Calculation method for the sandwich structure of the sandwich laminated floor slab

Abstract

The present invention discloses a calculation method for the sandwich structure of a sandwich laminated floor slab, comprising the following steps: First step, determine the values of the surface air convection heat transfer coefficients αi and αe according to the actual environment of the upper and lower surfaces of the sandwich laminated floor slab, and at the same time determine the thermal performance requirements that the sandwich laminated floor slab needs to meet, namely the structural heat transfer coefficient K'; Second step, select the basic structural parameters of the sandwich laminated floor slab; Third step, select the materials used for the sandwich structure layer, substitute the thermal conductivity λ3' of the used materials into the formula to obtain the recommended thickness δ3' of the sandwich layer; Fourth step, based on the recommended thickness δ3' of the sandwich layer, determine the thickness δ3 of the sandwich layer; Fifth step, use the heat transfer coefficient calculation formula of the sandwich laminated floor slab to calculate the structural insulation performance. The present invention has the characteristics of being able to accurately calculate the thickness of the sandwich layer and the actual heat transfer coefficient of the sandwich laminated floor slab.

Description

Technical Field

[0001] The present invention relates to the field of design of sandwich laminated floor slabs, and particularly to a calculation method for the sandwich structure of sandwich laminated floor slabs. Background Art

[0002] In view of the practice of laying thermal insulation materials on the surface of floor slabs in existing buildings to meet the mandatory thermal insulation requirements of codes, the concept of incorporating the thermal insulation performance of floor slabs into the structural system is proposed. Combining the manufacturing and construction characteristics of precast laminated floor slabs, a "sandwich" is formed by placing the thermal insulation layer between the top slab and the bottom slab of the laminated floor slab, as Figure 1 shown, which hinders the effective transfer of heat between floors. The sandwich layer is located in the neutral axis region of the overall structural force, and is connected by steel truss bars between layers. It bears a relatively small bending moment itself, and building fluid materials such as aerated concrete, foamed cement, thermal insulation mortar, etc. or building solid materials such as mineral wool boards, phenolic boards, polystyrene foam plastics, etc. can be selected. When the selected sandwich layer material has a certain stiffness, it can improve the stiffness and load-bearing capacity of the floor slab while exerting its own thermal insulation performance. The bottom slab of the laminated floor slab together with the sandwich layer is prefabricated in the factory and then transported to the construction site for pouring.

[0003] The existing heat transfer coefficient calculation methods only consider the interaction between material layers, and there is currently no accurate analysis method for the influence caused by the steel bars embedded in the layers. GB50176-2016 regards concrete and the steel bars embedded in it as a whole in the heat transfer analysis module, and separately defines the thermal conductivity of the reinforced concrete material layer. In the new sandwich laminated floor slab structure, the truss steel bars will form a complex heat bridge effect, resulting in the inapplicability of the traditional code formulas. Therefore, a supporting calculation and analysis method needs to be established. Summary of the Invention

[0004] The purpose of the present invention is to provide a calculation method for the sandwich structure of sandwich laminated floor slabs. The present invention has the characteristics of being able to accurately calculate the thickness of the sandwich layer and the actual heat transfer coefficient of the sandwich laminated floor slab.

[0005] The technical solution of the present invention: A calculation method for the sandwich structure of sandwich laminated floor slabs, comprising the following steps:

[0006] First step, determine the values of the surface air convection heat transfer coefficients α i and α e of the sandwich laminated floor slab according to the actual environments on the upper and lower surfaces of the sandwich laminated floor slab, and at the same time determine the thermal performance requirements that the sandwich laminated floor slab needs to meet, that is, the structural heat transfer coefficient K';

[0007] Second step, select the basic structural parameters of the sandwich laminated floor slab, including determining the specific values of the truss steel bar spacing S, the mortar thickness δ1, the top slab thickness δ2, the bottom slab thickness δ4, and the plastering thickness δ5;

[0008] Select the thermal conductivity of mortar, concrete and plastering materials;

[0009] Step 3: Select the materials used for the sandwich structure layer, and substitute the thermal conductivity λ3' of the materials used into the formula to obtain the recommended thickness δ3' of the sandwich layer;

[0010] Step 4: Based on the recommended thickness δ3' of the sandwich layer, determine the thickness δ3 of the sandwich layer;

[0011] Step 5: Use the calculation formula for the heat transfer coefficient of the sandwich-layer composite floor slab to calculate the structural insulation performance.

