Designing a truss is a right of passage for any Structural Engineer. Indeed it is usually covered during the first couple of years of university. So it is extremely important that a competent engineer knows how to design a truss.

Going through the design process of a truss can assist with load path understanding and general bending theory.

In this article we are going to go through forming a concept design for a roof truss. I will go through how I would approach this as a practicing Structural Engineer in the real world of consulting. This means we will need to come up with the geometry and the layout ourselves based on project constraints and client requirements.

Once you know how to use truss theory you can apply this to a range of applications including:

- Truss bridges
- Long span stadium roof structures
- Your very own homes timber roof trusses

The steps in designing a truss generally involve the following:

- Determine the truss span and spacing
- Determine the design loading on the truss
- Choose the structural depth of the truss to control deflection
- Choose the structural members which make up the truss
- Check the truss for strength requirements
- Verify your design to make sure it correct

Without further ado, lest jump straight into it…

## Setting the Scene for our Truss Design (Geometry and Span of Truss)

Ok, here is the scenario…

We have been engaged by an architect to provide structural advice on the framing arrangement for a multi-use sports facility at a local high-school.

The architect would like to gain an understanding on the structural depth requirements for the main roof trusses which span across the playing court of the facility. The dimensions of the facility are

**20**m Wide (**32.8**foot)**60**m Long (**196.85**foot)

The length of the building is divided by 6 grid bays each **10**m apart (**32.8** foot). This gives the following grid arrangement which has been provided by the architect…

The trusses will be spanning along the lettered grid lines making their full clear span **20**m (**65.6** foot).

## Loading Up our Roof Truss

Now that we have the span and spacing of the roof trusses sorted, the next piece of the puzzle is the loading. The loading assumptions we will be considering for this truss arrangement will include the following.

**Superimposed Dead Load:**This includes the roof sheeting, lighting and services (sprinkler pipes etc.) and any insulation on the roof (there will be no ceiling, the structure will be exposed).**Self Weight:**Self weight will be attributed to the weight of the truss member itself. Self weight is also imposed by the members spanning between the roof trusses. Considering that we haven’t designed the roof trusses or the secondary beams, we will need to make an educated guess on these elements. These assumptions can then be latter checked to make sure everything still works at the end of the process.**Live Load:**The roof will be non-trafficable with access only being granted to maintenance personnel.**Snow Loading:**The sports facility is located close to sea level in a region which does not receive snow fall, therefore the effects of snow loading do not need to be considered.**Wind Loading:**Given the roof construction is to be comprised of lightweight material, wind loading should be considered as a further load combination. However in the interests of keeping this article short enough to digest in one sitting, we will not be considering wind loading in this example. This can be a good topic for future discussion.

### Summarising our Primary Loads

Now that we have sorted out the scope of our loading, here is what we will have to work with as a summary for this example:

Element | Loading | Comments |

Roof Sheeting | 0.16 kPa( 3.34psf) | Sheeting specified by the architect |

Services (lighting, Sprinklers etc.) | 0.5 kPa( 10.4psf) | Services layout specified by the MEP consultant |

Live Load | 0.25 kPa( 5.2psf) | Taken from your local loading code for a non-trafficable roof, for Australia this is AS1170.1 Chapter 3, Table 3.2 |

Self Weight of Secondary Beams | 0.2 kPa( 4.2psf) | Educated “guess” based on past experience to be later verified |

Self Weight of Truss | 0.6 kN/m( 0.04kips/ft) | Educated “guess” based on past experience to be later verified |

Now that we have our primary loads, we need to convert them into a usable format for our truss design. You will notice that all but one of the loads in the table above are given in kPa (kilopascal or psf, pound per square foot). This is often referred to as an “area load”. Given that we are designing a “beam” element, its much easier to work with these loads if they a “line load”. A line load being force per unit length applied to the member. We know that the grid bay is **10**m (**32.8** foot) so we simply multiply the area loads by **10**m to convert these area loads to line loads (which is kilonewton per meter or kips/ft).

