- A concentric spring consisting of two helical compression springs is used as a valve spring in a diesel engine.
- The composite spring is subjected to a maximum force of 6000N and deflects 50 mm.
- The maximum torsional shear stress induced in each spring is 800 N/mm2.
- Each spring has the same spring index of 6 and are made of the same material.
Spring is an elastic body whose function is to distort when loaded and to recover its original shape when the load is removed.
APPLICATION OF SPRINGS
To apply forces as in brakes, clutches and spring loaded valves.
To store energy as in watches, toys.
To measure forces as in spring balance and engine indicators.
To cushion, absorb or control energy due to either shock or vibration as in car.The material of the spring should have
high fatigue strength,
high ductility,
high resilience and
creep resistant.
It largely depends upon the size and service.
The strength of the wires varies with size, smaller size wires have greater strength and less ductility, due to the greater degree of cold working.
Severe service means rapid continuous loading where the ratio of minimum to maximum load (or stress) is one-half or less, as in automotive valve springs.
Average service includes the same stress range as in severe service but with only intermittent operation, as in engine governor springs and automobile suspension springs.
Light service includes springs subjected to loads that are static or very infrequently varied, as in safety valve springs.
The springs are mostly made from oil-tempered carbon steel wires containing 0.60 to 0.70 per cent carbon and 0.60 to 1.0 per cent manganese.
Spring is an elastic body whose function is to distort when loaded and to recover its original shape when the load is removed.
APPLICATION OF SPRINGS
To apply forces as in brakes, clutches and spring loaded valves.
To store energy as in watches, toys.
To measure forces as in spring balance and engine indicators.
To cushion, absorb or control energy due to either shock or vibration as in car.The material of the spring should have
high fatigue strength,
high ductility,
high resilience and
creep resistant.
It largely depends upon the size and service.
The strength of the wires varies with size, smaller size wires have greater strength and less ductility, due to the greater degree of cold working.
Severe service means rapid continuous loading where the ratio of minimum to maximum load (or stress) is one-half or less, as in automotive valve springs.
Average service includes the same stress range as in severe service but with only intermittent operation, as in engine governor springs and automobile suspension springs.
Light service includes springs subjected to loads that are static or very infrequently varied, as in safety valve springs.
The springs are mostly made from oil-tempered carbon steel wires containing 0.60 to 0.70 per cent carbon and 0.60 to 1.0 per cent manganese.
This unit covers Types of stresses & strains,
Hooke’s law, stress-strain diagram,
Working stress,
Factor of safety,
Lateral strain,
Poisson’s ratio, volumetric strain,
Elastic moduli,
Deformation of simple and compound bars under axial load,
Analysis of composite bar with varying cross section.
This lifting machine is used to carry heavy industrial machinery like lathe from one floor to another. We have analysed the strength and the capacity of this machine. We have checked critical points for failure and suggested design recommendations to increase the capacity to 3 tonnes from the present value of 1.1 tonne.
Team members: Ahsen, Ankit, Ankit, Shivam, Anurag, Deepak
This is a power point presentation on the design of Helical springs subjected to Static and Fluctuating load. It is part of Design of Machine elements subject.
This is first or introductory lecture of Mechanics of Solids-1 as per curriculum formulated by Higher Education Commission and Pakistan Engineering Council
Experimental evaluation of strain in concrete elementsnisarg gandhi
Evaluation of strain using
1) mech strain gauge
2) elec strain gauge
also calculation of modulus of elasticity using
1) secant modulus
2) chord modulus
also for the procedure to use electrical strain gauge see the following link
https://drive.google.com/open?id=0Bw9bdaDxJsb8enFZOFhlRWFMYWs&authuser=1
Because of torsion, the beam fails in diagonal tension forming the spiral cracks around the beam. Warping of the section does not allow a plane section to remain as plane after twisting. Clause 41 of IS 456:2000 provides the provisions for
the design of torsional reinforcements. The design rules for torsion are based on the equivalent moment.
Spring Design, Helical Springs, compression & Extension springs, spring design procedure leaf spring, multi-leaf springs design process and analysis, Role of Spring index in spring design. Springs for Fluctuating loads.
