ch7answers

© Pearson Education, Inc. 7-1Pearson Prentice Hall, Pearson Education, Upper Saddle River, NJ 07458Ostwald and McLaren / Cost Analysis and Estimating for Engineering and Management

Chapter 7Answers to Problems

7.1. An operation has a setup of 1.5 hr and a cycle time of 0.50 min. If the lot quantity is 75, find the lothours and unit time expressed in hours.

Answer:Lot hr = 1.5 + 75 x 0.50/60 = 2.125 hrUnit time = 2.125/75 = 0.02833 hr = 1.7 min considering the prorata share of setup.Unit time without the influence of setup is 0.50/60 = 0.008 hr

7.2 An operation is estimated to require a setup of 10 hr and a cycle time of 1 min. If the productive hourcost is $75, and the run quantity is 210 units, find the operation cost.

Answer:Lot hr = 10 + 210 x 1/60 = 13.5Cdlo = $1,012.50

7.3 A collet-type turret lathe will rough turn, finish turn, drill and tap using an end turret, and cutoff withthe cross turret. The running of the 7-in. long stock includes the following metal cutting times:

Element MinRough turn 0.89Finish turn 1.26Drill 0.73Tap 0.47Cut off of stock 0.83

Estimate setup, pieces per hour, hr/100 units, and lot hours for a run of 710 units. (Hints: There are fivespeed changes for machining RPMs. Use Tables 7.2, 7.4 and 7.5.)

Answer:Setup Description Frequency Time HrPunch in and out 1 0.2 0.2First tool 1 1.3 1.3Additional tools 4 0.3 1.2Collet 1 0.2 0.2

Total 2.9

Element Description Frequency Time MinStart, stop 1 0.08 0.08Change speed 5 0.04 0.20Air clean 1 0.06 0.06Feed stock, 7 in. + 0.23 0.23End turret locations 4 0.08 0.32Cross slide location 1 0.09 0.09Rough turn 1 0.89 0.89Finish turn 1 1.26 1.26Drill 1 0.73 0.73Tap 1 0.47 0.47Cutoff stock 1 0.83 0.83

Cycle Total 5.16

N = 710 units. Setup = 2.9hr, Unit time = 5.16 min, pc/hr = 11.6, hr/100 = 8.600 hr, lot hr = 63.96


7.4 A slot milling operation is performed on a 13-lb gray-iron casting. The setup includes an angle platewith fixture mounting and a tolerance on the dimension being machined as ± 0.002 in. Machining time is2.24 min. The shop advises that air blowing and brushing chips will be necessary. Estimate setup, piecesper hour, hr/100 units, and lot hours for a run of 1783 units.

Answer:Setup Description HrPunch in and out 0.2Angle plate 1.4Slot milling cutter 1.6Tight tolerance 0.5

Total 3.7

Cycle Element MinStart and stop 0.08Air clean 0.06Brush chips 0.14Load casting 0.25Fixture clamping on angle plate 0.76Metal cutting 2.24

Cycle total 3.53

Setup = 3.7 hr; Pc/hr = 17.0; hr/100 units = 5.883; lot hr = 108.6

7.5 An operation is estimated for a Burgmaster NC turret drill press. Ten aluminum flat parts are stacked-drilled in 37 locations. Each part weighs 2.7 lb and is 1/8 in. thick and is loaded against a rail clamp-typefixture. Four turret stations are used. Drilling time through the stack is 0.19 min per hole. Estimate setup,unit time, pieces per hour, hr/100 units, and lot hours for a run of 983 units. (Hints: Hole sizes and locationsare not pertinent to the solution.)

Answer:Setup Description Hr

Punch in and out 0.2Fixture 0.1First turret 0.75Additional 3 turrets 0.21

Total setup 1.26

Cycle Elements Frequency Time MinStart & stop 0.08Air clean 0.04Move part 10 0.13 1.30Clamp with air 0.05Raise tool, move 10 0.06 0.60Index turret 4 0.03 0.12Drilling 37 0.19 7.03

Cycle time 9.22Unit time 1/10 0.922

Setup = 1.26 hr; unit time = 0.922 min, pc/hr = 65.1, hr/100 = 1.537; lot hr = 16.4


7.6 A press operation for an automobile body fender production line has a setup of 30 hr for the form die,which costs $72,500. The operation is estimated to require 2.173 min for loading, drawing to size and trim,and unloading to the conveyor, which completes the cycle. The run for this model year is 247,500 units.Material cost for the corrosion-resistant steel sheet-metal blank including all losses included is $7.12. Theproductive hour cost for the press is $192.50.

Find the operation lot hours. Each press is expected to operate 4000 annual hours. How manypresses are necessary? Find the unit cost. Find the proportion of productive hour cost, material cost, andtool cost. (Hint: It is necessary to find the number of machines and die sets that are required for theproduction quantity.)

Answer:Lot hr = 30 + 247,500(2.173/60) = 8994 hr. No. of presses = 8994/4000 = 2.3 or 3 conservatively. Thereare 3 form die sets necessary. A slight improved no. of lot hr = 3 x 30 + 247,500(2.173/60) = 9055 hr.Cdlo = 9054 x 192.50 =$1,742,895.Cu = 1,742,895/247,500 + 7.12 + (3 x 72,500)/247,500 = $15.04.

Component Proportion, %PHC 47Material 47Tooling 6

Note. The third press is idle a good part of the time for the requirement. Arguably, this idle time can bepicked up in the PHC for non-production of assets. The instructor can raise the question of a third shift andthe effect of this on the cost.

7.7 Permanent tooling is to be evaluated for an operation and product. Let the estimated unit savings indirect labor be $0.30 for a part if tooling is used, burden on labor saved = 40%, cost of each setup = $100,interest rate = 20%, allowance for taxes and insurance = 10%, allowance for depreciation and obsolescence= 50% and allowance for maintenance = 25%.(a) The cost of a fixture is $4000. With one run per year, how many pieces must be made per year to havethe fixture pay for itself?(b) Let depreciation be 100%, because the fixture must pay for itself within a year. How large must that runbe?(c) By using the initial data, how much money can we afford for a fixture for a single run of 15,000 units atan estimated savings of $0.30 per piece?(d) How many years for payback will a $4000 fixture require for an annual quantity of 20,000 units?

Answer:(a) C = $4000, N = 4000(I + T + D + M)/(a(1 + t) + SU) = (4000(.2 + .1 + .5 + .25) + 100)/(.3(1.4)) =

10,238 units are required for the fixture to pay for itself.(b) D = 100%. N = ((.2 + .1 + 1 + .25)4000 + 100)/.3(1.4) = 15,000 units(c) N = 15,000. C = (15,000(.3)(1.4) – 100)/(.2 + .1 + .5 + .25) = $5905(d) N = 20,000 units. C = (20,000 x .3((1.4) – 100)/(.2 + .1 + .5 + .25) = $7905. Years payback =

4000/7905 = 0.5 year


7.8 (a) An operator earns $15 per hour. Handling and other constant time elements total 1.35 min. What isthe cost for the element?(b) The machining length is 20 in. and the part diameter is 4 in. Standard velocity and feed for this materialis 275 fpm and 0.020 ipr. What is the time for turning?(c) The Taylor tool life equation is VT0.1 = 372. Find the average tool life for V = 275 fpm.(d) Tool changing time is 4 min and other conditions are VT0.1 = 372, V = 275 fpm, C0 = $0.25/min, L = 20in., D = 4 in., and f = 0.020 ipr. Determine the tool changing cost.(e) By using information in earlier parts of this problem and Ct = $5, find the tool cost per operation.

