1、一一. .超靜定結構的靜力特征和幾何特征超靜定結構的靜力特征和幾何特征The static and geometrical features of statically independent structures (SIS)靜力特征靜力特征: :僅由靜力平衡方程不能求出所有內力和反力僅由靜力平衡方程不能求出所有內力和反力. . Static features: Internal forces and reactions can not be determined only by Static equilibrium conditions 超靜定問題的求解要同時考慮結構的超靜定問題的求解要同時

2、考慮結構的“變形、本構、平變形、本構、平衡衡”. .Deformation and equilibrium must be considered simultaneously 幾何特征幾何特征: :有多余約束的幾何不變體系。有多余約束的幾何不變體系。Geometrical features: stable systems with redundant restraints. 與靜定結構相比與靜定結構相比, , 超靜定結構的超靜定結構的優點優點為為The advantages of SIS : 1.內力分布均勻內力分布均勻the distribution of internal forces a

3、re distributed more uniformly; 2.抵抗破壞的能力強抵抗破壞的能力強More resistance to fracture1.內力與材料的物理性質內力與材料的物理性質、截面的幾何形狀和尺寸有關截面的幾何形狀和尺寸有關。The internal forces are related with the properties of materials, the Geometrical form and sizes of the cross-section of the members二二. .超靜定結構的性質超靜定結構的性質The properties of SIS2.

4、2.溫度變化、支座移動一般會產生內力溫度變化、支座移動一般會產生內力。Temperature change, movements of supports usually cause internal forces. 1.1.力法力法force method-以多余約束力作為基本未知量以多余約束力作為基本未知量Take redundant forces as primary unknowns。2.2.位移法位移法displacement method-以結點位移作為基本未以結點位移作為基本未知量知量. . Take joint displacements as primary unknowns。

5、三三. .超靜定結構的計算方法超靜定結構的計算方法Computation methods3.3.混合法混合法combined method- -以結點位移和多余約束以結點位移和多余約束力作為基本未知量力作為基本未知量. . 4.4.力矩分配法力矩分配法moment distribution method-近似計算方法近似計算方法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.找出未知問題不能求解的原因找出未知問題不能求解的原因find the reasons whywe can not solve the

6、 problem, 2.將其化成會求解的問題將其化成會求解的問題 transform it into solvable problem, 3.找出改造后的問題與原問題的差別找出改造后的問題與原問題的差別find the differencebetween the transformed and original problems, 4.消除差別后消除差別后,改造后的問題的解即為原問題的解改造后的問題的解即為原問題的解removing the difference, the transformed problem becomessolvable.5.5.矩陣位移法矩陣位移法matrix disp

7、lacement method-結構矩陣分析法之一結構矩陣分析法之一.7-2 超靜定次數的確定超靜定次數的確定Determination of indeterminacy超靜定次數超靜定次數= =多余約束個數多余約束個數= =去除的多余約束數去除的多余約束數degrees of indeterminacy number of redundant restraintsnumber of released restraints. . 幾次超靜定結構幾次超靜定結構?X X比較法比較法: :與相近的靜定結構與相近的靜定結構 相比相比, , 比靜定結構比靜定結構 多幾個約束即為幾多幾個約束即為幾 次超靜

8、定結構次超靜定結構. .X X1 12 2X X1 1X X2 2力法基本體系不惟一力法基本體系不惟一Primary system of force method is not unique. .若一個結構有若一個結構有N個多余約束個多余約束,則稱其為則稱其為N次次超靜定結構超靜定結構 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、去除一個鉸支座或者去除一去除一個鉸支座或者去除一個單鉸相當于去掉個單鉸相當于去掉2個約束。個約束。Removing a hinged support or a single hinge is equivalent to releasing 2 restraints1.1.去掉一個鏈桿或切斷一個鏈去掉一個鏈桿或切斷一個鏈桿相當于去掉一個約束桿相當于去掉一個約束 Removing a roller support or cutting a link or two-force member is equivalent to releasing one restraint3.3.去掉一個固定端支座或去

