Understanding and Analysing Trusses

So what exactly is a truss?

It is essentially a rigid structure made up of a collection of straight members. But that’s not a complete definition.

There are two important assumptions we need to be able to make for a structure to be considered as a truss.

First, we need to be able to assume that all of the joints in the structure can be represented by a pinned connection, meaning that members are free to rotate at the joints.

The members of a truss are often rigidly connected using what is known as a gusset plate, but if the centre-lines of all of the members at a joint intersect at the same point, like they do here, it’s reasonable to assume that the joint behaves like a pinned connection.

The second assumption we need to be able to make is that loads are only ever applied at the joints of the truss.

We never have loads acting in the middle of a member, for example, because all joints are pinned, the members cannot carry bending moments – they can only carry axial loads.

This simplifies the analysis of a truss significantly.
Each member has to be in equilibrium, so the forces acting at each end of a member must be equal and opposite.
Each member is either in tension or in compression, these assumptions are what differentiate a truss from a frame, unlike trusses, frames don’t necessarily have pinned joints, and so members can carry bending moments.

A frame can also have loads applied directly to its members.
The base shape of a truss is three members connected to form a triangle.
If a load is applied, the angles of the triangle won’t be able to change if the length of each of the members stays the same.
This means that the triangle is a very stable shape which won’t deform when loads are applied to it, and so it is a great base from which to build a larger structure.
Joining four members together does not form a stable structure.
The angles between members can change without any change in the length of the members, and so using a four-sided shape as the base for building a truss would be a terrible choice.
An easy way to stabilize this configuration is to add a diagonal bracing member to split it into triangles.
We can start with our triangle and build it out to form a structure.
There are a lot of different ways to build a truss, but there are some particularly popular truss designs that you will see again and again, and so they are referred to by specific names.
The one shown here is a Fink roof truss, but there are many more, as you can see here.
Later on in this video I’ll cover how these different designs carry loads in different ways.
The members of these trusses are all located in the same plane.
These are called planar trusses, and we can analyse them as two-dimensional structures.
Even seemingly three-dimensional structures can often be analysed as planar trusses.
Take a look at this bridge for example.
The loads are transmitted from the horizontal floor beams to the two vertical trusses on each side of the bridge.
Each of these trusses only carries loads acting in its plane, and so we can analyse it as a two-dimensional structure.
To be able to design or analyse a truss we need to be able to determine the force in each of its members.
This allows us to check that the members can carry the loads without failing, or gives us the information we need to select the best cross-section for each of the members.
There are two main methods we can use to do this – the Method of Joints, and the Method of Sections.
Let’s look at the Method of Joints first, using the Fink roof truss we saw earlier.
The method is really simple.
First you draw a free body diagram, showing all of the external loads acting on the truss, and you use the three equilibrium equations to calculate the reaction forces.
Then you draw a free body diagram for every single joint, and work through them one by one to solve the unknown forces acting at each joint.
You solve the unknown forces using the equilibrium equations.
Since all of the joints are pinned connections, there are no moments, and so you only need to consider equilibrium of the horizontal and vertical forces.
Remember that we are calculating the forces acting at each joint, not the forces in the member.
If a member is in tension, the internal forces will be acting to make the member longer.
For every action, there is an equal and opposite reaction, which means that a member in tension will be exerting a force on the joint which is acting away from the joint.
For members in compression, the force will be acting towards the joint.
Let’s work through an example for a slightly simpler truss.
First let’s draw the free body diagram and determine the reaction forces using our three equilibrium equations.
Taking equilibrium of the horizontal forces, the horizontal force at joint A must be equal to zero, because it is the only force in the horizontal direction.
Taking equilibrium of the vertical forces, the reaction forces at joints A and E must sum up to 20 kN.
Both joints are located at the same distance from joint C, so taking equilibrium of the moments acting about joint C we can calculate that they both equal 10 kN.
Now let’s determine the forces acting on each joint.
Since we don’t know yet which members are in tension and which are in compression, it’s easiest to just assume that all of the members are in tension, and so we’ll draw the internal forces as pointing away from each joint.
If we end up with negative values for these forces, it just means that we guessed wrong, and the member is actually in compression.
Now we can work through each of the joints, starting with joint A.
Analysing trusses involves a lot of resolving forces to different angles, so if you want to be good at it you’re going to need to remember your trigonometry!
Here’s a quick reminder.
Back to our joint…
All we have to do is apply the equilibrium equations to determine the unknown forces at joint A.
Taking equilibrium of the vertical forces, the 10 kN reaction force must balance the vertical component of the force F-AB.
And so we can calculate F-AB as -10 divided by sine of the angle of 60 degrees.
Taking equilibrium of the horizontal forces, we get that the force F-AC must balance the horizontal component of the force F-AB, and so is equal to 5.8 kN.
That’s all of the forces acting on joint A calculated.
The force in each member is constant, and so we now also know the forces acting on the joints at the other ends of these two members.
We can then repeat the process for joint B.
We can start by considering equilibrium of the vertical forces, which allows us to calculate the force F-BC.
And then we can consider equilibrium of the horizontal forces to calculate F-BD.
We then need to work through all of the remaining joints.
But we can save a bit of time by noticing that the truss and the applied loads have an axis of symmetry, and so the forces on the other side of the truss must be identical.
That gives us all of the forces at the joints.
We can show which members are in tension and which are in compression, like this.
One thing you’ll notice as you analyze trusses is that some members don’t carry any loads at all.
We call these zero force members.
There are two main configurations where we have zero force members.
The first is where we have three members connected at a single joint, and two of the members are aligned.
Here, only one member has a component in the vertical direction, and so to maintain equilibrium of forces at the joint in the vertical direction, the force in this member must be zero.
