CN103970981B - Container house anti-side rigidity computational methods - Google Patents

Container house anti-side rigidity computational methods Download PDF

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CN103970981B
CN103970981B CN201410012570.2A CN201410012570A CN103970981B CN 103970981 B CN103970981 B CN 103970981B CN 201410012570 A CN201410012570 A CN 201410012570A CN 103970981 B CN103970981 B CN 103970981B
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查晓雄
刘乐
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Abstract

The present invention provides a kind of container house anti-side rigidity computational methods, for calculating the anti-side rigidity of container, comprises the following steps, step a, obtains detrusion and flexural deformation of the container by side plate during side force using the principle of virtual work;Step b, assume the shearing stress distribution of container upper beam and side sheet room, obtain the axial deformation of upper beam;Step c, container both ends deformation caused by being bent downwardly under concentrated force effect is obtained using symmetrical structure and antisymmetry structure;Step d, by the detrusion of the side plate of container, flexural deformation, the axial deformation of upper beam and it is bent downwardly the rigidity that deformation caused by deformation calculates container.The present invention combines container specific constructive form, upper beam axial deformation is added in container sidesway formula, so that container Rigidity Calculation accuracy has very big raising.

Description

Container house anti-side rigidity computational methods
Technical field
The present invention relates to a kind of container house anti-side rigidity computational methods.
Background technology
With the development of building construction, container is used in building structure more and more.In order to ensure container The safety and stablization of building, its anti-side rigidity should just be studied.At present, the research to container rigidity is also less, main It is that container side plate is equivalent into shear wall to want thinking, with method of equivalent brace approximate calculation, or directly uses european norm In calculating covering rigidity method it is approximate.
Wherein, method of equivalent brace is a kind of calculating mould for seeking assembled wall anti-side rigidity of extensive utilization in assembled wall Type, it can obtain the anti-side rigidity of equal value of assembled wall, and can also estimate assembled wall during rigidity is derived Sidesway.But it is built upon on the basis of certain hypothesis, it requires that the connection between each bar is be hinged in simplified model, This is extremely different with the construction of container, and last error is inevitable very big.Furthermore it does not account for container upper beam itself Axial deformation.The covering anti-side rigidity mainly connected on purlin with connector that method in european norm is directed to, but not It is applied to very much calculate container.Its reason has at following 2 points:First, its connected mode is different from container;2nd, because first point former Cause, formula is applied to not consider the situation of axial deformation in specification, is not particularly suited for container Rigidity Calculation.
In summary there is problems with and defect in existing computational methods:Simplified mathematical model and assumed condition are different from collection Vanning computation model, can cause larger calculation error;In container sidesway, axial deformation is very big, is accounted for mainly Position, and above method does not account for the axial deformation of container upper beam.
The content of the invention
In order to overcome the above-mentioned deficiencies of the prior art, the present invention provides a kind of container house anti-side rigidity computational methods, uses Calculate, comprise the following steps in the anti-side rigidity to container:
Step a, detrusion and flexural deformation of the container by side plate during side force are obtained using the principle of virtual work;
Step b, assume the shearing stress distribution of container upper beam and side sheet room, obtain the axial deformation of upper beam;
Step c, container is obtained under concentrated force effect using symmetrical structure and antisymmetry structure to draw due to being bent downwardly The both ends deformation risen;
Step d, by the detrusion of the side plate of container, flexural deformation, the axial deformation of upper beam and it is bent downwardly Deformation calculates the rigidity of container caused by deformation.
Further, in the step a, the detrusion and flexural deformation of the side plate of container meet below equation:
In formula:EyyFor container y directions modulus of elasticity;GeffFor the equivalent shear modulus in x/y plane;I is container lateral Plate is around strong equatorial moment of inertia;P is the concentrated force acted at container angle;H is container height;L is container longitudinal length;
In the step b, the axial deformation of the upper beam of container meets below equation:
In formula:IcThe moment of inertia for corner post of container to itself centre of form axle;A is container upper girder cross-section area.