[0012] In the above-mentioned calculation method for the sandwich structure of the sandwich-layer composite floor slab, in ordinary building interiors, α i = α e = 8.7 W / (m 2 ·K).

[0013] In the above-mentioned calculation method for the sandwich structure of the sandwich-layer composite floor slab, the thermal conductivity of the mortar λ1 = 0.93 W / (m·K), the thermal conductivities of the top and bottom concrete slabs λ2 = λ4 = 1.51 W / (m·K), and the thermal conductivity of the plastering λ5 = 0.87 W / (m·K);

[0014] In the above-mentioned calculation method for the sandwich structure of the sandwich-layer composite floor slab, the specific determination process of the thickness δ3 of the sandwich layer is as follows:

[0015] If 20 mm ≤ δ3' ≤ 40 mm, then determine the final parameters of the sandwich layer structure as λ3 = λ3', δ3 = δ3', and the final value of δ3 will be rounded up in 5 mm increments of δ3';

[0016] If δ3' < 20 mm, select materials that are cheaper and have a relatively higher thermal conductivity, and go back to Step 3 to recalculate;

[0017] If δ3' > 40 mm, then select materials with better heat insulation performance and go back to Step 3 to recalculate.

[0018] In the above-mentioned calculation method for the sandwich structure of the sandwich-layer composite floor slab, the calculation formula for the heat transfer coefficient K of the sandwich-layer composite floor slab is

[0019]

[0020] β = 0.24S - 2.53δ3 - 1.15δ4 + λ3 + 0.72;

[0021] where K is the structural heat transfer coefficient, W / (m 2 ·K); R is the thermal resistance, m 2 ·K / W; α i, α e is the convective heat transfer coefficient of air between the hot and cold surfaces of the structure, W / (m 2 ·K); S is the spacing of the truss reinforcement, m; δ1 is the thickness of the mortar, m; λ1 is the thermal conductivity of the mortar material, W / (m·K); δ2 is the thickness of the roof slab, m; λ2 is the thermal conductivity of the roof slab material, W / (m·K); δ3 is the thickness of the sandwich layer, m; λ3 is the thermal conductivity of the sandwich layer material, W / (m·K); δ4 is the thickness of the bottom slab, m; λ4 is the thermal conductivity of the bottom slab material, W / (m·K); δ5 is the thickness of the plaster, m; λ5 is the thermal conductivity of the plaster material, W / (m·K).

[0022] Compared with the prior art, the present invention first calculates the recommended thickness of the sandwich layer using a specific formula, and then determines the final thickness of the sandwich layer using a specific method, so as to obtain the thickness of the sandwich layer that meets the actual heat transfer performance requirements. Using the heat transfer coefficient calculation formula of the sandwich layer laminated floor slab, the actual heat transfer coefficient of the sandwich layer laminated floor slab can be calculated well, and its thermal insulation performance can be accurately evaluated during the structural design process. The calculation method is simple and fast, and the accuracy is also relatively high. At the same time, the above calculation method and calculation formula also have good applicability. It can provide guidance and reference for relevant personnel during the structural design and use stages. To sum up, the present invention has the characteristics of being able to accurately calculate the thickness of the sandwich layer and the actual heat transfer coefficient of the sandwich layer laminated floor slab. Description of the Drawings

[0023] Figure 1 is the structural view of the sandwich layer laminated floor slab;

[0024] Figure 2 is the calculation flow chart of the sandwich layer thickness;

[0025] Figure 3 is the calculation flow chart of the heat transfer coefficient K. Detailed Embodiments

[0026] The present invention will be further described below in conjunction with embodiments, but it shall not be used as a basis for limiting the present invention.

[0027] Embodiment. The calculation method for the sandwich structure of the sandwich layer laminated floor slab includes the following steps:

[0028] The first step is to determine the values of the surface air convective heat transfer coefficients α i and α e according to the actual environment on the upper and lower surfaces of the sandwich layer laminated floor slab, and at the same time determine the thermal performance requirements that the sandwich layer laminated floor slab needs to meet, that is, the structural heat transfer coefficient K'.

