Page: Helical Spring Terminology
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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

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

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$LS = dn + 2d$

$LS = 544 mm$

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

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.

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modified 23 days ago by gravatar for Sanket Shingote Sanket Shingote250 written 3 months ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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