1. What is the approximate ductility (%EL) of a cold worked brass that has a yield strength of 345 MPa (50,000 psi)1401201040 Steel1040 Steel120700600100500BrassBrass eoS400Brass300Copper40200Copper040 e0040Copper100O 10 20 30 40 50 60 70Percent cold work2000 10 20 30 40 50 60 70Percent cold workO 10 20 30 40 D0 60 70Percent cold work(a)(b)

Concept: From first graph-a, For brass(yellow graph) , at a yeild strength of 345 Mpa , the…

Answered: 1. What is the approximate ductility…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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The engineering stress-strain curve below was obtained for a precipitation hardened Aluminum alloy. What is the approximate Yield Strength for this alloy in psi? Engineering Stress Based on Original Area (psi) 50,000 45,000 40,000 35,000 30,000 25,000 20,000 15,000 10,000 5,000 0 O 0.02 0.04 0.06 Aluminum 6061-T6 0.08 0.1 0.12 Engineering Strain (in/in) 0.14 X 0.16 0.18
A cylindrical specimen of cold-worked steel has a Brinell hardness of 240. If the specimen remained cylindrical during deformation and its original radius was 10.4 mm, determine its radius after deformation. For steel, the dependence of tensile strength on percent cold work is shown in Animated Figure 7.19b. i mm
Cyclic Stresses (Fatigue); The S-N Curve 4. The fatigue data for a ductile cast iron are given as follows: Cycles to Failure Stress Amplitude [MPa (ksi)] 248 (36.0) 236 (34.2) 224 (32.5) 213 (30.9) 201 (29.1) 193 (28.0) 193 (28.0) 193 (28.0) 1 × 105 3 x 105 1 x 10⁰ 3 x 106 1 x 107 3 x 107 1 108 3 × 108 (a) Make an S-N plot (stress amplitude versus logarithm cycles to failure) using these data. (b) What is the fatigue limit for this alloy? (c) Determine fatigue lifetimes at stress amplitudes of 230 MPa (33,500 psi) and 175 MPa (25,000 psi). (d) Estimate fatigue strengths at 2 × 105 and 6 × 106 cycles.
6. Answer the following questions regarding the Lever Rule. [ a. Describe the purpose of the Lever Rule and write the equation. b. Using the lever rule for 52 wt% Zn-48 wt% Cu, cite the phases present and mass fractions at the following temperatures: 1000°C, 800°C, 500°C, and 300°C. Refer to the Cu-Zn diagram from problem 3 or use the book diagram on p. 326. 7. Returning to the Zn-Cu phase diagram, using the composition 52 wt% Zn-48 wt% Cu, draw schematics of the microstructures if you were to undergo a slow cooling process. Draw representations for 950°C, 860°C, 800°C, and 600°C. Make sure to properly label each phase and determine their approximate compositions.
Question 5: Following is tabulated data that were gathered from a series of Charpy impact tests on a tempered 4140 steel alloy. Temperature (°C) 100 75 50 25 0 -25 -50 -65 -75 -85 -100 -125 -150 -175 Impact Energy (J) 89.3 88.6 87.6 85.4 82.9 78.9 73.1 66.0 59.3 47.9 34.3 29.3 27.1 25.0 (a) Plot the data as impact energy versus temperature. (you can use Excel or other software). (b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (c) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is 70 J.
5.3) A cold-worked carbon steel was tested in a tensile test and the results are shown in Table 1 below: Table 1. Tensile test results Original diameter 10,00 mm Final diameter 7,60 mm Original gauge length 102,00 mm Final gauge length 114,30 mm Load at upper yield 31,21 kN Maximum force 36,25 kN Load at fracture 28,65 kN Factor of safety 2 For Questions 5.3.1 to 5.3.8, calculate the following tensile test properties based on the information provided in Table 1 above: 5.3.1) Stress at upper yield. ANSWER TO QUESTION 5.3.1 5.3.2) Ultimate tensile stress. ANSWER TO QUESTION 5.3.2
Answer the following questions based on the given information in the table and further calculation (if needed for any missing parameter, e.g. strain at yield point). Assume all metals started with the same shape and size before the tensile tests occurred. Material A B C D E Yield Strength (MPa) 105 600 430 355 The material that experienced the greatest percent reduction in area is material The hardest material is material The hardest material has Tensile Strength (MPa) The material that experienced the greatest percent reduction in area has 135 850 The material that experienced the largest change in shape before plastic deformation occurred is material The stiffest material is material The stiffest material has 535 365 Fracture before yielding The material that experienced the largest change in shape before plastic deformation occurred has Strain at Fracture [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] 0.45 0.15 0.20 0.25 Fracture Strength (MPa)…
Q2c) Listed in the table below is the tensile stress-strain data for different grades of steels. Utilizing the data given answer the three queries given below. Tensile Strain at Fracture Strength Strength Fracture Strength Modulus Material Yield Elastic (MPa) (MPa) (MPa) (GPa) А 410 1440 0.63 265 410 200 220 0.40 105 250 C 815 950 0.25 500 610 800 650 0.14 720 210 E Fractures before yielding 650 550 1) Which will experience the greatest percent reduction in area? Why? 2) Which is the strongest? Why? 3) Which is the stiffest? Why? B.
One metal sample is hot worked 40%; a second sample is hot worked 50%; a third sample is cold worked 40%; a fourth sample is cold worked 50%. Select all of the following that are true. Group of answer choices The 40% hot worked sample would have the highest tensile modulus. The two cold worked samples would have the highest tensile modulus and the cold worked moduli are equal. The 50% cold worked sample would have the highest tensile modulus. The 50% hot worked sample would have the highest tensile modulus. The 40% cold worked sample would have the highest tensile modulus. The two hot worked samples would have the highest tensile modulus and the hot worked moduli are equal. The tensile modulus of all the samples would be about the same.
Hall-Petch Constants Material Y; (MPa) k (MPa.ma) Copper 25 0.12 Titanium 80 0.40 Mild Steel 70 0.74 Using Hall-Petch equation, determine the Yield Stress (MPa) for a Mild Steel piece of metal having a grain size of 100 um. O 115 MPa O 101 MPa O 144 MPa O 130 MPa
O 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)
c). Interpret the following diagram? The effect of carbon and heat treatment on the properties of plain carbon steels. 160 140 Tensile 120 strength, 100 Impact energy 100 Annealed 80 80 Normalized 60 60 Annealed 40 `Yield strength 40 20-% Elongation Annealed 20 Normalized< 0.2 0.4 0.6 0.8 1.0 Weight percent carbon Figure-4 Yield and tensile strength (ksi) Normalized % Elongation or impact energy (It • Ib)

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