[0029] Step 2: Select the basic structural parameters of the sandwich laminated floor slab, including determining the specific values of the truss reinforcement spacing S, mortar thickness δ1, top plate thickness δ2, bottom plate thickness δ4, and plaster thickness δ5;

[0030] Select the thermal conductivity coefficients of the mortar, concrete, and plaster materials;

[0031] Step 3: Select the materials used for the sandwich structure layer, and substitute the thermal conductivity coefficient λ3' of the materials used into the formula to obtain the recommended thickness δ3' of the sandwich layer;

[0032] Step 4: Based on the recommended thickness δ3' of the sandwich layer, determine the thickness δ3 of the sandwich layer;

[0033] Step 5: Use the calculation formula for the heat transfer coefficient of the sandwich laminated floor slab to calculate the structural insulation performance.

[0034] In ordinary building interiors, α i = α e = 8.7 W / (m 2 ·K).

[0035] The thermal conductivity coefficient of the mortar λ1 = 0.93 W / (m·K), the thermal conductivity coefficients of the top and bottom plate concretes λ2 = λ4 = 1.51 W / (m·K), and the thermal conductivity coefficient of the plaster λ5 = 0.87 W / (m·K);

[0036] The specific determination process of the thickness δ3 of the sandwich layer is as follows:

[0037] If 20 mm ≤ δ3' ≤ 40 mm, then determine the final parameters of the sandwich layer structure λ3 = λ3', δ3 = δ3', and the final value of δ3 will be rounded up in 5 - mm increments from δ3';

[0038] If δ3' < 20 mm, select materials that are cheaper and have relatively higher thermal conductivity coefficients, and go back to Step 3 to recalculate;

[0039] If δ3' > 40 mm, then select materials with better heat insulation performance and go back to Step 3 to recalculate.

[0040] The calculation formula for the heat transfer coefficient K of the sandwich laminated floor slab is

[0041]

[0042] β = 0.24S - 2.53δ3 - 1.15δ4 + λ3 + 0.72;

[0043] where K is the structural heat transfer coefficient, W / (m 2 ·K); R is the thermal resistance, m 2 ·K / W; α i, α e is the convective heat transfer coefficient of air between the hot and cold surfaces of the structure, W / (m 2 ·K); S is the spacing of the truss bars, m; δ1 is the thickness of the mortar, m; λ1 is the thermal conductivity of the mortar material, W / (m·K); δ2 is the thickness of the roof slab, m; λ2 is the thermal conductivity of the roof slab material, W / (m·K); δ3 is the thickness of the sandwich layer, m; λ3 is the thermal conductivity of the sandwich layer material, W / (m·K); δ4 is the thickness of the floor slab, m; λ4 is the thermal conductivity of the floor slab material, W / (m·K); δ5 is the thickness of the plaster, m; λ5 is the thermal conductivity of the plaster material, W / (m·K).

[0044] Example 1

[0045] Now, a new type of sandwich laminated floor slab is to be designed for use as a household floor slab in civil buildings in hot summer and cold winter regions, with the requirement that the heat transfer coefficient K' ≤ 1.8 W / (m 2 ·K), that is, K' is taken as 1.8 W / (m 2 ·K). The truss spacing has been determined to be 600 mm, the mortar thickness is 30 mm, the roof slab thickness is 70 mm, the floor slab thickness is 60 mm, and the plaster thickness is 10 mm. All of the above are selected from commonly used engineering materials, and the design parameters of the floor slab sandwich layer structure are determined.

[0046] Design process:

[0047] It can be obtained from GB50176 - 2016 that the inner surface heat transfer coefficient α of the envelope structure under typical working conditions i = 8.7 W / (m 2 ·K), and this value of α can be taken for both the upper and lower surfaces of the household floor slab i = α e = 8.7 W / (m 2 ·K).

[0048] The thicknesses of each layer are given: δ1 = 0.03 m, δ2 = 0.07 m, δ4 = 0.06 m, δ5 = 0.01 m. The thermal conductivities of commonly used building materials: mortar λ1 = 0.93 W / (m·K), concrete λ2 = λ4 = 1.51 W / (m·K), plaster λ5 = 0.87 W / (m·K). The truss spacing S = 0.6 m, and the heat transfer coefficient K ≤ 1.8 W / (m 2 ·K).

