CN104408022A - Shallow tunnel surrounding rock stress and displacement explicit analytical solution solving method - Google Patents

Shallow tunnel surrounding rock stress and displacement explicit analytical solution solving method Download PDF

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CN104408022A
CN104408022A CN201410765074.4A CN201410765074A CN104408022A CN 104408022 A CN104408022 A CN 104408022A CN 201410765074 A CN201410765074 A CN 201410765074A CN 104408022 A CN104408022 A CN 104408022A
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mrow
msup
msub
msubsup
mfrac
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CN104408022B (en
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韩凯航
张成平
张顶立
王剑晨
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Beijing Jiaotong University
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Beijing Jiaotong University
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Abstract

The invention discloses a shallow tunnel surrounding rock stress and displacement explicit analytical solution solving method. The method is based on complex variables functions, an inverse mapping function is utilized to solve the series form of a z-plane analytical function, and the Cauchy-Riemann equation (C-R condition) is combined to conduct analytical function derivation on a shallow tunnel surrounding rock stress and displacement implicit analytical solution to obtain a series expression of the shallow tunnel stress and displacement function. Compared with the implicit analytical solution, the explicit analytical solution is visual and convenient to use by engineers. The programming calculation quantity is small, the actually measured stratum displacement of the engineering and the calculation result of the explicit analytical solution are high in consistency, and correctness and practicability of the explicit analytical solution are verified.

Description

Method for solving explicit analytic solution of shallow tunnel surrounding rock stress and displacement
Technical Field
The invention relates to urban underground tunnel engineering, in particular to a method for solving an explicit analytic solution of shallow tunnel surrounding rock stress and displacement.
Background
With the continuous development of urban traffic, the construction of urban tunnels represented by urban subways, underground roads and the like is greatly increased, the buried depth of the tunnels is shallow, the tunnels are subjected to the comprehensive action of complex loads such as buildings and the like on the ground and in the stratum, the stratum displacement influence factors in the construction process are many, and the accurate prediction of the stratum displacement is difficult. With the continuous enhancement of environmental awareness of people, stratum displacement caused by urban tunnel construction and the influence control problem on the surrounding environment are fully paid attention by related scholars, and accurate prediction of the stratum displacement before tunnel construction is a premise for effectively controlling the influence degree of the construction on the environment. In the prior art, a complex function method is used for analyzing the stress and displacement of surrounding rock around a shallow tunnel, the difficulty in calculation of a bipolar coordinate method is avoided, the defects of singularity assumption in a mirror image method are also avoided, and an accurate analytic solution of a displacement field is directly given while a stress field is given. However, the solution involves repeated mapping of two complex planes when solving, and the solution of stress and displacement cannot be directly represented by a function of coordinates (x, y) in the R domain before conformal mapping, and has the following inconveniences: firstly, the solution needs a certain professional theoretical knowledge, and if the solution is not a professional for researching a complex function, the solution cannot be directly used by engineering personnel like a Peck empirical formula, and for an explicit solution, the engineering personnel can directly input formation parameters to obtain stress and displacement, so that the engineering personnel do not need to spend much time on researching a complex mapping principle; the Verruijt solving process relates to the repeated mapping of two complex planes, the calculation efficiency is low, the solving process is complex, the programming calculation amount is large, the explicit analytic solution can directly program the stress and the displacement into a function of coordinates (x, y) in an R domain before conformal mapping, and the calculation amount is small; and thirdly, the Verruijt solution is not intuitive, and the superposition applicability is poor when the surrounding rock stress and displacement caused by tunnel construction under the condition of bearing stratum are solved.
Therefore, it is desirable to provide a method for solving an explicit analytic solution of the stress and displacement of the surrounding rock of the shallow tunnel, so as to overcome the above problems in the prior art.
Disclosure of Invention
The technical problem to be solved by the invention is to provide a method for solving an explicit analytic solution of the stress and displacement of the shallow tunnel surrounding rock, so as to overcome the problems in the prior art.
In order to solve the technical problems, the invention adopts the following technical scheme
The method for solving the explicit analytic solution of the stress and displacement of the shallow tunnel surrounding rock is characterized by comprising the following steps
S1, establishing an analytic function of shallow tunnel surrounding rock stress and displacement on a z plane without concrete expression forms based on a complex function methodAnd psi1(z) and a calculation formula of the stratum stress and displacement of the shallow tunnel:
wherein σxx,σyy,σxyIs formation stress, ux,uyIn order to displace the formation,and psi1(z) is an analytical function of two non-specific expression forms of a z plane, mu is a shear modulus, kappa is a poisson ratio correlation coefficient (kappa is 3-4 upsilon), upsilon is a poisson ratio, and i is an imaginary constant;
s2, utilizing conformal mapping functionWherein alpha is a parameter defined by the ratio of the radius r of the cavern to the buried depth h of the center of the cavern,analytical function of non-concrete expression form on z planeAnd psi1(z) conversion to an analytic function with a concrete representation on the zeta planeAnd psi (ζ), the relationship of the two sets of analytical functions beingAndthe specific form of the analytic functions phi (zeta) and psi (zeta) areAndwherein, a0,ak,bk,c0,ck,dkThe physical meaning of the method is a boundary condition coefficient of the tunnel cavern;
s3, analyzing zeta plane by inverse mapping functionAnd psi (ζ) into an analytical function having a specific form in the z-plane in a complex function methodAnd psi1(z) solving a series form of a z-plane analytical function;
s4, solving function with concrete expression form based on Cauchy-Riemann equationAnd psi1(z) performing derivation;
s5, an analytic function with a concrete formAnd psi1And (z) substituting the derivative function into the calculation formula of the shallow tunnel stratum stress and displacement to obtain a display analytical solution of the shallow tunnel stratum stress and displacement.
Preferably, the step S3 includes
To formulaInverse mapping is carried out to obtain an inverse mapping function <math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <msup> <mi>&omega;</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mi>z</mi> <mo>+</mo> <mi>ia</mi> </mrow> <mrow> <mi>z</mi> <mo>-</mo> <mi>ia</mi> </mrow> </mfrac> <mo>,</mo> </mrow> </math> Given z ═ x + iy, then this formula can be deformed as: <math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>-</mo> <msup> <mi>a</mi> <mn>2</mn> </msup> <mo>+</mo> <mn>2</mn> <mi>xai</mi> </mrow> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>y</mi> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>=</mo> <mi>U</mi> <mo>+</mo> <mi>Vi</mi> <mo>,</mo> </mrow> </math> wherein, U = x 2 + y 2 - a 2 x 2 + ( y - a ) 2 , V = 2 xa x 2 + ( y - a ) 2 ;
will be a formula <math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>-</mo> <msup> <mi>a</mi> <mn>2</mn> </msup> <mo>+</mo> <mn>2</mn> <mi>xai</mi> </mrow> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>y</mi> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>=</mo> <mi>U</mi> <mo>+</mo> <mi>Vi</mi> </mrow> </math> Triangular form converted to complex number zeta <math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mo>{</mo> <mi>cos</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>}</mo> <mo>,</mo> </mrow> </math> From the correlation law of complex power operations: <math> <mrow> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>=</mo> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mi>k</mi> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>;</mo> </mrow> </math>
will analyze the functionAnd <math> <mrow> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>c</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>d</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mo>[</mo> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>]</mo> </mrow> </math> converted into real and imaginary forms, let a0=a′0i,ak=a′ki,bk=b′ki,c0=c′0i,ck=c′ki,dk=d′ki. Then, there is,
<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>c</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>d</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> </mtd> </mtr> <mtr> <mtd> <mo>=</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mi>i</mi> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mi>i</mi> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mi>k</mi> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> </mtd> </mtr> <mtr> <mtd> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mi>i</mi> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> </mtd> </mtr> </mtable> </mfenced> <mo>;</mo> </mrow> </math>
after the process of simplification, the process of the method, <math> <mrow> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mo>*</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> <mo>,</mo> </mrow> </math> wherein,
W 1 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] k 2 cos [ k arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] ,
W 2 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] - k 2 cos [ ( - k ) arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] ,
W 3 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] k 2 sin [ k arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] ,
W 4 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] - k 2 sin [ ( - k ) arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] .
preferably, the step S4 includes
When f (z) ═ u (x, y) + iv (x, y) is differentiable at point z ═ x + iy, the pairAnd <math> <mrow> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mo>*</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> </mrow> </math> conducting derivation to obtain
<math> <mrow> <msubsup> <mi>&psi;</mi> <mn>1</mn> <mo>&prime;</mo> </msubsup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> <mo>,</mo> </mrow> </math>
Preferably, the step S5 includes
Will be provided with <math> <mrow> <msubsup> <mi>&psi;</mi> <mn>1</mn> <mo>&prime;</mo> </msubsup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> </mrow> </math> Andsubstitution formula To obtain
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>xx</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mn>2</mn> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mo>-</mo> <mn>2</mn> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>-</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <msup> <mi>x</mi> <mn>2</mn> </msup> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>yy</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mn>2</mn> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>-</mo> <mrow> <mo>(</mo> <mn>2</mn> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <msup> <mi>x</mi> <mn>2</mn> </msup> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>xy</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>u</mi> <mi>y</mi> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>&mu;</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>{</mo> <mi>&kappa;</mi> <msubsup> <mi>a</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>}</mo> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>u</mi> <mi>x</mi> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>&mu;</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>&kappa;</mi> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>&kappa;</mi> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>.</mo> </mrow> </math>
The invention has the following beneficial effects:
the technical scheme of the invention utilizes the inverse mapping function to analyze the Zeta planeAnd psi (ζ) into an analytical function with a specific representation of the z-planeAnd psi1(z) obtaining a derivative of an analytic function in a hidden analytic solution of the stress and displacement of the shallow tunnel based on a complex function method by combining a Cauchy-Riemann equation (C-R condition), and obtaining a series explicit expression form of the stress function and the displacement function; the explicit analytic solution is intuitive and convenient for engineering personnel to use, and has less programming calculation amount than the implicit analytic solution.
