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Distributed vs concentrated loads

  1. Dec 11, 2012 #1
    According to my book(mechanics of materials) the definition for concentrated load is when the area of contact of one body is small in comparison to the total area of the other body. So, small area in contact with big area equals concentrated load, but was watching an example on you tube that confused me. He was saying that when the foot touch the ground when walking it received a distributed load from the ground on the foot. Why? Area of foot < area ground

    Thanks
     
  2. jcsd
  3. Dec 11, 2012 #2

    rock.freak667

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    When you get down to it, all of your loads are really distributed loads. But say you are considering the load on a 60 foot long beam and you place a a 1 inch x 1 inch mass somewhere along it, in that case you can consider the mass to be a point load.

    If you were considering the a 2inch long beam, then mass would be distributed.

    In the video, the ground they were most like talking about was not the entire area of the country, but just the small piece surrounding the foot.
     
  4. Dec 11, 2012 #3
    Ok thanks
     
  5. Dec 12, 2012 #4
    I have a question with this one.

    1.Why they didnt considered Fcd(4/5) when calculating moment at A? Is it Because when multiplied by the distance is zero? like so, Fcd(4/5)(0)

    2. How do I determine the values of the triangle of Fcd? When I took static I never had to find them. BTW, they are h=5,y=3,x=4
     

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  6. Dec 12, 2012 #5

    SteamKing

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    1. Fcd (4/5) is the horizontal component of Fcd. Since it is inline with point A, there is no moment produced about A.

    2. Don't you know trigonometry? You can find the component ratios from the sides of the 3-4-5 triangle.
     
  7. Dec 12, 2012 #6
    no..........
     
  8. Dec 12, 2012 #7

    SteamKing

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    How about Pythagoras?
     
  9. Dec 12, 2012 #8
    how do I get an hypotenuse of 5 from square root of 1.5^2+2^2?
     
  10. Dec 13, 2012 #9

    SteamKing

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    You are mixing up different parts of the force Fcd diagram with the dimensions of the frame. The triangle attached to Fcd is 3-4-5, which satisfies the Pythagorean theorem.
    The diagonal support measures 2 meters from A to C, and 1.5 meters from A to D.
    The distance from D to C is not 5 meters but 2.5 meters. Therefore, triangle ACD is similar to a 3-4-5 triangle. Knowing the magnitude of Fcd (12,262.5 N), you can work out the components by using ratios.
     
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