# Life Prediction on a T700 Carbon Fiber Reinforced Cylinder with Limited Accelerated Life Testing Data

Authors: Ma Xiaobing; Zhang Yongbo
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

An accelerated life testing investigation was conducted on a composite cylinder that consists of aluminum alloy and T700 carbon fiber. The ultimate failure stress predictions of cylinders were obtained by the mixing rule and verified by the blasting static pressure method. Based on the stress prediction of cylinder under working conditions, the constant stress accelerated life test of the cylinder was designed. However, the failure data cannot be sufficiently obtained by the accelerated life test due to the time limitation. Therefore, most of the data presented to be high censored in high stress level and zero-failure data in low stress level. When using the traditional method for rupture life prediction, the results showed to be of lower confidence. In this study, the consistency of failure mechanism for carbon fiber and cylinder was analyzed firstly. According to the analysis result, the statistical test information of carbon fiber could be utilized for the accelerated model constitution. Then, rupture life prediction method for cylinder was proposed based on the accelerated life test data and carbon fiber test data. In this way, the life prediction accuracy of cylinder could be improved obviously, and the results showed that the accuracy of this method increased by 35%.


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## Body

## 1. Introduction

In recent years, T700 carbon fiber replacing T300 becomes the new general carbon fiber since densification gives the T700 higher tensile strength [1–6]. The cylinder of new special equipment consists of T700 carbon fiber composite materials, glass fiber composites, and aluminum alloy. The main role of aluminum alloy is to improve the axial modulus and corrosion resistance of cylinder, the glass fiber composite material is to guarantee the aluminum alloy being adapted to higher working because of large prestress, and the carbon fiber composite material is to improve the strength and modulus of the cylinder and it is also the main load-bearing material between the three-layer materials.The introduction of carbon fiber into the new generation cylinder improves the failure stress of the cylinder, but it also brings new problem to the reliability evaluation of the cylinder. The cylinder wound by T700 carbon fiber exhibits long life and high reliability in high-speed rotation mode, and the sufficient data cannot be obtained by the traditional life tests. Therefore, the accelerated test must be introduced, and the reliability index in the normal stress could be extrapolated by the statistical analysis using the high stress level data [7]. The maximum likelihood estimation method [8, 9] is very suitable for the censored data analysis, but this method only has good properties for the large sample. And it needs to iterate for the transcendental equations; sometimes the computation is difficulties and not convergence. The integral best linear unbiased estimation method [10, 11] solves problems by regression analysis based on linear transformation acceleration model. The information between different stresses is used comprehensively, and the estimation accuracy can be improved. But the evaluation accuracy is still unable to meet the actual engineering requirement of the cylinder structure with great life dispersion.This study found that the failure mode of the cylinder wound by T700 carbon fiber was carbon fiber breakage in the working mode with high-speed rotation. Therefore, there would be some relationship between the rupture life of fiber carbon and the cylinder. In this study, we analyzed the mechanical properties of the cylinder composite material and obtained the ultimate failure stress by mixing rules, which were verified by the static pressure burst test method. Then we designed accelerated life tests on the carbon fiber and the cylinder. According to the test results, we proposed the integral best unbiased estimate method for the cylinder life prediction. The evaluation accuracy was greatly improved by comprehensive evaluation using the information of T700 carbon fiber based on the test results of the accelerated equivalence and failure mechanisms.


## 2. Mechanical Analysis

### 2.1. Mechanical Analysis of the Composite Material

The new generation cylinder consists of aluminium alloy, glass fiber, and T700 carbon fiber, and its structure is illustrated in Figure1.Figure 1
Structure of the cylinder.The main role of the aluminium alloy is to improve the axial modulus and anticorrosion of the cylinder. The glass fiber is to impose prestress to the aluminium alloy according to winding, and so it could improve the maximum capacity. The carbon fiber is the main bearing materials and for their high strength and modulus the ultimate failure stress of cylinder is greatly improved. The mechanical properties of the glass fiber and the carbon fiber can be obtained by mixing rules according to tensile strength, modulus, and the volume content of each fiber. Thus, the overall mechanical properties of the cylinder could also be estimated by the mixing rule basing on the material properties and structures.According to the property test of T700 carbon fiber, the tensile strength is 4900 Mpa, the modulus is 230 GPa, and the strain is approximately 2%. Also, the matrix strength of the composite material is 85 MPa, the modulus is 2.8 Gpa, and the strain is 4%. The mix rule is(1)Xt=σfVf+σm1-Vf,where the Xt is tensile strength of composite, σf is the tensile strength of fiber, σm is the tensile strength of matrix, and Vf is the fiber volume. Therefore, for the fiber volume of composite material used in this study is 72%, the tensile strength of this composite would be calculated as 3551.8 MPa.For the glass fiber, the tensile strength is 3400 MPa, the modulus is 91 GPa, and the strain is approximately 3.7%. And the matrix is the same with the T700 carbon fiber reinforced composite. While the percentage of fiber volume is 76%, according to the mix rule, the tensile strength of this glass fiber reinforced composite would be calculated as 2604.4.The lining of the cylinder is aluminium alloy, its elastic modulus is 70 GPa, yield strength is 610 Mpa, tensile strength is 640 Mpa, and strain is greater than 4% [12].


