161

SpringsUNIT 13 SPRINGS

Structure 13.1 Introduction

Objectives

13.2 Close Coiled Helical Springs 13.2.1 Spring Subjected to Axial Load 13.2.2 Spring Subjected to Axial Couple

13.3 Open Coiled Helical Springs 13.3.1 Spring Subjected to Axial Load 13.3.2 Spring Subjected to Axial Couple

13.4 Compound Springs

13.5 Leaf Springs

13.6 Summary

13.7 Answers to SAQs

13.1 INTRODUCTION

Springs are generally used to control shocks and vibrations. These are adopted in automobiles and railway buffersete. They are classified as helical springs and leaf springs. Helical springs are formed by coiling a wire into a helix. When the helix angle is less than 18o, it is called as close coiled helical springs. When the helix angle is significant it is known as open coiled helical springs.

Objectives After studying this unit, you should be able to

• understand the different types of springs,

• find the stresses, deflection and stiffness of springs, and

• design the springs. Definitions

Proof Load The maximum load the spring can carry without failure.

Proof Stress The stress in the spring under proof load.

Proof Resilience The strain energy stored in the spring under proof load.

Spring Constant It is the stiffness of the spring. It is the load per unit deflection : units are N/m.

13.2 CLOSE COILED HELICAL SPRINGS

A helical spring is formed by winding a wire in the form of a circular cylindrical helix. These may be subjected to axial load or torsional loads. The angle made by the coil with the horizontal is known as helix angle (α). If α is less than 18, it is known close coiled helical springs.

162

Each coil can be regarded as lying in a plane at right angle to axis of the helix. The wire that makes close coiled helical spring is under torsion.

Stresses in Shafts and Shells

13.2.1 Spring Subjected to Axial Load Let us consider a close coil helical spring subjected to axial load W.

Let R = Mean coil radius,

D = Mean coil diameter,

d = Wire diameter,

n = Number of coil,

l = Length of the wire,

α = Helix angle, and

Δ = Axial deflection.

(b) (c) (d) (e)

(a)

Figure 13.1 : Close Coiled Helical Spring under Axial Load

Figure 13.1(a) shows a helical spring schematically. The force W acts along axis of the spring. Though in practice the axis can assume any position, for convenience it is regarded vertical AA represents a horizontal plane. α is the helix angle which is made by a coil with the horizontal plane. The angle is exaggerated in the figure. It is actually smaller. In practice, the close coiled helical carrying tensile load have their coils touching each other. The compression springs may have some clearance between the coils. Figure 13.1(b) shows the spring lower portion cut after second coil, thus, the spring wire section, which is circular is exposed. In a close coiled spring a normal on the section will be horizontal. It may be pointed out that if spring is open coiled then the section cut by a vertical plane will not be a circle but a larger section. The Figure 13.1(c) is free body diagram on which an upward force W is axial and is balanced by a downward force W on the section in its plane and a moment T = W R acting on the section about its axis. Moment, T = W . R . . . (13.1) From Figure 13.1(b), As explained earlier, T may have two components : cos and sinW R W Rα α

But since α is small sin α→ 0 and cos α → 1. Hence, T = W R . . . (i) But a direct shearing force W is also acting upon the wire section.

A A α

W

+ =

RT = WR

4 w π d2

16 T W π d3 τ1 = τ1 = τm = τ1 + τ2

163

Springs

2

Thus, the wire section is subjected to direct shearing stress τ1 and torsional shearing stress τ2. The distributions of τ1, τ2 and resultant stress 1mτ = τ + τ are shown in Figure 13.1(c).

1 24W

dτ =

π

2 316T

dτ =

π

∴ 316

4mWdWR

d⎛ ⎞τ = +⎜ ⎟

π ⎝ ⎠

316 1

4WR d

Rd⎛ ⎞= +⎜ ⎟

π ⎝ ⎠

or, 316 1

2mWR d

Dd⎛ ⎞τ = +⎜ ⎟

π ⎝ ⎠ . . . (13.2)

In such cases where d < < D, 2dD

may be neglected in comparison with 1. There

316

mWRd

τ =π

. . . (13.3)

Otherwise τm is expressed as

316

mWRKd

τ =π

. . . (13.4)

where K is stress concentration factor on the inside of the coil. K is also called Whal’s factor.

12dKD

⎛ ⎞= +⎜ ⎟⎝ ⎠

. . . (13.5)

Deflection

Under the axial load the free end of the spring will move vertically down. The amount by which the free end moves is called the deflection of the spring, denoted by Δ. Apparently the free end moves because of twist in the wire of the spring which can be regarded as a shaft of length l,

. . . (13.6) 2l R= π n

where n = Number of coils.

The amount by which the open section of wire will move down is R × twist.

The angle of twist of length l from T GJ l

θ= is Tl

GJθ = and substituting for l from

Eq. (13.6)

4 and32

J d T Wπ= = R

2 2

4 42 32 64WR n W R n

G d G dπ ×

θ = =π

∴ RΔ = θ

164

Stresses in Shafts and Shells or

3

464WR n

G dΔ = . . . (13.7)

Strain energy stored in spring

12

U W= Δ

or 2 3

432W R nU

G d= . . . (13.8)

It may be good to check that strain energy per unit volume

22

4u

Gτ

= . . . (13.9)

(Divide Eq. (13.8) by volume 2

4d lπ

= × ).

Stiffness is defined as load per unit deflection i.e.

WK =Δ

or 4

364G dK

R n= . . . (13.10)

13.2.2 Spring Subjected to Axial Couple Let us consider a close coiled helical spring subjected to a couple M shown in Figure 13.2(a). Two components maybe considered if α = angle of helix.

Mt = M cos α

T = M sin α

Out of these T → 0 as sin α → 0, since α is small. But Mt → M since cos α → 1. Hence, the spring is subjected to a torque about axis, of magnitude M. The effect of M will be to wind or unwind the coil of the spring, resulting into increase or decrease in radius of the coil. This effectively is the bending of the rod of the spring.

M

M

Figure 13.2

Let us say due to M acing on the spring, the radius of the coil changes from R to R′. The BM from theory of bending is given by

1 1M EIR R

⎛= −⎜ ′⎝ ⎠⎞⎟

R

. . . (13.11)

This twisting of the spring (not the twisting of spring rod) or bending of the rod of the spring will change the number of turns from n to n′ while length l will remain same.

Note 2 2l n R n′ ′= π = π

∴ 1 1 2 (n nR R l

)π ′− = −′

. . .

(i)

165

SpringsUse (i) in Eq. (13.11) to find

2 ( )EI n nMl

′π −= . . . (13.12)

You would remember that E = modulus elasticity of spring material and I = moment of inertia for rod section. The total twist in the spring or rotation

. . . (13.13) 2 ( )n n′φ = π −

Hence, from Eq. (13.12)

MlEI

φ = . . . (13.14)

or 44

2 128

64

M n R n M RE dE d

πφ = =

π⎛ ⎞⎜ ⎟⎝ ⎠

. . . (13.15)

The rod of the spring is subjected to BM

∴ Bending stress 332 M

dσ =

π . . . (13.16)

Strain energy in bending

2 1

2 2M lU

EI= = M φ . . . (13.17)

If U is divided by volume of the rod, the strain energy per unit volume.

2

8u

Eσ

= . . . (13.18)

(Check above equation by dividing).

Example 13.1

A close coiled helical spring is made of 6 mm diameter wire. The coil diameter is 80 mm, number of coil is 10. If the maximum stress in the spring is not to exceed 180 MPa, determine

(a) the proof load, and

(b) the extension of the spring.

Take G = 80 GPa.

Solution

Proof load is the load corresponding to maximum shearing stress in spring wire.

d = 6 mm, D = 80 mm, 40 mm2DR = = , n = 10, , G = 80 GPa 2

max 180 N/mmτ =

Check 6 0.01672 2 180dD= =

×

Hence, K = 1

(a) Proof load, 3

max .16

dWR

π= τ

3(6)180 191 N

16 40π

= × =×

166

Stresses in Shafts and Shells (b) Extension,

3

464 W R n

G dΔ =

3

3 464 191 40 10 75.5 mm

80 10 6× × ×

= =× ×

Example 13.2 For a close coiled helical spring subjected to an axial load of 200 N having 10 coils of wire of diameter 18 mm and made with coil diameter of 200 cm, find

(a) strain energy stored, (b) axial deflection, and (c) maximum shear stress in the wire.

Take G = 80 GN/m2. Solution

W = 200 N, n = 10, d = 8 mm, D = 20 cm = 200 mm, 100 mm2DR = =

Neglecting effect of direct shear since 12dD< < .

