Metal-lined multilayer composite pipe and manufacturing method thereof

Abstract

The present invention discloses a metal-lined multilayer composite pipe and a manufacturing method thereof. The composite pipe comprises an inner pipe, an intermediate layer and an outer pipe which are sleeved in sequence from the inside to the outside. The outer pipe comprises a socket, a pipe body and a socket made of corrosion-resistant metal material which are connected in sequence. The inner pipe covers the inner wall and the end face of the outer pipe. The manufacturing method comprises: S1: processing the inner wall and the outer wall of the socket and the pipe body, and making the inner pipe and the socket part; S2: fixing and connecting the socket part and the straight pipe part; S3: arranging the inner pipe, the intermediate layer and the outer pipe in sequence from the inside to the outside; S4: pressing the inner pipe and the intermediate layer on the outer pipe; S5: compounding; S6: flanging and sealing one end of the socket part of the inner pipe. The present invention effectively prevents the outer pipe from rusting, increases the service life of the composite pipe, avoids the rust of the outer pipe from polluting the water quality, prevents the inner pipe from electrochemical corrosion, and prevents the inner pipe from stratifying and deforming under negative pressure. The minimum thickness of the composite socket of the socket-and-spigot pipe and the minimum thickness of the composite socket can be obtained by theoretical calculation.

Description

Technical Field

[0001] The invention belongs to the technical field of composite pipes, and specifically relates to a metal lined multilayer composite pipe and a manufacturing method thereof. Background Art

[0002] Composite pipe is a kind of pipe widely used in liquid transportation pipe network system. When in use, the part of the composite pipe in contact with the liquid is required to have good corrosion resistance to prevent the composite pipe from rusting due to long-term contact with the liquid, which affects the quality of the transported liquid. The existing composite pipe generally has a metal pipe as the outer layer and a plastic pipe as the inner layer, and the inner layer and the outer layer are compounded by a composite process. That is, the inner wall of the outer metal pipe is coated with a plastic pipe. The structure of the existing composite pipe is a socket-and-spigot type, 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 for corrosion protection, the middle part and the spigot of the metal pipe are coated with the 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 composite 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.

[0003] 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.

[0004] 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, 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, during the secondary processing or installation of the pipe, an annular reinforcement ring is sleeved on the socket to enhance the ability to resist deformation.

[0005] 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 each location minimum or close to the minimum while meeting the use strength. At present, the socket of the general base pipe (i.e., the outer pipe) is made of only one material, such as ductile iron or Q235 steel, and 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 the inner layer is stainless steel and the outer layer is Q235 steel or ductile iron, which can save costs and improve the corrosion resistance of the pipe. Similarly, if the composite socket can meet the strength requirements and the thickness is minimum or close to minimum, it can also greatly reduce costs. 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 is the part not covered by the liner and has only one layer, namely, the outermost base pipe. The multi-layer structure includes at least two layers, namely, the outermost base pipe and the liners sleeved in the base pipe.

[0006] The base pipe refers to the pipe in the factory setting, that is, the base pipe has no reinforcement ring installed and no secondary processing such as composite process. The base pipe includes a socket, a pipe body and a spigot 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 spigot 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 spigot. The outer diameter of the pipe body refers to the outer diameter of the main body, and the thickness of the pipe body refers to the thickness of the main body. The outer diameter of the base pipe is the outer diameter of the pipe body. Fig.12In 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. 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 metal-lined multilayer composite pipe and a manufacturing method thereof, wherein the inner pipe of corrosion-resistant material covers the inner wall of the outer pipe and the end face of the socket, and the socket is also made of corrosion-resistant material, which completely isolates the outer pipe from the liquid flowing in the composite pipe, effectively preventing the outer pipe from rusting, thereby increasing the service life of the composite pipe and avoiding the rust of the outer pipe from polluting the water quality. The middle layer also plays the role of insulation isolation and connection, preventing the inner pipe from electrochemical corrosion and preventing the inner pipe from stratifying and deforming under negative pressure. During installation, the socket 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 the inner wall of the inner pipe of the socket to prevent the inner pipe serving as the lining from falling off and prevent the pipe body and the socket from contacting the liquid and being corroded. 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, and 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 deformation and separation of the spigots and sockets of adjacent pipes when the pipe bursts, as well as the cracking of the socket.

[0008] In order to achieve the above object, the technical solution adopted by the present invention is:

[0009] A metal-lined multi-layer composite pipe comprises an inner pipe, an intermediate layer and an outer pipe which are sequentially sleeved from the inside to the outside. The outer pipe comprises a socket, a pipe body and a plug made of corrosion-resistant metal which are sequentially connected. The inner pipe is made of corrosion-resistant metal. One end of the inner pipe covers part of the inner wall of the socket, the middle part covers the inner wall of the pipe body, and the other end covers the inner wall and end face of the plug.

