CN101900653A

CN101900653A - Method for determining anti-bending bearing capacity of stainless steel sandwich panel and application of stainless steel sandwich panel

Info

Publication number: CN101900653A
Application number: CN2010102154491A
Authority: CN
Inventors: 查晓雄; 宋新武
Original assignee: Shenzhen Graduate School Harbin Institute of Technology
Current assignee: Shenzhen Graduate School Harbin Institute of Technology
Priority date: 2010-06-30
Filing date: 2010-06-30
Publication date: 2010-12-01
Anticipated expiration: 2030-06-30
Also published as: CN101900653B

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 steel sandwich 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

Method for determining bending resistance bearing capacity of stainless steel sandwich plate and application

Technical Field

The invention relates to a method for determining the bending resistance bearing capacity of a sandwich plate and application thereof, in particular to a method for determining the bending resistance bearing capacity of a stainless steel sandwich plate and application thereof.

Background

The sandwich board for stainless steel heat insulation is formed by bonding or pouring an upper thin outer surface board (a bearing layer and a color steel plate) and a lower thin outer surface board (a core layer and novel straws) with high strength and a light and soft middle layer (a core layer and the like) through adhesives. It has obvious combination advantages, and can be used as ideal application material for wall plate and roof plate. The metal surface layer has a protection effect on the core layer, so that the core layer is prevented from being damaged by machinery, weathering is prevented, and water and steam are isolated; and the sandwich layer can be connected into whole with two surface course, bears the load jointly, when the surface course takes place the bucking under the load effect, the sandwich layer can also support the surface course, increases the ability that the surface course resisted the bucking, the sandwich layer still has effects such as adiabatic, give sound insulation. Can be respectively suitable for different building requirements, and comprises a plurality of building fields such as industrial factory buildings, public buildings, warehouses, combined houses, purification projects and the like. In the prior art, the determination of the bending resistance bearing capacity of the stainless steel sandwich plate still stays on the basis of practical and empirical formulas, and an accurate conclusion cannot be obtained, so that the accurate determination of the bending resistance bearing capacity cannot be performed, and the use and popularization of the stainless steel sandwich plate are greatly restricted.

Disclosure of Invention

The technical problem solved by the invention is as follows: the method for determining the bending resistance bearing capacity of the stainless steel sandwich plate and the application thereof are provided, and the technical problem that the bending resistance bearing capacity cannot be determined accurately in the prior art is solved.

The technical scheme of the invention is as follows: the method for determining the bending resistance bearing capacity of the stainless steel sandwich plate comprises the following steps:

collecting relevant parameters of the stainless steel sandwich plate: the span of the stainless steel sandwich panel, the width of the stainless steel sandwich panel, the stiffness of the stainless steel sandwich panel core, the stiffness of the stainless steel sandwich panel, the effective cross-sectional area of the stainless steel sandwich panel core, the effective shear modulus of the stainless steel sandwich panel core, the elastic modulus of the stainless steel sandwich panel face stock, the elastic modulus of the stainless steel sandwich panel core, the thickness of the stainless steel sandwich panel, and the thickness of the stainless steel sandwich panel core are collected.

Determining the deflection of the stainless steel sandwich plate: the determination of the deflection of the stainless steel sandwich panel comprises the deflection under concentrated load and the deflection under evenly distributed load,

the deflection of the stainless steel sandwich plate under the concentrated load is obtained by adopting the following formula:

the deflection of the stainless steel sandwich plate under uniform load is obtained by adopting the following formula:

wherein: w represents the total deflection of the sandwich panel; p represents the load value of the sandwich panel; l represents the span of the sandwich panel; e₁Denotes the modulus of elasticity of the face stock; i is₁Representing the moment of inertia of the upper and lower metal surfaces against the central axis of the sandwich panel; a. the_effRepresents the effective cross-sectional area of the sandwich panel; g_effRepresents the effective shear modulus of the sandwich panel; k denotes the shear stress non-uniformity coefficient, which can be taken to be 1.2 for common plate types.

Beta represents the shear distribution coefficient of the sandwich panel,

R₁、R₂、R₃、R₄the values are shown in Table 1.

Table 1: coefficient R₁、R₂、R₃、R₄Value taking table:

determining the bending resistance bearing capacity of the stainless steel sandwich plate: determining the bending resistance bearing capacity of the stainless steel sandwich plate: and determining the bending resistance bearing capacity of the stainless steel sandwich plate according to the relationship between the load and the bending resistance bearing capacity.

The further technical scheme of the invention is as follows: the stainless steel sandwich plate comprises a wall panel and a roof panel.

The further technical scheme of the invention is as follows: the core material of the stainless steel sandwich plate is formed by bonding or pouring rock wool through a bonding agent.

The further technical scheme of the invention is as follows: the core material of the stainless steel sandwich plate is formed by bonding or pouring polyurethane through a bonding agent.

The technical scheme of the invention is as follows: the method for determining the bending resistance bearing capacity of the stainless steel straw sandwich board is applied to determining the bearing capacity of the stainless steel straw sandwich board component.

The invention has the technical effects that: the method for determining the bending resistance bearing capacity of the stainless steel sandwich plate is characterized in that the bending resistance bearing capacity of the stainless steel sandwich plate is accurately determined by determining the deflection of the stainless steel sandwich plate under concentrated load and the deflection of the stainless steel sandwich plate under uniformly distributed load. The invention accurately determines the bending resistance bearing capacity of the stainless steel sandwich plate, and can more accurately evaluate the safety performance of the stainless steel sandwich plate, thereby promoting the application of the stainless steel sandwich plate.

