**Theory Questions and answers on Design of Springs**

**1) What is Spring ? What are its functions/objectives?**

**“A spring is an elastic object used to store mechanical energy. Spring acts as a flexible joint in between two parts or bodies.”**

**Objectives/functions of Spring:**

**A spring can is used for multiple purposes and functions in different devices, Some of them are listed below. Following are the objectives of a spring when used as a machine member:**

**1. Cushioning , absorbing , or controlling of energy due to shock and vibration.**

**Car springs or railway buffers acts as cushions.**

**2.Control of motion**

**Maintaining contact between two elements (cam and its follower)**

**Creation of the necessary pressure in a friction device (a brake or a clutch)**

**Restoration of a machine part to its normal position when force is withdrawn (a governor )**

**3.Measuring forces**

**Spring is used for measuring forces in Spring balances, gauges.**

**4. Storing of energy**

**In clocks,starters and toys spring is used to store energy and release it over time.**

**2) Give Classification of Springs?**

**Springs can be classified depending on how the load force is applied to them:**

**Tension/Extension spring – the spring is designed to operate with a tension load, so the spring stretches as the load is applied to it. examples helical tension spring,rubber spring.****Compression spring – is designed to operate with a compression load, so the spring gets shorter as the load is applied to it. Examples compression helical spring, Belleville spring****Torsion spring – unlike the above types in which the load is an axial force, the load applied to a torsion spring is a torque or twisting force, and the end of the spring rotates through an angle as the load is applied. examples torsion bars, clock spring etc.****Constant spring - supported load will remain the same throughout deflection cycle.****Variable spring - resistance of the coil to load varies during compression.**

**They can also be classified based on their shape:**

**Coil spring – this type is made of a coil or helix of wire****Flat spring – this type is made of a flat or conical shaped piece of metal.****Machined spring – this type of spring is manufactured by machining bar stock with a lathe and/or milling operation rather than coiling wire. example clip springs**

**3) What are commonly used materials used for springs ?**

**Following are the commonly used material for the springs**

**Hard-drawn wire: This is cold drawn, cheapest spring steel. Normally used for low stress and static load. The material is not suitable at subzero temperatures or at temperatures above 120C.**

**Oil-tempered wire:It is a cold drawn, quenched, tempered, and general purpose spring steel. However, it is not suitable for fatigue or sudden loads, at subzero temperatures and at temperatures above 180C.**

**Chrome Vanadium:This alloy spring steel is used for high stress conditions and at high temperature up to 220C. It is good for fatigue resistance and long endurance for shock and impact loads. When we go for highly stressed conditions then alloy steels are useful.**

**Music wire:This spring material is most widely used for small springs. It is the toughest and has highest tensile strength and can withstand repeated loading at high stresses. However, it can not be used at subzero temperatures or at temperatures above 120C.**

**4) What is the curvature effect in a helical spring? How does it vary with spring index?**

**Ans: For springs where the wire diameter is comparable with the coil diameter, the inside length of the spring segment is relatively shorter than the outside length. Hence shearing strain is more in the inner segment than the outer segment. This unequal shearing strain is called the curvature effect. Curvature effect decreases with the increase in spring index.**

**5) State the significance of Wahl’s correction factor in springs?**

**Ans :**

**The shear stress developed in the spring is given by,**

**Where ks is shear stress factor..**

**It is given by,**

**The curvature of the wire increases the shear stress on the inner surface of the spring and decreases it slightly on the outer surface. This curvature effect stress is localized and is significant only when fatigue load is present. In order to consider the effects of both direct shear stress as well as curvature effect, the shear stress correction factor Ks is replaced by another factor Kw known as Wahl’s factor.**

**..**

**Maximum shear stress produced is given by ,**

**The Wahl factor increases very rapidly as the spring index decreases. It signifies that the curvature stress also increases with increase in spring index.**

**6) What are the different stresses a helical spring is subjected to?**

**A helical spring is subjected to ,**

**1) Direct sheare stress : This is the stress induced in the wire as the effect load acting on spring.**

**2) Torsional shear stress : This stress is induced due to the torsional effect of load on coil.**

**From equation of torsion , this stress can be written as**

**3) Shear stress due to curvature: The curvature of the wire increases the shear stress on the inner surface of the spring and decreases it slightly on the outer surface. This is taken into account through Wahl’s correction factor.**

**7). Explain the following terms related to the spring with diagram, Spring index, solid length, free length, compressed length,pitch,stiffness.**

