The flexible shaft is a common transmission device used to either reduce tolerance requirements or develop otherwise impossible machines (e.g. mobile rotary hand tools driven by a stationary mounted motor). See, for example, this video. Is anyone aware of a standard "mechanics of materials" level model of one?

For example, imagine an ordinary shaft extending from the origin along the +X axis. If the driver at the origin applies a torque and rotation such that the right hand thumb points into the shaft (vector points along +X), then the loaded end of the shaft experiences a reactionary torque also pointing into the shaft, but at this end it would be in the -X direction. This is the standard explanation for how transmission of work along a shaft balances torques, so that the shaft rotates at a constant speed, and all work transmitted into the shaft by the source is extracted from the shaft by the load.

If we now imagine that the shaft has a section where its longitudinal axis curves with radius R a full 180 degrees, the far end of the shaft (the loaded end) is now facing the -X direction. However, the torque exerted on the shaft by the load still points "into" the shaft body, and this is now also in the +X direction. Both the source and the load are exerting the same +X torque on the shaft.

Clearly, this is not in equilibrium. In this case, it looks pretty intuitive that there would have to be vertical reactions, one at the source and one at the load that are equal and in opposing directions. The moment of these reactions would be the opposite of the total torque on the shaft, which is twice the torque that shaft is carrying, since both the source and the load are exerting the same torque (i.e. in the same direction). So, in the video linked above, if the hand tool is held pointed upwards (180 degrees from the downward driver motor), the craftsman's hand and the motor mount would both experience horizontal forces so that the net torque exerted on the system is zero.

I feel this portion should be uncontroversial. So then the question is, how is this carried through the length of the shaft? More specifically, when we are halfway through the radius (so at 90 degrees), how is the stress carried? The shaft should carry a shear flow to balance the reaction force described previously. The reaction force and the shear force half way through the turn will only make half the torque that the two reactions make (since the distance of this couple is now only half what it is for the two reaction forces). So there must be a bending moment, right? However, the whole idea of the "flexible shaft" is that it has virtually no stiffness in bending, this is what the word "flexible" really refers to in its name. So how does the flexible shaft carry a bending moment?

If my description is not clear, please let me know. If anyone knows of an analysis or can just provide an explanation, that would be great.

It is worth pointing out that my question is geometrically similar to a mechanics of materials example sometimes seen, where two opposite torques are exerted on two opposite sides of a torus (i.e. a "tire" or "doughnut"). The mid points at +/- 90 degrees of the torus carry bending moment. One can picture this by exerting these two torques on an O-Ring and seeing how it bends at the mid points between where your hands are twisting the O-Ring.

Edit: Just to further clarify, I would like to be able to model the shaft dynamics. So basically, if you move the load around, how does the shaft move? I am thinking about it, and for solid / continuous (i.e. non flexible) shafts, there is a relation $J^{2} = I_{x}^{2}+I_{y}^2$, where I is the 2nd moment of inertia and J the polar. A flexible shaft is basically assembled in a special way so that $I$ can be small while $J$ is large, the above relation does not hold. But in order to carry the bending moment created by sending torque through a turn, as I pointed out, there needs to be sufficient $I$. I guess that is the puzzle I am trying to get my head around.


2 Answers 2


Draw an FBD of a small portion of the flex drive and its casing. Basically the interaction at any point s along the curve re-aligns the torque vector to become parallel to the cable inner. This is analogous to developing the shape of a catenary.

  • $\begingroup$ When the axis of the shaft is no longer colinear with the axis at the mount, how can the two torque vectors cancel? The catenary uses three degrees of freedom (tension, distance, and angle) to balance three quantities (x, y force and z moment), with y force driven by cable weight. A cable will not make a catenary without weight along its length. If you fix the flex shaft input and output location and apply balancing torques at each end, what are your degrees of freedom? Especially if you are not allowed to resist bending moments. $\endgroup$ Dec 9, 2022 at 22:48
  • $\begingroup$ FWIW, I think that tensile forces need to develop in such a way that the shaft deflection causes these forces to balance the moments and reaction forces. $\endgroup$ Dec 9, 2022 at 22:57
  • $\begingroup$ Perhaps the catenary confused you. There's no such thing as colinear torques, they are either parallel or not. Statics 101. $\endgroup$ Dec 10, 2022 at 5:30
  • $\begingroup$ Yes, what I am saying is that when the axis of the shaft is not a straight line, then in its interior there must be places where the torque is not parallel to the torque applied at the boundary. They obviously cannot cancel. It looks to me like you need to use reaction forces (axial or transverse) and internal normal and shear stress and the curve of the shaft axis can be three dimensional (assuming no bending moments due to "flexibility"). It turns out that my semi circle example actually works, but the general case is harder. I know what a catenary is. $\endgroup$ Dec 10, 2022 at 10:21

So I just solved my example of the shaft bent into a half circle and it turns out that this is a bit of a bad example because it carries constant torque and shear with no issues. I will try to get a more general solution (for conditions where the circle is not possible) and post it here.