According to the technical scheme, based on an engineering example, the explicit analytical solution of the stress and displacement of the shallow tunnel surrounding rock is applied, the theoretical calculation result of the explicit analytical solution and the actually-measured stratum displacement of the engineering are compared and analyzed, and the correctness and the practicability of the explicit analytical solution are verified.
Drawings
The following describes embodiments of the present invention in further detail with reference to the accompanying drawings;
FIG. 1 is a schematic diagram illustrating a method for solving an explicit analytic solution of shallow tunnel surrounding rock stress and displacement;
FIG. 2 shows a schematic diagram of "black box theory";
FIG. 3-a is a graph showing a comparison of an analytic solution of san Francisco line No. 4 and an actually measured vertical displacement in an embodiment;
FIG. 3-b is a graph showing the analytical solution and the measured horizontal displacement comparison for line No. 4 of san Francisco in the example;
FIG. 4-a is a graph showing a comparison of an analytic solution of san Francisco line No. 2 and an actually measured vertical displacement in an embodiment;
fig. 4-b shows a comparison of the analytic solution of san francisco line No. 2 and the measured horizontal displacement in the example.
Detailed Description
The invention discloses a method for solving an explicit analytic solution of shallow tunnel surrounding rock stress and displacement, which is based on a complex function method, utilizes an inverse mapping function to solve a series form of a z-plane analytic function, and combines a Cauchy-Riemann equation (C-R condition) to derive an analytic function in an implicit analytic solution of shallow tunnel surrounding rock stress and displacement to obtain a series explicit analytic solution of a shallow tunnel stress field and a displacement function. And verifying the correctness and practicability of the explicit analytic solution by comparing and analyzing the engineering measured data.
Firstly, based on a complex function method, the problem of stress and displacement of the shallow tunnel is converted into a solution of a plane problem, and the solution can be expressed as an analytic function without concrete expression form analyzed in a position in a region R (a semi-plane y for removing holes is less than 0)And psi1(z), shallow tunnel stress solution and displacement solution can be determined according to the formula (1a), the formula (1b) and the formula (1 c).
Wherein σxx,σyy,σxyIs formation stress, ux,uyIn order to displace the formation,ψ1(z) is an analytical function of two non-specific expression forms of a z plane, mu is a shear modulus, kappa is a poisson ratio correlation coefficient (kappa is 3-4 upsilon), upsilon is a poisson ratio, and i is an imaginary constant;
due to the analytic function of the z planeAnd psi1Since (z) has no specific form, the solutions of equations (1a) to (1c) for calculating the stress solution and displacement of the shallow tunnel have no specific results, and it is necessary to map the solutions onto the ζ plane to obtain an analytic function with a specific expression formAnd psi (zeta), and converting the inverse mapping function back to the z-plane analysis function to obtain the analysis functionAnd psi1(z) specific expression forms.
