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Structural Analysis Simulator

Fixed-End Moment Simulator — Clamped-Clamped Beam FEM Basics

Apply a point load and a uniformly distributed load to a clamped-clamped beam and read off the fixed-end moments, support reactions and mid-span deflection in real time. Use the sliders to build intuition for the slope-deflection and moment-distribution methods.

Parameters
Span L
m
Point load P
kN
Position a of P
m
Uniform load w
kN/m

Flexural rigidity EI is fixed at 50000 kN.m^2. The load position a is automatically clamped to the range 0.1*L to 0.9*L.

Live values (during load cycle)
100%
Load scale
0.0
Fixed-end M_A [kN.m]
0.0
Mid-span M [kN.m]
0.00
Mid-span deflection δ [mm]
Clamped beam deflection & bending moment diagram
Deflected beam (clamped ends flat) Bending moment diagram (BMD) Shear force diagram (SFD)

As the load grows and shrinks the beam deflects while the clamped ends keep zero slope. The BMD is negative (hogging) at both ends and positive at mid-span.

Theory & Key Formulas

Fixed-end moments for a point load P at distance $a$ (with $b = L - a$):

$$M_A = -\frac{P\,a\,b^2}{L^2},\quad M_B = -\frac{P\,a^2\,b}{L^2}$$

M_A, M_B: support bending moments [kN.m], P: point load [kN], L: span [m].

Fixed-end moments for a uniformly distributed load $w$ over the full span:

$$M_A = M_B = -\frac{w\,L^2}{12}$$

w: distributed load [kN/m]. Both supports carry equal hogging moments under a UDL.

Reference mid-span deflections (UDL alone, point load at a = L/2 alone):

$$\delta_{c,w} = \frac{w\,L^4}{384\,EI},\quad \delta_{c,P} = \frac{P\,L^3}{192\,EI}$$

The simulator uses linear superposition and a closed-form expression for a point load at arbitrary a.

What is the Fixed-End Moment Simulator?

🙋
So a "fixed-end moment" is the bending moment that appears at the supports when both ends of a beam are clamped tight, right?
🎓
That is exactly right. When both ends cannot rotate, the supports have to push back with a moment to keep the beam from turning. That moment is called the fixed-end moment, or FEM. If you crank up "Point load P" on the slider above, you will see the two FEM cards M_A and M_B respond immediately.
🙋
Oh, but the values are negative. Is that a bug?
🎓
No, just a sign convention. This tool calls a sagging moment positive, meaning the bottom fiber is in tension under a downward load. At a clamped support the curvature flips, so you get a hogging moment, which is negative under this rule. In the BMD panel you can see the curve dip downward at both supports, which is what hogging looks like. Designers often call it a "negative moment" for the same reason.
🙋
And combining a point load with a UDL just means adding them?
🎓
Yes, linear superposition. The tool computes the FEMs from P alone using $-P a b^2 / L^2$ and the FEMs from w alone using $-w L^2 / 12$, then sums them. With the defaults (L=6, P=30, a=2, w=10) you should see roughly M_A = -57 and M_B = -43 kN.m. The same idea is how engineers combine dead, live, wind and seismic load cases in practice.
🙋
Why is a clamped-clamped beam so much stiffer than a simply supported one?
🎓
Because the support moments fight the curvature. For a UDL, the maximum deflection drops from $5 w L^4 / (384 EI)$ to $w L^4 / (384 EI)$, a factor of five. Try w=10 in the simulator and you should see delta_c near 0.7 mm. A simply supported beam under the same load would deflect about 3.4 mm. The trade-off is that the supports now have to be designed to carry that hogging moment.

Frequently Asked Questions

FEM stands for Fixed End Moment. The abbreviation is identical to that of the Finite Element Method, but in a structural mechanics context it almost always means the support moment of a fully clamped beam. Slope-deflection and moment-distribution (Hardy Cross) methods iteratively reference these moments, so practicing engineers commonly use "FEM tables" listing the values for standard load cases.
This tool focuses on FEMs and reactions, neither of which depends on EI. Only the mid-span deflection delta_c is sensitive to EI, and a representative fixed value keeps the educational focus clear. If you want to study how stiffness changes deflection, pair this tool with the "Beam Deflection and Stress" simulator.
When a is very close to 0 or L the point load sits almost on top of a support, which collapses the visualization and yields trivial moments. Clamping a to 0.1*L through 0.9*L keeps the diagrams legible while still covering the engineering range of interest. To explore boundary behavior, change L and use a proportional position rather than pushing a to its limits.
By linear superposition of two closed-form solutions. For the UDL we use $w x^2 (L-x)^2 / (24 EI)$ at $x = L/2$. For the point load we use the standard clamped-clamped beam deflection $y = P b^2 x^2 (3aL - 3ax - bx) / (6 L^3 EI)$ for $x \le a$ and the mirror form for $x \ge a$. When a = L/2 with only P, this collapses to the well-known $P L^3 / (192 EI)$.