Answer:(a) Handling cost = 15/60 x 1.35 = $0.34(b) Time to machine = LπD/12Vf = 20π4/(12 x 275 x 0.02) = 3.8 min(c) Average tool life = T = (372/275)10 = 20.5 min(d) Tool changing cost = 0.25(4)(3.8/20.5) = $0.19(e) Tool cost per operation = 5(3.8/20.5) = $0.93

7.9 (a) Operator wage and variable machine expenses are $60 per hour and handling is 1.65 min. Find thehandling cost for this element.(b) The length and diameter of a gray iron casting are 8.5 by 8.6 in. Surface rotational velocity of thecasting is 300 fpm and feed is 0.020 ipr. Find the turning time.(c) The Taylor tool life equation is VT0.15 = 500. Find the average tool life for 300 fpm.(d) The time to remove a square insert and index it to another new corner is 2 min, tool-life equation isVT0.15 = 500, V = 300 fpm, Co = $l/min, L = 8.5 in., D = 8.6 in., and f = 0.020 ipr. What is the cost tochange tools?(e) By using the information in part (d), and that an eight-corner insert for rough turning of cast iron is $24,find the tool cost per operation.

Answer:(a) Handling cost = $60/hr x hr/60 x 1.65 min = $1.65(b) Turning time = tm = LπD/12Vf = 8.5 π 8.6/12(300)(0.02) = 3.19 min(c) Average tool life = T = (500/300)1/0.15 = 30.13 min(d) Cost to change tools = ($1/min)(2 min)(3.19/30.13) = $0.21(e) Tool cost per operation = 24/8(3.19/30.13) = $0.32


7.10 (a) Stainless-steel material is to be rough and finish turned. Diameter and length of the bar stock are 4x 30 in. Recommended rough and finish cutting velocity and feed for tungsten carbide tool material are (V= fpm, f = ipr) or (350, 0.015) and (350, 0.007). Determine rough and finish cutting time.(b) Medium carbon steel is to be rough and finish turned by using high-speed steel-tool material. The partdiameter and cutting length are 4 in. and 20 in. Determine the total time to machine by using Table 7.2.(c) Gray cast iron is to be rough and finish turned with tungsten carbide tooling. Part diameter and cuttinglength are 8.6 in. and 8.5 in. What is the part RPM for the rough and finish? Find the total turning time.(Hint: Use Table 7.2 for the turning velocity and feed. To find the time, use the relationship tm = L/fN.)

Answer:(a) V = 350, f = 0.015) rough cuttingtm = 30π4/12(350).015) = 6 min for a rough passtm = 30π4/12(350).007) = 12.8 min for the finish pass machining of the stainless steel bar stock(b) High speed steel machining of medium carbon steel.Rough = (190, 0.015), tm = 20π4/12(190)0.015 = 7.3 minFinish conditions = 125, 0.007; tm = 20π4/12(125)0.007 = 23.9 minTotal time = 7.3 + 23.9 =31.2 min(c) Gray cast ironN = 12 V/πD; For roughing N = 12 (500)/ π (8.6) = 222 RPM,For finish, V, f = 675, 0.01; N = 300 RPM,tm = L/fN = 8.5/222 x .02 + 8.5/300 x 0.01 = 4.7 min

7.11 (a) Find the time to rough mill a stainless-steel work material having a total-machining 4-in.flatsurface length of 35 in. The plain milling cutter is high speed steel material, and is 4-in. O.D., 6-in. wideand has 16 teeth. (Hints: Assume only one pass. Refer to Table 7.2 for information.)(b) Find the time to finish mill a medium-carbon steel work material having a total-machining slot length of11.5 in. by 1-in. wide. The end milling cutter is high speed steel material, and is 1-in. O.D. and has 4 teeth.

Answer:(a) tm = LπD/12Vntft = 35π4/12(140)(16)(0.006) = 2.7 min(b) tm = LπD/12Vntft = 11.5π/12(95)4(0.002) = 4.0 min


7.12 (a) A gray cast-iron block has one hole that is drilled and tapped according to the drawing callout of5/16-16 for a length of 1.5 in. Standard time data retrieval for setup are 2 hours and all handling requires1.65 min. There are 93 units on the operation order. What is the cycle time? Find the lot hours for thisoperation.

(Hints: The final tap diameter is 5/16 in. and has 16 threads per in. The drill diameter is smallerthan 5/16 in. and thus _ in. is chosen, it being the only stated value in Table 7.2. The drilling length istypically longer than the tapped length by one thread length or so.)(b) A medium carbon steel part has one hole that is drilled and tapped according to the standard drawingcallout of 3/8-10 for a length of 0.75 in. Standard time data retrieval for setup is 1.22 hours and all handlingrequires 0.65 min. There are 273 units on the operation order. Find the lot hours for this operation.

Answer:(a) t(Time to drill _ hole) = Lfd = 1.6 x 0.20 = 0.32 min

t(time to tap 5/16-16 hole) = 1.5 x 0.21 = 0.32 minTotal cycle time = 1.65 + 0.32 + 0.32 = 2.29 minLot hr = 2.00 + 93(2.29/60) = 5.55 hr

(b) t(time to drill 5/16 in. hole) = 0.80 x 0.23 = 0.18 mint(time to tap 3/8 – 10 hole) = 0.75 x 0.32 = 0.24 minTotal cycle time = 0.65 + 0.18 + 0.24 = 1.07 minLot hr = 1.22 + 273(1.07/60) = 6.09 hr


7.13 (a) A 10-in. long block is to have a 1-in. wide slot machined on it. The HSS end mill is 1 in. O.D. andonly one finish pass is machined over the slot. Calculate the approximate length of cut for the tool if thesafety stock is 1/4 in. Repeat if the width of the slot is 2 in.(b) A slab milling cutter is 6 in. in diameter. The design length for metal removal is 15 in., and one roughmachining passes of 3/8 in. depth of stock removal is necessary. Safety stock is 1/4 in. Determine theapproximate length of cut for roughing.(c) A slab milling cutter is 6 in. in diameter. The design length for metal removal is 15 in., and two roughmachining passes of 3/8 in. depth of stock removal and one finish 0.015 in. depth of cut are necessary.Safety stock is 1/4 in. Determine the length of cut for roughing and finishing.(d) Find the length of cutting for a 1 in. dia drill in soft steel where the design length is 2 in.

