Question & Answer

Verify that the derivative of Eq. (24.6) results in Eq. (24.7)

Solve Problem 24.26 only using maximum production rate as the objective rather than minimum piece cost. Use of a spreadsheet calculator is recommended

In a turning operation on ductile steel, it is desired to obtain an actual surface roughness of 63 µ-in with a 2/64 in nose radius tool. The ideal roughness is given by Eq. (24.1) and an adjustment will have to be made using Figure 24.2 to convert the 63 µ-in actual roughness to an ideal roughness, taking into account the material and cutting speed. Disposable inserts are used at a cost of $1.75 per cutting edge (each insert costs $7.00 and there are four edges per insert). Average time to change each insert = 1.0 min. The workpiece length = 30.0 in and its diameter = 3.5 in. The machine and operator’s rate = $39.00 per hour including applicable overheads. The Taylor tool life equation for this tool and work combination is given by: vT0.23 f 0.55 = 40.75, where T = tool life, min; v = cutting speed, ft/min; and f = feed, in/rev. Solve for (a) the feed in in/rev that will achieve the desired actual finish, (b) cutting speed for minimum cost per piece at the feed determined in (a). Hint: To solve (a) and (b) requires an iterative computational procedure. Use of a spreadsheet calculator is recommended for this iterative procedure

As indicated in Section 23.4, the effect of a cutting fluid is to increase the value of C in the Taylor tool life equation. In a certain machining situation using HSS tooling, the C value is increased from C = 200 to C = 225 due to the use of the cutting fluid. The n value is the same with or without fluid at n = 0.125. Cutting speed used in the operation is v = 125 ft/min. Feed = 0.010 in/rev and depth = 0.100 in. The effect of the cutting fluid can be to either increase cutting speed (at the same tool life) or increase tool life (at the same cutting speed). (a) What is the cutting speed that would result from using the cutting fluid if tool life remains the same as with no fluid? (b) What is the tool life thatwould result if the cutting speed remained at 125 ft/min? (c) Economically, which effect is better, given that tooling cost = $2.00 per cutting edge, tool change time = 2.5 min, and operator and machine rate = $30/hr? Justify you answer with calculations, using cost per cubic in of metal machined as the criterion of comparison. Ignore effects of workpart handling time.

An NC lathe cuts two passes across a cylindrical workpiece under automatic cycle. The operator loads and unloads the machine. The starting diameter of the work is 3.00 in and its length = 10 in. The work cycle consists of the following steps (with element times given in parentheses where applicable): 1 - Operator loads part into machine, starts cycle (1.00 min); 2 - NC lathe positions tool for first pass (0.10 min); 3 - NC lathe turns first pass (time depends on cutting speed); 4 - NC lathe repositions tool for second pass (0.4 min); 5 - NC lathe turns second pass (time depends on cutting speed); and 6 - Operator unloads part and places in tote pan (1.00 min). In addition, the cutting tool must be periodically changed. This tool change time takes 1.00 min. The feed rate = 0.007 in/rev and the depth of cut for each pass = 0.100 in. The cost of the operator and machine = $39/hr and the tool cost = $2.00/cutting edge. The applicable Taylor tool life equation has parameters: n = 0.26 and C = 900 (ft/min). Determine (a) the cutting speed for minimum cost per piece, (b) the average time required to complete one production cycle, (c) cost of the production cycle. (d) If the setup time for this job is 3.0 hours and the batch size = 300 parts, how long will it take to complete the batch?

A vertical boring mill is used to bore the inside diameter of a large batch of tube-shaped parts. The diameter = 28.0 in and the length of the bore = 14.0 in. Current cutting conditions are: speed = 200 ft/min, feed = 0.015 in/rev, and depth = 0.125 in. The parameters of the Taylor equation for the cutting tool in the operation are: n = 0.23 and C = 850 (ft/min). Tool change time = 3.0 min, and tooling cost = $3.50 per cutting edge. The time required to load and unload the parts = 12.0 min, and the cost of machine time on this boring mill = $42.00/hr. Management has decreed that the production rate must be increased by 25%. Is that possible? Assume that feed must remain unchanged in order to achieve the required surface finish. What is the current production rate and the maximum possible production rate for this job?

Solve Problem 24.21 except that in parts (a) and (b), determine the cutting speeds and tool lives for maximum production rate. Use of a spreadsheet calculator is recommended

Three tool materials are to be compared for the same finish turning operation on a batch of 150 steel parts: high-speed steel, cemented carbide, and ceramic. For the high-speed steel tool, the Taylor equation parameters are: n = 0.130 and C = 80 (m/min). The price of the HSS tool is $20.00 and it is estimated that it can be ground and reground 15 times at a cost of $2.00 per grind. Tool change time is 3 min. Both carbide and ceramic tools are in insert form and can be held in the same mechanical toolholder. The Taylor equation parameters for the cemented carbide are: n = 0.30 and C = 650 (m/min); and for the ceramic: n = 0.6 and C = 3,500 (m/min). The cost per insert for the carbide is $8.00 and for the ceramic is $10.00. There are 6 cutting edges per insert in both cases. Tool change time is 1.0 min for both tools. The time to change a part is 2.5 min. The feed is 0.30 mm/rev, and depth of cut is 3.5 mm. The cost of machine time is $40/hr. The part is 73.0 mm in diameter and 250 mm in length. Setup time for the batch is 2.0 hr. For the three tooling cases, compare: (a) cutting speeds for minimum cost, (b) tool lives, (c) cycle time, (d) cost per production unit, (e) total time to complete the batch and production rate. (f) What is the proportion of time spent actually cutting metal for each tooling? Use of a spreadsheet calculator is recommended.

