warren truss experiment

Warren Truss: What is it? And How to Calculate it?

Warren Trusses are commonly used in railway bridges.

But why is that? Why is in some designs the Warren Truss favoured over the Pratt or Howe 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 Warren Truss is , its static syst em(s) , which members act in compression or tension and how to calculate the internal forces.

Without further ado. Let’s talk WARREN TRUSSES. 🚀🚀

What Is the Warren Truss?

The Warren truss is a structural system which can carry loads with relatively long spans compared to beams. It was patented by engineers James Warren and Willoughby Monzoni in 1946.

The Warren truss is characterized by having tension and compression members, which we’ll calculate later in this article.

Let’s have a look at the Warren Truss and its different members. In the picture, you can see the names of the members of a Warren Truss.

• Bottom chord
• Diagonals (sometimes called strut or tie depending on compression or tensions)

What is the Warren Truss used for?

The main application where Warren Trusses are used are Bridge Structures.

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

A mix of different elements is also an option. For example, the diagonals and top chord can be built in steel, while the bottom chord is integrated into the deck and built in concrete.

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

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” Warren truss with hinge connections.

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

The static system of the Warren 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

In most cases the Warren truss is used as a bridge where the load (dead and traffic load) is applied on the deck which distributes the load to the bottom chord

Support types

1 Pin and 1 Roller support

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.

Warren Truss with fixed connections

The Warren 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.🎉🎉

Warren Truss Analysis

Let’s say our Warren Truss is a bridge structure. Therefore, the truss is exposed to dead and traffic load mainly on the bridge deck, which transfers the loads to the bottom chord.

We also simplify and say that the design load ( Load combination of dead and traffic load) is 200 kN/m.

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

The Line load of 200 kN/m is approximated as point loads in the nodes, because otherwise the bottom chord members would be beams, which makes the calculation a lot more difficult.

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

$$200 kN/m \cdot 10m = 2000 kN$$

while the point loads applied to nodes (b) and (c) is calculated as

$$200 kN/m \cdot 20m = 4000 kN$$

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.

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 2000 \, \mbox{kN} – 2 \cdot 4000 \, \mbox{kN} = 0 $ -> $V_a = V_b = \frac{12000}{2} \mbox{kN} = 6000 \mbox{kN}$

$\sum M = 0: M_a = 0$

Calculation of the internal forces

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 , 2000 kN, N 1 and N 4 .

$$ \alpha = 45 ° $$

Vertical Equilibrium:

$$ \sum V = 0: A_v – 2000 kN + N_4 \cdot sin(\alpha) = 0$$

Let’s solve that for N 4 .

$$ N_4 = \frac{2000 kN – 6000 kN}{sin(\alpha)} = -5656.85 kN$$

Horizontal Equilibrium:

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

Let’s solve that for N 1 .

$$ N_1 = – N_4 \cdot cos(\alpha) = 4000 kN$$

$$ \sum V = 0: N_4 \cdot cos(\alpha) + N_5 \cdot cos(\alpha) = 0$$

Let’s solve that for N 5 .

$$ N_5 = – N_4 = 5656.85 kN$$

$$ \sum H = 0: -N_4 \cdot sin(\alpha) + N_5 \cdot sin(\alpha) + N_{10} = 0$$

Let’s solve that for N 10 .

$$ N_{10} = N_4 \cdot sin(\alpha) – N_5 \cdot sin(\alpha) = -8000 kN$$

$$ \sum V = 0: N_5 \cdot sin(\alpha) + N_6 \cdot sin(\alpha) – 4000 kN= 0$$

Let’s solve that for N 6 .

$$ N_6 = – N_5 + \frac{4000 kN}{sin(\alpha}= 0$$

$$ \sum H = 0: -N_1 + N_2 – N_5 \cdot cos(\alpha) = 0$$

Let’s solve that for N 2 .

$$ N_{2} = N_1 + N_5 \cdot cos(\alpha) = 8000 kN$$

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

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 9 , N 5 = N 8 , N 6 = N 7 , N 10 = N 11 .

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

• Cost-effective due to their efficient design
• Widely accepted around the world
• Easy to construct
• Very stable structure: Good buckling resistance as compression members are not too long
• Warren Truss doesn’t spread concentrated loads such as point loads evenly to all members. Most of the load is taken by the closest members. If this is considered in the design, the cross-section of the members taking up the concentrated load increases.
• Big deflection for long spans

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

• Snow load on a flat roof
• Wind load on a flat roof
• Wind load on walls

Because there are always multiple loads acting on Warren 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

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

Let us know in the comments below ✍️.

Warren Truss FAQ

The Warren Truss is mostly used for (railway) bridges of smaller and medium size, because trusses generally suffer from big deflections for longer spans.

Here are 3 reasons why the Warren truss is good! – Its efficient design makes it cost-effective – It’s a very stable structure – Easy to construct

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JSmol Viewer

Experimental study on the static behavior of reinforced warren circular hollow section (chs) tubular trusses.

1. Introduction

2. experimental study, 2.1. test specimens, 2.2. material properties, 2.3. test program, 3. test results, 3.1. failure mode, 3.2. load carrying capacity, 3.3. overall deflection, 3.4. strain intensity, 4. finite element analysis, 4.1. general, 4.2. material modelling, 4.3. finite element type and mesh size, 4.4. weld modeling and steel tube–concrete infill interface model, 4.5. loading and boundary conditions, 4.6. verification of finite element models, 5. conclusions and future work, author contributions, conflicts of interest.

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Share and Cite

Yang, W.; Lin, J.; Gao, N.-n.; Yan, R. Experimental Study on the Static Behavior of Reinforced Warren Circular Hollow Section (CHS) Tubular Trusses. Appl. Sci. 2018 , 8 , 2237. https://doi.org/10.3390/app8112237

Yang W, Lin J, Gao N-n, Yan R. Experimental Study on the Static Behavior of Reinforced Warren Circular Hollow Section (CHS) Tubular Trusses. Applied Sciences . 2018; 8(11):2237. https://doi.org/10.3390/app8112237

Yang, Wenwei, Jiankang Lin, Ni-na Gao, and Ruhao Yan. 2018. "Experimental Study on the Static Behavior of Reinforced Warren Circular Hollow Section (CHS) Tubular Trusses" Applied Sciences 8, no. 11: 2237. https://doi.org/10.3390/app8112237

Truss Series: Warren Truss

The Warren Truss is a very common design for both real and model bridges. It’s exact history and origination is a little muddled, however. James Warren patented a design in 1848 (in England), which many attribute the name “Warren Truss”. His patent was more about the methodology of building rather than a “design”. Regardless, the Warren Truss has been around a while and has been very popular. Examples of it can be found everywhere in the world.

The Warren Truss uses equilateral triangles to spread out the loads on the bridge. This is opposed to the Neville Truss which used isosceles triangles. The equilateral triangles minimize the forces to only compression and tension. Interestingly, as a load (such as a car or train) moves across the bridge sometimes the forces for a member switch from compression to tension. This happens especially to the members near the center of the bridge.

How the forces are spread out

Here are two diagrams showing how the forces are spread out when the warren truss is under a load. The first shows the load being applied across the entire top of the bridge. The second shows a localized load in the center of the bridge. In both cases the total load = 100. Therefore, you can take the numbers as a percentage of the total load.

Interestingly, there is a significant difference. When the load is concentrated on the middle of the bridge, pretty much all the forces are larger. The top and bottom chord are under larger forces, even though the total load is the same. Thus, if you want your school project bridge to be able to hold more weight then try to spread out the force across the top of the bridge.

For a real life Warren Truss bridge, the forces often will be very localized and not spread out along the bridge. Thus, engineers must calculate how strong to make each member of the bridge and build accordingly. Unfortunately, not many Warren bridges are made anymore.

Warren Truss for model bridges

I have definitely used the Warren truss design for many balsa and basswood bridges. I have also used for some popsicle stick bridges. In fact, you can get a learning kit using a Warren Truss from my store . I think the Warren is a very solid choice when designing a model bridge. If you do not know how to start designing your own bridge, I would recommend the Warren, or the Pratt or Howe trusses.

The Warren truss is easy to use with Lap Joints, which are very strong joints. Find out more on my Bridge Joints page. All you have to do is lay down your top and bottom chords, and glue on the truss members directly on top of the top and bottom chords. The example bridge that I build in my 5 Steps to Building a Model Bridge ebook is a Warren Truss design.

Additional Resources

Pictures of real Warren Truss Bridges In depth history of the Warren Truss

74 thoughts on “Truss Series: Warren Truss”

With a center load (which I have for my project), how am I supposed to distribute the force across the top?

I also have a center load project for Tech.and i also am wondering how to distribute the weight across the top.

Currently we have middle school project on Warren Truss Balsa bridge, How did you managed the load to spread across the top , bottom chords . We are only able to get 10 to 12 lbs most . How can we increase the load on to the bridge . Have to use only Balsa wood and did lap joints . Appreciate your help . We will be heading to states shortly . So need your inputs . Thanks .

