Deformation Processing - Rolling

Transcription

1 Deformation Processing - olling ver. 1 1

2 Overview Process Equipment Products Mecanical Analysis Defects 2

3 Process 3

4 Process 4

5 Process 5

6 ing olling 6

7 Equipment 7

8 Equipment 8

9 Products 9

10 Sapes Products I-beams, railroad tracks Sections door frames, gutters Flat plates ings Screws 10

11 Products A greater volume of metal is rolled tan processed by any oter means. 11

12 Objectives olling Analysis Find distribution of roll pressure Calculate roll separation force ( rolling force ) and torque Processing Limits Calculate rolling power 12

13 Flat olling Analysis Consider rolling of a flat plate in a 2-ig rolling mill V 0 V f (> V 0 ) b θ f Widt of plate w is large plane strain 13

14 Flat olling Analysis α V 0 N V f (> V 0 ) b Entry Zone Exit Zone f L N Friction plays a critical role in enabling rolling cannot roll witout friction; for rolling to occur eversal of frictional forces at neutral plane (NN) µ tanα 14

15 Flat olling Analysis Stresses on Slab in Entry Zone p µp φ σ x + dσ x σ x N µp b f p dx N Stresses on Slab in Exit Zone p µp σ x + dσ x σ x µp p 15

16 Equilibrium Appling equilibrium in x (top entry, bottom exit) ( σ + dσ ) ( + d) 2 p dφ sinφ ± 2µ p dφ cosφ σ = 0 x x x Simplifying and ignoring HOTs ( σ ) d x dφ = 2 p m ( sinφ µ cosφ ) 16

17 Simplifying Since α << 1, ten sinφ = φ, cosφ = 1 ( σ ) d x = 2 p dφ ( φ m µ ) Plane strain, von Mises p σ = Y Y x flow flow 17

18 Differentiating Substituting or d [( ) ] p Y d flow φ = 2p ( φ m µ ) d p Y flow 1 = 2 p dφ Yflow ( φ m µ ) 18

19 19 Differentiating ( ) ( ) µ φ φ φ m = + p Y d d Y p Y p d d Y flow flow flow flow 2 1 earranging, te variation Y flow. wit respect to φ is small compared to te variation p/ Y flow wit respect to φ so te second term is ignored ( ) µ φ φ m Y p Y p d d flow flow 2 =

20 = + 2 f Tickness ( 1 cosφ ) or, after using a Taylor s series expansion, for small φ 2 = + φ f from te definition of a circular segment b 2 φ φ cosφ = 1 + L 2! 4! 2 0 φ L 4 b f 2 f 2 20

21 Substituting and integrating p d Y p Y flow flow = f 2 + φ 2 ( φ m µ ) dφ In[1]:= 2 Hφ µl f + φ 2 φ Out[1]= 2 µ ArcTanB φ F f f + LogAf + φ2 E 2 ln p Y f = ln m 2µ f tan 1 φ f + lnc 21

22 Eliminating ln() p = C Yflow exp m µ ( H ) H = 2 tan 1 φ f f 22

23 Entry region at φ = α, H = H b, p = C Y flow exp µ C = exp( µ Hb ) b ( H ) ( Y σ ) exp [ H H ] p = flow xb µ b p = Y flow exp µ ( ) b b ( [ H H ]) b Wit back tension=(y flow σ xb ) H b = 2 f tan 1 α f H = 2 tan 1 φ f f 23

24 Exit region at φ = 0, H = H f =0, C = p = f p = ( Y ) exp( H ) flow µ ( Y σ ) exp( H ) flow xf µ f f Wit forward tension H = 2 tan 1 φ f f 24

25 Effect of back and front tension pressure maximum pressure Y Y σ xb Y Y σ xf distance 25

26 Flat olling Analysis esults witout front and back tension Stresses on Slab in Entry Zone p µp Stresses on Slab in Exit Zone p µp σ x + dσ x σ x σ x + dσ x σ x µp p µp p Using slab analysis we can derive roll pressure distributions for te entry and exit zones as: 0 and b are te same ting 2 p= Yf e 3 0 Entry Zone ( H H ) µ 0 H = 2 tan f φ f 1 H 0 = = α 2 p = Yf e 3 Exit Zone f µ H 26

27 Average rolling pressure per unit widt p ave, entry = φ n n 1 1 p d p = entry φ; ave, exit ( α φn) φ α n 0 φ p exit dφ 27

28 olling force F = p ave,entry x Area entry + p ave,exit x Area exit 28

29 Force An alternative metod F = α w p entry dφ + φ n w p exit dφ φ n 0 again, very difficult to do. 29

30 Force - approximation F / roller = L w p ave L = b - f p ave = f ave L 30

31 Derivation of L circular segment φ = + 2 f Taylor s expansion ( 1 cosφ ) 2 φ cosφ = 1 + 2! 4 φ 4! 0 L b 2 L b f 2 f 2 2 = + φ f φ = L 31

32 Derivation of L setting = b at φ = α, substituting, and rearranging b f = = L 2 or L = 32

33 Approximation based on forging plane strain von Mises p ave µ L = 1.15 Y flow 1 + 2ave average flow stress: due to sape of element 33

