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Sixth semester MechanicalDesign of Machine Elements -IIHelical springs are said to be closely coiled, when the helix angle is very small ( 10o),where as in open coil helical spring the helix angle is largeAdvantages.] These springs are easy to manufacture.] They are available in wide range.] They are highly reliable.] They have constant spring rates.] Their performance can be predicted more accurately.] There characteristics can be varied by changing dimensions.2. Conical and Volute springsThe conical and volute spring shown in the figure are used in special applications where thespring rate increases in increase in load. Another feature of these types of springs is thedecreasing number of coils results in an increasing spring rate. This characteristic issome times utilized in vibrations problems where springs are used to support to body thathave varying mass.3. Torsion springsThese springs may be of helical or spiral type asshown in figure. Helical types of springs are usedwhere the load tends to wind up the springs and areused in electrical mechanisms. Spiral type is usedwhere the loads tends to increase the number ofcoils and are used in watches and clocks.4. Laminated or Leaf springs.The laminated or leaf spring (also known as flatspring) consists of a number of flat plates (knownas leaves) of varying lengths held together bymeans of clamps and bolts. These types of springsare most used in automobiles.5. Disc springsThese springs consists of a number of conical discs heldtogether by a central bolt or tube as shown in figure. Thesesprings are used in applications where high spring rates andcompact spring units are requiredJagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IITERMS USED IN COMPRESSION SPRINGSSolid length: When the springs are compressed until the coils come in contact with eachother, then the spring is said to be solid. The solid length of a spring is the product of totalnumber of coils & the diameter of the wire. Mathematically,Solid length Free length Free length of acompression spring is the length ofthe spring in the free or unloadedcondition & is equal to the solidlength plus the maximum deflectionor compression of the spring & theclearance between the adjacentcoils.Free length - Lf solid length max. Deflection clearance between adjacent coils.Spring indexIt is defined as the ratio of the man diameter of the coil toSpring index C D/dthe diameter of the wire.Spring Rate spring rate (stiffness/spring constant) is the defined as the load required perunit deflection of the spring.Spring Rate, K F/ δF- load, Nδ - Defection, mmPitch: Pitch of the coils is defined as the axial distance between adjacent coils in uncompressed state.STRESS IN HELICAL SPRING AND CIRCULAR WIRE.Jagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IIConsider a helical compression spring made of circular wire & subjected to an axial load F,as shown in figure.Let,D Mean diameter of the coild Diameter of the spring wire,n number of active coils,G Modulus of Rigidity for the spring material,F Axial load on the spring,τ Max. Shear stress induced in the wire,C spring index D/dp pitch of the coilsδ deflection of the spring.Consider a point of the spring shown in fig (b). The load ‘F’ tends to rotate the wire & as aresult twisting moment (T) is developed in the wire, & thus torsional shear stress is induced in the wire.Let us consider that part of spring is in equilibrium under the action of two forces ‘F’ &twisting ‘T’.In addition to the torsional shear stress (τ1) induced in the wire, the following stresses alsoact on the wire. Direct stress due to the load. F, & Stress due to curvature of wireShear stress 8FD .K3πdWhere K 4C - 1 0.6154C - 4; Wahl stress Conc. FactorCJagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IIDeflection of helical springs of circular wire.Let l total active length of wire πD nθ Angular deflection of the wire due to Torque t.Therefore, Axial deflection of the spring δ θ D/2-----------(1)Also from the Torsion equation, we have,33Therefore δ 8FD n 8FC n4GdG.dwhere n is the number of active coils.Design Procedure for Helical Springs.Jagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -II1. Diameter of the wire:2. Mean diameter of the coil:(a) Outer diameter of the coil:(b) Inner diameter of the coil:3. Number of coils:4. Free length5. Stiffness or Spring Rate6. Pitch:PROBLEMS(1) Design a helical compression spring to support an axial load of N. The deflection underload is limited to 60 mm. The spring index is 6. The spring is made of chrome-vanadiumsteel & FOS 2.Spring specifications(i) Wire diameter(ii) Mean diameter(iii) Free length(iv) Total no of coils –(v) Style of ends squared & ground(vi) Pitch - p –(vii)Spring rate –(viii)Material – Chrome- vanadium(2) Design a helical compression spring for a max. load of 1000N for a deflection of 25mmusing the spring index as 5. the max permissible shear stress for spring wire is 420 N/mm2 &G 84 103 N/mm2Jagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -II(3) A railway carriage weighing 40KN & moving at 8km/hr is to be brought to rest by 2 buffersprings. The compression between the coils must be twice the wire diameter. Assume springindex as 8. And allowable shear stress for the spring material 450N/mm. Take G 0.8 * 10N/mm. Design the spring?(4) A railway carriage weighing 20KN & moving at 2.5 km/hr is to be brought to rest by buffersprings. Find how many springs each of 15 coils will be required to store the energy ofmotion during a compression of 150mm. The available size of wire is 20mm. The meanradius of coil is 100mm.