Shaft Size Calculator

Shaft design relies on classical mechanics of materials, specifically analyzing torsional shear stress ($ au$), bending stress ($Δ_b$), and equivalent twisting or bending moments ($T_e$ and $M_e$). For pure torsion, the transmitted power ($P$) relates to rotational speed ($n$) and torque ($T$) through the formula $P = 2͠ n T / 60$. When shafts experience combined loading—such as simultaneous bending and twisting—equivalent moments are calculated using fatigue and shock factors ($K_m$ and $K_t$). The ultimate sizing takes the maximum resulting diameter derived from both maximum shear stress theory (Guest's theory) and maximum normal stress theory (Rankine's theory), evaluating solid profiles against hollow profiles governed by the inner-to-outer diameter ratio ($k = d_i / d_o$).

Consider a solid steel transmission shaft required to deliver power under combined bending and torsional loads. Suppose the applied bending moment ($M$) is $400 \text{ N}⃆m$ and the twisting torque ($T$) is $300 \text{ N}⃆m$. Assuming fatigue factors $K_m = 1.5$ and $K_t = 1.0$, and an allowable shear stress ($ au$) of $50 \text{ MPa}$ ($50 \times 10^6 \text{ N/m}^2$). First, calculate the equivalent twisting moment: $T_e = \sqrt{(K_m M)^2 + (K_t T)^2} = \sqrt{(1.5 \times 400)^2 + (1.0 \times 300)^2} = \sqrt{360000 + 90000} = 670.82 \text{ N}⃆m$. Next, utilize the torsion equation for a solid shaft: $T_e = \frac{\pi}{16} \tau d^3$. Rearranging for diameter yields $d = \left(\frac{16 T_e}{\pi \tau}\right)^{1/3} = \left(\frac{16 \times 670.82}{\pi \times 50 \times 10^6}\right)^{1/3} \approx 0.0385 \text{ m}$ or $38.5 \text{ mm}$. Engineers would select the next standard stock size, typically $40 \text{ mm}$.

1. Account for Stress Concentrations: Keyways, steps, fillets, and cross-holes significantly reduce local fatigue strength; always apply appropriate stress concentration factors ($K_t$ and $K_m$).
2. Prioritize Rigidity Over Strength: Long shafts often fail due to excessive angular deflection or lateral vibration rather than direct material yielding, making torsional rigidity and critical speed primary constraints.
3. Hollow vs. Solid Selection: Hollow shafts offer a superior strength-to-weight ratio for aerospace and high-speed automotive applications, whereas solid shafts remain cost-effective for general industrial gearboxes.