1、6 Stress evaluation approach (slab method) (应力分析方法工程法),Assumption 1,Plane section remain plane.,Element cut from a plastic deformed material with sections parallel to the plane section.,Equilibrium equations,(1),Assumption 2,Take some special directions as the principal direction.,For example:,For p
2、lane strain:,Horizontal ( width ). ox,vertical ( thickness ). oy,y. c.,For axisymmetrical problem:,Axis of a work-piece oz,Radial direction . or,s1 s3 = 2k =Y,(2),(1) + (2),Stress boundary condition: for determining integration constants.,Assumption 3: Friction condition,1) Sliding friction, coulomb
3、 friction law: t =mp ( m = 0 , frictionless),2) Sticking friction (常摩擦应力区) (粘着摩擦),滑动摩擦, 遵守库仑摩擦定律(常摩擦系数区),t =k ( in hot working, rough tooling surface),3) Constant friction law （摩擦应力递减区）,t = mk (0m1),Assumption 4: Quasi- static deformation,The effect of the strain rate on the yield stress can be igno
4、red,6.2 Quasi- static plane strain compression between rigid parallel platens,6.2.1 Coulomb friction(库仑摩擦),1 Assumption,The vertical plane section remain plane.,2 Analysis of the stress acting on the element,Vertical sections,x: sx,x+dx: sx + dsx,Top interface and bottom interface,Shear stress txy=m
5、p,Normal pressure p= -sy,No shear stress acts on the sections,The normal stress is uniform on any sections,3 Equilibrium equation,In the ox direction,Assume that the ox and oy are principal directions, then sx = s1, sy = - p = s3,From yield criterion. (V-M. or Tr.),s1 s3 = K,ds1 = dsx = -dp,Therefor
6、e,After integration,sx + p = K,considering unit length of the block in the z direction,When,Determine integration constant C by means of stress boundary condition,At x = w,Therefore,For,At x = -w,Force exerted by the platens and mean platen pressure,6.2.2 Constant friction,Equilibrium equation,From
7、yield criterion,dsx = -dp,boundary condition,Yield criterion,When m=1, sticking friction,6.2.3 Transition between Coulomb friction and sticking friction,Coulomb friction txy = mp,txy attains k,sticking,A region in proximity to the vertical geometric axis of symmetry of the block is stiction.,At the
8、outer region of the block is Coulomb friction.,Unified equilibrium equation,or,When xd, t = mp k,When xd, t = k= K/2,When x=d, mp= K/2,Determine C,At x = w,When x=d, mp= k=K/2,When xd, t = k= K/2,When x=d, mp= K/2,6.3 Quasi- static plane strain drawing of sheet material through a wedge shaped die,1)
9、 Description of the process,Small semi-angle: a,t = m p (lubricate well),s longitudinal tensile stress from back tension,Problem : Drawing force t=?,2) Take element and analyze the stress,Assumption: vertical section remain plane.,Slab cut with vertical section,Normal stress: Uniform,Shear stress: n
10、o.,x , x + dx,Stress analysis,x , sx,x + dx sx+dsx,Top and bottom die-workpiece interface:,p , t = mp,Uniform on a section and there is no shear stress,Equilibrium equations. S x = 0 ( in horizontal direction),Neglect the second order of the term,The stresses in the vertical direction,Because m1 and
11、 a is very small.,Therefore,(A),3) The assumption of the principal direction,Principal direction,sx (- p) = K,(B),4) Drawing force at the exit,Let,Integrate (C),(D),Determine the integrate constant A by means of boundary condition .,At entrance,y = H,sx = s,(D),5) Maximum fractional reduction in a s
12、ingle pass （单道次最大压下率）,If t attains the yield stress in uniaxial tension (leaving the die), then the sheet yields. The limit condition is,When back tension is employed,If back tension is not employed,For example.,cota = 5.6713.,B = mcota = 0.2836.,(1+B ) / B = 4.53.,Therefore .,Hence .,Thus .,6) Norm
