Tools › Mechanical Design › Brakes, Clutches and Belts › Keys, Splines and Couplings › Example
2 in shaft with a 1/2 in square key under 3,000 lb-in: key shear and crushing, required length, keyseat geometry and the keyseated-shaft torque
| Tool | Keys, Splines and Couplings |
| Source | U.S. Army (1996). Fundamentals of Machine Tools. TC 9-524, pages 8-26 to 8-27 and Appendix A Table 8-5 (key proportions, length at least 1.5 D, the bottom-of-keyway dimension S); U.S. Navy (1990). Machinery Repairman. NAVEDTRA 12204, page 3-41, Table 3-4 (1/2 in square key for 1 3/4 to 2 1/4 in shafts, lengths 4 to 16 widths); Loewenthal, S. H. (1984). Design of Power-Transmitting Shafts. NASA RP-1123, Table V page 13 (keyway fatigue factors). |
| Level | Undergraduate |
Problem
A 2 in shaft carries 3,000 lb-in through a 1/2 in square key 3 in long (the TC 9-524 minimum of 1.5 shaft diameters; the 1/2 in key is the size the Navy and Army tables give for this shaft). Key, hub and shaft are a cold-drawn steel of 54,000 lb/in² yield (illustrative) and the design factor is 2. Find the force on the key, its shear and crushing stresses, the margins at 0.577 S_y/n in shear and S_y/n in bearing, the length the allowables require, the keyseat chord drop, depth and the shop dimension S, and the torque the keyseated shaft carries by the 0.75 rule with the RP-1123 fatigue factors of an annealed shaft with a profiled keyway.
Inputs to Enter
- d 2 in, T 3,000 lb-in, n 2, S_y shaft 54,000 lb/in²; parallel key w 0.5 in, h 0.5 in, L 3 in, S_y key 54,000 lb/in², hub yield blank (= key); keyseat factors annealed, profiled: k_e 0.63 bending, 0.77 torsion
Answers
- F = 2T/d = 3,000 lb; shear τ = 2T/(d w L) = 2,000 lb/in²; crushing σ = 4T/(d h L) = 4,000 lb/in² on each side; allowables 15,588 (shear) and 27,000 lb/in² (bearing, both sides): M.S. +6.79 shear, +5.75 bearing
- Required length: 0.385 in for shear, 0.444 in for bearing on either side, so 0.444 in governs; the machinist rules ask 3 in (1.5 d) and 2 to 8 in (4 w to 16 w)
- Keyseat: chord drop y = 0.03175 in, depth at the centre line 0.2818 in, S = d − h/2 − y = 1.718 in (TC 9-524 Table 8-5 prints 1.718 for the 2 in shaft)
- Shaft: nominal torsional stress 16T/(πd³) = 1,910 lb/in²; K_f = 1/0.63 = 1.587 bending, 1/0.77 = 1.299 torsion (RP-1123 Table V); plain-shaft torque at 0.577 S_y/n 24,486 lb-in, with the keyseat 0.75 × = 18,365 lb-in, M.S. +5.12
Notes
The tool's default inputs are this case; press Submit. The shaft, key and material are illustrative (stated on the page); the key sizes, the 1.5 D and 4 w to 16 w rules and the S dimension are the printed Navy and Army values, the keyway factors are RP-1123 Table V, and every number is checked in test_keys_splines_couplings.php (451 checks) against the independent numpy/scipy script References/Keys-Splines/checks/keys_check.py, which also reproduces all fifty S values of TC 9-524 Table 8-5 (one misprint found), the Woodruff table's geometry, the NASA-tested SSME preburner spline's pitch and base diameters and the AMCP 706-202 spline stresses.