CN101900653B - Method for determining anti-bending bearing capacity of stainless steel sandwich panel and application of stainless steel sandwich panel - Google Patents

Abstract

The invention relates to a method for determining the anti-bending bearing capacity of a stainless steel sandwich panel. The panel of the stainless steel sandwich panel is a stainless steel panel; and the core panel of the stainless steel sandwich panel is formed by sticking an interlayer with a binder or casting. The method for determining the anti-bending bearing capacity of the stainless steelsandwich panel comprises the following steps of: acquiring related parameters of the stainless steel sandwich panel; determining the deflection of the stainless steel sandwich panel; and determining the anti-bending bearing capacity of the stainless steel sandwich panel. The method for determining the anti-bending bearing capacity of the stainless steel sandwich panel precisely determines the anti-bending bearing capacity of the stainless steel sandwich panel by determining the deflection of the stainless steel sandwich panel under concentrated load and uniformly distributed load. The method for precisely determining the anti-bending bearing capacity of the stainless steel sandwich panel can precisely estimate the safety performance of the stainless steel sandwich panel so as to promote the application of the stainless steel sandwich panel.

Description

The stainless steel flexural capacity of sandwich plate is determined method and application

Technical field

The present invention relates to a kind of flexural capacity of sandwich plate and determine method and application, relate in particular to a kind of stainless steel flexural capacity of sandwich plate and determine method and application.

Background technology

Stainless steel is adiabatic to be to be bondd or pour into a mould by cementing agent by two bulk strengths are bigger up and down thin outer dash board (bearing bed, color steel) and light and middle layer (sandwich layer, novel stalk etc.) that matter is soft to form with battenboard.It has tangible advantageous combination, makes it become the ideal application material of wallboard and roof boarding.Metal faced have protective effect to sandwich layer, makes it avoid being mechanically damaged, and prevents weathering, isolated water and steam; And sandwich layer can connect into two surface layers integral body, common bearing load, and when surface layer under load action flexing took place, sandwich layer can also support surface layer, increased the ability of surface layer opposing flexing, and sandwich layer also has effects such as thermal insulation, sound insulation.Can be applicable to different building needs respectively, comprise a plurality of building fields such as industrial premises, public building, warehouse, assembled house, cleaning project.Prior art is determined still to rest on practice and the experimental formula for the stainless steel flexural capacity of sandwich plate, can not draw accurate conclusion, thus, cause to carry out accurate anti-bending bearing capacity and determine, restricted using and promoting of stainless steel battenboard greatly.

Summary of the invention

The technical matters that the present invention solves is: provide a kind of stainless steel flexural capacity of sandwich plate to determine method and application, overcome and can not accurately carry out the technical matters that anti-bending bearing capacity is determined in the prior art.

Technical scheme of the present invention is: provide a kind of stainless steel flexural capacity of sandwich plate to determine method, the panel of described stainless steel battenboard is corrosion resistant plate, the central layer of described stainless steel battenboard is that the middle layer forms by cementing agent bonding or cast, and the stainless steel flexural capacity of sandwich plate determines that method comprises the steps:

Gather the correlation parameter of stainless steel battenboard: the thickness of the elastic modulus of the effective cross-sectional area of the rigidity of the span of collection stainless steel battenboard, the width of stainless steel battenboard, stainless steel battenboard central layer, stainless steel battenboard panel rigidity, stainless steel battenboard core, effective modulus of shearing of stainless steel battenboard core, stainless steel battenboard plane materiel, the elastic modulus of stainless steel battenboard core, stainless steel battenboard plate thickness, stainless steel battenboard central layer.

Determine the amount of deflection of stainless steel battenboard: the amount of deflection of stainless steel battenboard determine to comprise amount of deflection under the load and the amount of deflection under the evenly load,

Amount of deflection under the stainless steel battenboard load adopts following formula to obtain:

$w_{\max }=\frac{{\mathrm{PL}}^3}{48E_1{\mathrm{<figure-callout\; id="1"\; label="I"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">I</figure-callout>}}_1}+\frac{\mathrm{k\&beta;PL}}{4A_{\mathrm{eff}}{G}_{\mathrm{eff}}}$

Amount of deflection under the stainless steel battenboard evenly load adopts following formula to obtain:

$w_{\max }=\frac{5p{L}^4}{384E_1{\mathrm{<figure-callout\; id="1"\; label="I"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">I</figure-callout>}}_1}+\frac{\mathrm{k\&beta;p}{L}^2}{8A_{\mathrm{eff}}{G}_{\mathrm{eff}}}$

Wherein: w represents the combined deflection of battenboard; P represents the payload values of battenboard panel; L represents the span of battenboard; E ₁The elastic modulus of expression plane materiel; I ₁Metal covering is to the moment of inertia of battenboard natural axis about the expression; A _EffThe net sectional area of expression battenboard; G _EffEffective modulus of shearing of expression battenboard; What k represented is the shear stress nonuniformity coefficient, can be taken as 1.2 for common template.

β represents the shearing partition factor of battenboard,

R ₁, R ₂, R ₃, R ₄Value sees Table 1.

Table 1: coefficients R ₁, R ₂, R ₃, R ₄List of values:

Determine the anti-bending bearing capacity of described stainless steel battenboard: the anti-bending bearing capacity of determining described stainless steel battenboard: the anti-bending bearing capacity of being determined the stainless steel battenboard by the relation of load and anti-anti-bending bearing capacity.

Further technical scheme of the present invention is: described stainless steel battenboard comprises shingle nail and roof boarding.

Further technical scheme of the present invention is: the core of described stainless steel battenboard is that rock wool forms by cementing agent bonding or cast.

Further technical scheme of the present invention is: the core of described stainless steel battenboard is that polyurethane forms by cementing agent bonding or cast.

Technical scheme of the present invention is: stainless steel straw flexural capacity of sandwich plate is determined that method is applied to stainless steel straw battenboard bearing of component and determines.

Technique effect of the present invention is: provide a kind of stainless steel flexural capacity of sandwich plate to determine method, by the amount of deflection under definite stainless steel battenboard load and the amount of deflection under the evenly load, accurately determine the anti-bending bearing capacity of stainless steel battenboard then.The present invention accurately determines the anti-bending bearing capacity of stainless steel battenboard, can more accurately assess the security performance of stainless steel battenboard, thereby advance the application of stainless steel battenboard.

Description of drawings

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

Fig. 2 is plane stainless steel battenboard cross-sectional view of the present invention.

Fig. 3 is the shallow die mould stainless steel of the present invention battenboard cross-sectional view.

Fig. 4 is the dark die mould stainless steel of the present invention battenboard cross-sectional view.

Fig. 5 is power, the stress of battenboard of the present invention and is out of shape a kind of synoptic diagram.

Fig. 6 is power, the stress of battenboard of the present invention and is out of shape another kind of synoptic diagram.

Fig. 7 is the plane battenboard structural representation under the evenly load of the present invention.

Fig. 8 is power and the distortion synoptic diagram of die mould stainless steel battenboard of the present invention.

Fig. 9 is divided into sandwich portion and edge of a wing structural representation for die mould stainless steel battenboard of the present invention.

Figure 10 is the stressed synoptic diagram of battenboard of the present invention under load.

Figure 11 is that surface stainless steel rock wool metope battenboard shearing partition factor of the present invention is with the thickness of slab change curve.

Figure 12 is the change curve of surface stainless steel rock wool metope battenboard shearing partition factor of the present invention with plate thickness.

Figure 13 is the change curve of surface stainless steel polyurethane metope battenboard shearing partition factor of the present invention along with thickness of slab.

Figure 14 is the graph of relation of surface stainless steel polyurethane metope battenboard shearing partition factor of the present invention with plate thickness.

Figure 15 is COHESIVE element stress figure of the present invention.

Figure 16 is COHESIVE of the present invention unit figure.

Figure 17 is the softening constitutive model of the linear elasticity-linearity of COHESIVE of the present invention unit.

Figure 18 is stainless steel rock wool roofing battenboard shearing partition factor of the present invention and plate thickness graph of relation.

Figure 19 is stainless steel rock wool roofing battenboard shearing partition factor of the present invention and plate thickness graph of relation.

Figure 20 is stainless steel glass silk flosssilk wadding roofing battenboard shearing partition factor of the present invention and plate thickness graph of relation.

Figure 21 is stainless steel glass silk flosssilk wadding roofing battenboard shearing partition factor of the present invention and plate thickness graph of relation.

Embodiment

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

As shown in Figure 1, the specific embodiment of the present invention is: provide a kind of stainless steel flexural capacity of sandwich plate to determine method, the panel of described stainless steel battenboard is corrosion resistant plate, and the central layer of described stainless steel battenboard is that the middle layer forms by cementing agent bonding or cast.Among the present invention, the concrete described stainless steel battenboard of stainless steel battenboard comprises shingle nail and roof boarding.

