能量法在超静定系统中应用

1、第二章第二章 能量法在超静能量法在超静 定系统中应用定系统中应用FFFFFFF1.1.找静定基（基本系统）找静定基（基本系统）2.2.将多余约束用支反力代替将多余约束用支反力代替3.3.加载荷（外载）得到相当系统加载荷（外载）得到相当系统4.4.变形与原结构一致变形与原结构一致FByF0ByBFBFByyyFABC2/ l2/ l2/FllFBy1l032216522211lllFllFlEIyByBFFBy165例题例题1 1ByF03221211aaaFaaaFaaFaEIyByByBFFBy83例题例题2 2F1aFByaFByaFaaABCaFa 已知平面刚架已知平面刚架的抗弯刚度的抗

2、弯刚度EIEI为常为常量，求解静不定量，求解静不定1.1.找静定基（基本系统）找静定基（基本系统）2.2.将多余约束用支反力代替将多余约束用支反力代替3.3.加载荷（外载）得到相当系统加载荷（外载）得到相当系统4.4.变形与原结构一致变形与原结构一致F1X01111FXFABC2/ l2/ l2/Fl1l例题例题3 301111FX11111XX1111FXEIllllEI332211311EIFlllFlEIF4856522211311653485331111FlEIEIFlXF例题例题4 4013132121111FXXXqaaqq1X2X3X111023232221212FXXX0333

3、32321313FXXX21123113322322qaaa1202221qxMMFF1x2x1x2xaMxM211111x2x222120 xMM1x2x112313MM例题例题4 4EIaaaaaEI343221132211EIaaaEI3322113222EIaaaEI2133EIaaaEI221322112EIaaaEI232112223113EIaaEI221223223EIqaaqaEIF62311431EIqaaqaEIF8432311432EIqaqaEIF612311333q22qa1a1a11202221qxMMFFaMxM21111222120 xMM112313MMF3

4、F1F22321938qaXaXaX2321312812qaXaXaX23211239qaXaXaX161qaX1672qaX 4823qaX 01313212111FXXX02323222121FXXX03333232131FXXX011212111FnnXXX022222121FnnXXX02211nFnnnnnXXXjiij对于高次超对于高次超静定问题：静定问题：例题例题5 5F已知平面桁架各杆的抗拉刚度已知平面桁架各杆的抗拉刚度EAEA为常量，求各杆轴力。为常量，求各杆轴力。Faa3124561X1XF1101111FX1111FX00F2F002/ 1112/ 12/ 12/ 1aa

5、a2aaa200Fa22/Fa00NjFNjFjljNjNjlFFjNjlF22/a2/aa22/a2/aa2例题例题5 500F2F002/ 1112/ 12/ 12/ 1aaa2aaa200Fa22/Fa00NjFNjFjljNjNjlFFjNjlF22/a2/aa22/a2/aa2EAalFEAjNj222161211EAFalFFEAjNjNjF2221611FXF42321111各杆轴力为各杆轴力为1XFFFNjNjNjFFFFN4231FFFN4232FFFN4223FFFN4214FFFN4235FFFN42236 如图结构受如图结构受F=10kN的外力作用的外力作用 。已知梁和

6、杆的材料相同，弹性模量。已知梁和杆的材料相同，弹性模量E =200GPa。梁为矩形截面，杆为圆形截面、尺寸如图，且梁的跨度。梁为矩形截面，杆为圆形截面、尺寸如图，且梁的跨度l=2m，试求梁的最大应力试求梁的最大应力443m1024112bhI242m1056.124dA解：解：结构为一次静不定，解除结构为一次静不定，解除D点约束，点约束，建立相当系统。建立相当系统。MPa24. 54Pa1024.451 . 005. 0210189. 06189. 066242maxmaxbhFlWMEIFlllFlF48565222131EAlEIl23311N3110311. 0/23/148/52111

7、1FAlIFXFF1X01111FXF2/Fl1l1例题例题6 6l2FABC2/ l2/ l4050100D443m1024112bhI242m1056.124dA解解2：结构为一次静不定，解除结构为一次静不定，解除B点约束，点约束，建立相当系统。建立相当系统。MPa24. 54Pa1024.451 . 005. 0210189. 06189. 066242maxmaxbhFlWMl2FABC2/ l2/ lEIFlllFlF48565222131EIl3311N3110311. 0/23/148/521FAlIFXF1XEAlXXF111112F2/Fl1X1l例题例题6 6laaEI=C

