Manufacturing Engineering & Technology
7th Edition
ISBN: 9780133128741
Author: Serope Kalpakjian, Steven Schmid
Publisher: Prentice Hall
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Chapter 4, Problem 38SDP
Figure 4.18b shows hardness distributions in end-quench tests, as measured along the length of the round bar. Make a simple qualitative sketch showing the hardness distribution across the diameter of the bar. Would the shape of the curve depend on the bar’s carbon content? Explain.
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Q3 contd.
(d) The yield strength values of pure aluminium (Al) and pure copper (Cu) are 25 MPa and 20
MPa, respectively; whereas the yield strength values of cold rolled Al-Mn-Mg alloy and cast
60-40 Brass (60% Cu, 40% Zn) are 200 MPa and 105 MPa, respectively. With aid of
schematics, explain the main mechanisms account for the increases in the strengths.
(e) A cylindrical tie rod with a diameter of 18.4 mm is subjected to cyclic loading. The stress range
is +/- 200 kN. Figure Q3.3 shows the S-N curve of the material of which the rod is made, how
many cycles will this rod survive?
Stress amplitude
O₂ (MPa)
1500
1400
1300
1200
1100
1000
900
800
700
600
500
400
10²
10³
104
4340 low-alloy steel
Stress ratio = -1
Fig. Q3.3
105
106
Number of cycles to failure, Nf
107
108
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)
part made from AISI 1212 steel undergoes a 20 percent cold-work operation.
(a) Obtain the yield strength and ultimate strength before and after the cold-work operation. Determine
the percent increase in each strength.
(b) Determine the ratios of ultimate strength to yield strength before and after the cold work operation.
What does the result indicate about the change of ductility of the part?
Chapter 4 Solutions
Manufacturing Engineering & Technology
Ch. 4 - Describe the difference between a solute and a...Ch. 4 - What is a solid solution?Ch. 4 - Prob. 3RQCh. 4 - Describe the difference between a single-phase and...Ch. 4 - What is an induction heater? What kind of part...Ch. 4 - Describe the major features of a phase diagram.Ch. 4 - What do the terms equilibrium and constitutional,...Ch. 4 - Prob. 8RQCh. 4 - What is tempering? Why is it performed?Ch. 4 - Explain what is meant by severity of quenching.
Ch. 4 - What are precipitates? Why are they significant in...Ch. 4 - Prob. 12RQCh. 4 - Prob. 13RQCh. 4 - Prob. 14RQCh. 4 - Prob. 15RQCh. 4 - You may have seen some technical literature on...Ch. 4 - Prob. 17QLPCh. 4 - What is the difference between hardness and...Ch. 4 - Prob. 19QLPCh. 4 - Prob. 20QLPCh. 4 - Prob. 21QLPCh. 4 - Describe the characteristics of (a) an alloy, (b)...Ch. 4 - Explain why carbon, among all elements, is so...Ch. 4 - Prob. 24QLPCh. 4 - In Section 4.8.2, several fluids are listed in...Ch. 4 - Why is it important to know the characteristics of...Ch. 4 - Explain why, in the abscissa of Fig. 4.16c, the...Ch. 4 - Prob. 28QLPCh. 4 - Prob. 29QLPCh. 4 - Prob. 30QLPCh. 4 - Design a heat-treating cycle for carbon steel,...Ch. 4 - Using Fig. 4.4, estimate the following quantities...Ch. 4 - Prob. 33QTPCh. 4 - Prob. 34QTPCh. 4 - Prob. 35SDPCh. 4 - Figure 4.18b shows hardness distributions in...Ch. 4 - Throughout this chapter, you have seen specific...Ch. 4 - Refer to Fig. 4.24, and think of a variety of...Ch. 4 - Inspect various parts in your car or home, and...
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- The 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_forwardAnswer this please importantarrow_forwardExplain the reason for the quenching cracking that may occur when hardening some parts. Describe the alternative methods applied to prevent this situation by showing them on a single figure.arrow_forward
- A sheet of a 70% Cu-30% Zn alloy is cold-rolled 20% to a thickness of 3.0mm. The sheet is then further coldrolledto 2.0 mm. What is the percentage cold work?arrow_forwardPlease explain ths graph from Materials Science and EngineeringAn Introductionarrow_forwardOrdinary sheets of borosilicate glass are tested in bending, and are found to fracture at an average stress of 72 MPa. After thermal tempering, the stress at failure increases by 90%. What are the sign and magnitude of the stress induced in the glass by the tempering operation?arrow_forward
- How does the carbon content in plane carbon steels affect the fracture toughness? Show this influence by A7 curves.arrow_forwardThe figure shows two work pieces of steel of 0.15 %C and 1.2 %C with different dimensions. Compare between them regarding:- 1- The existed phases and the expected mechanical properties of each. 2- Give 5 important applications for each alloy. 3- The stages of Normalizing treatments for each one. 900 800 1 Aj = 727°C 0.0218 0.77 700 Dimensions of (1.5*1.5*1.5) inch Dimensions of (1*1*1) inch 0.2 0.4 0.6 0.8 1.0 1.2 1.4 6.67 Weight percent carbon Temperature ("C)arrow_forwardQ1: 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 50arrow_forward
- An uncold-worked brass specimen of average grain size 0.006 mm has a yield strength of 178 MPa. Estimate the yield strength of this alloy (in MPa) after it has been heated to 600°C for 1000 s, if it is known that the value of ky is 11 MPa-mm1/2. The Animated Figure 7.25 may be helpful. Attached is the question and the figure referenced.arrow_forward(i) and (ii) are both separate questions. You must provide suitable material for borh cases (i) and (ii)arrow_forwardThe average grain diameter and yield strength for a brass material were measured as a function of time at 650°C. Given the following yield strengths for the two specimens, compute the heat treatment time required at 650°C to give a yield strength of 100 MPa. Assume a value of 2 for n, the grain diameter exponent. Time (min) 30 90 Yield Strength (MPa) 90 75 min Grain Diameter (mm) 3.9 x 10-2 6.6 x 10-2arrow_forward
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How to make metal stronger by heat treating, alloying and strain hardening; Author: Billy Wu;https://www.youtube.com/watch?v=7lM-Y4XndsE;License: Standard Youtube License