Machinery''''s Handbook 27th Episode 2 Part 5 potx

MACHINING ECONOMETRICS 1115 Number of parts before tool change = N ch = 90/3 = 30 parts. Cycle time before tool change = T CYC = 30 × (3 + 3) = 180 minutes Example 5: Given cutting time, t c = 1 minute, idle time t i = 1 minute, N ch = 100 parts, cal- culate the tool-life T required to complete the job without a tool change, and the cycle time before a tool change is required. Tool-life = T = N ch × t c = 100 × 1 = 100 minutes. Cycle time before tool change = T CYC = 100 × (1 + 1) = 200 minutes. Calculation of Cost of Cutting and Grinding Operations.—When machining data var- ies, the cost of cutting, tool changing, and tooling will change, but the costs of idle and slack time are considered constant. Cost of Cutting per Batch: C C = H R × T C /60 T C = cutting time per batch = (number of parts) × t c , minutes, or when determining time for tool change T Cch = N ch × t c minutes = cutting time before tool change. t c = Cutting time/part, minutes H R = Hourly Rate Cost of Tool Changes per Batch: where T = tool-life, minutes, and T RPL = time for replacing a worn edge(s), or tool for regrinding, minutes Cost of Tooling per Batch: Including cutting tools and holders, but without tool changing costs, Cost of Tooling + Tool Changes per Batch: Including cutting tools, holders, and tool changing costs, Total Cost of Cutting per Batch: Equivalent Tooling-cost Time, T V : The two previous expressions can be simplified by using thus: C CH H R 60 T C T RPL T ××= $ min min⋅ $= C TOOL H R 60 T C 60C E H R T ××= $ min min min hr $ hr $ ⋅⋅ min ⋅⋅ $= C TOOL C CH +() H R 60 T C × T RPL 60C E H R + T ×= C TOT H R 60 T C × 1 T RPL 60C E H R + T + ⎝⎠ ⎜⎟ ⎜⎟ ⎜⎟ ⎛⎞ = T V T RPL 60C E H R += C TOOL C CH +() H R 60 T C × T V T ×= Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 1116 MACHINING ECONOMETRICS C E = cost per edge(s) is determined using two alternate formulas, depending on whether tools are reground or inserts are replaced: Cost per Edge, Tools for Regrinding Cost per Edge, Tools with Inserts: Note: In practice allow for insert failures by multiplying the insert cost by 4/3, that is, assuming only 3 out of 4 edges can be effectively used. Example 6, Cost per Edge–Tools for Regrinding:Use the data in the table below to cal- culate the cost per edge(s) C E , and the equivalent tooling-cost time T V , for a drill. Using the cost per edge formula for reground tools, C E = (40 + 5 × 6) ÷ (1 + 5) = $6.80 When the hourly rate is $50/hr, Calculate economic tool-life using thus, T E = 9.17 × (1/0.25 – 1) = 9.16 × 3 = 27.48 minutes. Having determined, elsewhere, the economic cutting time per piece to be t cE = 1.5 min- utes, for a batch size = 1000 calculate: Cost of Tooling + Tool Change per Batch: Total Cost of Cutting per Batch: Example 7, Cost per Edge–Tools with Inserts: Use data from the table below to calculate the cost of tooling and tool changes, and the total cost of cutting. For face milling, multiply insert price by safety factor 4/3 then calculate the cost per edge: C E =10 × (5/3) × (4/3) + 750/500 = 23.72 per set of edges When the hourly rate is $50, equivalent tooling-cost time is T V = 2 + 23.72 × 60/50 = 30.466 minutes (first line in table below). The economic tool-life for Taylor slope n = 0.333 would be T E = 30.466 × (1/0.333 –1) = 30.466 × 2 = 61 minutes. When the hourly rate is $25, equivalent tooling-cost time is T V = 2 + 23.72 × 60/25 = 58.928 minutes (second line in table below). The