Manufacturing System Introduction

ME 346 ME 346 Manufacturing Manufacturing Processes IIProcesses II3-0 Cr Hr3-0 Cr Hr

by by

Gp Capt (R) Akhtar HusainGp Capt (R) Akhtar Husain

How to Contact the How to Contact the Instructor Instructor

Office: Room G16, IAA Building Office Phone: Ext 403 e-mail : akhtar.husain@ mail.au.edu.pk Office hours Monday Thu Friday:

0830 hrs - 1630 hrs

COURSE OBJECTIVECOURSE OBJECTIVE

ME 346 - Manufacturing Processes II is ME 346 - Manufacturing Processes II is continuation of ME 245 - Manufacturing continuation of ME 245 - Manufacturing Processes I which you have already covered in Processes I which you have already covered in 44thth Semester. These two courses aim at Semester. These two courses aim at familiarizing you with:-familiarizing you with:-– Manufacturing SystemsManufacturing Systems– Conventional Manufacturing Processes Conventional Manufacturing Processes – Latest Advancements In Manufacturing TechnologyLatest Advancements In Manufacturing Technology– Manufacturing AutomationManufacturing Automation– Familiarization with CNC Machine Tools Familiarization with CNC Machine Tools – Programming of CNC MachinesProgramming of CNC Machines– Manufacturing Aids Manufacturing Aids – Basics of Manufacturing managementBasics of Manufacturing management

ME 346 Course ContentsME 346 Course Contents

• Main Attributes of Manufacturing Processes• Material Removal

• Cutting Tools & Fluids

• Machine Processes For Cutting Various Shapes

• Abrasive Machining & Finishing Operations

• Non-Conventional Machining Processes

• Rapid Prototyping

• Computer Control of Machine Tools & CIM

• CNC Programming

• Electronics Fabrication

• Jigs & Fixtures

• Metrology & Precision Measurements

• Introduction to Process Planning

Books & ResourcesBooks & Resources

Text Books: – Manufacturing Engineering and Technology

4th Edition By Serope Kalpakjian & Steven Schmid , Pearson Education Inc.

– Fundamentals of Modern Manufacturing, 3rd Edition, by M P Groover, John Wiley & Sons

Reference Books:– Materials and Processes in Manufacturing By

E.P Degarmo, Wiley

EvaluationEvaluation

There will be at least 6 quizzes, one Mid Semester Exam, and One Final Exam. In addition, there will be at least 3 home Assignments a Project to be completed within specified due dates/timings. Grading will be relative. Marks distribution would be as under:-

WeightageQuizzes 7 %Assignments 6 %Project 7 %Mid Term Exam 35 %Final Exam 45 %

Total 100%

Main Attributes of Main Attributes of Manufacturing Manufacturing System System

Four Main Attributes of Four Main Attributes of Manufacturing System are:-Manufacturing System are:-

CostCostTime Time QualityQualityFlexibilityFlexibility

Cost ElementsCost Elements

Cost of Equipment (Capital Cost)Cost of Equipment (Capital Cost) Labour CostLabour Cost Material CostMaterial Cost Energy CostEnergy Cost Maintenance CostMaintenance Cost Training CostTraining Cost MiscMisc

For Capital cost -------------For Capital cost -------------Consider Following ExampleConsider Following Example

CNC LatheCNC Lathe Capital Cost – Rs 2.5 MilCapital Cost – Rs 2.5 Mil Estimated Life – 5 YearsEstimated Life – 5 Years Operation Based on :Operation Based on :

– 300 days/year300 days/year– 16 hours/day16 hours/day

Machine hourly rate:-Machine hourly rate:-

2.5x1000,000/5x300x162.5x1000,000/5x300x16

=Rs 1042 per hr=Rs 1042 per hr

Conventional LatheConventional Lathe Capital Cost – Rs 0.25 MilCapital Cost – Rs 0.25 Mil Estimated Life – 5 YearsEstimated Life – 5 Years Operation Based on :Operation Based on :

– 300 days/year300 days/year– 16 hours/day16 hours/day

Machine hourly rate:-Machine hourly rate:-

0.25x1000,000/5x300x160.25x1000,000/5x300x16

=Rs 104 per hr=Rs 104 per hr

Work Centre CostWork Centre Cost

It is generally taken as addition of a,b,c.It is generally taken as addition of a,b,c.

