Question

A cylindrical bar of metal having a diameter of 17.8 mm and a length of 196 mm is deformed elastically in tension with a force of 46400 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34, respectively, determine the following:
a. The amount by which this specimen will elongate (in mm) in the direction of the applied stress. Enter your answer for part (a) in accordance to the question statement 0.544.
b. The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answer 1

Answer:

A) ΔL = 0.544 mm

B) Δd = -0.0168 mm

Explanation:

We are given;

Elastic modulus; E = 67.1 GPa = 67.1 × 10^(9) Pa

Force; F = 46400 N

Diameter; d = 17.8 mm = 17.8 × 10^(-3) m

Radius; r = d/2 = 17.8/2 = 8.9 mm = 8.9 × 10^(-3) m

Length; L = 196 mm = 0.196 m

Poisson ratio; ν = 0.34

A) We know that formula for elastic modulus is;

E = σ/ε

Where;

σ = F/A

ε = ΔL/L

Thus;

E = (FL/ΔL•A)

ΔL is change in length. Making it the subject of the formula, we have;

ΔL = FL/AE

Now, A = πr²

A = π × (8.9 × 10^(-3))²

ΔL = [(0.196 × 46400)/(π × (8.9 × 10^(-3))² × 67.1 × 10^(9)]

ΔL = 0.544 × 10^(-3) m

ΔL = 0.544 mm

B) formula for Poisson ratio is given as;

ν = -ε_x/ε_z

Where;

ε_x is transverse strain = Δd/d

ε_z is longitudinal strain = ΔL/L

Thus;

ν = -Δd•L/d•ΔL

Making Δd the subject, we have;

Δd = -νdΔL/L

Δd = -(0.34 × 17.8 × 10^(-3) × 0.544 × 10^(-3))/0.196

Δd = -0.0168 × 10^(-3) m = -0.0168 mm