Rolling And Extrusion ME 482 - Manufacturing Systems Rolling Process Points: Significant shape change Capital intensive Large volume Usually hot worked (isotropic)* Oxide scale Tolerances difficult to hold * Can be followed by cold rolling to improve

tolerances and directional properties ME 482 - Manufacturing Systems Rolling Products Billet Blooms (> 6 x 6) Slab Ingots Billets (> 1.5 x 1.5) Slab (> 10 x 1.5) ME 482 - Manufacturing Systems

vr Rolling Model to R p vo vf tf L

Assumptions: Infinite sheet vr Uniform, perfectly rigid rollers Constant material volume: to wo Lo = tf wf Lf rate (to wo vo = tf wf vf ) where Lo = initial plate length Lf = final plate length ME 482 - Manufacturing Systems

R = roller radius p = roll pressure L = contact length = contact angle vr = roll speed to = initial plate thickness tf = final plate thickness vo = plate entry speed Vf = plate exit speed Rolling Model Define draft = d = to - tf Draft limit = dmax = 2R

= 0.1 cold = 0.2 warm = 0.4 1.0 hot Define forward slip = s = (vf vr)/vr Does it make sense that vr < vf? ME 482 - Manufacturing Systems Point of greatest contact pressure = no slip point Rolling Model stress, strain, force, power

Define true strain = = ln(to/tf) (Note: use to/tf to keep > 0) Apply average flow stress = Yf = K n /(1 + n) Approximate roll force = F = Yf w L where L = R(to tf) R Torque estimated by T = 0.5 F L Power = P = T = 2 F L (for two rollers) ME 482 - Manufacturing Systems L

R (totf)/2 Other rolling configurations ME 482 - Manufacturing Systems Example 21.1 in text Roll a 12 inch wide strip, that is 1 inch thick, to 0.875 inch thickness in one pass with roll speed of 50 rpm and radius = 10 inches. Material has K = 40,000 psi, n = 0.15 and = 0.12. Determine if feasible and calculate F, T, and power if so. Solution: Feasible since dmax = (0.12)2 (10) = 0.144 in. > d = 1.0 0.875 = 0.125 in. Contact length = L = 1.118 in.

= ln(1.0/0.875) = 0.134 Yf = (40,000)(0.134)0.15/1.15 = 25,729 psi Rolling force = (25,729)(12)(1.118) = 345,184 lb Torque = (0.5)(345,184)(1.118) = 192,958 in.-lb Power = P = (2)(50)(345,184)(1.118) = 121,238,997 in.-lb/min (306 hp) ME 482 - Manufacturing Systems Extrusion Limitation requires uniform cross-section vs length Advantages: Variety of shapes Control mechanical properties in cold and warm extrusion Little wasted material Good tolerances Types : Direct extrusion and indirect extrusion

Less friction! ME 482 - Manufacturing Systems Extrusion Model Assumptions: L Circular cross-section Uniform stress distribution p Do

p = ram pressure L = remaining billet length Do = chamber diameter Df = extrudate diameter ME 482 - Manufacturing Systems Df Extrusion Model stress and strain Define extrusion ratio = rx = Ao/Af Ao = billet (chamber) area Af = extrudate area

Frictionless model: ideal true strain = = ln rx ideal ram pressure = p = Yf ln rx With friction: Johnson eqn x = a + b ln rx ME 482 - Manufacturing Systems a = 0.8 1.2 b 1.5 Extrusion Model stress and strain Indirect extrusion ram pressure = p = Yf x (x is from Johnson eqn) and where Yf is found using the the ideal true strain = ln rx

In direct extrusion, difficult to predict the chamber/billet interactive friction, so use the shear yield strength ( about Yf /2 ) to estimate the chamber wall shear force as pf Do2/4 = Yf Do L/2 giving pf = 2 Yf L Do and where pf = additional pressure to overcome wall friction force Total ram pressure becomes p = Y ( + 2L Do ) ME 482 - Manufacturing Systems f x Extrusion Model non-circular sections Apply a shape factor Kx (experimental results): Kx = 0.98 + 0.02 (Cx / Cc)2.25

where Cx = perimeter of extruded shape Cc = perimeter of circle having same area of extruded shape Applies for 1.2 (Cx / Cc) 1.5 For complex extrudate: Indirect Direct p = K x Yf x p = Kx Yf (x + 2L Do ) ME 482 - Manufacturing Systems Extrusion Model forces and power

Ram force = F = pAo Power = P = Fv ME 482 - Manufacturing Systems Extrusion example A billet 3 long and 1 diameter is to be extruded as a round extrudate in a direct extrusion operation with extrusion ratio of r x = 4. Given die angle of 90, strength coefficient of 60,000 psi, and strain hardening exponent of 0.18, use the Johnson formula with a = 0.8 and b = 1.5 to estimate extrusion strain, and then determine the pressure applied to the end of the billet as the ram moves forward.

Solution: = ln rx = ln 4 = 1.3863 x = 0.8 + 1.5(1.3863) = 2.87945 Yf = 60,000(1.386)0.18/1.18 = 53,927 psi ME 482 - Manufacturing Systems Extrusion example A billet 3 long and 1 diameter is to be extruded as a round extrudate in a direct extrusion operation with extrusion ratio of r x = 4. Given die angle of 90, strength coefficient of 60,000 psi, and strain hardening exponent of 0.18, use the Johnson formula with a = 0.8 and b = 1.5 to estimate extrusion strain, and then determine the pressure applied to the end of the billet as the ram moves forward.

Solution continued: p = Yf (x + 2L/D) = 53,927 [2.87945 + (2)(3)/1] p = 478,842 psi ME 482 - Manufacturing Systems Rolling and Extrusion What have we learned? ME 482 - Manufacturing Systems