Drawings

FIG. 1 is a flow chart of the present invention.

FIG. 2 is a schematic cross-sectional view of a planar stainless steel sandwich panel according to the present invention.

FIG. 3 is a schematic cross-sectional view of a low-pressure stainless steel sandwich plate according to the present invention.

FIG. 4 is a schematic cross-sectional view of a deep-drawing stainless steel sandwich panel according to the present invention.

FIG. 5 is a schematic view of the force, stress and deformation of the sandwich panel of the present invention.

FIG. 6 is another schematic view of the force, stress and deformation of the sandwich panel of the present invention.

FIG. 7 is a schematic structural view of a planar sandwich panel under uniform load according to the present invention.

FIG. 8 is a schematic diagram of the forces and deformations of a profiled stainless steel sandwich panel of the present invention.

FIG. 9 is a schematic view of the division of a profiled stainless steel sandwich panel of the present invention into sandwich portions and flanges.

FIG. 10 is a schematic view of the sandwich panel under concentrated load according to the present invention.

FIG. 11 is a graph showing the variation of the shear force distribution coefficient of the rock wool wall surface sandwich panel with the stainless steel surface according to the invention along with the thickness of the panel.

FIG. 12 is a graph showing the variation of the shear force distribution coefficient of the rock wool wall sandwich panel with stainless steel surface according to the present invention.

FIG. 13 is a graph showing the variation of the shear force distribution coefficient of the stainless steel polyurethane wall sandwich panel with the thickness of the panel.

FIG. 14 is a graph showing the relationship between the shear force distribution coefficient of the stainless steel polyurethane wall sandwich panel and the thickness of the panel.

FIG. 15 is a stress diagram of a COHESIVE cell of the present invention.

FIG. 16 is a diagram of a COHESIVE unit of the present invention.

FIG. 17 is a linear elastic-linear softening constitutive model of the COHESIVE unit of the present invention.

FIG. 18 is a graph showing the relationship between the shear distribution coefficient and the thickness of the stainless steel rock wool roof sandwich panel of the present invention.

FIG. 19 is a graph showing the relationship between the shear distribution coefficient of the stainless steel rock wool roof sandwich panel and the thickness of the panel.

FIG. 20 is a graph showing the relationship between the shear force distribution coefficient and the sheet thickness of the stainless steel glass wool roof sandwich panel of the present invention.

FIG. 21 is a graph showing the relationship between the shear distribution coefficient and the panel thickness of the stainless steel glass wool roof sandwich panel of the present invention.

Detailed Description

The technical solution of the present invention is further illustrated below with reference to specific examples.

As shown in fig. 1, the specific embodiment of the present invention is: the method for determining the bending resistance bearing capacity of the stainless steel sandwich plate is provided, wherein the panel of the stainless steel sandwich plate is a stainless steel plate, and the middle layer of the stainless steel sandwich plate is formed by bonding or pouring through a bonding agent. In the invention, the stainless steel sandwich plate comprises a wall panel and a roof panel.

The method for determining the bending resistance bearing capacity of the stainless steel sandwich plate comprises the following steps:

step 100: collecting relevant parameters of the stainless steel sandwich plate: the span of the stainless steel sandwich panel, the width of the stainless steel sandwich panel, the stiffness of the stainless steel sandwich panel core, the stiffness of the stainless steel sandwich panel, the effective cross-sectional area of the stainless steel sandwich panel core, the effective shear modulus of the stainless steel sandwich panel core, the elastic modulus of the stainless steel sandwich panel face stock, the elastic modulus of the stainless steel sandwich panel core, the thickness of the stainless steel sandwich panel, and the thickness of the stainless steel sandwich panel core are collected.

Step 200: determining the deflection of the stainless steel sandwich plate: the determination of the deflection of the stainless steel sandwich panel comprises the deflection under concentrated load and the deflection under evenly distributed load,

wherein: the reference numerals are the same as those described above.

For confirming the shear force distribution coefficient, the shear force distribution coefficient is further analyzed by establishing a model. The specific process is as follows:

firstly, approximate processing of the rigidity of the sandwich plate.

The core and face sheets are firmly bonded together with cooperative deformation, and if the flexural rigidity of the sandwich panel is K, then if the sandwich panel is considered as a composite beam, since the sandwich panel is composed of the upper and lower face sheets and the core, the flexural rigidity to the central axis O-O is, according to the calculation formula of the mechanical rigidity of the material:

the first term is the stiffness of the panel itself; the second term is the stiffness of the panel relative to O-O; the third term is the stiffness of the core material itself.

In practical applications, the second term in equation (3) plays a major role. The values of the first and third terms are generally small, and the contribution of the rigidity of the sandwich plate per se to the overall rigidity can be ignored. The ratio of the third term to the second term is less than 1% for common soft core materials, such as polyurethane, EPS (polystyrene), rock wool, glass wool and the like, and the influence of the rigidity of the core materials can be ignored. Because the core materials of the stainless steel sandwich plates have different thicknesses, the stainless steel sandwich plates comprise a shallow-pressure stainless steel sandwich plate, a planar stainless steel sandwich plate and a deep-pressure stainless steel sandwich plate, the deep-pressure sandwich plate is generally adopted for a general roof panel for water resistance, and the planar sandwich plate or the shallow-pressure sandwich plate is generally adopted for a wall panel.

