The Fink Truss [All YOU Need to Know]

What is the fink truss and how does it work?

Last updated: June 6th, 2023

Fink Trusses are commonly used in roof structures of residential buildings, farms or garages.

But why is that? Why is in some designs the Fink Truss favoured over the King Post or Scissor Truss?

To get an answer to those questions, we need to understand the structural system and how the loads travel through its different members.

So in this post, we’ll show what the Fink Truss is, its static system(s), which members act in compression or tension and calculate the internal forces.

Without further ado. Let’s talk FINK TRUSSES. πŸš€πŸš€

What Is the Fink Truss?

The Fink truss is a structural system which can carry loads with relatively long spans compared to beams. It’s mainly used as roof structure carrying snow, wind, live and dead load and distributing it to its supports (walls or columns).

The Fink truss is characterized by its diagonal members, which are inclined with different angles.

Let’s have a look at the Fink truss and its members.

Fink truss member names.
Fink truss member names.
  • Top chord or Rafters
  • Bottom chord
  • Diagonals or also called webs

What is the Fink Truss used for?

Fink trusses are mainly used as roof structures of

  • residential buildings
  • farms
  • garages
  • etc.

In most cases, timber is the main material for Fink trusses, however steel can also be an option.

Let’s look at the static system to get a better understanding of the structural behaviour of the Fink Truss.πŸš€πŸš€

Static System

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 below, you can see the “normal” Fink truss with hinge connections.

We summarize the characteristics of the truss with fixed connections a bit further below.

Static system of the fink truss with all member names including top chord, bottom chord, web and hinge connections.
Static System of the Fink Truss

The static system of the Fink 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.
Characteristics
Loads

In most cases the Fink truss is used as a roof structure where the loads (dead, live, snow and wind load) are applied on the rafters.

Support types

1 Pin and 1 Roller support

Reactions

Pin support: Horizontal AH and vertical reaction force AV
Roller support: Vertical reaction force BV

Connection types

Hinge connections: Moment is 0 in hinge connections.

Fink Truss with fixed connections
Static system of fink truss with fixed connections.
Fink Truss with fixed connections.

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

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

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 truss.πŸŽ‰πŸŽ‰

Fink Truss Analysis

Let’s say our Fink Truss is a roof structure. Therefore, the truss is exposed to the snow load only on the rafters.

We also simplify and say that the load is 2 kN/m. This equals a snow load of 1 kN/m2 and a Fink truss spacing of 2m. Check out this article to learn more about the snow load.

$$1 kN/m^2 \cdot 2m = 2 kN/m$$

Static system of fink truss with dimensions and snow line load of 2 kN/m.
Static system with truss dimensions and snow line load of 2 kN/m.

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

Node πŸ”΅ and bar πŸ”΄ indices.

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

The Line load of 2 kN/m is approximated as point loads in the nodes, because otherwise the bottom chord members would be beams instead of bars, which makes the calculation a lot more difficult.

The point load applied to nodes (a) and (d) is calculated as

$$2 kN/m \cdot 0.75m = 1.5 kN$$

while the point loads applied to nodes (e) and (g) is calculated as

$$2 kN/m \cdot (\frac{1.5m}{2} + \frac{2.5m}{2}) = 4 kN$$

and the point load applied to (f) is calculated as

$$2 kN/m \cdot 2.5m = 5 kN$$

Line load approximated as point load in nodes.

Let’s calculate. πŸš€πŸš€

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 AH, AV and BV.

$\sum H = 0: A_H = 0$

$\sum V = 0: A_V + B_V \, – 2 \cdot 1.5 \, \mbox{kN} \, – 2 \cdot 4 \, \mbox{kN} – 5 \, \mbox{kN} = 0 $ -> $V_a = V_b = \frac{16}{2} \mbox{kN} = 8 \mbox{kN}$

$\sum M = 0: M_a = 0$

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 4.

