Question

A cylindrical bar has a length L = 5 m and cross section area S = 10 m². The bar is made of a linear elastic material with a density \(\rho\) = 2700 kg/m³ and Young's modulus E = 70 GPa. The bar is suspended as shown in the figure and is in a state of uniaxial tension due to its self-weight.
The elastic strain energy stored in the bar equals ............... J. (Rounded off to two decimal places)
Take the acceleration due to gravity as g = 9.8 m/s².

Hint:

For problems involving self-weight, remember that the stress and strain are functions of position. The most common mistake is to assume a uniform load equal to the total weight. Always set up an integral by considering a small element and the load acting on it. The formula for strain energy due to self-weight, \(U = \frac{W^2 L}{6AE}\) where \(W = \rho g A L\), can be memorized for faster calculation.

Answer 1

Step 1: Understanding the Concept:
The question asks for the elastic strain energy stored in a cylindrical bar hanging vertically under its own weight. The stress in the bar is not uniform; it varies linearly along the length. The stress is zero at the free end (bottom) and maximum at the fixed end (top). To find the total strain energy, we must integrate the strain energy of infinitesimal elements along the length of the bar.
Step 2: Key Formula or Approach:
The strain energy \(dU\) in a small element of length \(dx\) is given by \(dU = \frac{P(x)^2 dx}{2AE}\), where \(P(x)\) is the tensile force at a distance \(x\) from the free end.
The total strain energy \(U\) is the integral of \(dU\) over the entire length \(L\).
\[ U = \int_0^L \frac{P(x)^2}{2AE} dx \] The force \(P(x)\) at a distance \(x\) from the bottom is the weight of the portion of the bar below it.
\[ P(x) = (\text{Volume below x}) \times \text{density} \times g = (A \cdot x) \cdot \rho \cdot g \] Step 3: Detailed Explanation or Calculation:
Given data:
Length, \(L = 5\) m
Cross-section area, \(A = S = 10\) m²
Density, \(\rho = 2700\) kg/m³
Young's modulus, \(E = 70\) GPa = \(70 \times 10^9\) Pa
Acceleration due to gravity, \(g = 9.8\) m/s²
First, express the force \(P(x)\) at a section \(x\) from the free end:
\[ P(x) = A \rho g x \] Now, substitute this into the strain energy integral formula:
\[ U = \int_0^L \frac{(A \rho g x)^2}{2AE} dx = \int_0^L \frac{A^2 \rho^2 g^2 x^2}{2AE} dx \] \[ U = \frac{A \rho^2 g^2}{2E} \int_0^L x^2 dx \] \[ U = \frac{A \rho^2 g^2}{2E} \left[ \frac{x^3}{3} \right]_0^L = \frac{A \rho^2 g^2 L^3}{6E} \] Now, substitute the given values into the derived formula:
\[ U = \frac{10 \times (2700)^2 \times (9.8)^2 \times (5)^3}{6 \times (70 \times 10^9)} \] \[ U = \frac{10 \times 7290000 \times 96.04 \times 125}{420 \times 10^9} \] \[ U = \frac{875161500000}{420 \times 10^9} = \frac{8.751615 \times 10^{11}}{4.2 \times 10^{11}} \] \[ U \approx 2.0837 \text{ J} \] Step 4: Final Answer:
Rounding off to two decimal places, the elastic strain energy stored in the bar is 2.08 J.
Step 5: Why This is Correct:
The calculation is based on the correct principle of integrating the strain energy over the length of the bar, considering the varying axial force due to self-weight. The derived formula \(U = \frac{A \rho^2 g^2 L^3}{6E}\) is the standard result for this problem. The calculated value of 2.08 J falls within the specified correct answer range of 2.00 to 2.16.