CN101900648A - 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 the 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

Along with the building raising that span constantly increases and the no antidetonation of building requires as a result, make the research of structure shearing resistance torsional property seem important.And concrete filled steel tube is in to press and turns round in the practical structures, and press-bending is turned round, even presses the state of bending and twisting 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 and pressing the multiple serviceability under stressed of bending and twisting, determines the form of its bearing capacity dependent equation.Because existing multiple stressed dependent equation down research all is to draw fitting formula behind the correlation curve by FEM (finite element) calculation, be theoretically unsound, especially easily practice is carried out the definite of bearing capacity of concrete-filled steel tube under multiple stress, and this has restricted the 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 the prior art under the multiple 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 shearing and moment of torsion load, its security assessment method comprises the steps:

Gather the correlation parameter of concrete filled steel tube: steel and concrete resistance to compression, shearing strength design load in the collection concrete filled steel tube, the 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, moment of torsion load compound action, 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 that is subjected to of described concrete filled steel tube is combined as a 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;

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

Further technical scheme of the present invention is: the load that is subjected to of described concrete filled steel tube is combined as a power, moment of flexure 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 that is subjected to of described concrete filled steel tube is combined as a 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 the 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 the 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 of the concrete filled steel tube under the multiple load compound action and determines method.The present invention can determine bearing capacity of concrete-filled steel tube under the multiple stress simply and directly, greatly facilitates determining of bearing capacity of concrete-filled steel tube under the multiple stress, helps promoting the application and the popularization of concrete filled steel tube.

Description of drawings

Fig. 1 is a 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 a present invention isoboles.

Fig. 4 cuts the model of element equivalent schematic for rectangle round steel pipe concrete of the present invention.

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

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

Fig. 7 is the stressed synoptic 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 the complicated applied force of the present invention.

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

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

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

Embodiment

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

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 the 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; Eight limit 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

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.

Each parameter is taken by " concrete filled steel tube technical regulation " in the formula;

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

A′＝0.2B′(1-ψ)+0.05ψ+0.05

$B^{,}=\frac{E_c}{E_s}$

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

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

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

Each parameter is taken by " concrete filled steel tube technical regulation " each continuous item in the formula

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 shearing and torsional interaction, the concrete filled steel tube security assessment method comprises the steps:

Step 100: the correlation parameter of gathering concrete filled steel tube: gather steel and concrete crushing strength design load in the concrete filled steel tube, gather the sectional area and the concrete sectional area of steel pipe in the 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 shearing and the torsional interaction, the equation that its bearing capacity satisfies is:

The time:

The time;

Determine the bearing capacity of concrete filled steel tube under the combination stressed effect of load, elder generation's limit of utilization level theory theoretical derivation draws the form at the dependent equation of not considering the bearing capacity of concrete filled steel tube under the combination stressed effect of complicated load under steel pipe and the effect of concrete cuff, by studying bearing capacity under the independent bearing load condition and the press-bending load bearing capacity under making up dependent equation is revised branch again.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 square steel tube xoncrete structure 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, and ring draws and horizontal pressurized three-dimensional stress state, but increasing along with shaft pressuring load, the steel pipe axial compression stress reduces, and the hoop tension increases, and concrete axial compression stress is because the effect of contraction increase of steel pipe 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.

Considering that moment of flexure does the time spent, do not consider steel pipe to concrete cuff effect earlier, then xial feed is promptly 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 the moment of flexure effect.

(1) at rectangular elements.

