Fixed–fixed beam under uniformly distributed load\nFor a fixed beam of length l carrying a total load W uniformly distributed over the whole span, what is the maximum deflection?

Introduction / Context:
Deflection control is a key serviceability requirement in beams. End fixity significantly reduces midspan deflection compared to simply supported conditions. This problem asks for the maximum deflection of a fixed–fixed beam under a uniformly distributed load (UDL) expressed in terms of total load W.

Given Data / Assumptions:

• Span = l; flexural rigidity = E I (constant).
• Uniformly distributed load of intensity w (so total W = w l).
• Both ends fully fixed (zero rotation and zero deflection).

Concept / Approach:
Standard closed-form results from beam theory: for a fixed–fixed beam under UDL w, the maximum deflection at midspan is y_max = w l^4 / (384 E I). Converting using W = w l gives y_max = W l^3 / (384 E I).

Step-by-Step Solution:
Start with y_max = w l^4 / (384 E I) for fixed–fixed with UDL.Replace w by W / l (since W = w l).Therefore y_max = (W / l) * l^4 / (384 E I) = W l^3 / (384 E I).

Verification / Alternative check:
Compare with simply supported beam under the same total load: y_max,SS = 5 W l^3 / (384 E I). The fixed–fixed deflection is one-fifth of the simply supported value, consistent with increased restraint.

Why Other Options Are Wrong:
5 W l^3 / (384 E I) is the simply supported case; W l^3 / (48 E I) and W l^3 / (192 E I) correspond to other loading/support cases; W l^4 / (8 E I) has wrong dimensions and matches a cantilever UDL form when expressed in w and l.

Common Pitfalls:
Mixing up w and W; applying simply supported formulas to fixed beams; forgetting to convert intensity to total load.

Final Answer:
W l^3 / (384 E I)