e10-mse final report

Nick Renda MSE Module

Section 402B 12/6/11

Final Report

Figure 1. 1045 Steel, Martensite, 400x Magnification

Figure 2. 1045 Steel, Pearlite, 400x Magnification

2

Introduction In this lab, we examined the microstructures of several steel and brass samples to get a better understanding of how microscopy works, and to become better at analyzing grain boundaries and other features of engineering materials. By sanding and polishing the samples, we were able to photograph them and analyze them graphically as well as with programs like ImageJ. Looking on a microscopic level at commonplace materials like brass and steel enabled us to understand why they possess such unique attributes, and how different microscopic structures result in different overall attributes.

Methodology

Brass Sample Preparation:

After receiving my C26000 Alloy brass sample, I chose the side with the most surface area and ground it down slightly with the 240-‐grit sandpaper so that it would lie flat on the bottom of the mold. My instructor then mixed the epoxy with the hardener, and poured it into my mold. I positioned my brass sample inside the mold so that the selected side was face down, and waited an hour for the epoxy to harden. I removed my sample from the mold, and then moved on to the grinding phase.

In order to keep debris from building up on the sandpaper while I was

grinding my sample, I turned on the water by rotating the knob on the grinding mount, and started to grind my sample on the coarsest grit (240). I made sure to maintain steady pressure on my sample, keep it flat, and avoid rotating it or grinding it back and forth. Every ten or so grinds, I checked the bottom of the sample—as soon as I only saw parallel scratches (meaning the previous grit’s scratches were no longer visible), I moved up to the next grit (320). I then rotated my sample 90° so that I would be able to tell when the old scratches were gone and the new scratches were in my new grinding direction. I repeated this pattern—grind, move up in grit, rotate sample—as I moved to the 400 grit, then to the 600, and finally to the 1500.

After I finished grinding, I got a polishing board with a nylon sheet affixed to it, and rinsed it with water and to remove any extra debris left over from the previous user. I put a small amount of 6-‐micron diamond slurry on the sheet, added a couple drops of water, and began to polish the sample in a figure-‐eight pattern for about two minutes. I then washed the sample off, got a new nylon sheet, affixed it to the board, and washed it clean with water. I put a small amount of 1-‐micron slurry and a couple drops of water on the clean nylon sheet and polished using the same pattern as I did with the 6-‐micron sheet, except for only 15 seconds.

3

Etching the Brass Sample After checking to make sure I was wearing gloves and had the proper eye protection, I brought my sample to the fume hood for etching. I placed a couple drops of the brass etchant (10 mL H2O + 10 mL NH4OH + 3 mL H2O2) onto my sample, and swabbed it vigorously to keep the reaction moving along. I made sure not to let my sample turn black, and kept adding more etchant every fifteen seconds. After about 1 minute I put a couple extra drops on, swabbed for ten seconds, and then rinsed everything off in the sink.

Steel Sample Preparation

The procedure for preparing the steel sample is exactly the same as that of the brass sample, except the steel sample was given to me already mounted in epoxy. Also, the steel sample has two different pieces of steel inside, so it is important to make sure to keep the sample level and apply equal pressure when grinding to ensure that both pieces get ground evenly. Etching the Steel Sample The procedure for etching the steel sample is the same to the brass sample, except a different etchant is used (2% nitric acid and 98% methanol). Also, because there are two pieces of steel, I made sure that both were etched equally. Imaging Both Samples Both samples were imaged in the same way using optical microscopes, so they will be discussed in the same section. To begin with, I mounted the sample on a slide so it could be viewed in the microscope. I put the sample on the slide with a piece of clay in between them, covered the sample with a kimwipe to protect it, and put the entire arrangement on a press. I applied a small amount of pressure on the press so that the sample was pushed into the clay and leveled for optimal imaging. Next, I put the slide onto the microscope stage, affixed it in place with the clip, and centered it over the lens with the stage controls on the right side of the microscope. I rotated the lens so it was on the highest magnification (40x), and using the coarse adjustment knob, positioned the slide so that it was roughly one slide thickness away from the tip of the lens. This puts the sample at a position where it can be seen easily by the microscope. I then rotated the lens to the 5x magnification, adjusted the illumination with a knob on the light controls next to the microscope, and focused the lens with the fine adjustment knob until the image was as sharp as possible. To photograph the sample, I used a NikonTM Coolpix 4300 Camera with an attachment that allowed it to fit into one of the eyepieces. I verified that it had a memory card with enough space for my photos, and turned off the flash as it wasn’t being used. To attach the camera to the microscope, I unscrewed one of the

4

eyepieces and inserted the attachment into the hole where the eyepiece used to be. I adjusted the zoom on the camera so it was at the maximum optical zoom (on the screen the white zoom bar moved all the way up to the line between optical and digital zoom). When focusing the images, I made sure to use the screen as a reference and not the other eyepiece, because the camera has a different magnification than the eyepiece. Once focused the image using the fine adjustment knob, I photographed the sample. I took images from several different areas, making sure to check the focus before each picture. I tried to focus on areas with as few scratches as possible to produce the best quality images. After taking all of my pictures, I removed the sample from the stage and put on a graticule slide so I would have a reference frame to measure my images when I analyzed them later. To focus the graticule, I repeated the same steps from earlier: I moved the lens to the 40x magnification, and adjusted the graticule so it was 1 slide width away from the lens. I then switched the lens to the 5x magnification and used the stage controls to locate the 1mm increment. After focusing it with the fine adjustment knob, I took several pictures.

For the steel sample I followed the same procedures, except I took photos of both samples at several magnifications (5x, 10x, 20x, 40x). Brass Grain Counting Methodologies The first method I employed to count the grains of the brass sample and determine their size was the linear intercept method. This method involves drawing many lines on the image, counting how many grains they intercept, and turning this result into a grain diameter. To begin with, I drew enough lines on my image so that I intersected more than fifty grains over my three images (to ensure statistical relevance). Next, I counted how many times each line intersected a grain boundary. I used Photoshop, so I was able to mark each intersection with a black dot. Before calculating the average grain diameter, I needed to convert the line lengths to a real life distance. To do this, I used the graticule. I developed a conversion factor by dividing the graticule’s real length (1000 μm) by each line’s length in inches as measured in Photoshop. I plugged the values for each line into the equation

𝑑 = 𝐶𝐿𝑁

Where d is the average grain diameter, C is the constant 1.5, L is the length of the line in μm, and N is the number of intersections on that line. I multiplied by the constant C because the linear intercept traditionally underestimates the grain diameter. I then took an average of my d values for each line to find an average grain diameter for the image.

