Skip to main content

Closed Coiled helical springs

Springs are energy absorbing units whose function is to store energy and to restore it slowly or rapidly depending on the particular application.
A spring may be defined as an elastic member whose primary function is to deflect or distort under the action of applied load; it recovers its original shape when load is released.
Helical spring: They are made of wire coiled into a helical form, the load being applied along the axis of the helix. In these type of springs the major stresses is torsional shear stress due to twisting. They are both used in tension and compression.

Derivation of the Formula :
In order to derive a formula which governs the deflection and stress of closed coil helical springs, consider a closed coiled spring subjected to an axial load W. with the fallowing Assumptions:
1. The Bending and shear Force effects may be neglect
2. For the purpose of derivation of formula, the helix angle is considered to be so small that it may be neglected.



Let
W = axial load , D = mean coil diameter,
d = diameter of spring wire, n = number of active coils ,
l = length of spring wire, G = modulus of rigidity, X = deflection of spring
o = Angle of twist

when the spring is being subjected to an axial load to the wire of the spring gets be twisted like a shaft.
If o is the total angle of twist along the wire and x is the deflection of spring under the action of load W along the axis of the coil, so that
x = (D / 2) .O
[ consider ,one half turn of a close coiled helical spring ]

Any one coil of a such a spring will be assumed to lie in a plane which is nearly perpendicular to the axis of the spring. This requires that adjoining coils be close together. With this limitation, a section taken perpendicular to the axis the spring rod becomes nearly vertical. Hence to maintain equilibrium of a segment of the spring, only a shearing force F = W and Torque T =W.D/2 are required at any X – section.
In the analysis of springs it is customary to assume that the shearing stresses caused by the direct shear force is uniformly distributed and is negligible.
Using the torsion formula i.e
O = 2x / D
Twisting moment in spring coil,  T = WxD/2  

The SPRING DEFLECTION ' x' will be,

Shear stress in the spring:
Stiffness of Spring :
The stiffness is defined as the load per unit deflection therefore
Springs in Series:
If two springs of different stiffness are joined end on and carry a common load W, they are said to be connected in series and the combined stiffness and deflection are given by the following equation.
Springs in parallel:
If the two spring are joined in such a way that they have a common deflection ‘x' ; then they are said to be connected in parallel.In this care the load carried is shared between the two springs and total load W = W1 + W2

Comments

Post a Comment

Popular posts from this blog

Relation between Modulus of Elasticity and Modulus of Rigidity

Modulus of Elasticity (E)   It is the ratio between Normal stress to Normal strain within the elastic limit. Elastic Modulus E = Normal stress/Normal strain E =  s/e Modulus of Rigidity (G)  It is the ratio between Shear stress to Shear strain within the elastic limit. Rigidity Modulus G = Shear stress/ Shear strain G =    Ƭ / ø   Relation between Modulus of Elasticity and Modulus of Rigidity: Consider a solid cube  PQRS is  subjected to a shearing force F.  Let  Ƭ     be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure as Ƭ   Due to the pure shearing force, the cube is deformed PQRS to PQR'S' . The point   S moved to S' and point R moved to R' as shown in fig.  The shear strain = The angle of distortion  ø                          ø = RR'/ RQ ---(1) Shear strain = Shear stress /Rigidity modulus                

PORTAL METHOD and CANTILEVER METHOD

The behavior of a structure subjected to horizontal forces depends on its height to width ratio. The deformation in low-rise structures, where the height is smaller than its width, is characterized predominantly by shear deformations. In high rise building, where height is several times greater than its lateral dimensions, is dominated by bending action. To analyze the structures subjected to horizontal loading we have two methods. Portal method  and Cantilever method 1. PORTAL METHOD The portal method is an approximate analysis used for analysing building frames subjected to lateral loads such as Wind loads/ seismic forces.  Since shear deformations are dominant in low rise structures, the method makes simplifying assumptions regarding horizontal shear in columns.  Each bay of a structure is treated as a portal frame, and horizontal force is distributed equally among them. Assumptions in portal method   1. The points of inflection are located at the mid-height of each column above th

Shearing Stresses Distribution in Circular Section

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is  τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis.  Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us consider on