8、REI=CbbEI=Ch hFFaaFFaaMaaMaMaMMaaMaMaMFFF6060aaaqaFFaaFFaaFFaaFFaa力法正则方程力法正则方程 000333323213123232221211313212111FFFXXXXXXXXX 其中系数其中系数000232232112 F 力法正则方程简化为力法正则方程简化为 0033331311313111FFXXXX F(c)11(d)11(e)11M 图1M 图2M 图3FX31XX2FXX321X(a)(b)FFF(c)11(d)11(e)11M 图1M 图2M 图3FX31XX2FXX321X(a)(b)FFF(c)11(d)1

9、1(e)11M 图1M 图2M 图3FX31XX2FXX321X(a)(b)FF结论：结论：qqFFMMAA对称轴u=0=0MMF =0sFNee FABCDa2 a2 a2 a2 (a)(b)CDF/2ANX1X1CX1A(c)Gx1x2(d)Fa8(e)(f)FFFFNF/2F/2F/2AFa8Fa8Fa8Fa8Fa8F1XX1NFFNFAX1F/2F/2F/2F/2X1FABCDa2 a2 a2 a2 (a)(b)CDF/2ANX1X1CX1A(c)Gx1x2(d)Fa8(e)(f)FFFFNF/2F/2F/2AFa8Fa8Fa8Fa8Fa8F1XX1NFFNFAX1F/2F/2F/2F

10、/2X1FABCDa2 a2 a2 a2 (a)(b)CDF/2ANX1X1CX1A(c)Gx1x2(d)Fa8(e)(f)FFFFNF/2F/2F/2AFa8Fa8Fa8Fa8Fa8F1XX1NFFNFAX1F/2F/2F/2F/2X1FABCDa2 a2 a2 a2 (a)(b)CDF/2ANX1X1CX1A(c)Gx1x2(d)Fa8(e)(f)FFFFNF/2F/2F/2AFa8Fa8Fa8Fa8Fa8F1XX1NFFNFAX1F/2F/2F/2F/2X1封闭刚架受两对集中力作用，试作结构弯矩图。封闭刚架受两对集中力作用，试作结构弯矩图。力法正则方程力法正则方程01111 FX 单位载

11、荷法：单位载荷法：)20(, 1)(,2)(:1111axxMxFxMCG )20(, 1)(,42)(:2222axxMFaxFxMGA 故故,11220111 aadxEI 20202221118422aaFFadxFaFxdxFxEI所以所以81111FaXF FABCDa2 a2 a2 a2 (a)(b)CDF/2ANX1X1CX1A(c)Gx1x2(d)Fa8(e)(f)FFFFNF/2F/2F/2AFa8Fa8Fa8Fa8Fa8F1XX1NFFNFAX1F/2F/2F/2F/2X1例题例题7 7车床夹具如图所示，车床夹具如图所示，EI EI 已知，求夹具已知，求夹具A A 截面上的

12、弯矩。截面上的弯矩。F6060AX12XD(a)(b)1X =1(c)X =12(d)(e)解一：解一：0022221211212111 FFXXXX 曲杆选单位力法曲杆选单位力法1),cos1(),cos3(sin2)60sin(21 MRMFRFRMF 各个系数如下：各个系数如下：23)cos1(00232111RdRRdMEI 2120022112)cos1( EIRdRdsMMEI F6060AX12XD(a)(b)1X =1(c)X =12(d)(e)F6060AX12XD(a)(b)1X =1(c)X =12(d)(e)例题例题8 8RdRRdMEI 002222303311407

13、. 2)cos1)(cos3(sin2FRdFRRdMMEIFF 20332223)cos3(sin2FRdFRRdMMEIFF 代入正则方程得代入正则方程得0230407. 223221232213 FRRXXRFRXRXR 解得解得FRXFX0999. 0577. 021 （A A 截面轴力）截面轴力）（A A 截面弯矩）截面弯矩）解二：解二：相当系统关于载荷相当系统关于载荷F F 对称对称(f)2XX16060BX1X2FFXFXXXXF 30cos2, 0,12211FFX577. 031 问题简化为一次超静定问题，写出弯矩方程问题简化为一次超静定问题，写出弯矩方程1600),cos1