The stainless steel flexural capacity of sandwich plate determines that method comprises the steps:

Step 100: the correlation parameter of gathering the stainless steel battenboard: the thickness of the elastic modulus of the effective cross-sectional area of the rigidity of the span of collection stainless steel battenboard, the width of stainless steel battenboard, stainless steel battenboard central layer, stainless steel battenboard panel rigidity, stainless steel battenboard core, effective modulus of shearing of stainless steel battenboard core, stainless steel battenboard plane materiel, the elastic modulus of stainless steel battenboard core, stainless steel battenboard plate thickness, stainless steel battenboard central layer.

Step 200: the amount of deflection of determining the stainless steel battenboard: the amount of deflection of stainless steel battenboard determine to comprise amount of deflection under the load and the amount of deflection under the evenly load,

Amount of deflection under the stainless steel battenboard load adopts following formula to obtain:

$w_{\max }=\frac{{\mathrm{PL}}^3}{48E_1{\mathrm{<figure-callout\; id="1"\; label="I"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">I</figure-callout>}}_1}+\frac{\mathrm{k\&beta;PL}}{4A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(1\right)$

Amount of deflection under the stainless steel battenboard evenly load adopts following formula to obtain:

$w_{\max }=\frac{5p{L}^4}{384E_1{\mathrm{<figure-callout\; id="1"\; label="I"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">I</figure-callout>}}_1}+\frac{\mathrm{k\&beta;p}{L}^2}{8A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(2\right)$

Wherein: each label is with above-mentioned the same.

For the affirmation of shearing partition factor, the present invention further analyzes and draws by setting up model.Detailed process is as follows:

One, to the approximate processing of battenboard rigidity.

Core and panel secure bond together, cooperative transformation, and if the bendind rigidity of battenboard is K, if then regard battenboard as composite beam, because battenboard is made up of upper and lower panel and core, according to mechanics of materials stiffness Calculation formula, then it to central shaft O-O bendind rigidity is:

$K=2E_1{I}_3+E_1{\mathrm{<figure-callout\; id="1"\; label="I"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">I</figure-callout>}}_1+E_2{I}_2={\mathrm{<figure-callout\; id="1"\; label="E"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">E</figure-callout>}}_1\&times;\frac{{\mathrm{bt}}^3}{6}+{\mathrm{<figure-callout\; id="1"\; label="E"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">E</figure-callout>}}_1\&times;\frac{{\mathrm{btD}}_e^2}{2}+E_2\&times;\frac{{\mathrm{bD}}_c^3}{12}---\left(3\right)$

First is the rigidity of panel itself; Second is that panel is with respect to the rigidity of O-O; The 3rd is the rigidity of core itself.

In actual applications, second in the formula (3) plays a major role.The first, the three value is generally smaller, can ignore the rigidity of battenboard own to the contribution of integral rigidity.The 3rd with second ratio, for the ordinary soft core, all less than 1%, namely can ignore the influence of the rigidity of core own such as its ratios such as polyurethane, EPS (polystyrene), rock wool, glass silk flosssilk waddings.Because the core thickness of stainless steel battenboard differs, the stainless steel battenboard comprises shallow die mould stainless steel battenboard, plane stainless steel battenboard and dark die mould stainless steel battenboard, general roof boarding is for the general battenboard that adopts dark die mould of waterproof, and the general employing of shingle nail is plane battenboard or shallow die mould battenboard.

For shallow die mould, dark die mould, plane stainless steel battenboard, its rigidity can be expressed as:

$K_C=E_1{I}_1={\mathrm{<figure-callout\; id="1"\; label="E"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">E</figure-callout>}}_1\&times;\frac{{\mathrm{btD}}_e^2}{2}---\left(4\right)$

Two, to stainless steel battenboard effective cross-sectional area and the effective approximate processing of modulus of shearing.

(1) plane stainless steel battenboard.

As shown in Figure 2, the effective core thickness of plane stainless steel battenboard and core thickness approximately equal namely have following relation:

D _C≈D _eff G _eff＝GD _eff/D _C≈G A _eff＝bD _eff≈A _C (5)

(2) shallow die mould stainless steel battenboard.

As shown in Figure 3, the shallow effective core thickness of die mould stainless steel battenboard and core thickness approximately equal namely have following relation:

D _C≈D _eff G _eff＝GD _eff/D _C≈G A _eff＝bD _eff≈A _C (6)

(3) dark die mould stainless steel battenboard.

As shown in Figure 4, effective core thickness of dark die mould stainless steel battenboard and effective modulus of shearing are equal to core thickness, and core modulus of shearing relation is as follows:

D _eff＝D _C+d G _eff＝G _C(D _C+d)/D _C A _eff＝A _C(D _C+d)/D _C (7)

For common roofing die mould battenboard, d can be taken as 8.0mm.

Three, in the mechanics of materials about power and the distortion between relation

In the mechanics of materials, the relation of load and distortion can be expressed as follows.

Relation table between various power and the distortion:

Title	Expression formula	Title	Expression formula
				Amount of deflection	w	Moment of flexure	M＝-Kw″
Shearing	V＝-Dw″′	Even distributed force	q＝Kw iv

Four, the amount of deflection of plane stainless steel battenboard is calculated.

The distortion of stainless steel battenboard is considered respectively from two parts, the one, the flexural deformation of battenboard; Another one is exactly the detrusion of battenboard.And two-part distortion addition can be obtained the total deformation of battenboard.For the plane stainless steel battenboard, can ignore the factor that the shearing that brought by the bendind rigidity of panel itself distributes, can suppose that namely shearing born by core fully.For soft core, can suppose that moment of flexure born by panel fully;

As seen from the above table, power and distortion exist following relation, its synoptic diagram such as Fig. 5, Fig. 6:

M＝Kγ′ ₂＝K(γ′-w″) (8)

V＝A _effG _effγ (9)

Among Fig. 5, Fig. 6, γ ₁Be total strain, γ ₂The strain that the bending of serving as reasons causes, γ is shear strain.By Fig. 5, Fig. 6 as can be seen shear stress almost all born by core, namely shearing is born by core; And moment of flexure is provided by the normal stress of panel.

By moment of flexure, the differential relationship between shearing and the even distributed force can obtain following equation by the mechanics of materials:

$\frac{\mathrm{dM}}{\mathrm{dx}}-V=0---\left(10\right)$

$\frac{\mathrm{dV}}{\mathrm{dx}}+p=0---\left(11\right)$

With equation (8) with equation (9) is brought equation (10) into and (11) can get following relational expression:

K(γ″-w″′)-A _effG _effγ＝0 (12)

A _effG _effγ′＝-p (13)

The item that extracts about γ and w can obtain:

$w^{\&prime;\&prime;\&prime;\&prime;}=\frac{p}{K}-\frac{1}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}{p}^{\&prime;\&prime;}---\left(14\right)$

${\mathrm{\&gamma;}}^{\&prime;\&prime;}=-\frac{p^{\&prime;}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(15\right)$

Under actual conditions, moment of flexure and shearing are more handy usually.So can carry out integration to equation (14) and equation (15).Obtain following relational expression:

$w^{\&prime;\&prime;}=-\frac{M}{K}+\frac{V^{\&prime;}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(16\right)$

$\mathrm{\&gamma;}=\frac{V}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(17\right)$

Battenboard such as Fig. 7 for evenly load.

Therefore η=x/L wherein can obtain:

$M=\frac{p{L}^2}{2}(\mathrm{\&eta;}-{\mathrm{\&eta;}}^2)---\left(18\right)$

$V=\frac{\mathrm{pL}}{2}(1-2\mathrm{\&eta;})---\left(19\right)$

Can be obtained by equation (16):

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1^{\&prime;\&prime;}=-\frac{M}{K}=-\frac{p{L}^2}{2K}(\mathrm{\&eta;}-{\mathrm{\&eta;}}^2)---\left(20\right)$

Equation (20) is carried out integration twice, and m and n are constant term, can obtain:

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1=-\frac{{\mathrm{pL}}^4}{2K}(\frac{{\mathrm{\&eta;}}^3}{6}-\frac{{\mathrm{\&eta;}}^4}{12}+\mathrm{m\&eta;}+n)---\left(21\right)$

Boundary condition is: in the time of η=0 or 1, and w ₁=0; Therefore can calculate: m=-1/12, n=0.Therefore can obtain:

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1=\frac{p{L}^4\mathrm{\&eta;}}{24K}({\mathrm{\&eta;}}^3-2{\mathrm{\&eta;}}^2+1)=\frac{p{L}^4\mathrm{\&eta;}}{24K}(1-\mathrm{\&eta;})(1+\mathrm{\&eta;}-{\mathrm{\&eta;}}^2)---\left(22\right)$

By being deformed into that shearing causes:

$w_2^{\&prime;\&prime;}=\frac{V^{\&prime;}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}=-\frac{p}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(23\right)$

