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Introduction to FE Analysis

Tutorial CD

Introduction to FE Analysis

Tutorial CD

The nastran beam model has length L = 60 ins, center load W = 10,000 lbs.

I beam is 10 ins deep , 8 ins wide. Wall thickness for flange and web is 1 inch

(I is 386 ins^4)

E = 1.07e7 psi.

Central deflection in model = .01573 ins.

Checking with Roark, d max = Wl^3/(48EI) = 0.011428 ins

for the beam elements. This is also true whether the structure above is created using

shells or solids. (more later on relative accuracies).

Euler beam theory does not take account of shear stiffness. So if you used the

equations out of Roark or similar you will get a stiffer result. For example for a cantilever

beam Roark has the term:

d = PL^3/(3EI) for bending stiffness, but the additional term is PL/(GA) to account for

shear stiffness.

Full solution is d = PL^3/(3EI) + PL/(GA)

We can simulate the lack of a shear stiffness term in a Nastran Beam by putting the K

shear correction term to a large number (1e6). This effectively knocks out the shear

stiffness term and simulates Roark. The default value of k is the effective shear area

ratio.

Rerun and the nastran central deflection = .01143 ins. This exactly matches Roark

calculation.

You can download a demonstration Nastran file. It is called 'beam_shear_test.nas'.

Both beam property types (I and rectangular) are in the deck, plus annotated

comment/uncomment lines for k.

The experiment is repeated with a rectangular 10 by 10 beam. I is now 833 in^4.

Roark gives d = 0.00505 ins.

Nastran with default K term = 0.00549 ins

Nastran with high K term = 0.00505 ins.

So in this case the effect of the missing shear stiffness in Roark is less (10%)because

the flexural stiffness term dominates.

With regards to variations in element types:

Beams should give excellent results as long as there are sufficient to follow the required

curvature (high order buckling or vibration modes need more).

Shells will give good results, but I recommend two or three shells minimum in the web to

get a good through depth shear distribution. The end constraints are more complicated

however. A method here is to link all grids at each end with a rigid element. The single

independent grid is at the centroid of the cross section. This is simply supported, or

fixed dependent on the test. This is an attempt to model the 'theory' that all sections

remain plane.

Solid elements require many more elements through the web depth to get a good

answer. the performance varies amongst types, but assuming you have a standard

Nastran type brick ( 8 noded Hexa) then at least 8 elements through web of the I beam.

If you have a higher order brick then it would be a good idea. End constraint issues are

similar to the shell. You want to strap all the grids at each end face to a single

supported grid with rigid elements. Again it keeps sections planar and replicates what

the theory is assuming.

Don't use Tetrahedral elements unless you are really forced to, and then never used 4

noded Tets. These will give bad results. If you have to use 10 noded Tets then make

the mesh in the beam very fine and as regular as possible. Avoid 'jaggies' which

destroy accuracy (high aspect ratio icicle type shapes).

There always comes a point where trying to benchmark against 'theory' is a diminishing

return because of the limitations inherent in the theoretical assuptions versus real world

applications.