CN104018513B - The computational methods of sinking land in coalmining areas protective plate basis deflection deformation and section turn moment - Google Patents

Abstract

The invention discloses the computational methods of the deflection deformation on electric power pylon protective plate basis, the sinking land in coalmining areas and section turn moment, under the assumed conditions of elastic foundation, the four free composite plate foundations in limit are reduced to unidirectional Equivalent Beam; It is vertical distributed load by the vertical load equivalency transform of isolated footing and top steel tower thereof; Set up the synergism simplified mathematical model of depression ground and composite protective plate base in the sinking land in coalmining areas, it is proposed to basis deflection deformation differential equilibrium equations; In conjunction with the boundary condition that computation model need to meet, obtain the system of linear equations solved; Solve system of linear equations, obtain the function expression of deflection deformation and section turn moment; Determining the controlling sections of composite protective plate base, the arrangement of reinforcement carrying out Equivalent Beam cross section with section turn moment design load calculates, and finally determines the sectional reinforcement on basis. The invention has the beneficial effects as follows as the steel tower composite protective plate bases used a large amount of in the sinking land in coalmining areas, it is provided that a kind of theoretical calculation method about basis deflection deformation and section turn moment.

Description

The computational methods of sinking land in coalmining areas protective plate basis deflection deformation and section turn moment

Technical field

The invention belongs to the design of transmission line of electricity in the coal mining zone of influence and resist technology field, relate to the deflection deformation on electric power pylon protective plate basis, the sinking land in coalmining areas and the computational methods of section turn moment.

Background technology

A large amount of exploitations of coal resources, cause the substantial amounts of sinking land in coalmining areas in the coal main producing region of China. Meanwhile, along with the deep enforcement of " becoming defeated coal into transmission of electricity " energy policy, increasing power plant will build near colliery. Therefore, substantial amounts of newly-built transmission line of electricity is caused inevitably to build in the sinking land in coalmining areas. In recent years, for the unfavorable shortcoming to the deformation of resisting mining earth's surface in original steel tower independence long column isolated footing, power construction department of China and relevant colleges and universities propose this novel resistance to deformation basis of composite protective plate base, and have carried out successful Application in engineering practice, achieve good effect. But existing engineering practice depends on engineering experience, the internal force of composite protective plate base is calculated to the calculating of especially section turn moment, not yet propose the computational methods of scientific system. On the whole, China about minery electric power pylon composite protective plate base design and application still in the tentative stage, this has just buried huge potential safety hazard for the safety of built and newly-built work transmission line and steady in a long-term operation, and the major accident because the design bearing capacity deficiency on resistance to deformation basis causes steel tower deformation seriously even to be collapsed very easily occurs. Therefore, in the urgent need to proposing the computational methods of the section turn moment of composite protective plate base in a kind of sinking land in coalmining areas, design and engineering practice for China's minery electric power pylon composite protective plate base provide theoretical foundation.

Summary of the invention

The purpose of the present invention is in the computational methods of the deflection deformation and section turn moment that provide electric power pylon protective plate basis, the sinking land in coalmining areas.

The four free composite plate foundations in limit, under the assumed conditions of elastic foundation, are reduced to the unidirectional Equivalent Beam along subsidence basin moving direction by the present invention; The vertical load of isolated footing and top steel tower thereof is converted to vertical Equivalent Distributed load; Consider the combined effect of ground and basis, set up the synergism simplified mathematical model of depression ground and composite protective plate base in the sinking land in coalmining areas; In conjunction with the boundary condition that computation model need to meet, obtain the corresponding system of linear equations about model solution; The selection of composite protective plate base design con-trol cross section and respective cross-section control moment value and the method for Cross section Design are proposed. These computational methods, its theoretical foundation is clear and definite, and easy to use, result of calculation is entirely capable of meeting the requirement of engineering design, has wide applicability.

The technical solution adopted in the present invention is to carry out according to following steps:

The four free composite plate foundations in limit are reduced to unidirectional Equivalent Beam by step 1: under the assumed conditions of elastic foundation;

Step 2: be vertical distributed load by the vertical load equivalency transform of isolated footing and top steel tower thereof;

Step 3: consider the combined effect of ground and basis, sets up the synergism simplified mathematical model of depression ground and composite protective plate base in the sinking land in coalmining areas,

Step 4: the synergism simplified mathematical model according to depression ground in the above sinking land in coalmining areas Yu composite protective plate base, it is proposed to basis deflection deformation differential equilibrium equations;

Step 5: the boundary condition that need to meet in conjunction with computation model, obtains the system of linear equations solved;

Step 6: solve system of linear equations when meeting boundary condition, obtains the function expression of deflection deformation and section turn moment;

Step 7: determine the controlling sections of composite protective plate base, analyze the design load of subsidence basin moving process middle section moment of flexure, the arrangement of reinforcement carrying out Equivalent Beam cross section with section turn moment design load calculates, and in conjunction with the detailing requiments of concrete foundation, finally determines the sectional reinforcement on basis.

Further, in described step 2, each section of computing formula of segmentation evenly load is:

$q_{\mathrm{JC}}=q_1+{\mathrm{\γ}}_{m1}\·d\·1=\frac{F_{\mathrm{NV}}}{b}+{\mathrm{\γ}}_{m1}\·d\·1,$ q_KZ=��_m2��d��1��

Further, in described step 3, subsidence curve model w (x) can be represented by formula:

$w\left(x\right)=w_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\sin 2\mathrm{\π}\frac{x+s}{L}).$

Further, in described step 4, basis deflection deformation differential equilibrium equations sets up process, owing to composite plate load situation is divided into 3 kinds of situations from left to right, therefore is divided into 3 sections to solve, from x₀=0 arrives x₁The composite plate cross-sectional displacement of=b section is designated as y₁(x); x₁=b to x₂=l-b section is designated as y₂(x); And x₂=l-b to x₃=l section is designated as y₃(x); Simultaneously, it is assumed that the respectively q of line load in X direction caused by top steel tower, basis and deadweight of banketing in 3 sections of intervals₁+��_m1��d����_m2D and q₁+��_m1D, if it is assumed that the resistance coefficient of Subsidence Area foundation soil body is k, then in 3 sections, composite plate deformation differential equation on elastic foundation and general solution can be expressed as:

1)x₀To x₁Section:

$\mathrm{EI}\frac{d^4{y}_1}{d{x}^4}+k{y}_1=(q_1+{\mathrm{\γ}}_{m1}\·\mathrm{bd})+k{w}_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·\sin 2\mathrm{\π}\frac{x+s}{L}),$

Corresponding y₁General solution is:

$\begin{array}{c}{y}_1\left(x\right)=e^{-\mathrm{\αx}}(A_1\sin \mathrm{\αx}+A_2\cos \mathrm{\αx})+e^{\mathrm{\αx}}(A_3\sin \mathrm{\αx}+A_4\cos \mathrm{\αx})\\ +\frac{q_1+{\mathrm{\γ}}_{m1}d}{k}+w_0-\frac{w_0}{L}(x+s)+m\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array},$

2)x₁To x₂Section:

$\mathrm{EI}\frac{d^4{y}_2}{d{x}^4}+k{y}_2={\mathrm{\γ}}_{m2}\·\mathrm{bd}+k{\mathrm{\ω}}_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·\sin 2\mathrm{\π}\frac{x+s}{L}),$

Corresponding y₂General solution be:

$\begin{array}{c}{y}_2\left(x\right)=e^{-\mathrm{\αx}}(A_5\sin \mathrm{\αx}+A_6\cos \mathrm{\αx})+e^{\mathrm{\αx}}(A_7\sin \mathrm{\αx}+A_8\cos \mathrm{\αx})\\ +\frac{{\mathrm{\γ}}_{m2}d}{k}+w_0-\frac{w_0}{L}(x+s)+m\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array},$

3)x₂To x₃Section:

$\mathrm{EI}\frac{d^4{y}_3}{d{x}^4}+k{y}_3=(q_2+{\mathrm{\γ}}_{m1}\·\mathrm{bd})+k(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·\sin 2\mathrm{\π}\frac{x+s}{L}),$

Corresponding y₃General solution be:

$\begin{array}{c}{y}_3\left(x\right)=e^{-\mathrm{\αx}}(A_9\sin \mathrm{\αx}+A_{10}\cos \mathrm{\αx})+e^{\mathrm{\αx}}(A_{11}\sin \mathrm{\αx}+A_{12}\cos \mathrm{\αx})\\ +\frac{q_1+{\mathrm{\γ}}_{m1}d}{k}+w_0-\frac{w_0}{L}(x+s)+m\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array},$

Above-mentioned various in: A₁��A₁₂It is undetermined constant,

$\mathrm{\α}=\sqrt[4]{\frac{k}{4\mathrm{EI}}},m=\frac{k{w}_0}{2\mathrm{\π}(\frac{16{\mathrm{\π}}^4\mathrm{EI}}{L^4}+k)},$

Wherein, EI is the cross section bending rigidity of the protective plate of unit width.

The invention has the beneficial effects as follows as the steel tower composite protective plate bases used a large amount of in the sinking land in coalmining areas, it is provided that a kind of theoretical calculation method about basis deflection deformation and section turn moment.

Accompanying drawing explanation

Fig. 1 is the scope schematic diagram of composite protective plate base horizontal layout and Equivalent Beam Elements thereof;

Fig. 2 is that exemplary complex protective plate basis is vertically arranged schematic diagram;

Fig. 3 is the simplified mathematical model of composite protective plate base under Gravitative Loads;

Fig. 4 is the displacement before and after sedimentation of the protective plate basis and deformation schematic diagram;

Fig. 5 is the layout schematic diagram of protective plate computing nodes;

Fig. 6 is in the vertical displacement figure in protective plate each cross section during diverse location in basin;

Fig. 7 is the Bending moment distribution situation map in each cross section;

Fig. 8 is the sub-grade reaction distribution figure on protective plate basis.

Detailed description of the invention

Below in conjunction with the drawings and specific embodiments, the present invention is described in detail.

In view of prior art exist deficiency, the purpose of this fermentation patent be to provide a kind of have clear and definite theoretical foundation, application simplicity and meet the sinking land in coalmining areas electric power pylon composite protective plate base deflection deformation of engineering practice demand and the computational methods of section turn moment.

Sinking land in coalmining areas electric power pylon composite protective plate base deflection deformation involved in the present invention and the computational methods of section turn moment, including minery composite protective plate base Equivalent Calculation unit determination, consider Way on Soil-Foundation Interaction time the computation model on compound protective plate basis and the main contents such as determination of the differential equation, the determination of boundary condition, the solving of undetermined constant, the selection of controlling sections and section turn moment design load thereof. Patent of the present invention is best suited for design and the analysis of the electric power pylon composite protective plate base engineering that the sinking land in coalmining areas occurs limited continuous earth's surface to deform. The computational methods that patent of the present invention uses, it calculates theoretical according to fully, and process is simple and direct, reliable results.

The electric power pylon that the inventive method calculates is four isolated footings, the present invention is directed the situation of four isolated footings. Below in conjunction with accompanying drawing, the enforcement of patent of the present invention is further described, is specifically divided into following step:

The four free composite plate foundations in limit are reduced to unidirectional Equivalent Beam by step 1: under the assumed conditions of elastic foundation;

Being illustrated in figure 1 the scope schematic diagram of exemplary complex protective plate basic plane layout and Equivalent Beam Elements thereof, wherein dash area is the planar range of Equivalent Beam Elements; Fig. 2 is that schematic diagram (unit: mm) is arranged on exemplary complex protective plate basis, moves for subsidence basin along horizontal path direction; In Fig. 1 and Fig. 2, A, B respectively horizontal path and suitable line direction iron tower foundation root open value; The total length of E, F respectively horizontal path and the composite protective plate along line direction; C is at the bottom of long column isolated footing, top 0.5 times of rank width; H is the thickness of composite protective plate.

For concrete calculating object, it is determined that the numerical value of the physical dimension such as A, C and E in Fig. 1 and Fig. 2, wherein A is the distance between two independent long column base centers, and C is 0.5 times of the independent long column basis length of side, and E is the protective plate length of side;

The structure ledge of the end 500mm that inventive algorithm is left out in Fig. 1 and Fig. 2, with the l=E-500-500 Practical Calculation length being simplified model, and takes unit width and is calculated;

Step 2: be vertical distributed load by the vertical load equivalency transform of isolated footing and top steel tower thereof; Namely concentration power is scaled even distributed force.

The segmentation evenly load that vertical load equivalency transform is Equivalent Beam top that independent long column basis above protective plate basis in unit width is passed down with soil body deadweight and steel tower, the simplified mathematical model (unit width) being illustrated in figure 3 under Gravitative Loads composite protective plate base, each section of computing formula of segmentation evenly load is: $q_{\mathrm{JC}}=q_1+{\mathrm{\γ}}_{m1}\·d\·1=\frac{F_{\mathrm{NV}}}{b}+{\mathrm{\γ}}_{m1}\·d\·1,$ q_KZ=��_m2D 1.Wherein: ��_m1����_m2The respectively equivalent unit weight of each section internal upper part soil body, unit kN/m³; F_NVVertical active force (downforce) for top steel tower; q₁Equivalence all portion's line load, the unit kN/m of foundation bottom is acted on for steel tower; B is rank, the end width of long column isolated footing, top, unit m; D is isolated footing buried depth, m.

Step 3: consider the combined effect of ground and basis, sets up the synergism simplified mathematical model of depression ground and composite protective plate base in the sinking land in coalmining areas;

Only considering the effect of deadweight, the schematic diagram protective plate as shown in Figure 4 basis obtaining the displacement before and after settlement of foundation of the protective plate basis and deformation and coordinates computed is settling displacement and the deformation schematic diagram of front and back. In Fig. 4, if w (x) is subsidence curve; Y (x) is the vertical displacement value in arbitrary cross section after composite plate deformation; L is the computational length of composite plate Equivalent Beam, and b is rank, the end width of long column isolated footing, top.

With the calculating rectangular coordinate system shown in Fig. 4, theoretical according to classical subsidence forecast, it is assumed that after sufficient mining, subsidence curve model w (x) of moving basin can be represented by formula (1).

