Build a live free body diagram for a simply-supported beam, a cantilever beam, or a 2D truss joint. Adjust loads with the sliders and watch reaction forces solve in real time via static equilibrium: ΣFx = 0, ΣFy = 0, ΣM = 0.
This simulator builds a live free body diagram for three of the most common statics configurations — a simply-supported beam, a cantilever beam, and a two-member truss joint — and solves for the unknown reaction or member forces using the fundamental equations of static equilibrium: ΣFx = 0, ΣFy = 0, and ΣM = 0. Every input change redraws the FBD with correctly scaled, correctly directed force and moment arrows.
A free body diagram isolates a single body (or joint) from everything it touches, replacing every physical support and connection with the force (and, if applicable, moment) it exerts. Once isolated, a rigid body in static equilibrium must satisfy three scalar equations in two dimensions: ΣFx = 0 (horizontal forces balance), ΣFy = 0 (vertical forces balance), and ΣM = 0 about any point (moments balance). For a simply-supported beam (pin + roller), the pin provides two reaction components (but here loaded only vertically, so one is zero) and the roller provides one vertical reaction — exactly two unknowns, solved with ΣFy = 0 and ΣM = 0. A cantilever (fixed support) provides a vertical reaction and a reaction moment — again two unknowns from the same two equations. A truss joint modeled as a pin connecting two two-force members has two unknown member forces, solved directly from ΣFx = 0 and ΣFy = 0 (the method of joints).
The point load P is drawn as a single downward arrow scaled to its magnitude at its chosen location a along the span. The uniform load w is drawn as a row of smaller downward arrows spanning the full beam length, representing a distributed force in force-per-length units. Reactions are drawn as upward arrows at the supports — a pin symbol (triangle) at a simply-supported beam's left end and a roller symbol (circles) at its right end, or a fixed-wall hatch symbol plus a curved moment arrow at a cantilever's built-in end.
Two members meet at a pin joint at angles θ₁ and θ₂ measured from the positive x-axis, plus an externally applied load with components Px and Py. Because each truss member is a two-force member (loaded only at its two end pins, with no distributed load along its length), the internal force in each member must act along the member's own axis. The joint's two equilibrium equations (ΣFx = 0, ΣFy = 0) are solved simultaneously for the two unknown member forces F1 and F2 — this is the method of joints, the standard hand-calculation technique for statically determinate trusses.
A pin support resists translation in both the x and y directions but allows rotation, so it contributes two unknown reaction force components. A roller support resists translation in only one direction (typically vertical) and allows both rotation and horizontal sliding, contributing one unknown reaction force. A fixed (cantilever) support resists translation in both directions and rotation, contributing two force components plus a reaction moment — three unknowns total.
A simply-supported beam has two supports (pin and roller) spread along its length, so the two vertical reactions alone can balance any applied moment by acting at different distances from the load — the couple they form resists the tendency to rotate. A cantilever has only one support, so that single fixed connection must resist not just the vertical shear but the entire overturning moment on its own — hence the reaction moment.
By the sign convention used here, a positive member force means the member is in tension — it is being stretched, and it pulls the joint toward the member. A negative value means the member is in compression — it is being squeezed, and it pushes the joint away from the member. This distinction matters enormously in design because slender compression members can buckle, while tension members generally cannot.
Yes. With only vertical loading, the beam has two unknown reactions (R1 at the pin, R2 at the roller — the pin's horizontal reaction is zero with no horizontal load) and two independent equilibrium equations (ΣFy = 0 and ΣM = 0), so the problem is statically determinate and solvable by statics alone, without considering beam stiffness or deflection.
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