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Forces Due to Static Fluids

  1. Sep 8, 2016 #1

    CGI

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    1. The problem statement, all variables and given/known data
    Find height (h) of the water level which can result in failure of dam by tipping it over due to water pressure at U/S of dam. Also draw the free body diagram.

    Density of water to be = 1000 kg/m^3 and its specific weight=62.4 lb/ft^3
    Hints: The specific weight of concrete is = 150 lb/ft^3.
    Width of dam = 1ft.

    IMG_1005.JPG

    2. Relevant equations

    Fr = (Specific Weight)(Height to Centroid)(Area)


    3. The attempt at a solution

    I've thought about rearranging the Fr equation so that I get
    h = Fr/(Specific Weight * Area)
    But the part where I'm confused is where to get the Fr from.

    I thought that I could get Fr for the concrete by using
    150 lb/ft^3 for the specific weight
    (12ft * 1 ft*) As the Area
    If I were to go about it this way(which I'm not even sure if it is right) how would I go about finding the height
    to the centroid?

    I tried to draw the free body diagram and I know that the pressure due from the water is in a prism-like shape as shown, but this is all I have. The resultant force from the water is also shown on there.
    IMG_1006.JPG

    I hope I used the template right! If anyone could help me on this, I would really appreciate it!
     
  2. jcsd
  3. Sep 9, 2016 #2

    NascentOxygen

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    Staff: Mentor

    I think you'd be allowed to consult tables to get the centroid of that shape. Either the trapezium directly or work it out as the centroid of two combined simple shapes using a formula you can look up.
     
  4. Sep 9, 2016 #3

    CGI

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    Okay, so once I find the centroid, I use that to calculate Fr? And then from there I can use that to find the height to the center of pressure?
     
  5. Sep 9, 2016 #4

    NascentOxygen

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    Staff: Mentor

    The centroid of the concrete's cross-section shape is where you can assume all its weight is concentrated, for the purpose of determining moments, etc.
     
  6. Sep 13, 2016 #5

    rude man

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    Homework Helper
    Gold Member

    I would integrate the moment of the water pressure from top to bottom about the tipping axis A and equate that to the moment generated about the same axis by gravity. That is safer than assuming some sort of "center of pressure" since the moment arm of the water force is a function of depth, not only in length but also in direction.
     
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