Answer:
The steel is a candidate.
Explanation:
Given
P = 27,500 N
d₀ = 10.0 mm = 0.01 m
Δd = 7.5×10
⁻³ mm (maximum value)
This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is applied; this means that the stress corresponding to this load not exceed the yield strength of the material.
Upon computing the stress
σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²
⇒ σ = 350 MPa
Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.
Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10
⁻³ mm), it is necessary to calculate the change in diameter Δd for these four alloys.
Then we use the equation υ = - εx / εz = - (Δd/d₀)/(σ/E)
⇒ υ = - (E*Δd)/(σ*d₀)
Now, solving for ∆d from this expression,
∆d = - υ*σ*d₀/E
For the Aluminum alloy
∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm
0.0165 mm > 7.5×10
⁻³ mm
Hence, the Aluminum alloy is not a candidate.
For the Brass alloy
∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm
0.0118 mm > 7.5×10
⁻³ mm
Hence, the Brass alloy is not a candidate.
For the Steel alloy
∆d = - (0.3)*(350 MPa)*(10 mm)/(207*10³MPa) = - 0.005 mm
0.005 mm < 7.5×10
⁻³ mm
Therefore, the steel is a candidate.
For the Titanium alloy
∆d = - (0.34)*(350 MPa)*(10 mm)/(107*10³MPa) = - 0.0111 mm
0.0111 mm > 7.5×10
⁻³ mm
Hence, the Titanium alloy is not a candidate.