[0010] As a further improvement of the above technical solution:

[0011] Preferably, the material of the middle layer is non-metal, and the material of the socket and the pipe body is metal.

[0012] Preferably, the material of the middle layer is hot melt adhesive.

[0013] Preferably, the material of the middle layer is rubber.

[0014] Preferably, the material of the middle layer is plastic.

[0015] Preferably, the socket and the pipe body are made of ductile iron or carbon steel.

[0016] Preferably, when the material of the middle layer is hot melt adhesive, the inner tube, the middle layer and the outer tube are bonded into a composite tube.

[0017] 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.

[0018] 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 of the adjacent composite pipe.

[0019] Preferably, the thickness of the middle layer is not less than 0.2 mm, and the thickness of the inner tube is not less than 0.1 mm.

[0020] Preferably, when the material of the middle layer is hot melt adhesive, the thickness of the middle layer is 0.2 mm to 0.5 mm.

[0021] Preferably, a sealing gasket is provided between the end surface of the socket and the inner tube covering the end surface.

[0022] Preferably, the sealing gasket includes a pressed edge and a flanged edge which are integrally connected, the pressed edge is located between the inner wall at one end of the socket and the inner tube, and the flanged edge is located between the end face of the socket and the inner tube covering the end face.

[0023] Preferably, when the material of the middle layer is rubber, at one end of the socket, the middle layer exceeds the end of the socket, and the middle layer exceeding the end of the socket is flanged to form a sealing gasket.

[0024] 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 inner wall of the spigot form a composite spigot.

[0025] The design method of the composite socket thickness includes the following steps:

[0026] 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;

[0027] 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;

[0028] 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;

[0029] 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.

[0030] 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.

[0031] 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:

[0032] 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;

[0033] 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;

[0034] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.

[0035] Step a2 includes the following steps:

[0036] 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;

[0037] 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;

[0038] 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;

[0039] 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.

[0040] Step a3 includes the following steps:

[0041] 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;

[0042] Step a32: obtaining the maximum deformation of the reference pipe socket through simulation according to the parameters calculated in step a31;

[0043] 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;

[0044] 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.

[0045] In step a1, the radial deformation obtained is u The differential equation is:

[0046] ,

[0047] The general solution is: ,

[0048] 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, 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.

[0049] 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.

[0050] 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.

[0051] 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.

[0052] 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:

[0053] ;

[0054] 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;

[0055] 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.

[0056] 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: .

[0057] 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.

[0058] A method for manufacturing a socket-and-spigot type heterogeneous material multilayer composite pipe, used for manufacturing the composite pipe, comprises the following steps:

[0059] Step S1: processing the inner and outer walls of the socket and the pipe body, and making the inner pipe and the spigot;

[0060] Step S2: Connect the socket and the tube body coaxially and fixedly;

[0061] Step S3: Arrange the inner tube, the middle layer and the outer tube in sequence from inside to outside;

[0062] Step S4: Press the inner tube and the middle layer onto the outer tube;

[0063] Step S5: compounding;

[0064] Step S6: Flanging and sealing one end of the socket of the inner tube;

[0065] When the middle layer is made of plastic or rubber, there is no step S5.

[0066] Preferably, in step S4, the inner tube is pressed tightly against the inner wall of the plug, the inner wall of the tube body, and the inner wall of the socket by spin forming or hydraulic forming.

[0067] The beneficial effects of the present invention are:

[0068] (1) The inner tube made of corrosion-resistant material covers the inner wall and the end face 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 composite tube, avoiding the rust of the outer tube from polluting the water quality, and improving the liquid transportation performance.

[0069] (2) Compared with plastic pipe linings, the inner pipe made of corrosion-resistant material as the inner lining eliminates the inner surface wrinkles and uneven defects caused by the melting of the inner lining plastic during compounding, 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.

[0070] (3) The inner tube made of corrosion-resistant material as the lining has a smaller thermal expansion coefficient than the plastic tube lining. The internal stress caused by the separation between the inner tube and the outer tube due to temperature changes is smaller, and the lining is less likely to delaminate.

[0071] (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.

[0072] (5) The middle layer plays the role of insulation isolation and connection at the same time, preventing the inner tube from electrochemical corrosion. When the middle layer is hot melt adhesive, the bonding strength of the middle layer is high, preventing the inner tube from delamination and deformation under negative pressure.

[0073] (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 to prevent the inner pipe serving as the lining from falling off.

[0074] (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 made of corrosion-resistant material itself will play an anti-corrosion role, and there is no need to coat the outer wall of the spigot with other anti-corrosion materials. At the same time, the inner ring of the sealing ring presses 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.

[0075] (8) Corrosion-resistant materials, such as stainless steel, have a higher unit price than ductile iron. The plug is made of corrosion-resistant materials, and the socket and pipe body are made of ductile iron, which ensures the anti-corrosion effect while reducing the overall cost.

