In the previous note we reviewed the derivation of capstan equation that solves the tension developed in a rope (web, belt) undergoing kinetic friction on a capstan (roller, pulley). We set up a problem in which a 1-D belt is tensioned, constrained statically, and in contact for a wrap angle of 90° on a cylinder that is rotating at a constant angular velocity. So the belt is in equilibrium and the belt-cylinder contact interface is in complete slip condition. This setup avoids the complication that would be introduced by slip-stick condition. However, the slip-stick condition universally exists in similar contact problems. In this note we will review the slip-stick condition, its mechanism, and its location at the contact interface in 1-D belt-roller contact problems with various boundary conditions. We will use \(\{\)rope, belt, web, string \(\}\) and \(\{\)capstan, cylinder, roller, pulley \(\}\) interchangeable, respectively, since in 1-D cases elements in each set are equivalent to others in the physical sense. All vector quantities are treated as scalars in computation with their direction visually shown.

In the previous problem, the belt is constrained static at one end and the roller is rotating steadily—an obvious relative motion is undergoing between two bodies and the contact interface is in complete slip. A different set of boundary conditions may prohibit the macroscopic relative motion, and the contact interface will be in partial stick and partial slip, or still in complete slip. The slip when macroscopic relative motion is absent is called **micro-slip** or **creep** since two bodies only slip microscopically at their partial contact interface. Whether micro-slip or creep is possible is determined by boundary conditions. Depending on the state of each part in the belt-roller contact, the problem can be classified into following situations, which only in some the micro-slip is possible:

- when only one part is in motion: either the roller is rotating like in the previous problem, or the belt is transiting on a stationary roller, the contact interface must be in macroscopic complete slip due to the significant relative motion;
- when both parts are static, micro-slip is possible; micro complete slip is a critical state before parts start to move;
- when both parts are in motion, micro-slip and macro complete slip are both possible.

## A Constrained Belt on a Rotating Roller

This is the first situation and is exactly the problem in the last note. In this problem we applied an angular velocity on the cylinder to guarantee a complete slip condition at the contact interface. The magnitude of the reaction torque \(M\) was calculated by the capstan equation where the slip angle equals the wrap angle (\(\theta=\pi/2\)):

$$\frac{T_1}{T_0}=e^{\mu\frac{\pi}{2}},$$

$$M=(T_1-T_0)R=(e^{\mu\frac{\pi}{2}}-1)T_0R.$$

The magnitude of the reaction torque achieved its maximum value in all configurations where there is a quarter circle wrap angle and the down end tension \(T_0\) is maintained, since the slip angle, equal to the wrap angle, is at its maximum. We will use this value as a reference and so assign a new variable \(M_0\) to it,

$$M_0= (e^{\mu\frac{\pi}{2}}-1)T_0R.$$

## A Prestressed and Constrained Belt on a Torqued Roller

Now, we’ll discuss the second situation where both bodies are static macroscopically. We consider a purely theoretical case where we replace the angular velocity \(\omega\) in the last problem with a same direction torque \(M\), and maintain all others. The system is statically determinant still.

Although this set of boundary conditions is really basic in theory, it is not easy to apply in reality, because it actually involves two sequential states of equilibrium—the first in which the belt builds its balance on the roller under only tension; the second after the torque is applied. In reality the friction is always there, as the belt first builds its equilibrium, due to the extension under tension the interface will have to accommodate and slip in some areas. Then, as the torque is applied, the interface has to change its status due to the change of loadings for the second time. The final result of the status of interface can be seen as a superposition of two, which introduces unnecessary complexity for making the point. Thus, I’d like to use a special way of applying loadings to consider only one-time change of the status at interface. We assume the roller is frictionless during the process of applying tension. Since the roller is frictionless, the equilibrium state is that the belt has a constant tension (the applied \(T_0\)) along the whole contact arc. Then we somehow turn on the friction feature of the roller and apply torque \(M\) at last. This way of application ignores the change of status at interface in the pretension loading step and allow us to only consider slip due to the applied torque for a prestressed belt.

