A kind of multi-layer pipe with socket-and-spigot type and corrosion-resistant metal lining

Abstract

The present invention discloses a multi-layered bell-and-spigot pipe with a corrosion-resistant metal lining, which belongs to the technical field of pipes, and comprises an inner pipe, an intermediate layer and an outer pipe which are sleeved in sequence from the inside to the outside, the inner pipe, the intermediate layer and the outer pipe are bonded to form the pipe or pressed to form the pipe, the outer pipe comprises a socket, a pipe body and a spigot which are connected in sequence, the outer pipe is made of a metal material, the inner pipe is made of a material with corrosion-resistant function, each end of the pipe is provided with a sealing layer and a lining ring made of a corrosion-resistant metal material, the lining ring at one end of the socket covers one end of the inner wall of the inner pipe, one end face of the inner pipe, one end face of the intermediate layer, and the end face of the pipe body, and the lining ring at one end of the spigot covers the inner wall of the other end of the inner pipe, the end face of the other end of the inner pipe, the end face of the other end of the intermediate layer, and the end face of the spigot. The present invention can obtain the minimum or near minimum thickness of the spigot that meets the deformation requirements and the minimum or near minimum thickness of the socket that meets the strength requirements of bell-and-spigot pipes of different pipe diameters through theoretical analysis and calculation.

Description

Technical Field

[0001] The invention belongs to the technical field of pipes, and in particular relates to a socket-and-spigot type multi-layer pipe with a corrosion-resistant metal lining. Background Art

[0002] Composite pipes are widely used pipes, often used in liquid transportation pipe network systems. They are generally of a socket-and-spigot structure, including an outer metal pipe and an inner plastic pipe, that is, the plastic pipe is used as an inner lining, one end of the metal pipe is a spigot, and the other end is a socket. During installation, multiple composite pipes are connected in sequence, and the spigot of a composite pipe is inserted into the socket of the adjacent composite pipe. However, the thermal expansion coefficient of the plastic pipe lining is quite different from that of the metal outer pipe. The inner and outer pipes will produce a certain amount of internal stress that separates them from each other due to temperature changes, which will cause the lining to delaminate. In addition, liquids can easily penetrate from the pipe ends, causing the outer pipe to rust and contaminate the transported liquid, and causing the end faces of each layer to peel off and delaminate. The inner layer may even be washed away under long-term scouring.

[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 pipe, and the pressure in the pipe network will drop rapidly, even forming negative pressure. In addition, the pipe will also be subject to some external pressure, for example, the pipe buried in the soil will be subject to the gravity of the soil above it. Under the above external pressure and internal negative pressure, the pipe will deform radially. If the deformation of the socket is too large, the socket will separate from the socket connected to it, causing leakage at the connection between the socket and the socket, causing the inside of the water pipe to be contaminated. Therefore, it is necessary to ensure that the deformation of the socket of the pipe does not exceed a certain range under the above circumstances. To this end, for the socket, the minimum socket thickness that meets the deformation requirements must be obtained. If the socket thickness of the selected pipe is less than the minimum socket thickness that meets the deformation requirements, measures need to be taken to make it meet the deformation requirements. For example, when the pipe is secondary processed or installed, at least one layer of annular reinforcement ring is sleeved on the socket to form a composite socket to enhance the ability to resist deformation. However, the material of the reinforcement ring is often different from that of the base pipe, because the reinforcement ring is in direct contact with the water in the pipe, it needs to be made of corrosion-resistant materials such as stainless steel, while the base pipe is generally made of relatively cheap materials such as ductile iron and Q235. Therefore, how to ensure that the above-mentioned composite socket meets the maximum deformation requirements while the thickness is minimum or close to minimum becomes a problem to be solved.

[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 various locations minimum or close to the minimum while meeting the use strength. At present, the socket of a general base pipe is made of only one material, such as ductile iron or Q235 steel. This socket is a single-layer socket. However, in order to improve the performance of the pipe, the socket can be made into a composite socket, that is, a structure including at least two coaxial layers of different materials that are sequentially sleeved from the inside to the outside, such as an inner layer of stainless steel and an outer layer of Q235 steel, which can save costs while improving the corrosion resistance of the pipe. Similarly, if the composite socket can meet the strength requirements and the thickness is minimum or close to the minimum, the cost can also be greatly reduced. For the composite socket, the inner wall of the base pipe socket can be completely covered by the lining or not completely covered by the lining. When the inner wall of the base pipe socket is not completely covered by the liner, the composite socket includes a single-layer structure and a multi-layer structure. The single-layer structure 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 inside the base pipe. In addition, for the single-layer socket, in order to improve its corrosion resistance, the secondary processing manufacturer or user generally sleeves a liner ring made of corrosion-resistant material inside the socket. Obviously, the liner ring can also be used as a reinforcement ring. At this time, the liner ring and the socket of the base pipe form a composite socket.

[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. 9 In the figure, X is the length of the chamfer at the end of the socket along the axial direction, Y is the height of the chamfer at the end of the socket (i.e. the radial dimension), DE is the outer diameter of the socket, and T6 is the length of the socket. 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 socket-type multi-layer pipe with a corrosion-resistant metal lining, wherein the 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 spigot which are sequentially connected, the outer pipe is made of metal material, and the inner pipe is made of a material with corrosion-resistant function, each end of the pipe is provided with a sealing layer and a lining ring of a corrosion-resistant metal material, and the minimum or near-minimum thickness of a single-layer spigot and / or a composite spigot that meets the deformation requirements and the minimum or near-minimum thickness of a single-layer socket and / or a composite socket that meets the strength requirements of socket-type pipes of different diameters can be obtained through theoretical analysis and calculation.

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

[0009] A multi-layer socket-and-spigot pipe with a corrosion-resistant metal lining comprises an inner pipe, an intermediate layer and an outer pipe which are sequentially sleeved from the inside to the outside, the inner wall of the socket is provided with steps and / or grooves, the inner pipe, the intermediate layer and the outer pipe are bonded to form the pipe or compressed to form the pipe, the outer pipe comprises a socket, a pipe body and a spigot which are sequentially connected, the outer pipe is made of a metal material, the inner pipe is made of a corrosion-resistant material, each end of the pipe is provided with a sealing layer and a liner ring made of a corrosion-resistant metal material, the liner ring at one end of the socket covers one end of the inner wall of the inner pipe, one end face of the inner pipe, one end face of the intermediate layer and the end face of the pipe body, the liner ring at one end of the spigot covers the inner wall of the other end of the inner pipe, the end face of the other end of the inner pipe, the end face of the other end of the intermediate layer and the end face of the spigot.

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

[0011] The material of the middle layer is hot melt adhesive, plastic or rubber.

[0012] The material of the sealing layer is rubber or hot melt adhesive. When the material of the sealing layer is hot melt adhesive, the sealing layer bonds the liner ring and the inner tube together.

[0013] The liner ring at one end of the socket also extends to completely or partially cover the inner wall of the socket.

[0014] The length of the portion of the liner ring at one end of the socket extending into the interior of the pipe is not less than the length of the socket.

[0015] The liner ring at one end of the socket is turned outward to cover the outer wall of the socket.

[0016] At the socket end, the end face of the inner tube, the end face of the middle layer and the end face of the pipe body are all flush; at the spigot end, the end face of the inner tube, the end face of the middle layer and the end face of the spigot are all flush.

[0017] At one end of the socket, the sealing layer covers the end face of the pipe body, the end face of the middle layer, the end face of the inner pipe and the inner wall of one end of the inner pipe. At one end of the spigot, the sealing layer covers the end face of the spigot, the end face of the middle layer, the end face of the inner pipe and the inner wall of the other end of the inner pipe. The lining ring compresses the sealing layer. When in use, the lining ring and the sealing layer at one end of the spigot form a full seal with the end face of the spigot, and the lining ring and the sealing layer at one end of the socket form a full seal with the end face of the pipe body.

[0018] The sealing layer includes a sealing layer press edge and a sealing layer flange, the sealing layer press edge is cylindrical, the sealing layer flange is annular, the lining ring includes a lining ring press edge and a lining ring flange, the lining ring press edge and the lining ring flange are a whole, the lining ring press edge is cylindrical, the lining ring press edge presses the sealing layer press edge, the lining ring flange presses the sealing layer flange, and the length of the lining ring press edge is not less than 20mm.

[0019] When the length of the portion of the lining ring at one end of the socket extending into the interior of the pipe is less than the length of the socket, the socket of the pipe is a single-layer socket, and the design method of the thickness of the single-layer socket includes the following steps:

[0020] 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;

[0021] 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;

[0022] 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.

[0023] When the length of the portion of the lining ring at one end of the socket extending into the interior of the pipe is not less than the length of the socket, the socket and the lining ring at one end of the socket form a composite socket, and the design method of the composite socket thickness also includes the following steps:

[0024] 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.

[0025] Step a5: Determine the minimum thickness of the reinforcement ring. If the thickness of the reinforcement ring is not less than the minimum thickness, the thickness of the composite socket is the minimum thickness that meets the deformation requirement.

[0026] Step a2 includes the following steps:

[0027] 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;

[0028] 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;

[0029] 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;

[0030] 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.

