### Freebody Diagram (FBD) of a Pinned Truss

In this post SG test, I wanted to discuss the freebody diagram (FBD) of a simple pinned truss. Interestingly, I had to draw this FBD in an interview a long time ago. Additionally, to learn more about the FBD of an angle bracket, check out the “Classical Hand Calculations in Structural Analysis” blog post.

I deliberately left the question marks in the image above because I wanted you think about this problem before diving into the solution below.

First of all, the key to the freebody diagram (FBD) of this system is of course the fact that all the joints are pinned. Consequently, there are no moments constrained at the joints. And yet, this system is a perfectly stable and a statically determinate system. Furthermore, we have enough equations to solve for all the unknown forces in this system.

### Freebody Diagram (FBD) Logical Steps:

The first thing we need to determine is which truss member takes what kind of load. So, let us start with the angular member.

1. The angular member is constrained at the top via a pinned constraint.
1. So the top point takes no moments
2. It can react both a horizontal and a vertical load
2. Next, consider the bottom truss member.
1. This member is the tricky one – pinned at the left
2. While the right joint can move in plane, the two members can rotate relative to each other here
3. In addition, the main assumption here is that the deformations are very small, or small strain, maintaining inter-member relative stiffness compatibility
4. Therefore, the angular motion of the bottom member is considered negligible enough that it stays basically flat
5. Consequently, the only load it can react is a horizontal load (it being flat) in this case
6. Also, its load will always be ‘compressive’
7. Why? Because any motion of the top member bottom point results in an arc going down and left, therefore compressing the bottom flat member
8. While it stays flat, it is incapable of taking any vertical load
9. The dominant constraint for vertical load is therefore the top pinned constraint

So given what we know, we can now figure out the freebody diagram (FBD) of this system.

### Static Equilibrium Equations:

So, let us slap some numbers onto this problem and get going. But before we get into that, remember the basic static equilibrium equations of any system:

Sigma F = 0 (‘along’ a given direction) – Force Balance, and

Sigma M = 0 (‘about’ a given direction) – Moment Balance

In this case, the material and physical properties of the system constituents or truss elements are assumed to be the same. So, no need to worry about those.

Therefore, we will only focus on the reaction loads and the static equilibrium equations needed to calculate them.

### Freebody Diagram (FBD):

Figure 1 below shows the dimensions we will use for this problem. And, it also shows the freebody diagram (FBD) loads and reactions of the assembly.

You can see the reactions at the top and bottom as we discussed above.

### FBD Force Balance:

Sigma Fv = 0

Therefore, $Rtv src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=100 lb$

Sigma Fh = 0

Therefore, $Rth src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=Rbh$

### FBD Moment Balance:

First of all, taking moments about the bottom left pinned point (and assuming CCW moments positive):

$Rth*H - P*L src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=0$

Hence, $Rth src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=P*L/H$

Hence, $Rth src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=100*10/15 src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=66.67lb$

Therefore, $Rth src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=Rbh src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-=66.67lb$

$Rtv src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-= 100 lb$

$Rth src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-= Rbh src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-= 66.67lb$, forming a force couple

### Freebody Diagram (FBD) FEM Model:

Finally I built a test model of the same problem in FEMAP using rod elements with typical properties, and using the same dimensions and applied load.

Figure 2 below shows the Freebody Diagram (FBD) loads using the 'Freebody' tool (covered in detail in the Bronze level course) in the 'Post Processing' Window Pane in FEMAP.

Note that in the FEM, in order to keep things stable, all out of plane rotations were constrained, and as this is a 2D problem. Additionally, the translations at the left are pinned, and DOF 4,5 are also constrained. The joint at the right is allowed to translate in plane only.

In conclusion, looking at the freebody diagram (FBD) loads, Figure 2 reactions clearly match the hand calculations we performed.

## Aircraft Structures Modeling Course

### Problem Variations and Self Study:

Also, just for fun, cut the vertical distance H in half and see what happens with your hand calculations.

Additionally, use a slight angle for the bottom member making it a little shorter, and make the top member a little longer. As a result, see what happens to the FEM freebody diagram (FBD) loads. Try it out, it will be fun.