UNIT 15 SPRINGS

Structure 15.1 Introduction

Objectives

15.2 Close Coiled Helical Springs 15.2.1 Stresses in Springs 15.2.2 StrainEnergy 15.2.3 Deflection of Springs 15.2.4 Stiffness of Springs 15.2.5 Proof Load

15.3 Open Coiled Helical Springs 15.3.1 Spring Subjected to AxialLoad 15.3.2 Spring Subjected to Axial Couple 15.3.3 Spring Subjected to Moment along the Axis of theHelix 15.3.4 Stresses in Springs 15.3.5 Proof Load

1 5.4 Compound Springs

15.4 Leaf Springs 15.4.1 Stresses in Springs

. 15.4.2 Practical Applications

15.5 Summary

15.6 Answers to SAQs

15.1 INTRODUCTION

The primary function of a spring is to deflect or distort under load and to recover its original shape when the load is released. During deflection or distortion, it absorbs energy and release the same as and when required. Springs are used in many engineering applications such as automobiles and railway buffers in order to cushion, absorb or control energy due to shock and vibrations. Springs will suffer a sizeable change in form without being distorted permanently when the loads are applied. Springs are generally classified as leaf springs or helical springs. Leaf springs consist of a number of thin curved plates, each of same thiclcness and width but of different lengths, all bent to the same curvature. Helical springs are formed by coiling thick spring wire into a helix. Helical springs are classified into two groups. When the helix angle is less than about lo", it is named as close-coiled helical spring. In such springs, the wire experiences too little bending or direct shear stress and their effect is neglected. Torsional stresses are predominant in such springs. If, however the helix angle is significant, then the wire experiences both torsional and bending stresses. Such type of spring is termed as open-coiled helical spring.

Objectives After studying this unit, you should be able to

differentiate between close coiled, open coiled and leaf springs,

calculate the stresses in springs,

ci+lculate the deflection, proof load and stiffness of springs, and

identify the areas of application of springs to practical situations.

Definitions

The various term? used in this unit are as follows :

Proof load

It is the greatest load that the spring can carry without getting permanently dis totted.

Proof stress

It is the maximum stress in the spring when subjected to proof load. I

Stresses in Shafts & Shells Proof resilieqce and Thennal Stresw

It is the strain energy stored in the spring when it has been subjected to the maximum load i.e. proof load.

Spring constant (stiffness of the spring)

It is the load per unit deflection. It is expressed in Nlm or kN1m.

15.2 CLOSE COILED HELICAL SPRINGS

A helical spring is a wire wound in spiral form, which can undergo considerable deflection without getting permanently distorted. A helical spring is said to be close-coiled when the obliquity of the wire is small, i.e. the pitch of the coils is very small. Each turn can be regarded as practically lying in planes at right angles to the axis of the helix. Hence, closecoiled helical spring is a torsion spring under axial load it may be assumed that the spring is subjected to torsion only, neglecting the effects of bending and disect shear.

15.2.1 Stresses in Springs When Subjected to Axial Load

Figure 15.1 (a) shows a close-coiled helical spring subjected to axial load W.

Figure 15.1 : Close Coiled Helical Spring (Subjeded to Axial Load)

Let . D = mean coil diameter

d = wire diameter

n = number of coils

1 = lengthof the wire

6 = axial deflection

a = angle of helix

Moment at any point = W x [?I \ 1

The axis of the coil is inclined at an angle a. Resolving the moment, along and at right angles to the axis of the spring,

Vector component along'the axis of the spring = T = M cos cx

Vector component normal to the axis of the spring Mb = M sin a

The component normal to the axis of the spring causes bending and the helix angle a being small, it can be neglected and cos a tends to unity for small angles.

:. The vector component along the axis, Twill be equal to M.

'l'utal length of the spring = (7cD) n \ " I

Total volume of the spring = d2 (7cDn) ! 4 I

using torsion formula, -T = fS J r'

We get, shear stress, f , = ( f ) r

On substituting, we get T = W (9

Springs

Alternatively,

Considering the equilibrium of one part of the spring, the spring is subjected to two types of forces,

(i) due to direct shear force W which is assumed to be uniformly distributed over the cross-section of the wire and

(ii) the moment WR acting in the plane of the section, which will cause tonional stress.

W 4W - :. Direct shear stress = - - - d2 xd2

Torsional shear stress = \ ' I =16WR xd3

Total shear stress, i.e. maximum shear stress

The above equation is only for straight shafts, whereas in the case of springs, %wire is curved hence, the above equation is only approximate.

