Question

For a brass alloy, the stress at which plastic deformation begins is 345 MPa, and the

modulus of elasticity is 103 GPa.
a) What is the maximum load that can be applied to a specimen with a crosssectional area of 130 mm2 without plastic deformation?

Answer 1

To calculate the maximum load that can be applied to a specimen without plastic deformation, we need to use the stress-strain relationship and the given information.

Given:
Stress at which plastic deformation begins (σ) = 345 MPa = 345 N/mm²
Cross-sectional area of the specimen (A) = 130 mm²

We can use the formula for stress (σ) to calculate the maximum load (F):

σ = F / A

Rearranging the formula, we have:

F = σ * A

Substituting the given values into the formula:

F = 345 N/mm² * 130 mm²

Now, let's convert the units to a more convenient form:

1 N = 1 kg * m/s²
1 mm = 1 x 10^-3 m

F = (345 N/mm²) * (130 mm²) * (1 kg * m/s² / 1 N) * (1 m / 1000 mm) * (1 m / 1000 mm)

Simplifying the units:

F = (345 * 130 * 1 * 1 * 1) / (1000 * 1000)

F = 44.85 N

Therefore, the maximum load that can be applied to the specimen without plastic deformation is approximately 44.85 Newtons.