[0049] Inorganic thermal insulation mortar is initially selected as the sandwich layer material, and its thermal conductivity is λ3' = 0.14 W / (m·K). Substituting it into the formula, the recommended thickness of the sandwich layer δ3' = 0.0294 m = 29.4 mm is obtained.

[0050] 20 mm ≤ δ3' ≤ 40 mm, then it is determined that λ3 = 0.14 W / (m·K), δ3 = 30 mm.

[0051] The sandwich layer material is inorganic thermal insulation mortar with a thermal conductivity of 0.14W / (m·K). The sandwich layer thickness of 30mm can meet the design requirements.

[0052] Calculation example 2

[0053] Now we need to design a new type of sandwich composite floor slab for use as a household floor slab in civil buildings in mild climate areas, and the heat transfer coefficient K' is required to be ≤1.8W / (m 2 ·K), that is, K' is 1.8W / (m 2 ·K). The truss spacing has been determined to be 600mm, the mortar thickness is 30mm, the top plate thickness is 65mm, the bottom plate thickness is 55mm, and the plaster thickness is 10mm. The above materials are commonly used in engineering projects to determine the design parameters of the floor sandwich structure.

[0054] Design process:

[0055] According to GB50176-2016, the heat transfer coefficient α of the inner surface of the enclosure structure under typical working conditions is i =8.7W / (m 2 K), the upper and lower surfaces of the floor slabs can be taken according to this value α i =α e =8.7W / (m 2 ·K).

[0056] The thickness of each layer is given as: δ1 = 0.03m, δ2 = 0.065m, δ4 = 0.055m, δ5 = 0.01m.

[0057] Thermal conductivity of commonly used building materials: mortar λ1 = 0.93W / (m·K),

[0058] Concrete λ2=λ4=1.51W / (m·K), plaster λ5=0.87W / (m·K). Truss spacing S=0.6m, heat transfer coefficient K≤1.8W / (m 2 ·K).

[0059] The glass bead insulation mortar was initially selected as the sandwich layer material. Its thermal conductivity is λ3' = 0.08W / (m·K). Substituting it into the formula, the recommended thickness of the sandwich layer is δ3' = 0.0184m = 18.4mm.

[0060] If δ3' is less than 20mm, a cheaper material can be used for the core layer. Using aerated concrete as the core material, with a thermal conductivity of 0.1W / (m·K), the recommended core layer thickness is δ3' = 0.0225m = 22.5mm.

[0061] 20mm≤δ3'≤40mm, then λ3=0.1W / (m·K), δ3=25mm.

[0062] The sandwich layer material is selected as aerated concrete with a thermal conductivity of 0.1 W / (m·K). A sandwich layer thickness of 25 mm can meet the design requirements.

[0063] Example 3

[0064] Now, a new type of sandwich-layer composite floor slab is to be designed for use as a household floor slab in civil buildings in cold regions, with the requirement that the heat transfer coefficient K' ≤ 1.5 W / (m²·K), that is, K' is taken as 1.5 W / (m²·K). The truss spacing has been determined to be 600 mm, the mortar thickness is 25 mm, the top slab thickness is 75 mm, the bottom slab thickness is 65 mm, and the plaster thickness is 15 mm. All of the above use commonly used engineering materials, and the design parameters of the floor slab sandwich layer structure are determined.

[0065] Design process:

[0066] It can be obtained from GB50176-2016 that the inner surface heat transfer coefficient α of the envelope structure under typical working conditions i = 8.7 W / (m 2 ·K), and this value of α can be used for both the upper and lower surfaces of the household floor slab, i.e., α i = α e = 8.7 W / (m 2 ·K).

[0067] The thicknesses of each layer are given: δ1 = 0.025 m, δ2 = 0.075 m, δ4 = 0.065 m, δ5 = 0.015 m. The thermal conductivities of commonly used building materials: mortar λ1 = 0.93 W / (m·K), concrete λ2 = λ4 = 1.51 W / (m·K), plaster λ5 = 0.87 W / (m·K). The truss spacing S = 0.6 m, and the heat transfer coefficient K ≤ 1.5 W / (m 2 ·K).

