CN103324811A - Long-span bridge bending deformation handling method - Google Patents

Abstract

The invention relates to design of bridges, and discloses a long-span bridge bending deformation handling method which is a segmented and independent integrated integral method. The method includes the following solving steps by the utilization of a computer: selecting an original design scheme, determining a structural load diagram, segmenting a girder of a long-span bridge, building an independent skew curve differential equation with a derivative of the fourth order, performing four times of integration on each girder segment respectively to obtain a general solution, building a constraint equation and determining the integration constant by the utilization of boundary conditions and continuity conditions, solving the equation to obtain analytical expressions of the shearing force, the bending moment, the turn angle and the deflection, drawing up a shearing force diagram, a bending moment diagram, a turn angle diagram and a deflection diagram, determining the maximum value of the shearing force, the maximum value of the bending moment, the maximum value of the turn angle and the maximum value of the deflection, and checking the strength and the rigidity of the girder. The long-span bridge bending deformation handling method solves the problems existing in the prior art when the analytical expressions of the shearing force, the bending moment, the turn angle and the defection of the structure are solved to handle the bending deformation of the long-span bridge, wherein the method in the prior art is tedious and slow in computation speed, and can only obtain a numerical solution but can not obtain an integral analytical solution.

Description

The diastrophic disposal route of Longspan Bridge

Technical field

Technical scheme of the present invention relates to the design of bridge, specifically the diastrophic disposal route of Longspan Bridge.

Background technology

The development trend of bridge structure design is now: from simple structure to large and complex structure, the development of new structure and new material structures; From linearity to non-linear development; From static(al) to Power evelopment; Develop to uncertainty from determinacy; Conform to the principle of simplicity fractional analysis to the analysis development that becomes more meticulous.Longspan Bridge at first must satisfy intensity, rigidity, stability and shockproof requirements, and shearing, moment of flexure, corner and amount of deflection that these all need computation structure draw shear diagram, moment curve, corner figure and the amount of deflection figure of structure.But the existing mechanics of materials and calculation method of structure mechanics all are to adopt hand computation, and method is loaded down with trivial details and computing velocity is slow, can not get complete analytic solution (Sun Xunfang, Fang Xiaoshu, Guan Laitai. the mechanics of materials. Beijing: Higher Education Publishing House, 1985.; Li Liankun. structural mechanics. Beijing: Higher Education Publishing House, 2004.).For example, Sun Xunfang (Sun Xunfang, Fang Xiaoshu, Guan Laitai. the mechanics of materials. Beijing: Higher Education Publishing House, 1985.) enumerate: 1. method of section needs the equilibrium establishment equation, can obtain shearing and moment of flexure, can draw shear diagram and moment curve, but can not obtain corner and amount of deflection, adopt method of section to connect shearing and moment of flexure can not be tried to achieve for hyperstatic structure; 2. immediate integration and singularity functions method can obtain the analytical expression of corner and amount of deflection, and this is one of difficult point but at first need to know Bending Moment Equations, determine that integration constant is two of difficult point, calculate loaded down with trivial detailsly, and speed is slow; 3. adopt energy method, only can access corner and the amount of deflection of specified cross-section such as Moire technique, also need to know in advance Bending Moment Equations.Li Liankun (Li Liankun. structural mechanics. Beijing: Higher Education Publishing House, 2004.) enumerate: 1. force method can obtain the constraining force of hyperstatic structure, three moments euqation can obtain the constraining force of continuous beam, but at first need to select the static determinacy base, obtain constraining force and just statically indeterminate problem is transformed into statically determinate problem; 2. displacement method is found the solution hyperstatic structure take hyperstatic beams with single span as the basis, need to table look-up, and computing velocity is slow, can not get expression formula.Qiu Shundong (Qiu Shundong. science of bridge building software midas Civil application project example. Beijing: People's Transportation Press, 2011.) and Yuan team of four horses (Yuan team of four horses. the program structure mechanics. Beijing: Higher Education Publishing House, 2001.) although the Finite Element of enumerating can in the hope of the value of shearing, moment of flexure, corner and amount of deflection, can draw shear diagram, moment curve, corner figure and amount of deflection figure; But can only obtain numerical solution, can not obtain analytical expression, can only obtain approximate solution, can not obtain exact solution.

In a word, process the diastrophic problem of Longspan Bridge if adopt shearing, moment of flexure, corner and the amount of deflection analytical expression of solution structure of requiring of the prior art, need long time, even can not obtain whole analytic solution in several days, still do not have so far ripe fast resolving method.

Summary of the invention

Technical matters to be solved by this invention is: provide Longspan Bridge diastrophic disposal route, it is the independent integrated integral method of a kind of segmentation, only need the segmentation independent integral four times, by boundary condition and the condition of continuity, determine integration constant, just can obtain the analytical function of shearing, moment of flexure, corner and amount of deflection, overcome and adopted shearing, moment of flexure, corner and the amount of deflection analytical expression of solution structure of requiring of the prior art to process the diastrophic problem of Longspan Bridge, need long time, even can not obtain the defective of whole analytic solution in several days.

The present invention solves this technical problem the technical scheme that adopts: the diastrophic disposal route of Longspan Bridge is the independent integrated integral method of a kind of segmentation, and step is:

The first step is selected just design proposal, determines structural loads figure;

Second step carries out segmentation with the beam of Longspan Bridge

The beam of Longspan Bridge is divided into n (1≤n≤100) section, sets up independently Fourth-Derivative deflection differential equation:

$\frac{d^4{v}_i}{{\mathrm{dx}}^4}=\frac{q_i\left(x\right)}{E_i{I}_i}(i=\mathrm{1,2},\·\·\·,n)---\left(1\right)$

The 3rd the step, with each section beam respectively integration obtain shearing, moment of flexure, corner and deflection equation general solution for four times

Integration once obtains shearing equation general solution

$\frac{d^3{v}_i}{{\mathrm{dx}}^3}=\∫\frac{q_i\left(x\right)\mathrm{dx}}{E_i{I}_i}+C_{i,l}(i=\mathrm{1,2},\·\·\·,n)---\left(2\right)$

Integration twice obtains the Bending Moment Equations general solution

$\frac{d^2{v}_i}{{\mathrm{<figure-callout\; id="2"\; label="dx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">dx</figure-callout>}}^2}=\&Integral;(\&Integral;\frac{q_i\left(x\right)}{E_i{I}_i}\mathrm{dx})\mathrm{dx}+C_{i,l}x+C_{i,2}(i=1,2,\&CenterDot;\&CenterDot;\&CenterDot;,n)---\left(3\right)$

Integration three times obtains the equations of rotating angle general solution

$\frac{{\mathrm{dv}}_i}{\mathrm{dx}}=\&Integral;(\&Integral;(\&Integral;\frac{q_i\left(x\right)}{E_i{I}_i}\mathrm{dx})\mathrm{dx})\mathrm{dx}+\frac{1}{2}{C}_{i,1}{x}^2+C_{i,2}x+C_{i,3}(i=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,n)---\left(4\right)$

Integration four times obtains the deflection equation general solution

$v_i=\&Integral;(\&Integral;(\&Integral;(\&Integral;\frac{q_i\left(x\right)}{E_i{I}_i}\mathrm{dx})\mathrm{dx})\mathrm{dx})\mathrm{dx}+\frac{1}{6}{C}_{i,1}{x}^3+\frac{1}{2}{C}_{i,2}{x}^2+C_{i,3}x+C_{i,4}(i=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,n)---\left(5\right)$

In the 4th step, utilize displacement boundary conditions, force boundary condition and the condition of continuity to set up 4n boundary condition equation of constraint

f(C _i,j)=0(i=1,2,…,n,j=1,2,3,4) (6)

Simultaneous solution boundary condition equation of constraint (6) is determined 4n integration constant C _{I, j}(i=1,2 ..., n, j=1,2,3,4);

The 5th step is with integration constant C _{I, j}(i=1,2 ..., n, j=1,2,3,4) and substitution (2)～(5) formula just can obtain the analytical expression of shearing, moment of flexure, corner and amount of deflection;

The 6th step, with the shear diagram that computing machine draws, moment curve, corner figure and amount of deflection figure;

In the 7th step, ask shearing maximal value F _Smax, moment of flexure maximal value M _Max, corner maximal value θ _MaxMaximal value v with amount of deflection _Max

The 8th step is by the strength and stiffness of following formula check beam

${\mathrm{\&tau;}}_{\max }=\frac{F_{s\max }{S}_z^{*}}{I_{z}b}\&le;\&lsqb;\mathrm{\&tau;}\&rsqb;,{\mathrm{\&sigma;}}_{\max }=\frac{M_{\max }{y}_{\max }}{I_z}\&le;\&lsqb;\mathrm{\&sigma;}\&rsqb;,{\mathrm{\&theta;}}_{\max }\&le;\&lsqb;\mathrm{\&theta;}\&rsqb;,v_{\max }\&le;\&lsqb;v\&rsqb;---\left(7\right);$

Above-mentioned steps all adopts software Maple and Computer Processing.