Concrete constitutive relation is got σ-ε curve that concrete standard is insulted cylinder, 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 the figure, moment of flexure produces compressive stress on the top in cross section, and the bottom in outer cross section produces tension, because a little less than the concrete stretching resistance, the concrete of bottom, cross section will ftracture, tension is mainly born by steel pipe, and wherein d is the significant height in cross section;

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

3) be the member top edge concrete compressive stress distribution plan that do not ftracture, the height of compressive region is x _c, the lower limb of compressive stress figure is a 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 the steel pipe stress diagram, 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 is surrounded, 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: $C=\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{bdx}$

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

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

Differential has

The substitution following formula has:

$C=\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)b\frac{x_c}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}\&CenterDot;\mathrm{d\&epsiv;}=b\&CenterDot;x_c\&CenterDot;\frac{C_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}={\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HSA00000189286600021.png,HSA00000189286600031.png"\; state="\{\{state\}\}">k</figure-callout>}}_1\&CenterDot;f_c\&CenterDot;b\&CenterDot;x_c---\left(2\right)$

$y_c=\frac{\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{xdx}}{\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)\mathrm{dx}}=\frac{\underset{0}{\overset{x_c}{\&Integral;}}{\mathrm{\&sigma;}}_c\left({\mathrm{\&epsiv;}}_c\right)b{\left(\frac{x_c}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}\right)}^2\&CenterDot;{\mathrm{\&epsiv;}}_c\&CenterDot;\mathrm{d\&epsiv;}}{b\&CenterDot;x_c\&CenterDot;\frac{C_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}}}={\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HSA00000189286600033.png,HSA00000189286600034.png"\; state="\{\{state\}\}">k</figure-callout>}}_2\&CenterDot;x_c---\left(3\right)$

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

Because t is much smaller than the steel tube component diameter, thus approximate think on, the stress value on the same cross section of downside steel pipe is equal, is respectively σ _Draw, σ _Press, and side plate stress with top is-σ _PressThe bottom is σ _DrawHold linear distribution.

If x _c=kd then has

Pressure above the steel pipe:

Pressure below the steel pipe: T _Down=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 (neutral axis is asked square)

If the depth-width ratio in concrete-filled rectangular steel tube cross section is λ, then 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 moment of flexure formula abbreviation is had:

5) axle equilibrium equation

C+T _On+ T _Press=T _Down+ 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 the formula: k ₁, k ₂---the concrete material performance parameter, depend on the concrete compression stress-strain curve. $k_1{f}_c=\frac{C_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}},{\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HSA00000189286600033.png,HSA00000189286600034.png"\; state="\{\{state\}\}">k</figure-callout>}}_2=\frac{y_{\mathrm{cu}}}{{\mathrm{\&epsiv;}}_{\mathrm{cu}}};$

K---the ratio of compressive region concrete compression height and depth of section in the 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 the concrete filled steel tube.

Order $j=k_2(\frac{\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HSA00000189286600033.png,HSA00000189286600034.png"\; state="\{\{state\}\}">k</figure-callout>}}{2}\&CenterDot;\frac{1-2k}{1-k})+\frac{(2k^2-2k+1)+\mathrm{\&lambda;}(k^2-\mathrm{<figure-callout\; id="1"\; label="k\; +"\; filenames="HSA00000189286600021.png,HSA00000189286600031.png"\; state="\{\{state\}\}">k</figure-callout>}+\frac{1}{3})}{2\&CenterDot;(1+\mathrm{\&lambda;})(1-k)},$

As can be seen, j is one and depth of compressive zone x _cThe related amount of=kd, then the interior moment of flexure of the outer moment of flexure of opposing 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 then arranged:

M＝f _s·A _s·jd?(8)

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

J---equivalent height coefficient.

(2) other, as the round steel pipe concrete component, from all directions the shape concrete component draws by same derivation:

Concrete filled steel tube simple bending member balance equation:

M＝f _s·A _s·jd?(8)

In the 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 _sAmong the jd as jd is used as the arm of force, can think outer moment of flexure system by a pair of equal and opposite in direction, direction is opposite, the parallel to each other and power N of jd apart _tAnd N ₁Resist, so the concrete filled steel tube flexural member can carry out equivalence with following truss model.

