A multi-layer composite pipe with socket and spigot covering and its manufacturing method
By using corrosion-resistant inner tube and intermediate layer in composite tubes to bond with the outer tube, the problems of rust and electrochemical corrosion during use of composite tubes are solved, and higher anti-corrosion performance and negative pressure resistance are achieved, which extends the service life and improves the efficiency of liquid transportation.
Patent Information
- Application Number
- CN202411931608.6
- Authority / Receiving Office
- CN · China
- Patent Type
- Patents(China)
- Current Assignee / Owner
- Priority Date
- 2023-12-30
- Filing Date
- 2024-12-26
- Publication Date
- 2025-04-15
- Estimated Expiration
- 2044-12-26
AI Technical Summary
During use, existing composite pipes are prone to rust, electrochemical corrosion, inner lining layering and deformation and fall off of the inner stainless steel pipe under negative pressure, which affects the service life and the quality of the conveying liquid.
The inner tube made of corrosion-resistant material is used as the inner lining, covering the inner wall of the outer tube, the end face of the socket and the outer wall of the socket, and bonded to the outer tube through an intermediate layer (such as hot melt adhesive, plastic or rubber) to form a composite tube structure to improve corrosion resistance and negative pressure resistance.
Effectively prevent external pipe corrosion, improve service life, reduce operating energy consumption, enhance negative pressure resistance, and ensure the cleanliness and efficiency of liquid transportation.
Smart Images

Figure CN119353504B_ABST
Abstract
Description
Technical Field
[0001] The invention belongs to the technical field of composite pipes, and in particular relates to a multi-layer composite pipe with a socket coating and a manufacturing method thereof. Background Art
[0002] Composite pipe is a kind of pipe widely used in liquid transportation network system. When in use, the part of the pipe that contacts the liquid is required to have good corrosion resistance to prevent the pipe from rusting due to long-term contact with the liquid, which affects the quality of the transported liquid. Existing composite pipes generally have a metal pipe as the outer layer and a plastic pipe as the inner layer, or a metal pipe as the outer layer and a stainless steel pipe as the inner layer. When composite pipes with a plastic pipe as the inner layer are composited, the inner plastic pipe melts due to heat. When the temperature is too high, it is easy to produce an uneven inner surface or partial melting. The uneven inner surface will increase the resistance when transporting liquid. The partial melting of the inner plastic pipe will cause the product to be scrapped, and the production cost will increase. The inner plastic pipe has a large thermal expansion coefficient and shrinks greatly at low temperatures. The internal stress generated by mutual peeling with the outer metal pipe is large. There is a risk of lining delamination after long-term use. The composite pipe with a stainless steel pipe as the inner layer is prone to electrochemical corrosion due to the potential difference between the inner stainless steel pipe and the outer metal pipe, which affects the service life of the pipe. The inner lining stainless steel pipe and the outer metal pipe are not bonded, so the inner stainless steel pipe is prone to deformation and falling off under negative pressure, affecting the normal use of the pipe.
[0003] In order to solve the above problems, the solution of setting glue between the inner stainless steel pipe and the outer metal pipe is currently mainly adopted. However, when glue is used as the middle layer, the glue is easy to age. During bonding, due to the high fluidity of the glue, it is easy to produce a part of the inner stainless steel pipe directly contacting the outer metal pipe, which cannot achieve the effect of preventing electrochemical corrosion and resisting negative pressure. The structure of the existing composite pipe is a socket-and-spigot structure, with one end of the metal pipe being a spigot and the other end being a socket. When the pipe network is installed, multiple composite pipes are connected in sequence, and the spigot of a composite pipe is inserted into the socket of the adjacent composite pipe. The inner wall of the socket of the metal outer pipe is uneven and stepped. When the inner wall of the metal pipe is coated with anti-corrosion, the middle part and the spigot of the metal pipe are coated with an inner plastic pipe, and the inner wall of the socket and the outer wall of the spigot are generally directly painted with anti-corrosion paint, or sprayed / coated. However, this method does not make the anti-corrosion coating of the pipe form a whole. During use, the long-term scouring of the liquid in the pipeline will cause the liquid to penetrate into the inner wall of the metal pipe through the gaps between different materials, resulting in failure of anti-corrosion and corrosion of the pipe wall, affecting the service life of the pipeline and polluting the liquid transported in the pipeline.
[0004] For water supply pipes in normal use, they will be subjected to a large internal water supply pressure, which is a positive pressure, so the pipes must meet the specified strength requirements. Some existing standards (group standards and national standards) specify the nominal wall thickness under different pressure levels and different nominal diameters. Pipes that meet the standards can meet the strength requirements within the corresponding pressure level range. Therefore, the pipes produced by existing pipe manufacturers generally meet the corresponding standard requirements.
[0005] However, when a pipe network fails, for example, a pipe bursts in the pipe network, water in the pipe network will spray out from the burst pipe, and the pressure in the pipe network will drop rapidly, even forming negative pressure. In addition, the pipe will also be subject to some external pressure, for example, the pipe buried in the soil will be subject to the gravity of the soil above it. Under the above external pressure and internal negative pressure, the pipe will deform radially. If the deformation of the socket is too large, the socket will separate from the socket connected to it, causing leakage at the connection between the socket and the socket, causing the inside of the water pipe to be contaminated. Therefore, it is necessary to ensure that the deformation of the socket of the pipe does not exceed a certain range under the above circumstances. To this end, for the socket, the minimum socket thickness that meets the deformation requirements must be obtained. If the socket thickness of the selected pipe is less than the minimum socket thickness that meets the deformation requirements, measures need to be taken to make it meet the deformation requirements. For example, when the pipe is secondary processed or installed, at least one layer of annular reinforcement ring is sleeved on the socket to form a composite socket to enhance the ability to resist deformation. However, the material of the reinforcement ring is often different from that of the base pipe (i.e., the outer pipe), because the reinforcement ring is in direct contact with the water in the pipe, and it needs to be made of corrosion-resistant materials such as stainless steel, while the base pipe generally uses relatively cheap materials such as ductile iron and Q235. Therefore, how to ensure that the above-mentioned composite socket meets the maximum deformation requirements while the thickness is minimum or close to minimum becomes a problem to be solved. The base pipe refers to the pipe in the factory setting, that is, the base pipe is not installed with a reinforcement ring and has not undergone secondary processing such as composite process. The base pipe includes a socket, a pipe body and a socket connected in sequence. The pipe body can be divided into two parts, namely the main body and the transition part. The transition part is used for the transition connection between the main body and the socket, that is, the socket, the transition part, the main body and the socket are connected in sequence. The inner wall surface of the pipe body is a cylindrical surface, that is, the inner diameter of the transition part is equal to the inner diameter of the main body. The main body is cylindrical, and the outer wall surface of the transition part is a conical surface. In other words, the thickness of the main body is basically uniform everywhere, and the thickness of the transition part gradually decreases along the direction from the socket to the socket. The outer diameter of the tube body refers to the outer diameter of the main body, the thickness of the tube body refers to the thickness of the main body, and the outer diameter of the base tube is the outer diameter of the tube body. Fig.15 In the figure, X is the length of the chamfer at the end of the socket along the axial direction, Y is the height of the chamfer at the end of the socket (i.e. the radial dimension), DE is the outer diameter of the socket, and T6 is the length of the socket.
[0006] For the socket, the existing standards do not give its wall thickness requirements, but generally it is necessary to ensure that its thickness is sufficient to meet the strength requirements. It is best to make the thickness of the socket at various locations minimum or close to the minimum while meeting the use strength. At present, the socket of a general base pipe is made of only one material, such as ductile iron or Q235 steel. This socket is a single-layer socket. However, in order to improve the performance of the pipe, the socket can be made into a composite socket, that is, a structure including at least two coaxial layers of different materials that are sequentially sleeved from the inside to the outside, such as an inner layer of stainless steel and an outer layer of Q235 steel, which can save costs while improving the corrosion resistance of the pipe. Similarly, if the composite socket can meet the strength requirements and the thickness is minimum or close to the minimum, the cost can also be greatly reduced. For the composite socket, the inner wall of the base pipe socket can be completely covered by the lining or not completely covered by the lining. When the inner wall of the base pipe socket is not completely covered by the liner, the composite socket includes a single-layer structure and a multi-layer structure. The single-layer structure has only one layer, namely, the outermost base pipe, and the multi-layer structure includes at least two layers, namely, the outermost base pipe and the liners sleeved in the base pipe. Summary of the invention
[0007] In view of the above-mentioned problems existing in the prior art, the purpose of the present invention is to provide a multi-layer composite pipe coated with a socket and a manufacturing method thereof. The inner pipe of corrosion-resistant material is used as an inner lining to cover the inner wall of the outer pipe, the end face of the composite pipe socket and the outer wall of the socket. The transportation environment is highly clean, the composite pipe has good corrosion resistance, simple installation, stable operation and long service life.
[0008] In order to achieve the above object, the technical solution adopted by the present invention is:
[0009] A multi-layer composite pipe covered with a socket and a spigot, comprising an inner pipe, an intermediate layer and an outer pipe, wherein the inner pipe, the intermediate layer and the outer pipe are bonded together or compressed to form a composite pipe, the outer pipe comprises a socket, a pipe body and a spigot connected in sequence, the inner pipe is made of a corrosion-resistant metal material, the inner wall of the socket is provided with steps and / or grooves, one section of the inner pipe covers part of the inner wall of the socket and the inner wall of the pipe body, and the other section of the inner pipe completely covers the spigot.
[0010] As a further improvement of the above technical solution:
[0011] Preferably, the inner tube comprises an inner sleeve and a socket sleeve, and the inner sleeve, the middle layer and the outer tube are sequentially sleeved from the inside to the outside.
[0012] Preferably, the inner sleeve and the socket sleeve are integral.
[0013] More preferably, the socket sleeve is formed by spinning and turning one end of the inner tube outward.
[0014] More preferably, the socket sleeve and the inner sleeve are welded together.
[0015] More preferably, the welding point between the socket sleeve and the inner sleeve is located on the end face of the socket, and the inner tube area covering the end face of the socket is formed by flanging the socket sleeve and / or flanging the inner sleeve followed by welding.
[0016] Preferably, the material of the middle layer is non-metal.
[0017] More preferably, the material of the middle layer is hot melt adhesive.
[0018] More preferably, the material of the middle layer is plastic.
[0019] More preferably, the material of the middle layer is rubber.
[0020] Preferably, the outer tube is made of metal.
[0021] More preferably, the outer tube is made of ductile iron or carbon steel.
[0022] Preferably, the thickness of the middle layer is not less than 0.2 mm, and the thickness of the inner sleeve is not less than 0.1 mm.
[0023] Preferably, when the material of the middle layer is hot melt adhesive, the inner tube and the outer tube are processed into a composite tube through a composite process.
[0024] Preferably, when the material of the middle layer is plastic or rubber, the inner tube, the middle layer and the outer tube are pressed tightly to form a composite tube.
[0025] Preferably, when two adjacent composite pipes are connected, the socket of one composite pipe is inserted into the socket of the adjacent composite pipe and sealed by a sealing ring, the outer ring of the sealing ring contacts the inner wall of the inner pipe of the socket of one composite pipe, and the inner ring of the sealing ring contacts the outer wall of the socket sleeve on the outer wall of the socket of the adjacent composite pipe.
[0026] Preferably, at least one sealing ring is provided, and the sealing ring is installed in a groove on the inner wall of the socket or the sealing ring is limited by a limiting member.
[0027] The socket and the inner tube covering the inner wall of the socket form a composite socket, and the spigot and the inner tube covering the spigot form a composite spigot.
[0028] The design method of the composite socket thickness includes the following steps:
[0029] Step a1: Obtain the differential equation of radial deformation of the ring and the general solution of the differential equation according to the differential equation of the deflection curve of the ring and the moment balance equation;
[0030] Step a2: Calculate and obtain a calculation formula for the maximum radial deformation of the ring according to the general solution of the differential equation obtained in step a1;
[0031] Step a3: Calculating the minimum thickness of the socket of the base pipe that meets the deformation requirements according to the calculation formula of the maximum radial deformation obtained in step a2;
[0032] Step a4: for a base pipe whose socket thickness is less than the minimum thickness obtained in step a3, at least one layer of annular reinforcement ring is sleeved in the socket of the base pipe to form a composite socket. If the material of the reinforcement ring is the same as that of the socket of the base pipe, the thickness of the composite socket is not less than the minimum thickness obtained in step a3. If the material of the reinforcement ring is different from that of the socket of the base pipe, go to step a5.
[0033] Step a5: Determine the minimum thickness of the reinforcement ring. If the thickness of the reinforcement ring is not less than the minimum thickness, the thickness of the composite socket is the minimum thickness that meets the deformation requirement.
[0034] Assuming that the outermost layer of the composite socket is the base layer and the other layers are non-base layers, for the multi-layer structure in the composite socket, the design method includes the following steps:
[0035] Step c1: Based on the same strain of each layer of the composite socket, the calculation formula of the hoop stress of any non-base layer is obtained according to the non-single-layer material equilibrium equation;
[0036] Step c2: determining the relationship between the thickness of any non-base layer and the thickness of the base layer, the determination principle being that the ratio of the stress of the non-base layer to the yield strength of the material of the non-base layer is not greater than the ratio of the stress of the inner wall of the pipe body to the yield strength of the material of the pipe body;
[0037] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.