The conformal mapping function for the z plane to the ζ plane is:
<math> <mrow> <mi>z</mi> <mo>=</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>-</mo> <mi>ih</mi> <mfrac> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mn>1</mn> <mo>+</mo> <mi>&zeta;</mi> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mn>1</mn> <mo>-</mo> <mi>&zeta;</mi> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mi>ia</mi> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <mi>&zeta;</mi> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <mi>&zeta;</mi> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> </math>
where α is a parameter defined by the ratio of the chamber radius r to the chamber center burial depth h (r/h), and can be expressed as:
<math> <mrow> <mfrac> <mi>r</mi> <mi>h</mi> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mi>&alpha;</mi> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> </math>
analytical function of non-concrete expression form on z planeψ1(z) conversion to an analytic function with a concrete representation on the zeta planeAnd psi (ζ), the relationship of the two sets of analytical functions beingAndanalytic functionAnd ψ (ζ) has a specific form, which is determined by the following two formulae:
<math> <mrow> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>c</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>d</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mo>.</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mi>b</mi> <mo>)</mo> </mrow> </mrow> </math>
wherein, a0,ak,bk,c0,ck,dkThe physical meaning of (1) is the coefficients of the boundary condition of the tunnel cavern, and the coefficients are determined by the following recursion formula:
<math> <mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mn>1</mn> </msub> <mo>-</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>+</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <msub> <mi>b</mi> <mn>1</mn> </msub> <mo>=</mo> <msub> <mi>A</mi> <mn>0</mn> </msub> <mo>-</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mi>a</mi> <mn>0</mn> </msub> </mrow> </math>
<math> <mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>&kappa;</mi> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mn>1</mn> </msub> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <msub> <mi>b</mi> <mn>1</mn> </msub> <mo>=</mo> <msub> <mover> <mi>A</mi> <mo>&OverBar;</mo> </mover> <mn>1</mn> </msub> <mi>&alpha;</mi> <mo>+</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mn>0</mn> </msub> </mrow> </math>
<math> <mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>+</mo> <mi>&kappa;</mi> <msup> <mi>&alpha;</mi> <mrow> <mo>-</mo> <mn>2</mn> <mi>k</mi> </mrow> </msup> <mo>)</mo> </mrow> <msub> <mi>b</mi> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mi>k</mi> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mi>k</mi> </msub> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>&kappa;</mi> <msup> <mi>&alpha;</mi> <mrow> <mo>-</mo> <mn>2</mn> <mi>k</mi> </mrow> </msup> <mo>)</mo> </mrow> <msub> <mi>b</mi> <mi>k</mi> </msub> <mo>+</mo> <msub> <mi>A</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msub> <msup> <mi>&alpha;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mo>,</mo> <mi>k</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mrow> </math>
<math> <mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>&kappa;</mi> <msup> <mi>&alpha;</mi> <mrow> <mn>2</mn> <mi>k</mi> <mo>+</mo> <mn>2</mn> </mrow> </msup> <mo>)</mo> </mrow> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mi>b</mi> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>=</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>&kappa;</mi> <msup> <mi>&alpha;</mi> <mrow> <mn>2</mn> <mi>k</mi> </mrow> </msup> <mo>)</mo> </mrow> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mi>k</mi> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mi>k</mi> <msub> <mi>b</mi> <mi>k</mi> </msub> <mo>+</mo> <msub> <mover> <mi>A</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <msup> <mi>&alpha;</mi> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> <mo>,</mo> <mi>k</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mrow> </math>
<math> <mrow> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>=</mo> <mo>-</mo> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mn>0</mn> </msub> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msub> <mi>a</mi> <mn>1</mn> </msub> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msub> <mi>b</mi> <mn>1</mn> </msub> </mrow> </math>
<math> <mrow> <msub> <mi>c</mi> <mi>k</mi> </msub> <mo>=</mo> <mo>-</mo> <msub> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mi>k</mi> </msub> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mi>a</mi> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow> <mo>(</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mi>a</mi> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> </mrow> </math>
<math> <mrow> <msub> <mi>d</mi> <mi>k</mi> </msub> <mo>=</mo> <mo>-</mo> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mi>k</mi> </msub> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mi>b</mi> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow> <mo>(</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mi>b</mi> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> </mrow> </math>
wherein:
u0the physical meaning of (A) is as follows: value of uniform radial shrinkage of the tunnel cavern boundary.
udThe physical meaning of (A) is as follows: values of ovalization deformation of tunnel cavern boundaries.
However, as shown in fig. 2, the coordinates of each point in the region R before mapping may be represented by (x, y), and the coordinates of each point in the region γ after mapping may be represented by (ζ, η). The solving process involves the mapping of two complex planes, i.e. if the stress or displacement value at one point (x, y) is required to be solved, the mapping function is firstly used for mapping into the region gamma to obtain the corresponding point (zeta, eta), and the value is solved and then mapped back to the corresponding point (x, y). That is, the solution of stress and displacement during the solution process cannot be directly represented by the function of coordinates (x, y) in the R domain before conformal mapping, and is inconvenient for the engineer to use.