Whal gave a more exact relationship as

where K is called the Whal's correction factor.

Stresses in Shafts & Shells d and Thermal Stresses When the wire diameter is very small, compared to the mean radius of spring, - being too 4R

small and can be neglected.

When Subjected to Axial Couple

Figure 15.2 (a) shows a close coiled helical spring under an axial couple M.

\

I b l

( 0 )

Figure 15.2 : Close Coiled Helical Spring (Subjected to Axial Couple)

In this case, the component M sin a is negligible, as a is small and, therefore, the spring will be subjected to pure bending as can be seen from Figure 15.2 (b).

1 ~ ~ d s Strain Energy, U = -

0 2EI

Rotation of this spring,

15.2.2 Strain Energy When Subjected to Axial Load

Neglecting strain energy due to direct shear W, strain energy stored,

fs2 U = - x (Volume of the spring) 4G

where G = modulus of rigidity.

15.2.3 Deflection of Springs Let 6 is the axial deflection.

1 Then, work done by the load = 3 (W x 6)

Equating the work done to the strain energy stored in the spring,

15.2.4 Stiffness of Springs The stiffness of the spring is defined as the load required to produce unit deflection.

From Eq. (15.6a), for unit deflection, 6 = 1.

W Gd4 - :. Stiffness of the spring - - - - - 6 ~ R 3 n

This is also termed as spring constant k.

15.2.5 Proof Load Proof load is the maximum load carrying capacity of the spring, without getting permanently distorted.

From Eq. (15.2), we have

Proof Load, nd

Wmax = 8~ X VsImax

Springs

where (f,),, is the allowable shear stress of the material of the spring.

Hence, using Eq. (15.8), you can determine the proof load of the spring.

Example 15.1

A close coiled helical spring is made of 5 mm diameter wire. It is made up of 30 coils, each of mean diameter 75 mm. If the maximum stress in the spring is not to exceed 200 MPa, then determine

(a) the proof load

(b) the extension of the spring when carrying this load.

Take G = 80 GPa.

Solution

Here, we have d = 5 mm n = 30

D = 75mm, R = 37.5 mm

(f,),, = 200 MPa G = 80 GPa

Stresses In Shah & Shells Using Eq. (15.2), we get and Thermal Stresses

Thus, proof load

Deflection

Example 15.2

A helical spring in which the slope of the helix may be assumed small, is required to transmit a maximum pull of 1 kN and to extend 10 mm for 200 N load. If the mean diameter of the coil is to be the 80 mm, find the suitable diameter for the wire and number of coils required. Take G = 80 GPa and allowable shear stress as 100 MPa.

Solutlon

Shear stress, 8WD

fs = - 7cd

Here, we have W= 1OOON D = 8 0 m m

fs = 100 MPa

:. Diameter of spring wire = 12.68 mm.

Now 6 = 10 mmfor W= 200 N. / W 200 :. Spring constant, k = - = - = 20 Nlmm 6 10

- - - 2o = 2 x lo4 Nlm

We have, C = 80 GPa

Number of coils required = 25.28 say 26.

Example 15.3

A truck weighing 30 kN and moving at 6 km/hr has to be brought to rest by a buffer. Find how many springs each of 20 coils will be required to store the energy of motion during a compression of 200 mm. The spring is made out of 20 mm diameter steel rod coiled to a mean diameter of 200 mm. Take G = 100 GPa.

Solution

Weight of truck = 30 kN.

Mass of the truck = (30:8:03 ] kt?

Velocity of the truck = 6 k m h

1 Kinetic Energy of the truck = - mv2 2

= 4248 Nm

For the spring, let N ;: number of springs.

n = 20 S=200mm

d=20mm G = 100GPa

D=200mm R = 100mm

Using the relationship, 6 = 64 wIt3n G'f'

Springs

1 Strain energy stored in one spring - - WS - 2

Strain energy stored in N springs =

= 250 N

Equating the strain energy stored to kinetic energy of the truck,

250 N = 4248

:. Provide 17 springs.

Example 15.4

A close coiled helical spring of circular section has coils of 75 mm mean diameter. When loaded with an axial load of 250 N, it is found to extend 160 mm and when subjected toma twisting cou'ple of 3 N m, there is an angular rotation of 60" degrees. Determine the Poisson's ratio for the material.

