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Converting some data from a tensile testing device

by jmarcian
Tags: converting, data, device, tensile, testing
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jmarcian
#1
Mar3-09, 07:43 PM
P: 20
ok the device used is called a “Instron” electromechanical testing machine. not sure if anyone is familiar with this, but that isnt the real point. ok, so an aluminum beam, dimensions:

"ext. gauge length: 4in"
"Spec. gauge length: 2in"
"thickness: 0.037in"
"width: 0.5in"

now, the machine starts to pull the beam apart, and the first reading comes out as "-0.955460" at a load of 0.99328lbs. so ~1lb force on a aluminum beam should cause basically no extension or deformation. the next several entries look like this

displacement load
-0.955440 0.9664
-0.955260 1.10066
-0.955110 2.09395

between these there are about 2700 entries... but here are the last few

displacement load
-0.493260 894.494
-0.49310 893.1520
-0.492770 889.93
-0.492770 887.78
beam breaks.......

ok, i am kind of confused on how to interpret this data. first off i need to convert the displacement in to stress.... what i have so far is that i take the original displacement, (-0.95540) and subtract it from itself, leaving a displacement of 0. makes sense right?
now i go down the list a subtract that same number from the corresponding displacement, so the last entries look like this...

0.4622
0.4624
0.4625
0.4625
0.4629

now, does this means just before the beam broke it was displaced, or extended 0.4629in??? does that make sense for a 4in aluminum beam to extend that much? need some elite materials guys for that one! haha. all kidding aside, to find the stress would i then divide that number by the original beam length, 4in?


any help would be appreciated, maybe some of you have dealt with this before!


thanks
jared
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nvn
#2
Mar3-09, 08:48 PM
Sci Advisor
HW Helper
P: 2,121
jmarcian: Everything you described sounds like you are doing everything correctly. You did not mention the exact aluminum alloy, temper, and form you are testing, but the results you obtained sound entirely possible and reasonable for some aluminum alloys. To obtain the engineering stress, divide the applied force by the bar original cross-sectional area.
jmarcian
#3
Mar3-09, 09:27 PM
P: 20
thanks nvn, i guess i need some reassurance. plus measuring the displacement from the negative standpoint through me off... i converted all the data to strain and stress and came up with a graph... see attached. looks acceptable to me. i was just didnt know that aluminum could stretch that much before failing! i just under a load of 919lbf i could heheh...

check out the graph and let me know what you think...

thanks for your help!
jared
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nvn
#4
Mar4-09, 01:57 PM
Sci Advisor
HW Helper
P: 2,121
Converting some data from a tensile testing device

jmarcian: Your curve looks great, but something currently seems incorrect. The initial, straight-line portion of your stress-strain curve is not steep enough; it probably should have a slope of about E = 69 GPa (10 000 ksi). But state the aluminum alloy and temper you are testing. Ensure your test specimen is connected very tightly in the testing machine, and not slipping. It appears your test specimen might be slipping or something, before eps = 0.02, but I am not sure yet what is wrong. Perhaps show an example of the formula you used to compute the x and y coordinates of your plot, including the units. Another thing you need to do is, draw a straight line through the initial, straight-line, almost-vertical portion of your stress-stress curve, and see where this line intersects the x axis, which currently appears to be approximately eps0 = 0.003. Then subtract eps0 from the x coordinate of all of your current data points, to zero your stress-strain curve, shifting it horizontally to the left, so that the initial, straight-line, almost-vertical portion of your stress-strain curve would intersect the x axis at eps = 0.
jmarcian
#5
Mar4-09, 05:25 PM
P: 20
hi nvn,

the aluminum is 6061-T6, does the T6 mean temper? im not sure.

as for the formulas, to compute stress i used stress=(load)/(cross-sectional area)
and to compute the strain i used strain=(displacement)/(orignial length, 4in). now class mates measured the thickness, width, and length of the beam so i am guessing imprecise measuring could in hand effect my graph.....


thanks
jared
timmay
#6
Mar5-09, 07:59 AM
P: 122
Are you able to illustrate your test setup with a photograph?

Typically, standardised tensile specimens (either dumbbell or dogbone) are tested. Care is taken with the dimensions so that the clamping conditions do not affect the area that the strain change is being measured from. The problem with taking the crosshead displacement (the output from the Instron's internal position sensor) is that it will also take into account any slippage or deformation at the clamping positions, where applied stresses are much higher due to stress concentrations. As a result you're probably not getting a particularly accurate measure of how much the centre of the section is deforming under that load.

A reduced thickness or diameter section in the middle of the sample is generally either clamped in an extensometer (typically an LVDT) or monitored by a camera system (known as a video extensometer) to measure the actual strain that the section in the middle is experiencing at the stress that is being applied.

Aluminium is a particularly ductile metal, so it's not surpising that it has behaved in the way it has. If you get a decent materials or metals textbook and look at tensile curves for brittle and ductile metals, you can see that you're certainly seeing the right behaviour. Alloying changes a variety of different material properties but generally won't have enough effect on elastic modulus to account for the low modulus you appear to be seeing. I would suspect it's as nvn suggested, that your setup is letting the sample slip or that there is additional deformation at the clamping points.
nvn
#7
Mar5-09, 08:48 AM
Sci Advisor
HW Helper
P: 2,121
jmarcian: Yes, T6 is the temper. Plot the raw applied load versus displacement, and use the procedure described at the end of post 4 to zero the plot horizontally. Post the plot, or even better, post your raw data (.txt) output file, zipped, if possible.

Why did you or your classmates type into the Instron "extensometer gauge length" field "4 in"? Did you actually use an extensometer? It must be removed when the strain reaches 0.05. What was the actual gauge length of your extensometer? Why did you or your classmates type into the Instron "specimen gauge length" field "2 in"? What was the actual, initial length of the narrow portion of your test specimen? And what was the initial distance between the Instron jaws when you installed your test specimen? What is the Instron machine model number? Did you also look at the contents of your .rep or .rlt output file? What does it say? Did you press the "GL Reset" (gauge length reset) button, after adjusting the gripper jaws to the correct gauge length distance, but before installing your test specimen?


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