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Russian Journal of Building Construction and Architecture

Changing the integration order, we get the second formula:

 

 

 

(1

2)

 

 

 

 

 

r

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

2

 

2

 

 

 

C(s)

 

 

 

0

 

 

0

 

 

 

 

J0(sr)dr

dz

 

,

 

R r

 

 

z

 

.

 

 

 

4

 

 

e(R)R

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In the formula (7) in the internal integral r const . Thus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dz4 2z2dz2

2((z2

r2 ) r2 )d (z2

r2 ).

 

 

 

 

 

 

 

 

 

Replacing the variablez2 r2

R2,

 

0 z ,

r R by (7) we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

2

 

2

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C(s)

(1

 

)

0

r r

(R

 

r

 

)dR5

J0(sr)dr

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

e(R)R

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1

2

 

 

 

r

 

(R3 r2R)dR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

) 0

 

r

 

 

e(R)R

5

 

J0(sr)dr

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Similar transformations applied for the formula (5) result in the equation

 

 

 

 

 

 

 

 

1

2

(R3 r2R)dR

 

1 2

 

 

dR

 

 

2

 

 

 

dR

 

 

(r)

 

 

 

r

 

 

 

 

 

 

 

5

 

 

 

 

 

 

r

 

 

 

 

 

r

 

r

 

 

 

 

 

 

.

 

2

 

e(R)R

 

 

2

e(R)R

2

 

 

e(R)R

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let us change the order of integration in the formula (9) and repalce r tR 0 t 1:

(8)

(9)

(10)

C(s) (1

2

 

R rR3 r3R

J0

 

 

) 0

 

0

e(R)R

5

(sr)dr dR

 

 

 

 

 

 

 

 

 

1

 

4

 

3

 

4

 

 

 

1

(1 2) 0

 

0

tR

t

R5

(sRt)Rdt dR (1 2) 0

1

0 (t t3)J0(sRt

 

 

 

J0

 

 

 

 

e(R)

 

 

 

e(R)R

 

 

 

 

Let us use the formula (6.2) as suggested by V. G. Korenev [7, p. 23]:

zdzd (zk Jk (z)) zk 1Jk 1(z), according to which we get the known equation

0u zk 1Jk (z)dz uk 1Jk 1(u) Then assuming thatsR a , we get

01 tJ0(sRt)dt 01 tJ0(at)dt a12 01 atJ0(at)dat a12 zJ1(z) 0a J1a(a)

Integrating individual parts and using the previous formula we have

)dt dR

J1(sR) sR

 

 

 

 

0u t3J0(t)dt t2 0t

sJ0(s)ds u0 0u 2t 0t sJ0(s)dsdt

 

 

 

 

 

 

u2 0u tJ0 (t)dt 2 0u t2J1(t)dt u3J1(u) 2u2J2(u)

 

 

 

Hence

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

3

 

1

1

3

 

1

 

3

 

2

 

 

0

t

 

J0(sRt)dt

 

 

0 (sRt)

 

J0

(sRt)dsRt

 

((sR)

 

J1(sR) 2(sR)

 

J2

(sR)).

 

 

(sR)4

 

(sR)4

 

 

(11)

(12)

10

Issue № 1 (41), 2019

ISSN 2542-0526

Deducting this formula from (12) and considering (11) we get the expression we need

 

2(1 2)

J (sR)

 

 

C(s)

s2

0

2

dR.

(13)

e(R)R2

2.2. Representation of nuclei and quasi-transforms using a function of heterogeneity of a foundation for a power law of changeE(z). The main results regarding the nuclei determined using the formulas (2), (3) are related to the power claw of change of the elasticity modulus with the depthE(z) En zn n 0 (the Poisson coefficient const ). Hence in [6, 7] the following expressions for the nuclei and transforms were obtained:

 

(r)

1

2

,

c

(s)

sn 121 n (1 2) (1 n )

,

(14)

 

 

 

2

E rn 1

E (1 n )

n

 

 

n

 

 

 

 

 

 

n

 

 

 

n

2

 

 

where (z) is the Euler gamma-function.

