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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 \(S_{u t}=1 \mathrm{GPa}\).

Short Answer

Expert verified
The factor of safety, \(n\), for infinite fatigue life of the steel shaft is calculated based on the stress amplitudes, mean stress and using the Goodman Diagram. Substituting all the values in the respective equations provides the safety factor value.

Step by step solution

01

Determine the Equivalent Alternating Stress Amplitude

The first step is to calculate the equivalent alternating stress from bending (\( \sigma \)) and torsion (\( \tau \)). The equation to use is given by the von Mises stress hypothesis as: \(\sigma' = \sqrt{{3}\cdot( {\sigma_{a}}^2 )}\), where \(\sigma_{a}\) is the alternating stress amplitude. Substituting given values, we will have \(\sigma' = \sqrt{{3}\cdot( {100})^2 }\), which will provide the equivalent stress value.
02

Calculate the Principal Mean Stress

Next, as both stresses are 90 degrees out-of-phase, both mean stresses are equal. Therefore, the principal mean stress is equal to the stress amplitude. This then simplifies to \( \sigma_{m} = \sigma_{a} = 100 \mathrm{MPa}\).
03

Find the Axial Fatigue Strength Coefficient and Exponent

For steel as per given in the problem, we can use the standard material properties values for \(S_f'\) and \(b\). That is, \(S_f' = 0.9 \cdot S_{u t}\) and \(b = -0.087\). Substituting the given \(S_{u t} = 1 GPa\), we get \(S_f' = 0.9 \cdot 1 = 0.9 GPa\).
04

Determine the Fatigue Strength

By referring to the Goodman diagram, one can understand that the safe values of the stress amplitudes determine the fatigue strength. We compute this fatigue strength for infinite life for steel as \( \sigma_{f i} = \frac{{S_{f'}}}{{(1 + \sigma_{m}/S_{u t})^{-1/b}}}\). Plugging in our values we get the fatigue strength.
05

Calculate the Factor of Safety

Finally, find the safety factor, \(n\), for infinite life from the equation \( n = \frac{{\sigma_{f i}}}{{\sigma'}}\). Using the values of \(\sigma_{f i}\) and \(\sigma'\) calculated in the previous steps, this equation will give the safety factor.

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Key Concepts

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

Von Mises Stress
Understanding von Mises stress is vital in assessing the yield or failure of materials under complex loading conditions. Named after Richard von Mises, this is a theoretical method used to predict when a material will yield due to any combination of stress. For instance, the steel shaft in our exercise experiences both bending and torsional stresses.

The von Mises stress is calculated using the formula \( \sigma' = \sqrt{{3}\cdot( {\sigma_{a}}^2 )} \), where \( \sigma_{a} \) is the alternating stress amplitude from bending (in this exercise, 100 MPa). Essentially, this calculation takes multi-axial stresses—stresses from more than one direction—and translates them into an equivalent single-axis stress that would produce the same effect on the material. This single value can then be compared against the material’s yield stress to determine whether it is within safe limits.

In the solution provided, the von Mises stress is a crucial first step in calculating the safety factor for the shaft's infinite life.
Goodman Diagram
The Goodman diagram is an engineering tool used to visualize the relationship between mean stress and alternating stress in fatigue analysis. By plotting these two types of stress on a graph, we can predict the lifespan of a material under repetitive loading. This diagram incorporates the ultimate strength of the material (\( S_{ut} \)) and the material's performance in fatigue, often characterized by the material's endurance limit.

In the context of our exercise, the Goodman diagram helps to establish the fatigue strength of the steel shaft under the given conditions. Once the equivalent alternating stress and the mean stress have been calculated, you can use the Goodman diagram to determine a safe operating point. The fatigue strength for infinite life, \( \sigma_{f i} \), is a crucial value that tells us what level of alternating stress the material can withstand indefinitely without failure, considering the additional effect of mean stress.
Factor of Safety
The factor of safety (often abbreviated as FOS or NS) is a measure of how much stronger a system is than it needs to be for an intended load. Safety factors are applied in engineering to provide a margin for uncertainties in design and material properties. For the steel shaft undergoing fatigue analysis, the safety factor tells us how many times stronger the shaft is compared to the minimum requirement needed to avoid failure.

