Where do I place the bearings for a stable inclined rotating drum?

[Sook]

Homework Statement

I am basically designing a rotary drum dryer and need to know the positions of the bearings for the dryer to be stable.
The drum is of length 17 m with diameter 3.24 m, inclined at 2° rotating at 3.24 rpm. I need to find the distance X and OY. (Given that Y passes through the center of mass) Weight of the drum =280 kN

http://https://scontent-a-kul.xx.fbcdn.net/hphotos-ash3/t31.0-8/p417x417/10014962_10152301201595569_2033873904_o.jpg

Homework Equations

Moment= F x D = WD ;
Moment CW = Moment ACW

The Attempt at a Solution

I am trying to solve this using moments. But I am stuck with finding the centre of mass of the inclined cylinder.

According to me (and after reading alignments of rotary drum, they did not take into consideration the inclination), X can be found by : (17/2)/2 =4.25m (Is that possible?)

 

[CWatters]

Not sure I will be around over the next few days to help but I suggest you post a diagram providing more detail such as roughly where the bearings are.

Are these the bearings that the drum spins in OR the bearings that the drum pivots about when changing inclination to tip the contents out or ??

2 degrees is quite a small angle.. not far off horizontal..so why do you think it needs to be taken into account?

I am stuck with finding the centre of mass of the inclined cylinder.

Finding the centre of mass of an inclined cylinder is quite trivial. So I think you might be asking the wrong question?

Lets say that the drum is uniform and the centre of mass is mid way between the ends when viewed horizontally (eg 17/2 = 8.5m from one end). Then when inclined it will still be mid way between the ends when inclined. See diagram.

A = 0.5*17*Cos(0) = 8.5m
B = 0.5*17*Cos(2) = 8.4948m

Similarly if the centre of mass is say 1/3rd of the way from one end.

 

[CWatters]

PS. Just noticed you tried to post a diagram but I can't see it or open it.

 

[NascentOxygen]

This being the homework forum, PF rules stipulate that only assistance to understanding should be given, not a complete solution to any problem posted.

Sook's pic:

https://scontent-a-kul.xx.fbcdn.net/hphotos-ash3/t31.0-8/p417x417/10014962_10152301201595569_2033873904_o.jpg

 

[HalfLostAndSearching]

First of all, great job on designing a rotary drum dryer! In order to determine the positions of the bearings for a stable inclined rotating drum, we need to consider a few factors.

The first factor is the weight of the drum, which you have already provided as 280 kN. This is an important factor to consider because it will determine the amount of force that the bearings will need to support.

Next, we need to consider the inclination and rotation of the drum. The inclination angle of 2° will create a force that is perpendicular to the axis of rotation, which will need to be counteracted by the bearings. This will also affect the center of mass of the drum, as it will shift slightly due to the inclination.

To determine the center of mass, we can use the formula for the center of mass of a cylindrical object, which is given by:

Xcm = (L/2) * cosθ

Where Xcm is the center of mass, L is the length of the cylinder, and θ is the inclination angle. Plugging in the values from your problem, we get:

Xcm = (17/2) * cos2° = 8.48m

This means that the center of mass is located 8.48m from the left side of the drum.

Now, to determine the position of the bearings, we can use the concept of moments. The moment of a force is equal to the force multiplied by the distance from the point of rotation. Since we want the drum to be stable, we want the moments on both sides of the drum to be equal.

Let's assume that the left bearing is located at a distance of x from the left end of the drum, and the right bearing is located at a distance of y from the right end of the drum. The moment of the weight of the drum about the left bearing will be:

Ml = 280kN * (8.48m - x)

Similarly, the moment of the weight of the drum about the right bearing will be:

Mr = 280kN * (y)

Since we want the drum to be stable, we want these two moments to be equal. This means that:

Ml = Mr

280kN * (8.48m - x) = 280kN * (y)

Solving for x and y, we get:

x =