The next method I used to count the grains of my brass sample was the ASTM E112 method, which involves drawing a circle over the image and counting how many grains are inside. The ASTM E112 method provides the following formula to

5

calculate a parameter called the “grain size number” G, given a calculated value of grains per square inch area of micrograph “nA“, recorded at 100x magnification.

𝑛! = 2[!!!] However, to account for the fact that my sample was at 50x magnification and not

100x, I needed to multiple nA by a constant !"!""

!, squared because zooming

doubles both the x and y axes, and area is calculated from multiplying the two. Simplifying the constant yields:

𝑛!4 = 2[!!!]

Solving for my unknown, I got the following equation.

𝐺 =ln𝑛!4ln 2 + 1

The nA value is calculated with the equation:

𝑛! =𝑤 + 𝑝2𝐴

Where w is the number of grains completely inside the circle, p is the number of grains partially inside, and A is the area of the circle in inches2. Using the photograph of the graticule, the radius of the circle can be measured and converted to inches ! !"#!

!".! !!. This radius can then be turned into an area with the formula

𝐴 = 𝜋𝑟!. The finalized formula for determining the grain size number G is:

𝐺 =ln

𝑤 + 𝑝24𝜋𝑟!

ln 2 + 1 Where w is the number of grains completely inside the circle, p is the number of grains partially inside, and r is the radius of the circle in inches. Brass ImageJ Analysis

To begin with, I downloaded the ImageJ program at http://rsbweb.nih.gov/ij/. Once the program had installed, I opened it and opened the sample (File>Open). The first step I employed in processing the file was to enhance the grain boundaries by drawing over them with the brush tool. I right clicked the brush tool to change the diameter of the brush (I found that 15px was large enough for ImageJ to detect but small enough to not get in the way later).

Figure 3. Brush tool selected

6

If the boundaries don’t connect completely, ImageJ counts multiple grains as a single grain later on, so I made sure to completely connect them all. The enhanced image looked something like Figure 4.

Next, I adjusted the contrast (Image>Adjust>Brightness/Contrast) so ImageJ could more easily differentiate between grains and grain boundaries. I found that the “auto” button worked well, but if necessary, the sliders in the menu that pops up could be used. Any setting that made the boundaries more defined and the background less so was optimal. The image with adjusted contrast looked something like Figure 5.

Figure 4. Enhanced grain boundaries in ImageJ, 50x Magnification

7

Figure 5. Contrast applied, 50x Magnification

After adjusting the contrast, I converted the image into 8-‐bit (Image>Type>8-‐bit) to speed up processing. This was also critical for later processing stages, as certain steps would not work if the image wasn’t 8-‐bit.

I opened the threshold panel (Image>Adjust>Threshold) to view the various options for adjusting the threshold. There were two sliders on the panel—the top set the black threshold, and the bottom set the white. In general, setting the top slider all the way to the left and then slowly moving the bottom slider to the left was the best way to set a good threshold. Ideally, the image only had grain boundaries shown in black and everything else was in white. I avoided setting the white threshold so low that the image began to speckle and the grain boundaries faded or disappeared. The final image looked something like Figure 6.

8

Figure 6. Threshold applied, 50x Magnification

Something extra that I found enhances image analysis was the “Find Edges” feature (Process>Find Edges). ImageJ outlined the edges of the grain boundaries in the image, and I used this to check that there were no holes in the boundaries. Also, it made the images more aesthetically pleasing and easier to manipulate later. The image after the processing was similar to Figure 7.

Figure 7. Find Edges applied, 50x Magnification

9

Next, I began the analyzing phase by setting which measurements ImageJ should make (Analyze>Set Measurements). In addition to the three defaults, I made sure to check “Feret’s diameter” as it was used later to determine the grain diameter. The set measurements looked like Figure 8.

Next, I selected Analyze Particles (Analyze>Analyze Particles) and a window popped up with settings for the analysis. To ensure that extra particles were not recorded, I set the minimum pixel size to 100. Also, on the Show menu I selected Outlines, and made sure to check “Exclude on edges” and then clicked “OK.” The selections matched Figure 9.

Three windows popped up next: Summary (Figure 10), Results (Figure 11), and a Drawing of the grains (Figure 12). I scrolled through the Results window and looked at the Area column to make sure all of the areas are roughly the same size and there were no irregularities (e.g. mostly four-‐digit areas but 1 six-‐digit area). If there had been any problems, the Drawing window could have been used in conjunction with the number label in the Results window to identify which grains were responsible.

The most important value was located in the Summary window under “Feret.” The Feret value is the greatest distance between two points on the grain boundary, i.e. the greatest diameter. This value was a good approximation of the grain size, but it was in pixels, the default unit in ImageJ. To convert it to a meaningful unit, such as μm, I needed to know the length of an object in both pixels and μm. I knew the length of the graticule in μm, and I could measure it in pixels by loading it into ImageJ.

Figure 8. Set Measurements Window

Figure 9. Analyze Particles Window

10

Figure 10. Post-‐Analysis Summary window

Figure 12. Post-‐Analysis Drawing window

Figure 11. Post-‐Analysis Results window

11

After I opened the graticule file in ImageJ (File>Open), I used the line tool shown in Figure 13 to make a line that spanned the length of the graticule. Then, I clicked Analyze>Set Scale, and a window appeared with the length of the measured segment in pixels. Therefore, to convert Feret’s diameter from pixels to μm, I multiplied by the constant

1000 µμm𝑥 pixels

Where x was the length in pixels of the graticule. The resulting value was the average grain size.

Figure 13. Line tool

Figure 14. Measuring length of graticule

12

Steel ImageJ Analysis For this lab, the only steel image I needed to analyze was the pearlite to find a volume fraction. To prepare it for analysis, all of the steps were exactly the same. Once I had an image that was properly thresholded, I set which measurements ImageJ should make (Analyze>Set Measurements). In addition to the three defaults, I made sure to check “Area Fraction” as it was the value I was looking for. The set measurements looked like Figure 15.

