能量法在超靜定系統(tǒng)中應(yīng)用

1、第二章第二章 能量法在超靜能量法在超靜 定系統(tǒng)中應(yīng)用定系統(tǒng)中應(yīng)用FFFFFFF1.1.找靜定基（基本系統(tǒng)）找靜定基（基本系統(tǒng)）2.2.將多余約束用支反力代替將多余約束用支反力代替3.3.加載荷（外載）得到相當系統(tǒng)加載荷（外載）得到相當系統(tǒng)4.4.變形與原結(jié)構(gòu)一致變形與原結(jié)構(gòu)一致FByF0ByBFBFByyyFABC2/ l2/ l2/FllFBy1l032216522211lllFllFlEIyByBFFBy165例題例題1 1ByF03221211aaaFaaaFaaFaEIyByByBFFBy83例題例題2 2F1aFByaFByaFaaABCaFa 已知平面剛架已知平面剛架的抗彎剛度的抗

2、彎剛度EIEI為常為常量，求解靜不定量，求解靜不定1.1.找靜定基（基本系統(tǒng)）找靜定基（基本系統(tǒng)）2.2.將多余約束用支反力代替將多余約束用支反力代替3.3.加載荷（外載）得到相當系統(tǒng)加載荷（外載）得到相當系統(tǒng)4.4.變形與原結(jié)構(gòu)一致變形與原結(jié)構(gòu)一致F1X01111FXFABC2/ l2/ l2/Fl1l例題例題3 301111FX11111XX1111FXEIllllEI332211311EIFlllFlEIF4856522211311653485331111FlEIEIFlXF例題例題4 4013132121111FXXXqaaqq1X2X3X111023232221212FXXX0333

3、32321313FXXX21123113322322qaaa1202221qxMMFF1x2x1x2xaMxM211111x2x222120 xMM1x2x112313MM例題例題4 4EIaaaaaEI343221132211EIaaaEI3322113222EIaaaEI2133EIaaaEI221322112EIaaaEI232112223113EIaaEI221223223EIqaaqaEIF62311431EIqaaqaEIF8432311432EIqaqaEIF612311333q22qa1a1a11202221qxMMFFaMxM21111222120 xMM112313MMF3

4、F1F22321938qaXaXaX2321312812qaXaXaX23211239qaXaXaX161qaX1672qaX 4823qaX 01313212111FXXX02323222121FXXX03333232131FXXX011212111FnnXXX022222121FnnXXX02211nFnnnnnXXXjiij對于高次超對于高次超靜定問題：靜定問題：例題例題5 5F已知平面桁架各桿的抗拉剛度已知平面桁架各桿的抗拉剛度EAEA為常量，求各桿軸力。為常量，求各桿軸力。Faa3124561X1XF1101111FX1111FX00F2F002/ 1112/ 12/ 12/ 1aa

5、a2aaa200Fa22/Fa00NjFNjFjljNjNjlFFjNjlF22/a2/aa22/a2/aa2例題例題5 500F2F002/ 1112/ 12/ 12/ 1aaa2aaa200Fa22/Fa00NjFNjFjljNjNjlFFjNjlF22/a2/aa22/a2/aa2EAalFEAjNj222161211EAFalFFEAjNjNjF2221611FXF42321111各桿軸力為各桿軸力為1XFFFNjNjNjFFFFN4231FFFN4232FFFN4223FFFN4214FFFN4235FFFN42236 如圖結(jié)構(gòu)受如圖結(jié)構(gòu)受F=10kN的外力作用的外力作用 。已知梁和

6、桿的材料相同，彈性模量。已知梁和桿的材料相同，彈性模量E =200GPa。梁為矩形截面，桿為圓形截面、尺寸如圖，且梁的跨度。梁為矩形截面，桿為圓形截面、尺寸如圖，且梁的跨度l=2m，試求梁的最大應(yīng)力試求梁的最大應(yīng)力443m1024112bhI242m1056.124dA解：解：結(jié)構(gòu)為一次靜不定，解除結(jié)構(gòu)為一次靜不定，解除D點約束，點約束，建立相當系統(tǒng)。建立相當系統(tǒng)。MPa24. 54Pa1024.451 . 005. 0210189. 06189. 066242maxmaxbhFlWMEIFlllFlF48565222131EAlEIl23311N3110311. 0/23/148/52111

