6.4 Compression of Orthogonal Cylinders
59
c
a
0
a 2 − r 2 · 2πrdr = P ,
(6.16)
where P is the compressing force.
The condition (6.16) and the first of formulas (6.15) give
a =
1
2
3
6πρ 1 ρ 2 (k 1 + k 2 )P
ρ 1 + ρ 2
.
(6.17)
By substituting a and c into formula (6.14), we find contact stresses
p[r] =
3P
2πa 2
1 −
r 2
a 2 , (r a).
(6.18)
The approach of the adjoining bodies is obtained from formulas (6.15) and
(6.17):
δ =
1
2
3
9π 2 (ρ 1 + ρ 2 )(k 1 + k 2 )P 2
2ρ 1 ρ 2
.
(6.19)
6.4 Compression of Orthogonal Cylinders
6.4.1 Simplest Case
Two cylinders whose radii are (R 1 = R 2 = R) and whose axes are mutually
perpendicular are subject to mutual compression by the force P .
Let us designate the contact point of the cylinders as O and assume it to be the
origin of the rectangular coordinate system Oxyz. Let us direct the axis Oz inwards
the cylinder (1) along the common normal line to the adjoining bodies and axes
Ox and Oy along the formed first and second cylinders going through the point O
(Fig. 6.3).
In the selected coordinate system, the equations of cylinder surfaces can be
represented as follows
F 1 (x, y) =
1
2R
y
2 , F 2 (x, y) = −
1
2R
x
2 .
(6.20)
In the case under consideration, Eq. (6.8) looks as follows
(k 1 + k 2 )(x, y) = −
1
2R
(x
2
+ y
2 ) + δ.
(6.21)
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