22-Mec-A4 Design and Manufacture of Machine Elements · December 2019

Question 5 of 6: Minimum Shaft Diameter — Static Yield and Fatigue

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: Minimum Shaft Diameter — Static Yield and Fatigue (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The minimum allowable diameter of the 10-in section between the bearings, first from a static distortion-energy yield check and then from a completely-reversed-bending fatigue check, both at a design factor of 2.0.

Approach. Convert the transmitted torque into the gear tooth force, carry that overhung force back to bearing C to get the maximum bending moment on the 10-in section, and then size the shaft twice: once with the distortion-energy yield criterion applied to the steady stress state, and once with the DE-Goodman shaft equation for completely reversed bending combined with steady torsion.

1. Find the gear tooth force from the transmitted torque. The tangential (transmitted) component acts at the pitch radius, so

   $$W_t = \frac{T_A}{d_p/2} = \frac{3500}{3} = 1166.7\ \text{lbf}$$

   The radial component follows from the pressure angle, and the resultant is the force actually applied by the mating tooth:

   $$W_r = W_t\tan\phi = 1166.7\tan 20^\circ = 424.6\ \text{lbf}, \qquad F = \frac{W_t}{\cos\phi} = \frac{1166.7}{\cos 20^\circ} = 1241.5\ \text{lbf}$$

2. Locate the critical section and compute the bending moment there. The gear at D overhangs bearing C by 4 in, so the shaft between B and C carries a bending moment that is zero at B (a simple support) and rises linearly to a maximum at C. Taking moments of the overhung load about C,

   $$M = F\,e = 1241.5 \times 4 = 4966\ \text{lbf}\cdot\text{in}$$

   Because $W_t$ and $W_r$ act in perpendicular planes, working with the resultant $F$ gives the resultant moment magnitude directly — the same answer as combining $4W_t$ and $4W_r$ vectorially. The torque over this section is the full $T = 3500$ lbf·in, since it passes through from D to A. The critical section is therefore the right-hand end of the 10-in section, at the shoulder against bearing C, where bending moment and torque are both at their largest.

3. (a) Size the shaft for static yield by the distortion-energy theory. On the surface at the critical section, $\sigma = 32M/\pi d^3$ and $\tau = 16T/\pi d^3$, so the von Mises stress is

   $$\sigma' = \sqrt{\sigma^2 + 3\tau^2} = \frac{16}{\pi d^3}\sqrt{4M^2 + 3T^2}$$

   Setting $\sigma' = S_y/n$ and solving for the diameter,

   $$d = \left[\frac{16n}{\pi S_y}\sqrt{4M^2 + 3T^2}\right]^{1/3}$$

4. Evaluate the static result. The radical is

   $$\sqrt{4(4966)^2 + 3(3500)^2} = \sqrt{98.66\times10^6 + 36.75\times10^6} = 11\,636\ \text{lbf}\cdot\text{in}$$

   and therefore

   $$d = \left[\frac{16(2.0)(11\,636)}{\pi(65\,000)}\right]^{1/3} = (1.8237)^{1/3}$$

   $$\boxed{\,d_{static} = 1.22\ \text{in}\,}$$

   No stress-concentration factor is applied here: for a ductile material under static loading, local yielding at the fillet redistributes the stress, and Shigley's practice is to ignore $K_t$ in the static check.

5. (b) Set up the fatigue problem. The shaft rotates while the gear load stays fixed in space, so every surface element passes alternately through tension and compression: the bending stress is completely reversed with $M_a = 4966$ lbf·in and $M_m = 0$. The torque, by contrast, is transmitted steadily, so $T_m = 3500$ lbf·in and $T_a = 0$. This is the classic rotating-shaft loading, and Shigley's DE-Goodman shaft-diameter equation applies:

   $$d = \left\{\frac{16n}{\pi}\left[\frac{\sqrt{4(K_f M_a)^2 + 3(K_{fs}T_a)^2}}{S_e} + \frac{\sqrt{4(K_f M_m)^2 + 3(K_{fs}T_m)^2}}{S_{ut}}\right]\right\}^{1/3}$$

   which, with $M_m = T_a = 0$, reduces to

   $$d = \left\{\frac{16n}{\pi}\left[\frac{2K_f M_a}{S_e} + \frac{\sqrt{3}\,K_{fs}T_m}{S_{ut}}\right]\right\}^{1/3}$$

6. Build the endurance limit. Starting from the rotating-beam estimate for steel with $S_{ut} < 200$ kpsi,

   $$S'_e = 0.5\,S_{ut} = 0.5(75) = 37.5\ \text{kpsi}$$

   The surface factor for a machined finish is $k_a = aS_{ut}^{\,b}$ with $a = 2.00$, $b = -0.217$ for $S_{ut}$ in kpsi:

   $$k_a = 2.00(75)^{-0.217} = 0.784$$

   The size factor is $k_b = 0.879\,d^{-0.107}$ for $0.3 \le d \le 2$ in, which depends on the answer and so requires iteration. Loading, temperature and reliability factors are taken as unity ($k_c = 1$ for bending, room temperature, and no reliability adjustment specified).

7. Take the stress-concentration factors for a sharp fillet. The question directs us to assume sharp fillet radii at the bearing shoulder, which is Shigley's first-iteration case in Table 7-1:

   $$K_t = 2.7\ \text{(bending)}, \qquad K_{ts} = 2.2\ \text{(torsion)}$$

   and these are used directly as $K_f$ and $K_{fs}$, i.e. full notch sensitivity $q = 1$, which is the conservative first-iteration assumption.

8. Iterate to the fatigue diameter. Starting from a trial $d = 1$ in and cycling the size factor through the equation, the solution converges in a few passes to $d \approx 2.38$ in, at which

   $$k_b = 0.879(2.376)^{-0.107} = 0.801, \qquad S_e = k_a k_b S'_e = 0.784(0.801)(37.5) = 23.55\ \text{kpsi}$$

   The two bracket terms are then

   $$\frac{2K_f M_a}{S_e} = \frac{2(2.7)(4966)}{23\,550} = 1.139\ \text{in}^3, \qquad \frac{\sqrt{3}K_{fs}T_m}{S_{ut}} = \frac{1.732(2.2)(3500)}{75\,000} = 0.178\ \text{in}^3$$

   so that

   $$d = \left[\frac{16(2.0)}{\pi}(1.139 + 0.178)\right]^{1/3} = (13.41)^{1/3}$$

   $$\boxed{\,d_{fatigue} = 2.38\ \text{in}\,}$$

9. Select the governing diameter. The fatigue requirement is nearly twice the static one, so fatigue governs and the 10-in section must be at least 2.38 in in diameter. In practice one would specify the next preferred size, $d = 2\tfrac{1}{2}$ in, which restores a small additional margin.

The gap between the two answers is instructive: the static check sees a 4966 lbf·in moment applied once, while the fatigue check sees the same moment applied as a fully reversed cycle through a sharp fillet whose $K_f = 2.7$ multiplies it, against an endurance limit that is only about a third of the yield strength. Both effects push in the same direction, and together they account for the factor of roughly eight in required section modulus.