Actually, strain is defined as the ratio of that linear measurement (deformation) to the initial dimension of the material, so strain is unitless. (To help answer Stefano's question, the stress/strain relationship is typically linear up until the point at which the material yields, which is to say it won't return to its original shape when the load is removed.) The classic example is 10" long strip of metal that is loaded in tension such that it gets 0.01" longer. 0.01"/10" = 0.001. That's your strain. Or 0.1% if you prefer.
I still haven't found time to look up a photo of Biba's chair, so I'm not 100% certain what we're talking about here, but as far as shear stress goes, the shear strain is typically defined as the ratio of tangential deformation divided by the thickness of the material. So if you have two parts joined by a 0.05" thick layer of glue, and the applied load causes the parts to shift 0.001" relative to each other, your shear strain is 0.001"/0.05" = 0.02 or 2%. Of course, if one of those parts is a relatively flexible rubber block, it will undergo much greater strain. If it's an inch thick, and under load the glued surface moves 0.1" relative to the other surface (viewed from the side, that's a rectangle being deformed into a parallelogram with 84 and 96 degree interior angles), your shear strain in the rubber block is 0.1"/1.0" = 0.1 or 10%. To relate shear strain to the more common tensile strain in my first example (and the associated published material property data) you need to imagine a cube-shaped element of the material tilted 45 degrees from the shear plane. That element will see a tension load when the actual part is loaded in shear, with tensile stress and strain roughly 1.4 times (divide by the sine of 45 degrees) greater than the calculated values for shear stress and strain. So if you were imagining a cube shape element of the rubber block 0.1" wide in each dimension, oriented 45 degrees relative to the block, the shear load that deformed the block into the aforementioned parallelogram shape will have stretched your imaginary 45 degree 0.1" cube element into a rectangular solid 0.114" x 0.093" x 0.093" for a tensile strain of (0.114" - 0.1")/0.1" = 0.14 or 14% equivalent tensile strain. Joe Elliott, PE > Date: Fri, 16 Dec 2011 23:03:21 +0000 > From: ron holcomb <[email protected]> > Subject: RE: [alfa] RE: Little to zip Alfa glue question - Stress, Strain, > Shear > > Stress/Strain relationships are dependent upon the material involved. > Stress > will not be equal to strain since the units are different. Stress = force > per > area (i.e. pounds per square inch). Strain is a linear measurement (i.e. > one > eighth of an inch). > > > > Date: Fri, 16 Dec 2011 08:31:11 -0800 > > Subject: [alfa] RE: Little to zip Alfa glue question - Stress, Strain, > Shear > > From: [email protected] > > To: [email protected] > > > > >So, when a Shear Stress is applied to the glued bond between the metal > and > the > > >wood, the Strain will probably be great enough to separate the block > from > the > > >frame of the chair. > > > > So will the value of Shear Stress and Strain be equal at the time of > separation? > > Or are they not related in this way? > > > > And thanks to all that answered me about my Speedo cable question! > > > > Stefano > > Concord, CA > > -- > > to be removed from alfa, see http://www.digest.net/bin/digest-subs.cgi > > or email "unsubscribe alfa" to [email protected] -- to be removed from alfa, see http://www.digest.net/bin/digest-subs.cgi or email "unsubscribe alfa" to [email protected]