### 2.2. Failure Stress Analysis of the Cylinder

The tensile modulus of the glass fiber and the aluminium alloy is significantly lower than the outer carbon fiber composite material by the previous analysis. And according to the thickness of each laminate listed in Table1, it is obvious that the carbon fiber is the main load-bearing material in the entire cylinder. When the outer T700 carbon fiber reinforced composite is destructed, the glass fiber and aluminum alloy would be instantaneously destroyed due to the large load. The stress-strain curves of glass fiber and carbon fiber reinforced composite materials presented to be linear like the tensile stress-strain curves of glass fiber and carbon fiber that are shown in Figures 2 and 3.Table 1
Geometry parameters and properties of each laminate.
Aluminium alloy
Glass fiber
Carbon fiber
Thickness (mm)
1.20
0.85
1.50
Modulus (GPa)
71
70, 20 (90°)
163, 8 (90°)
Density (g/cm3)
2.8
2.15
1.58Figure 2
Tensile stress-strain curves of the glass fiber.Figure 3
Tensile stress-strain curves of the carbon fiber.The bearing stress of glass fiber is 380 Mpa that is the reverse stress to impose on the aluminum alloy; therefore, its bearing stress is 1777 Mpa when the conditions are the same with the outer carbon fiber. The tensile strength of aluminum alloy is 640 MPa, we consider that it imposed the reverse prestress of 380 MPa, its bearing stress is 260 Mpa when the breaking elongation is 2%, and its thickness is 1.2 mm. The thickness of glass fiber and the carbon fiber is 0.85 mm and 1.5 mm, respectively. Therefore, according to the mixing rule, the ultimate failure stress of the cylinder is theoretical as(2)σ=σl×tl+σg×tg+σf×tftl+tg+tf=2004MPa.The strip tensile test method [13] for ultimate failure stress of the cylinder cannot meet the requirements of test accuracy with a few samples. The NOL [13] ring stretching method is sensitive to the boundary effects and sample processing is very difficult. Therefore, we chose the blasting static pressure method [14] for the ultimate failure stress testing which dealt with the data as a whole and thus could reflect the mechanical properties better.The test samples were produced according to the national standard GB/T15560, and the processing technical, raw materials were the same with the cylinder. Figure4 shows the size of this test sample. As shown, two ends of the cylinder were strengthened by carbon fiber layers of 30 mm width. Table 2 shows the parameters of this static pressure burst test. Table 3 shows the test result.Table 2
Parameters of the static pressure burst test.
Aluminium alloy (mm)
Process conditions
Resin formula
Curing system
Glass fiber
Carbon fiber
Thickness: 1.2Length: 350
5 layers
10 layers
Proprietary formula
75°C/4 h + 80°C/12 hTable 3
Test result of the static pressure burst test.
Number
σ
b (MPa)
ε
b (%)
E (GPa)
1
2032.4
1.88
112.5
2
2087.6
1.87
113.3
3
2060.9
1.94
109.9
4
2180.9
2.02
112.3
5
2049.0
1.85
112.2
6
2110.5
1.93
112.3
7
1932.4
1.84
108.2
8
2086.7
1.91
112.6
Mean
2067.6
1.91
111.7
Dispersion
3.44%
3.08%
1.43%Figure 4
Size of the static pressure burst test.It can be seen from Table3 that the mean of the ultimate test failure stress is 2067.6 MPa, which is slightly higher than the theoretical prediction of 2004 Mpa. It may be the reason that the overall performance of the cylinder material would be slightly higher than fiber samples. The parameters used for theoretical prediction were the mean value of the tensile test results of fiber sample, and thus its volatility is relatively large. Another main reason may be that the length of the test sample in the static pressure burst test was required to be at least 5 times greater than the diameter in test standard of GB/T15560. But this long diameter ratio of our test sample is only 2.7. As the shorter sample is sensitive to the effects of the end portion in the testing process, this may be the main reason that the test results are slightly higher than the prediction result. In all, the test results of the static pressure burst test made a good agreement with the prediction result, and it is feasible to verify the overall performance of the cylinder. This test result would be taken as the important reference for the constant stress accelerated life test in the following research.


## 2.1. Mechanical Analysis of the Composite Material

The new generation cylinder consists of aluminium alloy, glass fiber, and T700 carbon fiber, and its structure is illustrated in Figure1.Figure 1
Structure of the cylinder.The main role of the aluminium alloy is to improve the axial modulus and anticorrosion of the cylinder. The glass fiber is to impose prestress to the aluminium alloy according to winding, and so it could improve the maximum capacity. The carbon fiber is the main bearing materials and for their high strength and modulus the ultimate failure stress of cylinder is greatly improved. The mechanical properties of the glass fiber and the carbon fiber can be obtained by mixing rules according to tensile strength, modulus, and the volume content of each fiber. Thus, the overall mechanical properties of the cylinder could also be estimated by the mixing rule basing on the material properties and structures.According to the property test of T700 carbon fiber, the tensile strength is 4900 Mpa, the modulus is 230 GPa, and the strain is approximately 2%. Also, the matrix strength of the composite material is 85 MPa, the modulus is 2.8 Gpa, and the strain is 4%. The mix rule is(1)Xt=σfVf+σm1-Vf,where the Xt is tensile strength of composite, σf is the tensile strength of fiber, σm is the tensile strength of matrix, and Vf is the fiber volume. Therefore, for the fiber volume of composite material used in this study is 72%, the tensile strength of this composite would be calculated as 3551.8 MPa.For the glass fiber, the tensile strength is 3400 MPa, the modulus is 91 GPa, and the strain is approximately 3.7%. And the matrix is the same with the T700 carbon fiber reinforced composite. While the percentage of fiber volume is 76%, according to the mix rule, the tensile strength of this glass fiber reinforced composite would be calculated as 2604.4.The lining of the cylinder is aluminium alloy, its elastic modulus is 70 GPa, yield strength is 610 Mpa, tensile strength is 640 Mpa, and strain is greater than 4% [12].


