Under what axial tensile load, the diameter of a steel bar will be reduced from 8 cm to 7.995 cm? Take E as t/cm2 and Poisson’s ratio as 0.3
To determine the axial tensile load that will result in the reduction of diameter from 8 cm to 7.995 cm, we can use the concept of axial strain and Hooke's Law.
Axial strain (ε) is defined as the change in length (ΔL) divided by the original length (L):
ε = ΔL / L
Hooke's Law states that stress (σ) is proportional to strain (ε) and is given by:
σ = E * ε
Here, E represents the modulus of elasticity (Young's modulus), which is given as t/cm², and Poisson's ratio (ν) is provided as 0.3.
We need to calculate the change in diameter (Δd) and convert it to strain. Since strain is typically expressed as a fractional change in length, we'll divide Δd by the original diameter (d).
Δd = 8 cm - 7.995 cm = 0.005 cm
d = 8 cm
ε = Δd / d = 0.005 cm / 8 cm = 0.000625
Now, we can calculate the tensile stress using Hooke's Law:
σ = E * ε = t/cm² * 0.000625
To find the tensile load that will cause this stress, we need to multiply it by the cross-sectional area (A) of the steel bar:
A = (π * d²) / 4
Substituting the value of d (8 cm), we get:
A = (π * (8 cm)²) / 4
Finally, we can calculate the tensile load (F):
F = σ * A = (t/cm² * 0.000625) * ((π * (8 cm)²) / 4)
Note: Please substitute the appropriate value of t/cm² (the modulus of elasticity) into the equation to get the final numerical result.