A design method for the thickness of single-layer spigot and composite spigot of spigot-and-socket pipes

Abstract

The present invention discloses a design method for the thickness of a single-layer spigot and a composite spigot of a spigot-and-socket type pipe. The design method for the thickness of a single-layer spigot includes: a1: obtaining a differential equation for radial deformation of a circular ring and a general solution of the differential equation; a2: calculating a calculation formula for the maximum radial deformation of a circular ring; a3: finding the minimum thickness of a single-layer spigot; assuming that the outermost layer of a composite spigot is a base layer and the other layers are non-base layers, for a multi-layer structure in a composite spigot, the design method includes: c1: obtaining a calculation formula for the hoop stress of any non-base layer; step c2: determining the relationship between the thickness of any non-base layer and the thickness of the base layer; c3: calculating the thickness of the non-base layer and / or the base layer. The present invention can obtain the minimum or near minimum thickness of a single-layer spigot that meets the deformation requirements and the minimum or near minimum thickness of a composite spigot that meets the strength requirements of spigot-and-socket type pipes of different pipe diameters through theoretical analysis and calculation.

Description

Technical Field

[0001] The invention belongs to the technical field of pipe design, and in particular relates to a design method for the thickness of a single-layer spigot and a composite spigot of a spigot-and-socket type pipe. Background Art

[0002] Most of the water pipes in the water supply and drainage systems for life and production are socket-and-spigot pipes, with a socket at one end and a spigot at the other. During installation, the spigot of one pipe is inserted into the socket of another pipe and sealed with a seal, thus achieving the connection of each pipe in the pipe network. Figure 6 In the figure, M is the socket and F is the socket. The base pipe includes a socket F, a pipe body and a socket M which are connected in sequence. The pipe body can be divided into two parts, namely a main body and a transition part. The transition part is used for the transition connection between the main body and the socket F, that is, the socket F, the transition part, the main body and the socket M 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 uniform or substantially uniform at all places, and the thickness of the transition part gradually decreases along the direction from the socket F to the socket M. The outer diameter of the pipe body refers to the outer diameter of the main body, the outer diameter of the base pipe is the outer diameter of the pipe body, and the thickness of the pipe body refers to the thickness of the main body. In the figure, X is the length of the chamfer at the end of the socket M along the axial direction, Y is the height of the chamfer at the end of the socket M (that is, the radial dimension), DE is the outer diameter of the socket M, and T6 is the length of the socket M. Base pipe refers to the pipe in factory settings, without reinforcing rings installed and secondary processing such as composite processes. 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 pipe 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.

[0003] However, when a pipe network fails, for example, a pipe bursts in the pipe network, water in the pipe network will spray out from the burst, and the pressure in the pipe network will drop rapidly, even forming negative pressure. In addition, the pipe will also be subject to some external pressure, for example, the pipe buried in the soil will be subject to the gravity of the soil above it. Under the above external pressure and internal negative pressure, the pipe will deform radially. If the deformation of the socket is too large, the socket will separate from the socket connected to it, causing leakage at the connection between the socket and the socket, causing the inside of the water pipe to be contaminated. Therefore, it is necessary to ensure that the deformation of the socket of the pipe does not exceed a certain range under the above circumstances. To this end, for the socket, the minimum socket thickness that meets the deformation requirements must be obtained. If the socket thickness of the selected pipe is less than the minimum socket thickness that meets the deformation requirements, measures need to be taken to make it meet the deformation requirements. For example, during the secondary processing or installation of the pipe, an annular reinforcement ring is sleeved on the socket to enhance the ability to resist deformation.

[0004] For the socket F, 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 F at various locations meet the use strength while the thickness is minimum or close to minimum. At present, the socket F of the general base pipe is made of only one material, such as ductile iron or Q235 steel. This socket F is a single-layer socket. However, in order to improve the performance of the pipe, the socket F 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 and improve the corrosion resistance of the pipe. Similarly, if the composite socket can meet the strength requirements and the thickness is minimum or close to minimum, the cost can also be greatly reduced. For the composite socket, the inner wall of the base pipe socket F can be completely covered by the lining or not completely covered by the lining. When the inner wall of the base pipe socket F is not completely covered by the liner, the composite socket includes a single-layer structure and a multi-layer structure. The single-layer structure is the part not covered by the liner, and has only one layer, namely, the outermost base pipe. The multi-layer structure includes at least two layers, namely, the outermost base pipe and the liners sleeved in the base pipe. In addition, for the single-layer socket F, in order to improve its corrosion resistance, the secondary processing manufacturer or user generally sleeves a liner ring made of corrosion-resistant material in the socket F. Obviously, the liner ring can also be used as a reinforcement ring. At this time, the liner ring and the socket F of the base pipe form a composite socket. Summary of the invention

[0005] In view of the above-mentioned problems existing in the prior art, the purpose of the present invention is to provide a design method for the thickness of a single-layer spigot and a composite socket of a socket-and-spigot pipe. The minimum or near-minimum thickness of a single-layer spigot that meets the deformation requirements and the minimum or near-minimum thickness of a composite socket that meets the strength requirements of socket-and-spigot pipes of different pipe diameters can be obtained through theoretical analysis and calculation, thereby saving materials and reducing the production, manufacturing and transportation costs of pipes. The method for designing the minimum wall thickness is theoretical calculation, which has low cost and high efficiency, provides strong support for experiments and pipe design, and provides a reference for secondary processing and installation. The pipe network system is more reliable to use, greatly reducing the leakage and water supply pollution caused by the deformation and separation of the spigots and sockets of adjacent pipes when a pipe bursts, as well as the cracking of the socket.