In the step c, container both ends deformation caused by being bent downwardly meets below equation:
In formula:Is is container side plate short side direction the moment of inertia
Further, for the container in case top applying power P, force end movement and free end meet below equation:
When the bottom girder of the container is fixed:
Exert a force end movement
Free end travel
When the bottom girder of the container is not fixed:
Exert a force end movement
Free end travel
Further, for individual layer m across container structure, when on case top, bottom girder is fixed during applying power P force side sidesway and Force side sidesway formula meets below equation when bottom girder is not fixed:
When bottom girder is fixed:
When bottom girder is not fixed:
N-layer single span container structure, the top displacement Δ at top during applying power Pn,n, meet below equation:
In above formula
Two layers of single span container structure, when pushing up applying power P for one layer, when both ends sidesway and bottom girder are not fixed when bottom girder is fixed Both ends sidesway formula, it is as follows:
When bottom girder is fixed:
When bottom girder is not fixed:
In formula:AUnderFor container underbeam area.
The displacement on n-layer single span container structure n-th layer case top during applying power P at the top of i-th (i≤n-1) layern,i, specifically Formula is as follows:
Wherein
For n-layer m across container structure, at top during applying power P, structural top force side displacement formula is as follows:
For n-layer m across container structure, it is assumed that horizontal concentrated force P mean allocations at each node, then each node The power of place's distribution is P/ (m+1), and in elastic range, every layer of displacement as caused by horizontal force is just Δ01'/(m+1), by its band Enter in n-layer single span structure lateral displacement formula, can obtain structural top displacement is:
I in formulamBe individual layer m across container structure to centre of form equatorial moment of inertia;Δ01' it is that the unfixed individual layer m of bottom girder exists across structure The lower force end movement of power effect in level set.
Further, the bottom girder of the container is fixed small when punching, and is punched when being less than 15% more than 3%, structure Lateral rigidity meets below equation:
Kh=α K
K is the rigidity of FCL in formula;α is the coefficient that punches,hfIt is the total height at hole;hsIt is Remove the residual altitude at hole, hs=h-hf;aiIt is the length at each hole.
Further, bottom girder is fixed middle through when symmetrically punching, and container total displacement meets below equation:
Further, when among when bottom girder is not fixed through symmetrical punch, below equation is met:
Further, non-through symmetrically punches sidesway, and hole top and the bottom equivalent bending stiffness is respectively:
The present invention is based on european norm covering rigidity formula, with reference to container specific constructive form, in container lateral Move and upper beam axial deformation is added in formula, so that container Rigidity Calculation accuracy can be used as the present by very big raising The theoretical foundation of container construction design afterwards.
Brief description of the drawings
Fig. 1 is equivalent perpendicular heterogeneous slab under container side plate local coordinate of the present invention;
Fig. 2 profiled sheet unit x directions force diagram
Fig. 3 is the container bottom girder fixed side deformation pattern in the present invention;
Fig. 4 is container bottom girder in the present invention not fixed side deformation pattern;
The unfixed container of Fig. 5 bottom girders is bent downwardly deformation pattern
Fig. 6 is container upper beam and side sheet room shearing stress distribution figure when the container bottom girder in the present invention is fixed;
Fig. 7 is container upper beam and side sheet room shearing stress distribution figure when the container bottom girder in the present invention is not fixed;
Fig. 8 is the small schematic diagram that punches of the container in the present invention;
Fig. 9 is the frame part schematic diagram of the container in the present invention;
Figure 10 is the Coupled Shear Wall partial schematic diagram of the container in the present invention;
Figure 11 is that the non-through of the container in the present invention symmetrically punches schematic diagram;
Figure 12 is the equivalent rear Coupled Shear Wall schematic diagram of the container in the present invention;
Embodiment
The present invention is further described for explanation and embodiment below in conjunction with the accompanying drawings.
The invention provides a kind of container house anti-side rigidity computational methods, for calculating the rigidity of container. The container side plate of container is equivalent to perpendicular heterogeneous slab by the container house anti-side rigidity computational methods of the present invention, is being counted Flexural deformation and detrusion are not only allowed for when calculating container sidesway, and is considered very big to container stiffness effect Container upper beam axial deformation, Rigidity Calculation basis is provided for container structure design.Container in the present invention is existing There is the container in technology, including the part such as side plate, top plate, bottom plate, upper beam, bottom girder, its structure will not be repeated here.
The container house anti-side rigidity computational methods of the present invention comprise the following steps:
By anisotropic plate it is assumed that obtaining the detrusion and bending when container receives side force using the principle of virtual work Deformation.Container stress end movement is made up of three parts:Axially become caused by detrusion, flexural deformation and container upper beam Shape.Container free end travel is made up of two parts:Detrusion and flexural deformation.
