Flat Truss: What Is It? And How To Calculate It?

Are you wondering how to calculate and design a flat truss? 🤔🤔

It can indeed be difficult, if you didn’t study structural engineering.

But in this blog post, we’ll try to explain the flat truss as simple as possible for everyone.👍👍

We’ll explain their key features, benefits, common uses and go through a calculation example, step-by-step.

Alright, let’s talk FLAT TRUSSES. 🚀🚀

What Is A Flat Truss?

The flat truss is a structural system commonly used in engineering projects like bridge and building structures. It’s characterized by a series of triangles that support the loads of a structure (wind, snow, dead, live load).

Let’s have a look at a visualization of the flat truss and its members.⬇️⬇️

• Top Chord
• Bottom Chord
• Diagonals

What is the flat truss used for?

Flat trusses are used as bridge structures but also for long spans as “beams” in buildings.

In most cases, steel is the main material, however timber could also be an option.

Let’s look at the static system to get a better understanding of the structural behaviour of the flat truss.🚀🚀

Static System

The static system of the flat truss is characterized by having

• hinge connections at all nodes
• a pin and roller support, which makes the system statically determinate. Therefore, the internal forces can be calculated with the 3 equilibrium equations.

Hinges

Most truss structures are designed with hinge connections, mainly due to 2 reasons:

• Easier to calculate: Trusses with hinge connections make the structure statically determinate, which means that the internal forces can be calculated by hand. Especially until advanced Finite element software programs weren’t widely available, this was the main reason for using hinge connections. If fixed connections are used, but no software is available, advanced methods like the method of consistent deformation can be used. However, these methods are complicated and susceptible to calculation failures.
• Cost: Hinge connections are cheaper to build than fixed connections.

It definitely makes sense to be aware of the differences of a truss with hinges and fixed connections. In the picture above, you can see the “normal” flat truss with hinge connections.

We also summarize the characteristics of the truss with fixed connections and a mix of fixed and hinge connections a bit further below.

Flat truss with fixed connections

The flat truss with fixed connections has the following differences to hinge connections:

• Bending moments in the connections
• More rigidity -> more robustness
• Smaller vertical deflection

Flat truss with fixed and hinge connections

The most realistic static system might actually be a mix of fixed and hinge connections.

So why is this the most realistic static system?

It’s because the top and bottom chord might be delivered in 1 piece, meaning that they are constructed continuous and the posts and diagonals are attached to them.

If the span is too long to deliver them in one piece, then moment stiff connections are design in areas where the moment is close to 0. This is cheaper and more sustainable.

The top chord basically turns into a 4-span continuous beam, which means that the static system is no longer statically determinate.

This means that advanced methods with FE programs need to be used to calculate the internal forces like Moment, Normal and Shear forces.

Alright, now that we have learned how to set up our static system, we are ready to calculate the internal compression and tension forces of the statically determinate truss.🎉🎉

Flat Truss Analysis

Let’s say our flat truss is supporting a floor in a building. Therefore, the truss is exposed to the dead and live load of the floor above on the top chord.

We also simplify and say that the design load (Load combination of dead and live load) is 30 kN/m.

Before we start with the calculations, let’s give the nodes 🔵 and bars 🔴 some indices, so the identification is easier later in the internal force calculation.

To calculate the compression and tension forces of the truss members with the 3 equlibrium equations, we do another approximation.

The line load of 30 kN/m is approximated as point loads in the nodes, because otherwise the top chord members would be beams, which makes the calculation a lot more difficult.

The point load applied to nodes (f) and (j) is calculated as

$$30 kN/m \cdot 1.5m = 45 kN$$

while the point loads applied to nodes (g) – (i) are calculated as

$$30 kN/m \cdot 3m = 90 kN$$

Let’s calculate. 🚀🚀

Calculation of Reaction Forces

As the structure is statically determinate, the reaction forces can be calculated with the 3 Equilibrium equations.

In our case, we are calculating the support forces A_H, A_V and B_V.

$\sum H = 0: A_H = 0$

$\sum V = 0: A_V + B_V \, – 2 \cdot 45 \, \mbox{kN} \, – 3 \cdot 90 \, \mbox{kN} = 0 $ -> $A_V = B_V = \frac{2 \cdot 45 + 3 \cdot 90}{2} \mbox{kN} = 180 \mbox{kN}$

$\sum M = 0: M_a = 0$

Calculation of the internal forces

Node a

Alright, now that we know the reaction forces, we can calculate the normal force of the first bar elements 1 and 5.

