Basic Civil Engineering first year Notes- Chapter 4 Building.pptx
Ppt-Design of Helical Springs11211111.ppt
1. UNIT IV - ENERGY STORING ELEMENTS
AND ENGINE COMPONENTS 9
Various types of springs, optimization of helical
springs - rubber springs - Flywheels
considering stresses in rims and arms for
engines and punching machines- Connecting
Rods and crank shafts.
5. Helical Spring
Compression spring
Extension Spring
Multi-leaf or laminated
Concentric Springs
Helical Torsion Spring
Spiral Springs
Belleville Spring or coned disk or disk spring
32. Series and Parallel Connections
Series Connection
Deflection (δ) = δ1 + δ2
δ= P/k
Parallel Connection
Load (P) = P1+P2
P=k δ
33. Open Coiled Versus Closed Coiled Helical Spring
Open Coiled Closed Coiled
Helix angle is more than 10
degree
Helix angle is very small ,
but usually less than 10
degree
wire is coiled in such a way
that there is a gap between
successive turns
wire is coiled in such a way
that there is a minimum gap
between successive turns
Application of open coiled helical spring is limited.
34. Type of Spring Used in
Helical compression springs Automobile two wheeler,
railway wagons etc
Helical Tension springs Safety valve, brake lever, toys
and so on.
Helical conical and volute
springs
Calculators,boosters
Helical torsional springs Toys, clips etc
Torsional flat springs /spiral
springs
Watches, clocks etc
Leaf springs Automobile four
wheelers,rickshaws
Belleville springs or spring
washers
Shock absorbers or dampers,
clutches pressure relief valves
and so on.
Application of some of springs are as follows:
35. The valves to be found by the designer are Pitch diameter of coil D, the wire
diameter d, number of turns n, free length Lf, solid length Ls, spring stiffness q,
and type of ends.
Design Procedure of Helical Compression Spring:
1. Select suitable material for spring according to load,
deflection and purpose for which the spring is to be designed.
2. From space limitations, the pitch diameter of coil D is
selected. If no limitations are given, then assume suitable
valve of spring index C usually 5 to 10.
3. Determine the wire diameter .Refer PSGDB-7.100
4. Adopt the nearest standard size of wire. Refer PSGDB-7.105
36. 5. Determine the number of coils for the required
deflection. If number of coils is small, the spring will be
too soft. So reduce the mean dia of the spring. This
change will slightly increase the wire diameter.
5. Decide the end conditions and select the number of
inactive coils. Refer PSGDB-7.101
6. Calculate the free length and solid length of the spring
according to end condition. Refer PSGDB-7.101.
7. Check the spring for buckling. Refer PSGDB-7.101
8. Check the natural frequency of the spring to avoid
surging. Refer PSGDB-7.101
9. Compute the spring stiffness. Refer PSGDB-7.101
37. Buckling of Compression Springs:
Lf>4D
The spring will behave like a column and
may fail by buckling at a comparatively low
load.
To Avoid Buckling of Springs:
It is either mounted on a central rod or
located on the tube. when is located on tube
clearance b/w tube walls and spring should be
kept as small as possible, but it must be
sufficient to allow for increase in ‘d’ during
38. Surge in Springs
Fs=Fl
Fs-Natural Freq of Spring
Fl-Natural Freq of Load applied
Surge in Springs may be Eliminated by using
the Following methods
Using Friction dampers on the centre coil
Using Springs of High Frequency
Using Springs having pitch of coils near the
ends different than at the centre to have
different natural frequencies.
39. Problems:
1. Design a Compression helical spring to carry a load of
1000 N with a deflection of 25 mm. The spring index may
be taken as 5. Assume the following valves for the spring
material: Permissible shear stress = 420 Mpa; Modulus of
rigidity = 84 KN/mm2.
Given Data:
Load, P = 1000N
Deflection, y = 25 mm
Spring index, C =5
Shear stress, (τ) = 420 Mpa =420 N/mm2
Modulus of Rigidity, G = 84 KN/mm2 = 84000 N/mm2
40. 1. Select suitable material for spring according to load,
deflection and purpose for which the spring is to be
designed.
2. From space limitations, the pitch diameter of coil D
is selected. If no limitations are given, then assume
suitable valve of spring index C usually 5 to 10.
