The projection of the whole stress on the inclined plane ab is Xn and YN respectively along with the shaft of x and y From the PAB equilibrium term 2F X,ds=o, ldS+t,mdS can get Divide ds and get XN=lox +mt Same from2F, =0and get: YN=moy +1 x The positive stress o N on the inclined plane AB, from the projection can get N IX +m=l-o + o +2lmT The shearing strength TN on the inclined plane AB, from the projection can get IN=LYN-XN=Im(,-0)+(--mt 21
21 The projection of the whole stress on the inclined plane AB is XN and YN respectively along with the shaft of x and y.From the PAB equilibrium term can get: Fx = 0 X N dS = x ldS + yxmdS Divide and get: dS N x m yx X = l + Same from and get: Fy = 0 N y xy Y = m + l The positive stress on the inclined plane AB,from the projection can get: N N N N x y xy lX mY l m 2lm 2 2 = + = + + The shearing strength on the inclined plane AB,from the projection can get: N N N N y x m xy lY mX lm( ) (l ) 2 2 = − = − + −
平的签论 斜面AB上全应力沿轴及轴的投影分别为和。由PB 的平衡条件∑F=0可得: Xds=o lds+t mdS 除以dS即得: XN=lox +mt 同样由∑F=0得出: Nmo +lT 斜面AB上的正应力ON,由投影可得: O=X tmy=lo tm o +lmt 斜面AB上的剪应力N,由投影可得: IN=lY=MXN=Im(o-o)+(L-m)t
22 斜面AB上全应力沿x轴及y轴的投影分别为XN和YN。由PAB 的平衡条件 Fx = 0 可得： X N dS = x ldS + yxmdS 除以 dS 即得： N x m yx X = l + 同样由 Fy = 0 得出： N y xy Y = m + l 斜面AB上的正应力 N ，由投影可得： N N N x y xy lX mY l m 2lm 2 2 = + = + + 斜面AB上的剪应力 N ，由投影可得： N N N y x m xy lY mX lm( ) (l ) 2 2 = − = − + −
3. Principal stress If the shearing stress of some inclined plane through point P is equal to zero then the positive stress of that inclined plane calls a principal stress of point p,but that inclined plane calls the main plane of the stress at point P, and the normal direction of that inclined plane calls the main direction of the stress at point P 1. The size of the principal stress +O +T 2 2 2.The direction of the principal stress o, is in the perpendicularity with o for each other 23
23 3.Principal stress If the shearing stress of some inclined plane through point P is equal to zero,then the positive stress of that inclined plane calls a principal stress of point P,but that inclined plane calls the main plane of the stress at point P,and the normal direction of that inclined plane calls 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 is in the perpendicularity with for each other. 1 2
平的签论 、主应力 如果经过P点的某一斜面上的切应力等于零,则该斜面 上的正应力称为P点的一个主应力,而该斜面称为P点的一个 应力主面,该斜面的法线方向称为P点的一个应力主向。 1.主应力的大小 土 x + ay 2 2 2 2.主应力的方向 σ1与σ2互相垂直。 24
24 二、主应力 如果经过P点的某一斜面上的切应力等于零，则该斜面 上的正应力称为P点的一个主应力，而该斜面称为P点的一个 应力主面，该斜面的法线方向称为P点的一个应力主向。 1.主应力的大小 2 2 2 1 ) 2 ( 2 xy x y x y + − + = 2.主应力的方向 1 与 2 互相垂直
82-4 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 oody, such as Fig, 2-5 show. After the elastic body suffers force, the point P,_A, B move to the point p'、A'、 Respectively The positive strain at point P au l!+ (u+ ou dx)=u ou sdx Ox -=dx Here because of small deformation B v+--d PA for causing stretch and shrink from B the y direction displacement v is the au small quantity of a high rank and this small quantity may be omitted Fig 2-5
25 §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,such as Fig.2-5 show.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 一、The positive strain at point P x u dx dx u x u u x = − + = ( ) Here because of small deformation, PA for causing stretch and shrink from the y direction displacement v is the small quantity of a high rank and this small quantity may be omitted