The reduction in computational effort-urgent

Peterz · April 18, 2011, 22:59

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 !

k_k · May 2, 2011, 04:39

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

April 18, 2011, 22:59	The reduction in computational effort-urgent	#1
Peterz 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, 04:39		#2
k_k Member Kailash Join Date: May 2011 Location: London, UK Posts: 45 Rep Power: 14	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