A cylindrical rod 120 mm long and having a diameter of 15.0 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 1.2 * 10-2 mm. Of the following materials listed, which are possible candidates? Justify your choice (s).

Material - Modulos of Elasticity (GPa) - Yield Strength (MPa) - Poisson′s Ratio

Aluminum Alloy - 70 - 250 - 0.33

Titanium Alloy - 105 - 850 - 0.36

Steel Alloy - 205 - 550 - 0.27

Magnesium Alloy - 45 - 170 - 0.35

The possible candidates that meet the listed criteria are Titanium Alloy and Steel Alloy as explained further as follows

Explanation:

The given variables are

length of rod = 120 mm

diameter of rod = 15 mm

applied tensile force = 35000 N

limit of diameter reduction = 1.2*10⁻² mm

To asses the listed metals if they meet the two criterion of

1. The metal selected must not experience plastic deformation from the applied load and

2. The reduction in diameter from the applied load should not exceed 1.2*10⁻² mm

For the first criteria we have

Stress = Force/Area where the area = π*r² and r = Diameter/2

We have area = π*0.0075² → and σ = (35000 N) / (π * (0.0075 mm) ²) = 198059484.74 Pa or 198.06 MPa

Hence from the listed metals, the ones that meet the first criteria are Aluminum Alloy, Titanium Alloy and Steel Alloy

For the second criterion, we apply Poisson's ratio as follows

v = - εx / εz = - (Δd/d₀) : (σ/E), rearranging to make Δd the subject of the formula we have Δd = - (v * σ * d₀) / E

For Aluminum Alloy we have Δd = - (0.33*198.06 MPa*15mm) / (70 GPa) = 1.4*10⁻²mm hence Aluminum Alloy should be excluded

For Titanium Alloy we have Δd = - (0.36*198.06 MPa*15mm) / (105 GPa) = 1.019*10⁻² mm, hence Titanium alloy meets the requirements

For the Steel Alloy we have Δd = - (0.27*198.06 MPa*15mm) / (205 GPa) = 3.9*10⁻³mm Hence the steel alloy also meets the requirement