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 (MPa) if the wire diameter is 0.2 cm.
Expand 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
Time Analysis
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