For this please explain the equation you formed ie sigma(2.5-x)/E + sigma b * x /E ...please explain how you formed this equation .
For this please explain the equation you formed ie sigma(2.5-x)/E + sigma b * x /E ...please explain how you formed this equation .
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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For this please explain the equation you formed ie sigma(2.5-x)/E + sigma b * x /E ...please explain how you formed this equation .
![Practice Problems
P2) A steel cylindrical rod 50 mm in diameter and 2.5 m long is subjected to
a pull of 100 kN. To what length the bar should be bored centrally so that
the total extension will increase by 15 % under the same pull, the bore
being 25 mm in diameter. Take E = 200 GN/m2.
Length of boring = 1.12 m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe3fe78da-ed7f-4acc-8090-c0216e6d800e%2Fb236eb70-597b-4f40-8578-9296988b5fb2%2Fthetjjb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Practice Problems
P2) A steel cylindrical rod 50 mm in diameter and 2.5 m long is subjected to
a pull of 100 kN. To what length the bar should be bored centrally so that
the total extension will increase by 15 % under the same pull, the bore
being 25 mm in diameter. Take E = 200 GN/m2.
Length of boring = 1.12 m
![Elongation of the rod &L = oL / E =
(50.92 x 106 x 2.5) / (200 × 10 °) =
0.000636m = 0.636 mm
%3D
Elongation after the rod is bored = 1.15
x 0.636 = 0.731mm
Area of the reduction section = (T/4)
(0.052 – 0.0252) = 0.001472 m²
Stress in the reduced section ob = (100
× 1000) / 0.001472 m² = 67.93 × 106
N/m2
Elongation of the rod X = o(2.5 – x)/E +
ob.x/E = 0.731 × 10-3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe3fe78da-ed7f-4acc-8090-c0216e6d800e%2Fb236eb70-597b-4f40-8578-9296988b5fb2%2Fuwcyvva_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Elongation of the rod &L = oL / E =
(50.92 x 106 x 2.5) / (200 × 10 °) =
0.000636m = 0.636 mm
%3D
Elongation after the rod is bored = 1.15
x 0.636 = 0.731mm
Area of the reduction section = (T/4)
(0.052 – 0.0252) = 0.001472 m²
Stress in the reduced section ob = (100
× 1000) / 0.001472 m² = 67.93 × 106
N/m2
Elongation of the rod X = o(2.5 – x)/E +
ob.x/E = 0.731 × 10-3
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