The diastrophic disposal route of above-mentioned Longspan Bridge, adopt the flow process of Computer Processing to be:

Beginning → selection is design proposal just, determine structural loads figure → beam is divided into the n section, set up independently the Fourth-Derivative deflection differential equation → with each section beam respectively integration obtain shearing four times, moment of flexure, corner and deflection equation general solution → utilize displacement boundary conditions, force boundary condition and the condition of continuity are set up 4n boundary condition equation of constraint → integration constant substitution general formula is obtained shearing, moment of flexure, the analytical expression of corner and amount of deflection → drafting shear diagram, moment curve, corner figure and amount of deflection figure → ask shearing maximal value, the moment of flexure maximal value, whether the strength and stiffness of the maximal value of corner maximal value and amount of deflection → check beam meet the demands

Return and select just design proposal, determine structural loads figure;

Finish.

The diastrophic disposal route of above-mentioned Longspan Bridge, described beam with Longspan Bridge is divided into the n section, wherein n=1～100.

The diastrophic disposal route of above-mentioned Longspan Bridge, used software Maple and computing machine are that those skilled in the art are known.

The diastrophic disposal route of above-mentioned Longspan Bridge can also be used for flexural deformation problem under other complex load effect.

The invention has the beneficial effects as follows: compared with prior art, the outstanding substantive distinguishing features of the diastrophic disposal route of Longspan Bridge of the present invention is: the diastrophic disposal route of Longspan Bridge of the present invention is the independent integrated integral method of segmentation, and a kind of computing machine combines with mechanics Quick look methodOnly need the segmentation independent integral four times, by boundary condition and the condition of continuity, determine integration constant, just can obtain the analytical function of shearing, moment of flexure, corner and amount of deflection, just can obtain support constraint forces without the column balancing equation, need not set up shearing equation and Bending Moment Equations, just can draw shear diagram, moment curve, corner figure and amount of deflection figure.

Compared with prior art, the significant progress of the diastrophic disposal route of Longspan Bridge of the present invention is:

(1) the inventive method is calculated simply, and step stylizes.And in the prior art, force method is found the solution statically indeterminate problem, needs to select the static determinacy base, judge degree of statical indeterminacy, displacement method is found the solution statically indeterminate problem, then needs to table look-up, these do not need and adopt the inventive method, and the step of processing for statically indeterminate problem and statically determinate problem is the same.

(2) the inventive method is easily programmed, and the independent integrated integral method of segmentation+Maple software has been accelerated computing velocity more.And prior art adopts hand computation, and speed is slow.

(3) the inventive method can be in the hope of analytic solution.And in the prior art, the limit elements method can only obtain numerical solution, can not obtain function expression.

(4) the inventive method can be in the hope of the integrated result of calculation of shearing, moment of flexure, corner, amount of deflection and support constraint forces.And prior art only can obtain partial results, and finding the solution a problem needs several method use in conjunction, calculates loaded down with trivial details.Only can obtain shearing, moment of flexure such as method of section; Figure multiplication, Moire technique only can obtain corner, amount of deflection; Force method only can be obtained the superfluous constraint power of hyperstatic structure.

Description of drawings

The present invention is further described below in conjunction with drawings and Examples.

Fig. 1 is the process flow diagram of the Computer Processing of the diastrophic disposal route of Longspan Bridge of the present invention.

Fig. 2 is Longspan Bridge structural loads figure among the embodiment 1.

Fig. 3 is the shear diagram that draws with computing machine among the embodiment 1.

Fig. 4 is the moment curve that draws with computing machine among the embodiment 1.

Fig. 5 is the corner figure that draws with computing machine among the embodiment 1.

Fig. 6 is the amount of deflection figure that draws with computing machine among the embodiment 1.

Fig. 7 is Longspan Bridge structural loads figure among the embodiment 2.

Fig. 8 is the shear diagram that draws with computing machine among the embodiment 2.

Fig. 9 is the moment curve that draws with computing machine among the embodiment 2.

Figure 10 is the corner figure that draws with computing machine among the embodiment 2.

Figure 11 is the amount of deflection figure that draws with computing machine among the embodiment 2.

Figure 12 is Longspan Bridge structural loads figure among the embodiment 3.

Figure 13 is the shear diagram that draws with computing machine among the embodiment 3.

Figure 14 is the moment curve that draws with computing machine among the embodiment 3.

Figure 15 is the corner figure that draws with computing machine among the embodiment 3.

Figure 16 is the amount of deflection figure that draws with computing machine among the embodiment 3.

Figure 17 is Longspan Bridge structural loads figure among the embodiment 4.

Figure 18 is the shear diagram that draws with computing machine among the embodiment 4,

Figure 19 is the moment curve that draws with computing machine among the embodiment 4.

Figure 20 is the corner figure that draws with computing machine among the embodiment 4.

Figure 21 is the amount of deflection figure that draws with computing machine among the embodiment 4.

Embodiment

Embodiment illustrated in fig. 1 showing, the flow process of the Computer Processing of the diastrophic disposal route of Longspan Bridge of the present invention is: beginning → selection is design proposal just, determine structural loads figure → beam is divided into the n section, set up independently the Fourth-Derivative deflection differential equation → with each section beam respectively integration obtain shearing four times, moment of flexure, corner and deflection equation general solution → utilize displacement boundary conditions, force boundary condition and the condition of continuity are set up 4n boundary condition equation of constraint → integration constant substitution general formula is obtained shearing, moment of flexure, the analytical expression of corner and amount of deflection → drafting shear diagram, moment curve, corner figure and amount of deflection figure → ask shearing maximal value, the moment of flexure maximal value, whether the strength and stiffness of the maximal value of corner maximal value and amount of deflection → check beam meet the demands

Return and select just design proposal, determine structural loads figure;

Finish.

Embodiment 1

Fig. 2 is Longspan Bridge structural loads figure in the present embodiment.

The diastrophic disposal route of Longspan Bridge shown in Figure 2 is processed the segmentation of the beam of this Longspan Bridge Longspan Bridge as the n=1 section.

The span of known this Longspan Bridge is L, and elastic modulus is E, and moment of inertia is I, uniformly distributed load q.

Utilize segmentation independent integral method to find the solution, step is as follows:

The first step: n=1, line of deflection approximate differential equation is as follows:

$\frac{d^{4}v}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(0\&le;x\&le;\frac{L}{2})---\left(8\right)$

Second step: the line of deflection approximate differential equation to (8) each section of formula is distinguished integration four times, obtains the general solution of shearing, moment of flexure, corner and amount of deflection.In general solution, include 4 integration constant C _i(i=1,2 ..., 4).

The 3rd step: utilize following displacement boundary conditions, force boundary condition and the condition of continuity

v(0)=0，EIv′′(0)=0 (9a)

v(L)=0，EIv′′(L)=0 (9b)

Simultaneous solution system of equations (9) formula draws 4 integration constant C _i(i=1,2 ..., 4).

The 4th step: with integration constant C _i(i=1,2 ..., 4) and the general solution of substitution shearing, moment of flexure, corner and amount of deflection obtains the analytical expression of shearing, moment of flexure, corner and amount of deflection.

The shearing function: $F_s=-\frac{q}{2}(2x-L),0\&le;x\&le;L;$

Bending moment functions: $M=-\frac{\mathrm{qx}}{2}(x-L),0\&le;x\&le;L;$

The corner function: $\mathrm{\&theta;}=-\frac{\mathrm{<figure-callout\; id="24"\; label="q"\; filenames="HDA00003483994000024.png"\; state="\{\{state\}\}">q</figure-callout>}}{24\mathrm{EI}}(2x-L)({2x}^2-2\mathrm{Lx}-L^2),0\&le;x\&le;L;$

Deflection functions: $v=-\frac{\mathrm{<figure-callout\; id="24"\; label="qx"\; filenames="HDA00003483994000024.png"\; state="\{\{state\}\}">qx</figure-callout>}}{24\mathrm{EI}}(x-L)(x^2-\mathrm{Lx}-L^2),0\&le;x\&le;L.$

The shearing maximal value: $F_{S^{,}\max }=\frac{\mathrm{qL}}{2}(x=0);F_{S,\max }^{(-)}=\frac{\mathrm{qL}}{2}(x=L).$

The moment of flexure maximal value: $M_{\max }=\frac{1}{8}{\mathrm{qL}}^2(x=\frac{L}{2});$

The corner maximal value: ${\mathrm{\&theta;}}_{\max }=\frac{1}{24}\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=L);{\mathrm{\&theta;}}_{\max }^{(-)}=\frac{1}{24}\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=0).$

The amount of deflection maximal value: $v_{\max }^{(-)}=\frac{5}{384}\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=\frac{L}{2}).$

Support constraint forces: $F_{\mathrm{Ax}}=0,F_{\mathrm{Ay}}=\frac{\mathrm{<figure-callout\; id="2"\; label="qL"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">qL</figure-callout>}}{2},F_B=\frac{\mathrm{<figure-callout\; id="2"\; label="qL"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">qL</figure-callout>}}{2}.$

Fig. 3 is the shear diagram that draws with computing machine in the present embodiment.