Next, I selected Analyze Particles (Analyze>Analyze Particles) and a window popped up with settings for the analysis. I wanted all pixels to be recorded, so I set the minimum pixel size to 0. The selections matched Figure 16.

Three windows popped up next: Summary (Figure 17), Results (Figure 18),

and a Drawing of the grains (Figure 19). The volume fraction is simply the number listed under the column “Area Fraction” in the summary window.

Figure 15. Set Measurements window



13



14

Original Images

Figure 20. C2600 Brass, 50x Magnification


15


Figure 23. Graticule, 50x Magnification

16



17



18

Figure 28. Graticule, 400x Magnification

19

Discussion

Question 1: What is the average grain diameter in your sample according to the linear intercept method?

In Figure 29, the length of the 1000μm strip on the bottom right is 2.89

inches (done by measuring in Photoshop), so we can find the conversion factor to get from inches to μm by dividing the two units, and the resulting value is 346 !!

!"#$.

For added clarification, I added the line numbers on the left side of the lines, black dots where grain boundaries occurred, and the number of boundaries per line on the right side. Description of Tables 1, 2, and 3

• The first row is has the column numbers that correspond to each column. • The second row has the symbol that corresponds to each column.

Figure 29. Micrograph A with linear intercept method applied, 50x Magnification

20

• Columns 1,2, and 4 are raw data obtained from the image. • Column 3 is obtained by multiplying the length of the line, L, by the

conversion factor 346 !!!"#$

.

• Column 5 is obtained with the equation 𝑛! =!!, where N is column 4 and L is

column 3. • Column 6 is obtained with the equation 𝑑 = !

!!, where C is the constant 1.5

and nL is column 5. Table 1. Raw and Calculated Data from Linear-‐Intercept Method Performed on Micrograph A

1 2 3 4 5 6 # L L N nL d

Line Number Length (in) Length (μm)

Number of Grain

Boundaries

Number of Boundaries per μm (1/μm)

Average Grain

Diameter (μm)

1 6.5 2250 12 .0053 281 2 6.5 2250 9 .0040 375 3 6.5 2250 10 .0044 337 4 6.5 2250 10 .0044 337 5 6.5 2250 10 .0044 337 6 6.5 2250 11 .0049 307 7 6.5 2250 10 .0044 337

The same descriptions of the pictures and graph apply to the next two images, so they will be omitted to save space.

21

Table 2. Raw and Calculated Data from Linear-‐Intercept Method Performed on Micrograph B

1 2 3 4 5 6 # L L N nL d

Line Number Length (inches)

Length (μm)

Number of Grain

Boundaries

Number of Boundaries per μm

Average Grain

Diameter (μm)

1 6.5 2250 10 .0044 337 2 6.5 2250 10 .0044 337 3 6.5 2250 8 .0036 422 4 6.5 2250 9 .0040 375 5 6.5 2250 7 .0031 482 6 6.5 2250 7 .0031 482 7 6.5 2250 6 .0027 562 8 6.5 2250 6 .0027 562

Figure 30. Micrograph B with linear intercept method applied, 50x Magnification

22

Table 3. Raw and Calculated Data from Linear-‐Intercept Method Performed on Micrograph C

1 2 3 4 5 6 # L L N nL d

Line Number Length (inches)

Length (μm)

Number of Grain

Boundaries

Number of Boundaries per μm

Average Grain

Diameter (μm)

1 6.5 2250 9 .0040 375 2 6.5 2250 6 .0027 562 3 6.5 2250 10 .0044 338 4 6.5 2250 9 .0040 375 5 6.5 2250 11 .0049 307 6 6.5 2250 10 .0044 337 7 6.5 2250 9 .0040 375

The average grain diameter of my brass sample by applying the linear intercept method to Figures 1, 2, and 3 is 388 μm. The standard deviation was 86 μm, and both values were calculated by inputting all data points into Excel.

Figure 31. Micrograph C with linear intercept method applied, 50x Magnification

23

Question 2: How precise is the linear intercept method? Use your data, citing the specific ways you executed the linear intercept method, and number of attempts, to support your conclusion. Remember "precision" means getting the same result each time; "accuracy" means getting the correct result. As with many engineering projects, no one knows the "correct" result for this exercise.

Simply put, precision is how consistent a given method is when it is repeated. For the linear intercept method to be perfectly precise, it needs to output the same average grain diameter for every image it is applied to. Because we are taking a cross section, we expect that the particles will not all be the same size. We can verify this by looking at the standard deviation for each image in Table 4. For micrographs B and C, the standard deviation was very high—within each image, there were a wide range of outputted grain diameters.

However, assuming each micrograph encompasses enough particles, the averages for each image should be somewhat similar to each other. From the bottom of Table 4, we can see this is not the case—the overall standard deviation of 88 μm was quite large. So what does this mean? It means that the averages for each micrograph varied significantly, and the method isn’t consistent. Each micrograph that the linear intercept method is applied to should give the same average: the material is the same throughout, and the images are all taken from the same cross-‐section.

The linear intercept method is inherently somewhat imprecise because it depends so much on human factors: choice of line placement, the length of line drawn, and determination of what qualifies as a grain boundary all significantly affect the results. Ideally lines are chosen arbitrarily and represent an “average” number of grains intersected, but in most cases there is some system or arrangement of lines, which can alter the data. In addition, the length of line drawn affects the number of intersections recorded: when lines are smaller, they intersect fewer grains and are less representative of the overall picture. The linear intercept method is also subject to whoever is using it—an inexperienced student may not be able to determine exactly which lines are grain boundaries, scratches, or twins.

I employed several tactics to try to eliminate these factors. First of all, I used

a grid method (as seen I Figure 29) when making my lines so I didn’t draw them arbitrarily. I tried to space them evenly throughout the image to obtain a result that reflected the contents of the image. I also standardized my line length so that I avoided ending lines in spots that I thought “looked good.” I may have made mistakes when determining grain boundaries, but that can’t be avoided.

As with any method that depends on averages, the result becomes more

precise with more data points. Perhaps one of my images was a fluke and the linear intercept method is actually quite precise. With 20 micrographs I would be able to

24

tell if one or two were off and the rest were similar, but with only three I can only conclude that the method itself is imprecise.

