Screw jack with square thread

Data

$W = 25\,000$ N, $d = 50$ mm, $p = 8$ mm (single start, lead $L = 8$ mm), $\mu = 0.12$, $R_c = 32.5$ mm, $\mu_c = 0.15$.

Mean diameter: $d_m = d - p/2 = 46$ mm. Core diameter: $d_c = d - p = 42$ mm.

Helix angle: $\alpha = \tan^{-1}\dfrac{8}{\pi\times46} = 3.169^\circ$. Friction angle: $\varphi = \tan^{-1}0.12 = 6.843^\circ$.

(a) Torque to raise

$$T_1 = W\tan(\alpha+\varphi)\frac{d_m}{2} = 25000\times\tan(10.012^\circ)\times0.023 = 101.5\ \text{N m}$$

Collar torque: $T_2 = \mu_c W R_c = 0.15\times25000\times0.0325 = 121.9$ N m.

$$T = T_1 + T_2 = 223.4\ \text{N m}$$

(b) Torque to lower

$$T_{l} = W\tan(\varphi-\alpha)\frac{d_m}{2} + T_2 = 25000\times\tan(3.674^\circ)\times0.023 + 121.9 = 36.9 + 121.9 = 158.8\ \text{N m}$$

(c) Efficiency and self-locking

Efficiency of the screw alone: $\eta_s = \dfrac{\tan\alpha}{\tan(\alpha+\varphi)} = \dfrac{0.05536}{0.17657} = 31.4\%$.

Overall efficiency including collar:

$$\eta = \frac{W L}{2\pi T} = \frac{25000\times0.008}{2\pi\times223.4} = 14.2\%$$

Since $\varphi = 6.84^\circ > \alpha = 3.17^\circ$, a positive torque is needed to lower the load: the screw is self-locking.

(d) Stresses in the screw

Direct compressive stress on core:

$$\sigma_c = \frac{4W}{\pi d_c^2} = \frac{4\times25000}{\pi\times42^2} = 18.0\ \text{MPa}$$

Torsional shear (thread torque $T_1$ acts on the screw core):

$$\tau = \frac{16T_1}{\pi d_c^3} = \frac{16\times101500}{\pi\times42^3} = 7.0\ \text{MPa}$$

Maximum shear $= \tfrac12\sqrt{18.0^2 + 4\times7.0^2} = 11.4$ MPa (small, safe). Buckling is checked as a column if the unsupported length is large.

(e) Nut

The bearing area of the threads is the projected annulus $\dfrac{\pi}{4}(d^2-d_c^2)$ per turn, so:

$$n = \frac{4W}{\pi(d^2-d_c^2)\,p_b} = \frac{4\times25000}{\pi(2500-1764)\times8} = 5.4 \;\Rightarrow\; n = 6$$

Nut height $H = n\,p = 6\times8 = 48$ mm. Shear stress in nut thread (thickness $t = p/2 = 4$ mm):

$$\tau_n = \frac{W}{\pi d\,t\,n} = \frac{25000}{\pi\times50\times4\times6} = 6.6\ \text{MPa}$$

(the nut thread shears at the major diameter; the stress is low for a bronze nut.)

Answer: $T_{raise} = 223$ N m; $T_{lower} = 159$ N m; $\eta = 14.2\%$ overall; self-locking; $\sigma_c = 18$ MPa, $\tau = 7$ MPa; nut with 6 threads, height 48 mm.