# The Fink Truss [All YOU Need to Know]

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 syst****, which members act in**

*em(s)***or**

*compression**and calculate the internal forces.*

**tension**Without further ado. Let’s talk FINK TRUSSES. ππ

## 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.

- 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.ππ

## Static System

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

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.*Easier to calculate:*- 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.

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 A_{H} and vertical reaction force A_{V}

Roller support: Vertical reaction force B_{V}

##### Connection types

Hinge connections: Moment is 0 in hinge 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.ππ

## 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/m^{2} and a Fink truss spacing of 2m. Check out this article to learn more about the snow load.

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

Before we start with the calculations, let’s give the nodes π΅ and bars π΄ 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 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

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

Let’s calculate. ππ

### 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 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 A_{V}, 1.5 kN, N_{1} and N_{4}.

$$ \alpha = atan(2m/4m) = 26.57Β° $$

**Vertical ***Equilibrium:*

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

Let’s solve that for N_{4}.

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

**Horizontal ***Equilibrium:*

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

Let’s solve that for N_{1}.

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

*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 N_{5}.

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

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 N_{5} 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 N_{5} 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 N_{9}.

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

**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 N_{2}.

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

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

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

Yes! But due to symmetry we know that N_{1} = N_{3}, N_{4} = N_{7}, N_{5} = N_{6}, N_{8} = N_{11}, N_{9} = N_{10}.

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

## 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.

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.π¦Ί

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 βοΈ.

## 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