CN101900648B - Method for determining bearing capacity of concrete-filled steel tube under various stressing conditions and application thereof - Google Patents

Abstract

The invention relates to a method for determining bearing capacity of a concrete-filled steel tube under various stressing conditions, which is characterized by solving the technical problems that the bearing capacity of the concrete-filled steel tube under various stressing conditions can not be determined simply, conveniently and directly, and providing a safety evaluation method of a concrete-filled steel tubular member under the complex loading combination of pressure, bending moment, shearing force and torsion moment. The concrete-filled steel tube is under the action of pressure, bending moment, shearing force and torsion moment, and the safety evaluation method comprises the following steps of: collecting the relative parameters of the concrete-filled steel tube; collecting the compressive strength design valves of steel and concrete in the concrete-filled steel tube, and collecting the cross sections of the steel tube and the concrete in the concrete-filled steel tube; and determining the bearing capacity of the concrete-filled steel tube: the bearing capacity of the concrete-filled steel tube under the action of pressure, bending moment, shearing force and torsion moment meets the equation shown in the specification.

Description

Bearing capacity of concrete-filled steel tube under multiple stressing conditions is determined method and application

Technical field

The present invention relates to a kind of bearing capacity of concrete-filled steel tube and determine method and application, the bearing capacity of concrete-filled steel tube that relates in particular under one kind of multiple stressing conditions is determined method and application.

Background technology

Span constantly increases and builds the raising that requires without antidetonation along with building as a result, makes the research of structure shearing resistance torsional property seem important.And in practical structures, concrete filled steel tube is in to press and turns round, and press-bending is turned round, and even the state of Subjected To Compression, Bending, Shear And Torsion is many, as the bridge pier as the shaped form bridge of the framework corner post of buildings, highway, the column of offshore production platform etc.Therefore it be necessary to further investigate concrete filled steel tube in the multiple serviceability under stressed of Subjected To Compression, Bending, Shear And Torsion, determines the form of its bearing capacity dependent equation.Because existing multiple dependent equation research under stressed is all to draw fitting formula after correlation curve by FEM (finite element) calculation, be theoretically unsound, especially easily practice is carried out determining of bearing capacity of concrete-filled steel tube under multiple effect of stress, and this has restricted application and the popularization of concrete filled steel tube greatly.

Summary of the invention

The technical matters that the present invention solves is: provide the bearing capacity of concrete-filled steel tube under one kind of multiple stressing conditions to determine method, overcoming can not be simply and directly to carrying out the technical matters that bearing capacity of concrete-filled steel tube is determined in prior art under multiple effect of stress.

Technical scheme of the present invention is: provide the bearing capacity of concrete-filled steel tube under one kind of multiple stressing conditions to determine method, described concrete filled steel tube is stressed, moment of flexure, and the combination in any effect of shear and torsion load, its security assessment method comprises the steps:

Gather the correlation parameter of concrete filled steel tube: steel and concrete anti-compression, shearing strength design load in the collection concrete filled steel tube, sectional area and the concrete sectional area of steel pipe in the collection concrete filled steel tube;

Determine the bearing capacity of concrete filled steel tube: described concrete filled steel tube is under pressure, bending, shearing, the effect of moment of torsion Load Combination, and the equation that its bearing capacity satisfies is:

The time:

The time;

Further technical scheme of the present invention is: described concrete filled steel tubular member comprises rectangular elements, round member and octagon member.

Further technical scheme of the present invention is: described concrete filled steel tubular member comprises solid concrete filled steel tubular member and concrete-filled steel tubular hollow member.

Further technical scheme of the present invention is: the Load Combination that is subject to of described concrete filled steel tube is axle power and moment of flexure, and when determining the bearing capacity of concrete filled steel tube, the equation that its bearing capacity satisfies is:

$\frac{N}{N_u}\≥0.2$ The time:

$\frac{N}{N_u}+\frac{{\mathrm{\β}}_{m}M}{1.071(1-0.4N/N_E)M_u}=1$

$\frac{N}{N_u}<0.2$ The time;

$-\frac{N}{7N_u}+\frac{{\mathrm{\&beta;}}_{m}M}{(1-0.4N/N_E)M_u}=1$

Further technical scheme of the present invention is: the Load Combination that is subject to of described concrete filled steel tube is axle power, bending and shearing, and when determining the bearing capacity of concrete filled steel tube, the equation that its bearing capacity satisfies is:

$\frac{N}{N_u}\&GreaterEqual;0.2[1-{\left(\frac{V}{V_u}\right)}^2]$ The time:

$(\frac{N}{N_u}+\frac{{\mathrm{\&beta;}}_{m}M}{1.071(1-0.4N/N_E)M_u})+{\left(\frac{V}{V_u}\right)}^2=1$

$\frac{N}{N_u}<0.2[1-{\left(\frac{V}{V_u}\right)}^2]$ The time;

$(-\frac{N}{7N_u}+\frac{{\mathrm{\&beta;}}_{m}M}{(1-0.4N/N_E)M_u})+{\left(\frac{V}{V_u}\right)}^2=1$

Further technical scheme of the present invention is: the Load Combination that is subject to of described concrete filled steel tube is axle power, moment of flexure and moment of torsion, and when determining the bearing capacity of concrete filled steel tube, the equation that its bearing capacity satisfies is:

$\frac{N}{N_u}\&GreaterEqual;0.2[1-{\left(\frac{T}{T_u}\right)}^2]$ The time:

$(\frac{N}{N_u}+\frac{{\mathrm{\&beta;}}_{m}M}{1.071(1-0.4N/N_E)M_u})+{\left(\frac{T}{T_u}\right)}^2=1$

$\frac{N}{N_u}<0.2[1-{\left(\frac{T}{T_u}\right)}^2]$ The time;

$(-\frac{N}{7N_u}+\frac{{\mathrm{\&beta;}}_{m}M}{(1-0.4N/N_E)M_u})+{\left(\frac{T}{T_u}\right)}^2=1$

Technical scheme of the present invention is: the bearing capacity of concrete-filled steel tube under multiple stressing conditions is determined that method is applied to concrete filled steel tubular member.

Technical scheme of the present invention is: the bearing capacity of concrete-filled steel tube under multiple stressing conditions is determined that method is applied to the safety assessment of concrete filled steel tubular member.

Technique effect of the present invention is: the bearing capacity of concrete-filled steel tube that the invention provides under one kind of multiple stressing conditions is determined method, by multiple stress resultant is considered, derives the bearing capacity determination methods of the concrete filled steel tube under multiple Load Combination effect.The present invention can determine bearing capacity of concrete-filled steel tube under multiple effect of stress simply and directly, greatly facilitates determining of bearing capacity of concrete-filled steel tube under multiple effect of stress, is conducive to promote application and the popularization of concrete filled steel tube.

Description of drawings

Fig. 1 is process flow diagram of the present invention.

Fig. 2 is concrete-filled rectangular steel tube simple bending member steel pipe of the present invention and concrete stress strain diagram.

Fig. 3 is the various section steel pipe concrete model of the present invention isoboles.

Fig. 4 is that rectangle circular steel tube concrete of the present invention is cut the model of element equivalent schematic.

Fig. 5 is the stressed schematic diagram of film unit body truss model of being cut of the present invention

Fig. 6 is the beam stress schematic diagram of being cut of the present invention.

Fig. 7 is the stressed schematic diagram of cell cube of being turned round of the present invention.

Fig. 8 is concrete filled steel tubular member space-truss model isoboles under complicated applied force of the present invention.

