<!doctype html public "-//W3C//DTD W3 HTML//EN"> <html><head><style type="text/css"></style><title>Re: Youngs Modulus for Piano wire</title></head><body> <div>At 9:48 pm +0200 27/8/06, Ric Brekne wrote:</div> <div><br></div> <blockquote type="cite" cite>Well, I had rather hoped to be able to avoid stretching a few samples of wire across the room and hanging weights to measure deflections and then figure the relevant numbers out... At least at this point.</blockquote> <div><br></div> <div>You mean you want someone else to do it for you :-)   Well here's a simple way to do it.  Hang a wire (preferably straightened) from as high a point as you have available, say the upstairs window-sill.  Rig up a rigid and immovable whatnot (technical term) as below so that the wire just touches the crossbar.  Attach a weight to the wire.  Tape a card to the crossbar and another card to the wire overlapping the first card. Carefully push a needle through the top card into the card behind it.  Remove the weight and replace it with just enough weight to hold the wire straight (just pull it tight).  Push the needle through the top card at the same point so that you now have two needle pricks in the back card.  Measure the wire from A to B.  Measure the distance between the two pin pricks.  End of practical.  Tidy up.</div> <div align="center"><img src="cid:p06240639c117ab3ee8ab@[10.0.0.1].1.0"></div> <div><br></div> <blockquote type="cite" cite>I just need a reliable ball park figure in order to proceed with my project.</blockquote> <div><br></div> <div>31,000 lbs per sq. mm.</div> <div><br></div> <div>Youngs Modulus is the force that would be required to stretch the wire to double its length if it were to continue obeying Hooke's law -- in other words if it were infinitely elastic.  Hooke's law states that if you pull a wire with F and it stretches L, then a force of 2F will stretch it 2L, 3F 3L etc.  In other words there is a linear relationship between the force and the extension.</div> <div><br></div> <div>Here are the data and the result in a Perl script, which is easy to understand even if you've never heard of Perl. </div> <div><br></div> <div><br></div> <div>#!/usr/bin/perl<br> $wire_number = 15;<br> $wire_diameter = 0.875; #mm.<br> $pi = 3.1415926;<br> $xsection_area = $wire_diameter**2 * $pi / 4;<br> $free_length = 2420; #mm.<br> $load = 42; #lbs.<br> $extension = 5.5; #mm.<br> $E = 1/$xsection_area * $free_length / $extension * $load;<br> printf "%0.0f\n", $E;</div> <div># --=> 30732</div> <div><br></div> <div>I used a No. 15 wire (not Ršslau in this case) at the end of which I made an English eye and passed the wire through a hole is a brass plate so that it would hang from the finishing coils of the eye and nothing would stretch at that point.  When the work was done I measured on the bench from the pin prick in the card (B) to the hanging point (A) to get 2,420 mm.  The distance between the two pin pricks in the back card was 5.5 mm.  If 42 lbs. stretched the wire 5.5mm. then to stretch it to double its length I would need 18,480 lbs and therefore to stretch a wire of 1 sq. mm. cross-section I would need 30,732 lbs. or, to be fussy, allowing for a generous error of 0.1mm in my measurement of the pin-pricks the figure will be somewhere between 30,184 and 31,301lbs.</div> <div><br></div> <div>ERGO, 31,000 lbs is the proper choice in this case.  There are likely to be variations according to the gauge number but I think the figure is more likely to go down than up as the wire gets thicker.  To test the fatter wires with any reasonable accuracy I would need a much higher house</div> <div><br></div> <div>JD</div> <div><br></div> <div><br></div> <div><br></div> <div><br></div> <div><br></div> </body> </html>