1. Chapter 16 Bulk Forming Processes (Part 3) EIN 3390 Manufacturing Processes Spring 2011

2. Swaging Also known as rotary swaging and radial forging Uses external hammering to reduce the diameter or produce tapers or points on round bars of tubes

3. Swaging Figure 16-21 (Below) Tube being reduced in a rotary swaging machine. (Courtesy of the Timkin Company, Canton, OH.) Figure 16-22 (Right) Basic components and motions of a rotary swaging machine. (Note: The cover plate has been removed to reveal the interior workings.) (Courtesy of the Timkin Company, Canton, OH.) Figure 16-23 (Below) A variety of swaged parts, some with internal details. (Courtesy of Cincinnati Milacron, Inc. Cincinnati, OH.)

4. Swaging

5. Drawing Force Estimation of Drawing Force required: F = Y avg A f ln (A 0 /A f ) Y avg = average true stress of material in the die gap; A 0 and A f are the original and final cross-sectional areas of the work. Assumptions: no friction.

6. Drawing Force (-continued) If considering the friction, the actual force is larger than provided by follows. F = Y avg A f ln (A 0 /A f ) In addition to the ratio A 0 /A f, other variables that influence draw force are die angle, and coefficient of friction at the work-die interface. A number of methods have been proposed for predicting draw force based on values of these parameters.

7. Drawing of Bar, Rod, or Wire

8. Drawing Force (-continued) Schey [1] suggested: F = Y avg A f (1 +  /tan  )  ln (A 0 /A f ) where F – drawing force, lb;  – die-work coefficient of friction;  – die angle (half-angle), degree;  – a factor that accounts for inhomogeneous deformation, which is determined as follows for a round cross section:  = 0.88 + D/L c Where D – average diameter of work during drawing, in; and Lc = contact length of the work with the draw die, in. D = 0.5 (D 0 + D f ), and L c = (D 0 – D f )/(2sin  The power required in a drawing operation is the draw force multiplied by exit velocity of the work. [1] Schey, j.A., Introduction to manufacturing Processes, 2 nd ed.., McGraw-Hill Book Co., New York, 1987, Chapter 4.

9. Flow stress Y avg The average flow stress (also called the mean flow stress) is the average value of stress over the stress- strain curve from the beginning of strain to the final (maximum) value that occurs during deformation. Stress-strain curve indication location of average flow stress y f in relation to yield strength Y and final flow stress Y f.

10. Flow stress Y avg (-continued) The average flow stress is determined by integrating the flow curve equation between zero and the final strain value defining the range of interest. Y avg = K  n /(1 + n) where Y avg – average flow stress, ib/in 2 ; K – the strength coefficient, lb/in 2,  - maximum strain value during the deformation process,  = ln (A 0 /A f ); and n – the strain hardening exponent. Finally, we have F = Y avg A f ln (A 0 /A f ) = [k  n /(1 + n)] A f ln (A 0 /A f )

11. Example to Calculate Drawing Force F force (0.00503) 70

12. Maximum Reduction per Pass for Drawing Work has to be done to overcome friction. Force increases with increasing friction. Cannot increase force too much, or material will reach yield stress. Maximum reduction in cross-sectional area per pass = 63%.

13. Maximum Reduction per Pass for Drawing (-continued) F/A f =

14. t 0 W 0 v 0 = t f w 0 v f v f = (t 0 v 0 )/t f v f > v r > v 0

15. Rolling Mill Configurations Smaller diameter rolls produce less length of contact for a given reduction and require less force to produce a given change in shape Smaller cross section provides a reduced stiffness ◦Rolls may be prone to flex elastically because they are only supported on the ends Figure 16-4 The effect of roll diameter on length of contact for a given reduction.

16. Rolling Mill Process X Y t0t0 txtx tftf v0v0 vfvf x y xfxf x0x0 Assume: Starting volume of rolling work is equal to the final volume of the rolling work: (Volume) 0 = (Volume) f t 0 W L 0 = t f W L f (t 0 W L 0 )T = (t f W L f )T (t 0 W v 0 ) = (t f W v f ) (t 0 v 0 ) = (t f v f ) t x = t f + (r – y) = t f + r – SQRT(r 2 – x 2 ) if x = 0, t x = t f if x = x 0, t x = t 0, where x 0 = SQRT(r 2 – y 0 2 ), and y 0 = r + t f – t 0 r y0y0

17. Rolling Mill Process v x = (t 0 v 0 )/t x = (t 0 v 0 )/[t f + r – SQRT(r 2 – x 2 )] Example: r = 10 in, v 0 = 1,000 in/min, t 0 - 1 in, and t f - 0.5 in. xVx 3.12251000.000 31041.007 2.751129.232 2.51223.179 2.251322.030 21424.418 1.751528.314 1.51630.949 1.251728.810 11817.767 0.751893.349 0.51951.190 0.251987.576 02000.000 V x X