Respuesta :
Answer:
The only suitable candidate is steel.
Explanation:
We are given;
Initial length;L_o = 120 mm = 0.12 m
Diameter;d_o = 15 mm = 0.015 m
Radius;r = 0.015/2 m = 0.0075 m
Applied force;F = 35000 N
Allowable Diameter reduction;Δd = 1.2 x 10^(-2) mm = 0.012 mm
Formula for Poisson’s ratio is;
v = ε_x/ε_z
Where;
ε_x is lateral strain = = Δd/d_o
ε_z is axial strain = σ/E
Thus;
v = (Δd/d_o)/(σ/E)
Making Δd the subject, we have;
Δd = (vσd_o)/E
Where σ is force per unit area
σ = F/A = 35000/πr² = 35000/(π0.0075)² = 198 × 10^(6) Pa
Now, let's find Δd for each of the metals given;
For Titanium alloy, E = 70 GPa = 70 × 10^(9) Pa and Poisson’s ratio (v) = 0.33
Thus;
Δd_ti = (0.33 × 198 × 10^(6) × 0.015)/70 × 10^(9) = 0.000014m = 0.014 mm
Similarly, for steel;
Δd_st = (0.36 × 198 × 10^(6) × 0.015)/(105 × 10^(9)) = 0.0000102 m = 0.0102 mm
For magnesium;
Δd_mn = (0.27 × 198 × 10^(6) × 0.015)/(205 × 10^(9)) = 0.0000039 m = 0.0039 mm
For aluminum;
Δd_al = (0.2 × 198 × 10^(6) × 0.015)/(45 × 10^(9)) = 0.0000132 m = 0.0132 mm
Since they must not experience a diameter reduction of more than 0.012 mm, then the only alloy that has a reduction diameter less than 0.012 is steel which has 0.0102.
Thus,the only possible candidate is stee.