As shown in Figure 3, in equivalent model:

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

2) the suffered tension of member is born with the form of girt by steel pipe in the BENDING PROCESS, 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 is produced is: M=f _sA _sJd, then the dv=jd item is the truss depth of equivalence.It can also be seen that by following formula go up lower beam during surrender and surrender simultaneously, its yield strength is N _t=N _b=f _yA _s, antitorque bearing capacity is: M=f _sA _sJd;

4) above-mentioned analysis has illustrated that concrete filled steel tube is subjected to curved theoretical ultimate principle, because the equilibrium condition of its satisfied 1. concrete and steel pipe 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, so think that to its equivalent reasoning be strict.

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

M＝f _s·A _s·jd(8)

In the 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 1, 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 the cell cube.Do not considering under 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 in the concrete filled steel tube steel pipe to the steel pipe mean force of whole core concrete.In equivalence, steel disc is divided into vertical steel disc and vertical steel disc, resists transverse shearing force and longitudinal shear stream q respectively.

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, the plane outside sweep takes place, 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 takes place, opposing longitudinal shear stream q.So vertical shear q can be born jointly by all steel pipes 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, owing to have only the steel pipe of side direction can both resist longitudinal shear, the steel pipe equivalent stress of horizontal steel disc is in the cell cube:

2) vertical steel disc: the opposing vertical shear, because four side steel pipes can be resisted vertical shear, so the steel pipe equivalent stress of vertical steel disc is in the cell cube:

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 the cell cube.Do not consider under 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 in the concrete filled steel tube steel pipe to the steel pipe mean force of whole core concrete.In equivalence, steel disc is divided into vertical steel disc and horizontal steel disc, resists vertical shear and transverse shearing force stream q respectively.

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 takes place, 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 the extraneous shear V effect is vertical, become angle with hoop, so need reduction, here the way of Cai Yonging is that circular equivalence is the direction steel pipe of homalographic, according to derivation, still think and have only the side direction steel pipe that transverse stress opposing longitudinal shear stream can be provided, so have: the rectangular steel tube length of side to rectangular steel pipe

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

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

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

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 _lIn following derivation, it is not carried out unfolding calculation, only use n _t=k _tf _sAnd n _l==k _lf _sExpression, k _tAnd k _lBe the 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 1, and thickness is b, and then the equivalent force of the vertical steel disc in the 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 bar that baroclines by crack segmentation behind the 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 the concrete bar that baroclines, steel pipe vertical and longitudinally steel disc constituted the truss that bears shear flow.

Cell cube steel disc power of being cut that triangle is firmly represented 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 bears jointly.(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 bears jointly.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: q=(σ by horizontal triangle _dB) sin α cos α.

So cell cube shearing balance equation is a 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 subjected 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 to resist the moment M that is produced by shearing, increases agent structure in addition and bears shear V.Horizontal by what cut, vertically the concrete brace behind steel disc and the cracking is formed and then is used to resist shearing.So just can divide two different mechanisms to come and analyze, selecting for use 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 d for length also 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, about the shearing of face all be V, the shear flow of generation is q, is a series of concrete bar that baroclines by crack segmentation behind the concrete cracking in the main body, 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 the 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 the truss model.

Based on above-mentioned model, adopted two supposition:

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

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

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

V＝(n _lb)·d _v·tanα

V＝(n _tb)·d _v·cotα

V＝(σ _db)d _vsinαcos?α(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 the unit of cells length of cutting of b _lB) multiply by the equivalent height dv of truss, i.e. whole longitudinal force N for causing 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.When design, can calculate the vertical and vertical tensile force that steel pipe produces, and then make up with the longitudinal stretching power that produces by bending by shear V.

Five, the pure member balance equation of turning round.

