The Basic Theoryof the Plane ProblemThe projection of the whole stress on the inclined plane AB is Xand Y along the axis of x and y, respectively. From theequilibrium condition of the PAB ZF =o, we can get:XrdS =o,lds +twmdsXn=lo,+mt,By dividing dS we get:VSimilarly, from F, =0 we can get: Y = mo, +ltThe normal stress on the inclined plane AB is on . Byprojecting we can get:O~ =IX~ +mY = I'o, +mo, +2lmTyThe shearing stress on the inclined plane AB is t From theprojection we can get:T = lY-mX= lm(α, -αx)+(I? -m2)tx11
11 The projection of the whole stress on the inclined plane AB is XN and YN along the axis of x and y, respectively. From the equilibrium condition of the PAB Fx = 0 , we can get: X N dS = x ldS + yxmdS By dividing we get: dS N x m yx X = l + Similarly, from we can get: Fy = 0 N y xy Y = m + l The normal stress on the inclined plane AB is . By projecting we can get: N N N N x y xy lX mY l m 2lm 2 2 = + = + + The shearing stress on the inclined plane AB is . From the projection we can get: N N N N y x m xy lY mX lm( ) (l ) 2 2 = − = − + −
The BasicTheoryofthe PlaneProblem2.Principal stressIf the shearing stress of some inclined plane through point P isequal to zero, then the normal stress of that inclined plane iscalled a principal stress of point P, and that inclined plane iscalled the main plane of the stress at point P. And the normaldirection of that inclined plane is called the main direction of thestress at point P.(1). The size of the principal stressOr6+0O1172202(2). The direction of the principal stressO, is perpendicular to 02.12
12 2.Principal stress If the shearing stress of some inclined plane through point P is equal to zero, then the normal stress of that inclined plane is called a principal stress of point P, and that inclined plane is called the main plane of the stress at point P. And the normal direction of that inclined plane is called the main direction of the stress at point P. (1). The size of the principal stress 2 2 2 1 ) 2 ( 2 xy x y x y + − + = (2). The direction of the principal stress 1 is perpendicular to . 2
The Basic Theory of the Plane ProblemS2-4 Geometrical EquationThe Displacement of the Rigid BodyIn plane problem, every point inside the elastic body can produce the arbitrarilydirectional displacement. Take an unit PAB through any point P inside the elasticbody (Fig.2-5). After the elastic body suffers force, the point P,A,B move to thepoint P'、A'、B',respectively.1. The normal strain of point P0xOudxu+ouC(u+dbxouax8rOvPdxdxaxaxA'BOBecause of small deformation,the12diayB'stretch and shrink of PA from the ydirection displacement v is a high rankauhusmall quantity, and this small quantityayis omitted.Fig.2-513
13 §2-4 Geometrical Equation. The Displacement of the Rigid Body In plane problem, every point inside the elastic body can produce the arbitrarily directional displacement. Take an unit PAB through any point P inside the elastic body (Fig.2-5). After the elastic body suffers force, the point P,A,B move to the point P′ 、A′ 、B′,respectively. P o x y A B P A B u v dx x u u + dy y v v + dy y u u + dx x v v + Fig.2－5 1. The normal strain of point P x u dx dx u x u u x = − + = ( ) Because of small deformation, the stretch and shrink of PA from the y direction displacement v is a high rank small quantity, and this small quantity is omitted
The Basic Theory of the Plane ProblemOvIn a similar way, we can get:ay2.Shearing strain at point PavThe angle of rotation of the(v+dxOvaxline segment PA:α=dxaxThe same can get the angle of rotation of theline segment PB:OuB=ayOvauThus=α+axay14
14 In a similar way, we can get: y v y = 2.Shearing strain at point P y u x v xy + = + = The angle of rotation of the line segment PA: x v dx dx v x v v = − + = ( ) The same can get the angle of rotation of the line segment PB: y u = Thus
The Basic Theoryof the Plane ProblemTherefore, we can get the geometrical equation of the plane problemou8XOxOvayavau2十axayFromtheabovegeometrical equations,when thedisplacement components of the object is completely determined.the deformation components is completely determined, but thedisplacement components can not be determined thoroughlyA15
15 Therefore, we can get the geometrical equation of the plane problem: + = = = y u x v y v x u xy y x From the above geometrical equations, when the displacement components of the object is completely determined, the deformation components is completely determined, but the displacement components can not be determined thoroughly