---
title: "Torque and efficiency of a power screw"
subject: "Machine Design II"
chapter: 8
key: "semester_7/machine_design_ii:power-screw-torque-efficiency-derive"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/power-screw-torque-efficiency-derive
site: BCT Engineering
---

# Torque and efficiency of a power screw

Machine Design II · Chapter 8: Power screw

- Asked in: year not given, 8 marks
- Video: [Power Screws - Torque to Force Relationships in Just Over 10 Minutes!](https://www.youtube.com/watch?v=BstZUC4tcOA) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/power-screw-torque-efficiency-derive
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-8

## Question

Derive the expressions for the torque required to raise and to lower a load with a square-threaded power screw, considering the screw as an inclined plane. Derive the efficiency and the condition for self-locking.

## Short answer

For a square thread, raising torque is (W d_m/2) tan(α+φ), lowering torque is (W d_m/2) tan(φ-α). Efficiency η = tan α / tan(α+φ). Self-locking requires φ ≥ α (μ ≥ tan α). Collar friction adds μ_c W R_c.

## Answer

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.

```text
                    /|
       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 $T = T_r + T_c$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The torque to raise the load is
   A. T = W (d_m/2) tan(alpha - phi)
   B. T = W (d_m/2) tan(alpha + phi)
   C. T = W d_m tan(alpha)
   D. T = W (d_m/2) sin(alpha + phi)
   Answer: B. Friction opposes raising, so the reaction is at alpha + phi from the normal: P = W tan(alpha + phi). (Key point: Torque to raise)

2. If phi < alpha when lowering, then
   A. Effort is needed to lower the load
   B. The screw is self-locking
   C. P is negative and the load falls by itself (overhauls)
   D. The efficiency is 50%
   Answer: C. tan(phi - alpha) is negative when phi < alpha, so the load overhauls. (Key point: Lowering behaviour)

3. The screw efficiency is
   A. tan(alpha + phi)/tan(alpha)
   B. tan(alpha)/tan(alpha + phi)
   C. tan(phi)/tan(alpha + phi)
   D. sin(alpha)/sin(alpha + phi)
   Answer: B. It is the ratio of torque without friction T_0 to torque with friction T_r. (Key point: Efficiency)

4. A screw is self-locking if
   A. phi >= alpha (mu >= tan alpha), and its efficiency is then below 50%
   B. phi <= alpha, with efficiency above 50%
   C. alpha = 45 degrees
   D. phi = 0
   Answer: A. A self-locking screw has phi >= alpha, giving efficiency at most (1 - tan^2 phi)/2, less than 50%. (Key point: Self-locking condition)

## Related in Machine Design II

- [Screw jack with square thread](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Ascrew-jack-square-thread-numerical) — Machine Design II, chapter 8
- [Power screw threads, standards and materials](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Ascrew-thread-types-standards-stress) — Machine Design II, chapter 8
- [Maximum efficiency and ball screws](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aself-locking-maximum-efficiency-ball-screw) — Machine Design II, chapter 8

## Related in other subjects

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- [200 V shunt motor power, torque and efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-200v-rotational-loss) — Electrical Machines I, chapter 4
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