Fig. 4 is the moment curve that draws with computing machine in the present embodiment.

Fig. 5 is the corner figure that draws with computing machine in the present embodiment 1.

Fig. 6 is the amount of deflection figure that draws with computing machine in the present embodiment 1.

Above-mentioned steps all adopts software Maple and Computer Processing.

Embodiment 2

Fig. 7 is Longspan Bridge structural loads figure in the present embodiment.

The diastrophic disposal route of Longspan Bridge shown in Figure 7 is processed the segmentation of the beam of this Longspan Bridge Longspan Bridge as the n=4 section.

It is L that known this Longspan Bridge has the span of four equal lengths, and elastic modulus is E, and moment of inertia is I, only strides the 3rd and loads q.

Utilize computer solving:

(i) draw shear diagram, moment curve, corner figure and amount of deflection figure;

(ii) if known q=50kN/m, L=70m, square-section deck-molding h=2.191m, wide b=4.564m, I=4m ⁴, A=10m ², [τ]=2MPa, [σ]=4MPa checks the intensity of this structure;

(iii) if known E=23GPa, [θ]=0.005rad,

Check the rigidity of this structure;

(iv) such as the discontented sufficient intensity of fruit structure or rigidity requirement, improved design project is proposed.

Step is as follows:

The first step is carried out segmentation with the beam of Longspan Bridge

The beam of Longspan Bridge shown in Figure 7 is divided into 4 sections, sets up independently Fourth-Derivative deflection differential equation:

$\frac{d^4{\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_1}{{\mathrm{dx}}^4}=0,(0\&le;x\&le;L)---\left(10a\right)$

$\frac{d^4{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">v</figure-callout>}}_2}{{\mathrm{dx}}^4}=0,(L\&le;x\&le;2L)---\left(10b\right)$

$\frac{d^4{v}_3}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(2L<x\&le;3L)---\left(10c\right)$

$\frac{d^4{v}_4}{{\mathrm{dx}}^4}=0,(3L<x\&le;4L)---\left(10d\right)$

Second step, with each section beam respectively integration obtain shearing, moment of flexure, corner and deflection equation general solution for four times

Integration once obtains shearing equation general solution

$\frac{d^3{v}_i}{{\mathrm{dx}}^3}=\&Integral;\frac{q_i\left(x\right)\mathrm{dx}}{E_i{I}_i}+C_{i,l}(i=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,4)---\left(11\right)$

Integration twice obtains the Bending Moment Equations general solution

$\frac{d^2{v}_i}{{\mathrm{<figure-callout\; id="2"\; label="dx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">dx</figure-callout>}}^2}=\&Integral;(\&Integral;\frac{q_i\left(x\right)}{E_i{I}_i}\mathrm{dx})\mathrm{dx}+C_{i,l}x+C_{i,2}(i=1,2,\&CenterDot;\&CenterDot;\&CenterDot;,4)---\left(12\right)$

Integration three times obtains the equations of rotating angle general solution

$\frac{{\mathrm{dv}}_i}{\mathrm{dx}}=\&Integral;(\&Integral;(\&Integral;\frac{q_i\left(x\right)}{E_i{I}_i}\mathrm{dx})\mathrm{dx})\mathrm{dx}+\frac{1}{2}{C}_{i,1}{x}^2+C_{i,2}x+C_{i,3}(i=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,4)---\left(13\right)$

Integration four times obtains the deflection equation general solution

$v_i=\&Integral;(\&Integral;(\&Integral;(\&Integral;\frac{q_i\left(x\right)}{E_i{I}_i}\mathrm{dx})\mathrm{dx})\mathrm{dx})\mathrm{dx}+\frac{1}{6}{C}_{i,1}{x}^3+\frac{1}{2}{C}_{i,2}{x}^2+C_{i,3}x+C_{i,4}(i=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,4)---\left(14\right)$

In general solution, include 16 integration constant C _{I, j}(i=1 ..., 4; J=1 ..., 4).

In the 3rd step, utilize displacement boundary conditions, force boundary condition and condition of continuity equation of constraint

v ₁(0)=0，EIv ₁′′(0)=0 （15a）

v ₁(L)=0，v ₂(L)=0 （15b）

v ₁′(L)=v′ ₂(L)，EIv ₁′′(L)=EIv ₂″(L) （15c）

v ₂(2L)=0，v ₃(2L)=0 （15d）

v ₂′(2L)=v ₃′(2L)，EIv ₂′′(2L)=EIv ₃′′(2L) （15e）

v ₃(3L)=0，v ₄(3L)=0 （15f）

v ₃′(3L)=v ₄′(3L)，EIv ₃′′(3L)=EIv ₄″(3L) （15g）

v ₄(4L)=0，EIv ₄″(4L)=0 （15h）

Simultaneous solution equation (15) formula draws 16 integration constant C _{I, j}(i=1,2 ..., 4, j=1,2 ..., 4);

The 4th step is with integration constant C _{I, j}(i=1,2 ..., 4, j=1,2 ..., 4) and substitution (11)～(14) formula just can obtain the analytical expression of shearing, moment of flexure, corner and amount of deflection;

The shearing function: $F_s=\left\{\begin{array}{cc}\frac{3\mathrm{<figure-callout\; id="0"\; label="qL"\; filenames="HDA00003483994000022.png,HDA00003483994000031.png"\; state="\{\{state\}\}">qL</figure-callout>}}{224},& 0\&le;x\&le;L\\ \frac{-15\mathrm{qL}}{224},& L<x\&le;2L\\ \frac{-q}{224}(224x-561L),& 2L<x\&le;3L\\ \frac{11\mathrm{qL}}{224},& 3L<x\&le;4L\end{array}\right.$

Bending moment functions: $M=\left\{\begin{array}{cc}\frac{3\mathrm{<figure-callout\; id="0"\; label="qLx"\; filenames="HDA00003483994000022.png,HDA00003483994000031.png"\; state="\{\{state\}\}">qLx</figure-callout>}}{224},& 0\&le;x\&le;L\\ \frac{-3q}{224}(5x-6L),& L<x\&le;2L\\ \frac{-q}{224}(112x^2-561\mathrm{Lx}+686L^2),& 2L<x\&le;3L\\ \frac{11\mathrm{qL}}{224}(x-4L),& 3L<x\&le;4L\end{array}\right.$

The corner function: $\mathrm{\&theta;}=\left\{\begin{array}{cc}\frac{\mathrm{qL}}{448\mathrm{EI}}({3x}^2-L^3),& 0\&le;x\&le;L\\ \frac{-\mathrm{qL}}{448\mathrm{EI}}(15x^2-36\mathrm{Lx}+{19L}^2),& L<x\&le;2L\\ \frac{-q}{1344\mathrm{EI}}(224x^3-1683{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+4116L^{2}x-3271L^3),& 2L<x\&le;3L\\ \frac{11\mathrm{qL}}{1344\mathrm{EI}}(3x^2-24\mathrm{Lx}+{47L}^2),& 3L<x\&le;4L\end{array}\right.$

Deflection functions: $v=\left\{\begin{array}{cc}\frac{\mathrm{qLx}(x-L)(x+L)}{448\mathrm{EI}},& 0\&le;x\&le;L\\ \frac{-\mathrm{qL}(x-L)(x-2L)(5x-3L)}{448\mathrm{EI}},& L<x\&le;2L\\ \frac{-q(x-2L)(x-3L)({56x}^2-281\mathrm{Lx}+317L^2)}{1344\mathrm{EI}},& 2L<x\&le;3L\\ \frac{11\mathrm{qL}(x-3L)(x-4L)(x-5L)}{1344\mathrm{EI}},& 3L<x\&le;4L\end{array}\right.$

The 5th step, with the shear diagram that computing machine draws, moment curve, corner figure and amount of deflection figure,

Fig. 8 is the shear diagram that draws with computing machine in the present embodiment.

Fig. 9 is the moment curve that draws with computing machine in the present embodiment.

Figure 10 is the corner figure that draws with computing machine in the present embodiment 2.

Figure 11 is the amount of deflection figure that draws with computing machine in the present embodiment 2.

