The composite bar BCD in the figure is fixed at the wall B (x=0), and has constant cross sectional area A_0. The bar is composed by joining, at section C, two segments (BC and CD) of equal length L. Segment CD is homogeneous, made of copper. Segment BC is obtained by joining two identical wedges, a copper wedge and a steel wedge, as indicated in the figure. Along segment BC the cross sectional areas of copper and steel are given, respectively, by:

A_c=A_0*x/L,
A_s=A_0(1-x/L),
which sum to give the constant cross-sectional area A_0 at each x within bar segment BC.

The bar is subjected to a uniform distributed load per unit length of magnitude f_0 in the direction indicated in the figure.

The Young’s modulus of copper is E_C=E_0, and the modulus of steel is E_S=2E_0.

NOTE: careful! The total length of the bar is 2L.

Find:

1. expression for the axial force resultant N(x) in terms of L, x and f_0.

2. expression for the axial strain in the bar, ϵ_a(x) in terms of x, L, E_0, A_0

3. expression for the normal stress in the steel at the midspan of segment BC, σ_n steel(x=L2), in terms of L, A0, and f0

4. expression for the displacement field in the bar, u_x(x) in terms of x, L, E_0, A_0 and f_0
for 0²x²L, ux(x)=
for L²x²2L, ux(x)=

5. expression for the elongation of entire bar BD, δ_BD, in terms of L, E_0, A_0 and f_0

To solve these problems, we'll use the principles of statics, mechanics of materials, and the equations for axial force, axial strain, normal stress, and displacement. Let's go through each question step by step.

1. Expression for the axial force resultant N(x) in terms of L, x, and f0:

To find the axial force resultant N(x), we'll consider the equilibrium of forces acting on the bar segment BC. The total axial force at any point x on segment BC is given by the integral of the distributed load:

N(x) = ∫[0 to x] f0 dx

So, the expression for N(x) is simply the integral of the distributed load f0 over the length from 0 to x.

2. Expression for the axial strain in the bar, ϵa(x) in terms of x, L, E0, and A0:

The axial strain ϵa at any point x can be found using Hooke's Law:

ϵa(x) = δ(x) / L

where δ(x) is the displacement at that point. To find δ(x), we can integrate the displacement field, which we will derive later. Once we have δ(x), we can substitute it into the equation for the axial strain.

3. Expression for the normal stress in the steel at the midspan of segment BC, σnsteel(x=L/2), in terms of L, A0, and f0:

To find the normal stress in the steel at the midspan of segment BC, we'll consider the equilibrium of forces acting on the steel section at x=L/2. At this point, the axial force N(x=L/2) will act on the steel section alone.

The normal stress σn in the steel can be calculated using the formula:

σn = N(x=L/2) / A(steel)

where A(steel) is the cross-sectional area of the steel at that point, given by:

A(steel) = A0 * (1 - x/L)

Substituting the expression for N(x=L/2) obtained from the first question, we can find σnsteel(x=L/2).

4. Expression for the displacement field in the bar, ux(x), for 0 ≤ x ≤ L and L ≤ x ≤ 2L:

To determine the displacement field, we need to analyze the deformation of the bar in both regions: 0 ≤ x ≤ L and L ≤ x ≤ 2L. In the first region, we have a combination of copper and steel, while in the second region, we have only copper.

For 0 ≤ x ≤ L:
In this region, the deformation is affected by both copper and steel segments. We need to determine the constants of integration for each segment and then join them at x = L.

For L ≤ x ≤ 2L:
In this region, the bar is purely copper and we can set the boundary condition as the displacement at x = L (where copper and steel segments join). We need to determine the constant of integration for this region.

Once we have the complete displacement field for both regions, we can write the final expression for ux(x).

5. Expression for the elongation of the entire bar BD, δBD, in terms of L, E0, A0, and f0:

The elongation of the entire bar BD can be obtained by integrating the axial strain ϵa(x) from x = 0 to x = 2L:

δBD = ∫[0 to 2L] ϵa(x) dx

Substituting the expression for ϵa(x) obtained in the second question, we can calculate δBD.

By following these steps and using the given equations and properties, you'll be able to find the expressions for the axial force resultant, axial strain, normal stress, displacement field, and elongation of the entire bar BD.