2D Statics Solver Back
Structural Mechanics

2D Statics Solver

Set beam length, support types, and loads to instantly compute reaction forces, shear force diagram (SFD), and bending moment diagram (BMD) in real time.

Beam Setup
Add Load
Reactions
Results
R_A (vertical)
5.00
kN
R_B (vertical)
5.00
kN
Max shear force
5.00
kN
Max bending moment
7.50
kNm
Free body diagram → Shear force diagram → Bending moment diagram
Fbd
Bmd
Theory & Key Formulas

$\sum F_y = 0$: $R_A + R_B = \sum F_i$
$\sum M_A = 0$: $R_B = \sum F_i \cdot a_i / L$
$\frac{dV}{dx}= -q$, $\frac{dM}{dx} = V$

What is Static Equilibrium & Bending?

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What exactly is "static equilibrium" for a beam? The simulator says it calculates reaction forces, but what does that mean?
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Basically, it means the beam isn't moving—it's perfectly still. For that to happen, all the forces pushing up must balance all the forces pushing down, and all the twisting effects (moments) must cancel out. In practice, the supports (like the pin and roller you can select above) provide the "reaction forces" needed to achieve this balance. Try adding a point load with the slider and watch the support reactions change instantly to keep ΣFy = 0.
🙋
Wait, really? So the shear force diagram (SFD) is just a picture of that balance? Why does it jump down at a load?
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Exactly! The SFD shows the internal vertical force at any point along the beam. Imagine making a cut just to the left and right of your applied load. The internal force must jump by the amount of the load to maintain equilibrium. That's why you see a sudden drop. A common case is a shelf bracket: the shear force is high right where the bracket attaches to the wall. Try moving the load position slider and see how the jump in the SFD travels with it.
🙋
Okay, and the bending moment diagram (BMD) is related to the SFD? Why is the moment highest where the shear is zero?
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Great observation! They are mathematically connected. The slope of the BMD at any point equals the shear force at that point ($\frac{dM}{dx}= V$). So, when the shear force crosses zero, the slope of the moment diagram is zero—that's a peak or a valley. For instance, in a simply supported beam with a central load, the maximum bending moment is right in the middle, where the SFD crosses zero. Change the support type to "cantilever" in the simulator and see how the relationship still holds, but the shapes are completely different.

Physical Model & Key Equations

The foundation of statics is the three equilibrium conditions. For a 2D beam, we use two force summations and one moment summation to solve for unknown reactions at the supports.

$$ \sum F_y = 0: \quad R_A + R_B = \sum F_i $$ $$ \sum M_A = 0: \quad R_B \cdot L = \sum (F_i \cdot a_i) $$

Here, $R_A$ and $R_B$ are the vertical reaction forces at the supports, $F_i$ are the applied point loads, $a_i$ is the distance of load $F_i$ from support A, and $L$ is the total beam length. These equations ensure the beam doesn't translate vertically or rotate.

The internal forces within the beam are described by differential relationships between the distributed load (q), shear force (V), and bending moment (M).

$$ \frac{dV}{dx}= -q \qquad \frac{dM}{dx} = V $$

Here, $q$ is the distributed load (force per length, which is zero for point loads), $V$ is the internal shear force, and $M$ is the internal bending moment. The second equation tells us the shear force is the slope of the moment diagram. Integrating these relationships (or using the graphical method) is how the simulator generates the SFD and BMD from your inputs.

Real-World Applications

Bridge Design: Engineers use these exact calculations to determine the size and material of girders in a bridge. The reaction forces tell them how much load the abutments must withstand, and the maximum bending moment dictates how deep the steel I-beams need to be to avoid failure.

Building Floor Joists: The wooden or steel joists supporting a floor are analyzed as beams. The shear force diagram helps locate where to place stiffeners or hangers, and the bending moment determines the required joist depth to prevent sagging under the weight of furniture and people.

Industrial Shelving & Crane Beams: Heavy-duty shelving units and the overhead beams for bridge cranes are classic examples of simply supported beams with point loads. Calculating the reaction forces ensures the uprights are strong enough, and the bending moment analysis prevents the beam from collapsing when a heavy load is moved to its center.

Aircraft Wing Spars: The main structural member of a wing (the spar) acts as a cantilever beam fixed at the fuselage. Bending moment analysis from aerodynamic lift forces is critical for determining material thickness and predicting fatigue life over thousands of flight cycles.

Common Misconceptions and Points to Note

First, understand that being able to calculate reactions does NOT mean the design is complete. The support reactions and maximum bending moment provided by this tool are merely "input values" for "selecting and designing" the member. For example, even if you determine the maximum bending moment is 500 kN·m, you still need separate calculations to decide what size H-beam can support it or how to arrange the rebar in a concrete beam.

Next, pay close attention to interpreting the "units for distributed loads". The tool requires input in "kN/m", which is "the force applied per meter length of the beam". For instance, when considering the weight transferred from a 5m wide floor slab to a single beam, you must multiply the total floor load (kN/m²) by the 5m width to convert it to "kN/m". Getting this wrong can lead to a major error, calculating with 1/5th or 5 times the actual load.

Finally, the practical limitation: "Simply supported beams are not a universal solution". While their calculation is simple, in practice, "deflection" and "vibration" often become problematic. For example, using a simply supported beam for a long office floor joist can cause excessive deflection at the center, leading to cracks, or create a noticeable bounce when people walk. In such cases, you need to increase stiffness by fixing both ends or adding intermediate supports (creating a statically indeterminate structure). Once you've experienced the "basic form" with this tool, start thinking about its "limitations" too.

How to Use

  1. Enter beam length (beamL) in meters, e.g., 5m for a standard industrial beam span
  2. Define support type (v_L): pin, roller, or fixed end conditions
  3. Input point load magnitude (loadF) in kN and its position (loadA) from left support
  4. Click solve to generate shear force diagram (SFD) and bending moment diagram (BMD) with numerical reaction values

Worked Example

Steel I-beam, 6m span, pinned at left, roller at right. Apply 25kN downward load at 2m from left support. Solver calculates: left reaction = 16.67kN upward, right reaction = 8.33kN upward. Maximum bending moment = 33.3kNm at 2m location. SFD shows linear change from +16.67kN to -8.33kN across span.

Practical Notes

  1. For cantilever beams, use fixed support (v_L=fixed) at one end; maximum moment occurs at fixed end, not load point
  2. Distributed loads require segmentation into equivalent point loads at centroid positions
  3. Verify moment equilibrium: sum of (reaction × distance) must equal applied moment for accuracy
  4. Use diagrams to identify critical sections for deflection analysis or material sizing