1. ME 482 - Manufacturing Systems RollingAndExtrusion

2. 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

3. ME 482 - Manufacturing Systems Rolling Products Billet Billet Blooms (> 6” x 6”) Blooms (> 6” x 6”) Slab Slab Ingots Billets (> 1.5” x 1.5”) Slab (> 10” x 1.5”) Slab (> 10” x 1.5”)

4. ME 482 - Manufacturing Systems Rolling Model Assumptions: Infinite sheet Infinite sheet Uniform, perfectly rigid rollers Uniform, perfectly rigid rollers Constant material volume: Constant material volume: t o w o L o = t f w f L f “rate”(t o w o v o = t f w f v f ) where L o = initial plate length L f = final plate length where L o = initial plate length L f = final plate length tftf toto vovo vfvf vrvr p R  L vrvr R = roller radius p = roll pressure L = contact length  = contact angle v r = roll speed t o = initial plate thickness t f = final plate thickness v o = plate entry speed V f = plate exit speed

5. ME 482 - Manufacturing Systems Rolling Model Define draft = d = t o - t f Draft limit = d max =  2 R  = 0.1 cold  = 0.2 warm  = 0.4 – 1.0 hot Draft limit = d max =  2 R  = 0.1 cold  = 0.2 warm  = 0.4 – 1.0 hot Define forward slip = s = (v f – v r )/v r Does it make sense that v r < v f ? Does it make sense that v r < v f ? Point of greatest contact pressure = no slip point

6. ME 482 - Manufacturing Systems Rolling – stress, strain, force, power Rolling Model – stress, strain, force, power Define true strain =  = ln(t o /t f ) (Note: use t o /t f to keep > 0) Apply average flow stress = Y f = K  n /(1 + n) Approximate roll force = F = Y f w L where L = R(t o – t f ) Torque estimated by T = 0.5 F L Power = P = T  = 2   F L (for two rollers) R R– (t o –t f )/2 L 

7. ME 482 - Manufacturing Systems Other rolling configurations

8. 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 d max = (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 Y f = (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)

9. 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!

10. ME 482 - Manufacturing Systems Extrusion Model Assumptions: Circular cross-section Uniform stress distribution p = ram pressure L = remaining billet length D o = chamber diameter D f = extrudate diameter L DoDo DfDf p

11. ME 482 - Manufacturing Systems Extrusion Model – stress and strain A o = billet (chamber) area A f = extrudate area a = 0.8 1.2  b  1.5 Define extrusion ratio = r x = A o /A f Frictionless model: ideal true strain =  = ln r x ideal ram pressure = p = Y f ln r x With friction: Johnson eqn  x = a + b ln r x

12. ME 482 - Manufacturing Systems Extrusion Model – stress and strain Indirect extrusion ram pressure = p = Y f  x (  x is from Johnson eqn) and where Y f is found using the the ideal true strain  = ln r x In direct extrusion, difficult to predict the chamber/billet interactive friction, so use the shear yield strength ( about Y f /2 ) to estimate the chamber wall shear force as p f  D o 2 /4 = Y f  D o L/2 giving p f = 2 Y f L  D o and where p f = additional pressure to overcome wall friction force Total ram pressure becomes p = Y f (  x + 2L  D o )

13. ME 482 - Manufacturing Systems Extrusion Model – non-circular sections Apply a shape factor K x (experimental results): K x = 0.98 + 0.02 (C x / C c ) 2.25 where C x = perimeter of extruded shape C c = perimeter of circle having same area of extruded shape Applies for 1.2  (C x / C c )  1.5 For complex extrudate: Indirect p = K x Y f  x Direct p = K x Y f (  x + 2L  D o )

14. ME 482 - Manufacturing Systems Extrusion Model – forces and power Ram force = F = pA o Power = P = Fv

15. 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 r x = ln 4 = 1.3863  x = 0.8 + 1.5(1.3863) = 2.87945 Y f = 60,000(1.386) 0.18 /1.18 = 53,927 psi

16. 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 = Y f (  x + 2L/D) = 53,927 [2.87945 + (2)(3)/1] p = 478,842 psi

17. ME 482 - Manufacturing Systems Rolling and Extrusion What have we learned?