The second configuration is when we have only two members connected at a joint, and the members are not aligned.
Only one member has a component in the vertical direction, and so both must be zero force members.
By the way, this is true regardless of how the members are oriented, because we can rotate the orientation of the coordinate system we are using to apply the equilibrium equations.
These two configurations only contain zero force members if there are no external loads acting at the joints.
If we have external loads, there will be components in the vertical direction, and so these will not be zero force members.
Let’s look at an example.
At this joint here, we have two connected members.
The members are not co-linear, and there are no external loads, so they must be zero force members.
And at this joint we have three members, of which two are co-linear.
The vertical member must be a zero force member.
We can remove these members, and so we have a much easier starting point for solving the truss.
You’ll notice that we haven’t removed the two members at this joint.
That is because there is an external load acting here, and so these can’t be zero force members.
You might be wondering why anyone would bother including zero force members in a truss if they carry no loads.
They are definitely not useless.
They are usually included to provide stability, for example to prevent buckling of long members which are under compression.
Or they may be used to make sure that unexpected loads won’t cause the structure to fail.
We’ve covered the Method of Joints.
Let’s look at the other method we can use to solve trusses, which is the Method of Sections.
The first step is the same as the Method of Joints.
We draw the free body diagram, and use the equilibrium equations to solve the reaction forces.
Next, we make an imaginary cut through the members of interest in our truss, and we draw the internal forces in the cut members.
The internal and external forces must be in equilibrium, and so we can apply the equilibrium equations to solve the internal forces.
When choosing how to cut your truss, remember that we only have 3 equilibrium equations.
If you cut through too many members, you will have too many unknowns and not enough equations.
You can choose which side of the cut you want to assess.
The left side looks easier to solve because there are less forces, but we could have chosen to solve the right side instead.
The Method of Sections is best used when you have a truss which has a lot of members, but you are only interested in the loading in a few specific members.
Let’s look at an example.
We need to determine the internal forces in these three members.
First let’s draw the free body diagram and apply the equilibrium equations to calculate the reaction forces.
The horizontal reaction force at joint A is the only force acting in the horizontal direction, so it must be equal to zero.
By considering equilibrium of forces in the vertical direction, and equilibrium of the moments acting about joint A, we can figure out that F-A is equal to 19 and F-H is equal to 21.
Next let’s make our imaginary cut through members F-D, F-E and G-E, and draw the internal forces.
Like we did earlier for the method of joints, we will assume that all unknown forces are tensile.
Next we just need to apply the equilibrium equations.
The force in member F-E is the only unknown force with a component in the vertical direction, so that’s a good place to start.
The diagonal members are all at 45 degree angles, so by considering equilibrium in the vertical direction we get that the force in member F-E is equal to 12.7 kN.
Now let’s consider equilibrium of moments acting about joint F.
This is a good joint to choose because 3 of the 5 forces in this free body diagram have a line of action passing through F, and so only the force in member G-E and the 21 kN reaction force generate a moment about this joint.
Both forces are located at a distance of 2 m from F, and so we can conclude that the force in member G-E is equal to 21 kN.
Finally, we can take equilibrium in the horizontal direction to calculate that the force in member F-D is equal to -30 kN.
And that’s it, we’ve calculated the internal forces in the three members we were interested in.
One member is in compression and two are in tension.
If it is possible to determine the reaction forces and the internal forces in the members of a truss by applying the equilibrium equations, the truss is said to be statically determinate.
Real-life structures sometimes contain more members than are needed for the structure to be stable, as this makes them safer.
This means we may not be able to apply the Method of Joints or the Method of Sections, because we have too many unknowns and not enough equilibrium equations, either to determine the reaction forces, or to determine the internal forces within the truss.
These trusses are said to be statically indeterminate, and would need to be solved using other methods like the force method or the displacement method, which I won’t get into in this video.
Now that we know how to calculate the loads in a truss, let’s explore some of the differences between truss designs.
Here we have three different bridge trusses, the Howe, Pratt and Warren trusses.
These trusses were all patented in the 1840s, at a time when new bridge designs were being developed to accommodate the expansion of the railroad industry.
They were typically constructed from a combination of wood and iron.
We can learn a lot about truss design by figuring out which members are in tension and which are in compression.
Let’s start with the Howe truss.
We can see that its vertical members are in tension, and its diagonal members are in compression.
Members in compression usually need to be thicker than members in tension, to reduce the risk of buckling.
This means that the Howe truss isn’t very cost effective, since the diagonal members, which need to be thicker, are quite long.
The Pratt truss addresses this issue.
Its vertical members are mostly in compression and its inner diagonal members are in tension.
This is more cost-effective than the Howe truss, since the longer diagonal members can be thinner.
Longer members are also more susceptible to buckling under compressive loading than shorter ones, so it’s a good idea for long members to be in tension.
The design of the Warren truss was based on equilateral triangles.
The fact that all of the members are the same length is an advantage for construction, and it uses less members overall than the Howe and Pratt trusses, so it is more efficient.
The diagonal members alternate between tension and compression, so it does have some quite long members in compression.
It can also be interesting to observe how the loading in members changes as a load moves across a bridge.
In this simplified model of a load moving across the Pratt bridge, we can see that some members alternate between tension and compression, and so will need to be designed accordingly.
The three-dimensional bridge we looked at earlier could be assessed as a collection of planar trusses.
But this won’t be possible for all structures, and sometimes a truss will need to be assessed in three dimensions.
This type of truss is called a space truss.
These can be analysed in essentially the same way as planar trusses, using the Method of Joints and the Method of Sections.
The only difference will be in the number of equilibrium equations.
We will have 6 equations instead of 3, and at each joint we will have 3 equations instead of 2.