Fig. 9 is the coefficient concrete filled steel tube equivalent truss of curved scissors of the present invention force diagram.

Figure 10 is the coefficient concrete filled steel tube equivalent truss of bending of the present invention force diagram.

Figure 11 is the coefficient concrete filled steel tube equivalent truss of bending and twisting of the present invention force diagram.

Embodiment

Below in conjunction with specific embodiment, technical solution of the present invention is further illustrated.

In the present patent application file, the meaning of correlation parameter is as follows:

Expression concrete filled steel tube long column stability factor;

A _sc=A _s+ A _cExpression concrete filled steel tube area;

A _sThe area of section of steel pipe in the expression concrete filled steel tube;

A _cBe expressed as concrete section area in concrete filled steel tube.

f _scBe expressed as concrete filled steel tube axial compression combined strength design load,

f _y, f _cExpression is respectively steel and concrete crushing strength design load.

Expression concrete filled steel tube cuff coefficient;

K represents cross section cuff adjustment coefficient.Quadrilateral, k=0.6879 ψ ²-0.8827 ψ+0.3285; Octagon k=-0.4545 ψ ²-0.3953 ψ+0.724

ψ represents hollow rate, ψ=A _k/ (A _c+ A _k);

A _kThe area of expression hollow parts;

The canonical slenderness ratio is

L ₀Be effective computational length of member, relevant with the constraint condition at two ends,

Be the turning radius.

Eular crippling stress ${\mathrm{\&sigma;}}_{\mathrm{Esc}}=\frac{{\mathrm{\&pi;}}^2{E}_{\mathrm{scm}}}{{\mathrm{\&lambda;}}^2},$

K=0.25 α ^N, α=A wherein _s/ A _cBe the steel ratio of concrete filled steel tube, N is the shape coefficient in cross section,

N is the limit number in cross section, for circle, N=1 is arranged; For octagon N=0.91; For square N=0.286.

In formula, each parameter is taken by " concrete filled steel tube technical regulation ";

$E_{\mathrm{sc}}=\frac{A^{\&prime;}{\mathrm{\&xi;}}^2+(A^{\&prime;}+2+k_e)\mathrm{\&xi;}+2}{A^{\&prime;}{B}^{\&prime;}{\mathrm{\&xi;}}^2+(A^{\&prime;}+2B^{\&prime;})\mathrm{\&xi;}+2}{E}_c$

A′=0.2B′(1-ψ)+0.05ψ+0.05

$B^{\text{'}}=\frac{E_c}{E_s}$

E _s=206000MPa

E _cBe calculated as follows: $E_c=\frac{{10}^2}{2.2+34.7/f_{\mathrm{cu},k}}$

M _u=k _mγ _mW _scf _sc

$T_u={\mathrm{\&gamma;}}_T{W}_{\mathrm{sc}}^T{f}_{\mathrm{sc}}^V$

$V_u={\mathrm{\&gamma;}}_V{A}_{\mathrm{sc}}{f}_{\mathrm{sc}}^V$

In formula, each parameter is taken by " concrete filled steel tube technical regulation " each continuous item

As shown in Figure 1, the specific embodiment of the present invention is as follows:

The bearing capacity of concrete-filled steel tube that the invention provides under one kind of multiple stressing conditions is determined method, it is characterized in that, described concrete filled steel tube is stressed, moment of flexure, and the shear and torsion effect, the concrete filled steel tube security assessment method comprises the steps:

Step 100: the correlation parameter that gathers concrete filled steel tube: gather steel and concrete crushing strength design load in concrete filled steel tube, gather sectional area and the concrete sectional area of steel pipe in concrete filled steel tube;

Step 200: the bearing capacity of determining concrete filled steel tube.Described concrete filled steel tube is at pressure, moment of flexure, and under the shear and torsion effect, the equation that its bearing capacity satisfies is:

The time:

The time;

Determine the bearing capacity of concrete filled steel tube under the stressed effect of Load Combination, first limit of utilization level theory is theoretical derives and draws form at the dependent equation of not considering the bearing capacity of concrete filled steel tube under the combination stressed effect of Complicated Loads under steel pipe and the effect of concrete cuff, then by studying bearing capacity under independent bearing load condition and the Compact-bending Load bearing capacity under making up, dependent equation is revised minute.The concrete analysis process is as follows:

One, the balance equation of the concrete filled steel tube under the effect of being stressed.

under the shaft pressuring load effect, common stressed Concrete Filled Square Steel Tubular is born load by steel pipe and pipe inner concrete, in the original upload stage, steel pipe and concrete acting in conjunction, but do not push between the two, along with axial strain increases, inside concrete generation microfracture, its transversely deforming surpasses the transversely deforming of steel pipe, at this moment, steel pipe is in axial compression, ring draws and horizontal pressurized three-dimensional stress state, but the increasing along with shaft pressuring load, the steel pipe axial compression stress reduces, the hoop tension increases, concrete axial compression stress is because the effect of contraction of steel pipe increases simultaneously, know and reach steel pipe three-dimensional stress ultimate limit state, bearing capacity value of reaching capacity of steel core concrete column.

The stress state of steel pipe can be reduced to vertical pressurized σ _1s, hoop tension σ ₂Two-dimensional state of stress, σ ₃=0 and concrete is in axial compression stress σ _1c, under the three-dimensional stress state of side direction compressive stress p, concrete filled steel tube is regarded as the structural system of steel pipe and two kinds of elements compositions of core concrete.

When considering Moment, first do not consider steel pipe to concrete confined effect, xial feed is namely got by steel pipe and concrete simple superposition, and the static balance condition equation is arranged:

N=A _cσ _c+A _sσ _s （1）

Two, only be subjected to the balance equation of the concrete filled steel tube under Moment.

(1) for rectangular elements.

Concrete constitutive relation is got concrete standard and is insulted the σ of cylinder-ε curve, and Fig. 2 is concrete-filled rectangular steel tube simple bending member steel pipe and concrete stress strain diagram.

1) on the xsect shown in figure, moment of flexure produces compressive stress on the top in cross section, and the bottom in outer cross section produces tension, a little less than the agent on crack resistance of concrete pulling force, cuts basifacial concrete and will ftracture, and tension is mainly born by steel pipe, and wherein d is the significant height in cross section;

2) figure is strain figure, much smaller than concrete filled steel tubular member cross section d, therefore the steel pipe thickness of thinking that can be similar to is that the strain value of the steel pipe of t equates, is all ε due to the steel pipe thickness t _s, concrete compressive strain ε _cFrom neutral axis 0 to the linear variation of the maximal value of top margin;

3) be not cracked concrete distribution of the compressive stress figure of member top edge, the height of compressive region is x _c, the lower limb of compressive stress figure is neutral axis, and stress is 0. to establish the C that makes a concerted effort to be of compressive region concrete compressive stress herein, and the C that makes a concerted effort is y to the distance of natural axis _c

4) figure is steel penstock stress figure, bottom tension upper portion pressurized, and steel pipe upper and lower side thickness is that the stress value of the steel pipe of t equates, is respectively σ _Draw, σ _Press, the stress of side direction steel pipe is linear distribution.

1) the interior moment M of concrete generation _cFind the solution (the centering profile shaft is got square)

Use the method for concrete specification, establish C _cuBe the area that concrete compressive stress-strain curve surrounds, y _cuThe centre of form of area is to the distance of coordinate axis y for this reason ^[45]

$C_{\mathrm{cu}}=\underset{0}{\overset{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)d{\mathrm{\&epsiv;}}_c$ $y_{\mathrm{cu}}=\frac{\underset{0}{\overset{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right){\mathrm{\&epsiv;}}_{c}d{\mathrm{\&epsiv;}}_c}{C_{\mathrm{cu}}}$

Order

The k here ₁And k ₂Only depend on the concrete compression stress-strain curve. be an amount relevant with the concrete material performance.

