11 Replies Latest reply on May 28, 2010 8:02 AM by 1-M4OM40

Water Pressure Distribution

Poovaiah Devenira Nov 20, 2009 12:50 PM

Hi,

I am analyzing a Vertical Solid Plate under water;

The height of the plate is 14 ft, Length 56 ft

The Maximum water head is 80 ft.

I can calculate the pressure, stress and deflection mathematically, however since the force distribution is trapezoidal, I was wonder if any could help me on how to apply this force in SolidWorks simulation, so that I can analyze it.

Re: Water Pressure Distribution

Jim Sculley Nov 20, 2009 2:34 PM (in response to Poovaiah Devenira)

Look at 'Loads and Restraints' in the Simulation help file. There is a non-uniform pressure load option.

You'll have a linear equation where the pressure is atmospheric at the surface of the water and increases linearly with distance below the surface, giving your the trapezoidal profile:

P = P_a+ ρgh

Switching to x-y nomenclature, assuming y represents depth into the water and x is along the surface of the water:

P_(x,y) = P_a+ ρgy

As we know, pressure depends only on the depth.

Simplify the constants:

P = A + By

So, in the non-uniform pressure distribution dialog, you set all the terms involving x to zero. You put P_afor the constant coefficient A, and ρg for the y coefficient B. Or zero for Pa if you don't care about atmospheric.

Jim S.
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- Re: Water Pressure Distribution
  
  1-M4OM40 Mar 23, 2010 10:24 AM (in response to Jim Sculley)
  
  Hi,
  Could you give me an example?
  The tank Im trying (and failing) to analyze is 96 inches tall, at the deepest part it is 3.6 psi. I built a coordinate system at the top surface of the water, but
  
  I have no idea what to put into the empty boxes in the nonuniform section.
  
  also, should the y direction point up or down?
  
  Do you know of any full fledged examples online where they go through it step by step?
  
  Thanks
  Doug
  
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  - Re: Water Pressure Distribution
    
    Poovaiah Devenira Apr 12, 2010 11:12 AM (in response to 1-M4OM40)
    
    Hi,
    The attached document should give you a clear idea; Make sure the u have the coordinate system set up properly.
    The first box, has the pressure value (40.26 psi) at the coordinate point.
    0.036 is specific weight of water in lb/in^3
    
    Pressure.pdf 134.2 KB
    
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    - Re: Water Pressure Distribution
      
      1-M4OM40 Apr 28, 2010 12:11 PM (in response to Poovaiah Devenira)
      
      Hi,
      Sorry if i already sent this question, It didnt show that It worked and i dont think it did.
      
      What do I put for the pressure value. It wont let me use zero even though I want to start at zero and add pressure as it gets deeper.
      
      Please see the image, do I use 0.00001 just to make it happy?
      
      Thanks
      
      Tank Pressure.bmp 4.1 MB
      
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      - Re: Water Pressure Distribution
        
        Derek Bishop Apr 28, 2010 8:21 PM (in response to 1-M4OM40)
        
        The equation for pressure distribution takes the mathematical form y=m.x+b. Where y is the pressure calculated at a given depth, the value of x is the pressure per metre of depth, and the value b is the pressure at the free surface. We are working in gauge pressures. The x value is the one entered as pressure. The equation simplifies to y=m.x since b, the pressure at the free surface is zero. If x is zero, the value of y is zero. ie. The pressure at the free surface is zero.
        
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        Re: Water Pressure Distribution
        
        1-M4OM40 Apr 29, 2010 8:53 AM (in response to Derek Bishop)
        
        Okay then for your example equation you need X pointing downwards on the coordinate system, and have the coordinate system positioned at the top of the volume of water?
        
        Solidworks says taht the coefficients (the rate of increase in pressure) go in the 6 boxes, not in the above pressure value box. What goes in the pressure value box if the coefficient goes somewhere in the 6 boxes below? 1? Zero? .00001 because it wont let you put zero?
        
        Thanks!!
        
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        Re: Water Pressure Distribution
        
        Derek Bishop Apr 29, 2010 9:41 AM (in response to 1-M4OM40)
        
        I have the coordinate system at the fluid surface. And sorry my mistake, the gradient m is the pressure increase per metre of water. The depth x is the variable.
        
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        Re: Water Pressure Distribution
        
        1-M4OM40 Apr 29, 2010 9:47 AM (in response to Derek Bishop)
        
        At the fluid surface and what axis is vertical? X, Y, or Z? Pointing up or down?
        
        You put the actual depth dimension in one of the 6 boxes?
        
        If the variable goes below in the 6 boxes what goes in the pressure value?
        
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      - Re: Water Pressure Distribution
        
        Poovaiah Devenira May 28, 2010 12:44 AM (in response to 1-M4OM40)
        
        0.036 is specific weight of water in lb/in^3
        
        Pressure = Specific weight X height
        
        The box for pressure input = 1 psi
        The box with no variable = 0 psi ( pressure at the co-ordinate point)
        The box for x input = 0
        The box for y input = 0.036 lb/in^3
        
        If your y coordinate is pointing down form the surface of the water
        
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        Re: Water Pressure Distribution
        
        1-M4OM40 May 28, 2010 8:02 AM (in response to Poovaiah Devenira)
        
        This is the clearest answer, thanks!!
        
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Re: Water Pressure Distribution

Anthony Botting Nov 20, 2009 2:39 PM (in response to Poovaiah Devenira)

Hi Poovaiah: since you know the distribution equation for your force, you can just use the "Nonuniform Distribution" option. If not already done, define a coordinate system to relate to the distribution equation. Then, right click on "External Loads" and choose the Force or Pressure option. In the PM there should be an empty option box labeled "Nonuniform Distribution". Check that, and it should open up and you can put-in your coefficients. There is an example in the Help file (click on the "?" in the top right corner of the PM). Good luck. Tony
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