Question

design a compression spring for a safety valve for the following data. Valve operating pressure =1 N/mm2 diameter of the valve seat= 100mm []=500N/mm2 G=0.5×10 5 N/mm2. The spring is kept to be in a casing of 115mm ID. Maximum lift of the spring is 5mm. When the pressure is 1.08N/mm2

Answer 1

Valve operating pressure$p=1\frac{N}{mm^2}$

G=0.5×10 5 N/mm2

D = 115

d =110

$W = 500\times \frac{\pi}{4}100^2 = 3926.99kN$

Spring index, $C = \frac{D}{d} =\frac{115}{110}=1.04$

1. Neglecting the effect of curvature

We know that the shear stress factor,

$Ks = 1 + \frac{1 } { 2C} = 1 + \frac{1 } { 2\times1.04}=1.48$

Maximum shear stress induced in the wire (τ),

$\tau = K_s (\frac{8WD} {πd^3}) =1.48(\frac{8\times 3926.99\times1000\times115}{\pi110^2})=140.66$ Mpa

Deflection per active turn,

$\frac{δ}{n} = \frac{8WD^3 }{Gd^4} =\frac{8\times3926.99\times1000\times115^3 }{0.5\times10^5\times110^4}=6.53$ mm

Number of turns of the coil

n = Number of active turns of the coil. We know that the deflection of the spring (δ),

$6.53= \frac{8WC^3n } {Gd} =\frac{8\times3926.99\times1000\times 1.04^3 n } {0.5\times10^5\times110}$

$n =1.01 = 2$ turns

For a spring having loop on both ends, the total number of turns, $n' = n + 1 = 2+1 =3$

Free length of the spring

 Taking the least gap between the adjacent coils as 1 mm when the spring is in free state, the free length of the tension spring,

$LF = n.d + (n – 1) 1 =3\times110+(3-1)=332$ mm