To: atm{at}shore.net
From: MLThiebaux 
Reply-To: MLThiebaux 


>We are supposed to be engineers with the ability to calculate things, so let
>us stop gossipping and let us do our jobs.  Beam formulas are not at all
>complicated.   For a cantilevered beam, the deflection is given in the
>Handbook of Engineering Fundamentals as y=p*l^3 / 3*E*I.  Oddly enough, this
>is the same as in Roark's table 3 situation 1a.  Hello?
>
>What is weird about the Handbook is that the guys who wrote it seem to be
>from the century before the last one:  they specify units of P in pounds,
>for l in inches, for E in psi, and for I in inches^4.    I guess that means
>the formula won't work with any other self-consistent unit system.   I
>wonder how my finite element software can do it: they only require that the
>units be self-consistent, but they don't tell you what the system must be.
>
>. . . Richard
>


If you are confident a formula is correct (e.g., from a reliable source)
but are concerned about units then check for DIMENSIONAL consistency using
ANY set of units.

For example, in the above formula replace all lengths by "inch",
all forces by "pound", and check whether the dimensions on both
sides of the resulting equation of dimensions are the same.  If they are
then the formula will work without any alteration for any other units of
length and force (such as meters and newtons, or whatever you like). 
"inch" and "pound" can be multiplied, divided,
cancelled, etc just as if they were algebraic quantities.

Your formula translates to

        inch = pound*inch^3/[(pound/inch^2)*inch^4]

which you see is correct after you carry out the appropriate cancellations
on the right hand side.

Even more simply and probably more to the point: just replace all lengths
by L and all forces by F, and carry out the check. The use of
"pound" can be annoying because the same word is used for both
mass and force.  The distinction should be clear from the context.  In such
a case it would be better to replace pound-mass by M and pound-force by
ML/T^2 (remember Newton's second law; T stands for time).

It is still possible that the formula is correct, yet the dimensional check
fails.  This is because the source has failed to mention that some
numerical constant (such as the 3 in the denominator of your formula) is
not just a dimensionless ratio but actually carries some physical
dimensions.  You will see cases like this in some empirical formulas that
attempt to fit formulas to poorly understood phenomena.  The formula can
still be fixed up to apply to any set of units, even if you are totally
ignorant of underlying physical principles, but some work is required and
the result varies from one set of units to another.  If anybody wants to
see how I would be glad to explain in another note.

Martial Thiebaux
Rawdon Hills, Nova Scotia

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