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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		.
C(s)				4		0				0		e(R)R			5	J0(sr)dr					dz			,		R r					z		.
				4								e(R)R
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
			(1					) 0			r		r			e(R)R			5		J0(sr)dr
																e(R)R
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							.
(r)		2		r		e(R)R						5				2			r		e(R)R				2	r		r		e(R)R				4	.
		2				e(R)R										2					e(R)R									e(R)R

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
C(s) (1		) 0	0	e(R)R	5	J0	(sr)dr dR
				e(R)R

	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

J0(sRt)dt

0 (sRt)

(sRt)dsRt

((sR)

J1(sR) 2(sR)

(sR)).

(sR)4

(11)

(12)

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		E rn 1			n		E (1 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

n 2

En Rn

e(R)

E(z)z

En z

n 3

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

1 2

(n 3)dR

n (r)

E Rn 2 r

E Rn 4

e(R)R2

e(R)R4

(1 2)(n 3)

(

)

(1 2)

2 E

rn 1n 1

rn 1

E rn 1

(n 1)

(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)

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)

((n 3)/ 2 1)

(1 2 )sn 1

2 n 1 ((1 n)/ 2)

((n

1)/ 2)(n 1)

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			,
																								e(R)																		n				3		Rn			,
																																	3									n				3
and using the formula (10) we get the expression for the quasi-nuclear
																										dR																				dR
						0n (r)												r								dR						r2							r							dR
						0n (r)												r					e(R)R								2	r2							r					e(R)R							4
																							e(R)R																					e(R)R																															(18)
									3																		dR																												dR																				(18)
									3																		dR																		2										dR
														r																										r						r																							,
										E				r				(1 (R /Q)n )R2																						r						r				(1 (R /Q)n )R4																			,
											0
where Q n			(n 3)E0				, а							1 v2									. At n 1 after the integration we get
where Q n				3E			, а									2							. At n 1 after the integration we get
					n
															3															r2										r /Q											2						r									1
							0	(r)							3					1										r2			ln							r /Q											2						r									1
						1
						1																									3																												2
															E								Q Q															1 r /Q 3r Q
															E			0					Q Q															1 r /Q 3r Q																												2Q
																		0
									3					2								3								1				r			2									r /Q										r										1
									3					2								3								1				r				3			ln					r /Q										r		2								1
																																						3			ln																	2
									E			3r E																		Q Q														1 r /Q Q																					2Q
									E			3r E																		Q Q														1 r /Q Q																					2Q
									0															0
																					3								1				r		2										r /Q									r											1
																					3								1				r												r /Q									r											1
							(r)																															ln																													,
							(r)																													3		ln																		2											,
											0										E							Q Q								3							1 r /Q Q													2							2Q
																					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
														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									lnr /Q																							(r)						4E										lnr			/Q					.
				1			E r										4E					2																	2								0								4E							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			3		2				r					1								r												Q																		3								r						1			r				Q
	2	(r)																									arctg																				(r)																										arctg	.
		(r)									2														3		arctg																				(r)																	2								3	arctg	.
				E		3r Q					2				Q							Q			3											r									0							E								Q				2			Q Q					3
				E		3r Q									Q							Q														r																E								Q							Q Q							r
				0																																																		0

Issue № 1 (41), 2019

ISSN 2542-0526

At r 0dropping the members converging to zero we get

0(r)

(r)

(1 2)

27E

2 .

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

n (r)

(R /Q)n )R2

(1 (R /Q)n )R4

r /Q

(1 tn )t2

(1 tn )t4

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

t n

(Q / r)n

t n 1

Therefore

3 1

r/Q

r /Q

(r)

(t n

1)tn 2

tn 4

r/Q

tn 2

(t n 1)tn 4

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

n 1

n 3

0n (r)

Q ((Q / r)

1)(n

1)r

n 1

(n 3)r

n 3

n 1

n 3

Q (n 1)r

n 1

((Q / r)

1)(n 3)r

n 3

3 Qn

(Q / r)n

n (r)

(n 1)(n 3)r

n 1

((Q / r)

1)(n 1)r

n 1

3 Qn

(Q / r)n

(n 1)(n

3)r

n 1

((Q / r)

1)(n

3)r

n 1

Ultimately considering that Qn (n 3)E0

, we conclude that at larger r

0 (r)

3 Qn

(r)

1)(n 3)rn 1

(n 1)rn 1

n 1

Now let be some small fixed number and r Q .

Russian Journal of Building Construction and Architecture

Then

(r)

r/Q

( Q),

(19)

(1 tn )t2

(1 tn )t4

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

(r)

r/Q

/Q dt2

r/Q dt4

r /Q

(1 t

(1 )t

Similarly, we get the estimate below

r/Q dt2

r/Q

dt4

r/Q

n (r).

t Q

(1 )t

After the integration we conclude that

(r)

3( Q)

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

(r)

(r) 0

( Q)

(r) o

I.e. at anyn

(r)

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

(r).

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)
0	(r)	(E r (n 1)E rn 1)		,	(20)
n		(E r (n 1)E rn 1)
		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).

Источник: https://studfile.net/preview/16566118/

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