answer option d. thanks Load(N)Al(mm)0.00005,00010,00015,00020,00025,00026,50027,00026,50025,0000.02960.05920.08880.150.510.901.50 (maximum load)2.102.79 (fracture)Pull test result data in Table 1. Obtained from the tensile testing of an ingredient. Specimendiameter A (A=11.75 mm), with gauge length = 30.00 mm. After breaking, the gauge lengthbecomes 32.61 mm and the diameter is broken B (B=11.51 mm). From the data of the tensiletest results:test-b.ppencd?d. If the material is loaded with a force of:• 560 x 158 (N• 280 x 158 (N)• 120 x 158 (N)What happened to the material?e. Calculate the modulus of resilience of the material.f. Calculate the ductility of the materialthe %Reduction in area

Answered: answer option d.

Engineering

Mechanical Engineering

answer option d.

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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5. The following data were collected from a standard 0.505-in.-diameter test specimen of a copper alloy (initial length lo= 2.0 in.). After fracture, the total length was 3.014 in. and the diameter was 0.374 in. Load (Ib) Al (in.) 00000 3,000 6,000 0.00167 0.00333 7,500 0.00417 9,000 10,500 0.0090 0.040 12,000 0.26 12,400 11,400 0.50 (maximum load) 1.02 (fracture) a) Plot the data as engineering stress versus engineering strain. b) Compute the modulus of elasticity. c) Determine the yield strength at a strain offset of 0.002. d) Determine the tensile strength of this alloy. e) What is the approximate ductility, in percent elongation? f) Compute the modulus of resilience. g) Compute from the data and plot true stress versus true strain diagram.
answer e and f only please..
In tensile test a plain carbon steel specimen has a (40mm) gauge length and the Final area (A final) of specimen after tensile test was 264.327. The load which caused fracture was (122.5 KN). After fracture, the final length was 47.516mm The % reduction in area is * O 15.819% 18.819 O 17.819 O 20.819
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Q19 and Q20