The composite bar BCD in the figure is fixed at the wall B (x=0), and has constant cross sectional area A0. 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:
Ac=A0xL,
As=A0(1−xL),
which sum to give the constant cross-sectional area A0 at each x within bar segment BC.
The bar is subjected to a uniform distributed load per unit length of magnitude f0 in the direction indicated in the figure.
The Young’s modulus of copper is EC=E0, and the modulus of steel is ES=2E0.
The given (known) quantities are: L[m], A0[m2], E0[Pa], and f0[Nm].
NOTE: careful! The total length of the bar is 2L.
1) Obtain a symbolic expression for the axial force resultant N(x) in terms of L, x and f0 (enter this as f_0):
N(x)=
2) Obtain an expression for the axial strain in the bar, ϵa(x) in terms of x, L, E0, A0 and f0 (enter E0, A0 and f0 as, respectively, E_0, A_0 and f_0):
for 0≤x≤L, ϵa(x)=
for L≤x≤2L, ϵa(x)=
3) Obtain a symbolic expression for the normal stress in the steel at the midspan of segment BC, σnsteel(x=L2), in terms of L, A0, and f0 (enter A0 and f0 as, respectively, A_0 and f_0).
σnsteel(x=L2)=
4) Obtain an expression for the displacement field in the bar, ux(x) in terms of x, L, E0, A0 and f0 (enter E0, A0 and f0 as, respectively, E_0, A_0 and f_0):
for 0≤x≤L, ux(x)=
for L≤x≤2L, ux(x)=
5) Obtain a symbolic expression for the elongation of entire bar BD, δBD, in terms of L, E0, A0 and f0 (enter E0, A0 and f0 as, respectively, E_0, A_0 and f_0):.
δBD=
Anyone knows?
To solve this problem, we need to use the equations provided and apply the principles of mechanics. Let's go step by step.
1) To obtain the axial force resultant N(x), we need to consider the forces acting on each segment of the bar separately.
First, let's look at the segment BC. The force acting on this segment is the force f0 distributed over the cross-sectional area Ac. So, the force in BC is N_BC = f0 * Ac.
Next, let's consider the segment CD. Since this segment is homogeneous and made of copper, the force in CD is simply N_CD = f0 * A0.
Now, to find the total axial force resultant N(x), we add the forces acting on BC and CD:
N(x) = N_BC + N_CD
Substituting the equations for Ac and A0, we get:
N(x) = f0 * (A0 * x * L + A0 * (1 - x * L))
= f0 * (A0 * x * L + A0 - A0 * x * L)
= f0 * A0
Therefore, the symbolic expression for N(x) is:
N(x) = f_0 * A_0
2) To find the axial strain ϵa(x), we know that strain is equal to the change in length divided by the original length.
For 0 ≤ x ≤ L, the original length is L and the change in length is x.
Therefore, ϵa(x) = x/L
For L ≤ x ≤ 2L, the original length is L and the change in length is L.
Therefore, ϵa(x) = L/L = 1
So, the symbolic expressions for the axial strain ϵa(x) are:
For 0 ≤ x ≤ L, ϵa(x) = x/L
For L ≤ x ≤ 2L, ϵa(x) = 1
3) To find the normal stress in the steel at the midspan of segment BC (x = L/2), we need to consider the forces acting on the steel wedge.
At the midspan of segment BC, the force in the steel is N_steel = f0 * As(x = L/2). Substituting the equation for As, we get:
N_steel = f0 * A0 * (1 - (L/2)*L)
So, the symbolic expression for the normal stress in the steel at x = L/2 is:
σnsteel(x=L/2) = N_steel / A0
4) To obtain the displacement field ux(x), we need to consider the deformation of the bar under the applied load.
For 0 ≤ x ≤ L, segment BC experiences both copper and steel deformations. The displacements in BC can be calculated using the equations for the strains and Young's modulus.
The displacement in the copper part is ux_copper(x) = ϵa(x) * L * E0.
The displacement in the steel part is ux_steel(x) = ϵa(x) * L * 2E0.
So, the total displacement in BC is the sum of the displacements in copper and steel:
ux(x) = ux_copper(x) + ux_steel(x)
Substituting the equations for ϵa(x), we get:
For 0 ≤ x ≤ L, ux(x) = (x/L) * L * E0 + (x/L) * L * 2E0
Simplifying:
ux(x) = (3x / L) * E0
For L ≤ x ≤ 2L, only segment CD (made of copper) contributes to the displacement. So, the displacement in this region is simply:
ux(x) = ϵa(x) * L * E0
= 1 * L * E0
= L * E0
So, the symbolic expressions for the displacement field ux(x) are:
For 0 ≤ x ≤ L, ux(x) = (3x / L) * E0
For L ≤ x ≤ 2L, ux(x) = L * E0
5) The elongation of the entire bar BD is the difference in displacements between the two ends of the bar.
δBD = ux(x = 2L) - ux(x = 0)
= L * E0 - 0
= L * E0
So, the symbolic expression for the elongation of entire bar BD is:
δBD = L * E0