λ=te averageVc =MDN1000T = S sinWhere λ is Chip Reduction CoefficientVc is cutting speedte is thickness of cut chipT is thickness of non-cut chipS is feed rateⒸ is cutting angle = 45⁰D is diameter of workpiece = 40mmResultsm/minDepth ofcut (a)mm22222Cuttingspeed(Vc)m/min500500500500500Feed rate(S)mm/revRequired:1.Complete the solution to the tablewith the given rules2-Draw chart between feed rate (S) and chipreduction coefficient2tc:tc₂tcs0.110.39 0.58 0.430.140.42 0.62 0.50.200.34 0.42 0.320.240.9 0.96 0.950.28 0.92 0.85 0.85te average(mm)T(mm)with discuss it.ChipReductionCoefficient(2)

Given data: The formula, T = S×sin θ

λ= te average Ve= "DN 1000 T = S sin e Where is Chip Reduction Coefficient Vc is cutting speed te is thickness of cut chip T is thickness of non-cut chip S is feed rate Ⓒ is cutting angle = 45° Dis diameter of workpiece = 40mm Results m/min Depth of cut (a) mm Cutting speed (Ve) m/min Feed rate (S) mm/rev Required: 1.Complete the solution to the table with the given rules 2-Draw chart between feed rate (S) and chip reduction coefficient 1 te: te tes te average (mm) T (mm) with discuss it. Chip Reduction Coefficient (2)

λ= te average Ve= "DN 1000 T = S sin e Where is Chip Reduction Coefficient Vc is cutting speed te is thickness of cut chip T is thickness of non-cut chip S is feed rate Ⓒ is cutting angle = 45° Dis diameter of workpiece = 40mm Results m/min Depth of cut (a) mm Cutting speed (Ve) m/min Feed rate (S) mm/rev Required: 1.Complete the solution to the table with the given rules 2-Draw chart between feed rate (S) and chip reduction coefficient 1 te: te tes te average (mm) T (mm) with discuss it. Chip Reduction Coefficient (2)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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A cemented carbide tool is used to turn a part with length = 18.0 in and diameter = 3.0 in. The parameters in the Taylor equation are: n = 0.27 and C = 1200. The rate for the operator and machine tool = $33.00/hr, and the tooling cost per cutting edge = $2.00. It takes 3.0 min to load and unload the workpart and 1.50 min to change tools. The feed = 0.013 in/rev. Determine: Cutting speed for maximum production rate, Tool life in min of cutting, and Cycle time and cost per unit of product.
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A cutting tool has a nose radius of 1.8 mm the feed rate for a theoretical surface roughness of R= 5 microns is (a) 0.268 mm/rev (b) 0 187 mm/rev (c) 0.036 mm/rev (d) 0.0187 mm/rev
Standard time & work sampling The following data resulted from a time study made on a horizontal milling machine: Mean manual effort time per cycle: 4.62 min Mean cutting time (power feed): 3.74 min Mean performance rating: 115 percent Machine allowance (power feed): 10 percent Fatigue allowance: 15 percent questions: 1.) What is the Basic Time for the manual effort? 2.) What is the Standard Time for the manual effort? 3.) What is the Standard Time for the machine?