Table 4. Select data compiled from Tables 1-‐3

Micrograph A Micrograph B Line

Number Number of Grain

Boundaries

Average Grain

Diameter (μm)

Line Number

Number of Grain

Boundaries

Average Grain

Diameter (μm)

1 12 281 1 10 337 2 9 375 2 10 337 3 10 337 3 8 422 4 10 337 4 9 375 5 10 337 5 7 482 6 11 307 6 7 482 7 10 337 7 6 562

Average (μm) 330 Average (μm) 445 Standard Deviation

(μm) 29.3 Standard

Deviation (μm) 91.7

Micrograph C Line

Number Number of Grain

Boundaries

Average Grain

Diameter (μm)

1 9 375 2 6 562 3 10 337 4 9 375 5 11 307 6 10 337 7 9 375

Average (μm) 381 Standard Deviation

(μm) 84.0

Average of the 3 averages (μm)

388

Standard Deviation (μm)

88

25

Question 3: What is the average ASTM grain size number of your sample according to ASTM E112? Use to data to establish the precision associated with this determination.

Using the ASTM E112 method, I inscribed a circle over the sample and marked the grains that were completely within the bounds with a black dot and those that were partially inside with a blue dot. All of my images were taken at 50x magnification, and for easy viewing I displayed the number of grains partially inside (blue) and completely inside (black) in the top left and right corners of the image, respectively.

The ASTM E112 method provides the following formula to calculate a

parameter called the “grain size number” G, given a calculated value of grains per square inch area of micrograph recorded at 100x magnification nA.

𝑛! = 2[!!!] However, to account for the fact that our sample is at 50x magnification and not

100x, we need to multiple nA by a constant !"!""

!, squared because zooming doubles

both the x and y axes, and area is calculated from multiplying the two. Simplifying the constant yields:

𝑛!4 = 2[!!!]

Solving for our unknown, we get the following equation.

𝐺 =ln𝑛!4ln 2 + 1

However, in order to determine the value of nA, we first need to determine the area of the circle. Using the graticule in the image, the radius of the circle can be measured to be 0.97mm. Converting this value to inches ! !"#!

!".! !! yields a radius of

0.038 in, and an area of 0.0046 in2.

26

Table 5. Raw and Calculated Data from ASTM E112 performed on Micrograph A

1 2 3 4 5 6 Number Weight Resultant

Value Combined Value

nA value (grains/inch2)

G value

Black 32 1 32 43 9385 12.196 Blue 21 .5 10.5

• Column 1 is raw data obtained from the image • Column 2 is the weight of whole versus partial grains determined by ASTM

E112 • Column 3 is obtained by multiplying Columns 1 and 2 • Column 4 is obtained by adding the values from column 3 together, and

rounded up to the nearest whole number • Column 5 is obtained by dividing column 4 by the area of the circle,

0.004582 in2 • Column 6 is obtained from the equation for G, where nA is column 5

𝐺 =ln𝑛!4ln 2 + 1

Figure 32. Micrograph A with ASTM E112 method applied, 50x Magnification

27

The calculations are repeated to obtain the data in Table 6 and 7

Table 6. Raw and Calculated Data from ASTM E112 performed on Micrograph B




G value

Black 23 1 23 35 7639 11.899 Blue 23 .5 11.5

Figure 33. Micrograph B with ASTM E112 method applied, 50x Magnification

28

Table 7. Raw and Calculated Data from ASTM E112 performed on Micrograph C

The average grain size number was 12.14, with a standard deviation of 0.22

as determined in Excel. At first glance, one may conclude that this method is precise because the standard deviation is so small, which means the G values are all relatively similar. However, we must consider the fact that the G value is the result of taking the natural log of a relatively large nA value. On an exponential scale, a standard deviation of 0.22 is much more significant than on a linear scale. For instance, if we take the standard deviation of the nA values from Table 5-‐7, we get 1334 grains/inch2, which is much larger and shows that the values vary significantly. The ASTM E112 method fails to generate consistent values when repeated, and with only three images to compare with, we can only conclude that it lacks precision.




G value

Black 38 1 38 47 10260 12.325 Blue 18 .5 9

Figure 34. Micrograph C with ASTM E112 method applied, 50x Magnification

29

Question 4: The linear intercept method gives a "grain size" in units of length while ASTM E112 gives a dimensionless "grain size number." Design a way to compare both methods on equal terms, explain it, and use your results to execute the comparison. What do you conclude? Explain your approach to comparing methods.

With the linear intercept method, we determined the average diameter of a single grain in our sample. Ideally, the linear intercept method accounted for the fact that all the grains had different sizes and shapes, and outputted a value for an average grain. We can assume that this average grain diameter is the same in every direction, which gives us a circular grain shape. We can then compute the average area of one grain. When we convert this value into inches, we can manipulate it to be similar to the nA value in Question 3. The nA value is simply a ratio of grains to area the grains take up, so if our 1 grain takes up x area, its nA value is simply !!. From this point, we can compute G value using the ASTM method from Question 3, and compare the two dimensionless grain size numbers. Use your results from the rest of the lab to execute the comparison.

The average grain diameter from Question 1 was 388 μm, or 0.388 mm. Dividing by 2 we get a radius of 0.194mm, or 0.00764in using ! !"#!

!".! !!. The area of

a circle is 𝜋r2, which gives an area of 0.000183 in2. Because we are only looking at one grain, the resultant nA value is 1 divided by this, or 5453. The resultant G value applying the same formulas from Question 3 is 11.413. State your conclusion about your comparison.

The G value of 11.413 (from manipulating the average grain diameter in

Question 1) was in the same range as the average G value from Question 3 of 12.14. Calculating the standard deviation of the two gives a value of 0.329—however, because G is the result of a logarithm, we can see that the values actually vary considerably (the same analysis is employed in greater detail in the conclusion of Question 3). From this data, we can conclude that the comparison is not very precise. There are many possible reasons for this—I assumed that the average grain was circular, when it may not be. Also, the same human sources of error for the linear intercept method apply to this comparison. If the linear intercept method wasn’t done accurately, then those mistakes would be reflected in the comparison.

