Stability of a skid

[Rajder]

Guys & Gals, I've got a question that I'm hoping someone here can point me in the right direction. I'm in the process of designing a skid that stores a large amount of cable and has the ability to rotate the cable drum in a CW and CCW direction. This skid happens to be constrained to a pretty small base but has a large diameter cable drum. So the cable drum is physically bigger than the base of the skid. I want to calculate the stability of the skid and specifically how high would the forces acting upon the skid have to be in order to cause the skid to tip over.

We have two scenarios for this skid that I want to evaluate. The first is that during operation we have to rotate this heavy cable drum and stop it from rotating. We rotate this cable drum at 15rpm and we utilize hydraulic braking through our gearbox to stop the rotating cable drum. I know through testing that our braking system applies a maximum torque of 136,000 in-lbs in the opposite direction of rotation while stopping the cable drum. This torque is applied along the centerline of the cable drum and is transferred through the skid. I want to be able to verify that machine is going to stay stable and not want to tip over while stopping the cable drum from rotating. I would also like to calculate what the maximum allowable torque along the centerline of the cable drum could be before the machine will tip over.

The other situation revolves around the problem that this skid is a mobile unit and will have to be transported via truck. During transportation we assume a maximum horizontal load of 40% of the total weight of the machine. The total weight of the machine 33,500 lbs. I would typically calculate stability by just finding the CG and making sure the force vector from the CG stays within the base of the machine. But since we are applying a horizontal load of 13,400 lbs in this case it's got me a little confused. If we apply a horizontal load to the machine I'm assuming we would apply it at the center of gravity of the machine. But applying the load at the center of a symmetrical machine doesn't end up moving the center of gravity away from the centerline of the machine and the CG doesn't move in the X direction.

My initial thought was to use the load cases to calculate a torque that acts upon the base of the machine. Assuming the torque acts along the centerline of the machine it would create a downward force on one side of the machine base and an upwards force on the other side of the machine base. Then as long as those forces didn't exceed the forces caused by the weight of the machine it would remain stable. Can someone help me out with this and point me in the right direction? Thanks!

 

[gruntguru]

Scenario 1. Draw a FBD of the skid at the point of tipping (All weight carried at one edge). This normal force and the mg acting vertically at the CG form a couple which resists your drum torque.

Scenario 2. The 13,400 lbs is applied at the bottom of the skid and reacted by an equal and opposite "inertial" force acting at the CG. These two form a couple which is trying to topple the skid.

je suis charlie

 

[MikeHalloran]

Do make sure to include a bunch of hole-type features in the skid so that it can be chained down,
in transport and in use.
... and some padeyes above the drum bearings so the assembly can be lifted safely.



Mike Halloran
Pembroke Pines, FL, USA

 

[Rajder]

I have already added lifting lugs above the CG of the machine to facilitate balanced lifting. Thanks for the looking out for me though Mike.

Gruntguru, just so I am understanding what you are saying correctly. In scenario 1 we would have the mg (the weight of the machine) acting downward on the CG and a normal force Fn acting upon only one edge of the machine. In this scenario Fn should equal mg which would be 33,500 lbs. Those two forces create a moment acting upon the machine and any moment force larger than that moment would cause the machine to become unstable and tip over. In my attached FBD I calculate a moment of 1,046,875 in-lbs so as long as any moment applied to the machine is less than that it would remain stable?

In scenario 2 we basically calculate the moment created by the 13,400 lb side load on the machine (at the CG) which would be 737,000 in-lbs. And since that moment is less than the 1,046,875 in-lbs previously calculated the machine would be stable then as well?

I've attached my quick FBD and hand calcs just to verify. This seems off to me simply because I would assume that the 40% side load on this particular machine would cause the machine to tip over if not properly restrained.

Does my calcs look correct? Thanks guys!

 

[zeusfaber]

Scenario 2, I'd not be comfortable with the assumption of maximum horizontal loads of just 40% of the weight of the machine. European practice these days is to restrain to 0.8g forwards (against forces from heavy braking), 0.5g towards the rear and 0.5g to either side (or 0.6g if there's any risk of the load tipping). There's a useful guide here.

Another way of looking at scenario 2 is to take the vector sum of the horizontal and gravitational forces, project those through the cg of the load and see if the resultant passes beyond the edge of the skid.

A.

 

[desertfox]

Hi Rajder

If the mg force is 33,500 lbs but it cannot act at one edge as you have assumed, mg acts at the centre of gravity therefore half of that figure acts on each side of the drum.
In case 1 if a vertical load of 16750lbs is applied at one corner as shown in your sketch then in theory the drum and frame will start to lift.

