Recent content by cmgames

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    Calculating 0.2% proof stress of aluminium

    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)...
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    Calculating 0.2% proof stress of aluminium

    i just rechecked, the graph measures 24.5 cm
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    Calculating 0.2% proof stress of aluminium

    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
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    Calculating 0.2% proof stress of aluminium

    i have done this now, is this correct?
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    Calculating 0.2% proof stress of aluminium

    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?
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    Calculating 0.2% proof stress of aluminium

    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?
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    Calculating 0.2% proof stress of aluminium

    hope i done it correctly, this is the graph i have now
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    Calculating 0.2% proof stress of aluminium

    thank you for that, i am however still confused as to how to work out the calculation
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    Calculating 0.2% proof stress of aluminium

    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...
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    Calculating 0.2% proof stress of aluminium

    Not quite sure how to calculate 0.2% proof stress of aluminium. The graph for which I need to read is below.
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    Measure the tensile strength using Young’s Modulus

    From my results i have found that wood and brass is not inherently stiff. What examples are there of using wood and brass, that by changing it's shape or distribution would make it stiffer?
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    Measure the tensile strength using Young’s Modulus

    Hi there, I am currently looking to measure the tensile strength using Young’s Modulus for steel and brass. My results I have obtained are comparable with published values of E. My question is regarding the formulae, the one I used was: E = F/x X l/a Where a is the original cross section...
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