The shearing stress distribution of reasonable assumption container upper beam and side sheet room, the axial direction that upper beam is obtained by direct integral become Shape.
By symmetrical structure and antisymmetry structure, obtain container both ends caused by being bent downwardly using layer control and become Shape.
It is bent downwardly by the detrusion of the side plate of container, flexural deformation, the axial deformation of upper beam and container Caused deformation calculates the rigidity of container.Single Container is fixed for bottom girder, stress end movement is made up of three parts:Shearing Deformation, flexural deformation and axial deformation caused by container upper beam.Free end travel is made up of two parts:Detrusion and curved Song deformation.Single Container is not fixed for bottom girder, stress end movement is made up of four parts:Detrusion, flexural deformation, packaging Axial deformation and container are bent downwardly caused deformation caused by case upper beam.Free end travel is made up of three parts:Shearing Deformation, flexural deformation and container are bent downwardly caused deformation.
Referring to Fig. 1 to Fig. 5, the side plate of container is profiled sheet, and profiled sheet is formed by flat board compacting.Suppress shape Into profiled sheet different directions modulus of elasticity and Poisson's ratio there is very big difference.According to prior art, an elastomer is such as The elasticity of its all directions of fruit is not fully identical, then referred to as anisotropic body.And with mutually orthogonal three elasticity it is main to, Then it is referred to as orthotropic solid.Therefore, profiled sheet can be equivalent to orthotropic plate, and then the performance of profiled sheet is carried out Judge.
Strain-stress relation after profiled sheet equivalent orthotropic plate can be written as shown equation:
In formula:Exx、EyyIt is x, the modulus of elasticity of y-axis under local coordinate;νxy、νyxIt is the Poisson's ratio in x, y direction;GeffIt is Equivalent shear modulus in x/y plane;σx、σy, be respectively equivalent perpendicular heterogeneous slab in level set power effect under x, y direction Direct stress;τxyFor shear stress of the equivalent perpendicular heterogeneous slab in level set under power effect in x/y plane, Fig. 1 is seen.
It can be obtained by equivalent stiffness principle:
The elastic modulus E in y directionsyy:Eyy=lE/a
Equivalent shear modulus G in x/y planexy
In formula:L is single ripple length of run;A is single ripple upright projection length, sees Fig. 2;E is the elasticity of steel Modulus, take 2.06 × 105(N/mm2);ν is the Poisson's ratio of steel, takes 0.3.
Flexural deformation and detrusion of the container structure under horizontal force action are asked using the principle of virtual work.Such as Fig. 1, packaging Shear stress and direct stress expression formula under the power P effects of case side plate in the horizontal direction is as follows:
In formula:σx、σy, be respectively equivalent perpendicular heterogeneous slab in level set power effect under x, y direction direct stress; τxyFor shear stress of the equivalent perpendicular heterogeneous slab in level set under power effect in x/y plane;I is container side plate around strong axle The moment of inertia;P is the concentrated force acted at container angle;H is container height;L is container longitudinal length, sees Fig. 1.
This structure Principle of Virtual Work formula is as follows:
Bring the stress formula tried to achieve above into above formula, obtain the flexural deformation and detrusion of structure.
In formula:EyyFor container y directions modulus of elasticity;GeffFor the equivalent shear modulus in x/y plane.
According to analysis, when beam interacts with plate, in force side, the shear stress between them is maximum, and the shearing of farthest Almost 0, it is therefore assumed that the shear stress between them is linear distribution, power q is cut in force sides, see Fig. 6, Fig. 7.Collection Power P suffered by vanning side plate is born by side plate, upper beam and corner post three parts respectively, by being distributed according to rigidity, can obtain upper beam institute The size for the power received.It can thus be concluded that the axial deformation Δ of container upper beam under concentrated force P effects3It is as follows:
Container deforms as shown in Figure 5 caused by being bent downwardly.By symmetrical structure Computing Principle and layer control, must can collect Vanning deforms Δ caused by being bent downwardly4It is as follows:
Using the computational methods of container anti-side rigidity of the present invention, rigidity numerical stability is ensure that, while avoids passing System method assumed condition deficiency and caused by error or the shortcomings that disregard some Zona transformans, be easy to structure design and analysis.
