A helical compression spring, made of circular wire, subjected to an axial force, which varies from 3.5 kN to 4.5 kN. Over this range of force, the deflection of the spring should be approximately 5 mm. The spring index can be taken as 5. The spring has square and ground ends. Take ultimate tensile strength as 1000 MPa and modulus of rigidity of 80000 MPa. The permissible shear stress for the spring wire should be half of the ultimate tensile strength.

Calculate

(I) Wire diameter

(II) Mean coil diameter

(III) Number of active coils

(IV) Total number of coils

(V) Solid length of the spring

(VI) Free length of the spring

(VIII) Required spring rate and

(VIII) Actual spring rate.

Subject :- Machine Design -I

Topic :- Spring

Difficulty :- Medium

1. Wire diameter:
   Let,

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}=\frac{4*5-1}{4*4-4}+\frac{0.615}{5}=1.3105$

$\tau=K\frac{8PC}{\pi d^2}=1.3105\frac{8*4500*5}{\pi d^2}=12.25=13mm$

1. Mean coil diameter:

$D=Cd=5*13=65mm$

1. Number of active coils:

$\delta=\frac{8PD^3N}{Gd^4}$

$5= \frac{8(4500-3500)56^3N}{80000*13^4}$

$N=8.13=9 coils$

1. Total number of coils:

For square and round ends, the total number of inactive coils is 2. Therefore,

$N_t$=N+2=9+2=11 coils

1. Solid length of the spring:

$N_1$d=1113=143mm

1. Free length of the spring:

Actual deflection of the spring under the maximum force of 4.5kN is given by,

$\delta=\frac{8PD^3N}{Gd^4}=\frac{8*4500*65^3*9}{80000*13^4}=38.94mm$

Assumed that there will be a gap of 0.5mm between the consecutive coils when the spring is subjected to the maximum force of 4.5kN. The total number of coils is 9. Therefore, total axial gap will be (9-1)*0.5=4mm

Free length =solid length+total axial gap+$\delta$ =143+4+38.94=185.94mm=186mm

1. Required spring rate:

$K=\frac{P1-P2}{\delta}=\frac{4500-3500}{38.94}=25.68 N/mm$

1. Actual spring rate:

$K=\frac{Gd^4}{8D^3N}=\frac{80000*13^4}{8*65^3*7}=148.57 N/mm$