Stiffness of Upholstery Springs

8.4.1 Stiffness of Cylindrical Springs

The comfort of the use of upholstered furniture is connected, to a large extent, with the softness of the spring layer. Its quality can be adjusted by choosing or designing the appropriate springs and spring units.

First, let us imagine a cylindrical spring stretched with two forces P acting in its axis (Fig. 8.22.). By cutting this spring in a plane perpendicular to the axis of the wire, which the spring is made of, a balance of cross sections will be conducted (Fig. 8.23). The transverse force P and the resulting torsion moment Ms = P • R must be differentiated by tangential stresses in this cross section, therefore

where

t1 tangential stresses resulting from transverse forces, t2 tangential stresses resulting from torsion moment,

P load stretching the spring,

Jo moment of inertia of the cross section at axial torsion, p fibres outward in relation to the torsion axis,

R radius of the spring coil and r radius of the wire cross section.

By adding both stresses t1 and t2 (Fig. 8.24), we obtain the value of the max­imum stress in the cross section of the spring wire in the form:

Fig. 8.22 Cylindrical spring subjected to stretching

However, since the share of the component P/(nr2) does not exceed 5 % of the Tmax, therefore the stresses in the wire section are expressed in a simple form:

2PR

Smax = 3 ■ (8.117)

nr3

Knowing the state of stresses in the spring coils, we can proceed to determine its deformations caused by external reasons. To this end, we cut from the spring a certain section of elementary length ds, using two planes perpendicular to the axis of the wire and passing through the axis of the cylinder (Fig. 8.25). Let us also assume that the R radiuses running from the axis of the cylinder to the centres of gravity of both cross sections before deformation lie in one plane. After defor­mation, these cross sections will turn in relation to one another by the angle d$> equal to:

and hence, the value of elementary deformation dl will amount to:

where

G shear modulus of the wire.

Taking into account the fact that all the elements with the length ds will deform identically, total deformation of the cylindrical spring l will be the sum of the elementary deformations dl:

(8.121)

therefore

which gives, for Jo = nr4/2,

4PR3 n

k —GiT

where

n number of the spring coils and 2nRn length of the spring wire.

A measurable indicator of the quality of the spring is its stiffness k, understood as the quotient of the load P to the displacement caused by this load:

Updated: October 14, 2015 — 3:33 am