Rope on a bollard, the capstan equation

Illustration of an elastic rope wrapped around a fixed item such as a bollard. The contact area is divided into stick and slip zones, depending on the loads exerted on both ends and on the loading history.

Consider a rope where equal forces (e.g., ${\displaystyle F_{\text{hold}}=400\,\mathrm {N} }$) are exerted on both sides. By this the rope is stretched a bit and an internal tension ${\displaystyle T}$ is induced (${\displaystyle T=400\,\mathrm {N} }$ on every position along the rope). The rope is wrapped around a fixed item such as a bollard; it is bent and makes contact to the item's surface over a contact angle (e.g., ${\displaystyle 180^{\circ }}$). Normal pressure comes into being between the rope and bollard, but no friction occurs yet. Next the force on one side of the bollard is increased to a higher value (e.g., ${\displaystyle F_{\text{load}}=600\,\mathrm {N} }$). This does cause frictional shear stresses in the contact area. In the final situation the bollard exercises a friction force on the rope such that a static situation occurs.

The tension distribution in the rope in this final situation is described by the capstan equation, with solution:

${\displaystyle {\begin{aligned}T(\phi )&=T_{\text{hold}},&\phi &\in \left[\phi _{\text{hold}},\phi _{\text{intf}}\right]\\T(\phi )&=T_{\text{load}}e^{-\mu \phi },&\phi &\in \left[\phi _{\text{intf}},\phi _{\text{load}}\right]\\\phi _{\text{intf}}&={\frac {1}{\mu }}\log \left({\frac {T_{\text{load}}}{T_{\text{hold}}}}\right)&\end{aligned}}}$

The tension increases from ${\displaystyle T_{\text{hold}}}$ on the slack side (${\displaystyle \phi =\phi _{\text{hold}}}$) to ${\displaystyle T_{\text{load}}}$ on the high side ${\displaystyle \phi =\phi _{\text{load}}}$. When viewed from the high side, the tension drops exponentially, until it reaches the lower load at ${\displaystyle \phi =\phi _{\text{intf}}}$. From there on it is constant at this value. The transition point ${\displaystyle \phi _{\text{intf}}}$ is determined by the ratio of the two loads and the friction coefficient. Here the tensions ${\displaystyle T}$ are in Newtons and the angles ${\displaystyle \phi }$ in radians.

The tension ${\displaystyle T}$ in the rope in the final situation is increased with respect to the initial state. Therefore, the rope is elongated a bit. This means that not all surface particles of the rope can have held their initial position on the bollard surface. During the loading process, the rope slipped a little bit along the bollard surface in the slip area ${\displaystyle \phi \in [\phi _{\text{intf}},\phi _{\text{load}}]}$. This slip is precisely large enough to get to the elongation that occurs in the final state. Note that there is no slipping going on in the final state; the term slip area refers to the slippage that occurred during the loading process. Note further that the location of the slip area depends on the initial state and the loading process. If the initial tension is ${\displaystyle 600\,\mathrm {N} }$ and the tension is reduced to ${\displaystyle 400\,\mathrm {N} }$ at the slack side, then the slip area occurs at the slack side of the contact area. For initial tensions between ${\displaystyle 400}$ and ${\displaystyle 600\,\mathrm {N} }$, there can be slip areas on both sides with a stick area in between.

Generalization for a rope lying on an arbitrary orthotropic surface

If a rope is laying in equilibrium under tangential forces on a rough orthotropic surface then three following conditions (all of them) are satisfied:

This generalization has been obtained by Konyukhov A.,^[12][13]

Sphere on a plane, the (3D) Cattaneo problem

Consider a sphere that is pressed onto a plane (half space) and then shifted over the plane's surface. If the sphere and plane are idealised as rigid bodies, then contact would occur in just a single point, and the sphere would not move until the tangential force that is applied reaches the maximum friction force. Then it starts sliding over the surface until the applied force is reduced again.

In reality, with elastic effects taken into consideration, the situation is much different. If an elastic sphere is pressed onto an elastic plane of the same material then both bodies deform, a circular contact area comes into being, and a (Hertzian) normal pressure distribution arises. The center of the sphere is moved down by a distance ${\displaystyle \delta _{n}}$ called the approach, which is equivalent to the maximum penetration of the undeformed surfaces. For a sphere of radius ${\displaystyle R}$ and elastic constants ${\displaystyle E,\nu }$ this Hertzian solution reads:

${\displaystyle {\begin{aligned}p_{n}(x,y)&=p_{0}{\sqrt {1-{\frac {r^{2}}{a^{2}}}}}&r&={\sqrt {x^{2}+y^{2}}}\leq a&a&={\sqrt {R\delta _{n}}}\\p_{0}&={\frac {2}{\pi }}E^{*}{\sqrt {\frac {\delta _{n}}{R}}}&F_{n}&={\frac {4}{3}}E^{*}{\sqrt {R}}\delta _{n}^{\frac {3}{2}}&E^{*}&={\frac {E}{2\left(1-\nu ^{2}\right)}}\end{aligned}}}$

Now consider that a tangential force ${\displaystyle F_{x}}$ is applied that is lower than the Coulomb friction bound ${\displaystyle \mu F_{n}}$. The center of the sphere will then be moved sideways by a small distance ${\displaystyle \delta _{x}}$ that is called the shift. A static equilibrium is obtained in which elastic deformations occur as well as frictional shear stresses in the contact interface. In this case, if the tangential force is reduced then the elastic deformations and shear stresses reduce as well. The sphere largely shifts back to its original position, except for frictional losses that arise due to local slip in the contact patch.