[0038] Step a2 includes the following steps:
[0039] Step a21: Assuming that the annular ring is subjected to at least two radial concentrated forces that are symmetrical relative to the center of the circle, based on the assumption, calculating the constant term in the general solution of the differential equation to obtain the first form of the differential equation;
[0040] Step a22: Based on the first form of the differential equation obtained in step a21, assuming that the annulus is subjected to only two radial concentrated forces that are symmetrical relative to the center of the circle, the first form of the differential equation is transformed into a second form of the differential equation;
[0041] Step a23: introducing the internal negative pressure of the annulus into the second form of the differential equation by introducing coefficients to obtain the third form of the differential equation;
[0042] Step a24: Obtain a calculation formula for the maximum change in the diameter of the circular ring with coefficients according to the third form of the differential equation obtained in step a23.
[0043] Step a3 includes the following steps:
[0044] Step a31: Calculate the parameters of a reference pipe according to an existing shear load formula, wherein the material and outer diameter of the reference pipe are respectively the same as the material and outer diameter of the base pipe;
[0045] Step a32: obtaining the maximum deformation of the reference pipe socket through simulation according to the parameters calculated in step a31;
[0046] Step a33: Substitute the maximum deformation of the socket obtained in step a32 into the calculation formula of step a2 to obtain the coefficient value in the calculation formula, thereby obtaining a calculation formula for the maximum radial deformation of the ring with a known coefficient;
[0047] Step a34: Calculate the minimum thickness of the base pipe socket that meets the deformation requirements based on the allowable deformation regulations and the formula obtained in step a33.
[0048] In step a1, the radial deformation obtained is u The differential equation is:
[0049] ,
[0050] The general solution is: ,
[0051] in, R is the average value of the outer circle radius and the inner circle radius of the ring; E is the elastic modulus of the pipe, Pa; I is the moment of inertia of the tube cross section, m 4 ; M is the bending moment borne by a certain section of the ring in the local area, in units of N*m; establish a two-dimensional coordinate system, φ The deformation is u The line connecting the point at and the origin of the coordinate system is the same as the line connecting the point at and the origin of the coordinate system in the two-dimensional coordinate system. x The angle between the positive directions of the axes, ; A1, A2, A3, B2, B3 are constants.
[0052] In step a2, it is assumed that the ring is subjected to at least two radial concentrated forces P symmetrically relative to the center of the circle, and the angle between two adjacent radial concentrated forces P is , the ring satisfies the following three boundary conditions:
[0053] First, in Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula: ;
[0054] Second, due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; that is, ;
[0055] Third, cut near point A, remove point A and the force P, and from the symmetry and equilibrium conditions, we know that the shear force on the section at the circular cut is , that is: ;
[0056] Combining the above three formulas with the general solution of the differential equation, we get the first form of the differential equation:
[0057] ;
[0058] Assuming that the annulus is subjected to only two radial concentrated forces P which are symmetrical relative to the center of the circle, then: , we get the second form of the differential equation:
[0059] ;
[0060] Introducing the coefficient m, we get the third form of the differential equation:
[0061] ;
[0062] Point of application of force P The point is the point of maximum deformation, and the change in the diameter of the ring is:
[0063] , where I = S 3 / 12, S is the wall thickness of the tube body.
[0064] In step a5, the minimum thickness of the reinforcement ring is obtained and determined by experiment, simulation or theoretical calculation. When the minimum thickness of the reinforcement ring is obtained and determined by theoretical calculation, the minimum thickness of the reinforcement ring is obtained based on the socket thickness of the base pipe and the bending strength calculation formula of the longitudinal composite beam.
[0065] For the same base pipe, along the direction from the socket to the spigot, the inner wall of the socket includes a first step, a second step, a third step, a fourth step, and a fifth step arranged in sequence, one end of the inner wall of the pipe body is a sixth step, the chamfer at the connection between the fourth step and the fifth step is a first chamfer, the chamfer at the connection between the fifth step and the sixth step is a second chamfer, the chamfer at the connection between the second step and the third step is a third chamfer, the chamfer at the connection between the first step and the second step is a fourth chamfer, and the chamfer at the connection between the third step and the fourth step is a fifth chamfer.
[0066] In step c1, the multilayer structure is regarded as a structure including at least one pressure vessel with a conical inner wall and at least one annular reinforcement ring installed on the inner wall of the pressure vessel. In step c1, the first cone is a pressure vessel with a conical inner wall, and the first chamfer and the second chamfer are located on the inner wall of the first cone.
[0067] In step c1, assuming that there are m layers of non-base layer and one base layer, the equilibrium equation of the multilayer structure is:
[0068] ;
[0069] in, p is the internal pressure of the conical container, D is the diameter of the conical container to be calculated, is the cone height, S1, S2, …, S m are the thickness of each non-base layer, S b is the thickness of the base layer, , , …, are the hoop stresses of each non-base layer, is the hoop stress of the base layer, α is the half cone angle of the conical container;
[0070] In step c1, the strain of the base layer The strain is the same as that of each non-base layer, satisfying: ,Right now , where E1, E2, …, E m are the elastic modulus of each non-base layer, E b is the elastic modulus of the base layer. Combined with the equilibrium equation, the calculation formula for the hoop stress of the nth non-base layer is: ,in, is the hoop stress of the nth non-base layer, n is 1~m, E n is the elastic modulus of the nth non-base layer.
[0071] In step c2, in order to make the composite socket meet the strength requirements, the internal stress of the nth non-base layer and the inner wall stress of the pipe body meet the following conditions: , , where Q0, Q n are the yield strengths of the pipe body and the nth non-base layer material respectively; is the hoop stress of the pipe body, S is the thickness of the pipe body, D0 is the outer diameter of the pipe body, and p is the pressure given by the fluid in the pipe. The calculation formula for the hoop stress of the nth non-base layer is combined to obtain the relationship between the thickness of the nth non-base layer and the thickness of the base layer: .
[0072] When the thickness of the base layer corresponding to the maximum stress point is known, the minimum thickness of the layer with the minimum yield stress is calculated based on the relationship between the thickness of the nth non-base layer and the thickness of the base layer. This layer is set as the minimum stress layer, and the thickness of other non-base layers is set to be not less than the minimum thickness of the minimum stress layer. When the thickness of the non-base layer is known, first, at least one point where the stress of the base layer is easily concentrated is selected. The selection principle is: after the strength of each selected point meets the requirements, the strength of other places of the base layer designed according to the outer wall design requirements also meets the requirements; then the thickness of each selected point is calculated based on the relationship between the thickness of the nth non-base layer and the thickness of the base layer.
[0073] A method for manufacturing a multi-layer composite pipe with a socket coating, for manufacturing the composite pipe, comprises the following steps:
[0074] Step S1: processing the inner wall and outer wall of the outer tube and making the inner tube;
[0075] Step S2: arranging the inner sleeve, the middle layer and the outer tube from inside to outside in sequence;
[0076] Step S3: Press the inner sleeve onto the outer tube and the socket sleeve onto the outer wall of the socket sleeve;
[0077] Step S4: compounding;
[0078] When the middle layer is made of plastic or rubber, there is no step S4.
[0079] Preferably, in step S3, the socket sleeve is obtained by spin-forming one end of the inner tube.
[0080] Preferably, in step S1, when making the inner tube, an inner sleeve and a socket sleeve are made separately, and the socket sleeve is put on the outer wall of the socket. Between step S3 and step S4, the inner sleeve and the socket sleeve are welded, and the welding point is located on the end face of one end of the socket of the composite tube.
[0081] Preferably, when the material of the middle layer is hot melt adhesive or plastic, in step S2, the middle layer is made into a round tube and then the inner sleeve, the middle layer and the outer tube are sequentially connected, or the powdered raw material of the middle layer is sprayed or coated on the outer wall of the inner sleeve and the outer wall of the socket, or the powdered raw material of the middle layer is sprayed or coated on the inner wall of the outer tube and the outer wall of the socket, and then the inner sleeve is inserted into the outer tube.
[0082] Preferably, in step S3, the inner sleeve is pressed tightly against the inner wall of the pipe body, the inner wall of the socket and the inner wall of the plug by spin forming or hydraulic forming.
[0083] The beneficial effects of the present invention are:
[0084] (1) The inner tube made of corrosion-resistant material covers the inner wall, the end face and the outer wall of the outer tube, completely isolating the outer tube from the liquid flowing in the composite tube, effectively preventing the outer tube from rusting, playing a good anti-corrosion role, and prolonging the service life of the pipe, avoiding the rust of the outer tube from polluting the water quality, and improving the liquid transportation performance.
[0085] (2) Compared with the plastic pipe lining, the inner pipe made of corrosion-resistant material as the lining improves the smoothness of the inner surface of the composite pipe and has a smaller lining thickness, which improves the water delivery efficiency of the pipeline and reduces operating energy consumption.
[0086] (3) The inner lining is made of corrosion-resistant material. Compared with the plastic pipe lining, it greatly reduces the internal stress caused by the mutual peeling between the inner pipe and the outer pipe due to the inconsistency of the material expansion coefficient, and the inner lining is not prone to delamination and deformation.
[0087] (4) The inner tube made of corrosion-resistant material is used as the inner lining. Compared with the plastic pipe lining, the pipe can be used to transport high-temperature liquid media. When the pipe is affected by high temperature such as external fire, the inner lining will not be damaged.
[0088] (5) The middle layer also acts as an insulator to prevent electrochemical corrosion between the outer tube and the inner tube.
[0089] (6) During installation, the spigot of the composite pipe is inserted into the socket of the adjacent composite pipe and sealed by a sealing ring. The outer ring of the sealing ring presses tightly against the inner wall of the inner pipe of the socket, and the inner ring presses tightly against the spigot sleeve of the outer wall of the spigot to prevent the inner pipe serving as the lining from falling off, and at the same time prevent the water flow from directly impacting the lining port and causing delamination of the composite pipe.
[0090] (7) After two adjacent composite pipes are connected, there is a gap between the end face of the spigot of one composite pipe and the end face of the pipe body of the adjacent composite pipe. Liquid will penetrate through this gap and contact the outer wall of the spigot. At this time, the spigot sleeve made of corrosion-resistant material will play an anti-corrosion role. At the same time, the inner ring of the sealing ring presses tightly against the outer wall of the spigot, completely isolating the pipe body and the socket from the liquid, preventing the pipe body and the socket from contacting the liquid and causing corrosion.
[0091] (8) When the middle layer is made of hot melt adhesive, the middle layer also plays a connecting role and has a high bonding strength, which prevents the inner tube from being delaminated and deformed under negative pressure. Using hot melt adhesive as the bonding material eliminates the inner lining delamination caused by the aging of ordinary glue, the uneven glue layer caused by the flow of ordinary liquid glue, the inner lining delamination and electrochemical corrosion caused by the direct contact between the inner lining and the outer tube without being bonded. It overcomes the technical problems of low bonding strength, poor hygienic performance, poor insulation effect and uneven inner wall of the outer tube resulting in the lining being suspended (i.e., there are areas in the corrosion-resistant lining that are not bonded to the outer tube).
[0092] (9) The socket sleeve and the inner sleeve are connected by welding, and the welding point is set on the end face of the socket, which is convenient for manufacturing and processing, and the weld has no effect on the installation of the pipe.
[0093] (10) The minimum or near minimum thickness of the composite spigot that meets the deformation requirements and the minimum or near minimum thickness of the composite socket that meets the strength requirements of socket-and-spigot pipes of different diameters can be obtained through theoretical analysis and calculation, which saves materials and reduces the production, manufacturing and transportation costs of pipes. The method for designing the minimum wall thickness is theoretical calculation, which is low in cost and high in efficiency. It provides strong support for experiments and pipe design and provides a reference for secondary processing and installation. The pipe network system is more reliable in use, greatly reducing the leakage and water supply pollution caused by the separation of the spigot and the socket of the adjacent pipe due to excessive deformation of the spigot when the pipe bursts, and the cracking of the socket, thereby improving the corrosion resistance of the pipe. BRIEF DESCRIPTION OF THE DRAWINGS
[0094] Figure 1 It is a schematic structural diagram of the case where the outer wall of the tube body and the outer wall of the socket sleeve are flush with each other according to the first embodiment of the present invention.
[0095] Figure 2 It is a structural schematic diagram of the first embodiment of the present invention when the thickness of the tube body and the thickness of the socket are equal.
[0096] Figure 3 It is a schematic diagram of connecting two adjacent composite pipes whose outer walls of the pipe body and the outer walls of the socket sleeve are flush with each other in embodiment 1 of the present invention.
[0097] Figure 4 It is a schematic diagram of connecting two adjacent composite pipes with equal pipe body thickness and socket thickness according to the first embodiment of the present invention.
[0098] Figure 5 It is a schematic structural diagram of the second embodiment of the present invention when the outer wall of the tube body and the outer wall of the socket sleeve are flush.
[0099] Figure 6 It is a structural schematic diagram of the second embodiment of the present invention when the thickness of the tube body and the thickness of the socket are equal.
[0100] Figure 7 It is a schematic diagram of connecting two adjacent composite pipes whose outer walls of the pipe body and the outer walls of the socket sleeve are flush with each other in embodiment 2 of the present invention.
[0101] Figure 8 It is a schematic diagram of connecting two adjacent composite tubes with equal tube body thickness and socket thickness according to the second embodiment of the present invention.
[0102] Fig. 9 Schematic diagram of the cross-section deformation before and after the socket is subjected to a pair of radially symmetrical concentrated forces.
[0103] Fig.10 for Fig. 9 Schematic diagram of local analysis.
[0104] Fig.11Schematic diagram of a circular ring subjected to multiple uniformly arranged radial concentrated forces.