Answer:(a)

1” width slot 2” width slotL = 1/4 + 1/2 + 10 + 1/2 = 11-1/4” L = 1/4 + 1/2 + 10 + 1/2 + 1 + 1/2

+ 10 + 1/4 + 1/2 + 1 = 24-1/2

(b) ( )dDdL ca −=

( ) 45.1375.06375.0 =−=L = 1/4 + 1.45 + 15 = 16.7

(c) from (b) L = 16.7for 0.015, La = 0.3

Total length of cut = 16.7 + 16.7 + 15.3 = 48.7

(d) L = 1/16 + 0 + 2 + 0.2 (1) = 2.26

f

f

f

1/2 1/

1/2

10

12

3 4

10


7.14 (a) A stainless steel surface 1 x 10 in. is to be end milled with a 1-in. high-speed steel four-tooth endmill for a depth of 0.0l5 in. The end mill will pass entirely over the material and have a safety stock of 1/8in. Velocity is 95 fpm, and chip load per tooth is 0.0015. What is the length of cut? How much time isnecessary for machining? What is the cutter RPM and feed rate in in. per min?(b) A 1-in. slot is to be end milled in gray cast iron for a depth of 1/4 in. for a design length of 20 in. Arough and finish pass is required for the HSS four-flute cutter. A safety stock of 1/8 in. is necessary What isthe machining length of cut for the rough or finish machining pass? Determine rough and finish machiningtime. Find cutter RPM and cutting rate in in. per min. If the non-machining time is 1.73 min per unit andthe setup is 3.24 hr, find the lot hours for 217 units for the operation.

(Hints: Use Table 7.2 for machining data, where data are cutting velocity and tooth load. Thecutter must pass entirely over the surface for the finish pass.)

Answer:(a) L = 1/8 + _ + 10 + _ = 11.125 in.tm = 11.125π/12(95)4(0.0015) = 5.1 minN = 12(95)/π = 363 RPMFeed rate in. per min = 11.125/5.1 = 2.18 in./min

(b) L(rough) = 1/8 + _ + 20 = 20.675 in.Rough tm = 21.125π/12(85)4(0.004) = 4.1 minRough cutter N = 12V/πD = 12(85)/3.14 = 325 RPMCutting rate = 20.675/4.1 = 5.0 in./minL(finish) = 1/8 + _ + 20 + _ = 21.125Finish tm = 21.125π/12(95)4(0.003) = 4.8 minN = 12(95)/3.14 = 363 RPMCutting rate (finish) = 21.125/4.8 = 4.4 in./minTotal cycle time = 4.1 + 4.8 + 1.73 = 10.63 minLot hr = 3.24 + 217(10.63/60) = 41.7 hr.

7.15 (a) A gray cast-iron surface 6 in. wide by 30 in. long is rough milled to a depth of 1/4 in. The 16-toothcemented-carbide face mill is 6 in. in diameter. Estimate the cutting time. Find the cutter RPM. (Hints: Thecomputer controlled milling machine is performing the operation as if the cutter is an end mill in a verticalposition. There is _ in. for safety space. Use Table 7.2 for machining data and apply Eq. (7.14). In exitingthe surface, the cutter does not pass entirely over the surface for a rough machining pass.)(b) A high-speed steel vertical milling cutter having 12 teeth is 4 in. in diameter and is used to mill a soft-steel surface 3 in. wide by 9 in. long with a depth of cut of 3/4 in. A cutting speed of 170 fpm and a feed of0.008 in per tooth-rev. are selected. Find the cutting time and cutter RPM. What is the operation lot hours ifthe setup is 3.25 hr and other cycle time is 14.17 min for 27 units?

Answer:(a) L = _ + 3 + 30 + 0 = 33.5

From the table, fpm = 170 and tooth load = 0.008. Tm = 33.5π6/12(170)0.008 = 2.42 min N = 12V/πD = 12(170)/3.14(6) = 108 RPM(b) Assume a rough pass over the surface, thus the cutter needs to only traverse over the end edge.L = _ + 2 + 9 + 0 = 11.5 approx.Tm = LπD/12Vnf = 11.5π4/12(170)12(0.008) = 0.74 minN = 12V/πD = 12(170)/π4 = 162 RPMLot hr = 3.25 + 27(0.75 + 14.17)/60 = 10 hr approx.


7.16 (a) A stainless-steel part is drilled 5/16 in. which is followed by a 3/8-16 N.C. tap for a depth of 7/8 in.Find the drilling length and the total of drilling and tapping time. (Hint: Use Table 7.2.)(b) Medium carbon steel is tap-drilled 1/4 in., which is followed with a tapping element of 3/8-24 N.F. for1.3 in. There are 26 holes. Find the drilling length and the total of drilling and tapping time.

There are 172 units in a lot and the setup is 0.7 hr and the cycle time exclusive of the drilling andtapping of holes is 7.13 min. What are the pieces per hour. Find the lot hours. (Hint: Pieces per hour isexclusive of setup time.)

Answer:(a) Stainless steel point included angle = 135°.

L = 1/16 in. safety stock + 5/16/2 tan 67.5 + 7/8 full length thread + 1/16 extra thread = 1.07 in.approx.Time to drill = 1.07 x .61 = 0.65 minTime to tap = 0.875 x 0.33 = 0.29 minTotal time = 0.65 + 0.33 = 0.94 min

(b) L = 1/16 safety stock + 1/4/2tan 59 + 1.3 1/24 extra thread for clearance = 1.42 approx.Time to drill = 1.42 x 0.20 = 0.284 minTime to tap = 1.3 x 0.33 (approx.) = 0.43 minTotal time per hole = 0.71 minFor 26 holes, total time = 18.54 minTotal cycle = 18.54 + 7.13 = 25.67 minPieces per hour = 2.3Lot hours = 0.7 + 172(25.67/60) = 74.29 hr

7.17 (a) Find the cutting time for a hard copper shaft 2 in. O.D. x 20 in. long. A surface velocity of 250 fpmis suggested with a feed of 0.009 in. per revolution. (Hint: This is a lathe turning operation.)(b) An end facing cut is required of a 10-in. diameter work piece. The revolutions per minute of the latheare controlled to maintain 400 surface feet per minute from the center out to the surface. Feed is 0.009.Find the time for the machining cut. (Hints: A facing cut is along the end of the bar stock, and the workpiece is held by a 3-jaw chuck allowing the tool to move free of interference. As the tool post moves fromthe outside to the inside the work piece RPMs will increase to maintain a constant 400 fpm. The length ofcut for a lathe facing operation is _ x diameter.)

Answer:(a) tm = LπD/12Vf = 20π2/12(250).009 = 4.7 min(b) D = 10 in.; V = sfpm; f = 0.009 ipr;. L = 5 in.

Tm = 5π10/12(400)0.009 = 3.6 min

7.18 If the tool material grade is Kennametal 3H carbide, work material is AISI 4140 steel, depth of cut is0.150 in., and the feed is 0.010 ipr, find the surface feet per minute for a 4-in. O.D. bar and a 6-min life ifthe tool life equation VT0.3723 = 1022. Also find the RPM.

Answer:VT0.3723 = 1022, then V = 1022/60.3723 = 525 fpm.N = 12V/πD = 12(525)/ π4 = 502 RPM


7.19 Find the ratio of tool life using tungsten carbide to high-speed steel tool materials for a rough- turningoperation of cast iron. (Hint: Use Tables 7.2 and 7.3.) What is the average tool life increase using tungstencarbide tool material?

Answer:Tool Material Work Material Taylor equation Fpm Average tool life

HSS Cast iron VT0.12 = 225 145 39 minCarbide Cast iron VT0.43 = 3000 500 65 min

Ratio = 65/39 = 1.7 or about 67% more life for the tungsten carbide.

7.20 Using the classical Taylor tool life model, and for the work and tool materials, find the cutting velocityfor a tool life of 20 min for each of these combinations.

Work Material Tool MaterialStainless steel Tungsten carbideMedium carbon steel Tungsten carbideMedium carbon steel High-speed steelGray cast iron Tungsten carbide

Answer:Work Material Tool Material K n V

Stainless steel Tungsten carbide 400 0.16 248Medium carbon steel Tungsten carbide 450 0.20 247Medium carbon steel High-speed steel 190 0.11 136Gray cast iron Tungsten carbide 3000 0.43 827*

* Probably above practical RPMs for equipment.


7.21 A company is planning to “turn” thousands of bars for a long production run. As one step in theplanning, it conducts a Taylor test and following a field study of the AISI 3140 work material, lathe, and anewly developed tool material, the tool-life curve in Fig. P7.21 is determined. The conditions of the test area feed of 0.013 ipr with a depth of cut of 0.50 in. Find the tool life for 100 fpm. What are the parameters forVTn = K ? (Hint: Use the curve to determine n and K.)

How much time is necessary to machine the new-product design AISI 3140 bar stock if thediameter and length are 2 in. O.D. and 27.125 in. for a life of 60 min if the feed is 0.0125 ipr? What are therevolutions per minute?

Figure P7.21

Answer:From graph, say 80 min = tool life for fpm = 100.Finding the parameters of the tool life equation: Using points (fpm, tool life); (150, 2); (100, 80).

Then V T V T Kn n n n1 1 2 2 150 2 100 80= = =; ( ) ( ) ; then log 1.5 = nlog 40; then n = 0.11, and K = 161.9

The equation: VT0.11 = 161.9 is found from the graph.At a life of 60 min, the equations gives V = 103 fpm, then tm = 27.125π2/12(103)0.0125 = 11 minN = 12V/πD = 12(103)/ π2 = 197 RPM


7.22 A Taylor tool-life performance test is conducted for a rough-turning operation of newly developedcarbon steel. The tool material is a recently developed grade of high-speed steel. The testing determinesVT0.1 = 372.

An operation requires planning for optimum performance. The work material diameter is 4 in.O.D. and the total cutting length is 20 in. The tool point costs $5 and time to change the tool after wear outis 4 min. Part handling is 2 min and the operator wage is $20 per hour. The feed of the turning operation is0.020 ipr for a depth of cut = 0.5 in.(a) Find the optimum cutting velocity analytically and calculate the RPMs.(b) Plot the handling, machining, tool-changing, and tool costs versus feet per minute to find the total costcurve. Find the optimum cutting velocity graphically and what is the cost?

(Hints: Transform the four equations (7.1, 7.3, 7.7. and 7.8) to have the cost expressed in units ofV, and then substitute values of V from 50 to 350 in steps of 50 units. For example, handling cost =$0.333/min x 2 min = $0.667, which is a constant. Again, machining cost = 349/V, and so on. You maywant to use a spreadsheet.)(c) Let the cost of Ct = 0 and compute Vmax , and Tmin.(d) Find the cycle time, units per hour, and hr/100 units.

Answer:

(a)Vmin .

.

.

.

=

−

× +

= =372

1

011

0 333 4 5

0333

222 2121

fpm, rpm =12(222)

3.14(4)

(b) Handling cost = Coth = ($0.333/min)(2) = $0.667, which is a constant line and independent of V.Machining cost = Cotm = 0.333 (20π4/12V(0.02) = 349/Vtm = 1047/Vaverage tool life T = (5.075 x 1025)/V10

Tool cost = Cttc/T = 2.310 x 10-19 V9

Tool changing cost = 2.0631 x 10-23V9

Cutting V Handling Cost Machining Cost Tool Cost Tool ChangingCost

TotalCost

50 0.333 6.98 0.000000 4.0E-08 7.313100 0.333 3.49 0.000103 2.1E-05 3.823150 0.333 2.33 0.003965 0.0008 2.665200 0.333 1.75 0.052813 0.0106 2.142222 0.333 1.57 0.135097 0.0270 2.068250 0.333 1.40 0.393486 0.0787 2.202300 0.333 1.16 2.030301 0.4061 3.933350 0.333 1.00 8.129833 1.6260 11.086

Answer continued on next page


7.22 Answer continued

Tmin = (1/n – 1)tc = (1/.1 – 1)4 = 36 minVmax = K/Tmax

n = 372/36.1 = 260 fpmtm = 20(3.14)4/12(212).02 = 4.9 minT = (372/212)10 = 272 minTu = th + tm + tc(tm/T) = 2 + 4.9 + 4(4.9/272) = 6.97 minPieces per hour = 60/6.97 = 8.6Hr/100 = 11.620

CHALLENGE PROBLEMS

7.23 Find the material cost and setup and cycle time for the fabrication of a pinion, part no. 4943806,similar in all respects to Fig. 7.7 except for the following changes. (Hints: Work each sub-problemseparately; complete the operations process sheet and find the unit cost as demonstrated by Table 7.6.Many work elements are similar to existing worksheets.)(a) Let the 18.750-in. dimension be 8.750 in., the 1.100-in. dimension be 1.40 in., and no holes. (Hint: The

1.40 in. O.D. dimension is reduced to 1,75 in. length. Check it out!)(b) Let the 4.75-in. flat dimension be 8.00 in. long, and the raw material be 2 in. O.D. instead of 1 3/4 in.

O.D.(c) Let the 6-in. wide milling cutter be 3 in. in width for operation 30, and operation 10 will use high-

speed tool material instead of tungsten carbide.(d) Let the material be medium carbon steel instead of stainless steel. Medium carbon steel costs $1.0/lb.(e) Let the lot quantity be 1000 instead of 200.

Answer on next page:

Taylor's Tool Life

0.000

2.000

4.000

6.000

8.000

10.000

12.000

50 100 150 200 222 250 300 350

Cutting velocity, fpm

Uni

t co

st, $

/pc

Handling Cost

Machining Cost

Tool Cost

Tool Changing Cost

Total Cost


7. 23 Answer

(a)Design length, in. 8.750Facing length, in. 0.015Cutoff length, in. 0.187

Total length, in. 8.952No. of units per 12 ft bar 15 –

16.1No. of bars per 200 lot quantity 14Cost per bar 8.178 x 12 x 1.50 =$147.20Lot cost for material 147.20 x 14 = $2061Cost per unit 2061/200= $10.30

Operations process sheet for part no. 4943806.Part no. 4943806 Material 430 F stainless steelPart name Pinion Size 1.750 ± 0.003Quantity 200 Length 12 ft bars

Material unitcost

$10.30

Workstation Op.no.

Description ofoperation

Setuphr

Cyclehr/100units

Lothr

PHC Lotcost

Unitcost

Turret lathe 10 Position 2.9 5.100 13.1 39.16 $513 $2.57Face 0.015Turn rough 1.45Finish turn 1.40Turn 1.735Cutoff to 18.75

Vertical mill 20 End mill 0.89 slotwith _ HSS endmill & collet

1.8 8.067 42.1 90.98 $3833 $7.67

Horizontalmill

30 Slab mill 4.75 x3/8 (Nesting vise)

1.3 1.467 8.6 90.98 $785 $1.57

Total PHC lot cost $5131Total PHC unit cost $11.81

Material, labor, and equipmentunit cost

$22.11

Worksheet for estimating setup and cycle for operation 10A. Setup elements HrPunch in and out, study drawing 0.2Collet 0.2First facing tool 1.3Additional 4 tools 1.2

Setup total, operation 10 2.9

Answer continued on next sheet



B. Handling and other equipment time elements MinStart and stop machine 0.08Advance stock through feed tube, 8.952 in. 0.23Place and remove oil guard 0.19Speed changes, assume 4 x 0.04 0.16End turret advance, return and index, 4 x 0.08 0.32Cross slide advance, return 0.09Inspect part with micrometer, irregular 1/5 x 0.30 0.06

Subtotal of handling and equipment elements 1.13

C. Calculation of machining timesElement Dim. Depth

of CutLengthof Cut,

Ld

Safetystock,

Ls

L,Length

D, Dia V,Velocity,

fpm

f, feed(ipr)

tm,, min

1.Face 0.015 0.875 1/32 0.906 1.750 350 0.007 0.172.Roughturn

1.45 0.15 6.5 1/32 6.53 1.750 350 0.015 0.57

3. Finishturn

1.40 0.025 6.5 1/32 6.53 1.45 350 0.007 1.01

5.Turn 1.735 0.0075 0.5 1/32 0.53 1.75 350 0.007 0.106.Cutoff

1.75 0.187 W 0.875 1/32 0.906 1.75 350 0.015 0.08

Subtotal of machining times 1.93Total cycle time for handling and

machining, min3.06

D. Entry values for operation sheet, operation 10Setup hr 2.9Hr/100 units 5.100

Worksheet for estimating setup and cycle for operation 20.Entry values for operation sheet, operation 20Setup hr 1.8Hr/100 units 8.067

Worksheet for estimating setup and cycle for operation 30

Entry values for operation sheet, operation 30Setup hr 1.3Hr/100 units 1.467

Operation 40 is not required.




(b) Let the 4.75-in. flat dimension be 8.00 in. long, and the raw material be 2 in. O.D. instead of 1 3/4 in.O.D.

Design length, in. 18.750Facing length, in. 0.015Cutoff length, in. 0.187

Total length, in. 18.952No. of units per 12 ft bar 7No. of bars per 200 lot quantity 29Lb per ft of 2 in. O.D. barstock

10.68

Cost per bar 10.68 x 12 x 1.50 =$192.17Lot cost for material 192.17 x 29 = $5572.87Cost per unit 5572.87/200= $27.86

Part no. 4943806 Material 430 F stainless steelPart name Pinion Size 2.000 ± 0.003Lot quantity 200 Length 12 ft bars

Material unitcost

$27.86

Workstation Op.no.

Description ofoperation

Setuphr

Cyclehr/100units

Lothr

PHC Lotcost

Unitcost

Turret lathe 10 Face 0.015 3.5 13.200 29.9 39.16 $1171 $5.85Turn rough 2 in.to 1.45 in.Turn rough 1.45Turn rough 1.15Finish turn 1.10Turn 1.735Cutoff to 18.75

Vertical mill 20 End mill 0.89 slotwith _ HSS endmill & collet

1.8 8.067 42.1 90.98 $3833 $7.67

Horizontalmill

30 Slab mill 4.75 x3/8 (Nesting vise)

1.3 1.467 8.6 90.98 $785 $1.57

N.C. turretdrill press

40 Drill 4-5/8 holes,tap 3/8-16 (collet)

1.12 5.217 27.2 39.16 $1065 $2.13

Total PHC lot cost $6854Total PHC unit cost $17.22

Material, labor, and equipmentunit cost

$45.08

Worksheet for estimating setup and cycle for operation 10.A. Setup elements HrPunch in and out, study drawing 0.2Collet 0.2First facing tool 1.3Additional 6 tools 1.8

Setup total, operation 10 3.5




B. Handling and other equipment time elements MinStart and stop machine 0.08Advance stock through feed tube, 18.952 in. 0.37Place and remove oil guard 0.19Speed changes, assume 5 x 0.04 0.20End turret advance, return and index, 6 x 0.08 0.48Cross slide advance, return 0.09Inspect part with micrometer, irregular 1/5 x 0.30 0.06

Subtotal of handling and equipment elements 1.47

Calculation of machining timesElement Dim. Depth

of CutLengthof Cut,

Ld

Safetystock,

Ls

L,Length

D, Dia V,Velocity,

fpm

f, feed(ipr)

tm,, min

1.Face 0.015 0.875 1/32 0.906 1.750 350 0.007 0.172.Roughturn

1.45 0.15 16.5 1/32 16.53 1.750 350 0.015 1.44

Another rough turning element is required.2.Roughturn

1.45 0.15 16.5 1/32 16.53 1.750 350 0.015 1.44

3.Roughturn

1.15 0.15 16.5 1/32 16.53 1.45 350 0.015 1.19

4. Finishturn

1.10 0.025 16.5 1/32 16.53 1.15 350 0.007 2.03

5.Turn 1.735 0.0075 0.5 1/32 0.53 1.75 350 0.007 0.106.Cutoff

1.75 0.187 W 0.875 1/32 0.906 1.75 350 0.015 0.08

Subtotal of machining times 6.45Total cycle time for handling and

machining, min7.92

Entry values for operation sheet, operation 10Setup hr 3.5Hr/100 units 13.200

Operations 20, 30, and 40 remain the same.

Parts (c), (d) and (e) are worked in a similar fashion.


7.24 Sometimes it is cheaper to design and construct one multipurpose tool that will serve two or moresimilar designs, that being less costly than an individual tool that will satisfy only one design. There arecomplications that can thwart this practice, such as simultaneously scheduling of parts or whenever thedesigns are so radically different that multipurpose tools are more expensive than if one tool were availablefor each design. Though multipurpose tooling may, at first, appear to offer cost benefits, the operation costfor different parts may cancel out many of those gains.

Two product designs are called Left Hand and Right Hand and are sufficiently similar to allowmultipurpose tools. Consider the following:

Left Hand Right Hand Both Hands,Multipurpose

Cost of tool $47,500 $32,500 $62,000Productivehour costper unit

$1.92 $2.17 $2.03 (LH)$2.65(RH)

Life timeunits

83,000 27,500

Should two tools or one multipurpose tool be designed and constructed? (Hints: Productive hour cost perunit includes direct labor and machine cost. Ignore material cost in the analysis.)

Answer:Left Hand Right Hand Both Hands,

MultipurposeCost of tool $47,500 $32,500 $62,000Totaloperationcost

$159,360 $59,675 $168,490$72,875

Total toolandoperationcost

$206,860 $92,175 $303,365

If two tools are built, it will cost $299,035 versus the cost for one tool of $303,365. That is a close call, andother conditions should be reviewed in the decision process. These quantities are the point of neutrality.


7.25 A malleable-iron casting having an as-cast nominal diameter of 9 1/4 in. is rough turned to 8.600 ±0.005 in. for a length of 8.500 ± 0.010 in. The feed for this turning operation is 0.020 ipr.

A renewable square carbide insert is used. The tungsten carbide insert has eight usable cornerssuitable for turning work and costs $24. The time for the operator to remove the insert and install anothernew corner and qualify the tool ready to cut is 2 min.

Operator and variable production costs are $60 per hour. Taylor's tool life equation for work andtool material is VT0.15 = 500. Robot handling time is 1.65 min for a casting mounted in a fixture.(a) Determine optimum cutting velocity analytically. Find the RPM.(b) Construct individual cost and total cost curves similar to Fig. 7.6. If the y axis is unit cost and x axis isfpm, then plot the curves to locate the minimum cost. What is the range of cost if shop performance can beexpected to vary ±10% from the optimum velocity.

(Hints: Transform the four equations (7.1, 7.3, 7.7. and 7.8) to have the cost expressed indimension of V, and then substitute values of V from 50 to 350 in jumps of 50 fpm units. A spreadsheetanalysis will be helpful. The diameter is the starting rough diameter of the casting.)(c) Find the unit time Tu, pieces per hour, and hr/100 units.(d) Assume that the cost of the tooling Cpt = 0 and find Vmax and Tmin..(e). Find the time-vs.-velocity curves if Cpt = 0.

Answer:(a)

VK

n

C t C

Co c t

o

nmin .

.

=

−

+

=

−

× +

=1

1

500

1

0151

1 2 3

1

303015

fpm

N = 12V/3.14(D) = 125 RPM. Note that the original diameter of 9.25 in. is used.

(b). Co = 1; th = 1.65; L = 8.5; D = 9.25 (use starting dia); V = TBD; f = 0.020; n = 0.15; K = 500; tc = 2; Ct

= 3.Handling cost = $1.65/unitMachining cost = 1029/VTool changing cost = 2.09 x 10-15V5.667

Tool cost per operation = 3.136 x 10-15 V5.667

Cutting V HandlingCost

MachiningCost

Tool Cost ToolChanging

Cost

TotalCost

Variation

100 1.65 10.29 0.001 0.000 11.94150 1.65 6.86 0.01 0.004 8.52200 1.65 5.15 0.03 0.023 6.85250 1.65 4.12 0.12 0.081 5.97273 1.65 3.77 0.20 0.134 5.75 -10%300 1.65 3.43 0.34 0.228 5.650303 1.65 3.40 0.36 0.241 5.649 optimum333 1.65 3.09 0.62 0.412 5.770 10%350 1.65 2.94 0.82 0.546 5.96400 1.65 2.57 1.75 1.164 7.13




(c) Tu = 1.65 + 3 + 2(3/28) = 4.86 min; pc/hr = 12.3, hr/100 = 8.107(d) Vmax = 500/[(1/.15 – 1)2].15 = 347 fpm

Tmin = (1/.15 – 1)2 = 11.3 mintm = L3.14D/12Vf = 8.5(3.14)9.25/12(347).02 = 3 min

The table can be recomputed, taking out the costs, and then the result gives optimum time. Also ignore thepurchase of the tool cost.

Cutting V Handling Machining tool change Total time50 2 20.9 4.0E-08 22.9

100 2 10.5 2.1E-05 12.5150 2 7.0 7.9E-04 9.0200 2 5.2 1.1E-02 7.2222 2 4.7 2.7E-02 6.7250 2 4.2 7.9E-02 6.3300 2 3.5 4.1E-01 5.9350 2 3.0 1.6E+00 6.6400 2 2.6 5.4E+00 10.0

(e)

Optimum velocity doesn’t change much once the perishable tooling cost is forced to zero dollars. Its about300 fpm, which is the same as above.

Taylor Tool Life for optimum time

0

5

10

15

20

25

50 100 150 200 222 250 300 350 400

Velocity, fpm

Cyc

le t

ime,

min

Handling time

Machining time

tool change time

Total time


7.26 Estimate the total lot and unit cost for the design given by Fig. P7.26 for material, labor, overhead andtooling. The quantity is 7500 units, which is produced in one lot. (Hints: The student will want to followthese steps.)

(a) How many bars are necessary? Find the material cost for the lot order and unit. Raw materialbar stock is 3 in. O.D., 8 ft long, and is $1.75/lb. Density is 0.285 lb/in.3 The cutoff and facing length is 1/8in. and 0.015 in.

(b) Find the setup, cycle min, hr/100 units for the two operations for entry to a operations processsheet. (Hint: Using a method similar to Table 7.6.) The operations area as follows:

Operation 10: Face, turn, chamfer, drill, and ream the 0.375 in. hole and cutoff with a turret latheOperation 20: Drill and counter bore three holes with a drill press.(c) Prepare an operations sheet and find the operation cost for the unit and lot. Productive hour

costs for turning and drilling are $75 per hour. (Hint: Use the following tabular approach:)Workstation Op.

No.Description Setup Hr/100 Lot Hr PHC Lot

CostUnitCost

(d) Find the total and unit tool cost. A small drill jig with pneumatic clamping is necessary for the0.25 in. drill and counterboring operation. Tool design is expected to cost $8000, and the PHC for toolbuilding is $100 per hour. (Hints: The tool cost is estimated using the part print, rather than the tool design,meaning visualize the elements of a tool design as suggested by Table 7.11. There are three removable slipbushings that come out for the counterboring. Expect about 4 lb for the jig. Make any material adjustmentsfor missing data based on local experience.)

(e) Using an operations sheet approach, estimate the total lot and unit cost of material, operationsand tooling.

