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 ro1 , and Young’s modulus E1 . The bottom half of the bar is made of material 2 with density ro2 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 p2 >> p1 [so you can neglect the gravity load on the top half of the bar].
1. Obtain a symbolic expression for .
For 0 <x< L/2, fx(x)=
For 0 <x< L, fx(x)=
2. Obtain a symbolic expression for .
For 0 <x< L/2, N(x)=
For 0 <x< L, N(x)=
3. Obtain a symbolic expression for
For 0 <x< L/2, ea(x)=
For 0 <x< L, ea(x)=
4. btain a symbolic expression for the total elongation of the bar
for part 4
(rho_2*g*L^2)/(4*E_1)+(rho_2*g*L^2)/(8*E_2)
To obtain the symbolic expressions for the given quantities, we need to consider the behavior of the bar and its response to the applied loads. Let's break down each question and explain the steps to find the symbolic expressions.
1. Symbolic expression for f(x):
The force acting on an infinitesimal element of the bar at position x is given by the weight of the material below it. Since we can neglect the gravity load on the top half of the bar, for 0 < x < L/2, the weight of the material below x is proportional to the mass density ro1 and the cross-sectional area A. Therefore, we have:
f(x) = ro1 * A * x * g,
where g is the acceleration due to gravity. Since the material is in a state of equilibrium, this force f(x) must be equal to the internal force within the material at x. However, the internal force depends on the deformation of the material, which we need to determine. So, for 0 < x < L/2, the expression for f(x) will be:
f(x) = E1 * A * u(x),
where u(x) is the displacement at x.
For L/2 < x < L, the weight of the entire bar needs to be considered. Therefore, the force is given by:
f(x) = ro1 * A * L/2 * g + ro2 * A * (x - L/2) * g.
Similarly, f(x) is also equal to E2 * A * u(x) for L/2 < x < L.
2. Symbolic expression for N(x):
Considering the equilibrium of forces acting on an infinitesimal element, the internal normal force N(x) at position x can be obtained by integrating f(x) over the length from 0 to x. So, for 0 < x < L/2, we have:
N(x) = ∫(0 to x) ro1 * A * x * g dx = (1/2) * ro1 * A * x^2 * g.
For L/2 < x < L, we integrate f(x) over that range:
N(x) = ∫(L/2 to x) (ro1 * A * L/2 * g + ro2 * A * (x - L/2) * g) dx = (1/2) * ro1 * A * L/2 * (x - L/2) * g + (1/2) * ro2 * A * (x - L/2)^2 * g.
3. Symbolic expression for ε(x):
The strain ε(x) at position x is defined as the deformation per unit length in the bar. In this case, since the bar is subjected to axial loading, the strain can be related to the displacement u(x) as:
ε(x) = du(x)/dx.
Therefore, for 0 < x < L/2, we differentiate the expression for u(x) to get ε(x):
ε(x) = du(x)/dx = N(x)/(E1 * A) = (1/2) * ro1 * x^2 * g / (E1 * A).
For L/2 < x < L, we have a different expression for ε(x) since E2 = 2E1:
ε(x) = du(x)/dx = N(x)/(E2 * A) = (1/2) * (ro1 * L/2 * (x - L/2) * g + ro2 * (x - L/2)^2 * g) / (E2 * A).
4. Symbolic expression for the total elongation:
The total elongation of the bar can be obtained by integrating the strain ε(x) from 0 to L. So, for 0 < x < L/2, we have:
ΔL = ∫(0 to L/2) ε(x) dx = ∫(0 to L/2) (1/2) * ro1 * x^2 * g / (E1 * A) dx.
Similarly, for L/2 < x < L, we integrate the expression for ε(x) over that range:
ΔL = ∫(L/2 to L) (1/2) * (ro1 * L/2 * (x - L/2) * g + ro2 * (x - L/2)^2 * g) / (E2 * A) dx.
Please note that these are the generalized symbolic expressions based on the given conditions for the materials, dimensions, and loads. To obtain numerical values, specific values for parameters need to be substituted into these expressions.