Draw a beam with pins, rollers or fixed supports, add point loads, distributed loads (uniform, triangular or trapezoidal) and moments, and instantly get the reactions, the axial, shear and bending moment diagrams, the elastic curve and the normal and shear stresses. It solves statically determinate and indeterminate beams, in 2D and 3D, with internal hinges.
With the equilibrium equations ΣF = 0 and ΣM = 0; indeterminate beams add deformation compatibility conditions. The calculator shows every equation.
Where the shear force crosses zero or at a concentrated load; the tool marks it on the diagram.
Yes, it solves continuous beams, fixed beams and beams with several supports.
A beam is a long member that carries loads perpendicular to its axis and takes them to its supports by bending. Analyzing it means answering four questions: what forces the supports exert (reactions), what internal forces act at each section (shear force V and bending moment M), how much it deforms (slope and deflection) and what stresses the material carries. This guide summarizes the theory the calculator uses and works a complete example by hand.
In the plane, each support prevents certain motions and therefore adds unknowns: a pin adds two (horizontal and vertical force), a roller one (the force perpendicular to its surface) and a fixed support three (two forces and a moment). The three planar equilibrium equations, ΣFx = 0, ΣFy = 0 and ΣM = 0, can find at most three unknowns; each internal hinge adds one more equation (the moment is zero there). If r is the number of reactions and h the number of internal hinges, the degree of indeterminacy is
n = r − (3 + h)
With n = 0 the beam is statically determinate and equilibrium is enough. With n > 0 it is statically indeterminate: n compatibility equations are needed (for example, zero deflection at an intermediate support). With n < 0 the beam is a mechanism and cannot be in equilibrium under arbitrary loads. The calculator shows this classification in the first step of the solution.
For overall equilibrium, a distributed load is replaced by a force equal to the area under its diagram, applied at the centroid of that area. A uniform load w over a length ℓ is equivalent to w·ℓ at the middle of the span; a triangular load going from 0 to w is equivalent to w·ℓ/2 applied ℓ/3 from the end with the larger intensity. This replacement is only valid for the reactions: to draw V and M the actual shape of the load must be kept.
Cutting the beam at a distance x and writing equilibrium of the left part gives two relationships that let you draw the diagrams without redoing equilibrium at every point (with w positive downward):
dV/dx = −w(x) dM/dx = V(x)
Practical consequences: over an unloaded segment V is constant and M is a straight line; under a uniform load V is a line and M a parabola; a point load causes a jump in V and a kink in M; an applied moment causes a jump in M. The moment is a maximum or minimum where V crosses zero or where V changes sign under a point load. The calculator uses the convention of many textbooks: M is positive when it compresses the top fiber (a "smiling" beam).
For slender beams of linear elastic material and small deformations, Euler–Bernoulli theory applies: plane sections remain plane and perpendicular to the deformed axis. The deflection v(x) satisfies
E·I · d²v/dx² = M(x)
where E is the modulus of elasticity of the material and I the moment of inertia of the section about the bending axis. It is integrated twice and the constants come from the support conditions: zero deflection at pins and rollers, plus zero slope at a fixed support. The product E·I is the flexural rigidity: doubling it halves the deflection. This theory leaves out shear deformation, which only matters in short, deep beams.
The bending moment produces normal stresses that vary linearly with the distance y from the neutral axis, and the shear force produces shear stresses, largest at the neutral axis:
σ = M·y / I τ = V·Q / (I·t)
Q is the first moment of the area above the point studied and t the thickness at that point (the web, in an I-shape). The largest normal stress is at the outermost fiber, c: σmax = M·c/I = M/S, where S = I/c is the section modulus. To compare with yielding when σ and τ act together, criteria such as von Mises or Tresca are used.
A 6 m beam with a pin at x = 0 and a roller at x = 6 m, a uniform load w = 10 kN/m over its whole length, steel IPE 300 section (I = 8356 cm⁴ = 83.56·10⁶ mm⁴, depth 300 mm) with E = 200 GPa and yield strength Fy = 250 MPa.
You can reproduce it in the calculator: length 6 m, pin at 0, roller at 6, uniform distributed load of 10 kN/m from 0 to 6 m, steel with E = 200 GPa and an IPE 300 section. It should give the same reactions, the same maximum moment and the same deflection.
Results are approximate and meant for learning and checking exercises; they do not replace a check against the applicable design code.
The theory in this guide is that of any strength of materials course, for example: J. M. Gere and B. J. Goodno, Mechanics of Materials (Cengage); R. C. Hibbeler, Mechanics of Materials (Pearson); F. P. Beer, E. R. Johnston, J. T. DeWolf and D. F. Mazurek, Mechanics of Materials (McGraw-Hill). Simuveliq is not affiliated with these authors or publishers.
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