Beam fixed at both ends with a load at midspan
When both ends of a beam are built in, the structure becomes the stiffest possible single span: neither end may move or rotate, so the beam carries the load with a fraction of the deflection a simply supported beam would show.
This is how real steel frames, machine housings and welded structures behave, and it is also what a beam embedded in concrete effectively looks like. The price is paid at the supports: they develop bending moments of their own, which the connection has to carry.
The calculator below models a stainless steel square tube, 40 mm across with a 2.5 mm wall, spanning 2000 mm between two fixed ends and loaded with 400 N at midspan.
Calculator
This calculator is already set up for the problem above. Change the length, the load, the profile or the material and it recalculates immediately; the address bar keeps a shareable link to your exact configuration.
Input data:
Beam:
Material: Construction steel (Generic)
Profile:
| Ax: 78.54mm2 | Jxx: 981.75mm4 |
| Asy: 78.54mm2 | Iyy 490.87mm4 |
| Asz: 78.54mm2 | Izz: 490.87mm4 |
Simulation results:
Displacements (mm):
| # | x | y | zz |
|---|
Reactions (newtons):
| # | x | y | zz |
|---|
Direct link to this setup: https://beamcalc.net/en/solve/fixed-fixed-mid-load#v1-ATUAAPpEAAAgQgAAIEIAACBAlQEBAgAAAAAAAMhDAAAAAA
How the beam is modelled
- Both ends are rigidly fixed: neither translation nor rotation is allowed at either support.
- A vertical point load of 400 N acts at midspan, pointing down.
- The section is a square hollow tube: 40 mm by 40 mm with a 2.5 mm wall.
- The span is 2000 mm and the material is stainless steel.
How to read the results
The midspan deflection is a quarter of what the same beam would show on simple supports — a fixed-fixed beam is four times stiffer in service, which is the whole reason structures are built in rather than simply supported.
The bending moment changes sign along the beam: it hogs over the supports and sags under the load. For a central point load on a uniform beam both magnitudes happen to be equal, F L / 8, which is a handy rule of thumb.
Each support carries half of the load, and since both ends are fixed the axial restraint is fully effective: a temperature change or a load applied along the beam axis will build up real axial force here, unlike in the pin-and-roller case.
Check the connection before trusting the numbers: the model assumes perfectly rigid fixings, and a real welded or bolted joint always rotates a little, which increases the deflection.
Formulas
| Formula | What it gives |
|---|---|
R_left = R_right = F / 2 | Support reactions; symmetry splits the load evenly. |
M_support = -F * L / 8 | Bending moment at each fixed end, hogging. |
M_mid = F * L / 8 | Bending moment under the load, sagging. |
delta_max = F * L^3 / (192 * E * I) | Deflection at midspan, the largest one. |
FAQ
How much stiffer is a built-in beam than a simply supported one?
Four times for a central point load: the midspan deflection is F L³ over 192 E I instead of F L³ over 48 E I.
Why is the support moment negative?
By convention the beam sags under the load and hogs over the supports. At the fixed ends the curvature reverses, so the bending moment changes sign; the connection has to be designed for that moment, not only for the vertical reaction.
Why are the support and midspan moments equal?
For a single central point load on a prismatic beam they both work out to F L / 8. With any other load pattern the two differ, so treat the equality as a special case rather than a rule.
What if the real supports rotate a little?
Then the beam behaves somewhere between the fixed-fixed and the simply supported cases. Since the deflection of the fixed-fixed model is the lower bound, the real deflection will be larger — a useful conservative check is to compare both models.
More beam examples
- Propped cantilever with a load at midspan A beam clamped at one end and propped by a roller at the other: the classic first step beyond statically determinate beams.