Calculate deflection with a varying inertia and material beam

[EspElement]

I have an application where I have a shaft being supported from end to end with bearings (self aligning) and a motor hard mounted driving it. The shaft starts at 1 3/4" RD to accept motor and bearings. Then is welded to a 2" x 1/4" square tube for the output. It allows for a wooden roller to be slid onto the shaft and then the machine wraps belt around this roller. Roller is made of wood chips glued together and is 5" in DIA with a 2 1/4" square down the center to accept 2" shaft. My question is, when I put a belt as heavy as 1000 lb over the span of 60" (bearing spacing is 83") what will be the result? Over the remaining 23" it will have the 1 3/4" RD supporting the load on each side. I can post pictures/drawings of this if required.

Since my motor is hard mounted I am accounting one side as fixed and other floating because of the self aligning bearings. Would I be able to consider this a fixed fixed application? Over the 60" span I believe my stress will be less because of the roller and belt wrapping will help keep area rigid. The at the points where I change from 2" square to 1 3/4" is where I am concerned about stress and deflection. I believe they would have the most. Would I be correct?

How would I go about insuring I will have no issues with this design?

Thanks guys.
Josh

 

[pongo38]

Maybe you should post a figure, but from what you have written, the span should be between SUPPORTS, and you say this is 83". I don't think you can have supports 'in the air'. The word 'fixed' is ambiguous. In its full glory it is fixed in position and in direction. But some people used 'fixed' to mean just 'fixed in position'. Without knowing more about 'self-aligning bearings' I can't help any more.

 

[EspElement]

Well, i have found a flaw in the design (A1) i just posted about requardless of this calc. The collars on each side will be impossible to remove once i wind belt. I will have to go with something like A4 and A5 with rollers inside a housing that the shaft sits on so they can remove the collar and then lift staft straight up with a forklift.

Now my inertia doesn't seem to change much from the bearing mounts and they have moved into around 63 now instead of 83 because i can with the new design. I don't see any issues with deflection or stress anymore.

Prongo, you seem to be the only one on here. Thanks for the posts.

Oh and what i mean by fixed is the ends are rigid, if i were to weld a bar to a rigid frame on both sides that is fixed on both sides. In a beam formula it says fixed or supported. Well since self aligning bearings allow for 1° angular i thought it would be considered more of a supported end because it can move and allow for deflection easier then if it was welded square to a frame. Does that make sense?

 

[EspElement]

now my delima is by using new method i lessen the gap from the bearings and load which allows me to instead of making frame adjustable with telescoping tube (i was in process of designing in as you see in new pictures) I can leave at a span of 2 meters (widest belt) and allow them to run any width belt with the frame. I have 2 size shafts on larger wider belts they use 3" square, on smaller less wide belts they use 2" square shafts. if i leave bearing span of around 82" and run a 36" wide belt in the center it would put me around 400-500 lbs of possible load on that belt. WIth it being in center 36" of it where it is winding should be very rigid because of the wooden roller and as belt accumlates it stiffens more. However, at the points outside of the 36" is where it gets hairy again. I figure a stress of 250 at those points and it being 23 inches from ends 23*250 is about 6000 lb*in moment and a max of around 10000 lb*in in center. I am back to point A with the deflection calcs from my last forum, you know well pongo which i have not investigated much because i havnt had a lot of time. By it being a single moment is there an easy why to determine max deflection @ 10000 with a .911in^4 inertia? When i take and put 500 lbs directly in center i get 0.094" using a fixed one end and other supported formula. My deflection would surely be less then that correct?

Thanks again, especially to you pongo