Question

A helical compression spring is to be made with oil-tempered wire ol 4 mm diameter with a spring index of C = 10 The spring is to operate inside a hole, so buckling is not o problem and the ends can be left plan The free length of the spring should be 80 mm. A force of 50 N should deflect the spring 15 mm (a) Determine the spring rate (b) Determine the minimum hole diameter tot the spring to operate in (c) Determine the total number of coils needed (d) Determine the solid length.

Answer 1

Answer:

a) the spring rate is 3.333 N/mm

b) the minimum hole diameter for the compression spring is 44 mm

c) the total number of coils needed is 11.6

d) the solid length is 50.4 mm

Explanation:

a)

to calculate the mean spring coil diameter, we take a look at the expression from the relation;

D = Cd

where C is the spring index ( 10 ) and d is the diameter of helical compression spring (4 mm)

so we substitute

D = 10 × 4 = 40 mm

Torsional stiffness G for the tempered wire with diameter 4 mm is 77.2 Gpa ( 77.2 × 10³ Mpa)  ( obtained from Table: Mechanical properties of spring wires).

so when the spring is compressed, the spring force is given by the following expression(realtion)

Fs = k × ys

where ys is the deflection of the spring (15 mm) and k is the spring rate, Fs is the force (50N)

so we substitute

50N = k × 15mm

k = 50N / 15mm

k = 3.333 N/mm

∴  the spring rate is 3.333 N/mm

b)

to calculate the minimum hole diameter for the compression spring

Now the entire spring is within a hole in the  ground, therefore the hole should have a diameter equal to the outer diameter of the spring.

so D₀ = D + d

and from our initial equations, the mean spring coil diameter D = 40mm and the diameter of the helical compression spring d = 4mm

we substitute

D₀ = 40 + 4

D₀ = 44 mm

the minimum hole diameter for the compression spring is 44 mm

c)

Consider the following relation to calculate the total number of coils needed

Na coils are actually working to support the springs structure and its all dependent on the cut at the edge (end). ( from the table, Nt elates to Na)

Na = (d⁴G) / 8D³k

where the mean spring coil diameter D = 40mm and the diameter of the helical compression spring d = 4mm, G is the torsional stiffness  (77.2 × 10³ Mpa), the the spring rate k is 3.333 N/mm

so we substitute

Na = (4⁴(77.2 × 10³)) / ( 8(40³)(3.333))

Na = 19,763,200 / 1,706,496

Na = 11.6

the total number of coils needed is 11.6

d)

As the number of active coils and total number of coils are the same, we get the following relation;

Na = Nt

Nt which is also total number of coils

Now to calculate the solid length

Ls = d ( Nt + 1 )

so we substitute

Ls = 4 ( 11.6 + 1 )

Ls = 50.4 mm

the solid length is 50.4 mm