You can spread the weight by extending the truss system above the single span pictured on this website, i.e. make the bridge a triangle. You can continue the design upwards, shortening each successive span by two equilateral triangles, until you have a central triangle which transfers forces downwards onto the base span. This will obviously increase the mass of the bridge and place extra load on the base span, but will spread the load 🙂 Hope this helps.

I have to do a science fair project to see what bridge designs hold the most weight. I want to test the best. 🙂 Can you tell me what are some of your favorites? Thanks!

truss bridge for sure

i agree that truss is the best

i agree as well

i love this i might get an a on my project

I like the suspension bridge.

I have a tech project and I’m trying to find the easiest bridge to build. Not that I’m trying to take the easy way out, but what do you think about this? Would it be a good choice?

I’m doing a tech project and we have to make a truss bridge and test it, are Warren trusses good??

hi, i have to do a tech project too, from what i have found, Warren trusses are the best.

yes they are great

I’ve done 2 bridge building project and have used the warren truss model for both. I think it is very effective and would definetly use it again

how do i calculate the forces on each member just using a calculator?

what grade are you in and what level of math can you do? some trigonometry is needed to calculate the forces. can you do trigonometry?

I am doing a project for school, we are making a bridge out of skewers i need some help and ideas! Thanks 😀

this web site is amazing! helped me out a ton 🙂

Hi guys! I have a project for science and we need to make a bridge out of skewers and straws. We only have 40 skewers and i need a strong design that can hold its weight x4! If anyone can help me find a design that includes warren trusses that would super duper helpful!! 😀 xx

having participated in a few model truss competitions while in college i can offer a few words of wisedom.

1, carefull clean construction with good joints goes a long way 2, most likely failure mode will be the joints or in compression. so make good joints since this is a science project think about bucking and do some simple tests. 3, shorter is better for compression, long and thin is good for tension. this will enable you to optimize the design. withadditional details on the span, width and materials I can provide more advise.

Hi, I will be doing this for a physics CP project that will be due in a week. From what my teacher has explained the weight will be held from the center of the bridge and not spread out equally throughout the bridge. Which design of a truss bridge would you most recommend for this project? We will be graded based on how much the bridge can hold, and how much the bridge islet weighs. I would appreciate any help you can give me.

Regan, use the Warren Truss bridge design, I have recently done an engineering assignment where we had to make a light weight bridge out of balsawood and see how much weight it could take then our strength/weight(bridge) ratio was calculated. the warren truss bridge I created weighted 120g and held 19kgs… I also did another type of bridge that was heavier and only held 15kgs. I would thoroughly reccomend an Warren Truss Bridge, hope this helps 🙂

We have to build a Warren Truss bridge out of tooth picks :O

We did too! It was so hard to design and to build!

yayv i hate stem

thank you for this!! well, we have a project too, we will make a bridge out of pasta 🙂 but i think this is easier than others :))

We have a requirement on our Steel Design, I’m going to design a bridge using warren truss, 2 lanes and 4 spans. What should I do first?

design the body of your bridge first and go from there

I was wondering if you had anymore info on the Neville truss

Hello, I have to do a basswood bridge project for school. I was planning on using a Warren Truss for my project but I am unsure, because the directions say that the load will be administered from a loading point that will hang down from the bridge. If the load is hanging down from the bridge should Is still make the bridge short and long, or try use a different design?

the designs that truss made are awesome and i agree to them in so many ways

Hi,Im doing a tech project aswell and I used the pratt.

This post was super awesome and the whole website in general was great. I’m a senior in high school and in my AP Physics class after the AP test we spent a week building bridges out of balsa wood. The bridge had to spam a 10 inch gap and was loaded from the top. Max height was 3 inches and max width was 2 inches. We built a warren truss with 6 bottom segments and 5 top segments. The load was placed in the middle of the middle top segment. We used Loctite glue for the joints. The bridge was 12 inches long and a little over 2 inches high and about an inch and three quarters wide. The bridge weighed 9.92 grams and held 20.2 pounds. The scoring was kilograms per gram and the score was 0.92 I believe. It was our first time building a bridge for a competition like this and we couldn’t have done it without this website. We used small gussets on our joints for both the truss and cross members. We also set the record for all time best bridge in the class. The balsa wood we had to use was 0.125 in by 0.125 in. The segment of the top that had the load used an extra piece of balsa to reinforce it because the load was not on a joint and we were afraid the balsa would break.

Great job, Patrick.

Pure awesomeness, Patrick

It is great it helped

This is a clear design and I have used it for many projects and is has not failed once

Could you tell me why do you always apply forces at the top joints? In real life, as I understand, only bottom joints are loaded.

Peter, this has to do with the ease of loading from the top verses loading from the deck of a model bridge. The theoretical load distribution doesn’t change based on top or bottom loading.

does the warrent bridge have a solid road bed

The Warren Truss is amazing!

How much weight can a warren truss bridge made out of popsicles hold?

A lot! A Warren truss is a great design and works well in model bridge made out of popsicle sticks and other types of materials.

It works well made out of balsa wood, too. Over 20 lb!

will the amount of trusses on a warrent bridge that is the smae length affect how much it can hold?

How much weight can a warren truss bridge hold if it is made out of straws?

what do all those numbers represent in the diagram and why are they of that value/ how do they add up??

we are trying to find something that will help with our science fair project. Our question is what size truss on a thirty centimeter bridge suits the bridge the most. if anyone could help we would appreciate it

hello cole, im glad to know that you are working so hard on your science fair project. To answer your question cole A thirty centimeter bridge should have a number that is easily divisible by thirty, in order to have a even amount of trusses that will cover the whole bridge. Thirty is a good number to pick, because all the numbers divisible by it are good picks for an experiment.

Hello there, I am doing some research for a TV programme about the history of the Warren Truss bridge.

I am looking for a standout example of a Warren Truss bridge that is still in existence. Do you know the oldest/first/longest/biggest Warren Truss bridge in the world? Or another notable one. The only stipulation is, it must be a Warren Truss to be exact.

Any advice greatly appreciated… thank you!

I don’t know off the top of my head what would be a good bridge for you to feature. However, you can browse this website for noteable Warren Truss bridges: http://bridgehunter.com/category/tag/warren-truss/

I just wanted to say that your web site has helped me a lot and probably always will

I think this information wasn’t that helpful. What sites can I use for background information about Warren Truss Bridge Designs?

I need help with some info fro the Warren Truss Bridge. Can you help me?

Hey I need some help on a project and I hope someone can help me out so my question is how much does an average warren truss bridge cost?

Hi there, I don’t understand what these numbers mean. Do they mean the newtons of force? And also why is there the number 139 on the bottom closest to centre?

The numbers in the truss analysis pictures represent a percentage of the total force. They are not associated with any specific unit, like Newtons. Instead, because they are showing a percentage, you can substitute any unit in their place as long as you are consistent.

So for instance, let’s say there is a 100 Newton force pushing down in the very center of the bridge. The bottom red member that says “139” now means that there is 139 Newtons of force pulling on that member (red in this case shows tension). How can this be if the total force applied at the top is only 100 Newtons? This is because the design of the truss sometimes multiples the original force in certain members.

Hi there, Because of you helping with this, the team I am in have won the bridge building competion. Our bridge, decided with the help of this website, got a paddle pop stick and hot glue bridge that held up 0.112kg per gram of bridge, and held up 20kg overall! We flattened the competition, who’s best was 0.29kg behind. So basically thank you.

Would the numbers be the same if the center load was supported from the base of the bridge construction?

Yes, it would be the same.

Hi, can anyone help with this question? Where is the longest single span Warren bridge in the uk? And how long is the span?

It will not be the railway bridge over the river Severn in the centre of Worcester which has a mid span support but it is an illustration of a classic Warren girder.

Why did James Warren make the Warren truss bridge?

I imagine that he thought it was a better way to build a bridge.

its because it uses less materials than the other designs that were made in the 18th century

I am using this site to help write a report on truss bridges and you have awesome information here. Is there any way I can find the resources that you had used through the making of this site? Thank you.

Where was the first Warren truss bridge build?

hi! I would like to know if Warren Truss Design is applicable in bridge 100meter long.

This is wrong, you should have only used the reaction forces of the individual forces. no wonder there is more compression and tension on the members. LOL

I am doing a science fair project for a bridge project with my friend and we are wondering if a howe, warren, or pratt truss would work best covering a 24 in. span.

I think that the warren would be better.

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FREE K-12 standards-aligned STEM

curriculum for educators everywhere!

Find more at TeachEngineering.org .

• TeachEngineering
• Doing the Math: Analysis of Forces in a Truss Bridge

Lesson Doing the Math: Analysis of Forces in a Truss Bridge

Grade Level: 11 (11-12)

Time Required: 3 hours

Lesson Dependency: None

Subject Areas: Algebra, Computer Science, Geometry, Number and Operations, Physics, Problem Solving

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Activities Associated with this Lesson Units serve as guides to a particular content or subject area. Nested under units are lessons (in purple) and hands-on activities (in blue). Note that not all lessons and activities will exist under a unit, and instead may exist as "standalone" curriculum.