Solution

Here, we have D = 75mm W = 250N

6 = 16Omm T = 3 N m

i$ = 60' 7 (60 x& lradiFS.

Let the Poisson's ratio be v.

Using the relationship, S = - 8WD3n for an axial load, cd4

Therfore,

64TDn For m axial torque, $ = - E$

Stresses in Shafts & Shells Thus, we get

MTDn and Thermal Stresses E = -

ti4

- 64 x 3 x (75 x n -

Therefore, E = 576 x75 x'109n

%a But we know the relationship E = 2G (1 + v)

Subsitituting the values of E and G from (A) and (B),

or :. v = 0.304

Example 15.5

Two close coiled helical springs are compressed between two parallel plates by a load of 1 kN. The springs have a wire diameter of 10 mm and the radii of coils are 5 0

and 75 mm. Each spring has 10 coils and is of the same initial length. If the s~iiallcr spring is placed inside the larger one such that both the springs are compressed by same amount, calculate

(a) the total deflection, and

(b) the maximum stress in each spring.

Take G = 40 GPa for both the springs.

Solution

Defections

Here, we denote spring 1 as larger and spring 2 as smaller.

dl = 10mm d2 = 1Omm

R1 = 75 mm R2 = 50 mm

W = 1OOON G = 40 x lo3 M P ~

Let W1 and W2 be the load carried by spring 1 and 2 respectively.

Since deflection of both the springs is same, we get

Also, we have W1 + W2 = 1000 N (B)

From (A) and (B), we get, W1 = 228.5 N, and W2 = 771.5 N.

Stresses

16W1R1 (fJrnax in spring 1 = -

xdi

16 W2R2 (fs)rnax in spring 2 = -

x (dd3

- . SAQ 1

A close coiled helical spring made out of 8 mm diameter wire has 18 coils. Each coil has 80 mnm mean dia. If the mnxin-urn allowable stress on the spring is 140 MPa. determine the maximum allowable l o d on the spring. the elongat.ion of the spriug aud the total strain energy stored in the spring at that load. Als:, deterrrline U1e sliffness of the spring. Take G = 82 GFa.

SAQ 2 .I clc~w coilctl Ilelic:~l fprlrig has to ahsorh 7(! N In ol energy wi1t.n comg)rccucrl t ~ > 58 m111. 'Thc toll dialnetor is ten llnles the wlre dlametel. If therc art. 12 toll,. cstlriiatc Lhe tliameters o C evil and wirc and Ole n~axli~iunl shew clross. Take 6' = 87000 ~/rnm'.

A close coiled helical spring i c ro he madc of 5 niln diameter wrle for which E - 200 GFa and G = 80 GPa. It is required to exteild 28 n ~ n l f(ir ;ul axial load !:

100 N and to twist 0.22 radian for an axi;il couple of 1 N m. Fintf rhe mean diami :r'r of tho coils and the number of coils required.

A close coiled helical spring of circular section extends 1 m when subjected t t i a:: ;!.ii;!l load of Wnewtons, a ~ v l therc is iln angular rolation of on(: ratliari when a t::! ;i.it:

11r' Tnewton metre is indepen(1ently applied about the axis of [lie spring. If Poisr.:!:':. r;ili!) is v and the mean dinmelor of the coil is 11. show that

Springs

Stresses in Shafts & Shells SAQ 5 and Thermal Stresses

A close coiled helical spring is subjected to an axial twist of 14 N m. If the spring has 14 coils with wire diameter of 12 mm and the mean coil diameter of 200 mm, find

(a) the strain energy stored,

(b) maximum bending stress in the wire, and

(c) axial twist.

~ a k e ' ~ = 200 GPa.

15.3 OPEN COILED HELICAL SPRINGS

In the case of open coiled helical springs, the coils are not so close together, hence, the helix angle a cannot be treated as small.

15.3.1 Spring Subjected to Axial Load Let W be the axial load acting at the end of the spring as shown in Figure 15.3 (a).

Flpro 15.3 a Open C d l d H d c d Spring (Subjected to Axld Ind)

Let d = wire diameter,

D = mean coil diameter,

1 = length of spring,

n = total number of turns, and

p = pitch of coils.

Moment of any point M = W x S F

WD The axial load W exerts a moment - about XX axis. 2

Resolving along the axis of the spring,

bending component = WD sin a . 2

Resolving normal to the spring axis,

WD twisting of wire = - cos a.

2

Let 1 = Total length of the spring

R 2 Total volume of the spring = - d x 1 = ( n e) 4 4

Strain Energy

Strain energy, U = Strain energy due to bending + Strain energy due to torsion

U = Wb12 x 1 f,2 + - (Volume of the spring )

2EI 4G

16T ' nb' Butweknow. f , = ~ and I = - 64

Deflection of Springs

The partial derivative of this total strain energy with respect to the applied load Wgives the deflection 6.

Mb = Msina = - wD sin a 2

Springs

Stresses inshafts & Shells Substituting in expression given in Eq. (1 5.1 I), and Thermal Stresses

D cos a x - cos a (15.12) 2

&E cos a l c d ~ cos a

Stiffness of Springs

W Stiffness of spring, k = - 6

:. k = d cos a cos2a 2 sin a [ 8 D 3 n i T + f )J

15.3.2 Spring Subjected to Axial Couple To find the angle $ by which the spring is wound up, apply a unit moment along the axis of helix as shown in Figure 15.4.

Figure 15.4 : Open Coiled Helical S p h g (Subjected to M a l Couple)

Strain Energy

Strain Energy, U = [ Z ) M ; + [ ~ ) ? ,

au Weknow, $ = -

aM

- 64 1 Mb -- 32 (sin a ) (- cos a ) + - Z ~ E lcdG

WD WD On substituting Mb = - sin a and T = - cos a, 2 2

:. $ = --- i s i n a c o s a [h - $) &

15.3.3 Spring Subjected to Moment along t4leAxis of the Helix Figure 15.5 (a) shows an open coiled helical spring subjected to an axial couple M.

I Axis of the colt wire

Figure 155 : Open Coiled Helical Spring (Subjected to Moment M along the Axis ofthe Helix)

Refer Figure 15.5 (b), we get

Component along the axis of the coil wire = M sin a causes twisting of the wire, and

Component at its right angles = M cos a causes bending of the wire

:. Mb = M cos a and T = M sin a

Strain energy, U = Strain energy due to bending + Strain energy due to torsion

- - - ~3 + - 6 x (Volume of the spring) 2EZ 4G

aMb M b = M c o s a - - " aM - cos a

a T T = M s i n a . . . - - aM - sin a

:. 9 = - ' (M cos a ) cos a + - 321 (Msina)s ina n d E ndc

nD On substituting 1 = n - cos a '

Springs

32 M . nnD sin2a 2 cos2a 9 = -x- [- cos a G E

Stresses in Shafts & Shells and Thermal Stresses

On putting (D = 2R), @ = 6 cos a E

where, R = radius of the coil.

Deflection of Springs

To frnd the deflection 6 of the spring, a unit moment is applied at the end as shown in Figure 15.6.

Figure 15.6 : Detldon Calcdatiom

Mb = Mcosa

WD . T = -sma = Msina 2

Strain energy, U =

a= aw ("Iw &E (%I+ n 6 ~

2Mcosa --sina + - = [ ) (n:t)-sina[+fcosa)

aU 16MDls inacosa - - aw - l~6 (15.19)

nnD Substituting for 1 = -

cos a

- - a' 16NiD MDsinaCosa($ - $1 aw - n d cos a

15.3.4 Stresses in Springs MD

(a) Bending stress, fb = - Z

where, WD nd3 MD = -sinol and Z = - 2 32 '

WD sina Thus,:. f - - -

- - 16 WD sin a nd3 16 T

(b) Sheai Stress, fs = - nd3

- L - nd3

- 8WD cos a. fs = d

(c) Principal stresses are as follov~ s :

(d) Maximum shear stress is,

Springs

15.3.5 Proof Load As def~ned earlier, it is the maximum load, the spring can take without getting permanently deformed. For theopen coiled helical spring, you can obtain this by equating the maxirraun allowable stress in shear to the stress equation or by equating the maximum principal stress equation to the allowable stress of the material.

Example 15.6

An open coiled helical spring consisting of 10 turns of 10 mm diameter wire wound to a coil of mean diameter 110 rnrn. The wire is making an angle of 60" to the axis of the coil which.is subjected to an axial load of 90 N. Find the extension of the coil. Take E = 210 GPa and v = 0.25.

Solution

Here, we have n = 10 d = lOmm

D = 110 mm W = 90N

The wire makes an angle of 60" with the axis of the coil.

