Boundary Condition for Underground Berried Vessel

niyamatniyamat Member Posts: 1

Dear All,

I am performing a Static Structural Analysis of mounded Bullets used for LPG gas Storage.

The vessel will be buried in the sand and the external loads will be acting on the shell due to sand load. I have applied load according to the above snap however I want to want to apply boundary conditions to prevent rigid body motion and have fixed end vertices on both dish end, and I am getting the error as "There is at least 1 small equation solver pivot term (e.g., at the UX  degree of freedom of node 2692). Please check for an insufficiently constrained model."

If I fix the node in x,y,z direction it performs analysis however maximum stress occurs at that location.

Kindly advise.

@peteroznewman

Comments

  • peteroznewmanpeteroznewman Posts: 11,404Member
    edited January 12

    @niyamat

    I can't see the triad, so I will assume the axis of the tank is along the X axis.

    Delete both Displacement BCs. As you have found, they are over-constraining the body.

    There are 6 rigid body DOF that have to be constrained.

    Since the geometry and loads are symmetric about the center plane (parallel to YZ), I suggest you slice the body on that plane, and delete one half.

    In mechanical, insert a Symmetry folder and insert a Region with the direction of X-axis, then select the cut edges. This constrains 3 of the 6 DOF. The remaining three DOF are Y, Z and rotation about X.

    Create a Deformable Remote Displacement on the circular edge at the dish end. Set Y, Z and rotation X to zero and leave the other DOF Free. Now all 6 DOF are constrained.

    If you don't want to use symmetry, create another Deformable Remote Displacement on the circular edge at the other end of the dish. Set X, Y, Z to zero and leave the other DOF Free. These two BCs exactly constrain the rigid body motion without overconstraint.

    You must change the Remote Displacement to Deformable because this adds no stiffness to the model. If the Remote Displacement behavior is Rigid, you will add stiffness to the model and get the wrong answer.

  • niyamatniyamat Posts: 4Member

    @peteroznewman

    Dear sir,

    Thanks for your reply.

    Sorry, I missed out triad in snapshot, yes the vessel axis is along the X-direction.

    Currently, I have used remote displacement on both ends. result appears to be feasible now.

    Hope this is OK.

    Thanks for your recommendation.

  • peteroznewmanpeteroznewman Posts: 11,404Member
    edited January 12

    @niyamat

    You scoped Remote Displacement to four faces, you can do four edges, which will add fewer equations to the solution, but faces will work fine also.

    You show the first remote displacement, but you don't show the second one. They are not the same! On the second one, set Y, Z and rotation X to zero and leave the other DOF Free.

    If you use the same XYZ set to zero at both ends you are overconstraining X again, but spreading the effect out over 8 faces so you won't even see that you have done that unless you Probe the Reaction Force on the remote displacements and see a large force in X that should be close to zero.

  • niyamatniyamat Posts: 4Member

    @peteroznewman

    Please refer below, scoped remote point to edges, Also the reaction Probe result attached as you pointed out Reaction in X direction is almost Zero.

    Thanks for your Valuable Support.



  • peteroznewmanpeteroznewman Posts: 11,404Member

    @niyamat

    It was good that you put the Force Reactions into your reply. The goal is that the forces should basically be zero, but you can see there is a large force in the -Y direction. This says that the pressure load on the top of the tank is too large and is not balanced by the pressure load on the bottom of the tank.

    I recommend you reduce the pressure on the top of the tank until the Y component of reaction force comes close to zero, within +/- 1000 N anyway.

  • niyamatniyamat Posts: 4Member
    edited January 14

    @peteroznewman

    Pressure on top of the tank cannot be reduced as it is buried in sand hence sand load will act on the tank.

    I have tried to check with symmetry region, due to stiffener location I cannot use symmetry in x-direction hence I have used symmetry in Z direction.

    Remote displacement With one end x, y, z direction zero and at other end Y, Z and rotation X to zero.

    Maximum Stress is at the junction of the face where remote displacement is applied. Which is at the same location without symmetry model. Also large force in the -Y direction.

    Can you advise how to overcome this large force apart from reducing top load?

  • peteroznewmanpeteroznewman Posts: 11,404Member
    edited January 14

    @niyamat

    The Y reaction force must be close to zero. If you can't reduce the pressure on the top, then increase the pressure on the bottom.

    Adding symmetry with a Z normal means you don't need some constraints on the Remote Displacements, but it won't hurt to leave them in. Remote displacement with one end X, Y is zero and other end Y is zero.

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