Thanks, Joe, for clearing that up.

@ Mr. Bayly:  Now, THAT is an engineer speaking!  The previous explanation
(mine) is merely an engineering geologist attempting to explain some physics
that he learned many years ago! ;-D

Ron Holcomb, CEG


> Date: Fri, 23 Dec 2011 11:31:16 -0500
> Subject: [alfa] RE: Little to zip Alfa glue question - Stress, Strain,
Shear
> From: [email protected]
> To: [email protected]
>
> 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
> > > --
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