S. F. Lebedev, mechanical engineer.

Rim Under Load According to an Exponential Law

The calculation of the rim of a pulley and of a flywheel is given by a whole series of authors. Grashof,¹ Föpphl, Grübler² consider the rim: the first two as a curvilinear beam; the third, as a closed ring, but in both cases under the action of a uniformly distributed load. Other authors (e.g., Prof. Timoshenko,³ Prof. Bobarykov) indicate the calculation of the rim for the case of a concentrated load (gear wheels).

The case of a nonuniformly distributed load has apparently not been investigated by anyone. Meanwhile, it is well known to all that the rim of a pulley, and likewise of a flywheel, almost always works under the conditions of the latter case, and moreover a load according to an exponential law occurs exceptionally often.

I therefore consider it of some interest to present here an investigation of the behavior of a rim for this load; all the more so since the formulas obtained make it possible to apply them without special complications to loads of every kind, i.e. they are quite general.

1. Acting forces and moments.

In Fig. 4 the rim is shown under a load according to an exponential law. The notation is clear from the figure. The reactions \(B\) and \(S\), as well as the reactive moment \(M_0\), will always be referred to the left-hand spoke bounding the given sector. The part of the rim “LOM” is considered as a curvilinear beam with a circular axis, the central angle of which is equal to the angle between the spokes.

A rim can always be placed with respect to the flexible body so that its leading end coincides with the direction of a known tangent and is thus perpendicular to some spoke; this spoke (the direction of the spoke) we take as the line of zero angles, measuring the latter clockwise.

The sector bounded by the zero spoke and the following spoke is also called the zero sector.

Then, with the notation of the drawing, for the \(n\)-th sector the expression for the bending moment in the section \(X—X\) will have the form:

\[ \text{(from the left) } 1.\quad M_x = M_0 + By_a - Sz_a - M_{px}, \quad \text{where} \]

\(M_{px}\)—the moment from the load by the belt (or other flexible body)—is composed in the following way. If \(dN\) is the pressure of the belt at the given place of the rim,

\(T\)—the tension of the belts » » »

then

\[ 2.\quad dN = Tdx \quad \text{and, further,} \]

\[ 3.\quad dM_{px} = yTdx \]

The total moment from the load is equal to

\[ 4.\quad M_{px} = \int_{n\beta}^{\alpha} yTdx. \quad \text{But} \]

\[ 5.\quad T=T_{\max} e^{f(\varphi_0-x)}{}^{1)} \]

[\(f\)—coefficient of friction] and

\[ 6.\quad y=r\sin(\alpha-x),\quad \text{therefore} \]

\[ 7.\quad M_{px}=\int_{n\beta}^{\alpha} T_{\max} r e^{-f(\varphi_0-x)}\sin(\alpha-x)\,dx \]
or, after integration,

\[ 8.\quad M_{px}= \frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} \left\{ e^{f\alpha} - \left[ \cos(\alpha-n\beta)+f\sin(\alpha-n\beta) \right]e^{fn\beta} \right\} \]

for the section \(X-X\) at an angle
\(\alpha<\varphi_0\) and within the limits from “\(n\beta\)” to “\(\varphi_0\)”.

For the section \(X-X\) with angle
\(\alpha>\varphi_0\)

\[ 9.\quad M'_{px}=\int_{n\beta}^{\varphi_0} T_{\max} r e^{-f(\varphi_0-x)}\sin(\alpha-x)\,dx= \]

\[ M'_{px}= \frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} \left\{ e^{f\varphi_0} \left[ \cos(\alpha-\varphi_0)+f\sin(\alpha-\varphi_0) \right] - e^{fn\beta} \left[ \cos(\alpha-n\beta)+f\sin(\alpha-n\beta) \right] \right\} \]

within the limits from “\(\varphi_0\)” to “\((n+1)\beta\)”.

[In the figure \(M_{px}\) is denoted by \(M'_{\alpha}\) and \(M'_{px}=M''_{\alpha}\).]

