Final answer:
The maximum force a 0.5 cm-diameter rod of Al2O3 can withstand without plastic deformation is approximately 4732 N, calculated using the provided engineering yield strength and the area of the rod's cross-section.
Step-by-step explanation:
To calculate the maximum force (F) that a 0.5 cm-diameter rod of Al2O3 can withstand without plastic deformation using its engineering yield strength, we use the formula:
F = σ₀ × A
Where:
- σ₀ is the yield strength (241 MPa or 241 x 106 N/m2).
- A is the cross-sectional area of the rod.
To find the cross-sectional area (A) of the rod, we use the formula for the area of a circle (A = πr2) given that a rod's cross-section is circular.
First, we convert the diameter into meters and find the radius:
Diameter = 0.5 cm = 0.005 m
Radius (r) = Diameter/2 = 0.0025 m
Now we calculate the area (A):
A = π(0.0025 m)2
A = π× 6.25 x 10-6 m2
A ≈ 1.9635 x 10-5 m2
Finally, we apply the yield strength to find the maximum force:
F = 241 x 106 N/m2 × 1.9635 x 10-5 m2
F ≈ 4732.135 N (rounded to three significant digits)
Thus, the maximum force that the rod can withstand without plastic deformation is approximately 4732 N.