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- 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?

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?