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Mechanical Engineering QUESTION #2500
Question 1
The total axial deformation \(\delta\) of a bar of length \(l\), cross-sectional area \(A\), modulus of elasticity \(E\), under axial force \(F\) is:
  • \(\delta = \frac{AE}{Fl}\)
  • \(\delta = \frac{Fl}{AE}\)✔️
  • \(\delta = \frac{FA}{El}\)
  • \(\delta = FAEl\)
Correct Answer Explanation
From Hooke's Law: \(\sigma = E\varepsilon\), where \(\sigma = F/A\) and \(\varepsilon = \delta/l\). Solving: \(\delta = Fl/(AE)\).