It is carried out twice integration gets:

$w_2=-\frac{{\mathrm{pL}}^2}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}(\frac{{\mathrm{\&eta;}}^2}{2}+\mathrm{m\&eta;}+n)---\left(24\right)$

Terminal conditions is: when η=0 and 1, and w ₂=0.Can get m=-0.5, n=0, so its value is:

$w_2=\frac{p{L}^2\mathrm{\&eta;}}{2A_{\mathrm{eff}}{G}_{\mathrm{eff}}}(1-\mathrm{\&eta;})---\left(25\right)$

So the total deformation of battenboard is:

$w={\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1+w_2=\frac{p{L}^4\mathrm{\&eta;}}{24K}(1-\mathrm{\&eta;})(1+\mathrm{\&eta;}-{\mathrm{\&eta;}}^2)+\frac{{\mathrm{pL}}^2\mathrm{\&eta;}}{2A_{\mathrm{eff}}{G}_{\mathrm{eff}}}(1-\mathrm{\&eta;})---\left(26\right)$

In the time of η=0.5, w can obtain maximal value as can be known:

$w_{\max }=\frac{5p{L}^4}{384K}+\frac{{\mathrm{pL}}^2}{8A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(27\right)$

Following formula can be turned to by equation (6):

$w_{\max }=\frac{5{\mathrm{pL}}^4}{384K}+\frac{{\mathrm{pL}}^2}{8A_C{G}_C}---\left(28\right)$

Former anti-bending bearing capacity formula need change into:

$f=\frac{{5\mathrm{pbl}}^4}{384\left(E_1{I}_1\right)}+\frac{{\mathrm{K\&beta;pbl}}^2}{8\mathrm{GA}}---\left(29\right)$

Wherein: E ₁---the elastic modulus of metal decking; Units MPa;

I ₁---go up lower steel plate with respect to the moment of inertia of neutral axis; The MIa of unit;

Last lower steel plate is to the moment of inertia I of battenboard natural axis ₁Approximate formula:

$I_1=\frac{A_u{A}_d}{A_u+A_d}{(D_C+d)}^2---\left(30\right)$

Wherein, A _uBoundary's area of section for upper steel plate; A _dArea of section for lower steel plate; D _CBe the sandwich thickness of slab; D be roof boarding upper steel plate centre of form axle to the distance of bottom surface, common roof boarding is taken as 8.0575mm.

The moment of inertia I of core itself ₂Approximate formula:

$I_2=\frac{b{(D_C+d)}^3}{12}---\left(31\right)$

Wherein, b is the battenboard width; D is battenboard thickness; D be roof boarding upper steel plate centre of form axle to the distance of bottom surface, common roof boarding is taken as 8.0575mm.

Five, the amount of deflection of die mould stainless steel battenboard is calculated.

(1) amount of deflection of die mould stainless steel battenboard under evenly load calculated.

When the used panel of battenboard is profiled sheet, need to consider the bendind rigidity of panel itself.Fig. 8 provided in this case power and the situation of distortion.Different with Fig. 5 is that it has considered the moment M that panel itself is born _F1And M _F2, the shear V that also has panel itself to bear _F1And V _F2

Relation by equation (8) and the resulting power of equation (9) and distortion does not change.And also had following relation:

M _F1＝-K _F1w″＝E ₁I _F1w″M _F2＝-K _F2w″＝E ₁I _F2w″ (32)

V _F1＝-K _F1w″′＝E ₁I _F1w″′ V _F1＝-K _F2w″′＝E ₁I _F1w″′(33)

Wherein, K _F1Be the rigidity of top panel, I _F1Be the moment of inertia of top panel own; K _F2Be the rigidity of lower panel, I _F2Moment of inertia for lower panel.Because last lower steel plate power is identical with the ratio of distortion, so following relational expression can be arranged:

M _D＝M _F1+M _F2 M＝M _D+M _C (34)

V _D＝V _F1+V _F2 V＝V _D+V _C (35)

K _D＝K _F1+K _F2 K＝K _D+K _C (36)

Above equation, battenboard has been divided into the part on sandwich portion and the edge of a wing, as shown in Figure 9.This hypothesis is practical in the application of reality.

From equation (8), equation (9), equation (32) and equation (33), can obtain following two differential equations in conjunction with equation (34), equation (35) and equation (3-36):

A _effG _effγ-K _Dw″′＝V (37)

K _Cγ′-K _Dw″＝M (38)

With V '=-p brings into, and cancellation γ can obtain one about the quadravalence differential equation of w:

$w^{\&prime;\&prime;\&prime;\&prime;}-{\left(\frac{\mathrm{\&lambda;}}{L}\right)}^2{w}^{\&prime;\&prime;}={\left(\frac{\mathrm{\&lambda;}}{L}\right)}^2\frac{M}{K}+\frac{1+\mathrm{\&alpha;}}{\mathrm{\&alpha;}}\frac{p}{K}---\left(39\right)$

Wherein L is the span of battenboard; α, δ and λ ²Value respectively as follows:

$\mathrm{\&alpha;}=\frac{K_D}{K_C},\mathrm{\&delta;}=\frac{K_C}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}{L}^2},{\mathrm{\&lambda;}}^2=\frac{1+\mathrm{\&alpha;}}{\mathrm{\&alpha;\&beta;}}---\left(40\right)$

Similarly, cancellation w can get in equation (37) and equation (38):

${\mathrm{\&gamma;}}^{\&prime;\&prime;}-{\left(\frac{\mathrm{\&lambda;}}{L}\right)}^2\mathrm{\&gamma;}=-\frac{\mathrm{\&delta;}{\mathrm{\&lambda;}}^{2}V}{K}---\left(41\right)$

When the moment of flexure of battenboard and shearing were known, the solution of equation (39) and (41) was:

$w={\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">m</figure-callout>}}_1\cosh \frac{\mathrm{\&lambda;x}}{L}+m_2\sinh \frac{\mathrm{\&lambda;x}}{L}+m_3+m_{4}x+w_p---\left(42\right)$

$\mathrm{\&gamma;}={\mathrm{<figure-callout\; id="1"\; label="n"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">n</figure-callout>}}_1\cosh \frac{\mathrm{\&lambda;x}}{L}+n_2\sinh \frac{\mathrm{\&lambda;x}}{L}+{\mathrm{\&gamma;}}_p---\left(43\right)$

W wherein _pAnd γ _pBe the integration particular solution relevant with load.Because solution must satisfy equation (8), can obtain following relational expression:

${\mathrm{<figure-callout\; id="1"\; label="n"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">n</figure-callout>}}_1=(1+\mathrm{\&alpha;})\frac{\mathrm{\&lambda;}}{\mathrm{<figure-callout\; id="1"\; label="L\; m"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">L</figure-callout>}}{m}_{}{}_1{,n}_2=(1+\mathrm{\&alpha;})\frac{\mathrm{\&lambda;}}{L}{m}_1---\left(44\right)$

Therefore the integration constant item coefficient of equation (42) and (43) has become four, and these integration constant item coefficients can obtain by boundary condition, and for the freely-supported battenboard, boundary condition is:

w(0)＝0 w″(0)＝0 w(L)＝0 w″(L)＝0 (45)

For evenly load, can get following relation by formula (18), formula (19):

$M=\frac{{\mathrm{pL}}^2}{2}(\mathrm{\&eta;}-{\mathrm{\&eta;}}^2),V=\frac{\mathrm{pL}}{2}(1-2\mathrm{\&eta;})$ η＝x/L (46)

Bringing equation (46) into formula (39) can obtain, and the particular solution that obtains in the formula (8) is

$w_p=\frac{{\mathrm{pL}}^4}{24K}({\mathrm{\&eta;}}^4-2{\mathrm{\&eta;}}^3-\frac{12}{\mathrm{\&alpha;}{\mathrm{\&lambda;}}^2}{\mathrm{\&eta;}}^2)---\left(47\right)$

Equation (46) is brought in the equation (41) and can be obtained, and the particular solution in the formula (43) is:

${\mathrm{\&gamma;}}_p=\frac{p{L}^3\mathrm{\&delta;}}{2K}(1-2\mathrm{\&eta;})---\left(48\right)$

Bring boundary condition (45) into equation (42) and equation (43) can obtain:

$\left.\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">m</figure-callout>}}_1+m_3+w_p=0\\ {\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">m</figure-callout>}}_1+w_p^{\&prime;\&prime;}=0\\ {\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">m</figure-callout>}}_1\cosh \mathrm{\&lambda;}+m_2\sinh \mathrm{\&lambda;}+m_3+m_{4}L+w_p=0\\ {\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">m</figure-callout>}}_1\frac{{\mathrm{\&lambda;}}^2}{L^2}\cosh \mathrm{\&lambda;}+m_2\frac{{\mathrm{\&lambda;}}^2}{L^2}\sinh \mathrm{\&lambda;}+w_p^{\&prime;\&prime;}=0\end{array}\right\}---\left(49\right)$