Making a concerted effort of cross section compressive region concrete compressive stress:

The distance of c to natural axis of making a concerted effort: $y_c=\frac{\underset{0}{\overset{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{bxd}{x}_c}{C}=\frac{{\underset{0}{\overset{x_c}{\&Integral;}}\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{xdx}}{\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{dx}}$

Represent neutral axis to the distance at strain stress place with x, have that triangle is similar to be had

Differential has

The substitution following formula has:

$C=\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)b\frac{x_c}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}\&CenterDot;\mathrm{d\&epsiv;}=b\&CenterDot;x_c\&CenterDot;\frac{C_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}={\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HSA00000189286600031.png"\; state="\{\{state\}\}">k</figure-callout>}}_1\&CenterDot;f_c\&CenterDot;b\&CenterDot;x_c---\left(2\right)$

$y_c=\frac{\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{xdx}}{\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{dx}}=\frac{\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)b{\left(\frac{x_c}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}\right)}^2\&CenterDot;{\mathrm{\&epsiv;}}_c\&CenterDot;\mathrm{d\&epsiv;}}{b\&CenterDot;x_c\&CenterDot;\frac{C_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}}={\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">k</figure-callout>}}_2\&CenterDot;x_c---\left(3\right)$

2) the interior moment M of concrete generation _cFind the solution (the centering profile shaft is got square)

M _c=C·y _c＝k ₁·k ₂·f _c·b·x _c ² (4)

3) the steel pipe each several part is made a concerted effort to find the solution

Due to t much smaller than the steel tube component diameter, therefore approximate think on, the downside steel pipe equates with the stress value on the cross section, is respectively σ _Draw, σ _Press, and side plate stress take on top as-σ _PressThe bottom is σ _DrawHold linear distribution.

If x _c=kd has

Pressure above steel pipe:

Pressure below steel pipe: T _Under=σ _DrawBt

Side direction steel pipe upper pressure:

Side direction steel pipe bottom pulling force:

4) the interior moment M of steel pipe each several part generation _sFind the solution (the centering axle is asked square)

If the depth-width ratio in concrete-filled rectangular steel tube cross section is λ, have

Again $A_s=2(b+d)\&times;t=2(\frac{d}{\mathrm{\&lambda;}}+d)\&times;t=2d\&times;t\&times;(1+\frac{1}{\mathrm{\&lambda;}})\&DoubleRightArrow;d\&times;t=\frac{1}{2}{A}_s\&CenterDot;\left(\frac{\mathrm{\&lambda;}}{1+\mathrm{\&lambda;}}\right)$

Following formula substitution bemding moment formula abbreviation is had:

5) axle equilibrium equation

C+T _On+ T _Press=T _Under+ T _Draw

Top various substitution is had:

6) moment of flexure balance equation

M=M _c+M _s

Formula (4) (5) (6) substitution following formula abbreviation there is simple bending concrete filled steel tubular member moment of flexure balance equation:

In formula: k ₁, k ₂---the concrete material performance parameter, depend on the concrete compression stress-strain curve.

$k_1{f}_c=\frac{C_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{ci}}},$ ${\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">k</figure-callout>}}_2=\frac{y_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}};$

K---the ratio of compressive region concrete compression height and depth of section in concrete, k=x _c/ d;

x _c---concrete compression district height;

λ---concrete-filled rectangular steel tube depth-width ratio, λ=d/b;

σ _Draw, σ _Press---the steel pipe tension, press side draw, compressive stress;

σ _c---concrete compressive stress;

A _s, A _c---steel pipe, concrete section area in concrete filled steel tube.

Order $j=k_2(\frac{\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">k</figure-callout>}}{2}\&CenterDot;\frac{1-2k}{1-k})+\frac{(2k^2-2k+1)+\mathrm{\&lambda;}(k^2-\mathrm{<figure-callout\; id="1"\; label="k\; +"\; filenames="HSA00000189286600031.png"\; state="\{\{state\}\}">k</figure-callout>}+\frac{1}{3})}{2\&CenterDot;(1+\mathrm{\&lambda;})(1-k)},$

Can find out, j is one and depth of compressive zone x _cThe related amount of=kd, the interior moment of flexure of resisting Moments can be written as: M=σ _sA _sDj.σ is arranged again _s=f _s, f _sFor the stress of concrete filled steel tubular member tensile region outermost layer steel pipe, work as f _s=f _yThe time can think member surrender.Concrete-filled rectangular steel tube simple bending member balance equation is arranged:

M=f _s·A _s·jd (8)

In formula: jd---the bending resistance equivalent height;

J---equivalent height coefficient.

(2) other, as the circular steel tube concrete member, from all directions the shape concrete component by same derivation, draws:

Concrete filled steel tube simple bending member balance equation:

M=f _s·A _s·jd （8）

In formula: jd represents the bending resistance equivalent height;

J represents the equivalent height coefficient.

(3) model equivalence.

As shown in Figure 3, M=f _sA _sIn jd as jd is used as the arm of force, can think Moments system by a pair of equal and opposite in direction, opposite direction, parallel to each other and at a distance of the power N of jd _tAnd N _lResist, therefore concrete filled steel tubular flexural member can carry out equivalence with following truss model.

As shown in Figure 3, in equivalent model:

1) in BENDING PROCESS, the suffered compressive stress of member is born with the form of horizontal struts by concrete and part steel pipe, is equivalent to the depression bar that winds up of truss, and the size of power is Nt=A _sF _s

2) in BENDING PROCESS, the suffered tension of member is born with the form of girt by steel pipe, is equivalent to the lower edge pull bar of truss, and the size of power is Nl=A _sF _s

3) the opposing moment of flexure that produces is: M=f _sA _sJd, the dv=jd item is the truss depth of equivalence.Be it can also be seen that by following formula, during surrender, upper lower beam is surrendered simultaneously, and its yield strength is N _t=N _b=f _yA _s, antitorque bearing capacity is: M=f _sA _sJd;

4) above-mentioned analytic explanation concrete filled steel tube be subjected to the ultimate principle of curved theory, because the equilibrium condition of its satisfied 1. concrete and steel penstock stress 2. Strain Distribution is obeyed Bernoulli Jacob's supposition, plane section keeps the plane the distortion successor, has satisfied the constitutive relation that has satisfied concrete and steel in the process that 3. deformation compatibility condition derive.Because it has satisfied above three conditions, therefore think that to its equivalent reasoning be strict.

Therefore concrete filled steel tube be subjected to curved balance equation gather for:

M=f _s·A _s·jd (8)

In formula: jd---the bending resistance equivalent height; J represents the equivalent height coefficient.

Three, cut the balance equation of film unit body.

(1) cut the model equivalence of cell cube.

Rectangle is cut the extraction of unit:

As shown in Figure 4, pair cross-section is wide, and height is respectively B, and the length of D is l, and steel pipe thickness is that the concrete-filled rectangular steel tube of t is got length, and wide, height is respectively 1 cell cube, studs with steel disc in cell cube.In the situation that do not consider steel pipe to concrete constraining force, can think that its steel pipe power to cell cube of distribution steel disc is equivalent to the steel pipe mean force of steel pipe to whole core concrete in concrete filled steel tube.Steel disc is divided into vertical steel disc and vertical steel disc in equivalence, resists respectively transverse shearing force and longitudinal shear stream q.