10、掉一個固定端支座或切斷一根彎曲桿相當于去切斷一根彎曲桿相當于去掉三個約束掉三個約束. .Removing a fixed support or cutting out a flexural member is equivalent to release 3 restraints 1X2X3X1X2X3X1X2X3X4.4.將剛結點變成鉸結點或將剛結點變成鉸結點或將固定端支座變成固定鉸將固定端支座變成固定鉸支座相當于去掉一個約支座相當于去掉一個約束束. .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幾何可變體系不能幾何可變體系不能作為基本體系作為基本體系一個無鉸封閉框有一個無鉸封閉框有三個多余約束三個多余約束. .Closed frame:3 redundant restraints1X2X3X基本結構基本結構指去掉多余約束后的結構指去掉多余約束后的結構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基本體系基本體系=基本結構基基本結構基本未知力外力本未知力外力Primary system=primary structure + unknown forces + external forces待解的未知問題待解的未知問題problem to be solved01變形條件變形條件 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. 確定基本體系確定基本體系 determine primary system2. 寫出位移條件寫出位移條件, ,力法方程力法方程 write out deformation Condition , develop canonical equations of force method 3. 作單位彎矩圖作單位彎矩圖, ,荷載彎矩圖荷載彎矩圖; construct Moment Diagrams

14、 caused by unit force and external forces4. 求出系數和自由項求出系數和自由項 find the coefficients and free terms5. 解力法方程解力法方程solve canonical equations 6. 疊加法作彎矩圖疊加法作彎矩圖 construct moment diagram by Superposition principle. qllEI2EIqllEI2EIX1X212變形條件變形條件:0021力法的典型方程力法的典型方程qllEI2EIqX1X212變形條件變形條件: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主系數主系數main coefficients0)(jiij付系數付系數secondary coefficientsiP自由項自由項free termsjiij位移互等位移互等law of reciprocal displacements柔度系數柔度系數flexibility coefficients, therefore force method-flexibility methodqllEI2EIqX1X212qX

16、1=11XX2=12X11212212P1P201212111PXX02222121PXXEIllEIllEI3321167132221M1lM2lMP22/qlEIlllEI32122121EIlllEI32212121EIlllEI3222313221EIqlP41169EIqlP424140320921/,/qlXqlXPMXMXMM2211202ql402/qlM內力分布與內力分布與剛度無關嗎剛度無關嗎? ?Is the distribution of inter-nal forces related to the rigidity? 荷載作用下超靜定結構內力分布與剛度荷載作用下超靜定

17、結構內力分布與剛度的絕對值無關只與各桿剛度的比值有關的絕對值無關只與各桿剛度的比值有關. . 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小結小結Summary:1.1.力法的典型方程是體系的變形協調方程力法的典型方程是體系的變形協調方程 Canonical equations of force method are equations of deformation compatibility conditions2.2.主系數恒大于零主系數恒大于零main coefficients are great than 0,副系數滿足位移互等定理副系數滿足位移互等定理secondary coefficients satisfy law of reciprocal displ

19、acements3.3.柔度系數是體系常數柔度系數是體系常數flexibility coefficients are constants of system4.4.荷載作用時荷載作用時, ,內力分布與剛度大小無關內力分布與剛度大小無關, ,與各桿剛度與各桿剛度比值有關比值有關. .荷載不變荷載不變, ,調整各桿剛度比可使內力重分布調整各桿剛度比可使內力重分布. . 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. 力法求解圖示結構的力法求解圖示結構的M M圖圖. 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. 力法解圖示結構力法解圖示結構,作作M圖圖Construct moment diagram by force method.solution:PllX1PX2X3000333323213123232221211313212111PPPXXXXXXXXX 0332233113P 023232333EAlGAskQEAsNEIsMd dd03X0022221211212111PPXXXX EIl 32211/EIl 62112/EIPlPP16221/882221/PlXplXPMX

23、MXMM221182/Pl例例3. 力法解圖示桁架力法解圖示桁架.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. 求作圖示梁的彎矩圖求作圖示梁的彎矩圖Draw moment diagram by forc

