Statistical Analysis Services, 4321 Kerve Fitting Ave. Regression, MI 48971

To: Ms. F. A. Sterprosesing, New products Division, Hardshell Container, Inc.

8765 Luggage St., Faraway, WA 98199

From: Siddhant Daga, Kevin Dulic, Elizabeth Daykin, Elise Machiele,

Engineers in Training, Lab 10, Team 01

Subject: Fiberglass Composite Material Testing and Analysis for Production Process Change

Date: 27 February 2014

Dist: Dr. L. O. Tofracture, Supervising Engineer, Statistical Analysis Services

Mr. T. M. N. Turtle, Vice President for Manufacturing, Hardshell Container Inc.

FOREWORD Hardshell Container, Inc. would like to increase the speed of manufacturing fiberglass-composites as you

have a back-log of orders which is increasing. You need to know whether fast production will lower

product quality or not. So you have asked us to perform stiffness and strengths tests on 3 types of

composites: (1) epoxy only, (2) 0° composite, and (3) 90° composite. For stiffness test, you asked us to

determine the density and isotropic or anisotropic elastic constants for each specimen with uncertainties at

95% confidence interval, compare these values with values obtained using the simple rule‐of‐mixtures

theory, and determine the extent to which a 10% void fraction in the epoxy would decrease the 0°

composite's stiffness. For strength test, you provided us with materials to make more composite

specimens, made carefully to reduce the occurrence of voids and quickly, and asked us to determine the

Weibull constants for both type of specimens, compare their strengths, estimate the production capacity

increase with the fast processes if 80% of your manufacturing time was devoted to mixing and mold-

filling, and find the cross-section with 99.5% probability for a 1 m long composite of each type to hold a

load of 0.25 metric tons assuming the mass fractions to be same as the tested specimens. You asked that

all measurements be reported in SI units, that we describe any difficulties encountered in the tasks, and

that we make recommendations. The purpose of this report is to provide our findings, conclusions,

supporting documentation, and our recommendations for the fast processes.

SUMMARY

We performed the required stiffness and strength tests on the composites to check for product quality after

increasing the speed of manufacturing. See Table 1 below and Table 5 for results. 10% void fraction in

epoxy would decrease the composite stiffness to E = 4.18 ± 1.10 GPa. The Weibull constants and the

cross-section for a 1.0 m long composite to hold a load of 0.25 metric tons with 99.5% probability are

listed in Table 2, below. We found that the quickly made specimens were stronger than the carefully

made specimens. With 80% of manufacturing time devoted to mixing and mold-filling, the production

capacity increase after transitioning to the fast processes was 430 ± 160 %. We faced difficulties in

production of specimens: some stuck to molds, and some did not harden and would not fracture.

Table 1: Results of Stiffness Testing and Comparison to Rule of Mixtures

Epoxy Longitudinal Transverse Rule of Mixtures

Density of Sample, ρ (kg/m3) 1.08 ± 0.01 1.59 ± 0.16 1.58 ± 0.16 N/A

Longitudinal Elastic Modulus, E (GPa) 1.90 ± 0.51 3.40 ± 0.58 4.30 ± 1.24 4.37 ± 1.25

Transverse Elastic Modulus, E (GPa) 1.90 ± 0.51 1.97 ± 0.56 2.08 ± 0.54 1.90 ± 0.51

Table 2: Results of Strength Testing and Calculated Cross-Sectional Area for Fast and Slow Samples

Weibull Parameters Cross-Sectional Area

(mm2) m ( ) (MPa) (mm

3)

Fast Manufactured Sample 6.62 ± 0.26 74.12 ± 2.90 10,750 100 ± 10

Slow Manufactured Sample 2.84 ± 0.15 65.68 ± 3.45 10,750 1300 ± 100

PROCEDUREWe conducted tensile tests on fiberglass composites with a hydraulic powered Instron 8516 with a 100kN load cell mounted to the crosshead. Two types of tests were performed: a stiffness test on the threeprefabricated test specimens, (1) epoxy only, (2) one longitudinal layer of glass fiber in the epoxy matrix(0° composite), and (3) one transverse layer of glass fiber in the epoxy matrix (90° composite), and astrength test on the composites we manufactured.

Stiffness TestingPre­test measurements: Before conducting the tensile tests, the three specimens you provided weremeasured four times for width W, thickness T, and length L, as shown in Figure 1, below, withmillimeter calipers of resolution 0.01 mm. The dimensions are listed in Table 3, below. Measurementswere taken at room temperature and at pressure of 1 atm.