The original table then becomes…

Element | Loading | Dead or Live Load |

Roof Sheeting | 1.6 kN/m( 0.11 kps/ft) | Dead |

Services (lighting, sprinklers etc.) | 5.0 kN/m( 0.32 kps/ft) | Dead |

Live Load | 2.5 kN/m( 0.17 kps/ft) | Live |

Self Weight of Secondary Beams | 2.0 kN/m( 0.14 kps/ft) | Dead |

Self Weight of Truss | 0.6 kN/m( 0.04 kps/ft) | Dead |

### Load Combinations

From the previous table, you will notice that I have added an additional column called “Dead or Live Load”. Our example is quite simple in that we are only dealing with Dead Loads and Live Loads. However you may encounter design examples which have wind, earthquake, snow, blast loading or collision loading. Each of these load categories have different safety factors which apply to them. These safety factors may also change depending on what combinations of primary loads and what design criteria you are adopting.

Given all of that, for our specific example, we wish to design the truss for two design conditions; Serviceability and Ultimate Design.

**Serviceability Design:**Serviceability Design is the consideration of deflection and vibration of a structural member. For our example we will be focusing on deflection as the roof is non-trafficable so vibration is not as much of a concern for this application.**Ultimate Design:**Ultimate Design (or strength design) is consideration of a structural members strength (bending, shear, torsion, tension or compression).

You will need to consult your local design code to determine what safety factors are require for Dead an Live loading for both Ultimate and Serviceability design. In Australia the safety factors can be found in AS1170.0 Chapter 4.

Here is a summary of our safety factors taken from AS1170.0 (design factors in different regions are quite similar to these):

Loading | Ultimate | Serviceability |

Dead Load | 1.2 | 1.0 |

Live Load | 1.5 | 0.7 |

OK, there’s been a lot of explanation to get to this point, but we are almost there. To finalise our load combinations, we simply add all the dead loads and live loads together then apply our safety factors and we are ready for the next step…

Loading | Working (Un-Factored) | Ultimate | Serviceability |

Dead Load | 9.2 kN/m( 0.63 kps/ft) | 11.04 kN/m( 0.76 kps/ft) | 9.2 kN/m( 0.63 kps/ft) |

Live Load | 2.5 kN/m( 0.17 kps/ft) | 3.75 kN/m( 0.26 kps/ft) | 1.75 kN/m( 0.12 kps/ft) |

Dead + Live | 11.7 kN/m( kps/ft)0.8 | kN/m14.79( kps/ft)1.02 | kN/m10.95( kps/ft)0.75 |

## How Deep Should my Truss be?

Now we get to the meat and potatoes of what we need to do. The depth of our truss is really what the architect would like to know at this point. So we need to make sure that we get this answer correct or they may never work with us again!

Before you jump on the computer to open up your ETABS, Strand7, SAP200, Robot, Microstran or Spacegass software, a word of warning….

**“Please step away from ****your FEA design software”**

Seriously, there is a time and a place for analysis software… concepting a simply supported truss is not the time nor the place. I’m all for using technology to assist you in your design (which you can tell from THIS article). However technology shouldn’t be used at the expense of coming up with the answer quickly and efficiently. This is a case where doing things the old fashioned way by hand is faster (trust me, you’ll see!).

Lets simplify our truss down to a conventional beam, much easier to manage at this point…

Nine times out of ten the depth of your spanning member is generally governed by the serviceability (deflection) design criteria. So we should start with our deflection criteria and work back from there.

