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Unformatted text preview: SOLUTION (3.213) Known: We are to determine the modulus of elasticity, ultimate tensile strength,
elongation at break, and density for certain carbon and alloy steels. Find: Search the materials property database at http'.£ﬁuﬂ}x,matweb.c0m and list the
(1) modulus of elasticity, E, (2) ultimate tensile strength, S". (3) elongation at break in
%, and (4) density in gI‘cc, for the following: (a) teels: [010 cold drawn,
1020 cold rolled, 1040 as roiled, 1050 as rolled, 1080 as rolled, and 1116 cold drawn;
and (b) Alloy steels: 4140 annealed, 4340 annealed, and 4620 annealed. Analysis: (a) From www.matweb.com we find for W: AISI 1010, cold drawn Density 7.87 gfcc Tensile Strength, Ultimate 365 MPa 52900 psi
Modulus of Elasticity 205 GPa 29700 ksi
Elongation at Break 20 % AISI 1020 Steel, cold rolled Density 7.87 g/cc Tensile Strength, Ultimate 420 MPa 60900 psi
Modulus of Elasticity 205 GPa 29700 ksi
Elongation at Break 15 % AISI 1040 Steel, as rolled Density 7.845 gz’cc Tensile Strength, Ultimate 620 MP3 89900 psi
Modulus of Elasticity 200 GPa 29000 ksi
Elongation at Break 25 % A151 1050 Steel, as rolled Density 7.85 gfcc Tensile Strength, Ultimate 725 MPa 105000 psi
Modulus of Elasticity 205 GPa 29700 ksi
Elongation at Break 20 % A151 1080 Steel, as rolled Density 7.85 g/cc Tensile Strength, Ultimate 965 MPa 140000 psi
Modulus of Elasticity 205 GPa 29700 ksi
Elongation at Break 12 % AISI 1116 Steel, cold drawn Density 7.85 glee Tensile Strength, Ultimate 530 MP3 76900 psi 
Modulus of Elasticity 205 GPa 29700 ksi
Elongation at Break 17 % (b) From www.matweb.eom we find for Aﬂoyggeels: AISI 4140 Steel, annealed at 815°C (1500°F) furnace cooled 11°C (20°F)lhour to
665°C (1230°F), air cooled, 25 mm (1 in.) round Density 7.85 glee Tensile Strength. Ultimate 655 MPa 95000 psi Elongation at Break 25.7 % Modulus of Elasticity 205 GPa 297’00 ksi AISI 4340 Steel, annealed, 25 mm round
Density 7.85 glee Tensile Strength, Ultimate 745 Mle 108000 psi
Elongation at Break 22 % Modulus of Elasticity 205 GPa 29700 ksi AISI 4620 Steel, annealed, 25 mm round
Density 7.85 glee Tensile Strength, Ultimate 510 MPa 74000 psi
Elongation at Break 31.3 % Modulus of Elasticity 205 GPa 29ml) ksi Comment: Matweb is an award winning web site. SOLUTION (3.3D)
Known: We are to determine the modulus of elasticity, ultimate tensile strength.
elongation at break, and density for certain carbon and alloy steels. Find: Search the materials property database at http',lfwww.matweb.com and list the
(1) modulus of elasticity, E. (2) ultimate tensile strength, S", (3) elongation at break in
%, and (4) density in gfcc, for the following: (a) Cast iron: ASTM class 20 and class
35; (0} W: 3003H12, 3003H18, 5052—H32, 50521138, 50520, 6061—
T4, 6061T9l. and 7075—0. Analysis: (a) From www.matwebaeom we ﬁnd for Castings Standard gray iron test bars, as cast, ASTM class 20 Density 7.15 gfcc
Tensile Strength, Ultimate 152 MPa 22000 psi Modulus of Elasticity 66  97 GPa 9570  14100 ksi Standard gray iron test bars, as cast, ASTM class 35 Density 7.15 gfcc
Tensile Strength, Ultimate 252 MPa 36500 psi Modulus of Elasticity 100  119 GPa 14500 — 17300 ksi (b) From www.matwebxom we find for Aluminum alloys: Aluminum 3003H12 Density 2.73 gfcc
Ultimate'l'ensile Strength 131 MPa 19000 psi Elongation at Break 10 % AA; Typical; 1(16 in. (1.6 mm) Thickness
Elongation at Break 20 % AA; Typical; 112 in. (12.7 mm) Diameter
Modulus of Elasticity 68.9 GPa 10000 ksi Aluminum 30031118 Density 2.?3 glcc Ultimate Tensile Strength 200 MPa 29000 psi Elongation at Break 10 % AA; Typical; 1K2 in. (12.7 mm) Diameter
Elongation at Break 4 % AA; Typical; 1/16 in. (1.6 mm) Thickness
Modulus of Elasticity 68.9 GPa Aluminum 5052H32 Density 2.68 g/cc Ultimate Tensile Strength 228 MP3 33000 psi Elongation at Break 12 % AA;Typical; 1! 16 in. (1.6 mm) Thickness
Elongation at Break 18 % AA; Typical; 1f2 in. (12.7 mm) Diameter Modulus of Elasticity 70.3 GPa Aluminum 5052H38 Density 2.68 glcc Ultimate Tensile Strength 290 MP3 42000 psi Elongation at Break 7 % AA; Typical; 1116 in. (1.6 mm) Thickness