## 2.2. Failure Stress Analysis of the Cylinder

The tensile modulus of the glass fiber and the aluminium alloy is significantly lower than the outer carbon fiber composite material by the previous analysis. And according to the thickness of each laminate listed in Table1, it is obvious that the carbon fiber is the main load-bearing material in the entire cylinder. When the outer T700 carbon fiber reinforced composite is destructed, the glass fiber and aluminum alloy would be instantaneously destroyed due to the large load. The stress-strain curves of glass fiber and carbon fiber reinforced composite materials presented to be linear like the tensile stress-strain curves of glass fiber and carbon fiber that are shown in Figures 2 and 3.Table 1
Geometry parameters and properties of each laminate.
Aluminium alloy
Glass fiber
Carbon fiber
Thickness (mm)
1.20
0.85
1.50
Modulus (GPa)
71
70, 20 (90°)
163, 8 (90°)
Density (g/cm3)
2.8
2.15
1.58Figure 2
Tensile stress-strain curves of the glass fiber.Figure 3
Tensile stress-strain curves of the carbon fiber.The bearing stress of glass fiber is 380 Mpa that is the reverse stress to impose on the aluminum alloy; therefore, its bearing stress is 1777 Mpa when the conditions are the same with the outer carbon fiber. The tensile strength of aluminum alloy is 640 MPa, we consider that it imposed the reverse prestress of 380 MPa, its bearing stress is 260 Mpa when the breaking elongation is 2%, and its thickness is 1.2 mm. The thickness of glass fiber and the carbon fiber is 0.85 mm and 1.5 mm, respectively. Therefore, according to the mixing rule, the ultimate failure stress of the cylinder is theoretical as(2)σ=σl×tl+σg×tg+σf×tftl+tg+tf=2004MPa.The strip tensile test method [13] for ultimate failure stress of the cylinder cannot meet the requirements of test accuracy with a few samples. The NOL [13] ring stretching method is sensitive to the boundary effects and sample processing is very difficult. Therefore, we chose the blasting static pressure method [14] for the ultimate failure stress testing which dealt with the data as a whole and thus could reflect the mechanical properties better.The test samples were produced according to the national standard GB/T15560, and the processing technical, raw materials were the same with the cylinder. Figure4 shows the size of this test sample. As shown, two ends of the cylinder were strengthened by carbon fiber layers of 30 mm width. Table 2 shows the parameters of this static pressure burst test. Table 3 shows the test result.Table 2
Parameters of the static pressure burst test.
Aluminium alloy (mm)
Process conditions
Resin formula
Curing system
Glass fiber
Carbon fiber
Thickness: 1.2Length: 350
5 layers
10 layers
Proprietary formula
75°C/4 h + 80°C/12 hTable 3
Test result of the static pressure burst test.
Number
σ
b (MPa)
ε
b (%)
E (GPa)
1
2032.4
1.88
112.5
2
2087.6
1.87
113.3
3
2060.9
1.94
109.9
4
2180.9
2.02
112.3
5
2049.0
1.85
112.2
6
2110.5
1.93
112.3
7
1932.4
1.84
108.2
8
2086.7
1.91
112.6
Mean
2067.6
1.91
111.7
Dispersion
3.44%
3.08%
1.43%Figure 4
Size of the static pressure burst test.It can be seen from Table3 that the mean of the ultimate test failure stress is 2067.6 MPa, which is slightly higher than the theoretical prediction of 2004 Mpa. It may be the reason that the overall performance of the cylinder material would be slightly higher than fiber samples. The parameters used for theoretical prediction were the mean value of the tensile test results of fiber sample, and thus its volatility is relatively large. Another main reason may be that the length of the test sample in the static pressure burst test was required to be at least 5 times greater than the diameter in test standard of GB/T15560. But this long diameter ratio of our test sample is only 2.7. As the shorter sample is sensitive to the effects of the end portion in the testing process, this may be the main reason that the test results are slightly higher than the prediction result. In all, the test results of the static pressure burst test made a good agreement with the prediction result, and it is feasible to verify the overall performance of the cylinder. This test result would be taken as the important reference for the constant stress accelerated life test in the following research.


## 3. Accelerated Life Test

### 3.1. Experiment Preparation

In the working conditions, the bearing stress of cylinder is about 913.8 Mpa, which is equivalent to 45.5% of the ultimate failure stress. We calculated the stress level of the cylinder by the finite element method, and the result shows that the bearing stress of the lining aluminum alloy is 240 MPa, the bearing stress of the glass fiber is 1060 Mpa, and the bearing of carbon fiber is 1350 MPa. The related parameters are shown in Table1.The cylinder used in the accelerated life test is the same as that used in the static pressure burst test, as shown in Figure5. Then, the cylinder is filled with hydraulic oil up to certain pressure and maintains this pressure for a long time so that the cylinder bears a uniform inner pressure in every direction. The test device is placed in one oven where working temperature could be maintained. The pressure can be tested by the pressure sensor. If the pressure of the cylinder drops significantly, the cylinder was considered to be failure and the failure time is automatically recorded.Figure 5
Diagram of the reliability test unit of the cylinder.


### 3.2. The Accelerated Test Plan of the Cylinder

We know the ultimate failure stress is 2067.6 Mpa given by the static pressure burst test and the working stress is 913.8 Mpa, which is equivalent to 45.5% of the ultimate failure stress of the cylinder. The stress-strain curve is close to linear in the scope of the ultimate failure stress of the cylinder.According to GB 2689.1-81, we divided the cylinders into five groups for constant stress accelerated life test. And we set the minimum stress level to be 1315 MPa (almost 63.6% of the ultimate failure stress) and the maximum stress level to be 1861 MPa (almost 90% of the ultimate failure stress). Also, other stress levels were set to be 72.7%, 80%, and 85% of the ultimate failure stress, respectively. According to the test standard of GB 2689.1-81, this test level setting could ensure that the failure mechanism of cylinders was the same. The working temperature of the cylinder will not exceed 40°C, so the test temperature is controlled at40±2°C.


## 3.1. Experiment Preparation

In the working conditions, the bearing stress of cylinder is about 913.8 Mpa, which is equivalent to 45.5% of the ultimate failure stress. We calculated the stress level of the cylinder by the finite element method, and the result shows that the bearing stress of the lining aluminum alloy is 240 MPa, the bearing stress of the glass fiber is 1060 Mpa, and the bearing of carbon fiber is 1350 MPa. The related parameters are shown in Table1.The cylinder used in the accelerated life test is the same as that used in the static pressure burst test, as shown in Figure5. Then, the cylinder is filled with hydraulic oil up to certain pressure and maintains this pressure for a long time so that the cylinder bears a uniform inner pressure in every direction. The test device is placed in one oven where working temperature could be maintained. The pressure can be tested by the pressure sensor. If the pressure of the cylinder drops significantly, the cylinder was considered to be failure and the failure time is automatically recorded.Figure 5
Diagram of the reliability test unit of the cylinder.


