2

pdf

School

University of Washington *

*We aren’t endorsed by this school

Course

501

Subject

Mechanical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by BailiffDragonfly2463

ME501-15 HW#2 Solutions Chapt 2 18.) An alloy of 95% B and 5% A, cooling from 3400 ° C to Room Temperature a) Superheat temperature = 3400 ° C - 2300 ° C = 1100 ° C b) Freezing range = Range liquidus and solidus line = 2300 ° C – 1500 ° C = 800 ° C c) Reactions as it cools from 3400 ° C 2200 ° C L 1 = L 2 + h 1500 ° C L = d + h 800 ° C e = d + h d) What phases are present at 1000 ° C = d and h e) % composition of each phase at 1000 ° C d = 58% h = 98% f) Section 3.9 1) Plot the engineering stress-strain curve using the data in Table 3.1 and compare the results to Fig. 3.1

From Table 3.1 Eng. Stress (MPa) Eng. Strain (m/m) 0 0 38.5 0.00016 76.9 0.00036 115 0.00054 154 0.00070 192 0.00090 230 0.00104 238 - 250 0.0102 269 0.0304 288 0.0406 327 0.0610 369 0.0914 385 0.132 392 0.254 369 0.314 350 0.376 307 0.448 Curves look similar although data from Table 3.1 are sparse and produce a rough approximation of Figure 3.1 Plotted data from Table 3.1 Figure 3.1 2) Plot the true stress-strain curve using the data in Table 3.1 and compare the results to Fig. 3.2 From Table 3.1 True Stress (MPa) True Strain (m/m) 0 0 38.5 0.00016 76.9 0.00036 115 0.00054

154 0.00070 192 0.00090 230 0.00104 - - 253 0.0101 277 0.0299 300 0.0398 347 0.0592 403 0.0875 436 0.124 490 0.226 565 0.273 771 0.319 1110 0.370 Curves look similar although data from Table 3.1 are sparse and produce a rough approximation of Figure 3.2 Plotted data from Table 3.1 Figure 3.2 3) Using the elongation data from Table 3.1 for the forces of 35, 42.5 and 48 kN, verify that the true strains are additive and that the engineering strains are not additive. From Table 3.1 Calculated using initial length of 50 mm Force (load) (kN) Elongation (mm) Instantaneous Length (mm) Engineering Strain (m/m) True Strain (m/m) 35 0 51.52 0.0304 0.0299 42.5 0.00016 53.05 0.0610 0.0592 76.9 0.00036 54.57 0.0914 0.0875

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

4) Plot appropriate true stress-strain data from Table 1 on log-log scale and obtain the strain hardening exponent and the strain hardening constant. Compare to Eq. 3.27 (i.e., e u =n) Discuss differences. The power relation only applies in the uniformly part (i.e. no necking) of the plastic portion of the true stress-strain curve. This part of the curve extends from the yeild point to the ultimate tensile strength as shown in the table. True Stress (MPa) True Strain (m/m) 253 0.0101 277 0.0299 300 0.0398 347 0.0592

403 0.0875 436 0.124 490 0.226 From the plot, K=787 MPa and n=0.293. Using the relation for e u =n, n=0.226. The difference in the two values is no doubt due to the paucity of the data points in the curve fit (although one could always question the general applicability of the power law curve fit of the plastic part of the stress-strain curve!), and/or the reliance on a single experimentally-measured data point for e u =n. 5) Using the empirical relations for Brinell hardness, estimate the tensile strength in ksi and MPa for a Brinell hardness of 180 kg/mm 2 .