Let us look at how this can be applied to the method based on the function of heterogeneity of the foundation. According to the formula (6) we have

 

 

 

 

 

 

 

 

 

1

R

 

 

 

 

 

 

 

2

 

1

 

R

 

n 2

 

En Rn

 

 

 

 

 

 

 

e(R)

 

 

 

0

E(z)z

dz

 

 

0

En z

 

dz

 

 

 

 

 

 

 

 

 

R3

 

R3

 

n 3

 

 

Hence using the formula (10) equivalent to the formula (5), we get

 

 

 

 

 

 

 

1 2

 

 

dR

 

2

 

 

 

dR

 

 

 

 

 

 

1 2

 

(n 3)dR

2

 

(n 3)dR

 

n (r)

2

r

 

r

 

r

 

 

 

 

 

 

2

 

 

r

 

 

E Rn 2 r

 

r

E Rn 4

 

 

e(R)R2

 

e(R)R4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

n

 

 

 

 

(1 2)(n 3)

(

 

1

 

 

 

 

 

 

 

1

 

 

)

 

 

 

(1 2)

 

 

 

 

 

 

 

 

2 E

 

 

rn 1n 1

 

rn 1

(n

3)

 

E rn 1

(n 1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

(15)

(16)

The comparison of the formulas (14) and (16) shows that in case when n 0 (homogeneous half-space) both methods yield the same result and at n 0 settling of a daytime surface of a foundation under a single load calculated using the formulas (5), (6) turns out to be n 1 times smaller. This indicates that in the foundation model designed using the function of heterogeneity of the foundation the distribution capacity of the soil turns out to be smaller than that of the model designed based on the laws of the theory of elasticity. Let us show that the quasi-transform calculated using the formula (13) is also n 1 times smaller. It is true that the integral emerging in (13) is a table integral for the Hankel transformation [11]:

 

 

2(1 2)

J2 (sR)

 

2(1 2)

(n 3)J2

(sR)

 

 

 

 

Cn (s)

s2

0

 

dR

 

 

s2

0

 

 

 

dR

 

 

 

e(R)R2

 

 

 

En Rn 2

 

 

 

 

2(1 2 )(n 3)

2 n 2 (1/ 2 (n 2)/ 2 1)

 

(1 2)(n 3)sn 1

2 n 1 ((1 n)/ 2)

 

(17)

s2En

s n 1 (1/ 2 (n 2)/ 2 1)

 

En

 

 

((n 3)/ 2 1)

 

 

 

 

 

 

 

 

 

 

(1 2 )sn 1

2 n 1 ((1 n)/ 2)

.

 

 

 

 

 

 

 

 

En

((n

1)/ 2)(n 1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

11

Russian Journal of Building Construction and Architecture

The last equation is obtained using the known property (z 1) z (z) two times [8].

2.3. Combined law of changes in the elastiticity modulus with evaluation of the behavior of a quasi-nuclear (settling) in zero and infinity. As at n 0 the elasticity modulus En zn turns to zero on a daytime surface of the half-space, which does not correspond with the actual properties of soil, we will look at a feasible case

E(z) E0 En zn ,

where En EH H E0 andEH is the elasticity modulus at the depthH . Тhen

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e(R)

 

 

E0

 

 

 

En

 

 

Rn

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and using the formula (10) we get the expression for the quasi-nuclear

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0n (r)

 

 

r

 

 

 

 

 

 

 

r2

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e(R)R

2

 

e(R)R

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(18)

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dR

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r

 

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

 

 

 

 

(1 (R /Q)n )R2

 

 

 

(1 (R /Q)n )R4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where Q n

(n 3)E0

, а

1 v2

 

. At n 1 after the integration we get

 

 

 

 

 

 

 

 

3E

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

r2

 

 

 

 

 

 

 

 

r /Q

 

 

 

 