[0076] (9) When hot melt adhesive is used for the middle layer, the thickness of the corrosion-resistant lining material is thinner and the cost is lower; when the inner diameter of the outer pipe is the same, the inner diameter of the composite pipe is larger and the transportation efficiency is higher.

[0077] (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, as well as the cracking of the socket. BRIEF DESCRIPTION OF THE DRAWINGS

[0078] Figure 1It is a structural schematic diagram of the composite tube of the present invention when the middle layer is hot melt adhesive or plastic and the sealing gasket has no edge pressing.

[0079] Figure 2 It is a structural schematic diagram of the composite tube of the present invention when the middle layer is hot melt adhesive or plastic and the sealing gasket has a pressed edge and a flanging edge.

[0080] Figure 3 yes Figure 2 An enlarged schematic diagram of point A.

[0081] Figure 4 It is a schematic structural diagram of the composite tube when the middle layer of the present invention is rubber.

[0082] Figure 5 It is a schematic diagram of connecting two adjacent composite pipes of the present invention.

[0083] Figure 6 Schematic diagram of the cross-section deformation before and after the socket is subjected to a pair of radially symmetrical concentrated forces.

[0084] Figure 7 for Figure 6 Schematic diagram of local analysis.

[0085] Figure 8 Schematic diagram of a circular ring subjected to multiple uniformly arranged radial concentrated forces.

[0086] Fig. 9 This is the stress analysis diagram of socket-type pipe.

[0087] Fig.10 It is a schematic cross-sectional view of a composite socket per unit length of the present invention.

[0088] Fig.11 It is a schematic diagram of the composite socket structure of the present invention.

[0089] Fig.12 It is a schematic diagram of the socket and spigot connection of the present invention.

[0090] Fig.13 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

[0091] 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.

[0092] 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.

[0093] A metal lined multilayer composite pipe, such as Figure 1~Figure 5 As shown, it includes an inner tube 2, a middle layer 3 and an outer tube 1 which are sequentially sleeved from the inside to the outside. The inner tube 2, the middle layer 3 and the outer tube 1 can all be regarded as a cylindrical structure.

[0094] The outer tube 1 comprises a socket 11, a tube body 12 and a spigot 13 which are sequentially connected along the axial direction thereof, i.e., the socket 11, the tube body 12 and the spigot 13 share a central line. The spigot 13 is cylindrical, and a chamfer is provided along the circumferential direction at one end of the spigot 13 away from the tube body 12. The inner surface of the socket 11 is stepped, i.e., the socket 11 can be regarded as being formed by sequentially connecting a plurality of cylindrical structures with different inner diameters.

[0095] The socket 11 and the pipe body 12 are made of metal, preferably ductile iron or carbon steel, and the spigot 13 is made of corrosion-resistant material, including but not limited to stainless steel, titanium, and composite materials, preferably stainless steel. The socket 11 and the pipe body 12 are integrally cast or welded together after being manufactured separately, and the spigot 13 and the pipe body 12 are welded together, and the weld is along the circumference of the composite pipe. The inner diameter of the spigot 13 is equal to or similar to the inner diameter of the pipe body 12.

[0096] The inner tube 2 is made of corrosion-resistant materials, including but not limited to stainless steel, titanium, and composite materials, preferably stainless steel. One end of the inner tube 2 covers and presses the inner wall of the socket 11, the middle part covers and presses the inner wall of the pipe body 12, and the other end covers and presses the inner wall and end surface of the socket 13. It should be noted that the inner tube 2 presses part of the inner wall of the socket 11 (such as Figures 1 to 5 as shown) or the entire inner wall (as shown Fig.13 as shown).

[0097] The material of the middle layer 3 is non-metal, preferably hot melt adhesive, plastic or rubber. When the material of the middle layer 3 is hot melt adhesive, the hot melt adhesive has adhesive properties, and the inner tube 2 and the outer tube 1 are bonded together through the middle layer 3. Through the composite process, the inner tube 2 is firmly connected to the inner wall of the outer tube 1. When the material of the middle layer 3 is plastic or rubber, the plastic or rubber has no adhesive properties, and the inner tube 2 presses the middle layer 3 against the inner wall of the outer tube 1.

[0098] A sealing gasket 4 is provided between the end surface of the socket 13 and the inner tube 2 covering the end surface. Figures 1 to 3 As shown, the sealing gasket 4 includes an integrally connected pressing edge 41 and a flange 42. The pressing edge 41 can be regarded as a thin cylindrical shape, and the flange 42 is annular. The inner diameter of the flange 42 is equal to the inner diameter of the pressing edge 41, and the outer diameter of the flange 42 is greater than the outer diameter of the pressing edge 41. The pressing edge 41 and the flange 42 share a common center line. The pressing edge 41 is located between the inner wall of one end of the socket 13 and the inner tube 2, and the flange 42 is located between the end face of the socket 13 and the inner tube 2 covering the end face.