Axial deflection, 3

464 W R n

G dΔ =

3

3 464 (200) (100) (10) 15.24 mm

(80 10 ) (18)= =

×

Strain energy, 12

W= Δ

1 15.24200 1.524 N-m2 1000

⎛ ⎞= × × =⎜ ⎟⎝ ⎠

Maximum shear stress, max 316 W R

dτ =

π

23

16 (200) (100) 17.5 N/mm(18)

= =π

Example 13.3 A close coiled helical spring has mean diameter of 80 mm, has spring constant of 100 kN/m. It has 10 coils. The maximum shear stress is 200 MPa. What is diameter of spring wire? What is the maximum load the spring can carry? Take G = 80 GPa.

Solution

D = 80 mm, 80 40 mm2

R = = , k = 100 kN/m, n = 10

3

464 W R n

G dΔ =

4

364W G d

R n=

Δ

167

Springs

3 4

3(80 10 )100 15 mm64 (40) (10)

d d×= ∴ =

max 316 W R

dτ =

π

316 40200 3313.4 N = 3.3 kN

(15)W W×

= ∴ =π

Example 13.4 A close coiled helical spring has a stiffness of 1 kN/m in compression with a maximum load of 50 N and a maximum shearing stress of 150 N/mm2. The solid length of the spring is 45 mm. Find the wire diameter, mean coil radius, and number of coils. Take G = 40 GPa.

Solution W = 50 N, k = 1 kN/m = 1 N/mm, τmax = 150 N/mm2, solid length = 45 mm, d = ?, R = ?, n = ? Use Eq. (13.10) for stiffness

4

364W G dk

R n= =Δ

or 3 4

3(40 10 )1

64d

R n×

=

∴ 3

4 625R nd

= . . . (1)

Use Eq. (13.3) for maximum shearing stress with K = 1

max 316 W R

dτ =

π

or 316 50150 R

d× ×

=π

∴ 3 0.6Rd

= . . . (2)

or R = 0.6 d3

Note solid length of spring is length when all coils are touching each other, so that 4n d 5=

45nd

= . . . (3)

Use R from Eq. (2) and n from Eq. (3) in Eq. (i)

3 34

45 1(0.6 ) 625dd d

⎛ ⎞ =⎜ ⎟⎝ ⎠

or 4 64.3d =

∴ 2.83 mmd =

From Eq. (3)

45 15.92.83

n = = (say 16)

3 30.6 0.6 (2.83) 13.6 mmR d= = =

168

Example 13.5 Stresses in Shafts and Shells

A close coiled helical spring has to extend by 120 mm under an axial force of 1200 N. If mean coil radius is 40 mm and maximum shear stress is 300 MPa, find the wire diameter, number of coils and length of the spring. Take G = 80 GPa.

Solution

Δ = 120 mm, W = 1200 N, R = 40 mm, max 300 MPaτ = , G = 80 GPa = 80 × 103 N/mm2, d = ?, n = ?, l = ?

Use Eq. (13.3) for maximum shearing stress

max 316W R

dτ =

π

or 316 1200300 R

d× ×

=π

∴ 3 0.049Rd

= or 30.049R d= . . . (1)

40 or 9.35 mm0.049

d = . . . (2)

From Eq. (13.7)

3

464 W R n

G dΔ =

∴ 3

3 464 1200120

80 10R nd

× ×=

× ×

or 3

4 125R nd

=

Using values of R and d,

3

440 1259.35

n×=

∴ 14.9 15n = ;

l = 2 π R n

= 2 π (40) (15)

= 3770 mm

Example 13.6

A close coiled helical spring has to absorb 100 N-m of energy when compressed to 10 cm. The coil diameter is 10 times the wire diameter. The number of coils is 12. Find the diameter of wire, mean radius and the maximum shear stress. Take G = 80 GPa.

Solution

Remember from Section 13.2.1

12

U W= Δ

169

Springs

∴ 3 1100 10 1002

W× = × ×

∴ W = 2000 N

D = 10 d or R = 5 d

G = 80 GPa = 80 × 103 MPa = 80 × 103 N/mm2

3

464 W R n

G dΔ =

∴ 3

3 464 2000 (5 ) 12100

(80 10 )dd

× × ×=

×

or d = 24 mm

5 5 24 120 mmR d= = × =

2max 3 3

16 16 2000 120 88.4 N/mm(24)

W Rd

× ×τ = = =

π π

Example 13.7

Design a close coiled helical spring which will deflect 100 mm under a load of 600 N. The radius of the coil is 6 times the wire diameter. The maximum shear stress is not to exceed 8 MPa. G = 80 GPa. What suddenly applied load will elongate the spring by 100 mm?

Solution

Δ = 100 mm, W = 600 N, R = 6d, , G = 90 GPa = 80 × 10

2max 80 N/mmτ =

3 N/mm2

max 316 W R

dτ =

π

⇒ 316 600 (6 )80 d

d× ×

=π

∴ d = 15.14 mm

R = 6d = 90.84 mm

3

464 W R n

G dΔ =

⇒ 3

3 464 600 90.84100 14.6

(80 10 ) (15.14)n n× × ×

= =×

=

A suddenly applied load will cause double the stress caused by same load if applied gradually. Hence, in this case if same stress is to be induced in spring wire

cross-section by suddenly applied load it will be 600 300 N2 2

W= = .

Example 13.8

A close coiled helical spring deflects 25 mm under a certain axial load. Find the deflection of the second spring under the same load if the effective length of the wire is same but the diameter of coils is 20% greater and that of the wire is 10% greater.

170

Solution Stresses in Shafts and Shells

3

464 W R n

G dΔ =

R2 = 1.20 R1, d2 = 1.1 d1

l1 = l2

G and W are same for two springs. If we call two springs 1 and 2 then they have the same length but different coil radius and number of coils ∴ 1 2 1 1 2or 2 2l l R n R n2= π = π

But R2 = 1.2 R1 (20% greater)

∴ 12 1 1 2

21.2 or 1.2nR n R n

n= =

Take the ratio of deflections of two springs

3 4

2 2 2 14

1 1 1 2. .R n d

R n d

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ= ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟Δ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⇒ 3 4 42

2 1 1

1 1

1.20 1 (1.2) 1. 0.25 1.20 1.1 (1.2) 1.1

R dR d

⎛ ⎞ ⎛ ⎞Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

984=

2

∴ 2 0.984 25 24.59 mmΔ = × =

Example 13.9 A close coiled helical spring has a stiffness of 10 N/mm. Its length when fully compressed, with adjacent coils touching each other is 400 mm. G = 80 GPa.

(a) Determine the wire diameter and the mean coil radius, if their ratio is 0.02.

(b) If the gap between any two adjacent coils is 2 mm, what maximum load can be applied before the adjacent coils touch?

(c) What is the corresponding maximum shear stress in spring? Solution

k = 10 N/mm, nd = 400 mm, , d = ?, R = ?, 380 GPa = 80 × 10 N/mmG =

0.02 5d R dR= ∴ = , Gap = 2 mm, W = ?, max ?τ =

(a) 4

364W G dk

R n= =Δ

∴ 4

364W G d

R n=

Δ

or 3 4

3

80 101040064 (5 ) .

d

dd

× ×=

⎛ ⎞⎜ ⎟⎝ ⎠

∴ d = 20 mm R = 5d = 5 × 20 = 100 mm

400 400 2020

nd

= = =

171

Springs(b) Gap between adjacent coils = 2 mm

Total gap = 2n = 2 × 20 = 40 mm, i.e. spring has to deflect 40 mm if coils are to touch each other.

3

464 W R n

G dΔ =

∴ 3

3 464 100 2040

80 10 (20)W× × ×

=×

or W = 400 N

∴ 2max 3 3

16 16 400 100 25.5 N/mm(20)

W Rd

× ×τ = = =

π π

Example 13.10 A vehicle weighing 28 kN and running at 2.5 m/sec is to be brought to rest by a buffer springs. Find the number of springs required to absorb the kinetic energy if

n = 20 Compression of each spring = 200 mm Diameter of wire = 30 mm Radius of coil = 100 mm G = 80 GPa

Solution

KE 3

2 21 1 28 10 2.5 8919.5 N-m2 2 9.81

mV⎛ ⎞×

= = × =⎜ ⎟⎜ ⎟⎝ ⎠

3

464 W R n

G dΔ =

⇒ 3

3 464 100 20200(80 10 ) (30)

W× × ×=

× ×

∴ W = 10125 N

For one spring, 1 1 20010125 1012.5 N-m2 2 1000

U W= Δ = × × =

∴ Numbers of springs, 8919.5 8.8 9.01012.5

KENU

= = = ;

Example 13.11 A close coiled spring is subjected to an axial moment of 16 N-m. If the spring has 15 coils with wire diameter of 10 mm and mean coil radius of 10 cm. Find :

(a) the strain energy stored, (b) maximum bending stress, and (c) axial twist.