**The spring index is defined as the ratio of mean coil diameter to wire diameter. Or C = D/d**

**In design of helical springs, the designer should use good judgment in assuming the value of the**

**spring index C. The spring index indicates the relative sharpness of the curvature of the coil.**

**Solid length:solid length is defined as the axial length of the spring which is so compressed, that the adjacent coils touch each other. In this case, the spring is completely compressed and no further compression is possible.The solid length is given by.**

**Solid length = ...mm {=total no of turns, d= diamter of wire}**

**Compressed length: Compressed length is defined as the axial length of the spring that is**

**subjected to maximum compressive force. In this case, the spring is subjected to maximum**

**deflection. When the spring is subjected to maximum force, there should be some gap or**

**clearance between the adjacent coils. The gap is essential to prevent clashing of the coils.**

**The clashing allowance or the total axial gap is usually taken 15% of the maximum deflection.**

**Free length: Free length is defined as the axial length of an unloaded helical compression**

**spring. In this case, no external force acts on the spring. Free length is an important dimension in**

**spring design and manufacture. It is the length of the spring in free condition prior to assembly.**

**Free length is given by, Free length = solid length + …...mm**

**The pitch of the coil is defined as the axial distance between adjacent coils in uncompressed state of spring. It is denoted by p. It is given by,**

**Stiffness: It is defined as the load required to produce the unit deflection of the spring.**

**8) Explain with sketch the construction and application of leaf spring.**

Leaf spring is built up of number of plates (known as leaves). The leaves are usually given as initial curvature or cambered so that they will tend to straighten under load. The leaves are held together by means of a band shrunk around them at the centre or by bolt passing through the centre. Since, the band exerts stiffening and straightening effect, therefore the effective length of the spring for bending will be overall length of spring minus width of band. In case of a centre bolt, two-third distance between centers of U-bolt should be subtracted from the overall length of spring in order to find effective length. The spring is clamped to the axle housing by means of U-bolt.

The longest leaf known as main leaf or master leaf has its ended formed in the shape of an eye through which the bolts are passed to secure the spring to its supports. Usually the eyes, through which the spring is attached to the hanger or shackle, are provided with bushings of some anti-friction material such as bronze or rubber. The other leaves of the spring are known as graduated leaves. In order to prevent digging in the adjacent leaves the ends of the graduated leaves are trimmed in various forms as shown in figure.

Applications:

1. It is used for mainly automobile suspension

2. For rail road spring

**9) What do you mean by Nipping of leaf spring?**

**Ans :The stress in the full length leaves is 50% greater than the stress in the graduated leaves. ****To distribute this additional stress from the full length leaves, pre-stressing is done, which is called “Nipping “ of leaves.**

**This is achieved by bending the leaves to different radii of curvature, before they are assembled with the centre bolt. The full length leaves are given in greater radii of curvature than the adjacent one. Due to the different radii of curvature, when the full length leaves are staked with the graduated leaves, without bolting, a gap is observed between them. This gap is called Nip The nip eliminated by tightening of the center bolt.**

**Numerical Problems on Design of Spring**

**1) Design a helical compression Spring for a maximum load of 1500 N and a deflection of 30 mm. Assume permissible shear stress for Spring material as 420 MPa andspring index as 5. Take G = 84 GPa.**

**2) Determine required number of coils and the allowable deflection in helical Spring made of 1.6 mm diameter wire. Assume thespring index as as 6 and a permissible shear stress as 345 MPa . The stiffness of the Spring is to be 1.8 N/mm. Take G = 80 GPa**

**3) Following data is refers to helical compression Spring.**

**Axial load=8000 N****Spring rate = 72 N/mm****Mean diamter of coil =125 mm****Modulus of rigidity for material (G) = 80 GPa****Tensile strength = 550 MPa****Allowable shear stress half that of tensile strength****Standard Spring wire dia 18,19,20,21,22,23,24,25,27,29,30**

** Find the wire diameter and number of active turns for the spring**

**4) A composite compression Spring has two closed coil. Outer Spring is of 15 mm longer than inner Spring. The outer Spring has 10 coils of mean diameter 36 mm & wire diameter 6 mm. The inner Spring has 8 coils of mean daimeter 30 mm & wire diameter 5 mm. When Spring is subjected to an axial load 1000 N, Modulus of rigidity for material (G) may be taken as 81370 MPa . Find. 1) Compression of each Spring 2) Load shared by each Spring 3) Shear stressindced in each Spring**

