Propped Cantilever (UDL)

stressmass-cost

Take a cantilever — fixed into a wall at one end, free at the other — and put a simple prop (a roller) under the free end so it can no longer sag. That single extra support changes the problem’s character, not just its numbers. A plain cantilever is statically determinate: two equilibrium equations, two unknown reactions, done. Add the prop and there are now three reactions to find — the wall’s vertical force $R_A$, the wall’s fixing moment $M_A$, and the prop’s force $R_B$ — but statics still gives only two equations. The beam is statically indeterminate to the first degree, and equilibrium alone cannot finish it.

This is the gap that half of a structures course exists to close, and it’s the first THING in this catalog that lives on the far side of it. You see propped cantilevers everywhere a stiff member is given one helper support: a shelf bracket with a diagonal tie to the wall, a balcony beam propped at its tip, a pipe run clamped at a wall and resting on a hanger, a machine bed bolted down at one end and shimmed at the other.

The missing equation is compatibility

The third equation does not come from force balance — it comes from geometry. Remove the prop and let the cantilever sag freely: the uniform load $w$ pushes the free end $B$ down by $w L^4/8EI$. Now push $B$ back up with the prop’s force $R_B$ alone: it lifts $B$ by $R_B L^3/3EI$. The real prop holds $B$ exactly on the wall’s line — its deflection is zero — so those two must cancel. That compatibility condition is the third equation, and with it the three reactions are determined:

$R_A = \tfrac{5}{8} w L, \qquad R_B = \tfrac{3}{8} w L, \qquad M_A = \tfrac{1}{8} w L^2$

The build does not hand you these formulas to trust. It certifies that the three relations (two equilibrium, one compatibility) form a system that is linear in the three unknowns, solves that $3\times3$ system exactly at build time, and then checks the solution back through every relation — the same total verification an ordinary closed form gets. This is the solveLinear capability, and the propped cantilever is its reference case, exactly as the eccentric column is for bracketed root-finding. What the machine proves and what still rests on a book is spelled out on the verification page.

The reactions do not care what the beam is made of

Look closely at the compatibility step in the derivation below. Both deflections carry the same flexural rigidity $EI$, and because the beam is prismatic (uniform section and material), $EI$ divides out completely — the line $\tfrac{1}{8}wL^4 = \tfrac{1}{3}R_B L^3$ has no $E$ and no $I$ in it. That cancellation is machine-checked, and it is the whole lesson: the reactions are material-blind. Switch the material in the widget from A36 steel to Ti-6Al-4V titanium and watch $R_A$, $R_B$, and $M_A$ not move at all, even as:

• Young’s modulus $E$ changes the midspan deflection $\delta_{mid} = w L^4/192EI$ — titanium is about half as stiff ($E \approx 110$ vs $200$ GPa), so it deflects nearly twice as far.
• Yield strength $\sigma_y$ changes the safety factor $\mathrm{SF} = \sigma_y/\sigma_{max}$ — titanium’s higher yield lifts the margin.
• Density $\rho$ changes the mass — titanium nearly halves it.

A redundant support redistributes load by geometry, not by material. That is a genuinely non-obvious result the first time you meet it: in a determinate structure the load path is fixed by statics; here the “sharing” between wall and prop is set purely by the shape of the deflection curves, which for a uniform beam are material-independent. (Make the beam non-prismatic — a tapered section, or two materials spliced together — and $EI$ no longer cancels; the reactions would then depend on the stiffness distribution. That is the composite-bar story, a later THING.)

What governs, and the sign convention

Throughout: $x$ runs from the wall $A$ to the prop $B$, downward load and deflection are positive, $R_A$ and $R_B$ are the upward reactions, and $M_A$ is the magnitude of the hogging fixing moment at the wall. The bending moment is largest at the wall: $|M|_{max} = M_A = wL^2/8$ (hogging) comfortably beats the sagging peak of $9wL^2/128$ at $x = 5L/8$, so the wall’s outer fiber is where first yield happens and what the safety factor guards. The clean midspan deflection $wL^4/192EI$ is quoted here; the true maximum is only a hair larger, $\approx wL^4/185EI$ near $x = 0.578L$.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget’s safety factor guards one thing — first yield at the wall’s outer fiber, where the hogging moment $M_A = wL^2/8$ peaks. A real propped cantilever has more exits, and the most interesting ones are specific to it being statically indeterminate:

• Prop settlement — the assumption that can quietly move every number. The whole solution rests on one geometric premise: the prop holds $B$ exactly on the wall’s line, zero deflection. Let the prop settle by even a small amount $\Delta$ and the compatibility equation changes to $wL^4/8EI - R_B L^3/3EI = \Delta$ — the reactions redistribute, $M_A$ can swing sharply, and the “material-blind” result stops being material-blind (now $EI$ does appear, multiplied by the settlement). In a determinate beam a support that settles just goes along for the ride; in an indeterminate one, support movement is a load case in its own right. This is why indeterminate structures are simultaneously stiffer and more temperamental.

• Yielding at the wall, then moment redistribution. Because the peak moment is at the built-in end, that’s where a plastic hinge forms first. Past that point the beam does not collapse — the wall hinge caps its moment and the span carries more, a redistribution the elastic formulas on this page cannot see. The σ_max > σ_y warning marks where they stop telling the truth.

• Loss of the prop. If the prop is a slender strut it can buckle, and if it is a tension tie it can yield or pull out of its anchor. Lose the prop entirely and the structure reverts to a plain cantilever: the free end that was held on the wall’s line now sags to the full cantilever value $wL^4/8EI$ (from zero), and the wall moment jumps from $wL^2/8$ to $wL^2/2$ — four times larger. The redundancy that makes the beam stiff is exactly what a single failed support removes.

• Fatigue at the fixed end. A propped cantilever that vibrates cycles its largest stress at the wall, at a re-entrant corner. Fatigue strength sits far below static yield, and nothing on this page addresses cyclic life.

• Excessive deflection or lateral-torsional buckling. A tall, narrow section ($h \gg b$) can twist and flop sideways before it yields — a stiffness failure, not a strength one — and a serviceability deflection limit often governs long before first yield does.

• Short, deep beams and brittle materials. For $L < 10h$ (the validity warning) shear deflection, neglected by Euler–Bernoulli theory, is no longer small. And a brittle material such as gray cast iron has no yield plateau at all — rate it against ultimate strength with a generous factor instead of the yield-based safety factor shown here.

Sources

• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — ch. 10 (Statically Indeterminate Beams): the force (flexibility) method, propped-cantilever reactions by superposition; §5.5 flexure formula; App. E section properties.
• Hibbeler, R. C., Mechanics of Materials, 10th ed., Pearson, 2017 — ch. 12 (Deflection of Beams and Shafts): statically indeterminate beams by the method of superposition; equilibrium of the propped cantilever.