• Trust in the Truss: Design a Wooden Bridge

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Engineering connection, learning objectives, worksheets and attachments, more curriculum like this, pre-req knowledge, introduction/motivation, associated activities, lesson closure, vocabulary/definitions, additional multimedia support, user comments & tips.

Determining the strength of structures is extremely important in civil and mechanical engineering. Engineers use sophisticated computer programs that solve all the equations resulting from a given problem solution. Engineers have to solve a wide variety of problems that requires finding the solution of one or many systems of linear equations. Because these systems may contain hundreds, if not thousands of equations, computers and software are used to solve them.

After this lesson, students should be able to:

• Identify the components of a truss bridge.
• Identify different type of trusses.
• Perform the analysis of forces at a truss’ nodes using free body diagrams.
• Use trigonometric ratios to find force magnitudes at a truss’ nodes.
• Use the method of joints to set up a system of linear equations to calculate the tensions and compressions on the truss elements.
• Use the substitution method to solve systems of linear equations larger than 3 x 3.

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Common Core State Standards - Math

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Students should have an understanding of the following mathematical concepts: systems of linear equations; solving systems of linear equations using the substitution method; writing a system of linear equations in matrix form; solving systems of linear equations in matrix form using the matrix inverse. right triangle trigonometry; sine and cosine ratios, the Pythagorean theorem, concept of force, and analysis of forces using free body diagrams. Students should also understand the basics of Google Sheets.

Start class with this question: What is the most efficient path to move to one point to another point, and why? (Expected answer: a straight line, because it is the shortest, the fastest, or easiest distance ).

One critical step in a bridge design process of is to determine if the proposed structure will be strong enough to withstand heavy traffic, strong winds or earthquakes. Civil engineers must perform precise calculations and select the proper materials and sizes for every bridge part. A mistake in these estimations may lead to fatal accidents (Figure 4).

You will work like civil engineers performing the basic calculations for a very simple bridge structure. But even though you will solve for a very simple structure, the same procedure is used for more complex structures. Therefore, by understanding the simple cases you will be able to understand and solve more complex cases.

This lesson will give you the background for your next job: the design, construction, and test of a wooden truss bridge (see the associated activity Trust in the Truss: Design a Wooden Truss Bridge ). You will also use math and technology to estimate the tensions-compressions on the truss’ elements, as well as the maximum strength of this bridge based on the wood and the thickness selected to build it. You will finish up by performing a strength test of your bridge. Your bridge must be able to support at least 90% of the estimated maximum strength.

Lesson Background and Concepts for Teachers

This challenging lesson is originally designed as an application of right triangle trigonometry and is directed to pre-calculus or trigonometry students. This section is divided into three parts. Parts 1 and 3 are core of this lesson, and the see the associated activity Trust in the Truss: Design a Wooden Truss Bridge and have to be taught during class . Teacher may or may not teach Part 2, depending on the students’ background. The Notetaking Sheet will help students follow along and easily annotate all of the explanations. There are also some video tutorials available (see Additional Multimedia Support below) that the teacher can assign to students to provide background or to reinforce the concepts in the lesson.

1. Truss Bridges

We are going to focus on a specific class of bridges: truss bridges. In civil engineering, a truss is a regular structure built with straight members with end point connections. No member is continuous through a joint. The straight elements usually form triangular units, because this is the most stable structure in this type of bridge. Truss bridges were widely used in the 19th century, because of their relative low cost and efficient use of materials. The truss design uses only tension and compression elements, which makes this structure strong and allows for simple analysis of forces on its structure.

Engineers have designed different kinds of truss bridges while searching for the optimal combination of strength, weight, span, and cost. (See Figure 5.)

This lesson and its associated activity will focus on four truss bridges: the Warren, the Warren with verticals, the Pratt, and the Howe (Figure 6).

Warren Truss.   A design distinguished by equal-sized components and the ability of some of the diagonals to act in both tension and compression. The type is generally characterized by thick, prominent, diagonal members, although verticals could be added for increased stiffness. This design was patented by the British engineer James Warren in 1848.

Warren truss bridges gained popularity in the United States after 1900 as American engineers began to see the structural advantages of riveted or bolted connections over those that were pinned. The design was well suited to a variety of highway bridge applications and was very popular from about 1900 to 1930.

Pratt Truss.  In this truss, its elements are arranged in right triangles. The United States railroad expansion in the 19th century required strong, dependable bridges to carry trains over ravines and rivers. In 1844, Caleb and Thomas Pratt developed a bridge that was built initially with wood and diagonal iron rods. Later, the bridge was built entirely of iron. This bridge had the advantage of low-cost construction, and could also be quickly erected by semi-skilled labor. This design became the standard American truss bridge for moderate spans from 7.62 meters (~25 feet) to 45.72 meters (~150 feet), well into the 20th century.

Howe Truss. Similar to the Pratt truss, elements of the Howe truss are also arranged in right triangles, but with different orientation. Designed by William Howe in 1840, it used mostly wood in construction and was suitable for longer spans than the Pratt truss. Therefore, it became very popular and was considered one of the best designs for railroad bridges back in the day. The diagonal structural beams slope toward the bridge center, while Pratt truss utilizes diagonal beams that slope outward from the center of the bridge. This approach makes diagonal members of Howe truss bridge in compression, while vertical web members are in tension.

2. Analysis of Forces. Basic Concepts

In physics, a force is any action that tends to maintain or alter the motion of a body or to distort it. When a bridge is loaded it is expected to be still, so all the forces applied on the bridge have to be absorbed by its structure. In other words, all forces are in equilibrium.

In a truss, it is assumed that the forces along the elements converge at the nodes of the structure. This fact allows us to use a free body diagram to find the acting forces values.

By definition, a free body diagram (FBD) is a representation of an object with all the forces that act on it. The external environment, as well as the forces that the object exerts on other objects, are omitted in a FBD. This allow us to analyze an object in isolation.

A FBD can be constructed in three simple steps: first, sketch what is happening on the body; second, identify the forces that act on the object; and third, represent the object as a point with the forces as arrows pointing in their acting direction: with an origin at the point representing the object, with a size proportional to their magnitude and a label indicating the force type. Figure 7 shows the FBD for a spherical object rolling on an incline.

When all the forces that act upon an object are balanced, then the object is said to be in state of equilibrium . This does not mean that the forces are equal, but that the sum of all forces add up to zero. FBD’s are used to decompose forces into their vertical (x) and horizontal (y) components; then the state of equilibrium can be stated as the sum of all these vertical and horizontal components equal to zero.

Using the sigma notation to represent sum, the equilibrium conditions can be mathematically written as:

Ʃ F x = 0       Ʃ F y = 0                                                              (1)

where F x represents all the forces’ components along the x -axis (horizontal), and F y all the components along the y -axis (vertical). These components are determined using the magnitude of the force, and the value of the sine or cosine of the angle the force make with the horizontal. These components will have a (+) sign when pointing in the positive direction of the axis, or (-) sign when pointing in the negative direction of the axis.  

Forces T and S make horizontal angles of 60° and 40° respectively. Force W works along the vertical axis. What should be the values for forces T and S to keep the systems static when W = 10 N?

T x = T ·cos60°                T y = T ·sin60°                                                 (2)

The equilibrium conditions (1) requires that all the forces’ x -components and y -components add up to zero. The T -force components ( T x , T y ) can be found using the trigonometric ratios sine and cosine of the 60° angle, because the force makes a right triangle with these axes (Figure 8(c)):

S x = S ·cos40°                S y = S ·sin40°                                                    (3)

The S-force components ( S x , S y ) can be found in a similar way (Figure 8(d)):

The W -force only has a y -component: W x = 0 , W y = W . Adding up the x -components and the y -components of the forces in equations (2) and (3), and making these additions equal to zero, we obtain the equilibrium equations for this problem:

Ʃ F x = T x + S x = T ·cos60° - S· cos40° = 0                                            (4)

Ʃ F y = T y + S y + W = T· sin60° + S· sin40° - W y = 0                             (5)  

where the minus signs in equations (4) and (5) are assigned because these components are pointing to the negative direction of the axis. Using now the given value for W = 10 N, and taking cos60° = 0.5, sin60° = 0.866, cos40° =0.766, and sin40° = 0.643, equations (4) and (5) can be used to determine the values for T and S to keep the system in equilibrium:

                                  0.5 T - 0.766 S = 0                                                                     (6)

                                  0.866 T + 0.643 S - 10 = 0                                                         (7)

The system of linear equations (6)-(7) can be easily solved. Using the substitution method, we can solve for T in equation (6)

 T = 1.532 S

and substituting this expression for T in equation (7), is possible to find a value for S :

0.866 (1.532 S ) + 0.643 S - 10 = 0

                   S = 5.077 N  

Using the found value for S , is now possible top find the value for T :

T = 1.532 (5.77) = 7.778 N  

3. The Joints Method                                                                                 

Note to Teacher: In this section, a step-by-step example of the simplest method used in civil engineering to solve for the unknown forces acting on members of a truss is presented. The bridge structure will be the smallest and simplest possible: The Warren truss with three equilateral triangles (Figure 9). Larger structures, or different like Pratt or Howe, are solved in the same way. Use the Notetaking Sheet to help students to follow you during this long process.