From Fig. 4 it is clear that

\[ 10.\quad Y_{\alpha}=r\left[1-\cos(\alpha-n\beta)\right] \]
and

\[ 11.\quad Z_{\alpha}=r\sin(\alpha-n\beta). \]

Hence the final expression for the bending moment is

\[ 12.\quad M_x=M_0+Br\left[1-\cos(\alpha-n\beta)\right] -Sr\sin(\alpha-n\beta)-M'_{\alpha} \quad \text{for } \alpha<\varphi_0, \]

\[ 13.\quad M_x=M_0+Br\left[1-\cos(\alpha-n\beta)\right] -S'r\sin(\alpha-n\beta)-M''_{\alpha} \quad \text{for } \alpha>\varphi_0. \]

Both expressions may be replaced by the single following one:

\[ 14.\quad M_x=M_0+Br\left[1-\cos(\alpha-n\beta)\right] -Sr\sin(\alpha-n\beta)-\delta M'_{\alpha}-E M''_{\alpha}, \]

where

\[ \begin{aligned} E&=0,\quad \delta=1 &&\text{for an arc loaded completely}\\ &&&\text{(in Fig. 4, sector “KOL”),}\\ E&=1,\quad \delta=0 &&\text{for an arc loaded partially}\\ &&&\text{(in Fig. 4, sector “LOM”).} \end{aligned} \]

2. Equations of equilibrium.

As the basis for the subsequent reasoning we adopt the usual assumption, which consists in this: 1) any sector of the rim (more precisely—its arc) is assumed to be in equilibrium under the action of the indicated forces, and 2) the points of application of the forces \(B\) and \(S\), and of the moment \(M_0\), do not move and do not rotate.

As a consequence of this assumption we immediately arrive at three equations with three unknowns, which makes it possible to determine \(M_0\), \(B\), and \(S\).

If \(A\) is the potential energy of the arc of sector “\(n\)” within the limits from “\(n\beta\)” to “\((n+1)\beta\),” then, by Castigliano’s theorem, we have

\[ 15.\quad \frac{dA}{dM_0}=0. \]

\(^{1)}\) See, for example: Hütte, I part, 1917.

\[ 16.\quad \frac{dA}{dB}=0 \quad \text{and} \]

\[ 17.\quad \frac{dA}{dS}=0,\quad \text{where } “A”, \text{ as is known, is equal to} \]

\[ 18.\quad A=\frac{1}{2EI}\int_{n\beta}^{(n+1)\beta} M_x^2\,ds \]

Here \(E\) and \(I\) are respectively the modulus of extension and the moment of inertia of the cross-section of the rim; \(ds=r\,d\alpha\) is the differential of the arc of the sector.

After substituting \(M_x\) (according to 14), differentiating and integrating, we obtain: [the coefficient “\(\delta\)” is taken equal to unity for the limits from “\(n\beta\)” to “\(\varphi_0\),” and \(=0\) from “\(\varphi_0\)” to “\((n+1)\beta\)”]

\[ \begin{aligned} 19.\quad \frac{dA}{dM_0}={}& M_0[\alpha-n\beta]+Br[\alpha-n\beta-\sin(\alpha-n\beta)]-Sr\,[1-\cos(\alpha-n\beta)]\\ &-\frac{T_{\max}re^{-f\varphi_0}}{1+f^2} \left\{ \frac{e^{f\alpha}-e^{fn\beta}}{f} -e^{fn\beta}\bigl[-f\cos(\alpha-n\beta)+f+\sin(\alpha-n\beta)\bigr] \right\}\\ &+E\Bigl\{ M_0[(n+1)\beta-\varphi_0] +Br[(n+1)\beta-\varphi_0-\sin\beta+\sin(\varphi_0-n\beta)]\\ &\qquad -Sr[-\cos\beta+\cos(\varphi_0-n\beta)] -\frac{T_{\max}re^{-f\varphi_0}}{1+f^2}\\ &\qquad\cdot \left\{ e^{f\varphi_0}\left[\sin((n+1)\beta-\varphi_0)-f\cos((n+1)\beta-\varphi_0)+f\right]\right.\\ &\qquad\qquad\left. -e^{fn\beta}\left[\sin\beta-\sin(\varphi_0-n\beta)-f\cos+f\cos(\varphi_0-n\beta)\right] \right\} \Bigr\}=0 \end{aligned} \]

\[ \begin{aligned} 20.\quad \frac{dA}{dB}={}& M_0\{\alpha-n\beta-\sin(\alpha-n\beta)\} +Br\left[(\alpha-n\beta)\frac{3}{2}-2\sin(\alpha-n\beta)+\frac{\sin 2(\alpha-n\beta)}{4}\right]\\ &-Sr\left[1-\cos(\alpha-n\beta)-\frac{\sin^2(\alpha-n\beta)}{2}\right]\\ &-\frac{T_{\max}re^{-f\varphi_0}}{1+f^2} \left\{ \frac{e^{f\alpha}-e^{fn\beta}}{f} -\left[-f\cos(\alpha-n\beta)+f-f\frac{\sin^2(\alpha-n\beta)}{2}\right]e^{fn\beta}\right.\\ &\qquad\left. -\frac{e^{f\alpha}}{1+f^2}\,[\sin(\alpha-n\beta)+f\cos(\alpha-n\beta)] +\frac{fe^{fn\beta}}{1+f^2}\\ &\qquad\left. -e^{fn\beta}\left[\sin(\alpha-n\beta)-\frac{\alpha-n\beta}{2}-\frac{\sin 2(\alpha-n\beta)}{4}\right] \right\}\\ &+E\Biggl\{ M_0\bigl[(n+1)\beta-\varphi_0-\sin\beta+\sin(\varphi_0-n\beta)\bigr]\\ &\qquad +Br\left[(n+1)\beta-\varphi_0-2\sin\beta+2\sin(\varphi_0-n\beta) +\frac{(n+1)\beta-\varphi_0}{2} +\frac{\sin 2\beta}{4}\right.\\ &\qquad\qquad\left. -\frac{\sin 2(\varphi_0-n\beta)}{4}\right] -Sr\left[-\cos\beta-\frac{\sin^2\beta}{2} +\frac{\sin^2(\varphi_0-n\beta)}{2} +\cos(\varphi_0-n\beta)\right]\\ &\qquad -\frac{T_{\max}re^{-f\varphi_0}}{1+f^2} \left\{ e^{f\varphi_0}\left[\sin((n+1)\beta-\varphi_0)-\right. \end{aligned} \]

\[ \begin{aligned} &-f\cos\bigl((n+1)\beta-\varphi_0\bigr) +f\left(\frac{1+f}{2}\right)\sin\bigl((n+2)\beta-\varphi_0\bigr)\\ &\quad+\left(\frac{1+f}{2}\right)\sin(\varphi_0-n\beta) +\left(\frac{1+f}{2}\right)\cos(n\beta-\varphi_0)\bigl((n+1)\beta-\varphi_0\bigr) \Bigg]\\ &\quad-e^{fn\beta}\Bigl[\sin\beta-f\cos\beta-\sin(\varphi_0-n\beta) +f\cos(\varphi_0-n\beta)\Bigr]\\ &\quad+\left[ \frac{(n+1)\beta-\varphi_0}{2} +\frac{\sin 2\beta}{4} -\frac{\sin 2(\varphi_0-n\beta)}{4} \right]\\ &\quad-f\left[ \frac{\sin^2\beta}{2} -\frac{\sin^2(\varphi_0-n\beta)}{2} \right]\Bigg\}=0 . \end{aligned} \]

\[ \begin{aligned} 21.\quad \frac{dA}{dS} &=M_0\Bigl[1-\cos(\alpha-n\beta)\Bigr] +Br\left[1-\cos(\alpha-n\beta)-\frac{\sin^2(\alpha-n\beta)}{2}\right]\\ &\quad-Sr\left[\frac{\alpha-n\beta}{2} -\frac{\sin 2(\alpha-n\beta)}{4}\right]\\ &\quad-\frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} \Bigg\{ \frac{e^{f\alpha}}{1+f^2}\Bigl[f\sin(\alpha-n\beta)-\cos(\alpha-n\beta)\Bigr]\\ &\qquad+\frac{e^{fn\beta}}{1+f^2} -e^{fn\beta}\left[ f\left(\frac{\alpha-n\beta}{2} -\frac{\sin 2(\alpha-n\beta)}{4}\right) +\frac{\sin^2(\alpha-n\beta)}{2} \right] \Bigg\}\\ &\quad+E\Bigg\{ M_0\Bigl[-\cos\beta+\cos(\varphi_0-n\beta)\Bigr]\\ &\qquad+Br\left[ -\cos\beta+\cos(\varphi_0-n\beta) -\frac{\sin^2\beta}{2} +\frac{\sin^2(\varphi_0-n\beta)}{2} \right]\\ &\qquad-Sr\left[ \frac{(n-1)\beta-\varphi_0}{2} -\frac{\sin 2\beta}{4} +\frac{\sin 2(\varphi_0-n\beta)}{4} \right]\\ &\qquad+\frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} \Bigg\{ e^{f\varphi_0}\Bigg[ -\frac{\cos\bigl((n+2)\beta-\varphi_0\bigr)}{2} +\frac{\cos(\varphi_0-n\beta)}{2}\\ &\qquad\qquad -\frac{\sin(n\beta-\varphi_0)}{2}\bigl((n+1)\beta-\varphi_0\bigr) +\frac{\sin\bigl((n+2)\beta-\varphi_0\bigr)}{2} +\frac{\sin(\varphi_0-n\beta)}{2}\\ &\qquad\qquad +f\cos(n\beta-\varphi_0)\left(\frac{(n+1)\beta-\varphi_0}{2}\right) \Bigg]\\ &\qquad\qquad -e^{fn\beta}\left[ \frac{\sin^2\beta}{2} -\frac{\sin^2(\varphi_0-n\beta)}{2} +f\left( \frac{(n+1)\beta-\varphi_0}{2} -\frac{\sin 2\beta}{4} +\frac{\sin 2(\varphi_0-n\beta)}{4} \right) \right] \Bigg\} \Bigg\}=0 . \end{aligned} \]

These equations are of an entirely general character and pertain to an arc of any \(n\)-th sector, where \(M_0\), \(B\), and \(S\) denote the moments and forces acting at the left-hand end of the sector. The solution of the equations presents no great difficulties and may be expressed in the following form:

\[ 22.\quad M_0=\frac{T_{\max}e^{-f\varphi_0}}{1+f^2}\,\frac{\Delta_{M0}}{\Delta}. \]

\[ 23.\quad B=\frac{T_{\max}e^{-f\varphi_0}}{1+f^2}\,\frac{\Delta_B}{\Delta}. \]

\[ 24.\quad S=\frac{T_{\max}e^{-f\varphi_0}}{1+f^2}\,\frac{\Delta_S}{\Delta}, \quad \text{where} \]

\(\Delta\) is the determinant customarily formed from the coefficients of \(M_0\), \(B\), and \(S\) in these equations, while \(\Delta_{M0}\), \(\Delta_B\), and \(\Delta_S\) are the same determinant with the corresponding vertical column replaced.

Having the quantities \(E\), \(n\), \(\alpha\), and \(\beta\) at one’s disposal in the sense indicated above, according to the data of each individual problem, we obtain the values of \(M_0\), \(B\), and \(S\), as well as the maximum values of the bending moment. Thus all the conditions necessary for checking the section of the rim will be found.

3. Application of the Equation.

Consider case I: the rim has three arms (Fig. 1).

For the sector \(KOL\):

\[ E=0, \]

\[ n=0, \]

\[ \alpha=\beta=\frac{2}{3}\pi . \]

Equations 21, 20, and 19 become the following:

\[ \begin{aligned} 25.\quad &M_0\beta -Br\left(\beta-\sin\beta\right) -Sr\left(1-\cos\beta\right) -\frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} \\ &\quad{}\times \left\{ \frac{e^{f\beta}-1}{f} -\left(-f\cos\beta+f+\sin\beta\right) \right\} =0 . \end{aligned} \]

\[ \begin{aligned} 26.\quad &M_0\left(\beta-\sin\beta\right) +Br\left(\frac{3\beta}{2}-2\sin\beta+\frac{\sin2\beta}{4}\right) -Sr\left(1-\cos\beta-\frac{\sin^2\beta}{2}\right) \\ &\quad{} -\frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} \left\{ \frac{e^{f\beta}-1}{f} -\left(-f\cos\beta+f-\frac{f\sin^2\beta}{2}\right) \right. \\ &\qquad\left. -\frac{e^{f\beta}}{1+f^2}\left(\sin\beta+f\cos\beta\right) +\frac{f}{1+f^2} -\left(-\frac{\beta}{2}+\sin\beta-\frac{\sin2\beta}{4}\right) \right\} =0 . \end{aligned} \]