Explicit analytic solution derivation of shallow tunnel surrounding rock stress and displacement
The invention takes the gamma domain after conformal mapping as a 'black box', as shown in fig. 2, tries to obtain an explicit expression function of stress and displacement field expressed by coordinates (x, y) in the R domain before conformal mapping, thereby providing a simple and practical form for engineering application personnel while keeping the prior art solution precision. To this end, zeta is leveled using an inverse mapping functionSurface analytical functionAnalytical function with embodied form for conversion of ψ (ζ) into z-planeψ1(z) carrying out derivation on an analytic function in hidden analytic solutions of stress and displacement of the shallow tunnel surrounding rock in the prior art by combining a Couchy-Riemann equation (C-R condition), and obtaining the analytic function with a specific formAnd psi1And (z) substituting the derivative function into the calculation formula of the shallow tunnel stratum stress and displacement to obtain a display analytical solution of the shallow tunnel stratum stress and displacement. The specific steps of explicit analytical solution derivation are as follows
(1) Solving for series form of z-plane analytical function
The corresponding inverse mapping function of equation (2) is:
<math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <msup> <mi>&omega;</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mi>z</mi> <mo>+</mo> <mi>ia</mi> </mrow> <mrow> <mi>z</mi> <mo>-</mo> <mi>ia</mi> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> </math>
if z is x + i y, equation (5) can be transformed as:
<math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>-</mo> <msup> <mi>a</mi> <mn>2</mn> </msup> <mo>+</mo> <mn>2</mn> <mi>xai</mi> </mrow> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>y</mi> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>=</mo> <mi>U</mi> <mo>+</mo> <mi>Vi</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein:
U = x 2 + y 2 - a 2 x 2 + ( y - a ) 2 , V = 2 xa x 2 + ( y - a ) 2
the triangular form of formula (6) to complex ζ is then converted as follows:
<math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mo>{</mo> <mi>cos</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>7</mn> <mo>)</mo> </mrow> </mrow> </math>
from the correlation law of complex power operations:
<math> <mrow> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>=</mo> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mi>k</mi> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mrow> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> </math>
according to the symmetry of the problem, all coefficients are pure imaginary numbers, and for the convenience (simplification) of derivation later, the analytic function is converted into the forms of a real part and an imaginary part, and the following can be set:
a0=a′0i,ak=a′ki,bk=b′ki,c0=c′0i,ck=c′ki,dk=d′ki。 (9)
the analytical functions (4a) and (4b) can be transformed into:
<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>c</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>d</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> </mtd> </mtr> <mtr> <mtd> <mo>=</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mi>i</mi> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mi>i</mi> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mi>k</mi> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mi>i</mi> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> </mtd> </mtr> <mtr> <mtd> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mi>b</mi> <mo>)</mo> </mrow> </mrow> </math>
simplifying the equations (10a) and (10b), and taking into account the equations (4a) and (4 b):
<math> <mrow> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mo>*</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mi>b</mi> <mo>)</mo> </mrow> </mrow> </math>
wherein:
W 1 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] k 2 cos [ k arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ]
W 2 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] - k 2 cos [ ( - k ) arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ]
W 3 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] k 2 sin [ k arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ]
W 4 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] - k 2 sin [ ( - k ) arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ]
(2) derivation of analytical functions according to Cauchy-Riemann equation
Because of the fact thatψ (ζ) is an analytic function, and it can be known from the cauchy-riemann equation (C-R condition): when f (z) is differentiable at point z x + iy, its derivative can be calculated by:
<math> <mrow> <msup> <mi>f</mi> <mo>&prime;</mo> </msup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>u</mi> </mrow> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <mo>+</mo> <mi>i</mi> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>v</mi> </mrow> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>v</mi> </mrow> <mrow> <mo>&PartialD;</mo> <mi>y</mi> </mrow> </mfrac> <mo>-</mo> <mi>i</mi> <mfrac> <mrow> <mo>&PartialD;</mo> <mi>u</mi> </mrow> <mrow> <mo>&PartialD;</mo> <mi>y</mi> </mrow> </mfrac> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> </math>
then can be aligned withAnd ψ 1(z) [ i.e., formula (11a) and formula (11b)]And (3) derivation, derivation is obtained:
<math> <mrow> <msubsup> <mi>&psi;</mi> <mn>1</mn> <mo>&prime;</mo> </msubsup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mi>b</mi> <mo>)</mo> </mrow> </mrow> </math>
(3) shallow tunnel surrounding rock stress and displacement solving method
The explicit expressions of the stress function and the displacement function are obtained by substituting equations (13a), (13b) and (13c) into equations (1a), (1b) and (1 c):
