# Solving truss member forces graphically

Is there a way to solve truss member forces of a complex truss like fink roof truss where the {Edit: external} forces and reactions are known, the angles of all the members and the total base length?

I have a text book problem in trying to solve. My difficulty is in how to resolve member forces, albeit with standard mathematical calculations. So this is why I'm wondering whether there a graphical method to estimate the individual member forces using just the mentioned givens and reactions?

Where the heck does the "ij" and "hi" come from? There is no labelled i or j joint.

Not a homework question but an educational experience.  • I am not sure so only posting as a comment. Since the forces are vectors with magnitude and direction, you should be able able to use vector geometry to figure it out. I could be wrong, its been a long time since I took that drafting class at university. Nov 1, 2022 at 0:35
• The problem is that I only have external forces. I can't figure out how to solve it mathematical as the answer given makes no sense to me at all! Nov 1, 2022 at 6:30
• @Rhodie You need to know both the applied forces and support reactions first, then draw them to scale to figure out the internal member forces. The article linked by "Dominik Kern" shows the way step by step.
– r13
Nov 1, 2022 at 12:45
• @r13 I think I will need to try it out. I'm just wondering whether it's possible to use just the angles to solve the problem knowing the actions and reactions and angles only given base length. Nov 2, 2022 at 9:23
• Yes. You can use the "method of joints", it is much easier than the graphic method for complex trusses.
– r13
Nov 2, 2022 at 13:12

So in order to figure this out graphically I first needed to figure out the reactions. This can be done either by taking the sum of the forces about A then Figuring out the Forces at E. Alternatively, since this is symmetrical a quick short cut is simply sum the vertical applied loads (30 kN), and dividing by the number of reactions (2) to get the reaction A both A and E since they will be equal due to the symmetry. Rya = Rye = 15 kN.

The next thing to consider with a truss is that there are only axial forces. So this means the geometry of the truss will control the direction of the force vector in each member. In this example you will note that the same coloured members are will carry the same vector magnitude because of the symmetrical loading and geometry. The next step is to draw you vectors to a an arbitrary scale. Whatever the scale you pick you will need to multiply the results by it to get your final force. In my example I used 1 kN as the same 1 m I used in the drawing the truss. Because I was in cad and had unlimited space this was not an issue for me. If you were doing it on paper you may need to use a smaller scale or get larger paper. The next I did was start at the reaction A. I did this as it had the fewest members at the joint. Since the sum of the forces at a node have to sum to zero we can use the sum of vectors to solve the unknown magnitude of the vector GA and BA. I started by drawing the reaction vector RyA to scale. You can then proceed clockwise or counterclockwise around the node and draw in the direction of unknown vectors. Note that each unknown vector is placed at either end of the known vector. Since we know that the vectors have to sum to zero, we know that they unknown vectors will need to intersect. Based on the direction of the RyA vector, the direction of the unknown vectors is determined. This process is repeated at node B. It is important to note that the direction of the vector at one member member end will be opposite at the other. For node B, The 2 of the 4 vectors are known. Draw in the FyB vector to scale and draw the same vector FAB that were determined at the previous at the same scale. You will know FCB will be attached to Fyb, and you know that FGB will be attached to FAB. Extend FCB and FGB until they intersect. When you have done this successfully you will wind up with something that looks like the following: Repeat this for node C and G. Note that in the next image vectors GF and GA are both horizontal and on top of one another. To get the magnitudes, measure the length of your vectors and multiply them by the scale you chose to use. Since this is a symmetrical system you can apply the magnitude to the corresponding members.

In my case I got the following results:

• AB=ED=27.0416 kN
• AG=EF=22.5 kN
• BC=DC=21.4946 kN
• GB=FD=8.3205 kN
• GC=FC=7.5 kN
• GH=15 kN

Vectors that point towards the node are in compression and the vectors that point away from the nodes are tension.

Alternatively normally loaded trusses like this have the top chord in compression and the bottom chord in tension. You can derive a convention from this to determine the interior member forces.

These values were also mathematically verified with a simple 2D online truss solver.

• Your answers are great but need to show relevant negative values where assumption of tension was incorrect. Thank you. Very comprehensive for anyone else coming to find educational answers. Jan 14 at 12:19

The method you are looking for is the Cremona Diagram

This is an example of solving by joint. I assume you know how to calculate the reactions, and only showing two joints to demonstrate the method.

First assume the positive direction of forces - +Fy upward and +Fx from left to right. Then draw the member forces arbitrarily as shown below, keep in mind that the member end reactions are in opposite direction. Joint A: Joint B: I think you can work out the other member forces by joint to joint. The resuly shall look like: The green shade represents compression on the respect member; the blue shade represents tension.

Good luck.

• The applied loads are symmetrical, and total 30 kN (3 x 10 kN). Due to the symmetry Rya=Ryb and Rya+Ryb= 30 kN. So Rya should equal 15 kN instead of 7.5? I also think you are missing a 10 kN applied load in your force diagram on the left side. Nov 6, 2022 at 4:19
• @ForwardEd Yes, I missed the force on the left. See my sketch.
– r13
Nov 6, 2022 at 15:52

The letters $$\pmb{h}$$, $$\pmb{i}$$, and $$\pmb{j}$$ mentioned in the answer, which gives the member forces of the shown truss , come from Bow’s notations. This notation is used to obtain a $$\pmb{\text{space diagram}}$$ of a truss. Each one of these letters corresponds to a $$\pmb{\text{space}}$$ in this $$\pmb{\text{space diagram}}$$ and has nothing to do with the joints of the truss. Two such letters together, (i.e., $$\pmb{hi}$$) denote a member that separates two adjacent spaces, such as $$\pmb{h}$$ and $$\pmb{i}$$. Please note that some engineers use numbers instead of letters to denote $$\pmb{\text{spaces}}$$. If you want I can draw and post the $$\pmb{\text{space diagram}}$$ of the given truss.