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Show that the efficiency of a self-locking screw is less than 50 percent.

Subject: Machine Design -I

Difficulty: Medium

md1(66) • 2.8k  views
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The efficiency of a square threaded screw is calculated by,

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

For a self-locking screw,

$\phi \geq \alpha$

Substituting the limiting value for $\phi=\alpha$ in eqn (a)

$\eta \leq\ \frac{tan\alpha}{tan(\phi+\phi}$

$\eta \leq\ \frac{tan\alpha}{tan2 \phi}$

$tan2 \phi= \frac{2tan\phi}{1-tan^2\phi}$

$\eta \leq\ \frac{tan\phi(1-tan^2\phi)}{2tan\phi}$

$\eta \leq\ [\frac{1}{2}-tan^2{\phi}/{2}]$

Therefore, efficiency of a self-locking square threaded power screw is less than 1/2 i.e less than 50%

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