In the radial fatigue test of the wheel, the wheel is rotated to withstand the constant radial load applied by the drum, and the load direction is connected along the center of the wheel and the drum. Therefore, according to the working principle of wheel radial load fatigue test, the wheel structure is fully restrained on the surface of the mounting disk and the bolt hole. The radial load of the cosine function distribution is applied over a certain range of rims. And exert an inflation pressure on the entire outer rim. In this paper, the tire pressure is 3 Bar.

According to the national standard with the following formula to determine the radial load, the unit is N:

F r = F v K (1)

where: F v is the maximum vertical static load on the wheel specified by the wheel or the automobile manufacturer or the rated load of the wheel, the unit is N; K is the reinforcement test coefficient.

Because the stress distribution of the rim and the tire contact is taken into account. When a radial load is applied to the hub, it is necessary to convert the cosine load into force, as follows [8] :

P = F r ( π 2 − B 2 ) cos ( a B π ) 2 R L B π cos B 2 (2)

where: R represents the radius, L represents the width, represents the angle, and B is the designated area angle.

In bending fatigue tests, full restraint is applied to the inside of the rim and eccentric force is applied to the hub at the other end of the shaft. The finite element analysis is simplified. Center to establish nodes. The node and the hub center for rigid connection. The bending of the nodes is applied.

Press the following formula determine the bending moment M:

M = ( μ R + d ) F v S (3)

where: μ is the coefficient of friction, R is the static load radius of the tire, d is the offset, F v is the maximum vertical static load on the wheel specified by the wheel or the automobile manufacturer or the rated load of the wheel.

The impact test is a transient response kinetic problem. In the process, the punch falls free and contacts the tire first, than touching the hub edge. In the whole process, Tires will be deformed. It is Involved a problem of energy absorption.

In order to simplify the problem, the punch from a certain height free fall impact of the wheel of this dynamic problem into a static problem. It can be based on momentum conservation that is equivalent to a force F. The action time is t₀. F is the time that can be calculated according to the following formula:

F = D v 0 / t 0 (4)

among them, D is the punch quality, v 0 = 2 g H , g is the acceleration of gravity, H is the punch height.

D v = ∫ 0 t 0 F max sin ( π t / t 0 ) d t (5)

among them: v = 2 ( 1 − η ) g H , η is the absorption of energy coefficient, about 20%. On the right side of the Formula (5) can be obtained:

F max = π 2 D ⋅ 2 ( 1 − η ) g H / t 0 (6)

Shang et al. [9] found that punch and flange from beginning to contact and at 34.5 ms, the drop velocity is zero and the stress is maximum. So it is considered that the punch and wheel contact time is about 70 ms.

The materials used in this paper are carbon fiber/epoxy (T300/5222) composites. The mechanical properties of the Table 1 below. The laying of the wheel is mainly [ 0 ° / 45 ° / 90 ° / − 45 ° ] s (s = 6). Pavement direction wheel axle. The pavement is 24 layers. Each layer thickness is 1mm. After the laying of the wheel compared to the weight of aluminum alloy wheels to reduce the 11.3%.

The boundary conditions and loads on the hub when the reference to the front of the wheel of the working conditions analysis. In order to form a contrast, in this paper, the same conditions of the aluminum alloy wheels for finite element analysis. Aluminum alloy material parameters, Elastic Modulus E = 7 × 10 4 MPa , density ρ = 2.7 × 10 9   t / mm 3 , Poisson’s ratio ν = 0.3 .

The strength check criteria chosen in this paper is a combination of multiple intensity criteria proposed by Tsai-Wu in the form of tensor. It is assumed that the damaged surface in the stress space has the following form:

F i σ i + F i j σ i σ j < 1 (7)

In the formula, the coefficients F i and F i j are called the strength parameters of the stress space. In the plane stress state, i , j = 1 , 2 , 6 .

Simplify:

F 11 σ 1 2 + 2 F 12 σ 1 σ 2 + F 22 σ 2 2 + F 66 τ 12 2 + F 1 σ 1 + F 2 σ 2 < 1 (8)

among them

F 1 = 1 X t − 1 X c , F 11 = 1 X t X c , F 2 = 1 Y t − Y c F 22 = 1 Y t Y c , F 66 = 1 S 2 (9)

The results of finite element analysis of composite wheel are compared with those of aluminum alloy wheel. At the same time, the maximum stress layer under each condition is replaced by (8) Tsai-Wu strength criterion for strength verification. It can be found that the strength of composite wheel hub can meet the requirement.

1) Radial load condition

Maximum stress layer σ 1 = 17.07 , σ 2 = 1.288 , τ 12 = 1.539 . By substituting Formula (8), the calculation result 0.0229 < 1, so the strength requirement can be satisfied under this condition. The stress distributions are shown in Figure 1 and Figure 2.

2) Bending load condition

Maximum stress layer σ 1 = 110.5 , σ 2 = 35.26 , τ 12 = 37.57 . By substituting Formula (8), the calculation result 0.8146 < 1, so the strength requirement can be satisfied under this condition. The stress distributions are shown in Figure 3 and Figure 4.

3) Impact load condition

Maximum stress layer σ 1 = 120.4 , σ 2 = 29.92 , τ 12 = 61.59 . By substituting Formula (8), the calculation result 0.991 < 1, so the strength requirement can be satisfied under this condition. The stress distributions are shown in Figure 5 and Figure 6.