How to Use the Method of Bar Conversions?

This post expands on our complex truss example by showing the complete solution of every member's internal axial bar force using the Method of Bar Conversions.

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Continuing from this example, let's learn how to solve this complex truss using the method of bar conversions.

The solution presented is in SI. The author will update the post soon to reflect English units.

Main Solution

Full Solution

The following section shows how to use the method of bar conversions:

For this method, we will break down the complex truss into two simple structures: a simple truss and a proxy truss. When we add these two structures together, you'll get the bar forces for each complex truss.

Convert to a Simple Truss

Convert complex truss example to simple truss

The first step is to convert our complex truss into a simple one by removing any member, say $A D$ ; however, in doing so, we made the truss unstable. To reverse that effect, we add another member, say $D G$ . There is no rule on what member to remove or add. Just remember that our goal here is to make our truss simple.

We load this simple truss with the supports and loads like that of our original complex truss. The result is a new structure with the same reactions and loads $S_{p}$ , serving as an element for the superposition process.

Like any other simple truss, the next step is to thoroughly analyze the reactions and bar forces of this simple truss using the method of joints or sections. It is up to the reader how to solve for such values.

At the end of the analysis, we will have the axial forces of each member summarised in the table shown. Remember that $A D$ does not exist in this truss, and $D G$ is a substitute member for stability purposes.

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Create Proxy Truss

The next step is to create a proxy truss for the superposition process $S_{x}$ . Remember that we could not solve for $A D$ in the simple truss because we removed it; hence, to counteract that effect, we must create the same simple truss as Sp and represent $A D$ with a tensile force $x$ .

Our goal here is to represent all bar forces in terms of this variable $x$ using either method of joints or section.

Let's consider joint A and solve for the unknowns in terms of $x$

$\sum F_{v} = 0] ↑_{+}$

$F_{A B} + \frac{1}{\sqrt{5}} x = 0$

$F_{A B} = - 0.447 x$

$\sum F_{h} = 0] \leftarrow_{+}$

$F_{A G} + \frac{2}{\sqrt{5}} x = 0$

$F_{A G} = - 0.894 x$

Continue doing this for all members. As shown in the table, we will get the forces in terms of $x$ for all members.

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Sign conventions for both $S_{p}$ and $S_{x}$ are critical, so double-check all signs before proceeding to the next part.

Apply Superposition

At this point, we have completely solved the bar forces of both structures $S_{p}$ and $S_{x}$ ; hence, we are ready to apply the idea of superposition.

Remember that we are solving for the bar forces of the complex truss $S_{T}$ . Equation-wise, it is:

$S_{T} = S_{p} + S_{x}$

We first apply superposition to solve for $x$ . With it, we can solve the bar forces of the complex truss. To find for $x$ , let's take an interest in $D G$ , the substitute member added to the simple truss for stability purposes.

We know that $D G$ does not exist in our truss example; hence, $S_{T}$ for the substitute member must be zero:

$S_{T} = S_{p} + S_{x}$

$0 = - 100.96 - 0.40 x$

$x = - 252.4 k N = F_{A D}$

When we solved for $x$ , we already got the axial force of the original member we removed which is $A D$ . With $x$ solved, we can complete our analysis by performing superposition to all members.

Summarize Your Results

After superimposing all bar forces, we have arrived at a complete solution! As one final step, it is always a good idea to summarise your results using a table. When summarizing, it's always great to categorize your results for easy reference (top chords, bottom chords, and web members).