Active Plywood Bending with Variable Stiffness

A few years ago I found makers cutting patterns into plywood and acrylic to weaken them and make them bendable. One early example of the technique is the folding wooden booklet by Snijlab.

Plywood with laser cut slits making it flexible. — *Folding Wooden Booklet by Snijlab.*

I had the idea that it should be possible to precisely control the stiffness of each part and so control its shape under stress.

Results

Physical model

After working out what stiffnesses I needed to match a given curve, full process below, I had a pattern ready for the laser cutter.

Laser cut plywood model with ends connected by string which buckles the plywood making a curve. — *Physical laser cut model.*

I used a curve as a target which I wanted the model to match when under stress. The manufactured model matches the target curve very well.

Laser cut plywood model compared to a curve printed on paper. The model is follows the curve perfectly. — *Shape of manufactured model compared to target curve.*

Interactive Demo

Try this interactive demo by moving the curve control points and see how the generated cutting path is adjusted when the curve is changed.

Segments

29 segments

Process

Calculating the spring stiffnesses

Let's make a simple schematic and name the components and important lengths.

Schematic of the laser cut paths. A series of bars alternately attached in the edges and in the middle. — *Schematic for laser cutting*

The structure is made up of torsion bars and connectors. The connectors are intended to be stiff enough to not significantly deform under bending whereas the bars are meant to be compliant and twist. We want to find the widths of the torsion bars so that when the edges are pushed together and the structure buckles it assumes a predetermined shape. For simplicity we set the torsion bar length $l$ and the connector length $L$ (measured between bar centre lines) to the same value for all elements and only vary the torsion bar width $w$ .

A series of rigid elements arranged in an arch. Each rigid element is connected to the next by a torsion spring. The ends of the arch are connected by a cable. — *Calculation model*

The calculation model we'll use is a number $n$ of rigid elements (the connectors) connected by torsion springs (the torsion bars). The first and last elements are connected by a cable. Let's isolate a single rigid element and the examine the forces acting on it.

One rigid element isolated. The rigid element is at the angle alpha and is L long. The left end is affected by the counter-clockwise moment M_i and positive horizontal force P. The right end is affected by the clockwise moment M_i+1 and negative horizontal force P. — *Isolated connector element with forces*

The horizontal force $P$ is equal to the tension of the cable and is transmitted through all the elements. The torsion springs apply moments to the ends of the elements. As each element is in equilibrium the sum of the moments acting on them is zero. The first and last element has one free end each as there is no $M_0$ or $M_n$ .

PL_i\sin(\alpha_i) + M_{i-1} - M_i = 0

The angular deflection from a straight line between two rigid elements is labelled phi_i. The angles of the elements are labelled alpha_i and alpha_i+1. — *Torsion spring deflection angle*

The moment at a torsion spring is proportional to the angular deflection $\varphi$ and the spring constant $k$ .

\begin{gather*} M = k\varphi\\ \varphi_i = \alpha_i - \alpha_{i+1} \end{gather*}

Here is an example of our system of equations if $n = 4$ .

\left\{ \begin{aligned} k_1 \varphi_1 &= PL_1\sin(\alpha_1)\\ k_2 \varphi_2 - k_1 \varphi_1 &= PL_2\sin(\alpha_2)\\ k_3 \varphi_3 - k_2 \varphi_2 &= PL_3\sin(\alpha_3)\\ -k_3 \varphi_3 &= PL_4\sin(\alpha_4) \end{aligned} \right.

The stiffnesses $k$ are proportional to the force $P$ . As we are only interested in the relative sizes of $k$ we can set $P = 1$ and solve for the stiffnesses.

Finding the widths

Now we just need to calculate the widths of each connector element. We first want to scale $k$ . There are a couple of factors to consider. First we do not want too thin and fragile bars as they may break easily. Too wide bars on the other hand might break from the increased shear stress. Wide bars also require longer connectors, making the structure larger. The difference between the largest and smallest $k$ can be made smaller by making the curve smoother or using more segments.

A rectangular bar of length l and section lengths a and b is under torsion by a moment M. The total twist is labelled phi. — *Bar under torsion*

\frac{M}{\varphi} = \frac{GJ}{l}

where $M$ is the applied moment, $\varphi$ is the angular deflection, $G$ is the shear modulus, $J$ the torsion constant and $l$ the length of the bar. We want find widths for the torsion bars so that the torsional stiffnesses $\frac{GJ}{l}$ equals the scaled spring stiffnesses $ck$ where $c$ is chosen so that the torsion bars are of acceptable width.

\frac{GJ_i}{l} = ck_i

For a rectangular section the torsion constant $J$ can be approximated with

J = \frac{ab^3}{3}F_1(\frac{a}{b})

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline a/b & 1 & 1.25 & 1.5 & 2 & 3 & 4 & 5 & 10 & \infty \\ \hline F_1 & 0.42 & 0.52 & 0.59 & 0.69 & 0.79 & 0.84 & 0.87 & 0.94 & 1 \\ \hline \end{array}

where $a$ and $b$ are side lengths and $a > b$ and $F_1(\frac{a}{b})$ is a value from the table. For the torsion bars the side lengths are the bar width $w$ and the material thickness.

Depending on whether we want to prescribe the minimum or maximum element width we can calculate the minimum or maximum $J$ which along with the maximum or minimum $k$ lets us solve for $c$ . Finally we need to find the remaining widths which is most easily done by a root finding algorithm. The secant method worked well for the demo as it is a quite well behaved function.

Now we have everything we need and it's time to generate the cutting lines and start fabrication.

Two curved plywood models arranged to form a heart. — *Valentine's greetings!*