MMAE557 Personal Consulting Project

Analysis of Ball Bearing Manufacturing

Li He A20358122

Xingye Dai A20365915

Instructor: John Cesarone

Apr.30, 2016

Section.1 Client Description

Our client is Weili bearing factory in China which we have ever visited. It’s a company located in

Hebei province and produces many models of bearings. It’s a private company so the factory

has not a large scale of manufacturing and their economic situation does not allow them to hire

a large amounts of workers. In that case, to improve their efficiency of manufacturing system

becomes essential.

In this factory, one of the production workshop was investigated. Totally three lines of work

stations are applied to the manufacturing process of bearings. The first line includes 4 steps

(Turning, heat treatment, grinding and testing) of manufacturing the inner and outer rings. The

second line includes 5 steps (shearing & heading, forging, heating treatment, lapping, testing)

of manufacturing the bearing balls. The third line includes 3 steps (blanking & punching,

forming and testing) of manufacturing the cages. All the machining processes are finished by

corresponding machining equipment and the heat treatments are processing in furnace. The

products are annealed at high temperature and quenched in oil for hardening. The inspections

are done by workers and operators. Each line has two operators, one for testing and the other

one for loading and unloading. The inspection for size and surface finish of balls is carried out

on a sample basis by means of microscopes and other precision equipment. The balls are then

cleaned and packed ready for bearing assembly operations. For inspections of cages, blanking,

punching, forming rivet holes and visual inspection is carried for any deformity. For inner and

outer rings, the diameters are measured. The finished parts of bearings are then manually

transport to assembling station after a batch of parts are produced. Then the assembling

station assembles the parts into bearings automatically. The finally products are then tested by

2 inspectors at next station. For each line, because all the steps are continuous and the

transport time between stations is accounted in load and unload time. the products are

transport so the multiple station cells are not applied to the manufacturing system. Two types

of products are recorded for the entire process and their average time for each procedure were

calculated and listed in following tables.

The classifications of manufacturing system of this factory belongs to ‘’Type II-A B. Type II-A

means it works on multiple automated stations and B means that the cell produces bearing

parts in batch production system.

Section 2: Client analysis

This factory is capable of manufacturing of large amount of bearings depends on the demands

and orders. In our investigation, processing times of two products are recorded and their cycle

time and production rates are calculated in following steps.

Line 1: Inner/Outer Ring Manufacturing Process and Step Time

Turning (s) Heat Treatment (S) Grinding (s) Testing (s) Load & Unload (s)

Product A 3 2 4 2 2.5

Product B 4 2.5 5 2 2.5

Note: The time of Heat Treatment is the average time of each ring.

Line 2: Ball Manufacturing Process and Step Time

Shearing & Heading (s)

Forging (s)

Heat Treatment (s)

Lapping (s)

Testing (s)

Load & Unload (s)

Product A 0.1 0.1 0.05 0.3 0.3 0.3

Product B 0.1 0.1 0.05 0.3 0.3 0.3

Note: Each of Product A needs 8 balls; Each of Product B needs 10 balls.

Line 3: Cage Manufacturing Process and Step Time

Blanking & Punching (s) Forming (s) Testing (s) Load & Unload (s)

Product A 0.5 2 1 1.5

Product B 0.8 3 1 1.5

Assembly:

Note: the Assembly & Testing time (Cycle Time) of Product A is 10s, the Assembly & Testing time (Cycle

Time) of Product B is 12s.

Plant Layout:

Cage

Rivet

Outer Ring

Inner Ring

Balls

Assembling (Pillow Wrapping Machine) Testing

Cycle Time and Production Rate

Line 1:

2 operators: one for Testing, one for loading & unloading

Turning Heating Grinding Testing

Breakdown Probability 0.005 0.003 0.008 0.007

Down Time (min) 5 10 6 3

𝑇𝐶 = 𝑇𝑀 + 𝑇𝑆

A: Tc1=3+2+4+2+2.5=13.5S

B: TC1=4+2.5+5+2+2.5=16S

𝑇𝑃 = 𝑇𝐶 + 𝐹 ∗ 𝑇𝑑

F*Td=0.005×5+0.003×10+0.008×6+0.007×3=0.124min=7.44s

A: Tp1=13.5+7.44=20.94s

B: Tp1=16+7.44=23.44s

𝑅𝑃 =3600

𝑇𝑃 (Parts/hr)

A: Rp1=3600

20.94=171.92 Parts/hr.

