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Shaft Design

Shaft design must consider both static and fatigue failure possibilities. While initially, focusing solely on fatigue may seem overly cautious, it’s not always the case. Despite predominantly experiencing uniform stresses, shafts can occasionally encounter significantly higher stress cycles. Cumulative damage plays a role in design. However, extreme loading events, like machine malfunctions, might occur rarely, if at all, with minimal impact on fatigue crack growth. Even during overloads or malfunctions, there’s usually a limit to the stresses on the shaft. Techniques like using keys, pins, or slip clutches help control loads. Preventing static failures is crucial, given shafts’ challenging serviceability, often leading to designs with large safety margins. Various loading conditions necessitate determining the minimum shaft diameter or safety factor, often through simplified solutions.

Static loading

Shaft under bending moment and torsion

Distortion-energy theory

The Distortion-Energy Theory (DET) criterion predicts the smallest diameter when failure will occur as,

\[d=\left(\frac{32 n_s}{\pi S_y} \sqrt{M^2+\frac{3}{4} T^2}\right)^{1 / 3} \nonumber\]

where;

$d$, diameter, [m]
$n_s$, safety factor, []
$S_y$, yield strength, [Pa]
$M$, moment [Nm]
$T$, torque [Nm]

Maximum Shear Stress

The Maximum-shear Stress Theory predicts the smallest diameter when failure will occur as,

\[d=\left(\frac{32 n_s}{\pi S_y} \sqrt{M^2+ T^2}\right)^{1 / 3} \nonumber\]

Static loading - shaft under bending moment, torsion, and axial loading

Distortion-Energy Theory

The Distortion-Energy Theory can be used to obtain an expression in $d$ for the smallest shaft diameter when failure will occur.

\[\frac{4}{\pi d^3} \sqrt{(8 M+P d)^2+48T^2}=\frac{S_y}{n_s} \nonumber\]

where;

$P$, normal force [N]

In this case an explicit form of the equation (in $d$) cannot be obtained, however numerical solutions can be applied to calculate $d$.

Maximum Shear Stress

The Maximum Shear Stress Theory can be used to obtain an expression in $d$ for the smallest shaft diameter when failure will occur.

\[\frac{4}{\pi d^3} \sqrt{(8 M+P d)^2+64T^2}=\frac{S_y}{n_s} \nonumber\]

Similarly in this case an explicit form of the equation (in $d$) cannot be obtained, however numerical solutions can be applied to calculate $d$

Cyclic loading

Ductile materials

Distortion-Energy Theory

Applying the Distortion–Energy Theory, the smallest safe diameter corresponding to a specific safety factor can then be expressed as,

\[d^3=\frac{32 n_s}{\pi S_y} \sqrt{\left(M_m+\frac{S_y}{S_e} K_f M_a\right)^2+\frac{3}{4}\left(T_m+\frac{S_y}{S_e} K_{f s} T_a\right)^2}\]

where;

$S_e$, modified endurance limit [Pa]
$K_f$, fatigue stress concentration factor
$K_{fs}$, fatigue (shear) stress concentration factor
$M_a$, moment (alternating) [Nm]
$M_m$, moment (mean) [Nm]
$T_a$, torque (alternating) [Nm]
$T_m$, torque (mean) [Nm]

Maximum Shear Stress

Applying the Maximum-Shear-Stress Theory (MSST):

\[d^3=\frac{32 n_s}{\pi S_y} \sqrt{\left(M_m+\frac{S_y}{S_e} K_f M_a\right)^2+\frac{3}{4}\left(T_m+\frac{S_y}{S_e} K_{f s} T_a\right)^2}\]

Brittle materials

Distortion-Energy Theory

\[d=\left\{\frac{16 n_s}{\pi S_u}\left[K_c \Psi+\sqrt{K_c^2 \Psi^2+K_{c s}^2\left(T_m+\frac{S_u}{S_e} T_a\right)^2}\right]\right\}^{1 / 3} \nonumber\]

where,

\[\Psi=M_m+\frac{S_u}{S_e} M_a \nonumber\]

$S_u$, ultimate strength [Pa]
$K_c$, stress concentration factor [Pa]
$K_{cs}$, stress (shear) concentration factor [Pa]\

Maximum Shear Stress

\[d=\left[\frac{32 n_s}{\pi S_y} \sqrt{\left(M_m+\frac{S_y}{S_e} K_f M_a\right)^2+\left(T_m+\frac{S_y}{S_e} K_{f s} T_a\right)^2}\right]^{1 / 3}\]