\[ \begin{aligned} 27.\quad &M_0\left(1-\cos\beta\right) +Br\left(1-\cos\beta-\frac{\sin^2\beta}{2}\right) -Sr\left(\frac{\beta}{2}-\frac{\sin2\beta}{4}\right) \\ &\quad{} -\frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} \left\{ \frac{e^{f\beta}}{1+f^2}\left(f\sin\beta-\cos\beta\right) +\frac{1}{1+f^2} \right. \\ &\qquad\left. -\left[ f\left(\frac{\beta}{2}-\frac{\sin2\beta}{4}\right) +\frac{\sin^2\beta}{2} \right] \right\} =0 . \end{aligned} \]

We determine the determinant equal to (omitting \(r^2\), which is common to all terms upon reduction)

\[ 28.\quad \Delta =- \left| \begin{array}{ccc} \frac{2}{3}\pi & \frac{2}{3}\pi-\frac{\sqrt{3}}{2} & 1+\frac{1}{2} \\[4pt] \frac{2}{3}\pi-\frac{\sqrt{3}}{2} & \pi-\sqrt{3}-\frac{\sqrt{3}}{8} & 1+\frac{1}{2}-\frac{3}{8} \\[4pt] 1+\frac{1}{2} & 1+\frac{1}{2}-\frac{3}{8} & \frac{\pi}{3}+\frac{\sqrt{3}}{8} \end{array} \right| =-0.06 . \]

In a similar manner,

\[ \Delta M_0=-0.08 . \]

\[ 29.\quad M_0 = \frac{T_{\max} r e^{-f\varphi_0}}{1+f^2} + \frac{-0.08}{-0.06} = +0.94\,T_{\max} r e^{-f\varphi_0} \qquad (f=0.25). \]

\[ 30.\quad B = \frac{T_{\max} e^{-f\varphi_0}}{1+f^2} \cdot \frac{-0.10}{-0.06} = 1.2\,T_{\max} e^{-f\varphi_0}. \]

\[ 31.\quad S = \frac{T_{\max} e^{-f\varphi_0}}{1+f^2} \cdot \frac{+0.02}{-0.06} = -0.25\,T_{\max} e^{-f\varphi_0}. \]

For \(\varphi_0 \simeq 0.9\pi\) (most commonly used)

\[ \begin{aligned} 32.\quad & M_0 = 0.48\,T_{\max}r,\\ 33.\quad & B = 0.6\,T_{\max},\\ 34.\quad & S = -0.13\,T_{\max}. \end{aligned} \]

For case II we have (see Fig. 2, sector \(LOD\)):

\[ E=1, \]

\[ n=1, \]

\[ \alpha=\varphi_0 \quad \text{and} \quad \beta=\frac{2}{3}\pi . \]

Equations 21, 20, and 19 take the form:

\[ \begin{aligned} 35.\quad &M_0'(\beta)+Br(\beta-\sin\beta)-Sr(1-\cos\beta) -\frac{T_{\max}re^{-f\varphi_0}}{1+f^2} \Bigg( \frac{e^{-f\varphi_0}-e^{-f\beta}}{f}\\ &\qquad +e^{-f\varphi_0}\big[\sin(2\beta-\varphi_0)-f\cos(2\beta-\varphi_0)+f\big] -e^{-f\beta}(\sin\beta-f\cos\beta+f) \Bigg)=0 . \end{aligned} \]

