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2 Form Cylinder 3 Diameter 12.8 mm 4 Gauge Length 5 Tensile Testing 50.8 mm To Failure 7 8 Load (N) Length (mm) 9 0 50.8 10 7330 50.851 11 15100 50.902 12 23100 50.952 13 30400 51.003 14 34400 51.054 15 38400 51.308 16 41300 51.816 17 44800 52.832 18 46200 53.848 19 47300 54.864 20 47500 55.88 AC10

2 Form Cylinder 3 Diameter 12.8 mm 4 Gauge Length 5 Tensile Testing 50.8 mm To Failure 7 8 Load (N) Length (mm) 9 0 50.8 10 7330 50.851 11 15100 50.902 12 23100 50.952 13 30400 51.003 14 34400 51.054 15 38400 51.308 16 41300 51.816 17 44800 52.832 18 46200 53.848 19 47300 54.864 20 47500 55.88 AC10

Elements Of Electromagnetics
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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From the data find the Young’s modulus in GPa
1 Material Aluminum 2 Form Cylinder ve 3 Diameter 12.8 mm 4 Gauge Length 5 Tensile Testing 50.8 mm To Failure 7 8 Load (N) Length (mm) 9 0 50.8 10 7330 50.851 11 15100 50.902 12 23100 50.952 13 30400 51.003 14 34400 51.054 15 38400 51.308 16 41300 51.816 17 44800 52.832 18 46200 53.848 19 47300 54.864 20 47500 55.88 21 46100 56.896 22 44800 57.658 23 42600 58.42 24 36400 59.182 25
Transcribed Image Text:1 Material Aluminum 2 Form Cylinder ve 3 Diameter 12.8 mm 4 Gauge Length 5 Tensile Testing 50.8 mm To Failure 7 8 Load (N) Length (mm) 9 0 50.8 10 7330 50.851 11 15100 50.902 12 23100 50.952 13 30400 51.003 14 34400 51.054 15 38400 51.308 16 41300 51.816 17 44800 52.832 18 46200 53.848 19 47300 54.864 20 47500 55.88 21 46100 56.896 22 44800 57.658 23 42600 58.42 24 36400 59.182 25
Expert Solution
Step 1 find youngs modulus

Given data:- 

Gauge diameter = 12.8 mm

Gauge length = 50.8 mm

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