1)In this , how did you multiply by 1.15 ..like from where did it come ? 2)Also pi/4( 0.052-0.0252) from where did 0.052 and 0.0252 come ...I didn't understand. 3)Also the equation which you formed , sigma ( 2.5-x)/E + sigma b ×x/E ..please explain how you formed this equation . 11:02O 37%You5 minutes agoPractice 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 11:44 bElongation of the rod SL = oL /E =(50.92 × 106 × 2.5) / (200 × 10 º) =0.000636m = 0.636 mmElongation 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 × 10-3II不lookN

x = Length of the bore D = Diameter of steel rod = 50 mm = 0.05 mL = Length of steel rod = 2.5 mP =…

Answered: Practice Problems P2) A steel…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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1)In this , how did you multiply by 1.15 ..like from where did it come ? 2)Also pi/4( 0.052-0.0252) from where did 0.052 and 0.0252 come ...I didn't understand. 3)Also the equation which you formed , sigma ( 2.5-x)/E + sigma b ×x/E ..please explain how you formed this equation .

11:02
O 37%
You
5 minutes ago
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

11:44 b
Elongation of the rod SL = oL /E =
(50.92 × 106 × 2.5) / (200 × 10 º) =
0.000636m = 0.636 mm
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
II
不
lookN

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