\[ \begin{aligned} 36.\quad &M_0'\,(\beta-\sin\beta) +Br\left(\frac{3\beta}{2}-2\sin\beta+\frac{\sin2\beta}{4}\right) -Sr\left(1-\cos\beta-\frac{\sin^2\beta}{2}\right)\\ &\quad -\frac{T_{\max}re^{-f\varphi_0}}{1+f^2} \Bigg\{ \frac{e^{-f\varphi_0}-e^{-f\beta}}{f} +e^{-f\varphi_0}\Bigg( -\frac{1}{1+f^2}\sin(\varphi_0-\beta) -\frac{f}{1+f^2}\cos(\varphi_0-\beta)\\ &\qquad +\sin(2\beta-\varphi_0)-f\cos(2\beta-\varphi_0)+f -\frac{1+f}{2}\sin(3\beta-\varphi_0)\\ &\qquad +\frac{1+f}{2}\sin(\varphi_0-\beta) +\frac{1+f}{2}\cos(\varphi_0-\beta)(2\beta-\varphi_0) +\frac{fe^{-f\beta}}{1+f^2} +e^{-f\beta}\\ &\qquad +\left( -\frac{2\varphi_0-3\beta}{2} +\sin\beta-f\cos\beta -\frac{\sin2(\varphi_0-\beta)}{2} +f\cos(\varphi_0-\beta)\right.\\ &\qquad\qquad\left. +\frac{\sin2\beta}{4} -f\frac{\sin^2\beta}{2} +f\frac{\sin^2(\varphi_0-\beta)}{2} \right) \Bigg\} \end{aligned} \]

\[ \begin{aligned} 37.\quad &M_0[1-\cos\beta] +Br\left(1-\cos\beta-\frac{\sin^2\beta}{2}\right) -Sr\left(\frac{\beta}{2}-\frac{\sin2\beta}{4}\right)\\ &\quad -\frac{T_{\max}re^{-f\varphi_0}}{1+f^2} \Bigg\{ e^{-f\varphi_0}\Bigg[ \frac{f}{1+f^2}\sin(\varphi_0-\beta) -\frac{1}{1+f^2}\cos(\varphi_0-\beta)\\ &\qquad -\frac{\cos(3\beta-\varphi_0)}{2} +\frac{\cos(\varphi_0-\beta)}{2} +\frac{\sin(\varphi_0-\beta)}{2}(2\beta-\varphi_0) -f\frac{\sin(3\beta-\varphi_0)}{2}\\ &\qquad +\frac{f\sin(3\varphi_0-\beta)}{2} +f\cos(\varphi_0-\beta)\left(\frac{2\beta-\varphi_0}{2}\right) \Bigg] -e^{-f\beta}\left[ f\left(\frac{\beta}{2}-\frac{\sin2\beta}{4}\right) +\frac{1}{2}\sin^2\beta \right] \Bigg\}=0 . \end{aligned} \]

For these equations the determinant will be equal to (\(r^2\) again omitted):

\[ 38.\quad \Delta = -\left| \begin{array}{ccc} \beta & \beta-\sin\beta & 1-\cos\beta \\[2mm] \beta-\sin\beta & \dfrac{3\beta}{2}-2\sin\beta+\dfrac{\sin2\beta}{4} & 1-\cos\beta-\dfrac{\sin^2\beta}{2} \\[3mm] 1-\cos\beta & 1-\cos\beta-\dfrac{\sin^2\beta}{2} & \dfrac{\beta}{2}-\dfrac{\sin2\beta}{4} \end{array} \right| =-0.06 . \]

Then for \(\varphi_0=0.9\pi\) and \(f=0.25\):

\[ \text{39.}\quad M_0=\frac{T_{\max} r e^{-f\varphi_0}}{1+f^2}\cdot \frac{-0.04}{-0.06}=0.6\,T_{\max}r \]

\[ \text{40.}\quad B=\frac{T_{\max} e^{-f\varphi_0}}{1+f^2}\cdot \frac{-0.01}{-0.06}=0.9\,T_{\max} \]

\[ \text{41.}\quad S=\frac{T_{\max} e^{-f\varphi_0}}{1+f^2}\cdot \frac{+0.03}{-0.06}=-0.2\,T_{\max} \]

For a uniformly distributed load (in particular also for centripetal forces), equations 19, 20, and 21, etc., are used in the following manner.

