Sometimes questions pops up around Inventor's calculation of beam deflections in the Stress Analysis environment.
Everyone knows the theoretical formulas from our school days. But not everyone knows how to correlate these schoolbook formulas to Inventor's way of defining loads and boundary conditions.
Let's take an example of a single point load in the middle of an Ibeam.
Fig 1: Schoolbook formula for a suspended beam with a central point force
The concrete data that we want to use in these formulas are:
 P= 32000 N
 L = 9000 mm
 E = 220 GPa = 220000 N/mm^2
 I = Ix = 182630000 mm^4
The Ix value can be found in the region properties of the Ibeam crosssection:
Fig 3: Identifying the moment of inertia values of the beam
With these numbers the theoretical deflection in the middle of the beam is:
d = 32000 x (9000)^3 / 48 / 220000/ 182630000 = 12 mm
Let's now verify if Inventor's Stress Analysis gives us an identical result.
Without too much thinking, the majority of our users will translate the schoolbook sketch as follows:
 A force of 32000 N applied to the top face
 A fixed constraint on two edges at both ends of the beam
Fig 4: Initial setup of the load and boundary conditions
If you run the simulation and look at the displacement results in the Y direction, you will see that the maximum displacement value obtained is 4.37 mm
Fig 5: Y displacement results with initial setup
Why do we have this large discrepancy between the calculated (4.37) and the theoretical value (12.0)?
Well the reason is that we made two fundamental mistakes in the way we applied the load and the boundary conditions.

First the force P we defined is not really a single point force but is a force applied to the entire face and becomes more of a distributed q force (hence the title of the article
J).
So instead of using a point load P=3200 N, we actually performed the calculation with a distributed load q = 3200 N/ 2.7 m^2 = 11851 N/m^2 (with 2.7 m^2 being to top surface area of the beam).
This can easily be verified by using a pressure force of 11851 N/m^2. You will get the exact same result as when using the P= 3200 N force on the top face.
It is too bad that Inventor does not change the load glyph to convey more clearly to users what is going on.
 Second mistake is that we clamped down both endpoints of the beam by adding fixed constraints.
If you look carefully in the schoolbook example, one endpoint of the beam can freely float in the longitudinal direction.
The first problem can be remedied by applying the force to a small circular split face in the middle of the beam's top face (note that you cannot apply a force to a sketch point or a work point):
Fig 6: Applying the force to a tiny centrally located circular split face (marked in brown)
The second problem can be fixed by defining a fixed constraint on one end and a frictionless constraint on the other end (but this again would require you to create a small split face near the beam's end)
A more convenient solution when the beam is suspended in horizontal fashion, is to simply allow the Zdirection of the fix constraint to float by unchecking the z checkbox in the advanced options
Fig 7: Allowing a fixed constraint to float in the Z direction (= longitudinal direction of the beam)
If we then repeat the calculation, we get a deflection of 12.46 mm which is much closer to the theoretical result of 12 mm.
This value can be improved upon by refining the mesh and by allowing more mesh refinement steps in the convergence settings.
Fig 8: Y displacement of 12.46 mm after modifying loads and constraints