Fundamental cncept metal rlling Assumptins 1) Te arc cntact between te rlls and te metal is a part a circle. v x x α L p y y R v 2) Te ceicient rictin, µ, is cnstant in tery, but in reality µ varies alng te arc cntact. 3) Te metal is cnsidered t derm plastically during rlling. 4) Te vlume metal is cnstant bere and ater rlling. In practical te vlume migt decrease a little bit due t clse-up pres. 5) Te velcity te rlls is assumed t be cnstant. 6) Te metal nly extends in te rlling directin and n extensin in te widt te material. 7) Te crss sectinal area nrmal t te rlling directin is nt distrted. Suranaree University Tecnlgy Tapany Udmpl
Frces and gemetrical relatinsips in rlling v x x α y y L p R v A metal seet wit a tickness enters te rlls at te entrance plane xx wit a velcity v. It passes trug te rll gap and leaves te exit plane yy wit a reduced tickness and at a velcity v. Given tat tere is n increase in widt, te vertical cmpressin te metal is translated int an elngatin in te rlling directin. b v bv bv Eq.1 Since tere is n cange in metal vlume at a given pint per unit time trugut te prcess, terere Were b is te widt te seet v is te velcity at any tickness intermediate between and. Suranaree University Tecnlgy Tapany Udmpl
Frm Eq.1 x R y b v b v v v Given tat b b x y v < v L t L t Ten we ave v v Wen >, we ten ave v < v Te velcity te seet must steadily increase rm entrance t exit suc tat a vertical element in te seet remain undistrted. v v Eq.2 Suranaree University Tecnlgy Tapany Udmpl
At nly ne pint alng te surace cntact between te rll and te seet, tw rces act n te metal: 1) a radial rce r and 2) a tangential rictinal rce F. I te surace velcity te rll v r equal t te velcity te seet, tis pint is called neutral pint r n-slip pint. Fr example, pint N. Between te entrance plane (xx) and te neutral pint te seet is mving slwer tan te rll surace, and te tangential rictinal rce, F, act in te directin (see Fig) t draw te metal int te rll. On te exit side (yy) te neutral pint, te seet mves aster tan te rll surace. Te directin te rictinal re is ten reversed and ppse te delivery te seet rm te rlls. x r F x N pint : v rll v seet N α θ β y y Frictin acts in ppsite directins Tapany Udmpl Suranaree University Tecnlgy
r is te radial rce, wit a vertical cmpnent (rlling lad - te lad wit wic te rlls press against te metal). Te speciic rll pressure, p, is te rlling lad divided by te cntact area. p Eq.3 bl p Were b is te widt te seet. L p is te prjected lengt te arc cntact. L p R ( ) ( ) 4 2 12 [ R ( )] 12 Eq.4 L p R Suranaree University Tecnlgy Tapany Udmpl
v p R N Te distributin rll pressure alng te arc cntact sws tat te pressure rises t a maximum at te neutral pint and ten alls. Te pressure distributin des nt cme t a sarp peak at te neutral pint, wic indicates tat te neutral pint is nt really a line n te rll surace but an area. A B Frictin ill in rlling Te area under te curve is prprtinal t te rlling lad. Te area in sade represents te rce required t vercme rictinal rces between te rll and te seet. Te area under te dased line AB represents te rce required t derm te metal in plane mgeneus cmpressin. Suranaree University Tecnlgy Tapany Udmpl
Simpliied analysis rlling lad Te main variables in rlling are: Te rll diameter. Te dermatin resistance te metal as inluenced by metallurgy, temperature and strain rate. Te rictin between te rlls and te wrkpiece. Te presence te rnt tensin and/r back tensin in te plane te seet. We cnsider in tree cnditins: 1) N rictin cnditin 2) Nrmal rictin cnditin 3) Sticky rictin cnditin Suranaree University Tecnlgy Tapany Udmpl
