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.

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

lhabdulsamirahmed lhabdulsamirahmed · Answer 2 · 2022-01-08T05:32:03+01:00

The spring rate is of the spring is 3.33N/mm, the minimum hole diameter required for the spring to operate in is 44mm while the number of coil needed is 11.6 or approximately 12 and the solid length of the spring is 50.4mm

Given data:

d=4mm
C=10mm; D/d = mean coil diameter / wire diameter = 10; D = 10 * 4 = 40mm.
Free length of the spring = 80mm
Force (f)= 50N
δ=15mm

a) Spring rate:

[tex]k = f/[/tex]δ = [tex]50/15=3.33N/mm[/tex]

b) Minimum hole diameter;

The minimum hole diameter(D[tex]_i[/tex]) is the sum of diameter of the wire + D

[tex]D_i=40+4=44mm[/tex]

c) Total number of coils needed:

To solve this, we need to use a constant known as the modulus of rigidity and it's given as G = [tex]77.2*10^3N/mm[/tex]

From δ=[tex]\frac{8fD^3N}{Gd^4}[/tex]

Making N the subject, we would have

[tex]N=\frac{15*77.2*10^3*4^4}{8*50*40^3}=11.58[/tex]≈11.6 coils

The number of coils needed is 11.6 coils or approximately 12 coils.

d) The solid length;

The solid length formula is given as

[tex]L_s=(N+1)d=(11.6+1)*4=50.4mm[/tex]

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