The equivalence of rectangular steel pipe steel disc:

In the concrete-filled rectangular steel tube member in shear, the steel pipe of upper and lower sides mainly bears moment of flexure, out-of plane bending occurs, produce normal stress and opposing vertical shear stream q, the steel pipe of left and right sides should bear moment of flexure generation plane inner bending, produce normal stress with opposing vertical shear stream q, also will bear vertical shear vertical sidesway occurs, opposing longitudinal shear stream q.Therefore vertical shear q can be by all steel pipe shareds of four sides, and longitudinal shear can only be born to steel pipe by left and right sides.

1) vertical steel disc: the opposing longitudinal shear, because the steel pipe that only has side direction can be resisted longitudinal shear, therefore in cell cube, the steel pipe equivalent stress of horizontal steel disc is:

2) vertical steel disc: the opposing vertical shear, because four side steel pipes can be resisted vertical shear, therefore in cell cube, the steel pipe equivalent stress of vertical steel disc is: $n_l=\frac{2(B+D)\&CenterDot;t\&CenterDot;l\&CenterDot;f_s}{B\&CenterDot;D\&CenterDot;l}=\frac{2(B+D)\&CenterDot;t}{B\&CenterDot;D}\&CenterDot;f_s$

Circle is cut the unit and is extracted:

As shown in Figure 4, be D to diameter, steel pipe thickness is that the concrete-filled rectangular steel tube of t is got length, wide, height is respectively 1 cell cube, studs with steel disc in cell cube.Do not consider in the situation of steel pipe to concrete constraining force, can think that its steel pipe power to cell cube of distribution steel disc is equivalent to the steel pipe mean force of steel pipe to whole core concrete in concrete filled steel tube.Steel disc is divided into vertical steel disc and horizontal steel disc in equivalence, resists respectively vertical shear and transverse shearing force stream q.

The equivalence of round steel pipe steel disc:

In the concrete-filled rectangular steel tube member in shear, steel pipe can produce meridional stress with opposing perimeter shear stream q, also will bear vertical shear vertical sidesway occurs, thereby can produce circumference stress with opposing longitudinal shear stream q.

1) vertical steel disc: because the circumference stress that round steel pipe produces is resisted longitudinal shear stream, and the shearing under extraneous shear V effect is vertical, become angle with hoop, therefore need reduction, here the way that adopts is that circular equivalence is the direction steel pipe of homalographic, according to the derivation to rectangular steel pipe, still think and only have the side direction steel pipe that transverse stress opposing longitudinal shear stream can be provided, therefore have: the rectangular steel tube length of side

Therefore in cell cube, the steel pipe equivalent stress of horizontal steel disc is:

$n_t=\frac{2B\&CenterDot;t\&CenterDot;l\&CenterDot;f_s}{B^{2}l}=\frac{2t}{B}\&CenterDot;f_s=\frac{2t}{0.9D}\&CenterDot;f_s=\frac{2.2t}{D}\&CenterDot;f_s.$

2) vertical steel disc: produce meridional stress opposing perimeter shear stream, because whole steel pipe can be resisted perimeter shear, therefore in cell cube, the steel pipe equivalent stress of vertical steel disc is:

In like manner, can extract the concrete filled steel tube member in shear of other section forms and be cut cell cube, and can calculate the equivalent stress value n of equivalent vertical and vertical steel disc _tAnd n _lIt is not carried out unfolding calculation in following derivation, only use n _t=k _tf _sAnd n _l==k _lf _sExpression, k _tAnd k _lThe amount relevant with cross sectional shape.

(2) cut the balance equation of film unit body.

Be illustrated in figure 5 as the film unit body of shear flow q effect, cell cube is square, and the length of side vertical and vertical both direction is l, and thickness is b, and the equivalent force of the vertical steel disc in cell cube is n _tB=k _tBf _s, vertically the equivalent force of steel disc is n _lB=k _lBf _s

As shown in Figure 5, be a series of concrete inclined strut by crack segmentation after concrete cracking, the trend of crackle becomes the α angle with the l axle, bears size and is σ _dCompressive stress, the tension f that bears with steel pipe _sThe shear stress τ that the extraneous shear flow q that produces of common opposing produces.What can be similar to thinks concrete inclined strut, steel pipe vertical and longitudinally steel disc consisted of the truss that bears shear flow.

The cell cube steel disc power of being cut that triangle firmly represents and the equilibrium relation of outside shear flow q be as shown in Figure 5: the equilibrium relation of cell cube left surface as shown in Figure 6, vertical shearing force q is mainly by vertical steel disc stress n _lB and oblique concrete (σ _dB) cos α shared.(σ _dB) cos α is b for acting on thickness, and width is cos α, and the concrete oblique stress is σ _dOblique concrete pressure.Draw shear flow q and steel pipe longitudinal force n by triangle of forces _lThe pass of b is: q=(n _lB) tan α.The equilibrium relation of cell cube upper side as shown in Figure 7, shear flow q is mainly by vertical steel disc stress n longitudinally _tB and oblique concrete (σ _dB) sin α shared.σ _dSin α is b for acting on thickness, and width is sin α, and the concrete oblique stress is σ _dOblique concrete pressure.Draw shear flow q and steel pipe vertical force n by triangle of forces _tThe pass of b is: q=(n _tB) cot α.Shear flow q and oblique concrete stress σ _dRelation, also can have by transverse triangle shape: q=(σ _dB) sin α cos α.

Therefore cell cube shearing balance equation is formula gather for:

q=(n _lb)tanα

q=(n _tb)cotα

q＝(σ _db)sinαcosα (9)

When concrete filled steel tube is surrendered, n _t=n _ty=k _lF _y, n _l=n _ly=k _tF _y, n herein _ly, n _tyBe respectively vertical, the yield force of vertical steel disc, f _yYield strength for steel pipe.By on can get:

α is the angle of concrete shear crack and l, can find out that by following formula carrying out of shear crack is relevant with the ratio of transverse and longitudinal steel disc yield force.

q _yShear flow when surrendering for concrete filled steel tube is the square root of the yield force product of steel disc.

Four, be subjected to the balance equation of shear-steel pipe concrete beam.

Cut the acting in conjunction that is subject to moment M and shear V from the beam element body that goes out, adopting the equivalent model of simple bending when the model equivalence is the basis, to wind up compression rod and lower edge tension member separated from unit main body, is used for opposing by the moment M that shearing produces, and increases in addition agent structure and bears shear V.Horizontal by what cut, vertically the concrete brace after steel disc and cracking forms for the opposing shearing.So just can separate two different mechanisms and analyze, selecting of the truss model that can also make has unitarity.

In truss model shown in Figure 6, the equivalent height of truss is d _v, in being cut analysis, we get is also d for length along the longitudinal direction _vThe beam infinitesimal, the moment of flexure on its left side is M, the moment of flexure on the right is M+Vd _v, the shearing of left and right face is all V, and the shear flow of generation is q, is a series of concrete inclined strut by crack segmentation after concrete cracking in main body, and the trend of crackle becomes the α angle with the l axle, bears size and is σ _dCompressive stress, be equivalent to the bar that baroclines in truss model.The transverse force that the equivalent steel disc of side direction steel pipe produces also can be resisted longitudinal shear, is equivalent to the perpendicular depression bar in truss model.

Based on above-mentioned model, adopted two supposition:

(1) the shear flow q on main body evenly distributes along height (namely vertical), because q is along being highly a constant, therefore V=qd is arranged _v

(2) the shear flow q on main body be also evenly distribution along length (namely vertical), thereby we think vertical steel disc stress n _tWith oblique concrete depression bar stress σ _dAlso even variation along its length.

