Analysis of cold rolling a more accurate method
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1 Analysis of cold rolling a more accurate method 1.1 Rolling of stri more accurate slab analysis The revious lecture considered an aroximate analysis of the stri rolling. However, the deformation zone in rolling rocess is very comlex and is curved. Therefore, we have to consider the various states of stresses acting, considering the curvature of the deformation zone. In cold rolling the work material is likely to undergo strain hardening as it comes out of the rolls. In the resent lecture we consider the analysis considering various stresses acting on an elemental stri. Slab method of analysis is alied in order to obtain the rolling load in terms of the geometry of the deformation zone and roll diameter. We assume that rolls are not undergoing any elastic deformation. Consider an elemental stri within the deformation zone, as shown below: We assume that the rolling is lane strain rocess, as there is little sread of material along the width of the stri. Further, the friction coefficient remains constant through the rolling rocess. dθ dθ h+dh h θ Element before neutral section After neutral section Fig : Elemental stri taken from the rolling deformation zone Joint Initiative of IITs and IISc Funded by MHRD Page 3 of 8
2 The element makes an angle of with the roll centre. Consider the element at an angle of from the line joining centres of the rolls The following forces act on the element: Normal roll ressure force: R Tangential friction force: R The comressive forces: and ][h+dh] The normal and tangential forces can be resolved along the direction of rolling x axis: R and R Making a force balance on the element shown above: ][h+dh] - - 2R 2 R = Ignoring the roducts of small quantities, dividing by and simlifying, we get: This equation is called von Karman equation. In cold rolling, under low friction conditions angle is small [6 degrees]. We can aroximately take; sin = and cos = 1. These aroximates were roosed by Bland and Ford. Now the above equation becomes: From von Mises yield criterion alied to lane strain we have: In rolling, for small angle, the two rincial stresses are: the roll ressure and Therefore, we have: - = = Y Substituting this in the above equation, Joint Initiative of IITs and IISc Funded by MHRD Page 4 of 8
3 Or Y h = The second term on left hand side can be ignored because, Y h is constant. That is, when h increases, Y decreases and vice versa. Now we have: we can aroximately write: h = h f +R Substituting this in 31 and integrating we get the general solution to the above differential equation as: where H = tan -1 [ ] A Alying the boundary conditions: At entry, and H = H o AT exit, and H = 0 We get the roll ressure as: = at the entry = at exit From the above exressions we note that the local rolling ressure deends on the angular osition of the section and the height of the work, h. It is also deendent on R/h f R/h f is equivalent of a/h in forging. As this ratio increases, the rolling ressure also increases. The total rolling force P can be evaluated by integrating the local rolling force over the arc of contact. P = Rb, where b is width of the stri Joint Initiative of IITs and IISc Funded by MHRD Page 5 of 8
4 Entry Exit /Y L Fig : Actual variation of roll ressure The above figure shows the variation of the non-dimensional roll ressure with resect to the coefficient of friction the friction hill. We observe that the roll ressure Neutral Point increases with increase in coefficient of friction. The area under the curves gives the total roll force. Further, we also observe that the neutral oint also shifts towards the exit as the coefficient of friction reduces. As the friction gets reduced, there is sliing between the rolls and the work. Hence the relative velocity between roll and stri is in the same direction. /Y 50% red ucti on 1 30% 10% L1 L2 Joint Initiative of IITs and IISc Funded by MHRD L3 Page 6 of 8 L
5 Fig.4.1.3: Roll Entry ressure versus reduction in thickness The above figure reresents the variation of roll ressure with resect to thickness reduction of the stri. As the reduction increases the roll ressure also increases. This is because, for larger reductions, the length of contact between roll and stri increases. 1.2 Determination of neutral oint: The neutral oint can be determined by equating the roll ressure before neutral oint to that after neutral oint. Equating the equations 33 and 34 and solving for Hn, H n = Substituting this in equation 32A and solving for, tan[ ] Examle: Determine the rolling ower required to roll low carbon steel stri, 250 mm wide, 12 mm thick, if the final thickness is 9 mm. Assume sliding friction between the rolls and work, with a coefficient of friction The 250 mm radius rolls rotate at a seed of 300 rm. Take k= 550 MPa, n = 0.26 for steel. Solution: We can take the average roll force for sliding friction condition as: (1+ ) True strain = = ln(ho/hf) = The average flow stress of the material = = MPa Plane strain flow stress Y = Average flow stress = MPa hav = (12+9)/2 = 10.5 mm Joint Initiative of IITs and IISc Funded by MHRD Page 7 of 8
6 L = = mm Rolling load F = 2.9MN Roll torque = FXL/2 Power = = 1,25 MW Joint Initiative of IITs and IISc Funded by MHRD Page 8 of 8
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