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Structural Analysis Simulator

Propped Cantilever Beam Simulator — One-Degree Indeterminate Beam

For a beam fixed at the left and pinned at the right, with UDL and a central point load, this tool visualizes the fixed-end moment, both support reactions, the maximum deflection, the SFD and the BMD in real time.

Parameters
Beam length L
m
Load type
Load magnitude w
kN/m
Flexural rigidity EI
kN·m²

Left end fully fixed, right end pinned. The load ramps up from 0 so you can watch the deflection, reactions, SFD and BMD develop in real time. In “Point P” mode you can also move the load position a.

Live results (updated with load factor λ)
—
Prop reaction R_B [kN]
—
Fixed-end moment M_A [kN·m]
—
Max deflection δ_max [mm]
—
Max sagging moment M+ [kN·m]
Deflection, SFD and BMD — real-time animation
Load factor λ = 0.00
Deflection No prop (comparison) SFD BMD Fixed-end M / load

Top = deflection curve (dashed = cantilever without prop) / Middle = SFD / Bottom = BMD. Fixed-end moment, max sagging moment and inflection point are annotated.

Theory & Key Formulas

For a beam fixed at the left and pinned at the right (one-degree statically indeterminate), the redundant reaction is found from the zero-deflection condition at the right support. For a UDL $w$ over the full length:

$$M_A = -\frac{wL^2}{8},\quad R_A = \frac{5wL}{8},\quad R_B = \frac{3wL}{8}$$

The maximum positive moment occurs at $x = 5L/8$ from the fixed end ($3L/8$ from the pin):

$$M_{+,\max} = \frac{9\,wL^2}{128}$$

The maximum deflection occurs at $x \approx 0.4215\,L$:

$$\delta_{\max} = \frac{wL^4}{185\,EI}$$

For a central point load $P$ at midspan:

$$M_A = -\frac{3PL}{16},\quad R_A = \frac{11P}{16},\quad R_B = \frac{5P}{16}$$

$w$ and $P$ can be combined by linear superposition. Compared with a simply supported beam, the fixed end reduces the maximum deflection by about 2.6 times and the maximum positive moment by about 1.8 times.

What is the Propped Cantilever Beam Simulator

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Is a "propped cantilever" really different from a regular cantilever? The names sound so similar.
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A propped cantilever is a regular cantilever with one extra support (a "prop") added at the free end. The left end is fully fixed and the right end is a pin (or roller). There are four unknown reactions (M_A, R_A, R_B and a horizontal reaction) but only three equilibrium equations, so we add one compatibility condition. That makes it a one-degree statically indeterminate beam. In the simulator the hatching on the left and the triangle on the right show those supports.
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Wait, indeterminate? Why can't equilibrium alone solve it once we add a support?
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Equilibrium gives only three equations (sum of vertical forces, sum of horizontal forces, sum of moments), and we have four unknown reactions. So we are one equation short. The fix is to add "the deflection at the right support equals zero". If you imagine removing the right support, the cantilever's free-end deflection under w is $wL^4/(8EI)$. Adding only R_B upward gives a deflection $R_B L^3/(3EI)$. Setting them equal gives $R_B = 3wL/8$ immediately. This is called the unit load method or compatibility condition.
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I see! With w=15 kN/m and L=8 m, M_A becomes -120 kN·m in the simulator. Is that a large number?
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For the same conditions, a simply supported beam has a maximum moment of $wL^2/8 = 120$ kN·m at midspan. The fixed-end moment of the propped cantilever has the same magnitude, -120 kN·m. The interesting thing is that the maximum positive moment drops to $9wL^2/128 \approx 67.5$ kN·m. Look at the red curve in the BMD: it dives down at the left and rises into a hump at $5L/8 = 5$ m from the fixed end (i.e. $3L/8$ from the pin). The design effectively trades a large negative moment at the fixed end for a smaller positive peak in the middle.
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And the deflection is only 4.15 mm — much smaller than a simply supported beam, right?
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The maximum deflection of a simply supported beam is $5wL^4/(384EI)$. With the same EI of 80000 kN·m² it works out to about 10.8 mm. The propped cantilever gives $wL^4/(185EI) = 4.15$ mm, about 2.6 times smaller. Because the right support shares the load, both reactions and deflections are reduced — that is the main benefit of going one-degree indeterminate. Add a point load P and the results change via superposition; try the P slider to feel it.

