Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 9.13, Problem 83AAP
An austenitized 40-mm-diameter 4340 steel bar is quenched in agitated water. Plot the Rockwell C hardness of the bar versus distance from one surface of the bar to the other across the diameter of the bar at the following points: surface,
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Q1: Austenitized 40 mm diameter 5140 alloy steel bar is quenched in agitated oil. Predict what is the
Rockwell hardness of this bar will be at (a) its surface and (b) its center (c) What do you think about
the difference in hardness number between the center and surface (d) Differentiate between hardness
and hardenability (e) Rank the steels in the figure below from lowest to highest hardenability and
explain why.
600-
Bar diameter (mm)
100
80
60
40
20
0
300
0
Cooling rate at 700°C (°C/sec).
-150
55
0
تنا
25
-------
5
S
10
12.5 8
M-R
L
1/2
34-R
Agitated oil
15
20
¼
¾
Distance from quenched end. De
(Jominy distance)
5,5
54
Car
Bar diameter (in.)
0
25 mm.
1 in.
Hardness (Rockwell C)
Where (C = center, S = surface, M-R = mid-radius)
2828 292
65
60-
55-
50
45
40
35
30
25
20
15
10
0
J
10
5140
30
20
Distance from quenched end (mm)
4340
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32
Distance from quenched end (sixteenths of an inch)
40
9840
4140
8640
50
Question-7. Steady-state creep data taken for an alloy at a stress level of 160 MPa are given below.
The stress exponent n for the alloy is 6.8. R is 8.3145 J/mol-K. Compute the steady-state creep
rate at 1000 °C and a stress level of 68 MPa.
Es (h-1)
6.8 × 10-5
T (°C)
800
8.6 x 10-3
900
A 2.0-inch diameter 4140 steel bar is austenized and tempered in oil without
Agitation, determine the resulting hardness at 0.6 inches below the surface, present all the
Procedure to arrive at the answer (include sketch of the piece and graph used with steps
Using word forms)
Chapter 9 Solutions
Foundations of Materials Science and Engineering
Ch. 9.13 - (a) How is raw pig iron extracted from iron oxide...Ch. 9.13 - (a) Why is the FeFe3C phase diagram a metastable...Ch. 9.13 - (a) What is the structure of pearlite? (b) Draw a...Ch. 9.13 - Distinguish between the following three types of...Ch. 9.13 - Prob. 5KCPCh. 9.13 - (a) Define an FeC martensite. (b) Describe the...Ch. 9.13 - (a) What is an isothermal transformation in the...Ch. 9.13 - How does the isothermal transformation diagram for...Ch. 9.13 - Draw a continuous-cooling transformation diagram...Ch. 9.13 - (a) Describe the full-annealing heat treatment for...
Ch. 9.13 - Describe the process-annealing heat treatment for...Ch. 9.13 - What is the normalizing heat treatment for steel...Ch. 9.13 - Describe the tempering process for a plain-carbon...Ch. 9.13 - (a) Describe the martempering (marquenching)...Ch. 9.13 - (a) Describe the austempering process for a...Ch. 9.13 - (a) Explain the numbering system used by the AISI...Ch. 9.13 - (a) What arc some of the limitations of...Ch. 9.13 - (a) What compounds docs aluminum form in steels?...Ch. 9.13 - Prob. 19KCPCh. 9.13 - (a) Define the hardenability of a steel. (b)...Ch. 9.13 - Prob. 21KCPCh. 9.13 - Prob. 22KCPCh. 9.13 - Prob. 23KCPCh. 9.13 - What is the difference between a coherent...Ch. 9.13 - Prob. 25KCPCh. 9.13 - Prob. 26KCPCh. 9.13 - Prob. 27KCPCh. 9.13 - (a) Describe the three principal casting processes...Ch. 9.13 - Prob. 29KCPCh. 9.13 - Prob. 30KCPCh. 9.13 - Prob. 31KCPCh. 9.13 - Prob. 32KCPCh. 9.13 - Prob. 33KCPCh. 9.13 - Prob. 34KCPCh. 9.13 - Prob. 35KCPCh. 9.13 - (a) What are the cast irons? (b) What is their...Ch. 9.13 - Prob. 37KCPCh. 9.13 - Prob. 38KCPCh. 9.13 - Prob. 39KCPCh. 9.13 - Prob. 40KCPCh. 9.13 - Prob. 41KCPCh. 9.13 - Prob. 42KCPCh. 9.13 - Prob. 43KCPCh. 9.13 - Prob. 44KCPCh. 9.13 - Prob. 45KCPCh. 9.13 - (a) Why arc titanium and its alloys of special...Ch. 9.13 - Prob. 47KCPCh. 9.13 - Prob. 48KCPCh. 9.13 - Prob. 49KCPCh. 9.13 - Prob. 50KCPCh. 9.13 - Prob. 51KCPCh. 9.13 - Prob. 52KCPCh. 