B: Rp1=3600

23.44=153.58 Parts/hr.

Line 1 Line 2 Line 3

Assembly

Testing

Line 2:

2 operators: one for Testing, one for loading & unloading

Shearing & Heading Forging Heat Treatment Lapping Testing

Breakdown Probability 0.0008 0.0003 0.0003 0.0005 0.0007

Down Time (min) 3 5 10 5 3

𝑇𝐶 = 𝑇𝑀 + 𝑇𝑆

Ball: Tc=0.1+0.1+0.05+0.2+0.3+0.3=1.05s

𝑇𝑃 = 𝑇𝐶 + 𝐹 ∗ 𝑇𝑑

F*Td=0.0008×3+0.0003×5+0.0003×10+0.0005×5+0.0007×3=0.0115min=0.69s

Ball: Tp=1.05+0.69=1.74s

𝑅𝑃 =3600

𝑇𝑃 (Parts/hr)

Ball: Rp=3600

20.91.744=2068.97 Parts/hr.

Tc

A: Tc2=8*Tc=8×1.05=8.4s

B: TC2=10* Tc=10×1.05=10.5s

Rp

A: Rp2=2068.97×1

8=258.62 Parts/hr.

B: Rp2=2068.97×1

10=106.90 Parts/hr.

Line 3:

2 operators: one for Testing, one for loading & unloading

Blanking & Punching Forming Testing

Breakdown Probability 0.005 0.003 0.004

Down Time (min) 5 8 3

𝑇𝐶 = 𝑇𝑀 + 𝑇𝑆

A: Tc3=0.5+2+1+1.5=5s

B: TC3=0.8+3+1+1.5=6.3s

𝑇𝑃 = 𝑇𝐶 + 𝐹 ∗ 𝑇𝑑

F*Td=0.005×5+0.003×8+0.004×3=0.061min=3.66s

A: Tp3=5+3.66=8.66s

B: Tp3=6.3+3.66=9.96s

𝑅𝑃 =3600

𝑇𝑃 (Parts/hr)

A: Rp3=3600

8.66=415.70 Parts/hr.

B: Rp3=3600

9.96=361.45 Parts/hr.

Assembly:

Breakdown Probability: 0.008; Down Time: 8min.

A: TPA=10+0.008×8×60=13.84s

B: TPA=12+0.008×8×60=15.84s

Ball Bearing:

Tc= max{Tc1, TC2, TC3}+TCA

A: TC=TC1+TCA=13.5+10=23.5s

B: TC=TC1+TCA=16+12=28s

Tp= max{TP1, Tp2, TP3}+TPA

A: Tp=TP1+TPA=20.94+13.84=34.78s

B: Tp=TP1+TPA=23.44+15.84=39.28s

RP=3600

𝑇𝑃 (Parts/hr)

A: RP=3600

34.78 =103.51 Bearing/hr.

B: RP=3600

39.28 =91.65 Bearing/hr.

Sections 3: optimize the system

Notice that the original system is not perfect efficient based on current process of

manufacturing, some improvement changes could be applied into the system like the number

of workstations and workers.

3.1 Optimum Number of Workstation—Assembling Workstation

Product A Product B

Annual Goal (Q) 800000 600000

Cycle Time without testing (TC) 8s 10s

Scrap Time (µ) 2%

Worker Efficiency (Ew) 95%

Machine Running Reliability (RM) 98%

Machine Setup Reliability (RS) 99.5%

Setup Time (TSP) 20min

Batch Size (N) 5000

Original Factory Operation Hours: 8 Hours/day; 5 Days/week; 50 Weeks/year.

Workload:

𝑊𝐿 =𝑄 ∗ 𝑇𝐶1 − µ

A: WL(A)=800000×8

1−2%=6530612s=1814.06hr.

B: WL(B)=600000×10

1−2%=6122449s=1700.68hr.