7、1FAlIFXFF1X01111FXF2/Fl1l1例題例題6 6l2FABC2/ l2/ l4050100D443m1024112bhI242m1056.124dA解解2：結(jié)構(gòu)為一次靜不定，解除結(jié)構(gòu)為一次靜不定，解除B點約束，點約束，建立相當系統(tǒng)。建立相當系統(tǒng)。MPa24. 54Pa1024.451 . 005. 0210189. 06189. 066242maxmaxbhFlWMl2FABC2/ l2/ lEIFlllFlF48565222131EIl3311N3110311. 0/23/148/521FAlIFXF1XEAlXXF111112F2/Fl1X1l例題例題6 6laaEI=C

8、REI=CbbEI=Ch hFFaaFFaaMaaMaMaMMaaMaMaMFFF6060aaaqaFFaaFFaaFFaaFFaa力法正則方程力法正則方程 000333323213123232221211313212111FFFXXXXXXXXX 其中系數(shù)其中系數(shù)000232232112 F 力法正則方程簡化為力法正則方程簡化為 0033331311313111FFXXXX F(c)11(d)11(e)11M 圖1M 圖2M 圖3FX31XX2FXX321X(a)(b)FFF(c)11(d)11(e)11M 圖1M 圖2M 圖3FX31XX2FXX321X(a)(b)FFF(c)11(d)1

9、1(e)11M 圖1M 圖2M 圖3FX31XX2FXX321X(a)(b)FF結(jié)論：結(jié)論：qqFFMMAA對稱軸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封閉剛架受兩對集中力作用，試作結(jié)構(gòu)彎矩圖。封閉剛架受兩對集中力作用，試作結(jié)構(gòu)彎矩圖。力法正則方程力法正則方程01111 FX 單位載

11、荷法：單位載荷法：)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例題例題7 7車床夾具如圖所示，車床夾具如圖所示，EI EI 已知，求夾具已知，求夾具A A 截面上的

12、彎矩。截面上的彎矩。F6060AX12XD(a)(b)1X =1(c)X =12(d)(e)解一：解一：0022221211212111 FFXXXX 曲桿選單位力法曲桿選單位力法1),cos1(),cos3(sin2)60sin(21 MRMFRFRMF 各個系數(shù)如下：各個系數(shù)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)例題例題8 8RdRRdMEI 002222303311407

13、. 2)cos1)(cos3(sin2FRdFRRdMMEIFF 20332223)cos3(sin2FRdFRRdMMEIFF 代入正則方程得代入正則方程得0230407. 223221232213 FRRXXRFRXRXR 解得解得FRXFX0999. 0577. 021 （A A 截面軸力）截面軸力）（A A 截面彎矩）截面彎矩）解二：解二：相當系統(tǒng)關(guān)于載荷相當系統(tǒng)關(guān)于載荷F F 對稱對稱(f)2XX16060BX1X2FFXFXXXXF 30cos2, 0,12211FFX577. 031 問題簡化為一次超靜定問題，寫出彎矩方程問題簡化為一次超靜定問題，寫出彎矩方程1600),cos1

14、(3)cos1()(21 MFRRXM 22302230222332)cos1(323212FRFRdFREIRRdEIF 故故A A 截面彎矩截面彎矩FRXF0999. 02222 例題例題8 8解三：解三：B60(g)CMBMCNBNC60F2yxoAA(h)CCMF2 3MAF2FFF2由對稱性可得由對稱性可得MMMFFFCBNCNB ,32)cos1(32sin2)( FRMFRM在在C C 截面加單位力矩截面加單位力矩 ，截面為對稱面，轉(zhuǎn)角為零，截面為對稱面，轉(zhuǎn)角為零1 M043634)cos1(32sin2)(2223030 FRFRMFRRdFRMFRRdMMEI 所以所以FRM