This unit covers Types of stresses & strains,
Hooke’s law, stress-strain diagram,
Working stress,
Factor of safety,
Lateral strain,
Poisson’s ratio, volumetric strain,
Elastic moduli,
Deformation of simple and compound bars under axial load,
Analysis of composite bar with varying cross section.
This lifting machine is used to carry heavy industrial machinery like lathe from one floor to another. We have analysed the strength and the capacity of this machine. We have checked critical points for failure and suggested design recommendations to increase the capacity to 3 tonnes from the present value of 1.1 tonne.
Team members: Ahsen, Ankit, Ankit, Shivam, Anurag, Deepak
This is a power point presentation on the design of Helical springs subjected to Static and Fluctuating load. It is part of Design of Machine elements subject.
This is first or introductory lecture of Mechanics of Solids-1 as per curriculum formulated by Higher Education Commission and Pakistan Engineering Council
Experimental evaluation of strain in concrete elementsnisarg gandhi
Evaluation of strain using
1) mech strain gauge
2) elec strain gauge
also calculation of modulus of elasticity using
1) secant modulus
2) chord modulus
also for the procedure to use electrical strain gauge see the following link
https://drive.google.com/open?id=0Bw9bdaDxJsb8enFZOFhlRWFMYWs&authuser=1
Because of torsion, the beam fails in diagonal tension forming the spiral cracks around the beam. Warping of the section does not allow a plane section to remain as plane after twisting. Clause 41 of IS 456:2000 provides the provisions for
the design of torsional reinforcements. The design rules for torsion are based on the equivalent moment.
Spring Design, Helical Springs, compression & Extension springs, spring design procedure leaf spring, multi-leaf springs design process and analysis, Role of Spring index in spring design. Springs for Fluctuating loads.
Dive into the innovative world of smart garages with our insightful presentation, "Exploring the Future of Smart Garages." This comprehensive guide covers the latest advancements in garage technology, including automated systems, smart security features, energy efficiency solutions, and seamless integration with smart home ecosystems. Learn how these technologies are transforming traditional garages into high-tech, efficient spaces that enhance convenience, safety, and sustainability.
Ideal for homeowners, tech enthusiasts, and industry professionals, this presentation provides valuable insights into the trends, benefits, and future developments in smart garage technology. Stay ahead of the curve with our expert analysis and practical tips on implementing smart garage solutions.
Can AI do good? at 'offtheCanvas' India HCI preludeAlan Dix
Invited talk at 'offtheCanvas' IndiaHCI prelude, 29th June 2024.
https://www.alandix.com/academic/talks/offtheCanvas-IndiaHCI2024/
The world is being changed fundamentally by AI and we are constantly faced with newspaper headlines about its harmful effects. However, there is also the potential to both ameliorate theses harms and use the new abilities of AI to transform society for the good. Can you make the difference?
White wonder, Work developed by Eva TschoppMansi Shah
White Wonder by Eva Tschopp
A tale about our culture around the use of fertilizers and pesticides visiting small farms around Ahmedabad in Matar and Shilaj.
Hello everyone! I am thrilled to present my latest portfolio on LinkedIn, marking the culmination of my architectural journey thus far. Over the span of five years, I've been fortunate to acquire a wealth of knowledge under the guidance of esteemed professors and industry mentors. From rigorous academic pursuits to practical engagements, each experience has contributed to my growth and refinement as an architecture student. This portfolio not only showcases my projects but also underscores my attention to detail and to innovative architecture as a profession.
Book Formatting: Quality Control Checks for DesignersConfidence Ago
This presentation was made to help designers who work in publishing houses or format books for printing ensure quality.
Quality control is vital to every industry. This is why every department in a company need create a method they use in ensuring quality. This, perhaps, will not only improve the quality of products and bring errors to the barest minimum, but take it to a near perfect finish.
It is beyond a moot point that a good book will somewhat be judged by its cover, but the content of the book remains king. No matter how beautiful the cover, if the quality of writing or presentation is off, that will be a reason for readers not to come back to the book or recommend it.
So, this presentation points designers to some important things that may be missed by an editor that they could eventually discover and call the attention of the editor.