[0105] Fig.12 This is the stress analysis diagram of socket-type pipe.
[0106] Fig.13 It is a schematic cross-sectional view of a composite socket per unit length of the present invention.
[0107] Fig.14 This is a schematic diagram of the composite socket structure of Example 1 of the present invention.
[0108] Fig.15 It is a schematic diagram of the spigot and socket connection of the socket type pipe of the present invention.
[0109] Fig.16 It is a schematic diagram of a composite socket in which the inner wall of the base pipe socket is completely covered by the liner according to the present invention. DETAILED DESCRIPTION
[0110] The specific implementation of the present invention is described in detail below in conjunction with the accompanying drawings. It should be understood that the specific implementation described here is only used to illustrate and explain the present invention, and is not used to limit the present invention.
[0111] For ease of description, spatially relative terms such as "above", "above", "on the upper surface of", "above", etc. may be used here to describe the spatial positional relationship between a device or feature and other devices or features as shown in the figure. It should be understood that spatially relative terms are intended to include different orientations of the device in use or operation in addition to the orientation described in the figure. For example, if the device in the accompanying drawings is inverted, the device described as "above other devices or structures" or "above other devices or structures" will be positioned as "below other devices or structures" or "below other devices or structures". Thus, the exemplary term "above" can include both "above" and "below". The device can also be positioned in other different ways (rotated 90 degrees or in other orientations), and the spatially relative descriptions used here are interpreted accordingly.
[0112] Embodiment 1
[0113] A multi-layer composite pipe with a socket coating, such as Figure 1~Figure 4 As shown, it includes an inner tube 2, an intermediate layer 3 and an outer tube 1. The inner tube 2 includes an inner sleeve 22 and a socket sleeve 21. The inner sleeve 22, the intermediate layer 3 and the outer tube 1 are sequentially sleeved from the inside to the outside.
[0114] The outer tube 1 comprises a socket 11, a tube body 12 and a spigot 13 which are connected in sequence along the axial direction thereof, i.e., the socket 11, the tube body 12 and the spigot 13 share a central line. The tube body 12 is cylindrical, and the spigot 13 is chamfered along the circumferential direction at one end away from the tube body 12. The inner surface of the socket 11 is provided with steps and / or grooves, i.e., the socket 11 can be regarded as being formed by connecting a plurality of cylindrical structures with different inner diameters in sequence.
[0115] The outer tube 1 is made of metal, preferably ductile iron or carbon steel, the socket 11, the tube body 12 and the spigot 13 are integrally cast, or the tube body 12 and the spigot 13 are integrally formed as the tube body, the socket 11 is made separately, and then the socket 11 and the tube body are welded. The inner diameter of the spigot 13 is equal to the inner diameter of the tube body 12.
[0116] The inner tube 2 is made of a material with corrosion resistance, including but not limited to stainless steel, titanium, and composite materials, preferably stainless steel. One end of the inner tube 2 covers part of the inner wall of the socket 11 or completely covers the inner wall of the socket 11, the middle part of the inner tube 2 covers the inner wall of the pipe body 12, and the other end of the inner tube 2 covers the inner wall of the socket 13, the end face of the socket 13, and the outer wall of the socket 13. In other words, the inner tube 2 can cover the entire socket 13. Specifically, the part of the inner tube 2 covering the outer wall of the socket 13 is the socket sleeve 21, and the part of the inner tube 2 covering the inner wall of the outer tube 1 is the inner sleeve 22. Obviously, the inner sleeve 22 is the inner lining of the composite pipe. The inner tube 2 formed by the inner sleeve 22 and the socket sleeve 21 is a whole, and there are no holes on the inner tube 2. It is a continuous medium and can completely cover at least most of the inner wall of the socket 11, the entire inner wall of the pipe body 12, the entire inner wall of the socket 13, the end face of the socket 13, and the outer wall of the socket 13.
[0117] The thickness of the socket sleeve 21 may be greater than, less than or equal to the thickness of the inner sleeve 22 .
[0118] The spigot sleeve 21 can be formed by spinning and turning one end of the inner tube 2 outward; or the spigot sleeve 21 and the inner sleeve 22 are welded together, preferably, the welding point is located on the end surface of the composite pipe spigot 13. Specifically, the stainless steel covering the end surface of the spigot 13 is formed by welding after the spigot sleeve 21 is flanging and / or the inner sleeve 22 is flanging.
[0119] The material of the middle layer 3 is non-metal, preferably hot melt adhesive, plastic or rubber, and the inner tube 2, the middle layer 3 and the outer tube 1 are pressed together or bonded together. When the material of the middle layer 3 is hot melt adhesive, the hot melt adhesive has bonding properties, and the middle layer 3 firmly connects the inner tube 2 to the outer tube 1 through the composite process. When the material of the middle layer 3 is plastic or rubber, plastic and rubber have no bonding properties, and the inner tube 2, the middle layer 3 and the outer tube 1 are pressed together.
[0120] Preferably, there is hot melt adhesive, plastic or rubber between the end surface of the socket 13 and the inner tube 2 and between the outer wall of the socket 13 and the socket sleeve 21 .
[0121] like Figure 2 As shown, the thickness of the socket 13 is equal to the thickness of the tube body 12. At this time, after the manufacturing is completed, the socket sleeve 21 protrudes from the outer wall of the tube body 12. Figure 1 As shown, the sum of the thickness of the socket 13, the thickness of the socket sleeve 21, and the thickness of the medium between the outer wall of the socket 13 and the socket sleeve 21 is equal to the thickness of the tube body 12. In this way, after manufacturing is completed, the outer wall of the tube body 12 and the outer wall of the socket sleeve 21 are flush, the appearance is more beautiful, and the socket 13 is not easily damaged during transportation.
[0122] like Figure 1 , 2 As shown, the socket 11 is sequentially provided with a first step 111, a second step 112, a third step 113, a fourth step 114 and a fifth step 115 from the end away from the pipe body 12 to the end close to the pipe body 12. The inner diameter of the first step 111 is smaller than the inner diameter of the second step 112, the inner diameter of the third step 113 is larger than the inner diameter of the first step 111 and smaller than the inner diameter of the second step 112, and the inner diameter of the fourth step 114 is substantially equal to or smaller than the inner diameter of the second step 112 and larger than the inner diameter of the third step 113. The inner wall of the fourth step 114 is a cylindrical wall. The inner wall of the fifth step 115 is the outer side surface of a cylinder or a truncated cone. When it is the outer side surface of a truncated cone, that is, from the fourth step 114 toward the pipe body 12, the inner diameter of the fifth step 115 gradually increases, which is equivalent to a bell mouth. Preferably, the maximum inner diameter of the fifth step 115 is smaller than the inner diameter of the fourth step 114. The minimum inner diameter of the fifth step 115 is greater than the inner diameter of the tube body 12 . Thus, a sixth step 116 is formed at one end of the tube body 12 close to the fifth step 115 .
[0123] The inner wall bell mouth of the fifth step 115 is provided to provide a deflection angle for the adjacent composite pipe connected thereto, i.e., to allow the center line of a composite pipe and the center line of the adjacent composite pipe connected thereto to be in an intersecting state rather than an overlapping state. This is to allow a certain installation angle deviation to facilitate actual operation, and to allow adjacent pipes to have a relative displacement of a certain angle when the soil moves naturally.
[0124] Preferably, the inner tube 2 covers the inner walls of the fourth step 114 and the fifth step 115 , and does not cover the inner walls of the first step 111 , the second step 112 , and the third step 113 .
[0125] In this embodiment, the outer diameter of the inner sleeve 22 is slightly smaller than the inner diameter of the outer tube 1, and the thickness of the inner sleeve 22 is not less than 0.1 mm.
[0126] The main function of the inner tube 2 is corrosion protection, so its thickness should not be too large. A large number of welding tests have shown that under the conditions of fiber laser welding, a stainless steel plate with a minimum wall thickness of 0.1 mm can obtain a weld with high welding strength and small deformation. Taking into account the complexity of the welding process equipment, the investment cost, and the fact that the middle layer 3 shrinks at low temperatures due to its large thermal expansion coefficient, and generates internal stress that separates it from the outer tube 1, it is necessary to give full play to the supporting role of the inner tube 2 on the middle layer 3 to offset the internal stress of the outer tube 1 and the middle layer 3. The preferred thickness of the inner sleeve 22 is 0.1 mm-2 mm.
[0127] The wall thickness of the middle layer 3 is not less than 0.2 mm. The inner diameter of the middle layer 3 is slightly larger than the outer diameter of the inner sleeve 22 , and the outer diameter of the middle layer 3 is slightly smaller than the inner diameter of the outer tube 1 . The length of the middle layer 3 is substantially the same as that of the inner sleeve 22 .
[0128] When the middle layer 3 is hot melt adhesive, it serves as an adhesive layer. When the inner wall roughness of the outer tube 1 is relatively large during compounding, the middle layer 3 is heated and melted to fill the pits on the inner wall of the outer tube 1. If the thickness of the middle layer 3 is too small, poor adhesion may occur in areas with large local roughness. At the same time, the middle layer 3 also serves as an insulating layer, and a certain thickness is required to produce an insulating effect between the outer tube 1 and the inner tube 2. The electric spark test and the peel strength test show that when the thickness of the middle layer 3 reaches more than 0.2mm, the peel strength and insulation with good performance can be stably obtained. The optimal thickness of the middle layer 3 is 0.2mm-2mm.
[0129] Compared with the thickness of the inner lining stainless steel required by the standard "Inner lining stainless steel composite steel pipe (CJ / T 192-2017)", when the middle layer 3 uses hot melt adhesive, the inner lining stainless steel thickness is thinner and the cost is lower. When the inner diameter of the outer pipe 1 is the same, the inner diameter of the composite pipe is larger and the transportation efficiency is higher.
[0130] The conclusions of the lining peel strength test (this test uses an outer tube with an inner diameter of 100-300mm and an inner lining stainless steel thickness of 0.1mm-0.3mm) are as follows:
[0131] When the rust removal grade of the inner wall surface of the metal outer tube 1 reaches Sa2.5, the surface roughness is Ra12.5~Ra25, and the thickness of the intermediate layer 3 is not less than 0.2mm, the peel strength between the hot melt adhesive and the outer tube can stably reach more than 70N / cm; the peel strength between the hot melt adhesive and the inner lining can stably reach more than 150N / cm;
[0132] When the rust removal grade of the inner wall surface of the metal outer tube 1 reaches Sa2.5, the surface roughness is Ra25~Ra50, and the thickness of the intermediate layer 3 is above 0.2mm, the peel strength between the hot melt adhesive and the outer tube can stably reach above 120N / cm; the peel strength between the hot melt adhesive and the inner lining can stably reach above 150N / cm;
[0133] When the rust removal level of the inner wall surface of the metal outer tube 1 reaches Sa2.5, the surface roughness is greater than Ra50, and the thickness of the intermediate layer 3 is above 0.2 mm, the peel strength between the hot melt adhesive and the outer tube can stably reach above 150 N / cm; the peel strength between the hot melt adhesive and the inner lining can stably reach above 150 N / cm.
[0134] The peel strength between the hot melt adhesive and the lining of this solution can reach 150N / cm, and the corresponding bonding strength is 15MPa, that is, the pipe of this solution can withstand a liquid lateral scouring force of 15MPa, which is much greater than the maximum liquid lateral scouring force in the existing pipe network system.
[0135] When the middle layer 3 is hot melt adhesive, the thickness of the middle layer 3 is preferably 0.2 mm to 0.5 mm. When the thickness of the middle layer 3 is greater than 0.5 mm, the peel strength remains substantially unchanged.
[0136] Based on the above structure, when two adjacent composite pipes are connected, Figure 3 and 4 As shown, the spigot 13 of a composite pipe is inserted into the socket 11 of the adjacent composite pipe and sealed by the sealing ring 4, the outer ring of the sealing ring 4 contacts the inner wall of the inner pipe 2 at one end of the socket 11 of the composite pipe, and the inner ring of the sealing ring 4 contacts the outer wall of the spigot sleeve 21 on the outer wall of the spigot 13 of the adjacent composite pipe. Preferably, the end face of one end of the sealing ring 4 contacts the end face of the first step 111, a part of the sealing ring 4 is accommodated on the second step 112, the third step 113 is stuck in the middle of the sealing ring 4, and the other part of the sealing ring 4 is located on the fourth step 114. At the same time, there is a gap between the end face of one end of the spigot 13 of a composite pipe and the end face of the pipe body 12 of the adjacent composite pipe, and the existence of this gap is also to allow the existence of a deflection angle when two adjacent composite pipes are installed.
[0137] As shown above, the length of the spigot 13 is greater than the length of the socket 11, so that when two adjacent pipes are connected by spigot, the inner and outer rings of the sealing ring 4 can contact the stainless steel, avoiding the metal pipe body 12 part extending between the sealing ring 4 and the end face of the spigot 13, causing this part of the pipe body 12 to contact the liquid and rust. At the same time, the inner tube 2 on the end face of the composite pipe prevents the liquid in the pipeline from contacting the middle layer of the end face of the spigot 13, and the tension of the water causes the composite pipe to be delaminated.
[0138] A method for manufacturing a multi-layer composite pipe with a socket coating, for manufacturing the composite pipe, comprises the following steps:
[0139] Step S1: processing the inner wall and the outer wall of the outer tube 1, making the inner sleeve 22, and making the socket sleeve 21.
