Solved Example: Hydrostatic Forces on a Vertical Gate
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- Опубликовано: 28 авг 2024
- MEC516/BME516 Fluid Mechanics: A simple solved exam problem of hydrostatic forces on a flat vertical gate. The solution includes locating the line of action i.e, the center of pressure.
All of the videos in this course, sample exams with solutions, and a copy (pdf) of this solution can be downloaded at www.drdavidnayl...
Minor Correction: There is a "typo" at 6:46. h_cg=2.8m (not 1.8m) . But the value of y_cp and the problem answers are all ok.
Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.
#fluidmechanics #fluiddynamics #fluidstatics
In 7 minutes I have learned more about Fluid Mechanics than I have in an entire semester so far with my professor. Thank you.
Thanks. Happy to help.
thx for the video-
little mistake at 06:48, hcg=2,8m !
Thanks. Yes That's a "typo". The value of y_cp is correct. I will make a note of this correction in the description. That's probably better than deleting this video. Thanks again.
true
OMG no way i just three hours until i just saw your comment.
thanks for the video though was very well explained, i understood it immediately as opposed to my lecturer that rambled for 2 hours straight
me attemping to learn this at 3 am and my exam is in 5hours 😪😪
Did you pass? Boutta take my final in 8 hours
Very clear explanation...thanks🙏
Can you please explain why the yp or the hp is different for vertical surfaces that protrude from the surface or whose tops are levelled with the surface compared to when the surface is below water?
I'm not sure I understand your question. The height of the centroid (h_cg) is the distance measured from the liquid surface. The centre of pressure (y_cp) is the distance below the centroid. A portion of the surface that protrudes above the liquid has no effect, because there is no hydrostatic pressure above the free surface.
Very well explained
Thanks. Glad to hear it was helpful.
Thanks for this, very clear explanation
Glad to hear it was helpful.
well explained, thank you.
Glad to hear it was helpful
At the end, Why did we write 1.8 intead of 2.8 ?
See the comment in the video description.
Thanks ✋🏻
For the last question why is the equation not : ycp = Ixx/hA +h
ycp is measured from the center of gravity, not the free surface. It's the the distance from the h_cg, with the negative sign indicating its below the cg. So, the distance from the free surface to the center of pressure is h_cg + abs(y_cp). I hope that helps.
How would I go about finding the reaction forces at points A and B?
Basic statics: Once you have the horizontal force at B, then set the sum of the forces in the x-direction equals zero to get F_Ax and set the sum of the forces in the y-direction equals zero to get F_Ay. Actually, F_Ay is zero by observation.
Well, I was expecting you to use the prism way to resolve it too. :)
Yes. That works too.
thanks . please share the link for the first culculation videos.thanks
All the videos are available at my website www.drdavidnaylor.net I hope that helps.
02:05 , free surface has Pa, a little correction i think??
No correction is needed because Pa also acts on the right side of the gate. So, the effect of atmospheric pressure cancels out.
I'm confused, why is the value of Ixx divided by 12?
The second moment of area of a rectangle about a horizontal axis through its centroid is the width*height^3/12. Google "second moment of area of a rectangle"
What is the distance 1.8 meters??
In the video description: Minor Correction: There is a "typo" at 6:46. h_cg=2.8m (not 1.8m) . But the value of y_cp and the problem answers are all ok.
I was looking for the correction. Thanks for the correction