13、al pressure on the die face （模具表面的正压力）,At exit,7) Compression with inclined platens,a,h,H,t1,o,t2,w2,w1,Neutral plane,On the left side of the neutral plane,Yield criterion,sx (- p) = K,sx = K- p,dsx = - dp,s.b.c. x = -w2,sx = t2,p = K t2,a,h,H,t1,o,t2,w2,w1,Neutral plane,On the right side of the neu
14、tral plane,Yield criterion,sx (- p) = K,sx = K- p,dsx = - dp,s.b.c. x = w1,sx = t1,p = K t1,Flow stress depends on strain rate ( strain rate sensitive) High friction coefficient.,6.4 Quasi- static plane strain cold rolling of plate and strip metal,1) Description of the process,hot rolling,stiction,c
15、old rolling,Flow stress is independent of strain rate Low friction coefficient.,Coulomb law,section material,plate, strip,elongation - rolling direction,spread - transverse direction,compression- vertical direction,3- Dimensional flow,elongation,spread 2%,compression,2- Dimensional flow,plane strain
16、,( w/ T) 10,The velocity of strip (vs) in passing through the roll gap,at the entry,vs = v vr cos f.,vr : the linear velocity of rolls (circumference),vs increase,at neutral plane,vs = vr cos fn.,vs increase,at exit,vs vr cos f.,backward zone,work-piece moves backward when referring to the rolls,neu
17、tral plane,exit,forward zone,work-piece moves forward when referring to the rolls,The relative velocity between rolls and strip,Direction of friction stress,Towards to the neutral point,From the incompressibility,HbvH = hxbvx = hbvh,Roll flattening,Assume :,The arc of contact is circular and of radi
18、us R.,RR ( R is the radius of the un-deformed rolls),Strain hardening,Tension - tensile stresses,The relation between hx and fx,2) Distribution of normal roll pressure,Assume :,Vertical plane section remain plane.,Take a slab,No shear stress on the sections and normal pressure is uniform,Stress anal
19、ysis :,f f +df,Section f : s x Section f +df : sx +dsx,Vertical section,y (hx) y+dy (hx+dhx),Top and bottom interface,Pressure p: normal to the interface. Friction shear stress t=mp: tangential to the interface.,Entry plane,exit,p,mp,f,df,s x,sx +dsx,h,H,t2,t1,R,x,dx,p,mp,s x,sx +dsx,mp,p,In forward
20、 zone,In backward zone,p,mp,s x,sx +dsx,mp,p,y,dy/2,Equilibrium equations :,In forward zone,In backward zone,(A),(B),(C),T.Karman equation（卡尔曼方程）1925,The stresses in the vertical direction,Because,Therefore,The assumption of the principal direction,Rolling direction,ox,oy,Yield criterion,s1 s3 = 2k,
21、sx (- p) = 2k,sx = 2k- p,dsx = d(2k- p),(D),Vertical direction,Integrate,Let,The positive sign applies to the forward zone and the negative sign applies to the backward zone.,(E),Determine the integral constant according to the stress boundary condition,At exit,f = 0,Q = 0,sx = t2,p+= 2k2 t2,y = h,k
22、 = k2,From (E),Sub. (E),At entry,f = f1,Q = Q1,sx = t1,p -= 2k1 t1,y = H,k = k1,From (E),Sub. (E),Position of the neutral plane,At neutral plane,f = fn,Q = Qn,y = hn,p+= p,Since,6.5 M. D. Stone Cold Rolling Equation,Assumption :,1) Simplify it as plane platens compression,2) Neglect wide spreading,3
23、) Coulomb friction tf = mp,4) Plane section remain plane,In forward slip zone, the equilibrium equation is,In backward slip zone, the equilibrium equation is,Because tf = mp,sx ( p) = 2k,dsx = dp,Therefore,Forward zone “+” Backward zone “”,In forward slip zone,Determine the integration constant C1 b
24、y means of stress boundary condition,When x = - l /2,sx = sf,p = K- sx = K sf = K(1-sf /K),In backward slip zone,Determine the integration constant C2 by means of stress boundary condition,When x = l /2,sx = sb,p = K- sx = K sb = K(1-sb /K),Rolling force P,Let,6.6 compression of thick plate with pla
