A method for establishing a small-diameter internal thread mechanical model based on intensity relation

Abstract

The application provides a method for establishing a small-diameter mechanical model of an internal thread based on an intensity relationship, a force analysis model of an internal and external thread and a screw rod is established, force analysis is performed on the internal and external thread and the screw rod, then a mathematical expression of the mechanical model of the small diameter D1 of the internal thread is obtained according to the yield limit of the rod part of the external thread: D 1实 ≤D‑0.7578P; a relationship model of the actual value D1 of the small diameter and the standard value D1 of the internal thread is established according to the mechanical model of the small diameter. Through the establishment of the small-diameter model of the internal thread, under the premise of ensuring the strength of the internal thread, the over pursuit of the fullness of the tooth profile of the internal thread due to the strength of the thread connection is avoided, the processing cost in the cold extrusion process of the internal thread is greatly reduced; the establishment of the small-diameter model also makes it necessary to only detect the small diameter of the internal thread after the detection by using a go-no-go gauge in the detection process of the internal thread, the qualified rate of the strength of the thread connection is greatly improved when the small diameter is qualified, the detection efficiency is improved, and the detection cost is reduced.

Description

Technical Field

[0001] This invention relates to a method for establishing a mechanical model of the minor diameter of an internal thread based on strength relationships. Background Technology

[0002] Cold-extruded internal threads often exhibit incomplete or recessed tooth profiles, which can affect the connection strength of the thread. Currently, the goal is to maximize the tooth height ratio during processing to ensure the connection strength of the internal thread. For example, CN115625281A discloses a method for improving the tooth height ratio in a cold-extruded internal thread process. This method utilizes the elastic rebound caused by multiple extrusions, where the volume recovered by the elastic rebound compensates for the volume of the tooth profile recess, thus increasing the tooth height ratio. Furthermore, slightly increasing the diameter of the pre-drilled hole reduces the torque on the tap during multiple extrusions, increasing the tap's lifespan. However, the multiple extrusions significantly increase the processing cycle, reduce processing efficiency, and lead to increased processing costs. Moreover, before processing a new batch of internal threads, trial processing of the new product and sample inspection are required. Existing inspection methods typically involve using go and no-go gauges and measuring the thread pitch diameter to test thread strength. For example, CN110567423B discloses a device and method for detecting large-pitch tapered internal threads. This method uses a thread detection component to measure the opening distance and longitudinal changes of the thread gauge at three different thread segments. The measurement data is then transmitted to a calculation component, which processes and calculates the measurement data to obtain the thread taper, the thread pitch diameter at a specified point, and the thread pitch. However, the thread pitch diameter alone cannot eliminate defects caused by thread concavity. When the thread concavity is too deep, the go / no-go gauge cannot be used, and the measurement of the thread pitch diameter cannot guarantee the thread connection strength. If the internal thread is formed by cutting, it is also necessary to perform go / no-go testing followed by minor diameter D1 testing to improve the thread connection strength. Summary of the Invention

[0003] To address the aforementioned technical problems, this invention provides a method for establishing a mechanical model of minor diameter internal threads based on strength relationships.

[0004] The present invention is achieved through the following technical solutions.

[0005] This invention provides a method for establishing a mechanical model of a minor diameter internal thread based on strength relationships, the steps of which are as follows:

[0006] S1. Establish a stress analysis model for the internal and external threads and the rod to determine that shear stress is the cause of the failure of the internal and external threads.

[0007] S2. Establish the model for the maximum shear stress;

[0008] S3. Based on the connection equation between the normal stress of the screw and the shear stress of the bolt's external thread. τ is used to verify the model for the maximum shear stress.

[0009] S4. When the shear stress is at its maximum value, establish a thread engagement depth model;

[0010] S5. Based on the verification conditions for the yield strength of the external thread shank, the mechanical model of the minor diameter D1 of the internal thread is obtained: D 1实 ≤D-0.7578P;

[0011] S6. Establish the actual value D of the minor diameter of the internal thread. 1实 The relationship model between the standard value D1 and the standard value;

[0012] S7. Establish the inequality equations for judging the failure modes of the bolt shank and the bolt external thread.

[0013] The process for establishing the maximum shear stress is as follows:

[0014] a. Establishing a threaded annular shear mechanical model:

[0015] b. Utilization Verification was performed; the nut thickness is ≥0.8D.

[0016] In the formula: Fs: shear force on a single thread, b: thread length, h: thread thickness, σ 杆 The normal stress A on the bolt / screw. 杆 τ represents the stress area of ​​the bolt / screw. max σ: Maximum shear stress. s : Yield limit of the material.

[0017] When the shear stress is at its maximum, and assuming the engagement depth of the internal and external threads is χ, find the value of χ:

[0018] Ensure that the shear stress is less than the bolt's yield strength.

