Introduction to Fatigue: Stress-Life Method, S-N Curve

Introduction to Fatigue in Mechanical Engineering

In this section, the instructor introduces the concept of fatigue in mechanical engineering and its significance for evaluating the suitability of equipment with moving parts.

Fatigue and its Importance for Mechanical Engineers

• Fatigue is a unique topic that is particularly relevant to mechanical engineers.

• Mechanical engineers are responsible for evaluating equipment with moving parts and cyclic processes.

• Examples include internal combustion engines where components like connecting rods experience fluctuating stresses.

• These components undergo cycles of compressive and tensile stress due to the engine's fast-paced operation.

Understanding Fatigue

• Fatigue occurs when parts are subjected to fluctuating stresses over time.

• Microscopic discontinuities within materials can lead to localized yielding under changing stress conditions.

• Yielding absorbs energy but also causes bonds to break around these discontinuities, leading to the formation of voids.

• As more cycles of load occur, these voids can coalesce and eventually result in part failure.

• Notably, parts tend to fail at stress levels much lower than their predicted yielding or ultimate strengths.

• The phenomenon of fatigue can occur even without observable macroscopic plastic deformation.

Challenges in Predicting Fatigue

• Quantifying fatigue remains challenging due to the stochastic nature of microscopic discontinuities within materials.

• Materials exhibit variations in properties and the manifestation of these discontinuities from piece to piece.

• Consequently, it is difficult to predict precisely how fatigue will occur in different scenarios.

Testing and Modeling Fatigue

This section discusses testing methods used to understand fatigue behavior and highlights the challenges associated with developing analytical models for predicting fatigue.

Testing Approaches for Fatigue

• Over time, tests have been developed to quantify fatigue behavior and understand its effects on materials.

• However, there is no perfect analytical model that can accurately predict the number of cycles at a specific stress level leading to failure.

Stochastic Nature of Fatigue

• The stochastic nature of fatigue arises from the microscopic discontinuities within materials.

• These discontinuities make it challenging to determine precise fatigue behavior and develop deterministic models.

• Materials are not completely isotropic, and variations in properties contribute to the complexity of fatigue prediction.

Conclusion

The instructor concludes by emphasizing the need for engineers to design with an understanding of fatigue, despite its complex and unpredictable nature.

Designing with Fatigue in Mind

• Despite the challenges in predicting fatigue, engineers must still consider it during the design process.

• Unlike disciplines such as mathematics, engineering deals with real-world materials that contain internal flaws and discontinuities.

• Understanding how these flaws affect fatigue behavior is crucial for designing reliable and durable mechanical systems.

This summary provides an overview of the transcript's content related to fatigue in mechanical engineering. It highlights key points discussed by the instructor while maintaining a clear and concise structure.

Understanding Failure Mechanisms through Experiments

The transcript discusses experiments designed to understand how failures occur in materials. These experiments involve subjecting a piece of material to a cyclic load, commonly done using a rotating bending test specimen.

Intentional Cyclic Loading Experiment

• Experiments are conducted to intentionally subject a material to a cyclic load.

• A rotating bending test specimen is commonly used for this purpose.

• The specimen consists of two forces applied at a distance, creating flexure and cyclic stress.

Describing Stress Cycles in Rotating Bending Specimens

This section explains the stress cycles experienced by rotating bending specimens and introduces different types of cyclic loading.

Stress Cycles in Rotating Bending Specimens

• Forces applied to the specimen create flexure, causing tension or compression on different parts of the material.

• As the specimen rotates, the stress on each element changes from tension to compression and vice versa.

• This results in fully reversed cyclic loading, where the stress magnitude is the same on both ends of the range but with opposite signs.

Other Types of Cyclic Loading

• Apart from rotating bending tests, other types of cyclic loading can occur.

• Repeated loads involve applying an axial load without reversing it.

• Different theories are used to analyze materials subjected to various types of cyclic loading.

Understanding Rotating Bending Test Results

This section focuses on understanding the outcomes of rotating bending tests using a fundamental chart that plots stress levels and number of cycles.

Fundamental Chart for Steel

• A chart is used to plot stress levels against the number of cycles taken by identical specimens made from steel.

• The shape of this chart remains fairly uniform for different types and strengths of steel.

• The chart helps predict the behavior of materials subjected to cyclic loading.

Unique Characteristic of Steel in Cyclic Loading

This section highlights a unique characteristic of steel in cyclic loading, where stress levels below a certain threshold do not affect the material's failure even with an increasing number of cycles.