:. a = (90 - 60) = 30"

For a = 30°, we get, cos a = 0.866 and sin a = 0.500

We have, E = 210 x lo3 MPa v = 0.25

u;irlg the relationship,

E = 2G (1 +v),

We get, G = 210Xl~3 = a x l o 3 M p a 2 (1 + 0.25)

:. Extension, 6 =

Stresses in Shafts & Shells Example 15.7 and Thermal Stresses

An open coiled helical spring of 50 mm mean diameter is made of steel of 6 mm, diameter. Calculate the number of turns required in the spring to give a deflection of 12 mm for an axial load of 250 N, if the angle of helix is 30". Calculate also, the rotation of one end of the spring relative to the other if it is subjected to an axial couple of 10 N m. Take E = 210 GPa and G = 84 GPa.

Solution

Here, we have, D = 50 mm d = 6 m

6=12mm W=250 N

a = 30" n = ?

E = 210 GPa G = 84 GPa

For a = 304, we get, sin a = 0.500, cos a = 0.866

Using the relationship, 6 = W L ? ~ (COP + 2sin2a) d cos a E

from which we get, the number of turns, n = 4.76

Here, axial couple M = 10 N m = 10 x lo3 N rnm

Angle of rotation, $ = 32 MDn sin2a + 2cos2a

E d c o s a (7 -1 :. $ = 0.6868 radian.

Example 15.8

An open coiled helical spring consisting of 12 turns of radius 100 mm and diameter of wire 12 mm and the angle of helix 15'. It is subjected to an axial load of 250 N. Determine the deflection under the load and also the angle of rotation of the free end. Take E = 21 0 GPa and G = 84 GPa.

Solution

Here, we. have, n = 12

d=12mm

For a = 15", we get cos a = 0.9659, sin a = 0.2588

Given ~ = 2 1 0 x l d ~ / m m ~ .

Using the relationship, 6 =

Angle of rotation, $ = 16 wn2n sin a (l - 2 ) d G E

Example 15.9 2

An open coiled helical spring is made of wire of 10 mm diameter. It has 10 coils of mean diameter 50 rnrn. What is the greatest axial load that can be applied to the ends of the spring if principal stress and maximum shear are not to exceed 100 MPa and 60 MPa respectively. Calculate for this load, the axial and angular deflection of the spring.

Take E = 200 GPa, and G = 80 GPa.

Springs

Axial Deflection, 6 = cos2a + 2 sin2 a 1

E

Angular Deflection, $I = 16 WD2n sin a d4 [a-s)

:. $I = 0.0314 radian.

Example 15.10

An o p n coiled spring consists of 10 coils, each of mean diameter of 50 mm, the wire formkg the coils being 6 mm diameter and making a constant angle of 30" with the planes perpendicular to the axis of the spring. What load would cause the spring to elongate by 12.5 mm and what are the magnitudes of bending and shearing stresses due to this load ? Take E = 210 GPa and G = 84 GPa.

Solution

Here, we have, n = 10 D = 5 0 m m

d = 5 m m 6 = 12.5 mm

E = 210 x l d MPa G = 84 x lo3 M P ~

For a = 30°, we get, cos a = 0.866, and sin a = 0.5

Solution

Here, we have, - d = 10 mm n = 10

D=5Omm R=25mm

fi -f2 8WD - Maximum shear stress, (A),, = - - - 2 nd3

Here a is unknown

Maximum principal stress, fi = - 8WD (sin a + 1 ) nd3

(sin a + 1 ) = 1.667

sin a = 0.667

Stream in Shafts & She& and Thermal S t r e s s Using the relationship, 6 = ----

E

.'. W = 124.1 N

Torque, T = WR cos a Bendii moment, Mb = WR sin a = M

Equivalent bending moment,

1 = 2 ( WRsina+ WR)

32 Me Maximum direct stress = 2

d3

Equivalent Torque. Teq =

= J(WR sin a12 + (WRCO~ a12

16 Te Maximum direct stress, f, = 2

ltd3

SAQ 6 In an open coiled helical spring having a = 20', if the inclination of the coils is ignored, calculate the percentage by which the axial extension is underestimated. Take E = 200 GPa and G = 80 GPa.

In an open coiled helical spring having a = 30", if the inclination of the coils is neglected. calculate the percentage error in the value obtained for the stiffness. TiLke,E = 200 GPa and G = 80 GPa.

