Bulk Deformation Forming Processes Amr Shehata Fayed, Ph.D. Assistant Professor of Mechanical Engineering Materials Engineering Department Faculty of Engineering Zagazig University amrfayed@yahoo.com

Bulk Deformation Processes in Metal Forming Metal forming operations which cause significant shape change by deformation in metal parts whose initial form is bulk rather than sheet "Bulk" refers to workparts with relatively low surface area-to-volume ratios. Starting forms: cylindrical bars and billets, rectangular billets and slabs, and similar shapes These processes work by stressing metal sufficiently to cause plastic flow into desired shape Performed as cold, warm, and hot working operations

Bulk Deformation Processes in Metal Forming Bulk deformation processes are performed as cold, warm and hot working operations. Cold and Warm working is appropriate when:- Shape change is less severe There is a need to improve mechanical properties Achieve good surface finish on the part Hot working is generally required when massive deformation of large work-parts is involved.

Importance of Bulk Deformation Processes When performed as hot working, they can achieve significant change in shape of the work-part. When performed as cold working operations, they can be used not only to shape the product, but also to increase its strength Deformation processes produce little or no waste. Some bulk deformation operations are near net shape or net shape processes They achieve final product geometry with little or no subsequent machining.

Products of Bulk Deformation Processes

Rolling Deformation process in which work thickness is reduced by compressive forces exerted by two opposing rolls. The rotating rolls perform two main functions: Pull the work into the gap between them by friction between workpart and rolls Simultaneously squeeze the work to reduce cross section.

Flat Rolling It involves the rolling of slabs, strips, sheets, and plates, workparts of rectangular cross section in which the width is greater than the thickness.

Flat Rolling Analysis Draft: The work is squeezed between two rolls so that its thickness is reduced by an amount called the draft. where d = draft; to = starting thickness; and tf = final thickness Reduction: The draft is sometimes expressed as a fraction of the starting stock thickness, called the reduction. When a series of rolling operations is used, reduction is taken as the sum of the drafts divided by the original thickness

Flat Rolling Analysis Spreading: In addition to thickness reduction, rolling usually increases work width. This is called spreading. Conservation of material is preserved, so the volume of metal exiting the rolls equals the volume entering: where: w0 & wf are the before and after work widths (mm) L0 & Lf are the before and after work lengths (mm)

Flat Rolling Analysis Similarly, before and after volume rates of materials flow must be the same, so the before and after velocities can be related: where:- v0 and vf are the entering and exiting velocities of the work

Flat Rolling Analysis Maximum Draft: There is a limit to the maximum possible draft that can be accomplished in flat rolling with a given coefficient of friction: where dmax : maximum draft (mm). μ : coefficient of friction R : roll radius (mm)

Flat Rolling Analysis Rolling Force: The rolling force can be calculated based on the average flow stress experienced by the work material in the roll gap: Where Y = average flow stress (MPa) w L = the roll-work contact area (mm2) Contact length can be approximated by

Flat Rolling Analysis The torque in rolling can be estimated by assuming that the roll force is centered on the work as it passes the rolls and that it acts with a moment arm of one-half the contact length L. The power required to drive each roll is the product of torque and angular velocity. P = power (kW) F = rolling force (N) N = rotational Speed (rev/min) L = contact length (m)

Example (1) A 12-in wide strip 1-in thick is fed through a rolling mill with two powered rolls each of 10-in radius. The work thickness is to be reduced to 0.875-in in one path at a roll speed of 50 rev/min. The work material has a flow curve defined by K = 40000 ib/in2 and n = 0.15, and the coefficient of friction between the rolls and the work is assumed to be 0.12. Determine if the friction is sufficient to permit the rolling operation to be accomplished, if so, calculate the roll force and torque. Given: R = 10 in, t0 = 1 in, tf = 0.875 in, m = 0.12

L = [ R (t0 – tf) ] 0.5 = ( 10 ( 1.0 – 0.875)) 0.5 = 1.118 in Example (1), Solution The draft attempted in this rolling operation is d = t0 – tf, then d = 1.0 – 0.875 = 0.125 in But, dmax = m2 R = (0.12)2 (10) = 0.144 in Since the max allowable draft exceeds the attempted reduction, the rolling operation is feasible. To compute the rolling force, we need the contact length L and average flow stress. The contact length is given by L = [ R (t0 – tf) ] 0.5 = ( 10 ( 1.0 – 0.875)) 0.5 = 1.118 in The flow stress is determined from the true strain: e = ln(t0/tf) e = ln( 1 / 0.875 ) = 0.134

Example (1), Solution, cont. Yf = k e n / (1 + n) Yf= 40000 (0.134)0.15 / (1+0.15) = 25729 ib/in2 Rolling force is determined from F = Yf w L F = 25729 (12) (1.118) = 345184 ib Torque required to derive each roll is given by: T = 0.5 F L = 0.5 (345184) (1.118) = 192.958 ib.in The power is obtained from P = 2p N F L P = 2p (345184)(1.118) = 121238997 in-lb/min Converting to horsepower (1 HP = 396000 in-lb/min) HP = 121238997 / 396000 = 306 HP