系数flexibility coefficients, therefore force method-flexibility methodqllEI2EIqX1X212qX

16、1=11XX2=12X11212212P1P201212111PXX02222121PXXEIllEIllEI3321167132221M1lM2lMP22/qlEIlllEI32122121EIlllEI32212121EIlllEI3222313221EIqlP41169EIqlP424140320921/,/qlXqlXPMXMXMM2211202ql402/qlM内力分布与内力分布与刚度无关吗刚度无关吗? ?Is the distribution of inter-nal forces related to the rigidity? 荷载作用下超静定结构内力分布与刚度荷载作用下超静定

17、结构内力分布与刚度的绝对值无关只与各杆刚度的比值有关的绝对值无关只与各杆刚度的比值有关. . the distribution of internal forces is relatedTo the relative rigidity of the members, but notTo the absolute rigidity.qllEI2EIqX1X212202ql402/qlM01212111PXX02222121PXX40320921/,/qlXqlX0021q1X2X40202221/,/qlXqlX01212111PXX02222121PXX00211X2X40203221/,/q

18、lXqlX01212111PXX02222121PXX0021小结小结Summary:1.1.力法的典型方程是体系的变形协调方程力法的典型方程是体系的变形协调方程 Canonical equations of force method are equations of deformation compatibility conditions2.2.主系数恒大于零主系数恒大于零main coefficients are great than 0,副系数满足位移互等定理副系数满足位移互等定理secondary coefficients satisfy law of reciprocal displ

19、acements3.3.柔度系数是体系常数柔度系数是体系常数flexibility coefficients are constants of system4.4.荷载作用时荷载作用时, ,内力分布与刚度大小无关内力分布与刚度大小无关, ,与各杆刚度与各杆刚度比值有关比值有关. .荷载不变荷载不变, ,调整各杆刚度比可使内力重分布调整各杆刚度比可使内力重分布. . the distribution of internal forces is related to the relative (stiffness) rigidity of the members, but not to the a

20、bsolute rigidity (stiffness). By regulating relative rigidity we can regulate the distribution of internal forces.例例1. 1. 力法求解图示结构的力法求解图示结构的M M图图. Construct moment diagram by force method017-5 算例算例Examplesl/2EIEIPl/2lX1PPX1=183/PlMP2/ lM1Solution:01111PX323/PlMEIl6311/EIPllPlllPllEIP9611442122324211

21、31)(4/Pl16111/PX PMXMM1101l/2EIEIPl/2lX1PPX1=183/PlMP2/lM1解解:01111PX323/PlMEIl6311/EIPllPlllPllEIP961144212232421131)(4/Pl16111/PXPMXMM1101解解:01111PXEIl 3211/EIPlPllEIP16214211213231/PlX PMXMM11PX14/PlMPP1M1X1=1另一解法另一解法 another solution03113000321PX1=1M1X2=1M2M3X3=1PMPX1PX2X3X1=1X2=1X3=1PM1M2M3MPPX1

22、X2X3000333323213123232221211313212111PPPXXXXXXXXX032PP例例2. 力法解图示结构力法解图示结构,作作M图图Construct moment diagram by force method.solution:PllX1PX2X3000333323213123232221211313212111PPPXXXXXXXXX 0332233113P 023232333EAlGAskQEAsNEIsMd dd03X0022221211212111PPXXXX EIl 32211/EIl 62112/EIPlPP16221/882221/PlXplXPMX

23、MXMM221182/Pl例例3. 力法解图示桁架力法解图示桁架.Calculate the truss by force method EA=constant.解解solution:Paa1XP0101111PXEAaEAlNN)(2141111EAPaEAlNNPP)(2121121/PXPNXNN11PP2P00P00NP11XN111111221XP-P/2-P/2P/2P/222 /22 /1X1XEAaX1101解解Solution：kXXP/11111 )(32251qlX例例 4. 4. 求作图示梁的弯矩图求作图示梁的弯矩图Draw moment diagram by forc

24、e method.PMXMM11)1(1111kXP ,310lEIk 当当k当当)(qlX451EIkX /11EIl6311EIPlP245310k当当01X解：解：01111PX 例例 5 5. . 求解图示加劲梁求解图示加劲梁。横梁横梁44m101IEIEAEIP3 .533,2 .1267.10111 当当kN .,m 944101123XA;,P11P11NXNNMXMM%./.3191925080415By changing relative rigidity We can regulate the distribution Of internal forces.当当kN .,m