Springs

Spring 2

SAQ 8 9

An ope11 coiled helical spring having 12 complete turns is made of 15 mnm diameter steel rod, the inearl diameter of the coil being 100 innl. The angle of hel~x, cx is 15'.

(a) Calculate the deflection under an axial load of 300 N.

(b) Also calculate the direct and shear stresses induced in the section of the wire.

(c) If, however, the axial load of 300 N is replaced by an axial torque of 8 N In, deterinhe lhe axial deflection and the angle or rotation about the axis of the coil. Takc E = 200 GPa and (; = 80 GEa.

((1) Also calculate the axial twist which will cause a bending stress of 10 MPa

SAQ Y An open coil sprlllg with a: = 30" has a vertical displacement of 16.8 Inn1 and an angular roration of the load end of 0.027 radi%iXntler an axial load of 100 N: The sprmg 1s fonned of a I0 nlnl diameter steel rod. Calculate the mean radius of the coil Tiike E = 21 0 <;Pa. (; = 84 GPa.

SAQ 10 If the close co:led sprillg Sormula 1s used in finding thz exterisioll of an open coiled 5prlng under rhe axial load. deterrmne thc maximum ariglc of helix for which the error 111 111e value oS the exlension is not to excccd 2 percent. Assume E = 2.5 G.

15.4 COMPOUND SPRINGS

More than one spring may have, at times, to be used to meet the specific requirements. The springs may be used either side by side (in parallel) or connected end to end (in series) as shown in Figures 15.7 (a) and (b). This type of composite system of springs is called as compound springs.

Springs in Parallel

Figure 15.7 (a) shows two springs connected in parallel.

Figure 15.7 (a) I Compound Springa (Spring in Parsllel)

Stresses in Shafts & Shells In this case, and Thermal Stresses

(a) The extension of both &elsprings is equal. ,

(b) The sum of the load carried by each spring is equal to the load W.

I w = w1 + w2

where, Wl = load shared by the spring 1, and

W2 = load shared by the spring 2.

Let S1 and S2 be the stiffnesses of these two springs respectively.

Spring 1 SS = 6S1 + 6S2

S = S1 + S 2 (15.25b)

Springs in Series

Figure 15.7 (b) shows two springs connected in series.

Each spring carries the same load W applied at the end and the total deflection is equal to the sum of the deflection in each spring.

Spring 2 :. 6 = h 1 + a 2 (15.26a)

W But, stiffness of the spring, S = -

6

Figure 15.7 (b) compound springs or (Springs in Series)

Example 15.11

A composite spring has two close coiled helical sprinzs connected in series, each spring has 12 coils at a mean diameter of 25 mm. Find the diameter of the wire in one of the springs if the diameter of the wire in the other spring is 2.5 mm and stiffness of the composite spring is 700 N/m. Estimate the greatest load that can be camed by the composite spriqg for a maximum shearing stress of 180 MPa. Take G = 80 GPa.

Solution

Springs are connected in series.

dl = 2.5 IIUII d2 = ?

For composite spring,

S = 700 N/m = 0.7 N/mm

W Using the relationship, S = - 6

For spring 1, !5-- - 1 - - 64# nl

W S1 Gd:

For spring 2, h 1 - 64 x (12.5)) x I%-- - - - - W - s2 (80 x lo3) (d214

For springs connected in series,

-- I - 0.05058 (d2)4

:. d2 = 2.109 mm

Given (f,),, = 180 MPa, then W = ?

Since both the springs carry the same load W,

180 x x (2.5)) = 44.17 For spring 1, W =

8 x 25

180 x n (2.109)) = 26.51 For spring 2, W = 8 x 2 5

Thus, the greatest load the spring can carry based on the maximum shear stress criteria is lesser of these two, i.e. 26.51 N.

Example 15.12

A rigid bar weighing 5 kN and carrying a load of 20 kN is supported by 3 springs a. shown in Figure 15.8, having spring constants S1 = 30 kNIm, S2 = 18 kN/m and S3 = 12 IdVIm. If the unloaded springs are all of the same length, find the distance x such that the bar is horizontal.

Springs

Figure 15.8 : Figure for Example 15.12

Stre~ses in Shafts &Shells Solution and Thennal Stresses

Here the springs are connected in parallel.

We know, spring constant, S = (."I 8 ~ d n Using the relationship, 6 = - cdl

W Here, Deflection of all the 3 springs will be equal, i.e. 6 = 7 = constant.