[0031] Step a3 includes the following steps:

[0032] 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;

[0033] Step a32: obtaining the maximum deformation of the reference pipe socket through simulation according to the parameters calculated in step a31;

[0034] 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;

[0035] 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.

[0036] In step a1, the radial deformation obtained is u The differential equation is: , its general solution is: ,

[0037] 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.

[0038] In step a2, it is assumed that the ring is subjected to at least two radial concentrated forces P symmetrically relative to the center of the circle, and the angle between two adjacent radial concentrated forces P is , the ring satisfies the following three boundary conditions:

[0039] First, in Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula: ;

[0040] Second, due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; that is, ;

[0041] Third, cut near point A, remove point A and the force P, and from the symmetry and equilibrium conditions, we know that the shear force on the section at the circular cut is , that is: ;

[0042] Combining the above three formulas with the general solution of the differential equation, we get the first form of the differential equation:

[0043] ;

[0044] Assuming that the annulus is subjected to only two radial concentrated forces P which are symmetrical relative to the center of the circle, then: , we get the second form of the differential equation:

[0045] ;

[0046] Introducing the coefficient m, we get the third form of the differential equation:

[0047] ;

[0048] Point of application of force P The point is the point of maximum deformation, and the change in the diameter of the ring is:

[0049] , where I = S 3 / 12, S is the thickness of the tube body.

[0050] 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.

[0051] When the liner ring at one end of the socket does not cover the inner wall of the socket, the socket of the pipe is a single-layer socket. When the liner ring at one end of the socket extends to completely or partially cover the inner wall of the socket, the socket and the liner ring at one end of the socket form a composite socket.

[0052] For the same 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, a fifth step, and a sixth step arranged in sequence. The chamfer at the connection between the fourth step and the fifth step is the first chamfer, the chamfer at the connection between the fifth step and the sixth step is the second chamfer, the chamfer at the connection between the second step and the third step is the third chamfer, the chamfer at the connection between the first step and the second step is the fourth chamfer, and the chamfer at the connection between the third step and the fourth step is the fifth chamfer.

[0053] The design method of single-layer socket thickness includes the following steps:

[0054] Step b1: Select multiple locations on the inner wall of the single-layer socket where stress is likely to be concentrated. The selection principle is: if the wall thickness of these locations meets the stress strength requirements, then the wall thickness of each location of the single-layer socket meets the strength requirements;

[0055] 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 socket 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 socket 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 socket 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.

[0056] In step b2, the minimum wall thickness of each selected point in step b1 is calculated by the stress calculation formula to obtain the outer wall of each selected point, and the single-layer socket is regarded as a pressure vessel including at least two coaxially connected pressure vessels with conical inner walls and at least one annular reinforcement ring installed on the inner wall of the pressure vessel. , ,in, is the hoop stress of the pipe body, is the hoop stress at the selected location, D0 is the outer diameter of the base pipe, D is the diameter of the selected location, p is the pressure of the fluid in the base pipe, S is the thickness of the base pipe, S0 is the wall thickness of the selected location, and α is the semi-cone angle of the cone at the selected location.

[0057] When the material of the single-layer socket is the same as that of the pipe body, the stress calculation formula is: .

[0058] When the material of the single-layer socket is different from that of the pipe body, the stress calculation formula is: , where Q0 is the yield strength of the pipe body material and Q1 is the yield strength of the material used for the single-layer socket.

[0059] The design method of composite socket thickness includes the following steps:

[0060] 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:

[0061] 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;

[0062] Step c2: Determine the relationship between the thickness of any non-base layer and the thickness of the base layer, and the determination principle is: the ratio of the stress of the non-base layer 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 material of the pipe body;

[0063] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.

[0064] 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.

[0065] 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:

[0066] ;

[0067] 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;

[0068] 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.

[0069] 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: .

[0070] 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.

[0071] The beneficial effects of the present invention are:

[0072] (1) The inner tube made of corrosion-resistant metal material is used as the inner lining to completely isolate the outer tube from the liquid flowing in the tube, effectively preventing the outer tube from rusting, playing a good anti-corrosion role, increasing the service life of the tube, avoiding the rust of the outer tube from polluting the water quality, and improving the liquid transportation performance.

[0073] (2) Compared with plastic pipe linings, the inner pipe made of corrosion-resistant metal 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 flatness of the inner surface of the pipe, and has a smaller lining thickness, which improves the water delivery efficiency of the pipeline and reduces operating energy consumption.

[0074] (3) The inner pipe made of corrosion-resistant metal material is used as the inner lining. Compared with the plastic pipe lining, the pipe can be used to transport high-temperature liquid media.

[0075] (4) The middle layer is made of insulating hot melt adhesive, plastic or rubber, which completely separates the inner tube from the outer tube to prevent the outer tube from contacting the inner tube and causing electrochemical corrosion to the inner tube. At the same time, it eliminates the uneven glue layer caused by the flow of ordinary liquid glue in the middle layer, and the electrochemical corrosion caused by the direct contact between the inner lining corrosion-resistant metal and the outer metal tube without bonding. It also overcomes the problems of easy aging and poor insulation of ordinary glue.

[0076] (5) When the middle layer is hot melt adhesive, the hot melt adhesive will bond the inner and outer layers together to prevent the inner lining from being delaminated due to negative pressure in the pipeline. When the middle layer is rubber, the elasticity of the rubber will cause the metal outer tube and the corrosion-resistant metal inner tube to form a mutual elastic force, which has good sealing performance for the inner wall of the metal tube and can prevent the liquid from contacting the inner wall of the metal outer tube and causing corrosion of the tube wall.

[0077] (6) A corrosion-resistant metal liner ring and a sealing layer are provided at the end of the pipe to prevent the bonding surface ends between the layers of materials from contacting water and peeling off due to the tension of water, thereby preventing the inner wall of the metal outer pipe from rusting and peeling off and causing water pollution; the liner ring and the sealing layer fix the inner pipe, and when the seal fails and the inner liner is completely delaminated, the inner liner can still be protected from being washed away by the scouring of water flow; the provision of the liner ring and the sealing layer enhances the ring stiffness of the pipe end, the plug is not easily deformed, and the installation of the pipe is facilitated, and the uniform sealing effect of the sealing layer and the sealing ring at the connection between the two pipes is improved, thereby reducing pipeline leakage. The sealing layer is made of rubber or hot melt adhesive, which improves the sealing performance of the pipe end face. When the sealing layer is made of hot melt adhesive, the sealing layer bonds the liner ring and the inner pipe together, further improving the installation stability of the liner ring.

[0078] (7) At one end of the socket, the length of the portion of the liner ring extending into the pipe can be set to be no less than the length of the socket, or the liner ring can be turned outward to cover the outer wall of the socket, so that the socket and the metal liner ring form a composite socket, which improves the strength and deformation resistance of one end of the pipe socket; at the socket end, the liner ring can be extended to partially cover or completely cover the inner wall of the socket, so that the socket and the metal liner ring form a composite socket, which improves the strength of one end of the pipe socket.

[0079] (8) The minimum or near minimum thickness of single-layer spigots and / or composite spigots that meet the deformation requirements and the minimum or near minimum thickness of single-layer sockets and / or composite sockets that meet 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, provides favorable support for experiments and pipe design, and provides a reference for secondary processing and installation. After the thickness of the base pipe spigot is known, the minimum thickness of the reinforcement ring of the composite spigot that meets the deformation requirements can be calculated. The pipes and pipe network systems installed in this way are more reliable to 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

[0080] Figure 1 It is a schematic diagram of the structure of the pipe described in Example 1 of the present invention.

[0081] Figure 2 yes Figure 1 An enlarged schematic diagram of point A.

[0082] Figure 3 It is a schematic diagram of connecting two adjacent pipes according to an embodiment of the present invention.

[0083] Figure 4 Schematic diagram of the cross-section before and after deformation when the socket is subjected to a pair of radially symmetrical concentrated forces.

[0084] Figure 5 for Figure 4 Schematic diagram of local analysis.

[0085] Figure 6 Schematic diagram of a circular ring subjected to multiple uniformly arranged radial concentrated forces.

[0086] Figure 7 This is the stress analysis diagram of socket-type pipe.

[0087] Figure 8 It is a schematic cross-sectional view of a composite socket per unit length of the present invention.

[0088] Fig. 9 Schematic diagram of socket and spigot connection.

[0089] Fig.10 It is a schematic diagram of the structure of the pipe according to the second embodiment of the present invention.

[0090] Fig.11 It is a schematic diagram of the structure of the pipe according to the third embodiment of the present invention.

[0091] Fig.12It is a schematic structural diagram of the pipe socket inner wall not covered by a liner ring in the fourth embodiment of the present invention.

[0092] Fig.13 It is a structural schematic diagram of the fourth embodiment of the present invention in which the inner wall of the pipe socket is not completely covered by the liner ring and there is no intermediate layer at the socket end.

[0093] Fig.14 It is a structural schematic diagram of the fourth embodiment of the present invention in which the inner wall of the pipe socket is completely covered by a liner ring and there is no intermediate layer at the socket end.

[0094] Fig.15 It is a schematic diagram of the structure in which the inner wall of the pipe socket of the fifth embodiment of the present invention is not covered by a liner ring.