The 6th goes on foot, and asks the maximal value of shearing maximal value, moment of flexure maximal value, corner maximal value and amount of deflection:

The shearing maximal value: $F_{S^{,}\max }=0.5045\mathrm{qL}(x=2L),F_{S,\max }^{(-)}=0.4955\mathrm{qL}(x=3L);$

The moment of flexure maximal value: $M_{\max }=0.07367q{L}^2(x=1.504L),M_{\max }^{(-)}=0.05357{\mathrm{qL}}^2(x=2L);$

The corner maximal value: ${\mathrm{\&theta;}}_{\max }=0.01900\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=2.888L),{\mathrm{\&theta;}}_{\max }^{(-)}=0.01871\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=2.121L);$

The amount of deflection maximal value: $v_{\max }=0.003150\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=3.423L),v_{\max }^{(-)}=0.006604\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=2.503L);$

Support constraint forces: $F_{\mathrm{Ax}}=0,F_{\mathrm{Ay}}=\frac{3\mathrm{qL}}{224},F_B=\frac{-9\mathrm{qL}}{112},F_C=\frac{4\mathrm{<figure-callout\; id="7"\; label="qL"\; filenames="HDA00003483994000031.png,HDA00003483994000041.png"\; state="\{\{state\}\}">qL</figure-callout>}}{7},F_D=\frac{61\mathrm{qL}}{112},F_E=\frac{-11\mathrm{qL}}{224}\text{.}$

In the 7th step, press the intensity that following formula is checked beam:

${\mathrm{\&tau;}}_{\max }=\frac{F_{s\max }{S}_z^{*}}{\mathrm{Ib}}\&le;\&lsqb;\mathrm{\&tau;}\&rsqb;,{\mathrm{\&sigma;}}_{\max }=\frac{M_{\max }{y}_{\max }}{I}\&le;\&lsqb;\mathrm{\&sigma;}\&rsqb;---\left(16\right)$

${\mathrm{\&tau;}}_{\max }=\frac{F_{s\max }{S}_z^{*}}{\mathrm{Ib}}=\frac{3}{2}\frac{F_{s\max }}{A}=\frac{3}{2}\frac{0.5045\mathrm{qL}}{A}=0.2649\mathrm{MPa}<\&lsqb;\mathrm{\&tau;}\&rsqb;=2\mathrm{MPa},$ Satisfy the bending shear stress requirement of strength;

${\mathrm{\&sigma;}}_{\max }=\frac{M_{\max }{y}_{\max }}{I}=\frac{0.07367{\mathrm{qL}}^2\mathrm{<figure-callout\; id="2I"\; label="h"\; filenames="HDA00003483994000051.png"\; state="\{\{state\}\}">h</figure-callout>}}{2I}=4.943\mathrm{MPa}>\&lsqb;\mathrm{\&sigma;}\&rsqb;=4\mathrm{MPa},$

The bending normal stresses requirement of strength.

In the 8th step, press the rigidity that following formula is checked beam:

θ _max≤[θ]，v _max≤[v] (17)

${\mathrm{\&theta;}}_{\max }=0.01900\frac{{\mathrm{qL}}^3}{\mathrm{EI}}=0.003542\mathrm{rad}<\&lsqb;\mathrm{\&theta;}\&rsqb;=0.005\mathrm{rad},$ Satisfy the rotation stiffness requirement;

$v_{\max }^{(-)}=0.006604\frac{{\mathrm{qL}}^4}{\mathrm{EI}}=0.08618m>\&lsqb;v\&rsqb;=\frac{L}{1000}=0.07m,$

The amount of deflection rigidity requirement.

In order to improve the strength and stiffness of beam, advise improved design proposal as shown in figure 12, specifically discussed by example 3.

Above-mentioned steps all adopts software Maple and Computer Processing.

Embodiment 3

Figure 12 is Longspan Bridge structural loads figure in the present embodiment.Be that with the difference of the Longspan Bridge structural loads figure shown in Figure 7 of embodiment 2 Longspan Bridge structural loads figure adds a rope at the 3rd mid point of striding in the present embodiment.

The diastrophic disposal route of Longspan Bridge shown in Figure 12 is processed the segmentation of the beam of this Longspan Bridge Longspan Bridge as the n=4 section.

Knownly add a rope, elastic modulus E at the 3rd mid point of striding ₁, area A ₁, length is H, wherein E ₁=10E, H=L/5, A ₁=10I/L ², other condition utilizes computing machine again to find the solution with embodiment 2:

(i) draw shear diagram, moment curve, corner figure and amount of deflection figure;

(ii) check the intensity of this beam;

(iii) check the rigidity of this beam;

(iv) consider the symmetry of structure, propose improved design project.

Step is as follows:

The first step is carried out segmentation with the beam of Longspan Bridge

The beam of Longspan Bridge is divided into 5 sections, sets up independently Fourth-Derivative deflection differential equation:

$\frac{d^4{\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_1}{{\mathrm{dx}}^4}=0,(0\&le;x\&le;L)---\left(18a\right)$

$\frac{d^4{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">v</figure-callout>}}_2}{{\mathrm{dx}}^4}=0,(L\&le;x\&le;2L)---\left(18b\right)$

$\frac{d^4{v}_3}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(2L\&le;x\&le;\frac{5L}{2})---\left(18c\right)$

$\frac{d^4{v}_4}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}\&le;x\&le;3L)---\left(18d\right)$

$\frac{d^4{\mathrm{<figure-callout\; id="5"\; label="v"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_5}{{\mathrm{dx}}^4}=0,(3L\&le;x\&le;4L)---\left(18e\right)$

Second step: the line of deflection approximate differential equation to (18) each section of formula is distinguished integration four times, obtains the general solution of shearing, moment of flexure, corner and amount of deflection.In general solution, include 20 integration constant C _i(i=1,2 ..., 20).

The 3rd step: utilize following displacement boundary conditions, force boundary condition and the condition of continuity

v ₁(0)=0，EIv ₁′′(0)=0 (19a)

v ₁(L)=0，v ₂(L)=0 (19b)

v ₁′(L)=v′ ₂(L)，EIv ₁′′(L)=EIv ₂″(L) (19c)

v ₂(2L)=0，v ₃(2L)=0 (19d)

v ₂′(2L)=v ₃′(2L)，EIv ₂′′(2L)=EIv ₃′′(2L) (19e)

v ₃(5L/2)=v ₄(5L/2)，v ₃′(5L/2)=v ₄′(5L/2) (19f)

EIv ₃′′(5L/2)=EIv ₄′′(5L/2)，EIv ₃′′′(5L/2)=EIv ₄′′′(5L/2) (19g)

v ₄(3L)=0，v ₅(3L)=0 (19h)

v ₄′(3L)=v ₅′(3L)，EIv ₄′′(3L)=EIv ₅′′(3L) (19i)

v ₅(4L)=0，EIv ₅′′(4L)=0 (19j)

$v_3(5L/2)=-\frac{F_{N}H}{E_1{A}_1}---\left(19k\right)$

Simultaneous solution system of equations (19) formula draws 20 integration constant C _i(i=1,2 ..., 20) and an axle power F _N

The 4th step is with integration constant C _i(i=1,2 ..., 20) and substitution general solution expression formula, just can obtain the analytical expression of shearing, moment of flexure, corner and amount of deflection;

The shearing function: ${\mathrm{<figure-callout\; id="1"\; label="F\; S"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=\frac{951\mathrm{qL}}{284008},0\&le;x\&le;L,$

${\mathrm{<figure-callout\; id="2"\; label="F\; S"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=-\frac{4755\mathrm{qL}}{284008},L\&le;x\&le;2L,$

$F_{S3}=-\frac{q}{284008}(284008x-639337L),2L<x\&le;\frac{5}{2}L,$

$F_{S4}=-\frac{q}{284008}(284008x-781337L),\frac{5}{2}L<x\&le;3L,$

${\mathrm{<figure-callout\; id="5"\; label="F\; S"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=\frac{3487\mathrm{qL}}{284008},3L<x\&le;4L.$

Bending moment functions: ${\mathrm{<figure-callout\; id="1"\; label="M"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">M</figure-callout>}}_1=\frac{951\mathrm{qLx}}{284008},0\&le;x\&le;L,$

${\mathrm{<figure-callout\; id="2"\; label="M"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">M</figure-callout>}}_2=-\frac{951\mathrm{qL}}{284008}(5x-6L),L<x\&le;2L,$

$M_3=-\frac{q}{284008}(142004x^2-639337\mathrm{Lx}+714462L^2),2L<x\&le;\frac{5}{2}L,$

$M_4=-\frac{q}{284008}(142004x^2-781337\mathrm{Lx}+1069462L^2),\frac{5}{2}L<x\&le;3L,$

${\mathrm{<figure-callout\; id="5"\; label="M"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">M</figure-callout>}}_5=\frac{3487\mathrm{qL}}{284008}(x-4L),3L<x\&le;4L$

The corner function: ${\mathrm{\&theta;}}_1=\frac{317\mathrm{qL}}{568016\mathrm{EI}}({3x}^2-L^2),0\&le;x\&le;L,$

${\mathrm{\&theta;}}_2=-\frac{317\mathrm{qL}}{568016\mathrm{EI}}({15x}^2-36\mathrm{Lx}+19L^2),L<x\&le;2L,$

