Torque on a square screw full with water

[Gh778]

Sorry for the confusion.

Thanks for your help :)

1. From message #63 on I thought the thread has no axis.

it can, I think this don't change the result. But for simplify the problem consider there is an axis like that all forces which are in the axis is canceled by axis.

2. I thought the water films are closed by gasket from bottom and top too.

Yes all is closed, but the film is always at the same place.

but perhaps you want to know the problem in your calculation.

Yes, I would like to calculate this ! I don't know how, I need help for this.

With or without axis, if the thread is forced to move with fixed distance from the fixed thread, the force any point is normal and the displacement in normal direction is zero.

For me, there is a torque and others forces in the axis of the square thread like you said. The pressure apply forces in 3 directions. Z axis which I see a torque. X and Y, but these axis can't turn due to the Z axis, could you explain how you see the torque is canceled by another forces ?regards

 

[Hassan2]

Seems here is your problem: The axis of the thread is not fixed by a bearing or something. It can't cancel any force or torque. In fact the torque is canceled by the fixed thread through the water films. The reaction forces from the fixed thread is normal to the fixed there ( and also to the square thread) , it has x and y components too.

 

[Gh778]

We can put the square thread like the drawing show. There is an axis but the thread can move free up/down. The sum up/down forces is 0. The X and Y axis can't turn.

 

[Hassan2]

But this has no effect. Since the water films are not compressible, they keep the axis just in the center of the bearing. No radial force then.

 

[Gh778]

Z axis is fixed with square thread. The square thread can only move up/down and turn around Z axis I think. The fixed threads are not fixed with Z axis. Where I wrote "fixed" in the drawing, supports of axis are fixed to fixed referential.

Since the water films are not compressible,

maybe it's the source of misunderstood. The film of water is not compressible, we have only gravity which create pressure.

 

[Hassan2]

Yes I understand you but again its the fixed thread which cancel the torque not the axis of the thread, because the axis of the tread is not pushing against the bearing.

 

[Gh778]

Since the water films are not compressible,

maybe it's the source of conflict. The film of water is not compressible, we have only gravity which create pressure.

 

[Hassan2]

That's why I prefer to fill the space between the threads with pressured gas.

 

[Gh778]

Ah, ok, but with gas, like I said before, I agree: there is no torque. For me, the torque appear with the change of pressure due to the gravity. I don't pressure the film of water, there is only pressure from gravity. And like you see with the last drawing, the axis of the square thread prevent the square thread to move X and Y axis and prevent to move around X and Y axis. The only liberty is move up/down and rotate around Z axis. Like that the square thread never move on X and Y, the film of water is always with gravity pressure only.

For me, now I see only one solution: the torque is the same everywhere in the thread but you said my calculation of the torque is good.

 

[Hassan2]

Another thing is that, for closed gasket, the pressure is not the gravity only. The pressures are determined by equilibrium equations. I think you consider the pressures as if their top was open.

 

[Gh778]

Another thing is that, for closed gasket, the pressure is not the gravity only.

But if you pressurized water, the pressure is added at each point and you add a fixed constant to the phase angle torque. The phase angle exist always I think.

But I give a drawing of the square thread with open gasket at top if you think it's important. The water can't move down.

 

[Hassan2]

You can even leave the lower gap filled with air. Only the upper film is filled with water. Yo can even fill it to a certain level. Again the axis cancels the torque.
Added: Sorry, in this case, the thread sits on the fixed thread.

 

[Gh778]

You can even leave the lower gap filled with air. Only the upper film is filled with water.

I don't understand can you explain ? Up and down gaps are filled with water not air.

Added: Sorry, in this case, the thread sits on the fixed thread

Not at all, there is no up/down force, so a mechanical system can easily adjust the altitude of the square thread. The mechanical system support the weight of the system (weight move up/down => energy=0). Again, films of water (up AND down) are always the same, the volume is constant. Water is under gravity pressure only.

Are you sure about the formula : dM=r*P*de*dl*sin(θ) ?

 

[Hassan2]

$dM=dF \times r$

It is a vector, when you integrate, you can't threat is like an scalar. the magnitude of this vector is

$|dM|=PdA r sinθ$ , where θ is the angle between the force and r , not the angle between the r and Z axis. Due to the curvature of the thread, the normal force is not along the Z axis.

dA: an infinitesimal area around the point.

 

[Gh778]

So, for you the formula is good ?

 

[Hassan2]

Seems fine. In this problem, θ seems to be constant at any phase angle.

 

[Gh778]

θ change with the radius, it is the slope of the thread. The pitch is the same at every radius but the length is bigger when the radius is increasing. With pitch = 1, when we pass from r=1 to r=3 we pass from sin(θ)=0.707 to sin(θ)= 0.316. The torque is not the same at each point of the thread, bigger if the radius is bigger.

 

[Hassan2]

I can't imagine the slope of the thread in radial direction correctly. What are the slopes at r=0 and r=infinity?

The torque is bigger mainly due to increase in r itself.

 

[Gh778]

r->0, the slope -> 90°
r->infinity, the slope ->0°
See last message for the drawing, the slopes are visible on the circular thread.

 

[Hassan2]

I don't get it. Normal to surface a any point makes an angle with Z axis. does the angle depends on r?

 

[Gh778]

does the angle depends on r?

for me yes, you can't see the drawing ?

 

[Gh778]

I'm working about the angle and the torque.

