The Method of Joints: Part 1

In one of my courses we’ve been tasked with building a truss bridge out of plastic tubes, slightly thicker than drinking straws, which must span 1 metre and hold 5kg of load. My team had a few basic designs that we wanted to start off with, which meant a whole bunch of calculation to see which one was best. Of course I’m not one to pass up an opportunity like that, and solving systems of linear equations is something that computers are kind-of alright at, so I wrote a python script that solves a given truss and returns the internal forces in each member. With the script complete, a new bridge design takes only about 2 minutes from data input to final results! The slowest step is sketching the bridge design…

01

 

The method of joints involves stepping through each joint in a truss and solving the equations of static equilibrium, i.e.

    \begin{align*} \sum{F_x} &= 0,\\ \sum{F_y} &= 0. \end{align*}

These two equations must be solved for each joint, giving 2n equations, where n is the number of joints present. The procedure is quite straightforward but having more than a few joints makes it get tedious very quickly. So, let’s make it faster.

A simple truss

Consider the frame above (with members 1 & 2 the same length). We have six unknowns to solve for: three reaction forces from the supports, and the three internal forces. We also have six equations describing the system, how convenient. These are:

    \begin{align*} 1:\sum{F_x} &= R_{1x} + F_3 \cdot \frac{1}{\sqrt{2}} = 0 \\ 1:\sum{F_y} &= R_{1y} - F_1 - F_3 \cdot \frac{1}{\sqrt{2}} = 0 \\ 2:\sum{F_x} &= R_{2x} + F_2 = 0 \\ 2:\sum{F_y} &= F_1 = 0 \\ 3:\sum{F_x} &= -F_2 - F_3 \cdot \frac{1}{\sqrt{2}} = 0 \\ 3:\sum{F_y} &= F_3 \cdot \frac{1}{\sqrt{2}} - 100N = 0 \\ &\Rightarrow F_3 \cdot \frac{1}{\sqrt{2}} = 100N \end{align*}

Now this system is simple enough to solve by inspection. Ignoring this though, we can represent this system as a matrix equation C \cdot F = P, where P is a vector of the applied (or known) forces, F is a vector of our unknowns, and C is the coefficient matrix. Thus:

    \begin{equation*} \begin{bmatrix} 0 & 0 & \frac{1}{\sqrt{2}} & 1 & 0 & 0 \\[0.3em] -1 & 0 & -\frac{1}{\sqrt{2}} & 0 & 1 & 0 \\[0.3em] 0 & 1 & 0 & 0 & 0 & 1 \\[0.3em] 1 & 0 & 0 & 0 & 0 & 0 \\[0.3em] 0 & -1 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 \\[0.3em] 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} F_1 \\[0.3em] F_2 \\[0.3em] F_3 \\[0.3em] R_{1x} \\[0.3em] R_{1y} \\[0.3em] R_{2x} \end{bmatrix} = \begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 100 \end{bmatrix} \end{equation*}

As long as the matrix C is non-singular, we can find it’s inverse and recover the solution F = C^{-1} \cdot P. In this case:

    \begin{equation*} \begin{bmatrix} F_1 \\[0.3em] F_2 \\[0.3em] F_3 \\[0.3em] R_{1x} \\[0.3em] R_{1y} \\[0.3em] R_{2x} \end{bmatrix} &= \begin{bmatrix} 0 & 0 & \frac{1}{\sqrt{2}} & 1 & 0 & 0 \\[0.3em] -1 & 0 & -\frac{1}{\sqrt{2}} & 0 & 1 & 0 \\[0.3em] 0 & 1 & 0 & 0 & 0 & 1 \\[0.3em] 1 & 0 & 0 & 0 & 0 & 0 \\[0.3em] 0 & -1 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 \\[0.3em] 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \end{bmatrix}^{-1} \cdot \begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 100 \end{bmatrix} = \begin{bmatrix} 0 \\[0.3em] -100 \\[0.3em] 141.4 \\[0.3em] -100 \\[0.3em] 100 \\[0.3em] 100 \end{bmatrix} \end{equation*}

That was easy! Now comes the question of somehow constructing the matrix for an arbitrary truss, without actually writing the equations by hand.

First, note that for a simple truss (constructed only of triangles) with n joints, there are 2n-3 members connecting them. If the truss is supported by three support forces, for example R_{1x}, R_{1y}, R_{2x} above, then the system is statically determinate: there are 2n equations describing the system and 2n unknowns to solve for.

An arbitrary member

Now consider an arbitrary member k in a truss, connected between two joints i and j. The equations

    \begin{equation*} i:\sum{F_x}, \quad i:\sum{F_y}, \quad j:\sum{F_x}, \quad j:\sum{F_y} \end{equation*}

are the only equations with any contribution from member k (as a member can only connect to two joints), and we can number these equations as 2i-1, 2i, 2j-1, and 2j respectively. The member k will appear in exactly these four equations, although the coefficient may be zero as with F_1 in the example above.

Looking at the x-component of forces at joint i, (equation number 2i-1):

    \begin{equation*} \dots + cos \theta \cdot F_k + \dots + P_{ix} = 0. \end{equation*}

We can simplify this a little by using ratios of the member’s dimensions rather than it’s angle. By defining

    \begin{align*} l_{ij} = cos \theta = \frac{\Delta x}{L_k} \\ m_{ij} = sin \theta = \frac{\Delta y}{L_k} \end{align*}

and by moving the known/applied force P_ix to the right hand side of the equation, we get the following:

    \begin{alignat*}{4} & \dots + l_{ij} \cdot F_k + \dots && = && -P_{ix} \quad && (eqn. \; 2i-1) \\ & \dots + m_{ij} \cdot F_k + \dots && = && -P_{iy} \quad && (eqn. \; 2i) \\ & \dots - l_{ij} \cdot F_k + \dots && = && -P_{jx} \quad && (eqn. \; 2j-1) \\ & \dots - m_{ij} \cdot F_k + \dots && = && -P_{jy} \quad && (eqn. \; 2j) \end{alignat*}

In the matrix C, the entire column k will be zero except for the rows corresponding to these equations. So the contributions to C from member k are:

    \begin{alignat*}{3} & C_{2i-1,k} \quad && = \quad && l_{ij} \\ & C_{2i,k} \quad && = \quad && m_{ij} \\ & C_{2j-1,k} \quad && = \quad && -l_{ij} \\ & C_{2j,k} \quad && = \quad && -m_{ij} \end{alignat*}

With C complete, we just construct the vector F as above, with F_k in order, and the three support reaction forces at the end. The vector of applied forces P is simple enough: for a load L applied at node i we just set

    \begin{alignat*}{3} & P_{2i-1} && = && -L_x \\ & P_{2i} && = && -L_y \end{alignat*}

And now we just find C^{-1}, solve F = C^{-1} \cdot P, and we’re done!

Leave a Reply

Your email address will not be published. Required fields are marked *