Transcript Two-Factor Anova Example (Cont`d)

Thermal-Fluid and Cost Analysis
of Injection Molded Gears
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Typical IM Process Cycle Basic Elements of an
Injection Molding Machine
3 ways to save $$
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Unsteady Temperature Distribution
Lumped Analysis
Analysis in mold
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Unsteady Heat Conduction
In 1D(!), heat equation becomes
 2T 1 T

, T ( x   L,0)  Tm , T ( x,0)  Ti
2
x
 t
Approximate Fourier series represented in Heisler chart
T ( x  0)  T
 f ( t / L2 , Bi  )
Ti  T
where subscripts i = injection, m = mold interface,
 = ambient (same as m for Bi), 0 = midline (x=0)
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Heisler Charts
2L= gear thickness
h = “h” 
 = k/(c)
To = midplane temp (x=0)
Ti = injection temp
T = ambient (mold) temp
Source: Incropera & DeWitt (1985)
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Pressure During Flow in Gates
Assumptions
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Newtonian fluid !! (consider power-law??)
Steady, incompressible flow
Isothermal
Laminar
Pipe (Poiseuille) Flow
L V2
hl  f
, f  64 / Re
D 2
or simply
 p D 4
Q
128L
where Q  volumetricflowrate and L, D  sprue length and diameter
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Properties of General Purpose Polystyrene
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Thermal conductivity
Specific gravity
Specific heat
Thermal expansion coef.
Elastic modulus
Tensile strength
Elongation to failure
k = .058 – .09 Btu/hr/ft/F
sg =1.04 – 1.07
c = 0.30 Btu/lb/F
3.3 – 4.8 10-5/F
E = 4 – 5 105 lb/in2
5 – 8 x 103 lb/in2
1.5 – 2.5 %
Source: Chem. Eng. Handbook,
Perry & Chilton (1973)
See also www.matweb.com
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Costs
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60 ton molder (Krauss Maffei Model 65C)
Footprint: 18’ x 6.5’
Availability: 86%, Capital costs $65k
Installation cost: 10% of capital cost
Expected life: 10 years
Leads to ….
• $0.60/min, includes the following:
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1 Direct Head 0.19
Indirect Labor 0.133
Fringe
0.164
MRO
0.023 (maintenance)
Overhead
0.019
• Compared to material costs of $1/lb
Source: Ford Motor Co.
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Thermal-Fluid/Cost Analysis
• Assuming uniform initial melt temperature set by controller and Tmold measured
initially by thermocouples, estimate time for gear center to drop to 80C  Tg.
• Estimate pressure drop within mold gates as function of injection rate. Estimate
gate diameter and length from part. Assume flow is Newtonian, fully
developed and isothermal (big assumptions). Estimate gear volume and
determine injection pressure and time required to fill mold as a function of melt
temperature using interpolation or curve fit to (polyethylene) viscosity data.
Compare to measured injection time.
• From your response to above questions, estimate minimum cycle time. Is this
consistent with actual cycle time? What is rate limiting step in molding
process? Based on strength vs. T data, analysis, and given cost data, can you
find optimal cycle time? Suggest design, process or material improvements.
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Statistical Analysis
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Statistical Distributions
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Normal Distributions
Most physical properties that are
continuous or regular in time or
space
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Log normal Distributions
a) Normal
Failure or durability projections; events
whose outcomes tend to be skewed
toward the extremity of the
distribution
b) Log Normal
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Statistical Distributions
• Poisson Distributions
Events randomly occurring in time
• Binomial Distributions
c) Poisson
Situations describing the number
of occurrences,n, of a particular
outcome during N independent
tests where the probability of any
outcome, P, is the same
d) Binomial
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Statistical Distributions
• Weibull Distributions
Fatigue tests, failure and durability projections;
similar to log normal applications
e) Weibull
*Theory and Design for Mechanical Measurements,
R. S. Figliola and D. E. Beasley, John Wiley & Sons, Inc., New York, 2000
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Weibull Distributions
• When there is a scatter in strengths, statistical methods are needed to
describe the strength of the material
• For a given applied stress there is a chance that a brittle material will
NOT fail. This probability decreases as the stress increases
• If there is a uniform state of stress and the test samples all have the
same volume, the probability of survival is
Ps(σ) = exp{-(σ/ σo)m}
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Weibull Distributions
• The Weibull parameters statistically portray the mechanical strength
characteristics of the load bearing parts.