[0006] In order to achieve the above object, the technical solution adopted by the present invention is:

[0007] A method for designing the thickness of a single-layer spigot and a composite spigot of a spigot-and-socket type pipe, wherein the method for designing the thickness of a single-layer spigot comprises the following steps:

[0008] 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;

[0009] 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;

[0010] Step a3: Calculating the minimum thickness of the single-layer spigot of the base pipe that meets the deformation requirements according to the calculation formula of the maximum radial deformation obtained in step a2;

[0011] 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:

[0012] 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;

[0013] 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;

[0014] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step a2.

[0015] As a further improvement of the above technical solution:

[0016] Step a2 includes the following steps:

[0017] 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;

[0018] 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;

[0019] 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;

[0020] 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.

[0021] Step a3 includes the following steps:

[0022] 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;

[0023] Step a32: obtaining the maximum deformation of the reference pipe socket through simulation according to the parameters calculated in step a31;

[0024] 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;

[0025] 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.

[0026] In step a1, the radial deformation obtained is u The differential equation is: , its general solution is: ,

[0027] in, R is the average value of the outer circle radius and the inner circle radius of the ring; E is the elastic modulus of the pipe, Pa; I is the moment of inertia of the tube cross section, m 4 ; M is the bending moment borne by a certain section of the ring in the local area, in units of N*m; establish a two-dimensional coordinate system, The deformation is u The angle between the line connecting the point at and the origin of the coordinate system and the positive direction of the x-axis in the two-dimensional coordinate system, ; A1, A2, A3、 B2 and B3 are constants.

[0028] 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:

[0029] First, in Because of symmetry, The rotation angle of the section at is 0, which satisfies the following formula: ;

[0030] Second, due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; that is, ;

[0031] 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: ;

[0032] Combining the above three formulas with the general solution of the differential equation, we get the first form of the differential equation:

[0033] ;

[0034] 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:

[0035] ;

[0036] Introducing the coefficient m, we get the third form of the differential equation:

[0037] ;

[0038] Point of application of force P The point is the point of maximum deformation, and the change in the diameter of the ring is:

[0039] , where I = S 3 / 12.

[0040] For the same base pipe, along the direction from the socket to the spigot, the inner wall of the socket includes a first step, a second step, a third step, a fourth step, and a fifth step arranged in sequence, one end of the inner wall of the pipe body is a sixth step, the chamfer at the connection between the fourth step and the fifth step is a first chamfer, the chamfer at the connection between the fifth step and the sixth step is a second chamfer, the chamfer at the connection between the second step and the third step is a third chamfer, the chamfer at the connection between the first step and the second step is a fourth chamfer, and the chamfer at the connection between the third step and the fourth step is a fifth chamfer.

[0041] 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. The pressure vessel with a conical inner wall is the first cone, and the first chamfer and the second chamfer are located on the inner wall of the first cone.

[0042] 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:

[0043] ;

[0044] 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;

[0045] 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.

[0046] 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: .

[0047] 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.

[0048] The beneficial effects of the present invention are as follows: the minimum or near minimum thickness of a single-layer spigot that meets the deformation requirements and the minimum or near minimum thickness of a composite spigot that meets the strength requirements of spigot-and-socket pipes of different diameters can be obtained through theoretical analysis and calculation, thereby saving materials and reducing the production, manufacturing and transportation costs of pipes. The method for designing the minimum wall thickness is theoretical calculation, which has low cost and high efficiency, provides strong support for experiments and pipe design, and provides a reference for secondary processing and installation. The pipe network system is more reliable in use, greatly reducing the leakage and water supply pollution caused by the separation of the spigot and the socket of adjacent pipes due to excessive deformation of the spigot when the pipe bursts, as well as the cracking of the socket. BRIEF DESCRIPTION OF THE DRAWINGS

[0049] Figure 1 Schematic diagram of the cross-section deformation before and after the socket is subjected to a pair of radially symmetrical concentrated forces.

[0050] Figure 2 for Figure 1 Schematic diagram of local analysis.

[0051] Figure 3 Schematic diagram of a circular ring subjected to multiple uniformly arranged radial concentrated forces.

[0052] Figure 4 This is the stress analysis diagram of socket-type pipe.

[0053] Figure 5 It is a schematic diagram of a composite socket in which the inner wall of the base pipe socket is not completely covered by the liner according to the present invention.

[0054] Figure 6 It is a schematic diagram of the socket and spigot connection of the present invention.

[0055] Figure 7 It is a schematic diagram of a composite socket in which the inner wall of the base pipe socket is completely covered by the liner according to the present invention. DETAILED DESCRIPTION

[0056] 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.

[0057] For ease of description, spatially relative terms such as "above", "above", "on the upper surface of", "above", etc. may be used here to describe the spatial positional relationship between a device or feature and other devices or features as shown in the figure. It should be understood that spatially relative terms are intended to include different orientations of the device in use or operation in addition to the orientation described in the figure. For example, if the device in the accompanying drawings is inverted, the device described as "above other devices or structures" or "above other devices or structures" will be positioned as "below other devices or structures" or "below other devices or structures". Thus, the exemplary term "above" can include both "above" and "below". The device can also be positioned in other different ways (rotated 90 degrees or in other orientations), and the spatially relative descriptions used here are interpreted accordingly.

