STATEMENT OF THE PROBLEM
It is desired to isolate a furnace, weighing 47,300lb from the surrounding by
mounting it on helical springs. Under the weight, the springs should deflect
approximately 1in, and at least 2in. before becoming solid. It has been decided to use
3
sprigs having a wire diameter of 1in. an outside diameter of 5 8 , 4.3 free coils.
Determine (a) the number of sprigs to be used, (b) the stress caused by the weight, (c) the
solid stress (d) what steel should be used?
Given:
= 47,300
= 1
= 1
3
= 5
8
= 4.3
Where :
d = Spring deflection
Dw = wire diameter of spring
Do = coil outside diameter
Nc = no. of active coils or live coils
Required :
a) The number of springs to be used
b) The stress caused by the weight, Ss
c) The solid stress
d) Steel should be used
DESIGN OF HELICAL SPRING Page 1
COMPLETE SKETCH OF THE ASSEMBLY
(NOT TO SCALE)
DESIGN OF HELICAL SPRING Page 2
Design Computation
Given:
= 47,300
= 1
= 1
3
= 5
8
= 4.3
Where :
d = Spring deflection
Dw = wire diameter of spring
Do = coil outside diameter
Nc = no. of active coils or live coils
Required :
a) The number of springs to be used
b) The stress caused by the weight, Ss
c) The solid stress
d) Steel should be used
Solution:
DESIGN OF HELICAL SPRING Page 3
Solving for coil mean Diameter, Dm
= (From Instructional Manual
by Engr. John Ike Uvas, pg.
81)
Given:
3
Do = 5 8
= 1
Where:
Do = coil outside diameter
Dw = wire Diameter
3
= 5 1
8
3
= 4
8
Solving for Spring Index, C
= (From Instructional Manual by Engr. John Ike
Uvas, pg. 82)
Given :
3
= 4
8
= 1
=
3
4
8
= 1
= .
DESIGN OF HELICAL SPRING Page 4
(a) The number of springs to be used
Solving for Axial Force, F
8 3
= (From Instructional Manual by Engr. John Ike Uvas,
pg. 82)
=
8 3
Given :
= 1.0
= 4.3
= 1
= 4.375
= 11.5 106
(From Table AT.7, DOME by Faires,
pg. 576)
Where :
d = Spring deflection
G = Shear modulus of elasticity or rigidity
= no. of active coils or live coils
= wire diameter of spring
F = Axial Force
=
8 3
(1.0)(11.5106 2 )(1.0)
F= 8(4.375)3 (4.3)
= .
DESIGN OF HELICAL SPRING Page 5
(a) The number of springs to be used
. = (From Instructional Manual by Engr. Uvas, Pg. 86)
Where:
= ;
= 47,300
= 3992.13
. =
47,300
=3992.135
. =
Solving for Wahls factor, K
41 0.615
= 44 + (From Instructional Manual by Engr. Uvas,
Pg. 82)
4(4.375) 1 0.615
= +
4(4.375) 4 4.375
= .
DESIGN OF HELICAL SPRING Page 6
Solving for the stress caused by the weight, Ss
=81 (From Instructional Manual by Engr. Uvas, Pg. 82)
Given :
= 1.3628
1 = 3942
= 1
= 4.375
Where :
K = Wahls Factor
F1 = axial load in each spring
Dw = Wire Diameter
C = Spring Index
=81
2
(1.3628)(8)(3942)(4.375)
=
(1.0)
= 59,850.38 2
= , .
DESIGN OF HELICAL SPRING Page 7
Solving for
8 3
=
(G)( )
= 8 3
(2.0)(11.5106 2 )(1.0)
= 8(4.375)3 (4.3)
=7989.27lb
Solving for solid stress,
=8
(1.3628)(8)(7984.27)(4.375)
= ()(1.0)2
=121,223.135psi
(d)steel should be use
= 0.6 where = =
121,223.135
=0.6 9(121,223.135psi)
= 72,733.88
Or
= 73
From Table AT7 DOME by Faires, = 75
Use AISI C1118, cold drawn
DESIGN OF HELICAL SPRING Page 8
AISI 1118 Carbon Steel Material Properties
Chemical Composition
Element Content (%)
Iron, Fe 98.03 - 98.48
Manganese, Mn 1.30 - 1.60
Carbon, C 0.14 - 0.20
Sulfur, S 0.080 - 0.13
Phosphorous, P 0.040
Physical Properties
Properties Metric Imperial
Density 7.85 g/cm3 0.284 lb/in
Melting point 1449C 2640F
Mechanical Properties
Properties Metric Imperial
Tensile strength 525 MPa 76100 psi
Yield strength 315 MPa 45700 psi
Bulk modulus (typical for steel) 140 GPa 20300 ksi
Shear modulus (typical for steel) 80.0 GPa 11600 ksi
Elastic modulus 190-210 GPa 27557-30458 ksi
Poisson's ratio 0.27-0.30 0.27-0.30
Elongation at break (in 50 mm) 32% 32%
Reduction of area 70% 70%
Hardness, Brinell 149 149
Hardness, Knoop (converted from Brinell hardness) 169 169
DESIGN OF HELICAL SPRING Page 9
Hardness, Rockwell B (converted from Brinell hardness) 80 80
Hardness, Vickers (converted from Brinell hardness) 155 155
Thermal Properties
Properties Metric Imperial
Thermal conductivity (typical steel) 49.8 W/mK 346 BTU in/hr.ft.F
Solving for Pitch and coil angle
= < 12
= + 2 (From table AT7 DOME by faires)
= (1.0)(4.3) + (2)(1.0)
= 6.3
= + +
= (1.0) + (2) + (6.3)
= 9.3
Use Lf=10in
Lf=P + 2
2
P=
10(2)(1.0)
= 4.3
P=1.86in
Tanl= < 12
1.86
= 3
(4 )
8
DESIGN OF HELICAL SPRING Page 10
l=7.71
Solving for Total No.Of Coils
= + 2 (From table AT16 DOME by
Faires,p. 589)
= . + = .3coils
materials Materails needed Materials cost Total cost
DESIGN OF HELICAL SPRING Page 11
Wire Dm by length Php 1600.00
AISI 1118 carbon 5.375in by 20in Php 80/Do by
steel length
Machine with labor Materials to operate Machine with labor Total cost
cost
Machine with labor Do by length Php 30/ Do by Php 600.00
5.375in by 20in length
Material cost Php 1600.00
machine with labor cost Php 600.00
Total cost of material and machine Php 2200.00
Number of spring Total cost of material and machine = total cost
12 2200.00 = 26400
Total cost=Php26400
DESIGN OF HELICAL SPRING Page 12
BIBLIOGRAPHY
Faires, Virgil Moring. Design of Machine Elements, 4th Edition, New York: Macmillan Co,
1969
Uvas, John Ike A. Instructional Manual in Machine Design I
Retrieved: October 2, 2017 https://www.azom.com/article.aspx?ArticleID=6563
Retrieved: October 2, 2017, www.google.com.ph
DESIGN OF HELICAL SPRING Page 13
