How to correct hourglass energy in Radioss
Altair EmployeeHello everyone,
As we all know, hourglass energy is not a physical energy and should be small. However, in my simulation, the hourglass energy was found to be large (figure as follow). So how to decrease the hourglass energy? Which options and parameters should I use?
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By the way, what's the possible reason for the dropping of total energy?
Best Wishes
Roy
Answers
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Hi Roy,
Normally hourglass occur in under integrated elements. Plot part wise energy plots and find out which part is responsible for higher hourglass energy. The recommended parameters for avoiding hourglass , for shell elements, are Ishell=24 (QEPH formulation) and keep N=5 (number of integration points) and for solid elements use Isolid=14.
Total Energy of the system should be somewhat constant throughout the run. Correct the hourglass energy, run the analysis and monitor the total energy..
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Hi,
Similar to Ishell=24 for shells, you can also use Isolid=24 (HEPH) which will run faster than the fully integrated Isolid=14.
Thanks,
Andy
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Altair Forum User said:
Hi Roy,
Normally hourglass occur in under integrated elements. Plot part wise energy plots and find out which part is responsible for higher hourglass energy. The recommended parameters for avoiding hourglass , for shell elements, are Ishell=24 (QEPH formulation) and keep N=5 (number of integration points) and for solid elements use Isolid=14.
Total Energy of the system should be somewhat constant throughout the run. Correct the hourglass energy, run the analysis and monitor the total energy..
Hi George P and Andy,
Isolid=24(HEPH) was tried, and I found the hourglass energy of corresponding solid part was zero. As far as I know, under integrated element will introduce hourglass energy, why the hourglass energy was zero when Isolid=24 (I think Isolid=24 HEPH element is under integrated element)?
The hourglass energy seems normal, but the total energy still drops (I think the total energy should be constant), what's the possible reason?
<?xml version="1.0" encoding="UTF-8"?>
Best Wishes
Roy
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Hi Roy,
HEPH is an improved reduced integration element with physical hourglass stabilization. The method of reduced integration with hourglass stabilization is one of the most powerful techniques in the field of finite element technology, which leads to very efficient finite element formulations. Please go through the theory manual of RADIOSS which gives details on physical hourglass formulation of HEPH elements.
Total energy is the sum of internal energy, kinetic energy, contact energy, hourglass energy..etc in the model.Taking into account the external works in the model, the total energy must remain constant or can decrease slightly. Please refer First Checks during/after Computation in the Help Menu where the energy balance in a model is explained.
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Hi George P,
I have read the content you mentioned (First Checks during/after Computation). However, I don't think I understand it.
In my opinion, the energy conservation law should be satisfied. That means, T=Ti + external work
(T is the total energy of the system at time t, Ti is the initial total energy of the system at time t=0, external work is the work introduced by external force)
If the external work is positive, the total energy increase. Otherwise, the total energy decrease.
So how to understand 'Taking into account the external works, the total energy must remain constant or decrease slightly' in the manual? And what's the total energy mean which included in T01 file?
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Thank you
Roy
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Hi Roy,
Whenever work is done upon an object by an external force, there will be slight changes in the total energy of the object. If only internal forces are doing work, then there is no change in the total energy and the total energy is said to be conserved.
Please check this post which is similar to your case:
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Hi George P,
I agree the points you mentioned. That is Whenever work is done upon an object by an external force, there will be slight changes in the total energy of the object. However, in the manual, it says 'Taking into account the external works, the total energy must remain constant or decrease slightly'. I don't think the total energy must remain constant or decrease, it also can increase if the work done by external force is positive. I express my understanding in my previous answers.
I did a simple free-fall example. One object with an initial velocity and only gravity load was been imposed. The total energy history showed it linear increased with time as I thought.
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So how to understand the energy balance and the saying in the manual?
Roy
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Is there anyone explain the question in the previous floor?
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Hi Roy,
This is a case of free fall and the components are not hitting the ground and no energy is transferred from the system to the surroundings.
When negative work, that is energy is transferred from the system to the outside, is done by the external force, then the energy of the system decreases. I presume that in that perspective it is written in the manual.
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