Based on above Computing Principle, below equation can be obtained according to the arrangement mode of multiple containers and architectural feature:
1st, two when two end movements and bottom girder are not fixed when for Single Container structure, in case top applying power P, bottom girder is fixed End movement formula is as follows:
When bottom girder is fixed:
Exert a force end movement
Free end travel
When bottom girder is not fixed:
Exert a force end movement
Free end travel
In formula:EyyFor container y directions modulus of elasticity;GeffFor the equivalent shear modulus in x/y plane;I is container lateral Plate is around strong equatorial moment of inertia;IsFor container side plate short side direction the moment of inertia;P is the concentrated force acted at container angle;H is Container height;L is container longitudinal length, sees Fig. 1;A is container upper girder cross-section area;Δ1Collection when being fixed for bottom girder Vanning exerts a force end movement under horizontal force action;Δ2Container free end travel under horizontal force action when being fixed for bottom girder;Δ3 Container upper beam axial deformation under horizontal force action, is shown in Fig. 3 when being fixed for bottom girder;Δ01Container exists when not fixed for bottom girder Exert a force end movement under horizontal force action;Δ02Container free end travel under horizontal force action when not fixed for bottom girder;Δ4For When bottom girder is not fixed the container caused by the horizontal force be bent downwardly and caused by both ends deform, see Fig. 4.
2nd, using displacement coordination condition, when to individual layer m, bottom girder is fixed when structure is in case top applying power P force side sidesway and Force side sidesway formula is derived when bottom girder is not fixed, and specific formula is as follows:
When bottom girder is fixed:
When bottom girder is not fixed:
3rd, using displacement superposed and iteration, top of n (n >=2) the layer single span container structure in top applying power P has been obtained Portion's displacementn,n, specific formula is as follows:
In above formula
In formula:Δ1,nOne layer of case top side is moved when acting on n-th layer for concentrated force P.
4th, by dynamic balance and displacement coordination condition, two layers of single span container structure are drawn when pushing up applying power P for one layer, bottom Both ends sidesway formula, as follows when both ends sidesway and bottom girder are not fixed when beam is fixed:
When bottom girder is fixed:
When bottom girder is not fixed:
In formula:AbIt is container underbeam area.
5th, using displacement superposed and iteration, n-layer single span container structure applying power at the top of i-th (i≤n-1) layer has been obtained The displacement on n-layer case top during Pn,i, specific formula is as follows:
Wherein
6th, for n-layer m across container structure, at top during applying power P, structural top force side displacement formula is as follows:
For n-layer m across container structure, it is assumed that horizontal concentrated force P mean allocations at each node, then each node The power of place's distribution is P/ (m+1), and in elastic range, every layer of displacement as caused by horizontal force is just Δ01'/(m+1), by its band Enter in n-layer single span structure lateral displacement formula, can obtain structural top displacement is:
I in formulamBe individual layer m across container structure to centre of form equatorial moment of inertia;Δ01' it is that bottom girder fixes individual layer m across structure in water The flat lower force end movement of concentrated force effect.
7th, sidesway when bottom girder fixes small punch:
When the area that punches is less than 3%, rigidity can be according to without the Rigidity Calculation that punches.Punch and be less than more than 3% as shown in Figure 8 When 15%, the lateral rigidity of structure can be calculated by following formula:
Kh=α K
K is the rigidity of FCL in formula;α is the coefficient that punches,hfIt is the total height at hole;hsIt is Remove the residual altitude at hole, hs=h-hf;aiIt is the length at each hole.
