Question

500-mm long aluminium billets of 250-mm diameter are separately forward extruded into bars of 125-, 50-, 25-, and 12.5-mm diameter at 500°C using a ram speed of 0.75 m/min. Flow rule is defined through C = 14 MPa, and m = 0.22 with Of = 35 MPa. The die angle is 2a= 140° whereas the Coulomb coefficient of friction is 0.08. a) Determine the extrusion force through the analytical approach for the four bar diameters, b) Determine the extrusion force through the empirical approach for the four bar diameters, c) Compute the exit velocities of the extrusions in the four cases, d) Figure out the ram speed that will give an analytical extrusion force of 4000 KN for the bar with 12.5 mm diameter, and e) Elaborate on whether the press is feasible enough for the economical extrusion of bars of these sizes.

Answer 1

given by the equation Ri The fore in analytical method is Fe 23 aogardy 20ft (22) Fer of a [Ri² _Ro? Ro Ro Ist extaution Riz 250 125mm 2 Roz 125 62.5 mm f 235MPa Fe 35xa [125²- 62.5%) 1288.54 KN and extantion Ri= 62.5mm Roz 50 2 25 mm Fe 2 35 XK [62.50 - 25²) 360.79 KN ord lextaution Riz 25mm Roe 25 2 12.5 mm 35 XÃ [ 5 - 10:53] 51.54 KW lth extrution Roz 6.25 Fe 2 Riz 12.5mm Fe. 35xK [ 12.5? – 6.25 12.885 KN b). The empirical equations for pressure is given for solid billet by Pavg 0.6+ 1.63 luk R-extaution ration F Pawg & T. [R - R2] = 00 [0.6 +1.63 ln R] x [R; ² - Ro?] 2 2 카 extsution Rz do C250 125 Z 4 Riz 125 Ro 2 62.5 60214 Af Fe = 14 (0.6+ 1.63 ln 4]*[125262158] - 1473.91 KN and extrusion R: (195) * = 6,25 Fe 2 2 14 (0.6+1631n 6.25)# [62154–2659] 2 515.41 KN Mod extaution R24 58.94 KN Fe, 40.03 x ñ [254-12.5?] - Ivath extrusion R24 Fe 2 40.03 XX [ 12.5² - 61252] - 14.13 KN

). we know by continuity equation. AV 2 A₂ V2 id, ²V, 2 d₂² V2 Le Given Vo be intial sam speed At d, diameter. V2 V3 V4 V5 be the velocity at end of est, und wird, with 1 extantion Vo 0.75m min 2 750 mm min. di = 250 d₂2125 ie, 2508x-15= 125² x V2 = V2 23 m/min V₂ x d² = V₂ xd ² - 3x 125² = Vzx50² » V₂ = 18.75 m/min 2 Vz xdz 2. V4 Xd4² - 18.75x50 2 25 X v1 = 75 m/min 2 V 4 x dx = Vs x d5 ² - 15 x 85² = 12.59 x V5 Vs 800 m/min