Question 8.1: Finding Wear Coefficients from Experimental Data A component......
Finding Wear Coefficients from Experimental Data
A component of a braking system consists of the unlubricated rounded end of a 2011-T3 wrought aluminum alloy pin being pushed with a force P against the flat surface of a rotating AISI-1095 HR steel disk (see Figure 8.5). The rubbing contact is at a radius R and the disk rotates at a speed n. Consequent to a t minutes test duration, the disk and pin are weighed. It is found that adhesive wear produced weight losses equivalent to wear volumes of V_a and V_s for the aluminum and steel, respectively.
Find: Compute the wear coefficients.
Data: The given numerical values are as follows:
Steel disk: 248 Brinell hardness (Table B.3) wear volume V_s=0.98 mm ^3
Aluminum pin: 95 Brinell hardness (Table B.6) wear volume V_a=4.1 mm ^3
Contact force: P=30 N at a radius R=24 mm
Test duration: t=180 min at a sliding speed n=120 rpm.
| TABLE B.3 Mechanical Properties of Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels |
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| UNS Number | AISI/ SAE Number | Processing | Ultimate
\text {Strength}^a S_u (MPa) |
Yield
\text {Strength}^a S_y (MPa) |
Elongation in 50 mm (%) | Reduction in Area (%) | Brinell Hardness (H_B) |
| G10060 | 1006 | HR | 300 | 170 | 30 | 55 | 86 |
| CD | 330 | 280 | 20 | 45 | 95 | ||
| G10100 | 1010 | HR | 320 | 180 | 28 | 50 | 95 |
| CD | 370 | 300 | 20 | 40 | 105 | ||
| G10150 | 1015 | HR | 340 | 190 | 28 | 50 | 101 |
| CD | 390 | 320 | 18 | 40 | 111 | ||
| G10200 | 1020 | HR | 380 | 210 | 25 | 50 | 111 |
| CD | 470 | 390 | 15 | 40 | 131 | ||
| G10300 | 1030 | HR | 470 | 260 | 20 | 42 | 137 |
| CD | 520 | 440 | 12 | 35 | 149 | ||
| G10350 | 1035 | HR | 500 | 270 | 18 | 40 | 143 |
| CD | 550 | 460 | 12 | 35 | 163 | ||
| G10400 | 1040 | HR | 520 | 290 | 18 | 40 | 149 |
| CD | 590 | 490 | 12 | 35 | 170 | ||
| G10450 | 1045 | HR | 570 | 310 | 16 | 40 | 163 |
| CD | 630 | 530 | 12 | 35 | 179 | ||
| G10500 | 1050 | HR | 620 | 340 | 15 | 35 | 179 |
| CD | 690 | 580 | 10 | 30 | 197 | ||
| G10600 | 1060 | HR | 680 | 370 | 12 | 30 | 201 |
| G10800 | 1080 | HR | 770 | 420 | 10 | 25 | 229 |
| G10950 | 1095 | HR | 830 | 460 | 10 | 25 | 248 |
| Source: ASM Handbook, vol. 1, ASM International, Materials Park, OH, 2020. | |||||||
| Note: To convert from MPa to ksi, divide given values by 6.895. | |||||||
| { } ^ {a} Values listed are estimated ASTM minimum values in the size range 18–32 mm. | |||||||
| TABLE B.6 Mechanical Properties of Some Aluminum Alloys |
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| Ultimate Strength S_u | Yield Strength S_y | |||||
| Alloy | (MPa) | (ksi) | (MPa) | (ksi) | Elongation in 50 mm (%) | Brinell Hardness ( H_B ) |
| Wrought: | ||||||
| 1100-H14 | 125 | (18) | 115 | (17) | 20 | 32 |
| 2011-T3 | 380 | (55) | 295 | (43) | 15 | 95 |
| 2014-T4 | 425 | (62) | 290 | (42) | 20 | 105 |
| 2024-T4 | 470 | (68) | 325 | (47) | 19 | 120 |
| 6061- T 6 | 310 | (45) | 275 | (40) | 17 | 95 |
| 6063- T 6 | 240 | (35) | 215 | (31) | 12 | 73 |
| 7075- T 6 | 570 | (83) | 505 | (73) | 11 | 150 |
| Cast | ||||||
| 201-T4^a | 365 | (53) | 215 | (31) | 20 | — |
| 295-T6^a | 250 | (36) | 165 | (24) | 5 | — |
| 355-T6^a | 240 | (35) | 175 | (25) | 3 | — |
| -T6^b | 290 | (42) | 190 | (27) | 4 | — |
| 356-T6^a | 230 | (33) | 165 | (24) | 2 | — |
| -T6^b | 265 | (38) | 185 | (27) | 5 | — |
| 520-T4^a | 330 | (48) | 180 | (26) | 16 | — |
| Sources: Materials Engineering, Materials Selector, Penton Publication, Cleveland, OH, 1991. | ||||||
| {} ^{a} Sand casting. | ||||||
| {} ^{b} Permanent mold casting. | ||||||
See Figure 8.5 and Equation (8.2).
V=K \frac{P L}{H} (8.2)
Total length of sliding is expressed as
\begin{aligned} L & =2 \pi R n t \\ & =2 \pi(24)(120)(180)=3.26 \times 10^6 mm \end{aligned} (b)
The values of hardness of pin and disk are
H_a=9.81(95)=932 MPa
H_s=9.81(248)=2433 MPa
Through the use of Equation (8.2), we have K=VH/PL . Therefore, introducing the numerical values, the wear coefficients for aluminum pin and steel disk are, respectively,
K_a=\frac{4.1(932)}{30\left(3.26 \times 10^6\right)}=3.91 \times 10^{-5}
K_s=\frac{0.98(2433)}{30\left(3.26 \times 10^6\right)}=2.44 \times 10^{-5}
Comments: Observe that the wear coefficient of the pin is about 1.6 times that of the disk wear coefficient. Interestingly, if the worn pin surface remains flat, for a given pin diameter, we approximately have
V_a=\frac{\pi d^2 h_p}{4} \quad V_s=2 \pi R d h_d (c)
Since wear volumes (as well as d and R) are known, then the linear pin wear depth h_p and the wear depth of h_d in the disk may readily be computed.