Overhung shaft bending moment ($M=Pe$)

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.