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>xx</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mn>2</mn> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mo>-</mo> <mn>2</mn> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>-</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <msup> <mi>x</mi> <mn>2</mn> </msup> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mi>a</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>yy</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mn>2</mn> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>-</mo> <mrow> <mo>(</mo> <mn>2</mn> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <msup> <mi>x</mi> <mn>2</mn> </msup> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mi>b</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>xy</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mi>c</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <msub> <mi>u</mi> <mi>y</mi> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>&mu;</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>{</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <msubsup> <mi>a</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <msubsup> <mi>a</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mi>&kappa;</mi> <msubsup> <mi>a</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> </mtd> </mtr> <mtr> <mtd> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mi>d</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>u</mi> <mi>x</mi> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>&mu;</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>&kappa;</mi> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>&kappa;</mi> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mi>e</mi> <mo>)</mo> </mrow> </mrow> </math>
as described above, in equations (14a) to (14e), the γ domain after conformal mapping is taken as a "black box", the stress and displacement functions are analyzed explicitly, and the left and right sides thereof are in the form of the R domain before mapping, so that engineers can directly obtain the values of the corresponding points from the coordinates (x, y) by directly inputting the parameters as in the Peck empirical formula without studying the theoretical principle. The input parameters contained in the formulas (14a) to (14e)The numbers can be divided into two groups: tunnel geometry and formation parameters: e, mu, nu, kappa, h, r, a. Formally, the geometry of the function is determined as it contains the part of the body function with all coordinates (x, y) as explicit analytical solution expressions; correlation coefficient of deformation boundary condition: alpha0,ɑk,bk,c0,ck,dkAnd the deformation condition of the tunnel boundary can be determined. Formally, it is the coefficient of the body function part that determines the geometric size of the function. Therefore, the explicit analytic solution can definitely research the influence rule of different stratum parameters and deformation boundary conditions on the analytic solution. In the formulas (14a) and (14e), the programming process only involves complex planes before mapping, the calculation efficiency is higher than that of an implicit analytic solution, the solving process is simple, and the obtained stress and displacement can be obtained by directly inputting stratum parameters. The explicit analytic solution can directly program the stress and the displacement into a function of coordinates (x, y) used in an R domain before conformal mapping, and has small calculation amount and high calculation efficiency.
The invention is further illustrated by the following set of examples:
to illustrate the utility of the explicit resolution solution, the following two sets of tunnel examples are referenced for analysis. Table 1 gives the geometrical dimensions, soil properties, etc. of the two sets of tunnels. In engineering, the displacement of the actual measuring point is usually measured by an extensometer and an inclinometer, and the horizontal displacement and the vertical displacement of the monitoring point on the horizontal measuring line and the vertical measuring line in the stratum can be obtained. By using the derived explicit analytic solution formula (14a) -formula (14e), a large amount of displacement data of each point of the stratum can be simply and conveniently provided, the measured engineering data and the analytic solution are compared, and the result analysis is as follows:
TABLE 1 Tunnel geometry and formation parameters
(1) N-2 mark section of san Francisco No. 4 line tunnel
And the N-2 section of the san Francisco No. 4 line tunnel is constructed by using an EPB shield machine with the diameter of 3.7m to construct a tunnel with the diameter of 3.56 m. During construction of the engineering, 23 measuring points are arranged by using an instrument to measure the vertical displacement and the horizontal displacement of the stratum, as shown in figure 3. The red dots are formula measured data, and the black solid lines are curves drawn by extracting data connecting lines every 1m for the explicit analytic solution.
(2) N-2 mark section of san Francisco No. 2 line tunnel
In order to avoid the universality of the analytical solution, the same method is used for researching the N-2 segment explicit analytical solution and the actually measured stratum settlement of the No. 2 linear tunnel in san Francisco, as shown in FIG. 4.
As can be seen from FIGS. 3 and 4, most of the actual measurement points are overlapped with the explicit analytic solution, which shows that the explicit analytic solution based on the complex function method derived by the invention can well predict the stratum displacement caused by tunnel construction.
The analysis shows that the advantage of displaying the analytic solution is that engineering personnel can conveniently make a plurality of earth surface and stratum internal measuring lines, so that the method is used for systematically researching the displacement change rule in the stratum, has good practicability and has important significance for the settlement prediction of urban tunnel construction. Moreover, if the stratum stress and the displacement under the load action of the building structures and the like on the ground and in the stratum are researched, various loads can be more conveniently superposed compared with an implicit analytical solution, and therefore the calculation efficiency of the stratum stress and the displacement of the tunnel construction under the complex condition is greatly improved.
In summary, the technical solution of the present invention utilizes the inverse mapping function to analyze the ζ plane as a functionConversion of sum psi (ζ) into z-plane analytic function in R-domain in complex function methodAnd psi1(z) obtaining a series explicit expression form of a stress function and a displacement function by derivation of an analytic function in a hidden analytic solution of the stress and displacement of the shallow tunnel surrounding rock by combining a Cauchy-Riemann equation (C-R condition); the explicit solutionThe analysis solution is visual and convenient for engineering personnel to use, and the programming calculation amount is less than that of the implicit analysis solution.