Can solve following relational expression:

$\left.\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">m</figure-callout>}}_1=\frac{{\mathrm{pL}}^4}{\mathrm{\&alpha;}{\mathrm{\&lambda;}}^{4}k};m_2=-\frac{{\mathrm{pL}}^4}{\mathrm{\&alpha;}{\mathrm{\&lambda;}}^{4}K}\frac{\cosh \mathrm{\&lambda;}-1}{\sinh \mathrm{\&lambda;}}\\ m_3=-\frac{{\mathrm{pL}}^4}{\mathrm{\&alpha;}{\mathrm{\&lambda;}}^{4}K};m_4=-\frac{{\mathrm{pL}}^4}{K}(\frac{1}{24}+\frac{1}{2\mathrm{\&alpha;}{\mathrm{\&lambda;}}^2})\end{array}\right\}---\left(50\right)$

Therefore final solution is:

$w=\frac{{\mathrm{pL}}^4}{K}[\frac{1}{24}\mathrm{\&eta;}(1-2{\mathrm{\&eta;}}^2+{\mathrm{\&eta;}}^3)+\frac{\mathrm{\&eta;}(1-\mathrm{\&eta;})}{2{\mathrm{\&alpha;\&lambda;}}^2}-\frac{\cosh \frac{\mathrm{\&lambda;}}{2}}{\mathrm{\&alpha;}{\mathrm{\&lambda;}}^4\cosh \frac{\mathrm{\&lambda;}}{2}}]---\left(51\right)$

Mid-span deflection can be got maximal value, and bringing η=0.5 into equation (48) can obtain:

$w_{0.5}=\frac{{\mathrm{pL}}^4}{K}(\frac{5}{384}+\frac{1}{8{\mathrm{\&alpha;\&lambda;}}^4}-\frac{\cosh \frac{\mathrm{\&lambda;}}{2}-1}{{\mathrm{\&alpha;\&lambda;}}^4\cosh \frac{\mathrm{\&lambda;}}{2}})---\left(52\right)$

As can be seen, following formula calculates very loaded down with trivial details, and application is not strong in the actual engineering.Therefore based on identical theory it is simplified below.

As seen from Figure 9, the die mould battenboard can be divided into two parts, a part is shearing and the moment of flexure that the rigidity of profiled sheet itself is born.A part is sandwich portion in addition, i.e. the shearing of being born by core and the moment of flexure of being born by axle power and the core itself of panel.Supposing that two parts are independently, is again cooperative transformation at contact point still.Two coefficients have been introduced here, i.e. the shearing partition factor of the distribution coefficient of bending moment ε of sandwich portion and interlayer (being approximately equal to the shearing partition factor of core) β.Therefore can obtain following relation by equation (8) and equation (9):

M _C＝K _C(γ′-w″) (53)

M _D＝-K _Dw″ (54)

V _D＝-K _Dw″′ (55)

V _C＝A _effG _effγ (56)

Can obtain following equation by equation (34) and (35):

$\mathrm{\&beta;}=\frac{V_C}{V_C+V_D},\mathrm{\&epsiv;}=\frac{M_C}{M_C+M_D}---\left(57\right)$

Can obtain following equation by equation (53)～(54):

$\mathrm{\&beta;}=\frac{A_{\mathrm{eff}}{G}_{\mathrm{eff}}\mathrm{\&gamma;}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}-K_D{w}^{\&prime;\&prime;\&prime;}}---\left(58\right)$

$\mathrm{\&epsiv;}=\frac{K_C({\mathrm{\&gamma;}}^{\&prime;}-w^{\&prime;\&prime;})}{K_C({\mathrm{\&gamma;}}^{\&prime;}-w^{\&prime;\&prime;})-K_D{w}^{\&prime;\&prime;}}---\left(59\right)$

And can obtain following relation:

K _C(γ′-w″)＝εM A _effG _effγ＝βV (60)

Can be got by equation (10) and equation (11):

K _C(γ″-w″′)/ε＝A _effG _effγ/β (61)

A _effG _effγ′/β＝-p (62)

Proposition can obtain following relation about the item of w and γ:

$w^{\&prime;\&prime;}=-\frac{\mathrm{\&epsiv;M}}{K_C}+\frac{V^{\&prime;}\mathrm{\&beta;}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(63\right)$

$\mathrm{\&gamma;}=\frac{\mathrm{\&beta;V}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(64\right)$

The amount of deflection that is caused by crooked and shearing is as follows respectively:

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1^{\&prime;\&prime;}=-\frac{\mathrm{\&epsiv;M}}{K_C},w_2^{\&prime;\&prime;}=\frac{{\mathrm{\&beta;V}}^{\&prime;}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(65\right)$

For evenly load, it is as follows respectively to obtain its moment and shearing:

$M=\frac{{\mathrm{pL}}^2}{2}(\mathrm{\&eta;}-{\mathrm{\&eta;}}^2),V=\frac{\mathrm{pL}}{2}(1-2\mathrm{\&eta;})---\left(66\right)$

Bring equation (66) into equation (65), and carry out twice integration and can obtain following equation:

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1=-\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^4}{2K_C}(\frac{{\mathrm{\&eta;}}^3}{6}-\frac{{\mathrm{\&eta;}}^4}{12}+\mathrm{m\&eta;}+n)---\left(67\right)$

$w_2=\frac{{\mathrm{\&beta;PL}}^2}{{2A}_{\mathrm{eff}}{G}_{\mathrm{eff}}}(\frac{{\mathrm{\&eta;}}^2}{2}+\mathrm{m\&eta;}+n)---\left(68\right)$

For the freely-supported battenboard, in the time of η=0 or 1, w ₁=0, w ₂=0; Obtain following equation:

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1=\frac{{\mathrm{\&epsiv;pL}}^4\mathrm{\&eta;}}{24K_C}(1-\mathrm{\&eta;})(1+\mathrm{\&eta;}-{\mathrm{\&eta;}}^2)---\left(69\right)$

$w_2=\frac{{\mathrm{\&beta;pL}}^2\mathrm{\&eta;}}{2A_{\mathrm{eff}}{G}_{\mathrm{eff}}}(1-\mathrm{\&eta;})---\left(70\right)$

So total distortion computing formula is as follows:

$w={\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1+w_2=\frac{{\mathrm{\&epsiv;pL}}^4\mathrm{\&eta;}}{24K_C}(1-\mathrm{\&eta;})(1+\mathrm{\&eta;}-{\mathrm{\&eta;}}^2)+\frac{\mathrm{\&beta;}{\mathrm{pL}}^2\mathrm{\&eta;}}{2A_{\mathrm{eff}}{G}_{\mathrm{eff}}}(1-\mathrm{\&eta;})---\left(71\right)$

Therefore in the time of η=0.5, mid-span deflection can be got maximal value:

$w=\frac{5\mathrm{\&epsiv;}{\mathrm{pL}}^4}{384K_C}+\frac{{\mathrm{\&beta;pL}}^2}{8A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(72\right)$

Shearing partition factor β for core can be definite with finite element analysis method, and introduced the inhomogeneous partition factor k of shear stress, and β is taken as k β in the formula (72), and wherein the concrete computing method of β are:

$\mathrm{\&beta;}=\frac{\stackrel{\&OverBar;}{\mathrm{\&tau;}}\&times;A}{Q}---\left(73\right)$

Wherein,

Average shearing stress near the sandwich material cross section bearing, A is the area in sandwich material cross section, Q is not equally distributed by the shear stress of battenboard in the actual conditions along the distribution of thickness of slab for total shear value at this place, can get the inhomogeneous partition factor k of a shear stress, can be similar to for the square-section and be taken as 1.2.

Distribution coefficient of bending moment for the sandwich part can be done as down conversion:

$\frac{K_C}{\mathrm{\&epsiv;}}=\frac{K_C({\mathrm{\&gamma;}}^{\&prime;}-w^{\&prime;\&prime;})-K_D{w}^{\&prime;\&prime;}}{K_C({\mathrm{\&gamma;}}^{\&prime;}-w^{\&prime;\&prime;})}\&times;K_C=\frac{K_C{\mathrm{\&gamma;}}^{\&prime;}-{\mathrm{Kw}}^{\&prime;\&prime;}}{{\mathrm{\&gamma;}}^{\&prime;}-w^{\&prime;\&prime;}}\&ap;K_C---\left(74\right)$

Therefore formula (72) can be converted into equation:

$w=\frac{5{\mathrm{pL}}^4}{384K_C}+\frac{{\mathrm{k\&beta;pL}}^2}{8A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(75\right)$

Can be got by equation (6) for shallow die mould battenboard:

$w=\frac{5{\mathrm{pL}}^4}{384K_C}+\frac{{\mathrm{k\&beta;pL}}^2}{8A_{C}G}---\left(76\right)$

K _CBendind rigidity for the sandwich portion of battenboard.

