Equation for a distributiong force on a half cylinder

In summary, the problem is to distribute a total force acting vertically upward over a half cylinder, represented by a set of nodes between 0 and 180 degrees. The force is maximum at 90 degrees and 0 at 0/180 degrees. The equation in cylindrical coordinates used is y=F*cos(theta)/n, where n is the number of nodes. The force remains constant into the page and is distributed circumferentially over 7 nodes and axially over 7 nodes, giving a total of 49 nodes. The nodes are spaced evenly around the circumference and the force is directed in the +z direction at all nodes. The formula for the force at any node is given by F= k*sin^2(2pi*j/
  • #1
bugatti79
794
1
Folks,

I have an FEA applicaton where I need to apply a total force acing vertically upward but to be distributed over half a cylinder (1st n 2nd quadrant). This half cylinder i representd by a set of nodes between [tex]0 \le \theta \le 180 [/tex]The force will be maximum at 90 degrees and 0 at 0/180 degrees. I thought that the equation in cylindrical coordinates would work like

y=F*cos(theta)/n where n is the number of nodes.

The force remains constant into the page as it acts over the area represented by the nodes.

Any suggestions?
 
Mathematics news on Phys.org
  • #2
Cool problem! I guess I have a few questions:

1. How long is the total cylinder?

2. Are you distributing the force only over the lateral area, or are you also putting some of it on the endcap?

3. You say the force is "acting upward", but you don't mention how the cylinder is oriented. How is the cylinder oriented? For example, if your cylindrical coordinates are $(\rho,\theta,z)$, then is the axis of the cylinder on the $z$ axis?

4. Suppose the cylinder is oriented with its axis on the $z$ axis. How is the force oriented to the cylinder, exactly? You mention that it's a maximum at $90^{\circ}$. How are you measuring that angle? Is it the usual cylindrical angle I mentioned before?

5. What is $y?$

If you could possibly draw a large, clear picture illustrating your setup, that would greatly help me visualize what you're trying to do.
 
  • #3
Ackbach said:
Cool problem! I guess I have a few questions:

1. How long is the total cylinder?

2. Are you distributing the force only over the lateral area, or are you also putting some of it on the endcap?

3. You say the force is "acting upward", but you don't mention how the cylinder is oriented. How is the cylinder oriented? For example, if your cylindrical coordinates are $(\rho,\theta,z)$, then is the axis of the cylinder on the $z$ axis?

4. Suppose the cylinder is oriented with its axis on the $z$ axis. How is the force oriented to the cylinder, exactly? You mention that it's a maximum at $90^{\circ}$. How are you measuring that angle? Is it the usual cylindrical angle I mentioned before?

5. What is $y?$

If you could possibly draw a large, clear picture illustrating your setup, that would greatly help me visualize what you're trying to do.

I cannot attached my sketch as its above max file size unfortunately. I will try again tomorrow
 
  • #4
Ackbach said:
Cool problem! I guess I have a few questions:

1. How long is the total cylinder?

2. Are you distributing the force only over the lateral area, or are you also putting some of it on the endcap?

3. You say the force is "acting upward", but you don't mention how the cylinder is oriented. How is the cylinder oriented? For example, if your cylindrical coordinates are $(\rho,\theta,z)$, then is the axis of the cylinder on the $z$ axis?

4. Suppose the cylinder is oriented with its axis on the $z$ axis. How is the force oriented to the cylinder, exactly? You mention that it's a maximum at $90^{\circ}$. How are you measuring that angle? Is it the usual cylindrical angle I mentioned before?

5. What is $y?$

If you could possibly draw a large, clear picture illustrating your setup, that would greatly help me visualize what you're trying to do.

I created a jpeg of size 145KB which is under 195KB for a jpeg but it still won't load up.

1) It is 7 nodes long
2) Distributing the force circumferentially over 7 nodes and axially over 7 nodes thereby giving total number of 49 nodes
3)Axis is z, correct
4)The total force is acting upwards, ie at 90degrees ACW. I want to distribute this force circumferentially so that at angles 0 and 180, the force is 0, then repeat these forces axially 7 times.
5)We could call y our total force Fr.
 