9.13 - Describe the structural changes that take place...Ch. 9.13 - Describe the structural changes that take place...Ch. 9.13 - If a thin sample of a eutectoid plain-carbon steel...Ch. 9.13 - If a thin sample of a eutectoid plain-carbon steel...Ch. 9.13 - (a) What types of microstructures arc produced by...Ch. 9.13 - A 0.65 % C hypoeutectoid plain-carbon steel is...Ch. 9.13 - A 0.25% C hypoeutectoid plain-carbon steel is...Ch. 9.13 - A plain-carbon steel contains 93 wt % ferrite7 wt%...Ch. 9.13 - A plain-carbon steel contains 45 wt% proeutectoid...Ch. 9.13 - A plain-carbon steel contains 5.9 wt%...Ch. 9.13 - A 0.90% C hypereutectoid plain-carbon steel is...Ch. 9.13 - A 1.10% C hypereutectoid plain-carbon steel is...Ch. 9.13 - If a hypereutectoid plain-carbon steel contains...Ch. 9.13 - A hypereutectoid plain-carbon steel contains 10.7...Ch. 9.13 - A plain-carbon steel contains 20.0 wt%...Ch. 9.13 - A 0.55% C hypoeutectoid plain-carbon steel is...Ch. 9.13 - A hypoeutectoid steel contains 44.0 wt% eutectoid...Ch. 9.13 - A hypoeutectoid steel contains 24.0 wt% eutectoid...Ch. 9.13 - A 1.10 % C hypereutectoid plain-carbon steel is...Ch. 9.13 - Draw timetemperature cooling paths for a 1080...Ch. 9.13 - Draw timetemperature cooling paths for a 1080...Ch. 9.13 - Thin pieces of 0.3-mm-thick hot-rolled strips of...Ch. 9.13 - Prob. 75AAPCh. 9.13 - Prob. 76AAPCh. 9.13 - Prob. 77AAPCh. 9.13 - Prob. 78AAPCh. 9.13 - Prob. 79AAPCh. 9.13 - Prob. 80AAPCh. 9.13 - Prob. 81AAPCh. 9.13 - Prob. 82AAPCh. 9.13 - An austenitized 40-mm-diameter 4340 steel bar is...Ch. 9.13 - Prob. 84AAPCh. 9.13 - Prob. 85AAPCh. 9.13 - Prob. 86AAPCh. 9.13 - Prob. 87AAPCh. 9.13 - Prob. 88AAPCh. 9.13 - Prob. 89AAPCh. 9.13 - Prob. 90AAPCh. 9.13 - Prob. 91AAPCh. 9.13 - Prob. 92AAPCh. 9.13 - (a) For a plain-carbon steel with 1 wt % carbon...Ch. 9.13 - Prob. 94SEPCh. 9.13 - Prob. 95SEPCh. 9.13 - Prob. 96SEPCh. 9.13 - Prob. 97SEPCh. 9.13 - Prob. 98SEPCh. 9.13 - Prob. 99SEPCh. 9.13 - Prob. 100SEPCh. 9.13 - Prob. 101SEPCh. 9.13 - Prob. 102SEPCh. 9.13 - Prob. 103SEPCh. 9.13 - Both 4140 and 4340 steel alloys may be tempered to...Ch. 9.13 - Prob. 105SEPCh. 9.13 - Aircraft fuselage is made of aluminum alloys 2024...Ch. 9.13 - Prob. 107SEPCh. 9.13 - Prob. 108SEPCh. 9.13 - Prob. 109SEPCh. 9.13 - (a) What makes austenitic stainless steels that...Ch. 9.13 - Prob. 111SEPCh. 9.13 - Prob. 112SEPCh. 9.13 - Prob. 113SEPCh. 9.13 - (a) Give examples of components or products that...Ch. 9.13 - Prob. 115SEPCh. 9.13 - Prob. 116SEPCh. 9.13 - Prob. 117SEP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- Determine the tensile yield strength (0.2% offset) and the maximum strength of a metal alloy having the following tensile stress-strain diagram. Select one: The tensile yield strength Sy = 100 Mpa and the maximum strength Smax = 250 Mpa. The tensile yield strength Sy = 170 Mpa and the maximum strength Smax = 250 Mpa. The tensile yield strength Sy = 150 Mpa and the maximum strength Smax = 200 Mpa. The tensile yield strength Sy = 240 Mpa and the maximum strength Smax = 250 Mpa. The tensile yield strength Sy = 80 Mpa and the maximum strength Smax = 250 Mpa.arrow_forwardQ1: 1.5 in diameter steel part is quenched in oil with no agitation. (H=0.20). If we use a 4340; 4140; 8640; 5140; 1040 steel, what would be the hardness number at the center of the part?arrow_forwardO The following engineering stress-strain data were obtained for a 0.2% C plain-carbon steel. (i) Plot the engineering stress-strain curve. (ii) Determine the ultimate tensile strength of the alloy. (iii) Determine the percent elongation at fracture. Engineering Engineering Engineering Engineering Stress Strain Stress Strain (ksi) (in./in.) (ksi) (in./in.) 76 0.08 30 0.001 75 0.10 55 0.002 73 0.12 60 0.005 69 0.14 68 0.010 65 0.16 72 0.020 56 0.18 74 0.040 51 0.19 75 0.060 (Fracture)arrow_forward
