Thermal Properties and Fracture Mechanics

Section 02, Sunday, April 12th, 2015

Authoring Student: Ben Schwartz, Gabriel A. T. da Silveira, Palloma M. Duarte, Wolney

M. Vasconcelos

Laboratory Group: Ben Schwartz, Gabriel A. T. da Silveira, Palloma M. Duarte, Wolney

M. Vasconcelos

Laboratory Teaching Assistants: Muhammad Sabry, Kazi M. Rahman

Part 1

Fracture Mechanics:

Abstract:

As it is known, a material can undergo two types of fractures: ductile or brittle depending on their ability to deform plastically. Mechanical fractures is more used to predict failures in brittle materials, since in these materials, failures happen quickly and without previous notice. The objective of this work is to characterize the strength of glass samples through a statistical study. In this experiment 20 samples of glass were tested. For each sample, the width and thickness were measured, then put in the Three Point Bending Tester to compute the minor and maximum load. With these data the study was conducted by the method of statistical Weibull and obtained values of m which is the Weibull modulus and σ0 is the reference strength. It was then found the value for m ≈ 4 which means a not very high variability in the fracture strength to glass slide.

Introduction:

Fracture mechanics has long been studied because of its importance in materials

design area. It is essential for a design engineer knowledge on this subject so as to

prevent structural failures. The study of fracture mechanics is related to quantitative

measurements among the stress level, material properties, crack length and crack

propagation [1].

A material can undergo two types of fracture: ductile and brittle, depending on

their capacity to suffer plastic deformation. In a ductile fracture, the material undergoes

a long process of plastic deformation before rupture, in the vicinity of an advancing

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crack. This process happens in a relatively slow manner. In a brittle fracture, the

material undergoes little or no plastic deformation before rupture. In this type of fracture,

the crack propagation happens quickly [2]. Because of this, brittle fracture is difficult to

predict.

Thus, the study of mechanical fractures is mainly used to predict catastrophic

failure in brittle materials. In addition to the above, brittle materials, as ceramics, has

high hardness, stiffness, poor toughness and low strength reliability. The last

characteristic is mainly due to the wide variation in distribution of shape crack, size, and

orientation with respect to the axis tensile loading [3].

It has been already observed experimentally that for the same samples and

under the same load conditions, the strength in ceramic materials ranging from

unpredictable way. And the average force depends on shape of test specimen, volume

of material stressed, and nature of loading [3]. Thus the importance of using

probabilistic methods as the Weibull method. Weibull proposed a two parameter

distribution function to characterize the strength of brittle materials. The generalized

strength distribution law has the following expression, Equation (1):

F (σ )=1−e−V g (σ )

V 0 (1)

where F(σ) is the probability of failure at a given stress level ‘σ’, V is the volume of the

material tested, V0 is the reference volume and g(σ) is the Weibull strength distribution

function Equation (2):

g (σ )=¿

where m is the Weibull modulus and σ0 is the reference strength for a given reference

volume V0.

As can be seen is of great importance to the study of mechanical fractures in

order to avoid structural failures. The objective of this experiment was to characterize

the strength of samples of glass through a statistical evaluation.

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Experimental Procedure:

Materials and equipment:

● Three Point Bending Tester;

● 20 Glass Slides;

● Dial Caliper;

● Pen and Paper.

Procedure:

Begin by obtaining the first glass slide to be tested. Using the dial caliper

measure the width and thickness of the slide and record the results. Place the slide in

Three Point Bending Tester so that it is supported on either end by the supports.

Record load that the Three Point Bending Tester is displaying (weight of the slide), this

will be the Minor load. All the steps are described in the flowchart of Fig. 2.

Next, it is time to perform the test. Push and hold the button that’s starts the test

on the Three Point Bending Tester (see Fig.1). This will cause the off center knob to

rotate, causing a force to be developed in the center of the glass slide. Hold button until

slide fractures. Display maximum load achieved during test on Three Point Bending

Tester, and record this result.