economic tool-life for Taylor slope n = 0.333 would be T E = 58.928 × (1/0.333 –1) =58.928 × 2 = 118 minutes. Time for cutter replacement T RPL , minute Cutter Price, $ Cost per regrind, $ Number of regrinds Hourly shop rate, $ Batch size Taylor slope, n Economic cutting time, t cE minute 1 40 6 5 50 1000 0.25 1.5 C TOT H R 60 T C × 1 T V T + ⎝⎠ ⎛⎞ = C E cost of tool number of regrinds cost/regrind×()+ 1 number of regrinds+ = C E cost of insert(s) number of edges per insert cost of cutter body cutter body life in number of edges += T V T RPL 60C E H R +1 60 6.8() 50 + 9.16minutes=== T E T V 1 n 1– ⎝⎠ ⎛⎞ ×= C TOOL C CH +() H R 60 T C × T V T × 50 60 1000 1.5×× 9.16 27.48 × $ 417== = C TOT H R 60 T C × 1 T V T + ⎝⎠ ⎛⎞ 50 60 1000 1.5×× 1 9.16 27.48 + ⎝⎠ ⎛⎞ × $ 1617== = Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY MACHINING ECONOMETRICS 1117 With above data for the face mill, and after having determined the economic cutting time as t cE = 1.5 minutes, calculate for a batch size = 1000 and $50 per hour rate: Cost of Tooling + Tool Change per Batch: Total Cost of Cutting per Batch: Similarly, at the $25/hour shop rate, (C TOOL + C CH ) and C TOT are $312 and $937, respec- tively. Example 8, Turning: Production parts were run in the shop at feed/rev = 0.25 mm. One series was run with speed V 1 = 200 m/min and tool-life was T 1 = 45 minutes. Another was run with speed V 2 = 263 m/min and tool-life was T 2 = 15 minutes. Given idle time t i = 1 minute, cutting distance Dist =1000 mm, work diameter D = 50 mm. First, calculate Taylor slope, n, using Taylor’s equation V 1 × T 1 n = V 2 × T 2 n , as follows: Economic tool-life T E is next calculated using the equivalent tooling-cost time T V , as described previously. Assuming a calculated value of T V = 4 minutes, then T E can be calcu- lated from Economic cutting speed, V E can be found using Taylor’s equation again, this time using the economic tool-life, as follows, Using the process data, the remaining economic parameters can be calculated as follows: Economic spindle rpm, rpm E = (1000V E )/(πD) = (1000 × 278)/(3.1416 × 50) = 1770 rpm Economic feed rate, F RE = f × rpm E = 0.25 × 1770 = 443 mm/min Economic cutting time, t cE = Dist/ F RE =1000/ 443 = 2.259 minutes Economic number of parts before tool change, N chE = T E ÷ t cE =12 ÷ 2.259 = 5.31 parts Economic cycle time before tool change, T CYCE = N chE × (t c + t i ) = 5.31 × (2.259 + 1) = 17.3 minutes. Time for replace- ment of inserts T RPL , minutes Number of inserts Price per insert Edges per insert Cutter Price Edges per cutter Cost per set of edges, C E Hourly shop rate T V min- utes Face mill 2 10 5 3 750 500 23.72 50 30.466 2 10 5 3 750 500 23.72 25 58.928 End mill 1 3 6 2 75 200 4.375 50 6.25 Turning 1 1 5 3 50 100 2.72 30 6.44 C TOOL C CH + ( ) H R 60 T C × T V T × 50 60 1000 1.5×× 30.466 61 × $ 624== = C TOT H R 60 T C × 1 T V T + ⎝⎠ ⎛⎞ 50 60 1000 1.5×× 1 30.466 61 + ⎝⎠ ⎛⎞ × $ 1874== = n ln V 1 V 2 ln T 2 T 1 ÷ ln 200 263 ln 15 45 ÷ 0.25== = T E T V 1 n 1– ⎝⎠ ⎛⎞ × 4 1 0.25 1– ⎝⎠ ⎛⎞ × 12 minutes== = V E1 T E () n × V 2 T 2 () n ×= V E1 V 2 T 2 T E ⎝⎠ ⎛⎞ n × 263 15 12 ⎝⎠ ⎛⎞ 0.25 × 278 m/min== = Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 1118 MACHINING ECONOMETRICS Variation Of Tooling And Total Cost With The Selection Of Feeds And Speeds It is a well-known fact that tool-life is reduced when either feed or cutting speed is increased. When a higher feed/rev is selected, the cutting speed must be decreased in order to