Forces for hydraulic driven radial gate 2

[Rockerdes]

I am designing a radial gate for coal discard bin that is going to be driven by a hydraulic cylider that is connect to a shaft arrangemnt. I am looking for the formulae (metric) to determing the force that needs to be exerted by the cylinder on the lever arm to rotate the radial gate 55 degrees. please find sketches attached and GA drgs of the bin

 

[desertfox]

hi Rockerdes

There is nothing in the attachment to see.

desertfox

 

[Rockerdes]

HI DESERT FOX
mY APOLOGIES PLEASE FIND ATTACHED THE SENARIO AS DESCIBED IN MY LAST THREAD

 

[desertfox]

Hi Rockerdes

Looking at the drawing I assume that the bin just rotates with the coal just sat on top, if that is the case you need only to overcome the friction between coal and drum so:-

50000 * coeff of friction = tangetial force on bin


tangential force on bin * 400mm (radius) = torque on bin

(torque on bin)/ (rad of drive link on cylinder)= cyl force

Now thats a very simplistic view, other things to consider is wind up in drive shaft, inertia of rotating parts, efficiency of system etc.
I have no idea from your post how fast this bin needs to move or whether it displaces coal which would all change the calculations.

Regards

desertfox

 

[hydtools]

Worst case will be to just open the bin. Coal will be at maximum weight and cylinder lever arm will be at minimum length. Once the bin begins to empty the weight of the coal will decrease and the cylinder lever arm will increase.

You may need to add some force margin to overcome the force of coal getting between the gate and bin clearance.

Limit the maximum force by setting the relief valve accordingly so no damage is done in the event of a serious gate jam.

Ted

 

[Rockerdes]

Thanks for the ideas, the bin is a fixed structure the radial gate is the moving parts, where could i possibly find a calculation to determine the forces required over the shaft lengh shown.

 

[desertfox]

hi Rockerdes

Change the word from "bin" to "gate" in my original post and the formula for the force is there.

desertfox

 

[Rockerdes]

Desert fox is the radius value in mm or m

 

[desertfox]

Hi Rockerdes

I have shown mm in my post, it doesn't matter so long as you use consistant units throughout.

desertfox

 

[hydtools]

Rockerdes,
You do not show a bearing supporting the shaft near the lever. There should be a support to react to the push/pull of the cylinder. Otherwise you will probably bend the shaft as you attempt to operate the gate. The force of the cylinder will act through a lever arm of approx. 2700mm creating a bending moment that the shaft and gate bearing must resist.

Ted

 

[Rockerdes]

Desert fox

The calculation you have shown i presume is for the torque directly below the load, I get 14000N/m torque my next question if my shaft point of force is approximately 2905mm from direct load applied will my shaft as i have calculated it increase from 125mm which is strong enough to support the load in torque and bending below the appplied load. If i apply a force at point b as shown in the sketches attached

 

[desertfox]

Hi Rockerdes

So to get to 14000Nm torque ( note 14000N/m in your post is incorrect)you used a coefficient of friction of 0.7 giving you a tangential force of 35000N.So stress on your 125mm diameter shaft is:-
tau = T*r/J

tau = 14000*0.0625m/(3.142*.125^4/32)

tau = 36501591.34N/m^2 or 36.5N/mm^2

so the stress in torsion is quite low however you need to check the torsional deflection over the 2905mm length.

angle of twist =32*T*L/(3.142*G*d^4) where G= modulus of
rigidity


32*14000*2.905/(3.142*80*10^9*.125^4)

angle of twist in radians is = 0.021207 which = 7.6 deg's

So your shaft will deflect 7.6 degs before any rotation of the gate takes place. I think thats quite a lot if you now connect your arm at B to your cylinder you will have
7.6 degs rotation of the arm before the gate moves.
I think you need to beef up the shaft to reduce deflection, also I don't know if the shaft you are proposing is solid or hollow.
One final point the shaft outside the bin which connects to the flange at A will have bending on it due to its own weight.

hope this helps

desertfox

 

[Rockerdes]

Desertfox

If i understand correctly my torque below the radial gate is correct(14000N/m)and my shaft below the is to small do you suggest i use a hollow shaft. What is the ideal(max angle of twist allowable).

 

[desertfox]

Hi Rockerdes

The way you write the units for torque is incorrect ie:-
torque = Nm not N/m.
You can use an hollow shaft to reduce the weight of shaft, however the diameter of the shaft at present in my opinion as to much deflection for example if your arm B is 400mm in radius then moving that via the cylinder it will move linearly by 52.9mm before it starts to rotate the gate, so almost 53mm of the cylinder stroke will be lost to shaft wind up.
What radius is the arm at point B? Oh its just trig to work this out so SIN 7.6 deg * radius arm B

desertfox

desertfox

 

[Rockerdes]

Desertfox

This type of shaft if i am not mistaken is classified as a line shaft which states that a max twist is 3deg/m or is it considered a machine tool shaft which has a max twist of 0.3deg/m.

 

[hydtools]

Rockerdes, I think you need to decide what you want the shaft to be. You need to decide what characteristics you want the shaft to have. Certainly, the shaft types you mentioned can be used as guides but you as the designer need to make the final decision on the performance of the shaft.

Ted

 

[desertfox]

Hi Rockerdes

The shaft deflection depends on two things:-

1/ Stroke of cylinder

2/ length of lever B driven by cylinder

If you allow the angular twist to be 3degs/m then the 7.6 deg calculated for your shaft is almost correct but your cylinder will need a stroke to go through the 55 degs operation of the gate plus the 7.6 degs that the shaft deflects so without allowing for any losses in terms of friction etc the operating angle for your gate 62.6 degrees, now if lever B as a radius arm of 400mm then the stroke of the cylinder would be:-

SIN 62.6 * 400mm = 355.126 mm


however if the shaft deflection is only 1 degree in total cylinder stroke is:-

SIN 56 * 400mm = 331.615mm

desertfox

 

[hydtools]

Also consider that when the gate unloads, the shaft will 'unload' and spring the gate forward. Limiting the windup will limit or eliminate the possibly sudden springing forward of the gate and shock load back to the cylinder and hydraulic system. Sudden shock to the system is not desirable.

Ted