Question

A 500mm-long rod has a diameter of 25mm. If an axial tensile load of 50kN is applied to it,please answer the following questions:______.

(a) Will the material undergoyielding? Why or Why not?
(b) Can Hooke’s law be applied to this problem? (Assumethe proportionallimitstress is same with the Yield Strength of the material.)
(c) What is the longitudinalstrain?
(d) What is the lateral strain?
(e) What isthe change in its diameter? (PleaseuseYoung’smodulus E=200GPa, Shear Modulus (or Modulus of Rigidity) G=75.2GPa, and Yield StrengthσY=400????Pa)

Answer 1

(a) Since, the stress is less than the yield strength. Therefore, the material does not go yielding.

(b) Yes

(c) Longitudinal Strain = 5.09 x 10⁻⁴

(d) Lateral Strain = 1.67 x 10⁻⁴

(e) Change in Diameter = 4.2 x 10⁻³ mm = 4.2 μm

Step-by-step explanation:

(a)

First we calculate stress:

Stress = Force/Area = F/πr²

Stress = 50 KN/π(0.0125 m)²

Stress = 101.86 MPa

Since, the stress is less than the yield strength. Therefore, the material does not go yielding.

(b)

Yes, Hooke's law can be applied to this problem because the material is within elastic limits.

(c)

Longitudinal Strain = Stress/E

Longitudinal Strain = 0.10186 GPa/200 GPa

Longitudinal Strain = 5.09 x 10⁻⁴

(d)

Poisson's Ratio = Lateral Strain/Longitudinal Strain = E/2G - 1

Lateral Strain/Longitudinal Strain = E/2G - 1

Lateral Strain/5.09 x 10⁻⁴ = 200 GPa/2*75.2 GPa - 1

Lateral Strain = 0.33*5.09 x 10⁻⁴

Lateral Strain = 1.67 x 10⁻⁴

(e)

Lateral Strain = Change in Diameter/Original Diameter

Change in diameter = (1.67 x 10⁻⁴)(25 mm)

Change in Diameter = 4.2 x 10⁻³ mm = 4.2 μm