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A cylindrical tank with hemispherical ends is made from hot rolled steel that has \(S_{u t}=\) \(450 \mathrm{MPa}\). The tank outside diameter is \(350 \mathrm{~mm}\) with \(25 \mathrm{~mm}\) wall thickness. The pressure fluctuates from 0 to \(15 \mathrm{MPa}\). Determine the infinite-life fatigue safety factor for \(99.9 \%\) reliability.

Short Answer

Expert verified
The infinite-life fatigue safety factor for 99.9% reliability is calculated using the formulas and principles outlined in the steps. Variables such as the cyclic stress, fatigue strength, reliability factor, and threshold endurance limit are considered in the solution.

Step by step solution

01

Calculate the cyclic stress

First, calculate the cyclic stress using the thin-walled pressure vessel theory formula for hoop stress or circumferential stress, \( \sigma = \frac{Pr}{t} \). Here, \( P \) is pressure, \(r\) is radius, and \( t \) is wall thickness. Since the pressure fluctuates from 0 to 15 MPa, \( P = 15 MPa \). Therefore, the cyclic stress, \( \Delta \sigma \) is equal to \( 2* \sigma \).
02

Calculate the fatigue strength

The next step is calculating the fatigue strength, \(Se\). This is equal to 0.5 * \(S_{ut}\) for hot rolled steel. Here, \( S_{ut} = 450 MPa \). Calculate \( Se \) using this formula.
03

Incorporate reliability factor

In this step, you need to correct for a reliability of 99.9%. The reliability factor, \( k_{a} \) can be found on a standard reliability chart for steels, or it may be given. For steel at 99.9% reliability, the usual value for \( k_{a} \) is about 0.868. The threshold endurance limit, \(Se'\) is then given by \( Se' = k_{a}Se \). Calculate \( Se' \) using this formula.
04

Determine infinite-life fatigue safety factor

Finally, determine the infinite-life fatigue safety factor. This is equal to \( \frac{Se'}{\Delta \sigma / 2} \). Use the previously calculated \( Se' \) and \( \Delta \sigma \) to find the fatigue safety factor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thin-Walled Pressure Vessel Theory
Thin-walled pressure vessels are commonly used in industry for storing and transporting liquids and gases under pressure. The thin-walled assumption is valid when the wall thickness is less than one-tenth of the vessel's radius. Under this assumption, the stresses can be considered uniformly distributed across the wall's thickness. This theory is particularly important for understanding how a tank, cylinder, or sphere responds to internal pressure.

To apply the theory to our cylindrical tank example with hemispherical ends, it's crucial to consider that as the internal pressure fluctuates, the material undergoes cyclic loading, which can lead to fatigue. Therefore, calculating the hoop stress — which is the circumferential stress in the tank wall due to the internal pressure — is a key step in assessing the vessel's ability to withstand repeated loads without failure.
Hoop Stress
Hoop stress is the force exerted circumferentially (perpendicular to the axis and parallel to the radius) in a material like a cylindrical tank or a pipe due to internal pressure. It's one of the principal stresses in cylinder walls, caused by forces working to elongate the material. The formula for hoop stress, given by \( \sigma = \frac{Pr}{t} \) where \( P \) is the internal pressure, \( r \) is the inside radius, and \( t \) is the wall thickness, highlights how pressure, radius, and thickness affect the overall stress state of the vessel.

In the case of our pressure vessel, hoop stress illustrates the direct relationship between pressure and vessel diameter — as either increases, so does the stress, which could potentially lead to failure if the material's strength is exceeded.
Fatigue Strength
Fatigue strength, denoted as \( S_e \) for metals, is the highest stress level that a material can withstand for an infinite number of cycles without leading to failure. This is a critical property taken into account when designing components that will undergo repeated loading, as is the case with pressure fluctuations in a tank.