Threshold Stress Level for Steel

• Steel exhibits a characteristic where, if the stress level is below a certain threshold, the material can withstand an unlimited number of cycles without failing.

• This phenomenon is observed in steel specimens at stress levels indicated on the chart.

• The threshold stress level for steel is approximately one million cycles.

The transcript does not provide further information beyond this point.

Understanding the Endurance Limit

In this section, we learn about the concept of endurance limit and its significance in designing with steel. The endurance limit is the maximum stress level at which a material can withstand cyclic loading indefinitely without failure.

Estimating the Endurance Limit

• The endurance limit can be estimated through fatigue tests specific to the material being used for design.

• However, for the purposes of solving problems, we will assume a value of one million cycles as the threshold for estimating the endurance limit.

• Designing with steel often involves aiming for infinite life by considering the endurance limit.

• Other materials may require a more finite design approach.

Using Ultimate Strength as a Constraint

• When designing with steel, we typically have access to data on ultimate tensile strength.

• A helpful chart in the textbook allows us to estimate the endurance limit based on the ultimate tensile strength.

• There is a trend observed in various materials where higher strength values tend to result in lower ductility and increased brittleness.

• Ductility plays a crucial role in withstanding cyclic loading and preventing bond failures.

Estimation Formula

• The book uses an estimation formula called "se Prime" (for US units) to calculate the estimated endurance limit.

• se Prime = 0.5 * Ultimate Strength (if Ultimate Strength ≤ 200 ksi)

• se Prime = 100 ksi (if Ultimate Strength > 200 ksi)

Considerations and Limitations

• It's important to note that these estimations are based on highly polished rotating bending specimens tested at room temperature.

• Surface conditions of test specimens significantly impact results, making proper polishing essential for accurate testing.

Impact of Surface Conditions on Test Results

This section discusses how surface conditions of test specimens can affect the results of fatigue tests.

• If the surface of a material is not perfectly polished during testing, it can rapidly alter the test results.

• Imperfections or roughness on the surface can introduce stress concentrations and lead to premature failure.

• Properly polished surfaces are crucial for obtaining reliable and consistent fatigue test results.

What to do with Material that's not Okay

In this section, the speaker mentions that they will address the question of what to do with material that is not okay in a future lecture. However, for now, they focus on discussing the endurance limit for a perfectly polished rotating bending specimen.

Endurance Limit and Log Scales

• The speaker explains that the chart used to represent the endurance limit has two axes on log scales.

• On log-log scales, the relationship between variables does not follow straight-line equations.

• The chart shows breaks at certain points, typically occurring around a thousand cycles and a million cycles.

Modeling the Curve

• The speaker notes that there are some discrepancies in how the breaks in the curve are represented in the figure.

• They clarify that they will use a curve that goes from one point to another instead of precisely aligning it with a specific number of cycles.

• The break point is set at a thousand cycles, which affects how different segments of the line are modeled.

Strain Life Method

• The speaker briefly mentions the strain life method but states it is not extensively used.

• They introduce a formula related to fatigue strength coefficient and number of cycles raised to power B.

• This formula represents the middle portion of the graph known as the high cycle region.

Designing for Parts with Ultimate Strength Data Only

In this section, the speaker discusses designing parts when only ultimate strength data is available. They introduce parameters such as s prime sub F and S sub u T and explain how to calculate F, which represents fatigue strength fraction.

Calculating F - Fatigue Strength Fraction

• F is calculated by taking s prime sub F (fatigue strength coefficient) at a thousand cycles divided by S sub u T (ultimate strength) and expressing it as a fraction.

• The value of F is typically around 80% to 90%.

• The challenge is determining F when only ultimate strength data is known.

Isolating F

• The speaker presents an equation where s prime sub F at a thousand cycles equals the fatigue strength coefficient times 2n raised to power B, with n being the number of cycles.

• By rearranging the equation, they isolate F as the fatigue strength coefficient divided by ultimate strength multiplied by two times a thousand raised to power B.

Determining the Value of B

• To determine the value of B, another piece of information is needed.

• The speaker mentions that this information is available but does not provide further details in this section.

New Section

In this section, the speaker discusses the fatigue strength coefficient and explores how to analyze the curve at a million cycles to determine the endurance number of cycles.

Analyzing the Curve at a Million Cycles

• The value for Sigma F prime represents the fatigue strength coefficient.

• Another expression can be written for S sub F Prime at a million cycles (n sub E), which is the endurance number of cycles.

