$x \le a:\; y = \dfrac{Pb\,x(L^2-b^2-x^2)}{6EIL}$
UDL w (full span):
$\delta_{max}= \dfrac{5wL^4}{384EI}$
Superposition:
$\delta_{total}(x) = \sum_i \delta_i(x)$
Apply up to 3 loads — concentrated force, uniformly distributed load, or concentrated moment — to a simply-supported beam. The deflection curve and bending moment diagram update in real time, along with max deflection, support reactions and end slopes.
The fundamental equation governing beam bending is the Euler-Bernoulli beam theory, which relates the beam's deflection $y(x)$ to its bending moment $M(x)$, flexural rigidity $EI$, and load.
$$EI \frac{d^2 y}{dx^2}= M(x)$$Here, $E$ is the material's Young's Modulus (stiffness), $I$ is the beam cross-section's second moment of area (shape stiffness), and $M(x)$ is the internal bending moment, which depends on the applied loads. Solving this differential equation for different load cases gives the deflection formulas.
For a simply-supported beam of length $L$, the deflection at any point $x$ due to a single point load $P$ applied at a distance $a$ from the left support (with $b = L - a$) is given by two piecewise equations. This is a classic solution you can superimpose with others.
$$ \text{For }x \le a: \quad y(x) = \frac{P b x}{6 E I L}(L^2 - b^2 - x^2) $$ $$ \text{For }x > a: \quad y(x) = \frac{P a (L-x)}{6 E I L}(L^2 - a^2 - (L-x)^2) $$The total deflection under $n$ loads is the linear sum: $\delta_{total}(x) = \sum_{i=1}^{n} \delta_i(x)$. This is the core of the superposition method visualized in the simulator.
Structural Engineering & Building Design: Engineers use superposition daily to check floor joist deflections under complex loading, such as the combined weight of furniture (distributed load), a heavy safe (point load), and partition walls. Ensuring deflections are within code limits (like L/360) prevents cracked ceilings and floor vibrations.
Bridge Design and Rating: When assessing a bridge's capacity, engineers model the simultaneous effect of the bridge's own dead load (UDL), the weight of vehicles (moving point loads), and wind pressure. Superposition allows them to calculate the worst-case deflection scenario efficiently for safety certifications.
Industrial Machine Design: The bed of a CNC machine or a conveyor support frame must resist deflection to maintain precision. Designers superimpose deflections from the motor's weight (point load), the weight of the workpiece (another point load), and the frame's own weight to ensure machining accuracy is not compromised.
Construction Planning & Scaffolding: Before placing a heavy concrete pump or storage containers on temporary supports, construction planners use superposition to calculate the combined deflection from these concentrated loads and the uniform load of the scaffolding planks. This ensures stability and safety for workers underneath.
First, let's firmly grasp the premise of this tool. "Linear elasticity and small deformations" is the fundamental principle. For example, it's dangerous to directly apply results calculated for steel members to rubber or plastic parts that undergo large deformations. If the material is non-linear, superposition no longer holds.
Next, the importance of the values for the "Second Moment of Area I" and "Young's Modulus E" is often overlooked. The tool provides default values, but in actual design, these are critical. For instance, the I for a rectangular cross-section 100mm wide and 200mm high is $$I = \frac{b h^3}{12} = \frac{100 \times 200^3}{12} = 66.7 \times 10^6 \, \text{mm}^4$$. However, the moment you orient the beam sideways (200mm wide, 100mm high), I plummets to about 16.7×10⁶ mm⁴. This would make the calculated deflection four times larger! Always cross-check input values against the actual part.
Also, there's a pitfall in the "maximum of 3 loads" setting. In practice, having four or more loads is common. In such cases, "equivalent transformation" is key. For example, if several small concentrated loads are close together, you can sum them into a single concentrated load or approximate them as a distributed load. Conversely, if you need to examine part of a long distributed load in detail, extract that section and create a separate model. Think of this tool as a way to get a feel for "how to combine basic parts."
Simply-supported steel beam: L = 4 m, E = 200 GPa, I = 7850 cm⁴ (HEB 200). Apply F₁ = 15 kN at midspan (x = 2 m) and distributed load q = 5 kN/m across full span. Superposition yields maximum deflection δmax ≈ 12.4 mm at x = 2.05 m. Individual contributions: point load δ = 10.7 mm, distributed δ = 6.8 mm; combined effect shows nonlinear interaction near support regions.