Question 10.23: A flat spiral spring, used in an electrical instrument, is r...

Given P = 5 N r = 75 mm b = 12.5 mm.

l=750 mm \quad \sigma_{b}=100 N / mm ^{2} .

E = 106 000 N/mm²
Step I Thickness of strip
M = Pr = 5 (75) = 375 N-mm
From Eq. (10.32),

\sigma_{b}=\frac{12 M}{b t^{2}}                 (10.32).

\sigma_{b}=\frac{12 M}{b t^{2}} \quad \text { or } \quad 100=\frac{12(375)}{12.5 t^{2}} .

∴              t = 1.897 or 2 mm               (i)
Step II Degrees of rotation
From Eq. (10.33),

\theta=\frac{12 M l}{E b t^{3}}             (10.33).

\theta=\frac{12 M l}{E b t^{3}}=\frac{12(375)(750)}{(106000)(12.5)(2)^{3}} \text { radians } .

\text { or } \quad \theta=\frac{12(375)(750)}{(106000)(12.5)(2)^{3}}\left(\frac{180}{\pi}\right) \text { degrees } .

\theta=18.24^{\circ}                 (ii).