• m – the Weibull slope, also known as the shape parameter. It indicates
the width of the fracture stress distribution (a small m indicates an
unreliable material with a wide range of failure stresses and a large m
indicates a reliable material with a well defined ultimate stress)
• σ o – is the characteristic failure stress. It is found when the probability
of survival is about 0.368 in a uniaxial tensile test. Note that it is NOT
the average failure stress.
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Weibull Constants
• Weibull constants are usually determined from a series of uniaxial tests
using many “identical” specimens
• The failure stresses are sorted from weakest to strongest
• If the strength index is “i” of the experimentally-determined
probability of survival, Ps(σ)i, of any sample up to the failure stress
endured by the ith sample is,
Ps(σ)i = (N+1-i)/(N+1)
where N is the total number if specimens tested
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Weibull Constants
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If a Weibull distribution can be used to describe the statistics of failure, a
double-log plot of the [Ps(σ)i, σi] data (with Ps(σ)i on the y-axis and σi on the xaxis) will give a straight line of the form
ln{ln[1/ Ps(σ)i]} = m ln[σ / σo]
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The slope and the intercept of the best-fit line through the data can then be
used to determine m and σo
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Solving for Weibull Parameters
• After collecting the data, sort it in ascending order
• Calculate the size of the array
• Calculate Ps(σ)
• Calculate y = ln [ln {1/Ps(σ)}]
• Plot the data y versus x = ln[σ]
• Fit a line to the data (least-squares)
• The Weibull parameters are the slope of this line (m) and
σo = exp(-b/m) where b is the y-intercept
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Weibull Parameters
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Useful commands in Matlab: POLYFIT, POLYVAL, SORT, SIZE, LOG
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In Matlab there are a series of functions beginning with ‘weib’ that relate to
the Weibull distribution including, weibplot, weibfit, weibcdf, weibpdf, and
weibstat. Make sure you understand the output from these functions
before using them.
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Useful commands in Excel: INDEX, LINEST, COUNT, LN
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Weibull Example
• Six bearings were tested to failure, and the cycles to failure were
determined. (Here, σ is number cycles, not strength)
• Sorted in order:
1.3x105
2.7x105
4.0x105
5.2x105
6.6x105
9.8x105
Ps(σ): 0.142 = 1/7
0.285
0.428
0.571
0.714
0.857 = 6/7
• Best fit through the data (slope and y-intercept) is
m = 1.28 and σ = 5.9x105 cycles
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Weibull Plot from Excel
y=ln{ln[1/Ps(σ)]}
1
0.5
0
11.5
-0.5
12
12.5
13
13.5
14
-1
y = 1.2859x - 17.087
-1.5
-2
-2.5
x=ln{σ}
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From Matlab
m = 1.2859
σ = 5.9x105
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Analysis of Variance
• Analysis of Variance breaks the total variance (NOT the
standard deviation) down into variations due to each major
factor, interacting factors, and residual (experimental)
error.
• Total variance = variance within the samples + variance
between the means of the samples
• For our lab, we might ask, “Are gear teeth near the weld
lines significantly weaker?”
