《弹性力学》课程教学课件（英文讲稿）Chapter 8 The Solution of the Space Problem

The Solution of the Space Problemp(1+μ) P(1-2μ)ppzuR22元ERR+z1+μ) PY2(1u.R?2元ERP(1-2μ) R3p2Z70R32元R2R+zR(1-2μ) PNO2元R2RR+z3Pz3O2元R53Ppzwhere R? = p? +z2元R516

16 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 3 2 3 5 2 5 1 1 2 2 1 2 1 2 1 2 3 2 1 2 2 3 2 3 2 z z z P z u ER R R z P z u ER R P z R R R z R P z R R R R z Pz R P z R                         + −   = −     + +   = − +       − = −     + −   = −     + = = − 2 2 2 R z = +  ρ where

The Solution ofthe Space ProblemFor the problems solved in this section, the stress distributionhas the following characteristics:(1) When R→o0, the stress →0; When R→0, the stress →00.(2) The stress on the horizontal section . and t=p is independentof the elastic constant(3) The total stress on the horizontal section points to the operatingpoint O of force P. Any subsidence at any point on the boundaryF (1-μ2)(f)n元Ep17

(2) The stress on the horizontal section and is independent of the elastic constant. 17 For the problems solved in this section, the stress distribution has the following characteristics: ( ) ( ) 2 0 1 ( ) z z F η u f E = −  = =   (3) The total stress on the horizontal section points to the operating point O of force P. Any subsidence at any point on the boundary: (1) When R ∞, the stress 0; When R 0, the stress ∞.  z z 

The Solution ofthe Space ProblemIf the unit force is uniformly distributed on the rectangulard d y, and integratingarea a × b, substituting the F for d FbaE, y , then we can obtain the formula in the book.a2IF1-xKyPdyo62dsvy8-318

18 If the unit force is uniformly distributed on the rectangular area , substituting the F for , and integrating , then we can obtain the formula in the book. a  b y ba F d d 1 d =   , y

The Solution of the Space ProblemS 8-4 Solving the Space ProblemAccording to the StressAdd the second derivative of the second eguation with respect to zand the second derivative of the third equation with respect to y,and we get:a?a"sa2a3va3wOvOw0z?Qy?Qyoz?(OzOzOy?OyOzayauSubstitute the fourth equation into the8axgeometric equation, and we get:av020a"y.8oy(a)0z?ay?OyOzOwCOz19

19 §8-4 Solving the Space Problem According to the Stress Add the second derivative of the second equation with respect to z and the second derivative of the third equation with respect to y, and we get: 2 2 3 3 2 2 2 2 2 y z v w v w z y y z z y y z z y            + = + = +               Substitute the fourth equation into the geometric equation, and we get: 2 2 2 2 2 y yz z z y y z       + =     （a） x y z u x v y w z     =   =   = 