2

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

(r)

 

1

 

 

 

 

 

ln

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

 

 

 

Q Q

 

 

 

 

 

 

 

 

 

 

1 r /Q 3r Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

2

 

 

 

 

 

 

 

 

 

3

 

 

 

 

1

 

 

 

 

 

 

 

r

2

 

 

 

 

 

 

 

 

r /Q

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

ln

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

 

 

3r E

 

 

 

 

 

Q Q

 

 

 

 

 

1 r /Q Q

 

 

 

 

 

 

 

 

2Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

1

 

 

 

 

 

 

r

2

 

 

 

 

 

 

 

 

 

r /Q

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(r)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

 

 

 

 

Q Q

 

 

 

 

 

1 r /Q Q

 

 

 

 

 

2Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where as previously

(r)

 

1

2

 

 

is the nuclear of a homogeneous space with the elasticity

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

E0r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

modulusE0 . At r 0 dropping the members converging to zero we get

 

 

 

 

 

 

 

 

 

 

 

 

0(r)

1 2

 

 

 

 

 

 

 

 

9E

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

9E

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

lnr /Q

 

 

 

 

 

 

 

(r)

 

 

 

 

 

 

 

 

 

1

 

 

 

lnr

/Q

 

 

 

.

 

 

 

 

 

E r

 

 

4E

 

 

 

 

 

 

 

 

 

4E

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hence at small r the quasi-nuclear is 0

(r) (r), but it has the same growth order as

(r).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

At n 2 after the integration we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

2

 

 

 

 

r

 

 

 

 

 

 

1

 

 

 

 

 

 

r

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

r

 

 

 

 

 

 

1

 

 

r

2

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

 

2

(r)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

arctg

 

 

 

 

 

 

 

 

(r)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

arctg

.

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

3

 

 

 

 

E

 

3r Q

 

 

 

 

 

Q

 

 

 

Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

E

 

 

 

 

Q

 

 

 

Q Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12

Issue № 1 (41), 2019

ISSN 2542-0526

At r 0dropping the members converging to zero we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0(r)

 

 

(r)

 

(1 2)

 

27E

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

0

 

 

 

 

 

 

4E0

 

 

 

 

5E0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let us now examine the behavior of 0n (r) at r and r 0 in a general case. Using the

formula (18) and replacing with t R / Q we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

3

 

 

 

 

 

 

 

dR

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

dR

 

 

 

 

 

 

 

 

 

 

n (r)

 

 

 

 

r

 

 

 

 

 

 

 

 

 

 

 

 

r

 

r

 

 

 

 

 

 

 

 

 

 

 

 

 

E

(1

(R /Q)n )R2

 

 

 

 

(1 (R /Q)n )R4

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

1

 

 

 

 

 

 

 

dt

 

 

 

 

 

r2

 

 

 

 

 

 

 

 

 

dt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r /Q

 

 

 

 

 

 

 

 

 

r /Q

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

Q

(1 tn )t2

Q3

(1 tn )t4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let r Q , then t r / Q 1 and thus

 

 

 

 

t n

 

 

 

 

 

 

 

 

 

 

t n

 

 

 

 

 

1

 

 

t

n

.

 

 

 

 

 

(Q / r)n

 

1

 

t n 1

 

1

tn

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Therefore

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3 1

 

 

dt

 

 

 

 

 

 

r2

 

dt

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r/Q

 

 

 

 

r /Q

 

 

 

n

(r)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

Q

(t n

1)tn 2

Q3

tn 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

1

 

 

 

 

dt

 

 

 

r2

 

 

dt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r/Q

 

 

 

 

r/Q

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

Q

tn 2

Q3

(t n 1)tn 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Replacing t n

with (r /Q) n , we will strengthen the inequality. Thus after the integration in

the left and right parts of the inequality we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

1

 

 

 

 

Q

n 1

 

 

 

 

 

 

r

2

 

Q

n 3

 

 

 

 

 