Can be got by equation (7) for dark profiled sheet:

$w=\frac{5{\mathrm{pL}}^4}{384K_C}+\frac{{\mathrm{k\&beta;pL}}^4}{8A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(77\right)$

K wherein _CBendind rigidity for sandwich portion.

(2) amount of deflection of die mould stainless steel battenboard under load calculated.

Its stressed form such as Figure 10 under the effect of load derive and adopt the method for simplification similar to the above, and shearing and the moment of flexure value of its arbitrary section that we can calculate for load are as follows:

M＝PL(1-δ)η-PL{η-δ} (78)

V＝P(1-δ)-P{η-δ} (79)

Wherein { η-ε } is taken as 1 in the time of η-ε＞0, otherwise is taken as zero.

Therefore can obtain following relational expression by equation (63) and equation (64):

Distortion for sweep has following relational expression:

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1^{\&prime;\&prime;}=-\frac{\mathrm{\&epsiv;M}}{K_C}=-\frac{\mathrm{PL}}{K_C}[(1-\mathrm{\&delta;})\mathrm{\&eta;}-\{\mathrm{\&eta;}-\mathrm{\&delta;}\left\}\right]---\left(80\right)$

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1=-\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^3}{K_C}[(1-\mathrm{\&delta;})\frac{{\mathrm{\&eta;}}^3}{6}-\frac{{\{\mathrm{\&eta;}-\mathrm{\&delta;}\}}^3}{6}+\mathrm{m\&eta;}+n]---\left(81\right)$

Its boundary condition is: w in the time of η=0 and η=1 ₁=0, therefore can obtain n=0, and:

$m=-\frac{1-\mathrm{\&delta;}}{6}(2\mathrm{\&delta;}-{\mathrm{\&delta;}}^2)---\left(82\right)$

Bringing equation (82) into can obtain:

${\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="HSA00000189235000052.png"\; state="\{\{state\}\}">w</figure-callout>}}_1=\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^3}{6K_C}[-(1-\mathrm{\&delta;}){\mathrm{\&xi;}}^3+{\{\mathrm{\&xi;}-\mathrm{\&delta;}\}}^3+\mathrm{\&xi;\&delta;}(1-\mathrm{\&delta;})(2-\mathrm{\&delta;})]---\left(83\right)$

Can obtain, the research point is in the load left end:

$w_{1L}=\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^3}{{6K}_C}(1-\mathrm{\&delta;})\mathrm{\&xi;}(2\mathrm{\&delta;}-{\mathrm{\&delta;}}^2-{\mathrm{\&xi;}}^2)---\left(84\right)$

(83) are put in order, can be obtained being deformed into of load right-hand member reference point:

$w_{1R}=\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^3}{{6K}_C}\mathrm{\&delta;}(1-\mathrm{\&xi;})(2\mathrm{\&xi;}-{\mathrm{\&xi;}}^2-{\mathrm{\&delta;}}^2)---\left(85\right)$

It is same that can to obtain detrusion as follows:

$w_2=\&Integral;\&Integral;\frac{\mathrm{\&beta;}{\mathrm{VL}}^2}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}\mathrm{d\&xi;d\&xi;}=\frac{\mathrm{PL}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}[\mathrm{\&xi;}(1-\mathrm{\&delta;})-\{\mathrm{\&xi;}-\mathrm{\&delta;}\}+\mathrm{m\&xi;}+n]---\left(86\right)$

By boundary condition η=0 or 1 o'clock, w ₂=0 can obtain: m=n=0 so detrusion can turn to:

$w_2=\frac{\mathrm{PL}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}[\mathrm{\&xi;}(1-\mathrm{\&delta;})-\{\mathrm{\&xi;}-\mathrm{\&delta;}\left\}\right]---\left(87\right)$

Its value of reference point for the load right-hand member is:

$w_{2R}=\frac{\mathrm{\&beta;PL}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}\mathrm{\&delta;}(1-\mathrm{\&xi;})---\left(88\right)$

Its value of reference point for the load left end is:

$w_{2L}=\frac{\mathrm{\&beta;PL}}{A_{\mathrm{eff}}{G}_{\mathrm{eff}}}\mathrm{\&xi;}(1-\mathrm{\&delta;})---\left(89\right)$

With the detrusion at load two ends and flexural deformation respectively addition can obtain expression formula in the distortion of load left end and right-hand member:

$w_L=\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^3}{6K_C}(1-\mathrm{\&delta;})\mathrm{\&xi;}(2m+2\mathrm{\&delta;}-{\mathrm{\&delta;}}^2-{\mathrm{\&xi;}}^2)---\left(90\right)$

$w_R=\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^3}{6K_C}\mathrm{\&delta;}(1-\mathrm{\&xi;})(2m-{\mathrm{\&delta;}}^2+2\mathrm{\&xi;}-{\mathrm{\&xi;}}^2)---\left(91\right)$

Wherein:

$m=\frac{3K_C\mathrm{\&beta;}}{\mathrm{\&epsiv;}{A}_{\mathrm{eff}}{G}_{\mathrm{eff}}{L}^2}---\left(92\right)$

For the situation of load in the span centre position.Can calculate this moment moment of flexure the maximal value maximal value appear at the span centre position, therefore can be calculated in conjunction with equation (92) by formula (90) or (91):

$w_{\max }=\frac{\mathrm{\&epsiv;}{\mathrm{PL}}^3}{48K_C}+\frac{\mathrm{\&beta;PL}}{4A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(93\right)$

Can be obtained by equation (74): $\frac{K_C}{\mathrm{\&epsiv;}}\&ap;K_C---\left(94\right)$

And equation (94) is relatively conservative

With the evenly load situation, introduce an inhomogeneous partition factor k of shearing, β becomes k β with the shearing partition factor.Then formula (93) can be reduced to following form:

$w_{\max }=\frac{{\mathrm{PL}}^3}{48K_C}+\frac{\mathrm{k\&beta;PL}}{4A_{\mathrm{eff}}{G}_{\mathrm{eff}}}---\left(95\right)$

Six, the stainless steel battenboard is as the shearing partition factor of shingle nail.

Do not have tangible dash board for the stainless steel shingle nail, core is peeled off situation, so it can not consider the bonding damage.Can set up the finite element model of rock wool and glass silk flosssilk wadding respectively studies respectively.

(1) set up surface stainless steel shingle nail model:

Suppose in the beam test that battenboard panel and core are in the linear elasticity stage.Panel and core there is no slippage, guarantee co-operation fully, employing be that panel is tied to core TIE.Bearing is used with panel and is set up with identical material, with the consistance of maintenance with plate.Its width is 50mm, and the distance between two steel discs is 1950mm.What panel adopted is three-dimensional shell unit, adopts S4R (4 nodes reduce integration) model.Core is the 3D solid unit, adopts C3D8R (8 nodes reduce integration) model.That analysis step adopts is static general, applies evenly load at the upper steel plate of battenboard, and bearing is constrained to freely-supported.

The rock wool material property parameter is obtained by the material property test, specifically arranges as follows.

(1) adopt isotropy for thin plate, what mainly consider is axial tension, so its parameter adopts rift grain to tensile modulus of elasticity: E=8.326MPa, v=0.13

What (2) adopt for slab is anisotropy, and main what consider is axial rift grain tension, and the against the grain pressurized of Width laterally be the rift grain pressurized, and concrete parameter is obtained by the material property experiment, and specifically parameter is set at:

E ₁＝8.326MPa，E ₂＝0.238MPa，E ₃＝3.29MPa，v ₁＝v ₂＝v ₃＝0.13，G ₁＝0.35MPa，

G ₂＝1.56MPa，G ₃＝0.35MPa

Glass fiber cotton material performance parameter is obtained by the material property experiment, specifically arranges as follows:

(1) adopt isotropy for thin plate, what mainly consider is axial tension, so its parameter adopts rift grain to tensile modulus of elasticity: E=7.7MPa, v=0.13

What (2) adopt for slab is anisotropy, and main what consider is axial rift grain tension, and the against the grain pressurized of Width laterally be the rift grain pressurized, and concrete parameter is obtained by the material property experiment, and specifically parameter is set at: E ₁=7.7MPa, E ₂=0.06MPa, E ₃=1.59MPa, v ₁=v ₂=v ₃=0.13, G ₁=0.23MPa, G ₂=1.5MPa, G ₃=0.23MPa.

Be along the long direction of plate for anisotropic material 1 direction; 2 directions are the direction along the plate width; 3 directions are the direction along plate thickness.Rock wool is against the grain to placement along the plate Width, is that rift grain is to placement along plate length and thickness direction.

(2) the stainless steel battenboard is determined as the shearing partition factor of shingle nail.

