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 ProblemsP2) A steel cylindrical rod 50 mm in diameter and 2.5 m long is subjected toa pull of 100 kN. To what length the bar should be bored centrally so thatthe total extension will increase by 15 % under the same pull, the borebeing 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%3DElongation after the rod is bored = 1.15x 0.636 = 0.731mmArea of the reduction section = (T/4)(0.052 – 0.0252) = 0.001472 m²Stress in the reduced section ob = (100× 1000) / 0.001472 m² = 67.93 × 106N/m2Elongation of the rod X = o(2.5 – x)/E +ob.x/E = 0.731 × 10-3

Answered: For this please explain the 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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Question

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

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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