On Thu, 23 Feb 2012 20:11:33 -0500, "Jim Wilkins"
wrote:


"Ned Simmons" wrote in message
.. .
On Thu, 23 Feb 2012 19:16:48 -0500, Ned Simmons
wrote:

On Thu, 23 Feb 2012 17:48:49 -0500, "Jim Wilkins"
wrote:...
If Young's Modulus is 30E6 and the load is 30E3, what's the
elongation?
jsw

Depends on the cross sectional area and length of the member. If,
for
example, the load is 30E3 lb on 1 in^2 of steel (Y=30E6 lb/in^2):
stress = Force/area = 30E3 lb / in^2
strain = stress/Young's modulus = 30E3 lb/in^2 / 30E6 lb/in^2 = 1E-3
in/in

Over a 10 inch length the change would be 10 in x 1E-3 in/in = .01
in

In other words, strain is a ratio that represents the change in
length
(length/length) of a body under stress.
Ned Simmons

And thus my millimeter per meter, which I think is too little to
measure accurately on a twisted wire cable with a catenary droop. It
might work with tie rods salvaged from a defunct hydraulic cylinder,
if you had a test rod that fit the gap between two collar clamps and
you measured the end clearance under load with feeler gauges.
http://image.made-in-china.com/2f0j0...ed-Eye-Nut.jpg

jsw


I agree with your conclusion, but the .2% number you used to get there
has nothing to do with elastic deformation, and it's only a
coincidence that our numbers are close. I could just as easily have
used .1 in^2 as the area in my example, and the resulting strain would
be 10mm per meter.

I suspect the practical limitations of tensile testing machines 100
years ago is why the transition from elastic to plastic was designated
at .2% elongation. In other words, that's what could be reliably
detected.

--
Ned Simmons