1, the stainless steel rockwool sandwich board is determined as the shearing partition factor of shingle nail

Because the distortion of battenboard is very big, often adopt f=L/200 as the ultimate design index in the design, i.e. use state restriction, in the analysis in order to obtain the relation of shearing partition factor value and thickness of slab and plate thickness, model has adopted 50,60,70,80,90 altogether, six kinds of thicknesss of slab of 100mm, with 0.4,0.5,0.6, four kinds of plate thicknesses of 0.7mm analyze, build together and found 48 finite element models.Analog result and theoretical result of calculation such as table 2.As can be seen from Table 2, theoretical result of calculation is compared the less basic controlling of error ratio in 10%, so it has certain applicability with the finite element analogy result.And the error of experiment value and calculated value control 9.3% and result of calculation relatively conservative, so result of calculation meets the requirements.

Table 2 rock wool shingle nail analog result and the contrast of theoretical result of calculation

Type	Trial value kPa	Simulation β	Analogue value kPa	Calculated value kPa	Calculate and test error	Calculate and simulation error
							YQ-0.4-50	-	0.261	2.58	2.74	0.062	-
YQ-0.5-50	-	0.258	2.73	3.06	0.121	-

YQ-0.6-50	-	0.254	2.94	3.32	0.129	-
							YQ-0.7-50	-	0.249	3.08	3.55	0.152	-
YQ-0.4-60	-	0.317	2.94	3.18	0.080	-
							YQ-0.5-50	-	0.313	3.35	3.48	0.039	-
YQ-0.6-60	-	0.309	3.54	3.72	0.051	-
							YQ-0.7-60	-	0.304	3.6	3.93	0.080	-
YQ-0.4-70	-	0.373	3.59	3.56	0.008	-
							YQ-0.5-70	-	0.368	3.65	3.79	0.038	-
YQ-0.6-70	-	0.364	3.71	4.01	0.081	-
							YQ-0.7-70	-	0.359	3.94	4.19	0.063	-
YQ-0.4-80	-	0.417	3.95	3.95	0	-
							YQ-0.5-80	-	0.411	4.07	4.22	0.037	-
YQ-0.6-80	-	0.405	4.18	4.44	0.062	-
							YQ-0.7-80	-	0.399	4.31	4.62	0.072	-
YQ-0.4-90	-	0.455	4.45	4.36	0.020	-
							YQ-0.5-90	-	0.449	4.60	4.46	0.030	-
YQ-0.6-90	-	0.443	4.74	4.62	0.025	-
							YQ-0.7-90	-	0.437	4.88	4.79	0.018	-
YQ-0.4-100	-	0.491	4.74	4.64	0.021	-
							YQ-0.5-100	-	0.489	5.03	4.86	0.034	-
YQ-0.6-100	5.83	0.487	5.26	5.01	0.048	-0.093
							YQ-0.7-100	-	0.485	5.45	5.14	0.057	--

For the relation of the thickness of studying shearing partition factor and stainless steel rockwool sandwich board thickness and panel, Figure 11, Figure 12 have provided the relation curve of shearing partition factor and stainless steel rockwool sandwich board thickness of slab and plate thickness.The shearing partition factor is along with the variation of thickness of slab is approximated to the secondary relation as can be seen by last figure, and changing value is bigger.Thickness with panel is approximated to linear relationship, and changing value is smaller.The shearing partition factor is as follows:

$\mathrm{\&beta;}=-0.2\&times;{\left(\frac{D}{100}\right)}^2+0.705\&times;\frac{D}{100}-0.063\&times;d+0.125---\left(96\right)$

Wherein, D is the sandwich thickness of slab, and d is plate thickness.

2, stainless steel glass silk flosssilk wadding battenboard is determined as the shearing partition factor of shingle nail.

The analog result of glass silk flosssilk wadding shingle nail and the contrast of theoretical result of calculation are as shown in table 3.

Table 3 glass silk flosssilk wadding shingle nail analog result and the contrast of theoretical result of calculation

Type	Trial value kPa	Simulation β	Analogue value kPa	Calculated value kPa	Calculate and test error	Calculate and simulation error
							BQ-0.4-50	-	0.495	1.73	1.702	-0.014	-
BQ-0.5-50	-	0.479	1.80	1.864	0.036	-
							BQ-0.6-50	-	0.463	1.87	2.007	0.073	-
BQ-0.7-50	-	0.447	1.94	2.139	0.101	-
							BQ-0.4-60	-	0.509	2.16	2.115	-0.018	-
BQ-0.5-50	-	0.494	2.23	2.295	0.030	-
							BQ-0.6-60	-	0.479	2.30	2.453	0.067	-
BQ-0.7-60	-	0.464	2.37	2.598	0.096	-
							BQ-0.4-70	-	0.524	2.59	2.516	-0.026	-
BQ-0.5-70	-	0.509	2.73	2.710	-0.006	-
							BQ-0.6-70	-	0.494	2.80	2.880	0.029	-
BQ-0.7-70	-	0.479	2.87	3.036	0.057	-
							BQ-0.4-80	-	0.542	3	2.893	-0.035	-
BQ-0.5-80	-	0.526	3.16	3.097	-0.019	-
							BQ-0.6-80	-	0.511	3.26	3.275	0.006	-
BQ-0.7-80	-	0.496	3.35	3.439	0.026	-
							BQ-0.4-90	-	0.559	3.46	3.254	-0.055	-
BQ-0.5-90	-	0.543	3.64	3.468	-0.046	-
							BQ-0.6-90	-	0.527	3.73	3.655	-0.020	-
BQ-0.7-90	-	0.512	3.83	3.827	0.001	-
							BQ-0.4-100	-	0.608	3.71	3.439	-0.073	-
BQ-0.5-100	-	0.5902	3.83	3.645	-0.048	-
							BQ-0.6-100	3.87	0.574	3.86	3.826	-0.031	-0.009
BQ-0.7-100	-	0.558	4.07	3.994	-0.018	-

As can be seen from Table 3, theoretical result of calculation is compared the less basic controlling of error ratio in 10%, so it has applicability with the finite element analogy result.And the error of experiment value and calculated value control 0.9% and result of calculation relatively conservative, so result of calculation meets the requirements.

For the relation between the thickness of studying shearing partition factor and thickness of slab and panel, Figure 13 is stainless steel glass silk flosssilk wadding battenboard shearing partition factor and thickness of slab graph of relation, and Figure 14 is stainless steel glass silk flosssilk wadding battenboard shearing partition factor and thickness of slab graph of relation.

Stainless steel-glass silk flosssilk wadding shingle nail shearing partition factor can be taken as secondary about the relation curve of thickness of slab, is taken as once about the relation of plate thickness.

The linear regression result of the shearing partition factor of stainless steel-glass silk flosssilk wadding shingle nail is as follows:

$\mathrm{\&beta;}=-0.1\&times;{\left(\frac{D}{100}\right)}^2+0.827\&times;\frac{D}{100}-0.046\&times;d-0.071---\left(97\right)$

Two, the stainless steel filled board is as the shearing partition factor of roof boarding.

(1) sets up surface stainless steel roof boarding model.

Suppose in the beam test that battenboard panel and core are in the linear elasticity stage.Bearing uses the material identical with panel to set up, and with the consistance of maintenance with plate, its width is 50mm, distance between two steel discs is 1950mm, every span length is 1950mm, and the bonding of its panel and core is unsatisfactory as can be seen for roof boarding, and particularly lower floor's steel plate and core are almost completely peeled off.So need consider its bonding damage, and can not arrive together by simple T IE.

The rock wool material property parameter is obtained by the material property test, specifically arranges as follows.

(1) adopt isotropy for thin plate, what mainly consider is axial tension, so its parameter adopts rift grain to tensile modulus of elasticity: E=8.326MPa, v=0.13

What (2) adopt for slab is anisotropy, and main what consider is axial rift grain tension, and the against the grain pressurized of Width laterally be the rift grain pressurized, and concrete parameter is obtained by the material property experiment, and specifically parameter is set at:

E ₁＝8.326MPa，E ₂＝0.238MPa，E ₃＝3.29MPa，v ₁＝v ₂＝v ₃＝0.13，G ₁＝0.35MPa，G ₂＝1.56MPa，G ₃＝0.35MPa

Glass fiber cotton material performance parameter is obtained by the material property experiment, specifically arranges as follows:

(1) adopt isotropy for thin plate, what mainly consider is axial tension, so its parameter adopts rift grain to tensile modulus of elasticity: E=7.7MPa, v=0.13

What (2) adopt for slab is anisotropy, and main what consider is axial rift grain tension, and the against the grain pressurized of Width laterally be the rift grain pressurized, and concrete parameter is obtained by the material property experiment, and specifically parameter is set at: E ₁=7.7MPa, E ₂=0.06MPa, E ₃=1.59MPa, v ₁=v ₂=v ₃=0.13, G ₁=0.23MPa, G ₂=1.5MPa, G ₃=0.23MPa.