Take E = 200 GN/m2. Solution

316 N-m = 16 × 10 N-mmM = , n = 15, d = 10 mm, R = 100 mm,

, 2 2 100 15 9434.8 mml R n= π = π × × = 4 4(10) 490.9 mm64 64

I dπ π= = = 4

172

Stresses in Shafts and Shells (a)

2 3 2

3(16 10 ) 9424.8 12287.4 N-m = 12.3 N-m

2 2 (200 10 ) (490.9)M lU

EI× ×

= = =× ×

(b) 3

23 3

32 32 16 10 163 N/mm10

Md

× ×σ = = =

π π ×

(c) 12

U M= φ

⇒ 112.3 162

= × × φ

∴ φ = 1.54 rad Example 13.12

A close coiled helical spring of circular section extends 1 mm when subjected to an axial load of W and then given an angular rotation of 1 radian when a moment

M is independently applied about the axis. Find the ratio MW

.

Solution For a close coiled spring :

3

464 W R n

G dΔ = (Δ = 1 mm)

∴ 4

364G dW

R n= . . . (1)

4128 M R n

E dφ =

∴ 4

128E dM

R n= . . . (2)

Eq. (2) divided by Eq. (1), we get

4 3

464

128M E d R nW R n G d

= ×

2

2E RG

= 2 (1 )E G= + ν

2 (1 )M RW

= + ν

Example 13.13 A circular rod of diameter d and length l is fixed at one end and a torque T applied at the other. Now, if a close coiled helical spring is made of the same rod and applied same moment T about the axis, determine ratio of strain-energy stored. E = 200 GPa; G = 80 GPa.

Solution Circular Rod

J = 2I

T lG J

θ =

173

Springs2

1 2 4T l T lUG J G I

= = . . . (1)

Helical Spring

M l T lE I E I

φ = =

2

212 2

T lU ME I

= φ = . . . (2)

1

2

200 1.252 2 80

U EU G

= = =×

SAQ 1 (a) A weight of 200 N falls freely from a height of 600 mm on a close coiled

helical spring which is compressed by 120 mm under the impact. Find the instantaneous stress produced by the impact and the number of coil if wire diameter is 25 mm and mean coil radius is 100 mm. G = 80 GPa.

(b) A trolley of weight 20 kN and moving at 3.6 kmph has to be brought to rest by a buffer spring. Find how many springs each of 12 coils will be required to store the KE for compression of 120 mm. Wire diameter = 20 mm, mean coil radius = 100 mm. G = 80 GPa.

(c) Find the maximum possible load a close coiled helical spring can carry if it is made out of 6 mm × 6 mm square root with 10 coils of 40 mm mean radius, if the maximum shearing stress is limited to 70 N/mm2. Find the deflection of the load. G = 80 GPa.

(d) A close coiled helical spring is subjected to an axial moment of 15 N-m. If the spring has 10 coils with wire diameter of 16 mm and mean coil radius of 150 mm, find

(i) the strain energy stored,

(ii) maximum bending stress in the wire, and

(iii) axial twist.

Take E = 200 GPa.

(e) Two close coiled helical springs are made from the same wire of a same diameter, one wound on a 30 mm radius core and the other on a 15 mm radius core. Calculate the ratio of spring coefficients of both the springs if each has 10 coils.

(f) A close coiled helical spring is made up of 6 mm diameter steel rod having 12 coils with 45 mm mean radius. If the spring is subjected to 1 N-m, determine the bending stress and increase in the number of turns.

Take E = 200 GPa.

(g) Determine the poisson’s ratio if mean radius of coil is 30mm, a load of 80 N extends the spring by 80 mm and a torque of 1.2 N-m produces an angular

rotation of 3π radians.

(h) A close coiled helical spring is required to just slide over a bar of 30 mm diameter. The spring is to carry a maximum load of 500 N and deflect under this load by 2 mm. The shear stress is limited to 25 MPa and G = 80 MPa. Find diameter of wire, mean coil radius, and number of coils.

174

Stresses in Shafts and Shells

(i) A close coiled helical spring is used to connect two shafts which transmit 2.4 kN of power at 200 rpm. Calculate the maximum normal stress and wind-up angle in the spring. The diameter of the wire = 10 mm. Mean radius = 30 mm. E = 200 GPa. Numbers of coils = 10.

(j) If two close coiled helical springs of same material are subjected to same axial load, find the ratio of deflection for the same length of wires for the two springs.

(k) Design a close coiled helical spring which coil deflect 100 mm under a load of 500 N. The mean radius is 5 times the wire diameter, maximum shear stress is not to exceed 60 MPa and G = 80 GPa. What suddenly applied load will elongate the spring by 100 mm?

13.3 OPEN COILED HELICAL SPRINGS

In these, the coils of the spring are not close to each other. The angle of helix is greater than 18.

13.3.1 Spring Subjected to Axial Load Let us consider an open coiled helical spring subjected to axial load as in Figure 13.3(a). The geometry is defined by R, d and α. These and other properties are defined below.

R = Mean radius of the coil.

d = Diameter of the wire,

n = Number of coils,

α = Angle of helix,

G = Modulus of rigidity,

Δ = Axial deflection of the spring, and

θ = Angle of twist.

An axial force, W acts along the axis. The lowest coil is cut by a vertical plane through the axis, exposing a section AB of the wire. Due helix angle α being considerable (> 180o) AB, which is shown enlarged in Figure 13.3(b) is not a circle. The circular cross-section is inclined at α to AB and shown as AO′C in Figure 13.3(b). Note that normal to AB makes an angle α with normal to AC. Thus, three lines

(a) normal to section AB,

(b) normal to cross-section AC, and

(c) vertical axis of cross-section AC passing through its centre O′, are in the same plane.

The effect of axial W is to cause a moment T = W R about normal to AC. This moment is broken into two components, shown in Figure 13.3(c), viz. :

(a) W R cos α about axis which is perpendicular to the cross-section of the wire, and

(b) moment W R sin α about AO′C, a vertical axis of the cross-section and parallel to spring axis.

Thus, apparently, an axial load W, on an open coil helical spring has dual effect of

175

Springs(a) twisting the spring wire and inducing torsional shearing stress due to torque

W R cos α, and

(b) bending the spring wire and inducing bending stress due to BM W R sin α. Both torque and BM will cause axial deflection.

WR Cos α

d

W R`

α

A

BC

αα

WR

O′

A

B CO

α WR WR

α

C

A

O′

WR Sin α

(a) (b) (c)

Figure 13.3 : Open Coiled Helical Spring Subjected to Axial Load

Moment, M = W . R . . . (13.19)

Moment about the axis of the spring, (bending moment)

Mb = WR sin α . . . (13.20)

Moment about normal to the wire cross-section, (Torque)

T = WR cos α . . . (13.21)

Total length of the spring,

2cos

R nl π=

α . . . (13.22)

Work done by axial load,

12

W Δ . . . (13.23)

Strain energy due to bending,

2

1 2bM lU

E I= . . . (13.24)

Strain energy due to torque,

2

2 2T lUG J

= . . . (13.25)

Total strain energy,

1 2U U U= +

2 2

2 2bM l T l

E I G J= +

2 2 2 2 2 2( sin ) ( cos

2 2W R l W R l

E I G Jα α)

= +

2 2 2 2sin cos2

W R lUE I G J

⎡ ⎤α α= +⎢ ⎥

⎢ ⎥⎣ ⎦ . . . (13.26)

This strain energy should be equal to the external work done.

176

Stresses in Shafts and Shells ∴

2 2 2 21 sin2 2

W R lW cosE I G J

⎡ ⎤α αΔ = +⎢ ⎥

⎢ ⎥⎣ ⎦

⇒ 2 2

2 sin cosW R lE I G J

⎡ ⎤α αΔ = +⎢ ⎥

⎢ ⎥⎣ ⎦

2 2

2 2 sin coscos

R nW RE I G J

⎡ ⎤⎛ ⎞π α α= +⎢ ⎥⎜ ⎟α ⎢ ⎥⎝ ⎠ ⎦

. . . (13.27) ⎣

4 4;64 32

I d J dπ π= =

Substituting the values of I and J into the Eq. (13.26), we can get

3 2

464 sec 2sin cosW R n

E Gd

2⎡ ⎤α α αΔ = +⎢ ⎥

⎢ ⎥⎣ ⎦ . . . (13.28)

Both W R cos α (T) and W R sin α (Mb) will cause rotation of the coil also.

13.3.2 Spring Subjected to Axial Couple Consider an open coiled helical spring subjected to an axial couple ‘T’ as shown in Figure 13.4.

T

T

Figure 13.4

Consideration similar to these in Section 13.2.2 will show that an axial torque will have two components T sin α and T cos α. Out of these the former is the torque which twists the spring wire, while the latter is the BM which bends the spring wire.

Bending moment component,

cosbM T= α . . . (13.29)

This changes the curvature of coils.

Torque component,

1 sinT T= α . . . (13.30)

This causes the twist in the wire.