For our particular example, the manufacturer of the roof sheeting has specified a deflection limit for our roof structure. Given the roof slope and the sheeting profile selected by the architect, a deflection limit of **SPAN/400** has been imposed to prevent water ponding on our roof. This gives us the following deflection limit we need to aim for…

This now gives us everything we need to determine how deep our truss needs to be… Enter the “Structural Engineers Secret Handshake”. What is the Structural Engineers Secret handshake you may ask? Well the less cool name would be a simply supported beam equation to determine the deflection given an applied uniform distributed load. Every Structural Engineer in my opinion should have this equation committed to memory, hence the Secret Handshake. If you want to find out who is an imposter Structural Engineer, simply ask them to recite this magical equation…

Lets break down this equation a little bit to see its components:

Factor | Explanation |

Deflection | We know this already, our deflection limit is 50mm or 1.97 inch |

W | Uniformly distributed load. This is also known and is 10.95 kN/m for our serviceability case (or 0.75 kps/ft) |

L | Length of the spanning member. We know this one also, 20m or 65.6 foot |

E | Young’s Modulus. This is a measure of the materials stiffness, we will be designing a steel truss, generally this is taken as 200,000MPa for steel (or 29,007 kip/sq in) |

I | Second Moment of Inertia (or the Second Moment of Area) this is a measure of the members stiffness based on its geometry. The deeper the cross-section of the member the stiffer is it. This factor we do not know because we have yet to determine its depth. |

Taking all that we now know and placing the values into the secret handshake, then doing a little bit of re-arranging, we can solve for “I”…

now that we solved for “I”, we need to back calculate our truss’ effective depth. We need to make another assumption here (this is part of concepting and early stages of design). We will choose our top and bottom chord members to allow us to progress further.

Today I feel like using **PFC** members for our truss (**P**arallel **F**lange **C**hannel). These member types are good for trusses as connection detailing is relatively straight forward for both welded connections as well as bolted connections.

Specifically, we will use a **200PFC** member. The 200 designation means the depth of the member is **200**mm (**7.87** inch). However we will be lying these PFC members flat within our truss. The top chord will have the flanges pointed downwards and the bottom chord will have the flanges pointed upwards as this cross section indicates…

To determine the “I” value of a built-up section such as this we use another equation which you should make yourself very familiar with as a Structural Engineer…

This means we need to break out our high-school algebra skills to solve for dimension “D” which will give us the depth of truss we are after. Because both the top and bottom chords are identical members with the same distance from the neutral axis, we can replace the summation sign in this equation with a multiplier of 2.

The “I” value of the PFC and the cross sectional area of the PFC we can take straight from the standard specifications book for a 200PFC section. I highly recommend that you have standard steel section booklets handy for your region. We don’t want to be calculating everything from scratch, we want to clock off at 5pm for the day (fingers crossed). Because the PFC members are laid flat, we are looking for the Iy value (bending about its weak axis) in this example.

Now placing all these values in our “I” equation and using some high-school algebra trickery, we can work out the dimension “D”…

From the previous truss cross section, we know that the dimension “D” is roughly half of the truss depth. So our full truss depth is **624 x 2 = 1248**mm (**49.1** inch). To account for the additional width of each of the top and bottom chord PFC members lets round up to **1300**mm (**51.2** inch)

## How to Design a Truss for Strength

So far we have determined that our truss has an outline of **20**m (**65.6** foot) length and **1.3**m (**51.2** inch) height making it look something like this….

Now we need to divide the truss into its separate bays. Since the truss is **20**m in length, it makes it clean and simply to make each bay length **1**m long…

Now this is starting to really look like a truss. Time to check that our top chord, bottom chord and diagonals are structurally adequate. To keep the truss simple to construct we will be using the 200PFC members for all sections of the truss.

The first step in the strength design of a truss is to determine the support reactions. For this, we are going to treat our truss as a beam again. So you don’t need to scroll back up, I’ll show you the summary of the beam and our loading here…

Because we are designing for strength, we will be using the ultimate load condition. Since the beam is symmetrical and its loading is consistent for its full length, the reaction at each support is determined by….

Now we can focus on one support to determine our worse case diagonal member within the truss. Lets blow up the left hand side support node to take a closer look…

Isolated view of the left hand side support node of our example truss

Here we have a couple of variables we need to solve. Firstly the geometry (the angle **Y** shown on the right-hand-side image), then the force within our worse case diagonal truss member (on the left-hand-side image indicated **X**).

