《弹性力学》课程教学课件（英文讲稿）Chapter 4 Polar Solutions for Planar Problems

POLARSOLUTIONSFORPLANEPROBLEMSIt can prove that these stress components can satisfy equilibriumdifferential equations when body force is zero.p.pa.1a.1pFrom (a)+(b), we can get:Or2r2 002ax?a?r OrThus from compatibility equation in rectangular coordinates.a2p=0xwe can get the compatibility equation in polar coordinates:a21?1a0=0rorr2a0Using polar coordinates to evaluate planar problems (body forceis negligible), we can find the stress functionfrom the(r,0)compatibility equation. Then we can get the stress components. Thestress components are checked to satisfy the boundary conditionand also satisfy displacement single valued conditions if it ismultiply connectedbody16

Using polar coordinates to evaluate planar problems (body force is negligible), we can find the stress function from the compatibility equation. Then we can get the stress components. The stress components are checked to satisfy the boundary condition, and also satisfy displacement single valued conditions if it is multiply connected body. (r, ) Thus from compatibility equation in rectangular coordinates, we can get the compatibility equation in polar coordinates： 2 2 2 2 2 2 2 2 2 1 1             +   +   + = x y r r r r ( ) 0 2 2 2 2 2 +  =     x y ) 0 1 1 ( 2 2 2 2 2 2 =   +   +    r r r r  It can prove that these stress components can satisfy equilibrium differential equations when body force is zero. From (a)+(b), we can get： 16

POLARSOLUTIONSFORPLANEPROBLEMS4-4Coordinates Conversionof Stress ComponentsIn a certain stress situation,if has known stress components inpolar coordinates,stress components in planar coordinates are foundby using simple relationship equation. Vice versa.Assuming that stress components O,, e, Tre have been known, tryto determine the stress components x, , , Txy in plane coordinates.xFig. 4-4, fetching a tiny triangle A in10e9Tre.caelastic body, stresses of each side areBdenoted as the figure. The thickness ofathe triangle takes one unit.0CrVCgFig.4-417

§4-4 Coordinates Conversion of Stress Components In a certain stress situation,if has known stress components in polar coordinates,stress components in planar coordinates are found by using simple relationship equation.Vice versa. Assuming that stress components , , have been known, try to determine the stress components , , in plane coordinates.  r    r  x  y xy   r   r  y yx   r   r   r  r  xy   x c a b  o y x A B Fig.4-4 Fig. 4-4, fetching a tiny triangle A in elastic body, stresses of each side are denoted as the figure. The thickness of the triangle takes one unit. 17

POLARSOLTIONSFORPLANEPROBLEMSThe length of bc is ds, thus the lengths of ab and ac are dssin 0and ds cose, respectively.According to equilibrium condition of the triangle A, ZF, =0we can get the equilibrium equation:0,ds-o,ds cos? 0-0edssin ?+treds cos0sin 0+Teds sin 0cos0 = 0Substituting tro for Ter, we have:O,=0, cos?0+0, sin?0-2tre sin 0cos0In a similar way, from equilibrium condition ZF, = O, we have :Tx, =(o, -0。)sin cos0+tre(cos?0-sin ?0)Fetching other tiny triangle B , Fig.4-4, in terms of equilibriumcondition F, =O, we can get:O, =0, sin?0+0。cos?0+2tre sin 0cos018

The length of bc is ds, thus the lengths of ab and ac are and , respectively. cos sin sin cos 0 cos sin 2 2 + + = − −               ds ds ds ds ds r r x r Substituting for , we have:   r r    cos   sin  2  sin  cos 2 2 x = r + − r In a similar way, from equilibrium condition , we have Fy = 0 : ( )sin cos (cos sin ) 2 2  xy =  r −   + r  −  Fetching other tiny triangle B , Fig.4-4, in terms of equilibrium condition , we can get: Fy = 0   sin   cos  2  sin  cos 2 2 y = r + + r dscos dssin According to equilibrium condition of the triangle A, , we can get the equilibrium equation: Fx = 0 18

POLARSOLUTIONSFORPLANEPROBLEMSCombining above solutions, we can obtain the transformation ofthe stress components from polar coordinates to rectangularcoordinates:,=0,cos?0+0。sin?0-2tresin 0cos00,=0, sin ?0+0。 cos?0+2tre sin 0cos0[Txy =(0, -0)sin 0cos+Tre(cos?-sin ?0)Using simple triangle formula, the above formulas can beoverwritten as:o.+o., -0e cos20 - Tre sin 20a22,+o。, -0e cos20 + Tre sin 20R22r-0gsin 20 + Tre cos20.2A19

       = − + − = + + = + − ( )sin cos (cos sin ) sin cos 2 sin cos cos sin 2 sin cos 2 2 2 2 2 2                               xy r r y r r x r r Using simple triangle formula, the above formulas can be overwritten as:          + − = + − − + = − − + + =                               sin 2 cos 2 2 cos 2 sin 2 2 2 cos 2 sin 2 2 2 r r xy r r r y r r r x Combining above solutions, we can obtain the transformation of the stress components from polar coordinates to rectangular coordinates： 19

POLARSOLUTIONSFORPLANEPROBLEMSS 4-5Axisymmetric Stress andIts DisplacementIf stress components are only the function of the radius, such ascircular ring with inside and outside pressure, it is called the axialsymmetry problemUsing inverse solution method, we assume that the stress function only is the function ofradial coordinate r :β=p(r)Simplifying the compatibility equation, and we have:-d?1d=0dr2rdrThis is a four-order ordinary differential equation, whose generalsolution isβ= Aln r + Br? n r +Cr? + D20

§4-5 Axisymmetric Stress and Its Displacement If stress components are only the function of the radius, such as circular ring with inside and outside pressure, it is called the axial symmetry problem. Using inverse solution method, we assume that the stress function only is the function of radial coordinate :  r  = (r) Simplifying the compatibility equation, and we have: 0 d 1 d d d 2 2 2 =         +  r r r This is a four-order ordinary differential equation, whose general solution is: = A r + Br r +Cr + D 2 2  ln ln 20