Solve Problem 24.19 except that in part (a), determine the cutting speeds for maximum production rate

Compare disposable and regrindable tooling. The same grade of cemented carbide tooling is available in two forms for turning operations in a certain machine shop: disposable inserts and brazed inserts. The parameters in the Taylor equation for this grade are: n = 0.25 and C = 300 (m/min) under the cutting conditions considered here. For the disposable inserts, price of each insert = $6.00, there are four cutting edges per insert, and the tool change time = 1.0 min (this is an average of the time to index the insert and the time to replace it when all edges have been used). For the brazed insert, the price of the tool = $30.00 and it is estimated that it can be used a total of 15 times before it must be scrapped. The tool change time for the regrindable tooling = 3.0 min. The standard time to grind or regrind the cutting edge is 5.0 min, and the grinder is paid at a rate = $20.00/hr. Machine time on the lathe costs $24.00/hr. The workpart to be used in the comparison is 375 mm long and 62.5 mm in diameter, and it takes 2.0 min to load and unload the work. The feed = 0.30 mm/rev. For the two tooling cases, compare (a) cutting speeds for minimum cost, (b) tool lives, (c) cycle time and cost per unit of production. Which tool would you recommend?

Solve Problem 24.17 except that in part (a), determine cutting speed for minimum cost.

A cemented carbide tool is used to turn a part with a length of 14.0 in and diameter = 4.0 in. The parameters in the Taylor equation are: n = 0.25 and C = 1000 (ft/min). The rate for the operator and machine tool = $45.00/hr, and the tooling cost per cutting edge = $2.50. It takes 2.5 min to load and unload the workpart and 1.50 min to change tools. The feed = 0.015 in/rev. Determine (a) cutting speed for maximum production rate, (b) tool life in min of cutting, and (c) cycle time and cost per unit of product.

Solve Problem 24.15 except that in part (a), determine cutting speed for minimum cost.

A high-speed steel tool is used to turn a steel workpart that is 300 mm long and 80 mm in diameter. The parameters in the Taylor equation are: n = 0.13 and C = 75 (m/min) for a feed of 0.4 mm/rev. The operator and machine tool rate = $30.00/hr, and the tooling cost per cutting edge = $4.00. It takes 2.0 min to load and unload the workpart and 3.50 min to change tools. Determine (a) cutting speed for maximum production rate, (b) tool life in min of cutting, and (c) cycle time and cost per unit of product.

A turning operation is to be performed on C1010 steel, which is a ductile grade. It is desired to achieve a surface finish of 64 µ-in, while at the same time maximizing the metal removal rate. It has been decided that the speed should be in the range 200 ft/min to 400 ft/min, and that the depth of cut will be 0.080 in. The tool nose radius = 3/64 in. Determine the speed and feed combination that meets these criteria.

A face milling operation is not yielding the required surface finish on the work. The cutter is a four-tooth insert type face milling cutter. The machine shop foreman thinks the problem is that the work material is too ductile for the job, but this property tests well within the ductility range for the material specified by the designer. Without knowing any more about the job, what changes in (a) cutting conditions and (b) tooling would you suggest to improve the surface finish?

A face milling operation is to be performed on a cast iron part to finish the surface to 36 µ-in. The cutter uses four inserts and its diameter is 3.0 in. The cutter rotates at 475 rev/min. To obtain the best possible finish, a type of carbide insert with 4/64 in nose radius is to be used. Determine the required feed rate (in/min) that will achieve the 36 µ-in finish.

The surface finish specification in a turning job is 0.8 µm. The work material is cast iron. Cutting speed = 75 m/min, feed = 0.3 mm/rev, and depth of cut = 4.0 mm. The nose radius of the cutting tool must be selected. Determine the minimum nose radius that will obtain the specified finish in this operation.

A part to be turned in an engine lathe must have a surface finish of 1.5 µm. The part is made of a aluminum. The cutting speed is 1.5 m/s and the depth is 3.0 mm. The nose radius on the tool = 1.0 mm. Determine the feed that will achieve the specified surface finish.

Solve previous Problem 24.8 except that the part is made of cast iron instead of aluminum and the cutting speed is reduced to 100 m/min

A part to be turned in an engine lathe must have a surface finish of 1.6 µm. The part is made of a free-machining aluminum alloy. Cutting speed = 150 m/min, and depth of cut = 4.0 mm. The nose radius on the tool = 0.75 mm. Determine the feed that will achieve the specified surface finish

A single-point HSS tool with a 3/64 in nose radius is used in a shaping operation on a ductile steel workpart. The cutting speed is 120 ft/min. The feed is 0.014 in/pass and depth of cut is 0.135 in. Determine the surface roughness for this operation.

A turning operation uses a 2/64 in nose radius cutting tool on a free machining steel with a feed rate = 0.010 in/rev and a cutting speed = 300 ft/min. Determine the surface roughness for this cut.

In a turning operation on cast iron, the nose radius on the tool = 1.5 mm, feed = 0.22 mm/rev, and speed = 1.8 m/s. Compute an estimate of the surface roughness for this cut.

Tool life turning tests have been conducted on B1112 steel with high-speed steel tooling, and the resulting parameters of the Taylor equation are: n = 0.13 and C = 225. B1112 is the base metal and has a machinability rating = 1.00 (100%). During the tests, feed = 0.010 in/rev, and depth of cut = 0.100 in. Based on this information, and machinability data given in Table 24.1, determine the cutting speed you would recommend for the following work materials, if the tool life desired in operation is 30 min (the same feed and depth of cut are to be used): (a) C1008 low carbon steel with 150 Brinell hardness, (b) 4130 alloy steel with 190 Brinell hardness, and (c) B1113 steel with 170 Brinell hardness.