3 Replies Latest reply on Nov 13, 2013 6:37 AM by Saravanamoorthy Mani

    Wind load on a structure

    Saravanamoorthy Mani

      I want to know correct method of analysis, velocity of air passes over the structure with height of 8m, when the velocity is converted  to pressure loading. How much accuracy we may get it?. or Any other mode of  transfer the load and find out max deflection of the structure within 4% of the height of structure.

        • Re: Wind load on a structure
          Jared Conway

          ideal method, fluid structure interaction. fluid flow problem that also bends your part. not possible with solidworks simulation tools.


          next best, decoupled FSI. run the fluid flow analysis to determine pressure loading, pass to solidworks simulation for analysis.


          next best after that. determine the pressure generated by your fluid at velocity, apply as a pressure or force to your components.


          to determine accuracy, write down the assumptions your method works and work out how different you think it may be. for example in #1 and #2 doesn't account for the change in the pressure when the structure changes. but that would potentially decrease the amount of deformation so you're actually working with a worst case loading with those methods.

          • Re: Wind load on a structure
            Rick McWilliams

            If your structure is extremely flexible like a fiberglass antenna you will have to interate the wind load calculation with the deformed shape. If your structure has smooth aerodynamic shapes, or shapes with thin wing like sections you will need to do a detailed aerodynamic analysis. The wind is never uniform with height. Most structures use constant velocity. The practical method is to assume the drag coefficeint of structural elements to be 1.0 for a round tube, and 1.4 for a square shape. Use the area of the element projected on a plane perpendicular to the wind direction as the area of each element. The pressure is simply the dynamic pressure 1/2 rho V^2 or at sea level in air in mph V^2/391 result is in pounds per square foot. Tubes that are angled 45 degrees are a bit less drag. Tubes do not hide behind each other unless they are almost touching.