Calculate the force exerted on a floor (unit issue)

[dpmecheng]

I am trying to calculate the pressure a machine exerts on a floor. My 3D cad reports the mass as 10,000 lbmass. The area of all the feet is 100 inches square. The floor mfg. reports the maximum allowable is 3,000 psi. Assume the load is evenly distributed over the feet.

Is the pressure exerted 10,000 lbmass / 100 inches square = 100 psi

Or do I need to multiply the 10,000 lbmass by 32.2 ft/sec^2 to get pounds force. Therefore 10,000 lbmass (32) / 100 inches square = 3,200 psi.

 

[MiketheEngineer]

My guess if it is in the US is the 100 psi

 

[hydtools]

In Imperial units 1 lb-mass = 1 lb-force, on Earth. Gotta love cad systems and the units used.

Ted

 

[Cockroach]

Yeah, typical uncertainty. As mentioned by HyTools, pounds mass is pounds force. The imperial system uses slugs to measure mass, not pounds. So pounds, whether it be mass or weight, is a force.

Regards,
Cockroach

 

[Twoballcane]

Well let’s get to the units. Lbm is different from Lbf on other planets and it depends on the gravitational constant. For earth, it is conveniently gc=32 (lbm s^2)/(lbf ft) thus old Newton equation will be lbf =(lbm g)/gc. Thus on earth it will be Lbm = Lbf

Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”
"People get promoted when they provide value and when they build great relationships"

 

[imcjoek]

The imperial system uses slugs to measure mass, not pounds. So pounds, whether it be mass or weight, is a force.

lbf = lbm * g / gc
N = kg * g / gc

In the first case g and gc happen to be numerically equal.
In SI gc happens to be 1, but it's still there, if only implied.
Neither is more complicated/correct/whatever than the other.

Slugs seem to be an anachronism in my experience. Never see the buggers.

 

[Cockroach]

I'm not sure where in left field you are standing, TwoBallCane, but I presume DPMechEng is somewhere on this Earth, not some other extra terrestial location. Just an assumption on my part, perhaps I'm wrong.

I also don't suscribe to your interpretation of "gravitational constant", you fail to reference the curvature of space-time as a continuum. Your statements are awfully Newtonian in nature. This would be important in terms of the general relativistic theory, the special counterpart refers to bodies approaching that velocity in the vicinity of light and the tensors generated in the mathematics thereof. Again, I further assume DPMechEng is not entering regions of increasing continuum curvature, frequency shifting due to gravity becoming important.

Just saying....but I digress....

Regards,
Cockroach

 

[chicopee]

What you have not stated is if there is any inertial force from the machine that could add additional pressure to the flooring

 

[chicopee]

Well looking at the responders I see only partial info. since units are omitted from several replies such as:
F=Lbf
m=Lbm
a=ft/sec^2
gc=32.2 bm-ft/lbf-sec^2

 

[chicopee]

Oops! Unit for gc=32.2 Lbm-ft/lbf-sec^2

 

[Twoballcane]

Well two things. My apologies that I had my units wrong where gc should have been lbm ft / lbf s^2 and had lbm and lbf reversed in my statement where as lbm is constant and lbf is different. I quickly wrote this post before going to a meeting.

Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”
"People get promoted when they provide value and when they build great relationships"