[0140] In this step, the method for processing the inner wall and outer wall of the outer tube 1 is to first grind the inner wall of the outer tube 1, and then sandblast or shot blast the inner wall and outer wall. Use an internal grinder to grind the inner wall of the outer tube 1 made of ductile iron, grind the inner wall of the tube flat, and remove surface oxide scale, scum and rust. Alternatively, after the inner wall is polished flat, use a laser cleaning machine to clean the inner wall of the tube to deeply remove the oxide scale and rust on the inner wall of the tube. When sandblasting or shot blasting, remove dust, oxide scale, and rust on the inner and outer walls, and sandblast the inner and outer walls of the tube into a frosted surface.
[0141] Making the inner sleeve 22 comprises the following steps:
[0142] S11: Process the stainless steel sheet into a ring or a thin cylinder to complete the rolling. In this step, the rolling is performed by a rolling machine (the technical solution of the rolling machine in Patent 2020114372537 can be used).
[0143] S12: Cut the rolled stainless steel sheet and weld it into a cylindrical shape. In this step, if a single roll is obtained in step S11, a straight seam welding is performed using a argon arc welder or a laser welder, and the weld extends from one end of the inner sleeve 22 to the other end. If multiple spirally connected rolls are obtained in step S11, a spiral seam welding machine is used to weld the multiple spirally connected rolls in sequence to form a thin cylinder.
[0144] Preferably, the length of the stainless steel round tube obtained in step S12 is greater than the length of the inner sleeve 22 of the composite tube that is finally processed. Specifically, the length of the stainless steel round tube obtained in step S12 is greater than the sum of the length of the socket 11, the length of the tube body 12 and the length of the spigot 13.
[0145] The steps of making the socket sleeve 21 are the same as the steps of making the inner sleeve 22. The thickness of the socket sleeve 21 is substantially equal to the thickness of the inner sleeve 22, and the length of the socket sleeve 21 is greater than the length of the socket 13.
[0146] Step S2: arrange the inner sleeve 22 , the middle layer 3 and the outer tube 1 in sequence from the inside to the outside, and put the socket sleeve 21 on the outer wall of the socket 13 .
[0147] In this step, when the material of the middle layer 3 is hot melt adhesive or plastic, step S3 can be completed in the following three ways:
[0148] The first way is as follows:
[0149] First, the hot melt adhesive or plastic is formed into a round tube by extrusion molding, vacuum sizing, pulling, and cutting. Extrusion molding, vacuum sizing, pulling, and cutting can be performed by an extruder, a vacuum box, a tractor, and a cutting machine in the prior art. The extrusion molding is a prior art and will not be described in detail here.
[0150] Then, the obtained shaped intermediate layer 3 is sleeved on the outer wall of the inner sleeve 22. In this step, the sleeve can be made manually or by a sleeve machine.
[0151] Next, the inner wall of the socket 11 and the outer wall of the plug 13 are sprayed. After the outer walls of the socket 11 and the plug 13 are heated, a powdered hot melt adhesive or plastic is rotomolded or sprayed on the inner wall of the socket 11 and the outer wall of the plug 13 by a roller coater or a plastic coating machine.
[0152] Finally, the intermediate layer 3 and the inner sleeve 22 are inserted into the outer tube 1 as a whole, and the socket sleeve 21 is inserted into the outer wall of the socket 13. Similarly, in this step, the sleeve can be completed manually or by a pipe threading machine.
[0153] The second way is as follows:
[0154] First, after the inner wall of the outer tube 1 and the outer wall of the socket 13 are heated, a roller coater or a plastic coating machine is used to roll-coat or spray-coat the powdered hot melt adhesive on the inner wall of the outer tube 1 and the outer wall of the socket 13. The thickness of the roll-coating or spray-coating is greater than 0.2 mm.
[0155] Then, the inner sleeve 22 is inserted into the outer tube 1 with the middle layer 3 on the inner wall, and the socket sleeve 21 is inserted into the outer wall of the socket 13 .
[0156] The third way is as follows:
[0157] First, after the outer wall of the inner sleeve 22 and the outer wall of the socket 13 are heated, a powdered hot melt adhesive is roto-molded or spray-molded on the outer wall of the inner sleeve 22 and the outer wall of the socket 13 made of stainless steel by a roller coating machine or a plastic coating machine. The thickness of the roto-molding or spray-molding is greater than 0.2 mm. Then, the inner sleeve 22 with the intermediate layer 3 on the outer wall is inserted into the outer tube 1, and the socket sleeve 21 is inserted into the outer wall of the socket 13.
[0158] When the material of the middle layer 3 is rubber, first roll the rubber sheet on the outer wall of the inner tube 2, then use adhesive to bond the rubber joints, and then insert the inner tube 2 with rubber into the outer tube 1. Alternatively, the rubber tube is sleeved on the outer wall of the inner tube 2 and then inserted into the outer tube 1 as a whole. The length of the rubber should be enough to cover the inner wall of the socket 11, the inner wall of the tube body 12, the inner wall of the plug 13, the end face of the plug 13 and the outer wall of the plug 13. The rubber tube that exceeds the end face of the plug 13 is turned outward and sleeved on the outer wall of the plug 13, and then the plug sleeve 21 is sleeved on the rubber pad on the outer wall of the plug 13.
[0159] Step S3: Press the inner sleeve 22 and the middle layer 3 onto the outer tube 1 , and press the socket sleeve 21 onto the outer wall of the socket sleeve 21 .
[0160] This step can be completed by spin forming or hydraulic forming. During spin forming, the inner sleeve 22 of the socket 11, the pipe body 12 and the socket 11 are respectively spin-formed by a pipe spinning machine. When the socket 11 is spun, the spinning is carried out from the inside to the outside, so that the inner sleeve 22 lined with stainless steel material, the middle layer 3 and the outer tube 1 are closely attached in sequence. During hydraulic forming, the two ends of the inner sleeve 22 of stainless steel material inserted into the outer tube 1 are sealed, and water is injected into the tube to pressurize it, so that the inner sleeve 22, the middle layer 3 and the outer tube 1 are closely attached in sequence.
[0161] Step S4: welding the inner sleeve 22 and the socket sleeve 21, wherein the welding point is located on the end surface of one end of the socket 13 of the composite pipe.
[0162] In this step, first, the inner sleeve 22 and / or the spigot sleeve 21 at one end of the spigot 13 are spun using a pipe mouth spinning machine, so that one end of the inner sleeve 22 and / or one end of the spigot sleeve 21 covers the end surface of the spigot 13 of the composite pipe. Then, the excess burrs are trimmed to facilitate subsequent welding. Finally, the inner sleeve 22 and the spigot sleeve 21 are welded together, and the welding point is located at the end surface of the spigot 13 of the composite pipe.
[0163] Step S5: Compounding.
[0164] In this step, the pipe is heated by a medium frequency inductor or a heating furnace, the hot melt adhesive is melted, and after cooling, the outer pipe 1 and the inner sleeve 22 are bonded together, and the socket sleeve 21 and the socket 13 are bonded together.
[0165] When the material of the middle layer 3 is plastic or rubber, there is no step S5.
[0166] After completing the above steps, the outer wall of the pipe is sprayed with paint or plastic coated or other methods are used for external anti-corrosion treatment.
[0167] Assuming that the portion of the inner sleeve 22 covering the socket 11 is a socket lining ring, and the portion of the inner sleeve 22 covering the socket 13 is a socket lining ring, then in this embodiment, the socket 11 of the outer tube and the socket lining ring can both improve the strength of the socket of the composite pipe, and the socket formed by the two is a composite socket. The inner wall and the outer wall of the socket 13 are completely covered by the socket lining ring, and the socket 13 of the outer tube and the socket lining ring can both improve the strength and deformation resistance of one end of the socket, and the socket formed by the two is a composite socket. Because the material of the middle layer 3 is hot melt adhesive or non-metal such as plastic or rubber, it has basically no effect on the strength and deformation resistance of the socket 13, and therefore does not belong to the category of composite sockets that can resist deformation.
[0168] Next, the thickness of the spigot end and the socket end of the composite pipe is optimized and designed.
[0169] The design method of the thickness of the composite socket includes the following steps:
[0170] Step a1: Obtain the radial deformation of the ring based on the differential equation of the deflection curve and the moment equilibrium equation of the ring Because the cross-sections of the tube body 12 and the socket 13 are circular rings, the deformation of the socket 13 is analyzed by analyzing the radial deformation of the circular ring. In addition, a cylinder of unit length can be regarded as a circular ring.
[0171] Step a2: Calculate and obtain a calculation formula for the maximum radial deformation of the ring based on the general solution of the differential equation obtained in step a1.
[0172] Step a3: The minimum thickness of the socket of the base pipe that meets the deformation requirements is obtained according to the calculation formula of the maximum radial deformation obtained in step a2.
[0173] Step a4: For a base pipe whose socket thickness is less than the minimum thickness obtained in step a3, at least one layer of annular reinforcement ring is sleeved in the socket of the base pipe to form a composite socket. If the material of the reinforcement ring is the same as that of the socket of the base pipe, the thickness of the composite socket is not less than the minimum thickness obtained in step a3. If the material of the reinforcement ring is different from that of the socket of the base pipe, go to step a5.
[0174] Step a5: When the material of the reinforcement ring and the material of the socket of the base pipe are different, the minimum thickness of the reinforcement ring is obtained based on the thickness of the socket of the base pipe and the calculation formula of the bending strength of the longitudinal composite beam. If the thickness of the reinforcement ring is not less than the minimum thickness, the thickness of the composite socket is the minimum thickness that meets the deformation requirements. In other words, the thickness of the composite socket of the pipe formed after the reinforcement ring is installed on the base pipe meets the deformation requirements and strength requirements.
[0175] The following is a detailed introduction to each step.
[0176] In step a1, the radial deformation obtained The differential equation is:
[0177] (1)
[0178] in ,R is half of the sum of the outer radius and the inner radius of the ring. In other words, R One quarter of the sum of the outer and inner diameters of the ring 。E is the elastic modulus of the pipe, Pa. I is the moment of inertia of the tube cross section, m 4 . M It is the bending moment borne by a certain section of the ring in the local area, in N*m. Establish a two-dimensional coordinate system, φ The deformation is u The line connecting the point at and the origin of the coordinate system is the same as the line connecting the point at and the origin of the coordinate system in the two-dimensional coordinate system. x The angle between the positive directions of the axes, In this embodiment, establish xy Coordinate system, point O is the coordinate origin and the center of the ring, such as Fig. 9 Point m 1 The radial deformation at u, but om 1 and x The angle between the positive directions of the axes is φ.
[0179] The above differential equation (1) is obtained as follows:
[0180] like Fig. 9 and 10 As shown in the figure, the dotted line is the ring that has not been deformed by external force, and the solid line is the shape of the ring after being deformed by two symmetrical radial external forces P. The center point of the ring before and after deformation is O The position remains unchanged. Fig. 9 midpoint O Too xy Vertex of the coordinate system , point m and n are two points on the ring before deformation; m 1 and n 1 are two points on the deformed shape of the ring, and the points m 1 and n 1 Located on the line segment om and on superior, φ The deformation is u The line connecting the point at and point O is x The angle between the positive directions of the axes, om and on The angle between φ, set up u Centripetal movement is positive. Fig.10 Middle, straight line m 1 t Passing point m 1 and perpendicular to the line segment om ,straight line m 1 t 1 For passing point m 1 A tangent line to n 1 t’ Passing point n1 and perpendicular to on ,straight line n 1 t 1 ’ For passing point n 1 tangent.
[0181] Arc before deformation mn The arc length is ds , whose curvature is k 1 , arc after deformation m 1 n 1 The curvature is k 2 , du Yes m and Point n The difference in radial deformation is The arc before deformation mn exist m Tangent line of point and n The angle between the tangent lines of a point, The deformation amount is u The radius corresponding to the point at after deformation is Is an arc mn Deform to Arc m 1 n 1 back ds The increment of The arc after deformation m 1 n 1 exist m 1 Tangent line of point and n 1 The angle between the tangent lines of the point is m 1 t 1 and The angle between , then:
[0182] (2)
[0183] (3)
[0184] (4)
[0185] (5)
[0186] m 1 t 1 andm 1 t The angle between , n 1 t 1 ’ and n 1 t’ The angle between .
[0187] m 1 section to n 1 The variable of the cross-section rotation is :
[0188] (6)
[0189] in, m The cross section is through om and perpendicular to the cross section of the ring, in other words, m The cross section is through the center axis and point of the corresponding pipe of the ring m The cross section of n The cross section is through on and perpendicular to the cross section of the ring, in other words, n The cross section is through the center axis and point of the corresponding pipe of the ring n cross section. m 1 t 1 and m 1 t The angle between m 1 The cross-sectional angle, n 1 t 1 ’ and n 1 t’ The angle between n 1 The corner of the cross section.
[0190] Substituting formula (5) and (6) into formula (4) yields: (7)
[0191] By series expansion we have:
[0192] Ignoring the higher-order terms, substituting the series expansion into (7) yields: (8)
[0193] Substituting (2) into (8), we obtain: (9)
[0194] Omitting the higher-order terms, we get: (10)
[0195] For a thin ring (i.e. a ring with a small difference between the outer and inner diameters relative to the diameter), we have: , by definition: ,therefore, , substituting into (10) we get, , combining formula (2) to obtain radial deformation The differential equation formula (1).