• This expression is given by Sigma F Prime times n sub E raised to the power of B.

New Section

In this section, the speaker explains how to solve for B in order to find F using known values such as fatigue strength coefficient and endurance limit.

Solving for B

• B can be determined by taking the negative log of the fatigue strength coefficient over the endurance limit divided by the log of two times the number of endurance cycles.

• The values needed to solve for B are defined, with n sub E generally assumed as one million.

• S sub e can be found based on a half relationship, while fatigue strength coefficient can be obtained from an SAE handbook.

• Once B is determined, it can be plugged into an expression to find F as a function of ultimate strength.

New Section

In this section, the speaker discusses plotting results based on ultimate strength and explains how to read and interpret curves for different levels of precision.

Plotting Results

• Since there is only one variable controlling this method, it is possible to plot results based on ultimate strength.

• Depending on desired precision, one can locate their material's ultimate strength on the curve and determine corresponding fraction f.

• The entire curve is generated using previously explained techniques.

New Section

In this section, the speaker addresses the low cycle region and explains how to determine the relationship between fatigue strength and number of cycles.

Low Cycle Region

• In the low cycle region, there is still a relationship between fatigue strength and number of cycles.

• This relationship can be represented by a log-log curve with a straight line.

• To determine the values of a and B in this region, two points are considered: one at one cycle and another at one thousand cycles.

• At one cycle, S sub u T (ultimate tensile strength) is equal to a. At one thousand cycles, F times S sub u T is equal to S sub u T times one thousand to the power of B.

• By canceling out S sub u T, we can solve for B as log base 10 of F over 3.

New Section

In this section, the speaker mentions that the previously identified endurance limit applies to perfectly polished rotating bending specimens at room temperature. They also mention that evaluating a fully corrected endurance limit can change the shape of the curve.

Fully Corrected Endurance Limit

• The previously identified endurance limit applies to perfectly polished rotating bending specimens at room temperature.

• Once a fully corrected endurance limit is evaluated, it can cause changes in the shape of the curve.

Formulas for Fatigue Strength Calculation

In this section, the speaker discusses the formulas used to calculate fatigue strength.

Formulas for Fatigue Strength Calculation

• The fatigue strength formula is of the form s sub F = a n to the power of B.

• The value of 'a' in the formula is given by F times S sub u T squared over S sub e, where S sub e is the fully corrected endurance limit.

• The value of 'B' in the formula is -1/3 log base 10 of F times S sub u T over S sub e, also considering the fully corrected S sub e.

• These formulas are derived based on factors that correct the endurance limit and relate strength to the number of cycles.

Correcting Endurance Limit

This section explains how to correct the endurance limit and introduces a new value for 'a' and 'B'.

Correcting Endurance Limit

• To correct the endurance limit, factors are applied that reduce its value.

• The corrected endurance limit is denoted as S sub e without a prime symbol.

• The new value for 'a' in the fatigue strength formula is obtained by using F times S sub u T squared over S sub e (fully corrected).

• The new value for 'B' in the fatigue strength formula is obtained by using -1/3 log base 10 of F times S sub u T over S sub e (fully corrected).

Example: Evaluating Number of Cycles

An example is provided to illustrate how to evaluate the number of cycles in relation to strength.

Example: Evaluating Number of Cycles

• When evaluating number of cycles in high cycle region, a new value for 'a' and 'B' is required.

• The example involves a shaft mounted in ball bearings, with a flywheel hanging on the end.

• The material of the shaft is polished 1141 steel, quenched and tempered at 1000 degrees Fahrenheit.

• The goal is to find the number of revolutions before the shaft breaks.

Example: Evaluating Number of Cycles (Continued)

The example continues with additional details and objectives.

Example: Evaluating Number of Cycles (Continued)

• In the example, it is assumed that no correction factors for endurance limit are applied yet.

• The flexural stress at the critical location is calculated using MC over I formula.

• The maximum bending moment occurs at the overhang point where weight is multiplied by eight inches.

• By calculating the stress, it is found to be 107.43 ksi.

Example: Evaluating Number of Cycles (Continued)

Further analysis and data about the material used in the example are discussed.

Example: Evaluating Number of Cycles (Continued)

• Data about the material can help determine if the stress value obtained is significant relative to its strength.

• The ultimate strength value for tempered 1141 steel at 1000 degrees Fahrenheit is given as 130 ksi.

• Additionally, the yield strength value for this material is stated as 111 ksi.

