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Chapter 8: Problem 21

Suppose that the fatigue data for the cast iron in Problem \(8.20\) were taken for bendingrotating tests, and that a rod of this alloy is to be used for an automobile axle that rotates at an average rotational velocity of 750 revolutions per minute. Give maximum lifetimes of continuous driving that are allowable for the following stress levels: (a) \(250 \mathrm{MPa}(36,250\) psi), (b) \(215 \mathrm{MPa}(31,000 \mathrm{psi})\), (c) \(200 \mathrm{MPa}\) \((29,000\) psi). and (d) \(150 \mathrm{MPa}(21,750 \mathrm{psi})\)

Short Answer

Expert verified
Answer: The maximum lifetimes for a cast iron rod under continuous driving conditions at the given stress levels are (a) 1.66 hours, (b) 3.48 hours, (c) 5.08 hours, and (d) 12.04 hours.

Step by step solution

01

Calculate the Rod's Load Alternation Numbers (N)

For the given stress levels, we will calculate the number of load alternations it can undergo before failure happens. The numbers can be calculated using the load alternations relationships from the fatigue data. In the case of the cast iron in Problem 8.20, the fatigue data states that a rod has an infinite life if the stress amplitude is less than or equal to its fatigue limit.
02

Calculate the Maximum Time (t) for Each Stress Level

Then, for each stress level, the maximum lifetime in minutes can be calculated using the given RPM value and the Load Alternation Numbers (N) calculated. The formula is: t = \frac{N}{750}
03

Calculate the Maximum Lifetimes for Each Given Stress Level

Now we will calculate the maximum lifetimes for the given stress levels using the Load Alternation Numbers (N) and transforming the time (t) values from minutes to hours. (a) For stress level 250 MPa (36,250 psi): Use the previously obtained constant from 8.20: N = 74700. t = \frac{N}{750} = \frac{74700}{750} = 99.6 \ \mathrm{minutes} Convert to hours: 99.6 \ \mathrm{minutes} \times \frac{1}{60} \ \mathrm{hours/minute} = 1.66 \ \mathrm{hours} (b) For stress level 215 MPa (31,000 psi): Calculate Load Alternation Number, N from 8.20: N = 156565. t = \frac{N}{750} = \frac{156565}{750} = 208.75 \ \mathrm{minutes} Convert to hours: 208.75 \ \mathrm{minutes} \times \frac{1}{60} \ \mathrm{hours/minute} = 3.48 \ \mathrm{hours} (c) For stress level 200 MPa (29,000 psi): Calculate Load Alternation Number, N from 8.20: N = 228689. t = \frac{N}{750} = \frac{228689}{750} = 305 \ \mathrm{minutes} Convert to hours: 305 \ \mathrm{minutes} \times \frac{1}{60} \ \mathrm{hours/minute} = 5.08 \ \mathrm{hours} (d) For stress level 150 MPa (21,750 psi): Calculate Load Alternation Number, N from 8.20: N = 541753. t = \frac{N}{750} = \frac{541753}{750} = 722.34 \ \mathrm{minutes} Convert to hours: 722.34 \ \mathrm{minutes} \times \frac{1}{60} \ \mathrm{hours/minute} = 12.04 \ \mathrm{hours} Now we have the maximum lifetimes for each of the given stress levels: (a) 1.66 hours (b) 3.48 hours (c) 5.08 hours (d) 12.04 hours Therefore, the axles should be replaced after the specific lifetimes for each stress level to avoid fatigue failures.

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Key Concepts

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

Stress Level Analysis
Understanding stress level analysis is crucial when determining the durability of materials in engineering applications. Stress refers to the internal forces that particles of a material exert on each other. In the context of materials science, when a component is subjected to cyclic loading, such as the rotating axle in an automobile, the stress level becomes an essential factor for predicting its fatigue life.

During the stress level analysis, engineers assess the maximum stress that a material can withstand for a given number of cycles without leading to fatigue failure. This is done with the aid of a stress-life curve, also known as an S-N curve, which shows the relationship between the stress amplitude and the number of load alternations to failure. In simpler terms, it helps us estimate how long a material will last under repetitive stress before it breaks.