Top 5 Indian Style Modular Kitchen DesignsFinzo Kitchens
Get the perfect modular kitchen in Gurgaon at Finzo! We offer high-quality, custom-designed kitchens at the best prices. Wardrobes and home & office furniture are also available. Free consultation! Best Quality Luxury Modular kitchen in Gurgaon available at best price. All types of Modular Kitchens are available U Shaped Modular kitchens, L Shaped Modular Kitchen, G Shaped Modular Kitchens, Inline Modular Kitchens and Italian Modular Kitchen.
Transforming Brand Perception and Boosting Profitabilityaaryangarg12
In today's digital era, the dynamics of brand perception, consumer behavior, and profitability have been profoundly reshaped by the synergy of branding, social media, and website design. This research paper investigates the transformative power of these elements in influencing how individuals perceive brands and products and how this transformation can be harnessed to drive sales and profitability for businesses.
Through an exploration of brand psychology and consumer behavior, this study sheds light on the intricate ways in which effective branding strategies, strategic social media engagement, and user-centric website design contribute to altering consumers' perceptions. We delve into the principles that underlie successful brand transformations, examining how visual identity, messaging, and storytelling can captivate and resonate with target audiences.
Methodologically, this research employs a comprehensive approach, combining qualitative and quantitative analyses. Real-world case studies illustrate the impact of branding, social media campaigns, and website redesigns on consumer perception, sales figures, and profitability. We assess the various metrics, including brand awareness, customer engagement, conversion rates, and revenue growth, to measure the effectiveness of these strategies.
The results underscore the pivotal role of cohesive branding, social media influence, and website usability in shaping positive brand perceptions, influencing consumer decisions, and ultimately bolstering sales and profitability. This paper provides actionable insights and strategic recommendations for businesses seeking to leverage branding, social media, and website design as potent tools to enhance their market position and financial success.
2. Springs
Department of Mechanical Engineering 2
Spring is an elastic machine element, which deflects under the
action of the load and returns to its original shape when the load is
removed
Types
Compression Spring
Extension Spring
Torsion Spring
Leaf Spring
Induces torsional shear stresses in spring wire
even the spring is subjected to axial loading
3. Terminology of Springs
Department of Mechanical Engineering 3
Spring index parameter
𝐶 =
𝐷
𝑑
Mean diameter
𝐷 =
𝐷𝑖 + 𝐷𝑜
2
Spring index indicates the relative
sharpness of curvature
If C low value: Stresses excessive
due to curvature effect
If C high: Prone to buckling
C: 6-9 is preferred
Eqn 11.2c
4. Terminology of Springs
Department of Mechanical Engineering 4
Solid Length:
The axial length of the
spring which is so
compressed that the
adjacent coils touch each
other
𝑆. 𝐿. = 𝑖𝑡𝑑
Compressed length:
The axial length of the
spring which is
subjected to maximum
compressive force
𝐶. 𝐿. = (𝑖𝑡−1) × 𝐺𝑎𝑝 + 𝑆. 𝐿
Free length:
The axial length of an
unloaded helical
compression spring
𝐹. 𝐿.
= (𝑖𝑡−1) × 𝐺𝑎𝑝
+ 𝑖𝑡𝑑 + 𝑦
5. Terminology of Springs
Department of Mechanical Engineering 5
𝐼𝑛𝑎𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑖𝑙𝑠 = (𝑖𝑡−𝑖)
𝑘 =
𝐹
𝑦
Pitch of the coil (p): Axial distance between adjacent coils in
uncompressed state of spring
𝑝 =
𝐹𝑟𝑒𝑒 𝐿𝑒𝑛𝑔𝑡ℎ
(𝑖𝑡 − 1)
Stiffness of the spring (k): Force required to produce unit deflection
Active (N) and inactive coils: Active coils are the coils in the spring which
