An Introduction to Graphic Statics

As I’m so fond of mentioning, engineering design required the use of a number of creative methods before the invention of calculators and computers. Some of the most important and widespread of these were graphic methods of analysis. Graphic methods essentially translate problems of algebra into geometric representations, allowing solutions to be reached using geometric construction (ie: drawing pictures) instead of tedious and error-prone arithmetic.

Unfortunately, these methods are slowly being forgotten. It’s extremely rare to ever see them used, outside of a select few occasionally taught in structural analysis courses. But understanding how, and more importantly why, they work unquestionably makes for a better engineer.

To remedy this, this post will lay out some of the basics of graphic statics. If there’s interest, more posts on more advanced methods will follow.

To start, consider a force vector.

vector-1

A vector can be split into component forces along any two axes. The initial force vector is the resultant of these two component forces.

vector-2

If the axes are perpendicular, the resultant can be constructed using the pythagoream theorem. But it can also be done by moving the beginning of one component force to the end of the other (making sure to scale the lengths properly).

vector-3

This allows the construction of a resultant for any two forces acting on a single point.

vector-4

And it’s not limited to just two forces. The resultant for any number of forces acting through a single point can be constructed this way. This construction is known as the force polygon.

vector-5

When the last force connects to the first force (ie: the polygon is closed), the system is in equilibrium.

Of course, typically forces aren’t all acting through a single point. In these cases we can still calculate a resultant, but we’ll also need to take a few extra steps to find it’s location.

beam

To start, consider a set of forces F1 through F4 acting on a beam. To find the resultant, first construct the force polygon (in this case, a straight line). Next, pick a random point “O” to the right side.

vector-6

Now, starting with F1, create a smaller force polygon, with a vertex at “O”.  F1 can be thought of as the resultant of these two new forces, which we’ll call F1A and F1B. Because this polygon is closed, it’s in equilibrium.

vector-7

Create a similar polygon for F2, F3 and F4.

vector-8

Now, starting with F1, copy force F1A and F1B over to the beam, such that each force’s endpoint is along F1’s line of action. Do this for force F2, F3 and F4, making sure that concurrent forces are kept concurrent (ie: F2A is on the same line as F1B).

funicular

This construction is known as the funicular polygon, as it forms the shape a string being deflected by forces acting on it. To find the resultant, we construct another funicular polygon that has identical end forces, but is acted on by only one force. Because a string can only carry tension, a string with identical end forces must have an identical deflected shape at each end. Thus, we can find the location of the resultant by simply extending the lines of the two outer forces until they meet.

The force and funicular polygons form the basis of a wide variety of graphic methods, from moment calculations to arch analysis. Their accuracy is only limited by the accuracy of the scale and straightedge used to construct them.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s