A cylindrical rod 120 mm long and having a diameter of 15 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2x10-2 mm. Of the materials listed below, which are possible candidates? Justify your choice(s). Material Modulus of Elasticity (GPa) Yield Strength (MPa) Poisson's ratio
Titanium alloy 70 250 0.33
Steel alloy 105 850 0.36
Magnesium alloy 205 550 0.27
Aluminium alloy 45 170 0.20

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.

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