## 3.2. The Accelerated Test Plan of the Cylinder

We know the ultimate failure stress is 2067.6 Mpa given by the static pressure burst test and the working stress is 913.8 Mpa, which is equivalent to 45.5% of the ultimate failure stress of the cylinder. The stress-strain curve is close to linear in the scope of the ultimate failure stress of the cylinder.According to GB 2689.1-81, we divided the cylinders into five groups for constant stress accelerated life test. And we set the minimum stress level to be 1315 MPa (almost 63.6% of the ultimate failure stress) and the maximum stress level to be 1861 MPa (almost 90% of the ultimate failure stress). Also, other stress levels were set to be 72.7%, 80%, and 85% of the ultimate failure stress, respectively. According to the test standard of GB 2689.1-81, this test level setting could ensure that the failure mechanism of cylinders was the same. The working temperature of the cylinder will not exceed 40°C, so the test temperature is controlled at40±2°C.


## 4. Results and Discussion

### 4.1. The Method on Rupture Life Evaluation of the Cylinder

We assume that there arenj samples prepared for type-I censored test in the vj  (j=1,2,…,s) stress level and the censored time is ξj∗. There are qj  (1≤qj≤nj) failures and the censored data is ξj1≤⋯≤ξjqj.Hypothesis 1.
The product life follows Weibull distributionW(mj,ηj) in the vj stress level.Hypothesis 2.
The failure mechanisms of the carbon fiber and the cylinder are the same in each stress level.Hypothesis 3.
The relationship between characteristic lifeηj and stress level vj follows the inverse power law model: ηj=Avj-c,  j=1,2,…s. And A,c are parameters that should be estimated.The linear expression can be obtained by logarithmic transformation on the accelerated model in Hypothesis3:(3)lnηj=a+bxj.Then the logarithmic life follows the extreme value distribution and the relationship between the logarithmic life and the logarithmic stress can be written as(4)yj=a+bxj+εjk,εjk~EV0,σ,j=1,2,…,s;k=1,2,…,qj+1,where yj=lnηj,  xj=lnvj,  a=lnA, and b=-c. Parameter a reflects the characteristic of test product; parameter b reflects the acceleration characteristic of the test; σ is the scale parameter of the extreme value distribution and the measure parameter for the consistency of the failure mechanism.In the condition of Hypothesis1, type-I censored data yj1≤⋯≤yjqj can be taken as the value of the former qj order statistics Yj1≤⋯≤Yjqj for the extreme value distribution with size nj. yj(qj+1)=yj∗ can be taken as the value of the qjth interval statistics Yj(qj+1) with the same sample.From literature [9, 10], we can obtain the estimations of a, b, and σ by partial derivative for Q(5)Q=∑j=1s∑k,l=1qj+1yjk-a-bxj-σujk×gjklyjl-a-bxj-σujl.The estimations of a, b, and  σ are(6)a^=y--b^x--σ^u-,b^=L22L1y-L12L2yL11L22-L122,σ^=L11L2y-L12L1yL11L22-L122,where(7)y-=1n∗∑j=1s∑k,l=1qj+1gjklyjk,u-=1n∗∑j=1s∑k,l=1qj+1gjklujk,x-=1n∗∑j=1s∑k,l=1qj+1gjklxj,n∗=∑j=1s∑k,l=1qj+1gjkl,L1y=∑j=1s∑k,l=1qj+1gjklxj-x-yjk-y-,L2y=∑j=1s∑k,l=1qj+1gjklujk-u-yjl-y-,L11=∑j=1s∑k,l=1qj+1gjklxj-x-2,L12=∑j=1s∑k,l=1qj+1gjklxj-x-ujk-u-,L22=∑j=1s∑k,l=1qj+1gjklujk-u-ujl-u-.And the covariance matrix ofa,b, and  σ is(8)cova^,b^,σ^=σ2C,C=∑j,k,lgjkl∑j,k,lgjklxj∑j,k,lgjklμjk∑j,k,lgjklxj∑j,k,lgjklxj2∑j,k,lgjklxjμjk∑j,k,lgjklμjk∑j,k,lgjklxjμjk∑j,k,lgjklμjkμjl-1,where G=(gjkl)qj+1×qj+1=V-1=(vjkl)qj+1×(qj+1)-1,  ujk(k=1,2,…,qj) is the mean of the kth order statistic for the standard extreme value distribution with size nj, vjkl(k,l=1,2,…,qj) is the covariance of the kth and lth order statistic for the standard extreme value distribution with size nj,  uj(qj+1) is the mean of the (qj+1)th order statistic for the standard extreme value distribution with size nj+1, and  vjk(qj+1)=vj(qj+1)k(k=1,2,…,qj+1) is the covariance of the kth and (qj+1)th order statistic for the standard extreme value distribution with size nj+1. These values could be all obtained by formula or table lookup [15].The reliability rupture life with reliability ofR and its upper and lower limits with confidence level γ could be calculated as follows:(9)yR=a^+b^x+σ^lnln1R,yRU=a^+b^x+σ^1-uγ2c33lnln1R+uγ2c13+c23x+uγωR,yRL=a^+b^x+σ^1-uγ2c33lnln1R+uγ2c13+c23x-uγωR,where cij are elements of the matrix C,  ωR=ω+c33(lnln1/R)2+2(c13+c23x)lnln1/R, and ω=uγ2(c13+c23x)2+(1-c33uγ2)(c11+2c12x+c22x2).The Failure Mechanism Consistency Analysis. This analysis focused on the consistency of failure mechanism and accelerated model parameter between the carbon fiber and cylinder in the accelerated life test. Denote the model parameters of the carbon fiber and the cylinder as a1, b1, and  σ1 and a2, b2, and  σ2, respectively. We assume the following.(1)
The distribution parametersσ1, σ2 are two independent normal populations. If the failure mechanism of the carbon fiber is the same with the cylinder, the mean and variance of the two normal populations are the same [16, 17].(2)