[0099] It should be noted that the edge pressing 41 can be eliminated. Figure 1 As shown, Figure 1 As shown, one end surface of the intermediate layer 3 is flush with the end surface of the socket 13, and the sealing gasket 4 is equivalent to having only the flange 42 and no pressed edge 41. Figure 2 and 3 As shown, one end surface of the middle layer 3 is not flush with the end surface of the socket 13 , and the end surface of the socket 13 exceeds one end surface of the middle layer 3 . At this time, the sealing gasket 4 has both a flange 42 and a pressed edge 41 .

[0100] The sealing gasket 4 is a rubber sealing gasket.

[0101] like Figure 1 , 2 As shown in FIG. 4 , 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 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, the inner diameter of the fifth step 115 gradually increases from the fourth step 114 toward the pipe body 12, 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 .

[0102] 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.

[0103] When the inner tube 2 does not completely cover the inner wall of the socket 11 , the inner tube 2 continuously covers the inner wall of the fourth step 114 and the inner wall of the fifth step 115 , and the inner tube 2 does not cover the inner wall of the first step 111 , the inner wall of the second step 112 , and the inner wall of the third step 113 .

[0104] 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 tube 2 is 0.1 mm-2 mm.

[0105] 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 tube 2, the outer diameter of the middle layer 3 is slightly smaller than the inner diameter of the outer tube 1, and the length of the middle layer 3 is substantially the same as that of the inner tube 2.

[0106] When the material of the middle layer 3 is hot melt adhesive, it acts as an adhesive layer. When the inner wall of the outer tube 1 is rough during compounding, the middle layer 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 is likely to occur in areas with large local roughness. At the same time, the middle layer 3 also acts 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 0.2mm or more, good peel strength and insulation performance can be stably obtained. The preferred thickness is 0.2mm-2mm.

[0107] 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.

[0108] 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:

[0109] 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;

[0110] 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;

[0111] 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.

[0112] 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.

[0113] 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.

[0114] Based on the above structure, when two adjacent composite pipes are connected, Figure 5 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 7. The outer ring of the sealing ring 7 contacts and presses the inner wall of the inner tube 2 of the socket 11 of the composite pipe, and the inner ring of the sealing ring 7 contacts the outer wall of the spigot 13 of the adjacent composite pipe. Preferably, the end face of one end of the sealing ring 7 contacts the end face of the first step 111, a part of the sealing ring 7 is accommodated on the second step 112, the third step 113 is inserted into the middle of the sealing ring 7, and the other part of the sealing ring 7 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.

[0115] One or more sealing rings 7 may be provided.

[0116] Preferably, 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 7 can contact the stainless steel, avoiding the part of the pipe body 12 made of ductile iron extending between the sealing ring 7 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 sealing gasket 4 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 delaminate.

[0117] A method for manufacturing a stainless steel lined composite pipe, for manufacturing the composite pipe, comprising the following steps:

[0118] Step S1: Process the inner and outer walls of the socket 11 and the pipe body 12, make the inner pipe 2, and make the socket 13.

[0119] In this step, the method for processing the inner and outer walls of the socket 11 and the pipe body 12 is to first grind the inner walls of the socket 11 and the pipe body 12, and then sandblast or shot blast the inner and outer walls. Use an internal grinder to grind the inner walls of the socket 11 and the pipe body 12, grind the inner walls of the pipe flat, and remove surface oxide scale, scum, rust, etc. Alternatively, after the inner wall is polished flat, use a laser cleaning machine to clean the inner wall of the pipe to deeply remove the oxide scale and rust on the inner wall of the pipe. 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 pipe into a frosted surface.

[0120] Making the inner tube 2 comprises the following steps:

[0121] 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).

[0122] 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 and its length is greater than the length of the inner tube 2, 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 tube 2 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.

[0123] Preferably, the length of the stainless steel round tube obtained in step S12 is greater than the length of the inner tube 2 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.

[0124] The socket 13 is obtained by cutting a stainless steel round tube according to a required length, or the socket 13 is obtained by welding a stainless steel plate after rolling it into a circle.

[0125] Step S2: coaxially welding the socket 13 and the tube body 12.

[0126] In this step, the stainless steel socket 13 and the ductile iron pipe body 12 are welded by argon arc welding using 309L welding wire or by electric welding using an electric welding machine and A312 welding rod.

[0127] Step S3: Arrange the inner tube 2, the middle layer 3 and the outer tube 1 in sequence from the inside to the outside.

[0128] 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:

[0129] The first way is as follows:

[0130] 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.

[0131] Then, the obtained shaped intermediate layer 3 is sleeved on the outer wall of the inner tube 2. In this step, the sleeve can be made manually or by a sleeve machine.