[0196] The general solution of the above differential equation (1) is:
[0197] (11)
[0198] Among them, A1, A2, A3, B2, B3 are constants. If the external force is symmetrical, the odd function terms in the solution can be deleted, and the deflection curve equation is:
[0199] (12)
[0200] Step a2: Calculating and obtaining a calculation formula for the maximum radial deformation of the ring according to the general solution of the differential equation obtained in step a1, comprising the following steps:
[0201] Step a21: Assume that the circular ring is subjected to at least two radial concentrated forces P that are symmetrical relative to the center of the circle. Based on this assumption, calculate the constant term in the general solution of the differential equation to obtain the first form of the differential equation.
[0202] Step a22: Based on the first form of the differential equation obtained in step a21, assuming that the annulus is subjected to only two radial concentrated forces P that are symmetrical relative to the center of the circle, the first form of the differential equation is transformed into the second form of the differential equation.
[0203] Step a23: Introduce the internal negative pressure on the annulus into the second form of the differential equation by introducing coefficients to obtain the third form of the differential equation.
[0204] Step a24: Obtain a calculation formula for the maximum change in the diameter of the circular ring with coefficients according to the third form of the differential equation obtained in step a23.
[0205] In step a21, it is assumed that the ring is subjected to at least two radial concentrated forces P symmetrically relative to the center of the ring. The radial concentrated force P means that the direction of the force P is radial to the ring and points to the center of the ring, and the angle between two adjacent radial concentrated forces P is ,like Fig.11 shown. Fig. 9 The circular ring shown in FIG. 1 is subjected to two radial concentrated forces P which are symmetrical relative to the center of the circle, and the two radial concentrated forces P are located on the y-axis.
[0206] The ring satisfies the following three boundary conditions:
[0207] (1) Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula:
[0208] (13)
[0209] (2) Due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; that is,
[0210] (14)
[0211] (3) Cutting near point A, removing point A and the acting force P, it can be seen from the symmetry and equilibrium conditions that the shear force on the cross section of the circular cut is , that is:
[0212] (15)
[0213] in, dM represents the differential with respect to M. It should be noted that point (3) is not a real cut of the ring, but an assumption.
[0214] By combining formulas (13), (14) and (15), we get:
[0215] (16)
[0216] (17)
[0217] (18)
[0218] Substituting formulas (16), (17), and (18) into formula (12), we obtain:
[0219] (19)
[0220] Formula (19) is the first form of the differential equation obtained in step a21. is the radial deformation of the ring.
[0221] In step a22, if only one pair of radial forces P is applied to the ring, then , the deflection curve equation (19) becomes:
[0222] (20)
[0223] Formula (20) is the second form of the differential equation obtained in step a22.
[0224] In step a23, considering that the pipe generally needs to withstand an internal negative pressure of 0.1 MPa during testing, the coefficient m is introduced, that is:
[0225] (twenty one)
[0226] For water supply pipes, if a fault occurs during actual use, such as a burst pipe, a certain negative pressure will be formed inside the pipe. Therefore, when testing the pipe, a negative pressure will be created inside the pipe to ensure that the pipe performance meets the standards.
[0227] Formula (21) is the third form of the differential equation obtained in step a23.
[0228] In step a24, the maximum deformation is At the point where the force P acts, the change in the diameter of the ring is the maximum change in the diameter of the ring. u max :
[0229] (twenty two)
[0230] Formula (22) is the third form of the differential equation obtained in step a24.
[0231] Step a3: Calculate the minimum thickness of the base pipe that meets the deformation requirements based on the calculation formula of the maximum radial deformation obtained in step a2. It includes the following steps:
[0232] Step a31: Calculate the parameters of a reference pipe according to the existing shear load formula. The material and outer diameter of the reference pipe are respectively the same as the material and outer diameter of the base pipe.
[0233] Step a32: Obtain the maximum deformation of the reference pipe socket through simulation based on the parameters calculated in step a31.
[0234] Step a33: Substitute the maximum deformation of the socket obtained in step a32 into the calculation formula of step a2 to obtain the coefficient value in the calculation formula, thereby obtaining the calculation formula for the maximum radial deformation of the ring with a known coefficient.
[0235] Step a34: Calculate the minimum thickness of the base pipe socket that meets the deformation requirements based on the national standard allowable deformation regulations and the formula obtained in step a33.
[0236] In step a31, the shear load formula used is derived from the group standard "Corrosion-resistant steel pipe with socket-type flexible joint for water supply and drainage" (T / CISA 108-2021), which is:
[0237] (twenty three)
[0238] like Fig.12 As shown, W is the shear load, in Newton (N); F is the shear force acting on the interface, in Newton (N); M P is the weight of the test assembly steel tube and the weight of the object in the steel tube, in kilograms (kg); a The distance between the shear load application point and the end face of the socket inserted into the socket, in millimeters (mm); b The distance between the center of gravity of the steel pipe at the spigot end and the end face of the spigot inserted into the socket, in millimeters (mm); c It is the distance between the support point of the pad and the end face of the socket that has been inserted into the socket, in millimeters (mm).
[0239] In this embodiment, the nominal diameter of the base pipe is DN=600mm, and the material is Q235 steel. The nominal diameter of the reference pipe is DN=600mm, and the material is Q235. Its wall thickness is selected according to the provisions of the group standard, and the wall thickness when the nominal pressure is 3.0MPa, that is, the nominal wall thickness S=7mm. At this time, the outer diameter of the base pipe D0=630mm, R =312mm. In addition, the shear force F It should be no less than 30 times of the nominal diameter DN. In this embodiment, 30 times is taken, that is, F = DN × 30 = 600 × 30 = 18000 N. Take the pipe length as 1m, c = 800mm.
[0240] In this embodiment, the parameters in formula (23) are M P and W The calculation process is as follows:
[0241]
[0242]
[0243] in, is the density of water, is the density of steel. M P The diameter used in the calculation is the outer diameter D0 of the base pipe, and the calculation error is within the allowable range. In addition, in this embodiment, the material of the base pipe socket is Q235 steel, the elastic modulus of the steel pipe material is E=200GPa, I=S 3 / 12. It should be noted that the calculation formula for the moment of inertia I of the circular tube section also needs to be multiplied by the length, but this solution considers a cylinder of unit length (i.e., the circular ring), so the length is 1.
[0244] Step a31 is thus completed.
[0245] In step a32, based on the parameters obtained by the above calculation, simulation software is used. In this embodiment, the maximum deformation of the socket of the reference pipe is simulated by finite element simulation. u max =12.19 mm. During simulation, the material parameter Poisson's ratio = 0.3. The simulation method is well known to those skilled in the art and will not be described in detail here.
[0246] In step a33, u max =12.19 Substituting into formula (22) we get m =1.18 (Note: When calculating, the length units are meters, and the unit of elastic modulus E is Pa), thus:
[0247] (twenty four)
[0248] From formula (24), we can see that u max The value is related to the pipe diameter and pipe thickness.
[0249] In this embodiment, in step a34, for a pipe with DN=600mm, the national standard stipulates that the allowable deformation must not be greater than 3.6% of the outer diameter of the socket DE=635mm, that is, u max =22.86 mm, and substituting it into formula (24), we can calculate that for a pipe with DN=600 mm, the minimum wall thickness of the socket is S=5.7 mm.
[0250] Table 1 Nominal wall thickness and absolute deformation of DN=600mm pipes at various nominal pressures
[0251]
[0252] In Table 1, each wall thickness is the nominal wall thickness under the corresponding nominal pressure specified in the industry standard, and the absolute deformation is calculated according to formula (24).
[0253] It can be seen from Table 1 that if the base tube purchased by the secondary processing enterprise or user is processed in accordance with the provisions of the industry standard, the base tubes with serial numbers 1, 2, and 3 in Table 1 have a socket thickness greater than 5.7 mm and an absolute deformation less than 22.86 mm, which meet the deformation requirements, while the base tubes with serial numbers 4, 5, and 6 have a socket thickness less than 5.7 mm and an absolute deformation greater than 22.86 mm, which do not meet the deformation requirements.
[0254] If the use requirement of the pipe in a certain occasion is to meet the nominal pressure within 2MPa, the general user will choose the pipe with serial number 4 in Table 1, and will not choose the pipe with serial number 1, 2 or 3 to meet the deformation requirement, because if one of these three is selected, although it will meet both the strength requirement and the deformation requirement, their thickness is larger and the cost will be much greater than the pipe with serial number 4. If the pipe with serial number 4 is selected, its strength meets the use requirement. For the deformation requirement, some measures can be taken, such as sleeve-fitting an annular reinforcement ring in the socket to increase the thickness of the socket and improve its ability to resist deformation, and this measure is low in cost.
[0255] Step a4: For a base pipe whose socket thickness is less than the minimum thickness obtained in step a3, at least one layer of annular reinforcement ring is sleeved in the socket of the base pipe to form a composite socket. If the material of the reinforcement ring is the same as that of the socket of the base pipe, the thickness of the composite socket is not less than the minimum thickness obtained in step a3. If the material of the reinforcement ring is different from that of the socket of the base pipe, go to step a5.
[0256] For example, in some occasions, the use requirement of the pipe is to meet the nominal pressure within 1.6MPa, and the selected base pipe is the pipe No. 5 in Table 1, with a wall thickness of 4mm. If the selected reinforcement ring and the base pipe are made of the same material, the thickness of the reinforcement ring is not less than 1.7mm to meet the deformation requirement, and the thickness of the composite socket is not less than 5.7mm.
[0257] Step a5: The material of the reinforcement ring is different from that of the socket of the base pipe. The minimum thickness of the reinforcement ring is obtained based on the socket thickness of the base pipe and the bending strength calculation formula of the longitudinal composite beam. The thickness of the reinforcement ring is not less than the minimum thickness. Then, the thickness of the composite socket is the minimum thickness that meets the deformation requirements.
[0258] In this embodiment, the calculation formula for the bending strength of the longitudinal composite beam is as follows:
[0259] (25)
[0260] in, C i The area is A i The distance from the centroid of the cross section to the bending center, I oi is the moment of inertia of the ith section about its own centroidal axis, E i is the elastic modulus of the corresponding part material. E is the elastic modulus of the base tube material, and I is the moment of inertia of the cross section corresponding to the minimum thickness obtained in step a3.
[0261] In this embodiment, the socket sleeve 21 is turned outward to the outer wall of the base pipe, and the turned outward length is not less than the length of the socket 13, that is, the outer wall of the entire socket 13 is covered by the socket sleeve 21, which is equivalent to the composite socket being provided with two layers of reinforcement rings, but the two layers of reinforcement rings are made of the same material and can be regarded as a whole.
[0262] The following is an example to illustrate this.
[0263] For the composite spigot of this embodiment, the use requirement of the pipe in some occasions is to meet the nominal pressure within 1.6MPa. The selected base pipe is the pipe No. 5 in Table 1, and its wall thickness is 4mm. The composite spigot includes a base pipe spigot and an annular reinforcement ring. In this embodiment, since the two layers of reinforcement rings are made of the same material, the two layers of reinforcement rings are regarded as a whole, and the thickness of the whole reinforcement ring is set to x ,like Fig.13 The figure shows the cross section per unit length of the composite socket or composite beam, where the plane of the cross section passes through the central axis of the base pipe. The base pipe is made of Q235, and the reinforcing ring is made of 304 stainless steel. Fig.13 The distance from the center point of the cross section shown (the center of the composite beam) to the outer or inner surface of the composite socket is (4+ x ) / 2, for the base tube socket (original socket): A i =4, E i =200, C i = x / 2; For the reinforcement ring: A i = x , E i =193, C i =2, substituting into (25) we get:
[0264]
[0265] Find: x =1.73mm.
[0266] Therefore, the total thickness of the reinforcement ring is not less than 1.73 mm, and the thickness of the inner tube 2 is not less than half of 1.73 mm, that is, 0.865 mm.
[0267] Based on the above, the design calculation of the minimum thickness of the composite socket is completed.
[0268] The design method of the thickness of the composite socket is as follows:
[0269] The inner wall shape of the pipe socket 11 is in accordance with the inner wall shape specified in the group standard "Corrosion-resistant steel pipe with socket and spigot flexible joint for water supply and drainage" (T / CISA108-2021) and GB / T 13295-2008, such as Fig.15 As shown. The inner wall of the socket 11 is stepped. For the same base pipe, along the direction from the socket 11 to the spigot 13, the inner wall of the socket 11 includes a first step 111, a second step 112, a third step 113, a fourth step 114, and a fifth step 115 arranged in sequence. One end of the inner wall of the pipe body is a sixth step 116, and the corresponding inner diameters are: D2, D1, (D3-2F0), D3, D4, (D0-2S), wherein the inner diameters of the sixth step 116, the first step 111, the fifth step 115, the third step 113, the fourth step 114, and the second step 112 increase in sequence, that is: (D0-2S) < D2 < D4 < (D3-2F0) < D3 < D1. The lengths of the first step 111, the second step 112, the third step 113, the fourth step 114, and the fifth step 115 are respectively: T3, T4, T5, (T2-T4-T5), (T1-T2-T3), that is, T1 is the sum of the lengths of the first step 111, the second step 112, the third step 113, the fourth step 114, and the fifth step 115, and T2 is the sum of the lengths of the second step 112, the third step 113, and the fourth step 114, and the lengths refer to the dimensions of the corresponding parts in the axial direction of the base pipe. S is the wall thickness of the pipe body, and F0 is the height of the third step 113 protruding from the fourth step 114, and F0 ≥ 0. The connection of each step is processed with chamfers, wherein the chamfer at the connection of the fourth step 114 and the fifth step 115 is the first chamfer R1, the chamfer at the connection of the fifth step 115 and the sixth step 116 is the second chamfer R2, the chamfer at the connection of the second step 112 and the third step 113 is the third chamfer R3, the chamfer at the connection of the first step 111 and the second step 112 is the fourth chamfer R4, and the chamfer at the connection of the third step 113 and the fourth step 114 is the fifth chamfer R5. Wherein, D0 is the outer diameter of the base pipe of the pipe.
