integration points

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lvy
Posts: 26
Joined: Tue Dec 14, 2004 8:22 pm
Location: Shanghai, China

integration points

Post by lvy » Mon Mar 14, 2005 6:18 am

Hello, everyone!

In the command, "element nonlinearBeamColumn $eleTag $iNode $jNode $numIntgrPts $secTag $transfTag <-mass $massDens> <-iter $maxIters $tol>". "$numIntgrPts" mean number of integration points along the element, and the help document told us two integration points at the element ends. For example, $numIntgrPts equal to 5, so it should be Fig b.
node— —1— —2— —3— —4— —5— —node
(Fig a)

(node)1— —2— —3— —4— —5(node)
(Fig b)
But I output the defomation of section 5 of element 1 and the defomation of section 1 of element 2. If Fig b is true, two section is same. In fact, it is different. I think the Fig a is right.

fmk, do you tell which is right? Do the specified number of integration point include the node?

berktaftali
Posts: 68
Joined: Fri Jul 02, 2004 6:10 am
Location: Computers and Structures, Inc.

Post by berktaftali » Mon Mar 14, 2005 7:10 am

nonlinearBeamColumn uses Gauss-Lobatto integration rule. Looking at the code below, you can see that there are always two points at the element ends (-1 and 1).

Code: Select all

void
LobattoBeamIntegration::getSectionLocations(int numSections, double L,
					    double *xi)
{
  switch(numSections) {
    
  case 2:
    xi[0] = -1.0;
    xi[1] =  1.0;
    break;
    
  case 3:
    xi[0] = -1.0;
    xi[1] =  0.0;
    xi[2] =  1.0;
    break;
    
  case 4:
    xi[0] = -1.0;
    xi[1] = -0.44721360;
    xi[2] =  0.44721360;
    xi[3] =  1.0;
    break;
    
  case 5:
    xi[0] = -1.0;
    xi[1] = -0.65465367;
    xi[2] =  0.0;
    xi[3] =  0.65465367;
    xi[4] =  1.0;
    break;
    
  case 6:
    xi[0] = -1.0;
    xi[1] = -0.7650553239;
    xi[2] = -0.2852315164;
    xi[3] =  0.2852315164;
    xi[4] =  0.7650553239;
    xi[5] =  1.0;
    break;
    
  case 7:
    xi[0] = -1.0;
    xi[1] = -0.8302238962;
    xi[2] = -0.4688487934;
    xi[3] =  0.0;
    xi[4] =  0.4688487934;
    xi[5] =  0.8302238962;
    xi[6] =  1.0;
    break;

  case 8:
    xi[0] = -1.0;
    xi[1] = -0.8717401485;
    xi[2] = -0.5917001814;
    xi[3] = -0.2092992179;
    xi[4] =  0.2092992179;
    xi[5] =  0.5917001814;
    xi[6] =  0.8717401485;
    xi[7] =  1.0;
    break;
    
  case 9:
    xi[0] = -1.0;
    xi[1] = -0.8997579954;
    xi[2] = -0.6771862795;
    xi[3] = -0.3631174638;
    xi[4] =  0.0;
    xi[5] =  0.3631174638;
    xi[6] =  0.6771862795;
    xi[7] =  0.8997579954;
    xi[8] =  1.0;
    break;

  case 10:
    xi[0] = -1.0;
    xi[1] = -0.9195339082;
    xi[2] = -0.7387738651;
    xi[3] = -0.4779249498;
    xi[4] = -0.1652789577;
    xi[5] =  0.1652789577;
    xi[6] =  0.4779249498;
    xi[7] =  0.7387738651;
    xi[8] =  0.9195339082;
    xi[9] =  1.0;
    break;
  }
  
  for (int i = 0; i < numSections; i++)
    xi[i]  = 0.5*(xi[i] + 1.0);
}
The reason you're having different responses for 1st and 5th sections may be due to the loading or the boundary conditions...
Berk Taftali
Georgia Institute of Technology
Ph.D. Candidate, Structural Engineering, Mechanics, and Materials
School of Civil and Environmental Engineering
Atlanta, GA 30332 USA
Email: gte994y@mail.gatech.edu

lvy
Posts: 26
Joined: Tue Dec 14, 2004 8:22 pm
Location: Shanghai, China

Post by lvy » Mon Mar 14, 2005 4:41 pm

Thank you! From the scripts, we know that Fig b is right. For example, $numIntgrPts equal to 5, and length of element is 2. We aquire the deformation of section 2.

fmk has said,
for the section it's the angle .. we try to update the documentation as users post questions like this .. keep them coming .. but place them in the documentation section.
berktaftali's scripts is
case 5:
xi[0] = -1.0;
xi[1] = -0.65465367;
xi[2] = 0.0;
xi[3] = 0.65465367;
xi[4] = 1.0;
break;
So the curvature is doformation/0.65465367. And the curvature of section 3 is doformation/1, the curvature of section 4 is doformation/1.65465367.......

If it is right, we couldn't find the curvature of section 1. Bcause the curvature of section 1 equal to doformation/0.

berktaftali
Posts: 68
Joined: Fri Jul 02, 2004 6:10 am
Location: Computers and Structures, Inc.

Post by berktaftali » Mon Mar 14, 2005 7:07 pm

Not necessarily. Those values represent the x coordinates of the integration points along the element. You don't need any of those to calculate the curvature at a section. I posted that for you to see that the two of the integration points are always at the ends of the element.

Use the element recorder to monitor section deformations (deformation option).

The following script will record the deformations at the 1st and the 5th sections of a beam-column element. For a fiber section, the recorded values will be axial deformation and curvature.

Code: Select all

recorder Element -file "section1.def" -ele 1 section 1 deformation
recorder Element -file "section5.def" -ele 1 section 5 deformation
Hope this helps...
Berk Taftali
Georgia Institute of Technology
Ph.D. Candidate, Structural Engineering, Mechanics, and Materials
School of Civil and Environmental Engineering
Atlanta, GA 30332 USA
Email: gte994y@mail.gatech.edu

lvy
Posts: 26
Joined: Tue Dec 14, 2004 8:22 pm
Location: Shanghai, China

Post by lvy » Mon Mar 14, 2005 8:00 pm

Thanks for your help, berktaftali.

I see. For a fiber section, the recorded values will be axial deformation and curvature. So what fmk said is wrong.

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