Figure 1: Dimension Locations of Composite Specimens

Table 3: Pre­fabricated dimensions of the three provided specimens

Tensile testing: The three test specimens, each with a 0­45­ strain gauge rosette attached at the09 o

center, were carefully installed one by one in the Upper Grip of Instron 8516 until snug, then the ramposition was adjusted to fully grip the end pad, and lastly the lower grip was tightened until snug but notover­tightened. The load cell was balanced (or zeroed) and the tensile test was conducted by one loadcycle per specimen up to 450N at a rate of 0.254 mm/min. LabVIEW 2010 collected load and strain dataduring these trials at a sampling frequency of 3 Hz.

Strength TestingFabrication of composite specimens: Two composite specimens, one unidirectional glassfiber­reinforced epoxy (0° composite) specimen made very carefully so as to minimize voids and oneunidirectional glass fiber­reinforced epoxy (0° composite) specimen made as rapidly as possible withoutthe concern for voids were manufactured using the molds, epoxy resin, hardener, and glass­fiber cloth youprovided. The molds were inspected, residual resin and debris were removed, and Slide Epoxease MoldRelease 40616T was liberally sprayed on both halves of each set of molds. The fiber­glass cloth was cutwith fibers oriented along the length of the fabric to the size of a 1.0” by 9.5” plastic template using arotary razor cutter. The cloth was laid in the lower mold with extra fibers tucked (or trimmed) extendingbeyond the cavity. The upper mold was installed and the dowel pins were aligned. Before the screwswere snugged, care was taken that the two halves fully mated to prevent any leaks. Nylon tape wasapplied to the two ends of the mold to seal the epoxy in the mold. A thin film of grease was applied with acotton swab to the inlet and outlet hole areas to prevent overflowing resin from bonding to the aluminum.

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Before the epoxy mixture was added, the mold was placed at a slight angle which allows the trapped air toescape as the mixture is being poured into the mold. We noted the time when the hardener first touchedthe resin and when the mold was fully filled. For the quickly made sample, two full pumps each of Resin(105) and Hardener (206) were mixed vigorously in a single cup with a resin­to­hardener ratio of 5:1; themixture was gently poured in the syringe with a red plug attached to the tip of it. The red plug wasremoved, tip of the syringe was inserted into the lower inlet hole of the mold and the plunger was used toinject the epoxy into the mold until it oozed out from the other end. The mold was flattened and kept asidefor one week.

For the carefully made sample, two full pumps each of Resin (105) and Hardener (206) were mixedvery slowly in a single cup with a resin­to­hardener ratio of 5:1; the mixture was gently poured in thesyringe with a red plug attached to the tip of it. The syringe was placed in a vacuum chamber to mitigateair bubbles. Then, the red plug was removed and the tip of the syringe was placed inside the lower inletand was left alone until gravity pulled the mixture inside and the epoxy oozed out from the other end. Themold was flattened and kept aside. The resins were allowed to cure for a week and tensile tests (refer p.xx Tensile testing) were conducted until failure on both the specimens.

Test measurements: The composites were manufactured and the specimens were filed to removerough edges. The masses of the specimens were measured using a weighing machine of resolution 0.01g.We measured the fast and the slow manufactured sample dimensions for L, gage length, W, width, and T,thickness as shown in Figure 1, three times with millimeter calipers of resolution 0.01mm. The dimensionsare listed in Table 4 below. Measurements were taken at room temperature and at pressure of 1 atm.

Table 4: Manufactured specimen dimensions

Data ReductionBefore beginning calculations, the recorded data was scrutinized and reduced to include only valid,relevant ranges. For all tensile tests, the data range was reduced to begin with the initiation of loading andend with the value after fracture, or one value after the maximum load. Load values were converted fromforce to stress using Stress = Force / Area, using the measured area of the specimen being tested. Forthe calculation of production capacity increase, we neglected two specimens from the 44 quicklymanufactured specimens as they did not harden completely.

FINDINGSBy conducting tensile tests on the prefabricated specimens and the specimens we fabricated, we wereable to assess the stiffness and associated parameters of the provided specimens, determine the strengthcomparisons of the two types of our manufactured specimens, verify the rule­of­mixtures theory, andprovide results to additional calculations based on these findings. All results are reported in SI units withuncertainties reported to the 95% confidence level.