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

25、 944101123XA23m107 . 1AqlX4598.4967.103 .5331當當,A梁的受力與兩跨梁的受力與兩跨連續梁相同。連續梁相同。（同例（同例4 4中中 ）k下側正彎矩為下側正彎矩為設基本未知力為設基本未知力為 X，則，則2)05. 04(5)05. 04)(5 . 040(XXXX跨中支座負彎矩為跨中支座負彎矩為80)5 . 040(4X根據題意正彎矩等于負彎矩，可得根據題意正彎矩等于負彎矩，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無彎矩情況判別無彎矩情況判別 the cases of zero mome

26、nts 不計軸向變形不計軸向變形, ,下述情況無彎矩下述情況無彎矩, ,只有軸只有軸力力. . Neglecting axial deformation, only axial forces arise in the following cases(1).(1).集中荷載沿柱軸作用集中荷載沿柱軸作用 concentrated forces act alone axis of membersP(2).(2).等值反向共線集中荷載沿桿軸作用等值反向共線集中荷載沿桿軸作用. .A pair of collinear, equal in magnitude, but opposite in direc

27、tion loads act alone axis of a member.PP(3).(3).集中荷載作用在不動結點集中荷載作用在不動結點 concentrated forces act on an unmovable jointP可利用下面方法判斷可利用下面方法判斷: :化成鉸接體系后化成鉸接體系后, ,若能若能平衡外力平衡外力, ,則原體系無彎矩則原體系無彎矩. . Method of determination: converting the structure into pin-connected system, if external forces can be balanced,

28、 then no moment arises.4.無彎矩情況判別無彎矩情況判別 the determination000333323213123232221211313212111PPPXXXXXXXXX 0321PPP齊次線性方程的系數組齊次線性方程的系數組成的矩陣可逆成的矩陣可逆, ,只有零解只有零解.The matrix composed by flexibility coefficients is inversable, therefore only zero solution exists. 0321XXXPMXMXMXMM332211超靜定拱的計算超靜定拱的計算The comput

29、ation of statically indeterminate archPPX1X1=111PP1dsGAQdsEANdsEIM2121211101111PX01dsEIMMPP11通常用數值積分方法或計算機通常用數值積分方法或計算機計算計算 Generally they can calculated by numerical method or by computer.7-6 對稱性利用對稱性利用Utilization of symmetrySymmetric structures: structures with geometry, supports and rigidity dist

30、ribution being symmetric with respect to symmetry axis.對稱結構對稱結構Symmetric structures非對稱結構非對稱結構Non-symmetric structures支承不對稱支承不對稱剛度不對稱剛度不對稱幾何對稱幾何對稱支承對稱支承對稱剛度對稱剛度對稱對稱荷載對稱荷載symmetric loads: :方向和作用點對稱于對稱軸、大方向和作用點對稱于對稱軸、大小相等的荷載小相等的荷載. .Loads with magnitudes, directions and action points being symmetric wi

31、th respect to symmetry axis反對稱荷載反對稱荷載 Antisymmetric loads: :作用在對稱結構對稱軸兩側作用在對稱結構對稱軸兩側, ,大小相等大小相等, ,作用點對稱作用點對稱, ,方向反對稱的荷載方向反對稱的荷載Loads with magnitudes, directions and action points being anti-symmetric with respect to symmetry axisPP對稱荷載對稱荷載Symmetric loadsPP反對稱荷載反對稱荷載Antisymmetric loadsPllMllPllEI=Cll

32、EI=CM選取對稱基本結構選取對稱基本結構, ,對稱基本未知量和反對稱基本未知量對稱基本未知量和反對稱基本未知量The selection of symmetric primary structure, symmetric and anti-symmetric primary unknowns1X2X3X11XM112XM213XM3MP 000333323213123232221211313212111PPPXXXXXXXXX 032233113 0003333P2222121P1212111PXXXXX 典型方程分為兩組典型方程分為兩組Force method equations are

