Recognitions:
Gold Member

## calculating 0.2% proof stress of aluminium

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 "

Recognitions:
Gold Member
 Quote by cmgames 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

 i thought the graph should be stress-strain graph? why is it force-elongation graph? is it the same/?

 Quote by dewking i thought the graph should be stress-strain graph? why is it force-elongation graph? is it the same/?
Stress is Force/Original Area
Strain is Elongation/Original test piece Length

The two graphs are not the same, but the shapes are identical, as the original area and original length are constant.

But that is a good point, why isn't your graph a Stress-Strain one?