Question

Determine the factor of safety at closure (ns) for the helical compression spring made from A228 music wire (G = 82.7 GPa) of diameter d 3 mm with an inner diameter of ID = 22 mm, a free length of Lo 68 mm, and 17 total coils. The ends of the spring are plain and ground and the torsional yield strength of the wire has been estimated to be Ssy = 848 MPa.

Answer 1

220.1

Step-by-step explanation:

To determine the factor of safety at closure, we need to find the maximum load the spring can support without exceeding its yield strength.

First, let's find the shear modulus of the A228 music wire:

G = 82.7 GPa

Next, we can use the formula for the maximum load of a helical compression spring:

Pmax = (π/8)Gd^4(n/ID + 1)(1 + 2c/d)

where:

d = wire diameter = 3 mm

n = total number of coils = 17

ID = inner diameter = 22 mm

c = spring index = D/d, where D is the mean coil diameter

D = (ID + d)/2

Lo = free length = 68 mm

We can calculate the spring index and mean coil diameter:

c = D/d = (ID + d)/2d = (22 + 3)/(2 × 3) = 4.17

D = (ID + d)/2 = (22 + 3)/2 = 12.5 mm

Then, we can plug in the values and simplify:

Pmax = (π/8)Gd^4(n/ID + 1)(1 + 2c/d)

= (π/8)(82.7 × 10^9)(3 × 10^-3)^4(17/22 + 1)(1 + 2(4.17)/3)

≈ 3,858.4 N

Finally, we can calculate the factor of safety at closure:

ns = Ssy/Pmax

= 848 × 10^6/3,858.4

≈ 220.1

Therefore, the factor of safety at closure for the given helical compression spring is approximately 220.1.