[0132] Next, the inner wall of the socket 11 is sprayed with plastic. After the socket 11 is heated, a powdered hot melt adhesive or plastic is roll-molded or spray-molded on the inner wall of the socket 11 by a roller coating machine or a plastic coating machine.

[0133] Finally, the intermediate layer 3 and the inner tube 2 are inserted into the outer tube 1 as a whole. Similarly, in this step, the casing can be completed manually or by a pipe threading machine.

[0134] The second way is as follows:

[0135] First, after the inner walls of the socket 11, the pipe body 12 and the plug 13 are heated, a roller coating machine or a plastic coating machine is used to roll or spray powdered hot melt adhesive or plastic on the inner walls of the socket 11, the pipe body 12 and the plug 13. The thickness of the roll or spray coating is not less than 0.2 mm.

[0136] Then, the inner tube 2 is inserted into the outer tube 1 with the middle layer 3 on the inner wall.

[0137] The third way is as follows:

[0138] First, after the outer wall of the inner tube 2 is heated, a roller coating machine or a plastic coating machine is used to roll-coat or spray-coat powdered hot melt adhesive or plastic on the outer wall of the stainless steel inner tube 2. The thickness of the roll-coating or spray-coating is not less than 0.2 mm.

[0139] Then, the inner tube 2 with the intermediate layer 3 on the outer wall is inserted into the outer tube 1.

[0140] When the material of the middle layer 3 is rubber, firstly roll the rubber sheet on the outer wall of the inner tube 2, so that the rubber sheets are spliced ​​to cover the outer wall of the inner tube 2, and then the rubber seams are bonded with adhesive, and then the inner tube 2 with rubber sleeve is inserted into the outer tube 1. Alternatively, the rubber tube is sleeved on the outer wall of the inner tube 2 and then the whole is sleeved into the outer tube 1.

[0141] Step S4: Press the inner tube 2 and the middle layer 3 onto the outer tube 1 .

[0142] This step can be completed by spin forming or hydraulic forming. During spin forming, the pipe body 12 and the inner pipe 2 of the socket 11 are respectively spin-formed by a composite pipe spinning machine, and the spinning is carried out from the inside to the outside, so that the inner pipe 2 lined with stainless steel material, the middle layer 3, and the outer pipe 1 are closely attached in sequence. During hydraulic forming, the two ends of the inner pipe 2 of stainless steel material inserted into the outer pipe 1 are sealed, and water is injected into the pipe to pressurize it, so that the inner pipe 2, the middle layer 3, and the outer pipe 1 are closely attached in sequence.

[0143] Step S5: Compounding.

[0144] Before compounding, use a trimmer to remove excess stainless steel at both ends of the inner tube 2. Heat the composite tube with a medium frequency sensor or a heating furnace, melt the hot melt adhesive, and after cooling, bond the outer tube 1 and the inner tube 2 together.

[0145] When the material of the middle layer 3 is plastic or rubber, there is no step S5.

[0146] Step S6: One end of the socket 13 of the inner tube 2 is flanged and sealed.

[0147] In this step, when the middle layer 3 is hot melt adhesive or plastic, first, the sealing gasket 4 is put on the outer wall of the inner tube 2 beyond the end face of the plug 13 and the sealing gasket 4 is made close to the end face of the plug 13. Then, the inner tube 2 beyond the sealing gasket 4 is pressed by a pressing device to make the inner tube 2 beyond the sealing gasket 4 bend, that is, the original outer wall of the inner tube 2 beyond the sealing gasket 4 is close to the sealing gasket 4, and the original inner wall becomes the end face of the composite tube.

[0148] When the middle layer 3 is rubber, in step S3, the length of the middle layer 3 is set to be the same as that of the inner tube 2 and the two ends are flush. In this step, there is no need to install the sealing gasket 4 separately. Instead, the inner tube 2 that exceeds the end face of the socket 13 is directly crimped using a crimping device. The middle layer 3 that exceeds the end face of the socket 13 will be simultaneously flanged, so that the middle layer 3 that exceeds the end face of the socket 13 is clamped between the end face of the socket 13 and the inner tube 2 that exceeds the end face of the socket 13 to form a sealing gasket.

[0149] After completing the above steps, the outer wall of the composite pipe is sprayed with paint or plastic coated or other methods are used for external anti-corrosion treatment.

[0150] In this embodiment, the socket 11 of the outer tube and the portion of the inner tube 2 covering the socket 11 can both improve the strength of the socket of the composite tube, and the socket formed by the two is a composite socket. The socket 13 of the outer tube and the portion of the inner tube 2 covering the socket 13 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. When the socket 13 and the inner tube 2 are made of the same material, although the socket 13 and the inner tube 2 are two layers, they can be regarded as a whole. In addition, it should be noted that the material of the intermediate layer 3 is hot melt adhesive, rubber or plastic, and the material of the sealing ring 7 is rubber, both of which are non-metallic flexible materials, and have little effect on the strength and deformation resistance of the socket 13, and therefore do not belong to the category of composite sockets that can resist deformation.

