Failure Analysis and Simulation Evaluation of an Al 6061 Alloy Wheel Hub

Abstract

This paper details the failure analysis of a wheel hub from a student designed Formula SAE^® race car that fractured at the roots of the rim finger attachment region. The wheel hub was identified to be manufactured from a rolled Al 6061 alloy. The experimental characterization included fracture surface analysis and microstructural analysis using scanning electron microscopy, as well as compressive stress–strain testing and micro-hardness testing to determine its mechanical properties. Analysis of the fractured surfaces of the hub revealed beach marks and striations, suggesting a fatigue failure. A kinematic model was developed to determine wheel hub loadings as defined by the car driving history. Detailed loads calculated from a kinematic equilibrium model and material properties obtained from the experiment results were used in a finite element model to simulate the stress distribution and fatigue life of the wheel hub. The wheel simulation results were consistent with the failure mode determined from the fractography study.

Appendix

The mathematical description of the quasi-static equilibrium state was developed by summing and setting equal to zero the forces in each global coordinate direction, and the moments about each global and local coordinate axis. From summation of forces in the global x, y, z coordinate direction separately, obtained Eqs 4–6.

$$\sum {F_{x} } = 0 = P_{\text{rl}} + P_{\text{rr}} - \mu_{\text{s}} \cdot W_{\text{fl}} \cdot { \sin }(\theta_{\text{fl}} ) - F_{\text{fr}} \cdot { \sin }(\theta_{\text{fr}} )$$

(4)

$$\sum {F_{y} = 0 = \mu_{\text{s}} \cdot W_{\text{fl}} \cdot { \cos }(\theta_{\text{fl}} ) + F_{\text{fr}} \cdot { \cos }(\theta_{\text{fr}} ) + \mu_{\text{s}} \cdot W_{\text{rl}} + F_{\text{rr}} - F_{\text{c}} }$$

(5)

$$\sum {F_{z} = 0 = W_{\text{fl}} + W_{\text{fr}} + W_{\text{rl}} + W_{\text{rr}} - M_{\text{c}} \cdot g}.$$

(6)

Summing the moments along global x, y, z coordinate axis resulted in the following Eqs 7–9.

$$\begin{aligned} \sum {M_{x} = 0 = \mu_{\text{s}} \cdot W_{\text{fl}} \cdot { \cos }(\theta_{\text{fl}} ) \cdot h + F_{\text{fr}} \cdot { \cos }(\theta_{\text{fr}} ) \cdot h + W_{\text{fl}} \cdot \Delta y} \hfill \\ + \mu_{\text{s}} \cdot W_{\text{fr}} \cdot h + F_{\text{rr}} \cdot h - W_{\text{fr}} \cdot \Delta y - W_{\text{rr}} \cdot \Delta y \hfill \\ \end{aligned}$$

(7)

$$\begin{aligned} \sum {M_{y} = 0 = W_{\text{fl}} } \cdot \Delta x_{\text{f}} + W_{\text{fr}} \cdot \Delta x_{\text{f}} + P_{\text{rl}} \cdot h - \mu_{\text{s}} \cdot W_{\text{fl}} \cdot { \sin }(\theta_{\text{fl}} ) \cdot h \hfill \\ - F_{\text{fr}} \cdot { \sin }(\theta_{\text{fr}} ) \cdot h - W_{\text{rl}} \cdot \Delta x_{\text{r}} - W_{\text{rr}} \cdot \Delta x_{\text{r}} \hfill \\ \end{aligned}$$

(8)

$$\begin{aligned} \sum {M_{z} = 0 = \mu_{\text{s}} \cdot W_{\text{fl}} \cdot { \cos }(\theta_{\text{fl}} ) \cdot \Delta x_{\text{f}} + F_{\text{fr}} \cdot { \cos }(\theta_{\text{fr}} ) \cdot \Delta x_{\text{f}} + \mu_{\text{s}} \cdot W_{\text{fl}} \cdot { \sin }(\theta_{\text{fl}} ) \cdot \Delta y} \hfill \\ + P_{\text{rr}} \cdot \Delta y - F_{\text{fr}} \cdot { \sin }(\theta_{\text{fr}} ) \cdot \Delta y - \mu_{\text{s}} \cdot W_{\text{rl}} \cdot \Delta x_{\text{r}} - F_{\text{rr}} \cdot \Delta x_{\text{r}} - P_{\text{rl}} \cdot \Delta y \hfill \\ \end{aligned}.$$

(9)

Summing the moments about the local X′, X″, Y′ coordinate axis obtained Eqs 10–12.

$$\sum {M_{{x^{\prime}}} = 0 = W_{\text{c}} \cdot \Delta y + F_{\text{c}} \cdot h - W_{\text{fl}} \cdot 2 \cdot \Delta y - W_{\text{rr}} \cdot 2 \cdot \Delta y}$$

(10)

$$\sum {M_{{x^{\prime\prime}}} = 0 = W_{\text{fl}} \cdot 2 \cdot \Delta y + W_{\text{rl}} \cdot 2 \cdot \Delta y + F_{\text{c}} \cdot h - W_{\text{c}} \cdot \Delta y}$$

(11)

$$\sum {M_{{y^{\prime}}} = 0 = W_{\text{c}} \cdot \Delta x_{\text{f}} - W_{\text{rl}} \cdot (\Delta x_{\text{f}} + \Delta x_{\text{r}} ) - W_{\text{rr}} \cdot (\Delta x_{\text{f}} + \Delta x_{\text{r}} )},$$

(12)

where $\mu_{\text{s}}$ is the dynamic frictional coefficient, and $h$ is the distance between the gravity center and the ground. These reaction forces corresponding to the hard turn conditions can be calculated from the above equations, and the unknown forces for hard left turn were solved and are shown in Table 3.

In the straight driving condition, since the gravity center is centrally located with respect to the wheels and there is no acceleration, it was assumed that the weight was evenly distributed to each wheel and no side loads imposed on the wheels.