Example: Evaluating Number of Cycles (Continued)

Objectives for different parts of the problem are outlined.

Example: Evaluating Number of Cycles (Continued)

• Three parts need to be addressed in this example:

• Part A: Finding how much weight reduction would make it last forever before breaking

• Part B: Ensuring it lasts an infinite number of revolutions without breaking

• Part C: Determining a new diameter to make it last 90,000 revolutions

Example: Evaluating Number of Cycles (Continued)

Calculating the flexural stress at the critical location is discussed.

Example: Evaluating Number of Cycles (Continued)

• The flexural stress at the critical location is calculated using MC over I formula.

• The maximum bending moment occurs at the overhang point where weight is multiplied by eight inches.

• The radius of the shaft is determined by dividing 0.8 inches by 2.

• By substituting values into the formula, a stress value of 107.43 ksi is obtained.

Example: Evaluating Number of Cycles (Continued)

Comparing the calculated stress with material strength values.

Example: Evaluating Number of Cycles (Continued)

• To determine if the calculated stress value is significant relative to material strength:

• Ultimate strength value for tempered 1141 steel at 1000 degrees Fahrenheit is given as 130 ksi.

• Yield strength value for this material is stated as 111 ksi.

• It appears that first cycle yielding is not a concern in this case.

Timestamps are approximate and may vary slightly depending on the source video.

Determining Fatigue Strength Coefficient and Endurance Limit

In this section, the speaker explains the steps to determine the fatigue strength coefficient and the uncorrected endurance limit.

Determining Fatigue Strength Coefficient

• The fatigue strength coefficient is calculated by adding 130 ksi to 50 ksi.

• The value of S sub e Prime (uncorrected endurance limit) is determined by multiplying 0.5 with 130 ksi, resulting in 65 ksi.

Calculating Exponent B

• Exponent B is calculated using the formula: B = -log base 10 of (180 ksi / S sub E Prime) / log base 10 of (2 * a million). If no information is given, assume a million as the basis.

• The calculated value for exponent B is approximately -0.072.

Formula for F

• The formula for F (fatigue strength) is derived using the fatigue strength coefficient and S sub UT.

• F = (180 ksi / S sub UT) * (2 * a thousand)^(-B).

• Corrected an error in notes: Use SWT instead of F in the formula.

• After correcting the error, the value of F is approximately 0.81205.

Break Point Calculation

• The break point from low cycle to high cycle occurs at S sub F Prime at a thousand cycles.

• Calculation: Point eight one two zero five times a hundred and thirty ksi equals approximately 105.6 ksi.

Determining Low Cycle or High Cycle Region

In this section, the speaker explains how to determine whether the stress is in the low cycle or high cycle region.

• The actual stress applied is 107.43 ksi, while the break point stress is 105.6 ksi.

• Since the applied stress is higher than the break point stress, we are in the low cycle region.

• The formulas for low cycle and high cycle regions differ based on stress levels.

Using Formulas for Low Cycle Region

In this section, the speaker discusses using formulas specific to the low cycle region.

• In the low cycle region, we use different formulas than in the high cycle region.

• The strength value S sub F can be calculated using S sub F = 130 ksi * n^B * log(F/3).

• Corrected an error in notes: Use S sub E Prime instead of S sub UT in the formula.

• After correcting the error, solving for F gives a value of not more than 408.4 ksi.

Reducing Weight to Prevent Failure

In this section, the speaker discusses reducing weight to ensure no failure occurs.

• Instead of a fixed weight of 675 pounds, it is now represented as a variable F.

• To prevent failure, we need to ensure that we do not exceed our endurance limit (65 ksi).

• Solving for F should result in a value not exceeding 408.4 ksi.

The transcript is already in English, so no language conversion is required.

Finding the Diameter for Predicted Lifespan

In this section, the speaker discusses the equations needed to find a diameter that would result in a predicted lifespan of 90,000 revolutions with a 675-pound weight.

Equations for Predicting Lifespan

• The equation used is: 675 pounds times 8 inches times D over 2 correct over PI over 64 times D to the fourth equals f zero point eight one two oh five x s sub u T squared over s sub e multiplied by n 90,000 raised to the minus one-third.

• The value of F is multiplied by S sub u T and divided by S sub e.

• The resulting equation can be solved for D.

: The diameter that satisfies these conditions is found to be 0.894 inches.

: To find the diameter required to hold a specific weight, you can plug in the endurance limit and solve for D. This method provides a more accurate result.