25、 944101123XA23m107 . 1AqlX4598.4967.103 .5331当当,A梁的受力与两跨梁的受力与两跨连续梁相同。连续梁相同。（同例（同例4 4中中 ）k下侧正弯矩为下侧正弯矩为设基本未知力为设基本未知力为 X，则，则2)05. 04(5)05. 04)(5 . 040(XXXX跨中支座负弯矩为跨中支座负弯矩为80)5 . 040(4X根据题意正弯矩等于负弯矩，可得根据题意正弯矩等于负弯矩，可得862915.46X有了基本未知力，由典型方程可得有了基本未知力，由典型方程可得23m 1072. 1A无弯矩情况判别无弯矩情况判别 the cases of zero mome

26、nts 不计轴向变形不计轴向变形, ,下述情况无弯矩下述情况无弯矩, ,只有轴只有轴力力. . Neglecting axial deformation, only axial forces arise in the following cases(1).(1).集中荷载沿柱轴作用集中荷载沿柱轴作用 concentrated forces act alone axis of membersP(2).(2).等值反向共线集中荷载沿杆轴作用等值反向共线集中荷载沿杆轴作用. .A pair of collinear, equal in magnitude, but opposite in direc

27、tion loads act alone axis of a member.PP(3).(3).集中荷载作用在不动结点集中荷载作用在不动结点 concentrated forces act on an unmovable jointP可利用下面方法判断可利用下面方法判断: :化成铰接体系后化成铰接体系后, ,若能若能平衡外力平衡外力, ,则原体系无弯矩则原体系无弯矩. . Method of determination: converting the structure into pin-connected system, if external forces can be balanced,

28、 then no moment arises.4.无弯矩情况判别无弯矩情况判别 the determination000333323213123232221211313212111PPPXXXXXXXXX 0321PPP齐次线性方程的系数组齐次线性方程的系数组成的矩阵可逆成的矩阵可逆, ,只有零解只有零解.The matrix composed by flexibility coefficients is inversable, therefore only zero solution exists. 0321XXXPMXMXMXMM332211超静定拱的计算超静定拱的计算The comput

29、ation of statically indeterminate archPPX1X1=111PP1dsGAQdsEANdsEIM2121211101111PX01dsEIMMPP11通常用数值积分方法或计算机通常用数值积分方法或计算机计算计算 Generally they can calculated by numerical method or by computer.7-6 对称性利用对称性利用Utilization of symmetrySymmetric structures: structures with geometry, supports and rigidity dist

30、ribution being symmetric with respect to symmetry axis.对称结构对称结构Symmetric structures非对称结构非对称结构Non-symmetric structures支承不对称支承不对称刚度不对称刚度不对称几何对称几何对称支承对称支承对称刚度对称刚度对称对称荷载对称荷载symmetric loads: :方向和作用点对称于对称轴、大方向和作用点对称于对称轴、大小相等的荷载小相等的荷载. .Loads with magnitudes, directions and action points being symmetric wi

31、th respect to symmetry axis反对称荷载反对称荷载 Antisymmetric loads: :作用在对称结构对称轴两侧作用在对称结构对称轴两侧, ,大小相等大小相等, ,作用点对称作用点对称, ,方向反对称的荷载方向反对称的荷载Loads with magnitudes, directions and action points being anti-symmetric with respect to symmetry axisPP对称荷载对称荷载Symmetric loadsPP反对称荷载反对称荷载Antisymmetric loadsPllMllPllEI=Cll

32、EI=CM选取对称基本结构选取对称基本结构, ,对称基本未知量和反对称基本未知量对称基本未知量和反对称基本未知量The selection of symmetric primary structure, symmetric and anti-symmetric primary unknowns1X2X3X11XM112XM213XM3MP 000333323213123232221211313212111PPPXXXXXXXXX 032233113 0003333P2222121P1212111PXXXXX 典型方程分为两组典型方程分为两组Force method equations are

33、decomposed into 2 sets:一组只含对称未知量一组只含对称未知量one involves only symmetric unknowns,另一组另一组只含反对称未知量只含反对称未知量 the second involves only anti-symmetric unknowns1X2X3X11XM112XM213XM3PMXMXMM2211MP3P=0，X3=0对称结构在正对称荷载作用下，对称结构在正对称荷载作用下，其弯矩图和轴力图是正对称的其弯矩图和轴力图是正对称的, ,剪力图反对称；变形与位移对称剪力图反对称；变形与位移对称. .对称荷载对称荷载symmetric lo