This is where our high-school trigonometry comes to save the day. I am referring to **SOH**–**CAH**–**TOA**. Did you learn this in a similar way? Please leave a comment down below…

Using **SOH**–**CAH**–**TOA**, lets first solve our angle “Y”…

We now have all the ingredients to solve for the forces entering the support node. The various segments of a truss meet at a point, these are referred to as nodes. Lets take a look at our support node one more time, this time totally cleaned it up to the bare essentials…

Structural Engineers often refer to “statics” or “equilibrium”. We intend for this truss to be a stable/solid support for our roof (and not move or fly off anywhere). This means that our support node needs to be stationary, or static at all times. Therefore the principal of equilibrium states that all forces entering and leaving our support node need to be balanced (cancel each other out) for the node to be in equilibrium (and therefore static)… if that doesn’t make sense, read it a few times then we will move onto what this looks like from a numbers perspective.

From our example support node, so far we know one force entering it, the reaction we determined earlier of **148**kN (**33.3** kips). This force is a vector, a vector has both a direction and a magnitude. The direction of the support force is upwards. In its current state this would mean that our node is wanting to sail off towards the top of our screen (not good!).

What is preventing the node from moving upwards is our diagonal member entering the node from above, or more accurately the vertical component of the fort within this member. Using geometry to obtain the diagonal component of our reaction force will give us the force within our diagonal member, lets give this a try…

We can add this force to our support node to see what it looks like now…

Its looking pretty good, however there is one problem. Our diagonal member force is entering the support node at an angle, this means that now our node wants to move across to the left of our screen (not good!). We’ve solved the vertical equilibrium, now we need to solve the horizontal equilibrium.

Can you guess what is going to prevent the node from moving away to the left of screen? You guessed it, a tension force now will develop in our bottom chord of the truss (the horizontal force exiting the node). We can use trigonometry again to determine the horizontal component of our diagonal member to work this out…

Putting it all together we can admire our lovely, stable node…

That nodes not going anywhere! Lets check that our diagonal 200PFC can handle that **186.8**kN (**42** kips) compression load. First we need to determine how long the member is. While we are on high-school mathematics, lets use some Pythagoras this time…

Now we have the members length and the force within that member we can determine if it is strong enough to handle what we are throwing at it. In a real world consulting environment, **I WOULD NOT** recommend working out the compressive capacity of a standard steel section by first principles, after all its nearly 5:00pm we need to get home! This is where our steel design book comes in handy once again, you should have one on your desk at all times (or a PDF version)…

The 200PFC will fail one of two ways in compression. It will either buckle about its strong axis (**X** Axis) or its weak axis (**Y** axis). In our truss, the 200PFC is a free spanning member and we are not restraining it in either the **x** or **y** axis so we choose the worse case buckling capacity to determine its adequacy (buckling about the weaker **y** axis).

The table from our steel design book indicates that we will have a capacity somewhere between **541**kN and **480**kN (**121** kips and **108** kips). Our calculated force in this member was **186.8**kN (**42** kips) which is lower than these capacities. So in summary our diagonal 200PFC has more capacity than the load we are asking for it to support, meaning it will not buckle.

Similarly, we can perform a similar check on our bottom chord 200PFC, however this time we are checking tension.

We are assuming that the truss will be fabricate using welded connections. This means we do not nee to reduce the 200PFCs tension capacity due to bolt holes. This gives us a tensile capacity of **788**kN (**117** kips) for our bottom chord. The calculated tension we derived earlier was **113.5**kN (**25.6** kips) so our bottom chord is much stronger than it needs to be. However in a simply supported member the bending moment is smallest near the supports, therefore we would expect to see the tension and compression in the top and bottom chords to be the smallest at these locations as well.

We need to go to the mid-span of the truss to obtain the maximum tension in the bottom chord and compression in the top chord. However to get there using vectors and truss theory means that we have a lot more nodes we need to check between our support node and the mid-point node.

