Sliding axle tension system

[minisimon]

In addition to using large screws, you can increase the width of your hexagonal puller so that it almost touches the two sliders. Such a configuration won't attempt to bend the screw; only shear it, and the screw should be able to handle that better. You probably won't be able to use a simple hex standoff for the puller any more, but you seem to be up for some machining.

[joshy1323]

Originally Posted by minisimon

In addition to using large screws, you can increase the width of your hexagonal puller so that it almost touches the two sliders. Such a configuration won't attempt to bend the screw; only shear it, and the screw should be able to handle that better. You probably won't be able to use a simple hex standoff for the puller any more, but you seem to be up for some machining.

i agree completely

Originally Posted by ZeroValue

how much should the "lips" go above and below the pocket in the aluminum to keep it from twisting?

that depends on what you are using it for. what kind of force do you plan on putting on it that would make it twist?

It would be good to increase the screw size, but it's not necessary. Keep in mind, this is only my opinion. If you're tightening the chain enough to bend a screw, your chain is probably too tight. I'd suggest using a #10 screw, then you should be fine. Nice design.

[MrForbes]

I agree that it should not bend a #8 screw if everyting is just right. Unfortunately our friend Murphy makes things go wrong...and making the design less vulnerable to mistakes like overtightening the tensioners, seems to me to be good design practice.

[AdamHeard]

Increasing the screw size increases the mechanical advantage which will help it stay in tension better, and will obviously have stronger threads. Stronger threads makes them less likely to wear out.

I'd just suggest 1/4-20 or 1/4-28, I'd also buy a solid rod end or clevis rod end and use that rather than the custom piece you have (lile 6071K41 on mcmaster). It'll save time and eliminate machining ops.

[=Martin=Taylor=]

Well ideally you could calculate the diameter of the bolt you need.

Stress is equal to Force over area. If you increase the area of contact with the bolt you decrease the stress. (as someone already suggested)

But what IS the force? I'm not really sure.

The term "tensioning" is rather ambiguous in chain systems. According to the Diamond Chain company the deflection at the center of the chain should be between 2-3% of the center-to-center distance of the sprockets. But if the chain can deflect, it isn't really under tension is it? If there is no tension there is no force...

Something doesn't make sense...

[Al Skierkiewicz]

Remember that the forces on the cross screws are greatest when the tensioner is receiving forces from the chain in motion. You may not bend the screws while tensioning but I bet you will in a pushing match with anotheer robot.

[IKE]

Originally Posted by Al Skierkiewicz

Remember that the forces on the cross screws are greatest when the tensioner is receiving forces from the chain in motion. You may not bend the screws while tensioning but I bet you will in a pushing match with anotheer robot.

Let's throw some numbers at this assume that while being pushedmost of the wieght is transferred to one wheel (100 pounds) and it is slipping give a coefficient of friction at 1.2(ish) and you have 120 pounds. for a 6 inch wheel this is 360 inch-pounds of torque in the system (120 pounds x 3 inch radius). If it is a 2 inch sprocket at the wheel, the chain force is then 1080 pounds. This gets added onto the tension that you put into the system. You can then run the calculation for the bending stress in a round beam constrained on both ends and then loaded in the middle (make sure you use the root diameter of the bolt if it is all thread).

We use two tensioners on each dead axle so that they are essentially in shear. Not only do they share the lods, but then you can directly connect them to the axle.

[Travis Covington]

Originally Posted by IKE

Let's throw some numbers at this assume that while being pushedmost of the wieght is transferred to one wheel (100 pounds) and it is slipping give a coefficient of friction at 1.2(ish) and you have 120 pounds. for a 6 inch wheel this is 360 inch-pounds of torque in the system (120 pounds x 3 inch radius). If it is a 2 inch sprocket at the wheel, the chain force is then 1080 pounds. This gets added onto the tension that you put into the system. You can then run the calculation for the bending stress in a round beam constrained on both ends and then loaded in the middle (make sure you use the root diameter of the bolt if it is all thread).

We use two tensioners on each dead axle so that they are essentially in shear. Not only do they share the lods, but then you can directly connect them to the axle.

It should be noted that the 2 bearing block screws provide a significant clamping force on the side rail. The friction between the bearing blocks and the side rail adds a ton of resistance and limits the amount of bending force the tensioning screw sees while driving. The clamping force alone will hold the bearing block in place better than you might imagine. The more serious concern is a high speed impact on a wheel, which will provide significantly more stress on that screw than anything our little motors can provide. Even still, you'll likely be able to replace a broken/bent screw if it does fail.

For reference, we used this style of tensioning in 2005 and 2006 and had no issues while using a #10 screw.

However, more importantly (in my eyes, and as Cory mentioned), the lack of axial alignment between the inner and outer bearings/blocks IS an issue and should get more attention than the load this screw sees from drivetrain forces.

[minisimon]

IKE: I think that your math might be a bit off. I don't see how a 2" sprocket would turn 360 in*lbs of torque into 1080 lbs of force. In fact, I think that a 2" diameter sprocket would result in 360 lbs of force in the chain. This is still a large number, but not quite as scary as 1080 lbs.

However, Travis has made a number of good points that mean any calculations we do only serve to give an idea of the forces the tensioning system (both the tightening screw and the clamp screws) will see.

[Al Skierkiewicz]

Ike,
I was trying to visualize the forces transferred to that part during a collision. (I am trained for failure analyzing) If both robots are traveling towards each other then for an instant we can believe the motors can be considered to be in locked rotor mode. That transfers all of the potential energy of both of the two robots to the drive train of each. It is in that moment of collision (with another robot or the wall) that would then transfer the maximum energy to the part in question. Still your 1080 number is pretty impressive when considering the forces involved.

[Al Skierkiewicz]

Remember that the forces on the cross screws are greatest when the tensioner is receiving forces from the chain in motion. You may not bend the screws while tensioning but I bet you will in a pushing match with another robot. BTW, nice looking design and CAD work.

[ChrisH]

Originally Posted by Al Skierkiewicz

Remember that the forces on the cross screws are greatest when the tensioner is receiving forces from the chain in motion. You may not bend the screws while tensioning but I bet you will in a pushing match with another robot. BTW, nice looking design and CAD work.

I agree with Al on this. But there is one other thing to consider. Screws need at least three threads worth of engagement to develop full strength, four is better. The hex shaped piece does not look to have enough thickness to have full engagement. You should also check the fastener lengths to ensure you are using a standard size.

Other than that it looks pretty good. By the way, it is customary in the design world to have others review your work as you are doing now. When your design decisions are questioned during the review process it is because the reviewers want to help you do a good job and have things turn out the way you want. So please do not be offended by any comments that you may disagree with.

Also, since all we are looking at is a picture, it is hard to tell the actual size of things. A few dimensions would help us evaluate just how much this design can take.

[Woody1458]

I'm confused as the where the wheel would fit. I see the axel going through but it seems the screw getting right in the way where a wheel would be.

[AndyB]

Originally Posted by Woody1458

I'm confused as the where the wheel would fit. I see the axel going through but it seems the screw getting right in the way where a wheel would be.

What you see goes inside a piece of 2"x1" aluminum. The wheel would be cantilevered off the side.