Design of shafts

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 principal shear stress (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 (about 1.0 to 2.0).

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.

Use the ASME code equation $d^3 = \dfrac{16}{\pi \tau_{allow}}\sqrt{(K_b M)^2 + (K_t T)^2}$.

Torque

$$T = \frac{60P}{2\pi n} = \frac{60 \times 25\,000}{2\pi \times 360} = 663.1\ \text{N.m}$$

Allowable shear stress

$$\begin{aligned} 0.30\,S_{yt} &= 0.30 \times 330 = 99\ \text{MPa} \\ 0.18\,S_{ut} &= 0.18 \times 560 = 100.8\ \text{MPa} \end{aligned}$$

The smaller value is 99 MPa. With a keyway the stress is reduced by 25 percent:

$$\tau_{allow} = 0.75 \times 99 = 74.25\ \text{MPa}$$

Equivalent moment

$$\begin{aligned} K_b M &= 1.5 \times 450 = 675\ \text{N.m} \\ K_t T &= 1.2 \times 663.1 = 795.8\ \text{N.m} \\ T_e &= \sqrt{675^2 + 795.8^2} = 1043.5\ \text{N.m} \end{aligned}$$

Diameter

$$d^3 = \frac{16 \times 1043.5 \times 10^3}{\pi \times 74.25} = 71\,572\ \text{mm}^3$$ $$d = 41.5\ \text{mm}$$

Adopt the next standard size, 42 mm.

Answer: d = 41.5 mm, so use a 42 mm shaft.

The endurance limit $S_e'$ found from a standard rotating-beam test on a small polished specimen is much higher than the endurance limit of an actual shaft. The actual value is obtained using the Marin equation:

$$S_e = k_a\, k_b\, k_c\, k_d\, k_e\, k_f\, S_e'$$

where, for steels, $S_e' \approx 0.5\,S_{ut}$ (for $S_{ut} \le 1400$ MPa, with an upper limit of 700 MPa).

Modifying factors

Stress concentration

Notches such as keyways, shoulders and grooves are included by the fatigue stress concentration factor $K_f = 1 + q(K_t - 1)$, where $q$ is the notch sensitivity. The nominal alternating stress is multiplied by $K_f$ (or the endurance limit is divided by it).

These corrected values are used in the Soderberg or Goodman equations when designing shafts for fatigue.

In a fatigue diagram the mean stress $\sigma_m$ is plotted on the horizontal axis and the alternating stress $\sigma_a$ on the vertical axis. A line is drawn to show the combinations that a material can bear for infinite life.

 sigma_a
   ^
 Se|*.
   |  *  .  Goodman
   |   *     .
   |     * Soderberg   .
   |       *      .      .
   |         *  .
   |           *.  .
   +-----------*-----*----> sigma_m
              Sy    Sut

(Soderberg line: from $S_e$ on the vertical axis to $S_y$ on the horizontal axis. Goodman line: from $S_e$ to $S_{ut}$. The Soderberg line is the lower one.)

Equations

Soderberg line (based on yield strength):

$$\frac{\sigma_m}{S_y} + \frac{\sigma_a}{S_e} = \frac{1}{n}$$

Modified Goodman line (based on ultimate strength):

$$\frac{\sigma_m}{S_{ut}} + \frac{\sigma_a}{S_e} = \frac{1}{n}$$

where $n$ is the factor of safety and $S_e$ is the corrected endurance limit. Stress concentration $K_f$ is applied to the alternating component (and, for brittle materials, also to the mean component).

Comparison

For a shaft with bending and torsion, the stresses are combined using the distortion energy theory: equivalent mean and alternating stresses are found and substituted in the line equation.

The bending stress is fully reversed ($\sigma_m = 0$) and the torsion is steady ($\tau_a = 0$). Using the Soderberg line with distortion energy equivalent stresses:

$$\frac{1}{n} = \frac{16}{\pi d^3}\sqrt{4\left(\frac{K_f M}{S_e}\right)^2 + 3\left(\frac{K_{fs} T}{S_y}\right)^2}$$

Corrected endurance limit

$$\begin{aligned} S_e' &= 0.5 \times 700 = 350\ \text{MPa} \\ S_e &= 0.77 \times 0.85 \times 0.814 \times 350 = 186.5\ \text{MPa} \end{aligned}$$

Stress concentration terms

$$\begin{aligned} K_f M &= 1.8 \times 500 = 900\ \text{N.m} = 900 \times 10^3\ \text{N.mm} \\ K_{fs} T &= 1.5 \times 700 = 1050\ \text{N.m} = 1050 \times 10^3\ \text{N.mm} \end{aligned}$$

Substituting

$$\begin{aligned} \frac{K_f M}{S_e} &= \frac{900 \times 10^3}{186.5} = 4826 \\ \frac{K_{fs} T}{S_y} &= \frac{1050 \times 10^3}{420} = 2500 \\ \sqrt{4(4826)^2 + 3(2500)^2} &= \sqrt{9.318\times10^7 + 1.875\times10^7} = 10\,580 \end{aligned}$$ $$d^3 = \frac{16\,n}{\pi}\times 10\,580 = \frac{16 \times 2}{\pi}\times 10\,580 = 1.078\times10^5\ \text{mm}^3$$ $$d = 47.6\ \text{mm}$$

The nearest standard size above this is 50 mm.