1) N rictin situatin In te case n rictin situatin, te rlling lad () is given by te rll pressure (p) times te area cntact between te metal and te rlls (bl p ). pbl p ' σ b R Eq.8 Were te rll pressure (p) is te yield stress in plane strain wen tere is n cange in te widt (b) te seet. Suranaree University Tecnlgy Tapany Udmpl
2) Nrmal rictin situatin In te nrmal case rictin situatin in plane strain, te average pressure p can be calculated as. σ p ' 1 Q Q ( e 1) Eq.9 Were Q µl p / te mean tickness between entry and exit rm te rlls. Frm Eq.8, pbl p We ave Q ( e ) b R 2 1 σ 1 3 Q Eq.10 Rll diameter Rlling lad Suranaree University Tecnlgy Tapany Udmpl
Terere te rlling lad increases wit te rll radius R 1/2, depending n te cntributin rm te rictin ill. Te rlling lad als increases as te seet entering te rlls becmes tinner (due t te term e Q ). At ne pint, n urter reductin in tickness can be acieved i te dermatin resistance te seet is greater tan te rll pressure. Te rlls in cntact wit te seet are bt severely elastically dermed. Small-diameter rlls wic are prperly stiened against delectin by backup rlls can prduce a greater reductin bere rll lattening becme signiicant and n urter reductin te seet is pssible. Backup rlls Example: te rlling aluminium cking il. Rll diameter < 10 mm wit as many as 18 backing rlls. Suranaree University Tecnlgy Tapany Udmpl
Frictinal rce is needed t pull te metal int te rlls and respnsible r a large prtin te rlling lad. Hig rictin results in ig rlling lad, a steep rictin ill and great tendency r edge cracking. Te rictin varies rm pint t pint alng te cntact arc te rll. Hwever it is very diicult t measure tis variatin in µ, all tery rlling are rced t assume a cnstant ceicient rictin. Fr cld-rlling wit lubricants, µ ~ 0.05 0.10. Fr t-rlling, µ ~ 0.2 up t sticky cnditin. Suranaree University Tecnlgy Tapany Udmpl
Example: Calculate te rlling lad i steel seet is t rlled 30% rm a 40 mm-tick slab using a 900 mm-diameter rll. Te slab is 760 mm wide. Assume µ 0.30. Te plane-strain lw stress is 140 Ma at entrance and 200 Ma at te exit rm te rll gap due t te increasing velcity. (40) ( ) x100 (40) 28mm x100 30% 30 (40) (28) 12mm µ L Q + 2 (40) + (28) 34mm 2 R (0.30) 450x12 (34) p µ 0.65 ' ' ' σ entrance+ σ exit 140+ 200 σ 170Ma 2 2 Frm Eq.10 ' 1 σ ( e Q 170 Q 1 (0.65) 1) b R 0.65 ( e 1(0.76) ) 0.45x0.012 13.4MN Suranaree University Tecnlgy Tapany Udmpl
3) Sticky rictin situatin Wat wuld be te rlling lad i sticky rictin ccurs? Cntinuing te analgy wit cmpressin in plane strain Frm Eq.8, Frm example; p ' a L ' p σ + + 1 1 σ0 2 4 pbl p σ ' R 4 + 1 b R 0.45x0.012 170 1 + (0.76) 4x0.034 14.6MN 0.45x0.012 Suranaree University Tecnlgy Tapany Udmpl
Example: Te previus example neglected te inluence rll lattening under very ig rlling lads. I te dermed radius R a rll under lad is given in Eq.11, using C 2.16x10-11 a -1, 13.4 Ma rm previus example. R ' ' C R 1 + Eq.11 b( ) 6 ( 13.4x10 ) 0. m 11 ' 2.16x10 R 0.45 1 + 464 0.76x0.012 Were C 16(1-ν 2 )/πe, Rlling lad based n te dermed rll radius. We nw use R t calculate a new value and in turn anter value R Q R '' '' µ R 170 1 0.66 0.45 1 + 0.30 0.66 ( e ) 464x12 34 10.76 11 2.16x10 (13.7x10 0.76x0.012 0.66 0.464x0.012 6 ) 0.465m 13.7MN Te dierence between te tw estimatins R is nt large, s we stp te calculatin at tis pint. Suranaree University Tecnlgy Tapany Udmpl