TRUSS BRIDGES

What is a beam bridge, truss bridge and how can you compute the forces at play.

In a THROUGH TRUSS bridge, the load ( people, trains, cars etc.) goes THROUGH the truss.

Here’s the Burnside Bridge in Portland OR. A DECK TRUSS, where the load (and the deck) is above the truss.

In a PONY TRUSS bridge, the load travels through the truss, even though there is no “top” to the truss. This 20ft long bridge
was built by a high school class.

This page will cover the different types of truss bridges, and how to use the Method of Joints and the Method of Sections to analyze the bridge.

What is a beam? A horizontal support. Here’s one

How to make 2 “trees” into a beam bridge

This beam bridge could easily support a couple hundred pounds.

How to turn a beam bridge into a truss.

A truss bridge is made of 2-force members that carry the load in either Tension or Compression. Here’s a simple model to illustrate that.

A truss is a beam made of many smaller parts (members) some of which form triangles. Here’s a 5 pound truss model that easily supports a 150 pound load.

More video of the bridge made using 1/4″ plywood.

METHOD OF JOINTS- a handy tool for computing the forces in a truss

This 3 member (piece) truss is made of the 2 angled legs and the connecting member- the black thread connecting the legs at the bottom.

Before we can analyze the truss, we need to understand the 3 equilibrium equations.

Now to find the External forces acting on the truss.

First the reactions at A and C. We use the sum of the forces in the Y direction and the sum of the Moments about point A.

I’ve been informed by engineers who know, that when computing the external forces, we DO NOT include the weight of the truss. So instead of 11.9 pounds for the “load” I should have used just the weight of the 2 bricks-9.4 pounds.

That means that when you use Sum forces in Y direction- the sum of the 2 reactions is not 11.9 but 9.4pounds. Next we use the Sum of the moments about point A-and use 9.4 instead of 11.9 pounds, and get 4.7pounds for the reaction @C and when substituting back into the earlier formula, we get 4.7 for the reaction @A.

Now for the Internal forces. We use the Free Body Diagram.

By convention, when analyzing the forces using the FBD, we always assume the forces are pulling away from the joint-so the arrow points away from A

Using the sum of the forces in the X (horizontal) direction we get an equation with 2 unknowns, so we have to continue. A diagonal force can be broken down into its vertical and horizontal components.

Now we use the sum of the forces in the Y (vertical) direction to compute the value of Fab.

But the reaction @A is not 5.95 but 4.7 so when we substitute 4.7 into the equation for Sum vertical forces- Fab=negative 6.89 (not 8.72)

Now we have to change Fab from 8.72 to 6.89. When you substitute 6.89into the formula, you’ll get Fac (the thread) = 5.02 pounds of tension.

Finally we can compute the value of Fac-(the thread), which is what was the goal from the beginning.