Parameters ofa1 and a2 reflect the life characteristic of carbon fiber and cylinder, and then a1 and a2 have no relation.(3)
The model parametersb1, b2 are two independent normal populations. If the acceleration of the carbon fiber is the same with the cylinder, the mean and variance of the two normal populations are the same.Based on the above assumptions,(b^1,σ^1) and (b^2,σ^2) can be taken as two bivariate normal populations. We can judge the consistency of the mean vector and covariance matrix of the two bivariate normal populations by hypothesis test.(1) The Consistency Judgment of the Mean Vector. The two independent normal populations are denoted by (b^1,σ^1)~N2(μ1,Σ1) and (b^2,σ^2)~N2(μ2,Σ2). We sample n,  m>2 specimens from them, respectively, and denote the mean vectors by X-,  Y-, respectively, and the variance matrix by Si(i=1,2). The hypothesis is(10)H0:μ1=μ2H1:μ1≠μ2.When Σ1=Σ2 and they were unknown, the test statistic(11)T2=nmn+mX--Y-TS-1X--Y-,where(12)S=n-1S1+m-1S2n+m-2.And F=(n+m-2)-1/2(n+m-2)T2~F(2,n+m-3).Then the rejection region with the significance levelα is(13)F>F1-α2,n+m-3.(2) The Consistency Judgment of the Variance Matrix. The hypothesis is(14)H0:Σ1=Σ2H1:Σ1≠Σ2.The amendatory likelihood ratio statistic is(15)λ∗=n+m-2n+m-2S1n-1/2S2m-1/2n-1n-1/2m-1m-1/2S1+S2n+m-2/2.Then,(16)-2lnλ∗~χ2f1-d,where(17)f=12pp+1k-1,d=2p2+3p-16p+1k-11n-1+1m-1-1n+m-kand k=p=2.Then the rejection region with the significance levelα is(18)χ2<χα2f1-d.The information fusion methodBased on the result of the integral best linear unbiased estimation and the consistency analysis of failure mechanism, we proposed an information fusion method for the covariance matrixes of the T700 carbon fiber and the cylinder. The fusion process can be accomplished by the two steps: firstly, fusing the information only related to parametersb and σ in the covariance matrix and then, secondly, to further improve the evaluation accuracy of the covariance matrix for model parameter of the cylinder, fusing the other information related to b and σ.Denote the point estimations of T700 carbon fiber and the cylinder bya^1, b^1, and  σ^1 and a^2, b^2, and  σ^2, respectively, and their covariance matrix is(19)cova^i,b^i,σ^i=vara^icova^i,b^icova^i,σ^icova^i,b^ivarb^icovb^i,σ^icova^i,σ^icovb^i,σ^ivarσ^i,i=1,2.Denote the covariance matrix of the bivariate normal distribution ofb1, σ1 and b2, σ2, respectively, by(20)V^i=varb^icovb^i,σ^icovb^i,σ^ivarσ^i,i=1,2.The elements of the above matrixes are one part of (19).The first step is fusing the information only related to parametersb and σ in the covariance matrix.If the covariance matrixesV1 and V2 are certified to be the same by the consistency analysis of the bivariate normal distribution b1,σ1 and (b2,σ2), the unbiased estimation of the covariance matrix of the bivariate normal distribution can be obtained by the following equation:(21)V^=n1-1V^1+n2-1V^2n1+n2-2.Compared to the small ratio of failure and high censored time of the cylinder life test, the data and failures of the carbon fiber are much greater and thus the evaluation result is more accurate. Therefore, when the matrixV^2 is replaced by V^ with fusing the information of parameters b and σ according to the above method, the prediction result of the cylinder is more accurate. The covariance matrix of b and σ based on fusion could be written as(22)V^=varb^12covb^12,σ^12covb^12,σ^12varσ^12.The covariance matrixcov(a^2,b^2,σ^2) for the cylinder with bottom right four elements replaced by V^ could be written as(23)cova^2,b^2,σ^2′=vara^2cova^2,b^2cova^2,σ^2cova^2,b^2varb^12covb^12,σ^12cova^2,σ^2covb^12,σ^12varσ^12.The second step is fusing the other elements related tob,σ of the covariance matrix.Based on the same correlation coefficient between the parameters, the value ofcov(a^2,b^12), cov(a^2,σ^12) with fusing the information of the T700 carbon fiber can be calculated as(24)cova^2,b^12=varb^12varb^2·cova^2,b^2,cova^2,σ^12=varσ^12varσ^2·cova^2,σ^2.According to the above two steps, the covariance matrixcov(a^2,b^12,σ^12) can be written as(25)cova^2,b^12,σ^12=vara^2cova^2,b^12cova^2,σ^12cova^2,b^12varb^12covb^12,σ^12cova^2,σ^12covb^12,σ^12varσ^12.Comparing with (25) and (19)  (i=2), it can be found that the elements of parameter covariance matrix of the cylinder have changed in addition to the variance va^r(a^2) by fusing the carbon fiber test information, thus making the evaluation result more reasonable. It should be noted that the parameter covariance matrix can be obtained by the integrated best linear unbiased estimation as follows:(26)cova^,b^,σ^=σ2C.In the calculation of the upper and lower limits for the reliability rupture life, the matrixC would be used. Therefore, in fusion process of the covariance matrix with T700 carbon fiber information and cylinder information, we can introduce the matrix C directly in the above method to make the calculation easier.