In case 2, a force of 19034.1 lbs acting at the horizontal position of fx would again start the frame and drum to tip.

I'm not sure why case 1 is being considered because I cannot see where a vertical force would manifest from.

Case 2 is a possibility but what about stability of the frame with a horizontal force at the drum centre acting at ninety degrees to the current fx?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[dhengr]

Rajder:
Think of these problems as righting moments (Mrtg.) vs. overturning moments (Motg.). You should also show another view of your cable reel system, at right angles to the view you’ve shown. You will likely have overturning moments about that orientation too. Also, it is likely that you could sill have some overturning issues when the system is almost empty of the cable loading, and thus the righting moments will be much smaller. You have to consider all of these conditions, not just the one you’ve shown. You can actually do this problem graphically if you do your drawings to a true/accurate scale, and then show the vert. load vector through the CG and pointing downward to some scale; then show the horiz. load vector from the lower end of the vert. load vector and pointing on to the right, and to the same scale (in lbs./unit length); then the vector connecting the start of this load diagram, at the CG, to the right end of the horiz. vector is the resultant load vector, and it can be scaled fairly accurately for a resultant load value. And, if its extended direction stays within your reaction point/line the system is stable. What does the cable reel weigh when its empty? What does the skid and its hydraulic equip. weigh and where is its CG? Given the trouble you are having with this stability problem, which is a pretty basic first year engineering problem, you may want to get someone to review the strength and stability of the skid structure itself. Zeusfaber’s link looks like it would really make good reading for you. It isn’t clear where your lateral load comes from, and you better determine those down properly. Then, this is probably the type of equipment which should be tied down by some means during use and transport.

 

[gruntguru]

Rajder. Your FBD's are correct.
desertfox. At the onset of toppling (which is the criterion under investigation) the entire weight is taken on one edge.

je suis charlie

 

[desertfox]

I would prefer to limit any forces and the resulting moments to that which just balance the the moment about the frame edge due to its mass and centre of gravity and that way we err on the safe side.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Rajder]

Thanks for the help on this guys. And thanks for the link Zeusfaber, that will be helpful. I was initially approaching the problem as desertfox has described. The reason why I was posted here was to get other opinions and I thank you guys for that. I didn't think of doing this problem graphically with the force vectors, I was only thinking about analyzing the overturning moments.

dhenger, don't worry the skid will obviously be chained down during transport. And it is also bolted to the floor during use. What I've put on here is obviously a very simplified version of the skid. I designed and built several versions of this type of equipment. It's just that I haven't looked at anything related to skid stability since college (which has been years ago) and wanted some other opinions on how they would approach the problem. This is a piece of equipment with a large rotating mass that has to be stopped relatively quickly, I really just wanted to calculate the point in which the moment created by stopping the rotating drum that would cause the unit to overturn and make sure we were within that figure (even though the machine is bolted down during use).

Before I posted here, I initially approached the problem as desertfox described. Then gruntguru gave me a different approach. I'll also quickly do a graphical force vector approach as well and take the most conservative case. Either way, thanks for the help guys. I've got what I need! Thanks!

 

[desertfox]

Hi Rajder

Taking the mass all on one edge is not correct, the only way the vertical force could equal mg is when the c of g is vertically in line with the edge. Now at this point the mg force would be passing through the pivot point and therefore produces zero moment to prevent tipping, in fact if you reach this point one could hit it with a feather and it would topple.





“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[gruntguru]

"the only way the vertical force could equal mg is when the c of g is vertically in line with the edge"

Have another look at the FBD. During cornering there is a large centrifugal force trying to roll the skid over. At the toppling limit, the entire weight (mg) is taken on one edge.

je suis charlie

 

[chicopee]

The second diagram (FB) is not adequate. You should have included a top view showing the effects of the cable being pulled from the sides of the reel as the cable is being unfurled and the position of the remaining cable on the reel moves back and forth on the reel. From side to side motion on the reel, there would be two resisting shear forces perpendicular to each other and two overturning moments perpendicular to each other. Also not shown on your second sketch, in the unwinding of the cable over the top of or under the reel which affect the overturning moment? In addition you have not taken into consideration the journal coefficient of friction which would increase the load line. You have taken into consideration the coefficient of friction between skid and ground if you wanted to have the skid unanchored which is proper. The last item deals with starting up the unwinding process when the reel is full and partially full which then you need to take into consideration angular acceleration and rotational inertias to determine the torques (or resisting moments) that would increase line loading.
zeusfaber reference to the European practice obviates the need to consider the information provided above but if you are a purist then indulge.