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Water invert house, fill it fail?

  1. May 17, 2009 #1
    Hello Every one,

    I am trying to design water vessel which shaped like an inverted house (i.e. a rectangle attached to a triangle). I want to find the pressure and force distribution along the walls of tank (it is open to atmosphere). I have tried solving this in excel using the equation

    P=P(atm) + (density)(gravity)(depth).

    I plotted the graph and it gave me a y=x graph (which makes sense to me)

    However, when i plotted the force curve,

    F=A*P=(length of tank)(dh)(pressure)=(density)(gravity)(length of tank)(depth)^2

    I graphed the function in excel (please attached photo) and got a curve[where y-pressure and x-depth. However it would seem to me that slope of the graph should change once the graph reaches the starting depth of the triangle (the top part of house) because the angle changes. Am i right?

    1Question: What would a force distribution curve (depth vs pressure) look like for an inverted house?

    2Question: In analyzing if an inverted house water tank might fail due to the weight of the water, what is the best way to approach this? (von mises stresses, ect...)

    NOTE: the tank is made of hard plastic. The red section of the curve is the point where the walls start tapering off to triangular point.

    ps. not sure if this is in right section, please move if necessary.

    Attached Files:

  2. jcsd
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