The joints method determine forces at the truss joints or nodes using FBD’s. The general assumptions to apply this method are:

(a). All truss elements are considered rigid, they never bend.

(b). A force applied to the truss structure will only produce compression or tension on the elements.

(c). Tension - compression forces’ directions are parallel to the elements.

(d). Any force on a truss element is transmitted to its ends.

(e). A truss structure in equilibrium means that every joint or node is at equilibrium.

(f). Once determined the value of a tension or compression force at one of the ends of an element, the complementary force at the other end of the element will be equal but in opposite direction. (equilibrium condition).

In this specific example (Figure 9) it is assumed that:

(g). Vertical and equal downward forces of 10 lbf are applied on the top nodes (nodes 2 and 4), and at the central node (node 3).

(h). The bridge stands only at its end bottom nodes (nodes 1 and 5).

(i). The equilateral triangles are 4 inches side.

Step 1. Identify the corresponding forces acting on each of the truss elements.

It is convenient to identify the forces on the truss elements making reference to the corresponding end nodes. For example, the force acting on the element joining nodes 1 and 2, is denoted F 12 , (Figure 10), and the force acting on element between nodes 4 and 3 is denoted F 43 . The vertical forces acting on nodes 2, 3, and 4, are denoted as F 2 , F 3 , and F 4 .

Step 2. Find the value of reaction forces on the end bottom of the truss (nodes 1 and 5)

Reactions are forces developed at the supports of a structure, to keep the structure in equilibrium. To find the reaction forces on the truss it is required to calculate the moments all the forces applied on the truss can produce, respect to every of the end bottom nodes.

By definition, the moment of a force is the product of the distance from the point to the point of application of the force and the component of the force perpendicular to the line of the distance:

The moment of a force M quantifies the turning effect or rotation the force can produce. In the case of a structure in equilibrium, the moment of all the forces applied has to be zero.

In this example (Figure 12), forces F 1 , F 2 and F 3 , tend to rotate the truss clockwise respect to node 1, but the reaction force R 5 on node 5 cancels this effect (Figure 12a), keeping the truss in equilibrium. This equilibrium condition is expressed mathematically as the sum of the moments of all forces and reactions equal to zero:

M 1 = Ʃ F 1 · x 1 = 0               (8) 

                                   = R 1  · 0 – F 2 · 2 - F 3 · 4 + F 4 · 6 + R 5   · 8 = 0

 where the minus signs indicate that the forces point in the negative direction.

These forces also tend to rotate the truss counter clockwise respect node 5, but the reaction force R 1 on node 1 cancels this effect (Figure 12b). This equilibrium condition can be also expressed as:

M 5 = Ʃ F 1 · x 1  = 0               (9) 

                                                                           =  R 1  · 8 –  F 2   · 2 -  F 3  · 4 +  F 4  · 6 +  R 5  · 0 = 0

  For this example, given F 1 = F 2 = F 3 = 10 lbf , the value of reaction R 5 can be found solving equation (8):

= R 1 · 0 – 10 · 2 - 10 · 4 + 10 · 6 + R 5 · 8 = 0

-120 +  R 5 · 8 = 0

R 5  = 15 lbf

and the value of reaction R 1 can be found solving equation (9):

=  R 1  · 8 – 10 · 6 - 10 · 4 + 10 · 2 +  R 5  · 8 = 0

R 1 · 8 - 120 = 0

R 1 = 15 lbf

Step 3. Analysis of Forces on Nodes using FBD and the equilibrium conditions ƩF y = 0 and  ƩF x = 0.

For those experienced solving systems of linear equations, is going to be odd to have an overdetermined system for a truss structure (more equations than variables). However, we know from Algebra that even though most of the times an overdetermined system of equations has no solution, there are overdetermined systems of equations that have a solution. This is the case for this example.

In structural analysis, a truss is statically determinate when all the forces on its elements can be found by equations of statics alone. This is what we have in this example. In Step 4 we can verify that our system has a solution.

Step 4. Solve the System of Equations. Identify the obtained forces as Compressions (-) or Tensions (+)

Note to teacher: Continue using the Notetaking Sheet to help students to follow you during this long process. Is recommended to model how to the first four or five equations and then ask students to work in small groups to obtain the solutions for the rest of the equations.

In this example, the substitution method is going to be used. The variables to solve are: F 12 , F 13 , F 23 , F 24 , F 34 , F 35 , and F 45 .  F 1 , F 2 , F 3 , R 1 and R 5 are known constants: F 1 = F 2 = F 3 = 10 lbf, and R 1 = R 5 = 15 lbf .

It is simple to verify that the obtained values for F 34 , F 35 and F 45 satisfy equations (17), (18), and (19):

Step 5.  Create the Matrix for the above System of Equations

Note: This simple concept is important in the computer graphic interface explained in Step 6. Continue using the Notetaking Sheet . Guide students to obtain the associated matrix for the system of equations (10)-(19):

Step 6. Spreadsheets to Calculate Trusses

The procedure shown in Steps 1 - 5 is also used to solve other type of trusses, larger trusses, isosceles triangles or right triangles trusses, or when different loads are applied on the nodes. However, a larger truss will produce a larger system of equations, which it will be more difficult to solve by hand. For example, a Warren truss with nine equilateral triangles will produce a 19 x 22 system of linear equations. Also, every change in the triangles angles or in the loads values will require to perform all the calculations again.

Because the procedure is always the same, no matter the truss type or the number of elements, it is possible to use Excel or Google Sheets and create a graphical interface to perform all these calculations automatically , giving the loads and the elements’ angles as entry values.

Note to teacher: It is not in the scope of this lesson that students write themselves such interface. But for those advanced or curious, the procedure used to write this interface is detailed in the Annex .

This section summarizes how to use a very friendly computation interface developed by the author in Google Sheets. This graphic interface:

• Calculates the tensions-compressions for the Warren truss, Warren with Vertical truss, Pratt truss, and Howe truss.
• Estimates maximum strength of the trusses considering the kind of wood used and element’s thickness.
• Gives solutions when the truss is supported only on its bottom end nodes, the truss’ diagonal elements are round or square, and the truss’ rails are square.

Visit this Trusses Calculations webpage for more information.

In this web page there are four interfaces (Figure 14(a)) corresponding to the four different trusses. Truss size is defined here by the number of triangles in the truss. The following table summarizes truss types and sizes:

To activate an interface on a PC, laptop, or tablet, click on the gray square at the right top of the window. (Figure 14(a)); click on the window when using a cellphone. (Note: these sheets are open to the public for distribution and to copy; the master will always remain locked.)

Once the spreadsheet is active, you can see its different sections (Figure 14(b)):

• A truss diagram with entries for the loads on each node.
• Two entries for the truss elements’ length and angle respect the horizontal.
• Cells to select the truss elements’ thickness and kind of wood they are made of.
• Section displaying the calculated truss elements’ tensions-compressions
• The operational section: matrix associated to the system of linear equations obtained from the FBD’s, the inverse matrix, and the solutions of the system of equations.

Students only have to input values in the cells in sections (1) and (2), select round or square diagonals (3), and select the elements’ wood type and thickness in the cells in section (4) (Figure 15). Sections (5) and (6), or other cells do not have to be altered. The tension-compression on the truss elements are automatically calculated once values in sections (1), (2), (3) or (4) are entered.

Next is suggested a guided practice to teach students the efficient use of these interfaces.

Important: Before this practice, reset the values in the Warren truss in the third sheet: 5 in the loads, 5 in the triangles’ base length, select round diagonals, and hardwood + 1/8 in the Diagonals entries, and also Hardwood + 1/8 in the Rails entries. There is a hardcopy of these instructions in the Notetaking Sheet , make enough copies for students. You will also need to be familiar with how to access and work with Google Drive and Google Sheets with a cellphone or computer. Use projections of these notes to support the practice steps.

Tell students: Because it is required in the associated activity to estimate the strength of the bridge to build, and this bridge is going to be larger than the one solved here, the number of nodes is going to increase, so the resulting system of equations will be larger, and the time to solve will be longer. For example, the analysis of a Warren truss with 11 equilateral triangles will produce a linear system with 26 equations and 23 variables!

But there is more. Any change in the loads values, or truss angles, will require to repeat all the analysis again. Repeating by hand the same process, even after the simplest change, is not really efficient.

You are going to learn how to use a graphic interface to easily perform all this calculations. You should have a basic knowledge of Google Sheets or Excel, a Google Drive account, and your cellphone or a computer with internet access.

