Respuesta :
Answer:
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