Frequently Asked Questions

As long as the beam is linearly elastic (Hooke's law), the deformations are small and the support conditions do not depend on the load, linear superposition holds rigorously even for indeterminate beams. For this propped cantilever, the solutions for UDL w and central point load P (M_A, R_A, R_B and deflection) can each be obtained separately and simply added to get the response to the combined load. Once the material yields, superposition no longer holds, so the assumption must always be checked when designing for ultimate limit states.
The propped cantilever is asymmetric (fixed on the left, pinned on the right), so the deflection curve is not symmetric either. The negative moment at the fixed end shifts the peak of the deflection toward the fixed end, landing at $x \approx 0.4215L$ rather than at the midspan. This value is obtained by differentiating the deflection function $w(x)$ and setting it to zero, and it is tabulated in standard beam deflection tables.
For a UDL, the differential relation $dV/dx = -w$ makes the shear decrease linearly from R_A = 5wL/8 at the left to -R_B = -3wL/8 at the right. The zero-crossing is at $x = R_A/w = 5L/8$ from the fixed end, which coincides with the location of the maximum positive moment in the BMD: this is the direct consequence of $dM/dx = V$. When a point load P is added at midspan, the SFD jumps vertically by P at $x = L/2$.
Typical examples include a cantilever canopy stiffened by an outrigger column at its tip, a bridge overhang propped by a temporary bent during erection, the base of a tower-crane jib with a tip pendant cable, and a cantilever balcony strengthened by an external diagonal brace. Because both deflection and stress are significantly smaller than a pure cantilever, this configuration is adopted for reinforcement, weight reduction or improved vibration performance.

Solving a Propped Cantilever (1st-Degree Indeterminate)

A propped cantilever is fixed at one end and simply supported (propped) at the other, making it statically indeterminate to the first degree (one more reaction than equilibrium alone can solve). A compatibility condition is needed.

Force method: choose the prop reaction $R_B$ as the redundant and first remove the prop, leaving a cantilever. The downward deflection $\delta_0$ at the prop due to the load and the upward deflection $\delta_R$ due to $R_B$ must cancel ($\delta_0 = \delta_R$, i.e. zero deflection at the prop), which gives $R_B$; equilibrium gives the rest.

Propped Cantilever Formulas (Fixed End + Support)

Representative values for span $L$, fixed end at the left, prop (simple support) at the right:

LoadFixed-end momentReactionsMax positive moment / deflection
Distributed load $w$$M_A=-wL^2/8$$R_A=5wL/8$, $R_B=3wL/8$$M^+=9wL^2/128$ (at $5L/8$), $\delta_{max}\approx wL^4/185EI$
Central point load $P$$M_A=-3PL/16$$R_A=11P/16$, $R_B=5P/16$$M^+=5PL/32$ at the load

For the same distributed load, the fixed-end moment is $0$ for a simply supported beam, $wL^2/2$ for a cantilever, and $wL^2/12$ for a fixed-fixed beam. The propped cantilever ($wL^2/8$) lies between these and greatly reduces deflection and moment versus a cantilever — the benefit of adding a prop.

Real-World Applications

Stiffening canopies and balconies in architecture: Cantilever canopies and balconies projecting from a building can show conspicuous deflection over time or under changed loading, and an auxiliary column (a "prop") is often installed at the tip to revive them. Going from a pure cantilever to a one-degree indeterminate propped cantilever cuts the tip deflection by about 2.6 times without increasing the fixed-end moment. The reduction in δ_max shown by this simulator illustrates that effect directly.

Continuous bridge supports and temporary erection bents: Continuous-span bridges use intermediate piers to limit midspan stresses. During cantilever erection, a temporary bent is sometimes placed under the projecting end to form a propped cantilever, allowing the overhang to be extended while stresses are controlled. The split between the fixed-end moment $-wL^2/8$ and the positive moment $9wL^2/128$ feeds directly into the construction stress check.