34、ads: Symmetric structures under symmetric loads: M-diagram, FN-diagram, deformation and displacements are symmetric; but FS-diagram is anti-symmetric.1X2X3X11XM112XM213XM3PMXMM33MPX1= X2 =0对称结构在反正对称荷载作用下，对称结构在反正对称荷载作用下，其弯矩图和轴力图是反正对称的其弯矩图和轴力图是反正对称的, ,剪力图对称；变形与位移反对称剪力图对称；变形与位移反对称.反正对称荷载反正对称荷载 Antisymm

35、etric loads: Symmetric structures under anti-symmetric loads: M-diagram, FN-diagram, deformation and displacements are anti-symmetric; but but F FS S-diagram is symmetric.-diagram is symmetric.取半结构计算取半结构计算 Select a half structure to analyzeA.A.奇数跨结构奇数跨结构 Frames with odd-number spans对称荷载作用对称荷载作用Under

36、 symmetric loads:半结构半结构反对称荷载反对称荷载Under anti-symmetric loads :半结构半结构B. 偶数跨结构偶数跨结构Frames with odd-number spansA.A.奇数跨结构奇数跨结构 Frames with odd-number spans对称荷载作用对称荷载作用Under symmetric loads:反对称荷载反对称荷载Under anti-symmetric loads :对称荷载作用对称荷载作用Under symmetric loads:反对称荷载反对称荷载Under anti-symmetric loads :练习练习:

37、练习练习:Pl/2l/2EI1X2X3XP/2P/201111PX11XM11MPP/2P/2Pl/4Pl/4EIl11EIPlP83181PlXPMXMM11MPPl/8Pl/8解解Solution：0P1 111 X11144EI 11800EIP 15 .12X P11MXMM例例: :求图示结构的弯矩图求图示结构的弯矩图EI=constant。M1MPM0P1 111 XEIplEIlP2331311,231PXPMXMM11例例1:1:求图示结构的弯矩图求图示结构的弯矩图EI=constant。0P1 111 XEIplEIlP1624731311,PX1431PMXMM11M1MP

38、M3Pl/28Pl/7Pl/73Pl/28Pl/73Pl/28Pl/73Pl/282Pl/73Pl/140P1 111 XEIplEIlP4322111,PlX831PMXMM11M1MPM3Pl/8Pl/8Pl/8Pl/8Pl/8Pl/83Pl/8例例4 4：求作图示圆环的弯矩图：求作图示圆环的弯矩图, , of the ringEIEI= =常数。常数。解：解：取结构的取结构的1/41/4分析分析11 MsinP2PRM ,dEIREIsM22111 ,dMPEIPREIsMP2211PRX 1)sin(P2111PRMXMM若只考虑弯矩对位移的影响，有：若只考虑弯矩对位移的影响，有：01

39、11P例例 5. 试用对称性对结构进行简化。试用对称性对结构进行简化。EI为常数。为常数。Simplify analytical modelsP /2P/2P/2P /2I/2I/2P /2P /2I/2方法方法 1PP /2P /2P /4P /4P /4I/2P /4P /4P /4P /4I/2P /4P/4P/4I/2P/4I/2P /4P方法方法 2PP /2P /2P /4P/2P /4P /4P /2P /4P /4P/2P /4P /4P /2P /4P /4P /4P /4P /4I/2P /4P/4P/4I/2P/4I/2P 7-7 7-7 超静定结构的位移计算超静定结构的

40、位移计算Computation of displacements of statically indeterminate structures思路思路ideaidea：将超静定位移计算转换为静定位移计算：将超静定位移计算转换为静定位移计算 Transform Transform the the computation of displacements of statically indeterminate structures into the computation of displacements of statically determinate structures 超静定位移状态超静

41、定位移状态 基本结构基本结构 外载外载 多余力多余力 引起的位移引起的位移状态状态TheThe displacement state of statically indeterminate structures is completely identical to the displacement state of the primary structure under external loads and redundant forces. 因此可以选因此可以选 基本结构（静定结构）为研究对象，而外载基本结构（静定结构）为研究对象，而外载 多余多余力力 统一看成外载。统一看成外载。We ca

42、n select primary structure (statically determinate structures) as object to analyze, and look at the original external loads and redundant forces as generalized external forces. 求求A A截面转角截面转角 Determine the rotation of section A0021qllEI2EIAX2X1Aq202ql402/qlM202ql402/qlM1Mi)()(EIqlqllqllEIA3228011402