[0095] Fig.16 It is a structural schematic diagram of the fifth embodiment of the present invention in which the inner wall of the pipe socket is not completely covered by the liner ring and there is no intermediate layer at the socket end.

[0096] Fig.17 It is a structural schematic diagram of the fifth embodiment of the present invention in which the inner wall of the pipe socket is completely covered by a liner ring and there is no intermediate layer at the socket end. DETAILED DESCRIPTION

[0097] 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.

[0098] 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" may include both "above" and "below". The device may also be positioned in other different ways (rotated 90 degrees or in other orientations), and the spatially relative descriptions used here are interpreted accordingly.

[0099] Embodiment 1

[0100] A multi-layered socket-and-spigot pipe with a corrosion-resistant metal lining, the pipe is mainly used for passing water and corrosive liquids. Figure 1 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.

[0101] The outer tube 1 is made of metal, preferably ductile iron or carbon steel. The outer tube 1 includes a socket 11, a pipe body 12 and a spigot 13 connected in sequence along its axial direction. The three are integrally cast, or the pipe body 12 and the spigot 13 are integrally formed, and the pipe body 12 and the socket 11 are welded. The spigot 13 is cylindrical, and the spigot 13 is chamfered along the circumferential direction. The inner diameter, outer diameter and thickness of the spigot 13 can be equal to or unequal to the inner diameter, outer diameter and thickness of the pipe body 12.

[0102] The inner tube 2 is made of a metal material with corrosion resistance, including but not limited to stainless steel and titanium, preferably stainless steel, and the inner tube 2 is cylindrical.

[0103] 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; when the material of the middle layer 3 is plastic or rubber, the plastic or rubber does not have adhesive properties, and the inner tube 2 presses the middle layer 3 against the inner wall of the outer tube 1.

[0104] The length of the inner tube 2 is equal to the length of the middle layer 3, which is equal to the sum of the lengths of the pipe body 12 and the socket 13. At the end of the socket 11, the ends of the inner tube 2, the middle layer 3 and the pipe body 12 are flush; at the end of the socket 13, the ends of the inner tube 2, the middle layer 3 and the socket 13 are flush.

[0105] The thickness of the intermediate layer 3 is not less than 0.2 mm, and the thickness of the inner tube 2 is not less than 0.1 mm.

[0106] 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.

[0107] 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.

[0108] 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 pipe is larger and the transportation efficiency is higher.

[0109] 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:

[0110] 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;

[0111] 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;

[0112] 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.

[0113] 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.

[0114] 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.

[0115] A lining ring 5 and a sealing layer 4 are provided at one end of the socket 13 of the pipe, and a lining ring 5 and a sealing layer 4 are also provided at one end of the socket 11. The lining ring 5 is made of corrosion-resistant metal, preferably stainless steel, and the sealing layer 4 is made of rubber. The sealing layer 4 at one end of the socket 13 is used to cover and compress the end face and inner wall of the socket 13 of the pipe, and the lining ring 5 is pressed on the sealing layer 4. The sealing layer 4 at one end of the socket 11 is used to cover and compress the end face and inner wall of the pipe body 12 of the pipe near the socket 11, and the lining ring 5 is pressed on the sealing layer 4.

[0116] like Figure 1 and 2 As shown, the sealing layer 4 includes a sealing layer edge 41 and a sealing layer flange 42, which are connected as one piece. The sealing layer edge 41 can be regarded as a thin cylindrical shape, and the sealing layer flange 42 is annular. The inner diameter of the sealing layer flange 42 is equal to the inner diameter of the sealing layer edge 41, and the outer diameter of the sealing layer flange 42 is greater than the outer diameter of the sealing layer edge 41. The sealing layer edge 41 and the sealing layer flange 42 share a common center line.

[0117] The structure of the liner ring 5 is similar to that of the sealing layer 4. The liner ring 5 includes an integrally connected liner ring pressure edge 51 and a liner ring flange 52. The liner ring pressure edge 51 can be regarded as a thin cylindrical shape, and the liner ring flange 52 is annular. The inner diameter of the liner ring flange 52 is equal to the inner diameter of the liner ring pressure edge 51, and the outer diameter of the liner ring flange 52 is greater than the outer diameter of the liner ring pressure edge 51. The liner ring pressure edge 51 and the liner ring flange 52 share a common center line.

[0118] In this embodiment, the length of the liner edge 51 at one end of the socket 13 is not less than the length of the socket 13. In other words, the length of the liner 5 at one end of the socket 13 extending into the pipe is not less than the length of the socket 13, so as to ensure that the socket 13 and the liner edge 51 form a composite socket. There is no special requirement for the length of the liner edge 51 at one end of the socket 11, which can generally be no less than 20 mm.

[0119] Figure 2 In the illustrated view, the upper end surface of the sealing layer flange 42 does not extend beyond the upper end surface of the backing ring flange 52 .

[0120] At one end of the socket 11, the outer surface of the sealing layer edge 41 of the sealing layer 4 fits the inner surface of the inner tube 2, and the inner surface of the sealing layer edge 41 fits the outer surface of the liner ring 5. One side of the sealing layer flange 42 fits the end face of the inner tube 2, the end face of the intermediate layer 3 and the end face of the pipe body 12, and the other side of the sealing layer flange 42 fits the liner ring 5. The liner ring edge 51 of the liner ring 5 fits and compresses the sealing layer edge 41, and the liner ring flange 52 fits and compresses the sealing layer flange 42.

[0121] At one end of the socket 13, the outer surface of the sealing layer edge 41 of the sealing layer 4 fits the inner surface of the inner tube 2, and the inner surface of the sealing layer edge 41 fits the outer surface of the liner ring 5. One side of the sealing layer flange 42 fits the end surface of the inner tube 2, the end surface of the intermediate layer 3 and the end surface of the socket 13, and the other side of the sealing layer flange 42 fits the liner ring 5. The liner ring edge 51 of the liner ring 5 fits and compresses the sealing layer edge 41, and the liner ring flange 52 fits and compresses the sealing layer flange 42.

[0122] A method for manufacturing a multi-layer bell-and-spigot pipe with a corrosion-resistant metal lining, for manufacturing the pipe, comprises the following steps:

[0123] Step S1: Process the inner wall and outer wall of the outer tube 1, make the inner tube 2, and make the liner ring 5.

[0124] In this step, the method for processing the inner and outer walls of the outer tube 1 is to first grind the inner wall of the outer tube 1, and then sandblast or shot blast the inner and outer walls. Use an internal grinder to grind the inner wall of the outer tube 1 made of metal, grind the inner wall of the tube flat, and remove surface oxide scale, scum and rust. Alternatively, after the inner wall of the outer tube 1 is polished flat, use a laser cleaning machine to clean the inner wall of the tube to deeply remove the oxide scale and rust on the inner wall of the tube. When sandblasting or shot blasting, remove dust, oxide scale, and rust on the inner and outer walls, and sandblast the inner and outer walls of the tube into a frosted surface.

[0125] Making the inner tube 2 comprises the following steps:

[0126] Step S11: Process the stainless steel sheet into a ring shape or a thin cylindrical shape to complete the rolling. In this step, the rolling is performed by a rolling machine.

[0127] Step S12: Cut the rolled stainless steel sheet and weld it into a cylindrical shape to obtain the tube blank of the inner tube 2. In this step, if a single roll is obtained in step S11, a straight seam welding is performed using a argon arc welder or a laser welder, and the weld extends from one end to the other end of the inner tube 2. 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 cylindrical shape.

[0128] 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 pipe material 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 tube body 12 and the length of the socket 13.

[0129] When making the liner ring 5, two sections can be cut from the stainless steel round tube obtained in step S12 as the tube blank of the liner ring 5. Alternatively, the liner ring 5 can be made separately. The method for making the liner ring 5 is as follows:

[0130] Step S11': using a flanging and rolling machine to process the stainless steel sheet into a rolled shape with a flanging, that is, into a structure with a liner ring pressing edge 51 and a liner ring flanging edge 52. In this step, the rolling is performed by a rolling machine.

[0131] Step S12 ′: using an argon arc welder or a laser welder to weld the rolled incision obtained in step S11 ′ to form the shape of the liner ring 5 .

[0132] Step S13': putting the sealing layer 4 on the outer wall of the liner ring 5 obtained in step S12'.

[0133] Step S2: Arrange the inner tube 2, the middle layer 3 and the outer tube 1 in sequence from the inside to the outside.

[0134] In this step, when the material of the middle layer 3 is hot melt adhesive or plastic, step S2 can be completed in the following three ways:

[0135] The first way is as follows:

[0136] 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.

[0137] 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.

[0138] 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 tube insertion can be completed manually or by a tube insertion machine.

[0139] The second way is as follows:

[0140] First, after the inner wall of the outer tube 1 is 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 wall of the tube body 12 and the inner wall of the socket 13 of the outer tube 1. The thickness of the roll or spray coating is greater than 0.2 mm.

[0141] Then, the inner tube 2 is inserted into the outer tube 1 with the middle layer 3 on the inner wall.

