Question

A cylindrical bar of metal having a diameter of 19.7 mm and a length of 206 mm is deformed elastically in tension with a force of 47500 N. Given that the elastic modulus and Poisson's ratio of the metal are 65.1 GPa and 0.35, 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.

Answer 1

a) ΔL = 4.93*10⁻⁴ m = 0.4931 mm

b) Δd = 0.0165 mm

Step-by-step explanation:

d = 19.7 mm = 0.0197 m

L = 206 mm = 0.206 m

P = 47500 N

E = 65.1 GPa = 65.1*10⁹ Pa

ν = 0.35

a) ΔL = ?

b) Δd = ?

Solution:

We get A as follows

A = 0.25*π*d² = 0.25*π*(0.0197 m)² = 3.048*10⁻⁴ m²

Then we can apply the following equation (Hooke's Law):

ΔL = P*L / (A*E)

ΔL = (47500 N)*(0.206 m) / (3.048*10⁻⁴ m²*65.1*10⁹ Pa)

ΔL = 4.93*10⁻⁴ m = 0.4931 mm

εa = ΔL / L = 0.4931 mm / 206 mm = 2.39*10⁻³

Now, we can use the formula

ν = εl / εa ⇒ εl = ν*εa = 0.35*2.39*10⁻³

⇒ εl = 8.378*10⁻⁴

If we know that

εl = Δd / d ⇒ Δd = εl*d

⇒ Δd = 8.378*10⁻⁴*0.0197 m = 1.65*10⁻⁵ m = 0.0165 mm