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Unformatted text preview: SOLUTION (3.1D)
Known: Deﬁnitions of the terms stress strength, yield strength, ultimate strength, elastic limit, proportional limit, modulus of elasticity, and yield point are given in
Section 3.2. Find: Write deﬁnitions of the above terms  see Section 3.2. Analysis: 1.
2.
3. The stress is the load divided by the cross—sectional area.
The strenth is the maximum value of stress a material will carry before failure. The yield strength, Sy, is the value of stress at which signiﬁcant plastic yielding
ﬁrst occurs. The ultimate strength is the maximum value of stress a material will carry
before fracture for nondynamic loading. The elastic limit is the highest stress the material can withstand and still return
exactly to its original length when unloaded. The proportional limit is the stress at which the stress—strain curve ﬁrst deviates
(ever so slightly) from a straight line. Below the proportional limit, Hooke's law applies. The modulus of elasticity (Young's modulus) E, is the constant of
proportionality between stress and strain (which is the slope of the curve between the origin and the proportional limit). The yield point of a material is a point for a material where appreciable yielding
occurs suddenly at a clearly deﬁned value of stress; for example, in soft steel.
In other materials the onset of appreciable yielding occurs gradually, and the
yield strength for these materials is determined by using the "offset method."
This is illustrated in Fig. 3.1; it shows a line, offset an arbitrary amount of 0.2
percent of strain, drawn parallel to the straightline portion of the original
stress—strain diagram. Point B is the yield mint of the material at 0.2 percent
offset. If the load is removed after yielding to point B, the specimen exhibits a
0.2 percent permanent elongation. Yield strength corresponding to a speciﬁed
(very small) offset is a stande laboratory determination, whereas elastic limit and proportional limit are not. 31 SOLUTION (3 .2D) Known: The materials to be selected have E greater than 207 GPa and Su greater than
1378 MPa. Find: Identify ﬁve materials with (a) modulus of elasticity greater than steel, and (b)
ultimate strength greater than 200 ksi. Analysis:
(a) From Appendix Cl, we attempt to select the following materials with higher E
values than steel, but inspection of the materials listed in the appendix of the textbook, reveals no materials with E values higher than that of steel. I
(b) From the textbook appendices, we select the following non—steel materials with Su
values greater than 1378 MPa. I Material Leaded beryllium copper 1379 MPa (max)
C 17300
(A n ndix C—13) Duranickel 301 1448 MPa (max)
CD aged bar
(A  ndix C15) Rene 95 1620 MPa
Su  rallo Comment: Inspection of the materials listed in the appendix of the textbook reveals
that steel is a relatively strong and stiff material. 32 SOLUTION (32D) —— alternate
Known: The materials to be selected have E greater than 207 GPa and Su greater than 1378 NIPa. Find: Identify ﬁve materials with (a) modulus of elasticity greater than steel, and (b)
ultimate strength greater than 200 ksi. Analysis:
(a) From www.matweb.com we select the following materials with higher E values than steel: I Modulus of Elastici , ksi, ; eater than 30 x 106 E (ksi) 30000 ksi
Be Ilium, Be 40020 ksi 35%" “1
Rec stallized Annealed Annealed Annealed Annealed Annealed Annealed Annealed 58000 ksi (b) We select the following non—steel materials with ultimate tensile strength Su
greater than 1378 MPa (200,000 psi). I Tensile Stren th, Ultimate, si, ; eater than 200 ksi AISI Grade 18Ni (300) Maraging 293625 psi
Steel Aged, round bar Tested lon itudinal, 75 mm Carpenter AerMet®forTooling
Tool Steel
Double A ed 468°C
BioDurTM 316LS Stainless Medical 223445 psi
90% Cold Worked
3 3 VascoMax® C—300 Specialty Steel 285070 psi
Heat Treatment: 927°C (1700°F) + A e
W25 Re Tungsten Rhenium Alloy 580000 psi
Deformed AISI A6, Type Tool Steel 345100 psi
Austenitized 830870°C (1525
1600°F) “900° F“
Temered at 500°C Titanium Ti15Mo52r 246500 psi
ST 730°C, Aed 400°C
AISI Type W2 Tool Steel 261000 psi
Water quenched at 775°C
(1425°F), and temered
AISI Type SS Tool Steel
Austenitized 855870°C (1575— 1600°F) Oil uenched to 55 HRC
Mo47.5 Re Molybdenum
Rhenium Alloy Deformed 290000?“
Cold—Drawn 30450” mi
Deformed 29986" “i
Hardened Technetium, Tc 218950 psi
As—Rolled
Pt20% Ni Alloy 250125 psi
Hard
79Pt15Rh—6Ru 300150 psi
A110 851
Pt8% W Alloy 300150 psi
Hard 323350 psi 464000 psi Comment: Matweb is an award winning web site. 34 SOLUTION (3.3)
Known: The critical location of a part made from a known steel is cold worked during fabrication.
