How to mind your P’s and q’s ?

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 I-beam.

 

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 I-beam cross-section:

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 articleJ).
  

  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 Z-direction 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

Comments

[J.D. Mather]

But where do we see this in the real world?
It seems there is a mis-guided attempt to validate software tools by showing that the results (if the only the problem is set up correctly) will match our "college try" hand calculations.

While setting up the problem correctly is certainly a requirement - since when is hand calc a validation or computer calc a validation?

The schematic simplification for hand calcs or the digitial model for computer calculations is only validated by physical experimentation and evidence.

Where, in the real world, do we set up a beam like this on infinitely thin edges and a point load (at least in your Inventor model you set up with a split face for a load of some area - whether the area used represents any real world application or not is open to discussion).

My thinking is we need to devise a whole new set of more realistic models for teaching/learning how to use modern analysis tools.

All this excercise says to me is that the initial Inventor model was not set up correctly if we expected it to show the same displacement as the equally flawed (in terms of real world application) schematic representation and simplified calculation.

Am I wrong? Where is the flaw in my thinking?

[Dan Margulius]

From this example we learn a lot of tools.
How to extract Moment of Inertia, how to apply forces and constraints. AD you are lucky that the sketch in the example is projected and there are no need for dimensions or else JD would send you the pdf tutorial he sends everybody in the discussion groups. Keep up the good work.

[Bob Van der Donck]

To J.D.'s point, I was not trying to asses the reality level of what we learn in school. I was just trying to make Inventor do what we learned. :-)
To Dan's point, yes I fully realize that I am playing here on J'D's home court and that he can deliver a home run any minute.
Besides it looks like J.D. has big plans to renovate the stadium.
Bob