Figure P7.26

Answer on next page


7.26 Answer

(a) Material in each piece:Cutoff length, in. 0.125Facing length, in. 0.015Length of finished piece, in. 1.000Total length, in. 1.148 ft/bar x 12 in./ft x 1 pc/1.14 = 84 pc/bar7500 pc/84 = 90 bars requiredMaterial cost for 90 bars.90 x π/4 (3)2 x 96 x 1.75 x 0.285 = $30,445Unit material cost $4.06

(b)Setup for operation 10 Freq. Hr

Punch in and out, study drawing 0.2 0.2First tool 1.3 1.3Each additional tool 5 0.3 1.5Collet fixture 0.2 0.2

Total setup 3.2

Cycle handling & non-machine time for operation 10 Freq. Time MinStart and stop machine 0.08Change speed 6 0.04 0.24Air clean part or fixture 0.06Brush chips 0.14Inspect dimension with micrometer 1/10 0.30 0.03End turret advance, return, and index 5 0.08 0.40Cross slide advance, return, and index 1 0.09Advance stock through feed tube 0 in. length < 6 0.18Place, remove oil guard 0.19

Subtotal of handling & other non-machining 1.41

Element Dim DepthOf cut

Lenof cut

Safety L Dia fpm ipr fdt tm

Face 3.00 0.015 1.5 1/32 1.53 3 400 0.007 0.43Turn 3.00 0.0625 1 1/32 1.03 3 400 0.007 0.28Chamf 3.00 .12 0.12 1/32 .15 3 400 0.007 0.04Drill 5/16 1.25 1/32 1.28 0.23 0.29Ream 0.375 1.05 1/32 1.08 0.23 0.25Cutoff W -1/8 1.45 1/32 1.48 2.875 400 0.007 0.40

Subtotal machining time 1.69Subtotal handling and other machine data 1.41

Total cycle time min 3.10Hr/100 5.167

Setup hr 3.2




Setup for operation 20 HrPunch in and out, study drawing 0.2Jig 0.1First turret for _ hole 0.75Additional turret for counterbore 0.07

Setup for op. 20 1.12

Cycle handling elements for operation 20 Freq Time MinStart and stop machine 0.08Air blow fixture 0.06Pick up part, move, and place; remove and lay aside 5 ≤ lb < 10 0.16Air cylinder, clamp part 0.05Remove bushing, re-install 3 0.07 0.21Change tool, each size 2 0.06 0.12Raise tool, position table to new location, lower toolready for drilling, per hole position for drill andc’bore

6 0.06 0.36

Index turret, each or change drill for c’bore 2 0.03 0.06Subtotal handling and other non-machining time 0.96

Cycle drilling, c’boring for operation 20 Freq L fdt MinDrill _ in. hole 3 1.2 0.20 0.723/8 in. c’bore 3 .4 0.25 0.30

Subtotal drilling and c’bore times 1.02Total cycle time min 1.98

Hr/100 3.300Setup hr 1.12

(c)Labor and equipment cost

Workstation No. Description Setup Hr/100 Lot Hr PHC Lot Cost Unit CostTurret lathe 10 Position 3.2 5.167 391 75 29,303 3.907

Face 0.015Turn rough1.45Turn finish1.436Drill 0.25Ream 0.375,1 in. deepCutoff tolength

Drill press 20 Position 1.12 3.300 248.6 75 18,646 2.49Drill 3-0.25holesC’bore 3holes

Total labor and equipment cost $47,949 $6.39Answer continued on next page



(d)Tool estimate for Fig. P7.26, drill fixture

Tool type: Small drill fixture with hinge top, threebushings

Tool material: Hardened high carbon steel alloy, 4 lbDescription of part: Flat block with three accurately located

holes, and removable slip bushings

Description of tool elements HrAngular base plate, C-angle 5.0Slip bushings, 3 6.0Hinge plate 12.0Feet, 3 7.5Pin locators, 3 5.4

Total 35.9Productive hour cost for tool making $100Cost of tool, 44.8 x 75 $3590Material cost, $25/lb 100Design cost $8000

Total cost of tool $11,690Unit cost of tool $1.56

(e)Cost of operations, material, tooling

Element Lot Cost Unit CostMaterial $30,445 $4.05Operations $44,485 $5.93Tooling $11,690 $1.56

Total $86,620.00 $11.54


7.27 Find the unit cost for flow line production which consists of three stations for an electronic productassembly. There are 1750 printed circuit boards produced. The flow line starts with a pre-drilled printedcircuit board obtained from a supplier costing $0.93 each. Designed tooling are holding fixtures, specialgrippers and a sled for wave soldering. Information is estimated as follows:

Station Machine ComponentInsertion

ManualStuffing

WaveSolder

Setup time, hr 1.6 0.5 1.2Cycle time, min/unit 3.75 5.13 0.65PHCR, $/hr $78.25 $27.25 $110.73Direct materials, $/unit-operation $17.62 $1.29 $0.85Designed tooling, $/operation $283 $18.46 $73.09

Answer:Station Machine Component

InsertionManualStuffing

WaveSolder

Calculation

Max setup, hr 1.6 1.6 1.6Max cycle, hr per unit 0.086 0.086 0.086Cost of setup $125.20 $43.60 $177.17 $345.97Flow line cost for cycle time,$/lot

.086 x 78.25 x 1750 =11,708

4,101 16,664 $32,473

Initial direct material, $/lot $1,627.50Added material, $/lot $30,825 $2,257.50 $1,487.50 $34,570.00Tooling cost $283 $18.46 $73.09 $374.55Total cost $69,362Unit cost $39.63


7.28 Can making of the 12 oz. beverage container is very large volume, and a can plant will make billionsper year. We are interested in finding the cost per can for material, labor, tooling, and equipment.

The description of 10 operations for the body part of the three-piece can is shown below:

Operation DescriptionMaterial Cost Per Can

For OperationYield OfOperation Min Per Can

Cost PerMinute

1. Blanking and cupping 0.0350 0.9800 0.00026 $5.422. Draw and iron 0.9750 0.00026 4.583. Trim top of can 0.9500 0.00500 2.334. Wash and dry 0.0001 0.9990 0.00020 3.005. Decorate and overcoat 0.0003 0.9988 0.00049 3.176. Oven cure for internal coating 0.9988 0.00020 2.087. Internal spray and bake 0.0002 0.9975 0.00167 3.008. Neck and flange 0.9400 0.00163 4.679. Inspection of cans 0.9990 0.00045 1.2510. Palletize and storage 0.9925 0.00020 8.33

Initially the can is blanked and cupped from aluminum coil stock, and material is added inoperations 4, 5, and 7.

Yield effects are given for each operation, and each operation reduces the quantity throughputbecause of waste losses due to processing. For example, a desired output of 1 million cans after operation10 is 1,007,557 units (= 1,000,000/0.9925), which are required as input. Because of yield losses foroperation 9, input is 1,008,000 units (= 1,007,557/0.9990). Each operation has yield losses, and thealgorithm is to work upwards to operation 1to find the incoming number of units to give a final output of1,000,000 units. The yield effect hurts the system as the waste means that earlier operations lose not onlythe material but also the productive cost to produce the unit.

The production is a “pull” system where queues are avoided, and successive operations havehigher throughput in order to prevent storage. This may mean additional and intermediate equipment tokeep the cans moving such as diverging and converging chains. The production rate for each operation isgiven by minutes per can

Conveyor, equipment, tooling and space cost is summarized by the cost per minute for eachoperation.

Find the cost of material and the cost of each operation. Find the total cost for each can and for 1million cans.

