# HW2_Sol - MAE 4500 Manufacturing Methods Assignment#2 1 The...

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Unformatted text preview: MAE 4500 Manufacturing Methods Assignment #2 1- The 302 stainless steel flange shown in figure 1 is to be machined out of a forged part. Both the central hole and the face of the flange required good surface finish (Ra = 0.8 µ m). Determine the process plan to finish the part and then for each operation give the following details: a) Sketch the process and determine tool geometry and material, b) Cutting speed, feed and machining time, c) Power requirement. Solution: Process plan: 1- Rough facing the diameter 120 with 1.5 mm depth of cut. 2- Finish facing the diameter 120 mm with .5 mm depth of cut. 3- Rough boring the part till hole diameter reaches 37 mm. 4- Finish boring the hole to the final demanded surface finish. 1- Rough facing with a depth of cut of 1.5 mm: Tensile strength of St. St. 302 is 600 MPa Hardness HB = 3x600/9.81 184 Kg/mm 2 From chart 16-45, using carbide tool Thus, V s = 1.8 m/s, fs = 0.5 mm/rev Using correction factor of 1.2 for carbide insert: V s = 1.8 x 1.2 = 2.16 m/s Using correction factor of 0.7 for stainless steel: V s = 2.16 x 0.7 = 1.512 m/s Depth of cut will be 1.5 mm From table 16-15 Z v = 1, Z f = 1 V = 1.512 m/s, f = 0.5 mm/rev N = V/ π D = 1.512/( π x (120/1000)) = 4 rev/sec Feed rate = f x N = 0.5 x 4 = 2 mm/sec V t = f x depth of cut x V = 0.5 x 1.5 x 1.512 x 1000 = 1134 mm 3 /s V = ( π /4) x ((120) 2 – (30) 2 ) x 1.5 = 15904 mm 3 Time for facing = V/V t = 14 sec E = E1(h/h ref )-a , a = -0.3, h ref = 1 mm From table 16-1, E1 = 2.3 w s/mm 3 E = 2.3 x (0.5)-0.3 = 2.83 w s/mm 3 η = 0.70 Power = (E x V...
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## This note was uploaded on 12/14/2009 for the course MAE 4500 taught by Professor El-gizawy during the Spring '06 term at Missouri (Mizzou).

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HW2_Sol - MAE 4500 Manufacturing Methods Assignment#2 1 The...

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