(5)A bumper spring consists of 2 helical steel springs of square cross section brings to rest arailway carriage weighing 50KN & moving at 1.5 km/hr. In doing so the springs arecompressed by 200mm. The mean diameter of coil is 6 times the side of square section.Design the spring if the permissible shear stress is 345 MPa and Modulus of rigidity is 78MPa.6) A loaded narrow cart weighing 7.5 kN and moving with a velocity of 1m/second is broughtto rest by a bumper consisting of 2 helical steel springs of square section. The mean coildiameter of spring is 6 times the side of square . the spring are compressed by 150mm.While bringing the car to rest. The permissible shear stress in spring is not to exceed 400MPa. Find the followingMean load on each springSide of square section of wireMean dia of coilNo of active coils, take G 82 Gpa.7) A load of 2KN is dropped axially on a helical spring from a height of 250mm. the springhas 20 turns, & it is made of 25mm diameter wire. The spring index is 8. Find the max.Shear stress induced in the spring 7 the amount of compression produced. Take G 82.7GN/mm8) Design a helical spring for a spring loaded safety valve for the following conditions:(i) Diameter of the valve 65mm(ii)Operating pressure 0.7N/mm2(iii) Max. Pressure on the valve 0.75N/mm2(iv) Max. lift of the valve when pressure 3.5mm rises from 0.7 to 0.75 N/mm2(v) Max. Allowable stress 550MPa(vi) Spring index 69) The valve spring of an I.C Engine is 40mm long, when the valve is open & 48mm longwhen the valve is closed . The spring loads are 250N when the valve is closed & 400n whenthe valve is open. The inside diameter of the spring is not to be less than 25mm & take FOSJagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -II 2. Assume spring index to be 6 &G 79.34 103MPa & yield shear stress 690N/mm2.Design the springAssignment Questions(1) A helical spring made from 6.3mm diameter steel wire has an outside diameter of57.3mm with squared & ground ends and has 12 coils. The allowable shear stress is827MPa. Determine the following(i) Spring rate(ii) Free length(iii) Pitch(2) The following data refers to the valve of a petrol engineLength of the spring when the valve is open – 40mmLength of the spring when the valve is closed – 48mmSpring load when the valve is closed – 350NSpring load when the valve is open – 220NSpring index – 6.8(3) The maximum shear stress allowed is 150MPa & the modulus of rigidity is 84GPa. Theends are squared & ground and the gap between the adjacent coils is 0.1 times the wirediameter. Determine the following(i) Wire dia(ii)Mean dia(iii) Number of coils(iv) Free length(v) PitchDesign of springs for fluctuating loadsForce acting on the spring in many applications varies in magnitude in time. Let us considera spring subjected to fluctuating force as shown in the figure. The load varies from Fmax toFminFmaxForceFminTIMEThe mean force Jagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IIA helical compression spring is subjected to purely compressive forces. In general, thespring wires are subjected to pulsating shear stresses. Let Sen be the endurance limit inshear. For cold drawn steel wires Sen 0.21x σ en and τ y 0.42σ u For oil hardenedtempered steel wires Sen 0.22xσenand τ y 0.45σ u whereσen is the ultimate tensilestrength is shown below:The mean stress τ m is plotted on the abscissa while the stress amplitudePoint A will co-ordinate (Sen Sen,2 2τa on the ordinate.) indicate the failure point of spring wire in a pulsatingstress cycle fatigue test. Point B indicates failure under static conditions i.e. when the stressreaches the torsional yield strength τ y . The line AB is called the line of failure. To considerthe effect of FOS a line of CD is drawn from point B in such a way thatτy OD. The lineNCD is parallel to AB. The line CD is called design line. Any point on CD such as X representsa situation with the same FOS.Considering the similar triangles XFD and AEB.This is called MODIFIED SODERBERG’S EQUATIONProblem : Design a helical coil spring for engine valve to exert force of 700 N in the openand 450 N in the closed position. The lift of the valve is 16mm. Keep the Outside spring coilJagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IIdiameter between 4 cm and 5cm. The spring ends are square and ground. The allowableshear stress of spring material is 450 MPa. For the material of the spring τ y 700 MPa.Sen 365 MPa. Take G 80 GPa.Helical tension springsHelical tension springs are usually wound with the coils closed and under initial tension.They are also called closed wound springs. The different styles of ends for the helicalsprings are shown in figure.When loops or end hooks are provided the small radius where the hook joins the 1st coil is aregion of highly local stress. The end should be designed in such a way that , the effect ofstress concentration at the bend is minimum. Sometimes the effect of stress concentrationin springs is so severe that the spring body becomes stronger that the end then the failureoccurs in the end coils.For helical extension springs, all coils are active, The number of active coils is same as thetotal number of coils. Figure 2 shows the relation between the external force and springelongation.External forceFiElongationHerethe forceF mustexceed the initial tension Fi before aJagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IIdeflection y is experienced. The free length lo of an extension spring is equal to the bodylength two times the hood distancelo d( i 1) body lengthFigure shows safety device operated by a lever and a tension spring. During the normalclosed position of the valve the pressure is 0.4 MPa. The maximum lift valve is 4mm at apressure of 0.55 Mpa. Diameter of the valve is 50mm. take allowable shear stress as 560MPa for the material of the spring. Spring Index C 7 and G 80 GPa. Design the spring.Spring surge ( critical frequency of helical springs)If one end of a compression spring is held against in a flat surface and the other end isdisturbed by repeated load, a helical compression wave is created that travels back and