$w\left(x\right)=w_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\sin 2\mathrm{\π}\frac{x+s}{L})---\left(1\right)$

Wherein: half basin of L subsidence basin is long, i.e. the horizontal range of basin boundary point and maximum lower sinker, m;

w₀The maximum sinking value of subsidence basin, m;

The s composite protective plate left end horizontal range from maximum lower sinker, m;

The arbitrary cross section of x composite protective plate is to the horizontal range in its left end cross section, m.

Step 4: the synergism simplified mathematical model according to depression ground in the above sinking land in coalmining areas Yu composite protective plate base, it is proposed to basis deflection deformation differential equilibrium equations.

Owing to composite plate load situation is divided into 3 kinds of situations from left to right, therefore it is divided into 3 sections to solve. From x₀=0 arrives x₁The composite plate cross-sectional displacement of=b section is designated as y₁(x); x₁=b to x₂=l-b section is designated as y₂(x); And x₂=l-b to x₃=l section is designated as y₃(x)��

Simultaneously, it is assumed that the respectively q of line load in X direction caused by top steel tower, basis (including composite plate) and deadweight of banketing in 3 sections of intervals₁+��_m1��d����_m2D and q₁+��_m1D. Wherein q₁The equivalent local line load of foundation bottom, kN is acted on for steel tower; B is equivalent strut width, m; Buried depth based on d, m.

If it is assumed that the resistance coefficient of Subsidence Area foundation soil body is k, then in 3 sections, composite plate deformation differential equation on elastic foundation and general solution can be expressed as:

1)x₀To x₁Section:

$\mathrm{EI}\frac{d^4{y}_1}{d{x}^4}+k{y}_1=(q_1+{\mathrm{\γ}}_{m1}\·\mathrm{bd})+k{w}_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·\sin 2\mathrm{\π}\frac{x+s}{L})---\left(2\right)$

In formula, y₁It is the deflection deformation function y of plate in the 1st section₁(x), lower same.

Corresponding y₁General solution is:

$\begin{array}{c}{y}_1\left(x\right)=e^{-\mathrm{\αx}}(A_1\sin \mathrm{\αx}+A_2\cos \mathrm{\αx})+e^{\mathrm{\αx}}(A_3\sin \mathrm{\αx}+A_4\cos \mathrm{\αx})\\ +\frac{q_1+{\mathrm{\γ}}_{m1}d}{k}+w_0-\frac{w_0}{L}(x+s)+m\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array}---\left(3\right)$

2)x₁To x₂Section:

$\mathrm{EI}\frac{d^4{y}_2}{d{x}^4}+k{y}_2={\mathrm{\γ}}_{m2}\·\mathrm{bd}+k{\mathrm{\ω}}_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·\sin 2\mathrm{\π}\frac{x+s}{L})---\left(4\right)$

In formula, y₂It is the deflection deformation function y of plate in the 2nd section₂(x), lower same.

Corresponding y₂General solution be:

$\begin{array}{c}{y}_2\left(x\right)=e^{-\mathrm{\αx}}(A_5\sin \mathrm{\αx}+A_6\cos \mathrm{\αx})+e^{\mathrm{\αx}}(A_7\sin \mathrm{\αx}+A_8\cos \mathrm{\αx})\\ +\frac{{\mathrm{\γ}}_{m2}d}{k}+w_0-\frac{w_0}{L}(x+s)+m\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array}---\left(5\right)$

3)x₂To x₃Section:

$\mathrm{EI}\frac{d^4{y}_3}{d{x}^4}+k{y}_3=(q_2+{\mathrm{\γ}}_{m1}\·\mathrm{bd})+k(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·\sin 2\mathrm{\π}\frac{x+s}{L})---\left(6\right)$

In formula, y₃It is the deflection deformation function y of plate in the 3rd section₃(x), lower same.

Corresponding y₃General solution be:

$\begin{array}{c}{y}_3\left(x\right)=e^{-\mathrm{\αx}}(A_9\sin \mathrm{\αx}+A_{10}\cos \mathrm{\αx})+e^{\mathrm{\αx}}(A_{11}\sin \mathrm{\αx}+A_{12}\cos \mathrm{\αx})\\ +\frac{q_1+{\mathrm{\γ}}_{m1}d}{k}+w_0-\frac{w_0}{L}(x+s)+m\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array}---\left(7\right)$

Above-mentioned various in: A₁��A₁₂It is undetermined constant;

$\mathrm{\α}=\sqrt[4]{\frac{k}{4\mathrm{EI}}},m=\frac{k{w}_0}{2\mathrm{\π}(\frac{16{\mathrm{\π}}^4\mathrm{EI}}{L^4}+k)}$

Wherein, EI is the cross section bending rigidity of the protective plate of unit width.

Step 5: the boundary condition that need to meet in conjunction with computation model, obtains the system of linear equations solved:

Consider boundary condition, the undetermined constant A in solving equation (3), (5), (7)₁��A₁₂. Have under boundary condition:

1) at x=0 place (composite plate foundation left end)

$M=0\⇒\mathrm{EI}{\frac{d^2{y}_1}{d{x}^2}|}_{x=0}=0---\left(8\right)$

$Q=0\⇒\mathrm{EI}{\frac{d^3{y}_1}{d{x}^3}|}_{x=0}=0---\left(9\right)$

Here, the moment in M, Q respectively cross section, corresponding x coordinate place and shear value, lower with.

2) at x=b place (inward flange of left isolated footing)

$y_1=y_2\⇒y_1{|}_{x=b}=y_2{|}_{x=b}---\left(12\right)$

${\mathrm{\θ}}_1={\mathrm{\θ}}_2\⇒\frac{{\mathrm{dy}}_1}{\mathrm{dx}}{|}_{x=b}=\frac{{\mathrm{dy}}_2}{\mathrm{dx}}{|}_{x=b}---\left(13\right)$

�� is the sectional twisting angle value at corresponding x coordinate place, lower same.

3) at x=x₂=l-b place (inward flange of right isolated footing)

$y_2=y_3\⇒y_2{|}_{x=l-b}=y_3{|}_{x=l-b}---\left(16\right)$

${\mathrm{\θ}}_3={\mathrm{\θ}}_3\⇒\frac{{\mathrm{dy}}_2}{\mathrm{dx}}{|}_{x=l-b}=\frac{{\mathrm{dy}}_3}{\mathrm{dx}}{|}_{x=l-b}---\left(17\right)$

4) at x=l place (right-hand member of composite plate foundation)

$M=0\⇒\mathrm{EI}\frac{d^2{y}_3}{d{x}^2}{|}_{x=l}=0---\left(18\right)$

$Q=0\⇒\mathrm{EI}\frac{d^3{y}_3}{d{x}^3}{|}_{x=l}=0---\left(19\right)$

Step 6: solve system of linear equations when meeting boundary condition, obtains the function expression of deflection deformation and section turn moment:

Formula (3), (5) and (7) is substituted into corresponding boundary condition formula (8)��formula (19) respectively, obtains about undetermined constant A₁, A₂..., A₁₂Order-1 linear equation group, see shown in formula (20)��(31).