Elastic Test ResultsWe determined the elastic modulus for each sample you provided us (Epoxy, Longitudinal, andTransverse) through tensile testing. We determined the density, longitudinal modulus, transverse modulus,major Poisson’s ratio, and minor Poisson’s ratio of each specimen. Our values for the material properties

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of the fabricated specimens are given in Table 5. Further discussion on these the modulus of the providedsamples, and the material properties of the fabricated samples are in the appropriate subsections.

Table 5: Composite Density and Elastic Properties of Specimens From Rule­of­Mixtures

Density: To determine the density of each composite specimen, , we applied the rule­of­mixturesρcexpression in Eq. 1 involving provided densities of the fiber and matrix, and respectively:ρf ,ρm

ρ 1 )ρρc = vf f + ( − vf m Eq. 1

where is the volume fraction of the fibers in the composite. The main contribution to error comes fromvfthe error in , the volume fraction of the fibers.vf

Elastic Modulus: To determine the modulus of each specimen, we used our reduced load and strain datafrom the tensile tests on the provided specimens, and converted those values to stress and principal axialstrain to produce Figure 2, below, where the slope of each data set is the composite’s Elastic Modulus,

, where is stress, and is strain. The error is determined by finding the error in the slope.E = εσ σ ε

Figure 2: Elastic Modulus for Each Composite Specimen from Elastic Test Data

To determine the longitudinal modulus, of each composite specimen, we applied the rule­of­mixtures:,EcL

E 1 )EEcL = vf f + ( − vf m Eq. 2

where is the volume fraction of fibers, is the modulus of the fibers, and is the modulus of thevf Ef Emmatrix. This value of longitudinal modulus found in Table 5, is the upper bound for the modulus of acomposite material. The main contributions to error come from , the volume fraction of the fibers.vf

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To determine the transverse modulus, of each composite specimen, we applied the rule­of­mixtures:,EcT

EcT =E Ef m

v E +(1−v )Ef m f fEq. 3

where is the volume fraction of fibers, is the modulus of the fibers, and is the modulus of thevf Ef Emmatrix. The value of transverse modulus found in Table 5 is the lower bound for the modulus of acomposite material. The main contributions of error come from the error in , the volume fraction of thevffibers.

Poisson’s Ratio: To determine the major Poisson’s ratio, of each specimen, we calculated the− ,νLT = εbεa

principal lateral and axial strains, and respectively, using Eq. 4, below:εa ,εb

εa,b = 2ε +ε1 3 ±√( 2

ε −ε1 3)2 + (2ε )2 − ε1 − ε32 Eq. 4

where are the measured strains at and from the strain gauge rosette, respectively. Theε1,2,3 ,0o ,45o 90o

major Poisson’s ratio was found by plotting the negative lateral strain vs axial strain and determiningεa εbthe linear relationship for each specimen, as showing in Figure 3, below. We determined the error byfinding the error in the slope.

Figure 3: Major Poisson’s Ratio for each composite specimen

To determine the minor Poisson’s ratio, we applied Eq. 5:,vTL

vTL = vLT EcLEcT Eq. 5

where is the major Poisson’s ratio, and are the elastic modulus of the composite in thevLT EcT EcL

transverse and longitudinal directions, respectively. We determined the error using error analysis todetermined the combined error of , , and .vLT EcT EcL

We determined that the epoxy sample is isotropic, while the longitudinal and transverse samples areanisotropic. The values for and must be in agreement for a material to be isotropic: only the epoxyvTL vLTsample satisfies this condition within uncertainty.

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Failure Test ResultsTo determine strength of specimens manufactured quickly and carefully, we manufactured single­cloth0o

composite specimens both ways with the supplied molds, epoxy resin, hardener and fiber­glass cloth andconducted tensile testing to failure on each sample as requested.

Manufacturing Times: As requested, we paid special attention during both manufacturing processes tothe amount of time that passed between the resin and hardener touching and the mold being fully filled. Asdiscussed in the procedure, we fabricated 41 “carefully made” samples in 21 17.9 minutes with special3. ±attention to reduce the occurrence of voids in the epoxy matrix, and 44 “quickly made” samples as fast aspossible in 4.0 5.1 / minutes, without regard for whether voids were created. The main contribution+ .0− 4to error is the broad distribution of manufacturing times; the negative error for quickly made samples wastruncated because time can not be negative. Each sample cured for one week before testing.

