Inclined Plane, Screw & Wedge

Load Standard Screw: a 3 mm pitch on a 10 mm radius, 1,000 N of load, μ = 0.15. The panel says an ideal advantage of 20.94 and an efficiency of 24.0%. Both come from the same triangle. Find it.

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1. Take one full turn of the thread and unroll it. The thread climbs by the pitch while it travels once round the shaft, so it is the hypotenuse of a right triangle whose height is the pitch and whose base is the circumference. A screw is an inclined plane that has been wrapped round a cylinder, and this is that sentence made arithmetic.

   $$\text{one turn unrolled: rise }p=3\mathrm{mm},\text{base }2\pi r=62.83\mathrm{mm}$$

2. The slope of that triangle is the helix angle. Under three degrees, which is why a screw feels nothing like a ramp even though it is one.

   $$\lambda =\arctan \frac{p}{2\pi r}=\arctan \frac{3}{62.83}=2.734^{\circ }$$

3. Ideal advantage is the distance you move over the distance the load moves: once round the circumference to advance one pitch.

   $${\mathrm{MA}}_{\text{ideal}}=\frac{2\pi r}{p}=\frac{62.83}{3}=20.94$$

4. Friction enters as an angle of its own. The friction angle is the slope at which a surface just begins to let go, and adding it to the helix angle is the standard way to fold friction into a screw.

   $$\phi =\arctan \mu =\arctan 0.15=8.531^{\circ }$$

5. The effort follows directly. Rather than 1,000 divided by 20.94, which is 47.8 N, you need four times that.

   $$E=L\tan (\lambda +\phi )=1000\tan (11.265^{\circ })=199.2\mathrm{N}$$

6. The advantage you actually get is the load divided by that effort, 1,000/199.2 = 5.02. So the efficiency is the ratio of what you got to what the geometry offered.

   $$\eta =\frac{{\mathrm{MA}}_{\text{actual}}}{{\mathrm{MA}}_{\text{ideal}}}=\frac{5.02}{20.94}=24.0\%$$

7. And the same comparison, written with a tangent instead of an angle, tells you whether the screw stays put with no hand on it.

   $$\tan \lambda =0.0477\le \mu =0.15\Rightarrow \text{self-locking}$$