[0068] Inorganic thermal insulation mortar is initially selected as the sandwich layer material, with its thermal conductivity λ3' = 0.14 W / (m·K). Substituting it into the formula, the recommended thickness of the sandwich layer δ3' = 0.0452 m = 45.2 mm is obtained.

[0069] Since δ3' > 40 mm, a material with better heat insulation performance needs to be selected for the sandwich layer. Aerated concrete is selected as the sandwich layer material, with its thermal conductivity of 0.1 W / (m·K). Substituting it into the formula, the recommended thickness of the sandwich layer δ3' = 0.0337 m = 33.7 mm is obtained.

[0070] Since 20 mm ≤ δ3' ≤ 40 mm, then λ3 = 0.1 W / (m·K) and δ3 = 35 mm are determined.

[0071] The sandwich layer material is selected as aerated concrete with a thermal conductivity of 0.1 W / (m·K). A sandwich layer thickness of 35 mm can meet the design requirements.

[0072] 2. Evaluation of Structural Thermal Insulation Performance

[0073] Calculation formula for the heat transfer coefficient of the new sandwich laminated floor slab:

[0074]

[0075] β = 0.24S - 2.53δ3 - 1.15δ4 + λ3 + 0.72 (2)

[0076] Where: K is the structural heat transfer coefficient, W / (m 2 ·K); R is the thermal resistance, m 2 ·K / W; α i 、α e are the air convection heat transfer coefficients on the hot and cold surfaces of the structure, W / (m 2 ·K); S is the spacing of the truss reinforcement, m; δ1 is the thickness of the mortar, m; λ1 is the thermal conductivity of the mortar material, W / (m·K); δ2 is the thickness of the top plate, m; λ2 is the thermal conductivity of the top plate material, W / (m·K); δ3 is the thickness of the sandwich layer, m; λ3 is the thermal conductivity of the sandwich layer material, W / (m·K); δ4 is the thickness of the bottom plate, m; λ4 is the thermal conductivity of the bottom plate material, W / (m·K); δ5 is the thickness of the plaster, m; λ5 is the thermal conductivity of the plaster material, W / (m·K).

[0077] The specific calculation process is as follows:

[0078] Determine the values of the air convection heat transfer coefficients α i 、α e on the upper and lower surfaces of the laminated floor slab according to the actual environment. The air convection heat transfer coefficient on the indoor surface of ordinary buildings is taken as 8.7 W / (m2·K).

[0079] Select the basic structural parameters of the new sandwich laminated floor slab, and determine the specific values of the spacing of the truss reinforcement S, the thermal conductivity of the sandwich layer material λ3, the thickness of the mortar δ1, the thickness of the top plate δ2, the thickness of the sandwich layer δ3, the thickness of the bottom plate δ4, and the thickness of the plaster δ5. The thermal conductivities of the mortar, concrete, and plaster materials can be selected according to the specification as λ1 = 0.93 W / (m·K), λ2 = λ4 = 1.51 W / (m·K), and λ5 = 0.87 W / (m·K).

[0080] Substitute the spacing of the truss reinforcement S, the thermal conductivity of the sandwich layer material λ3, the thickness of the sandwich layer δ3, and the thickness of the bottom plate δ4 into Equation (2) to obtain the value of the correction coefficient β.

[0081] Substitute the correction coefficient β and each variable into Equation (1) to obtain the overall thermal resistance R and heat transfer coefficient K of the structure.

[0082] Calculation example

[0083] There is a new type of sandwich laminated floor slab for household division. Given that the truss spacing is 600 mm, the mortar thickness is 20 mm, the top slab thickness is 70 mm, the sandwich layer material is vitrified microsphere thermal insulation mortar with a thickness of 30 mm, the bottom slab thickness is 60 mm, and the plastering thickness is 10 mm, please determine the structural heat transfer coefficient based on its basic parameters.

[0084] Evaluation process:

[0085] It can be obtained from GB50176 - 2016 that the heat transfer coefficient α of the inner surface of the envelope structure under typical working conditions i = 8.7 W / (m 2 ·K). This value of α can be used for both the upper and lower surfaces of the household floor slab. i = α e = 8.7 W / (m 2 ·K).