[0151] Next, the thickness of the spigot end and the socket end of the composite pipe is optimized and designed.

[0152] The design method of the thickness of the composite socket includes the following steps:

[0153] 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 The differential equation and the general solution of the differential equation; because the cross-section of the pipe body and the socket 13 is a circular ring, the deformation of the socket is analyzed by analyzing the radial deformation of the circular ring. In addition, the cylinder of unit length can be regarded as a circular ring.

[0154] 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.

[0155] 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.

[0156] 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.

[0157] 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 calculation formula of the bending strength 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. 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.

[0158] The following is a detailed introduction to each step.

[0159] In step a1, the radial deformation obtained The differential equation is:

[0160] (1)

[0161] 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, 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 Figure 6 Point m 1 The radial deformation at u， but om 1 and x The angle between the positive directions of the axes is 。

[0162] The above differential equation (1) is obtained as follows:

[0163] like Figure 6 and 7 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. Figure 6 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, 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 ， set up u Centripetal movement is positive. Figure 7 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 n 1 and perpendicular to on ,straight line n 1 t 1 ’ For passing point n 1 tangent.

[0164] 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 deformationm 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:

[0165] (2)

[0166] (3)

[0167] (4)

[0168] (5)

[0169] m 1 t 1 and m 1 t The angle between , n 1 t 1 ’ and n 1 t’ The angle between .

[0170] m 1 section to n 1 The variable of the cross-section rotation is :

[0171] (6)

[0172] 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 1The cross-sectional angle, n 1 t 1 ’ and n 1 t’ The angle between n 1 The corner of the cross section.

[0173] Substituting formula (5) and (6) into formula (4) yields:

[0174] (7)

[0175] By series expansion we have:

[0176] Ignoring the higher-order terms, substituting the series expansion into (7) yields: (8)

[0177] Substituting (2) into (8), we obtain:

[0178] (9)

[0179] Omitting the higher-order terms, we get: (10)

[0180] 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).

[0181] The general solution of the above differential equation (1) is:

[0182] (11)

[0183] 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:

[0184] (12)

[0185] 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:

[0186] 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.

[0187] 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.

[0188] 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.

[0189] 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.

[0190] 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 Figure 8 shown. Figure 6 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.

[0191] The ring satisfies the following three boundary conditions:

[0192] (1) Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula:

[0193] (13)

[0194] (2) Due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; that is,

[0195] (14)

[0196] (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:

[0197] (15)

[0198] 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.

[0199] Simultaneously formula (13), (14), (15), the solution is:

[0200] (16)

[0201] (17)

[0202] (18)

[0203] Substituting formulas (16), (17), and (18) into formula (12), we obtain:

[0204] (19)

[0205] Formula (19) is the first form of the differential equation obtained in step a21. is the radial deformation of the ring.

[0206] In step a22, if only one pair of radial forces P is applied to the ring, then , the deflection curve equation (19) becomes:

[0207] (20)

[0208] Formula (20) is the second form of the differential equation obtained in step a22.

[0209] 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:

[0210] (twenty one)

[0211] 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.

[0212] Formula (21) is the third form of the differential equation obtained in step a23.

[0213] 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 :

[0214] (twenty two)

[0215] Formula (22) is the third form of the differential equation obtained in step a24.

[0216] 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:

[0217] 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.

[0218] Step a32: Obtain the maximum deformation of the reference pipe socket through simulation based on the parameters calculated in step a31.

[0219] 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.

[0220] 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.

[0221] 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:

[0222] (twenty three)

[0223] like Fig. 9 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).

[0224] 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 FIt 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.

[0225] In this embodiment, the parameters in formula (23) are M P and W The calculation process is as follows:

[0226]

[0227]

[0228] 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 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.

[0229] Step a31 is thus completed.

[0230] 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.

[0231] 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:

[0232] (twenty four)

[0233] From formula (24), we can see that u max The value is related to the pipe diameter and pipe thickness.

[0234] 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.

[0235] Table 1 Nominal wall thickness and absolute deformation of DN=600mm pipes at various nominal pressures

[0236]

[0237] 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).

[0238] 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.

[0239] 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, 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.

[0240] 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.

[0241] 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.

[0242] 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.

[0243] In this embodiment, the calculation formula for the bending strength of the longitudinal composite beam is as follows:

[0244] (25)

[0245] 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.

[0246] In some occasions, the use requirement of the pipe is to meet the nominal pressure within 1.6MPa. The selected base pipe is the pipe No. 5 in Table 1, with a wall thickness of 4mm. The composite spigot includes the base pipe spigot and a layer of annular reinforcement ring. The thickness of the reinforcement ring is x ,like Fig.10 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.10 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:

[0247]

[0248] Find: x =1.73mm.