Based on the above-mentioned model of choosing and supposition, we have following model sketch 8, regard beam as one vertical, be all vertically the large cell cube of cutting of dv, and ditto have vertically and vertical two equilibrium states that triangle of forces represents, following three balance equations are arranged:

V=(n _lb)·d _v·tanα

V=(n _tb)·d _v·cotα

V=(σ _db)d _vsinαcosα (10)

Steel disc longitudinal force for being produced by shearing makes N _l=(n _lB) dv, thickness are the steel pipe longitudinal force (n on unit of cells length of cutting of b _lB) multiply by the equivalent height dv of truss, i.e. whole longitudinal force N for being caused by shearing action _lWhen concrete filled steel tube is surrendered, n _t=n _ty=k _lF _y, n _l=n _ly=k _tF _y, n herein _ly, n _tyBe respectively vertical, the yield force of vertical steel disc, f _yBe the yield strength of steel pipe,

V _yShearing when surrendering for concrete filled steel tube.Can calculate the vertical and vertical tensile force that steel pipe produces by shear V when design, and then make up with the longitudinal stretching power that is produced by bending.

Five, the pure member balance equation of turning round.

St.Venant St.Venant reverses theory to be had, and torsional deflection has following characteristics: after (1) is reversed, shape of cross section remains unchanged; (2) torsional deformation perpendicular to xsect is equal to along member length.

For concrete filled steel tubular member, getting vertical and horizontal in being turned round concrete filled steel tubular member is all d _vCut cell cube.Because reversing relational expression, Bredt all is suitable for for section form arbitrarily as shown in Figure 7, therefore for rectangle or circular steel tube concrete member, q=T/ ∮ rds=T/ (2A is arranged ₀).For above-mentioned cut cells body, it only is subjected to pure shear, and the area A that on xsect, the shear flow center line surrounds ₀=d _v ², the equivalent shear force stream q=T/ (2A that is produced by torque T ₀)=T/ (2d _v ²), the shear V=qd of equivalence _v=T/ (2d _v ²).Be inlaid with equivalent steel disc in the cut cells body, its equivalent stress is vertical n _l, vertical horizontal is to n _tTherefore, can use formula (9) expression by cell cube shearing balance equation, the V that replaces wherein with torque T has the following balanced type of being turned round:

T=(n _lb)(2A ₀)·tanα

T=(n _tb)(2A ₀)·cotα

T=(σ _db)(2A ₀)sinαcosα (11)

Steel disc longitudinal force for being produced by moment of torsion makes N _l=(n _lB) P ₀, thickness is the longitudinal force (n on unit of cells length of cutting of b _lB) multiply by the girth P of shear flow ₀Be whole longitudinal force N that twisting action causes _lWhen concrete filled steel tube is surrendered, n _t=n _ty=k ₁F _y, n _l=n _ly=k _tF _y, n herein _ly, n _tyBe respectively vertical, the yield force of vertical steel disc, f _yBe the yield strength of steel pipe,

T _ySuffered moment of torsion when surrendering for concrete filled steel tube.

Six, the equivalence of steel tube concrete soil model under multiple Load Combination effect.

Through several joints in front, concrete filled steel tube is born respectively pressure, curved, cut, the derivation of the balance equation when turning round load and the equivalence of model, we will be subjected under the Subjected To Compression, Bending, Shear And Torsion complex stress is by depression bar with the concrete filled steel tubular member equivalence, pull bar and the space-truss model of being cut the bar composition, and the thickness of truss is b _v, equivalent height is d _v, as shown in Figure 8: can think at the truss model of equivalence to be formed by two parts, namely bear the equivalent bending resistance truss element of Moment and bear the antitorque truss element of equivalent shearing resistance of shear stress effect.Equivalence bending resistance truss element is by being d at a distance of equivalent height _vOn press down pull bar and form, be used for bearing the longitudinal normal stress that moment of flexure and axle power produce.The antitorque truss element of shearing resistance is the main part of truss, and is horizontal by the equivalence of shearing resistance, and vertically the concrete inclined strut after steel disc and cracking forms, and jointly resists the shear stress that moment of torsion and shearing cause.According to above-mentioned division to equivalent truss model, the equivalent bending resistance truss that can be in computation process will be formed by wind up compression rod and lower edge tension member and being separated from unit main body respectively by the antitorque truss element of shearing resistance that the shearing resistance agent structure forms, be respectively used to resist normal stress and shear stress, so just can separate two different force mechanisms and analyze, simplified calculating, can make again selecting of truss model have unitarity simultaneously.

Seven, under different Load Combination effects, bearing capacity of concrete-filled steel tube is found the solution.

(1) at first curved Load Combination is cut in discussion

Concrete filled steel tubular member for being cut simultaneously with moment load has truss model as shown in Figure 9, and for top is depression bar, the bottom is the bending resistance truss that forms of pull bar and middlely is oblique and vertical depression bar, and height is d _vEquivalent model, moment M makes lower boom produce pulling force M/d _v, top boom produces the pressure-M/d that equates with it _v, shear V acts on is cut on the unit, and the vertical and vertical reinforcing bar power of its generation is respectively N _l=(n _lB) dv, N _t=(n _tB) dv, substitution beam are cut flat weighing apparatus equation (10) to be had: V=(n _lB) d _vTan α=N _lTan α has the upper lower beam column balancing equation of truss:

The pressure of top boom $N_{\mathrm{tl}}=-\frac{M}{d_v}+\frac{1}{2}{N}_l=-\frac{M}{d_v}+\frac{1}{2}V\cot \mathrm{\&alpha;},$

The pulling force of lower boom $N_{\mathrm{bl}}=\frac{M}{d_v}+\frac{1}{2}{N}_l=\frac{M}{d_v}+\frac{1}{2}V\cot \mathrm{\&alpha;},$

Pressure (the σ of brace _db)d _vCos α=V/sin α,

The equivalent force of vertical steel pipe is N _t=(n _tB) dv=Vtan α.

Owing to being subjected to top boom and the suffered pressure equal and opposite in direction of lower boom when curved in the concrete filled steel tube equivalent model, opposite direction, when being cut, top boom and lower boom are subject to equal-sized longitudinal force, lower edge is the effect of stack, wind up and be the effect of cancelling out each other, therefore must destroy for lower boom with vertical steel disc, surrender occurs in the equivalent model of the concrete filled steel tubular member under shearing and moment of flexure acting in conjunction, have therefore lower boom is carried out force analysis: N _bl=M/d _v+ (1/2) Vcot α, V=(n _tB) d _vCot α, the substitution α that divides out has

Abbreviation has: $\frac{M}{N_{\mathrm{bl}}\&CenterDot;d_v}+\frac{V^2}{\left(n_{t}b\right)\&CenterDot;\left(\frac{2N_{\mathrm{bl}}}{d_v}\right){{\mathrm{<figure-callout\; id="2"\; label="d\; v"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">d</figure-callout>}}_v}^2}=1$