[0142] The third way is as follows:

[0143] First, after heating the outer wall of the inner tube 2, a powdered hot melt adhesive or plastic is roto-molded or spray-molded on the outer wall of the stainless steel inner tube 2 using a roller coating machine or a plastic coating machine. The thickness of the roto-molding or spray-molding is greater than 0.2 mm. Then, the inner tube 2 with the intermediate layer 3 on the outer wall is inserted into the outer tube 1.

[0144] When the material of the middle layer 3 is rubber, the rubber sheet is first rolled onto the outer wall of the inner tube 2, and then the rubber seams are bonded with adhesive, and then the inner tube 2 with the rubber sleeve is inserted into the outer tube 1. Alternatively, the rubber tube is sleeved onto the outer wall of the inner tube 2 and then the entire tube is sleeved into the outer tube 1. The length of the rubber should be sufficient to cover the inner wall of the tube body 12 and the inner wall of the socket 13.

[0145] Step S3: Press the inner tube 2 and the middle layer 3 onto the outer tube 1 .

[0146] This step can be completed by spin forming or hydraulic forming. In spin forming, the inner tube 2 is spin-formed by a tube spinning machine so that the inner tube 2, the middle layer 3, and the outer tube 1 are closely attached in sequence. In hydraulic forming, the two ends of the inner tube 2 made of stainless steel that penetrates the outer tube 1 are sealed, and water is injected into the tube to increase pressure so that the inner tube 2, the middle layer 3, and the outer tube 1 are closely attached in sequence.

[0147] Step S4: Compounding.

[0148] In this step, the pipe is heated by a medium frequency inductor or a heating furnace, the hot melt adhesive is melted, and after cooling, the outer pipe 1 and the inner pipe 2 are bonded together.

[0149] When the material of the middle layer 3 is plastic or rubber, there is no step S4.

[0150] Step S5: Backing ring.

[0151] In this step, first, two liner rings 5 ​​with sealing layers 4 are manually placed at both ends of the pipe. Then, the liner rings 5 ​​are expanded by a diameter expander so that the liner rings 5 ​​press the sealing layers 4 against the inner wall and end surface of the pipe.

[0152] If the liner ring 5 tube blank is two sections cut from the stainless steel round tube obtained in step S12, then in this step, after the liner ring 5 with the sealing layer 4 is inserted into the tube end, it is also flanged to form a liner ring flange 52.

[0153] After completing the above steps, remove excess burrs and apply paint, plastic coating or other methods to the outer wall of the pipe for external anti-corrosion treatment.

[0154] In this embodiment, the socket 11 of the outer pipe is made of a single material and can be regarded as a single-layer socket.

[0155] The socket 13 of the outer tube 1, the portion of the inner tube 2 covering the socket 13, and the lining ring edge 51 can all improve the strength and deformation resistance of the corresponding parts, so the three form a composite socket of the pipe, which has a three-layer structure. For the socket, because the middle layer 3 is made of non-metallic flexible materials such as hot melt adhesive, rubber, plastic, etc., and the sealing layer 4 is made of rubber, it has basically no effect on the strength and deformation resistance of the socket 13, so it does not belong to the category of composite sockets that can resist deformation.

[0156] Next, the thickness of the spigot end and the socket end of the composite pipe is optimized and designed.

[0157] The design method of the thickness of the composite socket includes the following steps:

[0158] 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 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.

[0159] 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.

[0160] 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.

[0161] 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.

[0162] 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.

[0163] The following is a detailed introduction to each step.

[0164] In step a1, the radial deformation obtained The differential equation is:

[0165] (1)

[0166] in ，R is half of the sum of the outer radius and the inner radius of the ring. In other words, R One quarter of the sum of the outer and inner diameters of the ring 。E is the elastic modulus of the pipe, Pa. I is the moment of inertia of the tube cross section, m 4 . M It is the bending moment borne by a certain section of the ring in the local area, in N*m. Establish a two-dimensional coordinate system, φ The deformation is u The line connecting the point at and the origin of the coordinate system is the same as the line connecting the point at and the origin of the coordinate system in the two-dimensional coordinate system. x The angle between the positive directions of the axes, In this embodiment, establish xy Coordinate system, point O is the coordinate origin and the center of the ring, such as Figure 5 Point m 1 The radial deformation at u， but om 1 and x The angle between the positive directions of the axes is φ.

[0167] The above differential equation (1) is obtained as follows:

[0168] like Figure 4 and 5 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 4 midpoint O Too xy Vertex of the coordinate system ， point m and n are two points on the ring before deformation; m 1 and n 1 are two points on the deformed shape of the ring, and the points m 1 and n 1 Located on the line segment om and on superior, φ The deformation is u The line connecting the point at and point O is x The angle between the positive directions of the axes, om and on The angle between φ, set up u Centripetal movement is positive. Figure 5 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.

[0169] Arc before deformation mn The arc length is ds , whose curvature isk 1 , arc after deformation m 1 n 1 The curvature is k 2 , du Yes m and Point n The difference in radial deformation is The arc before deformation mn exist m Tangent line of point and n The angle between the tangent lines of a point, The deformation amount is u The radius corresponding to the point at after deformation is Is an arc mn Deform to Arc m 1 n 1 back ds The increment of The arc after deformation m 1 n 1 exist m 1 Tangent line of point and n 1 The angle between the tangent lines of the point is m 1 t 1 and The angle between , then:

[0170] (2)

[0171] (3)

[0172] (4)

[0173] (5)

[0174] m 1 t 1 and m 1 t The angle between , n 1 t 1 ’ and n 1 t’ The angle between .

[0175] m 1 section to n 1 The variable of the cross-section rotation is :

[0176] (6)

[0177] in, m The cross section is through om and perpendicular to the cross section of the ring, in other words, m The cross section is through the center axis and point of the corresponding pipe of the ring m The cross section of n The cross section is through on and perpendicular to the cross section of the ring, in other words, n The cross section is through the center axis and point of the corresponding pipe of the ring n cross section. m 1 t 1 and m 1 t The angle between m 1 The cross-sectional angle, n 1 t 1 ’ and n 1 t’ The angle between n 1 The corner of the cross section.

[0178] Substituting formula (5) and (6) into formula (4) yields:

[0179] (7)

[0180] By series expansion we have:

[0181] Ignoring the higher-order terms, substituting the series expansion into (7) yields: (8)

[0182] Substituting (2) into (8), we obtain: (9)

[0183] Omitting the higher-order terms, we can get: (10)

[0184] 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).

[0185] The general solution of the above differential equation (1) is:

[0186] (11)

[0187] 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:

[0188] (12)

[0189] 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:

[0190] 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.

[0191] 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.

[0192] 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.

[0193] 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.

[0194] 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 6 shown. Figure 4 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.

[0195] The ring satisfies the following three boundary conditions:

[0196] First, in Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula:

[0197] (13)

[0198] Second, due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; that is,

[0199] (14)

[0200] Third, cut near point A, remove point A and the force P, and from the symmetry and equilibrium conditions, we know that the shear force on the section at the circular cut is , that is:

[0201] (15)

[0202] in, dM represents the differential of M. It should be noted that the third point is not a real cut of the ring, but an assumption.

[0203] Simultaneously formula (13), (14), (15), the solution is:

[0204] (16)

[0205] (17)

[0206] (18)

[0207] Substituting formulas (16), (17), and (18) into formula (12), we obtain:

[0208] (19)

[0209] Formula (19) is the first form of the differential equation obtained in step a21. is the radial deformation of the ring.

[0210] In step a22, if only one pair of radial forces P is applied to the ring, then , the deflection curve equation (19) becomes:

[0211] (20)

[0212] Formula (20) is the second form of the differential equation obtained in step a22.

[0213] 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:

[0214] (twenty one)

[0215] 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.

[0216] Formula (21) is the third form of the differential equation obtained in step a23.

[0217] 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 :

[0218] (twenty two)

[0219] Formula (22) is the third form of the differential equation obtained in step a24.

[0220] 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:

[0221] 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.

[0222] Step a32: Obtain the maximum deformation of the reference pipe socket through simulation based on the parameters calculated in step a31.

[0223] 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.

[0224] 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.

[0225] 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:

[0226] (twenty three)

[0227] like Figure 7 As shown, W is the shear load, in Newton (N); F is the shear force acting on the interface, in Newton (N); M Pis 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).

[0228] In this embodiment, the nominal diameter of the base pipe is DN=600mm, and the material is Q235 steel. The nominal diameter of the reference pipe is DN=600mm, and the material is Q235. Its wall thickness is selected according to the provisions of the group standard, and the wall thickness when the nominal pressure is 3.0MPa, that is, the nominal wall thickness S=7mm. At this time, the outer diameter of the base pipe D0=630mm, R =312mm. In addition, the shear force F It should be no less than 30 times of the nominal diameter DN. In this embodiment, 30 times is taken, that is, F = DN × 30 = 600 × 30 = 18000 N. Take the pipe length as 1m, c = 800mm.

[0229] In this embodiment, the parameters in formula (23) are M P and W The calculation process is as follows:

[0230]

[0231]

[0232] 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.

[0233] Step a31 is thus completed.

[0234] 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.

[0235] 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:

[0236] (twenty four)

[0237] From formula (24), we can see that u max The value is related to the pipe diameter and pipe thickness.

[0238] 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.

