Question

square steel rod 20×20mm is section is to carry an axial load of 100 kn. calculate the shorthening in a length of 50mm.​

Answer 1

Explanation:

Given square steel rod 20 mm ×20 mm is section is to carry an axial load of 100 kn. calculate the shortening in a length of 50 mm.

• Now A = 20 mm x 20 mm = 0.02 x 0.02
•                                         = 0.0004 sq m
• Length l = 50 mm = 50 x 10^-3 = 0.05 m
• So load p = 100 KN
• Also E = 2.14 x 10^8 KN / m^2
• So Δ l is shortening of the rod.
• Stress Δ = P / A
•             = 100 / 0.0004
•           = 250000 KN / m^2
• Now E = Stress / Strain
• Or Strain = Stress / E
•               = 250000 / 2.14 x 10^8
• Therefore Δ l  = 250000 / 2.14 x 10^8 x l
•          Or Δ l = 250000 / 2.14 x 10^8 x 0.05
•                     = 5841.12 x 10^-8 m

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Answer 2

Answer:

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