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

POLARSOLUTIONSFORPLANEPROBLEMSAngle of rotation of radial line segment PA :ouedr)-ue(ug +ouarα =drOruAngle of rotation of hoop line segment PB:BrOugueThus shear strain is:r=α+β=OrrIf radial and loop displaces exist, from superposition method wehave:Ourar1 0ugur00rr1 ou,ougugYre+r a0arAThey are geometric functions in polar coordinates.11

If radial and loop displaces exist, from superposition method we have：          −   +   =   = +   = r u r u u r u r r u r u r r r r r           1 1 They are geometric functions in polar coordinates. Thus shear strain is: Angle of rotation of hoop line segment PB： r u r u r       −   = + = r u   = − Angle of rotation of radial line segment PA： r u dr dr u r u u   = −   + =      ( ) 11

POLARSOLUTIONSFORPTANEPROBLEMSII、Physical Equationsuoe(1) State of plane stress:E-uo)E12(1 + μ)2LreE(2) The situation of plane strain:Ea- for E and μin the above formulas,andSubstitute1urespectively.1-uLELuu-E一u2(1 + μ)YreTreE12

         + = − − − = − − − =                     r r r r r E E E 2(1 ) ) 1 ( 1 ) 1 ( 1 2 2 (2) The situation of plane strain: Substitute and for and in the above formulas, respectively. 2 E 1−  E    1− II、Physical Equations (1) State of plane stress：          + = = = − = −                 r r r r r r G E E E 1 2(1 ) ( ) 1 ( ) 1 12

POLARSOLUTIONSFORPLANEPROBLEMSS4-3 Stress Functions and CompatibilityEquations in Polar CoordinatesTo get the stresses and consistent equations denoted by stressfunctions in polar coordinates, the relationship between polarcoordinates and rectangular coordinates is used:r? = x + y2,3= arctan xx=rcos0, y=rsin0Then we have:ararxycosO,sin 0.Oxdyrra0a0cosasin 0xyOxayrr13

§4-3 Stress Functions and Compatibility Equations in Polar Coordinates To get the stresses and consistent equations denoted by stress functions in polar coordinates, the relationship between polar coordinates and rectangular coordinates is used：    cos , sin , arctan 2 2 2 x r y r x y r x y = = = + = Then we have： r r x r r y y x r y y r r x x r             cos , sin cos , sin , 2 2 = − = − = = = = = = 13

POLARSOLUTIONSFORPLANEPROBLEMS00dp arapsin papcosAaxarar ax0 ax00ra0op 0dp arcoso apsin e00ay ayOrar ayrapapsin ? 8 Qp2sin θcos0sin?8 a?p2sincos0ap0cos2A(a)r22002Q.2a2ar00aro0rrcos 0 dpapcos?0 a'p2sin0cos0ap2sin cos0 p(b)sin72r2.2002a2ar00aroorrcos?4-sin?8 ?p02sin cos ap0sincos6a.2araearaxayrr(c)cos?0-sin? apsin cos a*p12.200?a014

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2sin cos cos 2sin cos cos sin 2sin cos sin 2sin cos sin cos                                                 y r r r r r r r x r r r r r r r = + + − + = − + + + （a） （b）                                  +   = + = = + = − y y r r r y r x x r r r x r cos sin sin cos 2 2 2 2 2 2 2 2 2 2 2 2 cos sin sin cos cos sin sin cos sin cos                              −  −  − − − = + r r x y r r r r r （c） 14

POLARSOLUTIONSFORPLANEPROBLEMSWhen O-O, the components in polar coordinates equal to the ones inrectangular coordinates. Substituting these values into the equationsof stress components (constant body force) :a?paapax?apOxOyWe have:1 ag1 dpCr=(002r2arFa.2Odr00axaO15

We have： ) 1 ( ) ( ) ( ( ) ( ) 1 1 ( ) ( ) 0 2 0 2 2 2 0 2 0 2 2 2 0 2 2 0                                     x y r r x r y r r r r xy y r x = = − = − = = = = = = + = = = = = = When θ=0, the components in polar coordinates equal to the ones in rectangular coordinates. Substituting these values into the equations of stress components (constant body force) ： x y x y xy y x    = −   =   =       2 2 2 2 2 15