Mechanical Engineering 101 - UC Berkeley Department of...

© 2003‐2012, McMains, DornfeldME 101 lecturse 21‐22 1

Mechanical Engineering 101

University of California, Berkeley

Lecture #21

© 2003‐2012, McMains, Dornfeld ME 101 lecturse 21‐22 2

Today’s lecture

• Statistical Process Control– Process capability– Mean shift– Control chartsReading: pp. 373-383


.Precision


Process variation

• assignable causes– you know what caused variability– fix these!

• natural causes– inherent variability– or assignable, but cost/benefit doesn’t merit

fixing


+-

nominaldimension

L U

defects defectsgood parts

Scrap rate q

• from– natural variability of process– acceptable upper and lower spec. limits

• design specs

– deviation of actual from desired mean


Process capability

• cp• summary statistic for comparing design

tolerance to variation of (centered) process

6LUcp


Process capability

• Design spec is 1 .005– can make “setpoint” (i.e. = 1)– = .002 – cp = ?

6

LUcp


.Process capability

• If cp = 1, how many defective ppm?(ppm = parts per million)

6LUcp


.Defect rates from cp

Cp Tolerance Defect Defects(no. of std. dev.) rate, % ppm

0.333 ± 1 31.74 317,4000.667 ± 2 4.56 45,6001.000 ± 3 0.27 2,7001.333 ± 4 0.0063 631.667 ± 5 0.000057 0.572.000 ± 6 0.0000002 0.002

“acceptable”


Today’s lecture

• Statistical Process Control– Process capability– Mean shift– Control charts


.Mean shift

• process can’t make “setpoint” at nominal value• E.g. design spec is 1 .005

– process = 1.001, = .002 – fraction defective q=?


.Mean shift

• process can’t make “setpoint” at nominal value• design spec is 1 .005

– process = 1.001, = .002 – fraction defective q=?

upper limit +z = (1.005-1.001)/.002 = +2 std devs• F(2) = .9773

lower limit -z = (.995-1.001)/.002 = -3 std devs• F(3) = .9987

- +setpoint


Mean shift

• cp doesn’t reflect mean shift

6LUcp


Mean shift

• cp doesn’t reflect mean shift

• if can’t make setpoint, summary statistic is cpk

– calculated for side that’s worse

6LUcp

3,

3min LUcpk


.Mean shift

• Design spec is 1 .005– process = 1.001, = .002

3,

3min LUcpk

??,

??minpkc


.Mean shift

• Design spec is 1 .005– process = 1.000, = .002

• cp = 5/6

– process = 1.001, = .002 • cpk = 2/3• lower -- we lost capability with mean shift


Mean shift

• Some mean shift is expected– mean shift of 1.5 typical

• Set tolerances, reduce process variance accordingly– for tolerance of 6, mean shift of 1.5

• cpk = 1.5, for a defect rate of 3.4 ppm


Mean shift

Source: “Making war on defects,” IEEE Spectrum, September 1993, pp. 43-50


Today’s lecture


• X-bar and R control charts• Attribute control charts

Learning curve, defects• Reduction in manufacturing defects with time for two products• Testing if process in control only works at steady state


Ref: R. Mahoney, High Mix Low Volume Manufacturing, Prentice-Hall, 1997, p. 95

steady state value of defect rates

Statistical Process Control

• Especially once you’ve gotten defects low,may not be cost-effective to test all parts

• Continuous hypothesis testing– hypothesis: we’re (still) doing ok

• Two stages– determine capability of process– ensure process remains in control


Basic control charts


e.g.+3

e.g.-3

e.g.+3

e.g.-3

(UCL)

(LCL)

Control charts

• each “sample” on x-axis a group of n samples– “rational subgroup”– taken under same conditions

• hypothesis: we’re sampling a stationary process– so mean, variance of subgroups should be similar


Control chart types

• For quantitative, continuous variables(e.g. diameter or other dimension)

– chart (x-bar)• plots average of series of sample groups• indicates how process mean varies

– R chart• plots range of series of sample groups• indicates how variability of process changes


