1、2023-2-1632023-2-164第三章第三章 应力分析应力分析第一节第一节 柯西应力张量柯西应力张量第二节第二节 斜截面上应力分量斜截面上应力分量第三节第三节 应力张量坐标变换应力张量坐标变换第四节第四节 主应力和主方向主应力和主方向第五节第五节 八面体上的应力八面体上的应力第六节第六节 平衡微分方程平衡微分方程2023-2-165第一节第一节 柯西应力张量柯西应力张量一、柯西应力矢量一、柯西应力矢量v应力矢量定义应力矢量定义力矢量的分量力矢量的分量 dFj,微元面积微元面积 dA，外法线单位矢量的分量，外法线单位矢量的分量 niPdAdFjnidAdFt(n)ordAdFtjj)(n
2、2023-2-166v正六面体上的应力矢量正六面体上的应力矢量e1、e2、e3为沿坐标的单位矢量为沿坐标的单位矢量3132121111eeet)(e3232221212eeet)(e3332321313eeet)(ethat isjijeiet)(112233121321233231x1x2x3jje1jje2jje3ijejit)(2023-2-167v任意斜截面上的应力矢量任意斜截面上的应力矢量四面体为四面体为脱离体：脱离体：cos1dAdA 微元面积之微元面积之间的关系：间的关系：1ndAcos2dAdA 2ndAcos3dAdA 3ndAiindAdA2023-2-168dAt)(1n
3、331221111)(1nnntnjjiint)(niindAdA在在x1方向平衡：方向平衡：0331221111)(1ndAndAndAdAtn消去消去dA，得，得同理；在同理；在x2、x3方向平衡：方向平衡：jjn1jjnt2)n(2jjnt3)n(3111dA221dA331dA02023-2-169二、柯西应力张量二、柯西应力张量v应力张量应力张量 Since ti(n)and ni denote vectors,it follows from the quotient rule of Chapter 2 that the components ji are components of
4、 a second-order Cartesian tensor.This tensor is called Cauchy stress tensor.jjiint)(n2023-2-1610 Where 11,22,33 are normal stresses,and 12,13,21,23,31,32 are shear stresses.According to the theorem of conjugate shear stresses in Strength of Materials,we have333231232221131211jiijv二阶应力张量的矩阵表示二阶应力张量的矩
5、阵表示2023-2-1611三、应力张量分解三、应力张量分解 ijijkkijs31 The first term in the right-hand is called spherical stress tensor and the second term called deviator stress tensor.kkp31)(31332211)(31zyxv球形应力张量球形应力张量2023-2-1612ijijkkp31矩阵矩阵形式形式pppp000000Iv偏斜应力张量偏斜应力张量ijijijpsppp333231232221131211spppzyzzxyzyxyzxxyxijijk
6、kijs312023-2-1613第二节第二节 斜截面上应力分量斜截面上应力分量一、应力矢量的坐标分量一、应力矢量的坐标分量jjiint)(njijn313212111)(1nnntn323222121)(2nnntn333232131)(3nnntnnmlpzxxyxxnmlpyzyxyynmlpzyzzxz321333231232221131211nnnnmlzyzzxyzyxyzxxyxnt(n)2023-2-1614v总应力总应力t（total stress）)()(2nniittt)(kikjijnnkjikijnnv正应力正应力（normal stress）iiNnt)(njiij
7、nn)(2311323321221332322221121nnnnnnnnnzxyzxyzyxnlmnlmnml222222v剪应力剪应力（shear stress）22NNt二、应力矢量的法切分量二、应力矢量的法切分量2023-2-1615v例题例题3.130758075050805050v解解nt(n)321333231232221131211nnn2023-2-16162/15.05.030758075050805050(n)t7.180.286.106)()(2nniittt 222)7.18()0.28(6.10625.124978.111t2023-2-1617iiNnt)(n21