${\mathrm{\&theta;}}_3=-\frac{q}{170408\mathrm{EI}}(284008x^3-1918011L{x}^2+4286772L^{2}x-3166907L^3),2L<x\&le;\frac{5}{2}L,$

${\mathrm{\&theta;}}_4=-\frac{q}{170408\mathrm{EI}}(284008x^3-2344011L{x}^2+6416772L^{2}x-5829407L^3),\frac{5}{2}L<x\&le;3L,$

${\mathrm{\&theta;}}_5=\frac{3487\mathrm{qL}}{1704048\mathrm{EI}}(3x^2-24Lx+{47L}^2),3L<x\&le;4L.$

Deflection functions: ${\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_1=\frac{317\mathrm{qLx}(x-L)(x+L)}{568016\mathrm{EI}},0\&le;x\&le;L$

${\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">v</figure-callout>}}_2=-\frac{317\mathrm{qL}}{568016\mathrm{EI}}(x-L)(x-2L)(5x-3L),L<x\&le;2L,$

$v_3=-\frac{q(x-2L)}{1704048\mathrm{EI}}(71002x^3-497333{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+1148720L^{2}x-869467L^3),2L<x\&le;\frac{5}{2}L,$

$v_4=-\frac{q(x-3L)}{1704048\mathrm{EI}}(71002x^3-568331{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+1503393L^{2}x+1319228L^3),\frac{5}{2}L<x\&le;3L,$

${\mathrm{<figure-callout\; id="5"\; label="v"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_5=\frac{3487\mathrm{qL}}{1704048\mathrm{EI}}(x-3L)(x-4L)(x-5L),3L<x\&le;4L.$

The 5th step, with the shear diagram that computing machine draws, moment curve, corner figure and amount of deflection figure,

Figure 13 is the shear diagram that draws with computing machine in the present embodiment.

Figure 14 is the moment curve that draws with computing machine in the present embodiment.

Figure 15 is the corner figure that draws with computing machine in the present embodiment.

Figure 16 is the amount of deflection figure that draws with computing machine in the present embodiment.

The 6th goes on foot, and asks the maximal value of shearing maximal value, moment of flexure maximal value, corner maximal value and amount of deflection:

The shearing maximal value: $F_{S^{,}\max }=0.2511\mathrm{qL}(x=2L),F_{S,\max }^{(-)}=0.2489\mathrm{qL}(x=3L);$

The moment of flexure maximal value: $M_{\max }=0.01870{\mathrm{qL}}^2(x=2.751L),M_{\max }^{(-)}=0.01339{\mathrm{qL}}^2(x=2L);$

The corner maximal value: ${\mathrm{\&theta;}}_{\max }=0.004419\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=2.944L),{\mathrm{\&theta;}}_{\max }^{(-)}=0.004294\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=2.061L);$ The amount of deflection maximal value: $v_{\max }=0.0007876\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=3.423L),v_{\max }^{(-)}=0.001202\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=2.626L);$ Support constraint forces: F _Ax=0, F _Ay=0.003348qL, F _B=-0.02009qL,

F _C=0.2679qL，F _D=0.5000qL，F _E=1.249qL。

Suo Li: $F_N=\frac{17750\mathrm{qL}}{35501}.$

The 7th step, the intensity of check beam:

${\mathrm{\&sigma;}}_{\max }=\frac{M_{\max }{y}_{\max }}{I}=\frac{0.01870q{L}^2\mathrm{<figure-callout\; id="2I"\; label="h"\; filenames="HDA00003483994000051.png"\; state="\{\{state\}\}">h</figure-callout>}}{2I}=1.255\mathrm{MPa}<\&lsqb;\mathrm{\&sigma;}\&rsqb;=4\mathrm{MPa},$ Satisfied the bending strength requirement.

In the 8th step, press the rigidity that following formula is checked beam:

$v_{\max }^{(-)}=0.001202\frac{{\mathrm{qL}}^4}{\mathrm{EI}}=0.01568m<\&lsqb;v\&rsqb;=\frac{L}{1000}=0.07m,$ Satisfied rigidity requirement.

Consider symmetry, advise improved design proposal as shown in figure 17, specifically discussed by example 4.

Above-mentioned steps all adopts software Maple and Computer Processing.

Embodiment 4

Figure 17 is Longspan Bridge structural loads figure in the present embodiment.

Be that with the difference of the Longspan Bridge structural loads figure shown in Figure 17 of embodiment 2 Longspan Bridge structural loads figure in the present embodiment adds an identical rope at each mid point of striding.

The diastrophic disposal route of Longspan Bridge shown in Figure 17 is processed the segmentation of the beam of this Longspan Bridge Longspan Bridge as the n=4 section.

Longspan Bridge shown in Figure 17 has the span L of four equal lengths, and elastic modulus is E, and moment of inertia is I, two fixed ends, and whole beam loads distributed load q, adds an identical rope, known E at each mid point of striding ₁=10E, H=L/10, A ₁=100I/L ², utilize computer solving:

(i) draw shear diagram, moment curve, corner figure and amount of deflection figure;

(ii) if known q=50kN/m, L=70m, square-section deck-molding h=2.191m, wide b=4.564m, I=4m ⁴, A=10m ², [τ]=2MPa, [σ]=4MPa checks the intensity of this structure;

(iii) if known E=23GPa, [θ]=0.005rad,

Check the rigidity of this structure.

Step is as follows:

The first step is carried out segmentation with the beam of Longspan Bridge

The beam of Longspan Bridge is divided into 8 sections, sets up independently Fourth-Derivative deflection differential equation:

$\frac{d^4{\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_1}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(0\&le;x\&le;\frac{L}{2})---\left(20a\right)$

$\frac{d^4{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">v</figure-callout>}}_2}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}\&le;x\&le;L)---\left(20b\right)$

$\frac{d^4{v}_3}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(L\&le;x\&le;\frac{3L}{2})---\left(20c\right)$

$\frac{d^4{v}_4}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}\&le;x\&le;2L)---\left(20d\right)$

$\frac{d^4{\mathrm{<figure-callout\; id="5"\; label="v"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_5}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(2L\&le;x\&le;\frac{5L}{2})---\left(20e\right)$

$\frac{d^4{v}_6}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}\&le;x\&le;3L)---\left(20f\right)$

$\frac{d^4{\mathrm{<figure-callout\; id="7"\; label="v"\; filenames="HDA00003483994000031.png,HDA00003483994000041.png"\; state="\{\{state\}\}">v</figure-callout>}}_7}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(3L\&le;x\&le;\frac{7L}{2})---\left(20g\right)$

$\frac{d^4{\mathrm{<figure-callout\; id="8"\; label="v"\; filenames="HDA00003483994000054.png"\; state="\{\{state\}\}">v</figure-callout>}}_8}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}\&le;x\&le;4L)---\left(20h\right)$

Second step: the line of deflection approximate differential equation to (20) each section of formula is distinguished integration four times, obtains the general solution of shearing, moment of flexure, corner and amount of deflection.In general solution, include 32 integration constant C _i(i=1,2 ..., 32).

The 3rd step: utilize following displacement boundary conditions, force boundary condition and the condition of continuity

v ₁(0)=0，v ₁(0)=0 (21a)

v ₁(L/2)=v ₂(L/2)，v ₁′(L/2)=v ₂′(L/2) (21b)

EIv ₁′′(L/2)=EIv ₂′′(L/2)，EIv ₁′′′(L/2)=EIv ₂′′′(L/2) (21c)

v ₂(L)=0，v ₃(L)=0 (21d)

v ₂′(L)=v ₃′(L)，EIv ₂′′(L)=EIv ₃′′(L) (21e)

v ₃(3L/2)=v ₄(3L/2)’v ₃′(3L/2)=v ₄′(3L/2) (21f)

EIv ₃′′(3L/2)=EIv ₄′′(3L/2)’EIv ₃′′′(3L/2)=EIv ₄′′′(3L/2) (21g)

v ₄(2L)=0’v ₅(2L)=0 (21h)

v ₄′(2L)=v ₅′(2L)’EIv ₄′′(2L)=EIv ₅′′(2L) (21i)

v ₅(5L/2)=v ₆(5L/2)’v ₅′(5L/2)=v ₆′(5L/2) (21j)

EIv ₅′′(5L/2)=EIv ₆′′(5L/2)’EIv ₅′′′(5L/2)=EIv ₆′′′(5L/2) (21k)

v ₆(3L)=0’v ₇(3L)=0 (21l)

v ₆′(3L)=v ₇′(3L)’EIv ₆′′(3L)=EIv ₇′′(3L) (21m)

v ₇(7L/2)=v ₈(7L/2)’v ₇′(7L/2)=v ₈′(7L/2) (21n)

EIv ₇″(7L/2)=EIv ₈′′(7L/2)’EIv ₇′′′(7L/2)=EIv ₈′′′(7L/2) (21o)

v ₈(4L)=0，v ₈′(4L)=0 (21p)

$v_1(L/2)=-\frac{F_{N1}{H}^{,}}{E_1{A}_1}{v}_3(3L/2)=-\frac{F_{N2}H}{E_1{A}_1}---\left(21q\right)$

$v_5(5L/2)=-\frac{F_{N3}{H}^{,}}{E_1{A}_1}{v}_7(7L/2)=-\frac{F_{N4}H}{E_1{A}_1}---\left(21r\right)$

Simultaneous solution system of equations (21) formula draws 32 integration constant C _i(i=1,2 ..., 32) and four axle power F _Nk(k=1,2 ..., 4).

