**Terminology for helical spring:**

Length (LS): When spring is compressed until touch each other then length of spring is called solid length.

Free Length : The length of spring in a free unloaded condition is called free length spring stiffness/spring constant/spring rate(q). It is defined as load required per unit deflection spring.

q = load required = F/8

Pitch (P) : It is the axial dist bet adjacent at uncompressed condition.

Spring Index (C) : c = D/d

Factor of safety: For fluctuating loads F.O.S is used in between 1.5 to 2.

if helix angle is less than 10 it is cnsidered close coil spring.

if helix angle is more than it is considered open coil spring

**Helical Springs Design**

**Q.1. A closed coil helical spring is subjected to a load varying from 400 N to 1200 N continuouslly at a frequency 15 cycles/sec. The spring stiffness is 25 N for spring material wire.**

$S_{ut} = \frac{1735}{dw^{0.1}}$

$S_{ys} = \frac{875}{dw^{0.1}}$

$S_{no} = \frac{550}{dw^{0.1}}$

$G = 80000 N/mm^2$

making suitable assumption design spring considering fatigue. Draw dimensional sketch check for natural frequency of vibration.

$\sigma_{ut} = S_{ut} = ultimate \ tensile \ stress$

$\sigma_{ys} = \tau_y = torsional \ yeild \ stress$

$\sigma_{no} = \tau_o = endurance\ shear\ stress$

G = modulus of rigidity

$P_{min} = 400 N$

$P_{max} = 1200 N$

Freq applied = 15 cycles/sec = 15 Hz

$q = 25 N$

Amplitude load,

$P_a = \frac{P_{max} - P_{min}}{2} = 400 N$

Mean load

$P_a = \frac{P_{max} + P_{min}}{2} = 800 N$

Assuming spring index c = 6

wahls stress factor ... PSG 7.100

$K_s = \frac{4c-1}{4c-4}+ \frac{0.615}{c}$

$K_s = 1.2525$

$K_c = 1.15$ ...... PSG 7.102

$K_s = K_c \times K_{sh}$

$K_{sh} = 1.089130$

Amplitude shear stress ...... PSG 7.102

$t_a = \frac{8K_s P_a c}{\pi d^2} = \frac{765.71}{d^2} N/mm^2$

Mean shear stress ...... PSG 7.102

$\tau_m = \frac{8.K_{sh} \times P_m C}{\pi d^2} = \frac{13.2}{d^2}$

$\frac{1}{n} = \frac{6.4655}{d^{1.9}} + \frac{27.833}{d^{1.85}}$ ..... PSG 7.102

By assuming valve of d let us FDE which must be in bet 1.5 to 2 for fluct loads

$d = 5 mm \ N = 0.81$

$d = 7 mm \ N = 0.92$

$d = 9 mm \ N = 1.732$

calculation of mean coil dia (D)

$c = D/d$

$E = D/9$

$D = 54 mm$

Determination of no of active turns or coil

**PSG 7.100**

$q = \frac{Gd}{8c^3n}$

$n = 17$

Assuming square and ground end condition no of actual coils for total coil,

$n' = n+ 2$

Determination of Pitch (P)

assuming helix angle $\alpha = 4$

tan$\alpha = \frac{Pitch}{\pi D}$

Pitch = 11.36 mm

Determination of free length (LF)

For square and ground end condition

PSG(7.101)

LF = pn + 2d = 219.62 mm

calculation of solid length (LS)

For square and ground end condition

**PSG 7.101**

LS = dn + 2d = 171 mm

Determination of max stress developed in spring max stress developed in spring due to max PSG 7.100

$\tau = \frac{8K_s P.C}{\pi d^2}$

$\tau = 283.50 N/mm^2$

The design stress from PSG 1.102 for average service

$\tau_d = 0.324\times \sigma_u$

$\tau_d = 451.25 N/mm^2$ $\gt \tau =283.50 N/mm^2$

spring is safe

calculation of max and min deflection PSG 7.100

$y_{max} = \frac{8PC^3n}{Gd} = 48.90 mm$

$y_{min} = \frac{8PC^3n}{Gd} = 16.32 mm$

checking of spring in surge PSG 7.101

lowest natural frequency

f = $\frac{d}{\pi D^2 n}* (\frac{Gg}{8q})^{1/2}$

g = gravitational constant $= 9.81 m/sec^2$

v = specific weight of spring

Assuming spring material as $C = 50$

$v = 0.0783 N$

= 0.0783*$10^6 N/mm^3$

$d = 0.09 m$

$D = 0.054 m$

$G = 80000 N/mm^2$

On substituting all the values, we get

$f = 15 Hz$

Applied frequency is less than natural frequency hence spring is safe under surge or resonance.

