Fine wire tensile strengths?
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From: JosephKK on 5 Aug 2010 08:16 On Fri, 30 Jul 2010 22:49:58 -0400, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >On Fri, 30 Jul 2010 19:01:37 -0700 (PDT), the renowned "Ken S. Tucker" ><dynamics(a)vianet.on.ca> wrote: > >>I've been asked to 'light-up' aircraft models, such as, >> >>http://www.flickr.com/photos/35156618(a)N03/4754110575/ >> >>It is suspended from a rope. >>I need to use very fine 120V wire (like magnetic wire), >>the model uses 8# test monofilament right now and weighs >>1/2 # , but a strong wind requires that 8# test. >>Is there a table that gives wire gauge & tensile strength? >>The current will likely be a max of 1/2 amp. >>Ken > >You can pick the material and look up the yield strength. > >In Imperial units, maybe steel wire (like music wire) you could use >300,000 PSI, say. > >So for something that would break at 20lbs you would need >7E-5 in^2, which is about 0.01" diameter, if I did the sums correctly. > >You should calculate and test the strength and self-heating effect >before using it, of course. > > >Best regards, >Spehro Pefhany OK. diameter for wire gauges found here: http://www.interfacebus.com/Copper_Wire_AWG_SIze.html Yeild strength for several alloys can be found here: http://www.copper.org/resources/properties/144_8/144_8.html Using 60,000 psi for yeild strength we want 30,000 or 40,000 psi for test strength. Thus to get 8 lb test we need 8/30,000 ~= 266 E-6 square inches. Solved for diameter D=sqrt(A/(2*pi)) for A=sqrt(266/6.28) about 6.5 mils. Thus from the table, 2 #39 enameled magnet wire lightly twisted ought to provide the mechanical strength needed and a complete circuit and handle 18 mA without trouble.
From: JosephKK on 5 Aug 2010 08:23 On Sat, 31 Jul 2010 07:40:51 -0700 (PDT), "Ken S. Tucker" <dynamics(a)vianet.on.ca> wrote: >On Jul 31, 12:52 am, "Tim Williams" <tmoran...(a)charter.net> wrote: >> "Joe" <j...(a)somewhere.org> wrote in messagenews:i30cnt$d59$1(a)news.eternal-september.org... >> > Numbers in the ARRL handbook chart of breaking load for copper >> > antenna wire work out to about 39000 psi for soft drawn copper >> > wire and 65000 psi for hard drawn so perhaps use 2 strands of >> > varnished #24 or #26 copper magnet wire. Those strength numbers >> > are higher than Tim mentioned so run deadweight tests before >> > testing with model. >> >> 65 KSI is close to the ultimate tensile strength I saw; I specified yield strength, which is a little lower of course. >> >> The yield of dead soft copper is astonishingly low, it's like shiny pink clay. Only that, after you push it around a bit, it hardens up, and pretty soon it's so damn strong you can't bend it by hand anymore! Copper provides a particularly dramatic illustration of work hardening. > >This is a Good thread Gents, good of yo'all to be interested. >To Recap, >The models will be hung ~5' , the spec is to use 120V 5W & 7W >xmas bulbs, although I can run a 3rd line for 12V sissy lamps. >(Gnd, 120V, 12V). > >Using this ref >http://en.wikipedia.org/wiki/American_wire_gauge >I find 24 AWG to be ok. I used a Yield Tensile Strength of >Copper ~10,000psi, calculated area and found it to be ~4#. >(24 AWG has a dia 0.02" ) >Two wires gives a strength of 8# for the Gnd & 120V, and the >ohm loss calculated for 10'. >Regards >Ken > > I got to much finer wire #39. But have a current limit (~18 mA) that nearly forces an all LED solution.
From: JosephKK on 5 Aug 2010 08:45 On Sun, 1 Aug 2010 06:03:23 -0700 (PDT), "Ken S. Tucker" <dynamics(a)vianet.on.ca> wrote: >On Jul 30, 7:01 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: >> I've been asked to 'light-up' aircraft models, such as, >> >> http://www.flickr.com/photos/35156618(a)N03/4754110575/ >> >> It is suspended from a rope. >> I need to use very fine 120V wire (like magnetic wire), >> the model uses 8# test monofilament right now and weighs >> 1/2 # , but a strong wind requires that 8# test. >> Is there a table that gives wire gauge & tensile strength? >> The current will likely be a max of 1/2 amp. >> Ken > >We put a a/c display into operation yesterday, as you guys >can see (compared to the top photo) the suspension wire is >too heavy. > >http://www.flickr.com/photos/35156618(a)N03/4848740731/ > >The wire is ~24 gauge gray stranded speaker wire, very safe >and strong but too visible compared to 8# monofilament fish >line. >Regards all and thanks for your interest. >Ken That's why i went for much finer wire, 2 #39 enameled, complete with a very low current limit ~18 mA.
From: JosephKK on 5 Aug 2010 08:51 On Mon, 2 Aug 2010 09:47:30 -0700 (PDT), "Ken S. Tucker" <dynamics(a)vianet.on.ca> wrote: >On Aug 2, 2:13 am, Uwe Hercksen <herck...(a)mew.uni-erlangen.de> wrote: >> Ken S. Tucker schrieb: >> >> > I need to use very fine 120V wire (like magnetic wire), >> > the model uses 8# test monofilament right now and weighs >> > 1/2 # , but a strong wind requires that 8# test. >> > Is there a table that gives wire gauge & tensile strength? >> > The current will likely be a max of 1/2 amp. >> >> Hello, >> you might need a combination of steel wire for tensile strength and >> copper wire for low resistance to the electric current. To keep the >> weight low, replace copper with aluminium. >> Bye > >Hi >We did an experiment yesterday, we connected a AWG 36 wire >in series with a 100W bulb, dimmer and Current meter. >(The AWG 36 wire was carefully wrapped around a couple of >3/16" bolts and carefully snugged with nuts, a poor man's fuse). > >As we turned up the dimmer we expected the wire to fuse but it >didn't, it took the full current supplying a 100W bulb, I kid you not, >that's about 0.9 amps! >We got the wire from unwrapping a reed switch and checked the >gauge twice by tensile strength. >I invite anyone else to try the experiment. >Regards >Ken Alas, the temperature rise on the wire is far more than i care for. I get over 200 C. That could be enough to cause problems with plastics.
From: Tim Williams on 5 Aug 2010 10:27
"JosephKK" <quiettechblue(a)yahoo.com> wrote in message news:mj9l56dc2in9dj0qvi33ksnc40oo46r379(a)4ax.com... > Solved for diameter D=sqrt(A/(2*pi)) Just pi. But since it's diameter, not radius, it's 2*sqrt(A/pi). Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |