82-2 Differential Equation of Equilibrium Whether plane stress problem or plane strain problem, is the research problem in plane xy,all the physics quantity has nothing to do withz Discuss below the correlation between any point stress and volumetric force when the object is placed in the state of equilibrium, and lead an equilibrium differential equation from here. From the lamella shown in Fig. 2-1, we take out a small and positive parallelepiped PABC, and take for an unit length in the directional dimension in Establishing the function of the positive stress force in an unit on the left side ISO, =O,(x, y), the coordinate on the right side x gets the incrementer, the positive stress on Ox rdx the face is o,(+dx, y), spreading the formula above will be Taylors series dy ay (x+dx,y)=01(x,y) 001(x,y) dx t dy ao(x 1o1(x,y) Fig 2- (ax)2+ (dx) 2! ax n!. ax 11
11 §2-2 Differential Equation of Equilibrium Whether plane stress problem or plane strain problem, is the research problem in plane xy,all the physics quantity has nothing to do with z. Discuss below the correlation between any point stress and volumetric force when the object is placed in the state of equilibrium,and lead an equilibrium differential equation from here.From the lamella shown in Fig.2-1,we take out a small and positive parallelepiped PABC,and take for an unit length in the directional dimension in z. y o x y dy y y y + x dx x x x + xy dx x xy xy + yx dy y yx yx + P A B C X Y D Fig.2－3 (x, y) x = x dx Establishing the function of the positive stress force in an unit on the left side is ,the coordinate on the right side x gets the increment ,the positive stress on the face is ,spreading the formula above will be Taylor’s series: (x dx, y) x + n n x n x x x x dx x x y n dx x x y dx x x y x dx y x y ( ) ( , ) ! 1 ( ) ( , ) 2! 1 ( , ) ( , ) ( , ) 2 2 2 + + + + = +
平河的签论 §2-2平衡微分方程 无论平面应力问题还是平面应变问题,都是在y平面内研究问题, 所有物理量均与z无关。 下面讨论物体处于平衡状态时,各点应力及体力的相互关系,并 由此导出平衡微分方程。从图2-1所示的薄板取出一个微小的正平行 六面体PABC(图2-3),它在z方向的尺寸取为一个单位长度 设作用在单元体左侧面上的正 O 应力是ax=a2(x,y)右侧面上坐标x A 得到增量d,该面上的正应力为 dx x(x+dx,y),将上式展开为泰勒级 dx B 数: 01(x y +dy o (x+,y)=o (x,y)+=x,dx+ dy ay 10σ(x,y) 10:(x,y 图2-3 (ax)2 +∴ 2 ax n! ax 12
12 §2-2 平衡微分方程 无论平面应力问题还是平面应变问题，都是在xy平面内研究问题， 所有物理量均与z无关。 下面讨论物体处于平衡状态时，各点应力及体力的相互关系，并 由此导出平衡微分方程。从图2－1所示的薄板取出一个微小的正平行 六面体PABC(图2－3），它在z方向的尺寸取为一个单位长度。 y o x y dy y y y + x dx x x x + xy dx x xy xy + yx dy y yx yx + P A B C X Y D 图2－3 (x, y) x = x x dx 设作用在单元体左侧面上的正 应力是 ，右侧面上坐标 得到增量 ，该面上的正应力为 ，将上式展开为泰勒级 数： (x dx, y) x + n n x n x x x x dx x x y n dx x x y dx x x y x dx y x y ( ) ( , ) ! 1 ( ) ( , ) 2! 1 ( , ) ( , ) ( , ) 2 2 2 + + + + = +
After omitting small quantity of the two rank and above the two rank, can get a oo, (x) ax, at the same time, ay,Ty, r, are get the state of stress from the rawing show. While considering the volumetric force to the plane stress state still prove mutual and equal theory of shearing strength. Regard the center d and straight line in parallel with the shaft of z as the moment shaft, list the equilibrium equation of the moment shaft >MD=0 ax)ahy×1×+rxdy dx dx 2 (n+xd1×一,1x=0 The both sides of the formula above divide dxdy get 1 OT 17, N x 2 Ox 2 ay Cause x>0, dy ->0 Omitting small quantity isnt accounted can get xy 13
13 After omitting small quantity of the two rank and above the two rank,can get ,at the same time, , , are get the state of stress from the drawing show. dx x x y x y x x + ( , ) ( , ) y xy yx While considering the volumetric force to the plane stress state,still prove mutual and equal theory of shearing strength.Regard the center D and straight line in parallel with the shaft of z as the moment shaft, list the equilibrium equation of the moment shaft : MD = 0 0 2 1 2 ( ) 1 2 1 2 ( ) 1 − = − + + + dy dx dy dy dx y dx dy dx dx dy x yx yx yx xy xy xy The both sides of the formula above divide get: dxdy dy y dx x yx yx xy xy = + + 2 1 2 1 Cause dx → 0,dy → 0 ,Omitting small quantity isn’t accounted,can get: xy yx =
平河的签论 略去二阶及二阶以上的微量后便得σ(xy)+如(x卫同样叮、可、x 都一样处理,得到图示应力状态。 对平面应力状态考虑体力时,仍可证明剪应力互等定理。以通过中 心并平行于z轴的直线为矩轴,列出力矩的平衡方程∑M=0: (1+d×1x+×1x ax 2 2 -(x+dh)×1 2z,ax×1× 0 2 将上式的两边除以得到 1 0T 1at 7+ dx=t+ 2 OX 20 令→0→0,即略去微量不计,得:7x,=飞y 14
14 略去二阶及二阶以上的微量后便得 同样 、 、 都一样处理，得到图示应力状态。 dx x x y x y x x + ( , ) ( , ) y xy yx 对平面应力状态考虑体力时，仍可证明剪应力互等定理。以通过中 心D并平行于z轴的直线为矩轴，列出力矩的平衡方程 MD = 0 ： 0 2 1 2 ( ) 1 2 1 2 ( ) 1 − = − + + + dy dx dy dy dx y dx dy dx dx dy x yx yx yx xy xy xy 将上式的两边除以 dxdy 得到： dy y dx x yx yx xy xy = + + 2 1 2 1 令 dx → 0, dy → 0 ，即略去微量不计，得： xy yx =
Deduce the equilibrium differential equation of the plane stress problem below, list the equilibrium equation to the unit ∑ F =0 (ax+d)b×1-ax如×1+(x+dhy)d×1 ax Ov r·x×1+X,dxdy×1=0 ∑1 F.=0: 07 (a,+dy)x×1-a,x×1+(r+ax)dy×1 ax rndy×1+Y·dx:cy×1=0 15
15 Deduce the equilibrium differential equation of the plane stress problem below,list the equilibrium equation to the unit: 1 1 0 ( ) 1 1 ( ) 1 0 : − + = − + + + = dx X dx dy dy dx y dx dy dy x F yx yx x yx x x x 1 1 0 ( ) 1 1 ( ) 1 0 : − + = − + + + = dy Y dx dy dx dy x dy dx dx y F xy xy y xy y y y