[0058] A method for designing the thickness of a single-layer spigot and a composite spigot of a spigot-and-socket type pipe, wherein the spigot M of the pipe is a single-layer spigot and the spigot F is a composite spigot, and the thickness of the single-layer spigot and the thickness of the composite spigot are calculated respectively. The method for designing the thickness of the single-layer spigot comprises the following steps:

[0059] Step a1: According to the differential equation of the deflection curve of the circular ring and the moment balance equation, the differential equation of the radial deformation of the circular ring and the general solution of the differential equation are obtained; because the cross-sections of the pipe body and the socket of the pipe are circular rings, the deformation of the socket is analyzed by analyzing the radial deformation of the circular ring. In addition, a cylinder of unit length can be regarded as a circular ring.

[0060] 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.

[0061] 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.

[0062] The following is a detailed introduction to each step.

[0063] In step a1, the radial deformation obtained u The differential equation is:

[0064] (1)

[0065] in ，R is half of the sum of the outer radius and the inner radius of the ring. In other words, R One quarter of the sum of the outer and inner diameters of the ring 。E is the elastic modulus of the pipe, Pa. I is the moment of inertia of the tube cross section, m 4 . M It is the bending moment borne by a certain section of the ring in the local area, in N*m. Establish a two-dimensional coordinate system, The deformation is u The angle between the line connecting the point at and the origin of the coordinate system and the positive direction of the x-axis in the two-dimensional coordinate system, In this embodiment, establish xy Coordinate system, point O is the coordinate origin and the center of the ring, such as Figure 1 Point m 1 The radial deformation at u， but om 1 and x The angle between the positive directions of the axes is 。

[0066] The above differential equation (1) is obtained as follows:

[0067] like Figure 1 and 2 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 1 midpoint O Too xy Vertex of the coordinate system ， point m and n are two points on the ring before deformation; m 1 and n 1 are two points on the deformed shape of the ring, and the points m 1 and n 1 Located on the line segment om and on superior, The deformation is u The line connecting the point at and point O is x The angle between the positive directions of the axes, om and on The angle between ， set up u Centripetal movement is positive. Figure 2 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.

[0068] Arc before deformation mn The arc length is ds , whose curvature is k 1 , arc after deformation m 1 n 1 The curvature is k 2 , du Yes m and Point n The difference in radial deformation is The arc before deformation mn exist m Tangent line of point and n The angle between the tangent lines of a point, The deformation amount is u The radius corresponding to the point at after deformation is Is an arc mn Deform to Arc m 1 n 1 back ds The increment of The arc after deformation m1 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:

[0069] (2)

[0070] (3)

[0071] (4)

[0072] (5)

[0073] m 1 t 1 and m 1 t The angle between , n 1 t 1 ’ and n 1 t’ The angle between .

[0074] m 1 section to n 1 The variable of the cross-section rotation is :

[0075] (6)

[0076] 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.

[0077] Substituting formula (5) and (6) into formula (4) yields:

[0078] (7)

[0079] By series expansion we have:

[0080] Ignoring the higher-order terms, substituting the series expansion into (7) yields: (8)

[0081] Substituting (2) into (8), we obtain: (9)

[0082] Omitting the higher-order terms, we get: (10)

[0083] 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).

[0084] The general solution of the above differential equation (1) is:

[0085] (11)

[0086] 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:

[0087] (12)

[0088] 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:

[0089] 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.

[0090] 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.

[0091] 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.

[0092] 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.

[0093] 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 3 shown. Figure 1 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.

[0094] The ring satisfies the following three boundary conditions:

[0095] (1) Due to symmetry, the rotation angle of the section at this location is 0, which satisfies the following formula:

[0096] (13)

[0097] (2) Due to symmetry, the total tangential displacement of the arc in each equally divided segment is 0; that is,

[0098] (14)

[0099] (3) When the cut is made near point A, point A and the acting force P are removed. From the symmetry and equilibrium conditions, it can be seen that the shear force on the cross section at the circular cut is , that is:

[0100] (15)

[0101] in, dM represents the differential with respect to M. It should be noted that point (3) is not a real cut of the ring, but an assumption.

[0102] By combining formulas (13), (14) and (15), we get:

[0103] (16)

[0104] (17)

[0105] (18)

[0106] Substituting formulas (16), (17), and (18) into formula (12), we obtain:

[0107] (19)

[0108] Formula (19) is the first form of the differential equation obtained in step a21. is the radial deformation of the ring.

[0109] In step a22, if only one pair of radial forces P is applied to the ring, then , the deflection curve equation (19) becomes:

[0110] (20)

[0111] Formula (20) is the second form of the differential equation obtained in step a22.

[0112] In step a23, considering that the pipe generally needs to withstand an internal negative pressure of 0.1 MPa during testing, a coefficient m is introduced, that is,

[0113] (twenty one)

[0114] 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.

[0115] Formula (21) is the third form of the differential equation obtained in step a23.

[0116] 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 :

[0117] (twenty two)

[0118] Formula (22) is the third form of the differential equation obtained in step a24.

[0119] 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:

[0120] 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.

[0121] Step a32: Obtain the maximum deformation of the reference pipe socket through simulation based on the parameters calculated in step a31.

[0122] 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.

[0123] 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.

[0124] 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:

[0125] (twenty three)

[0126] like Figure 4 As shown, W is the shear load, in Newton (N); F n is the shear force acting on the interface, in Newton (N); M P is the weight of the test assembly steel tube and the weight of the object in the steel tube, in kilograms (kg); a The distance between the shear load application point and the end face of the socket inserted into the socket, in millimeters (mm); b The distance between the center of gravity of the steel pipe at the spigot end and the end face of the spigot inserted into the socket, in millimeters (mm); c It is the distance between the support point of the pad and the end face of the socket that has been inserted into the socket, in millimeters (mm).

[0127] 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 n It should be no less than 30 times of the nominal diameter DN. In this embodiment, 30 times is taken, that is, F n=DN×30=600×30=18000N. Take the pipe length as 1m, c=800mm.

[0128] In this embodiment, the parameters in formula (23) are M P and W The calculation process is as follows:

[0129]

[0130]

[0131] 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.

[0132] Step a31 is thus completed.

[0133] 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.

[0134] 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:

[0135] (twenty four)

[0136] From formula (24), we can see that u max The value is related to the pipe diameter and pipe thickness.

[0137] 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.

[0138] Table 1 Nominal wall thickness and absolute deformation of DN=600mm pipes at various nominal pressures

[0139]

[0140] 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).

[0141] 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.

[0142] If the use requirement of the pipe in a certain occasion is to meet the nominal pressure within 2MPa, the general user will choose the pipe with serial number 4, and will not choose the pipe with serial number 1, 2 or 3 to meet the deformation requirement, because if one of these three is selected, although it will meet both the strength requirement and the deformation requirement, their thickness is larger and the cost will be much greater than the pipe with serial number 4. If the pipe with serial number 4 is selected, its strength meets the use requirement. For the deformation requirement, some measures can be taken, such as sleeve-fitting an annular reinforcement ring in the socket to increase the thickness of the socket and improve its ability to resist deformation, and this measure is low in cost.

[0143] Based on the above, the design calculation of the minimum thickness of the socket is completed.

[0144] The design method of the thickness of the composite socket is as follows:

[0145] The inner wall shape of the pipe socket F is in accordance with the inner wall shape specified in the group standard "Corrosion-resistant steel pipe with socket and spigot flexible joint for water supply and drainage" (T / CISA108-2021) and GB / T 13295-2008, such as Figure 6As shown. The inner wall of the socket F is stepped. For the same base pipe, along the direction from the socket F to the spigot M, the inner wall of the socket F includes the first step 1, the second step 2, the third step 3, the fourth step 4, and the fifth step 5 arranged in sequence. One end of the inner wall of the pipe body is the sixth step 6, and the corresponding inner diameters are: D2, D1, (D3-2F0), D3, D4, (D0-2S), among which the inner diameters of the sixth step 6, the first step 1, the fifth step 5, the third step 3, the fourth step 4, and the second step 2 increase in sequence, that is: (D0-2S) < D2 < D4 < (D3-2F0) < D3 < D1. The lengths of the first step 1, the second step 2, the third step 3, the fourth step 4, and the fifth step 5 are respectively: T3, T4, T5, (T2-T4-T5), (T1-T2-T3), that is, T1 is the sum of the lengths of the first step 1, the second step 2, the third step 3, the fourth step 4, and the fifth step 5, and T2 is the sum of the lengths of the second step 2, the third step 3, and the fourth step 4. 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 3 protruding from the fourth step 4, and F0 ≥ 0. The connection of each step is processed with chamfers, wherein the chamfer at the connection of the fourth step 4 and the fifth step 5 is the first chamfer R1, the chamfer at the connection of the fifth step 5 and the sixth step 6 is the second chamfer R2, the chamfer at the connection of the second step 2 and the third step 3 is the third chamfer R3, the chamfer at the connection of the first step 1 and the second step 2 is the fourth chamfer R4, and the chamfer at the connection of the third step 3 and the fourth step 4 is the fifth chamfer R5. Wherein, D0 is the base pipe outer diameter of the pipe.

[0146] The above inner wall shape and size are generally for single-layer socket. The size and shape of the inner wall of the composite socket can also be designed according to the above single-layer socket. In this solution, the size and shape of the inner wall of the composite socket are also designed according to the above single-layer socket.

[0147] Assume that the outermost layer of the composite socket is the base layer F1, and the other layers are the non-base layer F2. The thickness of the base layer F1 is uneven, and the thickness of the non-base layer F2 is uniform. In other words, the thickness of the base layer F1 is different or not completely the same at different locations, and the thickness of the non-base layer F2 is the same at different locations. Since the third step 3 is the installation location of the sealing ring Z, the coverage range of the non-base layer F2 is generally from the fourth step 4 to the sixth step 6. Figure 5 shown.

[0148] For composite sockets, since the dimensions of the inner wall are fully given by the relevant standards, it is necessary to design some key dimensions of the outer wall and the thickness of each non-base layer F2 on this basis, in order to save material costs and meet strength requirements. In other words, for composite sockets, the purpose of this solution is to design the thickness of the composite socket so that it is the minimum or close to the minimum thickness while meeting the use strength, so as to reduce the production, manufacturing and transportation costs of pipes.