33、decomposed into 2 sets:一組只含對稱未知量一組只含對稱未知量one involves only symmetric unknowns,另一組另一組只含反對稱未知量只含反對稱未知量 the second involves only anti-symmetric unknowns1X2X3X11XM112XM213XM3PMXMXMM2211MP3P=0，X3=0對稱結構在正對稱荷載作用下，對稱結構在正對稱荷載作用下，其彎矩圖和軸力圖是正對稱的其彎矩圖和軸力圖是正對稱的, ,剪力圖反對稱；變形與位移對稱剪力圖反對稱；變形與位移對稱. .對稱荷載對稱荷載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對稱結構在反正對稱荷載作用下，對稱結構在反正對稱荷載作用下，其彎矩圖和軸力圖是反正對稱的其彎矩圖和軸力圖是反正對稱的, ,剪力圖對稱；變形與位移反對稱剪力圖對稱；變形與位移反對稱.反正對稱荷載反正對稱荷載 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.取半結構計算取半結構計算 Select a half structure to analyzeA.A.奇數跨結構奇數跨結構 Frames with odd-number spans對稱荷載作用對稱荷載作用Under

36、 symmetric loads:半結構半結構反對稱荷載反對稱荷載Under anti-symmetric loads :半結構半結構B. 偶數跨結構偶數跨結構Frames with odd-number spansA.A.奇數跨結構奇數跨結構 Frames with odd-number spans對稱荷載作用對稱荷載作用Under symmetric loads:反對稱荷載反對稱荷載Under anti-symmetric loads :對稱荷載作用對稱荷載作用Under symmetric loads:反對稱荷載反對稱荷載Under anti-symmetric loads :練習練習:

37、練習練習:Pl/2l/2EI1X2X3XP/2P/201111PX11XM11MPP/2P/2Pl/4Pl/4EIl11EIPlP83181PlXPMXMM11MPPl/8Pl/8解解Solution：0P1 111 X11144EI 11800EIP 15 .12X P11MXMM例例: :求圖示結構的彎矩圖求圖示結構的彎矩圖EI=constant。M1MPM0P1 111 XEIplEIlP2331311,231PXPMXMM11例例1:1:求圖示結構的彎矩圖求圖示結構的彎矩圖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：求作圖示圓環的彎矩圖：求作圖示圓環的彎矩圖, , of the ringEIEI= =常數。常數。解：解：取結構的取結構的1/41/4分析分析11 MsinP2PRM ,dEIREIsM22111 ,dMPEIPREIsMP2211PRX 1)sin(P2111PRMXMM若只考慮彎矩對位移的影響，有：若只考慮彎矩對位移的影響，有：01

39、11P例例 5. 試用對稱性對結構進行簡化。試用對稱性對結構進行簡化。EI為常數。為常數。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 超靜定結構的位移計算超靜定結構的

40、位移計算Computation of displacements of statically indeterminate structures思路思路ideaidea：將超靜定位移計算轉換為靜定位移計算：將超靜定位移計算轉換為靜定位移計算 Transform Transform the the computation of displacements of statically indeterminate structures into the computation of displacements of statically determinate structures 超靜定位移狀態超靜

41、定位移狀態 基本結構基本結構 外載外載 多余力多余力 引起的位移引起的位移狀態狀態TheThe displacement state of statically indeterminate structures is completely identical to the displacement state of the primary structure under external loads and redundant forces. 因此可以選因此可以選 基本結構（靜定結構）為研究對象，而外載基本結構（靜定結構）為研究對象，而外載 多余多余力力 統一看成外載。統一看成外載。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截面轉角截面轉角 Determine the rotation of section A0021qllEI2EIAX2X1Aq202ql402/qlM202ql402/qlM1Mi)()(EIqlqllqllEIA3228011402