[0019]

[0020] σ s =F / A 杆，

[0021]

[0022]

[0023] Substitute (2) into (1)

[0024]

[0025] A 纹 ≥A 杆 ,

[0026]

[0027] When τ reaches its maximum value, A 纹 =A 杆 ,

[0028] Let b = (d - 2χ)π (0 ≤ χ ≤ 0.54125P), h = 0.125P + 1.1547χ, and substitute into (3);

[0029] A 纹 =(d-2x)π(0.125P+1.1547x)×0.48114Z,

[0030] =0.48114π[0.125dP+(1.1547d-0.25P)x-2.3094x 2 Z,

[0031] According to GB / T3098.2 "Mechanical Properties of Fasteners and Nuts" standard, Z = 0.8d / P

[0032] A 纹 =

[0033] 0.48114π[0.125dp+(1.1547d-0.25P)x-2.3094x 2 Z =

[0034] π[0.048125d 2 +(0.4446d 2 / P-0.09625d)x-0.8891x 2 d / P],

[0035] According to HB6443 "General Specification for Nuts" (aviation standard) and GB / T3098.1 "Mechanical Properties of Fasteners - Bolts, Screws and Studs", the following is obtained:

[0036] Let A 纹 =A 杆 ,

[0037] △A=A 杆 -A 纹 =

[0038] π[0.25d 2 -0.6134dP + 0.3763P 2 -0.048125d 2 -(0.4446d) 2 / P-

[0039] 0.09625d)x+0.8891x 2 d / P] = 0,

[0040] When d = 6P,

[0041] 5.335x 2 -(16.006P-0.5775P)x+7.272P 2 -3.6804P 2

[0042] +0.3763P 2 =0

[0043] Solving for x, we get x1 = 0.285P and x2 = 2.6P. Since x2 > 0.54125P, we discard x2 and finally get x = 0.285P.

[0044] When d = 9P,

[0045] Solving for x, we get x1 = 0.347P and x2 = 4.046P. Since x2 > 0.54125P, we discard x2 and finally get x = 0.347P.

[0046] When d = 6P, x = 0.285P, A 杆 =A 纹 ;

[0047] When d = 9P, x = 0.347P, A 杆 =A 纹 Considering bolt manufacturing tolerances and strength safety factors, excessive torque increases the probability of bolt breakage, therefore x needs to be greater than 3 / 8H.

[0048] Take: x = 3 / 8H + 0.5 / 8H = 3.5 / 8H = 0.3789P;

[0049] H: Original triangle height of the thread, A 纹 A is the area of ​​equivalent shear stress on the external thread of the bolt. 杆 Bolt stress area, d nominal diameter of external thread, P pitch, Z number of turns of nut and bolt engagement, d1 basic minor diameter of bolt thread, d3 maximum minor diameter of bolt.

[0050] Verify x = 0.3789P:

[0051] A 纹 =π[0.048125d 2 +(0.4446d 2 / P-0.09625d)x-

[0052] 0.8891x 2 d / P],

[0053] =π(0.2166d) 2-0.1641dP);

[0054]

[0055] ΔA=A 纹 -A 杆 =π(0.2166d) 2 -0.1641dP)-π(0.25d 2 -

[0056] 0.6134dP + 0.3763P 2 )=π(-0.0334d 2 +0.4493dP -0.3763P 2 );

[0057] When d = 6P

[0058] ΔA=π(-1.2P 2 +2.6958P 2 -0.3763P 2 ) = 1.1195π > 0,

[0059] A 纹 >A 杆 ;

[0060] When d = 9P, ΔA = π(-2.7054P) 2 +4.0437P 2 -0.3763P 2 ) = 0.962π > 0,

[0061] A 纹 >A 杆 ;

[0062] Therefore, when the internal thread and bolt engagement depth x = 0.3789P, the screw will break first or fail due to yield deformation when there is excessive torque.

[0063] The minimum diameter that ensures the internal thread will not disengage when over-tightened:

[0064] D 1实 ≤D-2×0.3789P=D-0.7578P;

[0065] D is the nominal diameter of the internal thread.

[0066] Actual value of minor diameter D of internal thread 1实 The difference between the standard value D1 and the standard value D1 is calculated as ΔD1 = D 1实 -D1=D-0.7578P-(D-1.0825P)=0.3247P.

[0067] The beneficial effects of this invention are as follows: by establishing the minor diameter model of the internal thread, while ensuring the strength of the internal thread, it avoids enterprises from excessively pursuing the fullness of the internal thread profile due to thread strength, and greatly reduces the processing cost in the cold extrusion process of the internal thread.

