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Chapter 6: Problem 57

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 \(1550 \mathrm{MPa}\).

Short Answer

Expert verified
The corrected fatigue strength of the steel rod with 99.9% reliability is approximately 560.7 MPa.

Step by step solution

01

Determine size factor

- According to standards, for a rectangular section with \(h = 60\, mm\) and \(b = 40\, mm\), the size factor \(k_a\) is approximately 0.85. This can be found in relevant tables provided in most strength of materials textbooks.
02

Calculate surface finish factor

- Since the rod is ground, the surface finish factor \(k_b\) is found to be 0.87. This value is determined by referring to the relevant tables for surface finish factors.
03

Determine load factor

- Since the rod is subjected to repeated bending, the load factor \(k_c\) should be found to be 1. This is based on the load factor table in which bending loads correspond to a load factor of 1.
04

Calculate reliability factor

- The reliability factor \(k_d\) corresponding to 99.9% reliability is found to be 0.897. This value is usually obtained by referring to reliability factors tables.
05

Determine temperature factor

- Since the temperature is below \(450^{\circ}\mathrm{C}\), the temperature factor \(k_e\) is found to be 1.
06

Calculate fatigue strength

- The corrected fatigue strength \(S_e'\) is determined by the product of the ultimate tensile strength \(S_u\) and the factors determined in steps 1 through 6. Hence, \(S_e' = k_ak_bk_ck_dk_e(\frac{S_u}{\sqrt{2}}) = 0.85 \times 0.87 \times 1 \times 0.897 \times 1 \times \frac{1550}{\sqrt{2}} \approx 560.7\, MPa\)

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Key Concepts

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

Fatigue Strength
Fatigue strength is a crucial concept in mechanical engineering. When materials are subject to repeated stress cycles, their ability to withstand those stresses without failing is termed fatigue strength. In essence, it reflects the endurance limit of a material under cyclical loading conditions. This is especially important for components like rods, shafts, and beams that undergo repetitive loading during their operational life. Fatigue strength is not a fixed property and can be influenced by several factors, such as material composition, surface quality, and environmental conditions. Engineers often rely on empirical data from fatigue tests to predict the life expectancy of components. These predictions help in designing components that are not only efficient but also safe and reliable. To ensure accurate fatigue strength predictions, factors like the size, surface finish, load type, and reliability are carefully evaluated and incorporated in calculations.
Size Factor
The size factor is an adjustment made to the fatigue strength to account for the influence of a component's dimensions. Larger components tend to have a lower fatigue strength compared to smaller ones due to the presence of more material flaws and stress concentrations. In the problem provided, the size factor for a rod with a cross-section of 60 mm by 40 mm has been determined to be 0.85. This factor is typically derived from standard charts or tables that are available in engineering handbooks. To summarize, adjusting for size is critical as it ensures that the fatigue strength calculations reflect the actual behavior of larger components, helping to prevent premature failure.
Surface Finish Factor
Surface finish factor is a parameter that adjusts the fatigue strength to account for the effect of surface irregularities on a material's fatigue performance. A smoother surface typically means fewer stress concentrators, which can significantly enhance the component's ability to withstand cyclical stress. In the given problem, the rod is ground, which results in a surface finish factor of 0.87. This value, like the size factor, is obtained from standardized tables that relate different surface finishes to fatigue strength reductions. Surface finish is essential in fatigue analysis because even minute surface variations can cause concentrated stresses, leading to the initiation and propagation of cracks over time. Therefore, understanding and applying the correct surface finish factor ensures more accurate and reliable fatigue strength predictions.
Reliability Factor
The reliability factor considers the probability that a component will perform as expected or survive without failure for a certain period under specified conditions. The higher the reliability percentage, the greater the reduction applied to the fatigue strength to ensure safety. In the exercise, the reliability factor for 99.9% reliability is 0.897. This value means that there is a very high assurance level that the component will not fail under repeated stress cycles. Using the correct reliability factor is crucial as it provides an additional layer of safety. Engineers use this to adjust their design, ensuring components meet practical performance expectations and minimize the risk of unexpected failures in critical applications.

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Most popular questions from this chapter

Figure P6-19 shows a machined pivot pin that is a press-fit into part \(A\) and is a slip fit in part \(B\). If \(F=100 \mathrm{lb}, l=2\) in, and \(d=0.5\) in, what is the pin's safety factor against fatigue when made of SAE 1020 cold- rolled steel? The loading is fully reversed and a reliability of \(90 \%\) is desired. There is a bending stress concentration factor \(K_{t}=1.8\) at the section where the pin leaves part \(A\) on the right-hand side.

A paper machine processes rolls of paper having a density of \(984 \mathrm{~kg} / \mathrm{m}^{3}\). The paper roll is \(1.50\)-m outside dia \((O D) \times 0.22-\mathrm{m}\) inside dia \((I D) \times 3.23 \mathrm{~m}\) long and is on a simply supported, hollow, steel shaft with \(S_{u t}=400 \mathrm{MPa}\). Find the shaft \(I D\) needed to obtain a dynamic safety factor of 2 for a 10 -year life running three 8-hour shifts per day if the shaft \(O D\) is \(22 \mathrm{~cm}\) and the roll turns at \(50 \mathrm{rpm}\).

The paper rolls in Figure P6-17 are 0.9-m- \(O D \times 0.22-\mathrm{m}-I D \times 3.23 \mathrm{~m}\) long and have a density of \(984 \mathrm{~kg} / \mathrm{m}^{3}\). The rolls are transferred from the machine conveyor (not shown) to the forklift truck by the V-linkage of the off-load station, which is rotated through \(90^{\circ}\) by an air cylinder. The paper then rolls onto the waiting forks of the truck. The forks are 38 -mm-thick by 100 -mm-wide by \(1.2 \mathrm{~m}\) long and are tipped at a \(3^{\circ}\) angle from the horizontal and have \(S_{u t}=600 \mathrm{MPa}\). Find the infinite-life fatigue safety factor for the two forks on the truck when the paper rolls onto it under two different conditions (state all assumptions): (a) The two forks are unsupported at their free end. (b) The two forks are contacting the table at point \(A\).

A machine part is subjected to fluctuating, simple, multiaxial stresses. The fully corrected nonzero stress ranges are: \(\sigma_{x \min }=50 \mathrm{MPa}, \sigma_{x \max }=200 \mathrm{MPa}, \sigma_{y \min }=80\) \(\mathrm{MPa}, \sigma_{y \max }=320 \mathrm{MPa}, \tau_{x z \min }=120 \mathrm{MPa}, \tau_{x y \max }=480 \mathrm{MPa}\). The material properties are: \(S_{e}=525 \mathrm{MPa}\) and \(S_{u t}=1200 \mathrm{MPa}\). Using a Case 3 load line, calculate and compare the infinite-life safety factors given by the Sines and von Mises methods.

A \(42-\mathrm{mm}\)-dia steel shaft with a \(19-\mathrm{mm}\) 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 \(S_{u t}=1 \mathrm{GPa}\).
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