Answer on next page

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PRACTICAL APPLICATION

There are many opportunities for a practical experience in operations estimating. Your instructor maysuggest several. One possibility is to visit a contract production shop and interview the engineeringmanager responsible for cost analysis of the quotation procedure. Prepare a list of questions in anticipationof the visit. Be professional in the arrangements of the interview, and afterwards write a report detailing thelessons that you have learned. Send them a courtesy letter expressing your thanks after the interview.

CASE STUDY:

Estimate the unit cost of the design given by Fig. C7.1 for material, tooling and labor and overhead. Theannual quantity is 1500 units, and material is purchased and produced in three equal-sized lots.

Bar stock, stainless steel 430 Ferritic material, is 6-in. O.D., 6-ft long, and is $2.25/lb. Density is0.275 lb/in.3

Two operations are necessary:Operation 10: Face, turn the 5.88 in. dia, drill, ream and countersink the 0.375-in. hole, turn the

4.5000/4.995-in. dia, and cutoff and chamfer (Hint: Cutoff and chamfer are one tool.) in a lathe.Operation 20: Drill, tap, and counterbore six holes. A drill jig is necessary for the drill,

counterbore, drill-tap, and tap operation. (Hint: The following hole schedule is given for operation 20:)

Specification Tool No.holes

5/16 in. × 1 in. long 5/16 in. drill for cap screw 3_ in. socket head cap screw × 0.25 in. long 3/8 in. flat bottom c’bore for socket head 3

3/16 in. drill full-length hole 320 thread/in. tap 3

Productive hour cost rates for turning and drilling are $75 and $40 per hour. Tool design isexpected to cost $5000, and the PHC rate for tool building is $100 per hour.

After the analysis and cost estimate are complete, discuss the following: What are the higher costelements of the work, and where can cost reduction be suggested to the designer? What can be done toreduce the cost of the design? For this advice, what is the dollar and percentage reduction?

Case study continued on next page


Case study continued

Figure C7.1

Answer:(1) Material in each piece:

Cutoff length, in. 0.187Facing length, in. 0.015Length of finished piece, in. 1.000Total length, in. 1.2026 ft/bar x 12 in./ft x 1 pc/1.202 = 59 pc/bar500 pc/59 = 9 bars required

Material cost for 9 bars.9 x π/4 (6)2 x 72 x 2.25 x 0.275 = $11,331Unit material cost $22.66



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(2)Setup for operation 10 Freq. Hr

Punch in and out, study drawing 0.2 0.2First tool 1.3 1.3Each additional tool 7 0.3 2.1Collet fixture 0.2 0.2

Total setup 3.8

Cycle handling & non-machine time for operation 10 Freq. Time MinStart and stop machine 0.08Change speed 8 speed changes, one for each tool 9 0.04 0.32Air clean part or fixture 0.06Inspect dimension with micrometer, the .0005 dim 1/10 0.30 0.03End turret advance, return, and index 6 0.08 0.48Cross slide advance, return, and index 2 0.09 0.18Advance stock through feed tube 0 in. length < 6 0.18Place, remove oil guard 0.19

Subtotal of handling & other non-machining 1.52

Element Dim DepthOf cut

Lenof cut

Safety L Dia fpm ipr fdt tm

Face 6 0.015 3 1/32 3.03 6 350 0.007 1.94Turn 5.88 0.0625 1 1/32 1.03 6 350 0.007 0.66Drill 5/16 1.25 1/32 1.28 0.61 0.76Ream 0.375 1.05 1/32 1.08 0.61 0.66C’sink 0.25 1/32 0.28 0.61 0.17Turn 4.5000 0.25 1/32 0.28 5.88 350 0.007 0.18Cutoff &chamfer

W–3/16

2.25 1/32 2.28 4.5 350 0.007 1.10

Subtotal machining time 5.47Subtotal handling and other machine data 1.52

Total cycle time min 6.99Hr/100 11.650

Setup hr 3.8

Setup for operation 20 HrPunch in and out, study drawing 0.2Jig 0.1First spindle for _ hole 0.75Additional turret for counterbore,tap, and tap-drill

0.21

Probably need four tools. Sincethe _ tap will need a drill that issmaller than _.

Setup for op. 20 1.26

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Cycle handling elements for operation 20 Freq Time MinStart and stop machine 0.08Air blow fixture 0.06Pick up part, move, and place; remove and lay aside 5 ≤ lb < 10 0.16Remove & reinsert slip bushings 6 0.07 0.42Change tool, each size 4 0.06 0.24Raise tool, position table to new location, lower toolready for drilling, per hole position for drill andc’bore

12 0.06 0.72

Turn over for back side work, guess 0.15Index turret, each or change drill for c’bore 4 0.03 0.12

Subtotal handling and other non-machining time 1.81

Cycle drilling, c’boring, tapping for operation 20 Freq L fdt MinDrill 5/16 in. hole 3 1.2 0.61 2.20Drill 3/8 in. hole c’bore, but using drill data 3 0.3 0.65 0.593/16 in. drill 3 0.3 0.55 0.50Tap 0.250-20 x 0.67 3 0.67 0.30 0.60

Subtotal drilling and c’bore times 3.89Total cycle time min 5.70

Hr/100 9.500Setup hr 1.26

(3)Labor and equipment cost

Workstation

No. Description SU Hr/100 Lot Hr PHC Lot Cost Unit Cost

Turretlathe

10 Position 3.8 11.650 62.1 75 $4654 $9.31

FaceTurnDrillReamC’sinkTurnCutoff &chamfer

Drillpress

20 Position 1.26 9.500 48.8 40 $1950 $6.45

Total cost of labor andequipment

$6604 $13.20

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(4)Tool estimate for Figure C7.1, drill jig

Tool type: Small drill jig without hinge top, sixbushings

Tool material: Hardened high carbon steel alloy, 5 lbDescription of part: Round block with six accurately located

holes, and six removable slip bushings

Description of tool elements HrFlat Base plate 8.0Slip bushings, 6 12.0Feet, 3 x 7.5

Total 27.5Productive hour cost for tool making $100Cost of tool, 27.5 x 100 2750Material cost, $25/lb 100Design cost $5000

Total cost of tool $7850Unit cost of tool, divide by 1500 $5.23

(5)Cost of operations, material, tooling

Element Lot Cost Unit CostMaterial for500 Q

$11,331 $22.66

Operationsfor 500 Q

$6604 $13.20

Tooling onbasis of1500 Q

$7850 $5.23

Total $25,785.00 $41.09

Instructor discussion: This actual part design has overstated tolerances on hole locations, especially forsocket head screws and tapped holes, which can be reduced to 0.00x in. rather than 0.000x in. for the socketholes. The design is a conversion from fractional dimensions to decimal dimensions and the drafts-personmade accurate conversions, not realizing that the effect of tolerance on cost. (Once again, we are victims ofthe English system.) Indeed, the tolerance for this problem will encourage 6 slip bushings, but even that isfor teaching lesson, and I would recommend against a real design having bushings to guide the drill forsocket head screws for this crazy design. Further, it is also very possible that typical machine shops will beable to hold tolerances for thousandths place without any hard tooling. It is likely that the hard tool couldbe eliminated. You might want to see what happens when the design is changed to reflect this over-design.

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