0n (r)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

Q ((Q / r)

n

1)(n

1)r

n 1

 

Q

3

 

(n 3)r

n 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

1

 

 

 

 

Q

n 1

 

 

 

2

 

 

 

 

 

Q

n 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q (n 1)r

n 1

3

((Q / r)

n

1)(n 3)r

n 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

 

 

 

Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

or

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3 Qn

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

(Q / r)n

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n (r)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

(n 1)(n 3)r

n 1

((Q / r)

n

1)(n 1)r

n 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3 Qn

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

(Q / r)n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

 

 

(n 1)(n

3)r

n 1

 

((Q / r)

n

1)(n

 

3)r

n 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ultimately considering that Qn (n 3)E0

, we conclude that at larger r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3E

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0 (r)

 

 

3 Qn

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

(1

2)

 

 

 

 

(r)

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

(n

1)(n 3)rn 1

 

E

(n 1)rn 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

n 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Now let be some small fixed number and r Q .

13

Russian Journal of Building Construction and Architecture

Then

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

3

1

 

 

 

 

dt

 

 

 

 

 

 

r2

 

 

 

 

 

 

dt

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

(r)

 

 

 

 

 

 

 

 

 

r/Q

 

 

 

 

 

 

 

 

 

 

 

r/Q

 

 

 

 

n

( Q),

 

 

 

 

 

 

 

(19)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

Q

(1 tn )t2

 

Q3

 

(1 tn )t4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and for examining the behavior of 0n (r)

at r 0

 

all we have to do is to evaluate (r) that

is behind the bracket. We have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dt

 

 

 

2

 

 

 

 

 

 

dt

 

 

 

 

 

 

 

 

 

 

 

 

r

2

 

 

 

 

 

 

 

 

2

 

 

 

 

 

n

dt

 

 

 

 

(r)

1

 

 

r/Q

 

 

 

 

 

 

r

 

r/Q

 

 

 

 

 

 

 

 

 

1

 

 

r

/Q dt2

 

r/Q dt4

 

r

 

r /Q

 

 

 

 

.

 

 

 

 

 

n

)t

2

3

 

(1

n

)t

4

 

 

 

 

3

3

 

 

 

n

4

 

 

 

Q

 

 

(1 t

 

 

 

 

Q

 

 

 

 

 

t

 

 

 

 

Q

 

 

 

t

 

 

 

 

Q

 

 

 

 

t

 

 

Q

 

 

 

(1 )t

 

 

 

Similarly, we get the estimate below

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

r

2

 

 

 

 

 

1

 

 

 

 

 

 

 

 

n

dt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r/Q dt2

 

 

 

 

r/Q

dt4

 

 

r/Q

 

 

 

 

 

 

 

 

n (r).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

3

 

 

 

 

 

 

 

n

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

 

 

 

t

 

 

 

 

 

 

t Q

 

 

 

 

 

 

 

 

(1 )t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

After the integration we conclude that

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

1

 

 

 

 

r

2

 

 

 

 

 

 

n

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

2

 

 

 

 

1

 

 

 

 

 

 

r

2

 

 

 

 

 

 

n

 

 

1

 

 

 

r

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

(r)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

3( Q)

3

 

 

 

 

n

 

 

Q

3r

 

 

 

3( Q)

3

 

 

 

 

n

3r

 

3( Q)

3

 

 

3r

Q

 

 

1

 

 

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

 

 

 

1

 

 

 

 

 

 

 

As r 0and

is a randomly small fixed number,

 

we conclude that (r) is mostly made

up of 32r .

Therefore going back to the formula (19) we get the estimate

0

(r)

3

(r) 0

( Q)

3

2

o

1

 

0

( Q)

(r) o

1

.

 

 

 

 

 

 

 

 

 

 

n

 

 

 

n

 

 

3r

 

n

0

 

 

 

 

 

 

 

E0

 

E0

 

r

 

 

 

 

r

 

 

I.e. at anyn

0

(r)

of almost zero is mostly made up of the nuclear of the elastic half-

(r).

