To: "Richard Schwartz" 
From: MLThiebaux 
Cc: atm{at}shore.net
Reply-To: MLThiebaux 


Richard --
Just one more little thing (obvious I hope) regarding units.  I should have
mentioned that the author of a formula should specify CONSISTENT units in
order that ANY set of units applies.  A formula for example might specify a
length to be in feet and a diameter to be in inches.  Or power in
horsepower and stress in psi.  Mixing units this way is a common source of
inconsistency.  The formula should be fixed up for consistency in units AND
checked for dimensional consistency before it can be said to apply to any
set of units.

Martial Thiebaux
Rawdon Hills, Nova Scotia

>
>>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.
>

--- BBBS/NT v4.00 MP