$-A_1+A_3=\frac{m}{2{\mathrm{\α}}^2}\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\·\sin \frac{2\mathrm{\π}\·s}{L}---\left(20\right)$

$A_1+A_2+A_3-A_4=\frac{m}{2{\mathrm{\α}}^3}\·{\left(\frac{2\mathrm{\π}}{L}\right)}^3\·\cos \frac{2\mathrm{\π}\·s}{L}---\left(21\right)$

-e^-��bcos(��b)��A₁+e^-��bsin(��b)��A₂+e^��bcos(��b)��A₃-e^��bsin(��b)��A₄(22)

+e^-��bcos(��b)��A₅-e^-��bsin(��b)��A₆-e^��bcos(��b)��A₇+e^��bsin(��b)��A₈

=0

e^-��b(sin��b+cos��b)��A₁+e^-��b(-sin��b+cos��b)��A₂+e^��b(-sin��b+cos��b)��A₃(23)

-e^��b(sin��b+cos��b)��A₄-e^-��b(sin��b+cos��b)��A₅-e^-��b(-sin��b+cos��b)��A₆

-e^��b(-sin��b+cos��b)��A₇+e^��b(sin��b+cos��b)��A₈

=0

$\begin{array}{c}{e}^{-\mathrm{\αb}}\sin \left(\mathrm{\αb}\right)\·A_1+e^{-\mathrm{\αb}}\cos \left(\mathrm{\αb}\right)\·A_2+e^{\mathrm{\αb}}\sin \left(\mathrm{\αb}\right)\·A_3+e^{\mathrm{\αb}}\cos \left(\mathrm{\αb}\right)\·A_4\\ -e^{-\mathrm{\αb}}\sin \left(\mathrm{\αb}\right)\·A_5-e^{-\mathrm{\αb}}\cos \left(\mathrm{\αb}\right)\·A_6-e^{\mathrm{\αb}}\sin \left(\mathrm{\αb}\right)\·A_7-e^{\mathrm{\αb}}\cos \left(\mathrm{\αb}\right)\·A_8\\ =\frac{({\mathrm{\γ}}_{m2}-{\mathrm{\γ}}_{m1})d-q_1}{k}\end{array}---\left(2\right)$

$\begin{array}{c}{e}^{-\mathrm{\αb}}[-\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_1-e^{-\mathrm{\αb}}[\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_2\\ +e^{\mathrm{\αb}}[\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_3+e^{\mathrm{\αb}}[-\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_4\\ {-e}^{-{\mathrm{\αb}}_1}[-\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_5+e^{-\mathrm{\αb}}[\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_6\\ {-e}^{\mathrm{\αb}}[\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_7-e^{\mathrm{\αb}}[-\sin \left(\mathrm{\αb}\right)+\cos \left(\mathrm{\αb}\right)]\·A_8\\ =0\end{array}---\left(25\right)$

-e^-��(l-b)cos[��(l-b)]A₅+e^-��(l-b)sin[��(l-b)]��A₆+e^��(l-b)cos[��(l-b)]��A₇(26)

-e^��(l-b)sin[��(l-b)]��A₈+e^-��(l-b)cos[��(l-b)]��A₉-e^-��(l-b)sin[��(l-b)]��A₁₀

-e^��(l-b)cos[��(l-b)]��A₁₁+e^��(l-b)sin[��(l-b)]��A₁₂

=0

e^-��(l-b){sin[��(l-b)]+cos[��(l-b)]}��A₅+e^-��(l-b){-sin[��(l-b)]+cos[��(l-b)]}��A₆

+e^��(l-b){-sin[��(l-b)]+cos[��(l-b)]}��A₇-e^��(l-b){sin[��(l-b)]+cos[��(l-b)]}��A₈(27)

-e^-��(l-b){sin[��(l-b)]+cos[��(l-b)]}��A₉-e^-��(l-b){-sin[��(l-b)]+cos[��(l-b)]}��A₁₀

-e^��(l-b){-sin[��(l-b)]+cos[��(l-b)]}��A₁₁+e^��(l-b){sin[��(l-b)]+cos[��(l-b)]}��A₁₂

=0

$\begin{array}{c}{e}^{-\mathrm{\α}(l-b)}\sin \left[\mathrm{\α}(l-b)\right]\·A_5+e^{-\mathrm{\α}(l-b)}\cos \left[\mathrm{\α}(l-b)\right]\·A_6+e^{\mathrm{\α}(l-b)}\sin \left[\mathrm{\α}(l-b)\right]\·A_7\\ +e^{\mathrm{\α}(l-b)}\cos \left[\mathrm{\α}(l-b)\right]\·A_8-e^{-\mathrm{\α}(l-b)}\sin \left[\mathrm{\α}(l-b)\right]\·A_9-e^{-\mathrm{\α}(l-b)}\cos \left[\mathrm{\α}(l-b)\right]\·A_{10}\\ -e^{\mathrm{\α}(l-b)}\sin \left[\mathrm{\α}(l-b)\right]\·A_{11}-e^{\mathrm{\α}(l-b)}\cos \left[\mathrm{\α}(l-b)\right]\·A_{12}\\ =\frac{q_2+({\mathrm{\γ}}_{m1}-{\mathrm{\γ}}_{m2})d}{k}\end{array}---\left(28\right)$

e^-��(l-b)[-sin��(l-b)+cos��(l-b)]��A₅-e^-��(l-b)[sin��(l-b)+cos��(l-b)]��A₆(29)

+e^��(l-b)[sin��(l-b)+cos��(l-b)]��A₇+e^��(l-b)[-sin��(l-b)+cos��(l-b)]��A₈

-e^-��(l-b)[-sin��(l-b)+cos��(l-b)]��A₉+e^-��(l-b)[sin��(l-b)+cos��(l-b)]��A₁₀

-e^��(l-b)[sin��(l-b)+cos��(l-b)]��A₁₁-e^��(l-b)[-sin��(l-b)+cos��(l-b)]��A₁₂

=0

$\begin{array}{c}-e^{-\mathrm{\αl}}\cos \left(\mathrm{\αl}\right)\·A_9+e^{-\mathrm{\αl}}\sin \left(\mathrm{\αl}\right)\·A_{10}+e^{\mathrm{\αl}}\cos \left(\mathrm{\αl}\right)\·A_{11}-e^{\mathrm{\αl}}\sin \left(\mathrm{\αl}\right)\·A_{12}\\ =\frac{m}{2{\mathrm{\α}}^2}\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\·\sin \frac{2\mathrm{\π}(l+s)}{L}\end{array}---\left(30\right)$

$\begin{array}{c}{e}^{-\mathrm{\αl}}[\sin \left(\mathrm{\αl}\right)+\cos \left(\mathrm{\αl}\right)]\·A_9+e^{-\mathrm{\αl}}[-\sin \left(\mathrm{\αl}\right)+\cos \left(\mathrm{\αl}\right)]\·A_{10}\\ +e^{\mathrm{\αl}}[-\sin \left(\mathrm{\αl}\right)+\cos \left(\mathrm{\αl}\right)]\·A_{11}-e^{\mathrm{\αl}}[\sin \left(\mathrm{\αl}\right)+\cos \left(\mathrm{\αl}\right)]\·A_{12}\\ =\frac{m}{2{\mathrm{\α}}^3}\·{\left(\frac{2\mathrm{\π}}{L}\right)}^3\·\cos \frac{2\mathrm{\π}(l+s)}{L}\end{array}---\left(31\right)$

Adopt elimination solving equation (20)��(31), obtain A undetermined₁, A₂..., A₁₂Numerical value. Undetermined constant is substituted into respectively formula (3), (5) and (7), obtains the deflection deformation expression formula y of composite protective plate base in each section₁(x)��y₂(x) and y₃(x). The i.e. each cross-sectional displacement function of composite plate;

After obtaining each cross-sectional displacement function of composite plate, according to the differential relationship (32) of the section turn moment of the flat beam of Equivalent Elasticity Yu deflection deformation function, the Expression of Moment formula of respective cross-section can be tried to achieve in each interval of composite plate foundation respectively as shown in formula (33), (34) and (35).

