ASME code equation for shaft diameter

The ASME code equation gives the diameter of a shaft under combined bending and torsion by applying the maximum shear stress theory and multiplying the moment and torque by factors that allow for shock and fatigue.

Derivation

For a solid shaft of diameter $d$, the maximum bending stress and the torsional shear stress at the surface are

$$\sigma_b = \frac{32M}{\pi d^3}, \qquad \tau = \frac{16T}{\pi d^3}$$

The maximum shear stress at that point is

$$\begin{aligned} \tau_{max} &= \sqrt{\left(\frac{\sigma_b}{2}\right)^2 + \tau^2} \\ &= \sqrt{\left(\frac{16M}{\pi d^3}\right)^2 + \left(\frac{16T}{\pi d^3}\right)^2} \\ &= \frac{16}{\pi d^3}\sqrt{M^2 + T^2} \end{aligned}$$

The quantity $\sqrt{M^2+T^2}$ is the equivalent twisting moment $T_e$. In practice the loads are not steady, so the ASME code multiplies the bending moment by $K_b$ (combined shock and fatigue factor for bending) and the torque by $K_t$ (for torsion):

$$\tau_{max} = \frac{16}{\pi d^3}\sqrt{(K_b M)^2 + (K_t T)^2}$$

Putting $\tau_{max} = \tau_{allow}$ gives

$$d^3 = \frac{16}{\pi \tau_{allow}}\sqrt{(K_b M)^2 + (K_t T)^2}$$

For a hollow shaft with $k = d_i/d_o$, replace $d^3$ by $d_o^3(1-k^4)$.

Values of $K_b$ and $K_t$ (rotating shaft)

For a stationary shaft the values are smaller (1.0 for gradually applied loads, 1.5 to 2.0 for suddenly applied loads).

Allowable shear stress by the code

• $\tau_{allow} = 0.30\,S_{yt}$ or $0.18\,S_{ut}$, whichever is smaller.
• If the shaft has a keyway, reduce the value by 25 percent (multiply by 0.75) because of stress concentration.

The calculated diameter is rounded up to the nearest standard shaft size.