The equation of a helicoid is:

x=v*cos(u)
y=v*sin(u)
z=u

With u and v reals, v is the radius, u is the altitude

Perpendicular vector or normal vector (n) to the surface is :

n = v1 x v2 with "x" cross product and v1, v2 two vectors of the tangent plane

Vector=(v*cos(u), v*sin(u), u)

v1=dVector/du=(-v*sin(u), v*cos(u), 1)
v2=dVector/dv=(cos(u, sin(u), 0)

n = v1 x v2 = (sin(u), -cos(u), v)

We can see the angle of the perpendicular vector to the surface, it's like we said before.

I'm trying to do the full torque around Z axis.

 

[Hassan2]

Thanks for the equations. It like the formula of normals. I didn't know this formula and I usually use gradient method. For parametric curves your method is easier. Now I can imagine what surface the screw has.

In order to calculate the total torque, you need to use equations of equilibrium too.

 

[Gh778]

For you, these equations validate the torque dM=r*P*de*dl*sin(θ) ?

 

[Hassan2]

what are de and dl?

 

[Gh778]

dA=de*dl the small area

 

[Hassan2]

That's correct. But you can test your formula on a sphere in the water. the sphere haszero mass and is connected to a horizontal axis with a mass-less bar. the torque is expected to become ro*g*V*L , where the V is the volume of the sphere, ro is the density of water, and L is the distance from the center to the axis.

 

[Gh778]

It's very difficult to have the equation of the square thread for have the torque. I decided to find it with numerical solution. Here it's the program in C language. I hope you know this language but it's not difficult to read it with mathematical equations. I have take pressure in bar in local torque but done total torque in Pa.

The start of the helicoid is done by "d". If d=0 we have 214500 Nm but if d=-0.45 rd we have 217000 Nm for the total torque. Maybe something is wrong in my program or maybe there is a torque like I said before, in this case I don't find the contrary torque.#include <stdio.h>
#include <math.h>
#include <stdio.h>

int main(void)
{
double x=0.0,y=0.0,z=0.0; // each point of the circular thread, I use circular thread and after select only points which are on the square thread
double u=0.0,v=0.0; // u=altitude, v=radius
double a=0.0; // angle
double torque=0.0; // local torque
double torqueT=0.0; // total torque
double pas=0.001; // step for integration
double pi=3.1415927;
double vLimit=2*pi, uLimit=2*pi; // limit of integration, in meters
double c1=3, c2=4; // limit of the square thread in meters
double d=-0; // angle start for the helicoid

do
{
do
{
//calculation of each point of the helicoid
x=v*cos(u-d);
y=v*sin(u-d);
z=u;

// select only point for have the square thread
if( (x>c1 && x<c2 && y<c2 && y>-c2) || (x<-c1 && x>-c2 && y<c2 && y>-c2) || (y>c1 && y<c2 && x<c2 && x>-c2) || (y<-c1 && y>-c2 && x<c2 && x>-c2) )
{
a=atan(uLimit/2/pi/v);
torque=v*(uLimit-u)/10.0*sin(a)*pas*pas; // Here the pressure is (uLimit-u)/10.0 in bar //// pas*pas = surface
torqueT+=torque;
/*
// print part result
printf("\nu=%f, v=%f", u,v);
printf("\nx=%f, y=%f, z=%f",x,y,z);
printf("\na=%f, couple=%f",a, torque);
printf("\nT=%f", torqueT);
system("pause");
*/
}
v+=pas; // increase step of integration
}while(v<vLimit);
v=0;
u+=pas; // increase step for integration
}while(u<uLimit);

printf("The total torque is: %f",torqueT*100000.0); // * 100000 for pass from bar to Pa unity
}

 

[Hassan2]

I have checked the code by the result only and I think there is a problem in your code. I think if we make uLimit half, the torque becomes half too. in your code the relation is not linear. Also I have doubt about your differential volume. I don't think it simply becomes the product of the differential change in u and v. for example on the surface of a cylinder, ds=rdrdθ . For the helocoid, you can find a formula by relating the surface element to its projection on xy plane.

 

[Gh778]

For the helocoid, you can find a formula by relating the surface element to its projection on xy plane.

How can I do ?

Like the drawing ? maybe sin(a)+cos(a) ?

.

 

[Hassan2]

Imagine a differential surface of the surface between u and u+du, and v and v+dv. On the xy plane the projection of area is vdvdu ( as in polar coordinate). The surface on xy plane then becomes:

$ds_{xy}= ds cos\theta$

where θ is the angle between the normal and Z axis. so

$ds= \frac{vdvdu}{cos\theta}$

 

[Hassan2]

How can I do ?

Like the drawing ? maybe sin(a)+cos(a) ?

.

On xy plane it becomes much simpler. By the way, your software is great and of course you are so good in using it. Does it readily give you the projection of the object on the Cartesian planes?

 

[Gh778]

I confused, I don't understand your last message, what's dv, du ?

By the way, your software is great and of course you are so good in using it.

it's an internet drawing ;)

 

[Hassan2]

I confused, I don't understand your last message, what's dv, du ?

I used your own notation. u is the angle, so du is the differential angle which is "pas" in your code. The same for dv. So ds=v*pas*pas/cosθ .

 

[Gh778]

so for you the torque is torque=v*(uLimit-u)/10.0*(sin(a)/cos(a))*pas*pas ?