8、7.18210.28216.1061.2622NNt21.2625.124977.1082023-2-1618v例题例题3.2ppp000000v解解ijijpjijint)(njijnpipn)()()(2iiiipnpntttnniinnp22p2023-2-1619iiNnt)(niinpn)(iinpnp22NNt22pp 02023-2-1620三、球形张量的几何解释三、球形张量的几何解释v基本关系基本关系ijijp矩阵矩阵形式形式ppp000000任意斜截面上的应力大任意斜截面上的应力大小的平方小的平方)()()(2kikjijiinpnptttnniinnp22pjjiint)(
9、njijn或或2)n(3)n(3)n(2)(2)n(1)n(1pttttttn令令)n(1tx)n(2ty)n(3tz 2222pzyx球面方程球面方程2023-2-1621v应力球面应力球面,)n(1tx,)n(2ty)n(3tz 2222pzyx球面球面方程方程应力球张量应力球张量球形应力张量球形应力张量得名得名原因原因2023-2-1622第三节第三节 应力张量坐标变换应力张量坐标变换v新旧坐标系之间的关系新旧坐标系之间的关系x 3x 1x 2由第二章的定义，张量的分量满足坐标变换由第二章的定义，张量的分量满足坐标变换关系关系应力张量是二阶张量，应满足应力张量是二阶张量，应满足相应关系相
10、应关系2023-2-1623v应力分量之间的关系应力分量之间的关系pqjqipijaa333231232221131211332313322212312111333231232221131211332331232221131211aaaaaaaaaaaaaaaaaaTaa或或klljkiijaaaa T2023-2-1624v例题例题3.3v解：解：2023-2-1625333231232221131211332313322212312111333231232221131211332331232221131211aaaaaaaaaaaaaaaaaa1000cossin0sincos1000co
11、ssin0sincos333231232221131211333231231322122111231322122111cossincossincossinsincossincossincos1000cossin0sincos2023-2-1626333231231322122111231322122111cossincossincossinsincossincossincos1000cossin0sincos222122121111sincossincossincos2sinsincos122222112sincossin122222112233332sinsincos22xyyx2sinco
12、ssin22xyyxzz2023-2-1627cossincossinsincos2221222111122cos2sin2122211cossin233123sincos3231312cos2sin2xyyxsincoszxyzzcossinzxyzz333231231322122111231322122111cossincossincossinsincossincossincos1000cossin0sincos2023-2-16282sinsincos22xyyx2sincossin22xyyx2cos2sin2xyyx2sin2cos22xyyxyx2sin2cos22xyyxyx2c
13、os2sin2xyyx2023-2-1629第四节第四节 主应力和主方向主应力和主方向一、控制方程一、控制方程v主应力和主方向的概念主应力和主方向的概念iint)n(jijn可以证明，主应力是斜截面上正应力的极值可以证明，主应力是斜截面上正应力的极值2023-2-1630v控制方程控制方程0)(0)(0)(333232131323222121313212111nnnnnnnnnjijint)n(jjiint)(njijnjijjijnn0)(jijijn展开展开2023-2-1631032213III0ijij0333231232221131211其实，主应力就其实，主应力就是二阶应力张量是二
14、阶应力张量的主值，或应力的主值，或应力矩阵的特征值矩阵的特征值2023-2-1632First InvariantiiI1Second InvariantijijjjiiI21212231223212113333222211Third InvariantijkjiijkeIdet32132123323122223113123123322112332211zyx222zxyzxyxzzyyx2222xyzzxyyzxzxyzxyzyx试证明它们与坐标系无关试证明它们与坐标系无关2023-2-16333210000002023-2-16340)(0)(0)(33323213132322212131
15、3212111nnnnnnnnn1iinn1232221nnn2023-2-1635二、主应力的求解方法二、主应力的求解方法v三角函数法解一元三次方程三角函数法解一元三次方程131Ix 03qpxx21231IIp),0(3213131272IIIIq032213III2023-2-163633prrq2arccos3132pscossxi)120cos(sxj)240cos(sxk3/111Ix 3/122Ix 3/133Ix 3/1Ix2023-2-1637032213III32123III32122)(III22321III23)0(IIv迭代法求解主应力迭代法求解主应力2023-2-1