For a shallow, deep, flat stainless steel sandwich panel, the stiffness can be expressed as:

and secondly, approximate treatment of the effective cross-sectional area and the effective shear modulus of the stainless steel sandwich plate.

A plane stainless steel sandwich plate.

As shown in fig. 2, the effective core material thickness and the core material thickness of the planar stainless steel sandwich panel are approximately equal, i.e. the following relationship:

D_C≈D_eff G_eff＝GD_eff/D_C≈G A_eff＝bD_eff≈A_C (5)

and (II) shallow-pressure stainless steel sandwich plates.

As shown in fig. 3, the effective core material thickness and the core material thickness of the shallow stainless steel sandwich plate are approximately equal, i.e. the following relationship:

D_C≈D_eff G_eff＝GD_eff/D_C≈G A_eff＝bD_eff≈A_C (6)

and (III) deep-pressing stainless steel sandwich plates.

As shown in fig. 4, the core material thickness is equivalent to the effective core material thickness and the effective shear modulus of the deep-pressed stainless steel sandwich panel, and the core material shear modulus relationship 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 a common roof profiled sandwich panel, d can be 8.0 mm.

Third, the relation between force and deformation in material mechanics

In material mechanics, the relationship between load and deformation can be expressed as follows.

Table of relationships between various forces and deformations:

name (R)	Expression formula	Name (R)	Expression formula
				Deflection	w	Bending moment	M＝-Kw″
Shear force	V＝-Dw″′	Uniform force distribution	q＝Kw^iv

And fourthly, calculating the deflection of the plane stainless steel sandwich plate.

The deformation of the stainless steel sandwich plate is considered from two parts respectively, namely the bending deformation of the sandwich plate; the other is shear deformation of the sandwich panel. And the deformations of the two parts are added to obtain the total deformation of the sandwich plate. For a planar stainless steel sandwich panel, the factors of shear distribution due to the flexural stiffness of the panel itself can be ignored, i.e., it can be assumed that the shear is fully borne by the core material. For soft cores, it can be assumed that the bending moment is borne entirely by the panel;

as can be seen from the above table, the following relationship exists between the force and the deformation, and the schematic diagrams thereof are shown in fig. 5 and 6:

M＝Kγ′₂＝K(γ′-w″) (8)

V＝A_effG_effγ (9)

in FIGS. 5 and 6,. gamma.₁As total strain, γ₂Is the strain induced by bending and gamma is the shear strain. As can be seen from fig. 5 and 6, the shear stress is almost entirely borne by the core material, i.e., the shear force is borne by the core material; while the bending moment is provided by the positive stress of the panel.

From the differential relationship between bending moment, shearing force and uniform force, the following equation can be obtained from material mechanics:

\frac{dM}{dx} - V = 0 - - - (10)

\frac{dV}{dx} + p = 0 - - - (11)

substituting equations (8) and (9) into equations (10) and (11) yields the following relationships:

K(γ″-w″′)-A_effG_effγ＝0 (12)

A_effG_effγ′＝-p (13)

extracting terms for γ and w yields:

in practice, bending moments and shear forces are generally relatively easy to achieve. Equation (14) and equation (15) can be integrated. The following relationship is obtained:

for the uniformly loaded sandwich panel, fig. 7 is shown.

Where η ═ x/L, it is thus possible to obtain:

from equation (16), it can be found that:

by integrating equation (20) twice, m and n being constant terms, we can obtain:

the boundary conditions are as follows: when η is 0 or 1, w ₁0; thus, one can calculate: m-1/12, n-0. It is thus possible to obtain:

the deformation caused by shearing is:

it is integrated twice to obtain:

the boundary conditions are as follows: when η is 0 or 1, w ₂0. M is-0.5 and n is 0, so its value is:

the total deformation of the sandwich panel is therefore:

it is known that when η is 0.5, w can take a maximum value:

w_{\max} = \frac{5 p L^{4}}{384 K} + \frac{{pL}^{2}}{8 A_{eff} G_{eff}} - - - (27)

from equation (6), the above can be expressed as:

w_{\max} = \frac{5 {pL}^{4}}{384 K} + \frac{{pL}^{2}}{8 A_{C} G_{C}} - - - (28)

the original bending resistance bearing capacity formula needs to be changed into:

<math><mrow><mi>f</mi><mo>=</mo><mfrac><msup><mrow><mn>5</mn><mi>pbl</mi></mrow><mn>4</mn></msup><mrow><mn>384</mn><mrow><mo>(</mo><msub><mi>E</mi><mn>1</mn></msub><msub><mi>I</mi><mn>1</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>+</mo><mfrac><msup><mi>Kβpbl</mi><mn>2</mn></msup><mrow><mn>8</mn><mi>GA</mi></mrow></mfrac><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>29</mn><mo>)</mo></mrow></mrow></math>

wherein: e₁The modulus of elasticity of the metal panel; the unit MPa;

I₁the moment of inertia of the upper and lower steel plates with respect to the neutral axis; the unit MIa;

inertia moment I of the upper and lower steel plates to the central shaft of the sandwich plate₁Approximate calculation formula:

I_{1} = \frac{A_{u} A_{d}}{A_{u} + A_{d}} {(D_{C} + d)}^{2} - - - (30)

wherein A is_uThe boundary cross-sectional area of the upper steel plate; a. the_dThe cross-sectional area of the lower steel plate; d_CThe thickness of the sandwich plate is shown; d is the distance from the steel plate-shaped mandrel to the bottom surface on the roof panel, and the common roof panel is 8.0575 mm.