[0068] The establishment of the minor diameter model also means that during the internal thread inspection process, after using the go / no-go gauge, only the minor diameter of the internal thread needs to be inspected. If the minor diameter is qualified, the thread strength is qualified, thus improving inspection efficiency and reducing inspection costs. Attached Figure Description

[0069] Figure 1 This is a schematic diagram of the bolt and nut engagement of the present invention;

[0070] Figure 2 This is a schematic diagram of the threaded unfolded cantilever beam structure of the present invention;

[0071] Figure 3 This is a schematic diagram of the threaded cross-section structure of the present invention;

[0072] Figure 4 This is a schematic diagram of the shear stress distribution law of the present invention;

[0073] Figure 5 This is a schematic diagram illustrating the distribution law of normal stress during thread bending according to the present invention;

[0074] Figure 6 This is a schematic diagram of the shear mechanics model of the threaded ring of the present invention.

[0075] Figure 7 This is a schematic diagram of the shear force decomposition of the threaded annulus of the present invention;

[0076] Figure 8 This is a schematic diagram of the internal thread torque clamp structure of the present invention;

[0077] Figure 9 This is a graph showing the fastening force (tension) during the testing process of this invention;

[0078] Figure 10 This is a photograph of the actual bolt shank fractured during the test of this invention;

[0079] Figure 11 These are actual images showing bolt shank fracture, thread shearing, and changes in the internal thread of the nut during the test process of this invention.

[0080] Figure 12 This is a photograph of a bolt thread that slipped due to shear force during the test of this invention. Detailed Implementation

[0081] The technical solution of the present invention is further described below, but the scope of protection is not limited to what is described.

[0082] The strength checks for bolt external threads and nut internal threads both follow the Sopwith approach, treating the unfolded thread teeth as a cantilever beam structure for analysis. That is, bending normal stress... or It is based on the assumption that the thread teeth are cantilever beam structures, and is derived from the transverse bending theory of mechanics of materials. The transverse bending theory uses two assumptions in the calculation of pure bending normal stress: one is planar design; the other is that there is no stress on the longitudinal fiber surface.

[0083] First point This applies to actual beams where the beam section height (tooth thickness) is much smaller than the span (tooth height). Only such beams can satisfy the planar assumption and have a very small deviation from reality. The second assumption is that only when the height / span (tooth thickness / tooth height) ratio is very small (<1) will the ratio of longitudinal normal stress to bending normal stress be very small, and only then can the longitudinal normal stress be ignored.

[0084] In contrast, thread teeth have a thickness greater than their height. Secondly, consider a cantilever beam, open at one end and fixed at the other, while a thread is actually a closed loop. or It differs from reality.

[0085] Regarding the distribution law of shear stress on the cross section, mechanics of materials also made two assumptions: (1) the direction of shear stress at each point on the cross section is parallel to the shear force Fs; (2) the shear stress is uniformly distributed along the width of the cross section, and only when the height h of the cross section is greater than the width b is the accuracy sufficient, while the assumption of the threaded cantilever beam is exactly the opposite, b >> h. The results calculated using the assumptions of bending normal stress and shear stress distribution are... or This differs from reality, and the threaded cantilever beam is assumed to be in a state where one end is open and the other end is fixed, while the thread is actually closed.

[0086] The calculation process for the maximum thread shear stress in this application is as follows:

[0087] Establish a threaded annular shear mechanical model, see Figures 6-7 As shown, when the nut is tightened, the bolt shank is subjected to a uniform load q2, and the external thread is subjected to a uniform load q1, with q1 and q2 in opposite directions. The number of external thread teeth corresponds approximately one-to-one with the number of internal thread teeth on the nut. Each bolt thread tooth is a ring structure, similar to multiple overlapping flat washers. The internal thread teeth are similar to multiple dies, and the bolt shank is similar to multiple punches; each external thread punch corresponds continuously to each internal thread die. Because the tolerance between the minor diameter of the internal thread and the minor diameter of the external thread is very small, similar to the principle of punching, the circumference πd is subjected to bending and shearing, and ultimately, the shear stress τ causes metal fracture and separation.

[0088] Shear force Fs=bhτb Considering the safety factor of the punching machine and the high punching speed and large deformation resistance, the punching shear force is calculated using Fs = bhτ. b Multiplying by 1.3, Fs′=1.3Fs=1.3bhτ b The nuts (or bolts) are tightened slowly, so there is no risk of damage to the punch press equipment; therefore, the actual shear stress is... This model takes into account the closed loop of the thread, and the gap between the minor diameter of the bolt and the minor diameter of the nut is very small. From a mechanical point of view, it is a process of shear deformation to shear fracture. Therefore, the strength check only needs to check the shear stress.