30

Question 5: You've just been called in as a consultant on the failure of a water valve on the California aqueduct. The qualification specs indicate the valve should have been given a very specific heat treatment to generate a very specific grain size. The valve has an 18-‐inch throat and weighs over 700 lbs. In light of your new understanding of grain size analysis, what experimental procedures would you employ to determine if the proper heat treatment was done or not? Be specific.) Describe what experimental procedures you would employ to determine if the proper heat treatment was done. Be specific.

To start with, dealing with a 700 lb. sample is extremely impractical, so the first thing I would do is remove specific portions of the valve to examine. When choosing where to remove sample from, I would try to get at least three from near the point of failure to determine if it was an isolated irregularity, or if the entire valve is flawed. In addition, I would want to drill three holes in the valve where water flows through and remove samples so I could look at a cross section. For instance, maybe only the inside of the valve was heat-‐treated, and a cross section would reveal that the grain size was not constant throughout the valve.

To actually remove the samples, I would use a high-‐powered drill to cut out cylindrical sections of the valve. Depending on which faces I wanted to examine (planar versus cross sectional), I would grind the samples using the power grinder until I had a flat surface to work with. If the samples were too small to grip easily with my hands, I would mount them in epoxy to ensure they would be large enough to hold. From this point I would begin the grinding phase like we did in class. Starting at 240 grit, I would work my way up to the 1-‐micron polish until my samples were (mostly) scratch free. I would then etch them to reveal the grains, and examine them under a microscope.

From this point I would photograph the samples, and then photograph a graticule to have something to measure against. Looking at these photos in Photoshop, I would then apply both the Linear Intercept method and ASTM E112 as demonstrated in Questions 1 and 3 to determine what the average grain diameter and grain size number were, and compare them with what the actual specifications should be. If any of the values were more than one standard deviation away, I would take more samples from that area and analyze them in the same way to ensure that it was not just a fluke. If samples continued to have similar incorrect values, I would conclude that the heat treatment hadn’t occurred. If some samples had incorrect values but others did not, I would conclude that the heat treatment might have been done but only partially performed.

31

Question 6: What is the "grain size" of your sample determined by the image processing and analysis tools available in ImageJ? How do the grain sizes obtained in this lab compare to the grain sizes obtained in the first lab? Compare them using any metric of your choosing. Rationalize any differences.

Table 8. Raw Data from ImageJ Brass Sample Analysis

From Table 8 above, we get a Feret column, which represents the maximum diameter of a grain. We can use this value to compare to the values calculated in Question 1, but the current the values given are in pixels, and need to be converted into μm to be meaningful. To obtain our conversion factor, we need to find out how many pixel lengths correspond to a set distance. We can use an image of a graticule to find this.

Using the “Set Scale” tool in ImageJ, we are able to measure the length of the

graticule. We determine that 1000 μm is equal to 862.84 pixels, and obtain the following conversion factor:

1000 µμm862.84 pixels = 1.16

µμmpixels

When finding the average grain diameter in question 1, we multiplied by a constant 1.5 because it was found that the linear intercept method traditionally

Slice Count Total Area

Average Size

Area Fraction Feret Feret X Feret Y

Feret Angle

Min Feret

PS1.jpg 55 100977 1835.945 2.6 275.576 1183.455 818.091 95.906 178.585 PS2.jpg 40 89402 2235.05 2.3 327.037 1226.625 817.175 93.078 212.671

Figure 35. Calculating scale in ImageJ, 50x Magnification

32

underestimates grain diameter. Feret’s diameter is the greatest distance between two particles, so it won’t be an underestimate (if anything, it might be a slight overestimate). Multiplying the diameters from Table 9 by the conversion factor, we obtain values of 319 μm and 379 μm respectively, an average of 349 μm, and a standard deviation of 42.4 μm.

Table 9. Comparison of Linear Intercept and ImageJ results

For Micrograph A, the standard deviation between the two different methods

was extremely small. The methods both produced similar results. For Micrograph B, the standard deviation was much larger. However, when looking at which partial grains ImageJ didn’t consider, this makes sense. In Figure 29, the two large grains on the left boundary of the picture took up a substantial portion of almost every scribe line. Because these grains weren’t complete, ImageJ did not consider them, while the linear intercept method did. The resultant average Feret diameter was much lower than the linear intercept average, which makes sense given that ImageJ ignored the two largest grains in the micrograph.

By finding a ratio of pixels to μm, we can compare feret’s diameter calculated in ImageJ with the average grain diameter calculated in Question 1.

Micrograph Label

Linear Intercept

Average (μm)

Feret Diameter (μm)

Standard Deviation (μm)

A 330 319 7.8 B 445 349 67.9

33

Question 7: How precise is ImageJ for grain size determination? What pitfalls did you uncover in your use of its computational tools? Discuss the precision of determining grain size using ImageJ.

Initially, I had a lot of trouble getting precise grain sizes with ImageJ. The process usually fell apart in the threshold step. I could never adjust the sliders so I would get a clean outline around each grain—grains would always either fill in completely black or stay white. However, I received a helpful hint: manually draw in the grain boundaries so ImageJ can detect them better. With this in mind, ImageJ became incredibly easy to use. While going through with the brush tool and drawing in each boundary is time-‐consuming, it enables the program to easily pick up on the boundaries that may have been incredibly faint before.

When the user draws in the boundaries, ImageJ becomes extremely precise.

Assuming the lines have been drawn correctly, the program will consistently output very similar values every time an image is run through the various contrasts, thresholds, and analyzing stages. As I was learning how ImageJ worked I ran the same image through at least ten times, and by the end I was getting the exact same grain size every time.

Are the outcomes of your two images the same? Quantify how similar/different they are.

The outcomes of my two images are essentially the same, in terms of appearance. They both have all complete grains shaded black, with white grain boundaries. The boundaries are smooth, and have remained exactly as I first drew

Figure 36. Micrograph A, post-‐processing Figure 37. Micrograph B, post-‐processing

34

them for both images. When the lines are drawn first, the results are crisp, clear, and consistent. Discuss pitfalls of using ImageJ, relative to manually determining grain size.

One of the biggest pitfalls of using ImageJ is the time required to use it successfully. It simply isn’t possible to obtain a good image in a short amount of time. To ensure accuracy, the lines must be drawn on very carefully—if this step is rushed, the resultant data is useless. In addition, ImageJ analysis fails to account for grains that aren’t completely in the image. Often, this means losing a significant amount of data as well as reducing the sample size, as many grains are only partially present. In the case of Question 6, I hypothesized that this loss of data was the cause of a discrepancy between a manually measured diameter and ImageJ’s measured diameter.

