Simply supported beam

Anyone beyond the first year in engineering recognizes this sort of problem and can solve it in their sleep, as you’re reading this you’re already thinking about the reaction forces, one of the supports allows movement in the X-axis and the other doesn’t allow any movement. Shear and moment diagrams come to mind and so on. During statics course alone I have probably analyzed over 50 simply supported beams among others, and that’s without even mentioning Strength of materials course. These problems can take between one to 5 pages to solve manually, depending on the complexity of the problem.

I would like to explore the usage of ANSYS in the solution of these problems, extracting stress, strain and deformation values, which would normally take some formulas and some pages, instead, takes about 5 minutes of simulation setup. Which was an amazing realization for me, on top of that, because I’m a nerd, I’ll also look at the effect of mesh on the quality of the solution and compare it with the analytical solution.

This analysis has two parts, distributed and concentrated load

Let’s start with defining the problem:

fig-10-simply-supported-beam-with-uniformly-distributed-load-considered-in-example-4
Figure 1 – Mathematical representation of the problem

Let’s call the left support A and the right support B. Support A doesn’t allow displacement in both X and Y axes, while fixture B only allows movement in the X direction. And the load is distributed along the beam

moment of inertia square
Figure 2 – moment of inertia of square cross section

And for a 40 [mm] square this number will be I = 213333 [mm^4]
Beam length L = 1000 [mm]
The beam is made of structural steel with E = 210 [Gpa]
The distributed force P = 5 [N/mm]

The beam was created using the ANSYS modeling module, then exported to the mesh tool

Beam
Figure 3 – Beam geometry

 

As I’ve said before, the total length of the beam is 1000 [mm].

Next, the mesh was generated

 

Mesh.JPG
Figure 4 – Initial mesh

Element size is 20[mm], the elements are fairly large so that calculation time is reduced, I’ll include a finer mesh and the difference in results at the end of the article.

 

Boundary conditions are as follows:

 

bc.jpg
Figure 5 – Boundary conditions

 

A, as we’ve stated is a fixed support, in the current axial system not allowing movement in the Z or Y directions.

B, is a “displacement” B.C. which allows 0 displacement on the Y axis

C, Force applied normally to the top surface of the beam, the beam is 1000 [mm] long and the force is 5000 [N] which yields 5 [N/mm]

After setting up Boundary conditions and mesh the simulation is fully defined and can be calculated. The values I want ANSYS to calculated for me are:

  • Directional Deformation in the Y direction
  • Equivalent stress (Von Misses)
  • Normal stress
  • Force reactions in the supports

I’ve also added 2 line paths, one in the middle of the top surface and one in the center of the cross section, the latter represents the neutral axis while the first is for comparison with theory.

First things first, ANSYS didn’t like the mesh I made, which is why it auto-generated a finer mesh that looks like this:

 

Post mesh
Figure 6 – Calculation mesh

 

But I’ll fix it after the first round of results

Displacement: 1.4604 [mm] in the negative Y direction

 

Displacement
Figure 7 – Displacement, notice the direction and “min” / “max”

 

Equivalent stress: maximum 58.617 [Mpa]

 

Stress
Figure 8 – Equivalent stress

 

Normal stress: can see the maximum and minimum is the same magnitude but different “directions” this represents the compression and tension the different areas along the corss section

 

Normal stress.JPG
Figure 9 – Normal stress

 

Force reactions

Reaction 1

reaction 2

Reactions, as expected are the force\2 for each reaction, in this case 2500 [N]

 

 

 

b9780323068956000128_f012-002b-9780323068956
Figure 10 – Compression and tension magnitudes in a bending cross section and the neutral axis

 

Figure 10 is something you should be familiar with, the theory states that the neutral axis experiences little to no stress while under bending loads. let’s see what ANSYS says happens to the neutral axis.

 

Neutral axis
Figure 11 – Neutral axis stress

 

This is the path I’ve inserted along the beam so that I can extract the equivalent stress along the neutral axis (center of the cross section), it ranges from 5.4 [Mpa] at the supports to 0.068 [Mpa] in the middle, just to get an order of magnitude: the maximum stress experienced in the neutral axis is less than 10% of the maximum stress in the beam, and in the center it is close enough to 0 that we can round it to zero. One explanation I can stick to when considering these results is the method by which the program calculates the problem, which is through averaging over elements, I’ll check with a much finer mesh to see what happens in the neutral axis.

 

Top axis
Figure 12 – Path along the top surface of the beam

In figure 12 we can see a what stress a singular line on the top surface is experiencing, this correlates to what we’d expect from the theory and what we saw in the results earlier.

 

Now comes the comparison to what we know from the books and theory.

For displacement, we normally use this formula:

 

deflection formula
Equation 1 – Deflection formula

 

In our case:

(5*((1000)^4)*5)/(384*210000*213333) = 1.453 [mm]

(1.453 – 1.4604)/1.453 *100% =  0.5%

We can certainly live with a 0.5% error

 

Maximum stress from bending is calculated through:

 

Bending stress
Equation 2 – Bending stress

In our case:

M(max) = 1/2 *(M) * (L/2)  = 1/2 * P *L/2 * L/2 = (1/8)* 5*1,000^2 = 625,000 [N*mm]

y = 20 [mm]

I = 213333 [mm^4]

Max stress = (625,000 * 20 ) / (213333) = 58.5938 [Mpa]

(58.617 – 58.5938)/(58.5938) *100% = 0.03%

Which is definitely close enough for me.

As for shear force, it’s just:

P * L/2 = 5*1000/2 = 2500 [N]

Which is exactly the magnitude of the reactions.

 

One last thing I want to include is the extraction of a safety factor in ANSYS, which looks like this:

 

SF
Figure 13 – Saftey factor

The area that experiences the most stress is at a safety factor of 1, which would normally not be advised for actual use, as with errors in calculations and software errors this would mean the safety factor is -around- 1, so it could be a bit above or a bit below, and we want to avoid that.

 

 

 

Credit: DR. DALYO who is doing amazing work on reddit and youtube.

 

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