[0149] According to material mechanics knowledge and experience, the most likely places for stress concentration on the inner wall of the above-mentioned single-layer socket or composite socket are the first chamfer R1, the second chamfer R2 and the third chamfer R3. The analysis is as follows:

[0150] For the inner wall of the stepped socket F, according to mechanical knowledge, the stress concentration point is located at the connection between adjacent steps, that is, at each chamfer. Since the socket F is a trumpet-like shape, the outer diameter of the socket F tends to decrease from the socket F to the spigot M of the same base pipe.

[0151] Based on the above conditions, since the inner diameter at the fourth chamfer R4 is equal to the inner diameter at the third chamfer R3, both are equal to D1, the wall thickness at the fourth chamfer R4 must be greater than the wall thickness at the third chamfer R3 to meet the requirement that the socket F is shaped like a trumpet. Therefore, as long as the wall thickness at the third chamfer R3 meets the stress strength requirement, the wall thickness at the fourth chamfer R4 must meet the strength requirement. Similarly, the inner diameter at the fifth chamfer R5 is equal to the inner diameter at the first chamfer R1, both are D3, so the wall thickness at the fifth chamfer R5 must be greater than the wall thickness at the first chamfer R1 to meet the requirement that the socket F 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 places where stress is most likely to concentrate are the first chamfer R1, the second chamfer R2, and the third chamfer R3. Similarly, for the socket of a single-layer structure, as long as the thickness of the first chamfer R1, the second chamfer R2, and the third chamfer R3 meet the strength requirements, the base pipe socket manufactured according to the above requirements must also meet the strength requirements.

[0152] In the case where the inner wall of the base pipe socket F is not completely covered by the lining, since the third step 3 is the installation location of the sealing ring Z, the coverage range of the non-base layer F2 is generally from the fourth step 4 to the sixth step 6. Figure 5 At this time, for the composite socket, along the axial direction of the socket, one part is a single-layer structure, that is, there is no non-base layer F2 and only the base layer F1. In this solution, the first step 1, the second step 2, and the third step 3 are located in the single-layer structure part; the other part is a multi-layer structure, that is, there is a non-base layer F2 and a base layer F1. In this solution, the fourth step 4 and the fifth step 5 are located in the multi-layer structure part, as shown in FIG. Figure 5 shown.

[0153] For the composite socket, if both the single-layer structure part and the multi-layer structure part meet the strength requirements, then the entire composite socket meets the strength requirements.

[0154] 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.

[0155] For a single-layer structure of a composite socket, the design method comprises the following steps:

[0156] Step b1: Select at least one location where stress is likely to be concentrated in the single-layer structure. The selection principle is: if the wall thickness at each location meets the stress strength requirement, then the wall thickness at each location of the single-layer structure meets the strength requirement.

[0157] Step b2: Determine the minimum wall thickness of each selected location in step b1, and the determination principle is: when the material of the single-layer structure and the material of the pipe body are the same, the stress at each selected location is not greater than the stress of the inner wall of the pipe body; when the material of the single-layer structure and the material of the pipe body are different, the ratio of the stress at each selected location to the yield strength of the single-layer structure material is not greater than the ratio of the stress of the inner wall of the pipe body to the yield strength of the pipe body material.

[0158] In step b1, it can be seen from the above that the stress of the single-layer structure is most concentrated at the third chamfer R3, and the socket F 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.

[0159] For a thin-walled cylindrical pressure vessel, when the wall thickness is S, the internal pressure is p, and the diameter is D (the diameter D can be half of the sum of the outer diameter and the inner diameter of the cylinder, or the inner diameter of the cylinder, or the outer diameter of the cylinder), the hoop stress σ of the container wall is 环 for:

[0160] (25)

[0161] 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 can be half of the sum of the outer diameter and the 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 for:

[0162] (26)

[0163] in, α is the semi-cone angle of the conical shell, that is, half of the opening angle. In other words, in the section passing through the central axis of the cone, the angle between the cone wall and the central axis is the semi-cone angle. In step b1, the base layer F1 is regarded as consisting of at least one pressure vessel with a conical inner wall plus several internal reinforcement rings j, and the reinforcement rings are connected to the inner wall of the pressure vessel. In other words, the base layer F1 is regarded as consisting of at least one conical pressure vessel plus several internal reinforcement rings j. At this time, as long as the corresponding conical pressure vessel meets the strength requirements, the base layer F1 will definitely meet the strength requirements. Therefore, it is sufficient to ensure that the conical pressure vessel meets the strength requirements.

[0164] In step b1, take the second cone and the reinforcement ring on the inner wall of the second cone, such as Figure 6The 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 F 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 F is the annular reinforcement ring on the inner wall of the second cone.

[0165] exist Figure 6 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.

[0166] but:

[0167] (27)

[0168] In step b2, if the material of the base layer F1 is consistent with the material of the pipe body, the pipe body meets the strength requirements. To make the single-layer structure meet the strength requirements, the hoop stress at the third chamfer R3 Not greater than the hoop stress of the pipe body , that is:

[0169] (28)

[0170] Then we have:

[0171] (29)

[0172] Among them, S b3 is the wall thickness of the base F1 at the third chamfer R3, S is the wall thickness of the tube body, and we get:

[0173] (30)

[0174] 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 F 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 F is generally 1.4 times the fluid pressure. It can be obtained from formula (26) 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 (30) is corrected to:

[0175] (31)

[0176] That is, the minimum wall thickness at the third chamfer R3 is: .