[0239] Table 1 Nominal wall thickness and absolute deformation of DN=600mm pipes at various nominal pressures

[0240]

[0241] 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).

[0242] 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.

[0243] If the use requirement of the pipe in a certain occasion is to meet the nominal pressure within 2MPa, the general user will choose the pipe with serial number 4 in Table 1, and will not choose the pipe with serial number 1, 2 or 3 to meet the deformation requirement, because if one of these three is selected, although it will meet both the strength requirement and the deformation requirement, their thickness is larger and the cost will be much greater than the pipe with serial number 4. If the pipe with serial number 4 is selected, its strength meets the use requirement. For the deformation requirement, some measures can be taken, such as sleeve-fitting an annular reinforcement ring in the socket to increase the thickness of the socket and improve its ability to resist deformation, and this measure is low in cost.

[0244] 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.

[0245] 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.

[0246] 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.

[0247] In this embodiment, the calculation formula for the bending strength of the longitudinal composite beam is as follows:

[0248] (25)

[0249] 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.

[0250] For example, 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, whose wall thickness is 4mm. The composite spigot includes the base pipe spigot and two layers of annular reinforcement rings (i.e., the part of the inner pipe 2 covering the spigot 13 and the liner ring pressure edge 51). Since the inner pipe 2 and the liner ring pressure edge 51 are made of the same material, these two layers of annular reinforcement rings can be regarded as one layer. Assume that the thickness of the reinforcement ring is x ,like Figure 8 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. Figure 8 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:

[0251]

[0252] Find: x =1.73mm.

[0253] Therefore, the total thickness of the reinforcing ring is not less than 1.73 mm, that is, the total thickness of the inner tube 2 and the liner ring edge 51 is not less than 1.73 mm, and the specific thickness of the inner tube 2 and the liner ring edge 51 can be selected according to actual requirements. For example, if 1 mm thick stainless steel on the market is used as the liner ring material, that is, the thickness of the liner ring edge 51 is 1 mm, then the thickness of the inner tube 2 is not less than 0.73 mm, which falls within the thickness range of the inner tube 2 of the present solution (0.1 mm-2 mm).

[0254] If the materials of the socket 13, the inner tube 2 and the liner ring edge 51 are different, then there are two unknowns in this case, namely the thickness of the inner tube 2 and the liner ring edge 51, and there is only one equation (25). Therefore, it is necessary to limit the thickness of one layer according to the actual situation in order to calculate the thickness of the other. For example, if the minimum thickness of the inner tube 2 has been determined through a peel strength test, the thickness of the liner ring edge 51 can be calculated based on the thickness of the socket 13 and the inner tube 2 according to formula (25).

[0255] For another example, due to process or other requirements, the total thickness of the inner tube 2 and the liner ring edge 51 is known, that is, the total thickness of the reinforcement ring is known, and the required thickness of the original socket can be calculated by formula (25).

[0256] The above is the design method of the socket thickness.

[0257] The design method of the thickness of the single-layer socket is as follows:

[0258] The inner wall shape of the single-layer socket of the pipe material 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 / CISA 108-2021) and GB / T 13295-2008, such as Fig. 9 As shown. The inner wall of the single-layer socket is stepped. For the same base pipe, along the direction from the single-layer socket to the spigot, the inner wall of the single-layer 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 12 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.

[0259] Since the dimensions of the inner wall of the single-layer socket are fully given by the relevant standards, it is necessary to design some key dimensions of the outer wall on this basis, aiming to save material costs and meet strength requirements at the same time. In other words, the thickness of the single-layer socket is designed to be the minimum or close to the minimum while meeting the use strength requirements, so as to reduce the production, manufacturing and transportation costs of the pipe.

[0260] When the two pipes are connected, the socket 13 of one pipe is inserted into the socket 11 of the adjacent pipe and sealed by a sealing ring. In this embodiment, the inner wall of the fourth step 114 is a cylindrical wall. The inner wall of the fifth step 115 is the outer side of a truncated cone, that is, from the fourth step 114 toward the direction of the pipe body 12, the inner diameter of the fifth step 115 gradually increases, which is equivalent to a trumpet. 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 larger than the inner diameter of the pipe body 12. In this way, the sixth step 116 is formed at the end of the pipe body 12 close to the fifth step 115. The inner wall bell mouth of the fifth step 115 is provided to provide a deflection angle for the adjacent pipes connected thereto, i.e., to allow the center line of a pipe and the center line of the adjacent 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.

[0261] The design method of single-layer socket includes the following steps:

[0262] Step b1: Select multiple locations on the inner wall of the single-layer socket where stress is likely to concentrate. The selection principle is: if the wall thickness of these locations meets the stress strength requirements, then the wall thickness of each location of the single-layer socket meets the strength requirements.

[0263] Step b2: Determine the minimum wall thickness of each selected location in step b1, and obtain the outer wall of each selected location. The determination principle is: when the material of the single-layer socket 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 socket 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 socket 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.

[0264] Step b3: Connect the outer walls of the selected locations in sequence to obtain the outer wall shape and thickness of the single-layer socket.

[0265] In step b1, based on material mechanics knowledge and experience, several locations on the inner wall of the single-layer socket where stress is most likely to be concentrated are selected, namely the first chamfer R1, the second chamfer R2, and the third chamfer R3. The analysis is as follows:

[0266] For the inner wall of the stepped single-layer socket, according to mechanical knowledge, the stress concentration point is located at the connection between adjacent steps, that is, at each chamfer. Since the single-layer socket is in a trumpet shape, the outer diameter of the single-layer socket decreases from the single-layer socket of the same base pipe to the spigot.

[0267] Based on the above premise, 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 single-layer socket is shaped like a trumpet. Therefore, it is equivalent to the wall thickness at the fourth chamfer R4 being greater than the third chamfer R3. At this time, 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 single-layer socket is shaped like a trumpet. 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 stress concentration points finally selected are the first chamfer R1, the second chamfer R2, and the third chamfer R3. In other words, the analysis area of ​​the single-layer socket is the area between the third chamfer R3 and the second chamfer R2.

[0268] In step b2, the minimum wall thickness of each location selected in step b1 is calculated by the stress calculation formula. The specific method is as follows: the analysis area of ​​the single-layer socket is regarded as consisting of at least two coaxial pressure vessels with conical inner walls connected end to end, plus several internal reinforcement rings, and the reinforcement rings are connected to the inner wall of the pressure vessel. In other words, the analysis area of ​​the single-layer socket is regarded as consisting of at least two conical pressure vessels plus several internal reinforcement rings. At this time, as long as the corresponding conical pressure vessel meets the strength requirements, the single-layer socket must meet the strength requirements, such as Fig. 9 shown.

[0269] For a thin-walled cylindrical pressure vessel, its wall thickness is S, its diameter is D0, and its internal pressure is p. From material mechanics, we can get the axial stress σ of the pressure vessel wall: 轴 and hoop stress σ 环 They are as follows:

[0270] (26)

[0271] (27)

[0272] The hoop stress σ 环 That is, the circumferential stress. Wherein, D0 is 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.

[0273] Similarly, for a thin-walled conical pressure vessel, when the wall thickness is S and the internal pressure is p, the stress σ in the axial direction of the vessel wall at the diameter D is 轴 and hoop stress σ 环 They are:

[0274] (28)

[0275] (29)

[0276] in α It is the semi-cone angle of the conical pressure vessel, 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. Wherein, 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.

[0277] From the above, in this embodiment, the analysis area of ​​the single-layer socket includes two coaxially connected conical pressure vessels plus several internal reinforcement rings. The two coaxially connected conical pressure vessels are respectively the first cone and the second cone, which share the same central axis with the single-layer socket, wherein the first chamfer R1 and the second chamfer R2 are located on the inner wall of the first cone, such as Fig. 9 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 third chamfer R3 are located on the inner wall of the second cone. Fig. 9 The dotted line between the first chamfer R1 and the third chamfer R3 is the inner wall section of the second cone. For the area between the first chamfer R1 and the second chamfer R2, the portion between the inner wall of the first cone and the outer wall of the single-layer socket 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 single-layer socket is an annular reinforcement ring. Similarly, 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 single-layer socket 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 single-layer socket is another annular reinforcement ring.

[0278] exist Fig. 9 In the cross section shown, the angle between the straight line passing through the first chamfer R1 and the second chamfer R2 and the axis of the tube is the semi-cone angle of the first cone, set as α1. 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 semi-cone angle of the second cone, set as α2.

[0279] but:

[0280] (30)

[0281] (31)

[0282] When the pipe body meets the strength requirements and the material of the single-layer socket is consistent with that of the pipe body, in order to make the single-layer socket meet the strength requirements, the hoop stress at the first chamfer R1 is Not greater than the hoop stress of the pipe body , hoop stress at the second chamfer R2 Not greater than the hoop stress of the pipe body , Hoop stress at the third chamfer R3 Not greater than the hoop stress of the pipe body , that is:

[0283] (32)

[0284] (33)

[0285] (34)

[0286] Then we have:

[0287] (35)

[0288] (36)

[0289] (37)

[0290] Among them, S b1 , S b2 , S b3 are the wall thickness at the first chamfer R1, the second chamfer R2, and the third chamfer R3, respectively, and S is the wall thickness of the tube body, and we get:

[0291] (38)

[0292] (39)

[0293] (40)

[0294] Since there is a sealing ring z at the third chamfer R3, in order to achieve sealing, the pressure of the sealing ring z on the single-layer socket must be greater than the pressure p of the fluid in the pipe. According to existing research, the average pressure of the sealing ring z on the inner wall of the single-layer socket is generally 1.4 times the fluid pressure. It can be obtained from formulas (28) and (29) 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 (40) is corrected to:

[0295] (41)

[0296] So we have:

[0297] (42)

[0298] (43)

[0299] (44)

[0300] Among them, S 1m , S 2m , S 3m They are the minimum wall thickness at the first chamfer R1, the second chamfer R2, and the third chamfer R3 respectively.

