Stiffness and Stability of Chairs and Stools Stiffness of Chairs

Stiffness of construction of furniture is determined by both stiffness of joints and stiffness of components. Determining the dominant element or structural node, which decides about the stiffness of the product is not simple, because in furniture several parts at the same time are actively involved. We can, however, in the course of typical validation tests (Fig. 6.89), assess deformations, on the basis of which it is possible to formulate the criterion of stiffness of furniture correctly enough. In this case, coefficients k may be the measures of the stiffness of construction, calculated as follows:


Fl M

Ui Ui ’


F concentrated force corresponding to operational loads,

M bending moment,

SiF displacement of the point i in the direction of the force F, ф rotation of the node caused by the bending moment M.

When calculating stiffness of furniture, we assume that each cross section of each of the construction elements of the skeleton (beams) remains perpendicular to the deflected axis. So, it is tangential to the bending line and with the X-axis form the angle ф, equal to the angle of rotation of the cross section (without taking into account the effect of lateral forces on the bending). Assuming also constant stiffness of the beams (EJ) and structural nodes (EZJZ) between deflection of beams y(x), angle of rotation of cross sections ф(х), bending moment M(x), transverse force T(x) and load q(x), these are the following relations:

d, ,

(EJy) = EJ u, dx


d2 d

d? (EJy) = dxEJ u = M’


d3 d2 dM d? (EJy) = ^(EJ u)= ^ = T’


d4 , N d2 N d2M dT (EJy)= djj(EJ u) = – sr = d; = – q■


Displacements of bent beams can be specified with the use of several methods (Dyl^g et al. 1986; Nowacki 1976; Zielnica 1996): integration of differential equation of the bending line, virtual loads, energy method. Below, the energy method is presented, which assumes that elastic energy of linear elastic system under load of the forces F1, F2, F3, …, Fn, equals:

1 n

U = – J2 FidiF, (6.238)

2 i=1


SiF displacement of the point and applying the force F in the direction of this force (Clapeyron equation).

Elastic energy of the bent beam expressed as a function of internal forces amounts to:



df =——– (a F 3/),

F 3EJy

or from the Castigliano’s theorem:

a /

= CMcb Г МЛА dx;

F J 2EJ dF J 2EJ dF ’

0 0


Determining rotation ф requires loading the cross section C with the moment Mc. Bending moments and their derivatives in relation to Mc in individual cross sections amount to:

mcb _ EJ U a + 2l

Updated: October 6, 2015 — 10:03 am