Beer & Johnston statics problem 6.6 – FEM solution

I’m under the impression that statics and strength of materials is the base for machine design. It’s important to master these two topics in order to better understand and design more complex elements. I will be going into some iconic problems from Strength of Materials and statics courses and applying a FEM solution to it. So far from what I could see the solution is very neat and quick, as well as reliable.

I will be using the Beer & Johnston Vector Mechanics for Engineers Statics 9th edition book and will take problems and the theory from this book.

Chapter 6 deals with Structure analysis, it has different constructs that are usually made of bars, the crucial principle while calculating this manually is that the forces are transferred through the links between the bars. This way we can perform the truss analysis manually which, after a bit of practice, doesn’t take much thought but sometimes gets hairy. I set this problem up and solved it yesterday in less than five minutes which was both amazing and fun for me.

Let’s begin. The problem statement is as follows:6.6



Geometry sizes

Much like the manual solution, the simplification we use is of a linear nature, the whole body is made of lines which later on are attached a cross-section, the results we are looking for in this problem do not depend on the type of material nor the cross-section area. Or in other words, the forces in the bars are not a function of the young modulus or the cross-section area.

An arbitrary cross-section, the radius of 0.01[m] was assigned to all of the lines:

Cross section

And now we can move to the analysis window!

First the mesh, I just took something that would be quick and surely work, 10 [mm] minimum element size and we got this:

0Mesh is pretty straightforward, and you can also see the nice connection between the bars. The boundary conditions supplied were applied to points E and F. E is a fixed support meaning it doesn’t allow movement in any of the axes while F is a Roller support, allowing movement in the X direction (which is the right side in the following figure.


The “probe” tool was used to present these forces acting on the line elements, and of course, the sign tells the direction of the force, or in other words compression or tension in the bar.

next, let’s compare with the solution the manual calculation yields:

Force AB = Force BD = 0 [N]
Force AC = 675 [N]
Force AD = 1125 [N]
Force CD = 900 [N]
Force DF = 675 [N]
Force EF = 1800 [N]
Force CE = 2025 [N]
Force CF = 2250 [N]

The largest error I see is in Force CF, and it’s 55 [N] in magnitude, this is a rather small error which we could expect from different accuracy in calculators for example. I’m not sure why AB and BD do experience some force while it should be 0 according to the mathematical model, but I do assume this is because ANSYS calculates these things differently (FEM).

I deem this experiment a success and will continue to calculate other examples


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