Let φ = angle of twist at the free end of the spring, caused by axial torque, T

∴ Work done 1 T2

= φ . . . (13.31)

177

Springs

Strain energy due to 2

1,2

bM lBM UE I

=

or 2 2

1( cos )

2TU

E Iα

= . . . (13.32)

Strain energy due to torque, 2

12 2

T lUG J

=

or 2 2

2( sin )

2TU

G Jα

= . . . (13.33)

Total strain energy,

U = U1 + U2

2 2 2 2( cos ) ( sin )2 2

T TE I G

α α= +

J

2 2 2cos sin2

T lE I G J

⎡ ⎤α α= +⎢ ⎥

⎢ ⎥⎣ ⎦ . . . (13.34)

Equating the total strain energy to work done, we can get

∴ 2 2cos sinT l

E I G J⎡ ⎤α α

φ = +⎢ ⎥⎢ ⎥⎣ ⎦

. . . (13.35)

where 2 cosl R n= π α

4 4

and64 32d dI Jπ π

= =

2 2

464 sec 2cos sinT R n

E Gd

⎡ ⎤α αφ = +⎢ ⎥

⎢ ⎥⎣ ⎦

α . . . (13.36)

Bending stress,

332b b

bM Mz d

σ = =π

332 cosT

dα

=π

. . . (13.37)

Shear stress,

13

16Td

τ =π

316 sinT

dα

=π

. . . (13.38)

Principal stress,

2 2

1, 2 3 332 cos (32 cos ) (16 sin )

2 2T T T

d d 3dα α α

σ = ± +π π π

316 (cos 1)T

d= απ

± . . . (13.39)

178

Stresses in Shafts and Shells Maximum shear stress 1 2

2σ − σ⎛ ⎞= ⎜ ⎟

⎝ ⎠

316T

d=π

. . . (13.40)

Example 13.14

An open coiled helical spring is made having 10 turns of a mean radius of 60 mm. The wire diameter is 10 mm and coils make an angle of 38o with a plane perpendicular to the axis of the coil. Find

(d) the axial extension with a load of 200 N, and

(e) the angle of rotation of free end.

E = 200 GPa, G = 80 GPa.

Solution

n = 10, R = 60 mm, d = 10 mm, α = 38, E = 200 GPa, G = 80 GPa

(a) Axial Extension 3 2

464 sec cos 2sinW R

G Ed

⎛ ⎞2α α αΔ = +⎜ ⎟⎜ ⎟

⎝ ⎠

3 2 2

4 3 364 (200) (60) (10) sec 30 cos 30 2sin 30 36.7 mm

10 80 10 200 10

⎛ ⎞= +⎜ ⎟⎜ ⎟× ×⎝ ⎠

=

(b) Angle of Rotation at Free End 3

464 sin 1 2W R

G Edα ⎛ ⎞φ = −⎜ ⎟⎝ ⎠

3

4 3 364 (200) (60) (10) sec 30 1 2 0.0567 radian

10 80 10 200 10

⎛ ⎞= −⎜ ⎟⎜ ⎟× ×⎝ ⎠

=

Exampled 13.15

Find the mean radius of an open coiled spring of helix angle of 38o, to give a vertical displacement of 20 mm and an angular rotation of 0.02 radian at free end under an axial load of 30 N. The material available is 6 mm diameter steel bar. Take E = 200 GPa, G = 80 GPa.

Solution

α = 30o, Δ = 20 mm, φ = 0.02 radian, W = 30 N, E = 200 GPa, G = 80 GPa

3 2

464 sec cos 2sinW R

G Ed

⎛ ⎞2α α αΔ = +⎜⎜

⎝ ⎠⎟⎟ . . . (1)

3

464 sin 1 2W R

G Edα ⎛ ⎞φ = −⎜ ⎟⎝ ⎠

. . . (2)

Eq. (2) divided by Eq. (1), we get

∴

2 2cos 2sin

1 2sin cosG ER

G E

⎡ ⎤α α+⎢ ⎥

Δ ⎣ ⎦=φ α α ⎡ ⎤−⎢ ⎥⎣ ⎦

179

Springs

∴

2 2

3 3

3 3

cos 30 2sin 3080 10 200 1020

0.02 sin 30 cos 30 1 280 10 200 10

R

⎡ ⎤+⎢ ⎥

× ×⎣ ⎦=⎡ ⎤

−⎢ ⎥× ×⎣ ⎦

(0.115)10000.433 0.025

R=

×

∴ R = 94 mm

Example 13.16

In an open coiled spring of 10 coils the stresses due to bending and twisting are 120 N/mm2 and 150 N/mm2 respectively when the spring is loaded axially. Assuming the mean radius of the coil is 5 times the wire diameter, find the maximum permissible axial load and the wire diameter for a maximum extension of 20 mm.

E = 200 GPa, G = 80 GPa.

Solution

n = 10, R = 5d, Δ = 20 mm, , τ = 120 N/mm3200 GPa = 200 10 N/mmE = × 2

2

2,

, W = ?, d = ? 380 GPa = 80 10 N/mmG= ×

23

32 sin 120 N/mmbW R

dα

σ = =π

23

16 cos 150 N/mmW Rd

ατ = =

π

Taking ratio of b

τσ

1202tan150

α =

tan 0.4α =

2sin 0.37; sin 0.14α = α =

, sec α = 1.075 2cos 0.93; cos 0.86α = α =

3 2

464 sec cos 2sinW R

G Ed

⎛ ⎞2α α αΔ = +⎜ ⎟⎜ ⎟

⎝ ⎠

⇒ 3

4 364 (5 ) (10) 0.86 2 0.1420

0.93 80 10 200 10W dd

⎡ ⎤×= +⎢ ⎥

× × ×⎢ ⎥⎣ ⎦3

∴ 19.99Wd

= . . . (1)

332 (5 ) 0.37 120W d

d× × ×

=π

⇒ 2 6.37Wd

= . . . (2)

180

Eq. (1) divided by Eq. (2), we get Stresses in Shafts and Shells

d = 3.0 m 19.99 3 1145.43 NW = × =

Example 13.17

In an open coiled helical spring having α = 38o, if the inclination of the coil is ignored, calculate the percentage by which the axial extension is under estimated.

E = 200 GPa, G = 80 GPa.

Solution 3 2 2

464 cos 2sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

2 2cos 2sin

cosK

G E⎡ ⎤α α

= +⎢ ⎥α ⎢ ⎥⎣ ⎦

o 41

0.11538 , 10 0.01470.866

K K−α = Δ = × × = ×

If α is neglected,

20, 0.0125Kα = Δ = ×

% Error 1 2

1100Δ − Δ

= ×Δ

0.0147 0.0125 100 14.96%0.0147

−= × =

Example 13.18

An open coiled helical spring having 10 complete turns is made of 16 mm diameter steel rod, the mean radius of the coil being 50 mm. The angle of helix is 28o. Calculate the deflection under an axial load of 300 N. Also calculate the direct and shear stress induced in the section of the wire. If the axial load of 300 N is replaced by an axial moment of 9000 N-mm, determine the axial deflection and angle of rotation.

Take E = 200 GPa, G = 80 GPa.

Also, calculate the axial torque which will cause a bending stress of 12 N/mm2.

Solution

n = 10, d = 16 mm, R = 50 mm, α = 28o, W = 300 N

3 2 2

464 cos 2sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

3 2 2

4 364 300 50 10 cos 20 2sin 20

16 cos 20 80 10 200 103

⎡ ⎤× × ×= +⎢ ⎥

× × ×⎢ ⎥⎣ ⎦

∴ Δ = 4.74 mm

23 3

32 sin 32 300 50 sin 20 12.8 N/mm16b

W Rd

α × × ×σ = = =

π π ×

23 3

16 cos 16 300 50 sin 20 17.5 N/mm16

W Rd

α × × ×τ = = =

π π ×

181

SpringsPrincipal stresses,

2

21,2 2 2

b bσ σ⎛ ⎞σ = ± + τ⎜ ⎟⎝ ⎠

2

2 212.8 12.8 17.5 6.4 18.6 25 N/mm 12.2 N/mm2 2

⎛ ⎞= ± + = ± = −⎜ ⎟⎝ ⎠

2

Maximum shear stress 21 2 25 ( 1.22) 18.6 N/mm2 2

σ − σ − −= = =

M = 9000 N-m

2 2

464 2cos sin

cosM R n

E Gd

⎡ ⎤α αφ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

2 2

4 364 9000 50 10 2cos 20 sin 20

(16) cos 20 200 10 80 10

⎡ ⎤× × ×= +⎢ ⎥

× ×⎢ ⎥⎣ ⎦3

φ = 0.048 radians

2

464 sin 1 2M R n

G Edα ⎡ ⎤φ = −⎢ ⎥⎣ ⎦

2

4 364 9000 50 10 sin 20 1 2

(16) 80 10 200 103

⎡ ⎤× × × ×= −⎢ ⎥

× ×⎢ ⎥⎣ ⎦

= 0.19 mm

332 cos

bT

dα

σ =π

⇒ 332 cos 2012

16T ×

=π ×

∴ 5135.2 N-mmT =

SAQ 2 (a) The elongation of an open coiled helical spring is 1.8% greater than that of a

close coiled helical spring, otherwise they are similar in every respect. Calculate the helix angle.

