A4 - Tillverkningsteknik

Name the two main categories of cutting fluid according to function.

(1) coolants (2) lubricants.

Name three modes of tool failure in machining

fracture failure, temperature failure, gradual wear

In addition to cutting speed, what other cutting variables are included in the expanded version of the Taylor tool life equation?

Taylor equation include following: feed, depth of cut, and/or work material hardness.

2. Solve Problem 1 except that the tool life criterion is 0.50 mm of flank land wear rather than 0.75 mm.

v1 = 125 m/min, T1 = 13.0 min när FW = 0.50 mm v2 = 165 m/min, T2 = 5.6 min när FW = 0.50 mm. (c) Two equations: (1) 125(13.0)^n = C (2) 165(5.6)^n = C (1) and (2) ger 125(13.0)^n= 165(5.6)^n ln 125 + n ln 13.0 = ln 165 + n ln 5.6 4.8283 + 2.5649n = 5.1059 + 1.7228 n 0.8421 n = 0.2776, n = 0.3296 (1) C = 125(13.0)0.3894 = 291.14 (2) C = 165(5.6)0.3894 = 291.15 C = 291.15

Which of the following is not a common ingredient in cemented carbide cutting tools (two correct answers): (a) Al2O3, (b) Co, (c) CrC, (d) TiC, and (e) WC?

A) Al2O3 C) CrC

Two principal aspects of cutting-tool technology

Tool material and tool geometry

What are some of the new problems introduced by machining dry?

(1) overheating the tool (2) operating at lower cutting speeds and production rates to prolong tool life (3) absence of chip removal benefits that are provided by cutting fluids in grinding and milling.

Identify three desirable properties of a cutting-tool material.

(1) toughness to resist fracture failure, (2) hot hardness to resist temperature failure (3) wear resistance to prolong the life of the tool during gradual wear.

Identify mechanisms by which cutting tools wear during machining

Abrasion, adhesion, diffusion, and plastic deformation of the cutting edge.

Cast cobalt alloys typically contain which of the following main ingredients (three best answers): (a) aluminum, (b) cobalt, (c) chromium, (d) iron, (e) nickel, (f) steel, and (g) tungsten?

B) Cobalt C) Chromium G) Tungsten

Two principal locations on a cutting tool where tool wear occurs

The top face of the cutting tool and on flank of the tool, called flank wear.

What are some of the tool life criteria used in production machining operations?

Tool life criteria include: (1) complete failure of the tool (2) visual observation of flank or crater wear (3) fingernail test to feel flank (4) sound of the tool (5) chip disposal problems (6) degradation of finish (7) power increase (8) workpiece count (9) length of cutting time for the tool.

3. Tool life tests in turning yield the following data: (1) when cutting speed is 100 m/min, tool life is 10 min (2) when cutting speed is 75 m/min, tool life is 30 min. (a) Determine the n and C values in the Taylor tool life equation. Based on your equation, compute (b) the tool life for a speed of 110 m/min (c) the speed corresponding to a tool life of 15 min. v*T^n = C

a) Two equations: (1) 100(10)^n = C (2) 75(30)^n = C. 100(10)^n = 75(30)^n ln 100 + n ln10 = ln 75 + n ln 30 4.605 + 2.302 n = 4.317 + 3.401 n 4.605 - 4.317 = (3.401 - 2.302) n 0.288 = 1.099 n, n = 0.262 C = 100(10)0.262 , C = 182.8 C = 75(30)0.262, C = 182.8 b) 110 T^0.262 = 182.8 T^0.262 = 182.8/110 = 1.6618 T = 1.6618^1/0.262 = 1.6618^3.8167 = 6.948 min c) v*(15)^0.262 = 182.8 v = 182.8/(15)^0.262 = 182.8/2.033 = 89.916 m/min

PROBLEMS - Tool life and the Taylor Equation 1. Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.30 mm/rev and a depth of 4.0 mm. At a speed of 125 m/min, flank wear = 0.12 mm at 1 min, 0.27 mm at 5 min, 0.45 mm at 11 min, 0.58 mm at 15 min, 0.73 at 20 min, and 0.97 mm at 25 min a speed of 165 m/min, flank wear = 0.22 mm at 1 min, 0.47 mm at 5 min, 0.70 mm at 9 min, 0.80 mm at 11 min, and 0.99 mm at 13 min. The last value in each case is when final tool failure occurred. (a) On a single piece of linear graph paper, plot flank wear as a function of time for both speeds. Using 0.75 mm of flank wear as the criterion of tool failure, determine the tool lives for the two cutting speeds. (b) On a piece of natural log-log paper, plot your results determined in the previous part. From the plot. (c) Calculate the values of n and C in the Taylor equation solving simultaneous equations. Are the resulting n and C values the same?

