Not quite sure how to calculate 0.2% proof stress of aluminium. The graph for which I need to read is below.
Yes, there definitely appears to be a settling-in regime. The x-intercept of the best fit line to the linear regime could serve as the 0% extension. Having translated the x-axis accordingly, the standard procedure of drawing the parallel with intercept at 0.2% should now work.
i have the instructions on how to work it out but makes little sense to me: Measure the horizontal axis in mm. How many mm extension shown on the horizontal line is represented by your actual measurement? Do a proportional calculation to work out what 1mm in reality equates to on the graph Calculate what 0.2% of 25.25 is in mm (5.05mm). Multiply that by the answer to your proportion calculation. Offset a parallel line from your line of best fit by that distance. Note where your new line intersects with the curve of your graph. Drawn a horizontal line from that point to intersect with the vertical axis. Read off the force also when you talk about setling in what are you talking about ?
There are 2 common reasons for seeing the kind of "toe" that you have in the very beginning of your curve: 1. In porous materials, the initial part of the deformation involves the collapsing of pores and cracks, 2. There is a response lag in the testing machine - this is known as settling in.
The proof stress is measured by drawing a line parallel to the elastic portion of the stress/strain curve at a specified strain, this strain being a percentage of the original gauge length. In your example 0.2% proof is desired. Your post above (#4) sounds like the correct approach based on your graph (other than the mathematical error previously pointed out). Your second graph with the best fit line in the proportional area should now be shifted/offset with the same orientation by the indicated amount. Where ever the new line intersects the graph, draw a horizontal line over to the vertical axis and that will give you the proof stress for 0.2% (or whatever value you used). I've attached a graphic example. CS
Doesn't look quite right, the shift looks too far. If your measurement was 1mm and the graph's horizontal axis is 4mm in length what is this ratio? It shouldn't be a large shift.
1mm is the measurement, 0.6 mm is where the line of best fit meets, so i calculated 1-0.6 which is 0.4mm and x that by 0.505 mm. That gives me 0.202mm so if i add this to 0.6mm that gives me 0.8mm hence i have drawn a horizontal line and read off the force, is that not correct?
You need to shift the parallel line you had orginally at the point it crosses the horizontal axis by your calculated amount. It's hard to tell from the graph since it is not very clear but the original line appears to cross around 0.1mm. If so, shift it by whatever your distance is, so if your distance is .2mm then just add 0.1 + 0.2 = 0.3mm. That is where your new line will cross the horizontal axis. Which is not where it crosses in your current picture.
when you talk about 0.1mm are you talking about when the line of best fit starts? and the distance 0.2 mm is that from caculating 0.505 mm by 0.4 mm?
Yes the 0.1mm is where the best fit line appears to start so it will have to be shifted from that point by whatever amount you have calculated.
i havve just worked out on the graph the horizontal graph measures 245 mm, the extension is 4 mm. to get 1 mm i divide 245 by 4 = 61.25 0.02 x 61.25 gets the offset which is 1.225 0.505 x 1.2225 = 0.62 so do i add 0.1 and 0.62 to get 0.72
After thinking about this a bit more, I'm not sure the original instructions are correct. Based on those you end up with this... If the length of the graph is 245mm and the extension is 4mm then the proportion would be 245 / 4 = x / 1 So like you have already, 1mm extension would be equal to 61.25mm. 0.2% of 25.25 = 0.002 x 25.25 = 0.0505mm Which gives 0.0505 x 61.25 = 3.093mm of a shift. This added to the original 0.1 crossing of your best fit line is about 3.2mm on the graph which seems way too far. Normally to find the proof stress (the stress required to produce a small specified amount of plastic deformation in the test piece) from a stress/strain graph you follow this approach... If the specimen length is 25.25mm then 0.002 (0.2% proof) of that is 0.0505mm which is the specified permanent deformation. So you would shift the parallel line of the best fit by 0.0505mm to the right at the horizontal axis crossing. Then, where this new line intersects the curve, you draw a horizontal line to the vertical axis and read off the stress. That stress you read off is the specified proof stress.
i see your point however i am basing my calculations on what one of my lecturers emailed me~:~ "If, for example, your horizontal axis shows 5.0mm of extension, and you measure this as being say 220mm, it follows that 1mm in reality is represented by 44mm on your axis (220 divided by 5) Now multiply 0.2% of gauge length by 44 to obtain the amount of offset "
I don't see the point in scaling the graph down to a smaller scale. I must be missing something. Sorry I couldn't help more. CS