Why does the beam have bigger deflections than theory?.
The model set up is shown below.
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
It appears the model is 37% too flexible.
Answer:
The finite element model will include bending stiffness terms and shear stiffness terms 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.