Transcription of Deformation Processing - Rolling
1 Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton1 Deformation Processing -Rollingver. 1 Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton2 Overview Process Equipment Products Mechanical Analysis DefectsProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton3 ProcessProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton4 ProcessProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton5 ProcessProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton6 Ring RollingProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton7 EquipmentProf.
2 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton8 EquipmentProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton9 ProductsProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton10 Products Shapes I-beams, railroad tracks Sections door frames, gutters Flat plates Rings ScrewsProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton11 Products A greater volume of metal is rolled than processed by any other Ramesh Singh, Notes by Dr. Singh/ Dr. Colton12 Rolling Analysis Objectives Find distribution of roll pressure Calculate roll separation force ( Rolling force ) and torque Processing Limits Calculate Rolling powerProf.
3 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton13 Flat Rolling Analysis Consider Rolling of a flat plate in a 2-high Rolling millWidth of plate w is large plane strainhbhfRV0Vf(> V0) Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton14 Flat Rolling Analysis Friction plays a critical role in enabling Rolling cannot roll without friction; for Rolling to occur Reversal of frictional forces at neutral plane (NN)hbhfV0Vf(> V0) LEntry ZoneExit ZoneNNtan Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton15 Flat Rolling Analysishbhf NNdxpp p p x x+ d xStresses on Slab in Entry ZoneStresses on Slab in Exit Zonepp p p x x+ d xProf.
4 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton16 EquilibriumSimplifying and ignoring HOTs()() cossin2m =pRdhdx()()0cos2sin2= + +hdpRdpRdhhdxxx Appling equilibrium in x (top entry, bottom exit)Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton17 Simplifying Since << 1, then sin = , cos = 1 Plane strain, von Mises()() m =pRdhdx2flowflowxYYp = Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton18 Differentiating Substituting or()[]() m = pRdhYpdflow2() m = pRhYpYddflowflow21 Prof.
5 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton19 Differentiating()() m = + pRhYddYpYpddhYflowflowflowflow21 Rearranging, the variation Y with respect to is small compared to the variation p/ Y flowwith respect to so the second term is ignored() mhRYpYpddflowflow2= Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton20 Thickness() cos12 +=Rhhfor, after using a Taylor s series expansion, for small 2 +=RhhfL!4!21cos42 + =0from the definitionof a circular segmentRR 2fbhh L2bh2fhProf. Ramesh Singh, Notes by Dr.
6 Singh/ Dr. Colton21 Substituting and integratingChRhRRhYpffflntan2lnln1+ = m() dRhRYpYpdfflowflow += m22In[1]:= 2RH Lhf+R 2 Out[1]=2R ArcTanBR hfFhfR+LogAhf+R 2E2 RProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton22 Eliminating ln()()HRhYCpflow mexp = = ffhRhRH 1tan2 Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton23 Entry region at = , H = Hb, ()HRhYCpflow =exp()bbHhRC exp=()[]()HHhhYpbbxbflow = exp = ffhRhRH 1tan2 = ffbhRhRH 1tan2[]()HHhhYpbbflow = expWith back tension=(Y flow xb)Prof.
7 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton24 Exit regionat = 0, H = Hf=0, fhRC=()()HhhYpfxfflow exp = = ffhRhRH 1tan2()()HhhYpfflow exp =With forward tensionProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton25 Effect of back and front tensionYxbY xfY Ymaximum pressurepressuredistanceProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton26 Flat Rolling Analysis Results without front and back tensionpp p p x x+ d xStresses on Slab in Entry ZoneStresses on Slab in Exit Zonepp p p x x+ d xUsing slab analysis we can derive roll pressure distributions for the entry and exit zones as: h0 and hbare the same thing12tanffRRHhh = ()0023 HHfhpYeh =23 HffhpYeh =Entry ZoneExit Zone Ramesh Singh, Notes by Dr.
8 Singh/ Dr. Colton27 Average Rolling pressure per unit width = =nnRdpRpRdpRpexitnexitaveentrynentryave 0,,1;)(1 Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton28 Rolling force F = pave,entryx Areaentry+ pave,exitx AreaexitProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton29 Force An alternative method again, very difficult to do. + =nndRpwdRpwFexitentry 0 Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton30 Force - approximationF / roller = L w pave h = hb-hfhRL =LhfpaveaveProf.
9 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton31 Derivation of L () cos12 +=RhhfL!4!21cos42 + =02 +=RhhfLR= circular segmentTaylor s expansionRR 2fbhh L2bh2fhProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton32 Derivation of L setting h = hbat = , substituting, and rearranging2 = = RLRhhhfborhRL =Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton33 Approximation based on forgingplane strain von Mises + = average flow stress:due to shape of elementProf. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton34 Small rolls or small reductions friction is not significant ( -> 0)1>>= LhaveflowaveYp = + = 0 Prof.
10 Ramesh Singh, Notes by Dr. Singh/ Dr. Colton35 Large rolls or large reductions Friction is significant (forging approximation)1<< Lhave + = Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton36 Force approximation: low friction1>> LhaveflowYLwrollerF = Ramesh Singh, Notes by Dr. Singh/ Dr. Colton37 Force approximation: high friction1<< Lhave + = Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton38 Zero slip (neutral) point Entrance: material is pulled into the nip roller is moving faster than material Exit: material is pulled back into nip roller is moving slower than materialmaterialpull-invbvfvRvRvRmateria lpull-backProf.