a.a. Capital Cost (Machine Hourly Rate) say Capital Cost (Machine Hourly Rate) say 370/hr370/hr

b.b. Labour Rate say 120/hrLabour Rate say 120/hr

c.c. O/H say (40% of a) and (55% of b) aboveO/H say (40% of a) and (55% of b) above

Work Centre Cost = Work Centre Cost = 370+120+(0.4x370+0.55x120)370+120+(0.4x370+0.55x120)

Analyzing Cost per Analyzing Cost per Piece…………….Piece…………….Consider Following ExampleConsider Following Example A batch of 150 Part X is machined from A batch of 150 Part X is machined from

35mm bar stock costing Rs 1800/m. Work 35mm bar stock costing Rs 1800/m. Work centre hourly rate is Rs 705. The machine centre hourly rate is Rs 705. The machine setup time is 25 hrs, and each part requires setup time is 25 hrs, and each part requires 37 min of machining. The job requires a 37 min of machining. The job requires a single cutting tool costing Rs 610 which lasts single cutting tool costing Rs 610 which lasts for 7 workpieces. Tool change time is 20 min. for 7 workpieces. Tool change time is 20 min. The cost of indirect materials is Rs1184. The cost of indirect materials is Rs1184.

Analyzing Cost per PieceAnalyzing Cost per Piece

Material CostMaterial Cost– Direct Direct = 35/1000 x1800= 35/1000 x1800 =Rs63.00/piece=Rs63.00/piece– IndirectIndirect = 1184/150= 1184/150

=Rs7.90/piece=Rs7.90/piece– ToolTool = 610/7= 610/7 =Rs87.14/piece=Rs87.14/piece

Work Centre CostWork Centre Cost– SetupSetup = 25/150 x = 25/150 x 705 =Rs117.50/piece705 =Rs117.50/piece– MachiningMachining =37/60x705 = Rs 434.75/piece=37/60x705 = Rs 434.75/piece– Tool Change=20/60x1/7x705=Rs33.57/pieceTool Change=20/60x1/7x705=Rs33.57/piece

0.00

50.00

100.00

150.00

200.00

250.00

300.00

350.00

400.00

450.00

500.00

Dir Material IndirMaterial

ToolMaterial

Set Up Machining ToolChange

Analyzing Cost per Analyzing Cost per PiecePiece

Time!Time!

AvailabilityAvailability

Percentage of the total time the Percentage of the total time the equipment is available for equipment is available for production.production.

ReliabilityReliability

Probability that the equipment will Probability that the equipment will continue to remain operational for continue to remain operational for certain specified length of time.certain specified length of time.

Availability “A”Availability “A”

A = MTBF/(MTBF+MTTR)A = MTBF/(MTBF+MTTR)

MTBFMTBF =Mean Time Between Failures=Mean Time Between Failures

MTTRMTTR =Mean Time to Repair=Mean Time to Repair

5 11 4 7 4 7 35 11 4 7 4 7 3

42 44 87 23 22 21 26 2542 44 87 23 22 21 26 25

MTBF = (42+44+87+23+22+21+26+25)/7MTBF = (42+44+87+23+22+21+26+25)/7

= 41.43= 41.43

MTTR MTTR = (5+11+4+7+4+7+3)/7= (5+11+4+7+4+7+3)/7

=5.86=5.86

AA == 41.43/(41.43+5.86)41.43/(41.43+5.86) = 0.876 or = 0.876 or 87.6%87.6%

Failure Rate Failure Rate λλ = 1 / MTBF = 1 / MTBF

AvailabilityAvailability “A” “A”

Reliability “R”Reliability “R”

– λλ t t

RR = e= e

Where Where λλ = 1/MTBF ………(Failure = 1/MTBF ………(Failure Rate)Rate)

And t = Mission TimeAnd t = Mission Time

Here Here λλ = 1/41.43 = 1/41.43

= 0.0241= 0.0241

Let us assume 2 shift x 8 hour Let us assume 2 shift x 8 hour operation or mission time, thenoperation or mission time, then

- (0.0241x16)- (0.0241x16)

RR = e= e = 0.68 or 68 %= 0.68 or 68 %

Reliability “R”Reliability “R”

λ= .00023 λ= .00027λ= .0073λ= .00025

λ= .00023+.00025+.0073+.0027 = .00805Let t=16, Reliability R will be -λ tR = e

= 0.87987 or 87.9 %

Consider Following Example.Consider Following Example.