  • #5
bugatti79 said:
I created a jpeg of size 145KB which is under 195KB for a jpeg but it still won't load up.

1) It is 7 nodes long
2) Distributing the force circumferentially over 7 nodes and axially over 7 nodes thereby giving total number of 49 nodes

How are they spaced physically?

3)Axis is z, correct
4)The total force is acting upwards, ie at 90degrees ACW.

What does ACW stand for? Is the force, then, always in the $+z$ direction?

I want to distribute this force circumferentially so that at angles 0 and 180, the force is 0, then repeat these forces axially 7 times.
5)We could call y our total force Fr.

So, $1/7$ of your total force is at each of the seven $z$ values, correct? Then you have seven locations around the circumference, presumably equally distributed, where the force is located. Are you treating each location here as a point force? In other words, is the force only at the node, or is it continually varying through all the degrees? If the force is to be treated as only occurring at the node, then where are the nodes, precisely? Is there one at $\theta=0^{\circ}$? Are they evenly spaced around the circumference?

So, I think I know the answers to your questions above. Proceeding on these assumptions, I am picturing a cylinder, $7$ nodes high and $7$ nodes around. You must distribute a force $F_r$ amongst these nodes such that $F_r/7$ is distributed at each $z$ value, and that $F_r/7$ at each $z$ value is distributed circumferentially such that the force is $0$ at angles $0^{\circ}$ and $180^{\circ}$.

So the trig function that does this is the $\sin(\theta)$ function. At least, it's zero at the right places. The problem is that it goes negative in the third and fourth quadrants. One way to fix this is simply to square it. So, let's take $\sin^{2}(\theta)$ as our starting point for getting the angle correct. We now need to ensure that
$$A\sum_{j=0}^{6}\sin^{2}\left(\dfrac{2\pi j}{7}\right)=\dfrac17.$$
Mathematica says that
$$\sum_{j=0}^{6}\sin^{2}\left(\dfrac{2\pi j}{7}\right)=
2\left[\cos^2\left(\dfrac{\pi}{14}\right)+\cos^2\left(\dfrac{3\pi}{14}\right)+\sin^2\left(\dfrac{\pi}{7}\right)\right].$$
It follows that
$$A=\frac{1}{14\left[\cos^2\left(\dfrac{\pi}{14}\right)+\cos^2\left(\dfrac{3\pi}{14}\right)+\sin^2\left(\dfrac{\pi}{7}\right)\right]}.$$
So, taken with the fact that the force must be directed in the $+z$ direction everywhere, we have that the force at any node ($z$ shouldn't enter into it) $\vec{F}$ is given by
$$\vec{F}=\frac{\hat{\mathbf{k}}\sin^{2}\left(\dfrac{2\pi j}{7}\right)}{14\left[\cos^2\left(\dfrac{\pi}{14}\right)+\cos^2\left(\dfrac{3\pi}{14}\right)+\sin^2\left(\dfrac{\pi}{7}\right)\right]},$$
where $0\le j\le 6$ is the number of the angle corresponding to the node.

Would this work for you?
 
  • #6
Ackbach said:
How are they spaced physically?
What does ACW stand for? Is the force, then, always in the $+z$ direction?
So, $1/7$ of your total force is at each of the seven $z$ values, correct? Then you have seven locations around the circumference, presumably equally distributed, where the force is located. Are you treating each location here as a point force? In other words, is the force only at the node, or is it continually varying through all the degrees? If the force is to be treated as only occurring at the node, then where are the nodes, precisely? Is there one at $\theta=0^{\circ}$? Are they evenly spaced around the circumference?

So, I think I know the answers to your questions above. Proceeding on these assumptions, I am picturing a cylinder, $7$ nodes high and $7$ nodes around. You must distribute a force $F_r$ amongst these nodes such that $F_r/7$ is distributed at each $z$ value, and that $F_r/7$ at each $z$ value is distributed circumferentially such that the force is $0$ at angles $0^{\circ}$ and $180^{\circ}$.

