Question

zis 3. A Cylindrical billet is 75 mm long and 35 mm in diameter. It is reduced by direct extrusion to 20 mm diameter. The half die angle x 45°. In the flow curve for the work metal, strength coefficient 500 MPa and strain-hardening exponent -0.25, Coefficient of friction at the work- die interface 0.15. Given that total extrusion pressure P, is the sum of the die pressure P and the friction pressure P, between the container and the extruded billet interface, the equations obtained by the slab method of analysis, and the diagram of direct extrusion given below! p=P(BP -1 and P, 2 Where B = tamu Poming bilang a) (Ipt.) Determine the extrusion ratio, R Rampressure, b) (1 pt.) Determine the true strain, e) (2pts.) Determine the extrusion pressure. Pe and extrusion force, F. at the beginning of extrusion when L= 200 mm. d) (2pts.) Determine the extrusion pressure. P. and extrusion force, Fe. when L 0 mm. e) (Ipt.) Determine the extrusion pressure and extrusion force if the operation is frictionless (ideal case)

Answer 1

ExtHusion dio D 35 am R - R = R: 3.0625 True Stain ( 2 xfn

no.5 k 500Mfa o-15 y k( ntl 1o.25 Averax flow strey 41.42M1pa O-15 tan(5) 3o5 20-15 - 2D.15 35 P O-15 576.59 Mpa P Yx 2.L D b 4.42y 2 xLuD 35 gpMpaF09 MPa f Pe xA + F 5.07 MN Pe528. 5 Ma

Lihen 576. 5 35 o. 55 MN Fe Pe = 5765MPa 159L9TN. Extusion fa v29 F-kAo n R 500x TT (39), t-12 F o.538MN R= 3.0625 12 = 5278.5 Mpa Pe Fe = S09MN d) Pe= 576. 59 MPa o.55 MN e) F o.538 MN