This contact problem was solved approximately by Cattaneo using an analytical approach. The stress distribution in the equilibrium state consists of two parts:

${\displaystyle {\begin{aligned}p_{x}(x,y)&=\mu p_{0}\left({\sqrt {1-{\frac {r^{2}}{a^{2}}}}}-{\frac {c}{a}}{\sqrt {1-{\frac {r^{2}}{c^{2}}}}}\right)&0\leq {}&r\leq c\\p_{x}(x,y)&=\mu p_{n}(x,y)&c\leq {}&r\leq a\\p_{x}(x,y)&=0&a\leq {}&r\end{aligned}}}$

In the central, sticking region ${\displaystyle 0\leq r\leq c}$, the surface particles of the plane displace over ${\displaystyle u_{x}=\delta _{x}/2}$ to the right whereas the surface particles of the sphere displace over ${\displaystyle u_{x}=-\delta _{x}/2}$ to the left. Even though the sphere as a whole moves over ${\displaystyle \delta _{x}}$ relative to the plane, these surface particles did not move relative to each other. In the outer annulus ${\displaystyle c\leq r\leq a}$, the surface particles did move relative to each other. Their local shift is obtained as

${\displaystyle s_{x}(x,y)=\delta _{x}+u_{x}^{\text{sphere}}(x,y)-u_{x}^{\text{plane}}(x,y)}$

This shift ${\displaystyle s_{x}(x,y)}$ is precisely as large such that a static equilibrium is obtained with shear stresses at the traction bound in this so-called slip area.

So, during the tangential loading of the sphere, partial sliding occurs. The contact area is thus divided into a slip area where the surfaces move relative to each other and a stick area where they do not. In the equilibrium state no more sliding is going on.

Cylinder rolling on a plane, the (2D) Carter-Fromm solution

Consider a cylinder that is rolling over a plane (half-space) under steady conditions, with a time-independent longitudinal creepage ${\displaystyle \xi }$. (Relatively) far away from the ends of the cylinders a situation of plane strain occurs and the problem is 2-dimensional.

If the cylinder and plane consist of the same materials then the normal contact problem is unaffected by the shear stress. The contact area is a strip ${\displaystyle x\in [-a,a]}$, and the pressure is described by the (2D) Hertz solution.

${\displaystyle {\begin{aligned}p_{n}(x)&={\frac {p_{0}}{a}}{\sqrt {a^{2}-x^{2}}}&|x|&\leq a&a^{2}&={\frac {4F_{n}R}{\pi E^{*}}}\\p_{0}&={\frac {2F_{n}}{\pi a}}&&&E^{*}&={\frac {E}{2\left(1-\nu ^{2}\right)}}&\end{aligned}}}$

The distribution of the shear stress is described by the Carter-Fromm solution. It consists of an adhesion area at the leading edge of the contact area and a slip area at the trailing edge. The length of the adhesion area is denoted ${\displaystyle 2a'}$. Further the adhesion coordinate is introduced by ${\displaystyle x'=x+a-a'}$. In case of a positive force ${\displaystyle F_{x}>0}$ (negative creepage ${\displaystyle \xi <0}$) it is:

${\displaystyle {\begin{aligned}p_{x}(x)&=0&|&x|\geq a\\p_{x}(x)&={\frac {\mu p_{0}}{a}}\left({\sqrt {a^{2}-x^{2}}}-{\sqrt {a'^{2}-x'^{2}}}\right)&a-2a'\leq {}&x\leq a\\p_{x}(x)&=\mu p_{n}(x)&&x\leq a-2a'\end{aligned}}}$

The size of the adhesion area depends on the creepage, the wheel radius and the friction coefficient.

${\displaystyle {\begin{aligned}a'&=a{\sqrt {1-{\frac {|F_{x}|}{\mu F_{n}}}}},&{\mbox{for }}|F_{x}|\leq \mu F_{n}\\\xi &=-\operatorname {sign} (F_{x})\,{\frac {\mu (a-a')}{R}},&{\mbox{i.e. }}|\xi |\leq {\frac {\mu a}{R}}\\F_{x}&=-\operatorname {sign} (\xi )\,\mu F_{n}\left(1-\left(1+{\frac {R|\xi |}{\mu a}}\right)^{2}\right)\end{aligned}}}$

For larger creepages ${\displaystyle a'=0}$ such that full sliding occurs.