Following formula is M, and V correlationship equation owing to being lower boom and vertical steel disc generation yield failure, has N _bl=N _bly, n _t=n _yAs to get bending strength be M _u, shearing strength is V _u, by the balance equation of front, M is arranged when being subjected to curved surrender _u=N _blyd _vNow determine shearing strength V _u, when concrete filled steel tube is cut separately surrender, cause that by shearing the vertical yield force of steel disc is N in equivalent truss _ly=(n _lyB) dv.Owing to being that lower boom is first surrendered here, thus the lower edge stick force of the stick force of winding up during surrender during less than surrender, thereby according to top boom yield force N _tlyDetermine shearing strength V _uTo draw shearing strength V _uMinimum on the occasion of, by N _tll=1/2N _ly=1/2 (n _lyB) dv can derive $V_u=\mathrm{dv}\&CenterDot;b\sqrt{n_{\mathrm{ty}}\&CenterDot;n_{\mathrm{ly}}}=\sqrt{(n_{\mathrm{ty}}.b)\&CenterDot;(2N_{\mathrm{tly}}/\mathrm{dv}){\mathrm{<figure-callout\; id="2"\; label="dv"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">dv</figure-callout>}}^2}.$

With M _uAnd V _uThe relational expression substitution above the correlationship formula have:

$\frac{M}{N_{\mathrm{bl}}\&CenterDot;d_v}+\frac{V^2}{\left(n_{t}b\right)\&CenterDot;\left(\frac{2N_{\mathrm{bl}}}{d_v}\right){{\mathrm{<figure-callout\; id="2"\; label="d\; v"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">d</figure-callout>}}_v}^2}=1\&DoubleRightArrow;\frac{M}{M_u}+\frac{V^2}{{V^2}_u}\&CenterDot;\left(\frac{N_{\mathrm{tly}}}{N_{\mathrm{bly}}}\right)=1$

For concrete filled steel tubular member, because it has symmetry, therefore the yield force N of the upper lower beam of its equivalent truss _tly=N _blyTherefore, the dependent equation abbreviation is had:

$\frac{M}{M_u}+{\left(\frac{V}{V_u}\right)}^2=1---\left(12\right)$

(2) twist and warping Load Combination

Concrete filled steel tubular member for being cut simultaneously with moment load has truss model as shown in figure 10, and for top is depression bar, the bottom is the bending resistance truss that forms of pull bar and middlely is oblique and vertical depression bar, and height is d _vEquivalent model. in above-mentioned model, moment M makes lower boom produce pulling force M/d _v, top boom produces the pressure-M/d that equates with it _v, torque T acts on is cut on the unit, by the unit of being cut of deriving previously, the steel pipe of generation longitudinal force N is arranged _l=(n _lb)p ₀, because symmetrical, therefore wind up and the bar of lower edge respectively bears N _l/ 2=1/2 (n _lB) P ₀Pulling force.Laterally, torque T also will produce N in vertical steel disc _t=(n _tB) (2A ₀) transverse force.

Balance equation (3-36) is turned round in substitution to be had: T=(n _lB) (2A ₀) tan α=N _lTan α, T=(n _tB) (2A ₀) cot α, T=(σ _dB) (2A ₀) sin α cos α has the upper lower beam column balancing equation of truss:

The pressure of top boom $N_{\mathrm{tl}}=-\frac{M}{d_v}+\frac{1}{2}{N}_l=-\frac{M}{d_v}+T\&CenterDot;\frac{P_0}{\left(4A_0\right)}\&CenterDot;\cot \mathrm{\&alpha;},$

The pulling force of lower boom $N_{\mathrm{bl}}=\frac{M}{d_v}+\frac{1}{2}{N}_l=\frac{M}{d_v}+T\&CenterDot;\frac{P_0}{\left(4A_0\right)}\&CenterDot;\cot \mathrm{\&alpha;},$

Pressure (the σ of brace _dB) (2A ₀) cos α=T/sin α,

Laterally the equivalent force of steel pipe is $\left(n_{t}b\right)P_0=T\&CenterDot;\frac{P_0}{\left(2A_0\right)}\&CenterDot;\tan \mathrm{\&alpha;}.$

Owing to being subjected to top boom and the suffered pressure equal and opposite in direction of lower boom when curved in the concrete filled steel tube equivalent model, opposite direction, when being twisted and warped, top boom and lower boom are subject to equal-sized longitudinal force, lower edge is the effect of stack, wind up and be the effect of cancelling out each other, therefore must destroy for lower boom with horizontal steel pipe, surrender occurs in the equivalent model of the concrete filled steel tubular member under moment of torsion and moment of flexure acting in conjunction, have therefore lower boom is carried out force analysis:

T=(n is arranged again _tB) (2A ₀) cot α, the substitution α that divides out has

$N_{\mathrm{bl}}=\frac{M}{d_v}+T\&CenterDot;\frac{P_0}{\left(4A_0\right)}\&CenterDot;\frac{T}{\left(n_{t}b\right)\left(2A_0\right)}=\frac{M}{d_v}+\frac{T^2}{\left(4{A_0}^2\right)}\&CenterDot;\frac{p_0}{2\left(n_{t}b\right)}$

Abbreviation has: $\frac{M}{N_{\mathrm{bl}}\&CenterDot;d_v}+\frac{T^2}{\left(n_{t}b\right)\&CenterDot;\left(\frac{2N_{\mathrm{bl}}}{P_0}\right)\left(4{A_0}^2\right)}=1$

Following formula is M, and T correlationship equation owing to being lower boom and laterally steel pipe generation yield failure, has N _bl=N _bly, n _t=n _tyAs to get bending strength be M _u, twisting strength is T _u, by the balance equation of front, M is arranged when being subjected to curved surrender _u=N _blyd _v, in like manner according to top boom yield force N _tlyDetermine shearing strength V _uTo draw shearing strength V _u, with N _tly=1/2N _lySubstitution has $T_u=2A_{0}b\sqrt{n_{\mathrm{ty}}\&CenterDot;n_{\mathrm{ly}}}==2A_0\sqrt{\left(n_{\mathrm{ty}}b\right)\&CenterDot;(2N_{\mathrm{tly}}/P_0)}.$ With M _uAnd T _uThe relational expression substitution above the correlationship formula have:

$\frac{M}{N_{\mathrm{bl}}\&CenterDot;d_v}+\frac{T^2}{\left(n_{t}b\right)\&CenterDot;\left(\frac{2N_{\mathrm{bl}}}{P_0}\right)\left(4{A_0}^2\right)}=1$ Abbreviation has $\frac{M}{M_u}+{\left(\frac{T}{T_u}\right)}^2=1---\left(13\right)$

(3) cut the twist and warping Load Combination

Utilizing the equivalent model of front, can be depression bar for top with its equivalence, and the bottom is that bending resistance truss and the centre that pull bar forms is oblique and vertical depression bar, and height is d _vThe shearing resistance truss.First the bending resistance truss is considered Moment, moment M makes lower boom produce pulling force M/d _v, top boom produces the pressure-M/d that equates with it _vConsider again to cut and turn round acting in conjunction and create antagonism and cut the shear stress that antitorque truss produces, as shown in figure 10 for the last shear flow of the antitorque equivalent truss xsect of shearing resistance distribution situation, outer shear V only produces shearing stress flow q at the left and right sides of truss web _v, by the equivalent truss size, q is arranged _v=V/2d _vOuter torque T all produces shearing stress flow q in four sides of truss _t, by the equivalent truss size, q is arranged _t=V/2A ₀Shear stress line definition by the front has: A ₀=d _v* b _v, P ₀=2 (d _v+ b _v)

Both shear flows of causing of shear and torsion are superimposed in four sides of truss xsect, use subscript t, and b, l, r represent respectively, under, a left side, the shear flow on right side has the shear flow of four sides to be respectively:

$q_t=\frac{T}{2A_0},$ $q_b=\frac{T}{2A_0},$ $q_l=\frac{T}{2A_0}+\frac{V}{2d_v},$ $q_r=\frac{T}{2A_0}-\frac{V}{2d_v}$

