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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 µm). 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/mm2 From chart 16-45, using carbide tool Thus, Vs = 1.8 m/s, fs = 0.5 mm/rev Using correction factor of 1.2 for carbide insert: Vs = 1.8 x 1.2 = 2.16 m/s Using correction factor of 0.7 for stainless steel: Vs = 2.16 x 0.7 = 1.512 m/s Depth of cut will be 1.5 mm From table 16-15 Zv = 1, Zf = 1 V = 1.512 m/s, f = 0.5 mm/rev N = V/πD = 1.512/(π x (120/1000)) = 4 rev/sec Feed rate = f x N = 0.5 x 4 = 2 mm/sec Vt = f x depth of cut x V = 0.5 x 1.5 x 1.512 x 1000 = 1134 mm3/s V = (π/4) x ((120)2 – (30)2) x 1.5 = 15904 mm3 Time for facing = V/Vt = 14 sec E = E1(h/href)-a , a = -0.3, href = 1 mm From table 16-1, E1 = 2.3 w s/mm3 E = 2.3 x (0.5)-0.3 = 2.83 w s/mm3 η = 0.70 Power = (E x Vt /η) = 4.5 Kw Tool radius R = f2 / 32 Ra = 9.5 mm 2- Finish facing with a depth of cut of 0.5 mm: From chart 16-45, using carbide tool Vs = 1.8 m/s, fs = 0.5 mm/rev Using correction factor of 1.2 for carbide insert: Vs = 1.8 x 1.2 = 2.16 Using correction factor of 0.7 for stainless steel: Vs = 2.16 x 0.7 = 1.512 m/s Depth of cut will be 0.5 mm From table 16-15 Zv = 1.3, Zf = 0.5 V = 1.512 x 1.3 = 1.966 m/s, f = 0.5 x 0.5 = 0.25 mm/rev N = V/πD =1.966/(π x (120/1000)) = 5.2 rev/sec Feed rate = f x N = 0.25 x 5.2 = 1.3 mm/sec Vt = f x depth of cut x V = 189 mm3/s V = (π/4) x ((120)2 – (30)2) x 0.5 = 5301 mm3 Time for facing = V/Vt = 15 sec E = E1(h/href)-a , a = -0.3, href= 1 mm From table 16-1, E1 = 2.3 w s/mm3 E = 2.3 x (0.25)-0.3 = 3.49 w s/mm3 η = 0.70 Power = (E x Vt /η) = 0.94 Kw

Tool radius R = f2 / 32 Ra = 2.5 mm 3- Rough boring with a depth of cut of 3.5 mm:

Using carbide tool Vs = 1.8 m/s, fs = 0.5 mm/rev Using correction factor of 1.2 for carbide insert: Vs = 1.8 x 1.2 = 2.16 Using correction factor of 0.7 for stainless steel: Vs = 2.16 x 0.7 = 1.512 m/s Depth of cut will be 0.5 mm From table 16-15 Zv = 1, Zf = 1 V = 1.512 m/s, f = 0.5 mm/rev Vt = f x depth of cut x V = 2646 mm3/s V = (π/4) x ((37)2 – (30)2) x 160 = 58936 mm3 Time for facing = V/Vt = 22 sec E = E1(h/href)-a , a = -0.3, href= 1 mm From table 16-1, E1 = 2.3 w s/mm3 E = 2.3 x (0.5)-0.3 = 2.83 w s/mm3 η = 0.70 Power = (E x Vt /η) = 10.7 Kw Tool radius R = f2 / 32 Ra = 9.5 mm

4- Finish boring with a depth of cut of 0.5 mm: Using carbide tool Vs = 1.8 m/s, fs = 0.5 mm/rev Using correction factor of 1.2 for carbide insert: Vs = 1.8 x 1.2 = 2.16 Using correction factor of 0.7 for stainless steel: Vs = 2.16 x 0.7 = 1.512 m/s Depth of cut will be 0.5 mm From table 16-15 Zv = 1.3, Zf = 0.5 V = 1.512 x 1.3 = 1.966 m/s, f = 0.5 x 0.5 = 0.25 mm/rev Vt = f x depth of cut x V = 245.7 mm3/s V = (π/4) x ((37)2 – (30)2) x 160 = 9425 mm3 Time for facing = V/Vt = 38 sec E = E1(h/href)-a , a = -0.3, href= 1 mm From table 16-1, E1 = 2.3 w s/mm3 E = 2.3 x (0.25)-0.3 = 3.49 w s/mm3 η = 0.70 Power = (E x Vt /η) = 1.2 Kw Tool radius R = f2 / 32 Ra = 2.5 mm

2- A cast iron cylinder block (TS=500 MPa) is cleaned up by slab milling to a depth of 3mm. The surface subject to machining is 500 mm long and 200 mm wide. The tool used is a HSS milling cutter of 120 mm diameter with 8 cutting edges. a) Make a sketch of the process of down milling, showing the cutter rotation and the free direction. List the advantages and limitations of down milling,

b) Select the cutting speed, feed, feed rate and calculate time required, c) Calculate the power requirement, d) If the cylinder block is made out of Aluminum alloy instead of cast iron, would the motor power be less? (Justify your answer). Solution:

In down milling the feed motion is given in the direction of cutter rotation. Advantages: - Cut begins with a well-defined uniform chip thickness - The surface finish is good - Cutting direction holds down work piece, which means lower clamping force Limitations: - Requires more rigid machine tool - Machine tool must be free of backlash

Properties for cast Iron: TS = 500 Mpa Hardness (HB) = (3 x TS) / (9.81) = (3 x 500) / (9.81) = 153 kg/mm2 Using HSS milling cutter, from fig 8.44 Vs = 0.75 m/sec fs = 0.38 mm/rev Tool diameter = 120 mm Number of cutting edges = 8 Using correction factor of 0.8 for Hss tools: V = 0.8 x 0.75 = 0.6 m/sec = 600 mm/sec From the table for slab milling use Zv = 1.0 and Zf = 0.5/tooth V = Zv.Vs = 1.0 x 600 = 600 mm/sec f = Zf.fs = 0.5 x 0.38 x 8 = 1.52 mm/rev Depth of cut (d) = 3mm N = V/Π D = (600)/(Π x 120) = 1.59 rev/sec Feed rate = (feed)(rpm) = (1.52) (1.59) = 2.42 mm/sec Length of travel (L) = length of the work piece + (length of approach) Length of approach = (h(D-h))0.5 = (3(120 – 3))0.5 = 20 Time taken for facing = Length of travel/feed rate = 520/2.42 = 215 seconds Material removal rate (Vt) = (width)(depth of cut)(feed rate) = (200)(3)(2.42) = 1452 mm3 / sec Power requirements: E = E1(h/href)-a E1 = 2.14 W.sec/mm3 Power = (E x Vt /η) = 4.4 Kw If the part were made of aluminum alloy instead of cast iron, would the motor power be less? Aluminum is a softer material than cast iron. This hardness (HB) of aluminum alloy ranges from 30-150 Kg/mm2 while that of cast iron alloy ranges from 110-320 Kg/mm2. Also the specific energy (E1) of aluminum is lesser than that of cast iron. Hence for the same cutting speed and feed, power required to machine aluminum alloy be much less than the power required for machining cast iron. Also in case of aluminum alloys much higher cutting speed can be used.