So the trig function that does this is the $\sin(\theta)$ function. At least, it's zero at the right places. The problem is that it goes negative in the third and fourth quadrants. One way to fix this is simply to square it. So, let's take $\sin^{2}(\theta)$ as our starting point for getting the angle correct. We now need to ensure that
$$A\sum_{j=0}^{6}\sin^{2}\left(\dfrac{2\pi j}{7}\right)=\dfrac17.$$
Mathematica says that
$$\sum_{j=0}^{6}\sin^{2}\left(\dfrac{2\pi j}{7}\right)=
2\left[\cos^2\left(\dfrac{\pi}{14}\right)+\cos^2\left(\dfrac{3\pi}{14}\right)+\sin^2\left(\dfrac{\pi}{7}\right)\right].$$
It follows that
$$A=\frac{1}{14\left[\cos^2\left(\dfrac{\pi}{14}\right)+\cos^2\left(\dfrac{3\pi}{14}\right)+\sin^2\left(\dfrac{\pi}{7}\right)\right]}.$$
So, taken with the fact that the force must be directed in the $+z$ direction everywhere, we have that the force at any node ($z$ shouldn't enter into it) $\vec{F}$ is given by
$$\vec{F}=\frac{\hat{\mathbf{k}}\sin^{2}\left(\dfrac{2\pi j}{7}\right)}{14\left[\cos^2\left(\dfrac{\pi}{14}\right)+\cos^2\left(\dfrac{3\pi}{14}\right)+\sin^2\left(\dfrac{\pi}{7}\right)\right]},$$
where $0\le j\le 6$ is the number of the angle corresponding to the node.

Would this work for you?

Hi Ackbach,

This looks good, I think you are moving in the right direction! I have found a pictue online that should clarify my intention. From the picture that we can use as an example.



  • There are approximately 22 nodes circumferentially and 9 nodes axially giving a total of 198 nodes to which Fr must be distributed.
  • Force vectors are all vertical in the picture butI my case they will be normal to the surface at each point.
  • My distributed vectors are acting over half thecylinder, ie, the 1st and 2nd quadrant only and are 0 inmagnitude at 0 and 180 degrees ad maximum at 90 degrees (Obviously the applied force Fr at the centre of the bearing (not shown in picture) is acting upwards)

 

Attachments

  • Pressure Distribution.png
    Pressure Distribution.png
    6.5 KB · Views: 44
  • #7
Hmm. This is radically different from what you described before. If you had a cylinder with axis along the $z$ axis, and the force all pointing in the $+z$ direction, then the force would, of necessity, be parallel to the lateral surface of the cylinder. But you're saying it needs to be normal (perpendicular) at every point. That is, your force is going to be in the $\hat{\rho}$ direction.

Is it $7$ nodes by $7$ nodes? Or is it $22$ by $9$?

If you could please give a word-for-word, identical problem statement to that which you are solving, that would be much appreciated.
 

1. What is the equation for calculating the distribution of force on a half cylinder?

The equation for calculating the distribution of force on a half cylinder is F = 2πrσ, where F is the force, r is the radius of the cylinder, and σ is the surface tension.

2. How is this equation derived?

This equation is derived from the Laplace-Young equation, which relates the tension in a curved liquid surface to the radius of curvature and surface tension of the liquid. By taking the derivative of this equation and integrating, we arrive at the equation for distribution of force on a half cylinder.

3. What does this equation tell us about the distribution of force on a half cylinder?

This equation tells us that the force exerted on a half cylinder is directly proportional to the radius and the surface tension. It also tells us that the force is evenly distributed around the circumference of the cylinder, hence the 2π in the equation.

4. Can this equation be applied to other curved surfaces?

Yes, this equation can be applied to any curved surface that has a constant radius of curvature, such as a sphere or a cone. However, it may not accurately represent the distribution of force on more complex curved surfaces.

5. How can this equation be useful in practical applications?

This equation can be useful in understanding the forces acting on half cylinders in various situations, such as in fluid mechanics or in the design of cylindrical structures. It can also be used to calculate the stress and strain on the material of the cylinder, which is important in engineering and material science.

Similar threads

  • General Math
Replies
1
Views
7K
  • Advanced Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
2K
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
23
Views
2K
Replies
1
Views
584
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
954
Back
Top