To do that, we only look at node 1 and its point loads/normal forces AV, 1.5 kN, N1 and N4.

$$ \alpha = atan(2m/4m) = 26.57Β° $$

Vertical Equilibrium:

$$ \sum V = 0: A_v – 1.5 kN + N_4 \cdot sin(\alpha) = 0$$

Let’s solve that for N4.

$$ N_4 = \frac{1.5 kN – 8 kN}{sin(\alpha)} = -14.53 kN$$

Horizontal Equilibrium:

$$ \sum H = 0: N_4 \cdot cos(\alpha) + N_1 = 0$$

Let’s solve that for N1.

$$ N_1 = – N_4 \cdot cos(\alpha) = 13 kN$$

Node e

$$ \alpha = 26.57 Β° $$

$$ \beta = 36.9 Β° $$

Horizontal Equilibrium:

$$ \sum H = 0: N_5 \cdot cos(\alpha) – N_4 \cdot cos(\alpha) + N_8 \cdot cos(\beta) = 0$$

Let’s solve that for N5.

$$ N_5 = N_4 – \frac{N_8 \cdot cos(\beta)}{cos(\alpha)}$$

As this equation has 2 unknowns, we need another equation to solve the unknowns.

Vertical Equilibrium:

$$ \sum V = 0: N_5 \cdot sin(\alpha) – N_4 \cdot sin(\alpha) – N_{8} \cdot sin(\beta) – 4kN = 0$$

Let’s insert N5 in this equation.

$$(N_4 – \frac{N_8 \cdot cos(\beta)}{cos(\alpha)}) \cdot sin(\alpha) – N_4 \cdot sin(\alpha) – N_8 \cdot sin(\beta) – 4kN = 0$$

$$ – (\frac{cos(\beta)}{cos(\alpha)} \cdot sin(\alpha) + sin(\beta)) \cdot N_8 = 4 kN$$

$$N_8 = – \frac{4 kN}{\frac{cos(\beta)}{cos(\alpha)} \cdot sin(\alpha) + sin(\beta)} = -4 kN$$

And now let’s solve N5 with the horizontal equilibrium equation.

$$ N_5 = -14.53kN – \frac{-4 kN \cdot cos(\beta)}{cos(\alpha)} = -10.96 kN$$

Node b

$$ \beta = 36.9 Β° $$

$$ \gamma = 53.1 Β° $$

Vertical Equilibrium:

$$ \sum V = 0: N_8 \cdot sin(\beta) + N_9 \cdot sin(\gamma)= 0$$

Let’s solve that for N9.

$$ N_9 = – N_8 \cdot \frac{sin(\beta)}{sin(\gamma)} = 3 kN$$

Horizontal Equilibrium:

$$ \sum H = 0: -N_1 + N_2 – N_9 \cdot cos(\gamma) – N_8 \cdot cos(\beta) = 0$$

Let’s solve that for N2.

$$ N_{2} = N_1 – N_9 \cdot cos(\gamma) + N_8 \cdot cos(\beta) = 8 kN$$

Well done! Now we have calculated the normal forces of ALL bars.πŸŽ‰πŸŽ‰

What all? But we are missing half of the bars.

Yes! But due to symmetry we know that N1 = N3, N4 = N7, N5 = N6, N8 = N11, N9 = N10.

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

Normal force diagram

Normal force diagram of the Fink truss.
Normal force diagram of the Fink truss.

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 πŸ”΅, while a positive (+) normal force means that the member acts in tension πŸ”΄.

Compression and tension members of a Fink Truss due to point loads applied to the top chord nodes.
Compression and tension members of a Fink Truss due to point loads applied to the top chord nodes.

Advantages and Disadvantages

Pros

  • Can span much longer than beams
  • It’s a lightweight solution for longer spans
  • Easy to construct (can be prefabricated or assembled on site)
  • Very durable: Most truss members act mainly in compression or tension (depending on the static system used)

Cons

  • Takes up vertical space

Conclusion

Now, that you got an overview of the Fink 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 Fink 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!

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

Let us know in the comments below ✍️.

Fink Truss FAQ

What is a Fink roof truss?

The Fink truss is a structural system which can carry loads with relatively long spans compared to beams. It’s mainly used as roof structure carrying snow, wind, live and dead load and distributing it to its supports (walls or columns).

Why is a fink truss good?

Here are 3 reasons why the fink truss is good!
– It’s a very lightweight structure, with its members acting mainly in compression and tension
– It’s durable
– Can be prefabricated or assembled on-site

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