A uniform rod of cross-sectional area A and length L is subjected to an axial pull P. What is the change in length of the rod? (Assume that Young's modulus of elasticity E remains the same throughout the length.)

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DDA JE Civil 29 Mar 2023 Shift 2 Official Paper
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  1. \(\frac{PE}{AL}\)
  2. \(\frac{P}{AEL}\)
  3. \(\frac{PL}{AE}\)
  4. \(\frac{PA}{LE}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{PL}{AE}\)
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Detailed Solution

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Explanation:

We can find the change in length or longitudinal elongation by the formula:

\(\Delta L=\frac{Pl}{AE}\)

for this, we required Young Modulus of Elasticity E.

To find a decrease in the diameter of the rod, we need to know the Poisson's ratio (μ) defined as:

\(μ=-\frac{Lateral\;or\;diametral\;strain\;(ϵ_D)}{Longitudinal\;strain\;(ϵ_L)}\)

∴ Lateral deformation (ϵD) = -μ × Longitudinal deformation (ϵL)

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