16、638Example 3.45131623242023-2-1639Solution.155643322111I2312232121133332222112I2223124556646021233231222231131231233221132I2222536143122564545131623242023-2-1640054601523,91,73.4227.13 0325321nnn032321nnn043321nnn2023-2-16411333232221232221nnnnnn3/1321nnnSo,2023-2-1642v例题例题3.57060060802002050v解解0322
17、13III2023-2-164370600608020020503322111I2312232121133332222112I222060)20(50)70()70(808050910021233231222231131231233221132I060)20(2)70(805043200060708050222)20()70(08060502023-2-164421231IIp260319100103003213131272IIIIq)432000()9100(603160272323400033pr3310300 201175rq2arccos312011752234000arccos315
18、816.0arccos3185.412023-2-164532ps31030022.117cossxi85.41cos2.1173.87)120cos(sxj)12085.41cos(2.1174.111)240cos(sxk)24085.41cos(2.1171.2411131Ix 60313.873.10712231Ix 201.241.4413331Ix 204.1114.912023-2-1646910043200023II47.4722321III47)0(910047432000476022)1(9100432000602246.43910046.4343200046.436022
19、)2(19.44910019.4443200019.446022)3(05.44910005.4443200005.446022)4(08.442023-2-1647910008.4443200008.446022)5(07.44910007.4443200007.446022)6(07.44)(1.44(2cbcbcb1.44)1.44()1.44(23032213III11.44Ib 21.44Ibc2023-2-164811.44Ib9.15601.44bIc1.4429801)9.15(1.44910002cb242cbb2)9801(4)9.15(9.15226.1989.154.9
20、13.1072023-2-16490)(3132121111nnn70600608020020500)(3232122121nnn1232221nnn0203.5721nn0603.2720321nnn1232221nnn,314.01n,900.02n305.03n2023-2-1650v课堂练习题课堂练习题307580750508050502023-2-1651三、剪应力的极值三、剪应力的极值v主坐标下斜截面上的剪应力主坐标下斜截面上的剪应力jijint)n(332211nnniii321000000313212111)(1nnntn11n323222121)(2nnntn22n33323
21、2131)(3nnntn33n)n()(2iitttn2121n2222n2323niiNnt)(n211n222n233n2023-2-1652222NNt1232221nnn21212nt2222n2323nN211n222n233n2233222211232322222121nnnnnn它是截面法线方向余弦它是截面法线方向余弦n1、n2、n3的函数的函数v斜截面上剪应力的极值斜截面上剪应力的极值剪应力表达式中方向余弦剪应力表达式中方向余弦n1、n2、n3并不独立，它们满足条件并不独立，它们满足条件或或01232221nnn2023-2-1653剪应力的极值属于数学上的条件极值问题，可用剪
22、应力的极值属于数学上的条件极值问题，可用拉格朗日乘子法求解。拉格朗日乘子法求解。)1(2322212nnnFN2233222211232322222121nnnnnn 拉格拉格朗日乘子朗日乘子)1(232221nnn由驻值条件由驻值条件0inF和约束条件得到方程组和约束条件得到方程组2023-2-16542233222211232322222121nnnnnnF)1(232221nnn求剪应力极值的方程组求剪应力极值的方程组0inF解方程组得到解方程组得到ni，带回剪应力公式，即可得，带回剪应力公式，即可得到剪应力的极值到剪应力的极值0)(22332222111211nnnn0)(223322
