Re: Trouble with "flexure of beams" apparatus

[brian whatcott <inet@INTELLISYS.NET>]

At 16:41 12/11/00 -0600, Fred Lemmerhirt wrote:
> The apparatus instructions provide a formula:  (deflection) = [(Load)(beam
> length cubed)] / [4(beam width)(beam height cubed)(Young's modulus)].
> Graphing deflection vs load (for constant length) or deflection vs beam
> length cubed (for constant load) should then give a straight line from which
> Young's modulus is calculated as a factor of the slope.
> ______________________________________
> Fred Lemmerhirt

>                 /snip/
>        A simple beam with pinned supports and a fixed central point load of
> 1000 units       100 units long, deflection 0.718    /snip/


Let's call 'beamlength' span, to ease the writing.
It is given that  deflection = load (span cubed) / [4 W (Hcubed) E]
 for W beam width, H beam height, E Young's modulus.

Elsewhere *, one can read that max deflection = load (span cubed)/48EI
where E is Young's modulus, as before, and I is the second moment of the
particular beam cross section (which I don't yet know).

Assuming first, that both expressions are correct, we equate the
expressions for deflection to see that
4 W Hcubed = 48 I, and so I = W Hcubed/12

This matches the tabulated expression for a solid rectangular
cross section.

Now it happens, that the code example that I specified was a steel
square cross section of 12 inches on a side, and 100 ft long
(with no consideration of self weight) and a central load of 1000 lbs,
in the US customary units.

Let's see how this works out:

deflection = load (span cubed) / [4 W (Hcubed) E]


Plugging in my values:
0.718 in. = 1000 lb (100 ft  x 12 in/ft) cubed / 4  E 12 in^4
so E = 29E6 psi and indeed this is the default structural steel
modulus value in US customary units.

I conclude that the step that goes astray is the one that uses
the slope of the various deflections with load or span length.

Reviewing the meaning of the 'slope of the line' for various loads
 on various spans, we arrive at a meaning of
deflection/load in the one case, and in the other case
deflection/span cubed. We could possibly refine these two lines
 by ensuring the mean squared errors are minimized.
 Then we use
Slope = Deflection/load = span cubed/[4 W Hcubed E]
or
Slope = Deflection/span cubed = load/[4 W Hcubed E]

These two expressions should both give a reasonable value for E
as a factor of the inverse slope, with probably more error
in the varable span method, where length errors are cubed.
I wonder what experimental values were found in fact?

* Marks' Handbook, 'Beams' chapter.





brian whatcott <inet@intellisys.net>  Altus OK
                Eureka!