43、1120211qllEI2EIAX2X1Aq202ql402/qlM202ql402/qlM1Mi)()(EIqlqllqllEIA322801140211202111X2X202ql402/qlM1Mi)()(EIqlqllqllEIA3228012183232202121單位荷載法求超靜定結構位移時單位荷載法求超靜定結構位移時, ,單位力可加單位力可加在任意力法基本結構上在任意力法基本結構上When using unit load method to determine displacements, unit force may be applied on any primary stru

44、cture.求求A A截面轉角截面轉角 Determine the rotation of section A7-8 最后內力圖的校核最后內力圖的校核 The checking of final internal force diagrams1、平衡條件校核、平衡條件校核 The checking of equilibrium conditionsqllEI2EIA202ql402/qlM0202022qlqlMMMACABA從結構中取隔離體，驗證平衡條件是否滿足從結構中取隔離體，驗證平衡條件是否滿足 Isolate free bodies and verify whether or not

45、the equilibrium conditions are satisfied。取結點取結點A，驗證結點，驗證結點A的力矩平衡條件：的力矩平衡條件：202ql202qlA同樣可根據剪力圖和軸力圖驗證投影平衡條件是否滿足。同樣可根據剪力圖和軸力圖驗證投影平衡條件是否滿足。（同學可自己驗證）（同學可自己驗證）取結點，繪受力圖，取結點，繪受力圖，列平衡方程列平衡方程qllEI2EIAX2X1Aq202ql402/qlM202ql402/qlM011dsEIMM022dsEIMMX1=1M1lX2=1M2l2、位移條件校核、位移條件校核The checking of displacement con

46、ditions對有封閉無鉸框格的剛架，當只承受荷載時，利用框格上任意對有封閉無鉸框格的剛架，當只承受荷載時，利用框格上任意截面處的相對角位移為零的條件，則：截面處的相對角位移為零的條件，則：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或繞框格或繞框格M/EI面積之和

47、為零面積之和為零 The sum of the area of M/EI diagrams for the members of closed frame equals to zerodsEIM圖M圖KMt 1t2t1t1t2t1t1t1t2t10022221211212111ttXXXX0021htltNiiit)(0基本結構由于溫度變化引基本結構由于溫度變化引起的沿起的沿Xi方向的位移。方向的位移。位移條件位移條件 displacement conditions由疊加原理，又可寫成：由疊加原理，又可寫成：。的物理意義與前面相同的物理意義與前面相同ij7-9 溫度變化時超靜定結構內力的計算溫

48、度變化時超靜定結構內力的計算 Internal forces duo to temperature changes t1t1t2t1X1X2基本體系基本體系primary structure圖乘。與jijiijMMdsEIMM解解 Solution：01111 tX 例例Example. 求圖示剛架由于溫度變化引起的內力與求圖示剛架由于溫度變化引起的內力與K點點的位移。的位移。 t1=+250C t2=+350C,EI=常數常數,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溫度改變引起的內力與各桿的絕對剛度溫度改變引起的內力與各桿的絕對剛度 EIEI 有關。有關。The internal forces due to temperature change is related to t

50、he absolute rigidity EI of the members?dEIsMMiKyMi溫度低的一側受拉溫度低的一側受拉。?01CX111CX1C1CX10101111CXCRiiC7-10 支座移動時超靜定結構的計算支座移動時超靜定結構的計算 Internal forces duo to support settlements 計算原理與荷載作用時的計算相同計算原理與荷載作用時的計算相同 The computation principle is identical to the case under loads由基本體系位移與原結構位移相同建立基本方程由基本體系位移與原結構位移相

51、同建立基本方程, 計算多于未知力。計算多于未知力。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本例為方向的位移引起的沿移為基本結構由于支座位XCC注：方程的右端項可能不為零，與基本結構的選擇有關。注：方程的右端項可能不為零，與基本結構的選擇有關。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：什么情況下方程右端不為零？思考：什么情況下方程右端不為零？?01CX111CX1C1CX10101111CXCRiiC)(11本例為方向的位移引起的沿移為基本結構由于支座位XCCC1Is the displacement in the direction of X1 induced by the settle

