It is required to design a helical torsion spring for a window shade. The spring is made of patented and cold-drawn steel wire of Grade-4. The yield strength of the material is 60% of the ultimate tensile strength and the factor of safety is 2. From space considerations, the mean coil diameter is

Question

It is required to design a helical torsion spring for a window shade. The spring is made of patented and cold-drawn steel wire of Grade-4. The yield strength of the material is 60% of the ultimate tensile strength and the factor of safety is 2. From space considerations, the mean coil diameter is kept as 18 mm. The maximum bending moment acting on the spring is 250 N-mm.
The modulus of elasticity of the spring material is 207 000 N/mm². The stiffness of the spring should be 3 N-mm/rad. Determine the wire diameter and the number of active coils.

Accepted Answer

[latex] ext { Given } \quad S_{y t}=0.6 S_{u t} \quad(f s)=2 \quad D=18 mm [/latex].

[latex] M_{b}=250 N - mm \quad E=207000 N / mm ^{2} [/latex].

k = 3 N-mm/rad
Step I Wire diameter
The wire diameter is calculated by the trial and error method.
Trial 1 d = 1.4 mm
From Table 10.1,

Table 10.1 Mechanical properties of patented and cold-drawn steel wires

Minimum tensile strength (N/mm²)				Wire diameter d (mm)
Gr.4	Gr.3	Gr.2	Gr.1	Wire diameter d (mm)
2660	2460	2060	1720	0.3
2620	2430	2040	1700	0.4
2580	2390	2010	1670	0.5
2550	2360	1990	1650	0.6
2530	2320	1970	1630	0.7
2480	2280	1950	1610	0.8
2440	2250	1920	1590	0.9
2400	2240	1900	1570	1.0
2340	2170	1860	1540	1.2
2290	2090	1820	1500	1.4
2250	2080	1780	14700	1.6
2190	2030	1750	1440	1.8
2160	1990	1720	1420	2.0
2050	1890	1640	1370	2.5
1980	1830	1570	1320	3.0
1890	1750	1510	1270	3.6
1840	1700	1480	1250	4.0
1800	1660	1440	1230	4.5
1750	1600	1390	1190	5.0
1670	1530	1320	1130	6.0
1610	1460	1260	1090	7.0
1540	1400	1220	1050	8.0

[latex] S_{u t}=2290 N / mm ^{2} [/latex].

[latex] ext { The permissible stress }\left(\sigma_{t} ight) ext { is given by, } [/latex]

[latex] \sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{0.60 S_{u t}}{(f s)}=\frac{0.60(2290)}{(2)}=687 N / mm ^{2} [/latex].

[latex] C=\frac{D}{d}=\frac{18}{1.4}=12.857 [/latex].

[latex] K_{i}=\frac{4 C^{2}-C-1}{4 C(C-1)}=\frac{4(12.857)^{2}-(12.857)-1}{4(12.857)(12.857-1)} [/latex].

= 1.0616.

[latex] \sigma_{b}=K_{i}\left(\frac{32 M_{b}}{\pi d^{3}} ight)=(1.0616)\left(\frac{32(250)}{\pi(1.4)^{3}} ight) [/latex].

[latex] =985.18 N / mm ^{2} [/latex].

[latex] ext { Therefore, }\left(\sigma_{b} ight)>\left(\sigma_{t} ight) [/latex].

The design is not safe.
Trial 2 d = 1.6 mm
From Table 10.1,

[latex] S_{u t}=2250 N / mm ^{2} [/latex].

[latex] ext { The permissible stress }\left(\sigma_{t} ight) ext { is given by, } [/latex]

[latex] \sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{0.60 S_{u t}}{(f s)}=\frac{0.60(2250)}{(2)} [/latex].

[latex] =675 N / mm ^{2} [/latex].

[latex] C=\frac{D}{d}=\frac{18}{1.6}=11.25 [/latex].

[latex] K_{i}=\frac{4 C^{2}-C-1}{4 C(C-1)}=\frac{4(11.25)^{2}-(11.25)-1}{4(11.25)(11.25-1)} [/latex].

=1.071.

[latex] \sigma_{b}=K_{i}\left(\frac{32 M_{b}}{\pi d^{3}} ight)=(1.071)\left(\frac{32(250)}{\pi(1.6)^{3}} ight) [/latex].

[latex] =665.84 N / mm ^{2} [/latex].

[latex] ext { Therefore, }\left(\sigma_{b} ight)<\left(\sigma_{t} ight) [/latex].

The design is satisfactory and the wire diameter should be 1.6 mm.

Step II Number of active coils
From. (10.31),

[latex] k=\frac{E d^{4}}{64 D N} [/latex] (10.31).

[latex] N=\frac{E d^{4}}{64 D k}=\frac{(207000)(1.6)^{4}}{64(18)(3)}=392.53 [/latex].

or N = 393 coils.

Question 10.21: It is required to design a helical torsion spring for a wind...

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