M=EIy " (32)

Here, y is " for the second dervative of deflection deformation function y in each section.

$\begin{array}{c}{M}_1=\mathrm{EI}\·2{\mathrm{\α}}^2{e}^{-\mathrm{\αx}}(A_2\sin \mathrm{\αx}-A_1\cos \mathrm{\αx})+\mathrm{EI}\·2{\mathrm{\α}}^2{e}^{\mathrm{\αx}}(A_3\cos \mathrm{\αx}-A_4\sin \mathrm{\αx})\\ -\mathrm{EI}\·m\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array}---\left(33\right)$

$\begin{array}{c}{M}_2=\mathrm{EI}\·2{\mathrm{\α}}^2{e}^{-\mathrm{\αx}}(A_6\sin \mathrm{\αx}-A_5\cos \mathrm{\αx})+\mathrm{EI}\·2{\mathrm{\α}}^2{e}^{\mathrm{\αx}}(A_7\cos \mathrm{\αx}-A_8\sin \mathrm{\αx})\\ -\mathrm{EI}\·m\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array}---\left(34\right)$

$\begin{array}{c}{M}_3=\mathrm{EI}\·2{\mathrm{\α}}^2{e}^{-\mathrm{\αx}}(A_{10}\sin \mathrm{\αx}-A_9\cos \mathrm{\αx})+\mathrm{EI}\·2{\mathrm{\α}}^2{e}^{\mathrm{\αx}}(A_{11}\cos \mathrm{\αx}-A_{12}\sin \mathrm{\αx})\\ -\mathrm{EI}\·m\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\sin \frac{2\mathrm{\π}(x+s)}{L}\end{array}---\left(35\right)$

Step 7: determine the controlling sections of composite protective plate base, analyzes the design load of subsidence basin moving process middle section moment of flexure, carries out basis sectional reinforcement and calculates and Construction treatment:

After completing flexural function expression formula and section turn moment function expression solve, with spaning middle section for controlling sections, spaning middle section refers to the intermediate cross-section on protective plate basis, it is that to have changed the internal force in cross section be maximum that controlling sections refers in design, determine the Reinforcing Steel Bar Calculation value on whole protective plate basis, obtain the basic cross section Maximum bending moment in subsidence basin during diverse location of protective plate, here diverse location refers to plan-position, the value of namely different S, and to be in, when composite protective plate base, 0.25 times and 0.75 times that basin half is long, namely suppose that basin factory floor is the words of L, so the position of 0.25 times is the position of 0.25L of turning right from maximum lower sinker, the position of 0.75 times is the position of 0.75L of turning right from maximum lower sinker, in classical coal-mining subsidence theory, the two point is the application point of maximum curvature, the higher value of sagging moment is the control moment design load of composite protective plate base, the higher value of the moment of flexure in the higher value of sagging moment that is 0.25 times and 0.75 times of two position, certainly it is the maximum of whole plate, but moment of flexure has positive and negative, for two positions, positive and negative moment is respectively arranged with size, therefore take higher value in both and be calculated. the arrangement of reinforcement carrying out Equivalent Beam cross section with section turn moment design load calculates, and in conjunction with the detailing requiments of concrete foundation, finally determines the sectional reinforcement on basis.

Specific embodiment is set forth below, and the present invention will be described:

Embodiment 1: for verifying the feasibility of above-mentioned model and method, selects the composite protective plate of electric power pylon in certain sinking land in coalmining areas to be calculated. Maximum subsidence value w in this sinking land in coalmining areas₀=2m, the long L=200m in subsidence basin half basin. The length l=20m of the composite protective plate of electric power pylon, thickness h=0.6m, isolated footing, top side length b=5m, embedded depth of foundation d=3m. The equivalent line load q that top steel tower passes down₁=5kN/m. The weighted average unit weight �� of basis and earthing_m1=20kN/m³, banket unit weight ��_m2=17kN/m³. Base concrete elastic modulus E=2.55 �� 10¹⁰Pa. Coefficient of subgrade reaction takes k=3 �� 10⁴kN/m³��

For ease of analyzing and discuss when iron tower foundation is positioned at stress and the deformation of Subsidence Area diverse location, being divided into 12 sections (being respectively divided into again 4 segments in each section) along moving direction at the bottom of basin by big plate herein, its controlling sections is arranged and is numbered as shown in Figure 5. Calculate respectively when big plate centre distance maximum lower sinker distance respectively 0, L/8,2L/8,3L/8,4L/8,5L/8,6L/8,7L/8, L time the deflection deformation of each control point and moment. Wherein for distance for L in the case of, the situation of Practical Calculation is big plate low order end and Subsidence Area overlapping margins (now the maximum lower sinker horizontal range of actual range is L-l), and when being 0 for distance, the actual big plate left end s that overlaps with maximum lower sinker that takes is 0.Therefore, corresponding s value respectively 0m, 15m, 40m, 65m, 90m, 115m, 140m, 165m and 180m.

This protective plate basis is calculated by the above-mentioned model and the method for solving that adopt present invention proposition, obtains s and takes undetermined constant A1��A12 corresponding during different value, in Table 1, undetermined constant A₁��A₁₂(thickness of slab 0.6m).

Table 1

For verifying the correctness of above-mentioned solution, solving result during for s=0, corresponding undetermined constant is substituted into respectively formula (3), (5) and (7), obtain the deflection deformation function expression of composite plate, as shown in formula (36)��(38).

$y_1=\left\{\begin{array}{c}{e}^{-0.357530x}(-0.000034\sin 0.357530x+0.000131\cos 0.357530x)\\ +e^{0.357530x}(-0.000034\sin 0.357530x+0.000008\cos 0.357530x)\\ +2.4997\×{10}^{-3}+1.0-0.005x+0.159153\sin \frac{\mathrm{\πx}}{100}\end{array}\right\}---\left(36\right)$

$y_2=\left\{\begin{array}{c}{e}^{-0.357530x}(-0.001181\sin 0.357530x-0.000137\cos 0.357530x)\\ +e^{0.357530x}(-0.000034\sin 0.357530x+0.000131\cos 0.357530x)\\ +1.8333\×{10}^{-3}+1.0-0.005x+0.159153\sin \frac{\mathrm{\πx}}{100}\end{array}\right\}---\left(37\right)$

$y_3=\left\{\begin{array}{c}{e}^{-0.357530x}(-0.000034\sin 0.357530x+0.000008\cos 0.357530x)\\ +e^{0.357530x}(0.001181\sin 0.357530x-0.000137\cos 0.357530x)\\ +2.4997\×{10}^{-3}+1.0-0.005x+0.159153\sin 0.03142x\end{array}\right\}---\left(38\right)$

Then, counter-force (taking unit width 1m) at the bottom of the plate of big plate base is integrated summation, asks making a concerted effort of counter-force at the bottom of plate, as shown in formula (39).