14、(3)cos1()(21 MFRRXM 22302230222332)cos1(323212FRFRdFREIRRdEIF 故故A A 截面弯矩截面弯矩FRXF0999. 02222 例题例题8 8解三：解三：B60(g)CMBMCNBNC60F2yxoAA(h)CCMF2 3MAF2FFF2由对称性可得由对称性可得MMMFFFCBNCNB ,32)cos1(32sin2)( FRMFRM在在C C 截面加单位力矩截面加单位力矩 ，截面为对称面，转角为零，截面为对称面，转角为零1 M043634)cos1(32sin2)(2223030 FRFRMFRRdFRMFRRdMMEI 所以所以FRM

15、C18879. 0 FRFRMFRMCA0999. 0)60cos1(60sin2 例题例题8 8 轧钢机机架可以简化为封闭式矩形刚架，设刚架梁与立轧钢机机架可以简化为封闭式矩形刚架，设刚架梁与立柱的抗弯刚度均为柱的抗弯刚度均为EIEI。试作刚架的弯矩图，并求。试作刚架的弯矩图，并求A A、B B间的相间的相对位移。对位移。解：解：1 1）作弯矩图）作弯矩图FABCDaaaCDAXXFFaX1X122FF/2F/2Fa/2结构三次超静定结构三次超静定; 2/, 011FXXFy 22, 0XXMA 正则方程为：正则方程为：02222 FX 2122214412222FaFaaEIaaEIF 8

16、2222FaXF FABCDaaaCDAXXFFaX1X122FF/2F/2Fa/2FABCDaaaCDAXXFFaX1X122FF/2F/2Fa/2例题例题9 902222 FX 412212212222FaFaaEIaaEIF 81122FaXF 作弯矩图。作弯矩图。2 2）求）求AB 4852482232221213FaaaFaaFaaEIAB EIFaAB2453 Fa8Fa8Fa8Fa83Fa8DX2F/2F/2Fa/2FCFa/2FF/2DF/2F/2F/2Fa8Fa8a/2Fa8Fa8Fa81212DF/2Fa/2F/2Fa8aFa8Fa8Fa8Fa83Fa8DX2F/2F/2F

17、a/2FCFa/2FF/2DF/2F/2F/2Fa8Fa8a/2Fa8Fa8Fa81212DF/2Fa/2F/2Fa8a例题例题9 92 2）求）求AB 2452)2832221(3FaaaFaaFaaEIAB EIFaAB2453 Fa8Fa8Fa8Fa83Fa8DX2F/2F/2Fa/2FCFa/2FF/2DF/2F/2F/2Fa8Fa8a/2Fa8Fa8Fa81212DF/2Fa/2F/2Fa8a例题例题9 9 求图示刚架求图示刚架A A、B B 两截面间水平方向的相对位移。刚架抗弯两截面间水平方向的相对位移。刚架抗弯刚度均为刚度均为EIEI。轴力和剪力引起的变形忽略不计。轴力和剪力引起

18、的变形忽略不计。解：解：结构二次超静定结构二次超静定 结构上下对称，结构上下对称， A A 截面转角为零。截面转角为零。 B B 截面反对称内力剪力为零。截面反对称内力剪力为零。正则方程：正则方程：01111 FX 617235232022232222131311FaaFaaaFaaEIaaaaaaaEIF FaaaaaaaaAB(a)BAX(b)1AB(c)FaAX =11B(d)2a2a( )Aaa1BaaFFF40171111FXF EIFaaaFaaaFaaFaaaFaaEIAB40112240172401722122113 FaaaaaaaaAB(a)BAX(b)1AB(c)FaAX