[0270] The above inner wall shape and size are generally for single-layer socket. The size and shape of the inner wall of the composite socket can also be designed according to the above single-layer socket. In this solution, the size and shape of the inner wall of the composite socket are also designed according to the above single-layer socket.
[0271] Assume that the outermost layer of the composite socket is the base layer F1, and the other layers are the non-base layer F2. The thickness of the base layer F1 is uneven, and the thickness of the non-base layer F2 is uniform. In other words, the thickness of the base layer F1 is different or not completely the same at different places, and the thickness of the non-base layer F2 is the same at different places.
[0272] For composite sockets, since the dimensions of the inner wall are fully given by the relevant standards, it is necessary to design some key dimensions of the outer wall and the thickness of each non-base layer F2 on this basis, in order to save material costs and meet strength requirements. In other words, for composite sockets, the purpose of this solution is to design the thickness of the composite socket so that it is the minimum or close to the minimum thickness while meeting the use strength, so as to reduce the production, manufacturing and transportation costs of pipes.
[0273] According to material mechanics knowledge and experience, the most likely places for stress concentration on the inner wall of the above-mentioned single-layer socket or composite socket are the first chamfer R1, the second chamfer R2 and the third chamfer R3. The analysis is as follows:
[0274] For the inner wall of the stepped socket 11, according to mechanical knowledge, the stress concentration point is located at the connection between adjacent steps, that is, at each chamfer. Since the socket 11 is in a trumpet-like shape, the outer diameter of the socket 11 tends to decrease from the socket 11 to the spigot 13 of the same base pipe.
[0275] Based on the above conditions, since the inner diameter at the fourth chamfer R4 is equal to the inner diameter at the third chamfer R3, both are equal to D1, the wall thickness at the fourth chamfer R4 must be greater than the wall thickness at the third chamfer R3 to meet the requirement that the socket 11 is quasi-trumpet-shaped. Therefore, as long as the wall thickness at the third chamfer R3 meets the stress strength requirement, the wall thickness at the fourth chamfer R4 must meet the strength requirement. Similarly, the inner diameter at the fifth chamfer R5 is equal to the inner diameter at the first chamfer R1, both are D3, so the wall thickness at the fifth chamfer R5 must be greater than the wall thickness at the first chamfer R1 to meet the requirement that the socket 11 is quasi-trumpet-shaped. Therefore, as long as the wall thickness at the first chamfer R1 meets the stress strength requirement, the wall thickness at the fifth chamfer R5 must meet the strength requirement. Therefore, the places where stress is most likely to concentrate are the first chamfer R1, the second chamfer R2, and the third chamfer R3. Similarly, for the socket of a single-layer structure, as long as the thickness of the first chamfer R1, the second chamfer R2, and the third chamfer R3 meet the strength requirements, the base pipe socket manufactured according to the above requirements must also meet the strength requirements.
[0276] In the case where the inner wall of the base pipe socket 11 is not completely covered by the lining, since the third step 113 is the installation location of the sealing ring 4, the coverage range of the non-base layer F2 is generally from the fourth step 114 to the sixth step 116. Fig.14 Along the axial direction of the socket, one part is a single-layer structure, that is, there is no non-base layer F2 but only the base layer F1. In this solution, the first step 111, the second step 112, and the third step 113 are located in the single-layer structure part; the other part is a multi-layer structure, that is, there is a non-base layer F2 and a base layer F1. In this solution, the fourth step 114 and the fifth step 115 are located in the multi-layer structure part. Fig.14 shown.
[0277] For the composite socket, if both the single-layer structure part and the multi-layer structure part meet the strength requirements, then the entire composite socket meets the strength requirements.
[0278] For a single-layer structure, it is only necessary to ensure that the strength requirement is met at the point where the stress is maximum, and then the entire single-layer structure will meet the strength requirement.
[0279] For a single-layer structure of a composite socket, the design method comprises the following steps:
[0280] Step b1: Select at least one location where stress is likely to be concentrated in the single-layer structure. The selection principle is: if the wall thickness at each location meets the stress strength requirement, then the wall thickness at each location of the single-layer structure meets the strength requirement.
[0281] Step b2: Determine the minimum wall thickness of each selected location in step b1, and the determination principle is: when the material of the single-layer structure and the material of the pipe body are the same, the stress at each selected location is not greater than the stress of the inner wall of the pipe body; when the material of the single-layer structure and the material of the pipe body are different, the ratio of the stress at each selected location to the yield strength of the single-layer structure material is not greater than the ratio of the stress of the inner wall of the pipe body to the yield strength of the pipe body material.
[0282] In step b1, it can be seen from the above that the stress of the single-layer structure is most concentrated at the third chamfer R3, and the socket 11 is a trumpet-shaped structure. As long as the thickness of the third chamfer R3 meets the strength requirement, the entire single-layer structure meets the strength requirement.
[0283] For a thin-walled cylindrical pressure vessel, when the wall thickness is S, the internal pressure is p, and the diameter is D (the diameter D can be half of the sum of the outer diameter and the inner diameter of the cylinder, or the inner diameter of the cylinder, or the outer diameter of the cylinder), the hoop stress σ of the container wall is 环 for:
[0284] (26)
[0285] For a thin-walled conical pressure vessel, when the wall thickness is S, the internal pressure is p, and the diameter is D (the diameter D is half of the sum of the outer diameter and the inner diameter of the corresponding part of the cone, or the inner diameter of the corresponding part, or the outer diameter of the corresponding part), the hoop stress σ of the container wall is 环 for:
[0286] (27)
[0287] in, αis the semi-cone angle of the conical shell, that is, half of the opening angle. In other words, in the section passing through the central axis of the cone, the angle between the cone wall and the central axis is the semi-cone angle. In step b1, the base layer F1 is regarded as consisting of at least one pressure vessel with a conical inner wall plus several internal reinforcement rings j, and the reinforcement rings are connected to the inner wall of the pressure vessel. In other words, the base layer F1 is regarded as consisting of at least one conical pressure vessel plus several internal reinforcement rings j. At this time, as long as the corresponding conical pressure vessel meets the strength requirements, the base layer F1 will definitely meet the strength requirements. Therefore, it is sufficient to ensure that the conical pressure vessel meets the strength requirements.
[0288] In step b1, take the second cone and the reinforcement ring on the inner wall of the second cone, such as Fig.15 The dotted line between the first chamfer R1 and the third chamfer R3 is the inner wall section of the second cone. The first chamfer R1 and the third chamfer R3 are located on the inner wall of the second cone. For the area between the first chamfer R1 and the third chamfer R3, the portion between the inner wall of the second cone and the outer wall of the socket 11 is the entire side wall of the second cone, and the portion between the inner wall of the second cone and the inner wall of the socket 11 is the annular reinforcement ring on the inner wall of the second cone.
[0289] exist Fig.15 In the cross section shown, the angle between the straight line passing through the first chamfer R1 and the third chamfer R3 and the axis of the tube is the half cone angle of the second cone, which is set to α2.
[0290] but:
[0291] (28)
[0292] In step b2, if the material of the base layer F1 is consistent with the material of the pipe body, the pipe body meets the strength requirements. To make the single-layer structure meet the strength requirements, the hoop stress at the third chamfer R3 Not greater than the hoop stress of the pipe body , that is:
[0293] (29)
[0294] Then we have:
[0295] (30)
[0296] Among them, S b3 is the wall thickness of the base F1 at the third chamfer R3, S is the wall thickness of the tube body, and we get:
[0297] (31)
[0298] Since there is a sealing ring 4 at the third chamfer R3, in order to achieve sealing, the pressure of the sealing ring 4 on the socket 11 must be greater than the pressure p of the fluid in the pipe. According to existing research, the average pressure of the sealing ring 4 on the inner wall of the socket 11 is generally 1.4 times the fluid pressure. It can be obtained from formula (27) that the stress is proportional to the pressure and inversely proportional to the wall thickness. When the pressure increases to 1.4 times, in order to keep the stress unchanged, the wall thickness should be increased to 1.4 times at the same time. Then formula (31) is corrected to:
[0299] (32)
[0300] That is, the minimum wall thickness at the third chamfer R3 is: .
[0301] If the material of the base F1 is different from that of the pipe body, let the yield strength of the pipe body material be Q0, and the yield strength of the material used for the base F1 be Q1. Similarly, in order for the single-layer structure to meet the strength requirements, the stress at the third chamfer R3 and the stress on the inner wall of the pipe body should satisfy:
[0302] (33)
[0303] Then we have:
[0304] (34)
[0305] Considering the pressure of the sealing ring, the wall thickness at the third chamfer R3 is:
[0306] (35)
[0307] After the wall thickness at the third chamfer R3 is determined as above, the single-layer structure manufactured according to the above requirements meets the strength requirements.
[0308] For a multi-layer structure of a composite socket, the design method comprises the following steps:
[0309] Step c1: Based on the same strain of each layer of the composite socket and the non-single-layer material equilibrium equation, the calculation formula for the hoop stress of any non-base layer F2 is obtained.
[0310] Step c2: Determine the relationship between the thickness of any non-base layer F2 and the thickness of the base layer F1. The determination principle is that the ratio of the stress of the non-base layer F2 to the yield strength of the material of the layer is not greater than the ratio of the stress of the inner wall of the pipe body to the yield strength of the pipe body material.
[0311] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.
[0312] In step c1, for the multilayer structure, the multilayer structure is regarded as being composed of at least one pressure vessel with a conical inner wall and several internal reinforcement rings j, and the reinforcement rings are connected to the inner wall of the pressure vessel. In other words, the multilayer structure is regarded as being composed of at least one conical pressure vessel and several internal reinforcement rings j. At this time, as long as the corresponding conical pressure vessel meets the strength requirement, the multilayer structure will definitely meet the strength requirement. Therefore, it is sufficient to ensure that the conical pressure vessel meets the strength requirement.
[0313] In this solution, for the multilayer structure, the two places where the stress is most concentrated are the first chamfer R1 and the second chamfer R2. Fig.14 As shown. Therefore, the selected analysis area includes a conical pressure vessel plus an internal reinforcement ring, the conical pressure vessel is the first cone, the second cone and the first cone are two coaxially connected conical pressure vessels, and they share the same central axis with the socket 11. Fig.14 and 15 The dotted line between the first chamfer R1 and the second chamfer R2 is the inner wall section of the first cone. The first chamfer R1 and the second chamfer R2 are located on the inner wall of the first cone, the portion between the inner wall of the first cone and the outer wall of the socket 11 is the entire side wall of the first cone, and the portion between the inner wall of the first cone and the inner wall of the socket 11 (the innermost non-base layer F2) is an annular reinforcement ring j.
[0314] For the socket 11, the maximum load it bears is the tensile stress in the circumferential direction. The circumferential equilibrium equation for the thin-walled internal pressure conical shell is as follows:
[0315] (36)
[0316] in, p is the internal pressure of the conical shell, D is the diameter of the conical shell to be calculated, is the cone height, S is the side wall thickness of the conical shell to be calculated, σ 环 is the hoop stress of the conical shell, α is the semi-cone angle of the conical shell.
[0317] For a composite socket, assuming there are m layers of non-base layer F2 and one layer of base layer F1, then:
[0318] (37)
[0319] Among them, S1, S2, …, S m are the thickness of each non-base layer F2, S b is the thickness of the base layer F1, , , …, are the hoop stresses of each non-base layer, is the hoop stress of the base layer F1.
[0320] Since it is a thin-walled container, the hoop tensile stress of the composite socket can be regarded as the tensile stress of the composite beam, that is, the strain of the base F1 and the strain of each non-base layer F2 The same means that:
[0321] (38)
[0322] Right now:
[0323] (39)
[0324] Among them, E1, E2, …, E m are the elastic modulus of each non-base layer F2, E b is the elastic modulus of the base layer F1.
[0325] The stresses in formula (37) are all To express, is the hoop stress of the nth non-base layer F2, where n is 1~m, then E n is the elastic modulus of the nth non-base layer F2, S n is the thickness of the nth non-base layer F2. Based on formula (39), formula (37) is rearranged to obtain:
[0326] (40)
[0327] Formula (40) can be rearranged to obtain the hoop stress of the nth non-base layer F2:
[0328] (41)
[0329] Step c1 is thus completed.
[0330] In step c2, in order to make the composite socket meet the strength requirements, the internal stress of the nth non-base layer F2 and the inner wall stress of the pipe body satisfy:
[0331] (42)
[0332] (43)
[0333] Among them, Q0, Q n are the yield strengths of the pipe body and the nth non-base layer F2, respectively. The yield strength of the base layer F1 is Q b ;in is the hoop stress of the pipe body, S is the thickness of the pipe body, D0 is the outer diameter of the pipe body, and p is the pressure given by the fluid in the pipe.
[0334] Combining formulas (41), (42), and (43), we get:
[0335] (44)
[0336] Formula (44) is the relationship between the thickness of any non-base layer F2 and the thickness of the base layer F1.
[0337] The application of formula (44) can be carried out from two aspects: first, there is no limit on the thickness of the non-base layer F2; second, the thickness of the non-base layer F2 is limited.
[0338] When there is no limit on the thickness of the non-base layer F2, for a multi-layer structure, it is only necessary to ensure that the layer with the smallest yield stress meets the strength requirement. This layer is assumed to be the minimum stress layer, and the thickness of other layers is not less than that of the minimum stress layer. Then the entire multi-layer structure meets the strength requirement.