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Figure 1. Three Point Bending Tester.

The Three Point Bending Tester now needs to be reset to accommodate the next

slide. To do this push and hold the button to run the test, this will cause the off center

knob to rotate. Continue holding until the index on the knob is at its highest position,

this will allow enough clearance for the next slide to fit in the tester.

Obtain next slide to be tested and preform test following same procedure used

for first slide. Repeat this until all 20 slides have been tested. Record all results and

enter into a spread sheet.

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Figure 2. Flowchart of the experiment: Fracture Mechanics.

Results and Discussion:

Doing the procedures above, it was obtained the following raw data on Table 1.

After that, unit for load, which were in grams, was converted into American metric

system, pound-force, for the load. Then, it was ready to calculate modulus of rupture

(MOR), also known as bend strength or fracture strength, using the Equation (3):

MOR=3PL

2b t 2(3)

Where:

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P=load∈lbf L=bending fixturedistance (¿inches )between the cilindrical support centers

b=width of the glass slidet=thickness of the glass slide

Table 1. Raw data of the glass slide dimensions and net load needed to fracture.

Thickness T (inches) Width b (inches)

Load (grams)

0.9880 0.0430 22410.9990 0.0425 40050.9955 0.0425 19160.9985 0.0415 26521.0000 0.0415 22890.9945 0.0415 23730.9990 0.0425 28760.9945 0.0425 33680.9860 0.0425 47160.9970 0.0425 27110.9999 0.0435 24840.9987 0.0435 40430.9935 0.0420 21640.9975 0.0425 27680.9970 0.0425 15910.9990 0.0435 40201.0015 0.0430 23830.9900 0.0435 29881.0015 0.0415 28021.0005 0.0435 2695

The data obtained was then sorted to smallest to highest and the smallest and

the highest values were ignored in order to have a better and more concise datas.

The probability of survival, ps was then calculated using the Equation (4):

ps=1−n−0.5N

(4)

Where:

n=position∈the ascendent orderN=totalnumber of samples

Then, the remaining data in Table 2 was plotted in a graph (Fig. 3) below.

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Table 2. Relation between the MOR and probability of survival.

MOR probability of survival374.17 0.9250429.36 0.8750439.14 0.8250453.68 0.7750454.47 0.7250469.78 0.6750475.54 0.6250509.08 0.5750527.20 0.5250527.83 0.4750538.39 0.4250553.69 0.3750557.72 0.3250576.46 0.2750659.05 0.2250761.65 0.1750766.46 0.1250776.66 0.0750

0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.00000.00

100.00

200.00

300.00

400.00

500.00

600.00

700.00

800.00

900.00

f(x) = − 1250.95005426 x³ + 2347.95000704 x² − 1682.85361631 x + 926.319209594R² = 0.962391266280264

MOR vs. P(s)

P(s)

MO

R (p

si)

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Figure 3. A plot of MOR vs. Probability of survival.

So, in order to get the Weibull Exponent, the coefficients was used to calculate

the modulus of rupture corresponding to the probability of survival by the Equation (5).

P (so )=1e=0.368(5)

With these values, it was possible to plot a graph of ln(MOR/σo) on the x-axis

and ln(ln(1/p(s))) on the y-axis. These calculated values are shown on Table 3 and

plotted in the graph - Fig. 4 below:

Table 3. Table of ln(MOR/σo) and ln(ln(1/p(s))).

ln (MOR/ σo)

ln(ln(1/p(s))

-0.5969 -3.6762-0.4080 -2.5515-0.2705 -2.0134-0.2479 -1.6483-0.2154 -1.3669-0.2136 -1.1345-0.1805 -0.9338-0.1683 -0.7550-0.1001 -0.5917-0.0652 -0.4395-0.0640 -0.2951-0.0442 -0.1559-0.0161 -0.0194-0.0089 0.11680.0242 0.25540.1581 0.39990.3027 0.55560.3090 0.7321

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0.3223 0.95180.5119 1.3053

-0.8000 -0.6000 -0.4000 -0.2000 0.0000 0.2000 0.4000 0.6000

-4.0000

-3.0000

-2.0000

-1.0000

0.0000

1.0000

2.0000f(x) = 4.42282334307383 x − 0.348404142175833R² = 0.906538899417654

Weibull Exponent Calculation

ln(MOR/so)

ln(ln

(1/P

(s))

Figure 4. A plot of ln(MOR/σo) and ln(ln(1/p(s))).