[0086] The thicknesses of each layer are given: δ1 = 0.02 m, δ2 = 0.07 m, δ3 = 0.03 m, δ4 = 0.06 m, δ5 = 0.01 m. The thermal conductivity of common building materials: mortar λ1 = 0.93 W / (m·K), concrete λ2 = λ4 = 1.51 W / (m·K), plastering λ5 = 0.87 W / (m·K). The truss spacing S = 0.6 m, and the thermal conductivity of the sandwich layer material λ3 = 0.08 W / (m·K).

[0087] Substitute the truss bar spacing S, the thermal conductivity λ3 of the sandwich layer material, the thickness δ3 of the sandwich layer, and the thickness δ4 of the bottom slab into Equation (2) to obtain the correction coefficient β = 0.799.

[0088] Substitute the correction coefficient β and each variable into Equation (1) to obtain the heat transfer coefficient K of the new type of sandwich laminated floor slab as K = 1.542 W / (m 2 ·K).

Claims

1. Calculation method for the sandwich structure of a sandwich laminated floor slab, characterized in that, It includes the following steps: Step 1: Determine the values of the surface air convection heat transfer coefficients α i and α e according to the actual environment on the upper and lower surfaces of the sandwich laminated floor slab, and at the same time determine the thermal performance requirements that the sandwich laminated floor slab needs to meet, namely the structural heat transfer coefficient K'. Second step: Select the basic structure parameters of the sandwich laminated floor slab, including determining the specific values of the truss bar spacing S, mortar thickness δ1, top slab thickness δ2, bottom slab thickness δ4, and plaster thickness δ5; Select the thermal conductivity coefficients of the mortar, concrete, and plaster materials; Step 3: Select the materials for the sandwich structure layer, and substitute the thermal conductivity λ3' of the used materials into the formula to obtain the recommended thickness δ3' of the sandwich layer; Fourth step: Based on the recommended thickness δ3' of the sandwich layer, determine the thickness δ3 of the sandwich layer; Fifth step: Use the heat transfer coefficient calculation formula for the sandwich laminated floor slab to calculate the structural insulation performance; The heat transfer coefficient K calculation formula for the sandwich laminated floor slab is ， ； where, K is the structural heat transfer coefficient, W / (m 2 •K); R is the thermal resistance, m 2 ·K / W; α i , α e are the convective heat transfer coefficients of air on the hot and cold surfaces of the structure, W / (m 2 •K); S is the spacing of the truss bars, m; δ1 is the mortar thickness, m; λ1 is the thermal conductivity of the mortar material, W / (m•K); δ2 is the thickness of the roof slab, m; λ2 is the thermal conductivity of the roof slab material, W / (m•K); δ3 is the thickness of the sandwich layer, m; λ3 is the thermal conductivity of the sandwich layer material, W / (m•K); δ4 is the thickness of the floor slab, m; λ4 is the thermal conductivity of the floor slab material, W / (m•K); δ5 is the plaster thickness, m; λ5 is the thermal conductivity of the plaster material, W / (m•K).

2. The calculation method of the sandwich structure of the sandwich laminated floor slab according to claim 1, wherein: In ordinary building interiors, α i = α e = 8.7 W / (m 2 ·K).

3. The calculation method for sandwich structure of sandwich-layer laminated floor according to claim 1, characterized in that: The thermal conductivity coefficient of the mortar λ1 = 0.93 W / (m•K), the thermal conductivity coefficients of the top and bottom slab concretes λ2 = λ4 = 1.51 W / (m•K), and the thermal conductivity coefficient of the plaster λ5 = 0.87 W / (m•K).

4. The calculation method of the sandwich structure of the sandwich laminated floor slab according to claim 1, characterized in that, The specific determination process of the thickness δ3 of the sandwich layer is as follows: If 20mm ≤ δ3' ≤ 40mm, then determine the final parameter of the sandwich layer structure λ3 = λ3', δ3 = δ3', and the final value of δ3 is the rounded-up value of δ3' in 5mm increments; If δ3' < 20mm, select materials with a relatively higher thermal conductivity and lower price, and go back to the third step to recalculate; If δ3' > 40mm, then select materials with better heat insulation performance and go back to the third step to recalculate.

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