[0177] If the material of the base F1 is different from that of the pipe body, let the yield strength of the pipe body material be Q0, and the yield strength of the material used for the base F1 be Q1. Similarly, in order for the single-layer structure to meet the strength requirements, the stress at the third chamfer R3 and the stress on the inner wall of the pipe body should satisfy:

[0178] (32)

[0179] Then we have:

[0180] (33)

[0181] Considering the pressure of the sealing ring, the wall thickness at the third chamfer R3 is:

[0182] (34)

[0183] 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.

[0184] For a multi-layer structure of a composite socket, the design method comprises the following steps:

[0185] Step c1: Based on the same strain of each layer of the composite socket and the non-single-layer material equilibrium equation, the calculation formula for the hoop stress of any non-base layer F2 is obtained.

[0186] Step c2: Determine the relationship between the thickness of any non-base layer F2 and the thickness of the base layer F1. The determination principle is that the ratio of the stress of the non-base layer F2 to the yield strength of the material of the layer is not greater than the ratio of the stress of the inner wall of the pipe body to the yield strength of the pipe body material.

[0187] Step c3: Calculate the thickness of the non-base layer and / or base layer according to the relationship determined in step c2.

[0188] 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.

[0189] 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. Figure 5 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, and the second cone and the first cone are two coaxially connected conical pressure vessels, which share the same central axis with the socket F. Figure 5 and 6 The dotted line between the first chamfer R1 and the second chamfer R2 is the inner wall section of the first cone. The first chamfer R1 and the second chamfer R2 are located on the inner wall of the first cone, the portion between the inner wall of the first cone and the outer wall of the socket F 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 F (the innermost non-base layer F2) is an annular reinforcement ring j.

[0190] For the socket F, 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:

[0191] (35)

[0192] 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.

[0193] For a composite socket, assuming there are m layers of non-base layer F2 and one layer of base layer F1, then:

[0194] (36)

[0195] Among them, S1, S2, …, S m are the thickness of each non-base layer F2, S b is the thickness of the base layer F1, , , …, are the hoop stresses of each non-base layer, is the hoop stress of the base layer F1.

[0196] Since it is a thin-walled container, the hoop tensile stress of the composite socket can be regarded as the tensile stress of the composite beam, that is, the strain of the base F1 and the strain of each non-base layer F2 The same means that:

[0197] (37)

[0198] Right now:

[0199] (38)

[0200] Among them, E1, E2, …, E m are the elastic modulus of each non-base layer F2, E bis the elastic modulus of the base layer F1.

[0201] The stresses in formula (36) are To express, is the hoop stress of the nth non-base layer F2, where n is 1~m, then E n is the elastic modulus of the nth non-base layer F2, S n is the thickness of the nth non-base layer F2. Based on formula (38), (36) is rearranged to obtain:

[0202] (39)

[0203] Formula (39) can be rearranged to obtain the hoop stress of the nth non-base layer F2:

[0204] (40)

[0205] Step c1 is thus completed.

[0206] In step c2, in order to make the composite socket meet the strength requirements, the internal stress of the nth non-base layer F2 and the inner wall stress of the pipe body satisfy:

[0207] (41)

[0208] (42)

[0209] Among them, Q0, Q n are the yield strengths of the pipe body and the nth non-base layer F2, respectively. The yield strength of the base layer F1 is Q b ;in is the hoop stress of the pipe body, S is the thickness of the pipe body, D0 is the outer diameter of the pipe body, and p is the pressure given by the fluid in the pipe.

[0210] Combining formulas (40), (41), and (42), we get:

[0211] (43)

[0212] Formula (43) is the relationship between the thickness of any non-base layer F2 and the thickness of the base layer F1.

[0213] Based on formula (43), the thickness of the non-base layer F2 can be calculated when the thickness of the base layer F1 is known, and the thickness of the base layer F1 can be calculated when the thickness of the non-base layer F2 is known. Two embodiments are used to illustrate this.

[0214] Embodiment 1

[0215] In this embodiment, there is no limit on the thickness of the non-base layer F2. For a multi-layer structure, it is only necessary to ensure that the layer with the smallest yield stress meets the strength requirement, and this layer is set as the minimum stress layer, and the thickness of other layers is not less than the thickness of the minimum stress layer, then the entire multi-layer structure meets the strength requirement.

[0216] The yield stress of the base layer F1 is generally greater than the yield stress of the non-base layer F2, so the layer with the weakest yield stress is the non-base layer. Assuming that the yield stress of the nth layer of non-base layer F2 is the weakest, if yielding occurs, the nth layer should yield first. Based on this, if the thickness of the nth layer of non-base layer F2 meets the strength requirements, the minimum thickness of the base layer F1 and the thickness of other non-base layers F2 are not less than the thickness of the nth layer of non-base layer F2, then the strength of the base layer F1 and other non-base layers F2 will also meet the requirements. In addition, the thickness of the base layer F1 is generally greater than the thickness of the non-base layer. Therefore, it is necessary to calculate the thickness of the nth layer of non-base layer F2. Formula (41) is rewritten as:

[0217] (44)

[0218] in, is the maximum stress inside the nth non-base layer F2, that is From formula (40), we can see that in the analysis area, i.e., the first cone, Take the maximum value, then D takes the maximum value, and the thickness of other non-base layers F2 takes the minimum value, that is, The maximum point is located at the first chamfer R1, the maximum value of D in the first cone is D3, the thickness of other non-base layers F2 is not less than the thickness of the nth non-base layer F2, that is, the thickness of the non-base layer F2 is Sn, and the half cone angle is α 1. In Figure 5 and 6 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:

[0219] (45)

[0220] Substituting formula (42) and (45) into formula (44), we can obtain:

[0221] (46)

[0222] From formula (46), we get:

[0223] (47)

[0224] In this embodiment, S b Take the thickness of the base layer F1 at the first chamfer R1.