These interfaces are very easy to use. You have in your notes a hardcopy of these instructions in your notes for quick reference.

First you have to open the webpage where these interfaces are located. Open Google Chrome in your cellphones or on the computer and type the next address:

https://sites.google.com/gpapps.galenaparkisd.com/mramirez-math/courses-highlights/bridges/trusses-calculations

In this page select the Warren Truss interface. Click on this window to open the interface, and click on the Open button at the top of the document. (You will be required to use your Google account here.) Once the worksheet is open you con scroll this and identify the different parts of the graphic interface (Show to students Figure 15).

You can easily see the cells where you can input information. Do not type anything yet. Because this document is shared to every one of you, so every change made by one of you is going to be displayed in all other cellphones and computers. For this little practice it is necessary you work with your own copy of this document. So create your own copy in your Google Drive, using the commands Share & export + Make a copy, give a new name to the file and save it in your Drive. You will work now in your saved file.

Note to Teacher. It is very important you inform students that even though the original file cells that do not have to be modified are protected, the copies they save possibly will not be protected. So ask them to be very careful and not to modify or erase other cells but the indicated in the practices.

Your first assignment is to use the interface to calculate the Warren truss with three equilateral triangles we have solved in class. You know the correct solutions for this problem, so the interface, if correctly developed, will give you the same answers. Click on the cells next to the truss nodes (show Figure 16), and type the number 10 + ENTER in each one. Select now the cell for triangles’ base length and type 4 + ENTER. Verify the diagonal’s angle is 60°. Check the numbers below the legend “Tensions-Compressions on Truss Elements”, Are these the same as the previously obtained?   (Give some time to students to do this and then display the Warren truss interface with these values)

Once you verified that the interface gave the same values, you can use the interface for more complex calculations. Your next assignment is to overload the little truss. Change the 10 lbf load on the central bottom node (Node 3) by 65 lbf, and check the new tensions-compressions. What do you see now?

The expected answer is: four cells changed color . Ask to students now: What does this mean?  Check in the color code table. The answer is: the four diagonal elements break under this new load . The next question to students is: Assuming you need a bridge able to support this load, what could be the solution to this problem . The expected answer is: use a stronger material .

Continue saying: Let’s first change the shape of the diagonal elements from round to square, what happened now? The expected answer is: Only two diagonals are broken now. Ask students: Why is this possible? You only changed a round element of 1/8 inch dimeter by a square of 1.8 inch side? Draw on the board a square representing 1/8 inch and inside a circle with the same diameter to help students with their reasoning . It is expected they see that the square’s area is greater than the circle’s area, and therefore it is a thicker, and consequently, a stronger element.

Continue with the next: Change the diagonals to “round” again, and go now to the cells under the legend “Wood Compressive Strength – Truss Elements Strength. Here you have two options: Choose a different wood or choose a thicker element. Begin with the second option, click on the “Diameter” and “Side” cells and select thicker elements than 1/8. Select the next in size element. (Students will find 3/16 for Diagonals, and Rails) . What happened now? Students should answer something like: The thicker element of 3/16 inch resists this new load.

Ask students now to return Diagonals and Rails thicknesses to 1/8. Now ask them to change wood to Birch in both Diagonals and Rails, and ask students: What happened now? Students should answer: Birch is a stronger wood, 1/8 inch elements of this wood can resist the load 1/8 Hardwood elements canno t.

In the last part of this practice, students should independently manipulate the larger trusses. Assign problems from the last section of the Notetaking Sheet , four per student. Here is the first problem of this list:

Find the minimum diagonals and rails thicknesses for a hardwood Warren truss 20 inches long, made up with nine equilateral triangles that are able to resist a load of 50 lbf on every top node and 100 lbf on every bottom node (Figure 17).

The next table contains the approximate solutions for the practice problems at the end of the Notetaking Sheet . There is really more than one acceptable answer for every problem, depending on the wood type and thickness, or if round or square diagonals are used. The next table shows some possible answers to the proposed problems.

A last remark about the graphic interfaces. The wood options in these worksheets include only the most common dowels can be found in stores: hardwood, pine, basswood, poplar, oak, and birch.

During this assignment move around the room and check students’ work and provide the necessary help and support. It is very important they understand completely the use (and limitations) of these interfaces.

• Trust in the Truss: Design a Wooden Bridge - In this activity students design, construct, and test the strength of a wooden truss bridge and satisfy certain conditions like span, strength, and cost. Students perform the truss bridge strength estimation using a graphic interface that determine stress-compression on the truss elements using the method of joints.

To check the overall understanding of the lesson, ask students to write an exit note summarizing the steps used to solve a truss:

• Define the vertical loads magnitude and their location on the truss.
• Assume that the bridge is supported only at the bottom end nodes, and that the bridge and every node is at equilibrium.
• Perform the analysis of moments these loads produce respect the bottom ends to determine the reactions there.
• Perform on each node the analysis of forces using FBD’s to obtain the corresponding pair of equilibrium equations.
• Solve the obtained system of equations using the substitution method.

Ask some students to read aloud these notes. Be sure they have clear understanding of all these steps and their sequence. Make the necessary corrections.

bridge: A structure that is built over a railroad, river, or road so that people or vehicles cross from one side to the other in a safe way.

center of moment: The actual point about which a force causes rotation.

compression: Force that is transmitted through a solid when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the solid and pulls equally against the acting forces on the object’s opposite ends.

equilibrium: In Physics, the condition of a system when neither its state of motion nor its internal energy state tends to change with time. It is also defined as the condition in which the total force acting on an object is zero.

equilibrium conditions: Forty-components of all the acting forces be equal to zero: ∑Fx = 0, ∑Fy = 0

force: In Physics, a force is any interaction that, when unopposed, will change the motion of an object. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. When analyzed in two dimensions, a force F is separated in components along the x-y axes, using the force’s magnitude, its angle with the horizontal axis, and basic right triangle trigonometry.

free body diagram: In Physics and Engineering, a graphical illustration used to visualize the applied forces, movements, and resulting reactions on a body in a given condition. A tool used to solve the various forces acting on an object. It is used to minimize the complexity of any problem.

linear equation: An equation between two variables that gives a straight line when plotted on a graph.

method of joints: A method widely used by civil engineers to find the forces acting on a structure. It considers that every portion of the structure is in equilibrium and every joint with the bars connected to it as a free body. Truss members are assumed to be hinged together at the joints and all bars connecting at the joint do not pass through the joint. Bar forces do not produce any moments or rotation at the joint, then only two equations of statics are used to find the unknown forces: ∑F_x = 0, ∑F_y = 0

moment of a force: In Statics, it is a measure of the tendency of a force to cause a body to rotate about a specific point or axis. The magnitude of the moment is defined as the product of the magnitude of the perpendicular force (F), and the arm’s length (d).

rigid body: It is a solid body in which deformation is zero or so small that can be neglected. The distance between any two given points on a rigid body remains constant in time regardless of external forces exerted on it. A rigid body is usually considered as a continuous distribution of mass.

sigma notation: A convention used to write out a long sum in an abbreviated way, consisting in the capital letter S from the Greek alphabet, Sigma (∑), followed by an indexed term representing the elements or numbers to add: ∑x_i

substitution method: A sequence of algebraic steps to solve systems of equations, that for a 2x2 system with x-y variables follows the next way: Step 1: Solve one of the equations for either x = or y = . Step 2: Substitute the solution from step 1 into the other equation. Step 3: Solve this new equation. Step 4: Solve for the second variable in the first equation, using the value obtained in step 3

system of equations: A collection of two or more equations with a same set of unknowns. The equations in the system may be linear or non-linear.

tension: It is the force that is transmitted through a solid when it is pulled tight by forces F acting from opposite ends. The tension force is directed along the length of the solid and pulls equally against the acting forces on the object’s opposite ends.

truss: A regular structure or frame built with straight members with end point connections and forces that act only at these end points. No member is continuous through a joint.

truss bridge: A bridge whose load-bearing superstructure is composed of a truss, a structure of connected elements usually forming triangular units. The connected elements (typically straight) may be stressed from tension, compression, or sometimes both in response to dynamic loads.

Pre-Lesson Assessment

Pre-Test: Give students the Pre-Assessment to test students’ understanding of the use of the substitution method to solve linear equations.  

Lesson Summary Assessment

Check: Check students’ Notetaking Sheet and give a grade for completeness, clarity, and neatness. Make clear that these notes are going to be the most important reference during the associated activity, Trust in the Truss: Design, Construction, and Testing of a Wooden Truss Bridge .

There is no a homework for this lesson, but teacher may optimize class time assigning to students watch videos related with the content of this lesson. These videos are available at www.sophia.org .

There is available a set of video-tutorials for every one of the topics covered in this lesson. Make sure students know they have this on-line help. Bridges Project:  https://www.sophia.org/playlists/bridges-project

High school students learn how engineers mathematically design roller coaster paths using the approach that a curved path can be approximated by a sequence of many short inclines. They apply basic calculus and the work-energy theorem for non-conservative forces to quantify the friction along a curve...

