How to Analyze a Truss Bridge

Introduction:

This instructable will go over how to correctly analyze a truss bridge using the method of joints. This technique is common in civil engineering practice. The purpose of this analysis is to take into account the structure, supports, and loads, and calculate the forces within each member of the truss. This information is useful when planning a structure, as it will assist in deciding what materials to build with as well as how thick each member should be to withstand the load. This process will take about 60 minutes.

Supplies:

- Paper (2 sheets)

- Pencil

- Calculator (scientific is best)

The basic principle to understand when dealing with trusses has to do with the term "statics." This system is static, which means that it is not moving. Even though there are forces acting on the structure, there will always be other forces that will "cancel" them out. The sum of forces going in any direction (horizontally or vertically) will always equal zero.

The first step to any truss problem is to "detach" the structure from its surroundings and decide what forces are acting upon it. We will replace these supports with force arrows. In this case, the truss is sitting on two supports at D and F. These two different types of supports will contribute different kinds of forces back on the structure. At this point, redraw the diagram on the paper. It is very helpful to have a personal sketch to mark on.

The support at D is called a pin support. This means that point D will resist movement both vertically and horizontally, so you add two force arrows, one vertical and one horizontal. There is a force directed downward at point E, so you should assume that we need a force to oppose it. The vertical force at point D, call it Dy, should point up.

In the same way, the force at joint C directed to the right, indicating that you should point the horizontal force at point D, called Dx, to the left to oppose C.

The support at F is called a roller. This means that while it can resist movement in the vertical direction (y), it cannot in the horizontal direction (x). So here, you only need one force arrow, pointed up to oppose the force at E. Call this force Fy.

Your drawing of the truss should look like the second picture.

Now you treat the entire truss as one body and balance all of the external forces, making sure they balance to zero in both directions. Start with the x direction:

As seen in the picture, you define anything pointing to the right as positive in our equations. The "Σ" is the Greek letter sigma. It means to "add up." In this case, you will sum all of the forces in the x direction. As stated before, the key here is that all of these forces will add to zero.

The only external forces are a positive 25 kN (kilonewtons, a unit of force) and a "negative" unknown force Dx.

If you arrange those values as shown, you can solve for a value of Dx that will point to the left, just as you guessed before.

The next equation involves a concept called "moments". A moment is a measure of an objects tendency to rotate based on a perpendicular force applied a certain distance away from an axis. The moment equals the force multiplied by the perpendicular distance between it and the axis of rotation. Any force that goes directly through the axis will have a moment equal to zero because the "distance" from the axis will be zero. The handy thing about this moments equation is that you get to choose where you define the axis. This decision is important, as you can usually strategically eliminate some variables to leave only one unknown to solve for.

In this case, you want to choose your axis as point D. This will eliminate the forces Dx and Dy since they go directly through point D. Then you will only be left to worry about Fy, which is unknown, and the forces at C and E, which you do know.

Define a positive moment as anything that would tend to cause counter-clockwise rotation about point D. This direction of rotation can sometimes be a little tricky to visualize. Imagine holding down point D and applying a force elsewhere on the structure. Ask yourself which direction the structure would rotate.

Since Fy is pointed upwards, it would cause positive rotation about D (counter-clockwise), so you multiply Fy by the distance from D, which is 40 m. This distance is called the "moment arm".

The 100 kN force would cause negative rotation, so it gets multiplied by its distance of 20 m and subtracted from the first term.

The last force is at point C, which if looked at carefully, would also cause a negative rotation. This time, the 25 kN force gets multiplied by the moment arm of 15 m, as that is its perpendicular distance from D.

The resulting equation has only Fy as an unknown and can be solved.

Note: Remember the order of operations when solving this equation. Do all of the multiplying first and then add or subtract.

Next, take the same idea from step 2 and apply it in the y direction. Add up all of the forces going vertically and make sure they all balance to zero.

Defining upwards as the positive direction, see that there is the unknown force Dy which is pointed up (positive), and both Fy and the 100 kN force pointing down at E ( both negative). Use the value of Fy you found in the last step.

This equation should look similar to the picture above.

For this section, you will analyze each joint, where the members connect, individually. You will use the same concepts of balancing forces, but instead of using all of the external forces, you will only use forces acting directly on the joint.

It is important to start with the correct joint. Any joint that you analyze cannot have more than two unknown components. After a brief inspection, joint D looks like a good place to start.

Draw a free body diagram of point D separated from everything else. Include the values of Dx and Dy that you found earlier, pointing left and up respectively.

Forces AD and DE will be added in on top and on the right, but you need to choose which direction to point them. The logic here is as follows:

You know Dy is directed upward, so there must be an opposing force directed downward to make the sum of forces in the y direction equal zero. This means that force AD should point down.

The same idea applies to DE. Dx is pointed to the left, so there must be a force to the right to oppose it.

