Resultant force and centre of pressure in a tank

In summary, resultant force and centre of pressure in a tank refer to the overall force and point of application of that force on a liquid contained within a tank. Resultant force is determined by the sum of all the forces acting on the liquid, while the centre of pressure is the point where the resultant force acts. This is important in designing tanks and other structures to ensure stability and prevent tipping or collapse due to imbalanced forces. The location of the centre of pressure can be calculated using the principles of fluid mechanics. Additionally, changes in the shape or volume of the liquid can impact the resultant force and centre of pressure, making it crucial to consider these factors in tank design and operation.
  • #1
Guillem_dlc
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Homework Statement
The tank in the figure contains water and a uniform pressure ##p_a## acts on its surface. We wish to determine the resultant force ##F##, and its centre of pressure ##y_c-y_G##, on a flat gate of base ##3\, \textrm{m}## and height ##1,5\, \textrm{m}##. The damper is located on a wall of the tank with an inclination of ##30## degrees with respect to the horizontal. The upper base of the damper is located at ##3\, \textrm{m}## depth. Consider the following cases: a) ##p_a=p_{\textrm{atm}}## and b) ##p_a=125\, \textrm{kPa}##.
Relevant Equations
##I_{xx}=\iint y^2\, dA##
Figure:
5E13FCF2-1351-4777-88EC-B2DF9F5289A4.jpeg


a) CASE A ##\rightarrow p_a=101300\, \textrm{Pa}##
$$F_{\textrm{res}}?,\,\, y_c-y_{cg}=y_{cp}$$
We find ##h_{cg}\rightarrow h_{cg}=3+h##
$$h=0,75\cdot \sin (30)=0,375\, \textrm{m}\rightarrow h_{cg}=3,375\, \textrm{m}$$
$$p_{cG}=\rho_{H2O}gh_{cg}=33108,75\, \textrm{Pa}$$
We calculate ##A=1,5\cdot 3=4,5\, \textrm{m}^2##
$$\boxed{F=p_{CG}\cdot A=148989,375\, \textrm{N}}$$
$$I_{xx}=\iint y^2\, dA=\int_0^{1,5}\int_0^3 y^2\, dxdy=\int_0^{1,5}y^2\int_0^3 dxdy=3\int_0^{1,5}y^2\, dy=$$
$$=3\left[ \dfrac{y^3}{3}\right]_0^{1,5}=3,375$$
$$y_{CP}=-\dfrac{\rho_{H2O}g\sin \theta I_{xx}}{F_{CG}}=0,111$$
I don't get the ##I_{xx}## right and I don't know why? Is it calculated in this way?
$$I_{xx}=\iint y^2\, dA$$
It should give ##\dfrac{b\cdot L^3}{12}##.
 
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  • #2
You are supposed to be looking for the effective force acting on the gate, and it’s applied location(the center of pressure). The gate is in static equilibrium. The moment of inertia of the gate is not going to be a factor, it is not a dynamics problem.

Isolate the gate, assume it’s being held closed. Sketch the pressure distribution acting over its interior for the different gauge pressures acting over surface ##a##. Use that to determine the effective force on the gates interior, and it’s location using the proper formulas for those quantities.
 
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  • #3
The Second Moment of Area (Not Moment of Inertia o:) - Thanks @Lnewqban) is a factor! I'm sorry...I forgot it was in the formula for ##y_{cp}##. Anyhow, you don't have to do the integral, you have it written in the last part of the OP.
 
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  • #5
With the integral I have not done this in the end?
 
  • #6
For part (a) your force looks correct.

What they are asking you to find is:

$$ y_{cp} = \bar{y} + \frac{ \bar{I} }{ \bar{y} A} $$

$$ ( y_{cp} - \bar{y} ) = \frac{ \bar{I} }{ \bar{y} A} $$

Guillem_dlc said:
$$y_{CP}=-\dfrac{\rho_{H2O}g\sin \theta I_{xx}}{F_{CG}}=0,111$$

Where does the negative sign come from, and where did it go?

That expression is not equal to ##y_{cp}## (see above).

If you are going to use the Force and the density to calculate the area of the plate ##A## (which unnecessary and introduces computational error), label it correctly as ##F##. Its not ##F_{CG}##.

I feel like you have made a mess of things trying to be overly clever with substitutions. I think its correct (other than the minus sign and the labeling), but you have not made it easy to check.
 