maintain tool-life. However, a higher feed rate (feed rate = feed/rev × rpm, mm/min) can result in a longer tool-life if proper cutting data are applied. Optimized cutting data require accurate machinability databases and a computer program to analyze the options. Reason- ably accurate optimized results can be obtained by selecting a large feed/rev or tooth, and then calculating the economic tool-life T E . Because the cost versus feed or ECT curve is shallow around the true minimum point, i.e., the global optimum, the error in applying a large feed is small compared with the exact solution. Once a feed has been determined, the economic cutting speed V E can be found by calcu- lating the Taylor slope, and the time/cost calculations can be completed using the formulas described in last section. The remainder of this section contains examples useful for demonstrating the required procedures. Global optimum may or may not be reached, and tooling cost may or may not be reduced, compared to currently used data. However, the following examples prove that significant time and cost reductions are achievable in today’s industry. Note: Starting values of reasonable feeds in mm/rev can be found in the Handbook speed and feed tables, see Principal Speed andFeed Tables on page 1022, by using the f avg values converted to mm as follows: feed (mm/rev) = feed (inch/rev) × 25.4 (mm/inch), thus 0.001 inch/rev = 0.001× 25.4 = 0.0254 mm/rev. When using speed and feed Tables 1 through 23, where feed values are given in thousandths of inch per revolution, simply multiply the given feed by 25.4/1000 = 0.0254, thus feed (mm/rev) = feed (0.001 inch/rev) × 0.0254 (mm/ 0.001inch). Example 9, Converting Handbook Feed Values From Inches to Millimeters: Handbook tables give feed values f opt and f avg for 4140 steel as 17 and 8 × (0.001 inch/rev) = 0.017 and 0.009 inch/rev, respectively. Convert the given feeds to mm/rev. feed = 0.017 × 25.4 = 17 × 0.0254 = 0.4318 mm/rev feed = 0.008 × 25.4 = 8 × 0.0254 = 0.2032 mm/rev Example 10, Using Handbook Tables to Find the Taylor Slope and Constant:Calculate the Taylor slope and constant, using cutting speed data for 4140 steel in Table 1 starting on page 1027, and for ASTM Class 20 grey cast iron using data from Table 4a on page 1033, as follows: For the 175–250 Brinell hardness range, and the hard tool grade, For the 175–250 Brinell hardness range, and the tough tool grade, For the 300–425 Brinell hardness range, and the hard tool grade, For the 300–425 Brinell hardness range, and the tough tool grade, For ASTM Class 20 grey cast iron, using hard ceramic, n ln V 1 V 2 ⁄() ln T 2 T 1 ⁄() ln 525 705⁄() ln 15 45⁄() 0.27== =CV 1 T 1 () n × 1458== n ln V 1 V 2 ⁄() ln T 2 T 1 ⁄() ln 235 320⁄() ln 15 45⁄() 0.28== =CV 1 T 1 () n × 685== n ln V 1 V 2 ⁄() ln T 2 T 1 ⁄() ln 330 440⁄() ln 15 45⁄() 0.26== =CV 1 T 1 () n × 894== n ln V 1 V 2 ⁄() ln T 2 T 1 ⁄() ln 125 175⁄() ln 15 45⁄() 0.31== =CV 1 T 1 () n × 401== Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY MACHINING ECONOMETRICS 1119 Selection of Optimized Data.