Unit Load Method Simulator — Cantilever Beam Deflection and Rotation
Visualize the tip deflection delta and rotation theta of a cantilever beam with the unit load method (Mohr integral). Change the length, flexural rigidity, point load and uniform load to feel virtual work in action.
Parameters
Beam length L
m
Flexural rigidity EI
kN·m²
Tip point load P
kN
Uniform load w
kN/m
P and w act simultaneously; delta and theta are obtained by superposition. The deflected shape is drawn with exaggeration for clarity.
Results
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Deflection from P, delta_P = PL³/(3EI)
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Deflection from w, delta_w = wL⁴/(8EI)
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Total tip deflection delta_total
—
Total tip rotation theta_total
Cantilever Beam Model and Deflected Shape
Left end = fixed support / blue curve = deflected shape (exaggerated) / red arrow = tip point load P / orange arrows = uniform load w
Real Moment M(x) and Unit Virtual Moment m(x)
Solid blue = M(x) (real moment from P + w) / dashed yellow = m(x) = -(L-x) (virtual moment from a unit tip load)
Theory & Key Formulas
The unit load method places a unit virtual load at the point and direction of the desired displacement and integrates the product of its bending moment and the real bending moment over the beam.
Tip deflection delta (use a unit point load as the virtual load):
Cantilever with a tip point load P: real $M = -P(L-x)$, virtual $m = -(L-x)$, so
$$\delta_P \;=\; \frac{P L^3}{3 E I}$$
Cantilever with a uniform load w: real $M = -\tfrac12 w(L-x)^2$, so
$$\delta_w \;=\; \frac{w L^4}{8 E I}$$
The tip rotation theta uses a unit virtual moment $m = -1$:
$$\theta \;=\; \frac{P L^2}{2 E I} + \frac{w L^3}{6 E I}$$
When several loads act at once, the individual deltas and thetas simply add by superposition.
What is the Unit Load Method Simulator
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I had memorized beam deflection formulas like $PL^3/(3EI)$ and $5wL^4/(384EI)$, but where do they actually come from?
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One source is the "unit load method", also known as Mohr's integral. Roughly: at the place and direction where you want to know the deflection, put a fictitious unit load, multiply its bending moment $m(x)$ by the real bending moment $M(x)$, and integrate over the whole beam. As a formula, $\delta = \int_0^L M\,m / (EI)\,dx$. If you move P and w in the simulator above, you can feel how both classical formulas drop out of this integral.
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What does this "fictitious load" actually mean physically? It feels a bit creepy.
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I get that. It is a calculation trick that uses the principle of virtual work. The load is not really applied — you only invent it to create the "other half of a product" needed to extract one specific displacement. Want the tip deflection? Put a unit force at the tip. Want a rotation? Put a unit moment there. In the simulator, $m(x) = -(L-x)$ is exactly the virtual moment for a unit tip force.
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With the defaults the total deflection is 321 mm, which is huge. Is that realistic?
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Good catch. EI = 1000 kN·m² is a textbook value picked to make the numbers easy, and it is small for a real steel beam. A real beam over a 3 m span would have EI in the tens of thousands of kN·m² and the deflection would be only a few millimetres. Crank the EI slider up and watch the deflection collapse — $\delta$ is inversely proportional to EI.
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You compute P and w separately and just add them. Is that always allowed?
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As long as the response is linear elastic, the principle of superposition lets you. Both terms appear inside the same $M(x)$ which gets multiplied by the same $m(x)$ and integrated, so the result is automatically $\delta_P + \delta_w$. In practice we solve self-weight, live load and seismic load separately and add them later — the same idea. Once material yields or large deformations appear, superposition no longer holds, so keep this rule inside the elastic range.
Frequently Asked Questions
A simply supported beam works directly. For example, the mid-span deflection under a central point load gives the classical $PL^3/(48EI)$: both the real and the virtual moment diagrams are triangular, and the integral over 0 to L is straightforward. A fixed-fixed beam is statically indeterminate, so you first solve for the redundant reactions with the force method and then apply the unit load method to find any deflection or rotation.