35

Question 8: What changes would you suggest in the practice of optical metallography (sample selection, polishing, etching) to optimize the determination of grain size using ImageJ? Discuss changes to your SAMPLE PREPARATION to increase the utility of ImageJ processing. When grinding my sample, one of the first things I noticed was that the epoxy was much softer than the brass. It wears away much faster, and as a result often the sample tends to protrude more than the epoxy, and instead of being flat, the bottom of the sample is rounded. This creates problems because it becomes difficult to grind the edges of the brass—they often end up more scratched than the center because they aren’t actually being ground down. Most of the scratches in my sample were due to this, and a change I would propose is either using a harder mounting substance or having samples large enough to hold easily. This would result in more usable picture space, better quality images, and better ImageJ processing. Another problem I had with the sample preparation was that it was difficult to determine whether the camera had focused my images correctly. When using the camera with the microscope, the only way to manually focus the images was to look at the tiny display. The eyepiece has a slightly different magnification than the camera, so the only way to focus was to squint at the display and hope it was as sharp as possible. This has a direct effect on image processing—ImageJ works much better when it can see clear, distinct boundaries, versus the fuzzier boundaries that result from bad focus. Include a sample comparison. Include an image/portion of an image that was easier to analyze than the rest. Include an image/portion of an image that was harder to analyze. Discuss specific reasons why one was easier than the other.

The image on the left was

difficult to analyze for several reasons. First of all, several of the grains are almost the same color, and the boundaries are nearly impossible to see clearly. In addition, the image is slightly fuzzy, and the lack of sharp edges makes it difficult for ImageJ to process the grain boundaries. The scratch in the top left corner also adds another line into the picture that could generate an extra boundary. Figure 38. Brass sample, 50x Magnification

36

This image was much easier to process. Not only are the grain boundaries much clearer, but also the contrast is much sharper, and it is quite obvious where the grains touch each other. There are no visible scratches, and the twins are all well defined, so ImageJ won’t misinterpret them as extra boundaries. Also, the grain boundaries in this image seem to be slightly darker and wider, and as a result it is easier for ImageJ to register that they are boundaries.

Figure 39. Brass sample, 50x Magnification

37

Question 9: Design a method employing ImageJ for determining the grain size of a metallographic sample, and describe it in sufficient detail for a novice user of ImageJ to complete a successful grain size determination. In technical reports, this is known as the "experimental procedures" or "protocol" section.

To begin with, download the ImageJ program at http://rsbweb.nih.gov/ij/. Once the program has installed, open it and open the sample (File>Open). The first step in processing the file is to enhance the grain boundaries by drawing over them with the brush tool. Right click the brush tool to change the diameter of the brush, I found that 15px was large enough for ImageJ to detect but small enough to not get in the way.

Make sure that all of the grain boundaries connect! If the boundaries don’t connect completely, ImageJ will count multiple grains as a single grain later on. The enhanced image should look something like Figure 42.

Figure 41. Brush tool selected

Figure 42. Enhanced grain boundaries in ImageJ, 50x Magnification

38

Next, adjust the contrast (Image>Adjust>Brightness/Contrast) so ImageJ can more easily differentiate between grains and grain boundaries. I found that the “auto” button works well, but if necessary, the sliders in the menu that pops up can be used. Any setting that makes the boundaries more defined and the background less so is optimal. The image with adjusted contrast should look something like Figure 43.

Figure 43. Contrast applied, 50x Magnification

After adjusting the contrast, convert the image into 8-‐bit (Image>Type>8-‐bit) to speed up processing. This is also critical for later processing stages, as certain steps will not work if the image isn’t 8-‐bit.

Open the threshold panel (Image>Adjust>Threshold) to view the various options for adjusting the threshold. There are two sliders on the panel—the top sets the black threshold, the bottom sets the white. In general, setting the top slider all the way to the left and then slowly moving the bottom slider to the left is the best way to set a good threshold. Ideally, the image will only have grain boundaries shown in black and everything else will be white. Avoid setting the white threshold so low that the image begins to speckle and the grain boundaries fade or disappear. The final image should look something like Figure 44.

39

Figure 44. Threshold applied, 50x Magnification

Something extra that I found enhances image analysis is the “Find Edges” feature (Process>Find Edges). ImageJ outlines the edges of the grain boundaries in the image, and this can be used to check that there are no holes in the boundaries. Also, it makes the images more aesthetically pleasing and easier to manipulate later on. The image after the processing should be similar to Figure 45.

Figure 45. Find Edges applied, 50x Magnification

40

Next, begin the analyzing phase by setting which measurements ImageJ should make (Analyze>Set Measurements). In addition to the three defaults, make sure to check “Feret’s diameter” as it will be used later to determine the grain diameter. The set measurements should look like Figure 46.

Select Analyze Particles (Analyze>Analyze Particles) and a window will pop up with settings for the analysis. To ensure that extra particles are not recorded, set the minimum pixel size to 100. Also, on the Show menu select outlines, and be sure to check “Exclude on edges” and then click “OK.” The selections should match Figure 47.

Three windows should pop up: Summary (Figure 48), Results (Figure 49), and a Drawing of the grains (Figure 50). It is worth scrolling through the Results window and looking at the Area column to make sure all of the areas are roughly the same size and there are no irregularities (e.g. mostly four-‐digit areas but 1 six-‐digit area). If there are any problems, the Drawing window can be used in conjunction with the Numerical label in the Results window to identify which grains are responsible.

The most important value is located in the Summary window under “Feret.” The Feret value is the greatest distance between two points on the grain boundary, i.e. the greatest diameter. This value is a good approximation of the grain size, but it currently is in pixels, the default unit in ImageJ. To convert it to a meaningful unit, such as μm, we need to know the length of an object in both pixels and μm. We know the length of our graticule in μm, and we can measure it in pixels by loading it into ImageJ.

Figure 46. Set Measurements Window

Figure 47. Analyze Particles Window

41




42

After opening the graticule file in ImageJ (File>Open), use the line tool shown in Figure 51 to make a line that spans the length of the graticule. Then, click Analyze>Set Scale, and a window will appear with the length of the measured segment in pixels. Therefore, to convert Feret’s diameter from pixels to μm, multiply by the constant

1000 µμm𝑥 pixels

Where x is the length in pixels of the graticule. The resulting value is the average grain size.