[0339] The yield stress of the base layer F1 is generally greater than the yield stress of the non-base layer F2, so the layer with the weakest yield stress is the non-base layer. Assuming that the yield stress of the nth layer of non-base layer F2 is the weakest, if yielding occurs, the nth layer should yield first. Based on this, if the thickness of the nth layer of non-base layer F2 meets the strength requirements, the minimum thickness of the base layer F1 and the thickness of other non-base layers F2 are not less than the thickness of the nth layer of non-base layer F2, then the strength of the base layer F1 and other non-base layers F2 will also meet the requirements. In addition, the thickness of the base layer F1 is generally greater than the thickness of the non-base layer. Therefore, it is necessary to calculate the thickness of the nth layer of non-base layer F2. Formula (42) is rewritten as:
[0340] (45)
[0341] in, is the maximum stress inside the nth non-base layer F2, that is From formula (41), we can see that in the analysis area, i.e., the first cone, Take the maximum value, then D takes the maximum value, and the thickness of other non-base layers F2 takes the minimum value, that is, The maximum point is located at the first chamfer R1, the maximum value of D in the first cone is D3, the thickness of other non-base layers F2 is not less than the thickness of the nth non-base layer F2, that is, the thickness of the non-base layer F2 is Sn, and the half cone angle is α 1. In Fig.14 and 15 In the cross section shown, the angle between the straight line passing through the first chamfer R1 and the second chamfer R2 and the axis of the pipe is the semi-cone angle of the first conical pressure vessel, that is:
[0342] (46)
[0343] Substituting formula (43) and (46) into formula (45), we can obtain:
[0344] (47)
[0345] From formula (47), we get:
[0346] (48)
[0347] In this embodiment, S b Take the thickness of the base layer F1 at the first chamfer R1.
[0348] From the above, the number of layers m of the non-base layer F2 and the material of each layer are known, the material, thickness, diameter and other dimensions of the pipe body are also known, so the yield strength and elastic modulus of each layer are also known, and the internal shape and size of the socket 11 are in accordance with relevant standards. Therefore, the semi-cone angle α 1 is also known. In other words, in formula (48), only S n and S b Two unknowns. The thickness of the base layer F1 at the first chamfer R1 is given based on experience or other requirements. S b After that, we can find S n The minimum value of .
[0349] Therefore, according to formula (48), the minimum thickness of the non-base layer F2 with the smallest yield stress can be obtained, and the thickness of other non-base layers F2 should not be less than this value.
[0350] For the base layer F1, the thickness of the base layer F1 at the first chamfer R1 is S b1 It is given that the thickness S of the base layer F1 at the third chamfer R3 b3 It has also been calculated by formula (35), so we only need to give or calculate the thickness S of the base layer F1 at the second chamfer R2: b2 The thickness of the base layer F1 at each location can be determined to ensure that the multi-layer structure of the composite socket meets the strength requirements.
[0351] For the thickness S of the base layer F1 at the second chamfer R2 b2 , can be given directly or set to the thickness of the base layer F1 at the first chamfer R1 S b1 Equal, can also be obtained by using formula (44), specifically, formula (44) is rewritten as:
[0352] (49)
[0353] but:
[0354] (50)
[0355] The thickness S of the nth non-base layer F2 has been calculated by formula (48): n , and the thickness of other non-base layers is not less than S n , then formula (50) becomes:
[0356] (51)
[0357] The thickness S of the base layer F1 at the second chamfer R2 is obtained by formula (51): b2 .
[0358] From the above, the thickness of each non-base layer and the thickness of the first chamfer R1, the second chamfer R2 and the third chamfer R3 of the base layer F1 are obtained. According to the requirement of the trumpet shape of the outer wall of the socket 11, the outer walls corresponding to each chamfer are connected in sequence, and the thickness of each part of the base layer F1 and the thickness of each part of the composite socket are obtained. The outer diameter at the end face of the socket can be greater than the outer diameter at the third chamfer R3, and the specific value can be flexibly selected.
[0359] When the non-base layer F2 has only one layer, that is, the conical shell is a double-layer material, then formula (37) is in the following form:
[0360] (52)
[0361] Formulas (38) and (39) are organized into the following form:
[0362] (53)
[0363] For double-layer sockets, the material of the non-base layer F2 (i.e., inner lining layer) is generally stainless steel, and the material of the base layer F1 (i.e., outer layer) is generally Q235 steel or ductile iron. The yield stress of the non-base layer F2 is generally smaller than that of the base layer F1, that is, if yielding occurs, the non-base layer F2 should yield first. Based on this, the stress in formula (52) should be To express, based on formula (53), rearranging formula (52) we can get:
[0364] (54)
[0365] Formula (54) can be rearranged to obtain the hoop stress of the non-base layer F2:
[0366] (55)
[0367] In step c2, in this embodiment, formula (47) is in the following form:
[0368] (56)
[0369] The thickness of the non-base layer F2 is:
[0370] (57)
[0371] That is, the minimum thickness of the non-base layer F2 is: .
[0372] At this time, formula (50) is: (58)
[0373] A specific example is shown below.
[0374] Assume the nominal diameter of the purchased base pipe is DN600 mm, E b =200GPa, E n =193GPa, the base F1 material and the pipe body material are both Q235 steel, and the yield strength of Q235 steel is 235MPa, that is, Q1=Q0=235MPa. The non-base F2 material is 304 stainless steel, with a yield strength of Qn=205MPa, S=7mm, D0=630 mm, D1=673.4 mm, D3=668 mm, D4=642.6 mm, =8.21 o , =5.82 o Assume that the thickness of the first chamfer R1 of the base layer F1 is Sb=6 mm.
[0375] The thickness of the non-base layer F2 is calculated by formula (57) to be Sn≥2.38 mm, and the thickness of other non-base layers is not less than 2.38 mm.
[0376] For the thickness S of the base layer F1 at the second chamfer R2 b2 It can be set equal to the thickness Sb=6 mm at the first chamfer R1, or calculated according to formula (57). When formula (57) is used, Sn=2.38 mm is taken, and each parameter is substituted into formula (57) to obtain S b2 ≥5.68 mm.
[0377] For the thickness S of the base layer F1 at the third chamfer R3 b3 , obtained by formula (32), substituting each parameter into formula (32), and obtaining S b3 ≥10.53 mm.
[0378] When the inner wall of the entire base pipe socket 11 is completely covered by the lining, it is equivalent to the entire composite socket being a multi-layer structure. In this case, the design method of the thickness of the base layer F1 and each non-base layer F2 is the same as the design method when the inner wall of the base pipe socket 11 is not completely covered by the lining. The difference is that the base layer thickness S corresponding to the third chamfer R3 is b3At this time, the base thickness S corresponding to the third chamfer R3 b3 The method for calculating the thickness of the base layer F1 at the second chamfer R2 is the same as that for calculating the thickness of the base layer F1 at the second chamfer R2. It can be given directly, or it can be set equal to the thickness of the base layer F1 at the first chamfer R1. It can also be calculated using formula (44). Specifically, formula (49) gives:
[0379] (59)
[0380] In the case where the inner wall of the entire base pipe socket 11 is completely covered by the lining, there is generally no third step 113, or in other words, the inner diameter of the third step 113 is equal to the inner diameter of the fourth step 114, and the two are combined into a large step, such as Fig.16 shown.
[0381] When the thickness of the non-base layer F2 is limited, for example, in this embodiment, a socket liner made of corrosion-resistant material is sleeved in the socket 11, and the thickness of the liner is generally about 1 mm, which is equivalent to the thickness of the non-base layer F2 being known. In order to make the composite socket formed after the sleeve liner meet the strength requirements, it is necessary to calculate and design the thickness of the base layer F1 under this premise.
[0382] Based on this, formula (44) is rewritten as:
[0383] (60)
[0384] From the above analysis, it can be seen that in the multi-layer structure of the composite socket, the first chamfer R1 and the second chamfer R2 are the places where stress is most likely to concentrate, and the thickness of each non-base layer is uniform, that is, the thickness of each non-base layer is consistent. Therefore, it is only necessary to ensure that the thickness of the base layer F1 at the first chamfer R1 and the second chamfer R2 meets the strength requirements. The base layer F1 manufactured according to the above requirements and other parts of the multi-layer structure must also meet the strength requirements.
[0385] The thickness of the base layer F1 at the first chamfer R1 is S b1 , in formula (60), D is D3, and the half cone angle α for α 1, then formula (60) is:
[0386] (61)
[0387] The thickness of the base layer F1 at the second chamfer R2 is S b2 , in formula (60), D is D4, and the half cone angle α for α 1, then formula (60) is:
[0388] (62)
[0389] From the above, after knowing the thickness and material of the non-base layer F2, the minimum thickness of the base layer F1 at the first chamfer R1 and the second chamfer R2 can be calculated. If the outer wall of the composite socket is designed according to the above-mentioned trumpet-shaped requirement, the composite socket meets the strength requirement.
[0390] A specific example is shown below.
[0391] Assume a base pipe with a nominal diameter of DN600 mm, E b =200 GPa, the material of the tube body and the base tube F1 is the same, both are Q235 steel, yield strength Q1=Q0=235 MPa, S=7 mm, D0=630 mm, D1=673.4 mm, D3=668 mm, D4=642.6 mm, =8.21 o , =5.82 o A socket lining ring is sleeved inside, that is, there is a non-base layer F2, and the parameters of the lining ring are as follows: E n =193GPa, 304 stainless steel yield strength Q n =205 MPa, Q1=Q0=235 MPa, Sn=1 mm. Then, we can calculate from equations (61), (62), and (35) respectively:
[0392] S b1 ≥7.33 mm, S b2 ≥7.01 mm, S b3 ≥10.53 mm.
[0393] When the inner wall of the entire base pipe socket 11 is completely covered by the lining, it is equivalent to the entire composite socket being a multi-layer structure. In this case, the design method of the thickness of the base layer F1 and each non-base layer F2 is the same as the design method when the inner wall of the base pipe socket 11 is not completely covered by the lining. The difference is that the base layer thickness S corresponding to the third chamfer R3 is b3 At this time, the thickness of the base layer F1 at the third chamfer R3 is S b3 , in formula (60), D is D1, and the half cone angle α for α 2, then formula (60) is:
[0394] (63)
[0395] Embodiment 2
[0396] The difference between this embodiment and the first embodiment lies in the structure of the socket 11 and the number of sealing rings 4 .
[0397] like Figures 5 to 8As shown, from the direction away from the pipe body 12 toward the direction close to the pipe body 12, the socket 11 includes a sealing ring installation part 119 and a transition part 117 which are connected in an integral manner and arranged in sequence on the same center line. The sealing ring installation part 119 is a cylindrical structure, and the inner wall of the sealing ring installation part 119 is provided with at least two annular grooves arranged at intervals, the annular grooves and the composite pipe have the same center line, and the grooves are sealing ring installation grooves 118. The transition part 117 includes a cylindrical part and a truncated cone part, and the sealing ring installation part 119, the cylindrical part, the truncated cone part and the pipe body 12 are connected in sequence. The inner diameter of the cylindrical part is larger than the inner diameter of the pipe body 12.
[0398] In this embodiment, two sealing ring installation grooves 118 are provided on the inner wall of the socket 11. The inner tube 2 covers the inner wall of the cylindrical part of the socket 11, the inner wall of the truncated cone part, and the inner wall of a sealing ring installation groove 118 close to the cylindrical part, and the inner tube 2 does not cover the inner wall of a sealing ring installation groove 118 away from the cylindrical part.
[0399] Based on the above structure, Figure 7 , 8 As shown, when two adjacent composite pipes are connected, the spigot 13 of one composite pipe is inserted into the socket 11 of the adjacent composite pipe and sealed by at least two sealing rings 4, each of which is installed in each sealing ring installation groove 118. Similarly, the inner ring of at least one sealing ring 4 contacts the outer wall of the spigot sleeve 21 on the outer wall of the spigot 13 of one composite pipe, and the outer ring of at least one sealing ring 4 contacts the inner wall of the inner pipe 2 at one end of the socket 11 of the adjacent composite pipe. Compared with the first embodiment, the socket structure of this embodiment is simpler and easier to manufacture.
[0400] It should be noted that although the shape of the socket 11 of this embodiment is different from the shape of the socket 11 of the first embodiment, the minimum thicknesses at the first chamfer R1, the second chamfer R2 and the second chamfer R3 of the base layer F1 of the composite socket of the first embodiment can be calculated according to the method of the first embodiment, and the minimum thickness of the base layer F1 of this embodiment can be set to be not less than the minimum thickness of the base layer F1 of the composite socket of the first embodiment.
[0401] Embodiment 3
[0402] The difference between this embodiment and the first or second embodiment is the method of forming the socket sleeve 21 .
[0403] In this embodiment, the socket sleeve 21 and the inner sleeve 22 are not made separately, but the circular tubular inner tube 2 is made as a whole. After the inner tube 2 is inserted into the outer tube 1, one end of the inner tube 2 extends beyond the end face of the socket 13. The portion of the inner tube 2 that extends beyond the end face of the socket 13 is flanged by spin forming and pressed against the outer wall of the socket 13 to form the socket sleeve 21.
[0404] Finally, it is necessary to explain here that the above embodiments are only used to further illustrate the technical solution of the present invention in detail and cannot be understood as limiting the scope of protection of the present invention. Some non-essential improvements and adjustments made by technicians in this field based on the above content of the present invention all fall within the scope of protection of the present invention.