15、C18879. 0 FRFRMFRMCA0999. 0)60cos1(60sin2 例題例題8 8 軋鋼機機架可以簡化為封閉式矩形剛架，設(shè)剛架梁與立軋鋼機機架可以簡化為封閉式矩形剛架，設(shè)剛架梁與立柱的抗彎剛度均為柱的抗彎剛度均為EIEI。試作剛架的彎矩圖，并求。試作剛架的彎矩圖，并求A A、B B間的相間的相對位移。對位移。解：解：1 1）作彎矩圖）作彎矩圖FABCDaaaCDAXXFFaX1X122FF/2F/2Fa/2結(jié)構(gòu)三次超靜定結(jié)構(gòu)三次超靜定; 2/, 011FXXFy 22, 0XXMA 正則方程為：正則方程為：02222 FX 2122214412222FaFaaEIaaEIF 8

16、2222FaXF FABCDaaaCDAXXFFaX1X122FF/2F/2Fa/2FABCDaaaCDAXXFFaX1X122FF/2F/2Fa/2例題例題9 902222 FX 412212212222FaFaaEIaaEIF 81122FaXF 作彎矩圖。作彎矩圖。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例題例題9 92 2）求）求AB 2452)2832221(3FaaaFaaFaaEIAB EIFaAB2453 Fa8Fa8Fa8Fa83Fa8DX2F/2F/2Fa/2FCFa/2FF/2DF/2F/2F/2Fa8Fa8a/2Fa8Fa8Fa81212DF/2Fa/2F/2Fa8a例題例題9 9 求圖示剛架求圖示剛架A A、B B 兩截面間水平方向的相對位移。剛架抗彎兩截面間水平方向的相對位移。剛架抗彎剛度均為剛度均為EIEI。軸力和剪力引起的變形忽略不計。軸力和剪力引起

18、的變形忽略不計。解：解：結(jié)構(gòu)二次超靜定結(jié)構(gòu)二次超靜定 結(jié)構(gòu)上下對稱，結(jié)構(gòu)上下對稱， A A 截面轉(zhuǎn)角為零。截面轉(zhuǎn)角為零。 B B 截面反對稱內(nèi)力剪力為零。截面反對稱內(nèi)力剪力為零。正則方程：正則方程：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例題例題1010試作剛架的彎矩圖，剛架抗彎剛度為試作剛架的彎矩圖，剛架抗彎剛度為EIEI。解：解：結(jié)構(gòu)一次超靜定結(jié)構(gòu)一次超靜定llCABX1MPFl2MX12X l12X l1(a)(b)(d)(c)F =X1Ax(e)F =Ay(f)F2F =AyFM圖1(g)2lM圖C1X(h)720320Fl320(i)MX12lFFF2Fl2F2FlFlFABC2l方法方法1 1：不用對稱性：不用對稱性取圖取圖(b)(b)示相當系統(tǒng)示相當系統(tǒng)正

20、則方程：正則方程：01111 FX 31311222234022222322221FllFllEIlllllllEIF 4031111FXF 根據(jù)平衡方程根據(jù)平衡方程 0, 0yAFM2FFFByAy 0 xF)(403FFFBxAxllCABX1MPFl2MX12X l12X l1(a)(b)(d)(c)F =X1Ax(e)F =Ay(f)F2F =AyFM圖1(g)2lM圖C1X(h)720320Fl320(i)MX12lFFF2Fl2F2FlFlFABC2lX1MFFl2M2l2l12l2lFFFABC2lllCAB例題例題1111llCABX1MPFl2MX12X l12X l1(a)

21、(b)(d)(c)F =X1Ax(e)F =Ay(f)F2F =AyFM圖1(g)2lM圖C1X(h)720320Fl320(i)MX12lFFF2Fl2F2FlFlFABC2l結(jié)構(gòu)對稱、載荷對稱，結(jié)構(gòu)對稱、載荷對稱，中面中面C C 處轉(zhuǎn)角為零。處轉(zhuǎn)角為零。解解2 2：根據(jù)對稱性根據(jù)對稱性根據(jù)對稱性可知：根據(jù)對稱性可知：力法正則方程：力法正則方程：01111 FX BxAxByAyFFFFF,2222232022232222132/1311FllFllEIlllllllEIF 根據(jù)平衡方程根據(jù)平衡方程 0 xF)(403FFFAxBx4031111FXF 例題例題1111X31XX2XX321