Machine frames and sensor arms: Long sensor arms in measurement equipment or machine tools suffer from high stress at the cantilever root. Adding a tiny roller support at the tip turns the arm into a propped cantilever, raising the first natural frequency and reducing dynamic deflection during acceleration. Increase the EI slider and you can see the deflection δ_max drop inversely proportional to EI.

Tower crane jibs and stay cables: The horizontal jib of a tower crane is held up by counter-jib and pendant cables that effectively pull its tip upward. The geometry and loading are not exactly the simulator's model, but the idea of stiffening a cantilever with a remote support is the same, and this rationalization of support conditions is widely used to carry the self-weight of the jib together with the hoisted load.

Common Misconceptions and Cautions

The most common misconception is to overestimate the fixed-end moment compared with a simply supported beam. The fixed-end moment of a propped cantilever under UDL w is $-wL^2/8$, exactly the same magnitude as the maximum moment $wL^2/8$ of a simply supported beam with the same loading. People often think "fixing one end concentrates all the moment there", but in fact the positive moment is reduced to $9wL^2/128 \approx wL^2/14.2$ in exchange for the fixed end carrying the same magnitude — the total demand does not increase. The peak-to-trough ratio of the red BMD curve in the simulator makes this clear.

The next common error is to assume superposition does not apply to indeterminate beams. Under linear elasticity, small deformations and load-independent supports, superposition holds exactly even for indeterminate beams. Try moving the w slider and the P slider separately, and then compare with both loads applied together: each reaction, moment and deflection equals the sum of the individual solutions. On the other hand, when the material yields or a support lifts off, the problem becomes nonlinear and superposition breaks down.

Finally, take care not to assume that "because the right end is pinned, both deflection and rotation are zero there". A pin (roller) support constrains the vertical displacement only and leaves the rotation (slope) free. The BMD value at the right end x = L is exactly zero because the bending moment at a pin support is necessarily zero. The slope $dw/dx$ at the right end, however, is finite. This is different from a fully clamped end, so the distinction between support conditions must always be checked first.

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How to Use

  1. Set cantilever length L (meters) using the length slider; typical structural spans range 2–6 m.
  2. Input distributed load W (kN/m) and point load P (kN) via respective sliders; industrial beams often carry 10–50 kN/m.
  3. Adjust flexural rigidity EI (kN·m²); for steel IPE 300, EI ≈ 8,550 kN·m²; for concrete, EI ≈ 12,000–15,000 kN·m².
  4. Observe the three indeterminate reactions: fixed-end moment M_A, support reaction R_A, and prop reaction R_B displayed in real time.
  5. Review bending moment diagram, shear force diagram, and deflection curve to validate design constraints.

Worked Example

Consider a propped cantilever with L = 4 m, w = 12 kN/m, P = 8 kN at midspan, EI = 8,550 kN·m². With superposition the simulator calculates: M_A = −(wL²/8) − (3PL/16) = −24 − 6 = −30 kN·m; R_A = 5wL/8 + 11P/16 = 30 + 5.5 = 35.5 kN (fixed support); R_B = 3wL/8 + 5P/16 = 18 + 2.5 = 20.5 kN (prop reaction). Maximum deflection δ_max ≈ wL⁴/(185·EI) + 0.00932·P·L³/EI ≈ 2.5 mm, occurring near x ≈ 0.42L (the prop end deflection is zero).

Practical Notes

  1. Propped cantilevers are statically indeterminate to the first degree; removing the prop converts it to a simple cantilever with M_A = −wL²/2, doubling the bending moment and tripling deflection.
  2. For composite beams or timber joists, adjust EI based on modular ratio; laminated veneer lumber EI ranges 3,000–6,000 kN·m² depending on grade and section.
  3. Monitor shear force jump at point load P; if P exceeds R_B capacity, the prop lifts off and the structure becomes unstable.
  4. Deflection limits for floor systems typically cap δ ≤ L/360; for cantilever balconies, enforce δ ≤ L/180 due to visual perception.