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Single-Factor Anova Example
• Four brands of tires, A, B, C, and D are compared for tread loss after
20,000 miles of driving. The following tread loss in millimeters was
measured:
A B C D
14
14
12
10
13
14
11
9
17
8
12
13
13
13
9
11
• Calculate the totals for each column (brand), then calculate the squares
for each of these values, Tc2
• Calculate the total of all the measurements, then calculate the square of
this value, T2
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Single-Factor Anova Example (Cont’d)
• Calculate the sum of the squares of each measurement, Σx2
• Calculate the degrees of freedom for the columns (brands) and the
residual (experimental error)
DFc = c-1=3;
DFres = c*(r-1)=12;
where c is the number of columns and r is the
number of rows
• Calculate the variance among the columns (among the brands):
SSc=ΣTc2/r – T2/N=31
where N=r*c and equals the number of
measurements taken
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Single-Factor Anova Example (Cont’d)
• Calculate the total variance:
SStotal=Σx2 – T2/N=81
• Calculate the Residual variance (measure of experimental error or test
repeatability):
SSres= SStotal – SSc=50
• Next determine the Mean Square (MS) for the columns (brands) and
Residual:
SS/DF
MSc=SSc/DFc = 31/3=10.33
MSres=SSres/DFres=50/12=4.17
• Calculate the Mean-square ratio, MSR: MS/MSres
MSRc=MSc/MSres=10.33/4.17=2.48
• Finally, determine the minimum MSR required for factors to be
significant at the y % confidence level (you pick y)
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Single-Factor Anova Example (Cont’d)
• If MSR < minimum MSR, there is no significant difference between
the brands. Here, at y=95%, MSRmin equals 2.61 so the difference
between brands of tires is insignificant.
• MSRmin or (Fcrit) is determined from “F tables” using the cumulative
distribution function with DFc and DFres and y
• An Excel macro or a Matlab function will do all calculations for a
given test data, including MSRmin – the columns represent the
variation among the test samples
• Excel: Tools>DataAnalysis>Anova:Single-factor
and in Matlab the function is anova1(data)
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Two-Factor Anova Example
Our case has two variables we were changing –temperature and injection
rate. We need to do a 2-factor Anova.
The effect of sliding velocity and temperature on the wear-rate
characteristics of a material are the two variables:
Temperature B
B1 B2
A1
34
30
33
41
A2
43
37
50
44
Sliding Velocity, A
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Two-Factor Anova Example (Cont’d)
• In addition to the column (temperature), residual and total variance, we
must also calculate the row (sliding velocity) and interaction variances.
• Calculate Tc2, T2, and Σx2 just like for the single-factor case.
Tc=144 and 168
• In addition, calculate Tr2 (for the rows) where ‘rows’ refers to the
different temperatures: Tr=138 and 174
• And calculate Tcr2 (for the interaction between the columns and rows).
Tcr is the sum of the measurements for each case:
Tcr= 64, 74,
80, and 94
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Two-Factor Anova Example (Cont’d)
• Calculate the degrees of freedom for the columns (number of
temperatures), rows (number of sliding velocities), column-rows
interaction, and the Residual:
DFc = c – 1 = 1; DFr = r – 1 = 1;
DFcr = (c – 1)*(r – 1) = 1; Dfres = (c*r*(n – 1)) = 4
(n is the number of tests done for each combination
of factors – in this case n=2)
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Two-Factor Anova Example (Cont’d)
• The sums of squares for the various sources of variation are:
SSc=ΣTc2/(r*n) – T2/N=72
SSr=ΣTr2/(c*n) – T2/N=162
SScr=ΣTcr2/n – T2/N – SSc – SSr=2
SStotal=Σx2 – T2/N=312
SSres= SStotal – SSc – SSr – SScr =76
• Next determine the Mean Square (MS) for the columns , rows,
column-row interaction, and Residual: SS/DF
MSc=SSc/DFc = 72/1=72
MSr=SSc/DFr = 162/1=162
MScr=SSc/DFcr = 2/1=2
MSres=SSres/DFres=76/4=4.17
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Two-Factor Anova Example (Cont’d)
• Calculate the Mean-square ratio, MSR: MS/MSres
MSRc=MSc/MSres=72/19=3.8
MSRr=MSr/MSres=162/19=8.5
MSRcr=MScr/MSres=2/19=0.105
• For this case, the MSRmin=4.54; therefore, sliding velocity affects the
results, but the temperature is insignificant as compared to wear rate
due to chance. The interaction between the velocity and the
temperatures is insignificant, which means that the velocity is not
dependent on temperature.
•
Excel macro is found under Tools>DataAnalysis>Anova:Two-Factor with
Replication and in Matlab the function is anova 2(data,n)
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