89589540-Strength-of-Materials-by-S-K-Mondal-12.pdf

8/14/2019 89589540-Strength-of-Materials-by-S-K-Mondal-12.pdf

1/3

12. Spring

Theory at a Glance (for IES, GATE, PSU)1. A spring is a mechanical device which is used for the efficient storage

and release of energy.

2. Helical spring stress equation

Let us a close-coiled helical spring has coil diameter D, wire diameter d and number of turn n. The

spring material has a shearing modulus G. The spring index, D

Cd

= . If a force P is exerted in both

ends as shown.

The work done by the axial

force 'P' is converted into

strain energy and stored in

the spring.

( )

( )

U= average torque

angular displacement

T =

2

TLFrom the figure we get, =

GJ

PDTorque (T)=

2

4

2 3

4

length of wire (L)=Dn

dPolar moment of Inertia(J)=

32

4P D nTherefore U=Gd

According to Castigliano's theorem, the displacement corresponding to force P is obtained by

partially differentiating strain energy with respect to that force.

2 3 3

4 4

4 8UTherefore =

p D n PD n

P P Gd Gd

= =

Axial deflection3

4

8 =

PD n

Gd

Spring stiffness or spring constant ( )4

3

Pk =

8

Gd

D n=

Page 344 of 429


2/3

Chapter-12 Spring S K Mondals

The torsional shear stress in the bar,( )

1 3 3 3

16 / 216 8PDT PD

d d d

= = =

The direct shear stress in the bar, 2 2 324 8 0.5

4

P P PD d

Dd dd

= = =

Therefore the total shear stress, 1 2 3 38 0.5 81

sPD d PDK

Dd d

= + = + =

3

8s

PDK

d

=

Where0.5

1s

dK

D= + is correction factor for direct shear stress.

3. Wahls stress correction factor

38PDK

d

=

Where4 1 0.615

4 4

CK

C C

= +

is known as Wahls stress correction factor

Here K = KsKc; Where sK is correction factor for direct shear stress and Kc is correction

factor for stress concentration due to curvature.

Note:When the spring is subjected to a static force, the effect of stress concentration is neglected

due to localized yielding. So we will use,3

8s

PDKd

=

4. Equivalent stiffness (keq)

Spring in series 1 2( )e = + Spring in Parallel 1 2( = )=e

eq 1 2

1 1 1K K K

= + or 1 2

1 2

=+

eq

K KK

K K eq 1 2

K = +K K

Shaft in series ( 1 2 = + ) Shaft in Parallel( 1 2 = =eq )

Page 345 of 429


3/3

Chapter-12 Spring S K Mondals

eq 1 2

1 1 1K K K

= + or 1 2

1 2

=+

eq

K KK

K K eq 1 2

K = +K K

5. Important note

If a spring is cut into n equal lengths then spring constant of each new spring =nk

When a closed coiled spring is subjected to an axial couple M then the rotation,

4

64= c

MDn

Ed

6. Laminated Leaf or Carriage Springs

Central deflection,

3

3

3

8=

PL

Enbt

Maximum bending stress,max 2

3

2=

PL

nbt

Where P = load on spring

b = width of each plate

n = no of plates

L= total length between 2 points

t =thickness of one plate.

7. Belleville Springs

3

2 2

0

4 Load, ( )

(1 ) 2

= + f

EP h h t t

k D

Where, E = Modulus of elasticity

= Linear deflection

=Poissons Ratio

kf=factor for Belleville spring

Do= outside diamerer

h = Deflection required to flatten Belleville spring

t = thickness

Note:

Total stiffness of the springs kror =stiffness per spring No of springs

In a leaf springratio of stress between full length and graduated leaves = 1.5

Conical spring-For application requiring variable stiffness

Belleville Springs-For application requiring high capacity springs into small space

Do

t

P

ho

Page 346 of 429

Documents

89589540-Strength-of-Materials-by-S-K-Mondal-12.pdf