Then, it was found the value for the Weibull exponent m (the slope of the linear

trend line), that is m = 4.4228.

In the Fig. 3, the coefficient of determination R2 = 0.9624 > 0.9 indicates that the

polynomial trend line of 3rd order equation is a very good fit for the graph and the values

are concise enough for the project and also means the experiment was done in a

properly manner.

The second plot resulted a Weibull exponent equal 4.4228, which means the

glass slide does not have great variability in the strength. Materials with Weibull

exponent equal or less than one can be considered of huge variability in the strength

[4].

Conclusions and Recommendations:

Upon completing the statistical and data analysis for the glass slide, it was

possible to see a certain range in the load force for the glass slide, which varies from

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374 lbf to 776 lbf, which means it did not always have the same fracture strength in

every glass slide.

However, through Weibull coefficient analysis, it was possible to see this range is

a good range, since its Weibull exponent had a value of 4.4228, which means its

variability was not that greater than expected for glass slide.

However, there are ways to improve this experiment. One of them is increasing

number of samples to do the testing. Thus, it could have more data to analyze and

therefore, perform a better statistical analysis. Another good way would be just one

person managing the machine, because when many people manage the machine, the

operating errors increase since each operator has its intrinsic error.

Part 2

Thermal Properties:

Abstract:

Expansion and contraction are phenomena that often occur in materials. These phenomena are direct consequences of temperature changes and thermal properties of the materials. The thermal expansion coefficient (CTE) is one of the thermal properties and it is widely used to estimate how lengthy a materials can expand or contract when submitted to temperatures changes. The objective of this experiment is to show how this coefficient can be obtained quantitatively, compare different materials’ coefficients such as 1018 Steel and 8081 Aluminum, and explain why this parameter is important in design engineering. To do that, it was used cylindrical samples of each material. Each one was separately jacketed for heating, as well as the wire of the thermometer and the dial indicator were correctly positioned and adjusted. Thus, the data of temperature changes and expansion were collected. By plotting the data. it was possible to obtain the CTE of 10.0×10−6(℉−1) for the steel and 19.3×10−6(℉−1)19 for the aluminum. These results have a reasonable approximation of database engineering handbooks.

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Introduction:

In a materials design for application in engineering is essential to know both the

work required by the material as their working conditions by which it will be exposed.

Engineering materials are dynamic bodies, that is, they have some of their properties

directly affected by the environmental conditions, such as thermal property. The thermal

property of a material is defined as the response given from the material when exposed

to temperature variations. Among the most important thermal properties, thermal

expansion coefficient is a key parameter that must be know on projects, for which it is

possible to predict how a material will deform under heating or cooling [5].

The temperature of a material ismrelated to agitation of its molecules and the

average distance between molecules of the material. This relation is a direct

relationship, that is, an increase on temperature of material causes an increase on the

degree of thermal molecular agitation of the material, which consequently causes an

increase of the average distance between its molecules. This phenomenon is called of

thermal expansion [6].

Dilatation of materials can be classified based on the level of dimensions

changed on material. Thus, dilation may be linear (one-dimensional), surface (two-

dimensional), or volumetric (three dimensional), depending on the representativeness of

each dimension in the material [7]. The strain for a material exposed to a linear

expansion is expressed as Equation (6).

ƐT=l f−l0

l0

=α×ΔT (6)

where ƐT is the strain, ΔT is usually given in Fahrenheit, and α is the linear coefficient of

thermal expansion and has units as (ºF)-1.

