A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not experience either plastic deformation or a diameter reduction of more than 7.5×10-3 mm. Of the materials listed as follows, which are possible candidates? Justify your choice(s).

Material Modulus of Elasticity(GPa) Yield Strength(MPa) Poission's Ratio
Aluminum alloy 70 200 0.33
Brass alloy 101 300 0.34
Steel alloy 207 400 0.3
Titanium alloy 107 650 0.34

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.

AFOKE88 AFOKE88 · Answer 2 · 2022-01-08T15:06:59+01:00

From the materials listed , the only one that is a possible candidate that meets the two criterions is; Steel Alloy

We are given the ideal properties and dimensions as;

Tensile Load; P = 27,500 N

Diameter; d₀ = 10.0 mm = 0.010 m

Maximum change in diameter; Δd = 7.5 × 10 ⁻³ mm

Let us find the required yield strength from the formula for stress which is;

σ = P/A₀

where A₀ is area = πd₀²/4

Thus;

σ = P/(π*d₀²/4)

σ = 27500/(π × 0.01²/4)

σ = 350 × 10⁶ N/m²

σ = 350 MPa

Now, from the given table we see that;

Aluminium alloy yield strength = 200 MPa

Brass alloy yield strength = 300 MPa

Steel alloy yield strength = 400 MPa

Titanium alloy yield strength = 650 MPa

Since the required rod must not experience plastic deformation, then the possible candidates will have yield strength greater than 350 MPa.

Only Steel and Titanium alloy meet this condition.

Let us check the second condition which says diameter reduction must not be more than 7.5 × 10 ⁻³ mm. We will do this for only steel and titanium alloys since they both met the first criteria.

Formula for the change in diameter is;

∆d = υ × σ × d₀/E

where;

υ is poisson ratio

σ is yield strength

d₀ is diameter

E is modulus of elasticity

Change in diameter for steel Alloy;

∆d = (0.3 × 350 × 10)/(207 × 10³)

∆d = 5.1 × 10⁻³ mm

5.1 × 10⁻³ mm < 7.5 × 10⁻³ mm

Therefore, the steel is a candidate.

Change in diameter for Titanium Alloy;

∆d = (0.34 × 350 × 10)/(650 × 10³)

∆d = 11.1 × 10⁻³ mm

11.1 × 10⁻³ mm > 7.5 × 10⁻³ mm

Therefore, the Titanium is not a candidate.

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