LAMINATED SPRINGS CLASS NOTES FOR MECHANICAL ENGINEERING
LAMINATED SPRINGS CLASS
NOTES FOR MECHANICAL
ENGINEERING
This type of spring is very common in use.
There is an advantage over the coiled
spring. It acts as a structural member. In
addition it absorbs a shock. Used in cars,
trucks, buses and trains. This type of spring
has number of plates of constant
width and thickness. However, the length
of each leaf is different. These plates are
placed over one another and then
connected by a U-bolt in the center. The
plates are lamination’s. Thus, it is
called a laminated spring. There are two
types of laminated springs.
(i) Semi-elliptic type
Fig. Semi-elliptical Spring
Semi-elliptic type spring is simply supported at the ends. Bolts connects two eye ends to the vehicle body. In this case, load acts at the center. It is considered in simple bending.
(ii) Quarter elliptic type
Fig. Quarter Elliptical Spring
It is fixed at one. It acta as a cantilever. Load acts at the free end of the cantilever. These lamination’s have initial curvature. The load which can straighten these lamination’s is proof load. It is the maximum load, this type of spring can sustain. It absorbs the shocks due to unevenness of the road. Load acting on this spring tends to decrease the depth or try to straighten the initially curved plates. Hence design this type of spring in pure bending. Since maximum bending moment acts at the center, the other lamination’s decrease in length. Where a quarter elliptic acts as a cantilever, maximum bending moment acts at the fixed end. Thus, the maximum number of lamination’s are at the fixed end.
MATERIALS USED IN LEAF SPRINGS
Sr. No. |
Material |
Yield stressN/mm2 |
Endurance LimitN/mm2 |
1. |
Spring carbon steelSAE-1095 |
1170 |
670 |
2. |
Chrome Vanadium SteelsSAE -6140SAE-6150 |
1350 |
700 |
3. |
Silicon Manganese SteelsSAE-9250SAE-9260 |
13001350 |
700800 |
Commonly used plain carbon steels are with 0.9 to 1 % of carbon properly heat treated. Use Chrome-Vanadium and Silicon-Manganese steels for better performance. These alloy steels do not have strength more than that of carbon steels. But these have greater toughness and a higher endurance limit. These materials suit to springs subjected to rapidly fluctuating loads.
Semi-Elliptical Type
This spring is a beam of uniform strength. Load acts at the center. It is supported at the ends. Spring has number of lamination’s. One or two are of full length. Remaining plates reduce in length successively. Reduced length plates are truncated leaves. These are of constant width and constant thickness. Bending moment is different at different locations. Therefore, depth of the spring varies. Hence the section modulus varies.
Calculate the followings:
(a) Uniform stress
AT the center, bending moment is M = WL/4
AT the center, section modulus is n bt2/6
Uniform stress is = M/Z= (WL/4)/ n bt2/6= 6WL/4nbt2
σ =3WL/2nbt2
(b) Overlap
OVERLAP= a = (L/2) / n L/2n
(c) Number of plates
n = (L/2)/a= L/2a
(d) Central deflection
For an initially curved beam, deflection is
δ = ML2 / 8EI = (WL/4)L2/[8E(1/12) nbt3]
δ = 3WL3/8Enbt3
(e) Radius of curvature
σ/y = E/R
R = (Ex t/2)/σ= Et/2σ also = L2/8δ’
Where δ’ = R –(R2—L2/4)0.5
Where R is the initial radius of curvature
Δ’ is corresponding of proof load
Proof load is that load which can straighten the initially curved plate
(f) Strain energy and resilience
u = U/V= σ2/6E
We know U = (1/2) W δ= (W/2)( 3WL3/8Enbt3)
U = [(3/2) WL/nbt2 ]2 (n b t L/12E) = (σ2/6E)x( volume of the spring)
Therefore resilience u = σ2/6E
(g) Length of leaves
Full length = L
Firstly length of second leaf = L – 2a
Secondly length of third leaf= L-4a
Thirdly length of fourth leaf = L-6a
And so on
NOTE
In the above equations, assume originally straight spring. Plates are relatively thin and free to take lateral strains. Spring plates have initial curvature. Load acting at the center tries to straighten the curved plates.
https://www.mesubjects.net/wp-admin/post.php?post=7602&action=edit Salient features of springs