### 4.2. Examples

To verify the design level of the rupture life of a certain type of cylinder, a unit made accelerated life tests for T700 carbon fiber composite material and the cylinder structure, respectively. The test temperature is 40°C, and the test load conditions are shown in Table4.Table 4
Accelerated life test information for T700 carbon fiber and the cylinder.
Subjects
Stress level
T700 carbon fiber (N)
1700
1800
1900
—
—
Cylinder (the percentage of limit load %)
63.6
72.7
80
85
90The diagram of rupture life data for T700 carbon fiber and the cylinder structure from the test is as in Figures6 and 7.Figure 6
Rupture life of T700 carbon fiber.Figure 7
Ruptures life of the cylinder.(1) The Result of the Integral Best Unbiased Estimation. The results of model parameters and the matrix C obtained by the integrated best linear unbiased estimation are shown in Table 5.Table 5
Contrast result between T700 carbon fiber and the cylinder with the integrated best linear unbiased estimation.
Result
T700 carbon fiber
Cylinder
(
a
^
,
b
^
,
σ
^
)
(157.177, −19.872, 2.32)
(6.39, −18.41, 1.20)
C
[
495.06
-
66.08
0.46
-
66.08
8.82
-
0.06
0.46
-
0.06
0.012
]
[
0.53
2.70
-
0.06
2.70
21.49
-
1.05
-
0.06
-
1.05
0.13
](
2) The Test of Mean Vector and Covariance Matrix Equal to
(
b
^
,
σ
^
)
of the Carbon Fiber and the Cylinder. We consider parameters (b^,σ^) as a bivariate normal population and test whether the mean vectors and covariance matrixes of (b^,σ^) for the carbon fiber and the cylinder are equal or not, and the results are shown in Table 6.Table 6
The test results of mean vector and covariance matrix for a bivariate normal population(b^,σ^) of carbon fiber and cylinder.
Null hypothesis
Rejection region
Observation value
Critical valueα=0.025
μ
1
=
μ
2
{
F
>
F
1
-
α
(
2
,
n
+
m
-
3
)
}
8.09725
8.43
Σ
1
=
Σ
2
{
χ
2
<
χ
α
2
(
f
/
(
1
-
d
)
)
}
1.9468
0.831From Table6, the observed values of the test statistic are (27)F=8.09725<F0.9752,5=8.43,χ2=1.9468>χ0.02525.Then, we can receive the null hypothesis. The mean vectors and the covariance matrixes of (b^,σ^) for the carbon fiber and the cylinder are equal.(
3) The Information Fusion of the T700 Carbon Fiber and the Cylinder. By using the method in this paper, the matrix information of carbon fiber in Table 5 can be fused into the cylinder and the upper and lower limits of the reliability rupture life at confidence level γ=0.95 can be calculated. The comparison results are shown in Table 7.Table 7
Contrast result of cylinder after fusion information of T700 carbon fiber.
Result
The original data
Fusion of the information of T700
C
[
0.53
2.70
-
0.06
2.70
21.49
-
1.05
-
0.06
-
1.05
0.13
]
[
0.53
2.08
-
0.037
2.08
12.76
-
0.37
-
0.037
-
0.37
0.049
]
The interval estimation ofyR
[13.79, 27.35]
[14.29, 23.15]
Interval length
13.56
8.86From Table7, when the reliability is 0.9, the evaluation accuracy of the reliability life of the cylinder is increased by 35%.The curves of upper and lower limits of the logarithmic reliability rupture lifeyR changed with the reliability R are shown in Figure 7.Figure8 shows that the upper and lower limits of the reliability rupture life will be more accurate with the reliability changed after fusion of the information of T700 carbon fiber and the interval length is much shorter.Figure 8
Change curves foryR and upper and lower limits.