Wl w2 w3 . - - - - - - " Sl. s2 s3

where, W1, W2 and W3 are the loads carried by these 3 springs and S1, S2 and S3 are their respective stiffnesses.

Substituting in (i), Wl + W2 + W3 = 25 kN

Taking moments about the spring 3,

(Wl x 0.8) + (W2 x 0.4) = (5 x 0.4) + 20 (0.4 + X)

(12.5 x 0.8) + (7.5 x 0.4) = 2 + 20 (0.4 + x)

:. x = 0.150 m

The load of 20 kN must be placed 150 mm from the middle spring.

: ? i ' * .

. i t s -,o11 \pr?n:l\ who\e propcrt~cs are given below ale connected 111 series

0)

(ii)

Spring !i --+ coil diarrleter : 50 nim : nurnber of coils : 8

i f t l x wire dianierer of spril~g A be 8 nun and the stiffness of Lhc compound spring he 10 kYim, determine the wire diameter of s p r l ~ g B . What could k the safe load for ?.he spring so 1i:at the shear stress in the wire does not exceed 100 MPa. T6-ikkc G = 80 GI.'a.

SAQ 12 .e-

The following data applies to Lwo close colled helical spnog~ .

I Axial length 1 I Spring 1 I , uncornpreused 1

Spring If 1s placcd t~lsidc sprmg A and both art col~~pr:si~d hk fxrrr-en a pa:! (

pla!cs until LIlc dtslance between the plates measures hO mln.

Calculate

(a) tht3 Itjad applied lo thc plales. ant1

(b i the rnaxlrnunl lntcrlslty of shear stress m c;ch spn11;~

l'akc CI = 80 GPa.

15.5 LEAF SPRINGS

This type of springs are commonly used in carriages such as cars, railway wagons etc. and they are also termed as laminated or carriage springs. It is made up of a number of leaves of equal width and thickness, but varying length placed in laminations and loaded as a beam. The lengths of the plates are so adjusted that the maximum bending stress remains same in every plate and thereby it beh,aves like a beam of uniform strength.

It is assumed that each plate is free to slide relative to the adjacent plates as the spring deflects and the ends of each plate are tapered to provide a uniform change in effective breadth. Further it is assumed that the plates are bent to the same radius so that they contact only at their edges.

15.5.1 Stress in Springs L

1 Figure 15.9 shows a carriage spring carrying a central vertical load W, which is balanced by 1 W equal end reactions -.

2

Springs

,'

Figure 15.9 : Leaf Spring

Let W = load on the spring

R = initial radius of curvature of plates

6 = initial central deflection

Stresses in Shafts & Sheb b = width of each plate and Thermal Stresses

t = thickness of each plate

n = number of leaves (plates)

L = span of the spring

S = bending stress

b? Section modulus for a single plate or laminate = - 6 '

Section modulus for the whole spring (having n laminates) = n - [ b f l WL

Maximum bending moment, M = - 4

But !!!=f Y

Strain Energy

Resilience due to bending = 3 .6E

:. Total strain energy. U = f2 x (Volume of the equivalent place) 6E

volume = (9 L l )

.= Strain Energy U = $[%) Substituting for fi,

1 Work done by the load = - x W x 6

2

Equating these two,

Stiffness of Springs

It is defied as tile load required to produce unit deflection.

W :. Spring constant, S = - 6

Proof Load me-w-

If Wo is the load required to make the spring flat, it is known as the proof load.

If 60 is the deflection corresponding to proof load wo,

then,

8Enb? :. Proof Load, Wo = - 3L3

s,

15.5.2 Practical Applications Leaf springs are extensively used in railway carriages, railway wagons, trucks, trollies, buses and cars etc. the common purpose of all kinds of springs is to absorb energy and to release it as and when required. Carriage springs are used normally to absorb shock. In other words, they act as primarily shock absorbers.

Example 15.13

A leaf spring 0.8 nl long consists of 12 plates, each of them is 65 mm wide and 6 mm thick. It is simply supported at its ends. The greatest bending stress is not to exceed 180 MPa and the central deflection when the spring is fully loaded is not to exceed 20 mm. Estiinate the magnitude of the greatest central load that can be applied to the spring. Take E = 200 x lo3 MPa,

Solution

Here, we have, L = 0.8 m n = 12

b = 6 5 m m t = 6 m m

f < 21 0 ~ / m r n ~ 6 < 2 0 m m

3WL Using the relationship, f,, = -

2nb8

3 wL3 Using the relationship, 6 = - x -

8 Enb?

:. W = 3510N

Thus, the greatest central load that can be applied is lesser of these two, i.e. 3.51 kN.