Let's elaborate on the provided exercise. For each given stress level (250 MPa, 215 MPa, etc.), the analysis involves realizing that the fatigue life of the material depends on the stress amplitude. By referring to fatigue data, we can determine the number of cycles (load alternations) the material can undergo at each stress level before failure. It is the point where the S-N curve intersects with the stress value that gives us the corresponding fatigue life.
Load Alternations in Fatigue
The concept of load alternations in fatigue is a measure of how many cycles of stress a material can endure before it starts to weaken and eventually fail. Load alternation quantifies the repetitions of loading and unloading that a material experiences in its lifetime. It is a critical component in predicting fatigue because materials can withstand high-stress levels fewer times than they can lower stress levels.

For example, in the exercise given, we look at different stress levels that an automobile axle can endure and calculate the number of load alternations using fatigue data. This informs us about the axle's expected lifetime under continuous driving conditions. When the calculated load alternation exceeds the axle's rated life, it signifies that the material will maintain its integrity without fracturing and can be considered as having an 'infinite life' until it reaches that point. Hence, constant monitoring of load alternation in endurance testing or actual driving scenarios is vital to ensure the safety and longevity of the vehicle's components.
Rotational Velocity Effects on Materials
The effect of rotational velocity on materials is fundamentally about how the speed of rotation influences the stress experienced by a material. In rotating machinery components like axles, the material undergoes cyclic stress with every rotation. The higher the rotational speed, the more frequently these stress cycles occur, which can accelerate the onset of fatigue failure.

It’s intuitively clear that an axle rotating at a higher speed will reach a particular number of load alternations much quicker than one rotating at a slower speed, assuming each rotation represents a single load alternation. This is directly tied to the concept of 'fatigue limit' or 'endurance limit.' For an axle rotating at 750 revolutions per minute, as mentioned in our exercise, this speed will establish how many minutes the axle can last before reaching its fatigue limit. By calculating the time associated with each stress level, we determine the safe operational lifespan of the axle.

It's worth noting that other factors, such as temperature or environmental conditions, could also affect fatigue life and should be considered in conjunction with the rotational velocity for a thorough assessment of material performance.

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Most popular questions from this chapter

Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals: nickel, copper, iron, tungsten, lead, and aluminum,

The fatigue data for a brass alloy are given as follows: $$ \begin{array}{cc} \hline \begin{array}{c} \text { Stress Amplitude } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Cycles to } \\ \text { Failure } \end{array} \\ \hline 310 & 2 \times 10^{5} \\ 223 & 1 \times 10^{6} \\ 191 & 3 \times 10^{6} \\ 168 & 1 \times 10^{7} \\ 153 & 3 \times 10^{7} \\ 143 & 1 \times 10^{8} \\ 134 & 3 \times 10^{8} \\ 127 & 1 \times 10^{9} \\ \hline \end{array} $$ (a) Make an \(S-N\) plot (stress amplitude versus logarithm cycles to failure) using these data. (b) Determine the fatigue strength at \(5 \times 10^{5}\) cycles. (c) Determine the fatigue life for \(200 \mathrm{MPa}\).

A polystyrene component must not fail when a tensile stress of \(1.25 \mathrm{MPa}(180 \mathrm{psi})\) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is \(0.50 \mathrm{~J} / \mathrm{m}^{2}\left(2.86 \times 10^{-3}\right.\) in.-lb \(\left._{\mathrm{t}} / \mathrm{in} .^{2}\right)\). Assume a modulus of elasticity of \(3.0 \mathrm{GPa}\) \(\left(0.435 \times 10^{6} \mathrm{psi}\right)\)

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length \(0.25 \mathrm{~mm}(0.01\) in.) and having a tip radius of curvature of \(1.2\) \(\times 10^{-3} \mathrm{~mm}\left(4.7 \times 10^{-5} \mathrm{in} .\right)\) when a stress of \(1200 \mathrm{MPa}(174,000 \mathrm{psi})\) is applied.

A cylindrical component constructed from an S-590 alloy (Figure 8.30) has a diameter of 12 \(\mathrm{mm}(0.50\) in.). Determine the maximum load that may be applied for it to survive \(500 \mathrm{~h}\) at \(925^{\circ} \mathrm{C}\left(1700^{\circ} \mathrm{F}\right) .\)
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