No. of setup changes:

No.=𝑄(𝐴)+𝑄(𝐵)

𝑁=800000+600000

5000=280

Setup: WL(SP)=No. ×TSP=280×20=5600min=93.33hr.

Available Time:

AT=8×5×50=2000hr/yr.

Optimum number of workstations:

𝑁𝑜.=𝑊𝐿(𝐴) + 𝑊𝐿(𝐵)

𝐴𝑇 ∗ 𝐸𝑊 ∗ 𝑅𝑀+

𝑊𝐿(𝑆𝑃)

𝐴𝑇 ∗ 𝑅𝑆

𝑁𝑜.=1814.06+1700.68

2000×0.95×0.98+

93.33

2000×0.995=1.93452

So, 1 workstation 2 workstations OR double the operation hours.

Considering the high price of assembling machines, we choose doubling the operation hours as the

solution.

Then, the cycle time of assembling (with testing) changes from 8+2 to 4+2 for product A; 10+2 to 5+2 for

product B.

Ball Bearing:

Cycle Time:

A: TC=TC1+TCA=13.5+6=19.5s

B: TC=TC1+TCA=16+7=23s

Production Time:

A: TP=TP1+TPA=20.94+(6+0.008×8×60)=30.78s

B: TP=TP1+TPA=23.44+(7+0.008×8×60)=34.28s

Production Rate:

A: RP=3600

30.78=116.96 Bearing/hr.

B: RP=3600

34.28=105.02 Bearing/hr.

3.2 Machine Cluster Analysis (Determine the number of workers)

Line 1 Line 2 Line 3 Line 4

A (s) B (s) A (s) B (s) A (s) B (s) A (s) B (s)

Machine Time (TM) 6 7 0.5 0.625 1.5 2 6 7.5

Worker Service Time (Ts) 5 5 0.3 0.3 1.5 1.5 4 4.5

Repositioning Time (Tr) 5 5.5 0.15 0.1875 2.3 3.5 5 6

The number of machines could be operated by one person (N): 𝑁 =𝑇𝑀+𝑇𝑆

𝑇𝑆+𝑇𝑟

Line 1:

A: 𝑁 =6+5

5+5= 1.1 one worker could operate 1.1 line 1. So, in line 1, employee 1 worker.

B: 𝑁 =7+5

6+5.5= 1.14 1 worker

Line 2:

A: 𝑁 =0.5+0.3

0.3+0.15= 1.78 1 worker

B: 𝑁 =0.625+0.3

0.3+0.1875= 1.90 1 worker

Line 3:

A: 𝑁 =1.5+1.5

1.5+2.3= 0.79 2 workers

B: 𝑁 =2+1.5

1.5+3.5= 0.7 2 workers

Assembly:

A: 𝑁 =6+4

4+5= 1.11 1 worker

B: 𝑁 =7.5+4.5

4.5+6= 1.14 1 worker

Conclusion: Theoretically, set 1 worker in Line 1; 1 worker in Line 2; 2 workers in Line 3; 1 worker in

Assembly.

However, as to practice, some of the stuff could not be lifted by one worker, and there are a lot of

emergency situations which could cause the stop of whole factory if there is only one worker in a line,

such as sickness or vacation.

As to Line 2, one worker almost could operate 2 lines. The company could try to train the other guys,

and ask them to do both line 1 & line2, or both line 2 & line 3. Another method, the company could try

to buy standard Balls from other companies, just like buying Rivets.

Cost Analysis (Machine never idle):

Labor Cost ( CL): $2.5/hr. Machine/Line Cost (CM): $20/hr.

Cost/part=(𝐶𝐿

𝑛+ 𝐶𝑀) ∗ (𝑇𝑀 + 𝑇𝑆)

Original:

Line 1, Labor 2:

A: $=(2.5×2+20) ×13.5÷3600=0.09375

B: $=(2.5×2+20) ×16÷3600=0.1111

Line 2, Labor 2:

A: $=(2.5×2+20) ×8.4÷3600=0.0583

B: $=(2.5×2+20) ×10.5÷3600=0.0729

Line 3, Labor 2:

A: $=(2.5×2+20) ×5÷3600=0.0347

B: $=(2.5×2+20) ×6.3÷3600=0.0438

Assembly, Labor 2:

In order to meet the annual goal, we choose double the operator hours. And the Labor Cost at night is

$3/hr, which means the average Labor Cost is $2.75/hr.