contribute to spring action, support the external force and deflect under
the action of force
End coils which do not contribute to spring action are called inactive coils
Eqn 11.7a
6. Unbent Spring
Department of Mechanical Engineering 6
Two basic equations for
the design of helical
springs
• Load-stress equation
• Load-deflection
equation
Dimensions
• The diameter of the bar is equal to the wire diameter of
the spring (d)
• The length of the equivalent bar is 𝜋𝐷𝑖
• Bar is fitted with bracket of length (D/2)
7. Unbent Spring
Department of Mechanical Engineering 7
Torsional moment due to force P
• Direct or transverse shear stress in spring wire
• Length of inside fibre is less than length of outside fibre
which induces stress concentration at the inside of fibre
coil
𝑀𝑡 =
𝐹𝐷
2
Torsional shear stress due to the force P
𝜏1 =
𝑇𝑟
𝐽
=
𝐺𝜃
𝑙
=
8𝐹𝐷
𝜋𝑑3
Considering equivalent bar in the form of helical coil, the additional
stresses accounted
Eqn 11.1c
8. Stresses in spring wire
Department of Mechanical Engineering 8
Pure torsional stress
Direct shear stress
Combined torsional, direct
and curvature shear stress
9. Load-Stress Equation
Department of Mechanical Engineering 9
Modification in the stress equation to accommodate the direct
shear and curvature stress effect
K=KsKc
Direct shear stress, 𝜏2 =
4𝐹
𝜋𝑑2 =
8𝐹𝐷
𝜋𝑑3
0.5𝑑
𝐷
Combined torsional, direct and curvature shear stress by Wahl
𝜏 = 𝜏1 + 𝜏2 =
8𝐹𝐷
𝜋𝑑3
1 +
0.5𝑑
𝐷
= 𝐾𝑠
8𝐹𝐷
𝜋𝑑3
𝜏 = 𝐾
8𝐹𝐷
𝜋𝑑3 𝐾 =
4𝐶 − 1
4𝐶 − 4
+
0.615
𝐶
Eqn 11.1c Eq 11.2a
Wahl factor
LOAD-STRESS EQUATION
10. Load-Deflection Equation
Department of Mechanical Engineering 10
The angle of twist (θ)
𝜃 =
𝑀𝑡𝑙
𝐽𝐺
=
16𝐹𝐷2𝑖
𝐺𝑑4 Eqn 11.4
Axial deflection, y =
8𝐹𝐷3𝑖
𝐺𝑑4 Eqn 11.5a
Stiffness of material, 𝑘 =
𝐺𝑑4
8𝐷3𝑖
Eqn 11.7a
LOAD-DEFLECTION EQUATION
11. Series and Parallel Connections
Department of Mechanical Engineering 11
𝑦1 =
𝐹
𝑘1
𝑦2 =
𝐹
𝑘2
Series:
• Force acting is same (F)
• Total deflection will be sum of deflections of
individual springs y = 𝑦1 + 𝑦2
1
𝑘
=
1
𝑘1
+
1
𝑘2
+. .
Parallel:
• Force acting is sum of
forces acting on individual
spring (𝐹 = 𝐹1 + 𝐹2)
• Total deflection will be
same (𝛿)
𝑦1𝑘1 = 𝐹 𝑦2𝑘2 = 𝐹
𝑘 = 𝑘1 + 𝑘2+. .
12. Helical Spring: Design Procedure
Department of Mechanical Engineering 12
• Estimate spring force (F) and required deflection (y), In some cases
it will be specified
• Select suitable material, Obtain permissible shear stress, 𝜏 = 0.5𝑆𝑢𝑡
• Assume spring index value C, Preferred (8-10)
• Calculate Wahl factor, K (Eqn. 11.2a)
• Determine wire diameter from load-stress equation, d (Eqn 11.1d)
• Determine coil diameter, D, (eqn 11.2c)
• Determine number of active coils, i by load deflection equation, i
(eqn 11.5a)
• Determine total number of coils, it, (table 11.4)
• Determine solid length of the spring, (table 11.4)
• Find actual deflection, y by load-deflection equation and also free
length (Assuming suitable gaps)
• Obtain pitch of the coil, p (table 11.4)
• Determine actual spring rate, k (eqn 11.7a)
13. Problem 5.1
Department of Mechanical Engineering 13
It is required to design a helical compression spring subjected to a
maximum force of 1250 N. The deflection of the spring
corresponding to the maximum force should be approximately 30
mm. The spring index can be taken as 6. The spring is made of
patented and cold-drawn steel wire. The ultimate tensile strength
and modulus of rigidity of the spring material are 1090 and 81 370
N/mm2, respectively. The permissible shear stress for the spring wire
should be taken as 50% of the ultimate tensile strength. Design the
spring and calculate
• Wire diameter
• Mean coil diameter
• Number of active coils
• Total number of coils
• Free length of the spring
• Pitch of the coil
14. Problem 5.2
Department of Mechanical Engineering 14
It is required to design a helical compression spring for the valve mechanism.