Suppose the counterclockwise torque \(M\) we applied is initially small enough, \(M < M_0\), the roller remains static, both the belt and the roller are in equilibrium. Partial belt surface slips on the roller to produce counter moment that balances \(M\) for the roller to maintain equilibrium. Capstan equation holds for the belt on the slip contact arc. Since \(M < M_0\), the slip angle \(\theta<\pi/2\), there must be other partial interface that sticks. So the slip in this case is creep. To have the complete solution, we still need the distribution of slip and stick surfaces, in other words, the location of the slip.

### Location of slip and stick in static cases

For that a 1-D belt does not allow shear deformation (the assumption is detailed in the previous note) if the belt is going to bear any additional longitudinal load (tension or friction) from an unstressed state or an equilibrium state, its material points have to deform further longitudinally. Therefore, the material points that are going to deform longitudinally must slip on the roller until an new longitudinal equilibrium between tension and friction is built up. From the perspective of when and where those physical quantities of belt appear, **friction, change of tension, and slip happen simultaneously and in the same contact area.** Stick happens where the belt tension is constant, which is also where the friction is absent.

In this problem, the prestressed belt on the frictionless roller has a constant tension \(T_0\) everywhere at the beginning. Then, when the contact interface turns frictional and the torque is applied, since \((T_1-T_0)R=M\), tension at the upper end \(T_1 > T_0\) at the lower end, so the material at the right end first starts to elongate, thus, first slips on the roller; the slip interface climbs counterclockwise until the friction generated by the slip induces a counter torque that is equal to \(M\), then, a new equilibrium for the roller is achieved, so is for the belt. The conceptual slip-stick condition is shown in the figure below.

Below figure shows the change of tension and contact pressure along the belt when applying \(M=M_0 / 2\) in this theoretical situation.

We need to notice that the above results only hold for this theoretical setup which begins with a prestressed belt and a frictionless roller. This reminds us that if we want to validate those problems with tests, the way the boundary conditions are applied may affect results. In fact, if we do lab tests for the above case (\(M=M_0 / 2\)) just by first tensioning the belt on a typical (frictional) roller and then torquing the roller, the results will be a superposition of two events of loadings. In the tensioning stage, since the problem is symmetric about the 45° radial line in terms of geometry, boundary conditions, and loadings, the material points of belt at 45° stays where they are during the tensioning process. The slip-stick condition after the tension is shown below.

The tension at the middle point of the belt is lowest due to opposite slip in two contact zones at each side, which is a very different stress state for the torque be applied on. After the torque is applied a new equilibrium state will be achieved and the contact status will alter significantly. In the final equilibrium the locations of slip and stick arc will be very similar to the above theoretical results where slip locates at the higher tension end and stick locates at the lower, but the slip angle will be a bit greater than \(\pi/4\), and the distribution of belt tension on the roller will be quite different: it will have a small area of decreasing close to the lower end instead of being constant in the above case.

## A Moving Belt on a Rotating Roller

Next, we discuss the third situation—we change the boundary conditions to make the belt move on the roller and consider the possible slip-stick condition in moving cases. Two common and basic types of such problems are:

- a belt on a driving roller, where the applied (torque driving case) or reaction (velocity driving case) torque on the roller is in the same direction as the angular velocity of the roller. To achieve this condition, the constraint at the right end in the static problem is replaced with a tensile force \(T_1\), whose magnitude \(T_1 < T_0\) slightly; then a clockwise angular velocity \(\omega\) is applied to drive the belt moving. Since \(T_1 < T_0\), the reaction torque \(M\) is clockwise, which is in the same direction as angular velocity of the roller.
- a belt on a driven roller, where the applied torque on the roller is in the opposite direction of the angular velocity of the roller. The constraint at the right end in the static problem is replaced with a constant velocity \(v_1\), which drives the roller to rotate clockwise. The roller has a resistance torque \(M\), which acts in the counterclockwise direction. In reality, the resistance torque comes from the bearing resistance.