—Fig. 22 illustrates cutting time, cycle time, number of parts before a tool change, tooling cost, and total cost, each plotted versus feed for a con- stant tool-life. Approximate minimum cost conditions can be determined using the formu- las previously given in this section. First, select a large feed/rev or tooth, and then calculate economic tool-life T E , and the economic cutting speed V E , and do all calculations using the time/cost formulas as described previously. Fig. 22. Cutting time, cycle time, number of parts before tool change, tooling cost, and total cost vs. feed for tool-life = 15 minutes, idle time = 10 s, and batch size = 1000 parts Example 11, Step by Step Procedure: Turning – Facing out:1) Select a big feed/rev, in this case f = 0.9 mm/rev (0.035 inch/rev). A Taylor slope n is first determined using the Handbook tables and the method described in Example 10. In this example, use n = 0.35 and C = 280. 2) Calculate T V from the tooling cost parameters: If cost of insert = $7.50; edges per insert = 2; cost of tool holder = $100; life of holder = 100 insert sets; and for tools with inserts, allowance for insert failures = cost per insert by 4/3, assuming only 3 out of 4 edges can be effectively used. Then, cost per edge = C E is calculated as follows: The time for replacing a worn edge of the facing insert =T RPL = 2.24 minutes. Assuming an hourly rate H R = $50/hour, calculate the equivalent tooling-cost time T V T V = T RPL + 60 × C E /H R =2.24 +60 × 6/50 = 9.44 minutes 3) Determine economic tool-life T E T E = T V × (1/n − 1) = 9.44 × (1/ 0.35 − 1) = 17.5 minutes 4) Determine economic cutting speed using the Handbook tables using the method shown in Example 10, m/min = 280 / 17.5 0.35 = 103 m/min n ln V 1 V 2 ⁄() ln T 2 T 1 ⁄() ln 1490 2220⁄() ln 15 45⁄() 0.36== =CV 1 T 1 () n × 5932== 0.001 0.01 0.1 1 10 100 1000 0.01 0.1 1 10 f, mm/rev t c t cyc # parts C TOOL C TOT C E cost of insert(s) number of edges per insert cost of cutter body cutter body life in number of edges += 7.50 4 3⁄× 2 100 100 +$6.00== V E CT E n ⁄= Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 1120 MACHINING ECONOMETRICS 5) Determine cost of tooling per batch (cutting tools, holders and tool changing) then total cost of cutting per batch: C TOOL =H R × T C × (C E /T)/60 (C TOOL + C CH )=H R × T C × ((T RPL +C E /T)/60 C TOT =H R × T C (1 + (T RPL +C E )/T) Example 12, Face Milling – Minimum Cost : This example demonstrates how a modern firm, using the formulas previously described, can determine optimal data. It is here applied to a face mill with 10 teeth, milling a 1045 type steel, and the radial depth versus the cutter diameter is 0.8. The V–ECT–T curves for tool-lives 5, 22, and 120 minutes for this operation are shown in Fig. 23a. Fig. 23a. Cutting speed vs. ECT, tool-life constant The global cost minimum occurs along the G-curve, see Fig. 6c and Fig. 23a, where the 45-degree lines defines this curve. Optimum ECT is in the range 1.5 to 2 mm. For face and end milling operations, ECT = z × f z × ar/D × aa/CEL ÷ π. The ratio aa/CEL = 0.95 for lead angle LA = 0, and for ar/D = 0.8 and 10 teeth, using the formula to calculate the feed/tooth range gives for ECT = 1.5, f z = 0.62 mm and for ECT = 2, f z = 0.83 mm. Fig. 23b. Cutting time per part vs. feed per tooth Using computer simulation, the minimum cost occurs approximately where Fig. 23a indicates it should be. Total cost has a global minimum at f z around 0.6 to 0.7 mm and a speed of around 110 m/min. ECT is about 1.9 mm and the optimal cutter life is T O = 22 min- utes. Because it may be impossible to reach the optimum feed value due to tool breakage, 10 100 1000 0.1 1 10 ECT, mm V, m/min G-CURVE T = 5 T = 22 T = 120 0 0.1 0.2 0.3 0.4 0.5 0.6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 f z t c T = 5 T = 22 T = 120 Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY LIVE GRAPH Click here to view