Here are the results of my analysis. My assumption, as I stated in the comment to the other answer, is that the bending moment in the shaft is negligible. This is basically what makes it a "flexible shaft," it is "flexible" because you can snake it around with little to no resistive stiffness. This is similar, I guess, the the "cable assumption" seen in the catenary problem. In that problem, the assumption is that there is only tension (no shear or bending and torsion is not relevant in that problem). I am assuming three axes of force (tensile and two dimensions of shear flow) and a net torque in the shaft.

If you set up an input and an output to your flexible shaft and apply torques, the resultant torque is dependent upon the directions of the input and output. In an ordinary (straight) shaft, they obviously cancel, but in the case of the flexible shaft the input and output torque may not be anti-parallel (in the half circle, they are actually parallel). You therefore must assume reaction forces at the input and output that make a couple that cancels the net torque on the shaft. Additionally, the component of the reaction forces along the line between the input and output makes no net moment, and so is statically indeterminate. This component is a degree of freedom to the problem.

If you cut a section at an arbitrary location in the shaft and look at the portion from the input to the section cut, balancing forces necessitates that the net force at the section cut is equal and opposite to the reaction force at the input. Balancing moments requires more discussion.

Initially, I assumed that the torque in the shaft is a constant. If you make this assumption, the equation of static equilibrium is

$$\tau\frac{d\vec{r}}{ds}-\vec{r}\times\vec{F}_{R} = \tau\frac{d\vec{r}}{ds}\left(0\right)$$

where $\vec{r}(s)$ is parameterized to arc length, $\tau$ is the constant torque in the shaft, $F_{R}$ is the reaction force (which as stated contains a free component) and the derivative on the right side of the equation is evaluated at $s = 0$ as the boundary condition. This equation is linear and can be solved, the solutions are basically helices, and as can be verified, in addition to my circular arc example in my question, all regular helical (flex) shafts can carry a constant torque along their length.

The above equation is three coupled first order differential equations. Its general solution has 3 arbitrary constants. Along with the free force component, there are only 4 degrees of freedom with this method. Thus, the solution cannot satisfy arbitrary input location (3 DoF), output location (3 DoF), and shaft length (1 DoF).

If you perform a differential analysis on a segment of the shaft with these assumptions, the equation is:

$$\tau\frac{d^{2}\vec{r}}{ds^{2}}-\frac{d\vec{r}}{ds}\times\vec{F}_{R} = \vec{0}$$

While this may appear to have raised the order of the equation, and afforded us another 3 arbitrary constants, it has also lost the boundary torque information. It is evident that this equation can be integrated in $s$ directly to obtain the first equation, where the arbitrary vector constant of integration would be $\tau\frac{d\vec{r}}{ds}\left(0\right)$.

This method, then, still does not work in general. For this reason, I have determined that it is likely that the torque is not a constant in the shaft. I do not yet understand how this works exactly, but it seems to me that it is related to changes in curvature along the shaft length. Note that the circle and the helix both have constant curvature (the circle is a special case of a helix). If you allow the torque to vary, the differential analysis yields

$$\frac{d}{ds}\left(\tau\frac{d\vec{r}}{ds}\right)-\frac{d\vec{r}}{ds}\times\vec{F}_{R} = \vec{0}$$

where I am thinking that shaft torque $\tau$ is some function of the shaft axis curvature. Curvature is basically second order, so if this is the case, the derivative of torque with respect to arc length makes this a third order differential equation. 9 arbitrary constants along with the indeterminate component of the reaction force allow for 10 DoF, which should accommodate the input location (3), output location (3), input torque (3), and shaft length (1).

I am currently not sure of the relation between shaft curve geometry and the shaft torque. If anyone knows about this, or knows the solution to this problem (and I am wrong about what I have so far), please let me know. I will post an answer if I find it.

Edit 2:

It should be obvious that the third equation is still directly integrable in $s$, so that the equation can be reduced to

$$\tau\frac{d\vec{r}}{ds}-\vec{r}\times\vec{F}_{R} = \tau_{0}\frac{d\vec{r}}{ds}\left(0\right)$$

as before. However, this time one must remember the torque $\tau$ is not a constant factor, but rather, as I said I suspect, a function of the curvature. This is why the boundary condition on the right hand side is updated to include $\tau_{0}$, the torque at the boundary, rather than plain $\tau$. A second order system of 3 functions will have 6 arbitrary constants. In this case, the input torque has been built into the equation, so the boundary conditions are the six end points.

I have been thinking about whether the direction of the shaft at the end points will be guaranteed correct, i.e. parallel to the boundary condition torque. I believe this will fall out of the right hand side torque boundary condition and the component of the reaction force perpendicular to the line between the mounts (a portion of $\vec{F_{R}}$). In effect, the three parameters of torque can be thought of as two parameters of direction and one of magnitude, and the solution curve must start out parallel with the torque, since at the mount the reaction force can make no torque itself. At the endpoint, the solution curve must again be parallel with the ending torque vector, because the reaction at the initial point puts the whole system into equilibrium.

I'll post here again if I figure out the torque / curvature relation.


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