19、 =11B(d)2a2a( )Aaa1BaaFFFFaaaaaaaaAB(a)BAX(b)1AB(c)FaAX =11B(d)2a2a( )Aaa1BaaFFF例题例题1010试作刚架的弯矩图，刚架抗弯刚度为试作刚架的弯矩图，刚架抗弯刚度为EIEI。解：解：结构一次超静定结构一次超静定llCABX1MPFl2MX12X l12X l1(a)(b)(d)(c)F =X1Ax(e)F =Ay(f)F2F =AyFM图1(g)2lM图C1X(h)720320Fl320(i)MX12lFFF2Fl2F2FlFlFABC2l方法方法1 1：不用对称性：不用对称性取图取图(b)(b)示相当系统示相当系统正

20、则方程：正则方程：01111 FX 31311222234022222322221FllFllEIlllllllEIF 4031111FXF 根据平衡方程根据平衡方程 0, 0yAFM2FFFByAy 0 xF)(403FFFBxAxllCABX1MPFl2MX12X l12X l1(a)(b)(d)(c)F =X1Ax(e)F =Ay(f)F2F =AyFM图1(g)2lM图C1X(h)720320Fl320(i)MX12lFFF2Fl2F2FlFlFABC2lX1MFFl2M2l2l12l2lFFFABC2lllCAB例题例题1111llCABX1MPFl2MX12X l12X l1(a)

21、(b)(d)(c)F =X1Ax(e)F =Ay(f)F2F =AyFM图1(g)2lM图C1X(h)720320Fl320(i)MX12lFFF2Fl2F2FlFlFABC2l结构对称、载荷对称，结构对称、载荷对称，中面中面C C 处转角为零。处转角为零。解解2 2：根据对称性根据对称性根据对称性可知：根据对称性可知：力法正则方程：力法正则方程：01111 FX BxAxByAyFFFFF,2222232022232222132/1311FllFllEIlllllllEIF 根据平衡方程根据平衡方程 0 xF)(403FFFAxBx4031111FXF 例题例题1111X31XX2XX321

22、XM11111aalccMMMM1 1llaa1132力法正则方程力法正则方程 000333323213123232221211313212111FFFXXXXXXXXX 根据对称性根据对称性0, 0003132232112 FF 力法正则方程简化为力法正则方程简化为 0003331312222313111XXXXXF 故对称面上的对称内力为零。故对称面上的对称内力为零。X31XX2XX321XM11111aalccMMMM1 1llaa1132X31XX2XX321XM11111aalccMMMM1 1llaa113202222 FX 结论：结论：Fv=0对称轴FqM=0sF =0NAAFq

23、FseMeM求解图示超静定刚架，并作的弯矩图。求解图示超静定刚架，并作的弯矩图。Fx1(e)FFFF3Fa74Fa7EIEIEIaa1X(a)(b)Fa(c)X =11a2(d)3Fa74Fa7解：解：结构三次超静定结构三次超静定结构对称、载荷反对称，对称面上对称内力为零。结构对称、载荷反对称，对称面上对称内力为零。正则方程为：正则方程为：01111 FX 422124722232222121311FaaFaaEIaaaaaaaEIF 761111FXF Fx1(e)FFFF3Fa74Fa7EIEIEIaa1X(a)(b)Fa(c)X =11a2(d)3Fa74Fa7Fx1(e)FFFF3Fa

24、74Fa7EIEIEIaa1X(a)(b)Fa(c)X =11a2(d)3Fa74Fa7Fx1(e)FFFF3Fa74Fa7EIEIEIaa1X(a)(b)Fa(c)X =11a2(d)3Fa74Fa7例题例题1212求图示刚架，求图示刚架，弯矩图；弯矩图； A A、B B 两点间的相对位移。两点间的相对位移。AqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 图M 图E解：解：结构三次超静定结构三次超静定 结构对称、载荷反对称，结构对称、载荷反对称，对称面上对称内力为零。对称面上对称内力为零。取相当系统为图取相当系统为图(b)(b)，则，则 0,

25、0yxFFqaFFss 21作弯矩图为图作弯矩图为图(c)(c)。 取静定基如图取静定基如图(d)(d)，并加一对，并加一对竖直单位力竖直单位力1 1。对于水平轴，对于水平轴，M M 图反对称，而图反对称，而 图对称，故图对称，故 。M0 AB AqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 图M 图EAqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 图M 图E例题例题1313讨论讨论1.1.例中沿两条对角线是何种对称？例中沿两条对角线是何种对称？2.2.利用此对称性解体是否更简捷？利用此对称性解体是