6) From the force-elongation data for a tensile test of a test specimen with a 50 mm gage length and 12.8 mm in diameter determine a) ultimate tensile strength, b) the true ultimate tensile strength, c) the estimated strain hardening exponent, n, d) the strain hardening constant, K in the equation . Data Given Data Calculated Force (kN) Elongation (mm) Engineering Stress (MPa) Engineering Strain (m/m) Start 0 0 0.0000 0.0000 10 0.01 77.712 0.00020000 20 0.02 155.42 0.00040000 Yield 25 0.025 194.28 0.00050000 30 11.2 233.14 0.22400 Maximum 38 18.7 295.31 0.37400 Fracture 18 24.5 139.88 0.49000 a) By inspection, S uts (engineering)=295 MPa b) Because the strain state is uniform up to UTS, S uts (true)=S uts (eng)*(1+ e u (eng)) =295 (1+0.374)=406 MPa c) n= e u (true)=ln (1+ e u (eng))=ln(1+0.374)=0.317 d) Use the strain hardening relation at a known point, in this case, UTS such that

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Section 4.7 4.) 10 – 8 8 – 6 additional amount of cold work Total 10 – 6 \ Cold working is not additive 7.) R m = (R 0 + 2R 45 + R 90 )/4 = (0.9+2*1.2+1.1)/4 = 1.1 R p = (R 0 - 2R 45 + R 90 )/2 = (0.9 – 2*1.2 + 1.1)/2 = -0.2 Section 5.7 1.) The main alloy(s) and carbon content of the steels are: 1015 Carbon 0.15% C content 4140 Molybdenum with chromium and nickel 0.40% C content 8640 Nickel-chromium-molybdenum 0.40% C content The main alloy(s) of the aluminum alloys are: 1100 Aluminum 2024 Copper 7075 Zinc a.) ASTM 20 Gray Cast Iron Tensile Strength: 152 MPa Torsional Strength: 179 MPa Compressive Strength: 572 MPa Reversed bending fatigue: 69 MPa Hardness: 156 HB ASTM 50 Gray Cast Iron Tensile Strength: 362 MPa Torsional Strength: 503 MPa Compressive Strength: 1130 MPa Reversed bending fatigue: 148 MPa Hardness: 262 HB 60-40-18 Ductile Iron Tensile Strength: 414 MPa 0.2% offset YS: 290 MPa Elongation: 10%

Total Carbon: 3.6-3.8% Mn 0.15-1, Si 1.8-2.8, Cr 0.03-0.01 Ni 0.05-0.2, Mo 0.01-0.1, Cu 0.15-1 P 0.03, S 0.002, Mg 0.03-0.06 120-90-02 Ductile Iron Tensile Strength: 827 MPa 0.2% offset YS: 621 MPa Elongation: Total Carbon: 3.6-3.8% Mn 0.15-1, Si 1.8-2.8, Cr 0.03-0.01 Ni 0.05-0.2, Mo 0.01-0.1, Cu 0.15-1 P 0.03, S 0.002, Mg 0.03-0.06 32510 Malleable Iron Tensile Strength: 345 MPa YS: 224 MPa Hardness: 262 HB Elongation: 10% 50005 Malleable Iron Tensile Strength: 483 MPa YS: 345 MPa Hardness: 179-229 HB Elongation: 5% Section 6.8 b.) – a.) Material A: area = 140000*10/0.035*10 6 = 4 sq inch Or area = 140000/30000 = 4.667 sq inch C(u) = 6*0.066*10*4.667 = $18.48 Material B: area = 140000*10/0.035*15*10 6 = 2.667 sq inch Or area = 140000/50000 = 2.8 sq inch C(u) = 5*0.1*10*2.8 = $14 Material C: area = 140000*10/0.035*20*10 6 = 2 sq inch Or area = 140000/90000 = 1.555 sq inch C(u) = 3*0.275*10*2 = $16.5 b.) Minimum cost = $14/unit Material A: 6W* 0.066*10*4 = 14 Hence minimum weight W = 0.9 lb/unit Material B: 5W* 0.1*10*2.667 = 14

Hence minimum weight W = 1 lb/unit Material C: 3W* 0.275*10*2 = 14 Hence minimum weight W = 0.85 lb/unit c.) r = m/v v = 0.9/0.066 = 13.636 d.) X*0.066*10*4.667 = 14 Gives X = $ 4.545/lb Should be reduced by $ 1.455 i.e. 24.2%

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version