Propped cantilever with a load at midspan
A propped cantilever is a beam that is fixed at one end and simply supported at the other. It is the natural next step after the pin-and-roller beam, because equilibrium alone is no longer enough: the structure is statically indeterminate, and the load split between the two supports depends on how stiff the beam is.
This arrangement is everywhere in practice — a shelf bracket with an added stay, a machine shaft clamped at one end and carried by a bearing at the other, a wall-mounted arm with an end stop.
The calculator below is set to a 30 mm steel rod over 2000 mm, fixed at the left end, roller-supported at the right one, 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/propped-cantilever-mid-load#v1-AQkAAPpEAADwQZQBAQIAAAAAAADIQwAAAAA
How the beam is modelled
- The left end is rigidly fixed: it carries vertical force, horizontal force and a bending moment.
- The right end rests on a roller: it carries vertical force only and permits rotation and axial movement.
- A vertical point load of 400 N acts at midspan, pointing down.
- The beam is a solid steel rod, 30 mm in diameter and 2000 mm long.
How to read the results
The fixed end does most of the work: it takes eleven sixteenths of the load, while the roller takes only five sixteenths. The split is not equal, and it is not the simple lever rule either, because the fixed end also carries a moment.
The bending moment changes sign along the beam. It is negative — hogging, pulling the top fibres apart — at the fixed end, and positive — sagging — under the load. Both magnitudes matter for a real beam, since the steel has to cope with both.
Deflection is much smaller than for the same beam on two simple supports, and the maximum no longer sits exactly at midspan: it shifts slightly towards the roller, because the fixed end suppresses the rotation there.
Because the solution depends on stiffness, it also depends on the cross section and the elastic modulus: unlike a pin-and-roller beam, this one cannot be solved from geometry alone.
Formulas
| Formula | What it gives |
|---|---|
R_roller = 5 * F / 16, R_fixed = 11 * F / 16 | Support reactions. The fixed end carries the larger share. |
M_fixed = -3 * F * L / 16 | Bending moment at the fixed end, negative because it hogs. |
M_mid = 5 * F * L / 32 | Bending moment under the load at midspan. |
delta_mid = 7 * F * L^3 / (768 * E * I) | Deflection under the load. The true maximum is about two percent larger and sits just past midspan. |
FAQ
Why is this beam statically indeterminate?
Because there are more support reactions than equilibrium equations can determine: a vertical force and a moment at the fixed end plus a vertical reaction at the roller cannot be found from force and moment balance alone, so the deformation of the beam has to be used as an extra condition.
Why does the fixed end carry more than half the load?
Because it is also the stiffer end: it restrains rotation, so the beam leans on it more heavily than a simple lever rule would suggest.
What happens if I remove the roller?
You get a plain cantilever, and the deflection under the same midspan load increases several-fold: the beam loses both the end support and the rotation restraint that went with it.
Is the maximum deflection at midspan?
No. It is about two percent larger than the midspan value and sits slightly towards the roller side, where the beam is less restrained.
More beam examples
- Beam fixed at both ends with a load at midspan A built-in beam clamped at both supports: the stiffest single-span arrangement, and the one with the highest end moments.