Exercise 1: Statistical analysis
- a methodological experiment in MF2024, 2010
{Name of student #1}
{Name of student #2}
KTH Machine Design
School of Industrial Engineering and Management
MF2024 Robust and probabilistic design
School of Industrial Engineering and Management
October 2010
1 - PAPER CLIP FATIGUE TEST
Figure 1. The histogram
Normal distribution results:
Figure 2. Normal distribution plots


Normal probability density function:


1

f x;  ,   
e
 2
Our calculated parameter values for a normal distribution:
µ= ……………….
= ……………….
1
 x   2 
2 2 
Exponential distribution results:
Figure 3. Exponential distribution plots
Normal probability density function:
e x
f x;    
 0
,x 0
,x 0
Our calculated parameter value for an exponential distribution:
= ……………….
2
Weibull distribution results:
Figure 4. Weibull distribution plots
Weibull probability density function:
k x
f x; k ,     
 
, x  0;
k 1
e  x  
k
Our calculated parameter value for a Weibull distribution:
= ……………….
k= ……………….
Is there any reason to include a third Weibull parameter x0? If yes, what do you estimate x0
to be?
……………………………………………….
……………………………………………….
Conclusion:
State here what type of distribution that is the “best fit” to this set of test data, and also why
you came to that conclusion. Also comment on how the quality of the test could have been
improved,
3
2 - FASTENER SNAP FORCE TEST (N FASTENERS, N
TESTERS)
Figure 5. The histogram
Normal distribution results:
Figure 6. Normal distribution plots


Normal probability density function:


1
f x;  ,   
e
 2
Our calculated parameter values for a normal distribution:
µ= ……………….
= ……………….
4
 x   2 
2 2 
Exponential distribution results:
Figure 7. Exponential distribution plots
Normal probability density function:
e x
f x;    
 0
,x 0
,x 0
Our calculated parameter value for an exponential distribution:
= ……………….
5
Weibull distribution results:
Figure 8. Weibull distribution plots
Weibull probability density function:
k x
f x; k ,     
 
, x  0;
k 1
e  x  
k
Our calculated parameter value for a Weibull distribution:
= ……………….
k= ……………….
Is there any reason to include a third Weibull parameter x0? If yes, what do you estimate x0
to be?
……………………………………………….
……………………………………………….
Conclusion:
State here what type of distribution that is the “best fit” to this set of test data, and also why
you came to that conclusion. Also comment on how the quality of the test could have been
improved,
6
3 - FASTENER SNAP FORCE TEST (1 FASTENER, N
TESTERS)
Figure 9. The histogram
Normal distribution results:
Figure 10. Normal distribution plots


Normal probability density function:


1
f x;  ,   
e
 2
Our calculated parameter values for a normal distribution:
µ= ……………….
= ……………….
7
 x   2 
2 2 
Exponential distribution results:
Figure 11. Exponential distribution plots
Normal probability density function:
e x
f x;    
 0
,x 0
,x 0
Our calculated parameter value for an exponential distribution:
= ……………….
8
Weibull distribution results:
Figure 12. Weibull distribution plots
Weibull probability density function:
k x
f x; k ,     
 