Be along the long direction of plate for anisotropic material 1 direction; 2 directions are the direction along the plate width; 3 directions are the direction along plate thickness.Rock wool is against the grain to placement along the plate Width, is that rift grain is to placement along plate length and thickness direction.

The bonding damage can be simulated with the COHESIVE unit.Adopt thickness can be set to 0.0001 meter, to tally with the actual situation.It adopts isotropic model, and elastic modulus is taken as 9MPa and adopts the 3D solid unit, namely adopts the eight node linear units of COHESIVE.

(2) bonding element constitutive relation.

For tack coat is simulated, between panel and sandwich layer, add a new zone, set up a kind of new unit.In ABAQUS, claim that this unit is the COHESIVE unit, this new zone can be regarded as the superfluous zone of one deck resin bed.But different with simple resin zone, its main effect is to connect two individual layers up and down.It is connected by COHESIVE tractive force on two surfaces up and down, and this COHESIVEtraction is relevant with the spacing of upper and lower surface, and this relation is called " COHESIVE law " or " Traction-seperationlaw ".Acting force on the adhesive surface is divided into three kinds, and a kind of is the normal stress t of normal direction _n, two other is tangential shear stress t _sAnd t _t, Figure 15 is that COHESIVE element stress figure, Figure 16 are COHESIVE unit figure, Figure 17 is the softening constitutive model of the linear elasticity-linearity of COHESIVE unit.

If

Be the area that curve and coordinate axis surround among Figure 17, being called critical strain can release rate.That is:

$\left.\begin{array}{c}{\&Integral;}_0^{{\mathrm{\&delta;}}_n^{\max }}{t}_n\left(\mathrm{\&delta;}\right)d{\mathrm{\&delta;}}_n=G_C^n\\ {\&Integral;}_0^{{\mathrm{\&delta;}}_s^{\max }}{t}_s\left(\mathrm{\&delta;}\right)d{\mathrm{\&delta;}}_s=G_C^s\\ {\&Integral;}_0^{{\mathrm{\&delta;}}_t^{\max }}{t}_t\left(\mathrm{\&delta;}\right)d{\mathrm{\&delta;}}_t=G_C^t\end{array}\right\}---\left(98\right)$

Figure 17 is the softening constitutive model of the linear elasticity-linearity of COHESIVE unit, supposes that bonded areas has a thickness T _c, be the thickness (its value for 0.0001m) of the Part that builds in the model.So corresponding three strains are exactly:

${\mathrm{\&epsiv;}}_n=\frac{{\mathrm{\&delta;}}_n}{T_c};{\mathrm{\&epsiv;}}_s=\frac{{\mathrm{\&delta;}}_s}{T_c};{\mathrm{\&epsiv;}}_t=\frac{{\mathrm{\&delta;}}_t}{T_c}---\left(99\right)$

As δ＜δ ₀The time, being linear elasticity, have this moment:

$t=\left\{\begin{array}{c}{t}_n\\ t_s\\ t_t\end{array}\right\}=\left(\begin{array}{ccc}{K}_{\mathrm{nn}}& & \\ & K_{\mathrm{ss}}& \\ & & K_{\mathrm{tt}}\end{array}\right)\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_n\\ {\mathrm{\&epsiv;}}_s\\ {\mathrm{\&epsiv;}}_t\end{array}\right\}---\left(100\right)$

Work as δ ₀＜δ＜δ _MaxThe time, being the damage softening zone, have this moment:

$t=\left\{\begin{array}{c}{t}_n\\ t_s\\ t_t\end{array}\right\}=\left(\begin{array}{ccc}{(1-D)K}_{\mathrm{nn}}& & \\ & (1-D)K_{\mathrm{ss}}& \\ & & (1-D)K_{\mathrm{tt}}\end{array}\right)\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_n\\ {\mathrm{\&epsiv;}}_s\\ {\mathrm{\&epsiv;}}_t\end{array}\right\}---\left(101\right)$

Wherein, D is Damage coefficient, 0≤D≤1.When D=0, the expression material is not surrendered or has just been begun and surrenders; When D=1, the expression material damage loses load-bearing capacity.

As δ＞δ _MaxThe time, material has lost load-bearing capacity, is equivalent to bonded areas and destroys battenboard generation layering.

Also do not bear load at 0 among Figure 17,1 is in the elastic region, 2 surrenders, and 3 have entered the softened zone, and 4 are then just destroyed, and 5 have been destroyed layering.

Because the laminate layering caused by certain single cracking pattern, consider that separately a certain cracking pattern all can not simulate accurately to layering, so cracking criterion that must the consideration mixed mode.What I adopted in model is B-K cracking criterion, that is:

$G^C=G_n^C+(G_s^C-G_n^C){\left(\frac{G_S}{G_T}\right)}^{\mathrm{\&eta;}}---\left(102\right)$

G wherein _S=G _s+ G _t, G _T=G _n+ G _t, for compound polyurethane material, index η=2.When given

Behind the η, the critical strain release rate G of material ^CBe exactly G _S/ G _TThe function of determining.

(2) the stainless steel battenboard is determined as the shearing partition factor of roof boarding.

What adopt in this model is to add one deck COHESIVE solid element between lower floor's panel and core, simulates bonding.By checking, it is the complication experiment result relatively.Build together and found six kinds of thickness (50,60,70,80,90,100mm) (0.4,0.5,0.6,0.7mm) totally 48 finite element models are studied with four kinds of plate thicknesses.Table 4 is stainless steel rock wool roofing battenboard analog result and the contrast of theoretical result of calculation.

Table 4 stainless steel rock wool roofing battenboard analog result and the contrast of theoretical result of calculation

Type	Trial value kPa	Simulation β	Analogue value kPa	Calculated value kPa	Calculate and test error	Calculate and simulation error
							YW-0.4-50	-	0.334	2.65	3.02	-0.138	-
YW-0.5-50	-	0.329	3.08	3.29	-0.068	-
							YW-0.6-50	-	0.322	3.51	3.53	-0.004	-
YW-0.7-50	-	0.314	3.80	3.74	0.014	-
							YW-0.4-60	-	0.386	3.37	3.38	-0.004	-
YW-0.5-60	-	0.379	3.65	3.64	0.005	-
							YW-0.6-60	-	0.371	3.80	3.87	-0.018	-
YW-0.7-60	-	0.362	4.08	4.08	0.001	-
							YW-0.4-70	-	0.434	3.66	3.69	-0.010	-
YW-0.5-70	-	0.429	3.94	3.92	0.005	-
							YW-0.6-70	-	0.418	4.08	4.15	-0.016	-
YW-0.7-70	-	0.407	4.37	4.36	0.003	-
							YW-0.4-80	-	0.480	3.80	3.96	-0.042	-
YW-0.5-80	-	0.472	4.08	4.19	-0.026	-
							YW-0.6-80	-	0.462	4.37	4.39	-0.005	-
YW-0.7-80	-	0.450	4.65	4.59	0.013	-

YW-0.4-90	-	0.524	4.08	4.19	-0.027	-
							YW-0.5-90	-	0.515	4.37	4.41	-0.010	-
YW-0.6-90	-	0.5044	4.65	4.61	0.010	-
							YW-0.7-90	-	0.4919	4.94	4.80	0.029	-
YW-0.4-100	-	0.5652	4.37	4.41	-0.009	-
							YW-0.5-100	-	0.554	4.65	4.63	0.006	-
YW-0.6-100	5	0.5448	4.80	4.80	-0.0003	0.040
							YW-0.7-100	-	0.5318	5.08	4.98	0.019	-

As can be seen from Table 4, theoretical result of calculation is compared the less basic controlling of error ratio in 10%, so it has certain applicability with the finite element analogy result.And the error of experiment value and calculated value control 4.0% and result of calculation relatively conservative, so result of calculation meets the requirements.

Figure 18 is the graph of relation of stainless steel rock wool roofing battenboard shearing partition factor of the present invention and plate thickness, and Figure 19 is the graph of relation of stainless steel rock wool roofing battenboard shearing partition factor of the present invention and plate thickness.From Figure 18 and Figure 19 as can be seen the shearing partition factor be approximated to the secondary relation along with the variation of thickness of slab, and changing value is bigger.Along with the thickness of panel is approximated to linear relationship, changing value is smaller.In order to keep consistency, stainless steel-rock wool roof boarding shearing partition factor is taken as secondary about the relation curve of thickness of slab, seemingly is taken as once about the close of plate thickness.The fitting result of its shearing partition factor is as follows:

$\mathrm{\&beta;}=-0.1\&times;{\left(\frac{D}{100}\right)}^2+0.63\&times;\frac{D}{100}-0.092\&times;d-0.090---\left(103\right)$

Stainless steel glass silk flosssilk wadding roofing battenboard analog result and the contrast of theoretical result of calculation are as shown in table 5.

