The basic design of helical Springs

1. Mechanical Springs

2. Introduction
• Springs are used to exert forces or torques in a
mechanism or primarily to store the energy of
impact loads.
• These flexible members often operate with
high values for the ultimate stresses and with
varying loads.
• Helical springs are round or rectangular wire,
and flat springs (cantilever or simply supported
beams) are in widespread usage.

3. Torsion Bars
• A torsion bar is a straight hollow or solid bar
fixed at one end and twisted at the other, where
it is supported.
• Typical applications include counterbalancing
for automobile hoods and trunk lids.

4. Torsion bar springs: rod with bent ends and rod
with splined ends.

5. Note that in a passenger car, the bar may have about
0.75 m length, 25 mm diameter, and twist 30°–45°.

7. The angle of twist
TL
GJ
φ =
3
16PR
d
τ
π
=
Shear stress
Deflection
2
4
32TLR PLR
GJ d G
δ
π
=
The polar moment of inertia is
4
32
J d
π
=
The spring rate
4
32
d G
k
L
π
=

8. Helical Tension and Compression Springs

9. • wire diameter = d
• mean coil diameter =D
• coil pitch,
• pitch angle = λ.
• The ratio of the mean coil
diameter to wire diameter is
termed the spring index C
D
C
d
=
TERMINOLOGY OF HELICAL SPRINGS
( )1t
free length
p
N
=
−
The springs of ordinary geometry have C > 3 and λ < 12°. The
majority of springs, C varies from about 6 to 12.

10. Solid Length = tN d
Total gap = (Nt – 1) ×
Gap between adjacent
coils
Free length =
compressed length
+ δ = solid length +
total axial gap + δ

11. STRESS AND DEFLECTION
EQUATIONS
Torque
2
PD
T =

12. The torsional shear stress
( )
1 3 3
1 3
16 216
8
PDT
d d
PD
d
τ
π π
τ
π
= =
⇒ =
When the equivalent bar is bent in the form of helical
coil, there are additional stresses on account of following
two factors:
1. There is direct or transverse shear stress in the spring wire.
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 fibre. This
results in stress concentration at the inside fibre of the coil.

13. T=PD/2
P
P
Direct
shear
Pure
torsion
Combined
Torsional, Direct
and Curvature
Shear Stresses
In order to include the effect of direct shear stress and
stress concentration due to curvature effect, two factors:
KS = factor to account for direct shear stress
Kc = factor to account for stress concentration due to curvature
effect
K is the factor to account for the combined effect of two factors.
K=KS KC

14. Direct shear stress
( )
2 2 32
4 8 0.5
4
P P PD d
d d Dd
τ
π ππ
 
= = =  
 
Superimposing the two stresses
1 2 3
3
8 0.5
1
8 0.5
1
PD d
d D
PD
d C
τ τ τ
π
τ
π
 
= + = + 
 
 
⇒ = + 
 
The shear stress correction factor
0.5
1sK
C
 
= + 
 

15. AM Wahl derived the equation for resultant stress,
which includes torsional shear stress, direct shear stress
and stress concentration due to curvature.
3
8PD
K
d
τ
π
=
where K is called the stress factor or Wahl factor.
4 1 0.615
4 4
C
K
C C
−
= +
−