[0089] Example 1: The process of establishing the maximum shear stress model is as follows:

[0090] From the mechanics of materials, we know the relationship between shear stress and normal stress. Considering that the nut thickness follows the requirement of ≥0.8D, according to GB / T3098.2 "Mechanical Properties of Fasteners and Nuts", that is, when the nut thickness is ≥0.8D, the screw will basically break under over-tightening.

[0091] ZP≥0.8D, Z is the number of turns of the bolt and nut engagement. This is necessary to increase the probability of screw breakage under excessive torque. Now let's prove it. When K = what value? If the screw is over-twisted, it will break.

[0092] 1. First, the shear area A of the nut's internal thread. 母 =Dπ×0.75P, while the shear area A of the bolt external thread 栓 =d1π×0.75P. Because D>d1, therefore A 母 >A 栓 Therefore, we will only discuss the stress situation of the external thread of the bolt.

[0093] 2. The maximum shear stress should meet the following requirements.

[0094] make

[0095] make Pick

[0096]

[0097] From the formula, we get A. 纹 =A 杆 If A 纹 >A 杆 First, the screw undergoes plastic deformation under normal stress. If A 纹 <A 杆Then the external thread undergoes plastic deformation under shear force, A 纹 This represents the area of ​​the equivalent shear stress in the thread.

[0098] According to HB6443 "General Specification for Nuts" and GB / T3098.1 "Mechanical Properties of Fasteners - Bolts, Screws and Studs", the following standards were obtained:

[0099] According to GB / T197, EI G =+(15+11P)μm,EI H =0(1),

[0100] External thread major diameter grade 6 tolerance

[0101] For internal thread minor diameter grade 6, tolerance P ≥ 1 mm, TD1(6) = 230P 0.7 μm,

[0102] When 0.2mm≤P≤0.8mm, TD1(6)=(433P-190P) 1.22 μm(3); meshing depth 2x (limit case) = (d - Td) - (D1 + EI) G +TD1)(4);

[0103] x = [(d - Td) - (D1 + EI)] G +TD1)〕 / 2(5);

[0104] If we take formulas (1), (2), and (3) as Td, TD1, and EI, G Substitute (4) and (5)

[0105] Calculating x is very complex and there is no exact solution.

[0106] This invention uses specific values ​​from GB / T197 "Tolerances for Ordinary Threads" for calculation.

[0107] We selected the following: ① d = 6, P = 1, d / p = 6. ② d = 12, P = 1.75, d / P = 6.857. ③ d = 20, P = 2.5, d / P = 8. ④ d = 36, P = 4, d / P = 9.

[0108] A 纹 A 杆 、(A 纹 -A 杆 ) / A 纹 (Relative error %) is shown in Table 1.

[0109] Table 1

[0110]

[0111] Table 1 uses K = 1.2

[0112] Since a nut thickness ≥0.8D is specified by national standards, a nut thickness ≥0.8D will definitely not come off under normal circumstances, meaning it must be A. 纹 -A 杆 >0, if K is 1.1, A must be satisfied. 纹 -A 杆 When the relative error is greater than 0 and the relative error is less than 0.8%, then... This does not conform to the design principle of ensuring bolt breakage during overtightening, with a nut thickness ≥ 0.8D. By establishing a threaded annular shear mechanical model, the final mechanical model formula was determined:

[0113] Example 3: When the shear stress is at its maximum, what is the minimum engagement depth χ of the internal and external threads required to ensure that the shear stress is less than the bolt's yield strength of 0.5774σ? s :

[0114]

[0115] Take the maximum shear stress:

[0116]

[0117]

[0118] Substitute (8) into (7):

[0119]

[0120] A 纹 ≥A 杆 ,

[0121] When τ reaches its maximum value Therefore A 纹 =A 杆 ,

[0122] Given the thread circumference b = (d - 2χ)π (0 ≤ χ ≤ 0.54125P), as... Figure 2 As shown, tooth thickness h = 0.125P + 2y = 0.125P + 2tg30°χ = 0.125P + 1.1547χ. Substitute into (9);

[0123] A 纹 =(d-2x)π(0.125P+1.1547x)×0.48114Z

[0124] =0.48114π[0.125dP+(1.1547d-0.25P)x-2.3094x2 Z,

[0125] According to GB / T3098.2 "Mechanical Properties of Fasteners and Nuts", Z = 0.8d / P.