Example 15.14

A leaf spring is required to satisfy the following specification :

L = 0.75 m, W = 5 kN, b = 75 mm, maximum stress = 210 MPa,

Maximum deflection = 25 rnm, E = 200 GPa.

Find the number of leaves, their thicknesses and initial radius of curvature.

Solution

Here, we have, L = 0.75 m = 750 mm W = 5 kN = 5000 N

3 WL Maximum stress, f = - x - 2 rrb8

Springs

Stresses in Shafts & Shells and Thermal Stresses 3 wL3

Maximum deflection, 8 = - x - 8 ~ n b ?

':. n? = 2109 From (i) and (ii), t = 5.905 mm Thus, use 6 mm thick plates.

Adopt 10 leaves.

L L Using property of circle, - x - = 6 (2R - 6) 2 2

On account of 82 being very small, can be neglected.

and

(ii)

SAQ 13 :. Radius of curvature = 2.815 m.

A stcel carrlagc spring is hO0 mrn I:)nlz and carries a ceiitral load of 4.5 kN. Each plate is 75 rnm wide and 6 Irun thick. Thc strcss i b oot to exceed 170 MPa. C'alculate the required ~iun~ber of plates aid 152 deflection at thc cenuc of the spring. Take E = 210 GPa.

A laminated spring having a length 800 r r m is required to ciury .I cclilral polnt load of 8 kN. Calculate the thickness, width and the numQer of plates if the bcriding stress and central deflection are not to exceed 200 Nlm~n' and 20 inn1 respectively. Also calculate the rildius to which the plates should bt: curved. Take E = 200 GPa. Assume the width ol'lhe plate t c ~ be 12 tlmes ilx th~cklless.

I.* 15.6 SUMMARY

We conclude by summarising what we have covered in this unit. We have / !;I (a) Studied the definition of proof load, spring constant and proof stress, 1 (b) The defferent types of spring9namely close-coiled, open-coiled helical springs

and leaf springs. C.

(c) Obtained expressions for stresses in the springs, stiffness of springs and deflection of springs for the above three types of springs.

-, - (d) Also seen the compound springs i.e. springs in series and springs in parallel and

the situations where they are used.

(e) Studied some of the practical applications where springs are very often used.

15.7 ANSWERS TO SAQs

SAQ 1

W=351.8N,6=77.2mm, U = 13.59Nm,k=4.56N/mm.

SAQ 2 2 d = 45.92 mm, D = 459.22 mm, f, = 29.3 N/mm .

SAQ 3

D = 63.81 mm, n = 6.73.

SAQ 5

U = 4.23 N m, f,,= 82.52 MPa, @ = 34.7'.

SAQ 6

3.78%.

SAQ 7

9.7%.

SAQ 8

(a) 7.26 mm. (b) 28.49, -1 6.78,22.64 MPa.

(c) 0.1963 mm, 3.659'.

(d) 3.43 N m.

SAQ 9

56.7 mm.

SAQ 10

14.670.

SAQ 11

6.8 mm, 247 N.

SAQ 12

40.6 N, 19.1 MPa, 39.76 MPa.

SAQ 13

8.82, 11.9 mm. SAQ 14

t=8mm,b=96mm,n=7.81,R=4m.

Springs

1

Stresses in Shafts & Shells FURTHER READING and Thermal Stresses

(1) Aggarwal, S. K., and Gupta, P. K., Strengths of Materials, Metropolitan Book Company, New Delhi - 110 002.

(2) singh, Surendra, Strength of Materials, Vikas Publishing House, New Delhi - 110 002.

(3) Prasad, Jainti, Strength of Materials, CBS Publishers & Distributors.

(4) Timoshenko, Stephen, Strength of Materials - Part I & 11, CBS Publishers & Distributors.

(5) Schaum's OutlineSeries, (1989), Strength of Materials - Second Edition, McGraw Hill Book Company.

(6) Popov, E. P., (1993), Mechanics of Materials - Second Edition, Prentice Hall of India Private Limited.

(7) Ryder, G. H., (1993), Strength of Materials, Educational Low Priced Books Scheme.

(8) Case, John, Chilver, L., Ross, Carl T. F., (1993), Strength of Materials and Structures - Third Edition, Educational Low Priced Books Scheme.