【正文】 N’ dx dy u v Components of displacement and strain Plane strain state: for any given point on a member, displacements and strains occur on the same plane. In practical problems, the maximum strain usually occurs on the surface of the member and it is easy to obtain the surface strain. x o y dxxuu ???dxxv??dyyu??dyyvv ???167。 78 廣義虎克定律 在單向受拉（壓）時(shí)，若材料處于彈性階段。我們有公式 ?????????GE E???????在剪切時(shí)，我們有下面公式 定理：在小變形中，材料的線應(yīng)變只能產(chǎn)生正應(yīng)力。 剪應(yīng)力只能產(chǎn)生剪應(yīng)變。 證明：用反證法 若在正應(yīng)力作用下，材料發(fā)生 剪應(yīng)變。它繞 x軸轉(zhuǎn)動(dòng) 180。原來(lái)向外 變形現(xiàn)在向里。在同一力作用下，不 可能有二種變形形式。 同理可證明剪應(yīng)力只能產(chǎn)生剪應(yīng)變 generalized Hooke39。s law theorem Under the condition of small deformation, line strain of materials just results in normal stress, while shear strain corresponds to shear stress. Under the condition of uniaxial tension, materials deform elastically, we have the equation For shearing deformation verification apagoge Assume that under the action of normal stress, shear strain occurs in unit cube as shown in the figure. Turn the unit cube around x axis for 180 degree, then the original outward deformation is changed to inward deformation. It is impossible to exist different deformations under the action of same force. 在小變形情況下，線應(yīng)變可由各個(gè)方向的正應(yīng)力疊加而成 + + = )()]([)]([)]([,198111?????????????????????? E E EE E Eyxzzzxyyzyxxzxyxxx???????????????????????GGGyzyzxzxzxyxy?????????Under the condition of small deformation, line strain results from the superposition of normal stresses oriented from other orthogonal directions. 若單元體由三個(gè)主應(yīng)力產(chǎn)生。 )]([)]([)]([213331223211111????????????????????????EEE應(yīng)力通常由應(yīng)變的測(cè)量獲得。在平面應(yīng)變狀態(tài)分析中， 可導(dǎo)出和應(yīng)力相同的式子 ????????????????????????????????22222222222222222c o ss i nc o ss i ns i nc o ss i nc o sxyyxxyxxyyxyxxyxyx ???????????????????If unit cube is posed of three principal stresses, then the principal strains equal to Page 233, Eq. (, ) In the analysis of plane strain state, the same equations of strain can be derived as the stress. .,,xyxxyyxyyxyxxyxyyxxyxyx????????????????????????????求出后，代入平面虎克定律求出式。三向應(yīng)變值。代入上三向應(yīng)變片可測(cè)出????????????????222222229090045090450體積改變量 )1)(1)(1()1..1 zyxxdx d y dzVdxdxdx d y dzV?????????? 變成（受力后volume change Threeelement 45186。 rosette can measure the line strains along 0186。, 45186。 and 90186。, substitute the strains into the above equations, we have Plane Hooke’s law Page 235 3)21(33)21(3)218()(21)]([1)]([1)]([1)1()1)(1)(1(32132132121331232132111??????????????????????????????????????????????????????????????????????????mmzyxzyx EkkE E EE EVVVd x d yd z d x d yd zV體積改變量只能和平均主應(yīng)力有關(guān)。而和各個(gè)主應(yīng)力的比值無(wú)關(guān) 只要單元體的平均主應(yīng)力相同。它們的體積改變量相同。 The change in volume is just related to the average principal stress and has no relation with the ratio of each principal stress. Therefore if the average principal stresses (mean stress) is identical, the change in volume is the same. Bulk modulus Mean stress 材料超過(guò)彈性范圍。表示不能用虎克定律。若。小。體積就增大。反之就縮同號(hào)。當(dāng)物體受拉時(shí)，和。就一定。一定時(shí)。當(dāng)505000213..)(?????????????????kkEEkm例 75 剛性槽內(nèi)放入一立方體（長(zhǎng)為 1cm）的鋁塊。當(dāng)受壓力 p=6kN作用時(shí)。設(shè)它受力均勻。 E=70GPa。 =。 求它的三個(gè)主應(yīng)力的大小和相應(yīng) 的應(yīng)變。 y z x 分析： x方向沒(méi)有力 ,但有變形 . y方向應(yīng)變?yōu)榱恪５辛Υ嬖? Z方向受力均勻 ,有力又有變形 p ?bulk modulus Volumetric strain Materials exceeds elastic deformation range An aluminum cube with size 1cm is placed into a rigid slot, a force 6KN is applied to the cube. Assume the cube is stressed uniformly. Ask for the values of the three principal stresses and the corresponding strains. There are no deformation restriction in x direction and no deformation in y direction. y z x p 336943)]([110)(1070)]([1)60(0)]([16010106,0????????????????????????????????????yxzzzyxxzyzxyyzxE EM P aEM P aAp ????????????????????x z y p 1 2 M P aM P a M P a xEExApzEEzyxxzzxxzxzxxxzzyxzyxzzzyy00][1][10)(0)].([1)]([10212121222211112112122222111112121???????????????????????????????????????????????????????????方向：方向：方向：分析： 例 86 邊長(zhǎng)為 10 20 20mm的鋼塊 (1)和 鋁塊 (2)在 P力下同 步壓縮 . E1=200GPa. μ1=. E2=50GPa . μ2= P=6kN。 求主應(yīng)力。 Sizes of steel cube and aluminum cube are both 10 by 20 by 20mm and are subjected to a pressive force P simultaneously. Determine the principal stresses. 例 77 某一傳動(dòng)軸，直徑 d=40mm。 T= 377Nm。 P=。今測(cè)得 445445 1023210461 ??? ????? .,. ?? 求 E， G， ?p T 45 45 G P aEGG P aEEEEEM P aM P aM P aWTM P adpApxxxxxxnxx)1(2]2040[1][1]4020[1][140)90s i n ()90co s (2220301090s i n90co s223037716204454545445454545453232???????????????????????????????????????????????????????????????????????????，A driving shaft diameter torque x 例 78 圓柱形壓力容器，外直徑 D=3m t=24mm。 E=200GPa =。電測(cè) 軸向應(yīng)變 求內(nèi)壓力 , 外圓總伸長(zhǎng) , 厚度改變量 1?2?61050 ?????mmEpDEtttmmtEpDtpDtpDEDDSM P aDtEptE