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 0≤x≤L, ux(x)=
for L≤x≤2L, ux(x)=
5. expression for the elongation of entire bar BD, δ_BD, in terms of L, E_0, A_0 and f_0
1. To find the expression for the axial force resultant N(x), we need to consider the equilibrium of forces along the bar.
At any point x within segment BC, the axial force resultant N(x) can be found by summing up the differential forces acting on the segment BC and CD.
For the segment BC, the differential force can be calculated as:
dF_copper = σ_copper(x) * A_c * dx
= (E_0 * ε_copper(x)) * (A_0 * x/L) * dx
= (E_0 * ε_copper(x)) * A_0 * x/L * dx
dF_steel = σ_steel(x) * A_s * dx
= (E_0 * ε_steel(x)) * (A_0 * (1−x/L)) * dx
= (E_0 * ε_steel(x)) * A_0 * (1−x/L) * dx
The axial force resultant N(x) is the sum of these differential forces, so we have:
N(x) = ∫[0,L] (E_0 * ε_copper(x)) * A_0 * x/L * dx + ∫[0,L] (E_0 * ε_steel(x)) * A_0 * (1−x/L) * dx
2. The axial strain in the bar, ϵ_a(x), can be calculated based on the deformation of each segment (BC and CD).
For the segment BC, the strain can be calculated as:
ε_copper(x) = du_copper(x)/dx
By integrating the expression for A_c, we can relate the displacement du_copper to x:
du_copper(x) = u_copper(x) - u_copper(0) = ∫[0,x] ε_copper(x) dx = ∫[0,x] ε_copper(x) * L/x * dx
Using Hooke's Law, ε_copper(x) = σ_copper(x) / E_C = (E_0 * ε_copper(x)) / E_C, we can rewrite the above equation as:
du_copper(x) = ∫[(E_0 * ε_copper(x)) / E_C] * L/x * dx
Similarly, for the segment CD, the strain can be calculated as:
ε_steel(x) = du_steel(x)/dx
By integrating the expression for A_s, we can relate the displacement du_steel to x:
du_steel(x) = u_steel(x) - u_steel(L) = ∫[L,x] ε_steel(x) dx = ∫[L,x] ε_steel(x) * L/(L-x) * dx
Using Hooke's Law, ε_steel(x) = σ_steel(x) / E_S = (E_0 * ε_steel(x)) / E_S, we can rewrite the above equation as:
du_steel(x) = ∫[(E_0 * ε_steel(x)) / E_S] * L/(L-x) * dx
Therefore, the axial strain ϵ_a(x) is given by:
ϵ_a(x) = ε_copper(x) + ε_steel(x)
= ∫[(E_0 * ε_copper(x)) / E_C] * L/x * dx + ∫[(E_0 * ε_steel(x)) / E_S] * L/(L-x) * dx
3. The normal stress in the steel at the midspan of segment BC, σ_n steel(x=L/2), can be calculated by using the equilibrium condition.
At the midspan of segment BC, the total axial force must be zero. Therefore, we have:
N(L/2) = 0
Substituting the expression for N(x) from part 1, we can solve for σ_steel(x=L/2):
0 = ∫[0,L] (E_0 * ε_copper(x)) * A_0 * x/L * dx + ∫[0,L] (E_0 * ε_steel(x)) * A_0 * (1−x/L) * dx
Solving this equation for σ_steel(x=L/2) will give the desired expression.
4. The displacement field in the bar, u_x(x), can be calculated by integrating the strains along the bar.
For 0≤x≤L, the displacement field u_x(x) is given by:
u_x(x) = u_copper(x) = ∫[0,x] ε_copper(x) dx
For L≤x≤2L, the displacement field u_x(x) is given by:
u_x(x) = u_steel(x) - u_steel(L) = ∫[L,x] ε_steel(x) dx
5. The elongation of the entire bar BD, δ_BD, can be calculated by integrating the displacement u_x(x) over the entire length of the bar.
δ_BD = ∫[0,2L] u_x(x) dx = ∫[0,L] u_copper(x) dx + ∫[L,2L] (u_steel(x) - u_steel(L)) dx
1. To find the expression for the axial force resultant N(x), we can consider the equilibrium of forces on a small element of length dx at position x. The axial force on this element is given by N(x) = A(x) * σ(x), where A(x) is the cross-sectional area at position x and σ(x) is the normal stress at position x. The normal stress is related to the axial strain ε(x) by σ(x) = E(x) * ε(x), where E(x) is the modulus of elasticity at position x.