Moment of inertia I of the core material itself₂Approximate calculation formula:

I_{2} = \frac{b {(D_{C} + d)}^{3}}{12} - - - (31)

wherein b is the width of the sandwich plate; d is the thickness of the sandwich plate; d is the distance from the steel plate-shaped mandrel to the bottom surface on the roof panel, and the common roof panel is 8.0575 mm.

And fifthly, calculating the deflection of the profiling stainless steel sandwich plate.

And (I) calculating the deflection of the profiled stainless steel sandwich plate under uniformly distributed load.

When the face sheets used in the sandwich panel are profiled steel sheets, the bending stiffness of the face sheets themselves needs to be considered. The force and deformation conditions in this case are given in figure 8. Unlike in fig. 5, it takes into account the bending moment M borne by the panel itself_F1And M_F2And also the shear forces V to which the panel itself is subjected_F1And V_F2。

The relationship of force and deformation obtained from equations (8) and (9) does not change. Also, the following relationship is present:

M_F1＝-K_F1w″＝E₁I_F1w″M_F2＝-K_F2w″＝E₁I_F2w″ (32)

V_F1＝-K_F1w″′＝E₁I_F1w″′ V_F1＝-K_F2w″′＝E₁I_F1w″′(33)

wherein, K_F1For rigidity of the upper panel, I_F1Is the moment of inertia of the upper panel itself; k_F2For the rigidity of the lower panel, I_F2Is the moment of inertia of the lower panel. Since the ratio of the upper and lower steel plate forces to the deformation is the same, the following relationship is possible:

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)

the above equation divides the sandwich panel into a sandwich portion and a flange portion as shown in fig. 9. This assumption is practical in practical applications.

From equation (8), equation (9), equation (32), and equation (33), the following two differential equations can be obtained by combining equation (34), equation (35), and equations (3-36):

A_effG_effγ-K_Dw″′＝V (37)

K_Cγ′-K_Dw″＝M (38)

substituting V' ═ p in, and removing γ can yield a fourth order differential equation for w:

wherein L is the span of the sandwich panel; α, δ and λ²The values of (A) are respectively as follows:

similarly, eliminating w in equations (37) and (38) may result:

when the bending moment and shear of the sandwich panel are known, the solutions of equations (39) and (41) are:

wherein w_pAnd gamma_pIs an integral solution associated with the load. Because the solution must satisfy equation (8), the following relationship can be obtained:

the integral constant term coefficients of equations (42) and (43) thus become four, and these integral constant term coefficients can be derived from the boundary conditions, for a simple-supported sandwich panel:

w(0)＝0 w″(0)＝0 w(L)＝0 w″(L)＝0 (45)

for uniform load, the following relationship can be obtained from the equations (18) and (19):

η＝x/L (46)

substituting equation (46) into equation (39) can result in the special solution in equation (8) being

Substituting equation (46) into equation (41) yields the special solution in equation (43) as:

substituting the boundary condition (45) into equations (42) and (43) yields:

the following relationship can be solved:

the final solution is therefore:

the mid-span deflection may be taken to a maximum value, and substituting η of 0.5 into equation (48) may result:

it can be seen that the above formula is complicated to calculate, and the applicability in practical engineering is not strong. It is simplified below based on the same theory.

As can be seen from fig. 9, the profiled sandwich panel can be divided into two parts, one part being the shear and bending moments borne by the stiffness of the profiled steel sheet itself. The other part is the sandwich part, i.e., the shear force borne by the core material, the axial force of the panel, and the bending moment borne by the core material itself. The two parts are assumed to be independent, but co-deformed at the point of contact. Two coefficients are introduced, namely the bending moment distribution coefficient epsilon of the sandwich portion and the shear distribution coefficient beta of the sandwich (approximately equal to the shear distribution coefficient of the core). From equations (8) and (9), the following relationship can be obtained:

M_C＝K_C(γ′-w″) (53)

M_D＝-K_Dw″ (54)

V_D＝-K_Dw″′ (55)

V_C＝A_effG_effγ (56)

from equations (34) and (35), the following equation can be obtained:

from equations (53) to (54), the following equation can be obtained:

and the following relationship can be obtained:

K_C(γ′-w″)＝εM A_effG_effγ＝βV (60)

from equations (10) and (11), we can obtain:

K_C(γ″-w″′)/ε＝A_effG_effγ/β (61)

A_effG_effγ′/β＝-p (62)

it is proposed that terms for w and γ can yield the following relationship:

the deflection caused by bending and shearing forces are as follows:

for the uniform load, the moment and the shearing force can be obtained as follows:

substituting equation (66) into equation (65) and performing two integrations yields the following equation:

for a simply supported sandwich plate, w is when η is 0 or 1₁＝0，w ₂0; the following equation is obtained:

the overall deformation calculation is therefore as follows:

therefore, when η is 0.5, the mid-span deflection can take a maximum value:

the shear distribution coefficient beta of the core material can be determined by a finite element method, a shear stress non-uniform distribution coefficient k is introduced, and beta in the formula (72) is taken as k beta, wherein the specific calculation method of beta is as follows:

<math><mrow><mi>β</mi><mo>=</mo><mfrac><mrow><mover><mi>τ</mi><mo>&OverBar;</mo></mover><mo>×</mo><mi>A</mi></mrow><mi>Q</mi></mfrac><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>73</mn><mo>)</mo></mrow></mrow></math>

wherein,

is the average shear stress of the section of the sandwich material near the support, A is the area of the section of the sandwich material, Q is the total shear force value at the position and is determined by the actual shear forceThe distribution of the shear stress of the sandwich panel along the panel thickness in the case is not uniformly distributed, and a shear stress non-uniform distribution coefficient k may be taken, which may be approximately 1.2 for a rectangular section.

The bending moment distribution coefficient of the sandwich part can be transformed as follows:

equation (72) can therefore be converted to the following equation:

from equation (6) for a low pressure sandwich panel:

K_Cis the bending stiffness of the sandwich portion of the sandwich panel.

The deep-profiled steel sheet can be obtained from equation (7):

wherein K_CIs the bending stiffness of the sandwich portion.

And (II) calculating the deflection of the profiled stainless steel sandwich plate under the concentrated load.

The form of the stress under the action of the concentrated load is shown in fig. 10, and the simplified method similar to the above is deduced, and the values of the shearing force and the bending moment of any section of the stress under the action of the concentrated load can be calculated as follows:

M＝PL(1-δ)η-PL{η-δ} (78)

V＝P(1-δ)-P{η-δ} (79)

wherein { η - ε } is taken to be 1 when η - ε > 0 and zero otherwise.

From equations (63) and (64), the following relationship can be obtained:

the following relationship is given for the deformation of the bent portion:

the boundary conditions are as follows: when η is 0 and η is 1, w ₁0, so n can be obtained, and:

substituting equation (82) can result in:

it can be found that the study point is at the left end of the load:

and (83) arranging to obtain the deformation of the right end reference point of the load as follows:

the same shear deformation can be obtained as follows:

<math><mrow><msub><mi>w</mi><mn>2</mn></msub><mo>=</mo><mo>&Integral;</mo><mo>&Integral;</mo><mfrac><mrow><mi>β</mi><msup><mi>VL</mi><mn>2</mn></msup></mrow><mrow><msub><mi>A</mi><mi>eff</mi></msub><msub><mi>G</mi><mi>eff</mi></msub></mrow></mfrac><mi>dξdξ</mi><mo>=</mo><mfrac><mi>PL</mi><mrow><msub><mi>A</mi><mi>eff</mi></msub><msub><mi>G</mi><mi>eff</mi></msub></mrow></mfrac><mo>[</mo><mi>ξ</mi><mrow><mo>(</mo><mn>1</mn><mo>-</mo><mi>δ</mi><mo>)</mo></mrow><mo>-</mo><mo>{</mo><mi>ξ</mi><mo>-</mo><mi>δ</mi><mo>}</mo><mo>+</mo><mi>mξ</mi><mo>+</mo><mi>n</mi><mo>]</mo><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>86</mn><mo>)</mo></mrow></mrow></math>

when the boundary condition eta is 0 or 1, w ₂0 can result in: m-n-0 thus the shear deformation can be:

the reference point for the right end of the load has the value:

the reference point for the left end of the load has the value:

the expressions of the deformation at the left end and the right end of the load can be obtained by respectively adding the shear deformation and the bending deformation at the two ends of the load:

wherein:

for the case of a load in a mid-span position. It can be calculated that the maximum value of the bending moment at this time occurs at the mid-span position, and thus can be calculated by equation (90) or (91) in combination with equation (92):

from equation (74), one can obtain:

and equation (94) is more conservative

And introducing a shearing force uneven distribution coefficient k under the condition of uniformly distributing the load, and changing the shearing force distribution coefficient beta into k beta. Then equation (93) can be reduced to the following form:

and sixthly, taking the stainless steel sandwich plate as the shear force distribution coefficient of the wall panel.

The stainless steel wall panel has no obvious surface plate and core material stripping, so that the bonding damage can be not considered. Finite element models of rock wool and glass wool can be respectively established for research.

Establishing a stainless steel surface wall panel model:

assume that in the bending test, the sandwich panel face sheets and the core material are in the linear elastic phase. The panel and the core material do not slide, the joint work is completely ensured, and the panel and the core material TIE are constrained together. The stand-offs are constructed of the same material as the face plate to maintain conformance with the plate. The width was 50mm and the distance between the two steel sheets was 1950 mm. The panel uses three-dimensional shell elements, using the S4R (4-node reduced integral) model. The core material is a three-dimensional solid unit and adopts a C3D8R (8-node reduction integral) model. In the analysis step, static general is adopted, uniform load is applied to an upper steel plate of the sandwich plate, and the support is restrained to be simple.

The performance parameters of the rock wool material are obtained by a material performance test and are specifically set as follows.

(1) For the sheet, isotropy is used, and the main consideration is axial tension, so the parameters are the following modulus of elasticity: e is 8.326MPa and v is 0.13

(2) The thick plate adopts anisotropy, the main consideration is that the axial longitudinal grain is pulled, the width direction is reversely pressed, the transverse direction is pressed along the grain, the specific parameters are obtained by material performance experiments, and the specific parameters are set as follows:

E₁＝8.326MPa，E₂＝0.238MPa，E₃＝3.29MPa，v₁＝v₂＝v₃＝0.13，G₁＝0.35MPa，

G₂＝1.56MPa，G₃＝0.35MPa

the performance parameters of the glass fiber cotton material are obtained by material performance experiments and are specifically set as follows:

(1) for the sheet, isotropy is used, and the main consideration is axial tension, so the parameters are the following modulus of elasticity: e is 7.7MPa and v is 0.13

(2) The thick plate adopts anisotropy, the main consideration is that the axial longitudinal grain is pulled, the width direction is reversely pressed, the transverse direction is pressed along the grain, the specific parameters are obtained by material performance experiments, and the specific parameters are set as follows: e₁＝7.7MPa，E₂＝0.06MPa，E₃＝1.59MPa，v₁＝v₂＝v₃＝0.13，G₁＝0.23MPa，G₂＝1.5MPa，G₃＝0.23MPa。

The direction for the anisotropic material 1 is the direction along the plate length; the 2 direction is the direction along the width of the plate; the 3 direction is the direction along the thickness of the plate. The rock wool is placed along the width direction of the plate in the reverse grain direction and along the length direction and the thickness direction of the plate in the forward grain direction.

And secondly, determining the shear distribution coefficient of the stainless steel sandwich plate as the wall panel.

1. Determination of shear distribution coefficient of stainless steel rock wool sandwich board as wall panel

Because the deformation of the sandwich plate is large, f is L/200 which is often used as a limit design index in design, namely, state limitation is used, six plate thicknesses of 50mm, 60 mm, 70 mm, 80 mm, 90 mm and 100mm and four panel thicknesses of 0.4 mm, 0.5 mm, 0.6 mm and 0.7mm are used for analysis in the analysis in order to obtain the relation between the shearing force distribution coefficient value and the plate thickness and the panel thickness, and 48 finite element models are built. The simulation results and the theoretical calculation results are shown in table 2. As can be seen from Table 2, the theoretical calculation result has a smaller error compared with the finite element simulation result and is basically controlled within 10%, so that the method has certain applicability. And the error between the experimental value and the theoretical calculated value is controlled to be 9.3 percent, and the calculated result is slightly conservative, so the calculated result meets the requirement.

TABLE 2 comparison of rock wool shingle simulation results with theoretical calculation results

Type (B)	Test value kPa	Simulation of beta	Simulation value kPa	Calculated value kPa	Error of calculation and experiment	Calculation 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	--

In order to study the relationship between the shear distribution coefficient and the thickness of the stainless steel rock wool sandwich panel and the thickness of the panel, fig. 11 and 12 show the relationship between the shear distribution coefficient and the thickness of the stainless steel rock wool sandwich panel and the thickness of the panel. From the above graph, it can be seen that the shear distribution coefficient is approximately quadratic with the change in sheet thickness, and the change value is relatively large. The variation value is small along with the approximate linear relation of the thickness of the panel. The shear distribution coefficient is as follows:

wherein D is the thickness of the sandwich panel, and D is the thickness of the panel.

2. And determining the shear distribution coefficient of the stainless steel glass wool sandwich board as a wall panel.

The simulation results and theoretical calculation results of the glass wool shingle are shown in table 3.

TABLE 3 comparison of simulation results of glass wool wall panel with theoretical calculation results

Type (B)	Test value kPa	Simulation of beta	Simulation value kPa	Calculated value kPa	Error of calculation and experiment	Calculation 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, the theoretical calculation result has smaller error compared with the finite element simulation result and is basically controlled within 10%, so that the method has applicability. And the error between the experimental value and the theoretical calculated value is controlled to be 0.9 percent, and the calculated result is conservative, so the calculated result meets the requirement.

In order to study the relationship between the shear distribution coefficient and the thickness of the plate and the thickness of the panel, fig. 13 is a graph showing the relationship between the shear distribution coefficient and the thickness of the stainless steel glass wool sandwich plate, and fig. 14 is a graph showing the relationship between the shear distribution coefficient and the thickness of the stainless steel glass wool sandwich plate.

The relation curve of the shear distribution coefficient of the stainless steel-glass wool wall panel with respect to the thickness of the panel can be taken as two times, and the relation with respect to the thickness of the panel is taken as one time.

The linear regression results of the shear distribution coefficients of the stainless steel-glass wool shingle panels were as follows:

and secondly, the stainless steel sandwich plate is used as the shear distribution coefficient of the roof panel.

Firstly, establishing a stainless steel surface roof panel model.

Assume that in the bending test, the sandwich panel face sheets and the core material are in the linear elastic phase. The support is made of the same material as the face plate to maintain the consistency with the plate, the width of the support is 50mm, the distance between two steel sheets is 1950mm, the length of each span is 1950mm, the bonding between the face plate and the core material is not ideal for the roof plate, and particularly, the lower steel plate and the core material are almost completely peeled off. It is necessary to take into account the adhesive damage, but not simply TIE together.

E₁＝8.326MPa，E₂＝0.238MPa，E₃＝3.29MPa，v₁＝v₂＝v₃＝0.13，G₁＝0.35MPa，G₂＝1.56MPa，G₃＝0.35MPa

COHESIVE damage can be modeled using a cohesiveness cell. The thickness of the film may be set to 0.0001m to meet the practical situation. The method adopts an isotropic model, the elastic modulus is 9MPa, and three-dimensional solid units, namely eight-node linear units of COHESIVE are adopted.

(II) the bonding units are in constitutive relation.