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  • #7
You are correct, that whatever you are going to do after that should result in (for the rectangle):

$$I_{xx} = \frac{1}{12}bh^3$$
 
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  • #8
erobz said:
Where does the negative sign come from, and where did it go?
As ##y_{cp}## is a measure of the centre of gravity, I have given the result as an absolute value.

But maybe I should have left the sign to indicate where it goes.

For example, when you want to find the height of the centre of pressures then I do ##h_{cp}=h_{cg}+y_{cp}## with the sign I have been given from the formula so I know if it is higher or lower, but otherwise I put it in absolute value.
 
  • #9
The actual question you have is what you've done incorrectly calculating the second moment of area of a rectangular plate?

For starters the limits of integration are wrong. Put coordinate system through the centroid of the area. What are the limits of integration with respect to those axes?
 
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  • #10
Guillem_dlc said:
As ##y_{cp}## is a measure of the centre of gravity, I have given the result as an absolute value.
##y_{cp}## is the center of pressure, not the center of gravity. The depths are measured as positive downward from the surface. ##( y_{cp} - \bar{y}) > 0## for these types of problems. No need for absolute values.
 
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  • #11
erobz said:
The actual question you have is what you've done incorrectly calculating the second moment of area of a rectangular plate?

For starters the limits of integration are wrong. Put coordinate system through the centroid of the area. What are the limits of integration with respect to those axes?
##-0,75## to ##0,75## and ##-1,5## to ##1,5##?
 
  • #12
Guillem_dlc said:
##-0,75## to ##0,75## and ##-1,5## to ##1,5##?
ok, just leave it in parameters ##h## and ##b## for now. What are the limits for the integration w.r.t. ##y## in terms of ##h##, and ##x## in terms of ##b## respectively?
 
  • #13
erobz said:
ok, just leave it in parameters ##h## and ##b## for now. What are the limits for the integration w.r.t. ##y## in terms of ##h##, and ##x## in terms of ##b## respectively?
##-h/2## to ##h/2## and ##-b/2## to ##b/2##
 
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  • #14
Good. You can do that integration or (highly recommended), using the arguments of symmetry make life easier.

$$ I_{xx} = \int_{-b/2}^{b/2} \int_{-h/2}^{h/2} y^2 dy \, dx = 2 \int_{0}^{b/2} \int_{0}^{h/2} y^2 dy \, dx $$
 
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  • #15
erobz said:
Good. You can do that integration or (highly recommended), using the arguments of symmetry make life easier.

$$ I_{xx} = \int_{-b/2}^{b/2} \int_{-h/2}^{h/2} y^2 dy \, dx = 2 \int_{0}^{b/2} \int_{0}^{h/2} y^2 dy \, dx $$
So if I have a circle I will have to switch to polars, right?
 
  • #16
Guillem_dlc said:
So if I have a circle I will have to switch to polars, right?
I don't know that you have to... probably would be a good idea.
 
  • #17
Guillem_dlc said:
So if I have a circle I will have to switch to polars, right?

The problem is equivalent to finding the x coordinate of the mass centre of a vertical cylinder with a horizontal base but with a top sloped at some angle.
##\int_{-r}^r(a x+b)\sqrt{r^2-x^2}x.dx##
##\int_{-\pi}^\pi(ar\cos(\theta)+b)r\sin(\theta)r\cos(\theta).(-r)\sin(\theta)d\theta##
 
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FAQ: Resultant force and centre of pressure in a tank

1. What is the resultant force in a tank?

The resultant force in a tank is the net force acting on the tank due to the pressure of the fluid inside. It is the vector sum of all the individual forces acting on the tank.

2. How is the resultant force calculated in a tank?

The resultant force in a tank can be calculated by multiplying the density of the fluid by the volume of the tank and the acceleration due to gravity. This will give the weight of the fluid inside the tank, which is the resultant force.

3. What is the centre of pressure in a tank?

The centre of pressure in a tank is the point at which the resultant force acts on the tank. It is the point at which the tank will experience the same amount of pressure from the fluid on all sides.

4. How is the centre of pressure determined in a tank?

The centre of pressure in a tank can be determined by finding the centroid of the fluid's pressure distribution. This can be done by calculating the moment of the pressure forces about a reference point and then dividing by the total pressure force.

5. What factors can affect the resultant force and centre of pressure in a tank?

The resultant force and centre of pressure in a tank can be affected by factors such as the shape and size of the tank, the density and viscosity of the fluid, and the location and orientation of the tank. Changes in these factors can alter the distribution of pressure and thus affect the resultant force and centre of pressure.

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