Bar AB has length L, constant thickness d, and a width that tapers linearly from 3d at its fixed end (A) to d at its free end (B; see the figure below). The bar is homogeneous with mass density ρ and Young's modulus E. When hung vertically as shown in the figure, the bar stretches under its own weight. No other loads are applied. Assume that the x-axis is oriented downward with x=0 located at the ceiling (A).
Obtain a symbolic expression for N(x) in terms of L, d, g, x, and ρ
An interesting problem.
Would you kindly describe the figure so we can share how the bar was hung?
Is the bar tapering downwards, i.e. thin end at the bottom, and hung at the top (ceiling)?
Also, does N(x) represent the tensile stress at section x?
Assuming the wide end is at A, then the area is given by
a(x)=d²(3-2x/L)
and the tensile stress is given by the weight of material below divided by the area at each section.
i too have the same problem and
unable to paste the url for image
See if you can post by omitting "http://"
To obtain a symbolic expression for N(x) in terms of L, d, g, x, and ρ, let's break down the problem and consider the forces acting on the bar.
1. Weight of the bar:
The weight of the bar is given by the product of its mass and the acceleration due to gravity (g). Since the bar is hanging vertically, the weight acts downward. The weight can be expressed as W = m * g, where m is the mass of the bar.
2. Mass of the bar:
The mass of the bar can be calculated by finding its volume and multiplying it by the density (ρ). The volume of the bar can be approximated as a rectangular prism with varying width (from 3d to d) and constant thickness (d). The volume can be expressed as V = A * dx, where A is the cross-sectional area and dx is the differential length along the bar.
The cross-sectional area A can be determined by the width and thickness:
- At x = 0 (the fixed end A), the width is 3d.
- At x = L (the free end B), the width is d.
Since the width tapers linearly from 3d to d, we can express the width as a function of x: width = 3d - ((3d-d) / L) * x = 3d - (2d / L) * x.
The cross-sectional area can then be expressed as A = width * d = (3d - (2d / L) * x) * d.
Substituting the expression for A into the volume formula, we get the mass of the bar as: m = ρ * V = ρ * A * dx = ρ * ((3d - (2d / L) * x) * d) * dx.
3. Tension force:
Tension force (N) acts along the bar, counterbalancing the weight and preventing the bar from falling. At each differential length (dx), the tension force can be considered constant. We can express the differential tension force (dN) as equal to the weight acting on the element dx.
Therefore, we have dN = -W * dx = -m * g * dx.
Now, we can integrate dN from x = 0 to x = L to find the total tension force N(x):
∫dN = ∫(-m * g * dx)
N(x) = -∫(m * g * dx) [from x = 0 to x = L]
Substituting the expression for m into the integral, we get:
N(x) = -∫(ρ * ((3d - (2d / L) * x) * d) * dx) [from x = 0 to x = L]
Simplifying and integrating, we find the symbolic expression for N(x):
N(x) = -ρ * d * ∫((3d - (2d / L) * x) * dx) [from x = 0 to x = L]
N(x) = -ρ * d * [3dx - ((d / L) * x^2)] [from x = 0 to x = L]
N(x) = -ρ * d * [(3dL - (d / L) * L^2) - (0 - (d / L) * 0^2)]
Simplifying further gives us the final expression:
N(x) = -ρ * d * [(3dL - (d / L) * L^2)]
Therefore, the symbolic expression for N(x) in terms of L, d, g, x, and ρ is:
N(x) = -ρ * d * [(3dL - (d / L) * L^2)]