Tools › Mechanical Design › Fasteners and Power Screws › Power Screws and Ball Screws › Example
1 in - 5 Acme power screw lifting 2,000 lb: raising and lowering torque, efficiency, self-locking, thread stresses and column load
| Tool | Power Screws and Ball Screws |
| Source | U.S. National Bureau of Standards (1957). Screw-Thread Standards for Federal Services 1957, Part III. Handbook H28 (1957) Part III, U.S. Department of Commerce, Washington, DC (Section XII: Acme thread proportions, Table XII.3). U.S. Army Materiel Command (1964). Engineering Design Handbook: Carriages and Mounts Series, Elevating Mechanisms. AMCP 706-346, Washington, DC (eqs. 16 and 17 and the example of paragraph 48 for the efficiency of a screw-type drive). |
| Level | Undergraduate |
Problem
A single-start 1 in - 5 general-purpose Acme screw of 60 ksi steel raises 2,000 lb through a bronze nut 1.5 in long (yield 20 ksi), the thrust taken on a plain collar of 1.5 in mean diameter. Thread friction is 0.15 and collar friction 0.10. Find the torque to raise and to lower the load, the efficiency, whether the screw holds the load, the stresses at the thread root, the bearing pressure, and the buckling load of 12 in of screw standing free above its base, with a design factor of 2.
Inputs to Enter
- Thread: Acme (29°), major diameter 1 in, pitch 0.2 in, one start, minor-diameter allowance 0.020 in (H28: largest minor diameter of a general-purpose screw, 10 threads per inch and coarser)
- Load 2,000 lb, screw in compression; thread friction 0.15, collar friction 0.10 at 1.5 in mean diameter
- Nut length 1.5 in, first-thread share 0.38; S_y 60,000 psi screw, 20,000 psi nut; bearing limit 2,500 psi; design factor 2; E 30 Msi
- Speed 60 rev/min; column length 12 in, fixed - free
Answers
- Geometry: h = 0.5 p = 0.100 in, d_m = 0.900 in, d_r = 0.780 in, lead 0.200 in, lead angle 4.046° (H28 Table XII.3 prints 4°3′), flank angle normal to the helix 14.47°, thread thickness at the root 0.6293 p = 0.1259 in
- Thread torque raising T_R = (F d_m/2)(tan λ + μ sec α_n)/(1 − μ sec α_n tan λ) = 205.3 lb-in; lowering T_L = 74.94 lb-in; collar 0.10 · 2,000 · 1.5/2 = 150 lb-in
- Torque to raise 355.3 lb-in, to lower 224.9 lb-in; frictionless torque F l/(2π) = 63.66 lb-in; efficiency 0.3100 thread alone, 0.1792 with the collar
- Self-locking limit μ = tan λ cos α_n = 0.0685; the entered 0.15 is 2.19 times that: the thread holds the load
- Body: axial 4,186 psi, torsion 3,813 psi; first thread (760 lb): bending 5,874 psi, transverse shear 3,696 psi; von Mises at the root 10,965 psi, M.S. = +1.74; maximum shear 6,158 psi, M.S. = +1.44
- Nut: 7.5 threads engaged, bearing pressure 943.1 psi (M.S. +1.65 on 2,500 psi), nut thread bending 4,581 psi (M.S. +1.18), shear 2,883 psi (M.S. +1.00); required nut length 0.566 in
- Column: slenderness 61.5 (Euler above 49.7), P_cr = 9,340 lb, M.S. = +1.33
- At 60 rev/min: travel 12 in/min, rubbing speed 14.17 ft/min, PV 13,370 psi-ft/min, input power 0.338 hp for 0.0606 hp of useful work
Notes
The tool's default inputs are this case; press Submit. The case is illustrative, not a printed example: no public-domain document was found that works a complete power-screw problem. The thread proportions are those of NBS Handbook H28; the torque equation is checked against an independent numerical solution of the three-dimensional free body and against the printed efficiencies 0.422 and 0.408 of AMCP 706-346 paragraph 48 (test_power_screws_ball_screws.php, 800 checks). The first-thread share 0.38 and the 2,500 psi bearing limit are textbook figures entered as inputs. Choosing Ball screw in the Analysis select runs the second default, a 20 x 5 mm ball screw (illustrative ratings): L10 = 8.39e8 revolutions, 7,993 h, critical speed 8,907 rev/min fixed - supported over 24 in.