, x  0;
k 1
e  x  
k
Our calculated parameter value for a Weibull distribution:
= ……………….
k= ……………….
Is there any reason to include a third Weibull parameter x0? If yes, what do you estimate x0
to be?
……………………………………………….
……………………………………………….
Conclusion:
State here what type of distribution that is the “best fit” to this set of test data, and also why
you came to that conclusion. Also comment on how the quality of the test could have been
improved,
9
APPENDIX A: TEST DATA
Below is the data from the three tests that were performed at Lecture 1, August 27, presented
as Matlab vectors.
The number of cycles to fatigue for the paper clips (32 clips, 32 testers):
nclip=[5 10 6 5 12 6 5 6 7 4 9 8 5 5 7 6 7 6 8 10 4 7 4 8 8 10 13 9 3 12 10 7] ;
The measured force (N) for the snap fastener (22 fasteners, 22 testers):
nsnapnn=524 520 420 460 480 428 521 399 249 750 600 603 1002 1009 458 420 430
510 825 480 414 470]/1000*9.81
The measured force (N) for the snap fastener (1 fastener, 22 testers):
Nsnap1n=[489 879 1150 483 480 500 385 690 590 480 420 390 480 490 580 350 383
480 640 400 390 266]/1000*9.81
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APPENDIX B: MATLAB SOURCE CODE EXAMPLE
% Exercise.m
% Clip bending
%
%
Ulf Sellgren, 20101029
%
nclip=[5 10 6 5 12 6 5 6 7 4 9 8 5 5 7 6 7 6 8 10 4 7 4 8 8 10 13 9 3 12 10
7] ;
x0=0; % Weibull third parameter (threshold value)
nclip=nclip-x0;
%%
%bending cycles to failure
%
iplot=1 ;
figure(iplot)
hist(nclip,20)
iplot=iplot+1 ;
figure(iplot)
%subplot(2,2,1)
probplot('normal',nclip) % Normal distribution probability plot
%subplot(2,2,2)
iplot=iplot+1 ;
figure(iplot)
probplot('exponential',nclip) % Exponential distribution probability plot
%subplot(2,2,[3 4])
iplot=iplot+1 ;
figure(iplot)
probplot('weibull',nclip) % 2-param Weibull probability plot
x=0:0.1:30 ;
[ir,ic]=size(x) ;%
% ______________________________________________________
% normal distribution
iplot=iplot+1 ;
[nmean,nsigma]=normfit(nclip)
% alt nmean=mean(nc); nsigma=std(nc);
fnormal=normpdf(x,nmean,nsigma);
Fnormal=normcdf(x,nmean,nsigma)
f_norm=@(x)normpdf(x,nmean,nsigma);
Fnint5=quad(f_norm,-100,5) % probability of failure in the interval [0,5]
%
for ii=1:ic
Fnint(ii)=normcdf(x(ii),nmean,nsigma) ;
end
Rnint=1-Fnint ;
%
figure(iplot)
subplot(2,1,1)
plot(x,fnormal)
title('Normal probability density function')
xlabel('# of cycles')
ylabel('Frequency')
subplot(2,1,2)
plot(x,Fnint,x,Rnint)
title('Normal survival plot')
xlabel('# of cycles')
ylabel('Survival/death probability')
%
% ______________________________________________________
% exponental distribution
11
iplot=iplot+1 ;
mu=expfit(nclip)
lambda=1/mu
fexp=exppdf(x,1/lambda);
Fexp=expcdf(x,mu)
figure (iplot)
subplot(2,1,1)
plot(x,fexp)
title('Exponential probability density function')
xlabel('# of cycles')
ylabel('Frequency')
f_lambda=@(x)exppdf(x,1/lambda);
Feint5=quad(f_lambda,0,5) % probability of failure in the interval [0,5]
%
for ii=1:ic
Feint(ii)=expcdf(x(ii),1/lambda) ;
end
Reint=1-Feint ;
subplot(2,1,2)
plot(x,Feint,x,Reint)
title('Exponential survival plot')
xlabel('# of cycles')
ylabel('Survival/death probability')
%
%
% ______________________________________________________
% Weibul distribution
iplot=iplot+1;
parmhat=wblfit(nclip);
scale=parmhat(1)
shape=parmhat(2)
fwbl=wblpdf(x,scale,shape) ;
figure(iplot)
subplot(2,1,1)
plot(x,fwbl)
title('Weibull probability density function')
xlabel('# of cycles')
ylabel('Frequency')
f_wei=@(x)wblpdf(x,scale,shape);
Pw5=quad(f_wei,0,5) % probability of failure in the interval [0,5]
Fw5=wblcdf(5,scale,shape) % failure probability at a level of 5
Rw5=1-Fw5 % function reliability at a level of 5
for ii=1:ic
Fwbl(ii)=wblcdf(x(ii),scale,shape) ;
end
Rwbl=1-Fwbl ;
subplot(2,1,2)
plot(x,Fwbl,x,Rwbl)
title('Weibull survival plot')
xlabel('# of cycles')
ylabel('Survival/death probability')
% ______________________________________________________________
12
APPENDIX C: OUR MATLAB SOURCE CODE (M-FILE)
13