Study on the law of spring shrinkage during stress relief and tempering

Study on the law of spring shrinkage during stress relief and tempering
Abstract] In this paper, the empirical formula is obtained by studying the law of stress relief and tempering shrinkage of springs in order to reduce the number of tests and reduce costs.

[Descriptor] Tempering shrinkage regression equation

Preface

After the spring steel wire is made into a spring, it has to go through a stress relief tempering process. Generally speaking, the diameter of the spring should be reduced and the total number of turns should be increased. The shrinkage of the diameter is related to the winding ratio. The larger the winding ratio, the greater the shrinkage. Therefore, the first sample must be carried out before mass production, and mass production can only be carried out after the sample is confirmed. With the mass production of high-stress springs, oil tempered alloy spring materials are widely used, but there is currently no empirical formula in this regard. Our company’s years of experience in spring production have found that there is a certain law in spring stress relief and tempering shrinkage. Regression analysis of relevant data is now performed to obtain the following empirical formula: △D=3.188×10-6×C×D×T

The process of obtaining the experience formula is as follows:

1. The establishment of the equation:

1. Assuming that the law of stress relief and tempering shrinkage is a one-variable linear regression equation, namely

△D= a +Kt×C×D×T,

Among them, △D —diameter shrinkage after tempering, C—wound ratio,

D—spring diameter, T—tempering temperature.

△D is the dependent variable, C×D×T is the independent variable, a and Kt are undetermined parameters (regression parameters).

2. Collect samples: Collect data of 57 commonly used products of our company and summarize them in Table 1.

3. Calculate the values ​​of a and Kt in the equation:

According to the above table 1, the SPSS 12 software is used for statistical analysis, and the calculation results are shown in Table 2. The equation is as follows: △D=0.087+2.954×10-6×C×D×T

4. Is there really a linear relationship between △D and C×D×T? That is, who holds H0: Kt=0 and H1: Kt≠0? How much can the change of △D be changed by C×D×T Explained? Perform a significance test on the equation:

H0: Kt=0 equation is invalid

H1: Kt≠0 equation is valid

? Determine whether the equation is useful? Use the F test method. It can be seen from Table 2 ANOVAb that sig.<0.01 proves that the equation is useful;

? Determine whether the equation has a pitch term? Using the T test method, it can be seen from Table 2 Coefficientsa that sig.>0.05 in the (constant) term proves that there should be no constant term. The constant term is redundant and must be taken from the model Remove.

? Determine whether the equation has room for further simplification? Using the T test method, it can be seen from Table 2 Coefficientsa that sig.<0.05, the independent variable must exist in the model.

? In Table 2 Model summary R Square=64.9%<80%, it also shows that the explanatory ability of this equation is only 64.9%, the explanatory ability of this equation is poor, and this equation is not suitable.

5. Based on the above analysis, re-establish the unary regression equation: △D= Kt×C×D×T

6. Recalculate the Kt value and re-use the SPSS software for statistical analysis in Table 1 (without the constant term), and the calculation results are shown in Table 3. The equation is △D=3.188×10-6×C×D×T.

7. Perform a significance test on the equation:

H0 Kt=0 equation is invalid

H1 Kt≠0 equation is valid

? Determine whether the equation is useful? Use the F test method. It can be seen from Table 3 ANOVAb that sig.<0.01 proves that the equation is useful;

? Determine whether the equation has room for further simplification? Using the T test method, it can be seen from Table 3 Coefficientsa that the independent variable sig. <0.05 indicates that the variable has no room for simplification;

In the Model summary of Table 3, R Squarea=86.3%>80%, which also shows that the explanatory power of the equation is 86.3%, and the explanatory power is relatively strong, that is, the proportion of changes caused by C×D×T among the changes in △D Is 86.3%, the equation is valid.

The equation is △D=3.188×10-6×C×D×T

2. Application and promotion

for example:

An engine valve spring is made of Zhijin WOSC-V with oil tempered valve spring, the wire diameter is 3.2mm, the spring inner diameter is 16.9mm, the total number of turns is 7 turns, and the effective number of turns is 5 turns. Try to calculate the spring stress relief Shrinkage.

According to the above materials and steel wire diameter, the stress relief tempering after winding the spring is usually used: 420℃, 20 minutes. Known: d=3.2 D1=16.9 T=420 Nt=7

Calculation: D=d+D1=3.2+16.9=20.1

C=D/d=20.1/3.2=6.28 T=420

According to the formula of spring stress relief and tempering shrinkage

△ D=3.188×10-6×C×D×T=3.188×10-6×6.28×20.1×420=0.17mm

△ N=(ΔD×Nt)/(D+ΔD)=(0.17×7)/(20.1+0.17)=0.06(circle)

It can be seen that after the spring undergoes the stress relief and tempering process, the spring diameter is reduced by 0.17mm, and the total number of turns is increased by 0.06 turns. In order to ensure that the size of the spring product meets the drawings and technical requirements, it is necessary to consider the amount of spring shrinkage when winding the spring. Usually the winding spring process is designed as follows: the inner diameter of the winding spring = D1+ΔD=16.9+0.17=17.07mm, the total number of turns = Nt-ΔN=7-0.06=6.94 (turns)

Of course, in addition to the main factors such as the material used in the spring, the winding ratio, the diameter of the spring, and the temperature, the thread pitch and the tensile strength of the material must also be considered in the design of the winding spring process. In the end, it must be tested and verified before it can be used as a mass production process.

3. Conclusion:

The diameter shrinkage of the spring wound spring after tempering is △D=Kt×C×D×T,

Among them: Kt=3.188×10-6 C—wound ratio,

D—spring diameter, T—tempering temperature.

Scope of application:

a) Cold forming process

b) Unintentional rolling

c) Oil tempered alloy steel, such as domestic alloy steel wire such as 60Si2MnA, 55CrSi, 50CrV, SWOSC-V, SWOCV-V, SWOSC-B, SWOSM-B, OTEVA70, SWI-200 and other foreign alloy steel wires.

Through the above empirical formula, the number of tests can be reduced, the spring qualification rate can be improved, and the production cost can be reduced.