Claims
1. A multi-layer composite pipe with a socket coating, characterized in that: The composite pipe comprises an inner pipe (2), an intermediate layer (3) and an outer pipe (1); the inner pipe (2), the intermediate layer (3) and the outer pipe (1) are bonded together or pressed together to form the composite pipe; the outer pipe (1) comprises a socket (11), a pipe body (12) and a spigot (13) which are connected in sequence; the inner pipe (2) is made of a corrosion-resistant metal material; the inner wall of the socket (11) is provided with steps and / or grooves; a section of the inner pipe (2) covers a portion of the inner wall of the socket (11) and the inner wall of the pipe body (12); and another section of the inner pipe (2) completely covers the spigot (13); The socket (11) and the portion of the inner tube (2) covering the inner wall of the socket (11) form a composite socket, and the spigot (13) and the portion of the inner tube (2) covering the spigot (13) form a composite spigot. The design method of the composite socket thickness includes the following steps: Step a1: Obtain the differential equation of radial deformation of the ring and the general solution of the differential equation according to the differential equation of the deflection curve of the ring and the moment balance equation; Step a2: Calculate and obtain a calculation formula for the maximum radial deformation of the ring according to the general solution of the differential equation obtained in step a1; Step a3: Calculating the minimum thickness of the socket of the base pipe that meets the deformation requirements according to the calculation formula of the maximum radial deformation obtained in step a2; Step a4: for a base pipe whose socket thickness is less than the minimum thickness obtained in step a3, at least one layer of annular reinforcement ring is sleeved in the socket of the base pipe to form a composite socket. If the material of the reinforcement ring is the same as that of the socket of the base pipe, the thickness of the composite socket is not less than the minimum thickness obtained in step a3. If the material of the reinforcement ring is different from that of the socket of the base pipe, go to step a5. Step a5: determining the minimum thickness of the reinforcement ring. If the thickness of the reinforcement ring is not less than the minimum thickness, the thickness of the composite socket is the minimum thickness that meets the deformation requirement. Assuming that the outermost layer of the composite socket is the base layer (F1), and the other layers are non-base layers (F2), for the multi-layer structure in the composite socket, the design method includes the following steps: Step c1: Based on the same strain of each layer of the composite socket, the calculation formula of the hoop stress of any non-base layer (F2) is obtained according to the non-single-layer material equilibrium equation; Step c2: Determine the relationship between the thickness of any non-base layer (F2) and the thickness of the base layer (F1), and the determination principle is: the ratio of the stress of the non-base layer (F2) to the yield strength of the material of the non-base layer is not greater than the ratio of the stress of the inner wall of the pipe body to the yield strength of the material of the pipe body; Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.
2. The composite pipe according to claim 1, characterized in that: The inner tube (2) comprises an inner sleeve (22) and a socket sleeve (21). The inner sleeve (22), the middle layer (3) and the outer tube (1) are sleeved in sequence from the inside to the outside. The socket sleeve (21) is formed by spinning one end of the inner tube (2) outwardly, or by welding the socket sleeve (21) and the inner sleeve (22) together.
3. The composite pipe according to claim 2, characterized in that: When the socket sleeve (21) and the inner sleeve (22) are welded together, the welding point of the socket sleeve (21) and the inner sleeve (22) is located on the end surface of the socket (13), and the area of the inner tube (2) covering the end surface of the socket (13) is formed by welding after flanging the socket sleeve (21) and / or flanging the inner sleeve (22).
4. The composite pipe according to claim 2, characterized in that: When the material of the outer tube (1) is metal and the material of the middle layer (3) is hot melt adhesive, the inner tube (2) and the outer tube (1) are processed into a composite tube through a composite process; when the material of the middle layer (3) is plastic or rubber, the inner tube (2), the middle layer (3) and the outer tube (1) are pressed together to form a composite tube.
5. The composite pipe according to claim 4, characterized in that: The thickness of the middle layer (3) is not less than 0.2 mm, and the thickness of the inner sleeve (22) is not less than 0.1 mm.
6. The composite pipe according to claim 2, characterized in that: When two adjacent composite pipes are connected, the spigot (13) of one composite pipe is inserted into the socket (11) of the adjacent composite pipe and sealed by the sealing ring (4); the outer ring of the sealing ring (4) contacts the inner wall of the inner pipe (2) of the socket (11) of one composite pipe, and the inner ring of the sealing ring (4) contacts the outer wall of the spigot sleeve (21) on the outer wall of the spigot (13) of the adjacent composite pipe; at least one sealing ring (4) is provided, and the sealing ring (4) is installed in a groove on the inner wall of the socket (11) or the sealing ring (4) is limited by a limiting member.
7. The composite pipe according to claim 1, characterized in that: Step a2 includes the following steps: Step a21: Assuming that the annular ring is subjected to at least two radial concentrated forces that are symmetrical relative to the center of the circle, based on the assumption, calculating the constant term in the general solution of the differential equation to obtain the first form of the differential equation; Step a22: Based on the first form of the differential equation obtained in step a21, assuming that the annulus is subjected to only two radial concentrated forces that are symmetrical relative to the center of the circle, the first form of the differential equation is transformed into a second form of the differential equation; Step a23: introducing the internal negative pressure of the ring into the second form of the differential equation by introducing a coefficient, so as to obtain the third form of the differential equation; Step a24: Obtain a calculation formula for the maximum change in the diameter of the circular ring with coefficients according to the third form of the differential equation obtained in step a23.
8. The composite pipe according to claim 1, characterized in that: Step a3 includes the following steps: Step a31: Calculate the parameters of a reference pipe according to an existing shear load formula, wherein the material and outer diameter of the reference pipe are respectively the same as the material and outer diameter of the base pipe; Step a32: obtaining the maximum deformation of the reference pipe socket through simulation according to the parameters calculated in step a31; Step a33: Substitute the maximum deformation of the socket obtained in step a32 into the calculation formula of step a2 to obtain the coefficient value in the calculation formula, thereby obtaining a calculation formula for the maximum radial deformation of the ring with a known coefficient; Step a34: Calculate the minimum thickness of the base pipe socket that meets the deformation requirements based on the allowable deformation regulations and the formula obtained in step a33.
9. The composite pipe according to claim 1, characterized in that: In step a1, the radial deformation obtained is u The differential equation is: , The general solution is: , in, R is the average value of the outer circle radius and the inner circle radius of the ring; E is the elastic modulus of the pipe, Pa; I is the moment of inertia of the tube cross section, m 4 ; M is the bending moment borne by a certain section of the ring in the local area, in units of N*m; establish a two-dimensional coordinate system, φ The deformation is u The line connecting the point at and the origin of the coordinate system is the same as the line connecting the point at and the origin of the coordinate system in the two-dimensional coordinate system. x The angle between the positive directions of the axes, ; A1, A2, A3, B2, B3 are constants.
10. The composite pipe according to claim 9, characterized in that: In step a2, it is assumed that the ring is subjected to at least two radial concentrated forces P symmetrically relative to the center of the circle, and the angle between two adjacent radial concentrated forces P is , the ring satisfies the following three boundary conditions: First, in Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula: ; Second, due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; That is: ; Third, cut near point A, remove point A and the force P, and from the symmetry and equilibrium conditions, we know that the shear force on the section at the circular cut is , that is: ; Combining the above three formulas with the general solution of the differential equation, we get the first form of the differential equation: ; Assuming that the annulus is subjected to only two radial concentrated forces P which are symmetrical relative to the center of the circle, then: , we get the second form of the differential equation: ; Introducing the coefficient m, we get the third form of the differential equation: ; Point of application of force P The point is the point of maximum deformation, and the change in the diameter of the ring is: .
11. The composite pipe according to claim 1, characterized in that: In step a5, the minimum thickness of the reinforcement ring is obtained and determined by experiment, simulation or theoretical calculation. When the minimum thickness of the reinforcement ring is obtained and determined by theoretical calculation, the minimum thickness of the reinforcement ring is obtained based on the socket thickness of the base pipe and the bending strength calculation formula of the longitudinal composite beam.
12. The composite pipe according to claim 1, characterized in that: For the same base pipe, along the direction from the socket to the spigot, the inner wall of the socket includes a first step (111), a second step (112), a third step (113), a fourth step (114), and a fifth step (115) arranged in sequence, one end of the inner wall of the pipe body is a sixth step (116), the chamfer at the connection between the fourth step (114) and the fifth step (115) is a first chamfer (R1), the chamfer at the connection between the fifth step (115) and the sixth step (116) is a second chamfer (R2), the chamfer at the connection between the second step (112) and the third step (113) is a third chamfer (R3), the chamfer at the connection between the first step (111) and the second step (112) is a fourth chamfer (R4), and the chamfer at the connection between the third step (113) and the fourth step (114) is a fifth chamfer (R5).
13. The composite pipe according to claim 12, characterized in that: In step c1, the multilayer structure is regarded as a structure including at least one pressure vessel with a conical inner wall and at least one annular reinforcement ring installed on the inner wall of the pressure vessel. In step c1, the first cone is a pressure vessel with a conical inner wall, and the first chamfer (R1) and the second chamfer (R2) are located on the inner wall of the first cone.
14. The composite pipe according to claim 13, characterized in that: In step c1, assuming that there are m layers of non-base layer (F2) and one base layer (F1), the equilibrium equation of the multi-layer structure is: ,in, p is the internal pressure of the conical container, D is the diameter of the conical container to be calculated, l is the cone height, S1, S2, …, S m are the thickness of each non-base layer (F2), S b is the thickness of the base layer (F1), , , …, are the hoop stresses of each non-base layer, is the hoop stress of the base layer (F1), α is the half cone angle of the conical container; In step c1, the strain of the base layer (F1) The strain is the same as that of each non-base layer (F2), satisfying: ,Right now , where E1, E2, …, E m are the elastic modulus of each non-base layer (F2), E b is the elastic modulus of the base layer (F1), and the calculation formula for the hoop stress of the nth non-base layer (F2) is obtained by combining the equilibrium equation: ,in, is the hoop stress of the nth non-base layer, n is 1~m, E n is the elastic modulus of the nth non-base layer.
15. The composite pipe according to claim 14, characterized in that: In step c2, in order to make the composite socket meet the strength requirements, the internal stress of the nth non-base layer (F2) and the stress of the inner wall of the pipe body meet the following conditions: , , where Q0, Q n are the yield strengths of the pipe body and the nth non-base layer (F2) respectively; is the hoop stress of the pipe body, S is the thickness of the pipe body, D0 is the outer diameter of the pipe body, and p is the pressure given by the fluid in the pipe. The hoop stress calculation formula of the nth layer of non-base layer (F2) is combined to obtain the relationship between the thickness of the nth layer of non-base layer (F2) and the thickness of the base layer (F1): .
16. The composite pipe according to claim 15, characterized in that: When the thickness of the base layer (F1) corresponding to the maximum stress point is known, the minimum thickness of the layer with the minimum yield stress is calculated based on the relationship between the thickness of the n-th non-base layer (F2) and the thickness of the base layer (F1), and this layer is set as the minimum stress layer. The thickness of other non-base layers (F2) is set to be not less than the minimum thickness of the minimum stress layer; when the thickness of the non-base layer (F2) is known, first, at least one point where the stress of the base layer (F1) is easily concentrated is selected, and the selection principle is: after the strength of each selected point meets the requirements, the strength of other places of the base layer (F1) designed according to the outer wall design requirements also meets the requirements; then the thickness of each selected point is calculated based on the relationship between the thickness of the n-th non-base layer (F2) and the thickness of the base layer (F1).
17. A method for manufacturing a multi-layer composite pipe with a socket coating, used for manufacturing the composite pipe according to any one of claims 2 to 6, characterized in that: The steps include: Step S1: processing the inner wall and the outer wall of the outer tube (1) and manufacturing the inner tube (2); Step S2: arranging the inner sleeve (22), the middle layer (3) and the outer tube (1) in sequence from the inside to the outside; Step S3: Pressing the inner sleeve (22) onto the outer tube (1) and pressing the socket sleeve (21) onto the outer wall of the socket (13); Step S4: compounding; When the material of the middle layer (3) is plastic or rubber, there is no step S4.
18. The manufacturing method according to claim 17, characterized in that: In step S3, the spigot sleeve (21) is obtained by spin-forming one end of the inner tube (2), or when the inner tube (2) is manufactured, the inner sleeve (22) and the spigot sleeve (21) are separately manufactured, and the spigot sleeve (21) is sleeved on the outer wall of the spigot (13). Between step S3 and step S4, the inner sleeve (22) and the spigot sleeve (21) are welded, and the welding point is located on the end surface of one end of the spigot (13) of the composite tube.
19. The manufacturing method according to claim 17, characterized in that: When the material of the intermediate layer (3) is hot melt adhesive or plastic, in step S2, the intermediate layer (3) is made into a round tube and then the inner sleeve (22), the intermediate layer (3) and the outer tube (1) are sequentially sleeved, or the powdered raw material of the intermediate layer (3) is sprayed or coated on the outer wall of the inner sleeve (22) and the outer wall of the socket (13), or the powdered raw material of the intermediate layer (3) is sprayed or coated on the inner wall of the outer tube (1) and the outer wall of the socket (13), and then the inner sleeve (22) is sleeved into the outer tube (1); in step S3, the inner sleeve (22) is pressed tightly against the inner wall of the tube body (12), the inner wall of the socket (11) and the inner wall of the socket (13) by spin forming or hydraulic forming.
Citation Information
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