**check for buckling**

To avoid buckling

$\frac{LF}{P} \lt 3$ .... PSG 7.101

$\frac{219.62}{54} = 4.06 \gt 3$

Spring must be suitably guided.

**Q.2. A helical compression spring is subjected to the load varying from 0 to 9 kN, with a corresponding deflection of 90 mm. Consider the average service and a spring of steel material having ultimate tensile strength $\sigma_u = \frac{2000}{d^{0.16}}N/mm^2$**

Determine 1. Wire diameter 2. Total number of coil 3. Free length, pitch and helix angle 4. Check for solid stress

$Take \ G= 80000 N/mm^2$

**Solution:**

**Material selection**

Assume oil harderned and tempered prime carbon steel,

$S_{no} = \tau_o = 0.32 su = \frac{440}{d^{0.16}} N/mm^2$

$S_{ys} = \tau_y = 0.45\times \sigma_u = \frac{900}{d^{0.16}} N/mm^2$

$P_a = \frac{P_{max} - P_{min}}{2} = \frac{9}{2} = 4.5 KN$

$P_n = \frac{P_{max} + P_{min}}{2} = \frac{9}{2} = 4.5 KN$

Assume $c =6$

whals factor,

$K_s = \frac{4c -1}{4c - 4} + \frac{0.615}{c}$ .... PSG 7.100

= 1.2525

$K_c = curvature \ factor = 1.15$... PSG 7.102

$K_s = K_{sh}*K_c$

$K_{sh} = 1.0891$

$\tau_a = \frac{8K_sP_aC}{\pi d^2} = \frac{8615.5}{d^2}$ ... PSG 7.102

$\tau_m = \frac{8K_{sh}P_mC}{\pi d^2} = \frac{74881.00}{d^2}$ ... PSG 7.102

Factor of the safety

$\frac{1}{n} = \frac{\tau_n - \tau_a}{\tau_y} + \frac{2\tau_a}{\tau_o}$ ... PSG 7.102

Assume $n = 1.5,$

$d = 31.41$

$d = 32$

As above diameter is not available in PSG

The spring is manufactured for special purpose

$D = c\times d$

$= 172\ mm$

$q = \frac{P_{max}}{\delta} = 100 N/mm$

$q = \frac{Gd}{8 c^3 n}$

$n = 14.81$

take 15

$n^{'} = n + 2= 17$

$LS = dn + 2d$

$LS = 544 mm$

$LF\gt(544+90) = 634 mm$

Assume $LF = 644 mm$

$P = \frac{LF - 2d}{n} = 38.66 mm$

$tan\alpha = \frac{Pitch}{\pi D}$

$tan\alpha =\frac{38.66}{\pi \times 172}$

on solving,

$\alpha= 3.66^\circ$

Check for the solid stress

$Y_{solid} = L_F -L_S = 100 mm$

Let $P_{solid}$ be the force sequre to spring compress fully

$Y_{solid}=\frac{8P_sc^3n}{Gd}$......... PSG 7.100

$P_{solid} = \frac{Y_{soild} \times G \times d}{8 \times c^3 \times n}$

$P_{solid}= 9876.54 N$

$\tau_{solid} = \frac{K_s\times 8 \times P_{solid}\times c}{\pi d^2}$ ... PSG 7.100

$\tau_{solid} = \frac{1.2525 \times 8 \times 9876.54 \times 6}{\pi \times 32^2 }$

$\tau_{solid} = 184.57 N/mm^2$

For carbon steel wire **PSG 7.101**

$\tau_{solid}\lt0.5 \sigma_u$

Permissible stress $= 0.5 \times \sigma_u$

$= 0.5 \times \frac{2000}{d^{0.16}}$

$= 0.5 \times \frac{2000}{32^{0.16}}$

$= 574.34 N/mm^2$

As $\tau_{solid}\lt0.5 \sigma_u$

Therefore, spring is safe.