21.3 (SI units) A series of end-milling cuts is currently used to produce an alu

Question

21.3 (SI units) A series of end-milling cuts is currently used to produce an aluminum part that is an aircraft component. The purpose of the machining operation is to remove 95% of the part weight to create a structural frame. A total of 4.0 min is lost during the milling cycle due to tool repositioning. The part has a length = 1.6 m, width=0 m, and mm. The operation uses a four-tooth indexable-insert end mill with mm at a cutting speed 600 m/min, chip load 0.15 mm/tooth, and average cross-sectional area of cut-240 mm. High-speed machining has to replace the conventional milling process. The same chip load and average area of cut will be used, but the cutting speed will be increased to 3600 m/min (Table 21.1), and the time lost for tool repositioning will be reduced to 2.0 min. Determine (a) the cycle time of the current milling operation and (b) the cycle time of the proposed HSM operation. (c) Is this part a good candidate for high-speed machining? Explain. D 32mm0032m 005468.3 4min r 3S8 mm/min Feed

Explanation / Answer

(A) Given, L= 1.6m, W=0.5m, T=0.14m

No. of tooth on milling cutter (n)= 4

Diameter of cutter (D)= 0.032m

V= 600m/min

chip load (f) = 0.15mm/tooth

Assume, Approach Length(A) = Overtravel length (O) = D/2 = 0.016m

V = 3.14 *D*N

So, N= 5972 rpm

feed rate = N*n*f = 5972*4*(0.15/1000) = 3.58 m/min

Cycle time = Machining Time+Time lost in tool adjustment

Machining time = ( L+A+O/(N*feed rate) = 7.633*10^-5 min

Cycle time = 7.633*10^-5+4 = 4.00007633 min

(B) For HSM, V= 3600m/min

time lost = 2 min

N = 35828.02rpm

feed rate= 21.49m/min

maching time= 2.1196*10^-6

cycle time= 2.00000212min

(C) As the time lost is almost reduced to half so it is better to use HSM for the following process.

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