We can skip straight there using beam theory again with the second Structural Engineers Secret Handshake (bending moment in a simply supported beam with a uniformly distributed load)…

Bringing back our beam diagram and using this formula gives us…

Dividing the bending moment by the depth of the truss gives us the push/pull coupling forces in the top and bottom chords (the compression and tension we are looking for).

This is equivalent to the maximum compression in the top chord and maximum tension in the bottom chord at the mid-span of the truss. There are two things we need to consider when checking the top chord compression member.

- The top chord PFC is restrained by the truss itself so will not buckle about its minor axis
- The length of the member for buckling about its major axis is the full span of the truss unless we restrain it (that’s
**20**m or**65.6**foot!).

Lets look at the capacity table for a 200PFC in compression for buckling about its strong axis…

For one thing, our table doesn’t even reach **20**m. Secondly our capacity starts to be exceeded at around the **3**m mark (**9.8** foot). This means we need to provide some kind of restraint to the top chord at a maximum of **3**m intervals. We can achieve this by adding a single braced bay to our roof structure. Along with the braced bay, we can introduce struts across the whole width of the roof, this will allow us to collect the buckling loads from all of the truss top chords and support them by the single bracing bay. Lets see what this looks like in plan view for our example roof…

## Summary of our Roof Truss

We have now fully designed our roof truss. lets summarise what we have designed and how it all performed…

Element | Calculated Value | Limit |

Span | 20m (65.6 foot) | N/A |

Depth | 1.3m (4.26 foot) | N/A |

Max Deflection | 50mm (1.97 inch) | 50mm (1.97 inch) |

Truss Members | 200PFC Sections | N/A |

Top Chord Compression | 569kN (128 kips) | 734kN (165 kips) |

Bottom Chord Tension | 569kN (128 kips) | 788kN (177 kips) |

Diagonal Member Compression | 186.8kN (42 kips) | 550kN (124 kips) |

Believe it or not, to get to this point is around 10 minutes of work (tops) for a very experienced Structural engineer using the process outlined in this article. Through a mixture of muscle memory and having all the right reference equations and capacity tables at close reach allows these kind of elements to be designed very quickly.

But is this all correct? We need to verify our work before we can say that we are happy with our answer.

Here would be a fantastic opportunity to put together a very quick frame model in your favourite analysis software to at least check the deflection and member loads we just calculated earlier.

## Analysing a Truss using Structural Design Software

The analysis software I have chosen for this check is Microstran. Its getting a little bit dated these days however I have a sentimental attachment to this software package as it has been a trusty tool of mine ever since I graduated University.

I won’t get into detail on how to use Microstran, we can go through that on another article. however we will go through the high level workflow together.

First step is to draw all of the members with the correct geometry, assign the supports at each end, assign the cross section type as well as the material properties. Here is a snapshot of my framing and an isometric image taken from the model…

Next we assign our ultimate loading followed by our serviceability loading…

Hit the analyse button, fingers crossed you have no errors or warnings, then we read off the results. First lets take a look at the maximum deflection…

Now lets check strength, firstly we will go to the compressive force at the fist diagonal member near the left hand support…

Now lets look at the compression and tension at the top and bottom chords at mid-span…

Lets look at the hand based results and teh software based results side-by-side…

Hand Calculations | Software | Percentage Difference | |

Deflection | 50mm | 50mm | 0% |

Top Chord Compression | 569kN | 562kN | 1.2% |

Bottom Chord Tension | 569kN | 568kN | 0.17% |

Diagonal Member Compression | 186.8kN | 174kN | 7.3% |

There you have it, a good old fashioned hand base approach is plenty accurate. Now we can issue our findings to the architect and continue our concept design.

Did you find this article useful? Please feel free to leave a comment below if there are any specific topics you would like explored on **Sheer Force Engineering**.

A very useful article. Explained everything in a professional and easy to understand way. Thanks !

Hmm I am quite surprised that the additional deflections due to shear action of the long span member was not indicated in the FEM model (which should indicate a larger deflection value when compared to your hand calcs which only deal with deflections due to bending)