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8/12/2019 Nm 05 Matlab Case 2
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Nofrijon Sofyan, Ph.D.
8/12/2019 Nm 05 Matlab Case 2
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Case StudyLecture #05
Disclaimer: This lecture note was edited from various sources for the solely of teaching and learning purposes. It may contain copyrighte
owners; therefore, apart form teaching and learning purposes, this lecture note may not be reproduced, stored, or transmitted in any form or b
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Outline3
BackgroundFormula
MatLAB Solution
Exercise
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Case Study4
The strengths of 10 nominally identical ceramwere measured and found to be 387, 350, 3
400, 367, 410, 340, 345, and 310 MPa. (a)
Determine m and 0 for this material, (b) Calc
the design stress that would ensure a survivalprobability higher than 0.999.
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Background5
When we measure the strength of a series equivalent ceramic specimens we typically
considerable scatter results.
The reason is due to the size distribution of
that are responsible for failure.
This behavior is very different from that of
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Consequently we have to adopt different d
approaches when we use ceramics.
When we design components using metals w
determine the maximum stress that will be p
the component and then select a metal that
larger strength with safety margin is often i This approach is referred to as deterministi
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It does not work with ceramics because of t
scatter; rather we have to use a probabilist
approach in which we represent this scatterquantitative way so that these materials ca
used safely.
The most popular method is to use Weibulldistribution, which are based on the weake
approach.
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The Weibull distribution function gives the
probability of survival (S), or, alternatively,
probability of failure (SF), or Sn = 1 –
SF
The other way to determine survival proba
nth specimen is through the relation
= 1 + 1
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Since one is dealing with a strength
distribution, the random variable x
is defined as /0, where is thefailure stress and 0 is a
normalizing parameter, required to
render x dimensionless.
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Replacing x by /0 in the Weibull distribution, onthe survival probability, i.e., the fraction of samplewould survive a given stress level, is simply
= /
∝
or
= exp
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Rearrange and taking the natural log of bo
twice yields
lnln 1 = ln = ln ln Plotting – ln ln(l/S) versus ln yields a stra
with slope – m.
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The physical significance of 0 is now also o
It is the stress level at which the survival pro
is equal to 1/e, or 0.37.
Once m and 0 are determined from the seexperimental results, the survival probabilit
stress can be easily calculated
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Solution14
Weibull solution to the problem is by using the
= exp
To determine m and 0, the Weibull plot for th
data has to be made by ranking the specimenof increasing strength, 1 , 2 , 3 , . . . , n , n + 1
where N is the total number of samples.
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Determine the survival probability for the n
specimen
= 1 0.3 + 0.4 Plot - ln ln( 1 /S) versus ln , the least-squa
the resulting line is the Weibull moduluslnln 1 = ln = ln ln
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Summary of data needed to find m from a set of experimental resu
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The last two columns in
previous table are plotted
into a graph.A least-squares fit of the
data yields a slope of 9.5,
which is typical of manyconventional as-finished
ceramics.
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For statistical purposes, from the equation Y
BX
= Σ ΣΣΣ2 Σ 2
= Σ Σ
It can be shown that 0 381 MPa (i.e. wh1/S = 0).
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(b) To calculate the stress at which the surviprobability is 0.999, use equation:
= exp
or
0.999 = exp 381 .55from which = 184 MPa
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MatLAB solution20
%weibull1.m
clear all; clc;
dt = [387, 350, 300, 420, 400, 367, 410, 345, 310];
sigma = sort (dt, 'ascend'); %sort the datan=1:numel(sigma);
Sn=1-(n-0.3)/(numel(sigma)+0.4);
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ln_ln_Sn=-log(log(1./Sn));
ln_sigma=log(sigma);
disp(' Rank Sn sigma ln sigma ln lndisplay text above the table
disp([n' Sn' sigma' ln_sigma' ln_ln_Sn']) %create
p=polyfit(ln_sigma,ln_ln_Sn,1)
sigma_zero = exp(p(2)/-p(1))
sigma_max = sigma_zero*(log(1/0.999))^(1/-
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Manual graph fitting22
plot(ln_sigma, ln_ln_Sn)
A window will pop up
Click menu Tools, Basic
Fitting
Tick on Linear, then click
on the right Arrow
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You can also plot the graph in the forms of
model:
mdl=LinearModel.fit(ln_sigma,ln_ln_Sn)) %
command give simple statistic table
plot(mdl) % plot the graph and the linear m
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Interaction with Excel25
Data can be read from another source such
Excel with the format:
data=xlsread('filename', 'sheet', 'range');
Eg:
sigma=xlsread('data.xlsx','B3:B10');
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MatLAB m-file26
%Weibull2.m
clear all; clc %Clear all data from memory
data= xlsread('Ceramics data.xlsx', 'C1:L1'); %RExcel data
sigma = sort (data, 'ascend'); %Sort the dataN = numel(sigma); %Number of data
n =1:N;
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Sn = 1 - (n-0.3)/(N+0.4);
ln_ln_Sn = - log (log (1./Sn));
ln_sigma = log (sigma);
p = polyfit (ln_sigma, ln_ln_Sn,1) %1 refer to
plot
sigma_zero = exp(p(2)/-p(1))
sigma_max = sigma_zero*(log(1/0.999))^(1
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MatLAB m-file, interactive way29
%Weibull3.m
clear all; clc
dt = input ('Input the data: '); % data [x1, x2
sigma = sort (dt, 'ascend');
n=1:numel(sigma);
Sn = 1 - (n-0.3)/(numel(sigma)+0.4);
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ln_ln_Sn = - log (log (1./Sn));
ln_sigma = log (sigma);
p = polyfit (ln_sigma, ln_ln_Sn,1) % linear pl
sigma_zero = exp(p(2)/-p(1))
sigma_max = sigma_zero*(log(1/0.999))^(1
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disp('Rank Sn sigma ln sigma ln ln 1/
disp([n' Sn' sigma' ln_sigma' ln_ln_Sn'])
mdl=LinearModel.fit(ln_sigma, ln_ln_Sn);
plot(mdl), xlabel('ln sigma'), ylabel('-ln ln 1/S
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Quiz32
In an experiment, imagine that you need to d
the activation energy of Cu atom diffuses in A
versa. To carry out the experiment, firstly you
two bars of Al and Cu at certain temperature
time. After the joining process, you will examijoint to see the diffusion distance of the desire
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Knowing the diffusion distance and time of thprocess you will be able to find the diffusion
coefficient at a certain temperature in which. Next, to predict the activation energy (Qneed to do the experiment as a function oftemperature, after then you will be able to p
activation energy and temperature-independexponential (D0) based on the Arrhenius equa
= exp .
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Sketch your step in doing the experiment by w
on the flowchart, algorithm, and pseudocode.
that, write a small modeling program in MatLto determine the activation energy and tempe
independent pre-exponential factor from tho
experiments.
R f
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References35
W.Y. Yang, W. Cao, T.S. Chung, J. Morris: App
numerical methods using MATLAB, John WileyInc., Hoboken, New Jersey, 2005.
M.W. Barsoum: Fundamental of Ceramics, InsPhysics Publishing, Philadelphia, 2003.
D. A Porter, and K.E Easterling, Phase Transfoin Metals and Alloys, 2nd ed. Chapman and London, 1992.