Internal Force Analysis - How External Forces Create Internal Tension and Compression

Now that we have the external forces calculated, we can balance the internal forces. There are two methods that we can use for this: the method of sections and the method of joints. We’ll use the method of joints which analyzes bridge forces joint by joint across the structure. A limitation of the approach is that we must work from joints that have at most 2 unknown forces. In this bridge, A and D are the only joints like this. Here you see, for example, A has a force of 50 pounds pushing up in this direction and has two unknown forces: the force AB and the force AE. Since the bridge is symmetrical, we have the same situation occurring on the other side of the bridge. In order to do the calculation we are going to isolate joint A and figure out what AB and AE's forces are by doing vector analysis.

Joint "A"

Here we've isolated joint A. We're showing the force AB coming down the diagonal which will wind up compressing the member in the direction of the AB arrow.  AE is the tension member that pulls the joint in the direction of the AE arrow. Because the bridge isn't moving, we know that there is 0 pounds of force in the direction of the A_x arrow.  Because of the weight of 50 pounds coming in the corner, we know that there's 50 pounds of force pushing the bridge in the A_y direction. We now can take this diagram and decompose it into its X and Y components.

The force acting down on joint "A" is AB x sin(39.8 degrees). 39.8 degrees is the angle at corner "A". The tension force in the X direction has two components: The first is AB_x which is pushing to the left.  Its magnitude is AB x cos(39.8 degrees) The second is AE which has an unknown magnitude in the direction of the arrow. But AB x sin(39.8 degrees) equals 50 pounds (that’s the force pushing down on the corner support), so AB x sin(39.8 degrees) = 50 pounds. Solving for AB, we know AB (the force in that diagonal direction) is equal to 50 pounds divided by sin (39.8 degrees) which is equal to 78.11 pounds. Similarly, we know that the force on AE has to equal the force going in the other direction.  So AE is equal to AB x cos(39.8 degrees) which is equal to 78.11 x the cos(39.8 degrees) which is equal to 60 pounds. We’ve now fully solved the forces of corner A. we can now go to corner B and solve for the forces there.

Joint "B"

Here, we're looking at joint B. We've taken the 78.11 pounds that we know is the component of force in the AB direction and put it on this diagram. We’ve taken the 50 pounds of weight coming down onto point B and shown it at the top. We now need to figure out how much force BE exerts up and how much compression there is across BC. We're going to do this the same way we did before, by resolving these forces in terms of their X and Y components.

In the vector diagram's X direction, since the bridge is not moving, AB_x must equal BC.  But AB_x is equal to AB x cosine(39.8 degrees) which is equal to 78.11 x cos(39.8 degrees) which is equal to 60 pounds of force. Therefore, we have a 60 pound force component in the BC direction.

In the vector diagram's Y direction, we have 50 pounds coming down from the top, 50 pounds of force being transmitted down AB_y, and BE's force pushing up. This lets us solve for BE. BE is equal to 50 pounds minus the AB_y component which is 50 - AB x sin(39.8 degrees) which winds up being 50 – 50 or 0 pounds. In other words, the 50 pounds of force going in the Y direction is transmitted completely down the diagonal and is not supported by the "Vertical Hip" at all.  So BE is 0.

Joint "E"

We can now solve for the forces on joint E. Recall that there were 0 pounds of force on point E and 60 pounds of tension on cord AE. If there are 0 pounds coming on E and 60 pounds coming across AE that means that the 60 pounds of tension has to be transmitted all the way across.

By symmetry, we get exactly the same results for the other side.  The complete Free Body Diagram is shown below.

Summary

In summary, we've figured out if you have:

• 50 pounds of pressure coming down on point B

• 50 pounds of pressure going down on point C

then you get:

• "End Posts" AB, CD transmit 78.11 pounds of compressive force.

• "Bottom Chord" AD transmits 60 pounds of tensile force.

• "Vertical Hip" BE (surprisingly) transmits 0 pounds of force.

• "Top Chord" BC transmits 60 pounds of compressive force.

In our next section, we'll show you how to use these particular force calculations to better understand the materials one should use use in constructing this bridge.