Figure 51. Line tool

Figure 52. Measuring length of graticule

43

Question 10: Can you foresee any ethical dilemmas arising from the use of image processing in engineering analysis? Explain, using your own data if you can, how a "processed" image might deviate enough from an "actual" image to support different conclusions, and the implications of willfully choosing processing methodologies to achieve a specific goal. There is certainly a problematic ethical dilemma resulting from the use of image processing in engineering analysis. When the person doing the processing has a stake in the results of the analysis, he or she faces the choice of “altering” data to achieve a more favorable result. Altering can mean anything from being slightly biased towards a certain result and subconsciously influencing the data to outright fabrication of data or editing the images. Humans are inherently selfish, and if we can gain from slight manipulation of data, there’s a decent chance that our subconscious will steer us in that direction, even if we intend to be honest. Discuss. Use a concrete example, such as: imagine if you were guaranteed to get an “A” in this class, if your sample had the smallest average grain size. Even if you were trying to be honest, would the knowledge of this affect your adjustments/measurement?

Knowing this small fact would have a huge impact on my adjustments and measurement, as much I would like to think nothing would happen. If something as important as my grade in the class depended on something so small and arbitrary like rounding a number, my subconscious (and perhaps even my conscious mind) wouldn’t be able to help itself. Of course I wouldn’t fabricate data—but I would probably manipulate it to my advantage. For instance, I would definitely try both the Linear Intercept and ASTM E112 methods—and I would pick the one that yielded the smaller value. Say there was a blurry spot and it is unclear whether a grain boundary exists or not. As someone whose grade depended on getting a low value, I would probably “assume” that there was a boundary, splitting a grain in two and lowering my average grain size. Demonstrate by altering your image slightly to increase grain size and then again to decrease grain size. Show similar images that have very different outcomes.

In order to increase grain size, I decreased the pixel diameter of the brush used to draw the grain boundaries. The original image is Figure 53, and the first altered version is Figure 54. By decreasing the area between the grains, I effectively expanded them so that they would take up more space and output a larger resultant area and grain size. To decrease grain size, I did the opposite—I used a thicker brush, shrinking the grains and outputting a smaller area and grain size, as shown in Figure 55.

44

Table 10. Raw data from ImageJ analysis of different brush diameters

The ImageJ analysis results in Table 10 confirm that the large brush outputted the lowest average size and ferret diameter, while the small brush gave the largest of both.

The problem with having a personal stake in one’s own image analysis is that

there is no clear line between influencing the numbers with “human error” and outright fabrication. In addition, it is usually very easy to rationalize these tiny changes in data or methodology, as they are so small that they seem almost insignificant. But at what point does rounding a value up become falsifying data and not just abiding by a convenient choice of significant figures? Overall, one should refrain from choosing processes to meet a certain goal, and instead focus on doing a fair, unbiased analysis of the image.

Slice Count Total Area

Average Size

Area Fraction Feret

PS1 MED.jpg 56 93026 1661.179 2.4 253.225 PS1 SML.jpg 54 100204 1855.63 2.6 278.618 PS1 LRG.jpg 56 83800 1496.429 2.2 232.716

Figure 53. Medium brush diameter

Figure 55. Large brush diameter Figure 54. Small brush diameter

45

Question 11: Show and describe in detail the two microstructures associated with the two thermal treatments examined in this lab exercise. Note that the carbon concentration is the same for both specimens (both are 1045 steel), so any differences in microstructure are due exclusively to processing.

Both microstructures start out in a high temperature solid phase called austenite. When the temperature drops below 727°C, pearlite forms.

Pearlite has two major constituents: ferrite, which is composed primarily of iron, and cementite, which is an iron carbide. The ferrite and cementite form disk-‐like layers, called lamellae, and have a tendency to glitter under a microscope, as they do in Figure 56. Depending on how the pearlite is heat-‐treated when it is formed, the lamellae can be either coarse and thick or fine and thin. In Figure 57, the circle encloses a well-‐formed lamella, within lines drawn on to show the equiaxed grain structure. Figure 58 is also a particularly good image with interspersed lamellae of pearlite.

Figure 56. Pearlite, Optical Image, 400x Magnification

46

Figure 57. Pearlite, SEM Image, 5000x Magnification

Figure 58. Pearlite, SEM Image, 1000x Magnification

47

When the austenite is cooled off extremely quickly, a different microstructure called martensite forms. Martensite is characterized by acicular grain structure, which means its grains are much smaller and almost needlelike, like those within the circle in Figure 60. As shown in both Figures 59 and 60 below, the grains are almost imperceptible at 2000x magnification, and it takes zooming all the way to 10000x to get a close look at them. While martensite is generally stronger than pearlite, sometimes the rapid cooling process causes cracks to form. The process that causes the cracks, called quenching, is not ideal for big samples because large amounts steel cannot be cooled down quickly. However, martensite’s smaller grain size gives it superior strength when it can be created correctly.

Figure 60. Martensite, SEM Image, 10000x Magnification

Figure 59. Martensite, SEM Image, 2000x Magnification

48

Question 12: Pearlite is sometimes described as having a lamellar morphology. Can you identify pearlite in your images? Explain. Use ImageJ or other methods of analysis to determine the volume fraction of pearlite in your microstructure. Show your work.

Identify pearlite in your images, and explain why you have determined those regions are pearlite.

From Figure 57, we can see that pearlite is characterized by its lamellar structures, which in a cross section shows multiple layers on top of each other. Figure 58, at 1000x shows a zoomed-‐out view of the sample, and demonstrates how the pearlite regions are slightly darker than the other regions. Therefore, we can apply this to our sample: the pearlite regions are the darker regions, and since there are only white and black regions, the pearlite must be the black regions in our image. Determine the volume fraction of pearlite from two of your images of the steel that contain pearlite (not the martensite sample). Explain and illustrate your process for determining volume fraction.

