Question

The maximum deflection of a fixed beam of length \( l \) carrying a total load \( W \) being uniformly distributed over the entire length is

• A. \( \dfrac{Wl^3}{48EI} \)
• B. \( \dfrac{Wl^3}{96EI} \)
• C. \( \dfrac{Wl^3}{192EI} \)
• D. \( \dfrac{Wl^3}{384EI} \)

Hint:

Fixed beams have smaller deflection compared to simply supported beams under the same loading.

Answer 1

The Correct Option is C

Step 1: Understanding the beam condition.
The beam is fixed at both ends and carries a uniformly distributed load over its entire length. Fixed supports restrict both rotation and translation.
Step 2: Using standard deflection formula.
For a fixed beam under uniformly distributed load, the maximum deflection occurs at the centre of the beam.
Step 3: Writing the formula.
The maximum deflection is given by: \[ \delta_{\max} = \frac{Wl^3}{192EI} \]
Step 4: Conclusion.
The maximum deflection of the beam is \( \dfrac{Wl^3}{192EI} \).

Answer 2

A useful check is to compare a fixed (built-in) beam with the more familiar simply supported beam under the same uniformly distributed load \( W \). For a simply supported beam, the well-known maximum deflection is \( \dfrac{Wl^3}{48EI} \). Fixing both ends restrains rotation there, which makes the beam stiffer and reduces the deflection by a known factor of 4 compared to the simply supported case, giving \( \dfrac{Wl^3}{192EI} \). Let's test the options against this relationship.

1. \( \dfrac{Wl^3}{48EI} \): This is exactly the simply supported beam value; using it for a fixed beam would ignore the additional stiffness provided by the fixed-end restraints, so it is too large for this case.
2. \( \dfrac{Wl^3}{96EI} \): This is only twice as stiff as the simply supported case, but fixed-end restraint is known to reduce deflection by a factor of 4, not 2, so this undercounts the effect of both ends being fixed.
3. \( \dfrac{Wl^3}{192EI} \): This is exactly one-quarter of the simply supported deflection \( \dfrac{Wl^3}{48EI} \), consistent with the standard result that clamping both ends of a beam under a UDL reduces the maximum deflection by a factor of 4.
4. \( \dfrac{Wl^3}{384EI} \): This value corresponds to a beam fixed at both ends with a central point load, not a uniformly distributed load, so it belongs to a different loading case.

Comparing against the simply supported beam case with the known stiffening factor confirms the fixed beam's maximum deflection under a UDL is \( \dfrac{Wl^3}{192EI} \).

Therefore, the correct answer is \( \dfrac{Wl^3}{192EI} \).