International Journal of Engineering Trends and Technology (IJETT) – Volume 33 Number 4- March 2016
Re-Designing and Analysis of Leaf Spring
used In MX 400 Chasis
Rajeev V R #1, Vishnu A*2 , Ramjith Krishnan R #3
#
Assistant Professor, Archana College of Engineering Alappuzha, India
* Clerk, Traco Cable Company ltd, Pinarayi, India
#
Assistant Professor, Vidya Academy of Science and Technology Technical Campus, Trivandrum, India
Abstract There is a poor no load performance of the
existing leaf spring used in MX 400 chasis .On noload conditions, the passengers (only be driver and
a passenger) was finding it extremely uncomfortable
due to the low deflection of the spring. My aim is to
design a leaf spring such that passengers get more
comfort on no-load condition. There is a gap of 15
mm between 8th and 9th leaves in original design
and this creates the problem. In the proposed design
this gap is divided in to 3 equal divisions of 5 mm
each and analysis is carried out and concludes that
the passengers get more comfort from the proposed
design. The proposed design is to be carried out in
ANSYS. Then it is validated with numerical
equations.
Keywords— ANSYS, MX 400 chasis.
I. INTRODUCTION
This paper describes the working of leaf springs
and finding out the equation for calculating stress
and deflection for leaf spring. This paper also
mentioned the design theme of leaf spring. Here leaf
spring can be treated as cantilever beam and also as
simply supported beam.
Length of 1st leaf =638 mm
Length of 2nd leaf = 638 mm
Length of 3rd leaf = 580 mm
Length of 4th leaf = 560 mm
Length of 5th leaf = 540 mm
Length of 6th leaf = 475 mm
Length of 7th leaf = 420 mm
Length of 8th leaf = 365 mm
Length of 9th leaf = 310 mm
Length of 10th leaf= 265 mm
Length of 11th leaf= 200 mm
A; Softwares used
Leaf spring which is used in this
project has a gap between 8th and 9th leaves.
Therefore classical equation cannot be used for
finding out the stress and deflection of leaf spring. In
order to find out the stress and deflection of leaf
spring with this gap can be find out by analysis
software. That’s why ANSYS software is used in
this project. ANSYS software can give better results
when static load is applied.
II. EXISTING DESIGN ANALYSIS
Material as High Carbon Steel having
Young’s Modulus, E of 0.206 ×106
N/mm2
Maximum load on the spring i.e. leaf
spring on full load condition, 2F = 1460
kg
F = 730 kg
= 730 × 9.81
= 7161.3 N
No: of full length leaves, nf = 2
No: of graduated length leaves, ng = 9
Total Span length, 2L1= 638 mm
Distance between U bolt centres, l = 69 mm
Width of leaves, b = 40 mm
Thickness of leaves, h = 5 mm
ISSN: 2231-5381
B; Elements used
2-D modeling of leaf spring was
done in ANSYS software. In this project plane 42
element is used because plane 42 is mostly used as
2-D modeling of solid structure. The element can be
used either as a plane element (plane stress or plane
strain) or as an axisymmetric element. The element
is defined by four nodes having two degrees of
freedom at each node: translations in the nodal x and
y directions. The element has plasticity, creep,
swelling, stress stiffening, large deflection, and large
strain capabilities.
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International Journal of Engineering Trends and Technology (IJETT) – Volume 33 Number 4- March 2016
= 9×7161.3×296
40×52[(2×6) + (3×2)]
= 1096 N/mm2
Fig 1; Geometry of plane 42
Modeling is done by selecting key points. Then it
is connected by lines. Then area is assigned to these
lines. Free meshing is done. Leaf springs are
connected by applying glue option. So that load can
be distributed to entire leaf. In this project, leaf
spring is treated as cantilever. The main boundary
condition is that DOF of left portion of leaf spring
was arrested.
C; FE model of Existing Design
Fig 3;
condition
Deflection of leaf spring at full load
Fig 4; Stress of leaf spring at full load condition
At minimum load
Fig 2; FE model of existing leaf spring
Effective length, 2L = 2 L1- (2/3) × l
= 638 – (2/3) × 69
L = 296 mm
Deflection, y
y = 6×F×L3
b×h3×E[(2ng)×(3nf )]
= 6×7161.3×2963
40×53×0.206×106[(2×6) × (3×2)]
= 57.69 mm
Stress in the full length leaves,
σ = 9×F×L
b×h2[(2×ng)+(3×nf )]
ISSN: 2231-5381
Total rear axle weight be 300 kg
2F = 300×9.81
= 2943 N
F = 1471.5 N
Assume 30% as dynamic load
F = 1471.5 + 0.3×1471.5
= 1912.95 N
Deflection, y
y = 6×F×L3
b×h3×E[(2ng)×(3nf )]
= 6×1471.5×2963
40×53×.206×106[(2×6)× (3×2)]
= 12.35 mm
This deflection is not sufficient and the last
three
leaves
remaining
inactive
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International Journal of Engineering Trends and Technology (IJETT) – Volume 33 Number 4- March 2016
Fig 5; Deflection of leaf spring at no-load
condition
D; By Using ANSYS Software
Fig 6 ; Deflection of leaf spring at no-load
condition using Ansys software
Fig 7; Stress of leaf spring at no-load condition
using Ansys software
ISSN: 2231-5381
5 =
y
155
319
y = 39.12 mm
i.e., in order to activate all the leaf springs, a
deflection of 39.12 mm is needed
y = 6×F’×L3
b×h3×E[(2ng)×(3nf )]
39.12 = 6×F’×2963
40×53×0.206×106[(2×9) × (3×2)]
F’ = 3678.06 N
So, a force of 3678.06 N is needed to
activate all the leaf springs.
