Briefly explain the difference between fatigue striations and beachmarks in terms of both (a) size and (b) origin.

Short Answer

Expert verified
Answer: Fatigue striations are microscopic markings formed due to gradual growth of fatigue cracks caused by cyclic loading or stress, with a spacing ranging from several nanometers to micrometers. Beachmarks are macroscopic markings that form due to interruptions during cyclic stress, such as changes in loading conditions or stress levels, and have a spacing ranging from micrometers to several millimeters or more.

Step by step solution

01

(1. Define Fatigue Striations)

(Fatigue striations are microscopic lines or markings found on the fracture surface of materials that have experienced cyclic loading or stress. These lines represent progressive crack growth intervals and are parallel to the direction of crack propagation.)
02

(2. Define Beachmarks)

(Beachmarks, also known as clamshell marks or arrest lines, are macroscopic markings found on the fracture surface of a fatigue-cracked material. They are semi-circular or semi-elliptical in shape and indicate the successive positions of the crack front resulting from changing load conditions or interruptions in the cyclic loading.)
03

(3. Difference in Size)

(Fatigue striations are microscopic and are much smaller in size compared to beachmarks, which are visible to the naked eye or can be easily observed using a low-magnification tool like a hand lens. Fatigue striations typically have a spacing ranging from several nanometers to micrometers, while the spacing between beachmarks can range from micrometers to several millimeters or more.)
04

(4. Difference in Origin)

(The origin of fatigue striations is due to the gradual growth of fatigue cracks caused by cyclic loading or stress on the material. Each striation represents the incremental advance of the crack front. Beachmarks, on the other hand, form due to interruptions that occur during cyclic stress, such as changes in loading conditions or stress levels, and environmental variations. The arrest lines in beachmarks correspond to the points where the crack growth was temporarily halted or slowed down due to these interruptions.)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(3.0 \mathrm{GPa}\) \(\left(0.435 \times 10^{6} \mathrm{psi}\right)\)

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of \(55 \mathrm{MPa} \sqrt{\mathrm{m}}(50 \mathrm{ksi} \sqrt{\mathrm{in} .})\). If, during service use, the plate is exposed to a tensile stress of \(200 \mathrm{MPa}(29,000 \mathrm{psi})\), determine the minimum length of a surface crack that will lead to fracture. Assume a value of \(1.0\) for \(Y\).

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 cylindrical 1045 steel bar (Figure 8.34) is subjected to repeated tension- compression stress cycling along its axis. If the load amplitude is \(22,000 \mathrm{~N}\left(4950 \mathrm{lb}_{\mathrm{f}}\right)\), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of \(2.0\).

A specimen of a 4340 steel alloy having a plane strain fracture toughness of \(45 \mathrm{MPa} \sqrt{\mathrm{m}}\) (41 ksi \(\sqrt{\text { in. }})\) is exposed to a stress of 1000 MPa (145,000 psi). Will this specimen experience fracture if it is known that the largest surface crack is \(0.75 \mathrm{~mm}(0.03\) in.) long? Why or why not? Assume that the parameter \(Y\) has a value of \(1.0\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free