INTRODUCTION Springs are elastic members which distort under load and regain its original shape when load is removed. They are used in railway carriages, motor cars, scooters, motorcycles, rickshaws, governors etc. According to their uses, the springs perform the following function: To absorb shock or impact loading as in carriage springs. To store energy as in clock springs. To apply forces to and to control motions as in brakes and clutches. To measure forces as in spring balances. To change the variations characteristic of a member as in flexible mounting of motors.

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The springs are usually made of either high carbon steel (0.7 to 1.0%) or medium carbon alloy steels. Phosphor bronze, brass, 18/8 stainless steel and monel and other metal alloys are used for corrosion resistance springs.

HELICAL SPRINGS A helical springs is a length of wire or bar wound into helix. There are mainly two types of helical springs : (1) close-coiled, and (2) open-coiled

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THE CLOSE-COILED HELICAL SPRINGS Close-coiled helical spring with ‘Axial load’ Circular section wire springs: In fig 1 is shown a close-coiled helical spring loaded with an axial load W. Let R=Radius of the coil, d=Diameter of coil under the Load W, C= Modulus of rigidity, N=Number of coils or turns, θ = Angle of twist l=length of wire=2 π Rn

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τ =shear stress, and I = Polar moment of inertia = π d ⁴ 32 It may be noted that each section of the coil is under torsion but there are small bending and shearing stress which being small are usually neglected. Shear stress τ : From torsion equation, T = C θ = τ ; T = τ I l r I r Or, T= τ I = τ×π d ⁴ × 2 = τ . π d ³ r 32 d 16 P P P P

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τ = 16T π d ³ τ = 16WR (T=WR) …………(1) π d ³ Deflection, δ : Again, T = C θ I l θ = Tl = WR × 2 π Rn × 32 = 64WR ² n ……(2) CI C ×π d ⁴ Cd ⁴ δ = R ×θ …………….(3) δ = 64WR ³ n …………….(4) Cd ⁴

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Wahl’s correction factor: While deriving eqns.(1) and (2) the effect of curvature of spring and direct shear is neglected, Eqn. (1) is modified to include these effects by introducing a factor K called Wahl’s correction factor, τ = 16 WR K …………..(5) π d ³ Where, K is found from experiments and is given by K = 4S-1 + 0.615 …………..(6) 4S-4 S Where, S = D = spring index (where, D=mean d diameter of the coil)

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The deflection equating is not modified as the effect, if any, is considered to have been incorporated in the value of n by finding the effect on deflection due to end coils experimentally and the modified value of n is then called effective number of coils, Stiffness of the spring, K: K= W = W = Cd ⁴ δ 64WR ³ n 64R ³ n Cd ⁴ K= Cd ⁴ ………………(7) 64R ³ n

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Energy stored, U: Energy stored, U = 1 ×T× θ = 1 W.R × 64WR ² n 2 2 CD ⁴ = 1 . 1 16WR . 8WR ² n = 1 . 16WR . 16WR [2 π R n d ²× π ] 2 2 C d ³ d 4c π d ³ π d ³ 4 = 1 . τ²× volume of wire 4C i.e. U = τ² × volume of wire ……………..(8) 4C Again, energy stored, U= 1 . T . Θ = 1 . W.R. δ = 1 . W. δ ( δ =R θ ) 2 2 2 2 i.e. U= 1 W δ …………….(9) 2

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OPEN-COIL HELICAL SPRINGS With axial load Employing the same symbols as used in previous slides , the slope of coil α is introduced additionally. The wire length is now l=2 π Rsec α ×n ,where R is the radius of the coil. The couple applied to the material under the applied load W will be WR, and at each point along the centre line of this couple may be resolved into two components, one of torsion and one of bending . The couple producing torsion, T= Wrcos α The couple producing bending, M= Wrsin α .

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Refer to fig 1. The couple WR will act in a plane passing through the axes OY and OX, the centre line of the wire being at angle α to OX. Fig 1

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The bending moment will tend to wind the coils of the spring more tightly, and to a smaller radius of curvature. The axial extension of the spring may be most easily calculated by equating the work done by the load to the internal strain energy of the material. The strain energy due to bending= M²L 2EI The strain energy for twisting = T²L 2CI Where the deflection is δ , ½W. δ = T²L + M²L = W²R²cos² c.L + W²R²sin² α .L 2CI 2EI 2CI 2EI = W²R²L cos² α + sin² α 2 2CI 2EI P P P P

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= W²R²2 π Rnsec π cos² α + sin ² α 2 2CI 2EI Deflection, δ =2WR³n π sec α cos² α + sin ² α 2CI 2EI If α is as taken zero, δ = 2WR³n π CI = 64WR³n as before C d⁴ P P P

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Ψ = The resultant of the rotations θ₁ and α ₁ . Ψ = θ₁ sin α - ɸ ₁ cos α = TL sin α - ML cos α CI EI = WRcos α ×2 π Rsec α× n sin α - WRsin α × 2 π Rsec α× n cos α CI EI = 2WR²n π sin α 1 - 1 CI EI P P P

LEAF SPRING:

LEAF SPRING Sometimes it is also called as a semi-elliptical spring, as it takes the form of a slender arc shaped length of spring steel of rectangular cross section. The center of the arc provides the location for the axle,while the tie holes are provided at either end for attaching to the vehicle body. Heavy vehicles,leaves are stacked one upon the other to ensure rigidity and strenth . It provides dampness and springing function.

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It can be attached directly to the frame at the both ends or attached directly to one end,usually at the front,with the other end attched through a shackle,a short swinging arm. The shackle takes up the tendency of the leaf spring to elongate when it gets compressed and by which the spring becomes softer. Thus depending upon the load bearing capacity of the vehicle the leaf spring is designed with graduated and Ungraduated leaves. FABRICATION STAGES OF A LEAF SPRING

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