Problem 1 «A steel alloy has the following tensile properties and behavior in the plastic region:o Engineering strain, £g rs = 0.10 at the engineering tensile strength, TS; = 510 MPae...

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Problem 1 « A steel alloy has the following tensile properties and behavior in the plastic region: o Engineering strain, £g rs = 0.10 at the engineering tensile strength, TS; = 510 MPa e Strain-hardening exponent, n = 0.11 What is the alloy true strength (or stress), or, at an engineering strain, £; = 0.05? Sketch of stress-strain curve, Assumptions and Analysis sections are needed for this problem. Problem 2 A. A tensile stress (0) is applied to a single crystal. The angle between the direction of the tensile stress and the normal to the slip plane of the most favorable slip system is 37°. Slip (or yielding) occurs when the applied tensile stress increases to a magnitude of 2.3 times the critical resolved shear stress (T.ss). What is the angle between the direction of the tensile stress and direction of slip for this yielding condition? The Sketch and Analysis sections are needed for this part. B. For the conditions in Part A, the stress at yielding (oy) approaches 140 MPa as the crystal size (d) grows very large. If ky, = 28.4 MPa — mm®3 and the yield strength is 375 MPa, what is the grain size? The Analysis section is needed for this part. Problem 4 : Consider the magnesium-lead system and phase diagram region that includes the intermediate compound Mg; Pb and alloys with lesser compositions of Pb. Mg,Pb forms at the composition Cyyg, py, of about 80.7 wt% Pb. The maximum solubility of solid lead in solid magnesium, Cg, is about 41.2 wt% Pb at the eutectic temperature. The eutectic composition, Cg, is about 66.5 wt% Pb. Composition (at% Pb) 0 5 10 20 3040 70 100 T T T TTT 11 700 |— po % 1200 600 ” MgPb M 1000 500 © c = 800 = £ 400 g & § 2 2 § 600 E 2 300 2 200 400 100 200 o | | 0 20 40 60 80 100 (Mg) Composition (wt% Pb) (Pb) Adapted from Phase Diagrams of Binary Magnesium Alloys, A. A. Nayeb-Hashemi and J. B. Clark (Editors), 1988. Reprinted by permission of ASM International, Materials Park, OH Page dof 7 ba Name: . What is the eutectic temperature, Tg, within £5°C, in this region of interest? . What is (are) the phase(s) and/or intermediate compound(s) at the composition of 40 wt% Pb and temperature of 400°C? C. For a hypoeutectic (less than eutectic) alloy at a temperature just less than the eutectic temperature, the mass fraction of the eutectic microconstituent (alternating layers of @ and Mg, Pb), W,, is 0.602. What is the composition of the alloy, Co, in Wt% Pb? D. What is the mass fraction, Wy, of proeutectic (or primary) a for the composition C,? E. What is the mass fraction, Wyg, pp, of total Mg, Pb for the composition C,? wo» The Analysis section is needed for this problem. Problem 3 , A plate contains an internal flaw that extends through the plate thickness. The plate is to be loaded in tension with the direction of the tensile stress perpendicular to the directions of the flaw length and plate thickness. The plate is made of steel alloy 4340, temper heat treated at 425°C, and is 1-m wide and 100- mm thick. The flaw is 1.25-mm long with a radius of curvature of the flaw tip equal to 9.34x10~2 mm. The Y parameter for the given plate and flaw geometry equals 1.03. Assume the modulus of elasticity of the steel is not affected by tempering. A. What is the maximum tensile stress that can be applied without causing the flaw to grow? B. For the applied tensile stress in Part A, what is the stress at the flaw tip? Sketch, Assumptions and Analysis sections are needed for this problem.
Answered Same DayDec 17, 2022

Answer To: Problem 1 «A steel alloy has the following tensile properties and behavior in the plastic...

Dr Shweta answered on Dec 17 2022
30 Votes
Ans 1 Actual strength of alloy is calculated as:
Given: Engineering strain ϵ = 0.10
Engineering t
ensile strength = TS = σ = 510Mpa
Strain-hardening exponent n = 0.11
Now, firstly we’ll calculate strength coefficient K
K = σ/ ϵn -------1
Putting Values in equation 1 we get,
K = 510 *10^6/0.100.11
on solving we get, 657.2Mpa
Now, at an engineering strain ϵ = 0.05, the true strength of alloy is
σ = K ϵn --------2
on solving we get,
σ = 472.67Mpa.
Sketch of stress-strain curve
Stress
Ans 2. Part 1 Angle between direction of tensile stress and direction of slip is calculated as:
As we know that
σ =τcrss/m ------[1]
Given: σ = 2.3 times of τcrss
So, on solving equation 1 we get,
m = 0.435
now as we know that
m = cosφcosλ
here φ = angle angle between the normal of...
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