Power screw

Notation: $W$ = axial load, $d_m$ = mean diameter, $p$ = pitch (lead $L$ for single start), $\alpha$ = helix angle $= \tan^{-1}\dfrac{L}{\pi d_m}$, $\mu = \tan\varphi$ = thread friction coefficient, $\varphi$ = friction angle.

One turn of the thread unwrapped is an inclined plane of angle $\alpha$ carrying the load $W$; the nut is pushed by a horizontal force $P$ at the mean radius.

                    /|
       P ------>   / |  W (load)
          ______  /  |
                 /   |
        reaction R at friction angle

Raising the load

Friction opposes the motion and acts down the plane. Resolving forces on the load:

$$\begin{aligned} P &= R\sin(\alpha+\varphi) \\ W &= R\cos(\alpha+\varphi) \end{aligned}$$ $$P = W\tan(\alpha+\varphi) = W\,\frac{\tan\alpha + \tan\varphi}{1-\tan\alpha\tan\varphi}$$

Torque to raise:

$$T_r = P\,\frac{d_m}{2} = W\,\frac{d_m}{2}\tan(\alpha+\varphi)$$

Lowering the load

Now friction acts up the plane:

$$P = W\tan(\varphi - \alpha), \qquad T_l = W\,\frac{d_m}{2}\tan(\varphi-\alpha)$$

If $\varphi > \alpha$, $P$ is positive: an effort is needed to lower the load. If $\varphi < \alpha$, $P$ is negative: the load falls by itself (overhauls).

Self-locking

A screw is self-locking (holds the load without a brake) if

$$\varphi \ge \alpha \quad (\mu \ge \tan\alpha)$$

Efficiency

Without friction, $\varphi = 0$ and the torque is $T_0 = W\dfrac{d_m}{2}\tan\alpha$. Therefore

$$\eta = \frac{T_0}{T_r} = \frac{\tan\alpha}{\tan(\alpha+\varphi)}$$

Efficiency is $\dfrac{\text{useful work}}{\text{input}} = \dfrac{W L}{2\pi T_r}$. For a self-locking screw, $\alpha \le \varphi$ gives $\eta \le \dfrac{\tan\varphi}{\tan2\varphi} = \dfrac{1-\tan^2\varphi}{2} < 50\%$.

Collar friction

If a thrust collar of mean radius $R_c$ and friction coefficient $\mu_c$ is used, the additional torque is $T_c = \mu_c W R_c$, and total torque $T = T_r + T_c$.

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.

Power screws convert rotary motion into linear motion with high force (screw jacks, lathe lead screws, presses, valves).

Thread forms

Standards: metric trapezoidal threads Tr to ISO 2904 / IS 4694 (30 degrees); Acme to ASME B1.5 (29 degrees); buttress to the relevant IS standard or ASME B1.9; square threads follow a non-standard IS pitch table.

Stress concentration

At the thread root there are sharp corners and sudden change of section; stress concentration factors of 2 to 3 appear. Also the load is not shared equally by the threads in the nut: the first threads at the loaded face carry most of the load (often one third in the first thread). Failure of the nut thread or screw begins there and is more dangerous under fatigue and varying load. Remedies: root fillets, larger root radius, rolled threads (compressive residual stress), more uniform load distribution by tapering the nut.

Materials

• Screw: medium-carbon steel (C45, 40C8), alloy steels (e.g. 40Ni3, 40Cr1) for heavy duty, to resist wear and torsion; thread surface hardened or ground.
• Nut: a softer material to protect the costly screw, with low friction coefficient: phosphor bronze, gun metal, cast iron or aluminium bronze.
• Lubrication with oil or grease, and sometimes a second safety nut for jacks.

Maximum efficiency

$$\eta = \frac{\tan\alpha}{\tan(\alpha+\varphi)}$$

Using $\tan(\alpha+\varphi)$ and the trigonometric identity $\eta = \dfrac{\sin\alpha\cos(\alpha+\varphi)}{\cos\alpha\sin(\alpha+\varphi)} = \dfrac{\sin(2\alpha+\varphi)-\sin\varphi}{\sin(2\alpha+\varphi)+\sin\varphi}$.

The expression is largest when $\sin(2\alpha+\varphi)$ is largest, that is $2\alpha+\varphi = 90^\circ$:

$$\alpha_{opt} = 45^\circ - \frac{\varphi}{2}, \qquad \eta_{max} = \frac{1-\sin\varphi}{1+\sin\varphi}$$

Numerical ($\mu = 0.1$)

$\varphi = \tan^{-1}0.1 = 5.71^\circ$, $\sin\varphi = 0.0995$.

$$\alpha_{opt} = 45 - 2.86 = 42.1^\circ, \qquad \eta_{max} = \frac{1-0.0995}{1+0.0995} = 0.819 = 81.9\%$$

Why not used in practice

A helix angle of 42 degrees means a very large lead: the screw is not self-locking ($\alpha > \varphi$), has a small mechanical advantage and cannot hold a load. Practical power screws use $\alpha$ of 2 to 6 degrees (self-locking, large force), accepting efficiencies of 25 to 40%. Self-locking requires $\alpha \le \varphi$.

Acme thread

The normal force on a thread flank is increased by the thread angle, so the effective friction coefficient becomes

$$\mu' = \frac{\mu}{\cos\theta} = \frac{0.12}{\cos14.5^\circ} = 0.124$$

and $\varphi' = \tan^{-1}\mu'$ is used in the torque equation $T = W\dfrac{d_m}{2}\tan(\alpha+\varphi')$. The efficiency is slightly lower than for a square thread of the same lead.

Ball screw

The nut and screw have grooves with recirculating steel balls between them, replacing sliding friction by rolling friction ($\mu \approx 0.005$ to $0.01$).

• Efficiency 90% and above, and low starting torque (no stick-slip).
• Reversible: it is not self-locking, so brakes are required for vertical loads.
• Used in CNC machine feed drives, aircraft actuators and steering gear, where accuracy and efficiency matter more than cost.