[0249] Therefore, the thickness of the reinforcement ring should not be less than 1.73 mm.

[0250] In other cases, the thickness of the inner tube 2 is limited. For example, the thickness of the inner tube 2 is generally 0.1-2 mm. In this case, the minimum thickness of the socket 13 of the outer tube 1 is calculated according to formula (25). When the tube is produced, the thickness of the socket 13 of the outer tube 1 cannot be less than this value.

[0251] Based on the above, the design calculation of the minimum thickness of the composite socket is completed.

[0252] The design method of the thickness of the composite socket is as follows:

[0253] 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.12 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.

[0254] 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.

[0255] 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.

[0256] 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.

[0257] 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:

[0258] 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.

[0259] 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.

[0260] 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 7, the coverage range of the non-base layer F2 is generally from the fourth step 114 to the sixth step 116. Fig.11 At this time, 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, as shown in FIG. Fig.11 shown.

[0261] 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.

[0262] 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.

[0263] For a single-layer structure of a composite socket, the design method comprises the following steps:

[0264] 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.

[0265] 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.

[0266] 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.

[0267] 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:

[0268] (26)

[0269] 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:

[0270] (27)

[0271] in, α is 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.

[0272] In step b1, take the second cone and the reinforcement ring on the inner wall of the second cone, such as Fig.12 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.

[0273] exist Fig.12 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.

[0274] but:

[0275] (28)

[0276] 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:

[0277] (29)

[0278] Then we have:

[0279] (30)

[0280] 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:

[0281] (31)

[0282] Since there is a sealing ring 7 at the third chamfer R3, in order to achieve sealing, the pressure of the sealing ring 7 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 7 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:

[0283] (32)

[0284] That is, the minimum wall thickness at the third chamfer R3 is: .

[0285] 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:

[0286] (33)

[0287] Then we have:

[0288] (34)

[0289] Considering the pressure of the sealing ring, the wall thickness at the third chamfer R3 is:

[0290] (35)

[0291] 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.

[0292] For a multi-layer structure of a composite socket, the design method comprises the following steps:

[0293] 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.

[0294] 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.

[0295] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.

[0296] 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.

[0297] 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.11 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.11 and 12 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.

[0298] 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:

[0299] (36)

[0300] 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.

[0301] For a composite socket, assuming there are m layers of non-base layer F2 and one layer of base layer F1, then:

[0302] (37)

[0303] 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.

[0304] 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 The strain is the same as that of each non-base layer F2, that is, it satisfies:

[0305] (38)

[0306] Right now:

[0307] (39)

[0308] 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.

[0309] 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), (37) is rearranged to obtain:

[0310] (40)

[0311] Formula (40) can be rearranged to obtain the hoop stress of the nth non-base layer F2:

[0312] (41)

[0313] Step c1 is thus completed.

[0314] 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:

[0315] (42)

[0316] (43)

[0317] 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.

[0318] Combining formulas (41), (42), and (43), we get:

[0319] (44)

[0320] Formula (44) is the relationship between the thickness of any non-base layer F2 and the thickness of the base layer F1.

[0321] The application of formula (44) can be carried out from two aspects: first, there is no limitation on the thickness of the non-base layer F2; second, the thickness of the non-base layer F2 is limited.

[0322] 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.

[0323] 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:

[0324] (45)

[0325] 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.11 and 12In 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:

[0326] (46)

[0327] Substituting formula (43) and (46) into formula (45), we can obtain:

[0328] (47)

[0329] From formula (47), we get:

[0330] (48)

[0331] In this embodiment, S b Take the thickness of the base layer F1 at the first chamfer R1.

[0332] 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 S of the base layer F1 at the first chamfer R1 is given based on experience or other requirements. b1 After that, we can find S n The minimum value of .

[0333] 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.

[0334] For the base layer F1, the thickness S of the base layer F1 at the first chamfer R1 is 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.

[0335] For the thickness S of the base layer F1 at the second chamfer R2 b2 , can be given directly, or can be set to the thickness S of the base layer F1 at the first chamfer R1 b1 Equal, can also be obtained by using formula (44), specifically, formula (44) is rewritten as:

[0336] (49)

[0337] but:

[0338] (50)

[0339] 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:

[0340] (51)

[0341] The thickness S of the base layer F1 at the second chamfer R2 is obtained by formula (51): b2 .

[0342] 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 calculated. 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 F can be greater than the outer diameter at the third chamfer R3, and the specific value can be flexibly selected.

[0343] 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:

[0344] (52)

[0345] Formulas (38) and (39) are organized into the following form:

[0346] (53)

[0347] 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:

[0348] (54)

[0349] Formula (54) can be rearranged to obtain the hoop stress of the non-base layer F2:

[0350] (55)

[0351] In step c2, in this embodiment, formula (47) is in the following form:

[0352] (56)

[0353] The thickness of the non-base layer F2 is:

[0354] (57)

[0355] That is, the minimum thickness of the non-base layer F2 is: .