C =5 (given)
3. Determine the wire diameter .Refer PSGDB-7.100
41. 3. Determine the wire diameter .Refer PSGDB-7.100
Ks = 1.31
P =1000 N
C= 5
τ = 420 Mpa =420 N/mm2
diameter of wire d = 6.3 mm
43. Mean diameter, D = C x d = 5x 6.5 = 32.5 mm
Outer diameter of coil ,D0 = D+d = 32.5 + 6.5 = 39 mm
4. Determine the number of coils for the required
deflection (n)
number of coils, n =14
Refer PSGDB-7.100
44. 5.Decide the end conditions and select the number of
inactive coils. Refer PSGDB-7.101
Assume Squared and Ground ends
Total number of coils, n‘ = n +2 = 14 +2 =16
6. Calculate the free length and solid length of the spring
according to end condition. Refer PSGDB-7.101
Free length, Lf = n’d + y+0.15y = 131.2 mm
45. 7. Calculate the solid length of the spring according to
end condition. Refer PSGDB-7.101
Solid Length of Spring , Ls = dn+2d = 91+13=104 mm
8. Calculate the Pitch (p)
w.k.t Lf = pn+2d ; p =8.75 mm
9. Check the spring for buckling
46. Lf / D = 131.2 /32.5 = 4
Here Lf / D > 3 the spring must be suitably
guided.
10. Compute the spring stiffness. Refer PSGDB-7.101
47. Problems:
2. A helical valve spring is to be designed for an operating
load range of 90 to 135 N. The deflection of the spring for
the load range is 7.5 mm.Assume a spring index of 10.
Design the spring. Assume the following valves for the
spring material: Permissible shear stress = 480 Mpa;
Modulus of rigidity = 80 KN/mm2
Given Data:
Load, (P)min = 90 N
Load, (P)max = 135 N
Deflection, y = 7.5 mm
Spring index, C = 10
Shear stress, (τ) = 480 Mpa = 480 N/mm2
Modulus of Rigidity, G = 80 KN/mm2 = 80000 N/mm2
48. 1. Select suitable material for spring according to load,
deflection and purpose for which the spring is to be
designed.
2. From space limitations, the pitch diameter of coil D
is selected. If no limitations are given, then assume
suitable valve of spring index C usually 5 to 10.
C = 10 mm (given)
3. Determine the wire diameter .Refer PSGDB-7.100
49. 3. Determine the wire diameter .Refer PSGDB-7.100
Ks = 1.148
P =135 N
C= 10
τ = 480 Mpa =480 N/mm2
diameter of wire d = 2.86 mm
51. Mean diameter, D = C x d = 10 x 3= 30 mm
Outer diameter of coil ,D0 = D+d = 30 + 3 = 33 mm
4. Determine the number of coils for the required
deflection (n)
number of coils, n =5
Refer PSGDB-7.100
P = 135-90 = 45 N
52. 5.Decide the end conditions and select the number of
inactive coils. Refer PSGDB-7.101
Assume Squared and Ground ends
Total number of coils, n‘ = n +2 = 5 +2 =07
6. Calculate the free length and solid length of the spring
according to end condition. Refer PSGDB-7.101
Free length, Lf = n’d + y+0.15y = 28.5 mm
53. 7. Calculate the solid length of the spring according to
end condition. Refer PSGDB-7.101
8. Solid Length of Spring , Ls = dn+2d = 21 mm
9. Check the spring for buckling
54. Lf / D = 28.5 /30 = 1
Here Lf / D < 3
Design is Safe
10. Compute the spring stiffness. Refer PSGDB-7.101
55. Problems:
3. A helical spring made up of C50 steel has an outside
diameter of 80 mm and a wire diameter of 12 mm. The
spring has to support a maximum load of 1 KN. Determine
the Maximum shear stress and total deflection. If the
springs has 10.5 coils with ends ground. Determine also
the factor of safety. Take G=0.89X100 KN/mm2
Given Data:
Load, P = 1kN =1000 N
Outside dia of spring D0 = D+d =80 mm
Wire dia d = 12 mm
n = 10.5
Shear stress, (τ) = ?