Elongation at Break 8 % AA; Typical; 1/2 in. (12.7 mm) Diameter
Modulus of Elasticity 70.3 GPa Aluminum 50520 Density 2.68 glcc Ultimate Tensile Strength 193 MP3 28000 psi Elongation at Break 25 % AA; Typical; 1:16 in. (1.6 mm) Thickness
Elongation at Break 30 % AA; Typical; 1l2 in. (12.7 mm) Diameter
Modulus of Elasticity 70.3 GPa Aluminum 6061T4; 6061T451 Density 2.? glee Ultimate Tensile Stren th 24] MP3 35000 psi Elongation at Break 2 % AA; Typical; 1:16 in. (1.6 mm) Thickness Elongation at Break 25 % AA; Typical; 1:2 in. (12.7 mm) Diameter
Modulus of Elasticity 68.9 GPa Aluminum 6061T91
Density 2.7 gi’cc Tensile Strength, Ultimate 405 MPa 58700 psi
Elongation at Break 12 %
Modulus of Elasticity 69 GPa Aluminum 70750 Density 2.31 g/cc Ultimate Tensile Strength 228 MPa 33000 psi Elongation at Break 16 % AA; Typical; 1:2 in. (12.7 mm) Diameter
Elongation at Break 17 % AA; Typical; 1fl6 in. (1.6 mm) Thickness Modulus of Elasticity 71.7 GPa Comment: Matweb is an award winning web site. SOLUTION (3.?) Known: The critical location of a part made from 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
p01 nt 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 U 1.1 times its original area An. 0n the basis of the new area, the yield strength is Sy = 620.1) = 63.2 ksi. I
The ultimate strength is Su = 660. I} = 72.6 ksi. I At fracture, R increases to 2.5 On the graph.
R =2.5I1.1= 2.27 Using Eq. (3.3) and Eq. (3.2) PiﬁerP — __.1_= _——1 =
Ar‘l R l 2.27 0'56 I s=R—l=2.271=1.27or127% I SOLUTION {3.9)
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 o, 8, OT. 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, a = 66 ksi, E = 30%. I
2. For Figure 3.2, R = 1.3 at point H.
3. From Eq. (3.4), GT = UR = (66)(1_3) = 85.8 ksi. I
4. From Eq. (3.5), 812an + e) = in 1.30 = 0.26 = 26%. I
S. For the new specimen; 0 = 660.3) = 85.8 ksi. I
6. The new specimen behaves elastically, so 5 = GEE = 85.880300 = .00286. I
7. Within the elastic range, or  o and ﬂu 6. Therefore CT = 85.8 ksi and s1— = 0.29%. I Comment: Note also that 81‘ = ln(1 + e) = ln(l .0029) = 0.29%. SOLUTION (3.130)
Known: A steel is to be selected from Appendix (343. Find: Estimate Su and S); 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 C—4a shows that ANSI 1020 annealed steel has Su a 52.3, S}; = 42.8, and Bhn =111.
2. Using Eq. (3.11), we can estimate Su from the Brinell hardness using: Su = KBHB where KB as 500 for most steels. Su = 500011) = 55,500 psi I
3. S): can be estimated by using Eq. (3.12). 55: = 1.05 Su  30,000 psi = 1.0565500)  30,000 = 28,275 psi. I Comments: 1. Equation (3.12;) is a good estimate of the tensile yield strength of Stresswrelieved
(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 C—4a of Su 2 57.3 ksi. 2. Experimental data would be helpful to refine the above equations for Specific steels. SOLUTION (3.15D)
Known: A steel is to be selected from Appendix C4a. Find: Estimate Su and S), from the given values of Brinell hardness for the selected
Steel. Schematic and Given Data: AISI
1030
Steel
as—rolled 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, 33; = 50.0 psi, Bhn = 179; (2) normalized Su 2 75.5 psi, 85; =
32.0 psi, Bhn = 149; (3) annealed Sll = 67.3 psi, S), = 31.2 psi, Bhn = 126. 2. Using Eq. (3.11), we can estimate Su.
Su = KBHB
where KB =5 500 for most steels. 3. Sy can be estimated by using Eq. (3.12].
4. In asrolled condition: Su : 500(179) = 89,500 psi
Sy = 1.05 Su  30,000 psi = l.05(89,500)  30,000 = 63,975 psi. 5. In normalized condition:
Sn = 500(149) = 74,500 psi.
S3,, = 1.058”  30,000 psi = 48,225 psi. 6. In annealed condition:
Su = 500(126) = 63,000 psi.
8., = 1.058u — 30,000 psi = 36.150 psi. 7. Ratio of strength (Appendix C—4a values to Brinell hardness based values): i 80.0 _ 894 YB '
312 =.862 Su; 63.0 ‘sy 36.2 B i—ﬁlﬁ= 1.068 Comments: 1. Equation (3.12) is a good estimate of the tensile yield strength of stressrelieved
(not coldworked) steels. 2. Experimental data would be necessary to refine the above equations. ...
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 Spring '08
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