53、ment of support C 解解 Solution：例例Example. 求圖示梁由于支座移動引起的內力求圖示梁由于支座移動引起的內力. Find the internal forces induced by support settlementEIl1231121lClEI1X2X11X2/ lM10022221211212111CCXXXX002102112EIl22C212XM21216lEIX2211XMXMM12XM21lEIX2lEI4lEI2M支座移動引起的內力與各桿支座移動引起的內力與各桿的絕對剛度的絕對剛度 EI 有關。有關。練習練習: :寫出典型方程寫出典型方程,

54、,并求出并求出自由項。自由項。Exercise: Develop canonical equations and determine free termsaXXXXXXXXXCCC3333232131232322212113132121110 1 1C=b/l幾何法幾何法: 2 2C=-b/l 3 3C=0公式法公式法:1/l1/lCRiiC0lbblC/)/(11lbblC/)/(12001bCaXXXXXXXXXCCC3333232131232322212113132121110練習練習: :寫出典型方程寫出典型方程, ,并求并求出自由項。出自由項。Exercise: Develop ca

55、nonical equations and determine free termsCCCXXXaXXXbXXX333323213123232221211313212111 1 1C=0 2 2C=0 3 3C=01X3X2X1X3X2X000333323213123232221211313212111CCCXXXXXXXXX12X00111X10l00113X1CCCabllb3211)(支座移動時支座移動時, ,結構中的位移以及位移條件的校核公結構中的位移以及位移條件的校核公式如下式如下: :iiiiCiicREIsMMEIsMMdd制造誤差引起的內力計算制造誤差引起的內力計算: :1X3

56、X2XABAB桿造長了桿造長了1cm,1cm,如何作彎矩圖如何作彎矩圖? ?BA10m10m機動法作影響線機動法作影響線Construct influence lines by Mechanismic Method兩種方法：兩種方法：2 methods:2 methods:1 1、靜力法：按照力法求出影響線方程；、靜力法：按照力法求出影響線方程； Static method: determine influence lines by force method2 2、機動法：利用位移圖做影響線。、機動法：利用位移圖做影響線。Mechanismic Method: use displacement

57、 graph to construct influence lines1 1、靜力法、靜力法static method：3211112131111112363;30lxlxXEIxlxEIlXPPP求支座反力影響線求支座反力影響線construct IL of support reactions2、機動法、機動法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方方向的影響線向的影響線 X11 作用下在作用下在P=1方方向的影響線向的影響線超靜定結構的影響線問題化超靜定結構的影響線問題化為基本結構在固體荷載作用

58、為基本結構在固體荷載作用下的位移圖問題下的位移圖問題EIxlxP6321111設設則則11PX結果同前方法所得結果同前方法所得 由于由于P1P11 1向下為正，故上式負號表明向下為正，故上式負號表明X1X1影響線就是影響線就是X1=1X1=1作用下的桿的位移圖。作用下的桿的位移圖。 方法同靜定結構的機動法：方法同靜定結構的機動法：去掉所求未知力的相應約去掉所求未知力的相應約束后，體系沿著未知力方向產生單位位移，這時的體系位束后，體系沿著未知力方向產生單位位移，這時的體系位移圖就是該力影響線移圖就是該力影響線不同之處不同之處the differences： 靜定結構靜定結構去掉多余約束后變為機構

59、，體系去掉多余約束后變為機構，體系位移圖為位移圖為直線直線；Statically determinate structures transform into mechanisms after releasing redundant restraints, and the displacement graphs are straight lines. 而而超靜定結構超靜定結構去掉多余約束后認為幾何不去掉多余約束后認為幾何不變體系，體系位移圖為彈性撓曲線。變體系，體系位移圖為彈性撓曲線。 Statically indeterminate structures transform into geom

60、etrically stable systems after releasing redundant restraints, and the displacement graphs are elastic deflection lines.超靜定梁支座反力影響線超靜定梁支座反力影響線Reaction force IL of statically indeterminate beams彎矩、剪力影響線彎矩、剪力影響線IL of bending moment and shearing forces最不利荷載布置最不利荷載布置The most unfavorable position of load