ppt on springs (Mechanical)

1. A spring is defined as a elastic machine element which
deflects under the action of load and returns to its orignal
shape when the load is removed. It can take any shape and
form depending on its application.
IMPORTANT FUNCTIONS:
1)Springs are used to absorb shocks and vibrations.
2)Springs are used to store energy.
3)Springs are used to measure forces.
4)Springs are used to apply force and control motion.

2. Springs are classified according to their shape.
The most popular type of spring is helical spring.
There are two types of helical spring.
1. Compression spring
2. Extension spring

3. The helical springs are sometimes classified as closely-coiled
helical spring and open-coiled helical spring
The difference between them are as follows:
1) A helical spring is sais to be closely coiled spring when
the spring wire is coiled so close that the plane
containing each coil is almost at right angles to the axis
of the helix. The helix angle is very small ,it is usually less
than 10 degree.
2) A helical spring is said to be open coiled spring when the
spring wire is coiled in such a way that there are large gap
between adjacent coils. In other words the helix angle is
large . In other words it is more than 10 degree.

4. (i)They are easy to manufacture .
(ii)They are cheaper than other types of springs.
(iii)Their reliability is high.
(iv)The deflection of the spring is linearly propotional to the
force acting on the springs.
It is due to the above advantage that helical spring are
popular and extensively used in a number of application.
Helical torsional spring
The construction of the spring is similar to that of
compression or extension spring , the spring is loaded by a
torque about a axis of the coil. They are used to transmit
torque to a particular component in a machine or the
mechanism.

5. The helical torsion spring resists the bending moment(P x r),
Which tends to wind up the spring. The bending moment induces
bending stresses in spring wire. The term “torsional spring” is
somewhat misleading because the wire is subjected to bending
stresses, unlike torsional shear stresses induced in helical
compression or extension spring. It should be noted that although
the spring is subjected to torsional movement, the wire of the
helical torsion spring is not subjected to torsional shear stress. It is
subjected to bending stresses.
Terminology of helical spring:
The main dimension of helical spring subjected to compressive
force
d=wire diameter of spring (mm)
Di= inside diameter of the spring coil(mm)
Do= outside diameter of the spring coil(mm)
D= mean diameter coil(mm)

6. Therefore
D= Di+Do
2
There is an important parameter in spring design called spring index. It
is denoted by the letter C. The spring index is defined as the ratio of a
mean coil to the wire diameter . Or , C=D
d
A spring index from 4 to 12 is considered best from manufacturing
consideration close tolerance spring and those subjected to cyclic
loading.
There are three terms - Free length ,compressed length,& solid length
Solid length- defined as the axial length of the spring which is co
compressed that the adjacent coil touch each other.
Compressed length - defined as the axial length of the spring which is
subjected to maximum compressive force.
Free length – defined as the axial length of an unloaded helical
compression spring. No external forces act on the spring.

7. Solid length= Nt d
Where Nt= no. of coils
Compression length
Total gap=(Nt-1) x gap between adjacent coils
Free length =compressed length+ δ =solid length+total axial gap + δ
P=free length
(Nt-1)
The stiffness of the spring (k) defined as the force required to produce deflection .
Therefore k=P
δ
k=stiffness of the spring (N/mm)
P=axial spring force (N)
δ=axial deflection of the spring corresponding to the force P(mm)
The stiffness of the spring represents the slope of the load deflection line.

8. There are two terms related to the spring coils- active coils and
inactive coils
Active coils are the coils in the spring which contribute to
spring action, support the external force and deflect under the
action of force. A portion of the end coil which is in contact
with the seat , does not contribute to spring action and are
called inactive coil.The coil do not support the load and do not
deflect under the action of an external force. The no. of
inactive coils is given by,
inactive coils=Nt-N
Where N is the no. of active coils
STYLES OF END
Plain end :Nt
Plain end (ground) : (Nt-1/2)
Square end : (Nt-2)
Square end (ground) : (Nt-2)

9. The dimension of equivalent bar are as follows:
1)The diameter of the bar is equal to the wire dia of the spring(d)
.
2)The length of the coil of the spring is (πD). There is N such
alternative coils. Therefore the length of the equivalent
baris(πDN).
3)The bar is fitted with a bracket at each end. The length of the
bracket is equal to the mean coil radius of the spring(D/2).
The force P acting at the end of the bracket induces torsional
shear stress in the bar .
The torsional moment ,is given by
Mt =PD/2
The torsional shear stress of the bar is given by
τ1 = 16Mt/πd^3 = 16(PD/2)
πd^3
Or τ1=8PD/πd^3

10. When the equivalent bar is bent in the form of helical coil,
there is an additional in the form of the two factor:
1)There is a direct or transverse shear stress in the spring
wave
2)When the bar is bent in the form of coil the length of the
inside fibre is less than the length of the outside. The result
in stress concentration at the inside fibre of the coil. The
resultant consist of superimposition of torsional shear
stress direct shear stress and additional stresses due to the
curvature of the coil.

11. Stresses in spring wire:
(a) Pure torsional stress
(b) Direct shear stress
(c) Combined torsional , direct and curvature shear stresses
We will assume the following two factors to account for
these effects
Ks= factors to account for direct shear stress
Kc=factors to account for stress concentration due to
curvature effect
The combined effect of those two factor is given by
K=KsKc
Where k is the factor to account for the combined effect of
two factors

12. The shear stress correction factor (Ks)is defined as,
K=(1+0.5d/D) or K=(1+0.5/C)
τ2=p/(π/4 d²)=4P/πd²=8PD/πd^3 (0.5d/D)
Superimposing
τ= τ1+τ2 =8PD/πd^3+8PD/πd^3 (0.5d/D)=8PD/πd^3 (1+0.5d/D)
Substituting the above equation in the expression
τ=Ks(8PD/πd^3)
AM Wahl derived the equation for resultant stress, which includes
torsional stress, direct shear stress concentration due to curvature.
The equation is given by ,
τ=Ks(8PD/πd^3), where K is stress factor or wahl factor.
The wahl factor is given by,
K=4C-1 + 0.615
4C-4 C where Cis the spring index

13. The wahl factor provides a simple method to find out
resultant stresses in spring, the resultant shear stress is
maximum at the inside radius of the coil.
When the spring is subjected to fluctuating stresses, two
factor Ks & Kc are separately used.
The angle of twist(θ) for the equivalent bar,
θ= Mt l/JG
Where θ= angle of twist (radian)
Mt =torsional moment (PD/2)
l = length of the bar (πDN)
J=polar moment of inertia of the bar (πd^4/32)
G= modulas of elasticity
Substituting values
θ= (PD/2)(πDN) or θ =16PD²N
(πd^4/32)G Gd^4

14. The axial deflection