Fracture Loads: The fabricated specimens were put through tensile testing to fracture as discussed inthe procedure. The failure behavior of the two types of specimens was inconsistent, as evidenced byFigure 4, below. Specimens fabricated the same way had a wide variety of strengths and fracturestresses, so we cannot provide numbers for failure stress with any confidence. Additionally, many of thecarefully made specimens did not harden and so did not fracture. We used Weibull statistics to describefor you the specimen strength of the specimens that did fracture using probabilistic terms.

Figure 4: Fracture Plots of Stress Against Cross­head Displacement Illustrate Inconsistent Failure

Weibull Statistics: The two­parameter Weibull description of tensile strengths for both types ofmanufactured specimens were determined as requested using the results from the tensile testing of the 44fast samples and 41 carefully made samples. We assumed the tensile testing created a uniform stressstate in the constant gauge section of the specimen, so we used Eq. 6, below, which is the Weibullequation representing the probability of survival of a sample of volume, V, loaded uniformly with stress, σ.In Eq. 6, m is the Weibull modulus and is the Weibull constant for a reference volume Vo, which areσothe constants we sought to determine.

(V , ) xp P s σ = e −[ VV o ( σ

σ0) m] Eq. 6

To determine Weibull parameters from the results of our tensile testing, we first indexed the samples byfailure stress from weakest (lowest failure stress), i=1, to strongest (highest failure stress), i=N, where N

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is the number of samples tested. The probability of survival for the sample of volume, V, and failure stress,σi, was then calculated by

(V , )P s σi = N+1N+1−i Eq. 7

where N is the total number of samples tested, and i is the index value. Next, we used Microsoft Excel14.3.2 to create a linear log­log plot, derived from Eq. 6, where Ps(V,σi) is the calculated probability ofsurvival from Eq. 7. On the plot, displayed in Figure 5, below, we fit a linear trend line, Eq. 8, to the datapoints. The resulting trend line on this plot is equal to

n n n − n(σ )) l ln[ ( 1P (V ,σ )s i )] − l ( VV o) = m * l (σ) + ( m * l o Eq. 8

where the slope of the plot is m, the Weibull modulus, and a y­intercept is (­m* ln(σo)). Using the numbersfound with the best fit line, we solved for m, the Weibull modulus, and σo, the Weibull constant.

Carefully Made Samples: A total of 41 carefully made samples were produced in lab. Of the 41, 8samples did not cure properly so those samples were not considered in the data analysis; therefore, N=33for this data set. To fit a linear trend line to the data, the data points representing weakest 27% were notincluded. The two of lowest strengths lay so far away from the trend they are not shown on the graph.Samples of i=3 through i=9 were also outliers and are represented by X’s in Figure 5. Observationsrecorded in lab of the specimens after testing indicate that most outliers did not cure properly, and thus thefailure mechanism was not fracture, so they will be excluded from the trend­line. In some cases therewere exceptionally large voids near the grip pads, due to manufacturing errors, that caused the specimento fracture at a significantly lower stress. A linear trend line was fit to the remaining 22 samples, shown inFigure 5, to determine the Weibull Parameters. The main contribution to error for the Weibull Parametersis the slope linearity error. Returning the Weibull terms to Eq. 6, the probability of survival, , for aP scarefully made sample is given by Eq. 9, with the sample volume, V, in units of mm3, and applied stress, σ,in MPa.

(V , ) xp P s σ = e −[ V10,750 ( σ

65.68) 2.84] Eq. 9

Figure 5: Linear Fit for Carefully Made Samples Gives Weibull Parameters of m=2.84 ± 0.15and σo=65.68 ± 3.45 MPa for a reference volume, Vo , of 10,750 mm

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Quickly made samples: Although 44 quickly made samples were produced in the lab, three of thesesamples did not cure properly so those samples were not considered in the data analysis; therefore N=41

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for this set of data. To fit a linear trend line to the plotted data, the points representing weakest 27% werenot considered: the weakest lay too far from the rest of the data points it is not shown on the graph. Theother low­strength outliers are shown in Figure 6, represented by X’s. Based on in lab observations of thesamples after testing, we chose not to include theses samples in the trend line because it is likely thefailure mechanism for the outlier samples is different than the failure mechanism of those noted to be“clean fractures”, which the trend line is fit to. The trend line generated from the remaining 30 samples,shown on Figure 6, was used to determine Weibull parameters using Eq. 8. The main contribution to errorfor the Weibull Parameters is the slope linearity error. Returning the Weibull constants to Eq. 6, theWeibull probability of survival, , for a quickly made sample is given by Eq. 10 where the sample volumeP sV has units of mm3 and applied stress, σ, has units of MPa.

