Stiffness of Bottoms of Drawers and Containers

In the most common construction solutions, the bottom of a drawer is inserted into a groove on the entire perimeter without additional fastenings or inserted into grooves on three sides and fixed to the back of the drawer with staples. The first method of mounting the bottom corresponds to the theoretical scheme of the board supported on the entire perimeter in an articulated manner. The second is the scheme of a board of three sides supported elastically and one in a discontinuous manner. The deflection of such a board (Fig. 7.19), made from isotropic material (e. g. a thin particle board or fibreboard), can be expressed in the form of a double trigonometric series:

mnx nny

w(x, y)= Amn sin – sin b, (7.41)



a and b dimensions of the board.

By applying energy methods known in the theory of elasticity and using the equation for energy of bending an isotropic board,

1 f f ((d2w d2w2 ( @2w d2w d2w2

U = 2D дХ2 ~дУ2 — ( — v) дХ2 ~дХ2 — дХдУ y’ ( : )

0 0

We obtain the equation for the total energy of a bent board in the form

By providing the coefficient Amn with a small increment 3Amn, we obtain

= @Amn = P@w(x, y):


By entering the unit load 1 for P, we obtain the equation of the deflection of the board at a point of the coordinates x0 = f and y0 = n:

If a board (bottom of a drawer) is subjected to concentrated load (Fig. 7.19a) P(f, n) at the point of coordinates (f, n), then the deflection of this board (in accordance with the principle of superposition) will amount to

w(x, y) = PK(x, y, n, g): (7:46)

However, when an evenly distributed load works on the bottom of a drawer (Fig. 7.19b), then the deflection of that bottom at the point of the coordinates x and y can be presented in the following function:

where in all equations m and n are odd numbers.

A solution to this problem on the basis of elemental strength of materials requires determining the deflection of the board caused by cylindrical bending along every axis of symmetry of the board (Fig. 7.20). Presenting unit load in the
direction of the x axis as q(x) and the load in the direction of the y axis as q(y), the total load on both axes of symmetry of the board can be written in the form:

4a = 4(x) + 4(y), (7-48)

while the deformation at the intersection point of these axes amounts to

w = w(x) = w(y);


By comparing the deflections of the boards wx = wy, we obtain the equation

and transform it further


4(x) = 4a a4 + b4

The deflection of the bottom of the drawer of a depth h at the point of the coordinates (x = aJ2, y = b/2) shows the equation:

5 qVa4bh b4 W(a’i) _ 384 EJx a4 + b4 ’


qv volume load of the drawer,

a dimension of the bottom of the drawer in the direction of the x axis, b dimension of the depth of the drawer,

E linear deformations module and Jx moment of inertia of the cross section.

Updated: October 8, 2015 — 6:22 pm