From 2M=, can find equal relationship of shear stress From∑F=0, gives: ao dr(r+dr)de-o, rd 8-(0+ de)dr 06 2 de oodr-+ (To.+ oLer de)dr-tor dr+K, rear=0 2 06 From∑F=0,gves: (a+ 0oed0dr-aodr+( e+rdr)(r+ dr)d8 06 ar de de -trode+(te+ ate do)dr=+Ta.dr+ Kordadr=0 06 2 dede Because deis very micro, has sins 2 2, coc de ≈1and, substitutes tre for te into upper two formulas, thus 11
11 From M 0 ,can find equal relationship of shear stress： θr r FromFr 0 ,gives: ( ) 0 2 2 ( )( ) ( ) θr θr d dr dr K rd dr d dr d dr r dr d rd d dr r r r r r r From F 0 ,gives: 0 2 2 ( ) ( ) ( )( ) K rd dr d dr d rd d dr dr r dr d r d dr dr r r r r r r Because is very micro,has , and, substitutes for into upper two formulas, thus: 2 2 sin d d 1 2 cos d r r d
由∑M=0,可以得出剪应力互等关系:z=0 由∑F=0,有: (o, + dr(r+dr)de-o, rde-oe +be de)dr de 06 2 de -60 2+(a+o de)dr-ta dr+K rder==0 66 由∑F=0,有: (oB+ede)dr-opdr+(te+e dr)(r+dr)dB 06 ar de trerdo+(te +bde)dr+te.dr+Kordedr=0 因为d0很微小,所以取sina≈ de 22 d≈1,并用τ代替t, 整理以上两式,得: 12
12 由 M 0，可以得出剪应力互等关系： 由 Fr 0，有： 由 F 0，有： 0 2 2 ( ) ( ) ( )( ) K rd dr d dr d rd d dr dr r dr d r d dr dr r r r r r r 因为 很微小，所以取 ， ，并用 代替 ， 整理以上两式，得： d 2 2 sin d d 1 2 cos d r r ( ) 0 2 2 ( )( ) ( ) θr θr d dr dr K rd dr d dr d dr r dr d rd d dr r r r r r r θr r
00. 1 at. o 0-0 B+K=0 ar r ae 、0+<xr+K=0 1 a ra0 a These are differential formulas of equilibrium in polar coordinates Two differential formulas of equilibrium contain three unknown functions o, and oe, tre=te, so it is a statically determinate question. thus must consider deformation condition and physical relationship Above formulas differ from equilibrium equations in planar coordinates where stress components are expressed by partial derivative In polar coordinates, areas of which unit element is perpendicular to two side faces are not equal, and difference is increasing with radius reducing which can be seen from underline items in the formulas 13
13 0 1 2 0 1 K r r r K r r r r r r r r r These are differential formulas of equilibrium in polar coordinates. Two differential formulas of equilibrium contain three unknown functions and , , so it is a statically determinate question.thus must consider deformation condition and physical relationship. r r r Above formulas differ from equilibrium equations in planar coordinates where stress components are expressed by partial derivative.In polar coordinates, areas of which unit element is perpendicular to two side faces are not equal,and difference is increasing with radius reducing ,which can be seen from underline items in the formulas
o.+-0⊥0,-0+K=0 r00 100×0 0。2 +=+K6=0 这就是极坐标中的平衡微分方程。 两个平衡微分方程中包含三个未知函数a、o和0=, 所以问题是静不定的。因此必须考虑变形条件和物理关系。 上述方程和直角坐标系下的平衡方程有所不同,直角坐 标系中,应力分量仅以偏导数的形式出现,在极坐标系中, 由于微元体垂直于半径的两面面积不等,而且半径愈小差值 愈大,这些反映在方程里带下划线的项中 141
14 0 1 2 0 1 K r r r K r r r r r r r r r 这就是极坐标中的平衡微分方程。 两个平衡微分方程中包含三个未知函数 、 和 ， 所以问题是静不定的。因此必须考虑变形条件和物理关系。 r r r 上述方程和直角坐标系下的平衡方程有所不同，直角坐 标系中，应力分量仅以偏导数的形式出现，在极坐标系中， 由于微元体垂直于半径的两面面积不等，而且半径愈小差值 愈大，这些反映在方程里带下划线的项中
84-2 Geometric and Physical Formulas in polar coordinates I Geometric Formulas--Differential Relationship between Displacements and Deformation n polar coordinates, stipulate dr radial normal strain hoop normal strain B B B rre---shear strain(change of right angle between radial and hoop y lines segment) Fig. 4-2 ur, ---radial displacement ue ---hoop displacement 15
15 I、Geometric Formulas—Differential Relationship between Displacements and Deformation §4-2 Geometric and Physical Formulas in Polar Coordinates In polar coordinates , stipulate: r r r u u ---radial normal strain ---hoop normal strain ---shear strain(change of right angle between radial and hoop lines segment) ---hoop displacement ---radial displacement Fig.4-2 d r dr r u o