A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force of 50500 N. Given that the elastic modulus and Poisson's ratio of the metal are 65.5 GPa and 0.33, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

A) ΔL = 0.503 mm

B) Δd = - 0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π (d_o) ²) / 4

σ = 4F / (π (d_o) ²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F / (π (d_o) ²) = E (ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o) / (E•π (d_o) ²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 * 10^ (9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 * 50500 * 0.209) / (65.5 * 10^ (9) * π * (0.0202) ²)

ΔL = 0.503 * 10^ (-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = - (ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [ (Δd/d_o) / (ΔL/L_o) ]

v = - [ (Δd•L_o) / (ΔL•d_o) ]

Making Δd the subject, we have;

Δd = - [ (v•ΔL•d_o) / L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = - [ (0.33 * 0.503 * 20.2) / 209]

Δd = - 0.016 mm