St.Venant St.Venant reverses theory to be had, and torsional deflection has following characteristics: (1) is reversed the back shape of cross section and is remained 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 all is d _vCut cell cube.As shown in Figure 7 because Bredt reverses relational expression all is suitable for for section form arbitrarily, so, have for rectangle or round steel pipe concrete component

For above-mentioned cut cells body, it only is subjected to the pure shear effect, and the area A that the shear flow center line is surrounded on the xsect ₀=d _v ², then the equivalent shear force that is produced by torque T flows q=T/ (2A 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 ₁, perpendicular horizontal n _tSo, can use formula (9) expression by cell cube shearing balance equation, the V that replaces wherein with torque T then 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, then thickness is the longitudinal force (n on the 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 _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,

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

Six, the equivalence of steel tube concrete soil model under the multiple load compound action.

Through several joints in front concrete filled steel tube is born pressure respectively, curved, cut, the derivation of the balance equation when turning round load and the equivalence of model, we will be by depression bar with the concrete filled steel tubular member equivalence under the pressurized bending and twisting complex stress, pull bar and cut the space-truss model that bar is formed, 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 and to form, promptly bear the equivalent bending resistance truss element of moment of flexure effect and bear the antitorque truss element of equivalent shearing resistance of shear stress effect by two parts.Equivalence bending resistance truss element is by being d at a distance of equivalent height _vOn press down pull bar and form, be used to bear vertical normal stress that moment of flexure and axle power produce.The antitorque truss element of shearing resistance is the main part of truss, and by the equivalence of shearing resistance laterally, vertically the bar that baroclines of the concrete behind steel disc and the cracking is formed, and resists the shear stress that moment of torsion and shearing cause jointly.According to above-mentioned division to equivalent truss model, the equivalent bending resistance truss that can be in computation process will form by wind up compression rod and lower edge tension member and from unit main body, separate respectively by the antitorque truss element of shearing resistance that the shearing resistance agent structure is formed, be respectively applied for opposing normal stress and shear stress, so just can different separated by force mechanisms to analyze, simplified calculating, can make selecting for use of truss model have unitarity again simultaneously.

Seven, bearing capacity of concrete-filled steel tube is found the solution under the different load compound actions.

(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 formed of pull bar and middlely is oblique and vertical depression bar that 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) d _v, 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 last lower beam column balancing equation of truss:

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

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

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.

Because top boom and the suffered pressure equal and opposite in direction of lower boom when being subjected to bend in the concrete filled steel tube equivalent model, direction is opposite, top boom and lower boom are subjected to equal-sized longitudinal force when being cut, lower edge is the effect of stack, wind up and be the effect of cancelling out each other, so must destroy for lower boom with vertical steel disc surrender takes place in the equivalent model of the concrete filled steel tubular member under shearing and moment of flexure acting in conjunction, have so 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:

Following formula is M, and V correlationship equation owing to be lower boom and vertical steel disc generation yield failure, then has N _Bl=N _Bly, n _t=n _TyAs to get bending strength be M _u, shearing strength is V _u,, M is arranged when being subjected to curved surrender by the balance equation of front _u=N _Blyd _vNow determine shearing strength V _u, when concrete filled steel tube is cut surrender separately, in equivalent truss, cause that by shearing the vertical yield force of steel disc is N _Ly=(n _LyB) dv.Owing to be lower boom surrender earlier here, thus the lower edge stick force of stick force when surrendering of winding up during 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 _Tly=1/2n _Ly=1/2 (n _LyB) dv can derive

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

$\frac{M}{N_{\mathrm{bl}}\&CenterDot;d_v}+\frac{V^2}{\left(n_{t}b\right)\&CenterDot;\left(\frac{2N_{\mathrm{bl}}}{d_v}\right){{\mathrm{<figure-callout\; id="2"\; label="d\; v"\; filenames="HSA00000189286600033.png,HSA00000189286600034.png"\; state="\{\{state\}\}">d</figure-callout>}}_v}^2}=1\&DoubleRightArrow;\frac{M}{M_u}+\frac{V^2}{{V^2}_u}\&CenterDot;\left(\frac{N_{\mathrm{tly}}}{N_{\mathrm{bly}}}\right)=1$

