## Helical Spring

An applied force of 150 Newtons acts on a helical spring, with a spring diameter of 0.5 cm. Find the shear stress (Pascals) if the wire diameter is 0.2 cm.

Hint
The shear stress in a helical linear spring is:
$$\tau =\frac{K_{spring}(8)\cdot Force\cdot D_{spring}}{\pi d_{wire}^{3}}$$$where $$K_{spring}$$ is the spring constant, $$D_{spring}$$ is the mean spring diameter, and $$d_{wire}$$ is the wire diameter. Hint 2 To find the spring constant: $$K_{spring}=\frac{(2C+1)}{2C}$$$
where $$C=D/d$$
The shear stress in a helical linear spring is:
$$\tau =\frac{K_{spring}(8)\cdot Force\cdot D_{spring}}{\pi d_{wire}^{3}}$$$where $$K_{spring}$$ is the spring constant, $$D_{spring}$$ is the mean spring diameter, and $$d_{wire}$$ is the wire diameter. Spring Constant Equation: $$K_{spring}=\frac{(2C+1)}{2C}$$$
where $$C=D/d$$ :
$$C=\frac{.005\:m}{.002\:m}=2.5$$$Thus, the spring constant is: $$K_{spring}=\frac{(2(2.5)+1)}{2(2.5)}=\frac{6}{5}=1.2$$$
Finally solving for Shear Stress:
$$\tau =\frac{1.2(8)(150N) (0.005m)}{\pi (0.002m)^{3}}$$$$$\tau =\frac{7.2N}{(2.512\cdot 10^{-8})m^2}=2.87\cdot 10^{8}\:Pa=287\:MPa$$$
287 MPa