Each material deforms differently when subjected to temperature variations,

which means that each material has a different coefficient of thermal expansion. This

difference among the values of the thermal expansion coefficient for various materials is

related to the chemical bond energy among atoms or molecules.

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Due to this response given by the materials when exposed to temperature

variation, design engineers must consider it in their projects and have access to the

thermal properties of a wide variety of materials. The objective of this experiment was to

determine the coefficient of thermal expansion of 1018 steel and 6061 aluminum.

Experimental Procedure:

Materials and equipment:

Cylindrical 1018 Steel with 3.209 inches in length;

Cylindrical 6061 Aluminum with 3.201 inches in length;

Digital dial indicator;

One hinged heating sleeves with metal jacket;

One digital thermometer;

Safety clamp.

Procedure:

Begin by obtaining the 6061 aluminum cylinder.  Measure the total length from

one end to the next and record this measurement (See Fig. 5).  After this measurement

has been taken place cylinder in hinged heating sleeve.  Turn on digital thermometer,

and place sensor wire in the small hole on in top of cylinder.  Record the Initial

temperature reading from digital thermometer.

Next the dial indicator needs to be set up.  Position the dial indicator stand next

to cylinder and hinged heating sleeve so that the dial indicator is above cylinder with the

probe touching it (See Fig. 5 and Fig.6).  Adjust the dial on dial indicator so that it reads

zero.

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Figure 5. Cylinder Length Measurement and Dial Indicator Setup.

Figure 6. Photo showing the equipment and how they operated during the experiment.

Turn on the hinged heating sleeve and watch dial indicator.  Every time the dial

indicator reads an increase in length of 0.0005 inches record the temperature displayed

on the digital thermometer. Continue this until the digital thermometer reads 300 ºF.

After the cylinder has reached 300 ºF, turn off the hinged heating sleeve.

Carefully remove the digital thermometer sensor wire from top of cylinder.  Using tongs

and taking care not to be burnt, remove the cylinder from hinged heating sleeve and set

aside in a safe place.

Repeat this entire procedure using the 1018 steel cylinder.  Make sure to re-

measure and record the total length of the steel cylinder as it will not be the exact same

as the aluminum one. All the steps for this experiment are showed in the Fig. 7.

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Figure 7. Flowchart of the Thermal Expansion experiment.

Results and Discussion:

After the procedure above, the thermal expansion coefficient (α) was determined

by a plot of δT / l0vs.ΔT , since α can be calculated using the Equation (6), for both

materials –1018 Steel and 6061 Aluminium. The Fig. 8 represents this single plot.

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Figure 8. Thermal Expansion Coefficient (CTE) of ASIS 1018 and ASTM 6061 Aluminium.

Given the Equation (6), α is determined by the slope generated in Fig. 8.

Therefore, it was determined a value of 10.0×10−6(℉−1) for the 1018 Steel and

19.3×10−6(℉−1) for the 6061 Aluminium.

These data have found a reasonable approximation when they are compared

with the data previously available for consultation in database, such as handbooks of

engineering. According to The Engineering Toolbox web site [8], aluminum has a CTE

equal to 12.6×10−6(℉−1) while steel has half of its value, or more precisely

6.7×10−6(℉−1). These values make sense, since for the same average temperature

change and roughly same initial length, the 6061 aluminum had higher expansion than

1018 steel as showed on Table 4.

Table 4. Data obtained after experiment for 1018 Steel and 6061 Aluminum.