34 Small rolls or small reductions = ave L >>1 friction is not significant (µ -> 0) p p ave ave µ L = 1.15 Y flow ave = Y flow 0 34

35 Large rolls or large reductions ave L <<1 Friction is significant (forging approximation) p ave µ L = 1.15 Y flow ave 35

36 Force approximation: low friction ave L >>1 F roller = LwY flow 36

37 Force approximation: ig friction ave L <<1 F roller µ L = 1.15 LwY flow ave 37

38 Zero slip (neutral) point Entrance: material is pulled into te nip roller is moving faster tan material Exit: material is pulled back into nip roller is moving slower tan material v b v f v v v material material pull-in pull-back 38

39 System equilibrium Frictional forces between roller and material must be in balance. or material will be torn apart Hence, te zero point must be were te two pressure equations are equal. b f ( µ H ) exp = b = exp µ exp ( 2µ H ) n ( ( H 2H )) b n 39

40 40 Neutral point = f b b n H H ln µ = H f n f n 2 tan φ

41 Torque L = b - f p ave A L/2 Fy = 0 F = roller p ave A T = α φ n wµ p 2 entry dφ φ n 0 F roller wµ p 2 exit dφ Torque / roller = r L Froller = Froller = 2 F roller 2 L 41

42 Power Power / roller = Tω = F roller Lω / 2 ω = 2πN N = [rev/min] 42

43 Processing limits Te material will be drawn into te nip if te orizontal component of te friction force (F f ) is larger, or at least equal to te opposing orizontal component of te normal force (F n ). F f cosα F sinα n F f α α α /2 F n F = µ f F n tanα = µ µ = friction coefficient 43

44 Processing limits Also and << cosα = 2 = 1 2 sinα = 2 1 cos α sinα = sinα tanα = 2 44

45 Processing limits So, approximately 2 ( tan α ) = µ 2 = Hence, maximum draft 2 max = µ Maximum angle of acceptance φ = α = max tan 1 µ 45

46 Processing Limits α 0 f Max. reduction in tickness 2 ( ) =µ max Max. angle of acceptance φ = α = max tan 1 µ 46

47 Cold rolling (below recrystallization point) strain ardening, plane strain von 2 τ flow = 1.15 Y Mises flow = 1.15 K ε n n + 1 average flow stress: due to sape of element 47

48 Hot rolling (above recrystallization point) strain rate effect, plane strain - von Mises Average strain rate ε& = 2τ flow ε t = V L ln flow b f = 1.15 Y = C & ε m average flow stress: due to sape of element 48

49 Example 1.1 Cold roll a 5% Sn-bronze Calculate force on roller Calculate power Plot pressure in nip (no back or forward tension) 49

50 Example 1.2 w = 10 mm b = 2 mm eigt reduction = 30% ( f = 0.7 b ) f = 1.4 mm = 75 mm v = 0.8 m/s mineral oil lubricant (µ = 0.1) K = 720 MPa, n =

51 Example 1.3 Maximum draft: max = µ 2 = (0.1) 2 75 = 0.75 mm actual = b - f = = 0.6 mm 51

52 Example 1.4 Maximum angle of acceptance φ max = tan -1 µ =tan -1 (0.1) = 0.1 radian α = = ( ) b f ( 2 1.4) rad = = o

53 Example 1.5 oller force: F = L w p ave L = ( ) 0.5 = [75 x (2-1.4)] 0.5 = 6.7 mm w = 10 mm ave = ( b + f ) / 2 = 1.7 mm ave / L = 1.7 / 6.7 = 0.25 < 1 friction is important F µ L = 1.15 LwY flow 1 + roller 2 ave 53

54 Example 1.6 ε f = f 1.4 ln = ln = b τ flow = 1.15 = 1.15 Y = 1.15 ( 0.36) n Kε f n + 1 = 354 MPa 54

55 Example 1.7 F = 1.15 roller = LwY flow = 28,392 N = 3.2 tons 1 + µ L ave 55

56 Example 1.8 Power ( kw ) roller = T ω = F L V 2 Power ( kw ) / roll = = 3 28, kW / roll = 1.35p 56

57 Example 1.9 Entrance p ( ' ) Y exp ( H H ) = flow σ xb µ b ( ) b Exit p = ( ' ) Y σ exp ( H ) flow xf µ f ( ) 57

58 Example 1.10 φ = ( ) f H = 2 tan 1 φ f f 58

59 Example 1.11 pressure / 2Tflow Friction ill Exit Entrance φ φ / φ max 59

60 olling Normal Stress Sear stress 60

61 61

62 Widening of material φ Side view Top view 62

63 esidual stresses - due to frictional constraints a) small rolls or small reduction (ignore friction) b) large rolls or large reduction (include friction) 63

64 Defects a) wavy edges roll deflection b) zipper cracks low ductility c) edge cracks barreling d) alligatoring piping, inomogeniety 64

65 oll deflection olls can deflect under load olls can be crowned 65

66 oll deflection olls can be stacked for stiffness 66

67 Metod to reduce roll deflection 67

68 Summary Process Equipment Products Mecanical Analysis Defects 68

69 69