## 4.1. The Method on Rupture Life Evaluation of the Cylinder

We assume that there arenj samples prepared for type-I censored test in the vj  (j=1,2,…,s) stress level and the censored time is ξj∗. There are qj  (1≤qj≤nj) failures and the censored data is ξj1≤⋯≤ξjqj.Hypothesis 1.
The product life follows Weibull distributionW(mj,ηj) in the vj stress level.Hypothesis 2.
The failure mechanisms of the carbon fiber and the cylinder are the same in each stress level.Hypothesis 3.
The relationship between characteristic lifeηj and stress level vj follows the inverse power law model: ηj=Avj-c,  j=1,2,…s. And A,c are parameters that should be estimated.The linear expression can be obtained by logarithmic transformation on the accelerated model in Hypothesis3:(3)lnηj=a+bxj.Then the logarithmic life follows the extreme value distribution and the relationship between the logarithmic life and the logarithmic stress can be written as(4)yj=a+bxj+εjk,εjk~EV0,σ,j=1,2,…,s;k=1,2,…,qj+1,where yj=lnηj,  xj=lnvj,  a=lnA, and b=-c. Parameter a reflects the characteristic of test product; parameter b reflects the acceleration characteristic of the test; σ is the scale parameter of the extreme value distribution and the measure parameter for the consistency of the failure mechanism.In the condition of Hypothesis1, type-I censored data yj1≤⋯≤yjqj can be taken as the value of the former qj order statistics Yj1≤⋯≤Yjqj for the extreme value distribution with size nj. yj(qj+1)=yj∗ can be taken as the value of the qjth interval statistics Yj(qj+1) with the same sample.From literature [9, 10], we can obtain the estimations of a, b, and σ by partial derivative for Q(5)Q=∑j=1s∑k,l=1qj+1yjk-a-bxj-σujk×gjklyjl-a-bxj-σujl.The estimations of a, b, and  σ are(6)a^=y--b^x--σ^u-,b^=L22L1y-L12L2yL11L22-L122,σ^=L11L2y-L12L1yL11L22-L122,where(7)y-=1n∗∑j=1s∑k,l=1qj+1gjklyjk,u-=1n∗∑j=1s∑k,l=1qj+1gjklujk,x-=1n∗∑j=1s∑k,l=1qj+1gjklxj,n∗=∑j=1s∑k,l=1qj+1gjkl,L1y=∑j=1s∑k,l=1qj+1gjklxj-x-yjk-y-,L2y=∑j=1s∑k,l=1qj+1gjklujk-u-yjl-y-,L11=∑j=1s∑k,l=1qj+1gjklxj-x-2,L12=∑j=1s∑k,l=1qj+1gjklxj-x-ujk-u-,L22=∑j=1s∑k,l=1qj+1gjklujk-u-ujl-u-.And the covariance matrix ofa,b, and  σ is(8)cova^,b^,σ^=σ2C,C=∑j,k,lgjkl∑j,k,lgjklxj∑j,k,lgjklμjk∑j,k,lgjklxj∑j,k,lgjklxj2∑j,k,lgjklxjμjk∑j,k,lgjklμjk∑j,k,lgjklxjμjk∑j,k,lgjklμjkμjl-1,where G=(gjkl)qj+1×qj+1=V-1=(vjkl)qj+1×(qj+1)-1,  ujk(k=1,2,…,qj) is the mean of the kth order statistic for the standard extreme value distribution with size nj, vjkl(k,l=1,2,…,qj) is the covariance of the kth and lth order statistic for the standard extreme value distribution with size nj,  uj(qj+1) is the mean of the (qj+1)th order statistic for the standard extreme value distribution with size nj+1, and  vjk(qj+1)=vj(qj+1)k(k=1,2,…,qj+1) is the covariance of the kth and (qj+1)th order statistic for the standard extreme value distribution with size nj+1. These values could be all obtained by formula or table lookup [15].The reliability rupture life with reliability ofR and its upper and lower limits with confidence level γ could be calculated as follows:(9)yR=a^+b^x+σ^lnln1R,yRU=a^+b^x+σ^1-uγ2c33lnln1R+uγ2c13+c23x+uγωR,yRL=a^+b^x+σ^1-uγ2c33lnln1R+uγ2c13+c23x-uγωR,where cij are elements of the matrix C,  ωR=ω+c33(lnln1/R)2+2(c13+c23x)lnln1/R, and ω=uγ2(c13+c23x)2+(1-c33uγ2)(c11+2c12x+c22x2).The Failure Mechanism Consistency Analysis. This analysis focused on the consistency of failure mechanism and accelerated model parameter between the carbon fiber and cylinder in the accelerated life test. Denote the model parameters of the carbon fiber and the cylinder as a1, b1, and  σ1 and a2, b2, and  σ2, respectively. We assume the following.(1)
The distribution parametersσ1, σ2 are two independent normal populations. If the failure mechanism of the carbon fiber is the same with the cylinder, the mean and variance of the two normal populations are the same [16, 17].(2)
Parameters ofa1 and a2 reflect the life characteristic of carbon fiber and cylinder, and then a1 and a2 have no relation.(3)
The model parametersb1, b2 are two independent normal populations. If the acceleration of the carbon fiber is the same with the cylinder, the mean and variance of the two normal populations are the same.Based on the above assumptions,(b^1,σ^1) and (b^2,σ^2) can be taken as two bivariate normal populations. We can judge the consistency of the mean vector and covariance matrix of the two bivariate normal populations by hypothesis test.(1) The Consistency Judgment of the Mean Vector. The two independent normal populations are denoted by (b^1,σ^1)~N2(μ1,Σ1) and (b^2,σ^2)~N2(μ2,Σ2). We sample n,  m>2 specimens from them, respectively, and denote the mean vectors by X-,  Y-, respectively, and the variance matrix by Si(i=1,2). The hypothesis is(10)H0:μ1=μ2H1:μ1≠μ2.When Σ1=Σ2 and they were unknown, the test statistic(11)T2=nmn+mX--Y-TS-1X--Y-,where(12)S=n-1S1+m-1S2n+m-2.And F=(n+m-2)-1/2(n+m-2)T2~F(2,n+m-3).Then the rejection region with the significance levelα is(13)F>F1-α2,n+m-3.(2) The Consistency Judgment of the Variance Matrix. The hypothesis is(14)H0:Σ1=Σ2H1:Σ1≠Σ2.The amendatory likelihood ratio statistic is(15)λ∗=n+m-2n+m-2S1n-1/2S2m-1/2n-1n-1/2m-1m-1/2S1+S2n+m-2/2.Then,(16)-2lnλ∗~χ2f1-d,where(17)f=12pp+1k-1,d=2p2+3p-16p+1k-11n-1+1m-1-1n+m-kand k=p=2.Then the rejection region with the significance levelα is(18)χ2<χα2f1-d.The information fusion methodBased on the result of the integral best linear unbiased estimation and the consistency analysis of failure mechanism, we proposed an information fusion method for the covariance matrixes of the T700 carbon fiber and the cylinder. The fusion process can be accomplished by the two steps: firstly, fusing the information only related to parametersb and σ in the covariance matrix and then, secondly, to further improve the evaluation accuracy of the covariance matrix for model parameter of the cylinder, fusing the other information related to b and σ.Denote the point estimations of T700 carbon fiber and the cylinder bya^1, b^1, and  σ^1 and a^2, b^2, and  σ^2, respectively, and their covariance matrix is(19)cova^i,b^i,σ^i=vara^icova^i,b^icova^i,σ^icova^i,b^ivarb^icovb^i,σ^icova^i,σ^icovb^i,σ^ivarσ^i,i=1,2.Denote the covariance matrix of the bivariate normal distribution ofb1, σ1 and b2, σ2, respectively, by(20)V^i=varb^icovb^i,σ^icovb^i,σ^ivarσ^i,i=1,2.The elements of the above matrixes are one part of (19).The first step is fusing the information only related to parametersb and σ in the covariance matrix.If the covariance matrixesV1 and V2 are certified to be the same by the consistency analysis of the bivariate normal distribution b1,σ1 and (b2,σ2), the unbiased estimation of the covariance matrix of the bivariate normal distribution can be obtained by the following equation:(21)V^=n1-1V^1+n2-1V^2n1+n2-2.Compared to the small ratio of failure and high censored time of the cylinder life test, the data and failures of the carbon fiber are much greater and thus the evaluation result is more accurate. Therefore, when the matrixV^2 is replaced by V^ with fusing the information of parameters b and σ according to the above method, the prediction result of the cylinder is more accurate. The covariance matrix of b and σ based on fusion could be written as(22)V^=varb^12covb^12,σ^12covb^12,σ^12varσ^12.The covariance matrixcov(a^2,b^2,σ^2) for the cylinder with bottom right four elements replaced by V^ could be written as(23)cova^2,b^2,σ^2′=vara^2cova^2,b^2cova^2,σ^2cova^2,b^2varb^12covb^12,σ^12cova^2,σ^2covb^12,σ^12varσ^12.The second step is fusing the other elements related tob,σ of the covariance matrix.Based on the same correlation coefficient between the parameters, the value ofcov(a^2,b^12), cov(a^2,σ^12) with fusing the information of the T700 carbon fiber can be calculated as(24)cova^2,b^12=varb^12varb^2·cova^2,b^2,cova^2,σ^12=varσ^12varσ^2·cova^2,σ^2.According to the above two steps, the covariance matrixcov(a^2,b^12,σ^12) can be written as(25)cova^2,b^12,σ^12=vara^2cova^2,b^12cova^2,σ^12cova^2,b^12varb^12covb^12,σ^12cova^2,σ^12covb^12,σ^12varσ^12.Comparing with (25) and (19)  (i=2), it can be found that the elements of parameter covariance matrix of the cylinder have changed in addition to the variance va^r(a^2) by fusing the carbon fiber test information, thus making the evaluation result more reasonable. It should be noted that the parameter covariance matrix can be obtained by the integrated best linear unbiased estimation as follows:(26)cova^,b^,σ^=σ2C.In the calculation of the upper and lower limits for the reliability rupture life, the matrixC would be used. Therefore, in fusion process of the covariance matrix with T700 carbon fiber information and cylinder information, we can introduce the matrix C directly in the above method to make the calculation easier.