Find: Estimate Su, Sy and the ductility. Schematic and Given Data: Assumption: After cold working the stressstrain curve for the critical location starts
at point G. Analysis:
1. At point G in Fig. 3.2, the part has been permanently stretched to 1.1 times its
initial length. Hence, its area is 1/ 1.1 times its original area A0. 2. On the basis of the new area, the yield strength is Sy = 62(1.l) = 68.2 ksi. I
3. The ultimate strength is Su = 66(1.l) = 72.6 ksi. I
4. At fracture, R increases to 2.5 on the graph.
5. R=2.5/l.l =2.27 
6. Using Eq. (3.3) and Eq. (3.2)
— _ L: _ ; —_
Ar —1 R l 227 0.56 I
e=Rl=2.27l=l.27or127% I 3—5 SOLUTION (3.4) Known: The critical location of a part made from a known steel is cold worked
during fabrication. Find: Estimate Su, Sy and the ductility. Schematic and Given Data: Hot Rolled 1020 Steel Assumption: After cold working the stress—strain curve for the critical location starts
at point I. Analysis:
1. The area ratio at J is R = Ao/Af = 1.2. The initial area is thus 1/1.2.
2. The yield strength is Sy = 65(1.2) = 78 ksi. I
3. The ultimate strength is Su = 66(1.2) = 79.2 ksi.
4. At fracture, R increases to 2.5 on the graph.
5. R = 2.5/1.2 = 2.08
6. Using Eq. (3.3) and Eq. (3.2)
_ 1 _ 1 __ — _ =  =
Ar 1 R 1 2.08 0.519 8 R 1 2.08 1.08 or 108% I 3—6 SOLUTION (3.5)
Known: A tensile specimen of a known material is loaded to the ultimate stress, then
unloaded and reloaded to the ultimate stress point. Find: Estimate the values of 0', 8, 6T, ET for the ﬁrst loading and the reloading. Schematic and Given Data: Hot Rolled 1020 Steel Assumption: After unloading the stressstrain curve starts at point H for the new
specimen. Analysis:
1. For the initial sample, 0' = 66 ksi, 8 = 30%. I
2. For Figure 3.2, R = 1.3 at point H.
3. From Eq. (3.4), or: (SR = (66)(l.3) = 85.8 ksi. I
4. From Eq. (3.5), 8T=ln(1 + e) = In 1.30 = 0.26 = 26%. I
5. For the new specimen; 0' = 66(l.3) = 85.8 ksi. I
6. The new specimen behaves elastically, so a = 6/13 = 85.8/30,000 = .00286. I
7. Within the elastic range, CT = G and 8T 2 8. Therefore CT = 85.8 ksi and 81‘ = 0.29%. I Comment: Note also that 81‘ = ln(l + e) = ln(1.0029) = 0.29%. 3—7 SOLUTION (3.6)
Known: A tensile specimen of a known material is loaded to a known stress, then unloaded and reloaded to the same stress point. Find: Estimate the values of o, 8, 6T, ET for the ﬁrst loading and the reloading. Schematic and Given Data: Hot Rolled 1020 Steel Assumption: After unloading, the stress—strain curve starts at point G for the new
specimen. Analysis:
1. For the initial sample, 6 = 62 ksi, e = 10%. I
2 For Figure 3.2, R = 1.1 at point G.
3. From Eq. (3.4), O'T= OR = (62)(1.1) = 68.2 ksi. I
4. From Eq. (3.5), 8T: ln(l + e) = In 1.10 = 0.095 = 9.5%. I
5 For the new specimen; 6 = 62(1.1) = 68.2 ksi. I
6 The new specimen behaves elastically, so 8 = o/E = 68.2/30,000 = .00227. I
7 Within the elastic range, O'T z 6 and ET z 8. Therefore CT = 68.2 ksi and ET = 0.23%. I Comment: Note also that 81‘ = ln(l + 8) = ln(1.0023) = 0.23%. SOLUTION (3.7) Known: A tensile specimen of a known material is loaded to a known stress, then
unloaded and reloaded to the same stress point. Find: Estimate the values of 0', 8, GT, ET for the ﬁrst loading and the reloading. Schematic and Given Data: Hot Rolled 1020 Steel Assumption: After unloading the stress—strain curve starts at point J for the new
specimen. Analysis:
1. For the initial sample, (5 = 65 ksi, e = 20%. I
2 For Figure 3.2, R = 1.2 at point J.
3. From Eq. (3.4), CT = OR = (65)(1.2) = 78.0 ksi. I
4. From Eq. (3.5), 8T: ln(l + s)=ln1.20 = 0.18 = 18%. I
5 For the new specimen; 0' = 65( 1.2) = 78 ksi. I
6 The new specimen behaves elastically, so 8 = 6/13 = 78/30,000 = .0026. I
7 Within the elastic range, O'T z 6 and ET z 8. Therefore 61‘ = 78 ksi and 81‘ = 0.26%. I Comment: Note also that 81‘ = ln(l + e) = ln(1.0026) = 0.26%. 39 SOLUTION (3.8D)
Known: A steel is to be selected from Appendix C4a. Find: Estimate Su and Sy from the given value of Brinell hardness for the steel
selected. Schematic and Given Data: Decision: Select ANSI 1020 annealed. Assumptions: 1. The experimentally determined relationship of ultimate strength to hardness is
sufﬁciently accurate. 2. The experimentally developed relationship of yield strength to ultimate strength is
sufﬁciently accurate for our purposes. Analysis: 1. Appendix C4a shows that ANSI 1020 annealed steel has Su = 57.3, Sy = 42.8,
and Bhn =111. 2. Using Eq. (3.11), we can estimate Su from the Brinell hardness using:
Su 2 KBHB where KB z 500 for most steels. Su = 500(111) = 55,500 psi I
3. Sy can be estimated by using Eq. (3.12). Sy = 1.05 Su  30,000 psi = 1.05(55,500)  30,000 = 28,275 psi. I
Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stressrelieved
(not coldworked) steels. Note that the estimated value of Su from the Brinell
hardness of 55.5 ksi is close to the value given in Appendix C4a of Su z 57.3
ksi. 2. Experimental data would be helpful to reﬁne the above equations for speciﬁc
steels. 310 SOLUTION (3.9D)
Known: A steel is to be selected from Appendix C4a. Find: Estimate Su and Sy from the given values of Brinell hardness for the selected
steel. Schematic and Given Data: Decision: Select ANSI 1040 annealed steel. Assumptions:
1. The experimentally determined relationship of ultimate strength to hardness is sufﬁciently accurate.
2. The experimentally developed relationship of yield strength to ultimate strength is sufﬁciently accurate for our purposes. Analysis:
1. Appendix C—4a shows that ANSI 1040 annealed steel has Su = 75.3 ksi, Sy = 51.3 ksi, and Bhn = 149.
2. Using Eq. (3.11), we can estimate Su from the Brinell hardness value using: Su = KBHB where KB —~— 500 for most steels. SUB = 500(149) = 74,500 psi
3. Sy can be estimated by using Eq. (3.12). SyB = 1.05 Su  30,000 psi = 1.05(74,500)  30,000 = 48,225 psi.
4. Ratio of strength (Appendix C—4a values to Brinell hardness based values):
3“ — 753 — 1.011 s“B ‘ 74.5 Sy _ 51.3 _ 8y; 8.2 _ 1.064
Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stress—relieved (not coldworked) steels.
2. The Brinell hardness based strength values are slightly less than the Appendix C 4a strength values. 3—11 SOLUTION (3.10D)
Known: A steel is to be selected from Appendix C4a. Find: Estimate Su and Sy from the given values of Brinell hardness for the selected
steel. Schematic and Given Data: A181
1030
Steel
asrolled normalized
annealed Decision: Select ANSI 1030 steel. Assumptions: 1 . The experimentally determined relationship of ultimate strength to hardness is
sufﬁciently accurate. 2. The experimentally developed relationship of yield strength to ultimate strength is
sufﬁciently accurate for our purposes. Analysis: 1 . Appendix C4a shows that for ANSI 1030 the strengths are as follows: (1) as—
rolled Su = 80.0 psi, Sy = 50.0 psi, Bhn = 179; (2) normalized Su = 75.5 psi, Sy =
32.0 psi, Bhn = 149; (3) annealed Su = 67.3 psi, Sy = 31.2 psi, Bhn = 126. 2. Using Eq. (3.11), we can estimate 8“. Su = KBHB
where KB z 500 for most steels. Sy can be estimated by using Eq. (3.12).
In asrolled condition:
Su = 500(179) = 89,500 psi
Sy = 1.05 811 — 30,000 psi 2 1.05(89,500)  30,000 = 63,975 psi.
5. In normalized condition:
11 = 500(149) = 74,500 psi. Sy = 1.05811 — 30,000 psi = 48,225 psi.