[0126] A 纹 =

[0127] 0.48114π[0.125dp+(1.1547d-0.25P)x-2.3094x 2 Z =

[0128] π[0.048125d 2 +(0.4446d 2 / P-0.09625d)x-0.8891x 2 d / P],

[0129] According to HB6443 "General Specification for Nuts" (aviation standard) and GB3098.1 "Mechanical Properties of Fasteners - Bolts, Screws and Studs", the following is obtained:

[0130] Let A 纹 =A 杆 ,

[0131] △A=A 杆 -A 纹 =

[0132] π[0.25d 2 -0.6134dP + 0.3763P 2 -0.048125d 2 -(0.4446d) 2 / P-

[0133] 0.09625d)x+0.8891x 2 d / P] = 0,

[0134] When d = 6P,

[0135] 5.335x 2 -(16.006P-0.5775P)x+7.272P 2 -3.6804P 2

[0136] +0.3763P 2 =0

[0137] Solving for x, we get x1 = 0.285P and x2 = 2.6P. Since x2 > 0.54125P, we discard x2 and finally get x = 0.285P.

[0138] When d = 9P,

[0139] Solving for x, we get x1 = 0.347P and x2 = 4.046P. Since x2 > 0.54125P, we discard x2 and finally get x = 0.347P.

[0140] When d = 6P, x = 0.285P, A 杆 =A 纹 ;

[0141] When d = 9P, x = 0.347P, A 杆 =A 纹 Considering bolt manufacturing tolerances and strength safety factors, excessive torque increases the probability of bolt breakage, therefore x needs to be greater than 3 / 8H.

[0142] Take: x = 3 / 8H + 0.5 / 8H = 3.5 / 8H = 0.3789P; 0.3789P is greater than the median diameter (3 / 8H).

[0143] Verify x = 0.3789P:

[0144] A 纹 =π[0.048125d 2 +(0.4446d 2 / P-0.09625d)x-

[0145] 0.8891x 2 d / P],

[0146] =π(0.2166d) 2 -0.1641dP);

[0147]

[0148] ΔA=A 纹 -A 杆 =π(0.2166d) 2 -0.1641dP)-π(0.25d 2 -0.6134dP + 0.3763P 2 )=π(-0.0334d 2 +0.4493dP -0.3763P 2 ),

[0149] When d = 6P

[0150] ΔA=π(-1.2P 2 +2.6958P 2 -0.3763P 2 )=1.1195π>0

[0151] A 纹 >A 杆

[0152] When d = 9P, ΔA = π(-2.7054P) 2 +4.0437P 2 -0.3763P 2 )=0.962π>0

[0153] A 纹 >A 杆

[0154] Therefore, at an internal thread and bolt engagement depth x = 0.3789P, the bolt will either break first or fail due to yield deformation when subjected to excessive torque; A 纹 A is the area of ​​equivalent shear stress on the external thread of the bolt. 杆 Bolt stress area, d nominal diameter of external thread, P pitch, Z number of turns of nut and bolt engagement, d1 basic minor diameter of bolt thread, d3 maximum minor diameter of bolt;

[0155] When x = 0.3789P, the equivalent shear stress area of ​​the bolt's external thread is greater than the stress area of ​​the bolt shank. x = 0.3789P is the thread height of the nut's internal thread when the bolt and nut are engaged. Therefore, the maximum minor diameter is equal to D - 2 × 0.3789P. (D) 1实 ≤D-2X0.3789P can prevent the bolt threads from coming off during the connection process.

[0156] Establish a relationship model between the actual value D1 of the minor diameter of the internal thread and the standard value D1.

[0157] The difference between D1_actual and standard D1 is ΔD1 = D1_actual - D1 = D - 0.7578P - (D - 1.0825P) = 0.3247P.

[0158] Based on the actual value model of the minor diameter of the internal thread, D1actual = D1 + 0.3247P, 0.3247P is used to compare with the tolerance value of GB / T197, as shown in Table 2.

[0159] Table 2 Unit: mm

[0160]

[0161] The table shows that as DP increases, the differences 0.3247P-6H (tolerance value) and 0.3247P-(EI+6G tolerance value) increase. According to GB / T197 "Ordinary Threads, Tolerances", TD1(6) = 0.23P 0.7 EI: 0.015 + 0.011P, it can be proven that (0.015 + 0.011P + 0.23P) 0.7(P≥1)<0.3247P(P≥1)6G requirement, 0.23P 0.7 <0.3247P (P≥1)6H requirement. As long as the nut is manufactured according to the standard and meets the D1 standard value requirement, it can be fully guaranteed that the bolt will not disengage under over-tightening, and the internal thread connection is safe.

[0162] According to Appendix A of GB / T3098.2-2000 "Mechanical Properties of Fasteners and Nuts": "The breakage of the screw is sudden and relatively easy to detect, while the stripping occurs gradually, which is difficult to detect and increases the risk of accidents caused by fastener failure. Therefore, the design of threaded connections always aims for the failure to be screw breakage."

