RAM Frame distribution of wind loads in a rigid diaphragm accounting for changes in stiffness in shear walls caused by openings

I am trying to understand if RAM Frame takes into account the effect of openings in concrete shear walls on their stiffness when distributing story shears in a rigid diaphragm due to wind loading. To determine this I have modelled a square building, one side of which has shear walls with very large openings and on the opposite side of which it does not have openings. It has 3 floors all with rigid diaphragms. I then evaluate the shear induced on the walls by wind load in the direction of the shear walls with openings and without openings. Unfortunately, I have been alarmed to see that the shear forces are the same on each side... if there are 3 shear walls on 3 sides of the building. However, if I add a 4th shear wall, the walls with openings take on less shear than the walls without openings as expected! It is very strange. Adding another level of confusion is that the Center of Rigidity is very clearly in the right place and is taking into account this change in stiffness for both the 4 walled system and the 3 walled system. But, unless I specify a special load case for the CR, the button does not turn on, which indicates to me that RAM Frame does not compute torsional forces about the center of rigidity. Is it computing them about the COM? Essentially, I have two questions:

1. Does RAM Frame account for the change in stiffnesses due to wall openings when distributing lateral forces due to wind in systems with rigid diaphragms?

2. How does RAM Frame compute torsion for these systems? If not about the COR, why not?

  • 1. Yes, as you can see from the wall mesh, the openings are considered and the walls have a reduced stiffness when there are openings modeled. A wall with openings deflects more than a wall without openings under the same loads. 

    2. Torsion on the system is a natural product of the finite element analysis (whether the diaphragm is rigid or not). You can add a Center of Rigidity load case and then show the location of the COR, but it's optional to do so. For more see:  

    A point about your test, if you have a three walled open system like the image below, and assuming the out-of-plane stiffness of the walls is neglected, then there is a single load path for Y direction loads parallel. Consequently all the load goes to that wall regardless of it's stiffness. 



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    Figure 1: 3 walled model, reactions at base 

    Figure 2: Reactions at base of a 4 walled model 

    Figure 3: COR in 3 walled model and force case considered

      

    Figure 4: COR and load case considered for 4 walled building

    Figure 5: Roof of 3 walled structures with shears in each wall displayed

    Thanks for your reply. I now understand that RAM FRAME should be taking into account the change in stiffness per its documentation. However, the results from my experimental model are still very confusing and contradictory. My models (3 walled and 4 walled) are shown in Figures 1 and 2 with their respective Ry reactions at the base for the Y-load case. The walls with openings are along line 1 and should take on lower shear than that of the solid walls on line 4. However, surprisingly, in Figure 1 the walls with openings take on more load than the walls without openings. In Figure 2, with a 4th wall, for some reason the model behaves as expected and the solid walls do indeed take on more load than the walls with openings. 

    Figure 5 shows the resulting shears in the shear walls at the roof due to a Y-dir wind load case. It is very concerning to me as it appears that RAM Frame simply split the diaphragm load up equally among the 2 y-dir. lateral resisting systems and did not account for their differences in stiffness. 

    My initial thought was that shear due to torsion may have induced the perfect amount of shear to make the resulting total shears equally on both sides. However, I have tried changing building dimensions and with 3 walls, the amount of shear induced still remains the same on both sides. Additionally, Figures 3 and 4 show that the COR x coordinate remains very similar for both cases. 

    Another odd note is that both models have higher deflections in the shear walls with openings than the shear walls without openings, which indicates to me that RAM frame is appropriately lowering the stiffness of the wall panels with openings. 

  • I think you are on the right track with your comment about torsion, but think of it as an equilibrium check for moments about the z axis from the middle where the forces is applied like middle like so:

    As soon as you introduce a fourth wall the assumption that Fy=0 is no longer true. Another way to put it is like so - the 3 walled system has nearly zero resistance to torsion, any amount of applied torsion would result in huge diaphragm rotations.



  • Seth, you neglected to take into account that the center of rigidity is offset from center of mass significantly as shown in Figures 3-4 above. Here are my sample equilibrium calcs for distribution of forces in a rigid diaphragm:

    1. Solving for actual shear in each wall

    ∑Fy=V1+V2=0

    Then this shear is apportioned via proportion of stiffness of the overall stiffness. Ultimately, the overall shear is resisted and equilibrium is satisfied.

    2. Solving for additional shear induced by torsion
    The center of rigidity (as shown in previous pictures) is very far to the right, towards the rigid wall so the load is offset and with a rigid diaphragm the assumption is that the shear walls must resist this load

    Ex=71’-45’=26’
    Ey=0
    T=P*Ex

    This torsion is then distributed according to stiffness as well. Each wall would take the same amount of shear induced by torsion in a differing direction to create the resisting moment necessary so that equilibrium is satisfied for torsion.

    3. Total shear

    Total shear= Actual Shear + Shear induced by torsion

    I would not expect that to result in the same exact number for each member as each step of these computations accounted for distributing the shear based on stiffness of the members. Just in case, I have varied the building dimensions and still RAM is evenly dividing the shear force which indicates that it is not distributing the load according to relative stiffness and maybe that it is not accounting for torsion about the center of rigidity. Hence, my confusion and concern.
  • In my example the CR location is irrelevant, I'm doing an equilibrium check for a summation of moments about the vertical Z axis at the point of the applied load Fx.



    Answer Verified By: Gwen Carris