6. In annealed condition: Su = 500(126) = 63,000 psi. Sy = 1.058“ — 30,000 psi = 36,150 psi.
7. Ratio of strength (Appendix C—4a values to Brinell hardness based values): J>LM 312 asrolled: Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stress—relieved
(not cold—worked) steels. 2. Experimental data would be necessary to reﬁne the above equations. 313 SOLUTION (3.11) Known: An A181 4340 steel part is heat treated to 217 Bhn. A second A181 4340
steel part is heat treated to 363 Bhn. Find: Estimate values of Sn and Sy for both parts. Schematic and Given Data: Part 1, 217 Bhn Part 2, 363 Bhn Suz? Sy=? Su=? Sy=? Assumptions: 1. The experimentally determined relationship of ultimate strength to hardness is
sufﬁciently accurate. 2. The experimentally developed relationship of yield strength to ultimate strength is
sufﬁciently accurate for our purposes. Analysis:
1. Using Eq. (3.11), we can estimate Su.
Su = KBHB where KB 2 500 for most steels. For Part 1, Su 2 500(217) = 108,500 psi; for Part 2, Su = 500(363) = 181,500 psi I
2. Sy can be estimated by using Eq. (3.12). Sy = 1.05 Su — 30,000 psi. For Part 1, Sy = 1.05(108,500) — 30,000 = 83,925 psi. For Part 2, Sy = 1.05(181,500)  30,000 = 160,575 psi. I Comments:
1. Equation (3.12) is a good estimate of the tensile yield strength of stress—relieved
(not cold—worked) steels. 2. Experimental data would be helpful to reﬁne the above equations for speciﬁc
steels. 3~14 SOLUTION (3 . 1 2)
Known: An A181 1020 steel part is heat treated to 111 Bhn. Find: Estimate Sn and Sy. Schematic and Given Data: Assumptions: 1. The experimentally determined relationship of ultimate strength to hardness is
sufﬁciently accurate. 2. The experimentally developed relationship of yield strength to ultimate strength is
sufﬁciently accurate for our purposes. Analysis:
1. Using Eq. (3.11), we can estimate Su. Su 2 KBHB where K}; = 500 for most steels. u = 500(111) = 55,500 psi I 2. Sy can be estimated by using Eq. (3.12). Sy = 1.05 Su  30,000 psi = 1.05(55,500)  30,000 = 28,275 psi. I
Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stress—relieved
(not cold—worked) steels. 2. Experimental data would be helpful to reﬁne the above equations for speciﬁc
steels. 315 SOLUTION (3.13)
Known: An AISI 3140 steel part is heat treated to 210 Bhn. Find: Estimate Su and Sy. Schematic and Given Data: A181
3140
Steel Heat
Treated Assumptions:
1. The experimentally determined relationship of ultimate strength to hardness is
sufﬁciently accurate. 2. The experimentally developed relationship of yield strength to ultimate strength is
sufﬁciently accurate for our purposes. Analysis:
1. Using Eq. (3.11), we can estimate Su. Su = KBHB where KB z 500 for most steels. Su = 500(210) = 105,000 psi I
2. Sy can be estimated by using Eq. (3.12). Sy = 105 Sn — 30,000 psi = 1.05(105,000)  30,000 = 80,250 psi. I
Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stressrelieved
(not cold—worked) steels. 2. Experimental data would be helpful to reﬁne the above equations for speciﬁc
steels. 316 SOLUTION (3.14)
Known: The hardness versus distance curve for a Jominy endquench test of AISI 4340 steel is given. Find: Sketch similar curves for lowalloy steel and for plain carbon steel. Schematic and Given Data: . 4340 . Low Alloy . Plain Carbon 58 Rc 58 Rc 58 Rc / 4340 /_ Low Alloy O\\l
CO
——l——1 kl]
O Hardness (Rockwell C)
U.) 4:
O O \ Plain Carbon rd no
000 10 20 30 40 Distance From Quenched End (mm) Assumption: We assume that the hardness for the non heat treated end of the J ominy
test specimen properties are similar to 4340 steel. Analysis: From Appendix C—4a, for 1040 Normalized, HB = 170 and for 4340
Normalized, H}; = 363. We would expect at the end of the Jominy bar that the
hardness of the 1040 would be approximately 50% of the 4340 hardness. Comment: The J orniny curves show the effectiveness of the alloying elements in
imparting hardenability to steel. 317 SOLUTION (3.15D)
Known: Four given applications require steel. Find: Choose between (a) 0.1% C and 0.4% C, and between (b) plain carbon and
