Stiffness Influence Coefficient Fea For Plate

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SUMMARY

The discussion focuses on the calculation of the stiffness influence coefficient matrix (K(IJ)) for rectangular elements in Finite Element Analysis (FEA), specifically within Nastran and similar software. The stiffness matrix for rod elements is defined as AE/L, where A is the cross-sectional area, E is Young's modulus, and L is the length. For solid and plane elements, the stiffness matrix is derived using principles such as virtual work and the Galerkin method, leading to the formulation of the stiffness matrix as K^e = h_e ∫_Ve [B^e]^T [C^e] [B^e] dxdy. The discussion also addresses the implications of dividing a line into multiple rod or truss elements and the accuracy of using beam elements.

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chandran
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i have put this question in other forum also.

I have understood the stiffness matrix generation for a single rectangular element as below. I consider each node having 2 translational dofs

each in x and y coord system. There are 4 nodes in this rectangular plate element and hence 8 dofs overall.


K11U1+K12U2+K13U3+K14U4+K15U5+K16U6+K17U7+K18U8= F1
K21U1+K22U2+K23U3+K24U4+K25U5+K26U6+K27U7+K28U8= F2
K31U1+K32U2+K33U3+K34U4+K35U5+K36U6+K37U7+K38U8= F3
K41U1+K42U2+K43U3+K44U4+K45U5+K46U6+K47U7+K48U8= F4
K51U1+K52U2+K53U3+K54U4+K55U5+K56U6+K57U7+K58U8= F5
K61U1+K62U2+K63U3+K64U4+K65U5+K66U6+K67U7+K68U8= F6
K71U1+K72U2+K73U3+K74U4+K75U5+K76U6+K77U7+K78U8= F7
K81U1+K82U2+K83U3+K84U4+K85U5+K86U6+K87U7+K88U8= F8


where K(IJ) is the stiffness influence coefficient saying the force at ith dof due to a unit displacement at jth dof.
U1,U2,U3 ETC is the displacement at ith dof.
F1,F2,F3 ETC is the force at the ith dof.


Now i have a question. In nastran or other fea packages how this K(IJ) is calculated for rectangular elements. I know that for rod element it is

AE/L where A is the rod area,E youngs modulus and
L is the length of the rod. But for solid elements and plane elements how this is calculated.

One more question regarding rod or truss elements. There is a line of 10mm length horizontal to x axis. Why shouldn't i divide the
line into 10 rod or truss elements. What will be the error? Can i divide the line with 10 beam elements?
 
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chandran said:
One more question regarding rod or truss elements. There is a line of 10mm length horizontal to x axis. Why shouldn't i divide the
line into 10 rod or truss elements. What will be the error? Can i divide the line with 10 beam elements?

I'm not sure I'm getting this 'easy' part ... can't see why you couldn't, and basic finite elements such as rods, beams etc. typically produce in practise identical results with analytical theories (as long as we're considering typical bending theories and leave nonlinearities etc. out) == extremely accurate, work well with pretty crude meshes.

chandran said:
Now i have a question. In nastran or other fea packages how this K(IJ) is calculated for rectangular elements. I know that for rod element it is

AE/L where A is the rod area,E youngs modulus and
L is the length of the rod. But for solid elements and plane elements how this is calculated.

This goes into the "heart" of FEM ... you can work with simple element formulations using rod, beam and so forth analogies, but in general it's preferable to derive the balance equations using "valid" method of derivation, which in case of FEM means either principle of virtual work, minimum of potential energy, variational derivation or the "best" of all (personally) using weak forms, such as the Galerkin method.

Considering the bilinear 2D solid element, its stiffness matrix is typically derived as follows (the notation I'm using is similar to links I've added to the end):

*Assuming infinitesimal strain - displacement relations and linear - elastic constitutive relations (plane stress or strain), write the Cauchy's equation of motion (equilibrium equation) substituting the kinematic and constitutive relationship to it.
*Using virtual displacements, write for plane elastic body :

[tex] \int_\textrm{Ve}(\sigma_{ij}\delta\epsilon_{ij}+\rho\ddot{u_{i}}\delta u_{i})dV-\int_\textrm{Ve}f_{i}\delta u_{i}dV-\int_\textrm{Se}t_{i}\delta u_{i}dS=0[/tex]

... so write the virtual displacement energy balance law element - wise

*Introduce the FE approximation functions / interpolations, along the lines
[tex] u=\sum_{i=1}^n u_{i}^e \psi_{i}^e(x,y)[/tex]

*substitute them to above and you'll end up with a typical FEA matrix equation :

[tex] [M^e]({\ddot{\Delta^e}})+[K^e](\Delta^e)-(f^e)-(q^e)=0[/tex]

*where the stiffness matrix you're after is:

[tex] [K^e]=h_{e} \int_\textrm{Ve} [B^e]^T [C^e] [B^e] dxdy[/tex]

... get the specifics of the derivation for example here :

http://www.vector-space.com/TourPE.pdf
http://www.mathsoft.cse.clrc.ac.uk/felib/Docs/html/Intro/intro-node10.html

although looking at common FEA software manuals or any book is probably the best way to go.
 
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