[0225] From the above, the number of layers m of the non-base layer F2 and the material of each layer are known, the material, thickness, diameter and other dimensions of the pipe body are also known, so the yield strength and elastic modulus of each layer are also known, and the internal shape and size of the socket F are in accordance with relevant standards. Therefore, the semi-cone angle α 1 is also known. In other words, in formula (47), only S n and S b Two unknowns. The thickness S of the base layer F1 at the first chamfer R1 is given based on experience or other requirements. b1 After that, we can find S n The minimum value of .

[0226] Therefore, according to formula (47), the minimum thickness of the non-base layer F2 with the smallest yield stress can be obtained, and the thickness of other non-base layers F2 should not be less than this value.

[0227] For the base layer F1, the thickness S of the base layer F1 at the first chamfer R1 is b1 It is given that the thickness S of the base layer F1 at the third chamfer R3 b3 It has also been calculated by formula (34), so we only need to give or calculate the thickness S of the base layer F1 at the second chamfer R2: b2 The thickness of the base layer F1 at each location can be determined to ensure that the multi-layer structure of the composite socket meets the strength requirements.

[0228] For the thickness S of the base layer F1 at the second chamfer R2 b2 , can be given directly, or can be set to the thickness S of the base layer F1 at the first chamfer R1 b1 Equal, can also be obtained by using formula (43), specifically, formula (43) is rewritten as:

[0229] (48)

[0230] but:

[0231] (49)

[0232] The thickness S of the nth non-base layer F2 has been calculated by formula (47): n , and the thickness of other non-base layers is not less than S n , then formula (49) becomes:

[0233] (50)

[0234] The thickness S of the base layer F1 at the second chamfer R2 is obtained by formula (50): b2 .

[0235] From the above, the thickness of each non-base layer and the thickness of the first chamfer R1, the second chamfer R2 and the third chamfer R3 of the base layer F1 are calculated. According to the requirement of the trumpet shape of the outer wall of the socket F, the outer walls corresponding to each chamfer are connected in sequence, and the thickness of each part of the base layer F1 and the thickness of each part of the composite socket are obtained. The outer diameter at the end face of the socket F can be greater than the outer diameter at the third chamfer R3, and the specific value can be flexibly selected.

[0236] When the non-base layer F2 has only one layer, that is, the conical shell is a double-layer material, then formula (36) is in the following form:

[0237] (51)

[0238] Formulas (37) and (38) are organized into the following form:

[0239] (52)

[0240] For double-layer sockets, the material of the non-base layer F2 (i.e., inner lining layer) is generally stainless steel, and the material of the base layer F1 (i.e., outer layer) is generally Q235 steel or ductile iron. The yield stress of the non-base layer F2 is generally smaller than that of the base layer F1, that is, if yielding occurs, the non-base layer F2 should yield first. Based on this, the stress in formula (51) should be To express, based on formula (52), rearranging formula (51) we can get:

[0241] (53)

[0242] Formula (53) can be rearranged to obtain the hoop stress of the non-base layer F2:

[0243] (54)

[0244] In step c2, in this embodiment, formula (46) is in the following form:

[0245] (55)

[0246] The thickness of the non-base layer F2 is:

[0247] (56)

[0248] That is, the minimum thickness of the non-base layer F2 is: .

[0249] At this time, formula (50) is: (57)

[0250] A specific example is shown below.

[0251] Assume the nominal diameter of the purchased base pipe is DN600 mm, E b =200GPa, E n =193GPa, the base F1 material and the pipe body material are both Q235 steel, and the yield strength of Q235 steel is 235MPa, that is, Q1=Q0=235MPa. The non-base F2 material is 304 stainless steel, with a yield strength of Qn=205MPa, S=7mm, D0=630 mm, D1=673.4 mm, D3=668 mm, D4=642.6 mm, =8.21 o , =5.82 o Assume that the thickness of the first chamfer R1 of the base layer F1 is Sb=6 mm.

[0252] The thickness of the non-base layer F2 is calculated by formula (56) to be Sn≥2.38 mm, and the thickness of other non-base layers is not less than 2.38 mm.

[0253] For the thickness S of the base layer F1 at the second chamfer R2 b2， It can be set equal to the thickness Sb=6 mm at the first chamfer R1, or calculated according to formula (56). When formula (56) is used, take Sn=2.38 mm, substitute each parameter into formula (56), and calculate S b2 ≥5.68 mm.

[0254] For the thickness S of the base layer F1 at the third chamfer R3 b3 , obtained by formula (31), substituting each parameter into formula (31), and obtaining S b3 ≥10.53 mm.