[0301] From above, Fig. 9 As shown, the minimum outer diameter at the second chamfer R2 d 1. Minimum outer diameter at the first chamfer R1 d 2 , the minimum outer diameter at the third chamfer R3 d 3 are as follows:

[0302] (45)

[0303] (46)

[0304] (47)

[0305] The following example is used to calculate the wall thickness parameters of a single-layer socket.

[0306] In this embodiment, the material of the pipe body and the single-layer socket is Q235.

[0307] Table 2 is the socket size parameter table for the nominal pipe diameters of 600mm and 1200mm at a pressure of 3MPa as specified in the relevant standards:

[0308] Table 2 Dimensional parameters of sockets with nominal pipe diameters of 600 mm and 1200 mm as specified in relevant standards

[0309]

[0310] Based on this scheme, the wall thickness dimensions of the two single-layer sockets in Table 2 are calculated and shown in Table 3.

[0311] Table 3 Wall thickness, outer diameter and semi-cone angle of two single-layer sockets

[0312]

[0313] In step b3, the outer diameter of the end face of the socket only needs to be larger than the outer diameter at the third chamfer R3, and the specific value can be flexibly selected.

[0314] When the pipe body meets the strength requirements and the material of the single-layer socket is inconsistent with the material of the pipe body, in order to make the single-layer socket meet the strength requirements, the stress at the first chamfer R1, the second chamfer R2, and the third chamfer R3 and the stress on the inner wall of the pipe body should meet the following conditions:

[0315] (48)

[0316] (49)

[0317] (50)

[0318] Among them, Q0 is the yield strength of the pipe body material, and Q1 is the yield strength of the material used for the single-layer socket.

[0319] Then we have:

[0320] (51)

[0321] (52)

[0322] (53)

[0323] Similarly, considering the influence of the sealing ring at the third chamfer R3, the wall thickness of each chamfer is:

[0324] (54)

[0325] (55)

[0326] (56)

[0327] Therefore, the minimum wall thickness at the first chamfer R1, the second chamfer R2, and the third chamfer R3 are as follows:

[0328] (57)

[0329] (58)

[0330] (59)

[0331] Based on the above, the design calculation of the minimum thickness of the single-layer socket is completed.

[0332] Embodiment 2

[0333] Different from the first embodiment, in this embodiment, the inner tube 2 and / or the liner ring 5 at one end of the socket 11 extend to cover the inner wall of the socket 11, such as Fig.10The liner ring 5 or the inner tube 2 on the inner wall of the socket 11 can be directly attached to the socket 11, or an intermediate layer can be arranged between the socket 11 and the liner ring 5 (or the inner tube 2), and the material of the intermediate layer is hot melt adhesive, plastic or rubber.

[0334] In this embodiment, the liner ring 5 is attached to the inner wall of the socket 11, and the socket 11 and the portion of the liner ring 5 covering the inner wall of the socket 11 form a composite socket of the pipe. Because the middle layer is made of hot melt adhesive or flexible non-metal such as plastic or rubber, it has little effect on the strength of the socket 11, so it does not belong to the category of composite sockets with strength.

[0335] The design method of the thickness of the composite socket is as follows:

[0336] The shape and size of the inner wall of the socket 11 in the first embodiment are generally designed for a single-layer socket. The size and shape of the inner wall of the composite socket can also be designed according to the above-mentioned 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-mentioned single-layer socket.

[0337] The outermost layer of the composite socket is assumed to be the base layer (i.e., the socket 11 of the outer pipe 1), and the other layers are non-base layers. The thickness of the base layer is uneven, and the thickness of the non-base layer is uniform. In other words, the thickness of each part of the base layer is different or not completely the same, and the thickness of each part of the non-base layer is the same.

[0338] 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 on this basis, in order to save material costs and meet strength requirements. In other words, for composite sockets, the purpose of this scheme 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.

[0339] Similarly, the places where stress is most likely to concentrate on the inner wall of the composite socket are the first chamfer R1, the second chamfer R2 and the third chamfer R3.

[0340] In this embodiment, 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 Z, the coverage range of the non-base layer is generally from the fourth step 114 to the sixth step 116. Fig.10 As shown. 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 but only a base layer. 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 are non-base layers and base layers. In this solution, the fourth step 114 and the fifth step 115 are located in the multi-layer structure part. At this time, if both the single-layer structure part and the multi-layer structure part meet the strength requirements, the entire composite socket meets the strength requirements.

[0341] 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.

[0342] For a single-layer structure of a composite socket, the design method comprises the following steps:

[0343] Step d1: 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.

[0344] Step d2: Determine the minimum wall thickness of each selected location in step d1, 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.

[0345] In step d1, 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.

[0346] For a thin-walled cylindrical pressure vessel, when the wall thickness is S, the internal pressure is p, and the diameter is D0 (the diameter D0 can be half of the outer diameter and 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 Obtained by formula (27).

[0347] 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 outer diameter and inner diameter of the corresponding part of the cone, or the inner diameter at the corresponding part, or the outer diameter at the corresponding part), the hoop stress σ of the container wall is 环 Obtained by formula (29).

[0348] In step d1, the base layer 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 base layer 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 requirements, the base layer will definitely meet the strength requirements. Therefore, it is sufficient to ensure that the conical pressure vessel meets the strength requirements.

[0349] In step d1, take the second cone and the reinforcement ring on the inner wall of the second cone, such as Fig. 9The 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.

[0350] exist Fig. 9 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.

[0351] but:

[0352] (60)

[0353] In step d2, if the material of the base layer 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:

[0354] (61)

[0355] Then we have:

[0356] (62)

[0357] Among them, S b3 is the wall thickness of the base layer at the third chamfer R3, S is the wall thickness of the tube body, and we get:

[0358] (63)

[0359] Since there is a sealing ring z at the third chamfer R3, in order to achieve sealing, the pressure of the sealing ring z 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 z on the inner wall of the socket 11 is generally 1.4 times the fluid pressure. It can be obtained from formula (29) 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 (63) is corrected to:

[0360] (64)

[0361] That is, the minimum wall thickness at the third chamfer R3 is: .

[0362] If the material of the base layer is different from that of the pipe body, the yield strength of the pipe body material is Q0, and the yield strength of the material used for the base layer is 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:

[0363] (65)

[0364] Then we have:

[0365] (66)

[0366] Considering the pressure of the sealing ring, the wall thickness at the third chamfer R3 is:

[0367] (67)

[0368] 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.

[0369] For a multi-layer structure of a composite socket, the design method comprises the following steps:

[0370] 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 is obtained.

[0371] Step c2: Determine the relationship between the thickness of any non-base layer and the thickness of the base layer, and the determination principle is: the ratio of the stress of the non-base layer 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.

[0372] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.

[0373] 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.

[0374] 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. 9 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. 9The 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) is an annular reinforcement ring j.

[0375] For the socket 11, the maximum load it bears is the tensile stress in the circumferential direction. The annular equilibrium equation for the thin-walled internal pressure conical shell is as follows:

[0376] (68)

[0377] 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.

[0378] For a composite socket, assuming there are m layers of non-base material and one layer of base material, then:

[0379] (69)

[0380] Among them, 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.

[0381] 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 layer. The strain is the same as that of each non-base layer, that is, it satisfies:

[0382] (70)

[0383] Right now:

[0384] (71)

[0385] Among them, E1, E2, …, E m are the elastic modulus of each non-base layer, E b is the elastic modulus of the base layer.

[0386] The stresses in formula (69) are all To express, is the hoop stress of the nth non-base layer, where n is 1~m, then E n is the elastic modulus of the nth non-base layer, S n is the thickness of the nth non-base layer. Based on formula (71), (69) is rearranged to obtain:

[0387] (72)

[0388] Formula (72) can be rearranged to obtain the hoop stress of the nth non-base layer:

[0389] (73)

[0390] Step c1 is thus completed.

[0391] 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:

[0392] (74)

[0393] (75)

[0394] Among them, Q0, Q n are the yield strengths of the pipe body and the nth non-base material, respectively. The yield strength of the base material 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.

[0395] Combining formulas (73), (74), and (75), we get:

[0396] (76)

[0397] Formula (76) is the relationship between the thickness of any non-base layer and the thickness of the base layer.

[0398] Based on formula (76), the thickness of the non-base layer can be calculated when the thickness of the base layer is known, and the thickness of the base layer can be calculated when the thickness of the non-base layer is known. The following two cases are explained.

[0399] In the first case, there is no limit on the thickness of the non-base layer. In this case, 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 set as 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.