As is quite clear from the mnemonic scheme (p. 3),

\[ \text{42.}\quad T_{\max}=pr \]

Further, the coefficient of friction under conditions of radial loading, etc., is equal to 0, i.e., in all equations

\[ \text{43.}\quad f=0. \]

Then the expression

\[ \text{44.}\quad \left(\frac{e^{f\alpha}-e^{fn\beta}}{f}\right)=\frac{0}{0} \]

and, upon resolving the indeterminacy, is equal to

\[ \text{45.}\quad \alpha-n\beta. \]

The remaining expressions cause no doubt. As a result of the operations, with \(E=0\) and \(n=0\), in the equations:

\[ \text{46.}\quad M_0\beta+Br(\beta-\sin\beta)-Sr(1-\cos\beta)+pr^2(\beta-\sin\beta)=0 \]

\[ \text{47.}\quad M_0(\beta-\sin\beta)+Br\left(\frac{3\beta}{2}-2\sin\beta+\frac{\sin2\beta}{4}\right)-Sr\left(1-\cos\beta-\sin^2\frac{\beta}{2}\right)-pr^2\left(\frac{3\beta}{2}-2\sin\beta+\frac{\sin2\beta}{4}\right)=0. \]

\[ \text{48.}\quad M_0(1-\cos\beta)+Br\left(1-\cos\beta-\frac{\sin^2\beta}{2}\right)-Sr\left(\frac{\beta}{2}-\frac{\sin2\beta}{4}\right)-pr^2\left(1-\cos\beta-\frac{\sin^2\beta}{2}\right)=0. \]

Equations (46, 47, and 48) have exactly the form to which the equilibrium conditions for a uniformly distributed load lead.

Solving them by the method indicated above, we have, for \(\beta=\frac{2}{3}\pi\),

\[ \text{49.}\quad \Delta=-0.06 \]

\[ \text{50.}\quad M_0=0, \]

\[ \text{51.}\quad B=pr \]

\[ \text{52.}\quad S=0, \]

i.e., we obtain the case of simple compression, which arises in a uniformly compressed ring (Lamé problem), as of course was to be expected.

A study of the combined deformations of the rim and the handle\(^1\) leads to the conclusion that, at the point of connection of the rim with the handle, a moment \(M_0\) acts also in the case of a uniformly distributed load.

According to Berdov,\(^1\) for \(\beta=\frac{2}{3}\pi\)

\[ \text{53.}\quad B=0.98\,pr \]

\[ \text{54.}\quad M_0=0.01\,pr^2\quad \text{and} \]

\[ \text{55.}\quad S=-0.02\,pr \]

\(^1\) Berdov, Transmissions by Flexible Coupling, p. 87 (ed. 1911).

In a similar manner \(M_0\), \(B\), and \(S\) were computed for the case of 4 and 6 spokes. The results of the computations are brought together in Table I.

Table I¹

4. Conclusions

In comparing the data of Table I (see also Fig. 6) we arrive at the following conclusions:

1. The bending moments do not remain constant when the number of spokes is changed: as was to be expected, they decrease with a greater number of spokes.

2. The force perpendicular to the direction of the arm increases as the number of spokes is increased.

3. The force acting along the spoke likewise increases as the number of spokes is increased.

4. The change in these forces is weaker for small arcs of contact (column “a”); and the change in \(B\) is so insignificant under these conditions that it may be considered practically independent of the number of spokes.

5. Calculation of the rim for a uniformly distributed load gives smaller bending moments and exaggerated forces, as compared with the actual state (column “c”).

6. If certain average conditions are adopted for the operation of the pulley, with which one most often has to deal, an interesting conclusion is obtained. Let, for example, the pulley operate at an ordinary speed of from 5 to \(15\,\mathrm{m/s}\).

Then, under normal conditions (angle of contact \(=0.9\pi\), average pulley diameters), at rest the belt tension may be taken as equal to the circumferential force, multiplied by \(1\frac12\), i.e.

\[ 55.\quad T_{\max}=1.5P. \]

The circumferential force may be determined as a function of the load of the belt per one cm of its width. Namely, according to Gehrkens,²

\[ 56.\quad P=pa^2,\quad \text{where} \]

\(p\)—load per 1 cm of width,

\(a\)—width of the belt in cm.

Considering Gehrkens’s table, we see that for single belts at average speeds and average pulley diameters, \(p\) increases almost proportionally to the diameter of the pulley. It may therefore be taken for pulleys with diameter greater than 400 m/m and up to 1000 m/m that

\[ 57.\quad p=\frac{p_0 r}{r_0}, \]
here \(p_0\)—load

To the article by mechanical engineer S. F. Lebedev, “Rim Under Load According to an Exponential Law.”

