cantilever problem 1

[Anthony1780]

Hi,
I hope someone can help me with the following problem.
I need to know what the minimum length and diameter of fixed support would be required on one end of a solid steel shaft of 320mm diameter with a 500mm overhang, to prevent any deflection from the horizontal. There would be zero load applied in this situation as it would be in a rest state. The shaft is currently 1000mm long and can be machined to achieve the required size.
I have tried some cantilever formulae but I can't get my head around them as it is outside my field of experience.

Any guidance would be most appreciated.

 

[Lion06]

I am not really following your question. Can you post a sketch?

 

[Anthony1780]

Hope this clarifies the problem.

Regards

 

[Lion06]

What is the purpose of this? Why zero load? From a purely technical point of view, it will deflect slightly just under its own self weight. There is no way to get deflection to zero. Just design it to make the deflection meet whatever criteria you need (it just can't be 0).

 

[Anthony1780]

The shaft will be used to form a cylinder on a vertical press. Load will be applied during this process but a moving block will be engaged on the right side to provide full support. It is when the block is withdrawn to allow removal of the cylinder that the shaft will be subject to some degree of sag from the centre line. If this sag exceeds 1 to 2mm and perhaps increases with use then the engagement of the block will be severely hindered.
I just wanted to identify the minimum amount of material that would be required on the left side to enable rigid support of the shaft, the support blocks on the left side would be securly bolted to the machine platen.
I do realise that securing the shaft by half it's length whilst maintaining it's maximum diameter is the ideal solution, but due to press limitations I was hoping their would be a smaller ratio of left side to right side that would still work.

 

[swearingen]

As StructuralEIT said, you can never have truly zero deflection. However, if this thing is truly static and the only acceleration it experiences is gravity, the deflection would be as near to zero as makes no difference. I'm getting a bending deflection on the order of 470nm. Adding in shear deflection, the total won't even approach 1000nm, which is beyond the accuracy of most mechanical or electronic micrometers. I'll bet the sun heating up one side of it makes it move more than that...



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

 

[inertia4u]

Plus 10 points to swearingen for adding in the shear deflection as well!

-----
Nert

 

[rb1957]

i think that's assuming a fully fixed end ?

but back at the OP is the point it minimise weight ? (then keep the OD and machine out a bore). it sounds like you want to support the LH 1/2 on a couple of bearings. i would dtermine the weight of the RH (overhang) and the weight of the LH 1/2 separately (the LH 1/2 will vary as you "optimise" the section). self-weight appears to be the only load (as far as you've described things), easy enough to calculate. as a beam this is easy enough to determine the reactions (which willl size your support bearings) and the displacement of the end.

 

[Anthony1780]

Thanks to everyone for their help, it has been very useful.

Anthony