alloy steel. Schematic and Given Data: (a) Machine Frame (b) Round Rod 0: (c) Irregular Shaped Part (d) Rail CaI Wheel Eﬂaﬂr Assumption: Material is of usual quality. Analysis: (a) Machine frame. All steel materials provide approximately the same rigidity
(30 X 106 psi). In this application, since stresses are low and we also want the
cost to be low, we would select plain carbon because of low price and 0.1% C
because of low strength. (b) Small round rod. For high bending and torsional stresses we select 0.4% C and
to keep cost down we select plain carbon steel. (c) Large irregular shaped part. For high stress we select 0.4% C and for ease in heat
treating we select alloy steel. (d) Rail car wheel. For low cost we select carbon steel and for low interior strength
we choose 0.1% C. If we carburized the 0.1% C steel we can obtain a high
strength wear surface. Comment: In summary, (1) alloy steel is more expensive but it is more easily heat treated than plain carbon steel, and (2) 0.4% C steel has greater strength than 0.1% C
steel. 318 SOLUTION (3.16D)
Known: The site http://www.matweb.com provides the properties of aluminum 7075—0. Find: Compare the material properties for aluminum 707 50 from the web to those
given in Appendix C10 of the textbook. Analysis: 1 . From www.matweb.c0m we obtain for aluminum 70750: Key Words: Aluminum 70750; UNS A97075; ISO AlZn5.5MgCu(A);
Aluminium 70750; AA7075—O; PHYSICAL PROPERTIES Density, g/cc 2.81
Hardness, Brinell 60 500 kg load with 10 mm ball Hardness, Knoop 80 Converted from Brinell Hardness Value
Hardness, Vickers 68 Converted from Brinell Hardness Value MECHANICAL PROPERTIES Tensile Strength, Ultimate, MPa 220, 31,900 psi Tensile Strength, Yield, MPa 95, 13,775 psi Elongation %; break 17, In 5 cm; Sample 1.6 mm thick 17 % Modulus of Elasticity, GPa 72, Average of Tension and Compression. In
Aluminum alloys, the compressive modulus is typically 2% greater than the
tensile modulus, 10,440 ksi Poissons Ratio 0.33
Shear Modulus, GPa 26.9, 3,901 ksi
Shear Strength, MPa 150, 21,750 psi THERMAL PROPERTIES CTE, linear 20°C, um/mOC, 23.6, 20100°C, 13 pin/in~°F CTE, linear 250°C, um/m—°C, 25.2, Average over the range 20300°C, 14 pin/in—
°F Heat Capacity, J/g°C, 0.96, 0.23 BTU/lb°F Thermal Conductivity, W/m—K, 173, 1,201 BTUin/hrft2°F Melting Point, 0C, 477, Solidus, 891 °F Solidus, °C, 477, 891 °F Liquidus, 0C, 635, 1,175 °F ELECTRICAL PROPERTIES
Electrical Resistivity, Ohm—cm 0.0000038 I
From Appendix C—10 of the textbook for 7075—0 aluminum we ﬁnd: Aluminum 70750 Brinell Hardness 60 Tensile Strength, Ultimate, 38 ksi, MPa 230
Tensile Strength, Yield, 15 ksi, MPa 105 Elongation in 2 in %; 16 I 3—19 Comments: 1. The web wins again, providing a vast amount of information. 2. The tensile strength and the yield strength values from Appendix 010 are higher
than those from www.matweb.com. SOLUTION (3.17D)
Known: A designers favorite materials are 1020 steel, 1040 steel, 4340 steel, 2024
T4 aluminum, nylon 6/6, and acetal. Find: Compare the speciﬁc material properties for each material. Assumptions: The material properties presented in Appendix C are sufﬁcient for
comparison purposes. Decision: The material properties compared will be tensile strength, yield strength,
elongation, and hardness. Analysis: 1. From Appendix C4a, we obtain properties for normalized steel in 1 inch round
sections. 2. From Appendix C 10, we obtain properties for wrought aluminum alloy for 1/2
inch sizes. ' 3. From Appendix C18a, we obtain mechanical properties of plastics. MATERIAL Tensile Strength Yield Strength Elongation (%) Brinell
(W3) (MP 3) Hardness 1020 steel 441.3 346.5 35.8 131
1040 steel 589.5 374.0 28.0 170
4340 steel 1279.0 861.8 12.2 363
2024T4 470 325 19 120 4. Inspection of the table provides a comparison of the properties of the six
materials. Comments: The site http://www.matweb.com will provide additional material
property information. 3—20 ...
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This note was uploaded on 03/27/2009 for the course MECHENG EN.530.215 taught by Professor Wang during the Spring '09 term at Johns Hopkins.
 Spring '09
 Wang

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