22、XM11111aalccMMMM1 1llaa1132力法正則方程力法正則方程 000333323213123232221211313212111FFFXXXXXXXXX 根據(jù)對稱性根據(jù)對稱性0, 0003132232112 FF 力法正則方程簡化為力法正則方程簡化為 0003331312222313111XXXXXF 故對稱面上的對稱內(nèi)力為零。故對稱面上的對稱內(nèi)力為零。X31XX2XX321XM11111aalccMMMM1 1llaa1132X31XX2XX321XM11111aalccMMMM1 1llaa113202222 FX 結(jié)論：結(jié)論：Fv=0對稱軸FqM=0sF =0NAAFq

23、FseMeM求解圖示超靜定剛架，并作的彎矩圖。求解圖示超靜定剛架，并作的彎矩圖。Fx1(e)FFFF3Fa74Fa7EIEIEIaa1X(a)(b)Fa(c)X =11a2(d)3Fa74Fa7解：解：結(jié)構(gòu)三次超靜定結(jié)構(gòu)三次超靜定結(jié)構(gòu)對稱、載荷反對稱，對稱面上對稱內(nèi)力為零。結(jié)構(gòu)對稱、載荷反對稱，對稱面上對稱內(nèi)力為零。正則方程為：正則方程為：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例題例題1212求圖示剛架，求圖示剛架，彎矩圖；彎矩圖； A A、B B 兩點間的相對位移。兩點間的相對位移。AqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 圖M 圖E解：解：結(jié)構(gòu)三次超靜定結(jié)構(gòu)三次超靜定 結(jié)構(gòu)對稱、載荷反對稱，結(jié)構(gòu)對稱、載荷反對稱，對稱面上對稱內(nèi)力為零。對稱面上對稱內(nèi)力為零。取相當系統(tǒng)為圖取相當系統(tǒng)為圖(b)(b)，則，則 0,

25、0yxFFqaFFss 21作彎矩圖為圖作彎矩圖為圖(c)(c)。 取靜定基如圖取靜定基如圖(d)(d)，并加一對，并加一對豎直單位力豎直單位力1 1。對于水平軸，對于水平軸，M M 圖反對稱，而圖反對稱，而 圖對稱，故圖對稱，故 。M0 AB AqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 圖M 圖EAqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 圖M 圖E例題例題1313討論討論1.1.例中沿兩條對角線是何種對稱？例中沿兩條對角線是何種對稱？2.2.利用此對稱性解體是否更簡捷？利用此對稱性解體是

26、否更簡捷？3.3.如果求如果求E E、F F 的相對位移，應(yīng)如何加的相對位移，應(yīng)如何加 單位力？單位力？AqDBCaaa(a)CAFS1FS2q(c)qa2qa2qa2qa2(b)11aa(d)aM 圖M 圖EF F 作圖示剛架的彎矩圖，并求鉸鏈作圖示剛架的彎矩圖，并求鉸鏈 A A 處相鄰截面的相對轉(zhuǎn)處相鄰截面的相對轉(zhuǎn)角（不計軸力和剪力引起的變形）。角（不計軸力和剪力引起的變形）。BAqFsBsAF(a)(b)(c)(d)12qa2212qa1aqABaa解：解：結(jié)構(gòu)和載荷分別關(guān)于結(jié)構(gòu)和載荷分別關(guān)于AB AB 軸對稱和反對稱，軸對稱和反對稱，對稱面上對稱內(nèi)力為零。對稱面上對稱內(nèi)力為零。 取相當

