Question

Consider a rectangular bar of length L and cross-sectional area A suspended vertically. Which of the following is the appropriate expression for the deformation of the bar due to its own weight and a downward force P applied on its free end? The material has a specific weight γ (i.e. weight per unit volume) and a modulus of elasticity E.

a. δ= ⋎L²/2E − PL/AE ​
b. δ= ⋎L²/E + PL/AE
c. δ= ⋎L²/E + PL/AE ​
d. δ= ⋎L²/2E + PL/AE ​ ​

Answer 1

The correct expression for the deformation of a vertically suspended bar due to its own weight and an additional force is δ= γL2/2E + PL/AE. This considers the specific weight and modulus of elasticity of the material, as well as the bar's length and cross-sectional area.

Step-by-step explanation:

The student is asking about the deformation of a vertically suspended rectangular bar subjected to its own weight and an additional downward force. The deformation, or elongation, depends on the material's specific weight (γ), the modulus of elasticity (E), the length of the bar (L), and the cross-sectional area (A). The total deformation (δ) is the sum of the deformation due to the bar's own weight and the deformation caused by the external force (P) applied at its end.

Considering the specific weight, which is the weight per unit volume, the deformation due to the bar's own weight can be expressed as δweight = (γL2) / (2E), and the deformation due to the applied force P can be expressed as δP = (PL) / (AE). The total deformation δ is simply the sum of these two deformations:

δ = δweight + δP = (γL2) / (2E) + (PL) / (AE)

Based on this understanding, the correct answer is:

d. δ= γL2/2E + PL/AE