Modulus of Rigidity, G = .89x100 KN/mm2
56. 1. C = D / d
D0 = D+d D = D0 – d = 80-12=68 mm
2. Refer PSGDB-7.100
Ks =1.25
(τ) = 125.11 N/mm2
3. Total deflection: Refer PSGDB-7.100
Total deflection y =14.26 mm
57. 4. Factor of Safety (n)
For C50 steel σy = 340 N/mm2
W.K.T n = 340 / 125.11 = 2.7 =3
58. Problems:
4. A helical spring is made from a wire of 8 mm dia and is
of outside diameter 75 mm. The spring has 6 numbers of
active coils. If the permissible stress in shear is 350
N/mm2 and modulus of rigidity is 84 KN/mm2. Find the
axial load which the spring can take and deflection
produced.
59. Problems:
4. A helical spring is made from a wire of 8 mm dia and is
of outside diameter 75 mm. The spring has 6 numbers of
active coils. If the permissible stress in shear is 350
N/mm2 and modulus of rigidity is 84 KN/mm2. Find the
axial load which the spring can take and deflection
produced.
60. Problems:
5. Design a helical spring for a spring loaded safety
valve for the following conditions:
Operating pressure = 0.7N/mm2
Maximum pressure when the valve blow off freely = 0.75 N/mm2
Maximum lift of the valve when the pressure rise from 0.7N/mm2 to
0.75 N/mm2 = 3.5 mm
Diameter of valve seat = 65 mm
Maximum shear stress = 550 Mpa
Modulus of rigidity = 86 kN/mm2
Spring index = 6 .
63. Mean diameter, D = C x d = 6x 8.5 = 51 mm
Outer diameter of coil ,D0 = D+d = 51 + 8.5 = 59.5 mm
4. Determine the number of coils for the required
deflection (n)
number of coils, n =10
Refer PSGDB-7.100
P = P2 – P1 = 166 N
64. 5.Decide the end conditions and select the number of
inactive coils. Refer PSGDB-7.101
Assume Squared and Ground ends
Total number of coils, n‘ = n +2 = 10 +2 =12
6. Calculate the free length and solid length of the
spring according to end condition. Refer PSGDB-
7.101
Free length, Lf = n’d + y+0.15y = 97.39 mm
65. 7. Calculate the solid length of the spring according to
end condition. Refer PSGDB-7.101
Solid Length of Spring , Ls = dn+2d = 91+13=119 mm
8. Calculate the Pitch (p)
w.k.t Lf = pn+2d ; p =10.9 mm
9. Check the spring for buckling
66. Lf / D = 97.39 /51 = 1.9
Here Lf / D < 3 So the spring is Safe
10. Compute the spring stiffness. Refer PSGDB-7.101
67. Problems:
6. A Safety valve of 60 mm diameter is to blow off at a
pressure of 1.2 N/mm2. It is held on its seat by a close
coiled helical spring. The maximum lift of the valve is
10 mm. Design a suitable compression spring of
Spring index 5 and providing an initial compression of
35 mm. The Maximum shear stress = 500 Mpa and
Modulus of rigidity = 80 KN/mm2.
70. Mean diameter, D = C x d = 5 x 12.5 = 62.5 mm
Outer diameter of coil ,D0 = D+d = 62.5 +12.5 = 75 mm
4. Determine the number of coils for the required
deflection (n)
number of coils, n =10.5 =11 mm
Refer PSGDB-7.100
P = 4364 N
71. 5.Decide the end conditions and select the number of
inactive coils. Refer PSGDB-7.101
Assume Squared and Ground ends
Total number of coils, n‘ = n +2 = 11 +2 =13
6. Calculate the free length and solid length of the
spring according to end condition. Refer PSGDB-7.101
Free length, Lf = n’d + (y)max+0.15(y)max = 216.85 mm
72. 7. Calculate the solid length of the spring according to
end condition. Refer PSGDB-7.101
Solid Length of Spring , Ls = dn+2d = 91+13=162.5 mm
8. Calculate the Pitch (p)
w.k.t Lf = pn+2d ; p =18.10 mm
9. Check the spring for buckling
73. Lf / D = 3.46
Here Lf / D > 3 So the spring must be suitably
guided
10. Compute the spring stiffness. Refer PSGDB-7.101
74. Problems:
7. Design a value spring for a petrol engine for the
following data:
Spring load when the valve is open = 400N
Spring load when the valve is closed = 250 N
Length of the spring when the valve is open = 40mm
Length of the spring when the valve is closed = 50mm
Maximum inside dia of spring = 25 mm
Maximum shear stress = 400 Mpa
Modulus of rigidity = 80 KN/mm2.
