Shaft diameter for a rotary transfer device

[juddster94]

I am designing a rotary transfer device to transfer a 40ft tube from one conveyor to another. The device is 5 arms pivoting around a central shaft (I'm sizing the central shaft). See attached picture.
I am using the following equation to calculate the shaft diameter and I'd like any comments if anyone sees any errors in my calculations.

d^3=(16/(pi*Ss))*Sqrt((KbMb)^2+(KtMt)^2)
d = diameter of the shaft
Ss = allowable stress - I am using 6000 psi
Kt & Kb = Combined shock & fatigue factor applied to the torsional and bending moment. Kb=2.0 and Kt=1.5
Mt & Mb = torsional and bending moment

The entire assembly consists of 3 shafts, 174", 240" and 36" connected with a hub at each end. The arms and supports are at 0", 60", 180", 300" and 420". The pockets for the tube are on a 27" radius. The mass of the arm is 301 lbs each with a total of 5 arms with a total mass of 1505 lbs. The mass of the tube being transfered is 2136 lbs. The rotation starts with the arm horizontal at rest, picking up the tube and rotating through 180 degree's. The acceleration of the arm is from 0 to 6 rpm through pi/6 radians=.376 rad/s^2. The horizontal force due to acceleration I've calculated at Fn=mr'w = (3641 lb)(16/12 ft)(.376 rad/s^2)=1830 lbf or 367 lbf at each arm. The center of gravity of the arm is at the shaft and the shaft rotates at 6 rpm so I've calculated the HP required to rotate the tube through 180 degree's is

HP=F*r*N/63000=2136*27*6/63000=5.5 HP

I will put a 7.5HP motor on it. One question though, since the arm is loaded on one side only, should I include the mass of that half of the arm in my HP calculation?

I've calculated that my maximum bending moment as sqrt(Mbv^2+Mbh^2)=23068 lb-in. One question is should I calculate the bending moment on each shaft or should I calculate the system as a whole? I've done each shaft independently. I've positioned the drive near the 3rd arm/support assembly but attached to the 2nd shaft. So my bending moment free body diagram for the 174" shaft is 3 simple supports at 0", 60" and 174" (the third support is actually on the 2nd shaft). The vertical forces are two 750 lb loads acting down at 12" and 72" and two horizontal forces as a result of angular acceleration at 366 lbf at the same locations.

d=(16/pi*6000)*sqrt((2*Mb)^2+(1.5*Mt)^2))^(1/3)
= 0.000849*sqrt((2*8871)^2+(1.5*23068)^2)^(1/3)
= 3.207 = 3.5" dia shaft

Angle rotation = 584*Mt*L/(G*d^4)
= 584*23068*174/(12x10^6*3.5^4) = 1.3 degrees = 0.089 degrees / ft

Any assistance with this is greatly appreciated.
Best regards,
Judd

 

[desertfox]

Hi juddster94

Can put some dimensions on the sketch and some labels to the parts, where are the 5 arms, are they in line with the one arm I can see? the only 3 shafts I can see is one in the middle and two at each end with a grey v shape sat on them?
Another view might help and also can you enlighten us to how you got the Kb and Kt factors?

regards

desertfox

 

[juddster94]

Attached is a 2-D sketch of the design with dimensions

 

[desertfox]

Hi juddster94

Okay thats better but I only see 2 arms? anyway my next question is are you following a code for the shafting as I
see formula's in my machine design book but there not quite like yours? also on the 2d sketch I can't see the 3 shafts
at the dimensions you have posted please clarify.
Assuming that your formula are right you need to calculate the deflection of your shafting once the above formula have provided you with an intial shaft size.
You need to expand more were the formula came from and how you have determined the factor's you are using.
As regards the velocities and accelerations of the arm firstly you need the moment of inertia of the arms plus the tube your trying to manipulate, to work out the torque during acceleration, there are two components of acceleration during start up ie:- (tangential acceleration and centripetal acceleration) which you need to establish the resultant acceleration and its direction, from what I can see you have just simply added masses together and multiplied them by the angular acceleration and a radius of some ratio of 16/12 ft but I can't see where this is from again we need a lot more information if you want positive feedback.

regards

desertfox

 

[juddster94]

Hi desertfox,

This sketch I uploaded should clarify everything. This is a work in process and I have not generated all the drawings. The machine design book I'm using quotes the ASME standard for comercial steel and to use a factor of safety of 6 for a keyed shaft. For the modulus of rigidity use 12x10^6 psi. The forumula I used takes into account the bending moment (not stress) based upon the free body diagram. Once a shaft size has been determined I can certainly go back and check the bending stresses but it should be handled with the methods shown.

You are correct that I only used the angular acceleration to calculate the horizontal forces on the shaft. The 16 in/12 ft is the center of gravity of the mass I will be rotating with the shaft. So why would I need the moment of inertia of the arm if I only need to know the forces it is applying to the shaft? I would be interested in seeing the equations you have in your machine design book.


Juddster

 

[desertfox]

Hi juddster

Have a look at this sites regarding moment of inertia:-

http://www.cwu.edu/~acquisto/NLangular.htm

http://www.intmath.com/Applications-integration/6_Moments-inertia.php

The first paragraph of the second link states:-

The moment of inertia is a measure of the resistance of a rotating body to a change in motion.

The moment of inertia of a particle of mass m rotating about a particular point is given by:

Moment of inertia = md2

where d is the radius of rotation.

This should answer your question of moment of inertia the Torque is then found from :-

T= I[α]
where [α]= angular accel & I = moment of inertia


I have several formula like the one's you post one is based on the Max shear theory of failure, the other is on Max distortion energy theory I was trying to establish which one you are using.

When you post a question you have to remember we are not privy to all the information you have and your post needs to be clear not ambigous, how can I possibly know you haven't all the drawings available anymore than I would expect you to know my crystal ball failed its MOT yesterday.

 

[juddster94]

Thanks for the info Desertfox. I uderstand the complexities of sharing information, I just need to understand how much information is enough.

I've tried calculating my rotational moment of inertia but I've ran into some problems that I don't seem to be able to see through so here goes. My mass I am rotating is a tube 6 inches diameter x 480 in long and weight of 2136 lbs rotating around a 27 inch radius.

I=m(R^2+r^2)/2 + md^2

I=(2136 lb)(3 in)^2-(2 in)^2)/2 + (2136 lb)(27 in)^2=1571028 lb-in^2

My angular velocity is .628 rad/s. I have to accelerate from 0 to .628 rad/s through pi/6 radians.

Ang Acc=wf^2/2*theta = (.628 rad/s)^2/(2)(pi/6 rad) = .376 rad/s^2

T=I*alpha = (1571028 lb-in^2)*(.376 rad/s^2) = 590706 lb-in^2/s^2

The units don't seem to be working out. I've considered dividing my weight by the gravitational constent but the units still don't work out. I'm doing something wrong I just can't identify what it is. It is probably something simple that I've overlooked.

Thanks
Juddster

 

[desertfox]

Hi Juddster94

I'll try and post as soon as I can.

regards

desertfox

 

[desertfox]

Hi juddster94

You need to use the parallel axis rule for your tube at the 27" radius see this site:-

http://www.roymech.co.uk/Useful_Tables/Maths/M_of_Inertia.html

Regards

desertfox