The 4th step is with integration constant C _i(i=1,2 ..., 32) and substitution general solution expression formula, just can obtain the analytical expression of shearing, moment of flexure, corner and amount of deflection;

The shearing function: ${\mathrm{<figure-callout\; id="1"\; label="F\; S"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=-\frac{q}{2548}(2548x-649L),0\&le;x\&le;\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{<figure-callout\; id="2"\; label="F\; S"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=-\frac{q}{2548}(2548x-1899L),\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;L,$

$F_{S3}=-\frac{q}{2548}(2548x-3197L),L<x\&le;\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

$F_{S4}=-\frac{q}{2548}(2548x-4447L),\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;2L,$

${\mathrm{<figure-callout\; id="5"\; label="F\; S"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=-\frac{q}{2548}(2548x-5745L),2L<x\&le;\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

$F_{S6}=-\frac{q}{2548}(2548x-6995L),\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;3L,$

${\mathrm{<figure-callout\; id="7"\; label="F\; S"\; filenames="HDA00003483994000031.png,HDA00003483994000041.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=-\frac{q}{2548}(2548x-8293L),3L<x\&le;\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{<figure-callout\; id="8"\; label="F\; S"\; filenames="HDA00003483994000054.png"\; state="\{\{state\}\}">F</figure-callout>}}_S=-\frac{q}{2548}(2548x-9543L),\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;4L.$

Bending moment functions: ${\mathrm{<figure-callout\; id="1"\; label="M"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">M</figure-callout>}}_1=-\frac{q}{30576}({15288x}^2-7788\mathrm{Lx}+673L^2),0\&le;x\&le;\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{<figure-callout\; id="2"\; label="M"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">M</figure-callout>}}_2=-\frac{q}{30576}({15288x}^2-22788\mathrm{Lx}+8173L^2),\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;L,$

$M_3=-\frac{q}{30576}({15288x}^2-38364\mathrm{Lx}+23749L^2),L<x\&le;\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

$M_4=-\frac{q}{30576}({15288x}^2-53364\mathrm{Lx}+46249L^2),\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;2L,$

${\mathrm{<figure-callout\; id="5"\; label="M"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">M</figure-callout>}}_5=-\frac{q}{30576}({15288x}^2-68940\mathrm{Lx}+77401L^2),2L<x\&le;\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

$M_6=-\frac{q}{30576}({15288x}^2-83940\mathrm{Lx}+114901L^2),\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;3L,$

${\mathrm{<figure-callout\; id="7"\; label="M"\; filenames="HDA00003483994000031.png,HDA00003483994000041.png"\; state="\{\{state\}\}">M</figure-callout>}}_7=-\frac{q}{30576}({15288x}^2-99516\mathrm{Lx}+161629L^2),3L<x\&le;\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{<figure-callout\; id="8"\; label="M"\; filenames="HDA00003483994000054.png"\; state="\{\{state\}\}">M</figure-callout>}}_8=-\frac{q}{30576}({15288x}^2-114516\mathrm{Lx}+214129L^2),\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;4L$

The corner function: ${\mathrm{\&theta;}}_1=-\frac{\mathrm{qx}}{30576\mathrm{EI}}(2x-L)(2548x-673L),0\&le;x\&le;\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{\&theta;}}_2=-\frac{q}{30576\mathrm{EI}}(x-L)(2x-L)(2548x-1875L),\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;L,$

${\mathrm{\&theta;}}_3=-\frac{q}{30576\mathrm{EI}}(x-L)(2x-L)(2548x-3221L),L<x\&le;\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{\&theta;}}_4=-\frac{q}{30576\mathrm{EI}}(x-2L)(2x-3L)(2548x-4423L),\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;2L,$

${\mathrm{\&theta;}}_5=-\frac{q}{30576\mathrm{EI}}(x-2L)(2x-5L)(2548x-5769L),2L<x\&le;\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}.$

${\mathrm{\&theta;}}_6=-\frac{q}{30576\mathrm{EI}}(x-3L)(2x-5L)(2548x-6971L),\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;3L,$

${\mathrm{\&theta;}}_7=-\frac{q}{30576\mathrm{EI}}(x-3L)(2x-7L)(2548x-8317L),3L<x\&le;\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{\&theta;}}_8=-\frac{q}{30576\mathrm{EI}}(x-4L)(2x-7L)(2548x-9519L),\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;4L.$

Deflection functions: ${\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_1=\frac{{\mathrm{<figure-callout\; id="2"\; label="qx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">qx</figure-callout>}}^2}{61152\mathrm{EI}}({2548x}^2-25946\mathrm{Lx}+{673L}^2),0\&le;x\&le;\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}$

${\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">v</figure-callout>}}_2=\frac{{q(x-L)}^2}{61152\mathrm{EI}}({2548x}^2-2500\mathrm{Lx}+{625L}^2),\frac{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;L,$

$v_3=\frac{{q(x-L)}^2}{61152\mathrm{EI}}({2548x}^2-7692\mathrm{Lx}+{5817L}^2),L<x\&le;\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

$v_4=-\frac{{q(x-2L)}^2}{61152\mathrm{EI}}({2548x}^2-7596\mathrm{Lx}+{5673L}^2),\frac{3\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;2L,$

${\mathrm{<figure-callout\; id="5"\; label="v"\; filenames="HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_5=-\frac{{q(x-2L)}^2}{61152\mathrm{EI}}({2548x}^2-12788\mathrm{Lx}+{16057L}^2),2L<x\&le;\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

$v_6=-\frac{{q(x-3L)}^2}{61152\mathrm{EI}}({2548x}^2-12692\mathrm{Lx}+{15817L}^2),\frac{5\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;3L,$

${\mathrm{<figure-callout\; id="7"\; label="v"\; filenames="HDA00003483994000031.png,HDA00003483994000041.png"\; state="\{\{state\}\}">v</figure-callout>}}_7=-\frac{{q(x-3L)}^2}{61152\mathrm{EI}}({2548x}^2-17884\mathrm{Lx}+{31393L}^2),3L<x\&le;\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2},$

${\mathrm{<figure-callout\; id="8"\; label="v"\; filenames="HDA00003483994000054.png"\; state="\{\{state\}\}">v</figure-callout>}}_8=-\frac{{q(x-4L)}^2}{61152\mathrm{EI}}({2548x}^2-17788\mathrm{Lx}+{31507L}^2),\frac{7\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">L</figure-callout>}}{2}<x\&le;4L.$

The 5th step, with the shear diagram that computing machine draws, moment curve, corner figure and amount of deflection figure,

Figure 18 is the shear diagram that draws with computing machine in the present embodiment,

Figure 19 is the moment curve that draws with computing machine in the present embodiment,

Figure 20 is the corner figure that draws with computing machine in the present embodiment,

Figure 21 is the amount of deflection figure that draws with computing machine in the present embodiment;

The 6th goes on foot, and asks the maximal value of shearing maximal value, moment of flexure maximal value, corner maximal value and amount of deflection:

The shearing maximal value: $F_{S^{,}\max }=0.2547\mathrm{qL}(x=0)0,F_{S,\max }^{(-)}=0.2547\mathrm{qL}(x=4L);$

The moment of flexure maximal value: $M_{\max }=0.01043q{L}^2(x=0.2547L),M_{\max }^{(-)}=0.02201{\mathrm{qL}}^2(x=0);$

The corner maximal value: ${\mathrm{\&theta;}}_{\max }=0.001102\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=3.890L),{\mathrm{\&theta;}}_{\max }^{(-)}=0.001102\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=0.1103L);$ The amount of deflection maximal value: $v_{\max }^{(-)}=0.0001883\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=0.2641L);$

Support constraint forces: $F_{\mathrm{Ax}}=0,F_{\mathrm{Ay}}=\frac{649\mathrm{qL}}{2548},M_A=\frac{{673\mathrm{<figure-callout\; id="2"\; label="qL"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">qL</figure-callout>}}^2}{30576}{F}_B=\frac{649\mathrm{qL}}{2548},F_C=\frac{649\mathrm{qL}}{2548},$

$F_D=\frac{649\mathrm{qL}}{2548},F_{\mathrm{Ex}}=0,F_{\mathrm{Ey}}=\frac{649\mathrm{qL}}{2548},M_E=-\frac{\mathrm{<figure-callout\; id="2"\; label="673\; qL"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">673</figure-callout>}{\mathrm{qL}}^{}{}^2}{30576}.$