1a) Plotta en graf där y-axeln = Flank Wear och x-axeln = Tid, rita ett streck för Flank 0,75mm som går igenom y-axeln. 1b) Plotta en graf där cutting speed (mm/min) är y-axeln. x-axeln är tool life (min) at v1 = 125 m/min, T1 = 20.4 min using criterion FW = 0.75 mm, and at v2 = 165 m/min, T2 = 10.0 min using criterion FW = 0.75 mm. (c) Two equations: (1) 125(20.4)^n= C (2) 165(10.0)^n= C (1) and (2) 125(20.4)^n= 165(10.0)^n ln 125 + nln 20.4 = ln 165 + nln 10.0 4.8283 + 3.0155n = 5.1059 + 2.3026 n 0.7129 n = 0.2776 n = 0.3894 (1) C = 125(20.4)0.3894 = 404.46 (2) C = 165(10.0)0.3894 = 404.46 C = 404.46

Of the following cutting conditions, which one has the greatest effect on tool wear: (a) cutting speed, (b) depth of cut, or (c) feed?

A) Cutting speed

Which of the following are the two main functions of a cutting fluid in machining (two best answers): (a) improve surface finish on the workpiece (b) reduce forces and power (c) reduce friction at the toolchip interface (d) remove heat from the process (e) wash away chips?

C) Reduce friction at the toolchip interface D) Remove heat from the process

A series of turning tests are performed to determine the parameters n, m, and K in the expanded version of the Taylor equation, Eq. (23.4). The following data were obtained during the tests: (1) cutting speed = 1.9 m/s, feed = 0.22 mm/rev, tool life = 10 min; (2) cutting speed = 1.3 m/s, feed = 0.22 mm/rev, tool life = 47 min (3) cutting speed = 1.9 m/s, feed = 0.32 mm/rev, tool life = 8 min. (a) Determine n, m, and K (b) Using your equation, compute the tool life when the cutting speed is 1.5 m/s and the feed is 0.28 mm/rev. Formula: v*T^n*f^m = K*Tref^n*fref^m

Calculation: (1) K = v*Tn*fm = 114*10^n*0,22^m (2) K = v*T^n*f^m = 78*47^n*0,22^m (3) K = v*T^n*f^m = 114*8^n*0,32^m (1, 2) 114*10^n*0,22^m = 78*47^n*0,22^m ... (47/10)^n = (114/78)*(0,22^m/0,22^m) .. n = ln(114/78)/ln(47/10) = 0,245 (1, 3) 114*100,245*0,22^m = 114*80,245*0,32^m ...(0,32/0,22)^m = 114*100,245/(114*80,245) ..m=ln(114*100,245/(114*80,245))/ln(0,32/0,22) = 0,146 K = 114*10^0,245*0,22^0,146 = 161 Answer: n = 0,245, m = 0,146, K = 161 b) Calculation: v*T^n*f^m = K ... T^n = K/(v*f^m) ... T = K/(v*f^m)^1/n = 161/(1,5*60m/min*0,28^0,146)^1/0,245 Answer: T = 23 min

Dry machining is being considered by machine shops because of certain problems inherent in the use of cutting fluids. What are those problems associated with the use of cutting fluids?

Cutting fluids become contaminated over time with a variety of contaminants, including tramp oil, garbage and bacteria. In addition to causing odours and health hazards, contaminated cutting fluids do not perform their lubricating function as well as when they are fresh and clean.

4. In a production turning operation, the workpart is 125 mm in diameter and 300 mm long. A feed of 0.225 mm/rev is used in the operation. If cutting speed = 3.0 m/s, the tool must be changed every 5 workparts; but if cutting speed = 2.0 m/s, the tool can be used to produce 25 pieces between tool changes. Determine the Taylor tool life equation for this job. Formula: Tm = pi*Do*L/f*v, v*T^n = C

T1 = 5*pi*300*125/(0,225*180*10^3 (mm/min)) = 14,5 min T2 = 25*pi*300*125/(0,225*120*10^3 (mm/min)) = 109 min (1) C = v*T^n = 3*14,5^n (2) C = v*T^n = 2*109^n (1, 2), 3*14,5^n = 2*109^n n = 0,201 C = 180 (14.54)^0.2012 = 308,43