me495 laboratory ii-im- 1 thermal-fluid and cost analysis of injection molded gears

ME495 Laboratory II-IM- 1

Thermal-Fluid and Cost Analysisof Injection Molded Gears


Typical IM Process Cycle Basic Elements of an Injection Molding Machine

3 ways to save $$


Unsteady Temperature Distribution

Lumped Analysis Analysis in mold

smal

l ski

n ef

fect

in m

old

or T

m, s

low

ly in

cr. w

/tim

e


Unsteady Heat Conduction

In 1D(!), heat equation becomes

im TxTTLxTt

T

x

T

)0,(,)0,(,1

2

2

Approximate Fourier series represented in Heisler chart

),/()0( 2

BiLtf

TT

TxT

i

where subscripts i = injection, m = mold interface, = ambient (same as m for Bi), 0 = midline (x=0)


Heisler Charts

Source: Incropera & DeWitt (1985)

2L= gear thicknessh = “h” = k/(c)To = midplane temp (x=0)Ti = injection tempT = ambient (mold) temp


Pressure During Flow in Gates

Assumptions– Newtonian fluid !! (consider power-law??)

– Steady, incompressible flow

– Isothermal

– Laminar

Pipe (Poiseuille) Flow

L

DpQ

fV

D

Lfhl

128

Re/64,2

4

2

or simply

diameterandlengthsprueDLandrateflowvolumetricQwhere ,


Properties of General Purpose Polystyrene

• Thermal conductivity k = .058 – .09 Btu/hr/ft/F

• Specific gravity sg =1.04 – 1.07

• Specific heat c = 0.30 Btu/lb/F

• Thermal expansion coef. 3.3 – 4.8 10-5/F

• Elastic modulus E = 4 – 5 105 lb/in2

• Tensile strength 5 – 8 x 103 lb/in2

• Elongation to failure 1.5 – 2.5 %

Source: Chem. Eng. Handbook, Perry & Chilton (1973)

See also www.matweb.com


Costs

• 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:• 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/lbSource: Ford Motor Co.


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.


Statistical Analysis


Statistical Distributions• Normal Distributions

Most physical properties that are

continuous or regular in time or

space

• Log normal DistributionsFailure or durability projections; events

whose outcomes tend to be skewed

toward the extremity of the

distribution

a) Normal

b) Log Normal


Statistical Distributions

• Poisson DistributionsEvents randomly occurring in time

• Binomial DistributionsSituations describing the number

of occurrences,n, of a particular

outcome during N independent

tests where the probability of any

outcome, P, is the same

c) Poisson

d) Binomial


Statistical Distributions

• Weibull DistributionsFatigue tests, failure and durability projections;

similar to log normal applications

*Theory and Design for Mechanical Measurements,

R. S. Figliola and D. E. Beasley, John Wiley & Sons, Inc., New York, 2000

e) Weibull


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}


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.


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


Weibull Constants

• 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 x-axis) will give a straight line of the form

ln{ln[1/ Ps(σ)i]} = m ln[σ / σo]

• The slope and the intercept of the best-fit line through the data can then be used to determine m and σo


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


Weibull Parameters

• Useful commands in Matlab: POLYFIT, POLYVAL, SORT, SIZE, LOG

• 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.

• Useful commands in Excel: INDEX, LINEST, COUNT, LN


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 Ps(σ): 0.142 = 1/7

2.7x105 0.2854.0x105 0.4285.2x105 0.5716.6x105 0.7149.8x105 0.857 = 6/7

• Best fit through the data (slope and y-intercept) is

m = 1.28 and σ = 5.9x105 cycles


Weibull Plot from Excel

y = 1.2859x - 17.087

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

11.5 12 12.5 13 13.5 14

y=ln{ln[1/Ps(σ)]}

x=ln{σ}


From Matlab

m = 1.2859σ = 5.9x105


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?”


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

• Calculate the totals for each column (brand), then calculate the squares for each of these values, Tc

2

• Calculate the total of all the measurements, then calculate the square of this value, T2

14 14 12 10

13 14 11 9

17 8 12 13

13 13 9 11


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=ΣTc

2/r – T2/N=31

where N=r*c and equals the number of

measurements taken



• 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)



• 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)


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

Sliding Velocity, A

A2

34

30

33

41

43

37

50

44


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



• 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)



• 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



• 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)

me495 laboratory ii-im- 1 thermal-fluid and cost analysis of injection molded gears

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