LIVE GRAPH Click here to view MACHINING ECONOMETRICS 1121 the maximum practical feed f max is used as the optimal value. The difference in costs between a global optimum and a practical minimum cost condition is negligible, as shown in Figs. 23c and 23e. A summary of the results are shown in Figs. 23a through 23e, and Table 1. Fig. 23c. Total cost vs. feed/tooth When plotting cutting time/part, t c , versus feed/tooth, f z , at T = 5, 22, 120 in Figs. 23b, tool-life T = 5 minutes yields the shortest cutting time, but total cost is the highest; the min- imum occurs for f z about 0.75 mm, see Figs. 23c. The minimum for T = 120 minutes is about 0.6 mm and for T O = 22 minutes around 0.7 mm. Fig. 23d. Tooling cost versus feed/tooth Fig. 23d shows that tooling cost drop off quickly when increasing feed from 0.1 to 0.3 to 0.4 mm, and then diminishes slowly and is almost constant up to 0.7 to 0.8 mm/tooth. It is generally very high at the short tool-life 5 minutes, while tooling cost of optimal tool-life 22 minutes is about 3 times higher than when going slow at T =120 minutes. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 fz, mm C TOT, $ 0.01 0.06 0.11 0.16 0.21 0.26 0.31 T = 120 T = 22 T = 5 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 f z , mm Unit Tooling Cost, $ 0 10.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 T = 5 T = 22 T =120 Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY MACHINING ECONOMETRICS 1123 In the 1950’s it was discovered that cutting speed could be raised by a factor of 5 to 10 when hobbing steel with HSS cutters. This is another example of being on the wrong side of the Taylor curve. One of the first reports on high-speed end milling using high-speed steel (HSS) and car- bide cutters for milling 6061-T651 and A356-T6 aluminum was reported in a study funded by Defense Advanced Research Project Agency (DARPA). Cutting speeds of up to 4400 m/min (14140 fpm) were used. Maximum tool-lives of 20 through 40 minutes were obtained when the feed/tooth was 0.2 through 0.25 mm (0.008 to 0.01 inch), or measured in terms of ECT around 0.07 to 0.09 mm. Lower or higher feed/tooth resulted in shorter cutter lives. The same types of previously described curves, namely T–ECT curves with maximum tool-life along the H-curve, were produced. When examining the influence of ECT, or feed/rev, or feed/tooth, it is found that too small values cause chipping, vibrations, and poor surface finish. This is caused by inade- quate (too small) chip thickness, and as a result the material is not cut but plowed away or scratched, due to the fact that operating conditions are on the wrong (left) side of the tool- life versus ECT curve (T-ECT with constant speed plotted). There is a great difference in the thickness of chips produced by a tooth traveling through the cutting arc in the milling process, depending on how the center of the cutter is placed in relation to the workpiece centerline, in the feed direction. Although end and face milling cut in the same way, from a geometry and kinematics standpoint they are in practice distin- guished by the cutter center placement away from, or close to, the work centerline, respec- tively, because of the effect of cutter placement on chip thickness. This is the criteria used to distinguishing between the end and face milling processes in the following. Depth of Cut/Cutter Diameter, ar/D is the ratio of the radial depth of cut ar and the cutter diameter D. In face milling when the cutter axis