forthfrom one end to other end. Surging may occur in spring which is subjected to loads whosefrequency is close to the natural frequency of the spring. When the frequency of repeatedload is equal to natural frequency of spring resonance occur. To avoid this possible it isadvisable that the natural frequency of spring is atleast 20 times the frequency of theexternal repeated loads. If the frequency is not high enough, the spring should beredesigned to increase the k or decrease w.Natural frequency of spring fn 1 Fo g2π wHzProblem A disc cam 250mm diameter rotates off centre with eccentricity of 20 mm andoperates the roller follower that is carried by the arm as shown in the figure. The follower isheld against the cam by means of an extension spring. Assuming that the force between thefollower and the cam is 300 N at the lowest position and 500 N at the highest position offollower. The spring index is 6. determine the wire diameter, outside diameter of spring andno. of active coils. The maximum shear stress may be taken as 250 MPa and Modulus ofrigidity – 82.7 Gpa.Jagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IIProblem A helical coil spring made from 6.3 diameter steel wire has an outside diameter of57.3mm with square and ground ends has 12 coils. The length of the spring is such whatwhen it is compressed the torsional stress is 827 MPa. Determine the spring rate , Freelength , critical frequency . The density of material is 7800 kg/m3 and G 0.8x105 MPa.Concentric springsConcentric springs are closed coil helical springsforced one inside the other. They are used toobtain greater load carrying capacity or to obtaincertain load deflection characteristics. Figureshows two springs placed one inside the other.When springs are nested , the mechanismcontnues even if one of the springs break. Theconditions to be satisfied by the concentric springas follows.1. when the springs are of equal free lengthand made of same material, thenmaximum shear stress in the springs areequal.2. The deflection is same for both the springJagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalF1 d 1 F2 d 2Design of Machine Elements -II2also F F1 F2 , where F is the applied load. F1 is the load taken byspring 1 and F2 is the load taken by spring 2.Problem : The larger of the two concentric springs made of 38 mm diameter round bar has amean coil dia of 228mm and 6 active coils. The inner spring has a wire of 25mm, springindex 5 and no of active coils 9. Free height of outer spring is 19mm more than inner; Findthe deflection of each spring for a load of 90 kN. Take G 77000 MPa. Calculate the loadcarried by each spring.Problem One helical spring is nested inside the other . The dimensions of the spring aretabulated. Both springs have the same free length and carry a total maximum load of 2500 NOuter springInner springDescription61012.5mm9mm100mm70mmNumber of active coilsWire diameterMean coil diameterDetermine the maximum load carried by each spring. Total deflection of each spring.Maximum stress in 2 springs , take G 83 GPa.Jagadeesha T, Associate Professor, St Joseph Engineering College, Vamanjoor, Mangalore

Sixth semester MechanicalDesign of Machine Elements -IIProblem A concentric spring for an aircraft engine valve is to exert a maximum for of 5000Nunder an axial deflection of 40mm. Both springs have same free length and same solidlength and are subjected to equal maximum shear stress of 850 MPa. If the spring index forboth the spring is 6 find the load shared by each spring ii) maximum dimension of springs iii)No of active coils in each. Assume G 80 and dimensional clearance is equal to differencebetween the wire diameter.Leaf springsThe term flat spring is commonly applied to a wide variety of shapes made out of flat strip.The advantages of flat spring varies over the helical spring is that the end of the spring maybe guided along a definite path as it deflects . Thus the spring may act as structural memberas well as an emery absorbing device.Semi Elliptical leaf springs.A multi leaf spring used in automobile is shown in the figure. It is of semi elliptical form andhence the name semi elliptical leaf spring . It is built up of a number of plates known asleaves, The leaves are usually given an initial curvature or cambered.The top leaf is known as the master leaf. The eye is provided for attaching the spring withanother machine member. The amount of bend that is given to the spring from the centralline, passing through the eyes, is known as camber. The camber is provided so that even atthe maximum load the deflected spring should not touch the machine member to which it isattached. The camber shown in the figure is known as positive camber. The central clamp isrequired to hold the leaves of the spring. However, the bolt holes required to engage thebolts to clamp the leaves weaken the spring to some extent. Rebound clips help to share theload from the master leaf to the graduated leaIf the leaf spring has a shape of uniformly varying width (say Lozenge shape) then thebending stress at all section remains uniform. The situation is also identical as before in caseof varying thickness, the thickness should vary non-uniformly with length to make a beam ofuniform strength (L/h2 constant). These leaves require lesser material, have moreresilience compared to a constant width leaf. These types of springs are called leaf springsof uniform strength.In general the differential curvature between the master leaf and the next leaves is providedin a laminated spring, where, radius of curvature being more for the master leaf. Thisconstr

] To study and design stresses in closed rings. Curved Beam A beam in which the neutral axis in the unloaded condition is curved instead of straight. Or if the beam is originally curved before applying the bending moment, are termed as “Curved Beams Curved beams find a