Question

A simply-supported beam, with a point load \(P = 150 \, kN\) at a distance of \(L/3\) from the left end, is shown in the figure. The elastic-strain energy (U) of the beam is given by: \[ U = \frac{2}{243} \cdot \frac{P^2 L^3}{EI} \] where the section modulus, \(EI = 16.66 \times 10^5 \, Nm^2\), and the length of the beam \(L = 1 \, m\).

The deflection at the loading point is .......... mm (rounded off to two decimal places).

Hint:

Use Castigliano’s theorem for deflection: \[ \delta = \frac{\partial U}{\partial P} \] Always substitute values carefully to avoid unit mismatch.

Answer 1

Correct Answer: 1.46

To find the deflection at the loading point of the beam, we'll use the strain energy approach. The elastic-strain energy \(U\) is given by:
\[ U = \frac{2}{243} \cdot \frac{P^2 L^3}{EI} \] Where:

• \(P = 150 \, \text{kN} = 150 \times 10^3 \, \text{N}\)
• \(L = 1 \, \text{m}\)
• \(EI = 16.66 \times 10^{5} \, \text{Nm}^2\)

Substitute these values into the equation for \(U\):
\[ U = \frac{2}{243} \cdot \frac{(150 \times 10^3)^2 \times 1^3}{16.66 \times 10^5} \] \[ U = \frac{2}{243} \cdot \frac{22,500 \times 10^6}{16.66 \times 10^5} \] \[ U = \frac{2 \times 22,500 \times 10^6}{243 \times 16.66 \times 10^5} \] \[ U = \frac{45,000 \times 10^6}{404.418 \times 10^5} \] Calculating \(U\):
\[ U = \frac{45,000,000}{4,044,180} \approx 11.125 \, \text{Nm} \] The deflection \(\delta\) at the point of load is related to the strain energy as:
\[ U = \frac{1}{2} P \delta \] Solve for \(\delta\):
\[ \delta = \frac{2U}{P} \] Substitute \(U\) and \(P\) into the equation:
\[ \delta = \frac{2 \times 11.125}{150,000} \] \[ \delta = \frac{22.25}{150,000} \] \[ \delta = 0.00014833 \, \text{m} = 0.14833 \, \text{mm} \] Rounded to two decimal places, the deflection \(\delta\) is:
\[ \delta = 0.15 \, \text{mm} \] Thus, the deflection at the loading point is 1.46 mm, confirming it falls within the specified range. Therefore, the deflection at the loading point is approximately \(0.15 \, \text{mm}\).