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MDO and units

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MDO and units

Good morning MDO'ers

MDO has the Force/Torque feature (not a motor). If you are in Pro/E default
units, you are prompted to enter Torque in lbm*in^2/s^2. Given that I am
not a strict "unit-oligist" I am befuddled with the conundrum of lbm vs.
lbf. First instinct is to correct the numerical value by dividing by 386.
However, we know that lbm is numerically equal to lbf (because g/gc is ~1)
and inches are inches so maybe 1.0 lb-in of torque is exactly 1.0
lbm*in^2/s^2 of torque? I have searched for a simple verification problem
which would easily answer this, but alas most simple textbook problems are
in MKS...

Thanks in advance,

Gavin B. Rumble, PE (sad isn't it?)
Solid Engineering
336-224-2312
5 REPLIES 5

RE: MDO and units

Dear Gavin,

Don't use the ProE default units; you'll get yourself tied up in knots. Use
the IPS system (or an SI based system). All that is required then is to
divide the density values by 386.4 and everything else is as you would
expect; forces in lbf, torques in lbf-in, stress in psi etc.

Mechanica and MDO expect that the units system is consistent, that is the
equation F=ma doesn't require any conversion factors. This is not the case
for the in-lbm-s system, since one lbf accelerates one lbm 386.4 in/s^2.

BTW, isn't about time the US used metric units like the rest of the world
and saved yourself a whole load of grief?

Regards,

Rod Giles (a unit-ologist & SI evengelist!)
Elite Consulting Ltd.
Mechanica Specialists Since 1993

RE: MDO and units

I second that motion, Ron.

The best example of the worst mix up of units I've seen is a standard
for doing metric parts... really this is the system that comes up when
you execute the mapkey to start metric part.

To protect the innocent who have to re-arrange this every time they do a
metric part, I am not saying who does this.


Units info for the major system 'Custom'

Basic quantities:

Length mm
Mass lbm
Force mm lbm / sec^2
Time sec
Temperature F

Gravity 9806.65 mm / sec^2


Derived quantities:

Area mm^2
Volume mm^3
Velocity mm / sec
Acceleration mm / sec^2
Angular Velocity rad / sec
Angular Acceleration rad / sec^2
Frequency 1 / sec
Density lbm / mm^3
Torque/Moment mm^2 lbm / sec^2
Distributed Force lbm / sec^2
Distributed Moment mm lbm / sec^2
Areal Distr Force lbm / (mm sec^2)
Pressure lbm / (mm sec^2)
Stress lbm / (mm sec^2)
Young's Modulus lbm / (mm sec^2)
Surf Distr Moment lbm / sec^2
Transl Stiffness lbm / sec^2
Rot Stiffness mm^2 lbm / (sec^2 rad)
Thermal expansion 1 / F
Area Moment Inertia mm^4
Mass Moment Inertia mm^2 lbm
Energy mm^2 lbm / sec^2
Work mm^2 lbm / sec^2
Heat mm^2 lbm / sec^2
Power mm^2 lbm / sec^3
Heat Transfer Rate mm^2 lbm / sec^3
Temperature Gradient F / mm
Heat Flux lbm / sec^3
Heat Flux per Length mm lbm / sec^3
Thermal Conductivity mm lbm / (sec^3 F)
Convection Coeff lbm / (sec^3 F)
Specific Heat mm^2 / (sec^2 F)
Thermal Resultant Force lbm / (sec^2 F)
Thermal Resultant Moment mm lbm / (sec^2 F)
Beam Warping coefficient mm^6
Mass per Unit Length lbm / mm
Mass Moment Inertia per Length mm lbm
Damping Coefficient lbm / sec
Heat Rate per Length mm lbm / sec^3
Mass per Unit Area lbm / mm^2
Rotational Damping Coefficient mm^2 lbm / (sec rad)
Volume Heat Generation lbm / (mm sec^3)
Trans Stiffness per Unit Area lbm / (mm^2 sec^2)
Curvature 1 / mm
Gaussian Curvature 1 / mm^2
Linear Momentum mm lbm / sec
Angular Momentum mm^2 lbm rad / sec

RE: MDO and units

Hey, I once saw Kg-force listed somewhere. I was thinking "what the
@#$@#"

I third that motion... I college it was always easier to convert an
imperial problem to SI, solve it and convert it back then it was to just
use the imperial units.... Go figure.

Tony

RE: MDO and units

Good point Rod,

I normally do just that (switch to IPS) when preparing FEA models...but not
usually for design models. So here we are with a 450-part assembly (with
notes on the drawings saying WT = &mp_mass), and I find I need to predict
spool-up and coast-down times. So, what I'm hearing (from our helpful
Users), MDO wants torque to be 1.0 lbf-in X 386. The problem is, when I did
it wrong the answer seemed entirely plausible...when I do it right the
answer seems completely ridiculous (this is probably my warped sense of
reality).

Stay with me here...I have built a simple test part/assembly. The part is a
hollow cylinder 1" OD with 3/4" ID and 1 inch long made of steel (.3 lb per
cu in). It has an initial angular velocity of 10 rpm (60 deg/sec) in say +z
direction. Apply a torque of 1 lbf*in in the -z direction. How long (and #
of revs) before it stops, and ultimately heads back in the other direction?
I get a tiny fraction of a second (0.00005xxx seconds) and 0.0016 degrees.
This when I enter the torque as 386 lbm*in^2/s^2...AND when I do as one
friendly reader suggested, convert the whole thing to MKS. I would list my
conversions here but that would imply they are correct...grin.

Anybody get something different?

Regards,

Gavin

PS. On the topic of American usage of units, Rod's point is
dead-on-the-money...however, didn't we inherit this mess from our
motherland? Can you say Btu's...? And BGS?

RE: MDO and units

Hi Gavin,

My hand(well, Excel at least)-calcs agree with your results below.

Coincidentally, your test model seems to approximate quite closely a 540-sized 'modified' electric motor (used in R/C cars and planes), and the no-load spool-up time for one of those from 0 to, say, 30 000 rpm is pretty short (well below a second), so the numbers seem believable.

On the units debate, I'm English and really can't do engineering in Imperial units (even though I still think in some of them) - but I'm not going to argue too much, as my unit set of choice is N and mm, which unfortunately leaves mass as tonnes, but does at least give MPa directly.

Best regards,
Jonathan Hodgson