How to Calculate Leaf Springs
The
effortless, easy and comfortable experience produced while traveling in
vehicles may be largely due to the incorporation of these magnificent devices
called leaf springs. However designing these engineering marvels or calculating
leaf springs may involve many considerations and formulas.
A spring
is a multilayered arrangement
made up of specially seasoned metals which exhibit outstanding resilient
properties against all sorts of forced stresses and strains. These devices
therefore find their major applications as shock absorbers mostly in vehicles,
but actually you will find them being used everywhere, whether it’s a inside
the ball pens, gas lighters, or the toys they have today become almost an indispensible part
of our life.
Although
springs may be of many different kinds, the following two are most commonly
used in the engineering field: Laminated or Leaf Spring and Helical Spring.
Here we
are discussing leaf springs and the various expressions used for calculating
leaf springs.
From
the engineering point of view, designing a leaf spring can be critical,
requiring many considerations in order to produce the most suitable design that
is able to sustain the calculated loads and help absorb shocks optimally.
Fundamentally,
the parameters required for the calculations of leaf springs are:
○ Bending Moment,
○ Moment of Inertia,
○ Resisting Moment, and
○ Central Deflection
Let's
derive and learn each of the above expressions in a step-wise manner.
How to Measure Leaf Springs
A quick
peek under any four wheeler automobile will provide you with an instant view of
these important devices, or leaf springs, used extensively as suspensions for
the sole purpose of absorbing possible shocks due to accidental bumps and rough
roads. They are normally situated at the edges of the wheel axles.
But how
do you measure leaf springs that would guarantee an ideal and a safe journey
with the vehicles?
Before we
move into the technical details let’s first try to understand their basic
structure.
Referring
to the figure alongside we find that a leaf spring is basically comprised of a
number of parallel attached metal strips of decrementing lengths, but having
the same width and thickness. The strips are arranged in such a way that they
are free to slide one over the other.
Under no
load conditions, the shape of the strips of a leaf spring may appear bent over
a certain radius in the form of an arc. However in the presence of a load, the
orientations of the strips tend to straighten up, generating a cushioning
effect to the applied load. The arrangement of the integrated strips makes sure
that the load is distributed uniformly throughout the length of the entire leaf
spring.
In
vehicles and railway carriages, leaf springs are normally fitted such that the
wheel axle passes perpendicular and through the center of the spring
while the two ends of the spring are pinned to the body or the chassis of the
vehicle.
Let’s
assume the following parameters and denote them as follows:
l = Length of
the leaf spring,
t = thickness of
the strips,
b = width of the
strips,
n = number of
strips,
W = load acting
over the spring,
f = ultimate
bending stress produced over the strips,
δ = original
deflection of the topmost spring.
A little
analysis of the system shows that the load supported by the lowermost strip is
shared equally by the ends of the top most strip.
According
to the standard formula. the Bending
Moment (BM) developed at the center of the spring will be:
M
= Wl/4------------------------------------------------------------------------------------------------( i )
Similarly
the Moment of inertia developed over the strips may be presented as:
I
= bt^2/12
Also
since the Resisting Moment M is expressed through the relation:
M/I =
f/y,
Where M =
resisting moment,
I = Moment if
Inertia,
f = Bending Stress
developed in the used material,
y = distance of
the above stress from the neutral axis.
or M = f.I/y = f × bt^2/12 ÷ t/2 = f.bt^2/6
Therefore
the final expression indicating the Resisting Moment produced by n number
of pates can be written as:
M =
nfbt^2/6 -------------------------------------------------------------------------------------------( ii
)
Now since
the Bending Moment and the Resisting Moments act in the opposite directions and
are equal in magnitudes, equating the RHS of the respective expressions
i.e. ( i ) and (
ii ) gives:
Wl/4 = nfbt^2/6
i.e. f = 3Wl/2nbt^2
The above
equation may be used for calculating the maximum stress developed across the
plates of the spring.
Now the
shape of the leaf spring, or simply its geometry, indicates that the central
deflection must be:
δ
= l^2/8R ------------------------------------------------------------------------------------------------( iii
)
Also
according to another standard formula,
f/y = E/R
where E = Young’s Modulus and R =
Radius of curvature.
Therefore R
= E.y/f
= Et/2f, (since y = t/2).
Substituting
the above value of R in equation ( iii ) gives:
δ
= l^2 ÷ 8 × E.t/2f
or δ = fl^2/4Et.
This
final expression may be used to find out the central deflection while
calculating leaf springs.