8th, bottom girder is fixed middle through the symmetrical sidesway (punching greatly) that punches
Pass through structural mechanics knowledge and following three hypothesis made to calculating structure:(1) plane cross-section assumption is met;
(2) flexing of side plate is not considered, that is, does not consider out-of-plane deformation when calculating;(3) axial direction of post and beam is not considered Deformation, the container that obtains punching do not consider that the sidesway formula of upper beam axial deformation is:
The container upper beam axial deformation formula that punches of getting back is:
In summary, container total displacement formula is as follows:
9th, through the symmetrical sidesway (punching greatly) that punches among when bottom girder is not fixed
Now underbeam is deformed by the power handed down by side plate, and this integral rigidity on structure influences very big.Point Two parts are divided a structure into during analysis:A part is frame structure;Another part is that the connection limb being made up of side plate and upper beam is cut Power wall construction, as shown in Figure 9, Figure 10, the unfixed analysis of Container in bottom obtain, and structure lateral displacement formula is as follows:
Wherein
10th, non-through symmetrically punches sidesway, as shown in FIG. 11 and 12,
For punching for situation shown in Figure 11 and Figure 12, when calculating its deformation can with equivalent into the situation that punches straight up and down, Will in figure 2b length side plate it is equivalent on beam, then according to through the symmetrical sway calculation that punches.By can be calculated, on hole Lower part equivalent bending stiffness is respectively:
By finite element modelling result and mechanical knowledge, the deformation of container force side is mainly by upper beam axial deformation Cause.Therefore, the present invention is based on european norm covering rigidity formula, with reference to container specific constructive form, in container Upper beam axial deformation is added in sidesway formula, so that container Rigidity Calculation accuracy is by very big improve, can conduct The theoretical foundation of container construction design from now on.
Above content is to combine specific preferred embodiment further description made for the present invention, it is impossible to is assert The specific implementation of the present invention is confined to these explanations.For general technical staff of the technical field of the invention, On the premise of not departing from present inventive concept, some simple deduction or replace can also be made, should all be considered as belonging to the present invention's Protection domain.

Claims (4)

  1. A kind of 1. container house anti-side rigidity computational methods, for calculating the anti-side rigidity of container, it is characterised in that: Comprise the following steps,
    Step a, detrusion and flexural deformation of the container by side plate during side force are obtained using the principle of virtual work;
    Step b, assume the shearing stress distribution of container upper beam and side sheet room, obtain the axial deformation of upper beam;
    Step c, using symmetrical structure and antisymmetry structure obtain container concentrated force effect under caused by being bent downwardly Both ends deform;
    Step d, by the detrusion of the side plate of container, flexural deformation, the axial deformation of upper beam and it is bent downwardly deformation Caused deformation calculates the rigidity of container;
    In the step a, the detrusion and flexural deformation of the side plate of container meet below equation:
    <mrow> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mo>=</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>3</mn> </msup> </mrow> <mrow> <mn>3</mn> <msub> <mi>E</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mi>I</mi> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <msup> <mi>PhL</mi> <mn>2</mn> </msup> </mrow> <mrow> <mn>10</mn> <msub> <mi>G</mi> <mrow> <mi>e</mi> <mi>f</mi> <mi>f</mi> </mrow> </msub> <mi>I</mi> </mrow> </mfrac> </mrow>
    In formula:EyyFor container y directions modulus of elasticity, Eyy=lE/a, l are single ripple length of run;A is that single ripple is vertical Projected length, E are the modulus of elasticity of steel, take 2.06*105(N/mm2);GeffFor the equivalent shear modulus in x/y plane;I is collection Side plate case around strong equatorial moment of inertia;P is the concentrated force acted at container angle;H is container height;L indulges for container To length;
    In the step b, the axial deformation of the upper beam of container meets below equation:
    <mrow> <msub> <mi>&amp;Delta;</mi> <mn>3</mn> </msub> <mo>=</mo> <mfrac> <mi>P</mi> <mrow> <mn>3</mn> <msub> <mi>EI</mi> <mi>c</mi> </msub> <mo>/</mo> <msup> <mi>h</mi> <mn>3</mn> </msup> <mo>+</mo> <mn>3</mn> <mi>E</mi> <mi>A</mi> <mo>/</mo> <mi>L</mi> </mrow> </mfrac> </mrow>
    In formula:IcThe moment of inertia for corner post of container to itself centre of form axle;A is container upper girder cross-section area,
    In the step c, container both ends deformation caused by being bent downwardly meets below equation:
    <mrow> <msub> <mi>&amp;Delta;</mi> <mn>4</mn> </msub> <mo>=</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>2</mn> </msup> <mi>L</mi> </mrow> <mrow> <mn>16</mn> <msub> <mi>EI</mi> <mi>s</mi> </msub> </mrow> </mfrac> </mrow>
    In formula:IsFor container side plate short side direction the moment of inertia.