**5) A concentric Spring consists of two helical compression Spring having the same free length. The composite Spring is subjected to a maximum force of 2 kN. the wire and mean coil diameter of the outer Spring are 10mm and 80 mm respectively. The numbers of active coils in inner and outer Springs with G = 81370 MPa.Calculate :1) Force transmitted by each Spring,2) Maximum deflection of the Spring and3) maximum torsional shear stress induced in each Spring.**

**6) A railway wagon of mass 250kN moving with a velocity of 2.5m/sec is brought to rest by Springs of mean diameter 350mm. The maximum deflection of the Spring is 210mm. Find the wire diameter and number of turns. Take S s = 600MPa and G = 80 GPa.**

**7) it is required to design a Spring for the valve of IC Engine with following details:**

**Spring load = 80N, When valve is closed****Spring load = 100N, When valve is open**

**Space constraints for Spring fitment are:**

**Inside guide bush diameter = 24 mm****Outside recess diameter = 36mm.****Valve lift = 5mm**

**Spring steel has following properties:**

**Ultimate tensile strength = 710 MPa****Modulus of rigidity for material (G) = 8.0 ×10 4 MPa**

**Spring ends are square and ground. Permissible shear stress for Spring wire = 0.5 S ut .**

**Determine:**

**Wire diameter,****spring index as,****Total number of coils,****Solid length, Free length,****Pitch of the coil when additional 15% of working deflection is used to avoid complete closing of coil.**

**8) A torsional helical Spring is made from a music wire with wire diameter of 1.37 mm and mean coil diameter 22mm. The Spring has 400 turns. If the material of the Spring has ultimate tensile strength of 2076MPa and Yield point stress = 0.60 S ult with factor of safety of 2 based on yield point, compute maximum stress on the inside of the helix. Take modulus of elasticity of Spring material as 210 GPa. Consider the effect due to stress concentration and due to curvature. Find the torque that Spring can exert after unwinding 12 revolutions from the most highly stressed condition.**

**9) It is required to design a helical compression Spring of circular wire, subjected to an axial load, which varies from 2.5kN to 3.5kN. For this range of load, the deflection of the Spring should be limited to 5mm. The Spring index is 5 the Spring has square and ground ends. For Spring wire material, σ ut = 1050 MPa and G = 81370 MPa. The permissible shear stress for the Spring wire should be taken as 50% of the σ ut.**

** Calculate **

**i) Wire diameter and mean coil diameter. ii) Number of active coils & total number of coils. iii) Solid length of Spring. iv) Free length of Spring. v) Required Spring rate & vi) actual Spring rate.**

**10) A safety valve operated by a helical tension Spring through the lever mechanism is as shown in the figure below. The diameter of the valve is 50 mm. In normal operating conditions, The valve is closed and the pressure inside the chamber is 0.5 MPa. The valve is opened when the pressure inside the chamber increases to 0.6 MPa. The maximum lift of the valve is 5 mm. Thespring index as is 8. The Spring wire material has ultimate tensile strength of 1200 MPa and Modulus of rigidity for material (G) of 81370 MPa. The permissible shear stress for the Spring wire can be taken as 30% of the ultimate tensile strength.Calculate:i) Wire diameter.ii) mean coil diameter andiii) number of active coils.**

**11) A safety valve of 60 mm diameter is to blow off at a pressure of 1.2 MPa. It is held on its seat by closed coil helical Spring. The maximum lift of the valve is 10 mm. design a suitable compression Spring ofspring index as 5 and providing an initial compression of 35mm. the maximum shear stress in the material of the wire is limited to 500MPa . the Modulus of rigidity for material (G) of the Spring material is 80000 MPa .Calculate :a) diameter of Spring wire, b) mean coil diameter, c) number of active turns, and d) Pitch of the coil**

**Links to other Topics in Machine design I**

**Machine Design I - Introduction to Design : Theory Q&A**

**Machine Design -I -Design of joints : Theory Q&A**

**Knuckle Joint : Design Procedure,Problems and Questions**

**Design of turnbuckle : Design steps, Problems and Question**

**Design of Levers : Hand Lever, Foot Lever, Bell crank lever**

**Design Of Bolted and Welded Joints**

**Design of Shafts: Theory and Numerical Problems**

**Couplings : Design Procedure and Numerical problems**

**Design Of SPRINGS : Questions and Numerical problems**

**Power Screw Design**

**Belt drives:Theory Q&A and Selection of Flat and V belts**

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