Question

A spring made from ASTM A228 music wire with ends squared and ground, wire diameter d = 3 mm, outside diameter D. = 27 mm, 14 total coils, and free length 80 mm has been chosen for an application. Determine the static safety factor if the spring is subjected to a static load of 175 N.

Answer 1

The static safety factor of the spring is 7980.17, which means it can safely handle a load 7980 times greater than the static load of 175 N.

Step-by-step explanation:

The static safety factor of the spring can be calculated by dividing the maximum load the spring can handle by the load it experiences. The maximum load the spring can handle is determined by the tensile strength of the material used to make the spring. In this case, the spring is made from ASTM A228 music wire, which has a tensile strength of 2537 MPa.

The load the spring experiences is the static load of 175 N. To calculate the static safety factor, we divide the maximum load by the load:

Static Safety Factor = Maximum Load / Load

Maximum Load = Tensile Strength * Cross-sectional Area

The cross-sectional area of the spring is determined by its diameter, which is 27 mm. Using the formula for the cross-sectional area of a spring, A = (pi / 4) * (D^2 - d^2), we can calculate the cross-sectional area:

A = (pi / 4) * (27^2 - 3^2) = 550.07 mm^2

Now we can calculate the maximum load:

Maximum Load = 2537 MPa * 550.07 mm^2 = 1,396,730 N

Finally, we can calculate the static safety factor:

Static Safety Factor = 1,396,730 N / 175 N = 7980.17