Shear on bolts for a Decelarating rotating mass 4

[NewAtThis]

Hello all,

I have been asked to check some details for our customer.

We have an adapter on the spindle nose of a lathe, at a mass of around 335kg being held in place with 4 X M24 bolts and a drive dog.

The mass is rotating at 3300rpm. I want to check the force acting upon the bolts if the emergency stop is activated, which I assume is an immediate stop. To check the shear on the bolts and drive dog.

 

[flash3780]

No such thing as an immediate stop.

Perhaps you could determine the force by calculating a torsional deflection of your part based on the energy balance, and back out a bolt force. If you assume everything else is rigid, and your spinning thing is a torsional spring, you'll come out with a conservative number. Hard to say if that would work without seeing a picture.

 

[rb1957]

it's easy (isn't it) to calculate the energy in the lathe head. the difficulty in this, as in all "sudden" stops is "how sudden is sudden ?" 0.01 sec ? 0.001 sec ?? the shorter you make the time interval, the higher the deceleration becomes (average and peak), the higher the load becomes ...

good luck !

 

[NewAtThis]

Ok if we assume we have a time in seconds for the decelaration, wich we are getting from the machine tool comapny, and I have calculated the energy in joules for the rotating mass, how do i go about it from here?

Any help would be great.

 

[desertfox]

Hi NewAtThis

Put a sketch up with dimensions of the adaptor and on what pitch circle are the bolts acting.
Also you need to know the centre of gravity of the adaptor relative to the fixing bolts as there maybe some bending as well as shear.

desertfox

 

[NewAtThis]

Ok Heres the pickies, and my initial calcs. I'm not degree graduate, so dont be too harsh!!

The customer is just concerned about the shear forces acting on the bolts and drive dogs when a machine E-Stop is used. We are waiting on a decelaration value.

http://files.engineering.com/getfil...18-4407-830e-672de2ee3324&file=Image00207.pdf

http://files.engineering.com/getfil...44c2-95ca-555d3490c06f&file=Asm_For_Shear.PDF

 

[NewAtThis]

The decelaration is 12 seconds from 2500 rpm.

Hope this helps.

 

[hydtools]

Use impulse-momentum solution:
torque x delta time = I x delta omega I is mass moment of inertia. Delta omega is change in rotational speed. Delta t is the time interval.

Calculate torque. Figure shear on the bolts.

Ted

 

[rb1957]

this is the average (isn't it ?), you might want to double the answer for a peak.

 

[desertfox]

Hi NewAtThis

I have uploaded a file with the calculation for the shear of one bolt given the maximum torque at 3300rpm.
You need the power of the lathe motor and using this you can calculate the maximum torque for the above speed or different speeds,now if the sudden stop was an instanteous one, then I would double the torque obtained from the above information and ensure the bolts don't fail.
Bear in mind that if the speed of the adaptor is slower than 3300rpm then the torque and the shear on the bolts will be higher for a constant power.
I am not sure what your 175mm dimension is to the Centre of gravity of your device, do you mean 175mm from the bolt heads to the centre of gravity?

regards

desertfox

 

[BobM3]

desertfox -

You're assuming that an "e-stop" uses the motor to decelerate the load.

NewAtThis-

You mean the decel is 2500 rpm to 0 rpm in 12 seconds? If it is uniform, then the angular deceleration (alpha) is 2500/12. Use the deceleration in the formula T=I*alpha to get torque (careful with units) . Your inertia units (m/kg**2) don't look right in your calculation.

Are you making the assumption that the bolts have loosened up and will therefore have to react the entire torque? If the bolts remain tight then the torque reaction may be through the friction provided by the (4) bolts loaded in tension.

 

[desertfox]

Hi BobM3

I not intending that the motor deccelerates the load.
What I am saying is that if the lathe is turning and somebody runs the toolpost into it by accident and stops the chuck from turning, the bolts holding that adaptor will need to withstand the full motor torque which may be amplified through the gearbox depending on chuck rpm.
The motor won't know the chucks stopped and try to carry on turning, untill some protection as cut in.
If the bolts are designed on that basis and include a safety factor, then I see that as being conservative and you have no need to worry how long the emergency brake takes to slow the motor down so long as the brake does come into operation.

desertfox

 

[BobM3]

Even then, the torque to decelerate the motor rotor could exceed the rated torque of the motor.

 

[desertfox]

Yes I agree it could exceed the rated torque which is why I said put a good safety margin in.
Theres a good chance that the final torque at the chuck will be greater than that of the motor anyway, because of magnification through the gearbox.
Difficulty now, is if you work out the torque using the "moment of inertia" and angular deceleration, then you have to work it out again if the adaptor is used at different speeds.

desertfox

 

[NewAtThis]

Hello all,

Thanks for all the imput, all looking good.

As this is on a CNC lathe, we have a degree of collision control, and so feel that this isnt a problem. Also the customer feels that this is the worse case if the E-stop is used.

DesertFox-

I dont think the C of G is relevent in this case, just added just in case.

BobM3-

The moment of inertia is taken from the Cad data of Ixx=7.367 Kilograms/Square meter

I am assuming that the bolt connection is loose as I'm struggeling with this, let alone a clamped friction calc!!

I'm going to try the decelaration calc, as I believe this suits the application.

Any more ideas are welcome. Thanks again for all your help!!

 

[desertfox]

Hi NewAtThis

Well just a couple of thoughts,firstly if you think the braking situation is the worse case,I would ask what is the time taken to reach 3300rpm, because if it is less than 12 seconds then your bolts will see a greater shear load during start up due to a faster angular acceleration.
The centre of gravity is important, as it dicates the amount of tensile load in the bolts due to the centre of gravity being 175mm from the bolt heads.
However at 175mm from the from the bolt heads the additional tensile load caused by the 335kg mass is probably insignificant in comparison with the other loads but we couldn't say that untill we knew where the centre of gravity was.
Your moment of inertia should be in units of kgm^2 and not kg/m^2 as you have indicated.

desertfox

 

[NewAtThis]

I have had to assume a couple of things, I hope to get these calcs done out by someone qualifed.

I agree that the tensile load on the bolts is pretty much naff all, and I have assumed that acc=decel.

Yeah the inertia was a typo.

Thnaks again for all your help!!

 

[unclesyd]

One question.
Is there a possibility that the shear plane would encompass the threads?
Also the word "loose" kinda bothers me.

There reason for asking is that I recently witnessed the failure of a high speed balancing machine head where the shear failure was through the threads of the mounting bolts.

 

[NewAtThis]

Yes there is the possibility that the shear plane is thro the threads which effects the cross section!

I use the term loose, as to imply a simple shear effect rather than overcoming the friction force of the two plates clamped andd then the effect in shear.

 

[desertfox]

Hi NewAtThis

Well I have done the calculaton for the shear during an emergency stop based on torsion and the stress is very small.
Also included which surprised me, was the tensile load in the bolts due to the offset load at 175mm, it appears the static load for the latter generates a tensile stress roughly 4 times that of the torsional shear stress ue to inertia.
Note I haven't included any shear taken by the drive dogs which would make the torsional stress even less.

desertfox