  2. 2. container house anti-side rigidity computational methods according to claim 1, it is characterised in that:The container is applied on case top When reinforcing P, force end movement and free end meet below equation:
    When the bottom girder of the container is fixed:
    Exert a force end movement
    Free end travel
    When the bottom girder of the container is not fixed:
    Exert a force end movement
    Free end travel
  3. 3. container house anti-side rigidity computational methods according to claim 2, it is characterised in that:For individual layer m across container Structure, force side sidesway formula meets following when force side sidesway and bottom girder are not fixed when on case top, bottom girder is fixed during applying power P Formula:
    When bottom girder is fixed:
    <mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>&amp;prime;</mo> </msup> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>-</mo> <mfrac> <mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mn>2</mn> </msup> </mrow> <mrow> <mn>2</mn> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mrow> </mtd> <mtd> <mrow> <mi>m</mi> <mo>=</mo> <mn>2</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>-</mo> <mfrac> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mrow> <mn>2</mn> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <mn>4</mn> <msup> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mrow> <mo>(</mo> <mi>m</mi> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mn>2</mn> </msup> </mrow> <mrow> <mn>4</mn> <msup> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mrow> <mo>(</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mn>2</mn> </msup> </mrow> </mfrac> </mrow> </mtd> <mtd> <mrow> <mi>m</mi> <mo>&amp;GreaterEqual;</mo> <mn>3</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
    When bottom girder is not fixed:
    <mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>&amp;prime;</mo> </msup> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>-</mo> <mfrac> <mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mn>2</mn> </msup> </mrow> <mrow> <mn>2</mn> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> </mrow> </mfrac> </mrow> </mtd> <mtd> <mrow> <mi>m</mi> <mo>=</mo> <mn>2</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>-</mo> <mfrac> <mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mn>2</mn> </msup> </mrow> <mrow> <mn>2</mn> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> </mrow> </mfrac> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <mn>4</mn> <msup> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mrow> <mo>(</mo> <mi>m</mi> <mo>-</mo> <mn>3</mn> <mo>)</mo> </mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mn>2</mn> </msup> </mrow> <mrow> <mn>4</mn> <msup> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mn>2</mn> </msup> <mo>-</mo> <mrow> <mo>(</mo> <mi>m</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mn>2</mn> </msup> </mrow> </mfrac> </mrow> </mtd> <mtd> <mrow> <mi>m</mi> <mo>&amp;GreaterEqual;</mo> <mn>3</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
    N-layer single span container structure, the top displacement Δ at top during applying power Pn,n, meet below equation:
    <mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>n</mi> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>n&amp;Delta;</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>n</mi> </mrow> </msub> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>3</mn> </msup> </mrow> <mrow> <mn>2</mn> <mi>E</mi> <mi>I</mi> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>(</mo> <mi>n</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mn>2</mn> </mfrac> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> </mrow> <mn>2</mn> </mfrac> <mo>+</mo> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> </mrow>
    In above formula
    Two layers of single span container structure, when pushing up applying power P for one layer, both ends when both ends sidesway and bottom girder are not fixed when bottom girder is fixed Sidesway formula, it is as follows:
    When bottom girder is fixed:
    <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>1</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mfrac> <mrow> <mi>P</mi> <mi>L</mi> </mrow> <mrow> <msub> <mi>EA</mi> <mi>b</mi> </msub> </mrow> </mfrac> <mo>+</mo> <mn>2</mn> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> </mtd> </mtr> <mtr> <mtd> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>2</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mo>+</mo> <mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mfrac> <mrow> <mi>P</mi> <mi>L</mi> </mrow> <mrow> <msub> <mi>EA</mi> <mi>b</mi> </msub> </mrow> </mfrac> <mo>+</mo> <mn>2</mn> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> </mtd> </mtr> </mtable> </mfenced>
    When bottom girder is not fixed:
    <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>01</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mfrac> <mrow> <mi>P</mi> <mi>L</mi> </mrow> <mrow> <msub> <mi>EA</mi> <mi>b</mi> </msub> </mrow> </mfrac> <mo>+</mo> <mn>2</mn> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>02</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mo>+</mo> <mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mfrac> <mrow> <mi>P</mi> <mi>L</mi> </mrow> <mrow> <msub> <mi>EA</mi> <mi>b</mi> </msub> </mrow> </mfrac> <mo>+</mo> <mn>2</mn> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> </mtd> </mtr> </mtable> </mfenced>
    In formula:AbFor container underbeam area;