A: $=(2.75×2+20) ×10÷3600=0.0708

B: $=(2.75×2+20) ×12÷3600=0.085

Total:

A: ∑$=0.09375+0.0583+0.0347+0.0708=0.25755 $/bearing

B: ∑$=0.1111+0.0729+0.0438+0.085=0.3128 $/bearing

Optimization (theoretically)

Line 1, Labor 2:

A: $=(2.5+20) ×13.5÷3600=0.0844

B: $=(2.5+20) ×16÷3600=0.1000

Line 2, Labor 2:

A: $=(2.5+20) ×8.4÷3600=0.0525

B: $=(2.5+20) ×10.5÷3600=0.0656

Line 3, Labor 2:

A: $=(2.5×2+20) ×5÷3600=0.0347

B: $=(2.5×2+20) ×6.3÷3600=0.0438

Assembly, Labor 2:

A: $=(2.75+20) ×10÷3600=0.0632

B: $=(2.75+20) ×12÷3600=0.0758

Total:

A: ∑$=0.0844+0.0525+0.0347+0.0632=0.2348 $/bearing

B: ∑$=0.1000+0.0656+0.0438+0.0758=0.2852 $/bearing

Conclusion: After optimizing the number of workers, the cost per product A reduces from $0.25755 to

$0.2348, the cost per product B reduces from $0.3128 to $0.2852.

Section 4 Automate the system

Notice that the original layout of the system has different transportation time for each lines whereas

only the maximum Tr could be applied in to the calculation of calculation time. Then we assumed a new

lay out of the system and all the transportation will be carried out by conveyor belt automatically.

Original:

Tr=max{1, 1.5, 1, 1.5}=1.5s

Optimum:

0.5s

1s

1s

Ring Ball Cage

Assembly

Testing

Rivet

Same distance between each line to assembly workstation.

Tr=max{1.25, 1.25, 1.25, 1.25}=1.25s

Then optimized layout could save repositioning time from 1.5s to 1.25s

New Tc and Rp

Tc:

A: Tc=19.5-0.25=19.25s

B: Tc=23-0.25=22.75s

Rp:

Due to the Tp of line 1 which is the maximum can save 0.25s, the whole Tp could save 0.25s too.

A: Rp=3600

30.53=117.92 Bearing/hr.

B: Rp=3600

34.03=105.79 Bearing/hr.

1.25s

1.25s

Ring

Assembly

Ball

Rivet Cage

Testing

1.25s

Half Circle Layout

Section 5 Quality assurance

For any factory, quality is the most important specification for their reputations. Therefore, the

inspection procedure becomes essential. The statistical process and quality analysis is given by

the control charts. The mean and range charts are also plotted.

5.1 Product A

Y direction

Radius Testing Data AVG(x) Range

Y1 r 2.0032 2.0018 1.9958 1.9972 2.0081 2.0026 2.00145 0.0123

R 2.5092 2.5039 2.4909 2.4992 2.5041 2.4955 2.500467 0.0183

Y2 r 2.0071 2.0062 2.0021 1.9988 1.9943 2.0011 2.0016 0.0128

R 2.4989 2.509 2.5021 2.4991 2.4982 2.5033 2.501767 0.0108

Y3 r 1.9978 1.9954 1.9942 1.9933 1.9961 2.0006 1.996233 0.0073

R 2.5032 2.4929 2.4932 2.5077 2.4954 2.5031 2.49925 0.0148

Y4 r 1.9963 2.0085 1.9909 1.9959 2.0085 2.0044 2.00075 0.0176

R 2.5019 2.5055 2.5042 2.4959 2.5044 2.4951 2.501167 0.0104

Y5 r 2.0088 2.0105 2.0049 2.0088 2.0038 1.9992 2.006 0.0113

R 2.4982 2.4966 2.504 2.4921 2.5089 2.4906 2.4984 0.0183

1.988

1.99

1.992

1.994

1.996

1.998

2

2.002

2.004

2.006

2.008

2.01

1 2 3 4 5

Dem

ensi

on

of

Inn

er C

ircl

e (i

nch

)