[0400] The yield stress of the base layer is generally greater than the yield stress of the non-base layer, so the layer with the weakest yield stress is the non-base layer. Assuming that the yield stress of the nth non-base layer is the weakest, if yielding occurs, the nth layer should yield first. Based on this, if the thickness of the nth non-base layer meets the strength requirements, the minimum thickness of the base layer and the thickness of other non-base layers are not less than the thickness of the nth non-base layer, then the strength of the base layer and other non-base layers will also meet the requirements. In addition, the thickness of the base layer is generally greater than the thickness of the non-base layer. Therefore, it is necessary to calculate the thickness of the nth non-base layer. Formula (74) is rewritten as:

[0401] (77)

[0402] in, is the maximum stress inside the nth non-base layer, that is From formula (73), 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 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 is not less than the thickness of the nth non-base layer, that is, the thickness of the non-base layer is Sn, and the half cone angle is α 1. In Fig. 9 In the cross section shown, the angle between the straight line passing through the first chamfer R1 and the second chamfer R2 and the axis of the pipe is the semi-cone angle of the first conical pressure vessel, that is:

[0403] (78)

[0404] Substituting formula (75) and (78) into formula (77), we obtain:

[0405] (79)

[0406] From formula (79), we get:

[0407] (80)

[0408] S b Take the thickness of the base layer at the first chamfer R1.

[0409] From the above, the number of non-base layers m 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 (80), only S n and S bTwo unknowns. The thickness S of the base layer at the first chamfer R1 is given according to experience or other requirements. b1 After that, we can find S n The minimum value of .

[0410] Therefore, according to formula (80), the minimum thickness of the non-base layer with the smallest yield stress can be obtained, and the thickness of other non-base layers does not need to be less than this value.

[0411] For the base layer, the thickness S of the base layer at the first chamfer R1 b1 It has been given that the thickness S of the base layer at the third chamfer R3 b3 It has also been calculated by formula (67), so we only need to give or calculate the thickness S of the base layer at the second chamfer R2: b2 The thickness of the base layer at various locations can be determined to ensure that the multi-layer structure of the composite socket meets the strength requirements.

[0412] For the thickness S of the base layer at the second chamfer R2 b2 , can be given directly, or can be set to the thickness S of the base layer at the first chamfer R1 b1 Equal, can also be obtained by using formula (76), specifically, formula (76) is rewritten as:

[0413] (81)

[0414] but:

[0415] (82)

[0416] The thickness S of the nth non-base layer has been calculated from formula (80): n , and the thickness of other non-base layers is not less than S n , then formula (82) becomes:

[0417] (83)

[0418] The thickness S of the base layer at the second chamfer R2 is obtained by formula (83): b2 .

[0419] 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 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 and the thickness of each part of the composite socket are obtained. The outer diameter at the end face of the socket can be greater than the outer diameter at the third chamfer R3, and the specific value can be flexibly selected.

[0420] When the non-base layer has only one layer, that is, the conical shell is a double-layer material, then formula (69) is in the following form:

[0421] (84)

[0422] Formulas (70) and (71) are organized into the following form:

[0423] (85)

[0424] For double-layer sockets, the material of the non-base layer (i.e., inner lining layer) is generally stainless steel, and the material of the base layer (i.e., outer layer) is generally Q235 steel or ductile iron. The yield stress of the non-base layer is generally smaller than that of the base layer, that is, if yielding occurs, the non-base layer should yield first. Based on this, the stress in formula (84) should be To express, based on formula (85), rearranging formula (84) we can get:

[0425] (86)

[0426] Formula (86) can be rearranged to obtain the hoop stress of the non-base layer:

[0427] (87)

[0428] In step c2, in this embodiment, formula (79) is in the following form:

[0429] (88)

[0430] Thickness of the non-base layer after finishing:

[0431] (89)

[0432] That is, the minimum thickness of the non-base layer is: .

[0433] At this time, formula (83) is: (90)

[0434] A specific example is shown below.

[0435] Assume the nominal diameter of the purchased base pipe is DN600 mm, E b =200GPa, E n =193GPa, the base 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 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 , =5.82o Assume that the thickness of the base layer at the first chamfer R1 is Sb=6 mm.

[0436] The thickness of the non-base layer Sn is calculated by formula (89) to be ≥ 2.38 mm, and the thickness of other non-base layers must be no less than 2.38 mm.

[0437] For the thickness S of the base layer 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 (89). When formula (89) is used, take Sn=2.38 mm, substitute each parameter into formula (89), and calculate S b2 ≥5.68 mm.

[0438] For the thickness S of the base layer at the third chamfer R3 b3 , obtained by formula (64), substituting each parameter into formula (64), and obtaining S b3 ≥10.53 mm.

[0439] The second situation is that the thickness of the non-base layer is limited. For example, in order to improve the corrosion resistance of the pipe, the secondary processing manufacturer or user sleeves a liner ring made of corrosion-resistant material in the socket 11. The thickness of the liner ring is generally about 1mm, which is equivalent to the known thickness of the non-base layer. In order to make the composite socket formed after the sleeve liner ring meets the strength requirements, it is necessary to calculate and design the thickness of the base layer under this premise.

[0440] Based on this, formula (76) is rewritten as:

[0441] (91)

[0442] 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, the thickness of each non-base layer is uniform, and the thickness is consistent everywhere. Therefore, it is only necessary to ensure that the thickness of the base layer at the first chamfer R1 and the second chamfer R2 meets the strength requirements. The base layer and other parts of the multi-layer structure manufactured according to the above requirements will also meet the strength requirements.

[0443] The thickness of the base layer at the first chamfer R1 is S b1 , in formula (91), D is D3, and the half cone angle α for α 1, then formula (91) is:

[0444] (92)

[0445] The thickness of the base layer at the second chamfer R2 is S b2 , in formula (91), D is D4, and the half cone angleα for α 1, then formula (91) is:

[0446] (93)

[0447] From the above, after knowing the thickness and material of the non-base layer, the minimum thickness of the base layer 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 requirements, the composite socket meets the strength requirements.

[0448] A specific example is shown below.

[0449] 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 is sleeved inside, that is, there is a non-base layer, and 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 (92), (93), and (67) respectively:

[0450] S b1 ≥7.33 mm, S b2 ≥7.01 mm, S b3 ≥10.53 mm.

[0451] Embodiment 3

[0452] Different from the second embodiment, in this embodiment, the inner wall of the entire base pipe socket 11 is completely covered by the lining. Fig.11 As shown, the entire composite socket is a multi-layer structure. In this embodiment, the design method of the thickness of the base layer and each non-base layer is the same as the design method for the case where 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 The method of seeking.

[0453] For the case where the non-base thickness is not limited, the base thickness S corresponding to the third chamfer R3 b3 It can be given directly, or set to be equal to the thickness of the base layer at the first chamfer R1, or it can be obtained using formula (76). Specifically, formula (81) gives:

[0454] (94)

[0455] 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.11 shown.

[0456] For the case where the thickness of the non-base layer is limited, the thickness of the base layer at the third chamfer R3 is calculated using formula (91): S b3 , where D is D1, and the half cone angle is α for α 2, then:

[0457] (95)

[0458] Embodiment 4

[0459] Different from the first embodiment, the second embodiment or the third embodiment, in this embodiment, the liner ring 5 at one end of the socket 13 is extended and turned over to cover the outer wall of the socket 13, such as Figures 12-14 shown.

[0460] In this embodiment, the liner 5 portion located inside the pipe and the liner 5 portion covering the outer wall of the socket 13 can be regarded as a whole, which has a structure of two layers of the same material. The thickness of the whole is calculated and then divided by 2 to obtain the thickness of the single-layer liner 5.

[0461] Embodiment 5

[0462] Different from the first embodiment, the second embodiment or the third embodiment, in this embodiment, the liner ring edge 51 at one end of the socket 13 is only used to seal the end surface of the pipe, and is not used to form a composite socket. In this case, the length of the liner ring edge 51 at one end of the socket 13 is generally less than the length of the socket 13. Preferably, the length of the liner ring edge 51 at one end of the socket 13 is not less than 20 mm. Figures 15-17 In this embodiment, the most vulnerable part of the socket of the pipe is equivalent to only the outer pipe 1 having strength and being able to resist deformation, which is a single-layer socket.

[0463] The method for calculating the minimum thickness of the single-layer socket is steps a1 to a3 of the first embodiment.

[0464] Embodiment 6

[0465] Different from the first to fifth embodiments, in this embodiment, the material of the sealing layer 4 is hot melt adhesive. Obviously, the sealing layer 4 plays a sealing role and also plays a bonding role, bonding the liner ring 5 and the inner tube 2 together.

[0466] Correspondingly, step S5 in the method for manufacturing the pipe is different from that in the first embodiment. Specifically, in this embodiment, the specific process of step S5 (lining ring) is as follows:

[0467] In this step, first, the positions of the socket and the spigot of the inner tube 2 where the end lining ring is required are heated, and then hot melt adhesive powder is sprayed on the heated positions and cooled to form a hot melt adhesive layer, which is the sealing layer 4 of this embodiment. The thickness of the hot melt adhesive layer is not less than 0.2 mm.