λ= .00023 λ= .00027λ= .0073λ= .00025

λ= .00023+.00025+.001825 +.00027 = .002575Let t=16, Reliability R will be -λ tR = e

= 0.9596 or 95.96 %

λ= .0073

λ= .0073

λ= .0073

With Redundancy, With Redundancy, Notice the Notice the increase in R increase in R

QualityQuality

Conformance to standardsConformance to standards Quality is free!Quality is free! Quality is either….Quality is either….

– Specified Specified – ImpliedImplied

Mr. Deming! JapanMr. Deming! Japan

QualityQuality

Capability Index (Cp)Capability Index (Cp) Cp = UCL-LCL (say Tolerance) / 6σ Cp = UCL-LCL (say Tolerance) / 6σExplanation:Explanation: In this formula, USL and LSL In this formula, USL and LSL are the upper and lower specification are the upper and lower specification limits of the quality characteristic of the limits of the quality characteristic of the productproduct, respectively. The quantity in the , respectively. The quantity in the denominator, 6σ signifies the fact that denominator, 6σ signifies the fact that 99.73% of the values generated by the 99.73% of the values generated by the processprocess is contained within. is contained within.

Machine Accuracy

Product Tolerance

Cp = 14/6 = 2.33………Mfg Process too accurate for the product.

Product Tolerance

Machine Accuracy

Cp = 2/6 = 0.33 …… Mfg Process too rough / inaccurate for the product.

Product Tolerance

Machine Accuracy

Cp = 6.1/6 = 1.02 …. The mfg process is well suited for the job.

Quality : Quality : Consider Following Consider Following Example.Example.

Part Tolerance + 10 Part Tolerance + 10 µmµm Machine Accuracy (Following random error Machine Accuracy (Following random error

data available. All values in µm.)data available. All values in µm.)22 -3 -1 3 1 3 2 -2 1 -3 -1 0-3 -1 3 1 3 2 -2 1 -3 -1 0

Standard Deviation Standard Deviation σσ = = Cp = Tolerance/6Cp = Tolerance/6σσ = = = =

FlexibilityFlexibility

Flexibility of the Mfg System can be Flexibility of the Mfg System can be defined as defined as the ease with which the the ease with which the production can be changed…..production can be changed…..

Product, capacity or Process FlexibilityProduct, capacity or Process Flexibility– Product Type can be changed – Product flexibilityProduct Type can be changed – Product flexibility– Product Volume can be changed – Capacity Product Volume can be changed – Capacity

FlexibilityFlexibility– Manufacturing Process can be changed – Process Manufacturing Process can be changed – Process

FlexibilityFlexibility

………………..depending upon how easily ..depending upon how easily these parameters can be altered.these parameters can be altered.

Three manufacturing systems I, II & III Three manufacturing systems I, II & III are being considered for future are being considered for future production.production.

Marketing Dept has come up with Marketing Dept has come up with following forecast for new product line: following forecast for new product line: – The chances that our next product would beThe chances that our next product would be

Product A … 50 %Product A … 50 % Product B … 30 %Product B … 30 % Product C … 20 %Product C … 20 %

cont’dcont’d

Product Flexibility: Product Flexibility: Consider Following ExampleConsider Following Example

It would require following amounts to It would require following amounts to produce the product A,B & C on mfg systems produce the product A,B & C on mfg systems I, II & III respectively. I, II & III respectively.

ProductProduct Plant IPlant I Plant IIPlant IIPlant IIIPlant III– AA 3 Mil3 Mil 5 Mil5 Mil 7 Mil7 Mil– BB 0 Mil0 Mil 7 Mil7 Mil 2 Mil2 Mil– CC 6 Mil6 Mil 1 Mil1 Mil 1 Mil1 Mil

Cont’dCont’d

Flexibility: ExampleFlexibility: Example

Penalty of Change (POC) can be calculated for each Penalty of Change (POC) can be calculated for each mfg system as followsmfg system as follows

Product A Product B Product C Product A Product B Product C POCPOC

Sys I Sys I 3(.5)3(.5) 2(.3)2(.3) 6(.2)6(.2) 3.33.3

Sys II 5(.5)Sys II 5(.5) 7(.3)7(.3) 1(.2)1(.2) 4.84.8

Sys III 7(.5)Sys III 7(.5) 2(.3)2(.3) 1(.2)1(.2) 4.34.3

Since POC is minimum for Sys I (3.3), it is the most Since POC is minimum for Sys I (3.3), it is the most flexible. flexible.

Flexibility: ExampleFlexibility: Example

The EndThe End

Any Questions?Any Questions?