Question
Download Solution PDFA member of length 200 mm and diameter 25 mm is subjected to a tensile load of 20 kN. The final length of the member is found to be 220 mm. The percentage increase in the length of the member is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Percentage change in the length = \(\frac{L_f\;-\;L_i}{L_i}\times100\)
Where,
Lf = Final length of work-piece, Li = Initial length of work-piece.
Calculation:
Given:
Lf = 220 mm, Li = 200 mm
Then,
Percentage change in the length = \(\frac{L_f\;-\;L_i}{L_i}\times100\)
= \(\frac{220\;-\;200}{200}\times100\)
= 10%
Additional Information
Elongation in the bar under compressive and tensile load (P)
Elongation (δ) = \(\frac{PL}{AE}\)
Where,
P = Tensile or Compressive load, L = Initial length of bar, A = Cross-section area of bar, E = Modulus of elasticity.
Last updated on Jul 1, 2025
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