The bar in the figure has constant cross sectional area A. The top half of the bar is made of material "1" with mass density ρ1, and Young’s modulus E1. The bottom half of the bar is made of material "2" with density ρ2 and Young’s modulus E2=2E1. The total length of the bar is L. When the bar is hung from the ceiling, it stretches under its own weight. No other loads are applied. Assume ρ2>>ρ1[so you can neglect the gravity load on the top half of the bar].

Obtain a symbolic expression for N(x).

For 0≤x<L/2, N(x)=
For L/2<x≤ L, N(x)=

Obtain a symbolic expression for ϵa(x).

For 0≤x<L/2, ϵa(x)=
For L/2<x≤ L, ϵa(x)=

Obtain a symbolic expression for the total elongation of the bar, δ.

To obtain the symbolic expressions for N(x), ϵa(x), and the total elongation δ, we need to analyze the equilibrium and deformation of the bar.

1. For 0 ≤ x < L/2:
Since the gravity load on the top half of the bar is neglected, only the stretching due to the weight of the bottom half needs to be considered. In this region, the total force N(x) acting on a differential element of length dx is equal to the weight of the material below it. Thus, N(x) = (A * ρ2 * g * x), where g is the acceleration due to gravity.

2. For L/2 ≤ x ≤ L:
In this region, the stretching of the bar is due to the weight of both the top and bottom halves. The total force N(x) acting on a differential element of length dx is the sum of the weights of the top and bottom halves. It can be calculated as N(x) = (A * ρ1 * g * (L - x)) + (A * ρ2 * g * x).

3. For 0 ≤ x < L/2:
The axial strain ϵa(x) is the ratio of the change in length of a differential element to its original length. Since the bar is uniform in cross-section, the strain is the same throughout the element. Therefore, ϵa(x) = (∆L / L), where ∆L is the elongation of the differential element.

4. For L/2 ≤ x ≤ L:
Similar to the previous region, the axial strain ϵa(x) is the ratio of the change in length of a differential element to its original length (∆L / L).

5. The total elongation δ of the bar is obtained by integrating the axial strain over the whole length of the bar. Thus, δ = ∫[0,L] ϵa(x) dx.

By using the above explanations, you can now derive the symbolic expressions for N(x), ϵa(x), and δ based on the given conditions.