Under what axial tensile load, the diameter of a steel bar will be reduced from 8 cm to 7.995 cm? Take E as 2.0 ´ 103 t/cm2 and Poisson’s ratio as 0.3.
To find the axial tensile load required for the reduction in diameter of the steel bar, we can use the concept of stress and strain.
Stress is defined as force per unit area and is given by the formula:
Stress (σ) = Force (F) / Area (A)
Strain is defined as the change in length per unit length and is given by the formula:
Strain (ε) = Change in length (ΔL) / Original length (L)
The relationship between stress and strain is given by Hooke's Law:
Stress (σ) = Young's Modulus (E) × Strain (ε)
Poisson's ratio (ν) is the ratio of lateral strain to axial strain and is given as 0.3 in this case.
Now, let's solve the problem step by step:
1. Calculate the original area of the steel bar:
Area (A) = π × (Original diameter)^2 / 4
= π × (8 cm)^2 / 4
2. Calculate the original length of the steel bar:
Original length (L) = 2 cm
3. Calculate the change in length:
Change in length (ΔL) = Original length (L) × (1 - Strain (ε))
= 2 cm × (1 - Strain (ε))
4. Calculate the axial strain:
Axial strain = Change in diameter / Original diameter
= (Original diameter - Final diameter) / Original diameter
= (8 cm - 7.995 cm) / 8 cm
5. Calculate the axial stress:
Axial stress (σ) = Young's Modulus (E) × Axial strain
= 2.0 × 10^3 t/cm^2 × (8 cm - 7.995 cm) / 8 cm
Therefore, the axial tensile load required for the reduction in diameter from 8 cm to 7.995 cm can be calculated using the above steps.