Building on their understanding of graphs, students are introduced to random processes on networks. They walk through an illustrative example to see how a random process can be used to represent the spread of an infectious disease, such as the flu, on a social network of students.

Benson, Harris: University Physics. Revised Edition . Chapter 2 Vectors. John Wiley & Sons. 1996

Learn Civil Engineering. Structure Engineer Section Review / AM Section. Mechanics of Materials-Tension and Compression. http://www.learncivilengineering.com/wp-content/themes/thesis/images/structural-engineering/PE-reviewStructure-Mechanics-of-Materials-Tension-and-compression.pdf

Mr. Wayne’s Classroom. Free Body Diagrams. The Basics. http://www.mrwaynesclass.com/freebodies/reading/index01.html

NCDOT North Carolina Department of Transportation. Historic Bridges.

https://www.ncdot.gov/projects/ncbridges/historic/types/?p=17 

Arbabi, F. Structural Analysis and Behavior . McGraw-Hill Inc. 1991

Pinsdaddy. Steels Pin Connections. http://www.pinsdaddy.com/steel-pin-connections_PDE*7M5xqrsuQ099UuT1wYt2ME6U6BxL5UlfXt969ao/pxhere. Steel Rivet Connections. https://pxhere.com/en/photo/908147

University of North Carolina at Charlotte. Learning Activity #3. Analyze and Evaluate a Truss.

https://webpages.uncc.edu/~jdbowen/1202/learning_activities_manual/Learning_Activity_3.pdf  

University of North Carolina at Charlotte. Learning Activity #1. Build a Model of a Truss Bridge. https://webpages.uncc.edu/~jdbowen/1202/learning_activities_manual/Learning_Activity_1.pdf

University of N. C. Charlotte. Learning Activity #5. Design and Build a Model Truss Bridge.

https://webpages.uncc.edu/~jdbowen/1202/learning_activities_manual/Learning_Activity_5.pdf

The Historical Marker Database. “The Pratt Through-Truss Bridge.” Patuxent Branch Trail. https://www.hmdb.org/marker.asp?marker=20498

Mathalino. “Engineering Mathematics. Method of Joints, Analysis of Simple Trusses” https://www.mathalino.com/reviewer/engineering-mechanics/method-joints-analysis-simple-trusses

Journal of the American Institute of Architects . “Design Flaws Contributed to Fatal FIU Bridge Collapse.” Press Release. November 29, 2018

https://www.architectmagazine.com/practice/design-flaws-contributed-to-fatal-fiu-bridge-collapse_o

Contributors

Supporting program, acknowledgements.

The author expresses his thanks to Michelle Merritt, Galena Park ISD Director for Secondary Mathematics, Shameel Ali, Galena Park High School, Math Specialist.

Last modified: January 28, 2021

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The Warren Truss

Bridge historians and early textbooks generally call a truss with alternating compression and tension diagonals a Warren; however, sometimes it is called an equilateral truss since all panel lengths and diagonals are of equal length creating a series of equilateral triangles. When the panel lengths are shorter than the equal length diagonals, it was sometimes called an isosceles or isometric truss.

Figure 1. Commonly accepted Warren Truss.

When the span length increases and the height of the truss necessarily increases, the long compression members in the top chord need bracing to minimize buckling in the vertical direction. In this case, verticals are placed from the lower chord panel points up to the mid point of the chord member directly above. In addition, the deck structure stringers get longer requiring either heavier members or the addition of verticals from the top chord panel points dropping down to shorten panel lengths.

Figure 2. Warren Truss with verticals to support top chord and deck structure.

Neither of these truss styles are what James Warren and Willoughby Monzani patented in 1848 in England. They based their patent on similar trusses that were built in France by Alfred H. Neville and a patent that was granted in England to William Nash in 1839 on a similar design. Warren and Monzani were well known English engineers, and their design was for a truss that could be used as a deck or a through truss. They used cast iron for the top chord, and diagonals and wrought iron bars and links for the lower chord members. The top chord cast iron members were connected through cast iron junction blocks, and the cast iron diagonals and lower chord wrought iron members were connected with pins. The title of the patent application was Construction of Bridges and Aqueducts and was issued on August 15, 1848 with Patent #12,242. Their profile was rectangular. Even though Squire Whipple in the United States had published the method of determining loads in truss members under uniform and varying loads, this method had not made its way to England. It wasn’t until 1850 that W. B. Blood developed a method of analyzing triangular trusses, as Whipple had.

Figure 3. Warren and Monzani patent drawing showing deck at both levels.

Warren and Monzani’s patent stated,

The specification of this invention exhibits four different modes of building bridges, which it is stated may, with some slight modifications, be applied to the construction of aqueducts and roofing. 1) The bridge is built with cast iron side bands, rods, or plates, inclined towards each other, and combined so as to form a series of Vandykes [V shapes]. They are bolted at top to horizontal compression rods, and at bottom to horizontal tension rods, and carry a roadway at top or at bottom, or at both. 2) Or the bridge may be built of cast iron side angular frames (placed with the apices [point] downwards), which have their bases bolted together, end to end, and their apices bolted to horizontal rods. 3) Or, instead of the preceding modes of longitudinal construction, hollow cast-iron transverse frames may be employed, which are inclined, and bolted together at top, and are similarly attached at bottom to horizontal rods, bars, or plates. 4) Or wrought iron tie rods may be bolted at top to compression rods, and at bottom held together by the side of wooden girders, and the structure strengthened by means of stay rods. The angles of the plates are regulated by longitudinal screw rods and nuts.

It is clear they didn’t size their members or give details as to the load, either tension or compression in their diagonals. He didn’t even think of his web members as triangles but only connected VanDykes (V’s) between a compression member on top and a tension member on the bottom. They had four claims as follows,

1) The mode of constructing bridges, aqueducts, or roofing with iron rods, bars, or plates, inclined towards each other, and connected together at top by compression band, and at bottom by tension band, so as to carry a roadway at top or bottom, or at both. 2) The mode of constructing bridges with cast iron angular frames bolted together at their bases, and having their apices bolted to horizontal compression rods. 3) The mode of constructing bridges with transverse hollow cast iron frames inclined towards each other, and bolted together at top and at bottom to horizontal plates. 4) The mode of constructing bridges with wrought iron rods inclined towards each other, and attached at top and bottom, as described.

It appears their only claim to originality was in the use of triangles with top compression chords and lower tension chords. The first major bridge, built by Joseph Cubitt in 1852 roughly to the patent, was the Newark Dyke Railroad Bridge of the Great Northern Railroad. In it he used alternating cast iron compression and tension diagonals with cast iron upper chords and wrought iron links for the lower chord. At the middle panel he had opposing cast iron members.

Figure 4. Newark Dyke Bridge, cast iron A-frame on pier.

The bridge crossed the Dyke on a sharp angle, requiring a span of 240 feet 6 inches. Cubitt said the Warren design was brought to him by C. H. Wild. He wrote,

Each girder consists of a top tube, or strut of cast-iron, and a bottom tie of wrought-iron links, connected together by alternate diagonal struts and ties of cast and wrought iron respectively, dividing the whole length into a series of equilateral triangles, of 18 feet 6 inches length of side. These girders rest on the apices of cast-iron A-frames, placed on the masonry of the abutments (Figure 4). Each pair is connected by a horizontal bracing at the top and the bottom, leaving a clear width of 13 feet for the passage of the trains… The trusses are so arranged, that all compressive strains are taken by the cast-iron, and all tensile strains by the wrought iron; the strains, in all cases, in the direction of the length are of the respective parts, and all cross strain is avoided. The parts are so proportioned, that when loaded with a weight equal to one ton per foot run, which considerably exceeds the weight of a train of the heaviest locomotive engines in use on the Great Northern, or on any narrow-gauge line, no tensile, or compressive strain on any part, exceeds five tons per square inch of section.

It is clear that by 1852 Wild had taken the Warren configuration and, applying Blood’s method of analysis, calculated the load in each member so that it could be proportioned properly. An end view of the bridge showed, however, the hugeness of the members which was typical of English and European bridge design at the time. Over time, the bridge style was converted to all wrought iron with built up riveted members.

Figure 5. End View of one of the two parallel Newark Dyke Spans. Note massiveness of cast iron members as well as the verticals to support the deck at mid panel points.

In the United States, the Warren/Wild/Cubit design was known by our engineers. Many of them subscribed to the Proceedings of the Institution of Civil Engineers where Cubitt had published his article. Prior to 1848, Whipple had designed and built similar trusses on the New York and Erie Railroad and discussed them in his 1846/47 book on bridges. He included the plan shown in Figure 6.

Figure 6. Whipple plan for a bridge similar to the Warren Plan with inclined end posts.

Not only did he design this span, he built several for the New York and Erie Railroad in 1848, the same year Warren got his patent in England.