Sample Number,s

Chart of r

UCL

Center

LCL

r

Chart Range Chart

CL UCL LCL CL UCL LCL

2.00120667 2.0071282 1.995285 0.01226 0.024569 0

0

0.005

0.01

0.015

0.02

0.025

0.03

1 2 3 4 5

Ran

ge

Sample Number, s

Range Chart of r

Center

2.485

2.49

2.495

2.5

2.505

2.51

1 2 3 4 5

Dem

ensi

on

of

Ou

ter

Cir

cle

(in

ch)

Sample Number, s

Chart of R

UCL

Center

LCL

𝑋

5.2 Product B

Y direction

Radius Testing Data AVG(x) Range

Y1 r 3.0098 3.0044 3.0051 3.0026 2.9964 2.9956 3.002317 0.0142

R 3.9976 3.9985 4.0021 4.0016 4.0032 3.9925 3.99925 0.0107

Y2 r 2.9984 2.9973 3.0024 3.0036 2.9979 3.0035 3.000517 0.0063

R 4.0035 3.9946 3.9959 4.0031 3.9916 4.0035 3.9987 0.0119

Y3 r 3.0019 3.0026 2.9959 2.9967 3.0016 3.0025 3.0002 0.0067

R 4.0038 3.9958 3.9986 4.0025 4.0016 4.0035 4.000967 0.008

Y4 r 3.0026 3.0053 2.9935 2.9979 3.0019 3.0014 3.000433 0.0118

R 4.0009 3.9959 4.0016 3.9953 3.9929 4.0031 3.998283 0.0102

Y5 r 3.0049 3.0053 2.9959 2.9975 3.0041 3.0029 3.001767 0.0094

R 3.9989 4.0011 4.0021 4.0013 3.9989 4.0019 4.0007 0.0032

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

1 2 3 4 5

Ran

ge

Sample Number,s

Range Chart of R

UCL

Center

LCL

R

Chart Range Chart

CL UCL LCL CL UCL LCL

2.50021 2.5072232 2.493197 0.01452 0.029098 0

r

Chart Range Chart

CL UCL LCL CL UCL LCL

3.00104667 3.0057221 2.996371 0.00968 0.019399 0

2.99

2.992

2.994

2.996

2.998

3

3.002

3.004

3.006

3.008

1 2 3 4 5

Dem

ensi

on

of

Inn

er C

ircl

e (i

nch

)

Sample Number,s

Chart of r

UCL

Center

LCL

0

0.005

0.01

0.015

0.02

0.025

1 2 3 4 5

Ran

ge

Sample Number, s

Range Chart of r

Center

R

Chart Range Chart

CL UCL LCL CL UCL LCL

3.99958 4.0038304 3.99533 0.0088 0.017635 0

The charts show us the quality is quite well controlled during the process. While consider that the

company are using some low quality machining equipment. The way to make their defect rate better is

to purchase some new equipment with higher capability of controlling the precision and qualities, which

would also reduce the cost significantly. Assume the defect rate reduces from 5% to 3% after using the

new equipment, the quantity of required bearings to meet every 1 million will reduces from 1,052,632

3.99

3.992

3.994

3.996

3.998

4

4.002

4.004

4.006

1 2 3 4 5

Dem

ensi

on

of

Ou

ter

Cir

cle

(in

ch)

Sample Number, s

Chart of R

UCL

Center

LCL

𝑋

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02

1 2 3 4 5

Ran

ge

Sample Number,s

Range Chart of R

UCL

Center

LCL

to 1,030,928. Cost saved per million is $5,096.14 for A and $6,190.03 for B. Calculated as the following

steps.

Defect Rate=5%:

𝑁 =1,000,000

1 − 5%= 1,052,631.58 = 1,052,632

Defect Rate=3%:

𝑁 =1,000,000

1 − 3%= 1,030,927.84 = 1,030,928

Cost Saved (Price per Bearing is $0.2348 for A, $0.2852 for B):

CA=0.2348×(1,052,632-1,030,928)=$5,096.14/million

CB=0.2852×(1,052,632-1,030,928)=$6,190.03/million