Strength of Screw Joints
As an example of the strength of screw joints, let us consider the widely used method of fixing the rear wall to the side walls, bottoms and top (Fig. 7.63). On the number and spacing of connectors bx, ay, depends not only the stability of the rear wall, but also on stiffness and strength of the furniture body. In the construction loaded by horizontal force Q on the length of the edges of the rear wall, there are streams of contact intensities qyx = Q/b causing shearing of connectors. Depending on the number of connectors distributed on the length of edge a and b, the forces attributed to one connector can be calculated as
force acting on the connector located in horizontal rows, force acting on the connector located in vertical rows, appropriate streams of contact intensities, height of the rear wall,
b width of the rear wall,
n number of connectors on the edge of the length a and
m number of connectors on the edge of the length b.
In order for the constructor to provide the number and type of screws, it is necessary to specify the acceptable stresses on the screw and the greatest normal stresses caused by its bending. The static scheme of the connector with loading by shearing forces has been shown in Fig. 7.64.
In order to improve the calculation formulas, Korolew (1973) suggested the following simplifying assumptions:
• the intensity of the load in the plane of bending the screw changes according to the curve of the third degree qx = qox3. The beginning of the system of coordinates have been placed at a point of balance, which location is also the subject of calculation (Fig. 7.65),
• on the transverse cross section of the screw, the contact stresses are cosine-distributed and disappear only on side surfaces, in contact with the surface of the board (Fig. 7.66) where
– p < U < p, (7.196)
|
, . p M> 2 ■ |
(7.198) |
|
p(u) = o. |
(7.199) |
The dependence of the maximum values of contact stresses in the transverse cross section of screw x from the intensity of the attributed load is calculated as
p/2
У Po(x)Dx cos2 udu = q(x),
from which we can obtain
For the adopted system of coordinates, the diameter of the screw in the chosen cross section x amounts to
Dx = D(p +(1 – р)(П + По)), (7.202)
where
D largest screw diameter,
P = d/D reference coefficient of screw diameters,
f = x/l reference coefficient of the location of the transverse cross section of the
screw,
fo = xjl reference coefficient of the location of the beginning of the system of coordinates and
d smallest screw diameter
In order to determine unknown values qo and fo, the conditions of balance of the screw are used:
l xo
qo f x3dx = P
-xo
l-xo
qo = J x4dx = P(l — xo).
xo
After integrating equations and elementary transformations in order to determine qo and fo, we obtain a nonlinear system of algebraic equations:
‘ qo (1 — 4fo + 6f2 — 0= 4P,
qo( 1 — 5fo + 10f2 — 10fo + 5П4) = 5P (1 — no).
By calculating qo, from the above equation, the location of the origin of the system of coordinates is determined in the form:
According to Korolew’s (1973) calculations, = 0.38 determines the origin of the system of coordinates measured from the left end of the screw. Therefore, by solving the system of equations,
і qo ^ _ f + 6£8 = ^, (7.206)
we obtain that
The greatest intensity of load distribution on the length of the screw is formed in its largest cross section at f =1 – fo and can be expressed by the equation:
And the biggest value of stress force can be determined by
Strength calculations for the screw usually come down to determine the cross section, in which most normal stresses disappear. The bending moment forming in the cross section of the screw specified by the parameter f is determined from the following equations:
-По < П< 0
M(n) = 6.3PI n5 + nO)
0 < n < 1 – По
м(П) = 6.3PI [(1 – По)5-П5
Hence, the normal stresses arising in the transverse cross sections of the screw during its shearing are as follows:
for – По < П < 0
for 0 < n < 1 – no
(1 – По)5-П5
D3 (q + (1 – q)(П + По))l3′
The main cross section of the screw is the cross section located at the distance f = 0, so for this cross section, the largest normal stresses are as follows:
pi (1 – no)5
64.2 ( —^ 3.
D3 (p +(1 — р)П0)3
Eventually, the strength conditions of fixing a rear wall to the body using screws can be written in the form:
58pl kr
) D3(0.38+0.62p)3 – g f72l4i
9.6P kw ; ( ■ )
Dl — c
where
kg bending strength of the screw material and k’W compression strength of the board