$\mathrm{\Σ}\∫\mathrm{\σdx}=\underset{0}{\overset{b}{\∫}}{\mathrm{\σ}}_1\mathrm{dx}+\underset{b}{\overset{l-b}{\∫}}{\mathrm{\σ}}_2\mathrm{dx}+\underset{l-b}{\overset{l}{\∫}}{\mathrm{\σ}}_3\mathrm{dx}---\left(39\right)$

Wherein:WithCan be obtained by formula (40)��formula (42) respectively.

$\begin{array}{c}\underset{0}{\overset{b}{\∫}}{\mathrm{\σ}}_1\mathrm{dx}\\ =k[-\frac{A_1}{2\mathrm{\α}}{e}^{-\mathrm{\αx}}(\sin \mathrm{\αx}+\cos \mathrm{\αx})-\frac{A_2}{2\mathrm{\α}}{e}^{-\mathrm{\αx}}(\cos \mathrm{\αx}-\sin \mathrm{\αx})+\frac{A_3}{2\mathrm{\α}}{e}^{\mathrm{\αx}}(\sin \mathrm{\αx}-\cos \mathrm{\αx})\\ +\frac{A_4}{2\mathrm{\α}}{e}^{\mathrm{\αx}}(\sin \mathrm{\αx}+\cos \mathrm{\αx})+\left(\frac{q_1+{\mathrm{\γ}}_{m1}\·d}{k}\right)\·x-\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(x+s)}{L}+C]{|}_0^b\\ =k[-\frac{A_1}{2\mathrm{\α}}{e}^{-\mathrm{\αb}}(\sin \mathrm{\αb}+\cos \mathrm{\αb})-\frac{A_2}{2\mathrm{\α}}{e}^{-\mathrm{\αb}}(\cos \mathrm{\αb}-\sin \mathrm{\αb})+\frac{A_3}{2\mathrm{\α}}{e}^{\mathrm{\αb}}(\sin \mathrm{\αb}-\cos \mathrm{\αb})\\ +\frac{A_4}{2\mathrm{\α}}{e}^{\mathrm{\αb}}(\sin \mathrm{\αb}+\cos \mathrm{\αb})+\left(\frac{q_1+{\mathrm{\γ}}_{m1}\·d}{k}\right)\·b-\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(b+s)}{L}\\ +\frac{A_1}{2\mathrm{\α}}+\frac{A_2}{2\mathrm{\α}}+\frac{A_3}{2\mathrm{\α}}-\frac{A_4}{2\mathrm{\α}}+\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\πs}}{L}]\end{array}---\left(40\right)$

$\begin{array}{c}\underset{b}{\overset{l-b}{\∫}}{\mathrm{\σ}}_2\mathrm{dx}\\ =k[-\frac{A_5}{2\mathrm{\α}}{e}^{-\mathrm{\αx}}(\sin \mathrm{\αx}+\cos \mathrm{\αx})-\frac{A_6}{2\mathrm{\α}}(\cos \mathrm{\αx}-\sin \mathrm{\αx})+\frac{A_7}{2\mathrm{\α}}{e}^{\mathrm{\αx}}(\sin \mathrm{\αx}-\cos \mathrm{\αx})\\ +\frac{A_8}{2\mathrm{\α}}{e}^{\mathrm{\αx}}(\sin \mathrm{\αx}+\cos \mathrm{\αx})+\left(\frac{{\mathrm{\γ}}_{m2}\·d}{k}\right)\·x-\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(x+s)}{L}+C]{|}_b^{l-b}\\ =k\{-\frac{A_5}{2\mathrm{\α}}{e}^{-\mathrm{\α}(l-b)}[\sin \mathrm{\α}(l-b)+\cos \mathrm{\α}(l-b)]-\frac{A_6}{2\mathrm{\α}}{e}^{-\mathrm{\α}(l-b)}[\cos (l-b)-\sin \mathrm{\α}(l-b)]\\ +\frac{A_7}{2\mathrm{\α}}{e}^{\mathrm{\α}(l-b)}[\sin (l-b)-\cos (l-b)]+\frac{A_8}{2\mathrm{\α}}{e}^{\mathrm{\α}(l-b)}[\sin \mathrm{\α}(l-b)+\cos \mathrm{\α}(l-b)]\\ +\left(\frac{{\mathrm{\γ}}_{m2}\·d}{k}\right)\·(l-b)-\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(l-b+s)}{L}+\frac{A_5}{2\mathrm{\α}}{e}^{-\mathrm{\αb}}(\sin \mathrm{\αb}+\cos \mathrm{\αb})\\ +\frac{A_6}{2\mathrm{\α}}{e}^{-\mathrm{\αb}}(\cos \mathrm{\αb}-\sin \mathrm{\αb})-\frac{A_7}{2\mathrm{\α}}{e}^{\mathrm{\αb}}(\sin \mathrm{\αb}-\cos \mathrm{\αb})-\frac{A_8}{2\mathrm{\α}}{e}^{\mathrm{\αb}}(\sin \mathrm{\αb}+\cos \mathrm{\αb})\\ -\left(\frac{{\mathrm{\γ}}_{m2}\·d}{k}\right)\·b+\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(b+s)}{L}\}\end{array}---\left(41\right)$

