## Compression Spring

For a 200 N spring load, calculate the shear stress of a 20 mm diameter helical compression spring that has a 2.5 mm wire diameter.

##
__
__**Hint**

**Hint**

The shear stress in a helical linear spring is

$$$\tau =K_s\frac{8FD}{\pi d^3}$$$

where
$$d$$
is the wire diameter,
$$F$$
is the applied force,
$$D$$
is the mean spring diameter,
$$K_s$$
is the spring constant.

##
__
__**Hint 2**

**Hint 2**

The spring constant:

$$$K_s=(2C+1)/2C$$$

where
$$C=D/d$$
. (Note
$$D$$
is the mean spring diameter, and
$$d$$
is the wire diameter.)

The shear stress in a helical linear spring is

$$$\tau =K_s\frac{8FD}{\pi d^3}$$$

where
$$d$$
is the wire diameter,
$$F$$
is the applied force,
$$D$$
is the mean spring diameter,
$$K_s=(2C+1)/2C$$
is the spring constant, and
$$C=D/d$$

In our scenario:
$$C=20mm/2.5mm=8$$
and
$$K_s=[2(8)+1]/2(8)=1.0625$$

$$$\tau =(1.0625)\frac{8(200N)(0.02m)}{\pi (0.0025m)^3}=(1.0625)\frac{32N\cdot m}{4.9\times 10^{-8}m^3}$$$

$$$=(1.0625)(652\cdot 10^6)N/m^2=693\:MPa$$$

693 MPa