Table 5 stainless steel glass silk flosssilk wadding roofing battenboard analog result and the contrast of theoretical result of calculation

Type	Trial value kPa	Simulation β	Analogue value kPa	Calculated value kPa	Calculate and test error	Calculate and simulation error
							BW-0.4-50	-	0.4967	1.80	2.249	-0.251	-
BW-0.5-50	-	0.4924	2.087	2.393	-0.148	-
							BW-0.6-50	-	0.4864	2.371	2.513	-0.061	-
BW-0.7-50	-	0.4778	2.65	2.625	0.011	-
							BW-0.4-60	-	0.5779	1.941	2.440	-0.257	-
BW-0.5-60	-	0.5719	2.226	2.571	-0.155	-
							BW-0.6-60	-	0.5623	2.512	2.689	-0.070	-
BW-0.7-60	-	0.5526	2.798	2.792	0.002	-
							BW-0.4-70	-	0.6544	2.226	2.598	-0.167	-

BW-0.5-70	-	0.6466	2.512	2.719	-0.082	-
							BW-0.6-70	-	0.6363	2.798	2.826	-0.010	-
BW-0.7-70	-	0.6226	3.084	2.934	0.0485	-
							BW-0.4-80	-	0.7267	2.512	2.735	-0.089	-
BW-0.5-80	-	0.7131	2.798	2.863	-0.023	-
							BW-0.6-80	-	0.7052	3.084	2.950	0.043	-
BW-0.7-80	-	0.6894	3.369	3.057	0.0927	-
							BW-0.4-90	-	0.7956	2.798	2.855	-0.020	-
BW-0.5-90	-	0.7849	2.941	2.961	-0.007	-
							BW-0.6-90	-	0.7709	3.226	3.062	0.051	-
BW-0.7-90	-	0.7533	3.512	3.167	0.0983	-
							BW-0.4-100	-	0.8615	2.941	2.963	-0.008	-
BW-0.5-100	-	0.849	3.226	3.066	0.050	-
							BW-0.6-100	3.2	0.834	3.512	3.163	0.100	0.013
BW-0.7-100	-	0.8149	3.798	3.266	0.140	-

As can be seen from Table 5, except individual values, theoretical result of calculation is compared the less basic controlling of error ratio in 10%, so it has certain applicability with the finite element analogy result.And the error of experiment value and calculated value control 1.3% and result of calculation relatively conservative, so result of calculation meets the requirements.

Figure 20 is the relation curve of stainless steel glass silk flosssilk wadding roofing battenboard shearing partition factor and plate thickness, the graph of relation of Figure 21 stainless steel glass silk flosssilk wadding roofing battenboard shearing partition factor and plate thickness.Stainless steel-glass wool roof boarding shearing partition factor is taken as secondary about the relation curve of thickness of slab, seemingly can be taken as once about the close of plate thickness.The result of its curve match is as follows:

$\mathrm{\&beta;}=-0.2\&times;{\left(\frac{D}{100}\right)}^2+0.96\&times;\frac{D}{100}-0.113\&times;d-0.114---\left(104\right)$

In sum:

R ₁, R ₂, R ₃, R ₄Value sees Table 1.

Table 1: coefficients R ₁, R ₂, R ₃, R ₄List of values:

Step 300: the anti-bending bearing capacity of determining described stainless steel battenboard: the anti-bending bearing capacity of being determined the stainless steel battenboard by the relation of load and anti-anti-bending bearing capacity.

Because amount of deflection and the amount of deflection permissible value of battenboard have following relation: w _Max≤ [f]

[f] is referring to the limit value of amount of deflection in the national standard " the adiabatic battenboard of metal covering for building " (GB/T 23932-2009).Then:

Load is determined: $p\&le;\frac{\left[f\right]}{\frac{l^3}{48E_1{I}_1}+\frac{\mathrm{k\&beta;l}}{4A_{\mathrm{eff}}{G}_{\mathrm{eff}}}}$

Determining of local load: $p\&le;\frac{\left[f\right]}{\frac{5b{l}^4}{384E_1{I}_1}+\frac{\mathrm{kb\&beta;}{l}^2}{8A_{\mathrm{eff}}{G}_{\mathrm{eff}}}}$

For shallow die mould and plane battenboard (shingle nail):

A _eff≈A _C；G _eff≈G

A wherein _CArea of section for core; G is the modulus of shearing of core.

For dark die mould battenboard (roof boarding):

G _eff＝G _C(D _C+d)/D _C；A _eff＝A _C(D _C+d)/D _C

D wherein _CBe battenboard central layer thickness; D is taken as 8.0mm.

Thereby determine the anti-bending bearing capacity of stainless steel battenboard.

Stainless steel flexural capacity of sandwich plate of the present invention is determined method, by the amount of deflection under definite stainless steel battenboard load and the amount of deflection under the evenly load, accurately determines the anti-bending bearing capacity of stainless steel battenboard then.The present invention accurately determines the anti-bending bearing capacity of stainless steel battenboard, can more accurately assess the security performance of stainless steel battenboard, thereby advance the application of stainless steel battenboard.

The specific embodiment of the present invention is: the stainless steel flexural capacity of sandwich plate is determined that method is applied to stainless steel straw battenboard bearing of component and determines.

Above content be in conjunction with concrete preferred implementation to further describing that the present invention does, 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 (4)

1. a stainless steel flexural capacity of sandwich plate is determined method, it is characterized in that, the panel of described stainless steel battenboard is corrosion resistant plate, the central layer of described stainless steel battenboard is that the middle layer forms by cementing agent bonding or cast, and the stainless steel flexural capacity of sandwich plate determines that method comprises the steps:

Gather the correlation parameter of stainless steel battenboard: the thickness of the elastic modulus of the effective cross-sectional area of the rigidity of the span of collection stainless steel battenboard, the width of stainless steel battenboard, stainless steel battenboard central layer, stainless steel battenboard panel rigidity, stainless steel battenboard core, effective modulus of shearing of stainless steel battenboard core, stainless steel battenboard plane materiel, the elastic modulus of stainless steel battenboard core, stainless steel battenboard plate thickness, stainless steel battenboard central layer;

Determine the amount of deflection of stainless steel battenboard: the amount of deflection of stainless steel battenboard determine to comprise amount of deflection under the load and the amount of deflection under the evenly load,

Amount of deflection under the stainless steel battenboard load adopts following formula to obtain:

$w_{\max }=\frac{{\mathrm{pl}}^3}{{48E}_1{I}_1}+\frac{\mathrm{k\&beta;pl}}{{4A}_{\mathrm{eff}}{G}_{\mathrm{eff}}}$

Amount of deflection under the stainless steel battenboard evenly load adopts following formula to obtain:

$w=\frac{{5\mathrm{pl}}^4}{{384E}_1{I}_1}+\frac{{\mathrm{k\&beta;pl}}^2}{{8A}_{\mathrm{eff}}{G}_{\mathrm{eff}}}$

Wherein: w represents the combined deflection of battenboard;

P represents the payload values of battenboard panel;

L represents the span of battenboard;

E ₁The elastic modulus of expression plane materiel;

I ₁Metal covering is to the moment of inertia of battenboard natural axis about the expression;

A _EffThe net sectional area of expression battenboard;

G _EffEffective modulus of shearing of expression battenboard;

What k represented is the shear stress nonuniformity coefficient, is taken as 1.2 here,

β represents the shearing partition factor of battenboard, $\mathrm{\&beta;}=R_1{\left(\frac{D}{100}\right)}^2+R_2\frac{D}{100}+R_3\&times;d+R_4,$ D is battenboard thickness, and d is plate thickness, R ₁, R ₂, R ₃, R ₄Value sees Table 1,

Table 1: coefficients R ₁, R ₂, R ₃, R ₄List of values:

Determine the anti-bending bearing capacity of described stainless steel battenboard: determine the anti-bending bearing capacity of stainless steel battenboard by the relation of load and anti-bending bearing capacity, the relation of the amount of deflection of battenboard and amount of deflection limit value has following relation: w _Max≤ [f], the limit value of [f] expression amount of deflection

Then:

Load is determined: $P\&le;\frac{\left[f\right]}{\frac{l^3}{{48E}_1{I}_1}+\frac{\mathrm{k\&beta;l}}{{4A}_{\mathrm{eff}}{G}_{\mathrm{eff}}}}$

Determining of local load:

Namely determine the anti-bending bearing capacity of stainless steel battenboard.

2. stainless steel flexural capacity of sandwich plate according to claim 1 is determined method, it is characterized in that, described stainless steel battenboard comprises shingle nail and roof boarding.

3. stainless steel flexural capacity of sandwich plate according to claim 1 and 2 is determined method, it is characterized in that, the core of described stainless steel battenboard is that rock wool forms by cementing agent bonding or cast.

4. stainless steel flexural capacity of sandwich plate according to claim 1 and 2 is determined method, it is characterized in that, the core of described stainless steel battenboard is that polyurethane forms by cementing agent bonding or cast.

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