Example: Consider a metal whose strain hardeningbehavior follows: 25,000 0.25 psi.If an annealed bar of this metal is pulled in a tensiontest from a starting dia. of 12.7mm to a dia. of 11.5mm(uniform elongation, no friction, no distortion), what isthe work per unit volume required?If the bar is reduced by extruding and the deformationefficiency is 70%, what is the extrusion pressure?

Extrusion pressure Explanation:Step 1: Step 2: Step 3: Step 4:

Answered: Example: Consider a metal whose strain…

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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You are given a square rod of 6061-T6 aluminum (cross-section = 4mm X 4mm, length = 2.3m). The material has a Young’s modulus of 72 GPa. If a compressive load of 3 kN is applied parallel to the 2.3m dimension, what is the resulting engineering normal strain?
Suppose that the material in Q4 is fabricated in 25 mm (1 in) thick plate, estimate the largest valid K1c that can be measured on such a specimen.
Given a example of copper specimen that has had a tensile test conducted on it. Diameter of 6.4mm Max force of 11112.5N Max extension of 7.79mm Calculate: 1) modulus of elasticity 2) tensile strength 3) the ductility, in percent elongation 4) yield strength at a strain of 0.002
A ceramic part for a jet engine has a yield strength of 66,215 psi and a plane strain fracture toughness of 5000 psi (in) ¾. To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one-third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05 in. Iong. Assuming that f = 1.2 , does our non- destructive test have the required sensitivity? Type 1 for yes and O for no.
Correctly complete the following phrase below: "When a structure is subjected to general loading in two dimensions, Tresca and the Von-Mises criteria are used to... determine the maximum shear stress." Don't Know calculate the principle planes." calculate the deformation in a structure." determine failure of materials that behave plastically."
why use yield strength not tensile strength? You are drawing wire which you want the wire to elongate pernamently, and not return back to original shape .
Pls do this question at any cost
A material has a linear elastic perfectly plastic stress-strain. A cylindrical specimen of that material having an initial length of 35 mm and a tadius of 7 mm is pulled until it elongates 0.8mm. a. Calculate the stain and the corresponding stress at that elongation value. b. Afterward, the specimen is released to zero stress level, how does the final length of the specimen compare with its original length? How much plastic deformation remains if it exists?
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.
Question 4 Save Answer You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is ✓ MPa for 0.1 plastic strain, proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= MPa. Hence, we can predict that we need a prior plastic strain of approximately True Stress (MPa) True Stress (MPa) True Stress (MPa) 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 Ep = 0.1 0.01 Ep=0.2 0.01 Ep=0.3 0.01 0.02 True Strain 0.02 0.03 0.02 0.03 True Strain 0.03 0.04 0.04 0.04 True Stress (MPa) 0.05 0.000 0.001 True Stress (MPa) Ep=0.1 True…
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Question-8. . A cylindrical metal specimen with a diameter of 16 mm and length of 88 mm is to be subjected to a tensile stress of 80 MPa; at this stress the resulting deformation is elastic. (a) If the elongation is less than 0.058 mm, which of the metals in the table can be considered as candidates? Why ? (b) If the maximum permissible diameter decrease is 0.0012 mm when the tensile stress of 68 MPa is applied, which of the metals satisfy the criterion in question (a) are the suitable candidates? Why ? Metal Shear Modulus / GPa Modulus of Elasticity / GPa 69 Poisson's Ratio Al 25 0.33 Cu 110 46 0.34 Mg 45 17 0.29 Ni 207 76 0.31 Ti 107 6.5 0.34 W 407 23.2 0.28

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