Due to the q value four sidewall differences, due to q=(n _tB) cot α is therefore the inclination alpha of concrete compression bar is also different at four sidewalls, so have:

$\cot {\mathrm{\&alpha;}}_t=\frac{1}{\left(n_{t}b\right)}\frac{T}{2A_0},$ $\cot {\mathrm{\&alpha;}}_b=\frac{1}{\left(n_{t}b\right)}\frac{T}{2A_0},$

$\cos {\mathrm{\&alpha;}}_l=\frac{1}{\left(n_{t}b\right)}(\frac{T}{2A_0}+\frac{V}{2d_v}),$ $\cos {\mathrm{\&alpha;}}_r=\frac{1}{\left(n_{t}b\right)}(\frac{T}{2A_0}-\frac{V}{2d_v})$

Owing to being subjected to top boom and the suffered pressure equal and opposite in direction of lower boom when curved in the concrete filled steel tube equivalent model, opposite direction, turned round that when cutting, top boom and lower boom are subject to equal-sized longitudinal force, lower edge is the effect of stack, wind up and be the effect of cancelling out each other, therefore the side that must turn round the shear flow stack that causes by lower boom and cutting in the equivalent model of the concrete filled steel tubular member under moment of torsion and moment of flexure acting in conjunction is the horizontal steel disc surrender of left side wall and destroying, and has therefore lower boom is carried out force analysis:

Produce balance equation by shear flow q=(n is arranged _lB) tan α, the shearing that is produced by top and bottom sidewall shear flow q is V=qb _v, the longitudinal force N that causes _l=Vcot α=qb _vCot α, the shearing that is produced by left and right sidewall shear flow q is V=qd _v, the longitudinal force N that causes _l=Vcot α=qd _vCot α.

By the shearing in upper figure, it is as follows that longitudinal force and concrete brace pressure equilibrium triangle can get a longitudinal force of sidewall shear flow generation:

$N_{\mathrm{lt}}=q_t\&CenterDot;b_v\&CenterDot;\cot {\mathrm{\&alpha;}}_t=\frac{T}{2A_0}\&CenterDot;b_v\&CenterDot;\frac{1}{\left(n_{t}b\right)}\frac{T}{2A_0}=\frac{b_v}{\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2$

$N_{\mathrm{lb}}=q_b\&CenterDot;b_v\&CenterDot;\cot {\mathrm{\&alpha;}}_b=\frac{T}{2A_0}\&CenterDot;b_v\&CenterDot;\frac{1}{\left(n_{t}b\right)}\frac{T}{2A_0}=\frac{b_v}{\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2$

$N_{\mathrm{ll}}=q_l\&CenterDot;d_v\&CenterDot;\cot {\mathrm{\&alpha;}}_l=(\frac{T}{2A_0}+\frac{V}{2d_v})\&CenterDot;d_v\&CenterDot;\frac{1}{\left(n_{t}b\right)}(\frac{T}{2A_0}+\frac{V}{2d_v})=\frac{d_v}{\left(n_{t}b\right)}{(\frac{T}{2A_0}+\frac{V}{2d_v})}^2$

$N_{\mathrm{lr}}=q_r\&CenterDot;d_v\&CenterDot;\cot {\mathrm{\&alpha;}}_r=(\frac{T}{2A_0}-\frac{V}{2d_v})\&CenterDot;d_v\&CenterDot;\frac{1}{\left(n_{t}b\right)}(\frac{T}{2A_0}-\frac{V}{2d_v})=\frac{d_v}{\left(n_{t}b\right)}{(\frac{T}{2A_0}+\frac{V}{2d_v})}^2$

Lower boom is suffered make a concerted effort be:

$N_{\mathrm{bl}}=\frac{M}{d_v}+\frac{1}{2}{N}_{\mathrm{lt}}+\frac{1}{2}{N}_{\mathrm{lb}}+\frac{1}{2}{N}_{\mathrm{ll}}+\frac{1}{2}{N}_{\mathrm{lr}}=\frac{M}{d_v}+\frac{b_v+d_v}{\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2$

Arrangement has: $\frac{M}{N_{\mathrm{bl}}\&CenterDot;d_v}+\frac{P_0}{2N_{\mathrm{bl}}\&CenterDot;\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{N_{\mathrm{bl}}\&CenterDot;\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2=1$

Following formula is M, V, and T correlationship equation owing to being lower boom and laterally steel disc generation yield failure, has N _bl=N _bly, n _t=n _tyAs to get bending strength be M _u, shearing strength is V _u, twisting strength is T _uDerivation by the front has M _u=N _blyd _v, because be space-truss model, therefore shearing resistance yield strength V _UFor left and right two webbed side wall, therefore N _lBe twice, $V_u=\sqrt{(n_{\mathrm{ty}}.b)\&CenterDot;(4N_{\mathrm{tly}}/\mathrm{dv}){\mathrm{<figure-callout\; id="2"\; label="dv"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">dv</figure-callout>}}^2},$ $T_u=2A_0\sqrt{\left(n_{\mathrm{ty}}b\right)\&CenterDot;(2N_{\mathrm{tly}}/P_0)},$ With M _u, V _u, T _uThe relational expression substitution, concrete filled steel tube cut-turn round-curved correlationship formula has:

$\frac{M}{N_{\mathrm{bl}}\&CenterDot;d_v}+\frac{P_0}{2N_{\mathrm{bl}}\&CenterDot;\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{N_{\mathrm{bl}}\&CenterDot;\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2=1$

$\&DoubleRightArrow;\frac{M}{M_u}+a{\left(\frac{T}{T_u}\right)}^2+a{\left(\frac{V}{V_u}\right)}^2=1$

In like manner, abbreviation has: $\frac{M}{M_u}+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1---\left(14\right)$

(4) cut the twist and warping Load Combination

Axial compression is to produce axial compression stress on the impact of concrete filled steel tubular member, by steel pipe and concrete shared, therefore born by the bending resistance truss element of bearing moment of flexure in equivalent truss model in front, can not affect the internal balance mechanism of the antitorque truss element of shearing resistance that is formed by the shearing resistance main part, therefore only add an axial pressure in the equilibrium condition of top-bottom chord:

Pull bar: $N_{\mathrm{bl}}=\frac{M}{d_v}+\frac{P_0}{2\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2-\frac{\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">N</figure-callout>}}{2},$

Depression bar: $N_{\mathrm{tl}}=-\frac{M}{d_v}+\frac{P_0}{2\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2-\frac{\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">N</figure-callout>}}{2}$

Owing to having added axial pressure, therefore failure mode has two kinds, one is lower boom tension and vertical steel disc generation yield failure, and two is top boom pressurized and vertical steel disc generation yield failure, existing consideration respectively:

1) lower boom and vertical steel disc generation yield failure, have N _bl=N _bly, n _t=n _ty, the substitution following formula has: $\frac{M}{N_{\mathrm{bly}}\&CenterDot;d_v}-\frac{N}{2N_{\mathrm{bly}}}{+\frac{b_v+d_v}{N_{\mathrm{tly}}\&CenterDot;\left(n_{t}b\right)}\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{N_{\mathrm{tly}}\&CenterDot;\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2=1,$ As to get bending strength be M _u, shearing strength is V _u, twisting strength is T _u, had by the derivation of front: M _u=N _blyd _v, N _u=2N _bly $V_u=\sqrt{(n_{\mathrm{ty}}.b)\&CenterDot;\left(\frac{4N_{\mathrm{tly}}}{\mathrm{dv}}\right){\mathrm{<figure-callout\; id="2"\; label="dv"\; filenames="HSA00000189286600034.png"\; state="\{\{state\}\}">dv</figure-callout>}}^2},$ $T_u=2A_0\sqrt{\left(n_{\mathrm{ty}}b\right)\&CenterDot;\left(\frac{2N_{\mathrm{tly}}}{P_0}\right)}.$

