Question

A cantilever beam of length 0.3 m is subjected to a uniformly distributed load \( C = 10 \, \text{kN/m} \), as shown in the figure. The bending (flexural) rigidity of the beam is 5000 Nm\(^2\). Neglecting the self-weight of the beam, the magnitude of beam curvature in m\(^{-1}\) at the fixed end is 

• A. 1.10
• B. 0.02
• C. 0.09
• D. 0.05

Hint:

To calculate curvature in a beam under a uniform load, use \( \kappa = \frac{M(x)}{EI} \), where \( M(x) \) is the bending moment and \( EI \) is the bending rigidity.

Answer 1

The Correct Option is C

The beam is subjected to a uniformly distributed load, and we are tasked with finding the curvature at the fixed end of the cantilever beam. The curvature \( \kappa \) at any point on the beam is given by the equation: \[ \kappa = \frac{M(x)}{EI} \] where:
- \( M(x) \) is the bending moment at a point \( x \) on the beam,
- \( E \) is the Young's modulus of the material (we are given the bending rigidity \( C = EI \)),
- \( I \) is the second moment of area, which together with \( E \) gives the bending rigidity \( C \).
Step 1: Calculate the bending moment For a cantilever beam with a uniformly distributed load \( C \), the bending moment at a distance \( x \) from the fixed end is: \[ M(x) = \frac{C x^2}{2} \] Substituting \( C = 10 \, \text{kN/m} \) and \( x = 0.3 \, \text{m} \): \[ M(0.3) = \frac{10 \times (0.3)^2}{2} = \frac{10 \times 0.09}{2} = 0.45 \, \text{kN}\cdot\text{m}. \] Step 2: Calculate the curvature at the fixed end The bending rigidity is given as \( C = 5000 \, \text{Nm}^2 \). So, using the formula for curvature at the fixed end (\( x = 0 \)): \[ \kappa = \frac{M(0)}{C} = \frac{0.45}{5000} = 0.00009 \, \text{m}^{-1}. \] Thus, the magnitude of the curvature at the fixed end is \( 0.09 \, \text{m}^{-1} \), so the correct answer is (C).