(V , ) xp P s σ = e −[ V10,750 ( σ

74.12) 6.62] Eq. 10

Figure 6: Linear Fit for Quickly Made Samples Gives Weibull Parameters of m=6.62 ± 0.26 andσo=74.12 ± 2.90 MPa for a reference volume, Vo , of 10750 mm

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Strength Comparison: Based on the Weibull Statistics parameters displayed in Figure 7, below, thequickly and carefully made specimens do not have similar strengths. The quickly made specimens arestronger than the carefully made specimens.

Figure 7: Weibull Probability of Survival for Quickly and Carefully Made Samples

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Comparison and Evaluation of Rule­of­Mixtures TheoryTo evaluate the accuracy of the rule­of­mixtures, we compared the values of elastic modulus of thetransverse and longitudinal samples ( in Figure 2) with the modulus values we calculated using theE rule­of­mixtures found in Table 5. We have determined that the rule­of­mixtures gives us a validapproximation of elastic modulus in both longitudinal, and transverse directions. These values are inagreement because the rule­of­mixtures modulus lies within the uncertainty of the experimental datamodulus. Table 6, below, shows a comparison of these values.

Table 6: Longitudinal and Transverse Elastic Modulus Comparison Proves Rule­of­Mixtures Is Valid

Requested Engineering CalculationsTo determine the results for the additional calculations requested, we utilized the findings in the previoussections. We were asked to find the possible production capacity increase from adopting the fastmanufacturing process, the cross sectional area of composite specimens needed to meet the conditionsyou provided, and the extent to which a 10% void fraction in epoxy would decrease the composite0o

stiffness. This section details those findings.

Calculations of Production Capacity: To determine the production capacity increase possible with thefast process compared to the careful process, we took the data of time taken for mixing and filling themolds for the quickly and the carefully made specimens and calculated the increase using the differencebetween the time taken for the manufacturing of fast and slow specimens, divided by the time taken forthe manufacturing of slow specimens. The production increase was found to be 430 ± 160 %.

Calculation of Cross Sectional Area: We were requested to determine the cross sectional area of acarefully made and a quickly made specimen, assuming the mass fractions are the same as the specimenswe tested, that would be needed for an epoxy and fiber­glass composite of length 1.0 m to hold a load of0.25 metric tons, equivalent to a force of 2452.5 N, with 99.5% probability, or of 0.995. Using Eqs. 9P sand 10 and Volume = Area * Length and Stress = Force / Area, the cross sectional area using thesevalues that is required for a quickly made specimen is 100 and for a carefully made specimen is0± 1 m m 2

1,300 00± 1 m . m 2

Calculation of stiffness of a 0°composite with 10% void fraction in epoxy: By our previouslydiscussed confirmation of the accuracy of the rule­of­mixtures we were able to compute the stiffness of a0° composite with a 10% void fraction in the epoxy. The composite would be created with the samevolume fraction of fiber, because it is being cut to the same shape and placed in the same size mold. The10% void fraction would thus decrease the volume fraction of the epoxy matrix by 10%. Using Eq. 2, anda 10% decreased epoxy volume fraction, the stiffness, E = 4.18 ± 1.10 GPa.

CONCLUSIONS AND RECOMMENDATIONSThe requested tensile and strength tests were performed on composite specimens to check for the qualityof composites after increasing the speed of manufacturing. See Table 7 for results. The requestedcalculations were completed as well. We found that a 10% void fraction in the epoxy would decrease thecomposite’s stiffness to E= 4.18 ± 1.10 GPa. The Weibull constants and the cross­section for a 1.0 m0°

long composite to hold a load of 0.25 metric tons with 99.5% probability are listed in Table 8 below. With

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80% of manufacturing time devoted to mixing and mold­filling, the production capacity increase with thefast process compared to careful process was found to be 430 ± 160 %.

We have determined that the rule­of­mixtures is a valid approximation of longitudinal and transverseelastic modulus because, as shown in Table 7, the longitudinal and transverse modulus values using therule­of­mixtures overlap with the corresponding modulus values from our experimental data.

Based on Weibull Statistics parameters, the quickly made specimens were found to be stronger than thecarefully made specimens. Additionally, many more of the quickly made specimens were manufacturedsuccessfully. So, we recommend that you start increasing the speed of your manufacturing processes aswith our findings and results the quickly made specimens are stronger than the carefully made specimensand also the quickly made specimens do not compromise on product quality.

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