# Tainter Damper Figure: Analyzing Forces

• Guillem_dlc
In summary: Aha!The only potential issue I can see with this method of converting the problem (which, isn't an issue for the questions you had to answer), what would you do if you actually had to locate the center of pressure ##CP (x_{cp},y_{cp} )##? I have a feeling you would find it in a place where it doesn't seem like it should be w.r.t the actual scenario.
Guillem_dlc
Homework Statement
1. A "Tainter" damper has the shape of a partial cylinder as shown in the figure. The width of the damper is ##12\, \textrm{m}##. Note that the height of the water corresponds to the height of the pivot point of the gate. Calculate the horizontal force in ##\textrm{kN}##. Other data: ##\rho =1000\, \textrm{kg}/\textrm{m}##, ##I_{xx}=bh^3/12##

2. Calculate the vertical force in ##\textrm{kN}##

3. Calculate the total force in ##\textrm{kN}##

4. Calculate angle of total force in degrees

5. Calculate the momentum of the horizontal force in ##\textrm{kNm}##.
Relevant Equations
##F=pA, p=\rho h##
Figure:

Attempt at a solution:
$$b=12\, \textrm{m},\quad H=8\, \textrm{m}$$
a) $$F_H=p_{CG}A=3767040\, \textrm{N}=\boxed{3767,04\, \textrm{kN}}$$
$$A=8\cdot 12=96\, \textrm{m}^2$$
$$p_{CG}=\rho_g h_{cg}=39240\, \textrm{Pa}$$
b) $$F_V=mg=\rho_g V$$
We calculate ##\theta \rightarrow 8=10\cdot \sin \theta \rightarrow \dfrac{8}{10}=\sin \theta \rightarrow \theta =53,13\, \textrm{º}##
$$x=10\cdot \cos (\theta)=6\, \textrm{m}$$
We calculate ##A\rightarrow A=\dfrac{\pi r^2\cdot 53,13}{360}=\dfrac{6\cdot 8}{2}=22,3647\, \textrm{m}^2##
$$F_V=\rho_gV=\rho_g bA=\boxed{2632,77\, \textrm{KN}}$$
c) First of all we find ##\alpha## (angle between forces).
$$\alpha =\arctan \dfrac{F_V}{F_H}=34,95\, \textrm{º}$$
$$F=\dfrac{F_H}{\cos \alpha}=\boxed{4595,88\, \textrm{kN}}$$
d) $$\boxed{\alpha =34,95\, \textrm{º}}$$
e) First and foremost, we must locate the centre of pressure (CP).
$$y_{CP}=-\dfrac{\rho_g I_{xx}}{F_H}=-1,333\, \textrm{m}$$
$$M_o=F_H\cdot h_{cp}=F_H\cdot (4+1,333)=\boxed{20089,62\, \textrm{kN}\cdot \textrm{m}}$$
Does this one look good to you?

Think again about the volume displaced in part b.

haruspex said:
Think again about the volume displaced in part b.
Wouldn't this be it, or would it affect everything? I can see well how I've done it, can't I?

Okay, maybe like this?

Guillem_dlc said:
Figure:
View attachment 316081

e) First and foremost, we must locate the centre of pressure (CP).
$$y_{CP}=-\dfrac{\rho_g I_{xx}}{F_H}=-1,333\, \textrm{m}$$
Again... That is not the formula for ##y_{CP}##. What you are using is the formula for ##\bar y - y_{CP}##. They are not the same thing. It's obvious you understand that is not ##y_{cp}## from the remainder of your calculation, which is why it's confusing that you keep calling it that.

As for everything else, it seems to check out.

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Guillem_dlc
Guillem_dlc said:
Okay, maybe like this?
View attachment 316091
No.
It's a bit tricky. Suppose, instead of the sector of a circle given, the damper were to continue further to the right. Would that change the force? Would it change your calculation?
The area you have shaded would be correct if there were also water to the right. In that scenario, can you see a way to split the damper into two parts such that the water on the left is responsible for the buoyancy force on the left part and the water on the right responsible for that on the right part?

haruspex said:
No.
It's a bit tricky. Suppose, instead of the sector of a circle given, the damper were to continue further to the right. Would that change the force? Would it change your calculation?
The area you have shaded would be correct if there were also water to the right. In that scenario, can you see a way to split the damper into two parts such that the water on the left is responsible for the buoyancy force on the left part and the water on the right responsible for that on the right part?
And what I have marked in the figure at the beginning of everything, neither?

Guillem_dlc said:
And what I have marked in the figure at the beginning of everything, neither?
What is highlighted in post #1 is correct, but I looked at your working and you do not seem to have subtracted the triangular portion. I see ##A=\pi r^2\frac{\theta}{2\pi}##.

haruspex said:
What is highlighted in post #1 is correct, but I looked at your working and you do not seem to have subtracted the triangular portion. I see ##A=\pi r^2\frac{\theta}{2\pi}##.
They did. it's a typo.

Guillem_dlc said:
We calculate ##A\rightarrow A=\dfrac{\pi r^2\cdot 53,13}{360}=\dfrac{6\cdot 8}{2}=22,3647\, \textrm{m}^2##

Guillem_dlc said:
We calculate ##A\rightarrow A=\dfrac{\pi r^2\cdot 53,13}{360}-\dfrac{6\cdot 8}{2}=22,3647\, \textrm{m}^2##

Guillem_dlc and haruspex
erobz said:
They did. it's a typo. Should read:
Aha!

Guillem_dlc and erobz
The only potential issue I can see with this method of converting the problem (which, isn't an issue for the questions you had to answer), what would you do if you actually had to locate the center of pressure ##CP (x_{cp},y_{cp} )##? I have a feeling you would find it in a place where it doesn't seem like it should be w.r.t the actual scenario. The question: Is it just a simple transformation back to the actual scenario?

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## 1. What is a Tainter Damper Figure?

A Tainter Damper Figure is a type of mechanical device used to control the flow of water in a dam. It consists of a curved gate that can be rotated to adjust the amount of water passing through the dam.

## 2. How does a Tainter Damper Figure work?

The Tainter Damper Figure works by using the force of the water to rotate the curved gate. This gate can be adjusted to control the amount of water flowing through the dam, which helps to regulate the water level and flow downstream.

## 3. What forces are involved in the analysis of a Tainter Damper Figure?

The analysis of a Tainter Damper Figure involves the forces of gravity, water pressure, and the force required to rotate the curved gate. These forces must be carefully considered to ensure the proper functioning of the dam.

## 4. What is the purpose of analyzing forces in a Tainter Damper Figure?

The purpose of analyzing forces in a Tainter Damper Figure is to ensure the structural integrity and functionality of the dam. By understanding the forces at play, engineers can design and maintain the dam to withstand the pressure and flow of water, preventing potential failures.

## 5. What are the potential risks associated with a Tainter Damper Figure?

The potential risks associated with a Tainter Damper Figure include structural failure, damage to the surrounding environment, and disruption of water flow. Proper analysis and maintenance of the dam can help mitigate these risks and ensure the safety and functionality of the structure.

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