43、1120211qllEI2EIAX2X1Aq202ql402/qlM202ql402/qlM1Mi)()(EIqlqllqllEIA322801140211202111X2X202ql402/qlM1Mi)()(EIqlqllqllEIA3228012183232202121单位荷载法求超静定结构位移时单位荷载法求超静定结构位移时, ,单位力可加单位力可加在任意力法基本结构上在任意力法基本结构上When using unit load method to determine displacements, unit force may be applied on any primary stru

44、cture.求求A A截面转角截面转角 Determine the rotation of section A7-8 最后内力图的校核最后内力图的校核 The checking of final internal force diagrams1、平衡条件校核、平衡条件校核 The checking of equilibrium conditionsqllEI2EIA202ql402/qlM0202022qlqlMMMACABA从结构中取隔离体，验证平衡条件是否满足从结构中取隔离体，验证平衡条件是否满足 Isolate free bodies and verify whether or not

45、the equilibrium conditions are satisfied。取结点取结点A，验证结点，验证结点A的力矩平衡条件：的力矩平衡条件：202ql202qlA同样可根据剪力图和轴力图验证投影平衡条件是否满足。同样可根据剪力图和轴力图验证投影平衡条件是否满足。（同学可自己验证）（同学可自己验证）取结点，绘受力图，取结点，绘受力图，列平衡方程列平衡方程qllEI2EIAX2X1Aq202ql402/qlM202ql402/qlM011dsEIMM022dsEIMMX1=1M1lX2=1M2l2、位移条件校核、位移条件校核The checking of displacement con

46、ditions对有封闭无铰框格的刚架，当只承受荷载时，利用框格上任意对有封闭无铰框格的刚架，当只承受荷载时，利用框格上任意截面处的相对角位移为零的条件，则：截面处的相对角位移为零的条件，则：For any closed frame that is only subjected to external loads, using the condition that the relative rotation of any one of the cross sections is equal to zero, we obtain111KX0dsEIMdsEIMMKK或绕框格或绕框格M/EI面积之和

47、为零面积之和为零 The sum of the area of M/EI diagrams for the members of closed frame equals to zerodsEIM图M图KMt 1t2t1t1t2t1t1t1t2t10022221211212111ttXXXX0021htltNiiit)(0基本结构由于温度变化引基本结构由于温度变化引起的沿起的沿Xi方向的位移。方向的位移。位移条件位移条件 displacement conditions由叠加原理，又可写成：由叠加原理，又可写成：。的物理意义与前面相同的物理意义与前面相同ij7-9 温度变化时超静定结构内力的计算温

48、度变化时超静定结构内力的计算 Internal forces duo to temperature changes t1t1t2t1X1X2基本体系基本体系primary structure图乘。与jijiijMMdsEIMM解解 Solution：01111 tX 例例Example. 求图示刚架由于温度变化引起的内力与求图示刚架由于温度变化引起的内力与K点点的位移。的位移。 t1=+250C t2=+350C,EI=常数常数,矩形截面矩形截面,h=l/10. Determine the internal forces and the displacement of section K of

49、 the given rigid frame duo to temperature change. t1=+250C t2=+350C,EI=constant, rectangular section, h=l/10 10300tt , EIl35311230221030221)(llhltM1)(.)( dilhtltNEIsMMiiKy7534021138lEIX 11XMM M温度改变引起的内力与各杆的绝对刚度温度改变引起的内力与各杆的绝对刚度 EIEI 有关。有关。The internal forces due to temperature change is related to t

50、he absolute rigidity EI of the members?dEIsMMiKyMi温度低的一侧受拉温度低的一侧受拉。?01CX111CX1C1CX10101111CXCRiiC7-10 支座移动时超静定结构的计算支座移动时超静定结构的计算 Internal forces duo to support settlements 计算原理与荷载作用时的计算相同计算原理与荷载作用时的计算相同 The computation principle is identical to the case under loads由基本体系位移与原结构位移相同建立基本方程由基本体系位移与原结构位移相

51、同建立基本方程, 计算多于未知力。计算多于未知力。Using the condition that the displacements of the primary structure are identical to that of original structure to develop equation and further to determine unknown forces. )(11本例为方向的位移引起的沿移为基本结构由于支座位XCC注：方程的右端项可能不为零，与基本结构的选择有关。注：方程的右端项可能不为零，与基本结构的选择有关。Remark: The right side