points approximately to the middle of the work piece axis, eccentricity is close to zero, as illustrated in Figs. 3 and 4, page 1042, and Fig. 5 on page 1043. In end milling, ar/D = 1 for full slot milling. Mean Chip Thickness, hm is a key parameter that is used to calculate forces and power requirements in high-speed milling. If the mean chip thickness hm is too small, which may occur when feed/tooth is too small (this holds for all milling operations), or when ar/D decreases (this holds for ball nose as well as for straight end mills), then cutting occurs on the left (wrong side) of the tool-life versus ECT curve, as illustrated in Figs. 6b and 6c. In order to maintain a given chip thickness in end milling, the feed/tooth has to be increased, up to 10 times for very small ar/D values in an extreme case with no run out and otherwise perfect conditions. A 10 times increase in feed/tooth results in 10 times bigger feed rates (F R ) compared to data for full slot milling (valid for ar/D = 1), yet maintain a given chip thickness. The cutter life at any given cutting speed will not be the same, how- ever. Increasing the number of teeth from say 2 to 6 increases equivalent chip thickness ECT by a factor of 3 while the mean chip thickness hm remains the same, but does not increase the feed rate to 30 (3 × 10) times bigger, because the cutting speed must be reduced. How- ever, when the ar/D ratio matches the number of teeth, such that one tooth enters when the second tooth leaves the cutting arc, then ECT = hm. Hence, ECT is proportional to the num- ber of teeth. Under ideal conditions, an increase in number of teeth z from 2 to 6 increases the feed rate by, say, 20 times, maintaining tool-life at a reduced speed. In practice about 5 times greater feed rates can be expected for small ar/D ratios (0.01 to 0.02), and up to 10 times with 3 times as many teeth. So, high-speed end milling is no mystery. Chip Geometry in End and Face Milling.—Fig. 24 illustrates how the chip forming process develops differently in face and end milling, and how mean chip thickness hm var- ies with the angle of engagement AE, which depends on the ar/D ratio. The pertinent chip geometry formulas are given in the text that follows. Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY MACHINING ECONOMETRICS 1125 Table 2a. Variation of Chip Thickness and f z /f z0 with ar/D In Table 2a, a standard value f z0 = 0.17 mm/tooth (commonly recommended average feed) was used, but the f z /f z0 values are independent of the value of feed/tooth, and the pre- viously mentioned relationships are valid whether f z0 = 0.17 or any other value. In both end and face milling, hm = 0.108 mm for f z0 = 0.17mm when ar/D =1. When the f z /f z0 ratio = 1, hm = 0.15 for face milling, and 0.108 in end milling both at ar/D = 1 and 0.5. The tabulated data hold for perfect milling conditions, such as, zero run-out and accurate sharpening of all teeth and edges. Mean Chip Thickness hm and Equivalent Chip Thickness ECT.