To prepare the pearlite for analysis, all of the steps were exactly the same as for when we prepared the brass, except we don’t need to pre-‐draw lines around the grains as we are determining a volume fraction. The pre-‐analysis image should look something like Figure 61. Once we have an image that is properly thresholded, we set which measurements ImageJ should make (Analyze>Set Measurements). In addition to the three defaults, we make sure to check “Area Fraction” as it is the value we are looking for. The set measurements look like Figure 62.

Figure 61. Pre-‐analysis pearlite image

49

Next, we select Analyze Particles (Analyze>Analyze Particles) and a window pops up with settings for the analysis. We want all pixels to be recorded, so we set the minimum pixel size to 0. The selections match Figure 63.

Three windows popped up next: Results (Figure 64), a Drawing of the grains

(Figure 65), and Summary (Figure 66).




50

The only value we are interested in is the “Area Fraction” column in Figure 66. This value represents the percentage of black particles in our image. Because the black particles are pearlite, this value is the area fraction of pearlite. We assume that the area fraction of one planar cross-‐section is representative of the entire sample, so we assume the area fraction is the same as the volume fraction, and get a result of 37.2% pearlite.



51

Question 13: Show directly on a TTT curve the two cooling paths for the two specimens in your metallographic mount, supported by the data captured in your micrographs. Note that both thermal treatments began at time 0 in the austenitic region at 900°C, and finished at room temperature (assume 25°C).

Figure 67. TTT curve of martensite and pearlite samples

Figure 67 above shows the two cooling paths, where martensite is the black line and pearlite is the red line. This is supported both by our data and our knowledge of how the two samples form. Martensite forms when the steel is cooled very rapidly, and avoids the “knee” of the TTT curve at around 550° C. From the black line, we can see that steel drops to 25° C in less than a second, and does indeed avoid the protruding part of the TTT curve. Because this steel passes below the Ms formation temperature of 165° C before the TTT curve, it forms martensite. The black line is a general line—martensite can form from any line or curve that avoids the knee of the TTT curve and passes through the MS temperature before the curve does. From the red line, we can see that the pearlite cooling path first intersected the TTT curve at around 700° C. At this point, pearlite began to form and continued forming until around 610° C. This matches our knowledge of how pearlite forms: it is cooled in a furnace, so it cools over a longer timespan. The red line shown above is also a general line—pearlite will form from any cooling path that intersects the TTT curve at any point.

52

Question 14: Based on your results, what can you conclude about the effectiveness of light optical metallography for verification of the microstructures associated with steel components used in engineering applications? If you did obtain some scanning electron micrographs, are they any better for this application? Explain. Refer to specific images of optical micrographs and SEM micrographs in your discussion. Based on my results, I can conclude that light optical metallography has limited effectiveness for viewing and verifying microstructures of 1045 steel. Figures 24-‐27 show that there is not much detail to be seen on different steel microstructures at the optical microscope’s highest magnification, 400x. The smaller structures, such as the lamellar formations in the pearlite, cannot even be seen in these images. In addition, the martensite grains are barely visible because the magnification is so low—it is fairly difficult to do any sort of in-‐depth analysis on these images. SEM images are much better for analyzing microstructures of our 1045 steel samples. Figures 57-‐60 show much more detail about the samples, such as grain size and shape in both samples, and the lamellar structure in the pearlite. The scanning electron microscope we used had the ability to magnify to 30,000x, which is 75x what the optical microscopes can view. This extra zooming capability reveals a wealth of information about the samples that is hidden from us in the optical images. We conclude that SEM images are vastly superior to optical images for verifying the microstructures of our steel components.

53

Question 15: Imagine that you have been called in as a consulting engineer on a failed brake rotor that has been be implicated by an insurance agent in an automobile crash causing the deaths of three people. The design specification was for the rotor to be forged from SAE 1045 steel, cross drilled and slotted, austenitized at 900°C, then quenched into iced brine. The report of the insurance company agent claimed that the rotor "overheated because of faulty heat dissipation, causing the rotor to reach temperatures in excess of 1000°C, resulting in catastrophic brake fade, preventing the driver from stopping before impact with the concrete embankment." The report also noted that there was no fire after the crash. You reason that if optical metallography can verify microstructures associated with intentional heat treatments, it should also identify "unintentional" heat treatments. So you decide to perform metallographic analyses of the rotor to evaluate the claim of the insurance agent. What would you do? On what microstructural evidence would you be able to prove the insurance agent correct? On what evidence would you be able to prove the agent wrong? Be specific.

Using optical metallography, we need to determine if the insurance agent’s claims were correct or not. When the brake was forged, it was austenitized at 900°C then quenched into iced brine—from this description, we know a martensite structure formed. If the brake rotor did indeed heat up to 1000°C, it will have formed pearlite structures—martensite only forms when 1045 steel is quench-‐cooled; if allowed to cool naturally only pearlite will form.

To prove the insurance agent correct, we would need to see evidence of

pearlite structures in the brake rotor. This would mean that the rotor did heat up to 1000°C and cool back down to room temperature. In order to obtain microstructural images of the rotor, we should first take samples from several different areas of the rotor. Next, we would follow the steps listed in the Methodology section to prepare and image steel samples. Once we obtained our images, we would look for lamella, which would show us that pearlite had formed. We could use the techniques employed in Question 11 to locate and verify lamellar structure. If any evidence of lamellar structure was found in any of the samples, the insurance agent was correct and the sample did heat up to 1000°C and cool down naturally.

To prove the insurance agent wrong, we would need to see evidence of only

martensite structure in the brake rotor. Using the same techniques listed above, we would obtain images of our samples and examine them. Thin, acicular grains and lack of lamellar structure would prove the agent wrong—if this was the case, the brake rotor did not heat up to 1000°C, and the martensite structure would remain the same as when the rotor was forged.

54

Conclusion

By examining the microscopic structures of steel and brass samples, we learned more about microscopy and gained a better understanding of the formation and composition of engineering materials like steel and brass. In addition, we learned several techniques for preparing, imaging, and analyzing our images. We employed both manual analyses—such as the linear intercept method and ASTM E112—and used programs like ImageJ to determine grain size and volume fraction of our samples. Through this lab, we developed the skills necessary to look on a microscopic level at basic engineering materials, and understand how different microscopic structures resulted in different overall characteristics.

e10-mse final report

Documents

c26000 alloy brass sample

brass samples

brass etchant

micron sheet

new nylon sheet

new scratches

grit sandpaper

grinding mount