$\begin{array}{c}\underset{l-b}{\overset{l}{\∫}}{\mathrm{\σ}}_3\mathrm{dx}\\ =k[-\frac{A_9}{2\mathrm{\α}}{e}^{-\mathrm{\αx}}(\sin \mathrm{\αx}+\cos \mathrm{\αx})-\frac{A_{10}}{2\mathrm{\α}}{e}^{-\mathrm{\αx}}(\cos \mathrm{\αx}-\sin \mathrm{\αx})+\frac{A_{11}}{2\mathrm{\α}}{e}^{\mathrm{\αx}}(\sin \mathrm{\αx}-\cos \mathrm{\αx})\\ +\frac{A_{12}}{2\mathrm{\α}}{e}^{\mathrm{\αx}}(\sin \mathrm{\αx}+\cos \mathrm{\αx})+\left(\frac{q_1+{\mathrm{\γ}}_{m1}\·d}{k}\right)\·x-\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(x+s)}{L}+C]{|}_{l-b}^l\\ =k\{-\frac{A_9}{2\mathrm{\α}}{e}^{-\mathrm{\αl}}(\sin \mathrm{\αl}+\cos \mathrm{\αl})-\frac{A_{10}}{2\mathrm{\α}}{e}^{-\mathrm{\αl}}(\cos \mathrm{\αl}-\sin \mathrm{\αl})+\frac{A_{11}}{2\mathrm{\α}}{e}^{\mathrm{\αl}}(\sin \mathrm{\αl}-\cos \mathrm{\αl})\\ +\frac{A_{12}}{2\mathrm{\α}}{e}^{\mathrm{\αl}}(\sin \mathrm{\αl}+\cos \mathrm{\αl})+\left(\frac{q_1+{\mathrm{\γ}}_{m1}\·d}{k}\right)\·l-\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(l+s)}{L}\\ +\frac{A_9}{2\mathrm{\α}}{e}^{-\mathrm{\α}(l-b)}[\sin \mathrm{\α}(l-b)+\cos (l-b)]+\frac{A_{10}}{2\mathrm{\α}}{e}^{-\mathrm{\α}(l-b)}[\cos \mathrm{\α}(l-b)-\sin \mathrm{\α}(l-b)]\\ -\frac{A_{11}}{2\mathrm{\α}}{e}^{\mathrm{\αx}}[\sin \mathrm{\α}(l-b)-\cos \mathrm{\α}(l-b)]-\frac{A_{12}}{2\mathrm{\α}}{e}^{\mathrm{\α}(l-b)}[\sin \mathrm{\α}(l-b)+\cos (l-b)]\\ -\left(\frac{q_1+{\mathrm{\γ}}_{m1}\·d}{k}\right)\·(l-b)+\frac{L}{2\mathrm{\π}}(m-\frac{w_0}{2\mathrm{\π}})\cos \frac{2\mathrm{\π}(l-b+s)}{L}\}\end{array}---\left(42\right)$

By every known parameters �� of this paper example, l, b, m, s, q₁��q₂����_m1����_m2, d and k etc. substitutes into the summation of the Foundation pressure that protective plate can be tried to achieve in formula (40)��(42) and is:

�� �� �� dx=332.41+565.57+330.74=1229.72kN

And make a concerted effort (in the 1m width) of upper load is:

�� �� qdx=1224.98kN

Therefore, adopt the Foundation pressure that this method is tried to achieve to make a concerted effort and upper load ratio with joint efforts be:

$\frac{\mathrm{\Σ}\∫\mathrm{\σdx}}{\mathrm{\Σ}\∫\mathrm{qdx}}=\frac{\underset{0}{\overset{b}{\∫}}{\mathrm{\σ}}_1\mathrm{dx}+\underset{b}{\overset{l-b}{\∫}}{\mathrm{\σ}}_2\mathrm{dx}+\underset{l-b}{\overset{l}{\∫}}{\mathrm{\σ}}_3\mathrm{dx}}{\mathrm{\Σ}\∫\mathrm{qdx}}=\frac{1229.72}{1224.98}=1.004$

Both differ only by 4 ��, it is seen that above-mentioned solution meets static balance condition, and this shows that method for solving herein is correctly believable.

Solving result: Fig. 6��Fig. 8 respectively illustrates protective plate and is positioned at the distribution situation of (when s is from 0m to 180m) basic each cross section vertical displacement, each section turn moment value and Foundation pressure during the diverse location of Subsidence Area.

As shown in Figure 7, in the dynamic earth's surface deformation process of whole subsidence basin, the deformation process that protective plate basic stress will experience from negative cruvature to positive camber, wherein the negative cruvature stage should be designed with the big plate cross section of the isolated footing inside edge near pelvic floor hernia for controlling sections, and positive camber active phase should be designed with the spaning middle section of composite plate for controlling sections. Design internal force at whole dynamic earth's surface deformation stage should carry out value with the negative cruvature stage for least favorable situation. Meanwhile, in the deformation process of whole earth's surface, big plate moment of flexure experiences the process of positive and negative alternate, and therefore big plate should carry out basic constructional design by Dual-layer steel bar arrangement scheme, and is advisable with symmetric reinforcement cross section.

Conclusion: the analysis surface of above-mentioned example, the result of calculation of patent of the present invention is correctly believable, the computational analysis on the basis that is completely suitable in the sinking land in coalmining areas electric power pylon protective plate.

The foregoing is only presently preferred embodiments of the present invention, not in order to limit the present invention, all any amendment, equivalent replacement and improvement etc. made within the spirit and principles in the present invention, should be included within protection scope of the present invention.

Claims (1)

1. the computational methods of a sinking land in coalmining areas electric power pylon composite protective plate base deflection deformation and section turn moment, it is characterised in that carry out according to following steps:

Step 1: four limit free composite protective plate bases of sinking land in coalmining areas electric power pylon are reduced to the unidirectional Equivalent Beam on elastic foundation;

Step 2: be the vertical distributed load acting on composite protective plate base upper surface by the vertical load equivalency transform of top steel tower, namely concentration power is scaled even distributed force: the isolated footing above composite protective plate base in unit width and the soil body conducted oneself with dignity and vertical load that steel tower passes down, equivalency transform is the segmentation evenly load on unidirectional Equivalent Beam top, and each section of computing formula of segmentation evenly load is:

$q_{JC}=q_1+{\mathrm{\γ}}_{m1}\·d=\frac{F_{NV}}{b}+{\mathrm{\γ}}_{m1}\·d;$

q_kz=��_m2D;

Wherein: q_JCEvenly load in respective segments below isolated footing, unit: kN/m;

q_KZThe evenly load in span centre section between isolated footing, unit: kN/m;

��_m1����_m2The equivalent unit weight of each section internal upper part soil body, conducts oneself with dignity containing concrete isolated footing, unit: kN/m³;

F_NVThe vertical active force of top steel tower, unit: kN;

q₁Steel tower acts on the segmentation evenly load of foundation bottom, unit: kN/m;

Rank, the end width of isolated footing, b top, unit: m;

D isolated footing buried depth, unit: m;

Step 3: consider the combined effect of ground and composite protective plate base, sets up synergism simplified mathematical model subsidence curve model w (x) of depression ground and composite protective plate base in the sinking land in coalmining areas, and its expression formula is:

$w\left(x\right)=w_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}sin2\mathrm{\π}\frac{x+s}{L})$

Wherein: with the sinking on the earth's surface that horizontal range is x place of zero, the left end point of zero and composite protective plate, unit: m in w (x) subsidence basin direction of advance;

w₀The maximum sinking value of subsidence basin, unit: m;

Half basin of L subsidence basin is long, i.e. the horizontal range of basin boundary point and maximum lower sinker, unit: m;

The s composite protective plate left end horizontal range from maximum lower sinker, unit: m;

The arbitrary cross section of x composite protective plate is to the horizontal range in its left end cross section, unit: m;

Step 4: the synergism simplified mathematical model according to depression ground in the above sinking land in coalmining areas Yu composite protective plate base, it is proposed to basis deflection deformation differential equilibrium equations:

Basis deflection deformation differential equilibrium equations sets up process, owing to composite protective plate base load situation is divided into 3 kinds of situations from left to right, therefore is divided into 3 sections to solve: from x₀=0 arrives x₁The composite protective plate base cross-sectional displacement of=b section is designated as y₁(x); x₁=b to x₂=l-b section is designated as y₂(x); And x₂=l-b to x₃=l section is designated as y₃(x); The segmentation evenly load respectively q along the x-axis direction caused by top steel tower, isolated footing and deadweight of banketing in 3 sections of intervals₁+��_m1��d����_m2D and q₁+��_m1D;

