Power screws and helical springs

Why it matters

Power screws turn a small torque into a large axial force in screw jacks, presses, vices, machine-tool lead screws and valve stems. Helical springs store energy and control force in valves, clutches, brakes, suspensions and governors. Both are designed from a handful of formulas — torque, efficiency and self-locking for screws; shear stress, deflection and stiffness for springs — that appear constantly in university exams and GATE.

Key ideas

Power screw terms. Pitch p is the axial distance between adjacent threads; lead l is the axial advance per turn, l = n·p for an n-start thread. The mean diameter d_m lies midway between the major and minor (core) diameters; for a square thread d_m = d − p/2 and d_c = d − p. The helix (lead) angle α satisfies tan α = l/(π·d_m). The friction angle φ satisfies tan φ = μ.

Thread forms.

• Square: flanks perpendicular to the axis, highest efficiency, no radial (bursting) force on the nut; but hard to cut and wear of the nut cannot be compensated.
• Acme / trapezoidal (29° / 30° included angle): easier to make, can be used with a split nut to take up wear (lathe lead screws); slightly lower efficiency, because the flank angle raises the effective friction to μ/cos β, β being half the thread angle.
• Buttress: one flank nearly square, for large load in one direction only (vices, gun breeches).
• Ball screws: balls recirculate between screw and nut, giving efficiency above 90 %, but they are not self-locking and need a brake.

Raising and lowering. Unwrapping one turn of thread gives an inclined plane of angle α. Raising a load W is like pushing a block up the plane against friction, so the torque at the mean radius is W·(d_m/2)·tan(φ + α). Lowering needs W·(d_m/2)·tan(φ − α). If φ > α the lowering torque is positive: the load will not run down on its own, and the screw is self-locking. A self-locking square-thread screw always has an efficiency below 50 %.

Collar friction. In a screw jack the load rests on a collar or thrust washer that rubs as the screw turns, adding torque μ_c·W·R_c (R_c mean collar radius, uniform-wear assumption). Collar friction often consumes as much torque as the thread itself and must be included in the overall efficiency.

Screw stresses. The core carries direct compression (or tension) W/(π d_c²/4) together with torsional shear 16T/(π d_c³); combine them with the maximum shear stress theory. Long screws in compression are checked as columns. The nut threads are checked for bearing pressure and shear, which fixes the nut height.

Helical springs. A close-coiled helical spring under axial load is a wire in torsion: the load W at the coil radius D/2 produces a twisting moment W·D/2 on the wire. The spring index C = D/d (usually 4 to 12) measures how sharply the wire is curved. Direct shear and curvature make the inner fibre stress higher than the simple torsion value; the Wahl factor K accounts for both. Stiffness depends on the modulus of rigidity G (not E), on d⁴, and inversely on D³ and the number of active coils.

Arrangements and ends. Springs in series share the same force and add deflections; in parallel they share the same deflection and add forces. For squared-and-ground ends, the total coils are about n + 2 and solid length ≈ (n + 2)·d. Long, slender compression springs can buckle and are guided by a rod or tube.

Formulas

Lead and helix angle: l = n·p; tan α = l/(π·d_m); tan φ = μ.

Torque to raise (square thread): T = W·(d_m/2)·tan(φ + α) = W·(d_m/2)·(μ·π·d_m + l)/(π·d_m − μ·l).

Torque to lower: T = W·(d_m/2)·tan(φ − α); self-locking if φ > α (μ > tan α).

Trapezoidal or Acme thread: use μ' = μ/cos β in place of μ (β half thread angle).

Collar torque: T_c = μ_c·W·R_c.

Efficiency: η = W·l/(2π·T_total); thread only, square: η = tan α/tan(α + φ); maximum η_max = (1 − sin φ)/(1 + sin φ).

Core stresses: σ = 4W/(π·d_c²); τ = 16T/(π·d_c³).

W in N, T in N·m (or N·mm with lengths in mm), d_m, d_c, l, p in m or mm consistently.

Spring index: C = D/d. Wahl factor: K = (4C − 1)/(4C − 4) + 0.615/C.

Maximum shear stress: τ = K·8W·D/(π·d³) (N/mm² with W in N, D and d in mm).

Deflection: δ = 8W·D³·n/(G·d⁴); stiffness k = W/δ = G·d⁴/(8D³·n); n active coils, G modulus of rigidity.