X

X‐bar chart

• m subgroups, each of size n– Calculate X and s for each

• mean value all subgroups =• process standard deviation estimate =


Xs

Averages and sums of random variables

Assuming all variables have same variance,• Sum of n values

• Variance n2

• Since variance 2 for X+Y = x2

+ Y2

• Average of n values• Variance 2 /n


Correction for mean

• say we knew true V(Xi)=2– is standard deviation of variable– but we need standard deviation of mean

of n values


nLCL

nUCL

nXV i

3

3

)(2

Correction for small n

• for estimates from samples– correction factor from table 10.2


nLCL

nUCL

3

3

nncsXLCL

nncsXUCL

)(3

)(3

4

4

)(

)()(

4

22

ncs

sEsE

Small subgroup sizes

• sample range – almost as much info as std dev– easier to calculate– correlation with std dev known

• R-chart: X-bar chart:


RnDLCL

RnDUCL

R

R

*)(

*)(

3

4

RnAXLCL

RnAXUCL

X

X

*)(

*)(

2

2

Suspicious patterns


UCL (~3)

-3

2

-2

1

-1

Sample groupsqual

ity m

easu

re

.Example: is process in control?• Find center, UCL, LCL for X-bar and R charts• in control?


RnDLCL

RnDUCL

R

R

*)(

*)(

3

4

RnAXLCL

RnAXUCL

X

X

*)(

*)(

2

2

subgroup (n=5) 1 2 3 4 5 6 7 8x-bar 2.008 1.998 1.993 2.002 2.001 1.995 2.004 1.999R 0.027 0.011 0.017 0.009 0.014 0.02 0.024 0.018

Example: is process in control?• Find center, UCL, LCL for X-bar and R charts• in control? 1) yes 2) looks suspicious 3) no


RnDLCL

RnDUCL

R

R

*)(

*)(

3

4

RnAXLCL

RnAXUCL

X

X

*)(

*)(

2

2

subgroup (n=5) 1 2 3 4 5 6 7 8x-bar 2.008 1.998 1.993 2.002 2.001 1.995 2.004 1.999R 0.027 0.011 0.017 0.009 0.014 0.02 0.024 0.018

Example: is process in control?X-bar R


1.985

1.99

1.995

2

2.005

2.01

1 2 3 4 5 6 7 80

0.005

0.01

0.015

0.02

0.025

0.03

1 2 3 4 5 6 7 8

Control limits

• If we take more samples (n increases), control limits should be

1. Wider2. Narrower3. Unchanged



Today’s lecture


• X-bar and R control charts• Attribute control charts

Control chart types

• For quantitative, continuous variables – chart (x-bar)

• plots average of series of sample groups• indicates how process mean varies

– R chart• plots range of series of sample groups• indicates how variability of process changes

• For discrete attributes (good/bad; go/no-go)– p chart

• plots percentage defective in sample groups– c chart

• plots number (count) of defects in sample groups


X

Attribute control charts

• Examples– number of defects per automobile– fraction of non-conforming parts in a sample– presence or absence of flash in molded part– number of flaws in a sheet product– defects/flaws in a painted surface– shorts or “opens” in a printed circuit board


P‐charts

• Expected proportion defective = p• Sample size = n• Exponential distribution for number defective in sample – E(X)=np– V(X)=np(1‐p)

npppLCL

npppUCL

/)1(3

/)1(3


.P‐charts

• If average fraction defective = 1/6, samples of 4 units taken, UCL is


P‐charts derivation

• Proportion defective = p• For underlying population

– Mean = p– Variance = p(1-p)

derivation below

nXX 2

2 )(

22 )E( XX


222 )0)(1()1()E( ppppXX 22 )1()21( ppppp

)1(2 pppp

Averages and sums of random variables

Assuming all variables have same variance,• Sum of n values

• Variance n2

• Since variance 2 for X+Y = x2

+ Y2

• Average of n values• Variance 2 /n


P‐charts derivation

• Proportion defective = p• Sample size = n• Underlying population

– Mean = p– Variance = p(1‐p) {see prev. derivation}

• Number defective in sample X• Percentage defective X/n

– E(X/n)=expected percentage defective=p– V(X/n)=p(1‐p)/n

npppLCL

npppUCL

/)1(3

/)1(3


Feedback loop


Ref: Groover, p. 666

Mechanical Engineering 101 - UC Berkeley Department of...

Documents

Transcript of Mechanical Engineering 101 - UC Berkeley Department of...