Take E = 20 GPa, G = 80 GPa.

(b) In an open coiled helical spring having α = 38, if the inclination of the coils is neglected, calculate the percentage error in the value of stiffness.

E = 200 GPa, G = 80 GPa.

(c) An open coiled helical spring of wire 4 mm is diameter is coiled to a mean radius of 20 mm. It is loaded with an axial load only. Calculate the helix angle which would cause an error of 3% in deflection if the spring is assumed to be closely coiled.

Take μ = 0.25.

(d) The error in calculating the elongation of helical spring subjected to an axial load assuming the close coiled formula is not to exceed 1%. Calculate the angle of helix.

Take E = 2.5 G.

182

(e) An open coiled helical spring is made of steel wire having 3 mm diameter. It has 6 coils of 25 mm radius and 25 mm pitch. The spring is acted upon by a winding torque about the axis of the spring resulting in an increase of coils by 0.5, the length remaining 150 mm. Determine the torque required and the minimum elastic limit strength of steel to permit this amount of winding.

Take E = 200 GPa. G = 80 GPa.

Stresses in Shafts and Shells

13.4 COMPOUND SPRINGS

Sometimes, more than one spring may be used. The springs may be used in parallel or in series. This combination is called the compound springs.

Springs in Parallel

Figure 13.5 shows two springs connected in parallel. The springs are provided side by side.

W

W

1 2

Figure 13.5 : Springs in Parallel

In this case : (a) Sum of the individual loads carried by each spring is equal to the load

carried by compound spring W1 + W2 = W . . . (13.41)

(b) The deflections of both the springs as well as the combination are same

i.e. 1 2Δ = Δ = Δ . . . (13.42)

∴ 1 2K K K+ = . . . (13.43)

where K = Spring constant or stiffness of spring. Springs in Series

Figure 13.6 shows two springs connected in series. The springs are provided end to end.

1

2

Figure 13.6 : Springs in Series

In this case

183

Springs(a) The load carried by each spring is the same as that carried by the

compound spring

W1 + W2 = W . . . (13.44)

(b) The total deflection is equal to the sum of the deflection in each spring.

1 2Δ = Δ + Δ . . . (13.45)

i.e. 1 2W W W

Δ ΔΔ= +

∴ 1 2

1 1 1K K K

= + . . . (13.46)

Example 13.19

In a compound helical spring, the inner spring is arranged within and concentric with the outer one, but is 10 mm shorter. The outer spring has 10 coils of mean radius 12 mm, and the wire diameter is 3 mm. Find the stiffness of the inner spring if an axial load of 150 N causes the outer one to compress 20 mm.

If the radius clearance between the springs is 1.5 mm, find the wire diameter of the inner spring when it has 8 coils. G = 80 GPa.

Solution

Springs are in Parallel

Load carried by outer spring :

3

11 4

64 W R nG d

Δ =

3

13 4

64 12 1080 10 3

W× × ×=

× ×

∴ W1 = 117 kN

Load carried by inner spring, W2 = 150 –117 = 33 N

Compression, 2 20 10 10 mmΔ = − =

Stiffness, 233 3.30 N/mm10

K = =

3

22 4

64 W R nG d

Δ =

3

3 464 33 (9 0.5 ) 8

80 10d

d× × − ×

=× ×

⇒ 3 4(18 ) 379d d− =

Since d is small compared with 18, for a first approximation :

58304 1.98379

d = = mm

Second approximation

316.024 1.80

379d = = mm

Third approximation

184

Stresses in Shafts and Shells

316.24 1.83379

d = = mm

Fourth approximation

316.174 1.83

379d = = mm

∴ d = 1.83 mm Example 13.20

A composite spring has two close-coiled helical spring connected in series, each spring has 10 coils at a mean radius of 15 mm. Find the diameter of one if the other is 2.5 mm and the stiffness of the composite spring is 750 N/mm. Calculate the greatest load that can be carried by the composite springs, and the corresponding extension, for a maximum shearing stress of 200 N/mm2. G = 80 GPa.

Solution For spring in series, W is same. 1 2Δ = Δ + Δ . . . (1)

1 2

1 1 1K K K

= + . . . (2)

3

464 W R n

G dΔ =

3

1 3 464 15 1080 10 2.5

W × ×Δ =

× ×

11

1.45 N/mmW K= =Δ

. . . (3)

750 N/m = 0.75 N/mmK = . . . (4)

3

2 3 464 15 1080 10

Wd

× ×Δ =

× ×

42

2(0.037)W K d= =

Δ . . . (5)

Using (2), (3), (4) and (5), we get

41 1 1

0.75 1.45 0.037 d= +

×

∴ d = 2.55 mm

max 316 W R

dτ =

π

⇒ 316 15200

(2.5)W ×

=π

∴ W = 40.9 N

Total extension, 40.9 54.5 mm0.75

WK

Δ = = =

Example 13.21

185

SpringsTwo close-coiled helical springs are compressed between two parallel plates by a

load of 1000 N. The springs have a wire diameter of 10 mm and radii of the coils are 50 mm and 75 mm. Each spring has 10 coils and of the same initial length. If the smaller spring is placed inside the larger one,

calculate

(a) the total deflection, and

(b) the maximum stress in each spring.

Take G = 40 GPa.

Solution

In this case, the springs are connected in parallel.

1 2Δ = Δ

3

464 W R n

G dΔ =

3 3

1 24 4

64 50 64 75W n WG d G d× × × ×

=n

∴ W1 = 3.375 W2

W1 + W2 = 1000

⇒ 3.375 W2 + W2 = 1000

∴ W2 = 229 N

∴ W1 = 1000 – 229 = 771 N

3

1 2 3 464 771 50 10 6.168 mm

40 10 10× × ×

Δ = Δ = =× ×

2max1 3 3

16 16 771 50 196.3 N/mm10

W Rd

× ×τ = = =

π π ×

2max 2 3 3

16 16 229 75 87.5 N/mm10

W Rd

× ×τ = = =

π π ×

Example 13.22

Two concentric springs are subjected to an axial load of 60 kN. The maximum allowable deflection of the springs is 45 mm and the solid length is 55 mm. If the springs are of the material having G = 80 × 103 N/mm2 and the maximum allowable shear stress is 800 N/mm2, calculate

(a) load shared by the springs, and

(b) wire diameter and outer spring radius.

Take inner spring radius = 40 mm and radial clearance = 3 mm.

Solution

The springs are in parallel. Deflection of both the springs is same.

3

464 W R n

G dΔ =

3

1 13 4

1

64 404580 10

W nd

× × ×=

× ×

186

Stresses in Shafts and Shells 1 1

41

0.88W nd

= . . . (1)

max 316 W R

dτ =

π

131

16 40800 Wd×

=π

131

3.93Wd

= . . . (2)

11 1 1

1

5555 mmn d nd

= ⇒ = . . . (3)

From Eq. (1) and (3), we get

151

55 0.88Wd×

=

151

0.016Wd

= . . . (4)

From Eq. (2) to Eq. (1), we get

2 21 245.6 mmd =

∴ 1 15.67 mm 15.7 mmd = ;

31 3.93 15.7 = 15 129 N = 15.1 kNW = ×

1 2 60 kNW W+ =

∴ 2 44.9 kNW =

Outer Spring

22 2

80 15.7 2 3 = (50.85 0.5 ) mm2

dR d+ + × += +

max 316 W R

dτ =

π

3 2

32

16 44.9 10 (50.85 0.5 )800 dd

× × × +=

π ×

∴ 2 24.4 mmd =

∴ 2 (50.85 0.5 24.2) 63.05 mmR = + × =

Example 13.23

The table below gives particulars of a compound spring consisting of two co-axial close-coiled springs. If the spring is subjected to an axial load of 450 N, determine for each spring

(a) the change in length,

(b) the amount of load carried, and

(c) the maximum shearing stress.

Take G = 80 × 103 N/mm2.

187

Springs Mean Coil Radius

(mm) Diameter of Wire

(mm) n Free Length

(mm)

Outer Spring 25 5 10 100

Inner Spring 15 4 8 80

Solution The free length of outer spring is more than the inner spring by (100 – 80) = 20 mm. Load carried by the outer spring for a deflection of 20 mm is given by,

3

464 W R n

G dΔ =

⇒ 3

13 4

64 25 102080 10 5

W× × ×=

× ×

∴ 1 100 NW =

Now the springs are in parallel. Load to be shared by the two springs 450 100 350 N= − =

3

464 W R n

G dΔ =

⇒ 1 3

13 4

64 ( ) 25 1080 10 5

W ×Δ =

× ×

∴ 11 5W = Δ . . . (1)

1 32

3 464 ( ) 15 880 10 4

W ×Δ =

× ×

∴ 12 11.9W = Δ . . . (2)

1 11 2 350 NW W+ =

⇒ 5 11.9 350Δ + Δ =

∴ 20.7 mmΔ =

Change in length, 20.7 mmΔ =

11 5 5 20.7 103.5 NW = Δ = × =

12 11.9 11.9 20.7 246.5 NW = Δ = × =

Load taken by outer spring = 203.5 N. Load taken by inner spring = 246.5 N.

max 316 W R

dτ =

π

τmax is outer spring 23

16 203.5 25 207.3 N/mm5

× ×= =

π ×

τmax in inner spring 23

16 246.5 15 294.2 N/mm4

× ×= =

π ×

∴ Maximum shearing stress = 294.2 N/mm2 (inner spring).

