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

$\text { Given } \quad S_{y t}=0.6 S_{u t} \quad(f s)=2 \quad D=18 mm$.

$M_{b}=250 N – mm \quad E=207000 N / mm ^{2}$.

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

$S_{u t}=2290 N / mm ^{2}$.

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

$C=\frac{D}{d}=\frac{18}{1.4}=12.857$.

$K_{i}=\frac{4 C^{2}-C-1}{4 C(C-1)}=\frac{4(12.857)^{2}-(12.857)-1}{4(12.857)(12.857-1)}$.

= 1.0616.

$\sigma_{b}=K_{i}\left(\frac{32 M_{b}}{\pi d^{3}}\right)=(1.0616)\left(\frac{32(250)}{\pi(1.4)^{3}}\right)$.

$=985.18 N / mm ^{2}$.

$\text { Therefore, }\left(\sigma_{b}\right)>\left(\sigma_{t}\right)$.

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

$S_{u t}=2250 N / mm ^{2}$.

$\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{0.60 S_{u t}}{(f s)}=\frac{0.60(2250)}{(2)}$.

$=675 N / mm ^{2}$.

$C=\frac{D}{d}=\frac{18}{1.6}=11.25$.

$K_{i}=\frac{4 C^{2}-C-1}{4 C(C-1)}=\frac{4(11.25)^{2}-(11.25)-1}{4(11.25)(11.25-1)}$.

=1.071.

$\sigma_{b}=K_{i}\left(\frac{32 M_{b}}{\pi d^{3}}\right)=(1.071)\left(\frac{32(250)}{\pi(1.6)^{3}}\right)$.

$=665.84 N / mm ^{2}$.

$\text { Therefore, }\left(\sigma_{b}\right)<\left(\sigma_{t}\right)$.

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

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

$k=\frac{E d^{4}}{64 D N}$                (10.31).

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

or               N = 393 coils.