- A shear pin made from a 4340 steel (0.4% C) fails by ductile fracture. It was supposed to be initially normalised, austenitised, quenched and tempered to 100% martensite to give a hardness of 5.5 GPa. The hardness was measured to be 4.5 GPa. Which of the following summaries most accurately accounts for what may have gone wrong? O Insufficiently heating the steel would result in partial austenitisation, leading to only partial martensite after quenching and a reduction in final hardness. The temper could still occur at the correct conditions and still lead to an insufficient strength and encouraging ductile failure. O Performing a full austenitisation but then cooling slower than the critical cooling rate for the alloy would allow some ferrite/pearlite to form after the martensite reaction has finished. This would be softer than the tempered martensite and so reduce the final hardness. O Assuming a full austenitisation and then cooling faster than the critical cooling rate, the hardness…arrow_forward/ steel has zero mean stree of 320 Mpa ,mean stress of 200 Mpa,tensile stress of 740 Mpa,yield stress of 1.5 tensile stress 1- Calculate the fatique limit by using the three equations of fatique behavior 2-sketch the fatique behaviour for eachs 3- which of the three eaquatons is more suitable for calculations fatique of glasses materials and whyarrow_forward(ii) The turbine blade root is found to contain a surface fatigue crack of 3.3 mm during a routine maintenance inspection. The major stop-start load cycle is experienced twice per day with a minimum stress of 34 MPa and a maximum stress of 195 MPa. The turbine blade has a KIC of 75 MPa√m. Fatigue crack growth rate data for this alloy is given by: the Paris law constant, A = 4.5 x 10-11 and the Paris law exponent, m = 3.8. How many more stop-starts would you recommend be used? Explain your reasoning. You can assume the shape factor Q = 1.2, K is in MPaNm and a (crack length) is in marrow_forward
- A low-nickel steel in the heat-treated condition had an 'engineering' tensile strength of 708 N/mm2. The reduction in area of cross-section at the fracture was 44%. What was the true tensile strength of the steel?arrow_forward4. Assuming the following proportional relationship for steel alloys: TS = 3.45 x HB, where TS is in MPa. a. What is the Brinell hardness number (HB) for 1020 Steel that has been annealed at 785 "C? [Check Appendix B.4 for needed information] b. What diameter (d) indentation would you expect for a 10-mm-diameter indenter under 1000-kg load?arrow_forwardThe lower yield point for a certain plain carbon steelbar is found to be 135 MPa, while a second bar of the samecomposition yields at 260 MPa. Metallographic analysisshows that the average grain diameter is 50μm in the firstbar and 8μm in the second bar.a. Predict the grain diameter needed to cause a loweryield point of 205 MPa.b. If the steel could be fabricated to form a stablegrain structure of 500 nm grains, what strengthwould be predicted?c. Why might you expect the upper yield point to bemore alike in the first two bars than the lower yieldpoint?arrow_forward
- A mild steel material is subjected to a Brinell hardness test with an applied force of 7459 N using a hardened steel ball indentor of 11 mm and it is found that the Brinell Hardness Number as 954. Determine the surface area of indentation and indentation diameter. (NOTE: Please Solve the problem in a paper and upload in the separate submission link provided and also fill the answers without the unit in the box below) i) Surface Area of Indentation (in mm?) -- ( ii) Indentation Diameter (in mm)arrow_forwardFor a steel alloy, specify the amount of deformation (%CW) that is necessary to give a minimum Brinell hardness of 225 and at the same time provide a ductility of at least 12%EL. You may need to use Animated Figure 7.19b and Animated Figured 7.19c. There are 2 answers.arrow_forwardA. What are annealing purposes? B. Calculate the upper and lower critical temperature for steel composition as follow: %C %Si %Mn %P %S %Cr %Mo %Fe 0.15 1.00 0.60 0.025 0.025 9 1.10 balancearrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Introduction to Diffusion in Solids; Author: Engineering and Design Solutions;https://www.youtube.com/watch?v=K_1QmKJvNjc;License: Standard youtube license