[0468] Alternatively, the lining rings 5 ​​at both ends are heated, and hot melt adhesive powder is sprayed on the outer surface of the lining ring 5 and the flange connected to the outer surface, and then cooled to form a hot melt adhesive layer, the thickness of which is not less than 0.2 mm;

[0469] Two lining rings 5 ​​are manually placed at both ends of the pipe. Then, the lining rings 5 ​​are expanded by a diameter expander so that the lining rings 5 ​​press the hot melt adhesive layer against the inner wall and end face of the pipe.

[0470] If the liner ring 5 tube blank is two sections cut from the stainless steel round tube obtained in step S12, then in this step, after the liner ring 5 with the sealing layer 4 is inserted into the tube end, it is also flanged to form a liner ring flange 52.

[0471] The pipe sections where the lining rings 5 ​​at both ends are located are heated, and the hot melt adhesive layer 4 is melted, so that the two ends of the inner pipe 2 and the lining rings 5 ​​are bonded together.

[0472] After completing the above steps, remove excess burrs and apply paint, plastic coating or other methods to the outer wall of the pipe for external anti-corrosion treatment.

[0473] 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 design method for a multi-layer bell-and-spigot pipe with a corrosion-resistant metal lining, characterized in that: The pipe material comprises an inner pipe (2), an intermediate layer (3) and an outer pipe (1) which are sleeved in sequence from the inside to the outside; the inner wall of the socket (11) is provided with steps and / or grooves; the inner pipe (2), the intermediate layer (3) and the outer pipe (1) are bonded together to form the pipe material or pressed together to form the pipe material; the outer pipe (1) comprises a socket (11), a pipe body (12) and a plug (13) which are connected in sequence; the outer pipe (1) is made of a metal material; the inner pipe (2) is made of a corrosion-resistant material; Each end of the pipe is provided with a sealing layer (4) and a lining ring (5) made of a corrosion-resistant metal material. The lining ring (5) at one end of the socket (11) covers one end of the inner wall of the inner pipe (2), one end face of the inner pipe (2), one end face of the intermediate layer (3), and the end face of the pipe body (12). The lining ring (5) at one end of the spigot (13) covers the inner wall of the other end of the inner pipe (2), the end face of the other end of the inner pipe (2), the end face of the other end of the intermediate layer (3), and the end face of the spigot (13); When the length of the portion of the lining ring (5) at one end of the socket (13) extending into the interior of the pipe is less than the length of the socket (13), the socket of the pipe is a single-layer socket, and the design method of the thickness of the single-layer socket comprises 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; When the length of the portion of the liner ring (5) at one end of the socket (13) extending into the interior of the pipe is not less than the length of the socket (13), the socket (13) and the liner ring (5) at one end of the socket (13) form a composite socket, and the method for designing the thickness of the composite socket further includes the following steps: 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: Determine the minimum thickness of the reinforcement ring. If the thickness of the reinforcement ring is not less than the minimum thickness, the thickness of the composite socket is the minimum thickness that meets the deformation requirement.

2. The design method according to claim 1, characterized in that: The material of the middle layer (3) is hot melt adhesive, plastic or rubber.

3. The design method according to claim 1, characterized in that: The material of the sealing layer (4) is rubber or hot melt adhesive. When the material of the sealing layer (4) is hot melt adhesive, the sealing layer (4) bonds the liner ring (5) and the inner tube (2) together.

4. The design method according to claim 1, characterized in that: The lining ring (5) at one end of the socket (11) further extends to completely or partially cover the inner wall of the socket (11).

5. The design method according to claim 4, characterized in that: The length of the portion of the liner ring (5) at one end of the socket (13) extending into the interior of the pipe is not less than the length of the socket (13).

6. The design method according to claim 5, characterized in that: The liner ring (5) at one end of the socket (13) is turned outward to cover the outer wall of the socket (13).

7. The design method according to claim 1, characterized in that: At one end of the socket (11), the end face of the inner tube (2), the end face of the middle layer (3) and the end face of the tube body (12) are flush, and at one end of the spigot (13), the end face of the inner tube (2), the end face of the middle layer (3) and the end face of the spigot (13) are flush.

8. The design method according to claim 1, characterized in that: At one end of the socket (11), the sealing layer (4) covers the end face of the pipe body (12), the end face of the intermediate layer (3), the end face of the inner pipe (2) and the inner wall of one end of the inner pipe (2); at one end of the plug (13), the sealing layer (4) covers the end face of the plug (13), the end face of the intermediate layer (3), the end face of the inner pipe (2) and the inner wall of the other end of the inner pipe (2); the lining ring (5) presses the sealing layer (4) tightly; when in use, the lining ring (5) and the sealing layer (4) at one end of the plug (13) form a full seal with the end face of the plug (13); the lining ring (5) and the sealing layer (4) at one end of the socket (11) form a full seal with the end face of the pipe body (12); The sealing layer (4) comprises a sealing layer pressing edge (41) and a sealing layer flange (42), the sealing layer pressing edge (41) is cylindrical, and the sealing layer flange (42) is annular; the lining ring (5) comprises a lining ring pressing edge (51) and a lining ring flange (52), the lining ring pressing edge (51) and the lining ring flange (52) are an integral part, the lining ring pressing edge (51) is cylindrical, the lining ring pressing edge (51) presses the sealing layer pressing edge (41), and the lining ring flange (52) presses the sealing layer flange (42), and the length of the lining ring pressing edge (51) is not less than 20 mm.

9. The design method 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 change in the diameter of the circular ring with coefficients according to the third form of the differential equation obtained in step a23.

10. The design method 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.

11. The design method according to claim 9, 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.

12. The design method according to claim 11, characterized in that: In step a2, it is assumed that the ring is subjected to at least two radial concentrated forces P symmetrically relative to the center of the circle, and the angle between two adjacent radial concentrated forces P is , the ring satisfies the following three boundary conditions: First, in Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula: ; Second, due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; That is: ; Third, cut near point A, remove point A and the force P, and from the symmetry and equilibrium conditions, we know that the shear force on the section at the circular cut is , that is: ; Combining the above three formulas with the general solution of the differential equation, we get the first form of the differential equation: ； Assuming that the annulus is subjected to only two radial concentrated forces P which are symmetrical relative to the center of the circle, then: , we get the second form of the differential equation: ； Introducing the coefficient m, we get the third form of the differential equation: ； Point of application of force P The point is the point of maximum deformation, and the change in the diameter of the ring is: .

13. The design method according to claim 1, characterized in that: In step a5, the minimum thickness of the reinforcement ring is obtained and determined by experiment, simulation or theoretical calculation. When the minimum thickness of the reinforcement ring is obtained and determined by theoretical calculation, the minimum thickness of the reinforcement ring is obtained based on the socket thickness of the base pipe and the bending strength calculation formula of the longitudinal composite beam.

14. The design method according to any one of claims 1 to 8, characterized in that: When the liner ring (5) at one end of the socket (11) does not cover the inner wall of the socket (11), the socket of the pipe is a single-layer socket; when the liner ring (5) at one end of the socket (11) extends to completely or partially cover the inner wall of the socket (11), the socket (11) and the liner ring (5) at one end of the socket (11) form a composite socket.

15. The design method according to claim 14, characterized in that: For the same 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), a fifth step (115), and a sixth step (116) arranged in sequence; 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).

16. The design method according to claim 14, characterized in that: The design method of single-layer socket thickness includes the following steps: Step b1: Select multiple locations on the inner wall of the single-layer socket where stress is likely to be concentrated. The selection principle is: if the wall thickness of these locations meets the stress strength requirements, then the wall thickness of each location of the single-layer socket meets the strength requirements; Step b2: Determine the minimum wall thickness of each selected location in step b1, and obtain the outer wall of each selected location. The determination principle is: when the material of the single-layer socket 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 socket 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 socket 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.

17. The design method according to claim 16, characterized in that: In step b2, the minimum wall thickness of each selected point in step b1 is calculated by the stress calculation formula, and the single-layer socket is regarded as a pressure vessel including at least two coaxially connected pressure vessels with conical inner walls and at least one annular reinforcement ring installed on the inner wall of the pressure vessel. , ,in, is the hoop stress of the pipe body, is the hoop stress at the selected location, D0 is the outer diameter of the base pipe, D is the diameter of the selected location, p is the pressure of the fluid in the base pipe, S is the thickness of the base pipe, S0 is the wall thickness of the selected location, and α is the semi-cone angle of the cone at the selected location.

18. The design method according to claim 17, characterized in that: When the material of the single-layer socket is the same as that of the pipe body, the stress calculation formula is: .

19. The design method according to claim 17, characterized in that: When the material of the single-layer socket is different from that of the pipe body, the stress calculation formula is: , where Q0 is the yield strength of the pipe body material and Q1 is the yield strength of the material used for the single-layer socket.

20. The design method according to claim 15, characterized in that: The design method of composite socket thickness includes the following steps: 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: 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; 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; Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.

21. The design method according to claim 20, 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 and the second chamfer are located on the inner wall of the first cone.

22. The design method according to claim 21, characterized in that: 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: ,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; 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.

23. The design method according to claim 22, 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 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: .

24. The design method according to claim 23, characterized in that: 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.

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