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The alloy ti-6al-4v of table 9.1 is used to make a cylindrical pressure vessel having closed ends with an inner diameter of 250 mm and wall thickness of 2.5 mm. (a) What repeatedly applied pressure will cause fatigue failure in 10⁵ cycles? (Neglect the stress raiser effect of the end closure or other geometric discontinuities.) (b) For the pressure from (a), what is the safety factor against yielding?

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Final answer:

To cause fatigue failure in 10⁵ cycles, a pressure of 80,000 MPa/m needs to be applied to the cylindrical pressure vessel made of ti-6al-4v alloy. The safety factor against yielding for this pressure is 2.25.

Step-by-step explanation:

(a) To determine the repeatedly applied pressure that will cause fatigue failure in 10⁵ cycles, we need to use the endurance limit of the ti-6al-4v alloy. The endurance limit is the maximum stress that a material can withstand for an infinite number of cycles without experiencing fatigue failure. For ti-6al-4v alloy, the endurance limit is typically around 0.4 times the ultimate tensile strength (UTS) of the material. Assuming the UTS of ti-6al-4v alloy is 1000 MPa (megapascals), the endurance limit would be 0.4 * 1000 MPa = 400 MPa.

The formula to calculate the pressure is:

Pressure = Stress/Average Wall Thickness

Plugging in the values, we have:

Pressure = 400 MPa / (2 * 2.5 mm). Converting the wall thickness to meters, we get:

Pressure = 400 MPa / (2 * 0.0025 m) = 400 MPa / 0.005 m = 80,000 MPa/m.

(b) To calculate the safety factor against yielding, we need to compare the yield strength (or proportional limit) of the ti-6al-4v alloy to the stress caused by the repeatedly applied pressure. The yield strength of ti-6al-4v alloy is typically around 0.9 times the UTS of the material. Assuming the UTS of ti-6al-4v alloy is 1000 MPa, the yield strength would be 0.9 * 1000 MPa = 900 MPa.

The formula to calculate the safety factor is:

Safety Factor = Yield Strength / Stress

Plugging in the values, we have:

Safety Factor = 900 MPa / 400 MPa = 2.25

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