25、ne platens ( l / h 1),When without over end, the compression force increase with the increase of l/h.,When with over end,If l/h 1,The compression force decrease with the increase of l/h.,If l/h 1,The compression force is the same as that of without over end.,有外端,无外端,Assumption,On the interface of pl
26、atens and work-piece, the lubrication is very well, so that m = 0,tf = 0,On the interface between contact area and over end zone, txy = t e = k = K/2,Along x direction, txy is linear and in-dependent of y,From the equilibrium equations,(a),V-M yield criterion,(b),Differentiate the first equation of
27、(a) to y, and second equation to x.,(c),According to (b),(d),Sub. (d) into (c),(e),because txy is not related with y , therefore,Determine c1 and c2 according to the stress boundary condition,When x = 0, txy =0,c1 = 0,When x = l/2, txy = K/2,c2 = K/l,(f),(g),Sub. (g) into (a),(h),Integrate it,Sub. (
28、f), (h) into (d),Let,(i),Sub. (i), into (h) , and consider (f),Determine C,According to the condition of that the resulting force on the interface between over end and contacting area is equal zero, we can got that,When y = h/2, we got that on the contacting interface.,外端影响系数,6.7 Quasi- static axisy
29、mmetric upsetting of circular cylinder （轴对称圆柱体镦粗）,Description of the process,1) Coulomb friction at the interface, and m is lower.,2) h and a are the height and radius of the cylinder respectively.,3) Cylindrical coordinate.,Normal pressure distribution,Assumption,Cylindrical surface remain cylindri
30、cal.,Cylindrical surface : r.,Cylindrical surface : r + dr,Two meridian planes: dq is the angle between the two meridian planes.,On the cylindrical surface and meridian planes the normal stresses are uniform, and there are no shear stresses.,1) Take an element.,sr,sr + dsr,2) Analysis of stresses ex
31、erted on the element.,r,r+dr,meridian surface,h,p and mp,3) Equilibrium equation.,Sr = 0,sr,r dq h,sr +dsr,(r+dr) dq h,sq,dr h,r dq dr,Because dq is very small, therefore,Dividing the equilibrium equation by hdq , and neglect the second order of the term, we can get the following equation:,4) Yield
32、criterion.,(A),Take 0z, or and q as the principal directions. From the incompressibility law we have,Differentiate the above equation, we got,Because,Therefore,However,From L-M equation,therefore,Axisymmetrical stress state at any point of the cylinder,From v-M or Tr. Yield criterion,Y is the yield
33、stress in uniaxial stress state. (lode parameter md = -1 ),Because,(B),(C),Sub (B) and (C) into (A),(D),5) Determine A according to the prescribed stress boundary conditions.,At r = a.,From yield criterion.,Sub. Into (D).,Therefore,6) Force exerted by the platens and mean platen pressure.,r,dr,There
34、fore,6.8 Bar and wire drawing through a conical die with and without back tension,Description of the process,Small semi-angle of conical die a,Low friction coefficient m , t = m p,At entry r = R1 , tensile stress from back tension is S1,At exit r = R2 , drawing stress is S2,Assumption,All points on
35、a section normal to the drawing direction are in the same stress state, that is , axisymmetrical stress state.,Plane section normal to the drawing direction remain normal.,a,S1,o,R2,S2,R1,y,(At every point on a section),sz is uniform,There is no shear stress on the section plane.,Stress analysis,Neg
36、lect the second and higher order of the term,S z = 0,S r = 0,Yield criterion.,(A),(B),Let,(C),Determine A according to the boundary condition,At entry,Therefore,Hence,At exit,Therefore,6.9 Drawing and sinking of a thin-walled tube through a conical die,General consideration,Sinking of a thin-walled tube through a conical die 薄壁管减径拉拔,
