Question:
A cantilever beam with a uniform flexural rigidity \( EI = 200 \times 10^6 \, \text{N.m}^2 \) is loaded with a concentrated force at its free end. The area of the bending moment diagram corresponding to the full length of the beam is 10000 N.m². The magnitude of the slope of the beam at its free end is \(\underline{\hspace{1cm}}\) micro-radian (round off to nearest integer).
A cantilever beam with a uniform flexural rigidity \( EI = 200 \times 10^6 \, \text{N.m}^2 \) is loaded with a concentrated force at its free end. The area of the bending moment diagram corresponding to the full length of the beam is 10000 N.m². The magnitude of the slope of the beam at its free end is \(\underline{\hspace{1cm}}\) micro-radian (round off to nearest integer).
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The slope of the beam at the free end for a cantilever beam with a point load can be calculated using the bending moment and flexural rigidity.
Updated On: Dec 30, 2025
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Correct Answer: 48 - 52
Solution and Explanation
The slope at the free end of a cantilever beam subjected to a point load at the free end is given by:
\[
\theta = \frac{M_{\text{total}} \cdot L}{EI},
\]
where:
- \( M_{\text{total}} = 10000 \, \text{N.m}^2 \) is the total area of the bending moment diagram,
- \( L \) is the length of the beam, and
- \( EI = 200 \times 10^6 \, \text{N.m}^2 \) is the flexural rigidity.
We also know that the total moment at the free end is given by: \[ M_{\text{total}} = \frac{P \cdot L}{4}, \] where \(P\) is the point load applied at the free end. Thus, the magnitude of the slope at the free end is: \[ \theta = \frac{10000}{200 \times 10^6} = 50 \, \mu\text{rad}. \] Thus, the magnitude of the slope of the beam at its free end is: \[ \boxed{48 \, \text{to} \, 52 \, \mu\text{rad}}. \]
- \( M_{\text{total}} = 10000 \, \text{N.m}^2 \) is the total area of the bending moment diagram,
- \( L \) is the length of the beam, and
- \( EI = 200 \times 10^6 \, \text{N.m}^2 \) is the flexural rigidity.
We also know that the total moment at the free end is given by: \[ M_{\text{total}} = \frac{P \cdot L}{4}, \] where \(P\) is the point load applied at the free end. Thus, the magnitude of the slope at the free end is: \[ \theta = \frac{10000}{200 \times 10^6} = 50 \, \mu\text{rad}. \] Thus, the magnitude of the slope of the beam at its free end is: \[ \boxed{48 \, \text{to} \, 52 \, \mu\text{rad}}. \]
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