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April 18, 2011, 23:59 |
The reduction in computational effort-urgent
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#1 |
New Member
Peter
Join Date: Apr 2011
Posts: 1
Rep Power: 0 |
Hello everybody, I do have an urgent question..
I modeled a beam , using symmetry I just modeled one-fourth of the beam. I constrained the beam in cutting planes.. anyways. I just want to know by reducing the size of the model four time, ( the number of nodes , and number of elements) how much my matrix operation is reduced. how does it relate to the total computational effort ? I know the stiffness matrix will be one-fourth and the number of the components will be one/16th , but does this mean that the front matrix will be one/16th in each side !? noting that I just used normal solid elements with three DOFs at each node. please help me , your help in any extent is very much appreciated ! |
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May 2, 2011, 05:39 |
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#2 |
Member
Kailash
Join Date: May 2011
Location: London, UK
Posts: 45
Rep Power: 15 |
hi,
consider u have 'n' nodes. The size of your stiffness matrix will be 3n x 3n. If you reduce the number of nodes by 1/4 th, the size of the new stiffness matrix will be 3(n/4)x3(n/4). So you reduce n/16 numbers in the stiffness matrix. This implies a reduction in computational cost by (n/16)^2. Cheers |
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ansys, computational effort, matrix operations, symmetry |
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