Fig. 1.

In sector “KOL”

1. \(n = 0\)
2. \(\alpha \leq \beta\)
3. \(\varepsilon = 0\)

Moments \(M_\alpha\)

Fig. 2.

In sector “LOM”

1. \(n = 1\)
2. \(\alpha \leq \varphi_n\)
3. \(\varepsilon = 1\)

Moments \(M_\alpha\) and \(M_{\alpha'}\)

Fig. 3.

(\(N\) of the first formula for a uniformly distributed load)

Fig. 4.

(For the compilation of the general expression for the moment \(M_x\).)

Fig. 5.

\(M_\alpha,\ B^n,\ S\)

curve “\(M_\alpha\)”

curve “\(S\)”

curve “\(B\)”

number of spokes

for a belt with speed \(5\,\mathrm{m/s}\) and a pulley of diameter \(400\,\mathrm{m/m}\), \(r_0\) is the radius of the last pulley (\(240\,\mathrm{m/m}\)), and “\(p\)” and “\(r\)” refer to the driving pulley (within the indicated limits).

For our case, according to Gehrkens’ table,

\[ p_0 = 6\,\mathrm{kgr}. \]

\[ r_0 = 40\,\mathrm{cm}, \quad \text{and therefore} \]

\[ \tag{58} p = \frac{6r}{40} = 0.15\,r\,\mathrm{kgr}. \]

Neglecting the curvature of the rim, we have the strength condition for its bending:

\[ \tag{59} M_0 = \frac{b h^2}{6}\,k_b, \]

where

\(h\) is the height (mean) of the rim section (thickness),

\[ k_b = 300\,\mathrm{kg/cm^2} \]

for cast iron.

The mean value of \(M_0\), with the usual number of grooves, may, in accordance with Table 1, be taken as

\[ \tag{60} M_0 = 0.35\,T_{\max}\,r. \]

After substituting all quantities into equation 59, we obtain

\[ \tag{61} 0.35 \cdot 1.5 \cdot 0.15r \cdot b \cdot r = \frac{b h^2}{6}\,300, \]

and, after reduction and calculation,

\[ \tag{62} h = 0.03r. \]

“\(h\)” is the mean thickness of the rim; if one takes into account the slope from the middle to the edge of the rim \(\left(\frac{1}{25}\right)\), and also the camber of the rim, the coincidence may be considered quite sufficient with the usually adopted dimension at the edge:

\[ \tag{63} h = 0.01r. \]

The addition of \(3\,\mathrm{m/m}\), usually made, may also be justified.

For any number of grooves, the force “\(B\),” as has been shown, remains almost constant and equal to \(T_{\max}\), i.e.,

\[ \tag{64} B = T_{\max} = 1.5P. \]

Taking the limiting permissible stress for cast iron as \(350\,\mathrm{kg/cm^2}\) for the combined action of bending and compression, we obtain a residual stress, for compensation of the force \(B\), of \(50\,\mathrm{kg/cm^2}\).

Then, under the previous conditions, the required addition to the height of the rim will be equal to

\[ \tag{65} \frac{B}{bh} = k_2, \]

and after substitution

\[ \tag{66} h_{\mathrm{gov}} = \frac{1.5\,p\,b}{b \cdot 50} = \frac{1.5\,p}{50} = \approx 2\,\mathrm{m/m}, \]

and for

\[ p = 10\,\mathrm{kgr}, \quad h_{\mathrm{gov}} = 3\,\mathrm{m/m}, \]

i.e., entirely in agreement with the usual dimension.

Since “\(B\),” as shown, almost does not depend on the number of grooves and on the magnitude of the rim radius, whereas “\(M_0\)” varies considerably with the number of spokes and the magnitude of the radius, the general construction of the design formula (63) should be considered theoretically justified and entirely correct.

In any case, the actual work of the rim takes place under conditions other than those usually assumed in checking it, and it seemed important to us to draw attention to these circumstances.

1. Here \(T=T_{\max}\) in the formulas. ↩↩

2. Verov, Transmission by Flexible Coupling, p. 35. ↩↩

3. Timoshenko, S. M. 1916. ↩