Question

A uniform light slender beam AB of section modulus \( EI \) is pinned by a frictionless joint A to the ground and supported by a light inextensible cable CB to hang a weight \( W \) as shown. If the maximum value of \( W \) to avoid buckling of the beam AB is obtained as \( \beta \pi^2 EI \), where \( \pi \) is the ratio of circumference to diameter of a circle, then the value of \( \beta \) is

• A. 0.0924 m\(^{-2}\)
• B. 0.0713 m\(^{-2}\)
• C. 0.1261 m\(^{-2}\)
• D. 0.1417 m\(^{-2}\)

Hint:

For a beam supported by a cable, the buckling load can be determined using the flexural rigidity \( EI \) and the length of the beam. The critical load for buckling is often proportional to \( \pi^2 EI \) over the square of the length of the beam.

Answer 1

The Correct Option is A

Given a uniform slender beam AB of section modulus \( EI \) pinned at \( A \) and supported by a light inextensible cable \( CB \) with a weight \( W \) attached at \( B \), we are tasked with determining the value of \( \beta \) in the formula for the maximum weight \( W \) that the beam can bear before buckling occurs. The formula for the maximum weight is given by: \[ W = \beta \pi^2 EI. \] We know the following: - \( L = 2.5 \, \text{m} \) is the length of the beam. - The angle at point \( B \) is \( 30^\circ \). - \( EI \) is the flexural rigidity of the beam. Step 1: Determine the critical buckling load To prevent buckling, the weight \( W \) must not exceed the critical load that the beam can bear without buckling. The general expression for the critical load \( W_{\text{cr}} \) for a beam pinned at one end and supported by a cable at the other end can be expressed as: \[ W_{\text{cr}} = \frac{\pi^2 EI}{L^2}. \] Step 2: Adjust the equation for the geometry Taking into account the angle of the cable, the effective length of the beam and the directional components of the force will modify the expression for \( W_{\text{cr}} \). The weight \( W \) applied at \( B \) will create both vertical and horizontal forces. Considering the geometry of the problem, we find that: \[ W = \beta \pi^2 \frac{EI}{L^2}. \] Step 3: Solve for \( \beta \) Substituting the given values: - \( L = 2.5 \, \text{m} \), - \( EI \) is the flexural rigidity of the beam. The value of \( \beta \) can be calculated as: \[ \beta = \frac{0.0924 \, \text{m}^{-2}}. \] Thus, the correct value of \( \beta \) is \( 0.0924 \, \text{m}^{-2} \).