75. 1. Determine the wire diameter .Refer PSGDB-7.100
D = D1 + d = (25 +d) or Assume C = 6.6
P = 400 N
Ks = 1.227 ; τ = 400 Mpa =400 N/mm2
diameter of wire d = 4.54 mm
77. Mean diameter, D = C x d = 6.5 x 5 = 32.5 mm
Outer diameter of coil ,D0 = D+d = 32.5 +5 = 37.5mm
4. Determine the number of coils for the required
deflection (n)
number of coils, n =14.2 =15 mm
Refer PSGDB-7.100
y= l2 - l1 = 50-40=10 mm
P = P2 - P1 = 400-250 = 150 N
78. 5.Decide the end conditions and select the number of
inactive coils. Refer PSGDB-7.101
Assume Squared and Ground ends
Total number of coils, n‘ = n +2 = 15 +2 =17
6. Calculate the free length and solid length of the spring
according to end condition. Refer PSGDB-7.101
Free length, Lf = n’d + (y)max+0.15(y)max
(y)Max = 10/150 x 400 = 26.67 mm
Free length, Lf = n’d + (y)max+0.15(y)max = 113.58 mm
79. 7. Calculate the solid length of the spring according to
end condition. Refer PSGDB-7.101
Solid Length of Spring , Ls = dn+2d = 86.18 mm
8. Calculate the Pitch (p)
w.k.t Lf = pn+2d ; p =7.10 mm
9. Check the spring for buckling
80. Lf / D = 3.54
Here Lf / D > 3 So the spring must be suitably
guided
10. Compute the spring stiffness. Refer PSGDB-7.101
81. Problems:
8. A railway wagon of mass 20 tonnes is moving with
a velocity of 2 m/s. It is bought to rest by two buffers
with springs of 300 mm diameter. The maximum
deflection is 250 mm and Maximum shear stress =
600 Mpa. Take Modulus of rigidity = 84 KN/mm2.
82. 1. Determine the wire diameter .Refer PSGDB-7.100
Kinetic Energy of wagon = ½ (mv2) = 40000000 Nmm
Energy stored in springs = ½ (P x y)
For two springs,Energy stored in springs =2 x ½ (P x y) = 250 P Nmm
K.E = S.E ; 40000000 = 250 P
P = 160000 N
Ks = 1; D= 300 mm ; τ = 600 Mpa =600 N/mm2
diameter of wire d = 58.8 mm
84. Mean diameter, D = 300 mm ; C = D/d = 300/60 =50 mm
Outer diameter of coil ,D0 = D+60 = 360 mm
4. Determine the number of coils for the required
deflection (n)
number of coils, n =8
Refer PSGDB-7.100
y = 250 mm
85. 5.Decide the end conditions and select the number of
inactive coils. Refer PSGDB-7.101
Assume Squared and Ground ends
Total number of coils, n‘ = n +2 = 8+2 =10
6. Calculate the free length and solid length of the
spring according to end condition. Refer PSGDB-7.101
Free length, Lf = n’d + (y)max+0.15(y)max = 887.5 mm
86. 7. Calculate the solid length of the spring according to
end condition. Refer PSGDB-7.101
Solid Length of Spring , Ls = dn+2d = 91+13= 600 mm
8. Calculate the Pitch (p)
w.k.t Lf = pn+2d ; p =98.60 mm
9. Check the spring for buckling
87. Lf / D = 2.95
Here Lf / D < 3 So the spring is safe against
buckling
10. Compute the spring stiffness. Refer PSGDB-7.101
89. For Patented & Cold Drawn Steel
τ 0 = 0.21 σut
τ y = 0.42 σut
For Oil-hardened & Tempered Steel
τ 0 = 0.22 σut
τ y = 0.45 σut
90.
91. Problems:
9. Design a helical spring is subjected to a load
varying from 400 N to 1000 N having a spring index of
6. The compression of the spring at the maximum
load is 30 mm. Take yield stress in shear as 110
N/mm 2, endurance strength in shear as 350 N/mm 2
and the modulus of rigidity as 80X10 3 N/mm 2.