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

As seen from the above results, even at n 1,2 the expressions for the quasi-nuclear 0n (r)obtained based on the formula (18) are complex and not informative (as they contain growing and mutually destructive summands). A more simple but quite precise formula for the quasi-nuclear is certainly necessary.

This is the following formula:

0

(r)

1 2

 

,

(20)

(E r (n 1)E rn 1)

n

 

 

 

 

 

0

n

 

 

which obviously keeps 0n (r) at the same level of decrease for infinity and growth at zero. Tests show that at other arguments the formula proves more accurate for calculating 0n (r)as shown in the following figures obtained for n 1,2 at Q = 1 (Fig. 1, 2).

14

Issue № 1 (41), 2019

ISSN 2542-0526

Residual Error for Equation Formula

0,8

0,6

0,4

0,2

Residual

0,0

-0,2

-0,4

-0,6

 

 

 

 

 

 

 

 

 

 

 

0,0

10,0

20,0

30,0

40,0

50,0

60,0

70,0

80,0

90,0

100,0

110,0

Fig. 1. Absolute error of the formula (20) at n = 1 and Q = 1

Residual

Residual Error for Equation Formula2

0,4

0,3

0,2

0,1

0,0

-0,1

-0,2

-0,3

 

 

 

 

 

 

 

 

 

 

 

0,0

10,0

20,0

30,0

40,0

50,0

60,0

70,0

80,0

90,0

100,0

110,0

Fig. 2. Absolute error of the formula (20) at n = 2 and Q = 1

A somewhat lower accuracy at almost zero is due to the fact that there is not an asymptotic approximate equality 0n (r) 0(r)but an equality of the growth rate (equivalency of two infinitely large values).

15

Russian Journal of Building Construction and Architecture

Conclusions. The paper looks at a very important problem that has to do with mathematical modeling of the interaction between a base and a foundation which is also closely related to practical tasks facing foundation design and strength calculations of buildings and structures. The model of a soil foundation suggested by Aleynikov-Snitko has been considered in detail. By modifying the Hankel transformation the author was able to calculate quasi-transforms that correspond with the quasi-nuclear of the foundation. New formulas for the function of heterogeneity and a quasi-nuclear of an isotropic, heterogeneous linearly deformed foundations were obtained.

For the combined power law of changes in the elasticity modulus of an n-order foundation an asymptotic behavior of the quasi-nuclear at zero and infinity was studied.

A simple approximate formula describing a (quasi) nuclear in case of a combined power law was obtained.

The use of the method by Aleynikov-Snitko was found to show a large distribution capacity of soil compared to that by Winker and a smaller one than in that of the elastic half-space. The obtained formulas will be instrumental in calculating the strength of construction structures and solutions of spatial tasks of contact interaction of a base and a foundation.

References

1.Aleinikov S. M. Metod granichnykh elementov v kontaktnykh zadachakh dlya uprugikh prostranstvenno neodnorodnykh osnovanii [Boundary element method in contact problems for elastic spatially inhomogeneous bases]. Moscow, ASV Publ., 2000. 754 p.

2.Aleinikov S. M., Sedaev A. A. [Modeling of deformation of the surface of an elastic inhomogeneous base].

Trudy 6-i Vseros. nauch. konf. s mezhdunar. uchastiem “Matematicheskoe modelirovanie i kraevye zadachi”. Ch. 1 [Proc. of the 6th all-Russian conference with international participation “Mathematical modeling and boundary value problems“. Vol. 1]. Samara, SamGU Publ., 2009, pp. 13—15.

3.Aleinikov S. M. Prostranstvennaya kontaktnaya zadacha dlya zhestkogo fundamenta na uprugom neodnorodnom osnovanii [Spatial contact problem for a rigid Foundation on an elastic inhomogeneous base].

Izvestiya vuzov. Stroitel'stvo, 1997, no. 4, pp. 52—59.