—The basic formula for equivalent chip thickness ECT for any milling process is: ECT = f z × z/π × (ar/D) × aa/CEL, where f z = feed/tooth, z = number of teeth, D = cutter diameter, ar = radial depth of cut, aa = axial depth of cut, and CEL = cutting edge length. As a function of mean chip thickness hm: ECT = hm × (z/2) × (AE/180), where AE = angle of engagement. Both terms are exactly equal when one tooth engages as soon as the preceding tooth leaves the cutting section. Mathematically, hm = ECT when z = 360/AE; thus: for face milling, AE = arccos (1 – 2 × (ar/D) 2 ) for end milling, AE = arccos (1 – 2 × (ar/D)) Calculation of Equivalent Chip Thickness (ECT) versus Feed/tooth and Number of teeth.: Table 2b is a continuation of Table 2a, showing the values of ECT for face and end milling for decreasing values ar/D, and the resulting ECT when multiplied by the f z /f z0 ratio f z0 = 0.17 (based on hm = 0.108). Small ar/D ratios produce too small mean chip thickness for cutting chips. In practice, minimum values of hm are approximately 0.02 through 0.04 mm for both end and face milling. Formulas.— Equivalent chip thickness can be calculated for other values of f z and z by means of the following formulas: Face milling: ECT F = ECT 0F × (z/8) × (f z /0.17) × (aa/CEL) or, if ECT F is known calculate f z using: f z = 0.17 × (ECT F /ECT 0F ) × (8/z) × (CEL/aa) ar/D Face Milling End Milling (straight) ecentricity e=0 z=8 f z0 = 0.017 cos AE = 1 − 2 × (ar/D) 2 z=2 f z0 = 0.017 cos AE = 1 − 2 × (ar/D) AE hm/f z hm ECT/hm f z /f z0 AE hm/f z hm ECT/hm f z /f z0 1.0000 180.000 0.637 0.108 5.000 1.398 180.000 0.637 0.108 1.000 1.000 0.9000 128.316 0.804 0.137 3.564 1.107 143.130 0.721 0.122 0.795 0.884 0.8000 106.260 0.863 0.147 2.952 1.032 126.870 0.723 0.123 0.711 0.881 0.7355 94.702 0.890 0.151 2.631 1.000 118.102 0.714 0.122 0.667 0.892 0.6137 75.715 0.929 0.158 1.683 0.958 103.144 0.682 0.116 0.573 0.934 0.5000 60.000 0.162 0.932 0.216 0.202 90.000 0.674 0.115 0.558 1.000 0.3930 46.282 0.973 0.165 1.028 0.915 77.643 0.580 0.099 0.431 1.098 0.2170 25.066 0.992 0.169 0.557 0.897 55.528 0.448 0.076 0.308 1.422 0.1250 14.361 0.997 0.170 0.319 0.892 41.410 0.346 0.059 0.230 1.840 0.0625 7.167 0.999 0.170 0.159 0.891 28.955 0.247 0.042 0.161 2.574 0.0300 3.438 1.000 0.170 0.076 0.890 19.948 0.172 0.029 0.111 3.694 0.0100 1.146 1.000 0.170 0.025 0.890 11.478 0.100 0.017 0.064 6.377 0.0010 0.115 1.000 0.000 0.000 0.890 3.624 0.000 0.000 0.000 20.135 Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY [...]... 30 26 25 24 23 28 27 22 29 21 20 19 35 75 1 .5 18 17 40 16 15 14 75 1 .5 45 75 1 .5 50 800 900 1000 1 25 0 55 Inch 0 1 mm 14 18 14 18 14 18 2 3 5 10 15 20 25 10 14 8 12 10 14 10 14 6 10 4 6 8 4 6 9 1 11 4 1 .5 2. 5 9 2 3 75 8 2 3 5 8 1 1 2 11 10 1 .5 2. 5 3 4 5 8 7 12 150 100 4 6 6 10 6 10 13 1 .5 2. 5 500 450 400 350 300 25 0 20 0 50 5 8 8 12 8 12 4 5 6 700 600 7 2 3 3 4 6 5 3 4 13 4 2 1 4 1 23 4 3 31 4 3 2 2 2. .. 18-18 -2, 309, Ferralium 15- 5PH, 17-4PH, 17-7PH, 22 05, 310, AM 350 , AM 355 , Custom 450 , Custom 455 , PH13-8Mo, PH14-8Mo, PH 15- 7Mo 22 -13 -5, Nitronic 50 , 60 Speed (fpm) 50 0 Speed (m/min) 1 52 360 300 24 0 23 0 1 85 180 170 150 1 45 140 1 25 120 100 80 60 65 60 450 400 3 75 350 3 15 28 5 26 5 24 5 23 5 23 0 21 5 120 90 900 85 100 90 85 80 75 70 110 91 73 70 56 55 52 46 44 43 38 37 30 24 18 20 18 137 122 114 107 96 87 81 75. .. (50 0 05) , A536 (80 -55 -06) A48 (20 ksi) A536 (100-70-03) A48 (40 ksi) A 220 (60004) A436 (1B) A 220 (70003) A436 (2) A 220 (800 02) , A436 (2B) A536 ( 120 -90- 02) A 220 (90001), A48 (60 ksi) A439 (D -2) A439 (D-2B) WF-11 Astroloy M 356 , 360 353 187, 14 52 380, 54 4 173, 9 32, 934 330, 3 65 623 , 624 23 0, 26 0, 27 2, 28 0, 464, 6 32, 655 101, 1 02, 110, 122 , 1 72, 1 751 0, 1 82, 22 0, 51 0, 6 25 , 706, 7 15 630 811 Pyromet X- 15 A286,... Ti-3Al-8V-6Cr-4Mo-4Z, Ti-8Mo-8V-2Fe-3Al Ti-2Al-11Sn-5Zr-1Mo, Ti-5Al -2. 