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Topic: FEM: strain-displacement matrix for isoparametric elements
Replies: 2   Last Post: Jul 10, 2012 1:50 AM

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 Jörg Posts: 3 Registered: 7/8/12
FEM: strain-displacement matrix for isoparametric elements
Posted: Jul 8, 2012 6:45 AM
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Hi group,

I'm currently dealing with the basics of the Finite Element Method: I want
to compute the strains of an element based on the nodal displacements. The
nodal displacements are given from a FE-Program which is using isoparametric
elements.

In [1] (section 4.2.5) it is given how to compute the strain vector '{eps}'
using the strain-displacement matrix '[B]' and the nodal displacement vector
'{q}':
{eps} = [B] * {q}

To compute [B] numerically I have assemble it from several sub-matrices
[B_i] (given at page 23 in [1]). The several [B_i] have to be computed by
inverting the Jacobian matrix [J]. [J] itself is composed of the derivative
of the shape function Ni.

What I do not understand:

- How many [B_i] I have to provide? Is i ranging from 1 to the count of
nodes within the element?

- I can easily derive the shape functions Ni for the local coordinates 's'
or 't'. But in general Ni * d/ds or Ni * d/dt remain dependet on 's' and/or
't'. Therefore I have to provide a value for 's' and 't' when I numerically
calculate [J]. But what value of 's' and 't' I have to provide for the
current [B_i] ?

In other words: for me it is not clear how to assemble numerically the
matrix [B] for a general isoparametric element. Any help would be much
appreciated.

Thanks in advance,
Jörg

[1]
http://homepages.cae.wisc.edu/~suresh/ME964Website/M964Notes/Notes/introfem.pdf

Date Subject Author
7/8/12 Jörg
7/8/12 AMX
7/10/12 Jörg

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