23、22112222nnnn0)(22332222113233nnnn1232221nnn2023-2-16552/1,2/1,0321nnn23223maxN（1）2/1,0,2/1321nnn23113maxN（2）0,2/1,2/121nn22112maxN（3）最大剪应力最大剪应力23113max极值剪应力极值剪应力所在平面与所在平面与主平面夹主平面夹45 课外练习：试证明斜截面上课外练习：试证明斜截面上正应力的极值就是主应力。正应力的极值就是主应力。2023-2-1656四、莫尔应力圆四、莫尔应力圆v圆的方程圆的方程233222211nnnN232322222121222nnntNN主坐
24、标下主坐标下1232221nnn联立解得联立解得)()(312123221NNNn)()(123221322NNNn)()(231322123NNNn2023-2-16570)()(312123221NNNn斜截面法线方向余弦应满足条件斜截面法线方向余弦应满足条件0)(232NNN即即232223222NN应力点在应力点在C1圆以外圆以外022n由由应力点在应力点在C2圆以内圆以内023n由由应力点在应力点在C3圆以外圆以外以斜截面上的正应力以斜截面上的正应力 N为横坐标，剪应力为横坐标，剪应力 N为纵坐标，应力点为纵坐标，应力点在阴影部分内在阴影部分内2023-2-1658v应力圆应力圆 2
25、023-2-1659v应力圆的应用应力圆的应用),(212112),(213223)(213113)(213113max2023-2-1660一、正八面体构成一、正八面体构成v正八面体的定义正八面体的定义v任意一个面上的法线方向余弦任意一个面上的法线方向余弦321nnn第五节第五节 八面体上的应力八面体上的应力1232221nnn1321n31321nnn2023-2-1661二、正八面体上任意一个二、正八面体上任意一个 面上的应力面上的应力v坐标系坐标系v应力矢量的分量应力矢量的分量jijint)(n3/13/13/10000003213/3/3/3213210000002023-2-166
26、2)()(2nniittt 232221333)(3123222123222131t)(31321)(8iintn131I)(31zyx2023-2-16632828t2321232221)(91)(31213232221)()()(31)(6)()()(31222222zxyzxyxzzyyx2216231II21322321232可解释第四强度理论！可解释第四强度理论！2023-2-1664第六节第六节 平衡微分方程平衡微分方程一、主矢为零（力平衡）vEquilibrium Equationx1x2x3xixifiniti(n)dAdVAtAid)(nViVf d00ddViAjjiVfA
27、n2023-2-1665AjjiAn d Hence,the force equilibrium condition becomes0,ijjif we haveVjjiVd,0d)(,VijjiVf0,ijijf2023-2-1666000333322311323322221121331221111fxxxfxxxfxxxv Expanding0,ijjif0 xzxyxxfzyx0yzyyxyfzyx0zzyzxzfzyx2023-2-1667二、主矩为零（力矩平衡）The moment of force system acting on the body about any select
28、ed point,say the origin,must vanish,0dd)(VAVAfxtxnThat is,0dd)(VkjijkAkjijkVfxeAtxenppkknt)n(x1x2x3xixifiniti(n)dAdV2023-2-16680ddVkjijkAppkjijkVfxeAnxe0dVjkijkVpkjpijkVedVe0dd)(VkjijkAkjijkVfxeAtxenppkknt)n(VxeVppkjijkd),(VxexeVppkjijkpkpjijkd)(,0d)(,VfxexeVkppkjijkpkpjijkjp02023-2-1669 Therefore,t
29、he moment of forces equilibrium equation becomes0jkijkekjjk kkikkikkijkijkeeee33221113131212iiee23232121iiee32323131iiee032132231231eeejkjk3223031231132132eeejkjk3113021321123123eeejkjk21122023-2-1670v例题例题3.6,qxyx,0y224yhcxyv解解0 xzxxyxfzyx:02 cyqy2qc 2023-2-1671,qxyx,0y2242yhqxyqlh/2qlh/2qh2/8qh2/80yyzyxyfzyx0zzyzzxfzyx
