Shear Force 3

[windseaker1]

Help

Im looking for the shear force on any given object at 4' depth on the inside wall of a hopper of cement (density=90lb/ft3)? This seams like a dam problem??

 

[GregLocock]

"This seams like a dam problem" It may be a tricky problem but there's no need to swear.

I don't think there is a shear force.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[MintJulep]

It's a bin problem. Quite different from a dam problem.

 

[btrueblood]

To add to Greg's answer, "if the concrete is not in motion, then there will be no shear stresses". The object will see a hydrostatic pressure on all its surfaces, due to being submerged in a fluid (concrete), P = rho*g*h.

 

[MintJulep]

I had assumed that the cement was in it's dry powder form.

Solid material and bins and liquids in tanks behave differently.

 

[btrueblood]

Mint,

Yes, but I said "fluid", not liquid...but you're right, the equation would not be valid for dry powder.

 

[windseaker1]

OK, I think
The cement is not flowing , now.
And if dry power was flowing, just subsitute density values!

I have the engineer hat on now (there goes some variables)
At 90lb/ft3 and at 4', it would be a rough 360lb/ft3.
Now just a down force would be 360lb on a "surface"area of 1 sq ft., right?
If any object of say 1"x1"1" is present, them just convert from ft to in??

 

[MintJulep]

Wrong on just about all counts.

 

[Co]

Start with the equation btrueblood gave you. It will start you off on the right foot.

cheers,

 

[GregLocock]

So hang on, are you talking about a dry powder, bridging, or a wet slurry, hydrostatic forces, as in a dam?




Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[windseaker1]

Sorry everyone, I am talking about the forces that would be applied to an object in a ,say4'X4'X4' bin of cement dry power. The object is 2 or 3ft deep on the inside wall of the bin. The object could be a sensor or just a 2" round rod.

Btrueblood, is your equation good if there is a flow, 5cuyd/hr.??

 

[windseaker1]

Static, no motion.
P = (rho)x(g)x(h)
P = (p)x(g)x(h)
P = (90lb/ft3)x(32ft/sec)x(4ft)
P = 11,520lb of force on a object at 4' deep ??

If, and if this is right would it also be the shearing force equlivent on an object at that depth??

 

[btrueblood]

Well, no, the equation I gave is valid for static (non-flowing) fluids, where the only forces between particles are normal, and thus only hydrostatic pressure exists.

For dry powder, the problem gets stickier (pun intended), as Mint suggests. The powder can stick and clump, and even when dry it has an angle of repose...so voids may form below protrusions as the powder fills the hopper. Worst case is the full hydrostatic pressure of the fluid on one (typically the top) side of a protrusion, and zero pressure (void space) beneath. I think.

 

[FeX32]

How is the fluid flowing in the bin?

Fe

 

[FeX32]

windseaker1,
that is not a shearing force on an object under hydrostatic pressure.
Pressure_hydrostatic=rho*g*h (N/m^2 or lb/ft^2)
(as stated by the others)

Fe

 

[btrueblood]

" I think. " Which should give you great pause. A worser worst case would be during filling, when you have "dynamic forces" acting on the protrusion. These forces would depend on the height from which the powder is falling as it fills the hopper...as well as the density of the powder, and whether it has had time and moisture exposure to form "chunks"...

I did not look closely at your equations before. Even if you had a fluid, which you don't, the fluid equation gives a pressure

P = rho*g*h

So, you have P = (1440 kg/m3)*9.81 m/s2* 1.2 m = 17.2 kPa

Which you can convert back to psi at your leisure. There's a lesson to learn here: always convert to SI units when doing physics. Life is just easier that way.

To find the force acting on an object at that depth, you need to know the projected area of the object...

P*A = F (again, use the SI units and convert the force when you are done).

 

[GregLocock]

Well in the worst worst case then entire weight of the powder in the bin could be reacted through the obstruction. Or none of it could.

http://www.particles.org.uk/particle_technology_book/chapter_10.pdf

is at least a step forward

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[swearingen]

The benefit (in this case) of working with cement powder, is that it truly acts like a fluid when aerated. If the incoming stream is impinging directly on the protrusion, a simple equation giving force of a fluid mass flow would get you in the ball park.

If the flow is not falling on the object, and given the bin's size and that the cement is aerated, the forces would be only based on buoyancy, as they would be if you slowly filled the bin with water.

When the bin empties, you may get some build-up on the top of the sensor which you would need to account for.

Either way, a 2" diameter piece of pipe, 2' to 3' long should have no problems if properly secured at the fixed end.


If you "heard" it on the internet, it's guilty until proven innocent. - DCS

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[windseaker1]

The dry power is blown in from one corner of the top and trough out at bottom to tube.

This sensor could be a paddle wheel type and need shearing forces on it, to see if it will not snap off.

Here is bin design, sensor or object is at 4ft depth. say that the object is 3"X2"X1".
what would be shear forces on it?

 

[PeterCharles]

For the storage of bulk solids in hoppers, bins, bunkers, silos forget about formulae for fluids. Search out the work of Andrew Jenike the definitive source of knowledge on this subject!

If you are involved in handling and storage of bulk solids check out these specialist forums :-

http://www.bulk-online.com/Forum/index.php