Ppt springs

ppt on springs
(Mechanical)

1. SPRINGS A spring is defined as a elastic machine element which deflects under the action of load and returns to its orignal shape when the load is removed. It can take any shape and form depending on its application. IMPORTANT FUNCTIONS: 1)Springs are used to absorb shocks and vibrations. 2)Springs are used to store energy. 3)Springs are used to measure forces. 4)Springs are used to apply force and control motion.

2. TYPES OF SPRINGS Springs are classified according to their shape. The most popular type of spring is helical spring. There are two types of helical spring. Compression spring Extension spring

3. Sub classification of Helical spring The helical springs are sometimes classified as closely-coiled helical spring and open-coiled helical spring The difference between them are as follows: A helical spring is sais to be closely coiled spring when the spring wire is coiled so close that the plane containing each coil is almost at right angles to the axis of the helix. The helix angle is very small ,it is usually less than 10 degree. A helical spring is said to be open coiled spring when the spring wire is coiled in such a way that there are large gap between adjacent coils. In other words the helix angle is large . In other words it is more than 10 degree.

4. ADVANTAGES OF COMPRESSION & EXTENSION ( i )They are easy to manufacture . (ii)They are cheaper than other types of springs. (iii)Their reliability is high. (iv)The deflection of the spring is linearly propotional to the force acting on the springs. It is due to the above advantage that helical spring are popular and extensively used in a number of application. Helical torsional spring The construction of the spring is similar to that of compression or extension spring , the spring is loaded by a torque about a axis of the coil. They are used to transmit torque to a particular component in a machine or the mechanism.

5. The helical torsion spring resists the bending moment(P x r), Which tends to wind up the spring. The bending moment induces bending stresses in spring wire. The term “ torsional spring” is somewhat misleading because the wire is subjected to bending stresses, unlike torsional shear stresses induced in helical compression or extension spring. It should be noted that although the spring is subjected to torsional movement, the wire of the helical torsion spring is not subjected to torsional shear stress. It is subjected to bending stresses. Terminology of helical spring: The main dimension of helical spring subjected to compressive force d=wire diameter of spring (mm) Di= inside diameter of the spring coil(mm) Do= outside diameter of the spring coil(mm) D= mean diameter coil(mm)

6. Therefore D= Di+Do 2 There is an important parameter in spring design called spring index. It is denoted by the letter C. The spring index is defined as the ratio of a mean coil to the wire diameter . Or , C= D d A spring index from 4 to 12 is considered best from manufacturing consideration close tolerance spring and those subjected to cyclic loading. There are three terms - Free length ,compressed length,& solid length Solid length - defined as the axial length of the spring which is co compressed that the adjacent coil touch each other. Compressed length - defined as the axial length of the spring which is subjected to maximum compressive force. Free length – defined as the axial length of an unloaded helical compression spring. No external forces act on the spring.

7. Solid length= N t d Where N t = no. of coils Compression length Total gap=(Nt-1) x gap between adjacent coils Free length =compressed length+ δ =solid length+total axial gap + δ P= free length (N t -1) The stiffness of the spring (k) defined as the force required to produce deflection . Therefore k= P δ k=stiffness of the spring (N/mm) P=axial spring force (N) δ =axial deflection of the spring corresponding to the force P(mm) The stiffness of the spring represents the slope of the load deflection line.

8. There are two terms related to the spring coils- active coils 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. A portion of the end coil which is in contact with the seat , does not contribute to spring action and are called inactive coil.The coil do not support the load and do not deflect under the action of an external force. The no. of inactive coils is given by, inactive coils= N t -N Where N is the no. of active coils STYLES OF END Plain end : N t Plain end (ground) : (N t -1/2) Square end : (N t -2) Square end (ground) : (N t -2)

9. Stress and Deflection equation The dimension of equivalent bar are as follows: 1)The diameter of the bar is equal to the wire dia of the spring(d) . 2)The length of the coil of the spring is ( πD ). There is N such alternative coils. Therefore the length of the equivalent baris ( πDN ). 3)The bar is fitted with a bracket at each end. The length of the bracket is equal to the mean coil radius of the spring(D/2). The force P acting at the end of the bracket induces torsional shear stress in the bar . The torsional moment ,is given by M t =PD/2 The torsional shear stress of the bar is given by τ1 = 16Mt/πd^3 = 16(PD/2) πd^3 Or τ1=8PD/πd^3

10. When the equivalent bar is bent in the form of helical coil, there is an additional in the form of the two factor: 1)There is a direct or transverse shear stress in the spring wave 2)When the bar is bent in the form of coil the length of the inside fibre is less than the length of the outside. The result in stress concentration at the inside fibre of the coil. The resultant consist of superimposition of torsional shear stress direct shear stress and additional stresses due to the curvature of the coil.

11. Stresses in spring wire: Pure torsional stress Direct shear stress Combined torsional , direct and curvature shear stresses We will assume the following two factors to account for these effects Ks= factors to account for direct shear stress Kc =factors to account for stress concentration due to curvature effect The combined effect of those two factor is given by K= KsKc Where k is the factor to account for the combined effect of two factors

12. The shear stress correction factor (Ks)is defined as, K=(1+0.5d/D) or K=(1+0.5/C) τ2=p/(π/4 d²)=4P/πd²=8PD/πd^ 3 (0.5d/D) Superimposing τ= τ1+τ2 =8PD/πd^3+8PD/πd^ 3 ( 0.5d/D )=8PD/πd^ 3 ( 1+0.5d/D ) Substituting the above equation in the expression τ=Ks(8PD/πd^ 3 ) AM Wahl derived the equation for resultant stress, which includes torsional stress, direct shear stress concentration due to curvature. The equation is given by , τ=Ks(8PD/πd^ 3 ), where K is stress factor or wahl factor. The wahl factor is given by, K= 4C-1 + 0.615 4C-4 C where Cis the spring index

13. The wahl factor provides a simple method to find out resultant stresses in spring, the resultant shear stress is maximum at the inside radius of the coil. When the spring is subjected to fluctuating stresses, two factor Ks & Kc are separately used. The angle of twist(θ) for the equivalent bar, θ= M t l /JG Where θ= angle of twist (radian) M t = torsional moment (PD/2) l = length of the bar ( πDN ) J=polar moment of inertia of the bar (πd^4/32) G= modulas of elasticity Substituting values θ= (PD/2)( πDN ) or θ = 16PD²N (πd^4/32)G Gd^4

14. The axial deflection