1. Introduction

2. Design and Working Principle of Scratching Damper

3. Results and Discussions

3.1. Young’s Modulus of Extruded Aluminum Flat Bar

Upon scratch test of an extruded aluminum flat bar sample as shown in Figure 6 of size 15(L)x10(W)x4(T) in mm using scratching damper, the following sample tests in two to four divisions of the compression of spring pressure are conducted to investigate its material properties (such as Young’s modulus, hardness) in Figure 7.

These load curves indicate that loop energy increases linearly with the interval load [5,6,7,8]. Some of the results are complied with Akono and Ulm [7,8,9,10] that scratching energy release rate increases with test load in quadratic curve. From Table 1, one observes that normal force increases linearly with two to four dial divisions. However, the profile of loop damping energy per interval remains relatively constant.

Table 1. Extruded aluminum material property comparison from scratching test.

Table 1. Extruded aluminum material property comparison from scratching test.

The summation of loop energy per interval with pin conical rod is divisionals’ total energy expressed as

$$U_{pin}^j={\displaystyle \sum _{i=1}^N{U}_{pin}^{ji}\cdot j\cdot f_t}\cdot l_t/2=U_t^j$$

(17)

where $U_{pin}^{ji}$ is single loop energy with pin rod, i is number of test interval, j is dial division of the test, $f_t$=9 Hz is fixed-sine test frequency, $l_t$= 48s is the test period. It is interesting to observe that shape of loop is highly repeatable within the same test period with nearly 500 cycles. Meanwhile, the summation of loop damping energy per interval with round rod is expressed as

$${\mathsf{\Psi }}_{round}^j={\displaystyle \sum _{i=1}^N{\mathsf{\Psi }}_{round}^{ji}\cdot j\cdot f_t}\cdot l_t/2={\mathsf{\Psi }}_j$$

(18)

where $U_{round}^{ji}$ is single loop energy with round rod. By Eq.(1), scratching energy per interval with j divisions is

$${\wp }_j=U_{pin}^j-{\mathsf{\Psi }}_{round}^j$$

(19)

Divisional scratching energy release rate is

$${\wp }_j={\displaystyle \sum _{i=1}^{i=N}\frac{{\wp }_j^i}{N\cdot j}}$$

(20)

Meanwhile by Eq. (12), divisional Young’s modulus stands as

$$E_j=\frac{1-v^2}{{\wp }_j}\times \frac{Fe{q}_j^2}{2p_j{A}_j}$$

(21)

with notations in j divisions. Results from two to four divisions are input to Table 1. In general, their values decrease slightly with scratch performances and they tend to underestimate the actual E. High performance indicates that scratching process follows scratch energy release rate. Two divisions is the best at 29.5%. Thus, by two divisional test, its ith interval $E_2^i$ is generated. From its Nth data, ultimate Young’s modulus is obtained as $E_2^N=54.6\times {10}^{9}GPa$.

Tensile test is utilized to validate the Young’s modulus of extruded aluminum. Sample of nominal size in mm 57.4(L)x19.1(W)x2.76(T) is prepared conforming to the ASTM E 21 standards. Tensile testing machine is set to pull the sample as shown in Figure 8 until its fracture is exhibited (Figure 9). Stress and strain are calculated during the test by recording the applied tensile load and extension.

Stress-strain curve is generated by the machine as shown in Figure 10.

According to the analysis of Su and Young [9,10,11,12], Young’s modulus of extruded aluminum varies linearly in initial stage of this curve. In the linear elastic region, the material behaves elastically, and the value of the Young’s modulus remains relatively constant. It is important to note that that as the material gets closer to its yield point, the point at which plastic deformation starts, this linear connection could not hold true at greater stress levels. Young's modulus will go down as the substance experiences plastic deformation. By examining the change in stress and strain values at distinct positions within the linear section, insights into the material’s behaviour can be sought and Young’s modulus based on the ratio of the changes in stress and strain can be calculated. In our case, change in stress and strain at separate positions 1 and 2 of linear section are computed. Slope of the curve is used to calculate the Young’s modulus of this sample as

$$E^T=\frac{\Delta \sigma }{\Delta \delta }=\frac{{\sigma }_2^T-{\sigma }_1^T}{{\delta }_2^T-{\delta }_1^T}\times \frac{L}{WT}$$

(22)

The generated value is $E^T=57.2GPa$ which is +4.55% larger than$E_2^N$. Positive deviation arises at the elastic region estimation by tensile test. Thus, it tends to overestimate $E_2^i$ at plastic region. For extruded aluminum, it’s normalised dependence of Young’s modulus ratio $E_r=E_a/E_o$on the apparent density ${\rho }_r={\rho }_a/{\rho }_o$for flat bar sample [11,12,13,14] is

$$E_r=w{\rho }_r^u$$

(23)

where w = 0.974 is fitting constant, u is fitting exponent. Using digital weight test, experimental ${\rho }_r$ is found to be 0.91. Substituting ${\rho }_r$and$E^T=57.2GPa$ into this equation, u is determined to be 1.479. Now substituting the scratch $E_r=0.578$ into Eq. (23), scratch ${\rho }_r$ is computed as 0.844 which being -7.2% deviation from experimental value.

4. Conclusions

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