The substitution following formula has: $(\frac{M}{M_u}-\frac{N}{N_u})+a{\left(\frac{T}{T_u}\right)}^2+a{\left(\frac{V}{V_u}\right)}^2=1$

2) top boom and laterally steel disc generation yield failure, have N _tl=N _tly, n _t=n _ty, the substitution following formula has: $-\frac{M}{N_{\mathrm{tly}}\&CenterDot;d_v}-\frac{N}{2N_{\mathrm{tly}}}{+\frac{b_v+d_v}{N_{\mathrm{tly}}\&CenterDot;\left(n_{t}b\right)}\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{N_{\mathrm{tly}}\&CenterDot;\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2=1$ The substitution bending strength is M _u, shearing strength is V _u, twisting strength is T _u, various have abbreviation to have:

The surrender of lower boom pressurized $(-\frac{N}{N_u}+\frac{M}{M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1$

The surrender of top boom pressurized: $(-\frac{N}{N_u}-\frac{M}{M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1$

Have following formula to set up simultaneously when upper lower beam is surrendered simultaneously, two formula addition abbreviations have

Therefore concrete filled steel tubular compression bending is cut and is turned round common surrender form when stressed and determined by the N value, when $\frac{N}{N_u}\&GreaterEqual;1-{\left(\frac{T}{T_u}\right)}^2-{\left(\frac{V}{V_u}\right)}^2$ The time, for the pressurized surrender, dependent equation is arranged $(-\frac{N}{N_u}+\frac{M}{M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1,$ When $\frac{N}{N_u}<1-{\left(\frac{T}{T_u}\right)}^2-{\left(\frac{V}{V_u}\right)}^2$ The time, be the tension surrender, dependent equation is arranged: $(-\frac{N}{N_u}-\frac{M}{M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1.$

Therefore the dependent equation of concrete filled steel tube when bearing Subjected To Compression, Bending, Shear And Torsion can arrange as follows:

When $\frac{N}{N_u}<1-{\left(\frac{T}{T_u}\right)}^2-{\left(\frac{V}{V_u}\right)}^2$ The time: $(-\frac{N}{N_u}+\frac{M}{M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1$

When $\frac{N}{N_u}\&GreaterEqual;1-{\left(\frac{T}{T_u}\right)}^2-{\left(\frac{V}{V_u}\right)}^2$ The time: $(-\frac{N}{N_u}-\frac{M}{M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1---\left(15\right)$

The preferred embodiment of the present invention is: the stressed effect that also comprises shearing, moment of flexure, moment of torsion shearing of described concrete filled steel tube, when determining the bearing capacity of concrete filled steel tube, consider that simultaneously steel pipe is to concrete restraint stress.

Used limit equilibrium theory concrete filled steel tube has been carried out the equivalence of model and the derivation of bearing capacity dependent equation under stressed Subjected To Compression, Bending, Shear And Torsion is multiple.Each prerequisite of deriving is not all considered steel pipe to concrete restraint stress, and steel pipe and concrete are treated as two independently independent bearing loads of member.Because hindering the inside flexing of steel pipe to the constraint of core concrete and core concrete, steel pipe caused the concrete filled steel tube mechanical property to be strengthened, therefore the cuff effect of introducing now between steel pipe and concrete is revised the dependent equation of deriving before.The present invention will revise from two aspects:

Interaction between steel pipe and concrete not only is embodied in the independent raising of its ultimate bearing capacity when stressed of concrete filled steel tube, when concrete filled steel tube is subjected to various combined load, the raising of its bearing capacity is not only that the interaction between steel pipe and concrete can also be strengthened the combined effect of load to its independent stack that bearing capacity improves when stressed.Reflect the Subjected To Combined Axial Loading And Bending effect with equivalent bending resistance truss element when deriving multiple bearing capacity under stressed with limit equilibrium method, reflect to cut with equivalent shearing resistance truss element and turn round combined effect, bend and cut to turn round and revise respectively therefore dependent equation is divided into, consider that steel pipe and concrete effect of contraction are mainly reflected in the raising to the press-bending effect, to cutting not obvious that the combined effect turned round improves, therefore only considering it is bent, the present invention revises.

Revised Subjected To Compression, Bending, Shear And Torsion bearing capacity dependent equation

$\frac{N}{N_u}\&GreaterEqual;0.2[1-{\left(\frac{T}{T_u}\right)}^2-{\left(\frac{V}{V_u}\right)}^2]$ The time:

$(\frac{N}{N_u}+\frac{{\mathrm{\&beta;}}_{m}M}{1.071(1-0.4N/N_E)M_u}){+\left(\frac{T}{T_u}\right)}^2{+\left(\frac{V}{V_u}\right)}^2=1$

$\frac{N}{N_u}<0.2[1-{\left(\frac{T}{T_u}\right)}^2-{\left(\frac{V}{V_u}\right)}^2]$ The time;

$(-\frac{N}{7N_u}+\frac{{\mathrm{\&beta;}}_{m}M}{(1-0.4N/N_E)M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1---\left(16\right)$

The specific embodiment of the present invention is: the bearing capacity of concrete-filled steel tube under multiple stressing conditions is determined that method is applied to concrete filled steel tubular member.

The specific embodiment of the present invention is: the bearing capacity of concrete-filled steel tube under multiple stressing conditions is determined that method is applied to the safety assessment of concrete filled steel tubular member.

Above content is in conjunction with concrete preferred implementation further description made for the present invention, can not assert that concrete enforcement of the present invention is confined to these explanations.For the general technical staff of the technical field of the invention, without departing from the inventive concept of the premise, can also make some simple deduction or replace, all should be considered as belonging to protection scope of the present invention.

Claims (3)

1. the bearing capacity of concrete-filled steel tube under one kind of multiple stressing conditions is determined method, described concrete filled steel tube multiple stressed is stressed for described concrete filled steel tube, the combination in any effect of moment of flexure, shear and torsion load, and its security assessment method comprises the steps:

Gather the correlation parameter of concrete filled steel tube: steel and concrete anti-compression, shearing strength design load in the collection concrete filled steel tube, sectional area and the concrete sectional area of steel pipe in the collection concrete filled steel tube;

Determine the bearing capacity of concrete filled steel tube: described concrete filled steel tube is under pressure, bending, shearing, the effect of moment of torsion Load Combination, and the equation that its bearing capacity satisfies is:

The time:

The time;

Wherein: M represents moment of flexure, and T represents moment of torsion, and V represents shearing;

Mu represents the bending strength of concrete filled steel tubular member, and Tu represents the twisting strength of concrete filled steel tubular member, and Vu represents the shearing strength of concrete filled steel tubular member,

Expression concrete filled steel tube long column stability factor.

2. the bearing capacity of concrete-filled steel tube under multiple stressing conditions is determined method according to claim 1, it is characterized in that, described concrete filled steel tubular member comprises rectangular elements, round member and octagon member.

3. the bearing capacity of concrete-filled steel tube under multiple stressing conditions is determined method according to claim 1, it is characterized in that, described concrete filled steel tubular member comprises solid concrete filled steel tubular member and concrete-filled steel tubular hollow member.

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