Fixed-End Moment (FEM) Formulas

A fixed-end moment (FEM) is the bending moment produced at the ends of a fully fixed (clamped) beam under load. It is the starting point of the slope-deflection and moment-distribution (Hardy Cross) methods. For span $L$:

LoadFixed-end moment $|M|$Central deflection $\delta$
Distributed load $w$$wL^2/12$ (both ends)$wL^4/384EI$
Central point load $P$$PL/8$ (both ends)$PL^3/192EI$
Point load $P$ at $a$ ($b=L-a$)$M_{AB}=Pab^2/L^2$, $M_{BA}=Pa^2b/L^2$—

Under a distributed load both ends have the same magnitude $wL^2/12$, with $wL^2/24$ (positive) at midspan. For an off-center point load the FEM is larger at the nearer end (and both expressions reduce to $PL/8$ when $a=b=L/2$).

FEM in Slope-Deflection and Moment Distribution

The slope-deflection and moment-distribution (Hardy Cross) methods for indeterminate frames start from FEMs:

  1. Clamp all members: assume every joint is rotationally fixed and compute each member's FEM from the table above.
  2. Unbalanced moment: at each joint, sum the FEMs (the unbalanced moment). Real joints can rotate, so this is released.
  3. Distribute: share the unbalanced moment in proportion to each member's stiffness $K=I/L$ (distribution factor).
  4. Carry over: half of the distributed moment carries over to the far end.
  5. Iterate: repeat distribution and carry-over until the unbalanced moments are negligible, giving the final end moments.

FEM is thus the basic building block of indeterminate structural analysis. Varying the load and its position in this simulator builds intuition for frame analysis.

Real-World Applications

Rigid-frame structures: Reinforced concrete and steel moment frames model each beam as a clamped-clamped element to obtain initial FEMs, then iterate via slope-deflection or moment distribution to find the node rotations and full set of internal forces. The M_A and M_B values shown here are exactly the inputs that start that procedure.

Continuous bridge girders: Multi-span continuous bridges build an FEM table for each span, then redistribute moments across interior supports in proportion to stiffness. Watching how M_A and M_B change with the position a in this simulator gives a direct feel for how load location shifts the moment share between supports.

Machine shafts on two bearings: Spindle and transmission shafts supported by two bearings are treated as clamped-clamped beams. Self-weight enters as a UDL and pulleys or gears as point loads. The resulting bearing moments feed into bending stress checks $\sigma = M c / I$.

Sanity checks for commercial FE solvers: When setting up beam models in Ansys, Abaqus or SAP2000 it is standard practice to verify the boundary conditions by reproducing a simple clamped beam with this kind of hand-calculation. If the solver result and this tool disagree by more than about 5%, the boundary conditions or unit set are almost certainly wrong.

Common Misconceptions and Caveats

The most common pitfall is to assume that "clamped" means perfectly rigid in real construction. Even a steel beam embedded in concrete has some rotational compliance. Engineers introduce a fixity factor between 0 (pinned) and 1 (fully fixed) and interpolate between the pinned and clamped results. A safe design takes the smaller support moment and the larger mid-span moment of the two extremes, since neither idealization is exact.

The second pitfall is believing that FEMs depend on material strength. The expressions $M_A = -P a b^2 / L^2$ and $M_A = -w L^2 / 12$ contain neither E nor I. Fixed-end moments are geometric and load quantities only. Changing EI changes the deflection but not the moments, which is why this tool fixes EI without losing any information about the FEM itself.

Finally, do not mix up sign conventions. This tool calls a sagging moment positive, so the clamped-end moments come out negative. In the BMD panel, negative moments are drawn upward to match the visual intuition of a hogging curve. Other textbooks use the opposite convention, so when discussing results with colleagues always state which convention you are using to avoid confusion.

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How to Use

  1. Enter beam length L (mm) using slL slider, range 500–5000 mm typical for laboratory specimens.
  2. Set point load P (N) via slP and its position A (mm from left support) via slA; leave P=0 for distributed-load-only analysis.
  3. Input distributed load W (N/mm) with slW slider; simulator computes fixed-end moments M_A and M_B, reaction R_A, and mid-span deflection delta_c using 2-node Timoshenko beam elements.
  4. Read results in output stat labels; negative moments indicate clockwise rotation at clamped supports.

Worked Example

Steel I-beam (E=210 GPa, I=8.5e7 mm^4) clamped at both ends, length L=2000 mm, point load P=5 kN at A=1000 mm (mid-span). With W=0, FEM solver yields M_A=−1250 N·m, M_B=−1250 N·m (symmetric), R_A=2.5 kN (upward), and delta_c≈−0.85 mm (downward deflection). Adding distributed load W=2 N/mm increases |M_A| to −2417 N·m and reduces curvature symmetry.

Practical Notes

  1. For clamped-clamped beams, fixed-end moments always oppose applied loads; check sign convention: negative M indicates hogging (sagging if positive in your reference frame).
  2. Mid-span deflection becomes negligible (<0.1 mm) when I exceeds 1e8 mm^4 on steel; use reduced-section members for visible deflection in teaching labs.
  3. Distributed load W acts uniformly; for triangular or trapezoidal loading, decompose into equivalent point load and moment components.
  4. Reaction forces always sum to applied loads; verify R_A + R_B = P + (W × L) as FEM sanity check.