These arrow directions are just guesses and do not necessarily need to be correct. If you guess wrong, the numerical value for that force will come out negative. If that happens, just flip your arrow and make your answer positive.

Now apply the same concepts and create an equation on your own relating the two forces in the x direction.
Remember that they need to be equal to zero!
This equation should be solved for DE.
With each numerical answer for the force in a member, you also need to specify whether the member is in tension or compression. This is decided by which direction the arrow points in your free body diagram.
In this case, DE points away from the joint you are working with. This indications tension, denoted by (T) next to the numerical value.

Record DE=_____ kN (T) and draw a box around it. This will identify the value as a final answer.

The same idea applies to forces acting vertically. Write your own equation and solve for AD.

Force AD points towards the joint, meaning that it is in compression (C). Create your answer box.

Make a note and classify all members as in tension or compression as you solve:

Arrow points away from joint = Tension (T)

Arrow points towards joint = Compression (C)

Now that you know the value of member AD, you can move up to joint A where there are now only two unknowns. Draw a new free body diagram for joint A, just as you did at D.

A critical step here is the placement of the arrow AD on this joint. At joint D, member AD pointed downwards towards D because it was in compression. Here, AD will point upwards towards A to continue signifying compression.

After placing AD on the diagram, you need to predict the direction of the two unknown forces AB and AE also acting on joint A. The logic here is the same as before. You know for certain that AD will point up, so there must be another force contributing some force down to balance. The only other force at joint A that involves the y direction is AE, so it must be directed partially down and consequently also partially to the right. To balance this, AB must point to the left.

Your free body diagram should look like the picture.

Dealing with member AE is a little bit tricky, since it acts both in the x and y direction. You will use trigonometry to separate these components into two different values.

To do this, you first need to find angles alpha (α) and beta (β), as pictured. Use will use the tangent function on your calculator.

Start by identifying angle alpha in the picture. The equation you will use is tan(α)=(opposite/adjacent)

Look at the triangle created by AD, DE, and AE. The side length opposite to α is 20 m and the side length adjacent to α is 15 m. When inserted into the equation and rearranged, you get α=tan^-1(20/15) where tan^-1 is the "inverse tangent" found by pressing the green key followed by the T key on the TI-89 calculator.

Since all angles of triangles must add to 180 degrees and the angle between AD and AE is a right angle (90 degrees), angle β must equal 90-α. Make note of these two angles, as they will be used in future calculations.

Create equations to add up all forces in the x and y directions just as before. This time, however, you will use AEsin(α) as the x component of AE and AEcos(α) for the y component.

It is recommended here to start by balancing forces in the y direction first. This way you will have only AE as an unknown to solve for.

The equations should look like this:

Fy=0=AD - AEcos(α)

F=0= AEsin(α) - AB

Since you know the values of AD and alpha, plug those into equation one and solve for AE. Then plug that value into the second equation to solve for AB.

Continue moving clockwise around the structure to joint B. You can see there are now only two unknown components, so you are equipped to find both of them.

Remember that force AB pointed towards joint A, signifying compression, so now force AB should point towards joint B so it will still indicate compression.

This joint is fairly simple, as you will find that the force in member BC is equal in magnitude to member AB. Since the unknown force BE is the only force in the y direction, it has a value of zero.

This means that member BE has no force acting within it. This structure with the specific loads in this problem would function the exact same way if this member wasn't even there!

Joint C should be analyzed similarly to joint A. Draw another free body diagram and try to predict the directions of each force. Start by adding force BC and the external force of 25kN to the diagram first since you already know their directions. Then add CE and CF to oppose them.

When writing your equations, remember to use CEsin(α) for the x component and CEcos(α) for the y component of CE.

These equations should turn out as follows,

Fx=0=BC + 25 - CEsin(α)

Fy=0=CF - CEcos(α)

Again, solve the first equation for CE using the values of BC and alpha you already know. Then, using your newly found value of CE, use the second equation to solve for CF.

Hint: The magnitude of CF and AE should not be the same, even though they are symmetric on the structure.

If you have calculated correctly up to this point, joint F should be the easiest to analyze. It should turn out that force Fy that you found in the beginning will be equal to force CF. These forces are pointing opposite directions and therefore add to zero.

The only force in the x direction is EF, so this member also has zero force and is in neither tension or compression.

Now you should have values for all nine members along with indications of tension or compression. The last joint serves as a check.

Draw your free body diagram and include all of the values that you have found. Here is where angle β will be used. Enter ABcos(β) and CEcos(β) for x components and ABsin(β) and CEsin(β) for y components.

Neither equation will have any unknowns in them. Add up all of the terms and make sure both equations really do equal zero. If they both do, then you've done everything correctly.

At the end, gather all of the values you found (they should all be boxed) and place them all together at the end. Make sure to include units of kN and indicate T or C for each. This is your finished product.

All of the correct answers are shown above.