27、系統(tǒng)為圖取相當系統(tǒng)為圖(b)(b)，等效結(jié)構(gòu)關(guān)于水平，等效結(jié)構(gòu)關(guān)于水平軸載荷反對稱，所以軸載荷反對稱，所以qaFFsBsA21 取靜定基如圖取靜定基如圖(d)(d)，并加一單位力偶，并加一單位力偶1 1。對于水平軸，對于水平軸，M M 圖反對稱，而圖反對稱，而 圖對稱，圖對稱，故故 。M0 AB BAqFsBsAF(a)(b)(c)(d)12qa2212qa1aqABaaBAqFsBsAF(a)(b)(c)(d)12qa2212qa1aqABaa例題例題1414CBDAFAX1X1ABCDFFFFFFFFFX11XFX 45cos2121FX FX 45cos2121FX 求圖示結(jié)構(gòu)的內(nèi)力。求

28、圖示結(jié)構(gòu)的內(nèi)力。CBDAFAX1X1ABCDFFFFFFFFFX11XCBDAFAX1X1ABCDFFFFFFFFFX11XCBDAFAX1X1ABCDFFFFFFFFFX11X例題例題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)圖示剛架，抗彎剛度均為圖示剛架，抗彎剛度均為EIEI，F(xiàn) F = 80kN= 80kN。作彎矩圖。作彎矩圖。解：解：結(jié)構(gòu)三次超靜定結(jié)構(gòu)三次超靜定結(jié)構(gòu)對稱、載荷反對稱，對稱面上結(jié)構(gòu)對稱、載荷反對稱，對稱面上對稱內(nèi)力為零。對稱內(nèi)力為零。正則方程為：正則方程為

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)例題例題1616FABCDal2l2ExyzX1(a)(b)F2(c)(d)(e)X =11Fa2FFl41X =111l2(g)C(h)Ca1l2X1X1BDl2l2(i)(f)FFl4FF2 圓截面折桿，圓截面折桿，d d 2cm2cm，a a 0.2m0.2m

32、，l l 1m1m，F(xiàn) F 650N650N，E E 200GPa200GPa。求力。求力F F 作用點的垂直位移。作用點的垂直位移。解：解：正則方程為：正則方程為：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例題例題1717解：解：在在C C點加一垂直向下單位力點加一垂直向下單位力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 例題例題1717 圖示圓環(huán)，沿水平和垂直直徑各作用一對力圖示圓環(huán)，沿水平和垂直直徑各作用一對力F F ，求圓環(huán)，求圓環(huán)橫截面上的內(nèi)力。橫截面上的內(nèi)力。(d)FFFF2F2F2F2FF2FNFNFsFFaaX1NX1(a)1X(b)F2(c)X1( )M( )( )EI解：解：結(jié)構(gòu)三次超靜定結(jié)構(gòu)三次超靜定由平衡方程得：由平衡方程得：2,11FFFXXNN 取相當系統(tǒng)為圖取相當系統(tǒng)為圖(c),(c),寫出內(nèi)力方程

35、：寫出內(nèi)力方程：1,sin2)cos1(2)( MaFaFM )12(2)sincos1(2,2202212011 FadFaEIaadEIFFXF)212(1111 （與假設(shè)方向相反）（與假設(shè)方向相反）(d)FFFF2F2F2F2FF2FNFNFsFFaaX1NX1(a)1X(b)F2(c)X1( )M( )( )EI例題例題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例題例題1818022123

36、22)322(2232232223113222222112211 bbbEIEIbaaaabaaaEIabbbaEIEIbbbbEI X1X23XCX1X2X3CAqBCqa22qa22(b)(c)(d)1X =1bbX =1(e)2aaaa(f)aaaX =13aq1xx23x(a)ABqbbaaCCAqB 圖示剛架幾何上以圖示剛架幾何上以C C 為對稱中心。證明截面為對稱中心。證明截面C C 上軸上軸力和剪力皆等于零。力和剪力皆等于零。解：解：結(jié)構(gòu)三次超靜定結(jié)構(gòu)三次超靜定力法正則方程力法正則方程 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例題例題1919X1X23XCX1X2X3CAqBCqa22qa22(b)(c)(d)1X =1bbX =1(e)2aaaa(f)aaaX =13aq1xx23x(a)ABqbbaaCCAqBbqaqabqaaqaEIbqaqaabqaaqaaEIbqa