4.Gradshtein I. S., Ryzhik I. M. Tablitsy integralov, summ, ryadov i proizvedenii [Tables of integrals, sums, series and products]. Moscow, Gos. izd-vo fiz.-mat. lit., 1963. 1100 p.

5.Duraev A. E. Raschet konstruktsii na gruntovom osnovanii s vozrastayushchim po glubine modulem deformatsii [Calculation of structures on the ground base with increasing depth modulus of deformation]. Saransk, Izd-vo Mordovskogo un-ta, 1991. 192 p.

6.Klein G. K. [Taking into account the heterogeneity, discontinuity of deformations and other mechanical properties of the soil in the calculation of structures on a continuous basis]. Trudy MISI [MISI's writings], 1956, no. 14, pp. 168—180.

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Issue № 1 (41), 2019

ISSN 2542-0526

7.Korenev B. G. Vvedenie v teoriyu besselevykh funktsii [Introduction to the theory of Bessel functions]. Moscow, Nauka Publ., 1971. 287 p.

8.Korn G., Korn T. Spravochnik po matematike dlya nauchnykh rabotnikov i inzhenerov [Handbook of mathematics for researchers and engineers]. Moscow, Nauka, gl. red. fiz-mat. lit. Publ., 1973. 832 p.

9.Lebedev N. N. Spetsial'nye funktsii i ikh prilozheniya [Special features and their applications]. Moscow –– Leningrad, Fizmatgiz Publ., 1963. 358 p.

10.Novatskii V. Teoriya uprugosti [Theory of elasticity]. Moscow, Mir Publ., 1975. 872 p.

11.Perel'muter A. V., Slivker V. I. Raschetnye modeli sooruzhenii i vozmozhnost' ikh analiza [Design models of structures and the possibility of their analysis]. Kiev, Stal' Publ., 2002. 600 p.

12.Popov G. Ya. Kontaktnye zadachi dlya lineino deformiruemogo osnovaniya [Contact problems for a linearly deformable base]. Kiev –– Odessa, Vyshcha shkola Publ., 1982. 168 p.

13.Posadov M. I., Malikova T. A., Solomin V. I. Raschet konstruktsii na uprugom osnovanii [Calculation of structures on elastic base]. Moscow, Stroiizdat Publ., 1984. 679 p.

14.Sneddon I. Preobrazovaniya Fur'e [Fourier transformation]. Moscow, Inostran. lit. Publ., 1955. 667 p.

15.Snitko N. K. O deystvii sosredotochennoi sily na neodnorodnoe uprugoe poluprostranstvo [On the action of a concentrated force on an inhomogeneous elastic half-space]. Stroitel'naya mekhanika i raschet sooruzhenii, 1980, no. 2, pp. 76—78.

17

Russian Journal of Building Construction and Architecture

HEAT AND GAS SUPPLY,VENTILATION,AIR CONDITIONING,

GAS SUPPLY AND ILLUMINATION

DOI 10.25987/VSTU.2018.41.1.002

UDC 622.691.24

S. I. Astashev1, O. N. Medvedeva2, S. V. Chuikin3, K. A. Sklyarov4

THE OPTIMIZATION OF OPERATING MODES

OF TECHNOLOGICAL EQUIPMENT FOR UNDERGROUND GAS STORAGE

Yuri Gagarin State Technical University of Saratov

Russia, Saratov

1PhD student of the Dept. of Heat and Gas Supply, Ventilation, Water Supply and Applied Fluid Dynamics, e-mail: tester91@mail.ru

2D. Sc. in Engineering, Prof. of the Dept. of Heat and Gas Supply, Ventilation, Water Supply and Applied Fluid Dynamics, e-mail: medvedeva-on@mail.ru

Voronezh State Technical University

Russia, Voronezh

3PhD in Engineering, Assoc. Prof. of the Dept. of Heat and Gas Supply and Oil and Gas Business, tel.: (473)271-53-21, e-mail: ser.shu@vgasu.vrn.ru

4PhD in Engineering, Assoc. Prof., Dean of the Faculty of Construction and Technology, tel.: (473)271-59-26, e-mail: stf@vgasu.vrn.ru

Statement of the problem. The paper considers the process of regime adjustment of the gas preparation section of the Yelshansk underground gas storage. Instead of the obsolete gas preparation section, a new gas treatment complex was constructed and new equipment was commissioned. However, commissioning identified a number of unresolved problems requiring optimization.