Answer: d = 47.6 mm, so use a 50 mm diameter shaft.

        4 kN (down)     Fr=1.456 kN (horizontal)
            |
  A---------C--------------B----> coupling (torque)
  |<-300->|<-----500----->|
  |<----------800--------->|

Torque

$$T = F_t \times r = 4000 \times 0.1 = 400\ \text{N.m}$$

Vertical plane (load 4 kN at C)

$$\begin{aligned} R_{Av} &= 4 \times \frac{500}{800} = 2.5\ \text{kN}, \qquad R_{Bv} = 4 - 2.5 = 1.5\ \text{kN} \\ M_v &= R_{Av} \times 0.3 = 2.5 \times 0.3 = 0.75\ \text{kN.m} = 750\ \text{N.m} \end{aligned}$$

Horizontal plane (load 1.456 kN at C)

$$\begin{aligned} R_{Ah} &= 1.456 \times \frac{500}{800} = 0.91\ \text{kN}, \qquad R_{Bh} = 1.456 - 0.91 = 0.546\ \text{kN} \\ M_h &= 0.91 \times 0.3 = 0.273\ \text{kN.m} = 273\ \text{N.m} \end{aligned}$$

Check: in each plane $R_A + R_B$ equals the applied load, and moments about A balance: vertical $4 \times 0.3 = 1.2 = 1.5 \times 0.8$ kN.m; horizontal $1.456 \times 0.3 = 0.437 = 0.546 \times 0.8$ kN.m.

Resultant bending moment at C

$$M = \sqrt{750^2 + 273^2} = 798.1\ \text{N.m}$$

Diameter

$$\begin{aligned} T_e &= \sqrt{(K_b M)^2 + (K_t T)^2} = \sqrt{(1.5 \times 798.1)^2 + (1.0 \times 400)^2} \\ &= \sqrt{1197.2^2 + 400^2} = 1262.3\ \text{N.m} \end{aligned}$$ $$d^3 = \frac{16 \times 1262.3 \times 10^3}{\pi \times 40} = 1.607\times10^5\ \text{mm}^3$$ $$d = 54.4\ \text{mm}$$

Use the standard size 55 mm.

Answer: Reactions (vertical) 2.5 kN and 1.5 kN, (horizontal) 0.91 kN and 0.546 kN; M at C = 798 N.m; d = 54.4 mm, adopt 55 mm.

Torque

$$T = \frac{60P}{2\pi n} = \frac{60 \times 90\,000}{2\pi \times 250} = 3437.7\ \text{N.m}$$

Solid shaft

$$\begin{aligned} T &= \frac{\pi}{16}\tau d^3 \\ d^3 &= \frac{16 \times 3437.7\times10^3}{\pi \times 40} = 4.377\times10^5\ \text{mm}^3 \\ d &= 75.9\ \text{mm} \end{aligned}$$

Hollow shaft ($k = d_i/d_o = 0.6$)

For the same torque and stress,

$$\frac{\pi}{16}\tau d_o^3(1 - k^4) = T$$ $$d_o^3 = \frac{d^3}{1 - k^4} = \frac{4.377\times10^5}{1 - 0.6^4} = \frac{4.377\times10^5}{0.8704} = 5.029\times10^5$$ $$d_o = 79.5\ \text{mm}, \qquad d_i = 0.6 \times 79.5 = 47.7\ \text{mm}$$

Mass saving

Mass is proportional to cross-sectional area (same length and material):

$$\frac{A_h}{A_s} = \frac{d_o^2(1 - k^2)}{d^2} = \frac{79.5^2 \times 0.64}{75.9^2} = 0.702$$

Saving $= 1 - 0.702 = 0.298$, about 30 percent.

Hollow shafts save material because the inner core carries little shear stress. The outside diameter is only about 5 percent larger, but they are costlier to make.

Answer: Solid d = 75.9 mm; hollow $d_o$ = 79.5 mm, $d_i$ = 47.7 mm; mass saving about 29.8 percent.