A cylindrical rod of copper (Young’s modulus of 110 GPa) with a yield strength of 240MPa will be subjected to a load of 6660 N. If the length of the rod is 380 mm, whatmust be the diameter to allow an elongation of 0.50 mm?

To find the diameter of the cylindrical rod, we can use the formula for elongation:

ΔL = (F * L) / (A * E)

Where: ΔL is the elongation F is the force applied L is the length of the rod A is the cross-sectional area of the rod E is the Young's modulus of the material

First, let's convert the given values to the appropriate units: Yield strength = 240 MPa = 240 * 10^6 Pa Load = 6660 N Length = 380 mm = 0.38 m Elongation = 0.50 mm = 0.0005 m

Next, we can rearrange the formula to solve for the cross-sectional area:

A = (F * L) / (ΔL * E)

Substituting the given values:

A = (6660 N * 0.38 m) / (0.0005 m * 110 * 10^9 Pa)

Calculating the cross-sectional area:

A ≈ 0.000217 m^2

Finally, we can use the formula for the area of a circle to find the diameter:

A = π * (d^2) / 4

Rearranging the formula:

d^2 = (4 * A) / π

Substituting the value of A:

d^2 = (4 * 0.000217 m^2) / π

Calculating the diameter:

d ≈ 0.0166 m

Therefore, the diameter of the cylindrical rod should be approximately 0.0166 meters to allow for an elongation of 0.50 mm.