52、 of the equation may not equal to zero, whether or not the right side of the equation equals to zero depend on the primary structure selected.思考：什么情况下方程右端不为零？思考：什么情况下方程右端不为零？?01CX111CX1C1CX10101111CXCRiiC)(11本例为方向的位移引起的沿移为基本结构由于支座位XCCC1Is the displacement in the direction of X1 induced by the settle

53、ment of support C 解解 Solution：例例Example. 求图示梁由于支座移动引起的内力求图示梁由于支座移动引起的内力. Find the internal forces induced by support settlementEIl1231121lClEI1X2X11X2/ lM10022221211212111CCXXXX002102112EIl22C212XM21216lEIX2211XMXMM12XM21lEIX2lEI4lEI2M支座移动引起的内力与各杆支座移动引起的内力与各杆的绝对刚度的绝对刚度 EI 有关。有关。练习练习: :写出典型方程写出典型方程,

54、,并求出并求出自由项。自由项。Exercise: Develop canonical equations and determine free termsaXXXXXXXXXCCC3333232131232322212113132121110 1 1C=b/l几何法几何法: 2 2C=-b/l 3 3C=0公式法公式法:1/l1/lCRiiC0lbblC/)/(11lbblC/)/(12001bCaXXXXXXXXXCCC3333232131232322212113132121110练习练习: :写出典型方程写出典型方程, ,并求并求出自由项。出自由项。Exercise: Develop ca

55、nonical equations and determine free termsCCCXXXaXXXbXXX333323213123232221211313212111 1 1C=0 2 2C=0 3 3C=01X3X2X1X3X2X000333323213123232221211313212111CCCXXXXXXXXX12X00111X10l00113X1CCCabllb3211)(支座移动时支座移动时, ,结构中的位移以及位移条件的校核公结构中的位移以及位移条件的校核公式如下式如下: :iiiiCiicREIsMMEIsMMdd制造误差引起的内力计算制造误差引起的内力计算: :1X3

56、X2XABAB杆造长了杆造长了1cm,1cm,如何作弯矩图如何作弯矩图? ?BA10m10m机动法作影响线机动法作影响线Construct influence lines by Mechanismic Method两种方法：两种方法：2 methods:2 methods:1 1、静力法：按照力法求出影响线方程；、静力法：按照力法求出影响线方程； Static method: determine influence lines by force method2 2、机动法：利用位移图做影响线。、机动法：利用位移图做影响线。Mechanismic Method: use displacement

57、 graph to construct influence lines1 1、静力法、静力法static method：3211112131111112363;30lxlxXEIxlxEIlXPPP求支座反力影响线求支座反力影响线construct IL of support reactions2、机动法、机动法Mechanismic method：由位移互等定理得由位移互等定理得11PPP11111111PPX1PP=1作用下在作用下在X11方方向的影响线向的影响线 X11 作用下在作用下在P=1方方向的影响线向的影响线超静定结构的影响线问题化超静定结构的影响线问题化为基本结构在固体荷载作用

58、为基本结构在固体荷载作用下的位移图问题下的位移图问题EIxlxP6321111设设则则11PX结果同前方法所得结果同前方法所得 由于由于P1P11 1向下为正，故上式负号表明向下为正，故上式负号表明X1X1影响线就是影响线就是X1=1X1=1作用下的杆的位移图。作用下的杆的位移图。 方法同静定结构的机动法：方法同静定结构的机动法：去掉所求未知力的相应约去掉所求未知力的相应约束后，体系沿着未知力方向产生单位位移，这时的体系位束后，体系沿着未知力方向产生单位位移，这时的体系位移图就是该力影响线移图就是该力影响线不同之处不同之处the differences： 静定结构静定结构去掉多余约束后变为机构

59、，体系去掉多余约束后变为机构，体系位移图为位移图为直线直线；Statically determinate structures transform into mechanisms after releasing redundant restraints, and the displacement graphs are straight lines. 而而超静定结构超静定结构去掉多余约束后认为几何不去掉多余约束后认为几何不变体系，体系位移图为弹性挠曲线。变体系，体系位移图为弹性挠曲线。 Statically indeterminate structures transform into geom

60、etrically stable systems after releasing redundant restraints, and the displacement graphs are elastic deflection lines.超静定梁支座反力影响线超静定梁支座反力影响线Reaction force IL of statically indeterminate beams弯矩、剪力影响线弯矩、剪力影响线IL of bending moment and shearing forces最不利荷载布置最不利荷载布置The most unfavorable position of load