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0 50 100 150 200 2500.000

0.001

0.002

0.003

0.004

0.005

f(x) = 1.9266603011392E-05 x − 0.000112552794760166R² = 0.99757379454143

f(x) = NaN x + NaNR² = 0 Thermal Expansion Coefficient of AISI 1018 Steel and ASTM 6061

AluminiumASIS 1018 Steel

Linear (ASIS 1018 Steel)

ASTM 6061 Alu-minium

Linear (ASTM 6061 Alu-minium)

ΔT (ºF)

ΔL/L

0

1018 Steel 6061 Aluminum

Initial Length (in) ΔTT (℉) Initial Length (in) ΔTT (℉)

3.209 226 3.201 226

Final Length (in) ΔLT (in) Final Length (in) ΔLT (in)

3.2160 0.0070 3.2150 0.0140

There are several reasons why these numbers found does not exactly coincide

with those available in that web site. As a first consideration, a larger collection of

expansion data by temperature increase could promote greater accuracy. Moreover,

the temperature variation rate was higher than the rate at which the machine collected

the expansion promoted. Thus, there was an operational error in data collection. Finally,

the imprecision of equipment supplied may have contributed to the results found. The

equipment used to determine the expansion showed up to four decimal places, which

contributed to an inaccurate data collection process.

The knowledge and access to CTE values are essential for a design engineer,

because is from these values that is established estimates of expansion and contraction

of materials when heating or cooling respectively. An integrated circuit package, for

example, should be designed so that its integrity is maintained during thermal cycles of

heating/cooling. Another classic example is the construction of bridges and viaducts, in

which small slit are set so that the dilation occurs without causing damage to the

structure, such as cracks.

Conclusions and Recommendations:

This experiment allowed quantitatively determine the thermal expansion

coefficient (CTE) of the two materials very applied in design of engineering: the 1018

steel and 6061 aluminum. The knowledge of CTE allows design engineers to estimate

how much each material dilates/contracts when exposed to environments that provide

variations in temperature.

The results observed CTE for steel and aluminum was 10.0×10−6(℉−1) and

19.3×10−6(℉−1) respectively. These values are reasonably close to those in database

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of engineering handbooks. To make them more accurate from the data in handbooks

would be necessary an equipment more precise to quantify the dillatation of the

material, that is, an equipment with more than 4 decimal places. Also, it would be better

if it had been collected greatest amount of data to minimize errors.

References

1. Roylance, D. (2001). Introduction to Fracture. Mechanics.Department of

Materials Science and Engineering: Massachusetts Institute of Technology.

Available from:

http://ocw.mit.edu/courses/materials-science-and-engineering/3-11-mechanics-

of-materials-fall-1999/modules/frac.pdf

2. William, D., Callister, Jr., Rethwisch, David G. (2014). Material Science and

Engineering “An Introduction”, (9th ed.), Wiley.

3. Basu, B.; Tiwari, D.; Kundu, D.; Prasad, R. Is Weibull Distribution the Most

Appropriate Statistical Strength Distribution for Brittle Materials? Indian Institute

of Technology Kanpur. Available from:

http://home.iitk.ac.in/~kundu/paper132.pdf

4. Engineering Statistics. Available from:

http://www.itl.nist.gov/div898/handbook/apr/section1/apr162.htm

5. Kailas, S. V.. Chapter 15. Thermal properties, Indian Institute of Science.

Available from:

http://nptel.ac.in/courses/Webcourse-contents/IISc-BANG/Material

%20Science/pdf/Lecture_Notes/MLN_15.pdf

6. Murthy, V. S. R., et all (2003). Structure and Properties of Engineering

Materials, Tata McGraw-Hill. Available from: https://books.google.com/books?

id=fi_rnPJeTV8C&pg=PA410&lpg=PA410&dq=thermal+expansion+increase+dis

tance+between+atoms&source=bl&ots=cxTthvKBPb&sig=vUZQmX4WoqgrIUpZ

xRgkp_1OQzs&hl=pt-

BR&sa=X&ei=4topVeHvII3loASkkoCAAw&ved=0CDUQ6AEwAw#v=onepage&q

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%20atoms&f=false

7. ASM International. Thermal Properties of Metals. Available from:

http://www.owlnet.rice.edu/~msci301/ThermalExpansion.pdf

8. The Engineering ToolBox. Available from:

http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html

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