Figure 7. Whipple’s Brandywine Creek Bridge, New York and Erie Railroad, 1848.

In an article in Appleton’s Magazine and Engineers Journal in January, 1851, he described some of his New York and Erie bridges and wrote,

These were wrought-iron skeleton girders upon the triangular plan, such as have since been called Warren girders, and by some regarded as a newly invented combination. But they are merely trusses with parallel chords and diagonals, or rather, oblique members, with only one series of obliques, and without verticals, except to concentrate weight upon the obliques from intermediate points along the upper or lower chord, according as the girder is loaded at such upper or lower chord.

Whipple did not think there was anything new with what was being called a Warren Truss. In fact, in his 1846/47 book he wrote of trusses without verticals. He called this a “cancelled truss which dispenses with vertical pieces, except perhaps at the ends, or at the first bearing points from the ends.” He found, in fact, that a truss, his trapezoidal without verticals, used 8% less iron.

Figure 8. Whipple Plan 1846 but bridge historians call it a Double Warren Truss.

Several trusses that were patented in the United States incorporated the alternating tension and compression diagonals associated with the Warren Truss. The first was a wood and iron rectangular truss by A. D. Briggs in 1858 (#20,987) followed by Alber Fink in 1867 (#62,714) with a combination wood and iron trapezoidal truss with equilateral triangles with verticals dropping down to support the deck at mid panel points. He wrote, “I adopt the triangular, system of bracing between the two chords, both because this system best avoids evils arising from the unequal expansion of a wrought-iron bottom and wooden top chord, and because it is the system of bracing having the least amount of material for equal strength with other systems.” In the same year, J. Dutton Steele (#63,666) received a patent for an Isometric Truss. He had been building them since 1863 and called it an isometric plan, as the diagonals were of equal length with a shorter panel length. He had Charles Macdonald write a long report comparing all of the standard bridge designs, including the Pratt, Howe, Whipple and Warren trusses. Macdonald concluded the only cost savings in a truss bridge are in the web members, as the top and bottom chord requirements were the same for most bridges. For a standard bridge span length of 165 feet, he determined the Howe trusses needs 54% more iron in the web and the Pratt needs 31% more iron than the Isometrical truss. He then compares the Isometrical truss with Linville’s double intersection truss and determines the isometrical uses 19% less iron in the web. He presents the results of a study by C. Shaler Smith in 1865 where he compared the Fink, Bollman, Triangular (Warren) and Murphy Trusses. Smith determined the Triangular and Murphy were more efficient than either the Fink or Bollman trusses for both through and deck trusses. His conclusion stated the Isometrical Truss required less iron in the web system than any other trusses considered. In addition, he found the Isometrical Truss was, especially in wood, much easier to adjust in the event of wood shrinkage.

Figure 9. J. Dutton Steele patent drawing for an isometric plan.

In 1872, Whipple, in an article in the Transactions ASCE entitled “On Truss Bridge Building,” wrote that he had objections to Macdonald’s pamphlet and how he used the Whipple Double Intersection truss in his comparison, stating: “Now, Mr. Macdonald represents what he designates as the ‘Whipple Truss,’ with diagonals inclining only 30° from the vertical. I desire here to enter my emphatic protest against the imputation of ever having tolerated any such practice.” He then got into the Isometric Truss (and the Warren style), writing:

But what of the Isometric? The name, at least, as applied to bridge trusses, is new, and euphonic [pleasing to the ear] withal. This is a truss with parallel chords without vertical members in the web: one of the general types discussed and compared in my publication of 1847 with reference to Fig. A., page 14… I am not aware that there had existed any examples of the parallel chord truss without verticals, prior to their construction by me over 20 years ago, with the important exception of the plank lattice bridge. This was first known to me under the name of “Town’s Lattice Bridge,’” and it was a very cheap and serviceable bridge when properly constructed… But somehow it occurred to me…that a plan in which every member of the web system should do something in the way of advancing the weight toward the abutments, might possess advantages over one having vertical members merely to transfer the action of weight directly from chord to chord without advancing it at all horizontally… The Trapezoidal truss, with and without verticals, though depending upon combinations so old that “the memory of man” (especially the present generation) “runneth not to the contrary” still, perhaps, owes something to me for economical form and proportions… These gentlemen [Macdonald and Merrill] are pleased to term ‘The Whipple Truss;’ and considering that the Isometric and the Post [with inclined posts] trusses are merely modifications (and not very favorable modifications either) of a type of truss first used and thoroughly discussed by me.

It is clear that Whipple believed that the Warren or Isometric trusses were simply extensions of trusses he wrote about in the 1840s, and built in the 1840s and 1850s. In an article on the Pratt Truss (STRUCTURE, May 2015), a case was made that the trusses called Howe and Pratt should really be called Whipple Trusses. A similar case is made here that the Warren Truss should really be called a Whipple Truss. The reasons are that Warren, when he developed his truss, did not know how to size his members nor could he distinguish between tension and compression in his web members. He never designed or built a truss with an inclined end post nor a truss with verticals. A truss as he patented it was never built. Whipple on the other hand had analyzed, designed and built trusses with various web members and inclined end posts prior to Warren’s patent.

Figure 10. Little Juniata Bridge, Pennsylvania RR, cast and wrought iron with verticals, Pony Truss ~1870.

Figure 11. Bell’s Bridge, Delaware, Lackawanna & Western RR 1872, Double Warren or Whipple.

It is probably too late to change what most people call the various trusses, but it should be at least recognized that most of the truss patterns used in the late 19 th and 20 th century had their origins in the United States and Squire Whipple between 1841 and the 1880s. What were called the Warren trusses were built in the thousands as short span pony trusses with no verticals, longer spans with verticals, even longer spans with double intersections, and still longer spans with subdivided panels. They were originally built with cast and wrought iron members with pins and, later, with wrought iron members and cast iron joints with pins, and later fully riveted in steel. Polygonal top chords were also added in many trusses to extend the span length. J. A. L. Waddell used the pattern in many of his lift spans after the turn of the century. Several examples of the bridge style are shown in Figures 10, 11 and 12.▪

Figure 12. Warren, Isometric, Truss, Polygonal top chord, with verticals, all riveted steel bridge for BNSF Railroad over Verdigris River, Oklahoma~1960.

About the author  ⁄  Frank Griggs, Jr., Dist. M. ASCE, D. Eng., P.E., P.L.S.

Dr. Frank Griggs, Jr. specializes in the restoration of historic bridges, having restored many 19th Century cast and wrought iron bridges. He is now an Independent Consulting Engineer. ([email protected])

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5.6: Methods of Truss Analysis

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• Page ID 42966

• René Alderliesten
• Delft University of Technology via TU Delft Open

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There are several methods of truss analysis, but the two most common are the method of joint and the method of section (or moment).

5.6.1 Sign Convention

In truss analysis, a negative member axial force implies that the member or the joints at both ends of the member are in compression, while a positive member axial force indicates that the member or the joints at both ends of the member are in tension.

5.6.2 Analysis of Trusses by Method of Joint

This method is based on the principle that if a structural system constitutes a body in equilibrium, then any joint in that system is also in equilibrium and, thus, can be isolated from the entire system and analyzed using the conditions of equilibrium. The method of joint involves successively isolating each joint in a truss system and determining the axial forces in the members meeting at the joint by applying the equations of equilibrium. The detailed procedure for analysis by this method is stated below.

Procedure for Analysis

•Verify the stability and determinacy of the structure. If the truss is stable and determinate, then proceed to the next step.

•Determine the support reactions in the truss.

•Identify the zero-force members in the system. This will immeasurably reduce the computational efforts involved in the analysis.

•Select a joint to analyze. At no instance should there be more than two unknown member forces in the analyzed joint.

•Draw the isolated free-body diagram of the selected joint, and indicate the axial forces in all members meeting at the joint as tensile (i.e. as pulling away from the joint). If this initial assumption is wrong, the determined member axial force will be negative in the analysis, meaning that the member is in compression and not in tension.

•Apply the two equations \(\Sigma F_{X}=0\) and \(\Sigma F_{Y}=0\) to determine the member axial forces.

•Continue the analysis by proceeding to the next joint with two or fewer unknown member forces.

Example 5.2

Using the method of joint, determine the axial force in each member of the truss shown in Figure 5.10a.

\(Fig. 5.10\). Truss.