Then in 3 sections, composite protective plate base deformation differential equation on elastic foundation and general solution can be expressed as:

1)x₀To x₁Section:

$EI\frac{d^4{y}_1\left(x\right)}{{\mathrm{dx}}^4}+{\mathrm{ky}}_1\left(x\right)=(q_1+{\mathrm{\γ}}_{m1}\·bd)+{\mathrm{kw}}_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·sin2\mathrm{\π}\frac{x+s}{L})$

Corresponding y₁X () general solution is:

$\begin{array}{c}{y}_1\left(x\right)=e^{-\mathrm{\α}x}\left(A_1\sin \mathrm{\α}x+A_2\cos \mathrm{\α}x\right)+e^{\mathrm{\α}x}\left(A_3\sin \mathrm{\α}x+A_4\cos \mathrm{\α}x\right)\\ +\frac{{\mathrm{\γ}}_{m2}d}{k}+w_0-\frac{w_0}{L}\left(x+s\right)+m\sin \frac{2\mathrm{\π}\left(x+s\right)}{L}\end{array}$

2)x₁To x₂Section:

$EI\frac{d^4{y}_2\left(x\right)}{{\mathrm{dx}}^4}+{\mathrm{ky}}_2\left(x\right)={\mathrm{\γ}}_{m2}\·bd+{\mathrm{kw}}_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·sin2\mathrm{\π}\frac{x+s}{L})$

Corresponding y₂X the general solution of () is:

$\begin{array}{c}{y}_2\left(x\right)=e^{-\mathrm{\α}x}\left(A_5\sin \mathrm{\α}x+A_6\cos \mathrm{\α}x\right)+e^{\mathrm{\α}x}\left(A_7\sin \mathrm{\α}x+A_8\cos \mathrm{\α}x\right)\\ +\frac{{\mathrm{\γ}}_{m2}d}{k}+w_0-\frac{w_0}{L}\left(x+s\right)+m\sin \frac{2\mathrm{\π}\left(x+s\right)}{L}\end{array}$

3)x₂To x₃Section:

$EI\frac{d^4{y}_3\left(x\right)}{{\mathrm{dx}}^4}+{\mathrm{ky}}_3\left(x\right)=(q_2+{\mathrm{\γ}}_{m1}\·bd)+{\mathrm{kw}}_0(1-\frac{x+s}{L}+\frac{1}{2\mathrm{\π}}\·sin2\mathrm{\π}\frac{x+s}{L})$

Corresponding y₃X the general solution of () is:

$\begin{array}{c}{y}_3\left(x\right)=e^{-\mathrm{\α}x}\left(A_9\sin \mathrm{\α}x+A_{10}\cos \mathrm{\α}x\right)+e^{\mathrm{\α}x}\left(A_{11}\sin \mathrm{\α}x+A_{12}\cos \mathrm{\α}x\right)\\ +\frac{q_2+{\mathrm{\γ}}_{m1}d}{k}+w_0-\frac{w_0}{L}\left(x+s\right)+m\sin \frac{2\mathrm{\π}\left(x+s\right)}{L}\end{array}$

Above-mentioned various in:

The arbitrary cross section of x composite protective plate is to the horizontal range in its left end cross section, unit: m;

x₀The x coordinate value in the left end point cross section of composite protective plate, unit: m;

x₁The x coordinate value of the composite protective plate computing nodes corresponding with the right hand edge of isolated footing, left side, left side i.e. close maximum lower sinker side, unit: m;

x₂The x coordinate value of the composite protective plate computing nodes corresponding with the left hand edge of isolated footing, right side, right side i.e. close subsidence basin boundary point side, unit: m;

x₃The x coordinate value in the right endpoint cross section of composite protective plate, unit: m;

Half basin of L subsidence basin is long, i.e. the horizontal range of basin boundary point and maximum lower sinker, unit: m;

L composite protective plate is along the length on subsidence basin moving direction, unit: m;

The s composite protective plate left end horizontal range from maximum lower sinker, unit: m;

A₁��A₁₂It is undetermined constant;

��, m middle coefficient,

The representative section bending rigidity of the composite protective plate base of EI unit width, unit: 10⁶N��m²;

The resistance coefficient of k foundation soil body, unit: kN/m³;

��_m1����_m2The equivalent unit weight of each section internal upper part soil body, conducts oneself with dignity containing concrete isolated footing, unit: kN/m³;

F_NVThe vertical active force of top steel tower, unit: kN;

q₁Steel tower acts on the segmentation evenly load of foundation bottom, unit: kN/m;

Rank, the end width of isolated footing, b top, unit: m;

D isolated footing buried depth, unit: m;

Step 5: the boundary condition that need to meet in conjunction with the synergism simplified mathematical model of depression ground in the sinking land in coalmining areas Yu composite protective plate base, obtains the system of linear equations solved;

Step 6: solve system of linear equations when meeting boundary condition, obtains the function expression of deflection deformation and section turn moment:

Deflection deformation expression formula in each section of composite protective plate is y₁(x)��y₂(x) and y₃(x);

In each interval of composite protective plate, respective cross-section Expression of Moment formula is respectively as follows:

$\begin{array}{c}{M}_1=EI\·2{\mathrm{\α}}^2{e}^{-\mathrm{\α}x}\left(A_2\sin \mathrm{\α}x-A_1\cos \mathrm{\α}x\right)+EI\·2{\mathrm{\α}}^2{e}^{\mathrm{\α}x}\left(A_3\cos \mathrm{\α}x-A_4\sin \mathrm{\α}x\right)\\ -EI\·m\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\sin \frac{2\mathrm{\π}\left(x+s\right)}{L}\end{array}$

$\begin{array}{c}{M}_2=EI\·2{\mathrm{\α}}^2{e}^{-\mathrm{\α}x}\left(A_6\sin \mathrm{\α}x-A_5\cos \mathrm{\α}x\right)+EI\·2{\mathrm{\α}}^2{e}^{\mathrm{\α}x}\left(A_7\cos \mathrm{\α}x-A_8\sin \mathrm{\α}x\right)\\ -EI\·m\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\sin \frac{2\mathrm{\π}\left(x+s\right)}{L}\end{array}$

$\begin{array}{c}{M}_3=EI\·2{\mathrm{\α}}^2{e}^{-\mathrm{\α}x}\left(A_{10}\sin \mathrm{\α}x-A_9\cos \mathrm{\α}x\right)+EI\·2{\mathrm{\α}}^2{e}^{\mathrm{\α}x}\left(A_{11}\cos \mathrm{\α}x-A_{12}\sin \mathrm{\α}x\right)\\ -EI\·m\·{\left(\frac{2\mathrm{\π}}{L}\right)}^2\sin \frac{2\mathrm{\π}\left(x+s\right)}{L}\end{array}$

Step 7: determine the controlling sections of composite protective plate base, analyze the design load of subsidence basin moving process middle section moment of flexure, the arrangement of reinforcement carrying out Equivalent Beam cross section with section turn moment design load calculates, and in conjunction with the detailing requiments of concrete foundation, finally determines the sectional reinforcement on basis.