According to the technical scheme, based on an engineering example, the explicit analytical solution of the stress and displacement of the shallow tunnel surrounding rock is applied, the theoretical calculation result of the explicit analytical solution and the actually-measured stratum displacement of the engineering are compared and analyzed, and the correctness and the practicability of the explicit analytical solution are verified.
It should be understood that the above-mentioned embodiments of the present invention are only examples for clearly illustrating the present invention, and are not intended to limit the embodiments of the present invention, and it will be obvious to those skilled in the art that other variations or modifications may be made on the basis of the above description, and all embodiments may not be exhaustive, and all obvious variations or modifications may be included within the scope of the present invention.

Claims (4)

1. The method for solving the explicit analytic solution of the stress and displacement of the shallow tunnel surrounding rock is characterized by comprising the following steps
S1, establishing an analytic function of shallow tunnel surrounding rock stress and displacement on a z plane without concrete expression forms based on a complex function methodAnd psi1(z) and a calculation formula of the stratum stress and displacement of the shallow tunnel:
wherein σxx,σyy,σxyIs formation stress, ux,uyIn order to displace the formation,and psi1(z) is an analytical function of two non-specific expression forms of a z plane, mu is a shear modulus, kappa is a poisson ratio correlation coefficient (kappa is 3-4 upsilon), upsilon is a poisson ratio, and i is an imaginary constant;
s2, utilizing conformal mapping functionWherein alpha is a parameter defined by the ratio of the radius r of the tunnel cavern to the buried depth h of the circle center of the tunnel cavern,analytical function of non-concrete expression form on z planeAnd psi1(z) conversion to an analytic function with a concrete representation on the zeta planeAnd psi (ζ), the relationship of the two sets of analytical functions beingAnd psi (ζ) ═ psi1(ω(ζ))=ψ1(z), analytic functionAnd psi (zeta) are each in the specific formAnd <math> <mrow> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>c</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>d</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mo>,</mo> </mrow> </math> wherein, a0,ak,bk,c0,ck,dkThe physical meaning of the method is a boundary condition coefficient of the tunnel cavern;
s3, analyzing zeta plane by inverse mapping functionAnd psi (ζ) into an analytical function having a specific form in the z-plane in a complex function methodAnd psi1(z) solving a series form of a z-plane analytical function;
s4 based on Cauchy-Riemann equationFor analytical functions with concrete expression formAnd psi1(z) performing derivation;
s5, an analytic function with a concrete formAnd psi1And (z) substituting the derivative function into the calculation formula of the shallow tunnel stratum stress and displacement to obtain a display analytical solution of the shallow tunnel stratum stress and displacement.
2. The solution method according to claim 1, wherein the step S3 includes
To formula <math> <mrow> <mi>z</mi> <mo>=</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>-</mo> <mi>ih</mi> <mfrac> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msup> <mi>&alpha;</mi> <mn>2</mn> </msup> </mrow> </mfrac> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <mi>&zeta;</mi> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <mi>&zeta;</mi> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mi>ia</mi> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <mi>&zeta;</mi> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <mi>&zeta;</mi> </mrow> </mfrac> </mrow> </math> Inverse mapping is carried out to obtain an inverse mapping function <math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <msup> <mi>&omega;</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mi>z</mi> <mo>+</mo> <mi>ia</mi> </mrow> <mrow> <mi>z</mi> <mo>-</mo> <mi>ia</mi> </mrow> </mfrac> <mo>,</mo> </mrow> </math> Given z ═ x + iy, then this formula can be deformed as: <math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>-</mo> <msup> <mi>a</mi> <mn>2</mn> </msup> <mo>+</mo> <mn>2</mn> <mi>xai</mi> </mrow> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>y</mi> <mo>-</mo> <mi>a</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>=</mo> <mi>U</mi> <mo>+</mo> <mi>Vi</mi> <mo>,</mo> </mrow> </math> wherein, U = x 2 + y 2 - a 2 x 2 + ( y - a ) 2 , V = 2 xa x 2 + ( y - a ) 2 ;
will be a formulaTriangular form converted to complex number zeta <math> <mrow> <mi>&zeta;</mi> <mo>=</mo> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mo>{</mo> <mi>cos</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>}</mo> <mo>,</mo> </mrow> </math> From the correlation law of complex power operations: <math> <mrow> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>=</mo> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mi>k</mi> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>;</mo> </mrow> </math>
will analyze the functionAnd <math> <mrow> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>c</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>d</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mo>[</mo> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>]</mo> </mrow> </math> converted into real and imaginary forms, let a0=a′0i,ak=a′ki,bk=b′ki,c0=c′0i,ck=c′ki,dk=d′ki. Then, there is,
<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>0</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>c</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mi>k</mi> </msup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msub> <mi>d</mi> <mi>k</mi> </msub> <msup> <mi>&zeta;</mi> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> </mtd> </mtr> <mtr> <mtd> <mo>=</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mi>i</mi> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mi>i</mi> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mi>k</mi> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mi>k</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> </mtd> </mtr> <mtr> <mtd> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mi>i</mi> <msup> <mrow> <mo>(</mo> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mo>{</mo> <mi>cos</mi> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>+</mo> <mi>i</mi> <mi>sin</mi> <mo>[</mo> <mrow> <mo>(</mo> <mo>-</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>arccos</mi> <mfrac> <mi>U</mi> <msqrt> <msup> <mi>U</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>V</mi> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> </mtd> </mtr> </mtable> </mfenced> <mo>;</mo> </mrow> </math>
after the process of simplification, the process of the method,
<math> <mrow> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mo>*</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> <mo>,</mo> </mrow> </math> wherein,
W 1 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] k 2 cos [ k arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] ,
W 2 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] - k 2 cos [ ( - k ) arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] ,
W 3 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] k 2 sin [ k arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] ,
W 4 = [ x 2 + ( y + a ) 2 x 2 + ( y - a ) 2 ] - k 2 sin [ ( - k ) arccos ( x 2 + y 2 - a 2 x 4 + y 4 + a 4 + 2 a 2 x 2 + 2 x 2 y 2 - 2 a 2 y 2 ) ] .