To simulate the tie layer, a new region is added between the face sheet and the core layer, creating a new cell. In ABAQUS, where such units are referred to as COHESIVE units, this new area can be considered as the area where one resin layer is in excess. But unlike the simple resin region, it has the main function of connecting the upper and lower monolayers. The upper and lower surfaces are connected by a cohesiveness Traction force, which relates to the distance between the upper and lower surfaces, and is referred to as the "cohesiveraw" or "Traction-continuity raw". On the adhesive surfaceThe applied force is divided into three types, one is normal stress t_nThe other two being tangential shear stresses t_sAnd t_tFig. 15 is a stress diagram of a COHESIVE cell, fig. 16 is a stress diagram of a COHESIVE cell, and fig. 17 is a linear elastic-linear softening constitutive model of a COHESIVE cell.

Is provided with

The area enclosed by the curve and the coordinate axis in fig. 17 is called critical strain energy release rate. Namely:

<math><mrow><mfenced open='' close='}'><mtable><mtr><mtd><msubsup><mo>&Integral;</mo><mn>0</mn><msubsup><mi>δ</mi><mi>n</mi><mi>max</mi></msubsup></msubsup><msub><mi>t</mi><mi>n</mi></msub><mrow><mo>(</mo><mi>δ</mi><mo>)</mo></mrow><mi>d</mi><msub><mi>δ</mi><mi>n</mi></msub><mo>=</mo><msubsup><mi>G</mi><mi>C</mi><mi>n</mi></msubsup></mtd></mtr><mtr><mtd><msubsup><mo>&Integral;</mo><mn>0</mn><msubsup><mi>δ</mi><mi>s</mi><mi>max</mi></msubsup></msubsup><msub><mi>t</mi><mi>s</mi></msub><mrow><mo>(</mo><mi>δ</mi><mo>)</mo></mrow><mi>d</mi><msub><mi>δ</mi><mi>s</mi></msub><mo>=</mo><msubsup><mi>G</mi><mi>C</mi><mi>s</mi></msubsup></mtd></mtr><mtr><mtd><msubsup><mo>&Integral;</mo><mn>0</mn><msubsup><mi>δ</mi><mi>t</mi><mi>max</mi></msubsup></msubsup><msub><mi>t</mi><mi>t</mi></msub><mrow><mo>(</mo><mi>δ</mi><mo>)</mo></mrow><mi>d</mi><msub><mi>δ</mi><mi>t</mi></msub><mo>=</mo><msubsup><mi>G</mi><mi>C</mi><mi>t</mi></msubsup></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>98</mn><mo>)</mo></mrow></mrow></math>

FIG. 17 is a linear elastic-linear softening constitutive model of COHESIVE units, assuming that the bonded region has a thickness T_cI.e. the thickness of the Part built in the model (the value is0.0001 m). Then the three strains that correspond are:

when delta < delta₀When in use, the elastic thread is elastic, and the elastic thread has the following characteristics:

when delta₀＜δ＜δ_maxIn this case, the soft zone is damaged, and there are:

wherein D is a damage coefficient, and D is more than or equal to 0 and less than or equal to 1. When D ═ 0, it means that the material did not yield or just started to yield; when D ═ 1, it indicates that the material is broken and loses its load-bearing capacity.

When delta > delta_maxWhen the material loses its load-bearing capacity, the bond area is destroyed and the sandwich panel delaminates.

In fig. 17 point 0 has not yet taken the load, point 1 is in the elastic zone, point 2 yields, point 3 has entered the softening zone, point 4 has just broken, and point 5 has broken the delamination.

Since laminate delamination is often not caused by a single cracking mode, and the delamination cannot be accurately simulated by considering a single cracking mode, mixed mode cracking criteria must be considered. I used in the model is the B-K cracking criterion, namely:

wherein G is_S＝G_s+G_t，G_T＝G_n+G_tFor polyurethane composites, the index η is 2. When is given

η, critical strain relief G of the material^CIs exactly G_S/G_TA determined function.

And (II) determining the shear distribution coefficient of the stainless steel sandwich plate as the roof plate.

In the model, a layer of COHESIVE entity unit is added between a lower layer panel and a core material to simulate bonding. By verification, the method compares the composite experimental results. A total of 48 finite element models of six thicknesses (50, 60, 70, 80, 90, 100mm) and four panel thicknesses (0.4, 0.5, 0.6, 0.7mm) were built for study. And table 4 shows the comparison between the simulation result of the stainless steel rock wool roof sandwich panel and the theoretical calculation result.

Table 4 comparison of simulation results of stainless steel rock wool roof sandwich panel with theoretical calculation results

Type (B)	Test value kPa	Simulation of beta	Simulation value kPa	Calculated value kPa	Error of calculation and experiment	Calculation 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, the theoretical calculation result has a smaller error compared with the finite element simulation result and is basically controlled within 10%, so that the theoretical calculation result has certain applicability. And the error between the experimental value and the theoretical calculated value is controlled to be 4.0 percent, and the calculated result is conservative, so the calculated result meets the requirement.

Fig. 18 is a graph showing the relationship between the shear force distribution coefficient and the panel thickness of the stainless steel rock wool roof sandwich panel of the present invention, and fig. 19 is a graph showing the relationship between the shear force distribution coefficient and the panel thickness of the stainless steel rock wool roof sandwich panel of the present invention. It can be seen from fig. 18 and 19 that the shear distribution coefficient is approximately quadratic with the change in sheet thickness, and the change value is relatively large. The variation is small as the thickness of the panel is approximately linear. In order to keep consistency, the relation curve of the shear distribution coefficient of the stainless steel-rock wool roof panel with respect to the thickness of the panel is taken as the second order, and the relation with respect to the thickness of the panel is taken as the first order approximately. The fitting results of the shear distribution coefficients are as follows:

the simulation results and the theoretical calculation results of the stainless steel glass wool roof sandwich panel are shown in table 5.