188

Example 13.24 Stresses in Shafts and Shells A rigid bar of negligible weight transmits a load W to a combination of 3 springs

as shown in Figure 13.7. The three springs are made of the same material and are of equal diameters. They are of the same length before loading. The number of coils in three springs are 10, 12 and 15 respectively, while the mean radii of coils are in the proportion 1 : 1.2 : 1.5 respectively. Find the distance x such that the rigid bar remains horizontal after the application of the load.

W

x

1

Figure 13.7

Solution Since the springs are in parallel, Δ is same. 1 2 3W W W W+ + = . . . (1)

G is same, d is same. 2 11.2R R=

3 11.5R R=

1 2 310; 12; 15n n n= = =

3

464 W R n

G dΔ =

⇒ 4

364G dWR n

Δ=

3KW

R n=

1 3 31 1

0.110

K KWR R

= =×

2 3 31 1

0.048(1.2 ) 12

K KWR R

= =×

3 3 31 1

0.02(1.5 ) 15

K KWR R

= =×

1 2 3 31

0.168 KW W W WR

= + + =

The FBD of bar shown in Figure 13.8.

Figure 13.8

2 3

Rigid Bar

W

D B C A

X

W1 W2 W3

l l

189

SpringsNow, taking moments about A,

2 3. 2W x W l W l= × + ×

⇒ 0.168 0.048 0.02 2x l l× = × + ×

∴ x = 0.52 l

SAQ 3 (a) The following data refers to two close-coiled helical springs A and B.

Spring n R (mm) d (mm) Uncompressed Axial Lengths (mm)

A 8 50 6 70

B 10 40 5 80

Spring B is placed inside A and both are compressed between a pair of parallel plates until the distance between the plates measures 60 mm. Calculate

(i) the load applied to the plates, and

(ii) the maximum shear stress in the spring.

Take G = 80 GPa.

(b) Two close-coiled springs A and B are connected in series to form a composite spring of stiffness 1000 N/m. The spring A has 20 effective turns with a wire diameter 2 mm while B has 30 effective turns. If the spring index of the springs A and B is 8, find the wire diameter of spring B. Also calculate the maximum tensile load that can be applied to the composite spring and the maximum extension of the composite spring under this load. τmax = 200 N/mm2, G = 80 GPa. Neglect the effect of spring index on shear stress.

(c) A close-coiled helical spring held with axis vertical is built in at both ends. If an axial load W is applied at an intermediate point on it such that the number of coils above the point of application of load is 16 and below this point is 12, calculate how the load W will be shared between top and bottom of application point.

13.5 LEAF SPRINGS

These are known as laminated, carriage or built-up springs. These are used in cars, lorries, automobiles and railway wagons.

These are made of number of leaves of equal width and thickness, but varying length. The number of leaves decreases towards the end of the spring. The spring is designed such that the maximum bending stress is same at all sections.

Types of leaf springs :

(a) Semi-elliptic type, and

(b) Quarter-elliptic type.

190

Semi-elliptic Type Leaf Springs Stresses in Shafts and Shells

Figure 13.9 shows a carriage spring carrying a central load W.

y

W

W / 2 W / 2

l / 2 l / 2

Figure 13.9

Let l = Span, b = Width of leaves, t = Thickness of leaves, W = Load, y = Rise of crown, n = Number of plates, and R = Initial radius of curvature of plates.

Section modulus of single plate 2

6bt

= .

Z for the whole spring 2

6nbt⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

.

Maximum BM, 4

W lM =

.bM Z= σ

2234 .2

4

b

W lM W lZ nbtbtn

⎛ ⎞⎜ ⎟⎝ ⎠σ = = =⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

. . . (13.47)

From geometry of circles, because each leaf is assumed to be bent in form of a part of a circle.

(2 )2 2l ly R y− = ×

∴ 28lyR

= . . . (13.48)

0

1 1ME I R R

= −

⇒ 02384 ( )

.12

W l

y ylnbtE

−⎛ ⎞⎜ ⎟⎝ ⎠ = −⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

191

Springs

∴ Deflection, 3

33

8W l

nbt EΔ = . . . (13.49)

2 3

31 32 16

W lU Wnbt E

= Δ = . . . (13.50)

Spring constant 3

38

3W E nbt

l= =Δ

. . . (13.51)

Proof Load : Load required to make the spring flat.

Quarter-elliptic Type Leaf Springs

Figure 13.10 shows a quarter elliptic spring carrying an end load W at free end

2

6nbtZ =

M W l= −

26

bM W lZ nb

σ = = . . . (13.52)

y

l W

Figure 13.10

From geometry of circles,

(2 )y R y l l− = ×

22lyR

=

0

1 1ME I R R

= −

⇒ 023( ) 2 ( )

12

W l y ylnbtE

−= −

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

∴ 3

36W lnbt E

Δ = . . . (13.53)

2 3

31 32

W lU Wnbt E

= Δ = . . . (13.54)

Example 13.25

A laminated steel spring, simply supported at the ends and centrally loaded, with a span of 0.8 m; is required to carry a proof load of 8 kN; and the central deflection is not to exceed 50 mm; the bending stress must not exceed 400 kN/mm2, plates

192

are available in multiples of 1 mm for thickness and 4 mm for width. Determine suitable values of width, thickness and number of plates, and calculate the radius to which the plates should be formed. Assume width = 12 × thickness. E = 200 GPa.

Stresses in Shafts and Shells

Solution Simply Supported Spring

l = 0.8 m W = 8 kN = 800 N Δ = 50 mm σb = 400 N/mm2

b = 12 t E = 200 GPa = 200 × 103 N/mm2

3

33

8W l

nbt EΔ =

⇒ 3

3 33 8000 80050

8 (12 ) (200 10 )n t t× ×

=× ×

∴ . . . (1) 4 12,800nt =

23

2W lnbt

σ =

⇒ 23 8000 800400

2 (12 )n t t× ×

=

∴ . . . (2) 3 2,000nt =

From Eq. (1) to Eq. (2), we get ⇒ t = 6.4 mm = 7 mm 12 12 7 84 mmb t= = × =

3 2,000nt =

⇒ 37 20n × = 0

∴ 5.8 6n = ;

Actual deflection under the proof load,

3

0 3 33 (8000) (800) 44.4 mm

8 6 84 7 200 10Δ = =

× × × × ×

Initial radius of curvature,

2 2

00

(800) 1802 mm8 (8 44.4)lR = = =Δ ×

Example 13.26 A laminated spring of the quarter elliptic type, 0.6 m long, is to provide a static deflection of 80 mm under an end load of 2000 N. If the leaf material is 60 mm wide and 5 mm thick, find the number of leaves required and the maximum stress.

From what height can the load be dropped on to the undeflected spring to cause of maximum stress of 8000 N/mm2? E = 200 GPa.

Solution

193

SpringsQuarter Elliptic Leaf Spring

l = 0.6 m

Δ = 80 mm

W = 2000 N

b = 60 mm

t = 5 mm

n = ?

σb = ?

3

36W lnbt E

Δ =

⇒ 3

3 33 2000 6008060 5 200 10n× ×

=× × × ×

∴ 21.6n = (say 22 leaves)

26

bW l

nbtσ =

22

6 2000 600 369 N/mm22 60 5× ×

= =Δ × ×

For maximum stress of 800 N/mm2

26 60080022 60 5

W× ×=

× ×

⇒ 7333.3 NW =

Corresponding deflection 3

3 36 7333.3 600 288 mm

22 60 5 200 10× ×

= =× × × ×

Loss of PE = Gain of KE

12000 ( 288) 7333.3 2882

h + = × ×

∴ 240 mmh =

Example 13.27

A carriage spring, centrally loaded and simply supported at its ends, has 12 steel plates each 60 mm wide by 6 mm thick. If the longest plate is 700 mm, find the initial radius of curvature if the maximum stress is 150 N/mm2 and the plates are finally straight.

Neglecting loss of energy at impact, determine the height from which 200 N can be dropped centrally on the spring. E = 200 GPa.