This classification leans to reality since the perfect idle roller (no bearing resistance) is not included. If the belt moves in a steady state on a perfect idle roller, it sticks to the roller at the whole interface.

## Location of slip and stick in moving cases

In moving cases, we need to switch to Eulerian framework rather than staying in Lagrangian framework, since in steady state the region of slip and stick will be fixed in space but the material points in those regions are constantly changing. Furthermore, the deformation in moving cases will be directly related with velocity rather than the displacement in static cases; so we should focus on analysis of velocity. We’ve assumed that the roller is rigid, it will have one linear velocity; while the belt is deformable, its velocity can very along the contact interface. **Those material points of the belt and roller touching together with the same velocity will stay together as long as the velocity and contact condition remain, those touching together with velocity difference will separate apart with time**. **Slip locates where the belt and roller have a velocity difference, stick locates where there is none**.

The mass conservation law gives us a relation between the longitudinal strain and velocity at any spatial location in a moving web in steady state:

$$\frac{\epsilon+1}{v}=C,$$

where \(C\) is a constant.

To determine the location of slip, we need three ingredients:

- the strain relation (which is greater) at two ends of the belt, which combined with the mass conservation law, can determine the velocity relation at two ends (which end is greater) ;
- the type of roller (driving or driven), or in other words, the direction of friction forces, that can determine if the belt should run faster or slower on the roller;
- then the location of velocity difference, or at which end of the belt, the roller should match velocity with can be decided.

The slip starts from end of velocity difference, the stick locates at the other.

### A moving belt on a driving roller

For the case of a driving roller that drives the belt transporting,

- The equilibrium of the whole system tells \( (T_0-T_1)R=M\), so \(T_0 > T_1\), thus, \(\epsilon_0 > \epsilon_1\), based on which the mass conservation equation draws \(v_0 > v_1\).
- Since the roller drives the belt to move on its surface, the friction is clockwise on the belt surface pushing the belt forward; bearing that the friction points opposite to relative motion, the belt moves counterclockwise (backward) relative to the roller surface, meaning the belt is slower than the driving roller. Actually the word ‘driving’ have packaged all those middle reasoning to give us the intuition to draw that the roller is faster.
- So the roller’s peripheral speed has to match the higher speed in the belt, \(\omega R=v_0\), which guarantees that the roller runs faster than the belt in some contact area, thus, drives the belt, and there exists stick zone that synchronizes the movements two parts. Therefore, the speed difference happens at the lower speed \(v_1\) end, which is the exit of the roller.

### A moving belt on a driven roller

For the case of a driven roller, a moving belt drives a roller to rotate, the resistance torque \(M\) is counterclockwise, opposite to the direction of rotation. Repeat above analysis:

- \( (T_1-T_0)R=M \Longrightarrow T_1 > T_0 \Longrightarrow \epsilon_1 > \epsilon_0 \Longrightarrow v_1 >v_0\);
- a driven roller (since the belt drives the roller, friction \(q\) is counterclockwise on the belt surface) \(\Longrightarrow\) the belt moves faster than the driven roller;
- \(\Longrightarrow \omega R=min(v_0, v_1)=v_0\) \(\Longrightarrow\) the slip starts from the location of \(v_1\), which is also the exit of the roller.

### A moving belt on two pulleys

For the case of a belt that is moving on a set of pulleys—a right driving pulley and a left driven, we can directly apply results of above two cases and find the location of slip-stick condition. In fact, this problem is from Johnson’s book of contact mechanics, from which the above two cases are merely disassembled. Details refer to Johnson^{1}.

## Conclusions (tl;dr)

When a 1-D belt contacts a rigid roller,

- in static cases of which an equilibrium is built, the friction, the change of belt tension, and the slip happen simultaneously and in the same contact arc between the belt and roller;
- in steady moving cases, the slip zone always locates close to the exit of the roller and the stick zone locates close to entry;

A visual summary of slip-stick condition with the special theoretical static case and steady moving cases is shown below.

- K.L. Johnson, “Contact Mechanics”, p246 ↩