For our hot rolled steel tank, the estimated fatigue strength is 50% of the ultimate tensile strength \( S_{ut} \) due to its material properties. These values come from standard material tests and tables that provide average properties for common materials. Engineers use these properties, along with theories of failure and safety factors, to design safer, more reliable structures.
Reliability Factor
The reliability factor, often represented as \( k_a \) in engineering equations, is a multiplier used to adjust the fatigue strength of a material to reflect the desired level of reliability in the face of statistical variability inherent in material properties. A high reliability requirement, such as 99.9%, means that only one out of a thousand identical pressure vessels is expected to fail under specified conditions.

For components where failure could result in significant safety or economic consequences, engineers use reliability factors like \( k_a = 0.868 \) to provide an additional margin of safety. This ensures that even with variations in material quality, manufacturing processes, or real-world loading conditions, the component is still very likely to perform as intended.
Infinite-Life Fatigue
Infinite-life fatigue describes the behavior of materials that are expected to withstand an unlimited number of load cycles without failing, provided the loading is kept below a certain stress level. This concept is integral to safe design for parts that experience cyclic stress, such as the walls of our pressure vessel, because it ensures that these parts will be durable over their operational lifetime.

Using the thin-walled pressure vessel theory, the fatigue strength, and the reliability factor, one can calculate the fatigue safety factor—a measure of how much stress a material can handle relative to what it will experience during service. Within the context of infinite-life design, the goal is to maintain stress levels within the material's endurance limit to avoid fatigue failure.

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Most popular questions from this chapter

A cylindrical tank with hemispherical ends has been built. It was made from hot rolled steel that has \(S_{u t}=380 \mathrm{MPa}\). The tank outside diameter is \(300 \mathrm{~mm}\) with \(20 \mathrm{~mm}\) wall thickness. The pressure may fluctuate from 0 to an unknown maximum. For an infinitelife fatigue safety factor of 4 with \(99.99 \%\) reliability, what is the maximum pressure to which the tank may be subjected?

A nonrotating, ground, steel rod has a rectangular section with \(h=60 \mathrm{~mm}\) and \(b=40\) \(\mathrm{mm}\). It is loaded in repeated bending. Determine its corrected fatigue strength with \(99.9 \%\) reliability if it is used in an environment that has a temperature that is below \(450^{\circ} \mathrm{C}\) and has an ultimate tensile strength of \(1550 \mathrm{MPa}\).

A 42-mm-dia steel shaft with a 19-mm transverse hole is subjected to a combined loading of \(\sigma=\pm 100 \mathrm{MPa}\) bending stress and an alternating torsion of \(\pm 110 \mathrm{MPa}\), which are \(90^{\circ}\) out-of- phase. Find its safety factor for infinite life if \(S_{u t}=1 \mathrm{GPa}\).

A 42-mm-dia steel shaft with a 19-mm transverse hole is subjected to a sinusoidal combined loading of \(\sigma=\pm 100 \mathrm{MPa}\) bending stress and a steady torsion of \(110 \mathrm{MPa}\). Find its safety factor for infinite life if \(S_{u t}=1 \mathrm{GPa}\).

A closed, thin-wall cylinder is made from an aluminum alloy that has a fracture toughness of \(38 \mathrm{MPa}-\mathrm{m}^{0.5}\) and has the following dimensions: length \(=200 \mathrm{~mm}, O D=84 \mathrm{~mm}\), and \(I D=70 \mathrm{~mm}\). A 2.8-mm-deep semicircular crack is discovered on the inner diameter away from the ends, oriented along a line parallel to the cylinder axis. If the cylinder is repeatedly pressurized from 0 to \(75 \mathrm{MPa}\), how many pressure cycles can it withstand? The values of the coefficient and exponent in equation \(6.4\) for this material are \(A=5 \mathrm{x}\) \(10^{-12}(\mathrm{~mm} / \mathrm{cycles}) /\left[\mathrm{MPa}^{n}(\sqrt{\mathrm{m}})^{n}\right]\) and \(n=4\). (Hint: The value of the geometry factor for a semicircular surface flaw is \(\beta=2 / \pi\) and the crack grows in the radial direction.)

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