To calculate the safety factor for infinite life, we follow the final step in the given solution where \( n = \frac{{\sigma_{f i}}}{{\sigma'}} \). We are essentially comparing the fatigue strength of the material under the given stress conditions to the von Mises stress to ensure that the shaft will last indefinitely under the specified dynamic loads. A higher factor of safety implies a greater assurance against unexpected failures but might also indicate an overdesigned and potentially costlier system.
Fatigue Strength
Finally, let's talk about fatigue strength. This is the stress level at which a material will fail after a specified number of cycles. For materials like steel, which don't have a clearly defined endurance limit, the fatigue strength is typically defined for a large number of cycles, such as 1 million or 1 billion, to represent 'infinite life'.

In our example, we consider infinite life because the shaft is expected to operate for a very long time without failure. The fatigue strength, \( \sigma_{f i} \), is calculated using the material's ultimate tensile strength, \( S_{ut} \), material constants, and the mean stress, \( \sigma_{m} \). The calculation provided in the solution properly considers these variables to ensure the steel shaft will withstand the iterative loads without experiencing fatigue failure. Incidentally, this number goes hand-in-hand with the safety factor, as it is essentially the denominator in our safety factor calculation.

Fatigue strength is a critical design consideration for all engineering components subjected to cyclic loading, ensuring they meet the safety and longevity expected in their application.

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Most popular questions from this chapter

Determine the size of the clevis pin shown in Figure P6-8 needed to withstand an applied repeated force of 0 to \(130000 \mathrm{lb}\) for infinite life. Also determine the required outside radius of the clevis end to not fail in either tearout or bearing if the clevis flanges each are \(2.5\) in thick. Use a safety factor of \(3 .\) Assume \(S_{u t}=140 \mathrm{kpsi}\) for the pin and \(S_{u t}=80 \mathrm{kpsi}\) for the clevis.

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.

A cylindrical tank with hemispherical ends is to be fabricated from cold- rolled steel that has \(S_{u t}=550 \mathrm{MPa}\). The tank outside diameter is to be \(250 \mathrm{~mm}\). The pressure will fluctuate from 0 to \(15 \mathrm{MPa}\). For an infinite-life fatigue safety factor of 2 with \(99 \%\) reliability, determine the minimum wall thickness that may be used.

A component in the shape of a large sheet is to be fabricated from SAE 4340 steel, which has a fracture toughness \(K_{c}=98.9 \mathrm{MPa}-\mathrm{m}^{0.5}\). The sheets are inspected for crack flaws after fabrication, but the inspection device cannot detect flaws smaller than \(5 \mathrm{~mm}\). Determine the minimum thickness required for the sheet to have a minimum cycle life of \(10^{6}\) cycles (using fracture-mechanics criteria) if its width is \(400 \mathrm{~mm}\) and the load normal to the crack varies from 20 to \(170 \mathrm{kN}\). The values of the coefficient and exponent in equation \(6.4\) for this material are \(A=4 \times 10^{-9}(\mathrm{~mm} / \mathrm{cycles}) /\left[\mathrm{MPa}^{n}(\sqrt{\mathrm{m}})^{n}\right]\) and \(n=3\).

A machine part is made from 2024-T351 aluminum plate. For each cycle of the machine the part undergoes a tensile stress that varies from 0 to \(1 / 2\) of the material's tensile yield strength. The machine cycles 20 times per minute, 24 hours per day, 365 days per year. At the last inspection of the machine the part was found to have a central crack 1.7 mm long. The machine is inspected every six months. Estimate the expected crack length at next inspection. The values of the coefficient and exponent in equation \(6.4\) for this material are \(A=4.8 \times 10-11(\mathrm{~mm} /\) cycle \()\) and \(n=3.5\).

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