Solution

Simply supported carriage spring :

n = 12

b = 60 mm

t = 6 mm

194

l = 700 mm Stresses in Shafts and Shells

σb = 150 N/mm2

E = 200 GPa = 200 × 103 N/mm2

23

2bW lnbt

σ =

∴ 3703 NW =

23

8W l

nbt EΔ =

3

3 33 3703 700 15.3 mm

8 12 60 6 200 10× ×

= =× × × × ×

2 2

0700 4003 mm = 4 mm

8 8 15.3lR = = =Δ ×

U = Change in PE

1 3703 15.3 200 ( 15.3)2

h× × = × +

∴ h = 126 mm

Example 13.28 A leaf spring having a span of 1.40 m consists width and thickness of leaves to be 100 mm and 12 mm respectively. The maximum bending stress is 150 N/mm2 and the spring must absorb 125000 N-mm when straightened. Calculate the number of leaves and initial curvature. Take E = 200 GPa.

Solution d = 1400 mm b = 100 mm t = 12 mm

σb = 150 N/mm2

U = 125000 N-mm n = ? R0 = ?

E = 200 × 103 N/mm2

23

2bW lnbt

σ =

⇒ 23 1400150

2 (100) (12)W

n×

=

∴ W = 1029 n . . . (1)

3 3

3 3 33 3 (1029 ) (1400) 30.6 mm

8 8 (200 10 ) (100) (12)W l n

n E bt nΔ = = =

×

12

U W= Δ

195

Springs

⇒ 30.6125000 10292

n= ×

∴ 79 8n = ;

2 2

0(1400) 8007 mm

8 8 30.6lR = = =Δ ×

SAQ 4 A quarter elliptic leaf spring of length 450 mm has 4 leaves of thickness 9 mm. An end load of 2500 N causes a deflection of 40 mm and E = 200 GPa. Calculate the width of the leaves.

13.6 SUMMARY

Close Coiled Helical Springs

Under axial load

416W R

dτ =

π

3

464W R n

G dΔ =

Open Coiled Helical Springs

Under axial load

3 2 2

464 cos 2sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

Semi-Elliptic Springs

23

2W lnbt

σ =

3

23

8W l

nbt EΔ =

Quarter-Elliptic Springs

23

2W lnbt

σ =

3

36W lnbt E

Δ =

196

Stresses in Shafts and Shells 13.7 ANSWERS TO SAQs

SAQ 1 (a) P = 200 N d = 25 mm

h = 600 mm R = 100 mm

Δ = 120 mm G = 80 GPa = 80 × 103 N/mm2

1( ) . .2

p h W+ Δ = Δ

∴ 1200 (600 120) 1202

W+ = × ×

∴ 2,400 NW =

2max 3 3

16 16 2400 100 78.2 N/mm(25)

W Rd

× ×τ = = =

π π

3

464W R n

G dΔ =

∴ 3

3 464 2400 100120

80 10 (25)n× × ×

=× ×

∴ 24.4 25n = ;

(b) Weight of trolley = 20 kN = 20,000 N

3.6 53.6 kmph = 1 m/sec.18

V ×= =

21KE2

mv=

21 20000 1 1019.4 N-m2 9.81⎛ ⎞= × =⎜ ⎟⎝ ⎠

3

464W R n

G dΔ =

∴ 3

3 464 (100) 1212080 10 20

W ×=

× ×

∴ W = 2,000 N

12

U W= Δ

31 2000 120 120 10 N-mm2

= × × = ×

∴ 3

31019.4 10 8.5 9

120 10KENU

×= = =

×;

(c) d = 6 mm, n = 10,

R = 40 mm, τmax = 70 N/mm2

G = 80 GPa = 80 × 103 N/mm2

W = ?, Δ = ?

197

Springs

max 316W R

dτ =

π

∴ 316 4070

(6)W× ×

=π

∴ W = 74.2 N

3

464 W R n

G dΔ =

3

3 464 74.2 40 10 29.3 mm

80 10 6× × ×

= =× ×

(d) M = 15 N-m, n = 10, d = 16 mm, R = 150 mm, E = 200 GPa

2 2 150 10 9424.8 mml R n= π = π × × =

4 4(16) 3217 mm64 64

I dπ π= = = 4

3

3(15 10) (9424.8) 0.22 rad(200 10 ) (3217)

M lE I

×φ = = =

×

1 1 15 0.22 1.65 N-m2 2

U M= φ = × × =

32

3 332 32 15 10 37.3 N/mm

16bMd

× ×σ = = =

π π ×

(e) 3

464 W R n

G dΔ =

4

364W G dK

R n= =Δ

d is same. G is same.

n is same.

1 31

AKR

=

2 32

AKR

=

23

1 23

2 1

15 130 8

K RK R

⎛ ⎞= = =⎜ ⎟⎝ ⎠

(f) d = 6 mm, n = 12,

R = 45 mm, M = 1 N-m

E = 200 GPa

3

23 3

32 32 1 10 47.2 N/mm16b

Md

× ×σ = = =

π π ×

198

2 2 45 12 3392.9 mml R n= π = π × × = Stresses in Shafts and Shells

4

464 63.6 mm64 64dI π π ×

= = =

3

3(1 10) (3392.9) 0.27 rad(200) 10 63.6

M lE I

×φ = = =

× ×

2 ( )nπ Δ = φ

0.043nΔ =

Here Δ n = increase in number of turn.

(g) R = 30 mm, M = 1.2 N-m,

W = 80 N, radians3π

φ = ,

Δ = 80 mm

3

464 W R n

G dΔ =

n is same.

d is same.

⇒ 3

464 80 3080 n

G d× × ×

=

4

3 31728 10 1.73 10G dn

= × = × . . . (1)

M lE I

φ =

⇒ 3

4(1.2 10 ) (2 30)

364

ndE

π × × π × ×=

⎛ ⎞π ×⎜ ⎟⎜ ⎟⎝ ⎠

4

64.4 10E dn

= × . . . (2)

Dividing Eq. (2) by Eq. (1), we get

⇒ 2.5EG

=

2 (1 ) 2.5GG+ μ

=

∴ μ = 0.25 (h) Let d = diameter of wire.

Diameter of bar = 30 mm

15 mm2dR ⎛ ⎞= +⎜ ⎟

⎝ ⎠

W = 500 N

Δ = 2 mm

199

Springs τmax = 25 MPa

G = 80 GPa = 80 × 103 N/mm2

d = ?

R = ?

n = ?

max 316W R

dτ =

π

⇒ 3

16 500 15225

d

d

⎛ ⎞× × +⎜ ⎟⎝ ⎠=

π

⇒ 3 50.9 1527.9 0d d− − =

By trial and error,

d = 13 mm

1315 15 21.5 mm2 2dR = + = + =

3

464 W R n

G dΔ =

⇒ 3

3 464 500 21.52

80 10 (13)n× × ×

=× ×

∴ 14.4 15n = ;

(i) Shaft

260N TP π

=

⇒ 3 2 2002.4 1060

Tπ × ×× =

∴ T = 115. N-m

Spring

M = T = 11.5 N-m

d = 10 mm

R = 30 mm

E = 200 GPa

n = 10

2 2 30 10 1885 mml R n= π = π × × =

4 4

410 491 mm64 64dI π π ×

= = =

3

23 3

32 32 11.5 10 117.1 N/mm10b

Md

× ×σ = = =

π π ×

3

3(11.5 10) (1885) 0.22 rad(200 10 ) (491)

M lE I

×φ = = =

× ×

200

(j) G, W, d and R1 n are same. Stresses in Shafts and Shells

3

464 W R n

dΔ =

π

1 24 41 2

,A Ad d

Δ = Δ =π π

∴ 4

1 2

2 1

dd

⎛ ⎞Δ= ⎜ ⎟Δ ⎝ ⎠

(h) Δ = 100 mm, W = 500 N, R = 5 d τmax = 60 MPa, G = 80 GPa = 80 × 103 N/mm2

d = ? R = ? n = ?

max 316 W R

dτ =

π

⇒ 316 500 (5 )60 d

d× ×

=π

∴ 14.6 mm 15 mmd = ;

