Question

A copper rod with a diameter of 0.5 in., modulus of 16 x 10⁶ psi, and a Poisson's ratio of 0.35 is subjected to a tensile load within the elastic region. Determine the force that will produce a reduction of 1 x 10⁴ in. in the diameter.

Answer 1

To determine the force required to produce a reduction in diameter of a copper rod, we can use formulas for tensile strain, stress, and force. By calculating strain, stress, and the cross-sectional area of the rod, we can determine the force. The force is approximately 25.12 x 10¹⁰ psi.

Step-by-step explanation:

To determine the force required to produce a reduction in diameter, we can use the formula for tensile strain:

Strain = (original diameter change / original diameter)

From the given values, the original diameter change is 1 x 10⁴ in. and the original diameter is 0.5 in.

Substituting these values into the formula, we get:

Strain = (1 x 10⁴ / 0.5)

Strain = 2 x 10⁴

Now, we can use the formula for stress to calculate the force:

Stress = (modulus x strain)

From the given values, the modulus is 16 x 10⁶ psi.

Substituting the values into the formula, we get:

Stress = (16 x 10⁶ x 2 x 10⁴)

Stress = 32 x 10¹⁰ psi

To convert psi to force, we need to multiply the stress by the cross-sectional area.

The cross-sectional area can be calculated using the formula:

Area = (π x (diameter/2)²)

Substituting the given values into the formula, we get:

Area = (π x (0.5/2)²)

Area = (π x 0.25)

Area ≈ 0.785 in²

Finally, we can calculate the force:

Force = (Stress x Area)

Substituting the values into the formula, we get:

Force = (32 x 10¹⁰ psi x 0.785 in²)

Force ≈ 25.12 x 10¹⁰ psi