Energy stored: U = ½·W·δ. Series: 1/k = 1/k₁ + 1/k₂; parallel: k = k₁ + k₂.

Worked examples

Example 1 (standard). A screw jack has a single-start square thread, nominal diameter 50 mm and pitch 8 mm. It lifts 20 kN. Thread friction μ = 0.1; collar friction μ_c = 0.12 at a mean collar radius of 30 mm. Find the torque to raise the load, the efficiency and whether the screw is self-locking.

1. d_m = d − p/2 = 50 − 4 = 46 mm; l = p = 8 mm.
2. tan α = l/(π·d_m) = 8/(π × 46) = 0.05536, α = 3.17°; φ = tan⁻¹0.1 = 5.71°.
3. T_thread = W·(d_m/2)·tan(φ + α) = 20 000 × 23 × tan 8.88° = 71 860 N·mm = 71.9 N·m.
4. T_c = μ_c·W·R_c = 0.12 × 20 000 × 30 = 72 000 N·mm = 72.0 N·m.
5. T_total = 143.9 N·m.
6. η = W·l/(2π·T_total) = 20 000 × 8/(2π × 143 860) = 0.177; thread alone η = tan α/tan(α + φ) = 0.354.
7. T ≈ 143.9 N·m, overall η ≈ 17.7 %; φ = 5.71° > α = 3.17°, so the screw is self-locking.

Example 2 (GATE level). A close-coiled helical compression spring has wire diameter 8 mm, mean coil diameter 48 mm, 10 active coils and G = 80 GPa. It carries 600 N. Find the maximum shear stress (with Wahl factor), deflection, stiffness and energy stored. What is the stiffness if it is put in series with a spring twice as stiff?

1. C = D/d = 48/8 = 6.
2. K = (4 × 6 − 1)/(4 × 6 − 4) + 0.615/6 = 23/20 + 0.1025 = 1.2525.
3. τ = K·8W·D/(π·d³) = 1.2525 × 8 × 600 × 48/(π × 512) = 1.2525 × 143.24 = 179.4 N/mm².
4. δ = 8W·D³·n/(G·d⁴) = 8 × 600 × 48³ × 10/(80 000 × 8⁴) = 16.2 mm (G = 80 000 N/mm²).
5. k = 600/16.2 = 37.04 N/mm; U = ½ × 600 × 16.2 = 4860 N·mm = 4.86 J.
6. Series with 74.07 N/mm: k = 1/(1/37.04 + 1/74.07) = 24.69 N/mm.
7. τ_max ≈ 179 MPa, δ = 16.2 mm, k ≈ 37.0 N/mm, U ≈ 4.86 J, series stiffness ≈ 24.7 N/mm.

Common mistakes

• Using the pitch instead of the lead for a multi-start thread.
• Using the nominal diameter instead of the mean diameter in tan α and the torque, or the mean diameter instead of the core diameter in the stress.
• Forgetting the collar friction torque in a screw jack.
• Writing the self-locking condition the wrong way round (it is φ > α).
• Using E instead of G in the spring stiffness, or forgetting to convert GPa to N/mm² (80 GPa = 80 000 N/mm²).
• Using total coils instead of active coils for deflection.
• Adding stiffnesses of springs in series (it is the compliances that add).
• Omitting the Wahl factor when the question asks for the maximum shear stress.

For GATE PI

Expect torque to raise or lower a load, efficiency and self-locking, maximum efficiency of a square thread, the effect of a multi-start thread on lead, spring stiffness and deflection, Wahl-corrected shear stress, springs in series and parallel, and energy stored. Practise keeping N·mm and N·m apart and checking φ against α.

Quick check

1. A double-start screw has pitch 5 mm. What is its lead?
2. State the condition for a square-thread screw to be self-locking.
3. What is the stiffness of a spring with d = 10 mm, D = 100 mm, n = 10 and G = 80 GPa?
4. Two identical springs of stiffness k are in series. What is the combined stiffness?
5. Why is a self-locking square-thread screw less than 50 % efficient?

Answers: 1. 10 mm; 2. φ > α, i.e. μ > tan α; 3. 10 N/mm (10 000 N/m); 4. k/2; 5. because with φ ≥ α, η = tan α/tan(α + φ) ≤ tan α/tan 2α, which is below 0.5.