## 4.2. Examples

To verify the design level of the rupture life of a certain type of cylinder, a unit made accelerated life tests for T700 carbon fiber composite material and the cylinder structure, respectively. The test temperature is 40°C, and the test load conditions are shown in Table4.Table 4
Accelerated life test information for T700 carbon fiber and the cylinder.
Subjects
Stress level
T700 carbon fiber (N)
1700
1800
1900
—
—
Cylinder (the percentage of limit load %)
63.6
72.7
80
85
90The diagram of rupture life data for T700 carbon fiber and the cylinder structure from the test is as in Figures6 and 7.Figure 6
Rupture life of T700 carbon fiber.Figure 7
Ruptures life of the cylinder.(1) The Result of the Integral Best Unbiased Estimation. The results of model parameters and the matrix C obtained by the integrated best linear unbiased estimation are shown in Table 5.Table 5
Contrast result between T700 carbon fiber and the cylinder with the integrated best linear unbiased estimation.
Result
T700 carbon fiber
Cylinder
(
a
^
,
b
^
,
σ
^
)
(157.177, −19.872, 2.32)
(6.39, −18.41, 1.20)
C
[
495.06
-
66.08
0.46
-
66.08
8.82
-
0.06
0.46
-
0.06
0.012
]
[
0.53
2.70
-
0.06
2.70
21.49
-
1.05
-
0.06
-
1.05
0.13
](
2) The Test of Mean Vector and Covariance Matrix Equal to
(
b
^
,
σ
^
)
of the Carbon Fiber and the Cylinder. We consider parameters (b^,σ^) as a bivariate normal population and test whether the mean vectors and covariance matrixes of (b^,σ^) for the carbon fiber and the cylinder are equal or not, and the results are shown in Table 6.Table 6
The test results of mean vector and covariance matrix for a bivariate normal population(b^,σ^) of carbon fiber and cylinder.
Null hypothesis
Rejection region
Observation value
Critical valueα=0.025
μ
1
=
μ
2
{
F
>
F
1
-
α
(
2
,
n
+
m
-
3
)
}
8.09725
8.43
Σ
1
=
Σ
2
{
χ
2
<
χ
α
2
(
f
/
(
1
-
d
)
)
}
1.9468
0.831From Table6, the observed values of the test statistic are (27)F=8.09725<F0.9752,5=8.43,χ2=1.9468>χ0.02525.Then, we can receive the null hypothesis. The mean vectors and the covariance matrixes of (b^,σ^) for the carbon fiber and the cylinder are equal.(
3) The Information Fusion of the T700 Carbon Fiber and the Cylinder. By using the method in this paper, the matrix information of carbon fiber in Table 5 can be fused into the cylinder and the upper and lower limits of the reliability rupture life at confidence level γ=0.95 can be calculated. The comparison results are shown in Table 7.Table 7
Contrast result of cylinder after fusion information of T700 carbon fiber.
Result
The original data
Fusion of the information of T700
C
[
0.53
2.70
-
0.06
2.70
21.49
-
1.05
-
0.06
-
1.05
0.13
]
[
0.53
2.08
-
0.037
2.08
12.76
-
0.37
-
0.037
-
0.37
0.049
]
The interval estimation ofyR
[13.79, 27.35]
[14.29, 23.15]
Interval length
13.56
8.86From Table7, when the reliability is 0.9, the evaluation accuracy of the reliability life of the cylinder is increased by 35%.The curves of upper and lower limits of the logarithmic reliability rupture lifeyR changed with the reliability R are shown in Figure 7.Figure8 shows that the upper and lower limits of the reliability rupture life will be more accurate with the reliability changed after fusion of the information of T700 carbon fiber and the interval length is much shorter.Figure 8
Change curves foryR and upper and lower limits.


## 5. Conclusion

(1)
Since the carbon fiber bears the main load at work, the acceleration and failure mechanism of T700 carbon fiber and the cylinder are the same. And it can be proved by the structure analysis and the statistic test of the test data.
(2)
When the acceleration and failure mechanism are the same, the evaluation accuracy of the reliability life for the cylinder can be improved by fusion of the information of the carbon fiber.
(3)
The method in this paper is based on Weibull distribution and the inverse power law model for structured products. It can be applied to other location-scale family distribution and acceleration models.



---

*Source: 902157-2015-06-25.xml*