I need the answer quickly Task #7 The yield strength of mild steel with an average grain size of 0.05 mm is 20,000 psi. The yield stressof the same steel with a grain size of 0.007 mm is 40,000 psi. What will be the average grain size of the samesteel with a yield stress of 30,000 psi? Assume the Hall-Petch equation is valid and that changes in theobserved yield stress are due to changes in dislocation density.

Given Case 1 Average grain size = 0.05 mm Yield strength = 20000 psi Case 2 Grain size = 0.007 mm…

Answered: Task #7 The yield strength of mild…

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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Consider dislocations blocked with an average spacing of in a coppercrystal. If the flow stress is controlled by the stress necessary to operate aFrank--Read source, compute the dislocation density ρ in this crystal when itis deformed to a point where the resolved shear stress in the slip plane is 42MPa. Take G = 50 GPa.
Draw two schematic graphs using pencil showing a typical stress-strain curve for aluminum. The first graph should show engineering stress vs engineering strain, and the second graph should show true stress vs true strain. Label the showing: (i) elastic modulus (ii) proportional limit (iii) yield stress (iv)yield strain (v) fracture stress (vi) fracture strain on each graph. You may showboth graphs on one plot. Explain the difference between engineering stress and true stress.
2.3 A dislocation is pinned between two obstacles spaced 1micron apart. Calculate (i) the magnitude of stress to bow the dislocation into a semi circle. (ii) its yield strength. Given G = 100 GPa, Burgers vector = 0.25nm.
Samples 1 and 2 are identical except for the grain size. Which sample has the smaller grains (clearly mention 1 or 2)?
A hypothetical face-centred cubic metal has a lattice parameter a = 0.400 nm. A singlecrystalline bar made from the metal has a length of 20 mm and a diameter of 2 mm. TheBurgers vector of dislocations in the metal is <110>/2. Assuming only one slip system isactivated during uniform tensile straining and the angle between the direction of the Burgersvector of the dislocations and the tensile direction is 60°. Calculate the number of dislocationsneeded to achieve an engineering plastic strain of 10%. (pls give full handwritten solution to this)
A hypothetical face-centred cubic metal has a lattice parameter a = 0.400 nm. A singlecrystalline bar made from the metal has a length of 20 mm and a diameter of 2 mm. TheBurgers vector of dislocations in the metal is <110>/2. Assuming only one slip system isactivated during uniform tensile straining and the angle between the direction of the Burgersvector of the dislocations and the tensile direction is 60°. Calculate the number of dislocationsneeded to achieve an engineering plastic strain of 10%. (pls provide full calculations)
6.31 The flow stress for an alloy is 100 MPa when its dislocation density is 106 cm-2, and 150 MPa when its dislocation density is 108 cm-². When the flow stress is 190 MPa, what is the dislocation density?
6.27 Compute the dislocation density in tungsten if the flow stress is controlled by the stress necessary to operate a Frank--Read source. The shearstress in the slip plane is 50 MPa. Take G = 166 GPa.
Problem 5. Starting with a 2-inch diameter rod of brass, we would like to process 0.2-inch diameter rods that possess minimum yield strength of 40 ksi and a minimum elongation to fracture of 40%, (see the following figure). Design a process that achieves that. 120 70% Cu-30% Zn 100 Tensile strength 80 60 60 Yield strength 40 40 20 20 Elongation 10 20 30 40 50 60 Percent cold work Tensile strength and yield strength (ksi) Elongation percent
The yield stress of AISI 1020 steel with a grain size of 200 μm is 200 MPa.Estimate the yield stress for a grain size of 10 μm if the Hall--Petch constantk = 0.8 MN/m3/2.
Please help with the attached problem
A 3105 aluminum plate is reduced from 4.38 cm to 2.88 cm. Determine the final properties of the plate. See attached figure

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