5Sn, Ti-6Al-2Sn-4Zr-2Mo Ti-6Al-4V Ti-7Al-4Mo, Ti-8Al-1Mo-1V Speed (fpm) 4 25 400 340 330 320 310 300 29 0 28 0 27 5 27 0 25 0 24 0 22 5 22 0 21 0 20 0 190 1 85 180 1 75 160 140 1 25 110 1 05 100 90 70 80 75 70 65 Speed (m/min) 130 122 104 101 98 94 91 88 85 84 82 76 73 69 67 64 61 58 56 55 53 49 43 38 34 32 30 27 21 24 23 21 20 The speed figures given... Rev 50 110 0.004 30 60 70 150 0.0 05 40 75 70 150 0.004 40 75 70 150 0.003 40 75 70 150 0.0 02 40 75 70 150 0.00 15 40 75 70 150 0.006 40 75 50 110 0.001 30 75 50 110 0.0 02 30 75 80 150 0.0004 50 85 80 150 0.001 50 85 80 150 0.001 50 85 80 150 0.00 05 50 85 30 … … 14 … 30 40 … 16 20 40 60 0.0006 30 45 40 60 0.0008 30 45 40 60 0.00 12 30 45 40 60 0.0016 30 45 40 75 0.0 02 30 60 40 75 0.003 30 60 40 75 0.003... 70 150 0.008 40 85 70 150 0.008 40 85 70 150 0.006 40 85 70 150 0.00 45 40 85 70 150 0.008 40 85 150 … 0.010 1 05 … 150 … 0. 0 25 1 05 … 150 … 0.0 02 1 05 … 150 … 0.003 1 05 … 150 … 0.0 02 1 05 … 150 … 0.004 1 05 … 70 150 0.00 05 40 80 70 150 0.0008 40 80 70 1 05 0.006 – 0.004 40 60 70 1 05 0.006 – 0.008 40 60 70 150 0.0004 40 75 70 150 0.0 02 40 75 70 1 05 0.00 15 40 60 70 1 05 0.0004 40 60 70 150 0.00 05 40 85 70 150 ... 1 0 25 10 35 1018, 1 021 , 1 022 , 1 026 , 151 3, A2 42 Cor-Ten A 1137 1141, 1144, 1144 Hi Stress 41L40 1040, 4130, A2 42 Cor-Ten B, (A36 Shapes) 10 42, 154 1, 4140, 41 42 86 15, 8 620 , 8 622 W-1 1044, 10 45, 1330, 4340, E4340, 51 60, 8630 13 45, 41 45, 6 150 1060, 4 150 , 8640, A-6, O-1, S-1 H-11, H- 12, H-13, L-6, O-6 10 95 A -2 E9310 300M, A-10, E 52 1 00, HY-80, HY-100 S -5 S-7 M-1 HP 9-4 -20 , HP 9-4 - 25 M -2, M- 42, T1 D -2 T- 15. .. 0.3 32 0.303 0 . 25 2 0 .24 3 0 .20 6 0.089 0. 051 0. 026 0.0 12 0.004 0.0 02 End Milling (straight) ECT0 corrected for fz/fz0 0 .57 5 0.410 0.344 0.303 0 .24 2 0 .23 1 0.1 92 0.080 0.046 0. 023 0.011 0.004 0.0 02 hm 0.108 0. 122 0. 123 0. 121 0.116 0.1 15 0.108 0.076 0. 059 0.0 42 0. 029 0.017 0.0 05 fz/fz0 1.000 0.884 0.880 0.8 92 0.934 0.9 45 1.000 1. 422 1.840 2. 57 4 3.694 6.377 20 .1 35 ECT 0.103 0.093 0.083 0.076 0.063 0.061 0. 051 ... 72 70 66 37 27 27 4 26 30 27 26 24 23 21 65 60 20 18 50 190 150 140 120 1 15 110 100 95 90 80 15 58 46 43 37 35 34 30 29 27 24 60 18 Copyright 20 04, Industrial Press, Inc., New York, NY Machinery's Handbook 27 th Edition 11 42 BAND SAW BLADES Bimetal Band Saw Speeds for Cutting 4-Inch Material with Coolant (Continued) Material Steel Titanium Category (AISI/SAE) 12L14 121 3, 121 5 1117 1030 1008, 10 15, 1 020 ,... 100 90 80 70 60 50 40 30 20 10 0 ft/min 40 m/min 12 Click here to view 80 24 120 37 Starting Feed Rate 160 49 20 0 61 24 0 73 28 0 85 320 98 360 110 Break-In Area Band Speed (Machinability) LIVE GRAPH Total Break-In Area Required in .2 Click here to view 100 90 80 70 60 50 40 30 20 10 0 ft/min 40 80 120 160 20 0 24 0 28 0 m/min 12 24 37 49 61 73 85 cm2 6 45 580 51 5 450 3 85 320 26 0 1 95 130 65 0 320 98 360 110 . material. 700 900 800 600 50 0 450 400 350 300 25 0 20 0 150 100 75 50 25 20 15 10 5 1 25 0 1000 mm 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 35 40 45 50 55 Inch 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 1 10. 12 6 10 8 12 6 10 5 8 4 6 5 8 5 8 4 6 4 6 3 4 3 4 2 3 6 10 3 4 2 3 1 .5 2. 5 1 .5 2. 5 . 75 1 .5 . 75 1 .5 . 75 1 .5 1 4 2 2 3 1 4 3 1 4 1 1 2 2 1 2 3 1 2 1 3 4 2 3 4 3 3 4 8 12 10 14 14 18 2 3 1 .5 2. 5 Machinery's. mill 2 10 5 3 750 50 0 23 . 72 50 30.466 2 10 5 3 750 50 0 23 . 72 25 58. 928 End mill 1 3 6 2 75 20 0 4.3 75 50 6 . 25 Turning 1 1 5 3 50 100 2. 72 30 6.44 C TOOL C CH + ( ) H R 60 T C × T V T × 50 60