5 5 15 75 mR d m= = × =

3

464 W R n

G dΔ =

⇒ 3

3 464 500 75100

(80 10 ) (15)n× × ×

=×

∴ n = 30

12

W P× × Δ = × Δ

⇒ 1 1 500 250 N2 2

P W= × = × =

SAQ 2

(a) 3 2 2

1 464 cos 2sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

3

2 464 W R n

G dΔ =

1 21.018Δ = × Δ

⇒ 2 21 2cos sin 1.018cos

GE

⎡ ⎤α + α =⎢ ⎥α ⎣ ⎦

⇒ 2 2cos 0.8 sin 1.018 cosα + α = α

⇒ 2 2cos 0.8 (1 cos ) 1.018 cosα + − α = α

⇒ 2cos 5.09 cos 4 0α − α + =

∴ cos 0.97α =

∴ o14.07α =

201

Springs

(b) 3 2 2

464 cos sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

For α = 38,

3

4 41 4

64 0.133 10 0.133 10W R n Ad

− −⎛ ⎞Δ = × × = ×⎜ ⎟⎜ ⎟

⎝ ⎠

For α = 8, Δ2 = 0.125 × 10– 4 A

41

17.5 10W WK B

A⎛ ⎞= = × =⎜ ⎟Δ ⎝ ⎠

B

42

28 10WK B−= = ×

Δ

% error 1 2

1100K K

K−

= ×

7.5 8 100 6.67%7.5−

= × = −

(c) d = 4 mm, R = 20 mm, μ = 0.25

E = 2G (1 + μ) = 2.5 G

3 2 2

1 464 cos 2sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

3

2 464 W R n

G dΔ =

2 10.97Δ = Δ

⇒ 2 20.97 21 cos sincos

GE

⎡ ⎤= α +⎢ ⎥α ⎣ ⎦α

⇒ 2 2cos 0.97 cos 0.78 sinα = α + α

⇒ 20.19 cos cos 0.78 0α − α + =

∴ cos 0.952α =

∴ o17.82α =

(d) 3 2 2

1 464 cos 2sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

3

2 464 W R n

G dΔ =

1 2

1100 1Δ − Δ

× =Δ

⇒ 1 2 0.01Δ − Δ = Δ1

⇒ 2 10.99Δ = Δ

⇒ 2 20.99 21 cos sincos

GE

⎡ ⎤= α +⎢ ⎥α ⎣ ⎦α

202

Stresses in Shafts and Shells

⇒ 2 2cos 0.99 cos 0.79 sinα = α + α

⇒ 20.2 cos cos 0.79 0α − α + =

∴ cos 0.983α =

∴ o10.6α =

(e) d = 3 mm, n = 6, R = 25 mm

p = 25 mm, 3 2200 GPa 200 10 N/mmE = = ×

3 280 GPa 08 10 N/mmG = = ×

25tan2 2

pR

α = =π π × 25

sin 0.16 cos 0.99α = α =

2 2sin 0.02 cos 0.97α = α =

Deflection due to axial load, W

3 2 2

1 464 cos 2sin

cosW R n

G Ed

⎡ ⎤α αΔ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦ . . . (1)

Deflection due to axial couple, M

2

2 464 1 2sinM R n

G Ed⎡ ⎤Δ = α −⎢ ⎥⎣ ⎦

. . . (2)

1 2 0Δ + Δ =

⇒ 1 2Δ = − Δ

⇒ 3 2 2

464 cos 2sin

cosW R n

G Ed

⎡ ⎤α α= +⎢ ⎥

α ⎢ ⎥⎣ ⎦

2

464 sin 1 2M R n

G Edα ⎡ ⎤= − −⎢ ⎥⎣ ⎦

⇒ 2 2. cos 2sin

sin cosW R

G E⎡ ⎤α α

= +⎢ ⎥α α ⎢ ⎥⎣ ⎦

1 2MG E⎡ ⎤= − −⎢ ⎥⎣ ⎦

⇒ 3 325 0.97 2 0.02

0.16 0.99 80 10 200 10W ⎡ ⎤× ×

+⎢ ⎥× × ×⎢ ⎥⎣ ⎦

3 31 2

80 10 200 10M −

⎡ ⎤= − −⎢ ⎥

× ×⎢ ⎥⎣ ⎦

⇒ 19.45 0.025W M= −

778M W= − . . . (3)

0.5 coil 0.5 2 radiansφ = = × π = π

203

SpringsAngle of rotation due to axial load, W

2

1 464 sin 1 2W R n

G Edα ⎡ ⎤φ = −⎢ ⎥⎣ ⎦

2

4 364 (25) (6) (0.16) 1 2

3 80 10 200W

310

⎡ ⎤= −⎢ ⎥

× ×⎢ ⎥⎣ ⎦

31 102 10 W−φ = ×

Angle of rotation due to axial couple, M

2 2

2 464 2cos sin

cosM R n

E Gd

⎡ ⎤α αφ = +⎢ ⎥

α ⎢ ⎥⎣ ⎦

4 3 364 ( 778 ) (25) (6) 2 0.97 0.16 1.09

3 (0.99) 200 10 80 10W W

⎡ ⎤− ×= +⎢ ⎥

× ×⎢ ⎥⎣ ⎦= −

1 2φ + φ = π

⇒ 31.2 10 1.09W W−× − = π

2.9 NW = −

778 2245 N-mmM W= − =

23 3

32 32 2245 847 N/mm3b

Md

×σ = = =

π π ×

SAQ 3

(a) Springs are in parallel

A BW W W= +

max 316 W R

dτ =

π

3

464 W R n

G dΔ =

For Spring A 70 60 10 mmΔ = − =

3

3 464 (50) 810

(80 10 ) (6)AW× ×

=×

∴ 16.2 NAW =

For Spring B 80 60 20 mmΔ = − =

3

3 464 40 1020

(80 10 ) 5BW× × ×

=× ×

∴ 24.4 NBW =

∴ Total laod, 16.2 24.4 40.6 NA BW W W= + = + =

204

Stresses in Shafts and Shells τmax for spring A 2

316 16.2 50 19.1 N/mm

6× ×

= =π ×

τmax for spring B 23

16 24.4 40 39.8 N/mm5

× ×= =

π ×

∴ Maximum shear stress = 39.8 N/mm2

(b) For spring in series.

Spring index = 8

8Dd=

R = 4d

A BW W W= = . . . (1)

A BΔ = Δ + Δ . . . (2)

3

464 W R n

G dΔ =

1 N/mmW WK

Δ = ⇒ =Δ

WΔ = . . . (3)

3

3 464 ( ) (4 2) 20 0.512

(80 10 ) (2)AW W× ×

Δ = =×

. . . (4)

3

3 464 ( ) (4 ) 30 1.54

(80 10 ) ( )B

BBB

W d Wdd

×Δ = =

× . . . (5)

Eq. (3) = Eqs. (4) + (5)

⇒ 1.540.512B

WW Wd

= +

∴ 3.2 mmBd =

Maximum Load

max 316 W R

dτ =

π

316 8200

2W× ×

=π ×

∴ 39.3 NW =

316 12.8200

3.2W× ×

=π ×

∴ 100.5 NW =

Maximum load = 39.3 N (less)

39.3 mmΔ =

205

W2

W2

n = 12

Springs(c) W1

W1

n = 12

W

(a) (b) (c)

Figure

For springs, 1 2 0Δ + Δ =

⇒ 3 3

1 1 1 2 2 24 41 2

64 64 0W R n W R nd d

− =π π

⇒ 1 1 2 2W n W n=

⇒ 1 216 12W W× = ×

1 20.75W W= . . . (1)

1 2W W W+ = . . . (2)

1 0.43W W=

2 0.57W W=

SAQ 4

Quarter-elliptic Leaf Spring

l = 450 mm, n = 4, t = 9 mm

W = 2500 N 3 2200 GPa 200 10 N/mmE = = ×

Δ = 40 mm b = ?

3

36 W ln E bt

Δ =

⇒ 3

3 36 2500 45040

4 (200 10 ) 9b× ×

=× × × ×

∴ b = 58.6 mm

206

Stresses in Shafts and Shells FURTHER READINGS

Aggarwal, S. K., and Gupta P. K., Strengths of Materials, Metropolitan Book Company, New Delhi.

Singh, Surendra, Strength of Materials, Vikas Publishing House, New Delhi.

Prasad, Jainti, Strength of Materials, CBS Publishing and Distributors.

Timoshenko, Stephen, (1989), Strength of Materials – Part I and II, CBS Publishers and Distributors.

Schaum’s Outline Series, (1989), Strength of Materials, Second Edition, McGraw Hill Book Company.

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

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

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

SpringsSTRESSES IN SHAFTS AND SHELLS

Engineering materials, used singly or in combinations, are subjected to stresses and strains under practical working conditions. It is, therefore, necessary to visualise and compute these parameters for appropriate design of structural members. This block deals with these aspects of Strength of Materials.

This block contains five units.

Unit 9 deals with deflection of beams. This unit introduces different methods to calculate slopes and deflection of beams. It also discusses slopes and deflection of different types of beams like simply supported, continuous and overhanging and cantilever beams. It also outlines the application of deflection of beams.

Unit 10, in relation to torsional forces, discusses deformations undergone by both solid and hollow circular shafts while transmitting power. It also outlines the criteria for their design; and discusses the behaviour of stepped and non-circular shafts. Lastly, the analogy between membrane problem and torsion problem, as well as calculations for fully plastic torque in various structural sections are presented.

Unit 11 presents concepts of thin and thick cylinders; deals with stresses, strains and deformation in thin cylinders; and discusses stresses in a wire-bound pipe. Lastly, it analyses the behaviour of a thick cylinder as well as a compound cylinder.

Unit 12 analyses thin shells (spherical, doubly curved and conical) in respect of stresses and strains.

Unit 13 deals with various types of spring, i.e. close coiled, open coiled and leaf springs. It also presents the methodology of computation of stresses, deflection, proof load, and stiffness for different types of spring.

207