Answer:
A) ΔL = 0.503 mm
B) Δd = -0.016 mm
Explanation:
A) From Hooke's law; σ = Eε
Where,
σ is stress
ε is strain
E is elastic modulus
Now, σ is simply Force/Area
So, with the initial area; σ = F/A_o
A_o = (π(d_o)²)/4
σ = 4F/(π(d_o)²)
Strain is simply; change in length/original length
So for initial length, ε = ΔL/L_o
So, combining the formulas for stress and strain into Hooke's law, we now have;
4F/(π(d_o)²) = E(ΔL/L_o)
Making ΔL the subject, we now have;
ΔL = (4F•L_o)/(E•π(d_o)²)
We are given;
F = 50500 N
L_o = 209mm = 0.209m
E = 65.5 GPa = 65.5 × 10^(9) N/m²
d_o = 20.2 mm = 0.0202 m
Plugging in these values, we have;
ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)
ΔL = 0.503 × 10^(-3) m = 0.503 mm
B) The formula for Poisson's ratio is;
v = -(ε_x/ε_z)
Where; ε_x is transverse strain and ε_z is longitudinal strain.
So,
ε_x = Δd/d_o
ε_z = ΔL/L_o
Thus;
v = - [(Δd/d_o)/(ΔL/L_o)]
v = - [(Δd•L_o)/(ΔL•d_o)]
Making Δd the subject, we have;
Δd = -[(v•ΔL•d_o)/L_o]
We are given v = 0.33; d_o = 20.2mm
So,
Δd = -[(0.33 × 0.503 × 20.2)/209]
Δd = -0.016 mm