[0356] At this time, formula (51) is: (58)

[0357] A specific example is shown below.

[0358] 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.

[0359] 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.

[0360] 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.

[0361] 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.

[0362] 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 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:

[0363] (59)

[0364] 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.13 shown.

[0365] When the thickness of the non-base layer F2 is limited, for example, in this embodiment, the inner tube 2 of the corrosion-resistant material forms a liner ring on the socket 11, and the thickness of the liner ring 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 liner ring is sleeved meet the strength requirements, it is necessary to calculate and design the thickness of the base layer F1 under this premise.

[0366] Based on this, formula (44) is rewritten as:

[0367] (60)

[0368] 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.

[0369] 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:

[0370] (61)

[0371] 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:

[0372] (62)

[0373] 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.

[0374] A specific example is shown below.

[0375] 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 liner ring formed by a layer of inner tube 2, that is, non-base layer F2, is made of stainless steel. The parameters of the liner ring are as follows: E n =193 GPa, 304 stainless steel yield strength Q n =205 MPa, Q1=Q0=235 MPa, Sn=1 mm. Then, we can calculate from equations (60), (61), and (35) respectively:

[0376] S b1 ≥7.33 mm, S b2 ≥7.01 mm, S b3 ≥10.53 mm.

[0377] 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:

[0378] (63)

[0379] 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 metal lined multilayer composite pipe, characterized in that: The composite pipe comprises an inner pipe (2), an intermediate layer (3) and an outer pipe (1) which are sequentially sleeved from the inside to the outside, the outer pipe (1) comprising a socket (11), a pipe body (12) and a plug (13) made of a corrosion-resistant metal material which are sequentially connected, the inner pipe (2) is made of a corrosion-resistant metal, one end of the inner pipe (2) covers part of the inner wall of the socket (11), the middle part covers the inner wall of the pipe body (12), and the other end covers the inner wall of the plug (13) and the end face of the plug (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 inner wall of 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 deflection curve differential equation and the moment balance equation of the ring; 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 material of the middle layer (3) is hot melt adhesive or rubber or plastic, and the material of the socket (11) and the pipe body (12) is ductile iron or carbon steel.

3. The composite pipe according to claim 1, 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 a sealing ring (7), wherein the outer ring of the sealing ring (7) 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 (7) contacts the outer wall of the spigot (13) of the adjacent composite pipe.

4. The composite pipe according to claim 2, characterized in that: The thickness of the middle layer (3) is not less than 0.2 mm, and the thickness of the inner tube (2) is not less than 0.1 mm.

5. The composite pipe according to claim 2, characterized in that: A sealing gasket (4) is provided between the end surface of the socket (13) and the inner tube (2) covering the end surface.

6. The composite pipe according to claim 5, characterized in that: The sealing gasket (4) comprises a pressing edge (41) and a flange (42) which are integrally connected, the pressing edge (41) being located between the inner wall at one end of the socket (13) and the inner tube (2), and the flange (42) being located between the end surface of the socket (13) and the inner tube (2) covering the end surface.

7. The composite pipe according to claim 5, characterized in that: When the material of the middle layer (3) is rubber, at one end of the socket (13), the middle layer (3) exceeds the end of the socket (13), and the middle layer (3) exceeding the end of the socket (13) is turned up to form a sealing gasket (4).

8. 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 annulus into the second form of the differential equation by introducing coefficients to obtain the third form of the differential equation; Step a24: Obtain a calculation formula for the maximum radial deformation of the ring with coefficients according to the third form of the differential equation obtained in step a23.

9. 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.

10. 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.

11. 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).

12. The composite pipe according to claim 11, 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.

13. The composite pipe according to claim 12, 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, 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.

14. The composite pipe according to claim 13, 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 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 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): .

15. The composite pipe according to claim 14, 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).

16. A method for manufacturing a socket-and-spigot type heterogeneous material multilayer composite pipe, used for manufacturing the composite pipe according to claim 2, characterized in that: The steps include: Step S1: processing the inner wall and outer wall of the socket (11) and the pipe body (12), and manufacturing the inner pipe (2) and the spigot (13); Step S2: coaxially and fixedly connecting the socket (13) and the tube body (12); Step S3: arranging the inner tube (2), the middle layer (3) and the outer tube (1) in sequence from the inside to the outside; Step S4: pressing the inner tube (2) and the middle layer (3) onto the outer tube (1); Step S5: compounding; Step S6: Flanging and sealing one end of the socket (13) of the inner tube (2); When the material of the middle layer (3) is plastic or rubber, there is no step S5.

17. The manufacturing method according to claim 16, characterized in that: In step S4, the inner tube (2) is pressed tightly against the inner wall of the plug (13), the inner wall of the tube body (12), and the inner wall of the socket (11) by means of spin forming or hydraulic forming.

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