Support reactions. By applying the equations of static equilibrium to the free-body diagram shown in Figure 5.10b, the support reactions can be determined as follows:

\(\begin{array}{ll} +\curvearrowleft \sum M_{A}=0 \\ 20(4)-12(3)+(8) C_{y}=0 \\ C_{y}=-5.5 \mathrm{kN} & C_{y}=5.5 \mathrm{kN} \downarrow \\ +\uparrow \sum F_{y}=0 \\ A_{y}-5.5+20=0 \\ A_{y}=-14.5 \mathrm{kN} & A_{y}=14.5 \mathrm{kN} \downarrow \\ +\rightarrow \sum F_{x}=0 \\ -A_{x}+12=0 \\ A_{x}=12 \mathrm{kN} & A_{x}=12 \mathrm{kN} \leftarrow \\ \end{array}\)

Analysis of joints. The analysis begins with selecting a joint that has two or fewer unknown member forces. The free-body diagram of the truss will show that joints \(A\) and \(B\) satisfy this requirement. To determine the axial forces in members meeting at joint \(A\), first isolate the joint from the truss and indicate the axial forces of members as \(F_{A B}\) and \(F_{A D}\) , as shown in Figure 5.10c. The two unknown forces are initially assumed to be tensile (i.e. pulling away from the joint). If this initial assumption is incorrect, the computed values of the axial forces will be negative, signifying compression.

Analysis of joint \(A\).

\(\begin{array}{l} +\uparrow \sum F_{y}=0 \\ F_{A B} \sin 36.87^{\circ}-14.5=0 \\ F_{A B}=24.17 \\ +\rightarrow \sum F_{x}=0 \\ -12+F_{A D}+F_{A B} \cos 36.87^{\circ}=0 \\ F_{A D}=12-24.17 \cos 36.87^{\circ}=-7.34 \mathrm{kN} \end{array}\)

After completing the analysis of joint \(A\) , joint \(B\) or \(D\) can be analyzed, as there are only two unknown forces.

Analysis of joint \(D\).

\(\begin{array}{l} +\uparrow \sum F_{y}=0 \\ F_{D B}=0 \\ +\rightarrow \sum F_{x}=0 \\ -F_{D A}+F_{D C}=0 \\ F_{D C}=F_{D A}=-7.34 \mathrm{kN} \end{array}\)

Analysis of joint \(B\).

\(\begin{array}{l} +\rightarrow \sum F_{x}=0 \\ -F_{B A} \sin 53.13+F_{B C} \sin 53.13+15=0 \\ F_{B C} \sin 53.13=-15+24.17 \sin 53.13= \\ F_{B C}=5.42 \mathrm{kN} \end{array}\)

5.6.3 Zero Force Members

Complex truss analysis can be greatly simplified by first identifying the “zero force members.” A zero force member is one that is not subjected to any axial load. Sometimes, such members are introduced into the truss system to prevent the buckling and vibration of other members. The truss-member arrangements that result in zero force members are listed as follows:

1.If noncollinearity exists between two members meeting at a joint that is not subjected to any external force, then the two members are zero force members (see Figure 5.11a).

2.If three members meet at a joint with no external force, and two of the members are collinear, the third member is a zero force member (see Figure 5.11b).

3.If two members meet at a joint, and an applied force at the joint is parallel to one member and perpendicular to the other, then the member perpendicular to the applied force is a zero force member (see Figure 5.11c).

\(Fig. 5.11\). Zero force members.

5.6.4 Analysis of Trusses by Method of Section

Sometimes, determining the axial force in specific members of a truss system by the method of joint can be very involving and cumbersome, especially when the system consists of several members. In such instances, using the method of section can be timesaving and, thus, preferable. This method involves passing an imaginary section through the truss so that it divides the system into two parts and cuts through members whose axial forces are desired. Member axial forces are then determined using the conditions of equilibrium. The detailed procedure for analysis by this method is presented below.

Procedure for Analysis of Trusses by Method of Section

•Check the stability and determinacy of the structure. If the truss is stable and determinate, then proceed to the next step.

•Make an imaginary cut through the structure so that it includes the members whose axial forces are desired. The imaginary cut divides the truss into two parts.

•Apply forces to each part of the truss to keep it in equilibrium.

•Select either part of the truss for the determination of member forces.

•Apply the conditions of equilibrium to determine the member axial forces.

Example 5.3

Using the method of section, determine the axial forces in members \(CD\), \(CG\), and \(HG\) of the truss shown in Figure 5.12a.

\(Fig. 5.12\). Truss.

Support reactions. By applying the equations of static equilibrium to the free-body diagram in Figure 5.12b, the support reactions can be determined as follows:

\(\begin{array}{l} A_{y}=F_{y}=\frac{160}{2}=80 \mathrm{kN} \\ +\rightarrow \Sigma F_{x}=0 \quad A_{x}=0 \end{array}\)

Analysis by method of section. First, an imaginary section is passed through the truss so that it cuts through members \(CD\), \(CG\), and \(HG\) and divides the truss into two parts, as shown in Figure 5.12c and Figure 5.12d. Member forces are all indicated as tensile forces (i.e., pulling away from the joint). If this initial assumption is wrong, the calculated member forces will be negative, showing that they are in compression. Either of the two parts can be used for the analysis. The left-hand part will be used for determining the member forces in this example. By applying the equation of equilibrium to the left-hand segment of the truss, the axial forces in members can be determined as follows:

Axial force in member \(CD\). To determine the axial force in member \(CD\) , find a moment about a joint in the truss where only \(CD\) will have a moment about that joint and all other cut members will have no moment. A close examination will show that the joint that meets this requirement is joint \(G\). Thus, taking the moment about \(G\) suggests the following:

\(\begin{array}{l} +\curvearrowleft \sum M_{G}=0 \\ -80(6)+80(3)-F_{C D}(3)=0 \\ F_{C D}=-80 \mathrm{kN} & 80 \mathrm{kN} (C) \end{array}\)

Axial force in member \(HG\).

\(\begin{array}{l} +\curvearrowleft \sum M_{C}=0 \\ -80(3)+F_{H G}(3)=0 \\ F_{H G}=80 \mathrm{kN} & 80 \mathrm{kN} (T) \end{array}\)

Axial force in member \(CG\). The axial force in member \(CG\) is determined by considering the vertical equilibrium of the left-hand part. Thus,

\(\begin{array}{l} +\uparrow \sum F_{y}=0 \\ 80-80-F_{C G} \cos 45^{\circ}=0 \\ F_{C G}=0 \end{array}\)

Chapter Summary

Internal forces in plane trusses: Trusses are structural systems that consist of straight and slender members connected at their ends. The assumptions in the analysis of plane trusses include the following:

1.Members of trusses are connected at their ends by frictionless pins.

2.Members are straight and are subjected to axial forces.

3.Members’ deformations are small and negligible.

4.Loads in trusses are only applied at their joints.

Members of a truss can be subjected to axial compression or axial tension. Axial compression of members is always considered negative, while axial tension is always considered positive.

Trusses can be externally or internally determinate or indeterminate. Externally determinate trusses are those whose unknown external reactions can be determined using only the equation of static equilibrium. Externally indeterminate trusses are those whose external unknown reaction cannot be determined completely using the equations of equilibrium. To determine the number of unknown reactions in excess of the equation of equilibrium for the indeterminate trusses, additional equations must be formulated based on the compatibility of parts of the system. Internally determinate trusses are those whose members are so arranged that just enough triangular cells are formed to prevent geometrical instability of the system.

The formulation of stability and determinacy in trusses is as follows:

\(\begin{array}{l} m+r<2 j \quad \text { structure is statically unstable } \\ m+r=2 j \quad \text { structure is determinate } \\ m+r>2 j \quad \text { structure is indeterminate } \end{array}\)

Methods of analysis of trusses: The two common methods of analysis of trusses are the method of joint and the method of section (or moment).

Method of joint : This method involves isolating each joint of the truss and considering the equilibrium of the joint when determining the member axial force. Two equations used in determining the member axial forces are \(\Sigma F_{X}=0\) and \(\Sigma F_{y}=0\). Joints are isolated consecutively for analysis based on the principle that the number of the unknown member axial forces should never be more than two in the joint under consideration in a plane trust.

Method of section: This method entails passing an imaginary section through the truss to divide it into two sections. The member forces are determined by considering the equilibrium of the part of the truss on either side of the section. This method is advantageous when the axial forces in specific members are required in a truss with several members.

Practice Problems

5.1 Classify the trusses shown in Figure P5.1a through Figure P5.1r.

\(Fig. P5.1\). Truss classification.

5.2 Determine the force in each member of the trusses shown in Figure P5.2 through Figure P5.12 using the method of joint.

\(Fig. P5.2\). Truss.

\(Fig. P5.3\). Truss.

\(Fig. P5.4\). Truss.

\(Fig. P5.5\). Truss.

\(Fig. P5.6\). Truss.

\(Fig. P5.7\). Truss.

\(Fig. P5.8\). Truss.

\(Fig. P5.9\). Truss.

\(Fig. P5.10\). Truss.

\(Fig. P5.11\). Truss.

5.3 Using the method of section, determine the forces in the members marked X of the trusses shown in Figure P5.13 through Figure P5.19.

\(Fig. P5.13\). Truss.

\(Fig. P5.14\). Truss.

\(Fig. P5.15\). Truss.

\(Fig. P5.16\). Truss.

\(Fig. P5.17\). Truss.

\(Fig. P5.18\). Truss.

\(Fig. P5.19\). Truss.