Suo Li: ${\mathrm{<figure-callout\; id="1"\; label="F\; N"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">F</figure-callout>}}_N=\frac{625\mathrm{qL}}{1274},{\mathrm{<figure-callout\; id="2"\; label="F\; N"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">F</figure-callout>}}_N=\frac{625\mathrm{qL}}{1274},F_{N3}=\frac{625\mathrm{qL}}{1274},F_{N4}=\frac{625\mathrm{qL}}{1274}$

The 7th step, the intensity of check beam:

${\mathrm{\&tau;}}_{\max }=\frac{F_{s\max }{S}_z^{*}}{\mathrm{Ib}}=\frac{3}{2}\frac{F_{s\max }}{A}=\frac{3}{2}\frac{0.2547\mathrm{qL}}{A}=0.1337\mathrm{MPa}<\&lsqb;\mathrm{\&tau;}\&rsqb;=2\mathrm{MPa},$ Satisfy the bending shear stress requirement of strength;

${\mathrm{\&sigma;}}_{\max }=\frac{M_{\max }^{(-)}{y}_{\max }}{I}=\frac{0.02201{\mathrm{qL}}^2\mathrm{<figure-callout\; id="2I"\; label="h"\; filenames="HDA00003483994000051.png"\; state="\{\{state\}\}">h</figure-callout>}}{2I}=1.477\mathrm{MPa}<\&lsqb;\mathrm{\&sigma;}\&rsqb;=4\mathrm{MPa},$ Satisfy the bending normal stresses requirement of strength.

The 8th step, the rigidity of check beam:

${\mathrm{\&theta;}}_{\max }=0.001102\frac{{\mathrm{qL}}^3}{\mathrm{EI}}=0.0002054\mathrm{rad}<\&lsqb;\mathrm{\&theta;}\&rsqb;=0.005\mathrm{rad},$ Satisfy the rotation stiffness requirement;

$v_{\max }^{(-)}=0.0001883\frac{{\mathrm{qL}}^4}{\mathrm{EI}}=0.002457m<\&lsqb;v\&rsqb;=\frac{L}{1000}=0.07m,$ Satisfy the amount of deflection rigidity requirement.

Above-mentioned steps all adopts software Maple and Computer Processing.

Embodiment 5

The segmentation of the beam of the present embodiment Longspan Bridge Longspan Bridge is processed as the n=50 section.

Known the present embodiment Longspan Bridge has the span L of 50 equal lengths, and elastic modulus is E, and moment of inertia is I, pin-ended, and whole beam loads distributed load q, utilizes computer solving:

(i) analytical expression of shearing, moment of flexure, corner and amount of deflection;

(ii) determine maximum shear, maximal bending moment, hard-over and maximum defluxion;

(iii) determine each support constraint forces.

Step is as follows:

The first step is carried out segmentation with the beam of Longspan Bridge

The beam of Longspan Bridge is divided into 50 sections, sets up independently Fourth-Derivative deflection differential equation:

$\frac{d^4{v}_k}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(k=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,50)$

Second step: the line of deflection approximate differential equation to (22) each section of formula is distinguished integration four times, obtains the general solution of shearing, moment of flexure, corner and amount of deflection.In general solution, include 200 integration constant C _i(i=1,2 ..., 200).

The 3rd step: utilize following displacement boundary conditions, force boundary condition and the condition of continuity

v ₁(0)=0，EIv ₁′′(0)=0 (23a)

v _k(kL)=0，v _k(kL)=0(k=1,2,…,49) (23b)

v _k′(kL)=v _k+1′(kL)，EIv _k′′(kL)=EIv _k+1″(kL)(k=1,2,…,49) (23c)

v ₅₀(50L)=0，EIv ₅₀′′(50L)=0 (23d)

Simultaneous solution system of equations (23) formula draws 200 integration constant C _i(i=1,2 ..., 200).

The 4th step is with integration constant C _i(i=1,2 ..., 200) and substitution general solution expression formula, just can obtain the analytical expression of shearing, moment of flexure, corner and amount of deflection;

Shearing function: F _S1=-qx+0.3943qL, 0≤x≤L,

F _S2=-qx+1.528qL,L<x≤2L,

…………………………………………

F _S25=-qx+24.50qL,24L<x≤25L

F _S26=-qx+25.50qL,25L<x≤26L,

…………………………………………

F _S50=-qx+49.61qL,49L<x≤50L。

Bending moment functions: M ₁=-0.5qx ²+ 0.3943qLx, 0≤x≤L,

M ₂=-0.5qx ²+1.528qLx-1.134qL ²,L<x≤2L，

…………………………………………

M ₂₅=-0.5qx ²+24.50qLx-300.1qL ²,24L<x≤25L，

M ₂₆=-0.5qx ²+25.50qLx-325.1qL ²,25L<x≤26L，

…………………………………………

M ₅₀=-0.5qx ²+49.61qLx-1230qL ²,49L<x≤50L

The corner function: ${\mathrm{\&theta;}}_1=-\frac{q}{\mathrm{EI}}(0.1667x^3-0.1972L{x}^2+0.02406L^3),0\&le;x\&le;L$

${\mathrm{\&theta;}}_2=-\frac{q}{\mathrm{EI}}(0.1667x^3-0.7342L{x}^2+1.134L^{2}x-0.5429L^3),L<x\&le;2L$

…………………………………………

${\mathrm{\&theta;}}_{25}=-\frac{q}{\mathrm{EI}}(0.1667x^3-12.25L{x}^2+300.1L^{2}x-2450L^3),24L<x\&le;25L$

${\mathrm{\&theta;}}_{26}=-\frac{q}{\mathrm{EI}}(0.1667x^3-12.75L{x}^2+325.1L^{2}x-2762L^3),25L<x\&le;26L$

…………………………………………

${\mathrm{\&theta;}}_{50}=-\frac{q}{\mathrm{EI}}(0.1667x^3-24.80L{x}^2+1230L^{2}x-20340L^3),49L<x\&le;50L$

Deflection functions: ${\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_1=-\frac{\mathrm{qx}}{\mathrm{EI}}(0.04167x^3-0.06572L{x}^2+0.02406L^3),0\&le;x\&le;L$

${\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">v</figure-callout>}}_2=-\frac{q}{\mathrm{EI}}(0.04167x^4-0.2547L{x}^3+0.5670L^2{x}^2-0.5429L^{3}x+0.1890L^4),L\&le;x\&le;2L$

…………………………………………

$v_{25}=-\frac{q}{\mathrm{EI}}(0.04167x^4-4.083L{x}^3+150.0L^2{x}^2-2450L^{3}x+15000L^4),24L\&le;x\&le;25L$

$v_{26}=-\frac{q}{\mathrm{EI}}(0.04167x^4-4.250L{x}^3+162.5L^2{x}^2-2762L^{3}x+17600L^4),25L\&le;x\&le;26L$

…………………………………………

$v_{50}=-\frac{q}{\mathrm{EI}}(0.04167x^4-8.268L{x}^3+{615.1L}^2{x}^2-20340L^{3}x+252200L^4),49L<x\&le;50L$

The 5th goes on foot, and asks the maximal value of shearing maximal value, moment of flexure maximal value, corner maximal value and amount of deflection:

The shearing maximal value: $F_{S^{,}\max }=0.528\mathrm{qL}(x=L),F_{S,\max S^{,}\max }^{(-)}=0.6057\mathrm{qL}(x=L);$

The moment of flexure maximal value: $M_{\max }=0.007774q{L}^2(x=0.3943L),M_{\max }^{(-)}=0.1057q{L}^2(x=L);$

The corner maximal value: ${\mathrm{\&theta;}}_{\max }=0.02406\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=50L),{\mathrm{\&theta;}}_{\max }^{(-)}=0.02406\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=0);$

The amount of deflection maximal value: $v_{\max }=0.0001424(x=1.073),v_{\max }^{(-)}=0.006550\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=0.4411L);$

In the 5th step, ask support constraint forces:

F _A1x=0，F _A1y=0.3943qL，

F _A2=1.134qL，

…………………………………………

F _A25=qL，

F _A26=qL，

…………………………………………

F _A50=0.3943qL。

Above-mentioned steps all adopts software Maple and Computer Processing.

Embodiment 6

The segmentation of the beam of the present embodiment Longspan Bridge Longspan Bridge is processed as the n=100 section.

Known the present embodiment Longspan Bridge has the span L of 100 equal lengths, and elastic modulus is E, and moment of inertia is I, pin-ended, and whole beam loads distributed load q, utilizes computer solving:

(i) analytical expression of shearing, moment of flexure, corner and amount of deflection;

(ii) determine maximum shear, maximal bending moment, hard-over and maximum defluxion;

(iii) determine each support constraint forces.