[0163] The actual value model of the minor diameter of the internal thread shows that as long as D1actual ≤ D1 + 0.3247P, it can be guaranteed that the screw will break under over-torsion and the thread will not disengage.

[0164] Compare the shear areas of external threads (bolts) and internal threads (nuts):

[0165] Let the shear area of ​​the bolt's external thread be A1, and the shear area of ​​the nut's internal thread be A2.

[0166] Analyze the relationship between the shear area of ​​the external thread, the shear area of ​​the internal thread, and χ.

[0167] The area of ​​the external thread A1 = π(d - 2χ)(0.125P + 1.1547χ) = π(d - 2.3094χ) 2 =π〔0.125Pd﹢(1.1547d-0.25P)χ-2.3094χ 2 〕,dA1 / dχ=π(1.1547d-0.25P-4.6188χ), when: d=6P, dA1 / dχ=π(6.6782P-4.6188χ)>0, (0≤χ≤0.54125P).

[0168] When d = 9P, dA1 / dχ > 0, (0 ≤ χ ≤ 0.54125P).

[0169] This indicates that A1 = π(d - 2χ)(0.125P + 1.1547χ) is an increasing function of χ. τ1 = Fs / A1, τ1 is a decreasing function of χ.

[0170] Shear area A2 of the nut's internal thread:

[0171] A2=π(D1﹢2χ)(0.25P﹢1.1547x),

[0172] A2 is an increasing function of x, and τ2 = Fs / A2 is a decreasing function of x. When the bolt, nut, and thread dimensions meet the requirements of GB / T192~193.196~197 standards, the shear surface of the internal thread is at the major diameter D. The area at D, A2 = πDh = πD × 0.875P, is naturally larger than the area at d1 of the external thread, A1 = 0.75Pπd1. When d = 6P~9P:

[0173] A2-A1=(0.125d﹢0.812P)πP>0.

[0174] The results show that the shear area of ​​the internal thread is greater than that of the external thread (A2 > A1). Therefore, as long as the minor diameter D1 of the nut's internal thread meets the standard size requirements, the shear area of ​​the internal thread being greater than that of the external thread will always hold true. When the mechanical properties of the external and internal threads are matched, they meet GB3098.2. From a strength analysis perspective, the internal thread (nut) will not be sheared off (disengaged) under shear stress; instead, the external thread will either break or disengage.

[0175] When a nut is over-twisted on a bolt, from a safety perspective, it is desirable for the fracture to occur in the bolt shank rather than due to thread stripping. For verification testing, we conducted tests using: a) aircraft blind hole internal threads of M3 material, 1Cr18Ni9T, HRC 24-27; and b) through hole M3 internal threads of carbon steel. M3 and 1Cr18Ni9T are representative because they constitute a very large proportion of aircraft blind hole internal threads. Test data and failure states are shown in Table 3 (see...). Figures 10-12 ).

[0176] Threaded connection test table 3

[0177]

[0178]

[0179]

[0180] The above analysis leads to the following conclusions: When the bolt, nut, and thread manufacturing meet the requirements of GB / T192~193 and GB / T196~197, the bolt is subjected to axial load during engagement with the nut, and the bolt shank will fracture due to excessive torque. The shear stress of the nut's internal thread is less than that of the bolt's external thread, and the nut is safer when the bolt and nut have the same strength σs.

[0181] A tightening test was conducted using the torque + yield point control method to ensure a tensile force of 3400 N for HB6443 M3 nuts with a strength of 900 MPa.

[0182] Torque T = KDF, T: tightening torque, K: torque coefficient, D: nominal thread diameter, F: axial clamping force, F is the tensile force on the bolt in this test due to reaction force, see... Figure 8 .

[0183] If the bearing surface is smooth and lubricated, K is 0.1; if no lubricant is applied, K is 0.2.

[0184] T = (0.1 ~ 0.2) × 3 × 10 -3 ×3400=(1.02~2.04)Nm.

[0185] To achieve greater precision, the experiment employed the torque + yield point tightening method. First, the internal thread (nut) was tightened with a certain torque so that the lower end face of the bolt head was in close contact with the end face of the internal thread and the interlayer plane.

[0186] Torque clamp dimensions are shown in the image. Figure 8 :

[0187] The elongation-axial tightening force (tensile force) curve using the yield point tightening method is generally horizontal. Figure 9 .

[0188] Some experimental results are shown below, and all experimental data are shown in Table 3.

[0189] Experiment 1: The cold extrusion pre-drilled hole d0 is 2.80. T and Z are qualified, but D1 is unqualified. The torque T = 2.44~2.96Nm (>2.04Nm) does not break after one torsion, indicating that the bolt is subjected to a tensile force greater than 3400N, which meets the standard requirements.