26、否更简捷？3.3.如果求如果求E E、F F 的相对位移，应如何加的相对位移，应如何加 单位力？单位力？AqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 图M 图EF F 作图示刚架的弯矩图，并求铰链作图示刚架的弯矩图，并求铰链 A A 处相邻截面的相对转处相邻截面的相对转角（不计轴力和剪力引起的变形）。角（不计轴力和剪力引起的变形）。BAqFsBsAF(a)(b)(c)(d)12qa2212qa1aqABaa解：解：结构和载荷分别关于结构和载荷分别关于AB AB 轴对称和反对称，轴对称和反对称，对称面上对称内力为零。对称面上对称内力为零。 取相当

27、系统为图取相当系统为图(b)(b)，等效结构关于水平，等效结构关于水平轴载荷反对称，所以轴载荷反对称，所以qaFFsBsA21 取静定基如图取静定基如图(d)(d)，并加一单位力偶，并加一单位力偶1 1。对于水平轴，对于水平轴，M M 图反对称，而图反对称，而 图对称，图对称，故故 。M0 AB BAqFsBsAF(a)(b)(c)(d)12qa2212qa1aqABaaBAqFsBsAF(a)(b)(c)(d)12qa2212qa1aqABaa例题例题1414CBDAFAX1X1ABCDFFFFFFFFFX11XFX 45cos2121FX FX 45cos2121FX 求图示结构的内力。求

28、图示结构的内力。CBDAFAX1X1ABCDFFFFFFFFFX11XCBDAFAX1X1ABCDFFFFFFFFFX11XCBDAFAX1X1ABCDFFFFFFFFFX11X例题例题1515(c)qq/2q/2q/2=+F(a)CMalM2l(b)F2CM2laaM2l(c)aaCM2l=+aE(a)=(b)+(c)EEBCCBBCF2F2F2FF2F2F2F2(a)C(b)(c)qq/2q/2q/2=+F(a)CMalM2l(b)F2CM2laaM2l(c)aaCM2l=+aE(a)=(b)+(c)EEBCCBBCF2F2F2FF2F2F2F2(a)C(b)(c)qq/2q/2q/2=+

29、F(a)CMalM2l(b)F2CM2laaM2l(c)aaCM2l=+aE(a)=(b)+(c)EEBCCBBCF2F2F2FF2F2F2F2(a)C(b)(c)qq/2q/2q/2=+F(a)CMalM2l(b)F2CM2laaM2l(c)aaCM2l=+aE(a)=(b)+(c)EEBCCBBCF2F2F2FF2F2F2F2(a)C(b)图示刚架，抗弯刚度均为图示刚架，抗弯刚度均为EIEI，F F = 80kN= 80kN。作弯矩图。作弯矩图。解：解：结构三次超静定结构三次超静定结构对称、载荷反对称，对称面上结构对称、载荷反对称，对称面上对称内力为零。对称内力为零。正则方程为：正则方程为

30、：01111 FX mkN162042)2212(m31242)424(2)222322221(212311 FbaaFbaEIabbbabbbbEIF kN2 .391111 FXF(a)(b)(c)=+2FFa2Fa2(f)11b2b278.4kNm78.4kNm101.6kNm101.6kNmFF2F2F2F2b=4ma=4.5mF2(d)2FX1X12F(e)F(a)(b)(c)=+2FFa2Fa2(f)11b2b278.4kNm78.4kNm101.6kNm101.6kNmFF2F2F2F2b=4ma=4.5mF2(d)2FX1X12F(e)F(a)(b)(c)=+2FFa2Fa2(

31、f)11b2b278.4kNm78.4kNm101.6kNm101.6kNmFF2F2F2F2b=4ma=4.5mF2(d)2FX1X12F(e)F(a)(b)(c)=+2FFa2Fa2(f)11b2b278.4kNm78.4kNm101.6kNm101.6kNmFF2F2F2F2b=4ma=4.5mF2(d)2FX1X12F(e)例题例题1616FABCDal2l2ExyzX1(a)(b)F2(c)(d)(e)X =11Fa2FFl41X =111l2(g)C(h)Ca1l2X1X1BDl2l2(i)(f)FFl4FF2 圆截面折杆，圆截面折杆，d d 2cm2cm，a a 0.2m0.2m

