13.8k views
4 votes
Calculate the maximum force that a 0.5 cm-diameter rod of Al2O3, having an engineering yield strength of 241 MPa, can withstand with no plastic deformation. Express your answer in N.

1 Answer

7 votes

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.

User Cwhelms
by
8.9k points