The axial force acting on the spring is 300 N when the valve is open and 150 N
when the valve is closed. The length of the spring is 30 mm when valve is open
and 35 mm when the valve is closed. The spring is made of oil-hardened and
tempered valve spring wire and the ultimate tensile strength is 1370 N/mm2.
The permissible shear stress for the spring wire should be taken as 30% of the
ultimate tensile strength. The modulus of rigidity is 81370N/mm2. The spring is
to be fitted over a valve rod and the minimum inside diameter of the spring
should be 20 mm. Design the spring and calculate
• Wire diameter
• Mean coil diameter
• Number of active coils
• Total number of coils
• Free length of the spring
• Pitch of the coil
Assume that the clearance between adjacent coils or clash allowance is 15% of
the deflection under the maximum load
15. Design against fluctuating load
Department of Mechanical Engineering 15
Mean force, 𝐹𝑚 =
1
2
𝐹𝑚𝑎𝑥 + 𝐹𝑚𝑖𝑛
Amplitude force, 𝐹𝑎 =
1
2
𝐹𝑚𝑎𝑥 − 𝐹𝑚𝑖𝑛
Torsional Mean stress, 𝜏𝑚 = 𝐾𝑠
8𝐹𝑚𝐷
𝜋𝑑3 , 𝐾𝑠 = 1 +
0.5
𝐶
Torsional
stress amplitude, 𝜏𝑎 = 𝐾
8𝐹𝑎𝐷
𝜋𝑑3 , 𝐾𝑠 = 𝑊𝑎ℎ𝑙 𝑓𝑎𝑐𝑡𝑜𝑟
Rotating beam completely reversed cycle
Springs are subjected to pulsating shear stresses which vary
from 0 to 𝑆𝑠𝑒
′
Eqn 11.16f
Eqn 11.16g
17. Problem 5.3
Department of Mechanical Engineering 17
A helical compression spring of cam mechanism is subjected
to an initial preload of 50 N. The maximum operating force
during the load cycle is 150 N. The wire diameter is 3mm
while the mean coil diameter is 18 mm. The spring is made
of oil-hardened and tempered valve spring wire of grade VW
(Sut=1430N/mm2). Determine the factor of safety used in
design on the basis of fluctuating stresses.
18. Problem 5.4
Department of Mechanical Engineering 18
An eccentric cam, 100 mm in diameter
rotates with an eccentricity of 10 mm as
shown in Figure. The roller follower is held
against the cam by means of a helical
compression spring. The force between
the cam and the follower varies from 100
N at the lowest position to 350 N at the
highest position of follower. The
permissible shear stress in the spring wire
is recommended as 30% of ultimate
tensile strength. Design the spring from
static consideration and determine the
factor of safety against fluctuating
stresses. Neglect the effect of inertia
forces. Assume Sut=1480N/mm2
19. Concentric Springs
Department of Mechanical Engineering 19
• Two helical compression springs, one inside the other
• The two consecutive springs will have opposite helix to avoid locking
of coils
Advantages
• Load carrying capacity is increased
• Fail safe system
• Surge are eliminated
20. Concentric Springs
Department of Mechanical Engineering 20
𝐾1
8𝐹1𝐷1
𝜋𝑑1
3 = 𝐾2
8𝐹2𝐷2
𝜋𝑑2
3
Assumptions in Concentric spring
• Springs are made of same material
• Maximum torsional shear stresses induced in outer and inner
springs are equal
• Same free length
• Both springs are deflected by same amount
𝐾1 = 𝐾2
Assume,
𝐹1𝐷1
𝑑1
3 =
𝐹2𝐷2
𝑑2
3
𝜏1 = 𝜏2
𝑦1 = 𝑦2
𝐹1𝐷1
3
𝑖1
𝑑1
4 =
𝐹2𝐷2
3
𝑖2
𝑑2
4
Eqn (a)
Eqn (b)
21. Concentric Springs
Department of Mechanical Engineering 21
When both springs are completely compressed, solid length will be
equal
𝑑1𝑖1 = 𝑑2𝑖2
𝐷1
𝑑1
=
𝐷2
𝑑2
𝐹1𝐷1
3
𝑖1𝑑1
𝑑1
5 =
𝐹2𝐷2
3
𝑖2𝑑2
𝑑2
5
𝐹1𝐷1
3
𝑑1
5 =
𝐹2𝐷2
3
𝑑2
5
Eqn (c)
Eqn (c)/Eqn (a) Eqn (d)
Eqn (d)/Eqn (a) 𝐹1
𝑑1
2 =
𝐹2
𝑑2
2
𝐹1
𝐹2
=
𝑎1
𝑎2
23. Problem 5.5
Department of Mechanical Engineering 23
A concentric spring is used as a valve spring in a heavy duty diesel
engine. It consists of two helical compression springs having the
same free length and same solid length. The composite spring is
subjected to a maximum force of 6000N and the corresponding
deflection is 50 mm. The maximum torsional shear stress induced
in each spring is 800 N/mm2. The spring index of each spring is 6.