We can start by finding the expression for A(x) in segment BC. From the given information, A(x) = Ac(x) + As(x) = A0 * (x/L) + A0 * (1 - x/L) = A0. Therefore, the cross-sectional area is constant within segment BC and equal to A0.
Next, we can find the expression for E(x) in segment BC. Since segment BC is composed of copper and steel, we need to consider the contribution from each material. The modulus of elasticity for copper is EC = E0, and for steel is ES = 2E0. Therefore, we can define the modulus of elasticity as a piecewise function:
E(x) = EC = E0, for 0 <= x <= L
E(x) = ES = 2E0, for L < x <= 2L
Now, we can substitute A(x) and E(x) into the equation for N(x) to get the expression:
N(x) = A(x) * σ(x) = A0 * σ(x)
Since σ(x) = E(x) * ε(x), we substitute E(x) with the corresponding values for copper and steel:
N(x) = A0 * E(x) * ε(x) =
A0 * E0 * ε(x), for 0 <= x <= L
A0 * 2E0 * ε(x), for L < x <= 2L
Therefore, the expression for the axial force resultant N(x) in terms of L, x, and f0 is:
N(x) = A0 * E0 * ε(x), for 0 <= x <= L
N(x) = A0 * 2E0 * ε(x), for L < x <= 2L
2. To find the expression for the axial strain εa(x) in the bar, we need to relate it to the displacement field. The axial strain is defined as the change in length per unit length, which can be expressed as ε(x) = du(x)/dx, where du(x) is the change in displacement in the x-direction at position x.
Given that the bar is subjected to a uniform distributed load per unit length f0, the displacement field u(x) can be found by integrating the distributed load. We can divide the calculation into two segments: 0 <= x <= L and L < x <= 2L.
For 0 <= x <= L, the distributed load acts only on the copper segment and has a magnitude of f0. Therefore, the displacement field is given by:
u(x) = ∫f0 dx = f0x, for 0 <= x <= L
For L < x <= 2L, the distributed load acts on both the copper and steel segments, but the magnitude for the steel segment is double (2f0) since its modulus of elasticity is twice that of copper. Therefore, the displacement field is given by:
u(x) = ∫(f0 + 2f0) dx = 3f0(x - L), for L < x <= 2L
Therefore, the expression for the displacement field in the bar, u(x) in terms of x, L, E0, A0, and f0 is:
For 0 <= x <= L: ux(x) = f0x
For L < x <= 2L: ux(x) = 3f0(x - L)
3. To find the normal stress in the steel at the midspan of segment BC, σnsteel(x=L/2), we need to consider that the steel contributes to the total area at that point. The cross-sectional area equation A(x) = A0 * (x/L) + A0 * (1 - x/L) gives us the cross-sectional area at any point in segment BC.
When x = L/2, we substitute x into the equation for A(x):
A(L/2) = A0 * (L/2L) + A0 * (1 - L/2L) =
A0 * 1/2 + A0 * 1/2 =
A0 * 1 = A0
Therefore, at the midspan of segment BC, the cross-sectional area is A0, and the normal stress in the steel can be determined by substituting A(x) into the expression for N(x):
N(L/2) = A(L/2) * σ(L/2) = A0 * σ(L/2)
Therefore, the 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 is:
σnsteel(x=L/2) = A0 * σ(L/2)
4. To find the expression for the displacement field in the bar, u(x), in terms of x, L, E0, A0, and f0 at different intervals, we can refer back to the previous solution for part 2.
For 0 <= x <= L, we found that u(x) = f0x. Therefore, for this interval:
ux(x) = f0x
For L < x <= 2L, we found that u(x) = 3f0(x - L). Therefore, for this interval:
ux(x) = 3f0(x - L)
5. To find the elongation of the entire bar BD, δBD, in terms of L, E0, A0, and f0, we need to consider the displacement of the bar at both ends.
At the wall B (x = 0), the displacement is zero, so we can set u(0) = 0. Therefore, u(0) = 0 = f0(0), which implies f0 = 0.
For the free end D (x = 2L), we can substitute x = 2L into the displacement field equation for the interval L < x <= 2L:
ux(2L) = 3f0(2L - L) =
3f0(L)
Therefore, the expression for the elongation of the entire bar BD, δBD, in terms of L, E0, A0, and f0 is:
δBD = u(2L) = ux(2L) = 3f0(L)