[0190] Experiment 2: d0 = 2.83~2.85, T and Z are qualified, D1 2.61~2.63, D1 is unqualified, engagement length 2.90~3.0mm. Torque T 2.7~2.79Nm, the screw broke at the failure point in the torsion test, which meets the ideal design requirements of nut in GB / T3098.2.

[0191] Experiment 3: d0 = 2.82~2.85, T and Z are qualified; D1 = 2.62~2.64, engagement length 4.70~4.80 > 0.8D; torsional fracture test, torque T 2.38~2.87 Nm, the screw broke at the failure point, which meets the ideal design requirements of GB / T3098.2 nut.

[0192] Experiment 4, d0=2.90D1=2.81T, Z is qualified, D1 is unqualified, the bolt thread is stripped (slipped) with an engagement length of 2.90~3.0.

[0193] The bolts used in the above tests were grade 12.9, with internal thread material of stainless steel 1Cr18Ni9Ti and a hardness of HRC24-27.

[0194] Experiment 5, d0 = 2.80~2.85, see test serial numbers 1, 2, 4, 5, D1 is unqualified, but the strength is qualified when the engagement length is >0.8D.

[0195] Similarly, 2.60≤D1≤2.63 is unqualified, but if the engagement length is greater than 0.8D, the strength can still be guaranteed to be qualified.

[0196] Experiment 6: Nut material: carbon steel, hardness HRA42-50, HRC<18, engagement length 2.26mm<0.8d (d=3). Test bolts: grade 8.8, serial numbers 23-25, T and Z are qualified. Indications: hardness is too low, thickness is too thin, D1 is qualified, bolt stripping.

[0197] Experiment 7, Table 2, Serial No. 19, M3 nut, T, Z, D1 qualified, nut material carbon steel HRA48-52, HRC < 18. Engagement length 2.40, T = 2.23 (> 2.04 Nm), meeting standard requirements. Bolt broken.

[0198] Experiment 8: Nut M3 T and Z are qualified, D1 = 2.82~2.85 is unqualified, H = 2.40 (see test numbers 20~22), torque T < 1.40 Nm is unqualified, bolt thread is stripped (slipped).

[0199] Experiment 10T, Z is qualified, D1 = 2.73-2.80 is unqualified, H = 2.26, see test numbers 26-30, torque < 1.22 Nm is unqualified. Bolt is loose.

[0200] The bolt thread stripping (slippage) indicates that the engagement length is short (<0.8d) and D1 is not qualified, failing to meet the standard force value requirement, and the bolt thread stripping (slippage) is located at the failure point.

[0201] In the experiment, the nut had a low hardness: HRC < 18 (HRC 40-52). Due to the stress characteristics of the internal thread structure, A 栓 (shear area of ​​bolt thread) < A 母 (Shear area of ​​nut thread), no stripping (slippage) was found in the nut. See Table 3, serial numbers 19-30.

[0202] Therefore, as nut D1 increases, formula X decreases. For the same engagement length of 2.40 (H = 0.8D), the torque decreases (see Table 2, serial numbers 19-22), proving the correctness of the minor diameter mechanical transverse type. b = (d - 2χ)π, h = (0.125P + 1.1547χ), if χ increases, τ decreases; if χ decreases, τ increases.

[0203] Stainless steel has a hardness of HRC24-27, which is lower than that of bolts (grade 12.9), but the bolt still broke even with D1≤2.63, proving that when χ=3.5 / 8H, A 杆 <A 纹 See Table 3, Serial No. 4.

[0204] In the tripping (slipping) test, the macroscopic fracture surface clearly shows that the clips are sheared. Figures 11-12 That is, the shear stress is the maximum! Strength verification can be done based on shear stress. (Allowable shear stress).

[0205] Macroscopic inspection of the broken bolts (grades 8.8 and 12.9) revealed bending deformation in the threads. The bending under stress began from the second thread (counting from the fracture point). Because the thread at the chamfered end of the first thread of the nut was incomplete, the engaged threads of the bolt all bent, indicating that each thread was subjected to uniform stress. Furthermore, A... 纹 >A 杆 The bending deformation indicates that the thread teeth have yielded, and eventually the bolt shank fractures. See Figures 10-12 .

[0206] Experimental conclusion:

[0207] D 1实 ≤D-0.7578P=3-0.3789=2.6211, for all cases where D1≤2.63 and nut thickness≥2.4, the torque is qualified, but the screw breaks. This proves that the formula D1 ≤D-0.7578P is correct.

[0208] 1. Through analytical analysis, it is theoretically proven that as long as the shear stress τ is adjusted, That's fine; numerous experiments have verified that the theoretical conclusions are correct. The χ dimension (D1 dimension) directly affects the fastener connection strength and the thread failure fracture mode of the fastener connection. These are not independent but related.