TABLE 5 comparison of simulation results of stainless steel glass wool roof sandwich panel with theoretical calculation results

Type (B)	Test value kPa	Simulation of beta	Simulation value kPa	Calculated value kPa	Error of calculation and experiment	Calculation 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, apart from the individual values, the error of the theoretical calculation result is basically controlled within 10% compared with the finite element simulation result, so that the method has certain applicability. And the error between the experimental value and the theoretical calculated value is controlled to be 1.3 percent, and the calculated result is conservative, so the calculated result meets the requirement.

FIG. 20 is a graph showing the relationship between the shear force distribution coefficient and the panel thickness of the stainless steel glass wool roof sandwich panel, and FIG. 21 is a graph showing the relationship between the shear force distribution coefficient and the panel thickness of the stainless steel glass wool roof sandwich panel. The relation curve of the shear distribution coefficient of the stainless steel-glass wool roof panel with respect to the thickness of the panel is taken as the second order, and the relation with respect to the thickness of the panel can be taken as the first order approximately. The results of the curve fitting are as follows:

in summary, the following steps:

R₁、R₂、R₃、R₄the values are shown in Table 1.

Table 1: coefficient R₁、R₂、R₃、R₄Value taking table:

step 300: determining the bending resistance bearing capacity of the stainless steel sandwich plate: and determining the bending resistance bearing capacity of the stainless steel sandwich plate according to the relationship between the load and the bending resistance bearing capacity.

The deflection of the sandwich panel is related to the allowable deflection value as follows: w is a_max≤[f]

[f] See the limit value of deflection in the national standard metal surface heat insulation sandwich plate for buildings (GB/T23932-2009). Then:

concentrated load determination:

determination of local load:

for shallow and flat sandwich panels (shingles):

A_eff≈A_C；G_eff≈G

wherein A is_CIs the cross-sectional area of the core material; g is the shear modulus of the core material.

For deep-pressed sandwich panels (roof panels):

G_eff＝G_C(D_C+d)/D_C；A_eff＝A_C(D_C+d)/D_C

wherein D_CIs the thickness of the sandwich core plate; d is taken to be 8.0 mm.

Thereby determining the bending resistance bearing capacity of the stainless steel sandwich plate.

The method for determining the bending resistance bearing capacity of the stainless steel sandwich plate accurately determines the bending resistance bearing capacity of the stainless steel sandwich plate by determining the deflection of the stainless steel sandwich plate under concentrated load and the deflection of the stainless steel sandwich plate under uniformly distributed load. The invention accurately determines the bending resistance bearing capacity of the stainless steel sandwich plate, and can more accurately evaluate the safety performance of the stainless steel sandwich plate, thereby promoting the application of the stainless steel sandwich plate.

The specific implementation mode of the invention is as follows: the method for determining the bending resistance bearing capacity of the stainless steel sandwich plate is applied to determining the bearing capacity of the stainless steel straw sandwich plate component.

The foregoing is a more detailed description of the invention in connection with specific preferred embodiments and it is not intended that the invention be limited to these specific details. For those skilled in the art to which the invention pertains, several simple deductions or substitutions can be made without departing from the spirit of the invention, and all shall be considered as belonging to the protection scope of the invention.

Claims

1. The method for determining the bending resistance bearing capacity of the stainless steel sandwich plate is characterized in that a panel of the stainless steel sandwich plate is a stainless steel plate, a core plate of the stainless steel sandwich plate is formed by bonding or pouring an intermediate layer through a bonding agent, and the method for determining the bending resistance bearing capacity of the stainless steel sandwich plate comprises the following steps:

wherein: w represents the total deflection of the sandwich panel;

p represents the load value of the sandwich panel;

l represents the span of the sandwich panel;

E₁denotes the modulus of elasticity of the face stock;

I₁representing the moment of inertia of the upper and lower metal surfaces against the central axis of the sandwich panel;

A_effrepresents the effective cross-sectional area of the sandwich panel;

G_effrepresents the effective shear modulus of the sandwich panel;

k denotes the shear stress non-uniformity coefficient, which can be taken to be 1.2 for common plate types. Beta represents the shear distribution coefficient of the sandwich panel,R₁、R₂、R₃、R₄the values are shown in Table 1.

Table 1: coefficient R₁、R₂、R₃、R₄Value taking table:

2. The method of determining a bending resistance bearing capacity of a stainless steel sandwich panel according to claim 1, wherein the stainless steel sandwich panel comprises a wall panel and a roof panel.

3. The method for determining bending resistance and bearing capacity of a stainless steel sandwich panel according to claim 1 or 2, wherein the core material of the stainless steel sandwich panel is rock wool bonded by an adhesive or cast.

4. The method of claim 1 or 2, wherein the core material of the stainless steel sandwich panel is formed by bonding polyurethane with an adhesive or by casting.

5. A stainless steel sandwich panel applying the method for determining the bending resistance bearing capacity of the stainless steel sandwich panel is characterized in that the method for determining the bending resistance bearing capacity of the stainless steel straw sandwich panel is applied to the stainless steel straw sandwich panel.