    The displacement on n-layer single span container structure n-th layer case top in i-th layer of top applying power Pn,i, specific formula is as follows:
    <mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>n</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mi>n</mi> <mo>&amp;lsqb;</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mn>1</mn> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>-</mo> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>2</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> <mtd> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>&amp;lsqb;</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>-</mo> <mo>(</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>01</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>02</mn> </mrow> </msub> <mo>)</mo> <mo>&amp;rsqb;</mo> <mo>-</mo> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>)</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> <mo>,</mo> <mi>i</mi> </mrow> </msub> </mrow> </mtd> <mtd> <mrow> <mn>2</mn> <mo>&amp;le;</mo> <mi>i</mi> <mo>&amp;le;</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>
    Wherein
    <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>i&amp;Delta;</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>j</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>3</mn> </msup> </mrow> <mrow> <mn>2</mn> <mi>E</mi> <mi>I</mi> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>(</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mn>2</mn> </mfrac> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> </mrow> <mn>2</mn> </mfrac> <mo>+</mo> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>p</mi> <mn>01</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>=</mo> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>i</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <msup> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>3</mn> </msup> </mrow> <mrow> <mn>2</mn> <mi>E</mi> <mi>I</mi> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mn>2</mn> </mfrac> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;</mi> <mn>02</mn> </msub> </mrow> <mn>2</mn> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;</mi> <mrow> <mn>1</mn> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&amp;Delta;</mi> <mn>1</mn> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;</mi> <mn>2</mn> </msub> </mrow> <mn>2</mn> </mfrac> <mo>+</mo> <mrow> <mo>(</mo> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>3</mn> </msup> </mrow> <mrow> <mn>2</mn> <mi>E</mi> <mi>I</mi> </mrow> </mfrac> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>,</mo> <mn>2</mn> <mo>&amp;le;</mo> <mi>i</mi> <mo>&amp;le;</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow>
    For n-layer m across container structure, at top during applying power P, structural top force side displacement formula is as follows:
    For n-layer m across container structure, it is assumed that horizontal concentrated force P mean allocations at each node, then each node punishment The power matched somebody with somebody is P/ (m+1), and in elastic range, every layer of displacement as caused by horizontal force is just Δ01'/(m+1), carry it into n-layer In single span structure lateral displacement formula, can obtain structural top displacement is:
    <mfenced open = "" close = ""> <mtable> <mtr> <mtd> <mrow> <mi>&amp;Delta;</mi> <mo>=</mo> <mi>n</mi> <mfrac> <mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>&amp;prime;</mo> </msup> </mrow> <mrow> <mo>(</mo> <mi>m</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mfrac> <mo>+</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>3</mn> </msup> </mrow> <mrow> <mn>2</mn> <msub> <mi>EI</mi> <mi>m</mi> </msub> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mo>(</mo> <mi>n</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </mrow> <mn>2</mn> </mfrac> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msup> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>&amp;prime;</mo> </msup> </mrow> <mrow> <mo>(</mo> <mi>m</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mfrac> <mo>+</mo> <msup> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>&amp;prime;</mo> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </munderover> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <msup> <mi>Ph</mi> <mn>3</mn> </msup> </mrow> <mrow> <mn>2</mn> <msub> <mi>EI</mi> <mi>m</mi> </msub> </mrow> </mfrac> <mo>+</mo> <mo>&amp;lsqb;</mo> <mfrac> <mrow> <mo>(</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> <mo>(</mo> <mi>n</mi> <mo>+</mo> <mn>2</mn> <mo>)</mo> </mrow> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <mi>m</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>&amp;rsqb;</mo> <msup> <msub> <mi>&amp;Delta;</mi> <mn>01</mn> </msub> <mo>&amp;prime;</mo> </msup> </mrow> </mtd> </mtr> </mtable> </mfenced>
    In formula:ImBe individual layer m across container structure to centre of form equatorial moment of inertia;Δ01' it is that bottom girder does not fix individual layer m across structure in water The flat lower force end movement of concentrated force effect.
  4. 4. container house anti-side rigidity computational methods according to claim 1, it is characterised in that:The bottom girder of the container is consolidated It is fixed small when punching, and punch when being less than 15% more than 3%, the lateral rigidity of structure meets below equation:
    Kh=α K
    K is the rigidity of FCL in formula;α is the coefficient that punches,hfIt is the total height at hole;hsIt is to remove The residual altitude at hole, hs=h-hf;aiIt is the length at each hole.
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