[0255] When the inner wall of the entire base pipe socket F is completely covered by the lining, it is equivalent to the entire composite socket being a multi-layer structure. At this time, the design method of the thickness of the base layer F1 and each non-base layer F2 is the same as the design method when the inner wall of the base pipe socket F is not completely covered by the lining. The difference is that the base layer thickness S corresponding to the third chamfer R3 is b3 At this time, the base thickness S corresponding to the third chamfer R3 b3 The method for calculating the thickness of the base layer F1 at the second chamfer R2 is the same as that for calculating the thickness of the base layer F1 at the second chamfer R2. It can be given directly, or it can be set equal to the thickness of the base layer F1 at the first chamfer R1. It can also be calculated using formula (43). Specifically, formula (48) yields:

[0256] (58)

[0257] In the case where the inner wall of the entire base pipe socket F is completely covered by the lining, there is generally no third step 3, or in other words, the inner diameter of the third step 3 is equal to the inner diameter of the fourth step 4, and the two are combined into a large step, such as Figure 7 shown.

[0258] Embodiment 2

[0259] In this embodiment, the thickness of the non-base layer F2 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 F. The thickness of the liner ring is generally about 1 mm, which is equivalent to the known thickness of the non-base layer F2. 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 F1 under this premise.

[0260] Based on this, formula (43) is rewritten as:

[0261] (59)

[0262] From the above analysis, it can be seen that in the multi-layer structure of the composite socket, the first chamfer R1 and the second chamfer R2 are the places where stress is most likely to concentrate, and the thickness of each non-base layer is uniform, that is, the thickness of each non-base layer is consistent. Therefore, it is only necessary to ensure that the thickness of the base layer F1 at the first chamfer R1 and the second chamfer R2 meets the strength requirements. The base layer F1 manufactured according to the above requirements and other parts of the multi-layer structure must also meet the strength requirements.

[0263] The thickness of the base layer F1 at the first chamfer R1 is S b1 , in formula (59), D is D3, and the half cone angle α for α 1, then formula (59) is:

[0264] (60)

[0265] The thickness of the base layer F1 at the second chamfer R2 is S b2 , in formula (59), D is D4, and the half cone angle α for α 1, then formula (59) is:

[0266] (61)

[0267] From the above, after knowing the thickness and material of the non-base layer F2, the minimum thickness of the base layer F1 at the first chamfer R1 and the second chamfer R2 can be calculated. If the outer wall of the composite socket is designed according to the above-mentioned trumpet-shaped requirement, the composite socket meets the strength requirement.

[0268] A specific example is shown below.

[0269] 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 F2, 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 (60), (61), and (34) respectively:

[0270] S b1 ≥7.33 mm, S b2 ≥7.01 mm, S b3 ≥10.53 mm.

[0271] When the thickness of the base pipe meets the above requirements, the composite socket formed by sleeve-fitting the liner ring inside meets the strength requirements.

[0272] When the inner wall of the entire base pipe socket F is completely covered by the lining, it is equivalent to the entire composite socket being a multi-layer structure. At this time, the design method of the thickness of the base layer F1 and each non-base layer F2 is the same as the design method when the inner wall of the base pipe socket F is not completely covered by the lining. The difference is that the base layer thickness S corresponding to the third chamfer R3 is b3 At this time, the thickness of the base layer F1 at the third chamfer R3 is S b3 , in formula (59), D is D1, and the half cone angle α for α 2, then formula (59) is:

[0273] (62)

[0274] 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 method for designing the thickness of a single-layer spigot and a composite spigot of a spigot-and-socket pipe, characterized in that: The design method of the single-layer socket thickness includes the following steps: Step a1: Obtain the differential equation of radial deformation of the ring and the general solution of the differential equation according to the differential equation of the deflection curve of the ring and the moment balance equation; Step a2: Calculate and obtain a calculation formula for the maximum radial deformation of the ring according to the general solution of the differential equation obtained in step a1; Step a3: Calculating the minimum thickness of the single-layer spigot of the base pipe that meets the deformation requirements according to the calculation formula of the maximum radial deformation obtained in step a2; 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; 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.

2. 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 ring into the second form of the differential equation by introducing a coefficient, so as to obtain the third form of the differential equation; Step a24: Obtain a calculation formula for the maximum change in the diameter of the circular ring with coefficients according to the third form of the differential equation obtained in step a23.

3. 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.

4. The design method according to claim 1, 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: ； Where I = S 3 / 12, S is the wall thickness of the tube body.

5. The design method according to claim 1, characterized in that: For the same base pipe, along the direction from the socket to the spigot, the inner wall of the socket includes a first step, a second step, a third step, a fourth step, and a fifth step arranged in sequence, one end of the inner wall of the pipe body is a sixth step, the chamfer at the connection between the fourth step and the fifth step is a first chamfer, the chamfer at the connection between the fifth step and the sixth step is a second chamfer, the chamfer at the connection between the second step and the third step is a third chamfer, the chamfer at the connection between the first step and the second step is a fourth chamfer, and the chamfer at the connection between the third step and the fourth step is a fifth chamfer.

6. The design method according to claim 5, 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.

7. The design method according to claim 6, 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.

8. The design method according to claim 7, 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 materials of the pipe body and the nth non-base layer F, 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: .

9. The design method according to claim 8, 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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