Yes. The generalized unit load method reads $\delta = \int (Mm/EI + Vv/(GA_s) + Nn/(EA))\,dx$, summing the bending, shear and axial contributions. For slender beams with span-to-depth ratio above 10 the bending term dominates, so the bending-only formulation used in the simulator is sufficient. For trusses, short beams or Timoshenko beams the other terms cannot be ignored.
Both belong to the family of energy methods but the viewpoint differs. Castigliano's second theorem says the partial derivative of the strain energy with respect to a load equals the displacement at the load's point, $\delta = \partial U / \partial P$. The unit load method comes from the principle of virtual work and is written as the integral $\delta = \int Mm/(EI)\,dx$. They are mathematically equivalent and often lead to the same formula for a cantilever; you pick whichever is more convenient.
Inside commercial FEA codes the stiffness equation $[K]\{u\} = \{F\}$ is solved directly, so users never crank out a unit load integral by hand. But the spirit of the method underlies influence lines, sensitivity analysis and adjoint methods for structural optimization. As a sanity check against an FEA result, comparing with a hand calculation by the unit load method is still one of the most reliable verifications available.
Real-World Applications
Hand checks of bridge and building structures: Bridge girders, building beams, columns and frames often need a hand-calculated cross-check of their displacements in design practice. The unit load method gives the deflection or rotation at any chosen point directly by superposing self-weight, live load and seismic load, which makes it perfect for verifying FEA output. It is also a standard tool for checking deflection limits such as span over 300.
Stress analysis of indeterminate structures: Continuous beams, frames and truss bridges are solved by combining the force method with the unit load method. Treat the redundants as unknowns, write the compatibility equations for their displacement using the unit load method, solve the simultaneous system, and the reactions and section forces follow. Before computers this was the dominant approach to structural design and it is still the backbone of textbooks on railway and highway bridge design.
Construction of influence lines: In the design of bridges under moving loads, influence lines show the position of the load that produces the maximum stress at each point. The variation of section forces or displacements as a unit load travels across the bridge is itself the influence line, and the unit load method is its theoretical foundation. The concept is still essential for railway bridges and crane runways.
Core of structural mechanics education: In university courses on structural and material mechanics, the unit load method is the canonical example of an energy method. The chain — principle of virtual work, unit load method, Castigliano's theorems, matrix displacement method — leads naturally to the theoretical basis of the finite element method. Doing it once by hand keeps FEA from being a black box.
Common Misconceptions and Cautions
The most common misconception is to think that "the unit load method is a special-purpose formula for the tip deflection of a cantilever". In reality, as long as you place the right virtual load at the right point and direction, you can compute the displacement, rotation or relative displacement of any point in any statically determinate structure. For a mid-span deflection put a unit point force at mid-span; for an end rotation put a unit moment there. The simulator only shows tip displacement for clarity, but the principle is far more general.
The next common reaction is to feel that "delta_total is too large, the calculation must be wrong". With the default values, delta_total is about 321 mm, more than 10 percent of $L = 3$ m — catastrophic for a real structure. The arithmetic is correct, however; the reason is that EI = 1000 kN·m² is deliberately tiny, a value used in textbooks to keep the numbers easy to read. A real steel beam (an H-shape like H-400) has EI in the tens or hundreds of thousands of kN·m² and the deflection under the same loads is orders of magnitude smaller. Move the EI slider up to 10000 and watch $\delta$ drop to one tenth.
Finally, beware the trap of getting the sign of $m(x)$ wrong. You must enforce a single, consistent sign convention for both the real moment $M$ and the virtual moment $m$ — for example "tension on the bottom is positive". For a cantilever with a downward tip force, $M = -P(L-x)$ and $m = -(L-x)$ are both negative, but their product $M\cdot m = +P(L-x)^2$ is positive, and the integral is also positive, giving a downward deflection. Switching the convention mid-way through flips the sign and yields the impossible result of an upward deflection. Pick a convention at the start and stick with it to the end.