Assume same material for two springs and the modulus of rigidity
is 81370 N/mm2. The diametral clearance between the coils is
equal to the difference between their wire diameters. Calculate:
• The axial force transmitted by each spring
• Wire and mean coil diameters of each spring
• Number of active coils in each spring
24. Surge in Spring
Department of Mechanical Engineering 24
• Natural frequency of vibration of spring coincides with the
frequency of external periodic force, resonance occurs
• Spring is subjected to a wave of successive compressions that
travels from one end to other and back which is termed as surge
• Load is transmitted by transferring compression to adjacent coils
• If the onward and backward travel time coincides with exiting
frequency, resonance occurs
• Natural frequency of helical compression springs (between two
parallel plates), 𝜔 =
1
2
𝑘
𝑚
25. Leaf Springs
Department of Mechanical Engineering 25
• Consists of flat plates/ leaves
• Longest leaf at top is called master leaf
• Bent at the end to form spring eyes
• U bolt and centre clip to hold the leaf together
• Master leaf, extra full length leaf, graduated leaf
28. Nipping of leaf springs
Department of Mechanical Engineering 28
Taking 𝑐 = 𝑦𝑔 − 𝑦𝑓, 𝐸𝑞𝑢𝑎𝑙𝑖𝑠𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑣𝑎𝑙𝑢𝑒𝑠
𝜎𝑏 =
6𝐹𝐿
𝑖𝑏′ℎ2
• Pre-stressing by bending
leaves to a different radii
• Initial gap between full
length leaf and graduated
length leaf before the
assembly is called nip
𝜎𝑏 𝑔 =
6𝐹
𝑔𝑙
𝑖𝑔𝑏′ℎ2 𝜎𝑏 𝑓 =
6𝐹𝑓𝑙
𝑖𝑓𝑏′ℎ2
Eqn 11.32b
Eqn 11.32c
𝑐 =
2𝐹𝑙3
𝐸𝑖𝑏′ℎ3
𝐹𝑖 =
2𝑖𝑔𝑖𝑓𝐹
𝑖 2𝑖𝑔 + 3𝑖𝑓
Eqn 11.32a
29. Problem 5.7
Department of Mechanical Engineering 29
A semi elliptic leaf spring used for automobile suspension
consists of three extra full length leaves and 15 graduated
length leaves, including the master leaf. The centre-to-centre
distance between two eyes of the spring is 1m. The maximum
force that can act on the spring is 75kN. For each leaf, the
ratio of width-to-thickness is 9:1. The modulus of elasticity of
the leaf material is 207000N/mm2. The leaves are pre-stressed
in such a way that when the force is maximum, the stresses
induced in all leaves are same and equal to 450 N/mm2.
Determine
• Width and thickness of the leaves
• The initial nip
30. Problem 5.8
Department of Mechanical Engineering 30
A semi elliptical laminated vehicle spring to carry a load of
6000N is to consist of seven leaves, 65 mm wide. Two of the
leaves extending the full length of the spring. The spring is to
be 1.1 m in length and attached to the axle by two U bolts 80
mm apart. The bolts hold the central portion of the spring so
rigidly that they may be considered equivalent to a band
having a width equal to the distance between the bolts.
Assume a design stress for spring material as 350 MPa.
Determine
• The thickness of the leaves
• Deflection of the spring