Results. The issues arising in the operational mode were eliminated by means of introducing a new section of complex gas treatment. In order to improve the quality of the gas prepared for transportation, analytical and technical-economic comparison of different modes of operation of the fire regenerator was carried out.

Conclusions. Based on the results of the research, an optimal method of adjusting the technological mode of operation of the gas preparation section was set forth. It allows material and financial resources to be saved.

Keywords: underground gas storage, equipment, gas preparation, absorber, fire regenerator, fuel gas.

Introduction. The objective of designing underground gas storages is to increase the reliability of gas distribution system by controlling seasonal consumption of gas fuel by different categories of consumers [10, 13, 22, 24, 29].

© Аstashev S. I., Medvedeva О. N., Chuykin S. V., Sklyarov К. А., 2019

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Issue № 1 (41), 2019

ISSN 2542-0526

There have been a lot of research attempts here and abroad at studying existing technologies of gas transportation preparation and optimization of storages in order to improve the quality and efficiency of fuel [2, 3, 11, 15, 23, 25, 27, 28, 30, etc.]. As research results suggest, disruptions of technological operation have a negative impact on gas-pumping aggregates, separation devices as they cause a 10 to 15 % reduction in their efficiency factor [5, 9, 17, 20].

Search for new ways of optimizing industrial safety maintenance costs as well as lower rates of faulty decision-making regarding the operation of all underground gas storages is crucial for economic efficiency of energy complexes [1, 4, 6—8, 14, 17—19, 21, 26, 31 и др.].

The Yelshano-Kyurdyumskoe underground gas storage is designed to control seasonal gas consumption in the depleted gas and oil field of the same name [12]. Since 1966 depleted fields of the Tula region and Bobrikovsk-Kyzelevsk complex have been operated as independent gas storages. The Yelshano-Kyurdyumskoe gas and oil field is located in the Leninskiy District of Saratov, Saratov and Tatischevskiy districts of Saratov region. Orographically the horizon lies in the right bank of the water reservoir of the Volga River. The surface of the area of the storage has a permanent bend from the South West to the North East. The relief of the area is quite scattered with significant differences at the absolute points (from 50 to 125 m).

One of the features of the operation of the storage is an extremely even distribution of the wells throughout the structure. The main causes are as follows: the roof of the field is occupied by gardens, in the South there are residential areas, in the West there are no collectors in the Bobrikovsk (main) region and the collectors of the Kyzelevsk region have low filtration and capacity characteristics and thus provide no high well flow rates. Therefore the operating wells are located in the eastern, south-eastern and north-western parts of the structure.

1. A brief history of the opening and operation of the Yelshano-Kurduymskiy field. The Yelshano-Kurdumskiy structure was discovered in 1940 as part of a geological survey. In 1941 seismic work was performed and welling got underway resulting in an industrial gas presence of Vereian horizons (an open fountain was opened). Following the 1941—1948 geological surveys gas deposits of the Vereian, Melekessk, Cheremshano-Prikamsk complex, gas and oil fields of the Tula horizon, the Bobrikovsk-Cherepetsk-Kyzelevsk complex. The gas presence of deposits of the Bobrikovsk, Kyzelovsky and Cherepetskiy horizons was identified in 1943 with the well № 12. As the well was being prepared to be examined, some washingout liquid was resulted which resulted in an open fountain. The mine was surveyed and its contouring continued accompanied by its development into a field. Oil fringe was found in 1945, in 1946 oil started being mined from the oil fringe with gas extracted from a gas cap.

19

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