Overhung idler loading

Overhung idler loading occurs when an idler gear's mesh resultant acts outside the axial span of its shaft supports. The overhang creates a lever arm, so the nearer mechanical bearing can carry a larger transverse load than the mesh resultant and the farther support can have the opposite reaction direction. This geometry is separate from whether the idler's two transmitted torques cancel.^[ref]

Place ideal radial supports A and B at $z=0$ and $z=L$. An upward mesh resultant $P$ at $z=L+e$, with overhang $e>0$, gives static equilibriumHenceFor $P=200\mathrm{N}$, $L=100\mathrm{mm}$ and $e=30\mathrm{mm}$, A acts upward with60N while B acts downward with260N. Their signed sum balances the applied200N. The larger reaction at B is a lever effect, not an additional external mesh force.

The overhung shaft bending moment immediately next to the near support has magnitude $Pe=6\mathrm{N}\mathrm{m}$ in this ideal model. That bending moment can exist while the net shaft-axis torque on the steady idler is zero. Doubling overhang at fixed load doubles the near-support bending moment and changes the reaction magnitudes to120 and320N.

Idler bearing load sharing must therefore include the gear's axial position, not just its pitch radius and speed ratio. Loads in independent transverse directions can be balanced separately and then combined as vectors. A support model with more constraints requires compatible deformations as well as force balance. Shaft bending and support flexibility can also alter the gear face load distribution, linking the overhang geometry to tooth loading.

An overhung shaft bending moment is the internal bending moment produced when a transverse load acts beyond the shaft's support span. For a single end load $P$ at overhang distance $e$, its magnitude immediately next to the nearer support isin the ideal two-support shaft model. The relevant lever arm is the distance beyond that support, rather than the distance between supports. Manufacturer bearing-load guidance distinguishes this arrangement from a load located between bearings.^[ref]

Place ideal supports A and B at axial positions $z=0$ and $z=L$, and apply an upward load $P$ at $z=L+e$. With upward reactions positive, static equilibrium givesChoosing the internal moment sign consistently with the left-hand force sums, the bending moment along the shaft isIt is continuous at B, reaches $Pe$ there and vanishes at the free end. An ideal bearing need not apply a concentrated couple for the shaft's internal moment at its location to be nonzero: the shaft continues beyond that support.

For original values $P=200\mathrm{N}$, $L=100\mathrm{mm}$ and $e=30\mathrm{mm}$, the reactions are $+60$ and $-260\mathrm{N}$ and the peak moment is $6\mathrm{N}\mathrm{m}$. Doubling $e$ at fixed $P$ doubles that moment. Changing $L$ while keeping $P,e$ fixed changes the reactions but leaves $Pe$ unchanged.

For a uniform solid circular shaft in simple elastic bending, overhung shaft bending stress follows from ${\sigma }_{\max }=Mc/I$ and $I=\pi d^4/64$:At diameter $d=20\mathrm{mm}$, the example gives about $7.64\mathrm{MPa}$. This excludes stress concentrations and other loads; a shaft's allowable load requires more than this nominal estimate. Bending can remain even when opposing gear torques give zero net shaft-axis torque, because transverse-force moments and axial twisting are different components of the mechanical loading.

Overhung shaft bending stress is the longitudinal normal stress caused by the bending moment from a transverse load beyond the shaft's supports. It differs from twisting due to shaft-axis torque. In the ideal two-support model, the largest overhung shaft bending moment has magnitude $M=Pe$, where $P$ is the end load and $e$ the overhang.^[ref]

For a homogeneous linear elastic beam, plane sections in the beam model give longitudinal strain $ϵ(y)=\kappa y$ relative to the neutral axis. With Young's modulus $E$,where $I$ is the second moment of area. Thus $\sigma (y)=My/I$. For a solid circular shaft of diameter $d$, $I=\pi d^4/64$ and the outermost distance is $d/2$, givingAn original example uses $P=200\mathrm{N}$, $e=30\mathrm{mm}$ and $d=20\mathrm{mm}$. Then $M=6000\mathrm{N}\mathrm{mm}$ and ${\sigma }_{\max }\simeq 7.64\mathrm{MPa}$. Reducing diameter to $16\mathrm{mm}$ multiplies nominal stress by $(20/16{)}^3\simeq 1.953$, raising it to about $14.9\mathrm{MPa}$. The cubic diameter dependence makes a diameter change different from an equal percentage change in overhang.

A measurement check can place opposed axial strain gauges at one shaft station. In a simple axial-plus-bending model, readings ${ϵ}_{+}=44$ and ${ϵ}_{-}=-32$ microstrain give axial strain $({ϵ}_{+}+{ϵ}_{-})/2=6$ microstrain and bending strain $({ϵ}_{+}-{ϵ}_{-})/2=38$ microstrain. This separates two contributions that a single gauge would mix.

The nominal formula excludes local geometric concentrations, multiaxial loading and yielding. It is not a fatigue-life prediction. Under a stationary bending direction, a rotating material point can experience alternating stress, motivating rotating shaft bending fatigue even when the external load is steady. A permissible load requires the actual geometry, material and load history.