Step is as follows:

The first step is carried out segmentation with the beam of Longspan Bridge

The beam of Longspan Bridge is divided into 100 sections, sets up independently Fourth-Derivative deflection differential equation:

$\frac{d^4{v}_k}{{\mathrm{dx}}^4}=-\frac{q}{\mathrm{EI}},(k=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,100)---\left(24\right)$

Second step: the line of deflection approximate differential equation to (24) each section of formula is distinguished integration four times, obtains the general solution of shearing, moment of flexure, corner and amount of deflection.In general solution, include 400 integration constant C _i(i=1,2 ..., 400).

The 3rd step: utilize following displacement boundary conditions, force boundary condition and the condition of continuity

v ₁(0)=0，EIv ₁′′(0)=0 (25a)

v _k(kL)=0，v _k(kL)=0(k=1,2,…,99) (25b)

v′ _k(kL)=v _k+1′(kL)，EIv _k′′(kL)=EIv _k+1′′(kL)(k=1,2,…,99) (25c)

v ₅₀(100L)=0, EIv ₅₀' ' (100L)=0 (25d) simultaneous solution system of equations (25) formula draws 400 integration constant C _i(i=1,2 ..., 400).

The 4th step is with integration constant C _i(i=1,2 ..., 400) and substitution general solution expression formula, just can obtain the analytical expression of shearing, moment of flexure, corner and amount of deflection;

Shearing function: F _S1=-qx+0.3943qL, 0≤x≤L,

F _S2=-qx+1.528qL,L<x≤2L,

…………………………………………

F _S50=-qx+49.50qL,49L<x≤50L

F _S51=-qx+50.50qL,50L<x≤51L,

…………………………………………

F _S100=-qx+99.61qL,99L<x≤100L。

Bending moment functions: M ₁=-0.5qx ²+ 0.3943qLx, 0≤x≤L,

M ₂=-0.5qx ²+1.528qLx-1.134qL ²,L<x≤2L，

…………………………………………

M ₅₀=-0.5qx ²+49.50qLx-1225qL ²,49L<x≤50L，

M ₅₁=-0.5qx ²+50.50qLx-1275qL ²,50L<x≤51L，

…………………………………………

M ₁₀₀=-0.5qx ²+99.61qLx-4961qL ²,99L<x≤100L

The corner function: ${\mathrm{\&theta;}}_1=-\frac{q}{\mathrm{EI}}(0.1667x^3-0.1972{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+0.02406L^3),0\&le;x\&le;L$

${\mathrm{\&theta;}}_2=-\frac{q}{\mathrm{EI}}(0.1667x^3-0.7642{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+1.134L^{2}x-0.5429L^3),L<x\&le;2L$

…………………………………………

${\mathrm{\&theta;}}_{50}=-\frac{q}{\mathrm{EI}}(0.1667x^3-24.75{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+1225L^{2}x-20210L^3),49L<x\&le;50L$

${\mathrm{\&theta;}}_{51}=-\frac{q}{\mathrm{EI}}(0.1667x^3-25.25{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+1275L^{2}x-21460L^3),50L<x\&le;51L$

…………………………………………

${\mathrm{\&theta;}}_{100}=-\frac{q}{\mathrm{EI}}(0.1667x^3-49.80{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+4961L^{2}x-164700L^3),99L<x\&le;100L$

Deflection functions: ${\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="HDA00003483994000024.png,HDA00003483994000025.png"\; state="\{\{state\}\}">v</figure-callout>}}_1=-\frac{\mathrm{qx}}{\mathrm{EI}}(0.04167x^3-0.06572{\mathrm{<figure-callout\; id="2"\; label="Lx"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">Lx</figure-callout>}}^2+0.02406L^3),0\&le;x\&le;L$

${\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="HDA00003483994000022.png,HDA00003483994000024.png"\; state="\{\{state\}\}">v</figure-callout>}}_2=-\frac{q}{\mathrm{EI}}(0.04167x^4-0.2547{\mathrm{Lx}}^3+0.5670L^2{x}^2-0.5429L^{3}x+0.1890L^4),L<x\&le;2L$

…………………………………………

$v_{50}=-\frac{q}{\mathrm{EI}}(0.04167x^4-8.250{\mathrm{Lx}}^3+612.5L^2{x}^2-20210L^{3}x+250100L^4),49L<x\&le;50L$

$v_{51}=-\frac{q}{\mathrm{EI}}(0.04167x^4-8.417{\mathrm{Lx}}^3+637.5L^2{x}^2-21460L^{3}x+270900L^4),50L<x\&le;51L$

…………………………………………

$v_{100}=-\frac{q}{\mathrm{EI}}(0.04167x^4-16.60{\mathrm{Lx}}^3+2480L^2{x}^2-164700L^{3}x+4.101\&times;{10}^6{L}^4),99L<x\&le;100L$

The 5th goes on foot, and asks the maximal value of shearing maximal value, moment of flexure maximal value, corner maximal value and amount of deflection:

The shearing maximal value: $F_{S^{,}\max }=0.528\mathrm{qL}(x=L),F_{S,\max }^{(-)}=0.6057\mathrm{qL}(x=L);$

The moment of flexure maximal value: $M_{\max }=0.007774q{L}^2(x=0.3943L),M_{\max }^{(-)}=0.1057q{L}^2(x=L);$

The corner maximal value: ${\mathrm{\&theta;}}_{\max }=0.02406\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=100L),{\mathrm{\&theta;}}_{\max }^{(-)}=0.02406\frac{{\mathrm{qL}}^3}{\mathrm{EI}}(x=0);$

The amount of deflection maximal value: $v_{\max }=0.0001424(x=1.073),v_{\max }^{(-)}=0.006550\frac{{\mathrm{qL}}^4}{\mathrm{EI}}(x=0.4411L);$

In the 5th step, ask support constraint forces:

F _A1x=0，F _A1y=0.3943qL，

F _A2=1.134qL，

…………………………………………

F _A50=qL，

F _A51=qL，

…………………………………………

F _A100=0.3943qL。

Above-mentioned steps all adopts software Maple and Computer Processing.

Software Maple and computing machine used in above-described embodiment are that those skilled in the art are known.

Claims (3)

1. the diastrophic disposal route of Longspan Bridge is characterized in that: be the independent integrated integral method of a kind of segmentation, step is:

The first step is selected just design proposal, determines structural loads figure;

Second step carries out segmentation with the beam of Longspan Bridge

The beam of Longspan Bridge is divided into n (1≤n≤100) section, sets up independently Fourth-Derivative deflection differential equation:

The 3rd the step, with each section beam respectively integration obtain shearing, moment of flexure, corner and deflection equation general solution for four times

Integration once obtains shearing equation general solution

Integration twice obtains the Bending Moment Equations general solution

Integration three times obtains the equations of rotating angle general solution

Integration four times obtains the deflection equation general solution

In the 4th step, utilize displacement boundary conditions, force boundary condition and the condition of continuity to set up 4n boundary condition equation of constraint

f(C _i,j)＝0(i＝1,2,…,n,j＝1,2,3,4) (6)

Simultaneous solution boundary condition equation of constraint (6) is determined 4n integration constant C _{I, j}(i=1,2 ..., n, j=1,2,3,4);

The 5th step is with integration constant C _{I, j}(i=1,2 ..., n, j=1,2,3,4) and substitution (2)～(5) formula just can obtain the analytical expression of shearing, moment of flexure, corner and amount of deflection;

The 6th step, with the shear diagram that computing machine draws, moment curve, corner figure and amount of deflection figure;

In the 7th step, ask shearing maximal value F _Smax, moment of flexure maximal value M _Max, corner maximal value θ _MaxMaximal value v with amount of deflection _Max

The 8th step is by the strength and stiffness of following formula check beam

Above-mentioned steps all adopts software Maple and Computer Processing.

2. the diastrophic disposal route of said Longspan Bridge according to claim 1 is characterized in that above-mentioned steps all adopts the flow process of Computer Processing to be:

Beginning → selection is design proposal just, determine structural loads figure → beam is divided into the n section, set up independently the Fourth-Derivative deflection differential equation → with each section beam respectively integration obtain shearing four times, moment of flexure, corner and deflection equation general solution → utilize displacement boundary conditions, force boundary condition and the condition of continuity are set up 4n boundary condition equation of constraint → integration constant substitution general formula is obtained shearing, moment of flexure, the analytical expression of corner and amount of deflection → drafting shear diagram, moment curve, corner figure and amount of deflection figure → ask shearing maximal value, the moment of flexure maximal value, whether the strength and stiffness of the maximal value of corner maximal value and amount of deflection → check beam meet the demands

Return and select just design proposal, determine structural loads figure;

Finish.The diastrophic disposal route application documents of Longspan Bridge word format text.

3. the diastrophic disposal route of said Longspan Bridge according to claim 1, it is characterized in that: described beam with Longspan Bridge is divided into the n section, wherein n=1～100.

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