3. the solution method according to claim 2, wherein the step S4 includes
When f (z) ═ u (x, y) + iv (x, y) is differentiable at point z ═ x + iy, the pair
And
<math> <mrow> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&psi;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&omega;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>&psi;</mi> <mrow> <mo>(</mo> <mi>&zeta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mo>*</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> </mrow> </math> conducting derivation to obtain
<math> <mrow> <mrow> <msubsup> <mi>&psi;</mi> <mn>1</mn> <mo>&prime;</mo> </msubsup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>k</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> </mrow> <mo>,</mo> </mrow> </math>
4. The solution method according to claim 3, wherein said step S5 includes
Will be provided with
<math> <mrow> <msubsup> <mi>&psi;</mi> <mn>1</mn> <mo>&prime;</mo> </msubsup> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>}</mo> <mi>i</mi> <mo>-</mo> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>k</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>}</mo> </mrow> </math> And
substitution formula
To obtain
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>xx</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <msubsup> <mrow> <mo>-</mo> <mn>2</mn> <mi>a</mi> </mrow> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mrow> <mo>-</mo> <mn>2</mn> <mi>b</mi> </mrow> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>-</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>yy</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>2</mn> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msubsup> <mrow> <mn>2</mn> <mi>b</mi> </mrow> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>&sigma;</mi> <mi>xy</mi> </msub> <mo>=</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <msup> <mo>&PartialD;</mo> <mn>2</mn> </msup> <msup> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> <mn>2</mn> </msup> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>u</mi> <mi>y</mi> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>&mu;</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>{</mo> <mrow> <mo>(</mo> <msubsup> <mi>&kappa;a</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mn>0</mn> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <mo>+</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mo>[</mo> <mrow> <mo>(</mo> <msubsup> <mi>&kappa;a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mi>&kappa;b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>,</mo> </mrow> </math>
<math> <mrow> <msub> <mi>u</mi> <mi>x</mi> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mn>2</mn> <mi>&mu;</mi> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>{</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>&infin;</mo> </munderover> <mrow> <mo>(</mo> <msubsup> <mrow> <mo>-</mo> <mi>&kappa;a</mi> </mrow> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>c</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mrow> <mo>-</mo> <mi>&kappa;b</mi> </mrow> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>+</mo> <msubsup> <mi>d</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mo>)</mo> </mrow> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>+</mo> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>3</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>4</mn> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>y</mi> <mrow> <mo>(</mo> <msubsup> <mi>a</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>1</mn> </msub> <mo>+</mo> <msubsup> <mi>b</mi> <mi>k</mi> <mo>&prime;</mo> </msubsup> <mfrac> <mo>&PartialD;</mo> <mrow> <mo>&PartialD;</mo> <mi>x</mi> </mrow> </mfrac> <msub> <mi>W</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>]</mo> <mo>}</mo> <mo>.</mo> </mrow> </math>
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CN110377990A (en) * 2019-07-08 2019-10-25 北京交通大学 Containing empty stratum shallow tunnel surrouding rock stress and Methods of Solving Displacement Problems under gravity

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CN107194136A (en) * 2017-07-31 2017-09-22 中国水利水电第七工程局成都水电建设工程有限公司 A kind of pressure from surrounding rock computational methods suitable for many stratum shallow tunnels
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