Stress in the full length leaves,
σ = 9×F’×L
b×h2[(2×ng)+(3×nf )]
= 9×3678.06×296
40×52[(2×9) + (3×2)]
= 465.26 N/mm2
Fig 8; Deflection of leaf spring at when a load of
3678.06 N using Ansys software
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International Journal of Engineering Trends and Technology (IJETT) – Volume 33 Number 4- March 2016
Fig 9; Stress of leaf spring at when a load of
3678.06 N using Ansys software
III PROPOSED DESIGN
The gap between eighth and ninth leaves is 15
mm in original leaf spring.
Proposed design is that divide this gap into 3
equal divisions of 5 mm each between 2nd & 3rd, 5th
& 6th, and 8th & 9th leaves respectively.
290
319
x = 5.5 mm
i.e. a deflection of 5.5 mm is needed to activate
first 5 springs
Deflection, x = 6×F1×L3
b×h3×E[(2ng)×(3nf )]
5.5 = 6×F1×2963
40×53×0.206×106[(3×2)]
1
F = 218.44 N
Stress in the full length leaves,
= 9×F1×L
b×h2[(2×ng)+(3×nf )]
= 9×218.44×296
40×52[(3×2)]
= 96.9 N/mm2
B; Deflection between 5th and 6th leaves
5
= y
237.5
319
y = 6.175 mm
i.e. a deflection of 6.175 mm is needed to
activate first 8 springs.
Fig 10; FE model of proposed design
A; Deflection between 2nd and 3rd leaves
Deflection, y = 6×F2×L3
b×h3×E[(2ng)×(3nf )]
6.175 = 6×F2×2963
40×53×0.206×106[(2×3) + (3×2)]
F2 = 533.33 N
A total force,F2’= F1+F2 is needed for the
deflection of 10 mm
= 218.44 + 533.33
= 751.177 N
Stress in the full length leaves, σ
= 9×F2’×L
b×h2[(2×ng)+(3×nf )]
= 9×751.77×296
40×52[(2×3) + ((3×2)]
= 166.89 N/mm2
5 = x
ISSN: 2231-5381
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International Journal of Engineering Trends and Technology (IJETT) – Volume 33 Number 4- March 2016
C; Deflection between 8th and 9th leaves
5
= z
155
319
z = 10.29 mm
i.e. a deflection of 10.29 mm is needed to
activate all springs.
Deflection, z = 6×F3×L3
b×h3×E[(2ng)×(3nf )]
3
10.29 = 6×F ×2963
40×53×0.206×106[(2×6) + (3×2)]
F3 = 1126.02 N
A total force, F3’= F1+F2+F3 is needed
for the deflection of 15 mm
= 218.44 + 533.33 + 1126.02
= 1877.79 N
Stress in the full length leaves,
= 9×F3’×L
b×h2[(2×ng)+(3×nf )]
= 9×1877.79×296
40×52[(2×6) + (3×2)]
= 209.32 N/mm2
=
9×F×L
b×h2[(2×ng)+(3×nf )]
= 9×7161.3×296
40×52[(2×9)+((3×2)]
= 854.6 N
Fig 12; Stress of existing leaf spring design at
full load condition.
IV
From the analysis of existing design, it is a
observed that the deflection obtained at no-load
condition is very much smaller than the required
deflection. So the task was to implement a suitable
design which will give the required deflection,
which is it will give smooth running at no-load
condition. Therefore the solution obtained is to
propose a leaf spring design so that stress obtained is
very low when compared to the existing design. All
the leaves become active at the minimum load. So,
passengers get enough comfort from the proposed
design
REFERENCES
[1]
[2]
[3]
[4]
[5]
Fig 11; Stress of existing leaf spring design at
minimum load.
The maximum load acting on the leaf spring is
7161.3N
[6]
[7]
[8]
[9]
[10]
Stress in the full length leaves, σ
ISSN: 2231-5381
CONCLUSIONS
[11]
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