For concrete filled steel tubular member, because it has symmetry, so the yield force N of the last lower beam of its equivalent truss _Tly=N _BlySo, 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 formed of pull bar and middlely is oblique and vertical depression bar that 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 symmetry, so the bar that winds up with 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 last lower beam column balancing equation of truss:

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

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

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

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

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

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

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

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 be lower boom and laterally steel pipe generation yield failure, then has N _Bl=N _Bly, n _t=n _TyAs to get bending strength be M _u, twisting strength is T _u,, M is arranged when being subjected to curved surrender by the balance equation of front _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

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 combination of twist and warping load

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 is formed is oblique and vertical depression bar, and height is d _vThe shearing resistance truss.Earlier the bending resistance truss is considered the moment of flexure effect, moment M makes lower boom produce pulling force M/d _vTop 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, q is arranged by equivalent truss size _v=V/2d _vOuter torque T then all produces shearing stress flow q in four sides of truss _t, q is arranged by equivalent truss size _t=V/2A ₀Shear stress line definition by the front has: A ₀=d _v* b _v, P ₀=2 (d _v+ b _v)

The shear flow that shearing and moment of torsion cause is superimposed in four sides of truss xsect, uses subscript t, b, and 1, r represents respectively, descends, a left side, the shear flow on right side then has the shear flow of four sides to be respectively:

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

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

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

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

Because top boom and the suffered pressure equal and opposite in direction of lower boom when being subjected to bend in the concrete filled steel tube equivalent model, direction is opposite, turned round that top boom and lower boom are subjected to equal-sized longitudinal force when cutting, lower edge is the effect of stack, wind up and be the effect of cancelling out each other, so 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 so lower boom is carried out force analysis:

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

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

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

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

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

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

Then 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 be lower boom and laterally steel disc generation yield failure, then has N _Bl=N _Bly, n _t=n _TyAs to get bending strength be M _u, twisting strength is V _u, twisting strength is T _uDerivation by the front has M _u=N _Blyd _v, because be space-truss model, so shearing resistance yield strength V _UFor about two webbed side wall, so N _lBe twice,

With M _u, V _u, T _uThe relational expression substitution, then concrete filled steel tube cut-turn round-the correlationship formula of bending 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 combination of twist and warping load

Axial compression is to produce axial compression stress to the influence of concrete filled steel tubular member, bear jointly by steel pipe and concrete, so bear by the bending resistance truss element of bearing moment of flexure in the equivalent truss model in front, can not influence the internal balance mechanism of the antitorque truss element of forming by the shearing resistance main part of shearing resistance, so only in the equilibrium condition of last lower boom, add an axial pressure:

Pull bar: $N_{\mathrm{bl}}=\frac{M}{d_v}+\frac{P_0}{2\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2-\frac{\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="HSA00000189286600033.png,HSA00000189286600034.png"\; state="\{\{state\}\}">N</figure-callout>}}{2},$

Depression bar: $N_{\mathrm{tl}}=-\frac{M}{d_v}+\frac{P_0}{2\left(n_{t}b\right)}{\left(\frac{T}{2A_0}\right)}^2+\frac{d_v}{\left(n_{t}b\right)}{\left(\frac{V}{2d_v}\right)}^2-\frac{\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="HSA00000189286600033.png,HSA00000189286600034.png"\; state="\{\{state\}\}">N</figure-callout>}}{2}$

Owing to added axial pressure, so 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 then have N _Bl=N _Bly, n _t=n _Ty, the substitution following formula has:

As to get bending strength be M _u, twisting strength is V _u, twisting strength is T _u, have by the derivation of front: M _u=N _Blyd _v, N _u=2N _Bly $V_u=\sqrt{(n_{\mathrm{ty}}.b)\&CenterDot;\left(\frac{4N_{\mathrm{tly}}}{\mathrm{dv}}\right){\mathrm{<figure-callout\; id="2"\; label="dv"\; filenames="HSA00000189286600033.png,HSA00000189286600034.png"\; state="\{\{state\}\}">dv</figure-callout>}}^2},$ $T_u=2A_0\sqrt{\left(n_{\mathrm{ty}}b\right)\&CenterDot;\left(\frac{2N_{\mathrm{tly}}}{P_0}\right)}.$

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 then have N _Tl=n _Tly, n _t=n _Ty, the substitution following formula has: The substitution bending strength is M _u, twisting 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 last lower beam is surrendered simultaneously, two formula addition abbreviations have To turn round common surrender form when stressed definite by N value so concrete filled steel tubular compression bending is cut, when

The time, for the pressurized surrender, dependent equation is arranged

When

The time, be the tension surrender, dependent equation is arranged:

So the dependent equation of concrete filled steel tube when bearing the pressure bending and twisting can be put in order 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: $(-\frac{N}{N_u}-\frac{M}{M_u})+{\left(\frac{T}{T_u}\right)}^2+{\left(\frac{V}{V_u}\right)}^2=1$ (15)

Preferred implementation 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 is being pressed bending and twisting multiple stressed carried out the down equivalence of model and the derivation of bearing capacity dependent equation.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 steel pipe has caused the concrete filled steel tube mechanical property to be strengthened to the constraint and the inside flexing of core concrete obstruction steel pipe of core concrete, so the cuff effect of introducing now between steel pipe and the concrete is revised the dependent equation of deriving before.The present invention will revise from two aspects:

Interaction between steel pipe and the 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.When deriving multiple bearing capacity under stressed with limit equilibrium method, reflect the press-bending combined effect with equivalent bending resistance truss element, reflect to cut with equivalent shearing resistance truss element and turn round combined effect, so dependent equation is divided into bends and cut to turn round and revise respectively, 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, so the present invention only considers its item that bends is revised.

Revised pressure bending and twisting 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;

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

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

Above content be in conjunction with concrete preferred implementation to further describing that the present invention did, 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 (8)

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, shearing and moment of torsion load, and its security assessment method comprises the steps:

Gather the correlation parameter of concrete filled steel tube: steel and concrete resistance to compression, shearing strength design load in the collection concrete filled steel tube, the 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, moment of torsion load compound action, and the equation that its bearing capacity satisfies is:

The time:

The time;

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

3. determine method according to the bearing capacity of concrete-filled steel tube under the described multiple stressing conditions of 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.

4. determine method according to the bearing capacity of concrete-filled steel tube under the described multiple stressing conditions of claim 1, it is characterized in that, the load that is subjected to of described concrete filled steel tube is combined as a power and moment of flexure, and when determining the bearing capacity of concrete filled steel tube, the equation that its bearing capacity satisfies is:

$\frac{N}{N_u}\&GreaterEqual;0.2$ The time:

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

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

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

5. determine method according to the bearing capacity of concrete-filled steel tube under the described multiple stressing conditions of claim 1, it is characterized in that, the load that is subjected to of described concrete filled steel tube is combined as a power, moment of flexure 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;

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

6. determine method according to the bearing capacity of concrete-filled steel tube under the described multiple stressing conditions of claim 1, it is characterized in that, the load that is subjected to of described concrete filled steel tube is combined as a 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;

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

7. a bearing capacity of concrete-filled steel tube of using under the multiple stressing conditions is determined the concrete filled steel tubular member of method, it is characterized in that, the bearing capacity of concrete-filled steel tube under the multiple stressing conditions is determined that method is applied to concrete filled steel tubular member.

8. a bearing capacity of concrete-filled steel tube of using under the multiple stressing conditions is determined the safety assessment of the concrete filled steel tubular member of method, it is characterized in that, the bearing capacity of concrete-filled steel tube under the multiple stressing conditions is determined that method is applied to the safety assessment of concrete filled steel tubular member.

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