To do that, we only look at node (a) and its point loads/normal forces A_V, A_H, N₁ and N₅.

Vertical Equilibrium:

$$ \sum V = 0: A_v + N_5 = 0$$

Let’s solve that for N₅.

$$ N_5 = -180 kN$$

Horizontal Equilibrium:

$$ \sum H = 0: A_H + N_1 = 0$$

Let’s solve that for N₁.

$$ N_1 = A_H = 0$$

Node f

$$ \alpha = 45° $$

Vertical Equilibrium:

$$ \sum V = 0: N_{5} + 45 kN + N_6 \cdot sin(\alpha) = 0 $$

Let’s solve that for $N_6$

$$N_{6} = – \frac{N_5 + 45 kN}{sin(\alpha)} = 190.9 kN$$

Horizontal Equilibrium:

$$ \sum H = 0: N_{14} + N_6 \cos{\alpha} = 0$$

Let’s solve that for N₁₄.

$$ N_{14} = -N_6 \cdot cos(\alpha) = -135 kN$$

Node b

Horizontal Equilibrium:

$$ \sum H = 0: N_{1} – N_2 + N_6 \cdot cos(\alpha) = 0$$

Let’s solve that for N₂.

$$ N_{2} = N_{1} + N_6 \cdot cos(\alpha) = 135 kN$$

Vertical Equilibrium:

$$ \sum V = 0: N_{7} + N_6 \cdot sin(\alpha) = 0$$

Let’s solve that for N₇.

$$ N_{7} = -N_6 \cdot sin(\alpha) = -135 kN$$

Node g

$$ \alpha = 45° $$

Vertical Equilibrium:

$$ \sum V = 0: N_{7} + 90 kN + N_{8} \cdot sin(\alpha) = 0$$

Let’s solve that for N₈.

$$ N_{8} = \frac{-N_7 – 90 kN}{sin(\alpha)} = 63.6 kN$$

Horizontal Equilibrium:

$$ \sum H = 0: N_{14} – N_{15} – N_8 \cdot cos(\alpha) = 0$$

Let’s solve that for N₁₅.

$$ N_{15} = N_{14} – N_8 \cdot cos(\alpha) = -180 kN$$

Alright, I think you understood the principle. This can be continued for nodes h and c to calculate the normal forces of the other bars.

Now, a way to speed up the calculation is utilizing a FE program, which I did for this flat truss.

I used the free analysis software Ftool to do it.

So, to summarize it, a normal force diagram helps to understand how the loads travel through the truss.

Normal force diagram

Compression and Tension Members

Now, as you can see in the normal force diagram, some members have a positive (+) and some have a negative (-) normal force.

A negative (-) normal force means that the member is under compression 🔵, while a positive (+) normal force means that the member acts in tension 🔴.

Advantages and Disadvantages

Pros

• Flexibility: Flat trusses can be designed to span a wide range of lengths and widths, making them suitable for a variety of applications, from small residential buildings to large commercial or industrial structures.
• Simplicity: Flat trusses have a simple design. The diagonals always have the same angle to the chords. This can result in lower construction costs and faster installation times compared to other trusses. This also means that there is not a big variety of different connections.

Cons

• Aesthetics: Even though I like the design of the flat truss, there are other truss structures that have a better architectural appearance, such as the scissors truss or the king post truss.

Conclusion

Now, that you got an overview of the Flat Truss, and how we calculate its internal forces, you can learn about loads, because every truss is exposed to loads.

1. Snow load on a flat roof
2. Wind load on a flat roof
3. Wind load on walls

Because there are always multiple loads acting on Flat Trusses, considering these different loads in the structural design is done by setting up Load Combinations with safety factors.🦺🦺

Once all load cases and combinations are set up, the structural elements can be designed. We have already written a guide on how to design a timber truss. Check it out!

• Timber truss roof design

If you want to learn more about trusses, make sure to read our guide on the different types of trusses.

I hope that this article helped you understand the Flat Truss and how to go further from here. In case you still have questions.

Let us know in the comments below. ✍️✍️

Flat Truss FAQ