PSI - Issue 40
Kirill E. Kazakov et al. / Procedia Structural Integrity 40 (2022) 201–206 Kirill E. Kaz kov / Structural Int grity Procedia 00 (2022) 00 – 000
205
1
( ( ), ( )) f z f z
( ) ( ) . f z f z dz
1
2
1
2
1
Without going into details of the conclusions, we give the final formula to calculate the contact stresses
* *
1
( ) ( )
c t g z
0 k
* * *
( ) ( ) ( * * * z f t
) * 1 in
*
*
I V
( , )
,
[ 1,1],
1,
(4)
q z t
z
t
k
k
* *
* *
( )
m z
where
( ) * z
( ) k
( ), *
p z m
k
m
0
m
,
)( I V I V ( * out
) * 1 * * in c t
( , )( * *
)( I V I V * out
) * 1 * * in
*
( )
( ) c d
g R t k k
t
* *
( )
f t k
g
k
* *
* *
(5)
( )
( )
c t
c
1
k
k
( ) ( ) ( ) * * * * *
( ) ( ) ( , ) ( ) ( ) * * * * * * * c m z m p z p k z l i * *
m p g l
1
1
1
( ) R il k i
1
, d R *
* *
,
g
dz d
k
il
1
1
0
0
i
l
, , i k l
0,1,2, ,
p k ( z *) are polynomials, which can be found, for example, from the formulas
0 J J J J
J
0 J J J J
J
2 1
1
i
j
1
J
J
( ) *
i
1
1
2
1
i
1
1
j
*
( ) *
,
,
,
p z
d
J
d
i
j
i
* *
( )
m
1 d d j
J
1
j
*
( ) * z
j
1
J J
z
1
2
i
i
i
kernels R k ( t *, *) are the resolvents of the kernels
( ) ( , ) * * * * * in * * c t K t
* * * out
( , )
K t
k
,
( )
c t
k
( ) k i can be found from solution of the following problem
coefficients k and
0 ( ) k l l il R
. ( ) k
k i
Using formulas (2), (4), (5), we can obtain an expression for the contact pressures in dimensional form
~ ( ) f t k
( ) ( ) ( ) h z c t r h z g z ( ) in in 0
a
a z
0 k
( , )
,
[ , ],
,
(6)
q z t
z a a
t
k
0
( )
in h z
in
where
,
( E t E t ( out
) )
( E t out
)
( E t out
)
~ ( ) f t k
t
) * 1 in in
*
I V
out
out
out
(
.
( )
f
0 c t
k
2 in
2 in
1
1
in
0
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