[0209] 2. Under static axial force (with a small frequency of change) on threaded connections, the engagement and interchangeability checks are usually performed using T and Z. Based on the above analysis and verification, the minor diameter D1 should be added to the check, and the major diameter d of the external thread should also be added and passed. When D1 meets the standard requirements, the nut thickness H ≥ 0.8D, and the internal thread meets the connection strength requirements.

[0210] Considering the safety and reliability of fasteners, and taking into account factors such as the material selection for internal threads and the hardness of heat treatment, cracks and permanent deformation can be inspected after unloading according to the tensile force value specified in HR6443 (aerospace products).

[0211] Based on the above analysis and verification, if there is a slight indentation at the crest of the extruded internal thread, and T and Z are qualified, an additional D1 dimension check is performed. If D1 is qualified, the connection strength is also qualified. This avoids reducing the diameter of the reserved hole in pursuit of thread integrity, which can lead to premature wear and breakage of the extrusion tap.

Claims

1. A method for establishing a mechanical model of minor diameter internal threads based on strength relationships, comprising the following steps: S1. Establish a stress analysis model for the internal and external threads and the rod to determine that shear stress is the cause of the failure of the internal and external threads. S2. Establish the model for the maximum shear stress; The process for establishing the maximum shear stress is as follows: a. Establishing a threaded annular shear mechanical model: τ max = , b. Using τ max ≤ The nut thickness must be checked and verified to be ≥0.8D; S3. Based on the connection equation between the normal stress of the screw and the shear stress of the bolt's external thread. 杆 = τ 纹 The maximum shear stress is checked; S4. When the shear stress is at its maximum value, establish a thread engagement depth model; S5. Based on the verification conditions for the yield strength of the external thread shank, the mechanical model of the minor diameter D1 of the internal thread is obtained: D 1实 ≤D-0.7578P; S6. Establish the actual value D of the minor diameter of the internal thread. 1实 The relationship model between the standard value D1 and the actual value D of the minor diameter of the internal thread. 1实 The difference between the calculated relationship and the standard value D1 D1 = -D1=D-0.7578P-(D-1.0825P)=0.3247P; S7. Establish the inequality equations for judging the failure modes of the bolt shank and the bolt external thread. A = A 纹 -A 杆 0; In the formula, τ max : Maximum shear stress, Fs: Shear force on a single thread, b: Thread length, h: Thread thickness 杆 τ is the normal stress on a bolt or threaded rod. 纹 : Thread shear stress, D 1实 D1: Actual value of minor diameter of internal thread; D: Standard value of minor diameter of internal thread; D: Nominal diameter of internal thread. : Material yield strength A: Equivalent shear stress area of ​​bolt external thread 杆 : Bolt stress area, P: Pitch.

2. The method for establishing a mechanical model of minor diameter internal threads based on strength relationships as described in claim 1, characterized in that: When the shear stress is at its maximum value, and assuming the engagement depth of the internal and external threads is χ, find the value of χ: Ensure that the shear stress is less than the bolt's yield strength. : =F / A 杆 (1), =F / A 杆 , ， （2）， Substitute (2) into (1) ， ， （3）； when When taking the maximum value, , Let b = (d - 2χ)π (0 ≤ χ ≤ 0.54125P), h = 0.125P + 1.1547χ, and substitute into (3); ， According to GB / T3098.2 "Mechanical Properties of Fasteners and Nuts" standard Let Z = 0.8d / P ， According to HB6443 "General Specification for Nuts" (aviation standard) and GB / T3098.1 "Mechanical Properties of Fasteners - Bolts, Screws and Studs", the following is obtained: = , , make , ， when , ， Solving for: , ,because Therefore, abandon Finally, the solution was obtained. ; when , Solving for: , , because Therefore, abandon Finally, the solution was obtained. ; when hour, ; when hour, Considering bolt manufacturing tolerances and strength safety factors, excessive torque increases the probability of bolt breakage. It needs to be greater than 3 / 8H. Pick: ; H: Original triangle height of the thread A: Equivalent shear stress area of ​​bolt external thread 杆 : Bolt stress area, d: Nominal diameter of external thread, P: Pitch, Z: Number of turns of engagement between nut and bolt, d1: Basic dimension of minor diameter of bolt thread : Bolt minimum diameter, Fs: Shear force on a single thread, b: Thread length, h: Thread thickness, τ max : Maximum shear stress.

3. The method for establishing a mechanical model of minor diameter internal threads based on strength relationships as described in claim 2, characterized in that: right Verification required: ， ; ; when hour, ， ; when hour, ， ; Therefore, the engagement depth of the internal thread and the bolt When the screw is over-torsed, it will either break or fail due to yielding deformation.

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