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 the following expression:
\[ U = \frac{2}{243} \frac{P^2 L^3}{EI}, \] where the section modulus, \( EI \), is \( 16.66 \times 10^5 \, {Nm}^2 \) and the length of the beam \( L \) is 1 m.
The deflection at the loading point is ............ mm (rounded off to two decimal places).

Hint:

For a beam under a point load, the elastic-strain energy is related to the deflection by \( U = \frac{P \delta}{2} \). Use this relationship to solve for deflection when the strain energy is given.

Answer 1

We are given the expression for the elastic-strain energy \( U \) and asked to calculate the deflection at the loading point. We can use the relationship between strain energy and deflection for a beam under a point load. The deflection \( \delta \) at the loading point is related to the strain energy \( U \) by the formula: \[ U = \frac{P \delta}{2}. \] Step 1: Calculate the Strain Energy \( U \)
Substitute the given values into the expression for \( U \): \[ U = \frac{2}{243} \times \frac{(150 \times 10^3)^2 \times (1)^3}{16.66 \times 10^5}. \] Simplifying: \[ U = \frac{2}{243} \times \frac{(150^2) \times 10^6}{16.66 \times 10^5} = \frac{2 \times 22500 \times 10^6}{243 \times 16.66 \times 10^5}. \] Solving: \[ U = \frac{45000 \times 10^6}{4047.78 \times 10^5} \approx 11.12 \, {J}. \] Step 2: Calculate the Deflection \( \delta \)
Now, use the relation \( U = \frac{P \delta}{2} \) to solve for \( \delta \): \[ 11.12 = \frac{150 \times 10^3 \times \delta}{2}. \] Solving for \( \delta \): \[ \delta = \frac{2 \times 11.12}{150 \times 10^3} = \frac{22.24}{150 \times 10^3} = 0.000148 \, {m} = 1.48 \, {mm}. \] Thus, the deflection at the loading point is approximately 1.48 mm, which lies between 1.46 mm and 1.50 mm, as expected.