A flat spiral spring, used in an electrical instrument, is required to exert a maximum force of 5 N at the free end against the retaining drum. The line of action of this force is 75 mm from the centre of gravity of the spiral. The spring is made of brass strip (E = 106 000 N/mm^2) and the maximum

Question

A flat spiral spring, used in an electrical instrument, is required to exert a maximum force of 5 N at the free end against the retaining drum. The line of action of this force is 75 mm from the centre of gravity of the spiral. The spring is made of brass strip (E = 106 000 N/mm²) and the maximum bending stress should not exceed 100 N/mm². The width and length of the strip are
12.5 and 750 mm. Calculate
(i) the thickness of strip; and
(ii) the number of degrees of rotation through which the arbor should be turned to produce the required force.

Accepted Answer

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

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

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

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

[latex] \sigma_{b}=\frac{12 M}{b t^{2}} \quad ext { or } \quad 100=\frac{12(375)}{12.5 t^{2}} [/latex].

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

[latex] heta=\frac{12 M l}{E b t^{3}} [/latex] (10.33).

[latex] heta=\frac{12 M l}{E b t^{3}}=\frac{12(375)(750)}{(106000)(12.5)(2)^{3}} ext { radians } [/latex].

[latex] ext { or } \quad heta=\frac{12(375)(750)}{(106000)(12.5)(2)^{3}}\left(\frac{180}{\pi} ight) ext { degrees } [/latex].

[latex] heta=18.24^{\circ} [/latex] (ii).

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

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