32、，l l 1m1m，F F 650N650N，E E 200GPa200GPa。求力。求力F F 作用点的垂直位移。作用点的垂直位移。解：解：正则方程为：正则方程为：01111 FX ppFppGIFlaEIFlFlaGIFllEIGIaEIlaGIlEI416141142211211111212111 mNFlaEGlaEGlXF 3 .108)(821111 FABCDal2l2ExyzX1(a)(b)F2(c)(d)(e)X =11Fa2FFl41X =111l2(g)C(h)Ca1l2X1X1BDl2l2(i)(f)FFl4FF2FABCDal2l2ExyzX1(a)(b)F2(c)(

33、d)(e)X =11Fa2FFl41X =111l2(g)C(h)Ca1l2X1X1BDl2l2(i)(f)FFl4FF2FABCDal2l2ExyzX1(a)(b)F2(c)(d)(e)X =11Fa2FFl41X =111l2(g)C(h)Ca1l2X1X1BDl2l2(i)(f)FFl4FF2例题例题1717解：解：在在C C点加一垂直向下单位力点加一垂直向下单位力1 1。mlaXaFlGIlXFaFlEIlXalFlaGIlXlaFlalFllEIppC3122133111086. 4)28(1)8648(1)224(1)42322212324221(1 FABCDal2l2ExyzX

34、1(a)(b)F2(c)(d)(e)X =11Fa2FFl41X =111l2(g)C(h)Ca1l2X1X1BDl2l2(i)(f)FFl4FF2pDBFGIaFlaXX4211111 例题例题1717 图示圆环，沿水平和垂直直径各作用一对力图示圆环，沿水平和垂直直径各作用一对力F F ，求圆环，求圆环横截面上的内力。横截面上的内力。(d)FFFF2F2F2F2FF2FNFNFsFFaaX1NX1(a)1X(b)F2(c)X1( )M( )( )EI解：解：结构三次超静定结构三次超静定由平衡方程得：由平衡方程得：2,11FFFXXNN 取相当系统为图取相当系统为图(c),(c),写出内力方程

35、：写出内力方程：1,sin2)cos1(2)( MaFaFM )12(2)sincos1(2,2202212011 FadFaEIaadEIFFXF)212(1111 （与假设方向相反）（与假设方向相反）(d)FFFF2F2F2F2FF2FNFNFsFFaaX1NX1(a)1X(b)F2(c)X1( )M( )( )EI例题例题1818在任意截面上在任意截面上)2cossin2()(),cos(sin2)(),cos(sin2)( FaMFFFFsN(d)FFFF2F2F2F2FF2FNFNFsFFaaX1NX1(a)1X(b)F2(c)X1( )M( )( )EI例题例题1818022123

36、22)322(2232232223113222222112211 bbbEIEIbaaaabaaaEIabbbaEIEIbbbbEI X1X23XCX1X2X3CAqBCqa22qa22(b)(c)(d)1X =1bbX =1(e)2aaaa(f)aaaX =13aq1xx23x(a)ABqbbaaCCAqB 图示刚架几何上以图示刚架几何上以C C 为对称中心。证明截面为对称中心。证明截面C C 上轴上轴力和剪力皆等于零。力和剪力皆等于零。解：解：结构三次超静定结构三次超静定力法正则方程力法正则方程 000333323213123232221211313212111FFFXXXXXXXXX X

37、1X23XCX1X2X3CAqBCqa22qa22(b)(c)(d)1X =1bbX =1(e)2aaaa(f)1X =13q1xx23x(a)111ABqbbaaCCAqBX1X23XCX1X2X3CAqBCqa22qa22(b)(c)(d)1X =1bbX =1(e)2aaaa(f)1X =13q1xx23x(a)111ABqbbaaCCAqB例题例题1919X1X23XCX1X2X3CAqBCqa22qa22(b)(c)(d)1X =1bbX =1(e)2aaaa(f)aaaX =13aq1xx23x(a)ABqbbaaCCAqBbqaqabqaaqaEIbqaqaabqaaqaaEIbqa