Module 16 · Lesson 16.2
Fixed-ended beams and fixed-end moments
The building blocks every hand method for frames is assembled from.
Why this matters
Fixed-end moments are the currency of every hand method for indeterminate frames. Slope-deflection starts from them, moment distribution starts from them, and even a matrix analysis uses the same quantities as equivalent nodal loads. Get comfortable with a short list of them and a great deal becomes straightforward.
By the end of this lesson you should be able to
- Analyse a fixed-ended beam under a UDL
- Compare fixed-ended and simply supported moment distributions
- Quote and use fixed-end moments for the standard load cases
What you should already know
- The force method and compatibility (this module, lesson 1)
- Bending moment diagrams (Module 3)
A beam built in at both ends has six reaction components against three equations, so it is indeterminate to the third degree. In the common case of vertical loading only, the horizontal reactions vanish and it reduces to degree 2 — but symmetry usually cuts that down further.
For a full UDL on a fixed-ended beam the results are worth memorising:
- End moments: wL²/12 at each end, hogging
- Mid-span moment: wL²/24, sagging
- Reactions: wL/2 at each end, unchanged from the simply supported case
The reactions are unchanged because the end moments are equal and opposite; they form a couple that does not disturb the vertical balance.
Compare with the simply supported beam, where the maximum moment is wL²/8 = 3wL²/24. Fixing the ends redistributes the bending: the peak moves to the supports and drops from wL²/8 to wL²/12 — a reduction of a third. The mid-span moment falls much further, to a third of its simply supported value.
Predict first
A simply supported beam under a UDL has a maximum moment of wL²/8. Both ends are then built in. What happens to the maximum moment in the beam?
Try it
How end restraint shares out the moment
A single span under a full UDL. Dial the end fixity from a pin (simply supported) to a full clamp (fixed ends) and watch the bending-moment diagram morph — but its total depth never changes. Restraint moves the moment; it cannot remove it.
Jump to
- End fixity r
- 0.60
- Spring stiffness SL/EI
- 3.00
- Hogging at each end
- 43.2 kNm
- Sagging at mid-span
- 64.8 kNm
- Their sum (free moment)
- 108.0 = wL²/8
- Reaction at each support
- 72.0 kN
Things worth trying
- Drag the fixity to 0 and 100%: the diagram flips from all-sagging to the wL²/12 : wL²/24 split, but the envelope height never moves.
- Watch the sum row — hogging plus sagging is always the free moment, whatever the fixity or the load.
- Change w or L: every moment scales, yet the reactions stay at wL/2 because the end moments form a couple.
- At full fixity the peak has fallen by a third — but it has moved to the supports, where the connections now carry it.
The standard fixed-end moments (FEMs) — the moments developed at the ends of a member with both ends fully fixed against rotation — are the inputs to slope-deflection and moment distribution. The three worth knowing:
- Full UDL of intensity w: wL²/12 at each end
- Point load P at mid-span: PL/8 at each end
- Point load P at distance a from the left end (b = L − a): Pab²/L² at the left, Pa²b/L² at the right
The general point-load case reduces to the central one when a = b = L/2: Pab²/L² becomes P(L/2)(L/2)²/L² = PL/8 ✓
Note how the general case puts the larger moment at the nearer end. A load close to the left support produces a large moment there and a small one at the far end — which matches the intuition that the near support does most of the work.
Worked example
Fixed-ended beam under a UDL, and the effect of the fixity
Given
- Fixed-ended beam, span L = 6.00 m
- Uniformly distributed load w = 24.0 kN/m over the whole span
Find
The end moments, the mid-span moment and the reactions, compared with the simply supported case.
Practice
A fixed-ended beam spans 6.00 m and carries a UDL of 24.0 kN/m over its whole length. What is the magnitude of the moment at each end, in kNm?
Practice
For the same beam, what is the sagging moment at mid-span, in kNm?
Practice
A member of span 8.00 m has both ends fixed and carries a single point load of 80.0 kN at 3.00 m from the left-hand end. What is the fixed-end moment at the left-hand end, in kNm?
Summary
- Fixed-ended beam with a full UDL: wL²/12 at the ends, wL²/24 at mid-span
- Reactions stay at wL/2 under symmetric loading, because the end moments form a couple
- The free moment wL²/8 is fixed by statics; restraint only decides how it is shared
- Standard FEMs: wL²/12 for a UDL, PL/8 for a central point load, Pab²/L² and Pa²b/L² in general
- Restraint reduces span moments but transfers moment into the supports and connections
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint