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.
for problem 1,part 4
(rho_2*g*L^2)/(4*E_1)+(rho_2*g*L^2)/(8*E_2)
To obtain the symbolic expressions for N(x), ϵa(x), and the total elongation of the bar δ, we need to analyze the equilibrium and deformation of the bar under its own weight. Let's go step by step:
1. Equilibrium Analysis:
Since there are no other loads applied apart from the weight of the bar itself, the bar will be in static equilibrium. This means that the internal forces in the bar must balance the gravitational force acting on it.
Considering the bar is hung from the ceiling, the top half of the bar experiences no gravitational load due to the neglect of the weight. Hence, the internal force N(x) in the top half must balance out the force exerted by the bottom half.
2. Deformation Analysis:
The deformation of the bar can be analyzed using Hooke's Law, which states that the stress (σ) in a material is proportional to the strain (ε) applied to it. The relation is given as σ = Eε, where E is the Young's modulus of the material.
Since the cross-sectional area of the bar is constant, the stress and strain will also be constant across the bar for each material. However, the strain will differ in the two halves of the bar due to their different Young's moduli.
3. Symbolic Expressions:
For 0 ≤ x ≤ L/2:
- In this region, the top half of the bar experiences no gravitational load, so N(x) = 0.
- The strain εa(x) for the top half is given by εa(x) = 0, as there is no deformation present.
For L/2 < x ≤ L:
- In this region, the bottom half of the bar experiences the full gravitational load, balanced by N(x).
- Since the bottom half is made of material "2" with a Young's modulus of E2 = 2E1, the internal force N(x) can be given as N(x) = ρ2 * g * A * (L - x), where ρ2 is the density of material "2," g is the acceleration due to gravity, and A is the cross-sectional area of the bar.
- The strain εa(x) for the bottom half is given by εa(x) = -(L - x) * (ρ2 * g * A) / (E2 * A).
For the total elongation of the bar, δ:
- We can find the elongation by integrating the strain over the length of the bar.
- Integrating εa(x) from 0 to L/2 gives δ1 = 0, as the strain is zero in the top half.
- Integrating εa(x) from L/2 to L gives δ2 = -(ρ2 * g * A / E2) * [(x(L - x) - (L^2)/4)].
- Therefore, the total elongation of the bar is δ = δ1 + δ2 = -(ρ2 * g * A / E2) * [(x(L - x) - (L^2)/4)].
By following these steps, you can obtain the symbolic expressions for N(x), εa(x), and the total elongation of the bar, δ, for different regions along the length of the bar.