Engineering · Simulator 094
A beam bends under load. Choose how it is supported, where the loads go, how long it is, what it is made of and the shape of its cross-section. See the deflected shape, the shear force and bending moment along the beam, the peak stress, and whether the sag stays within a typical limit.
In the Euler–Bernoulli model, valid for small deflections, the curvature of the beam is proportional to the bending moment M(x):
EI d²v/dx² = M(x)
E is the material's Young's modulus and I the cross-section's second moment of area; for a rectangle of width b and depth h, I = b h³ / 12. Depth matters most: doubling h makes the beam 8 times stiffer. The simulator finds M(x) from the loads and support reactions, then integrates twice, using the support conditions (zero deflection at both supports, or zero deflection and slope at a fixed end).
Standard results it reproduces:
simply supported, central load P: δ = P L³ / (48 E I)
simply supported, uniform load w: δ = 5 w L⁴ / (384 E I)
cantilever, end load P: δ = P L³ / (3 E I)
The largest bending stress is σ = M c / I with c = h/2. Building codes often limit live-load deflection to about span/360.
Linear elastic, small deflections, no self-weight unless you add it as a uniform load, no shear deformation (accurate for slender beams), and no buckling or yielding checks beyond the stress readout.
A timber joist (E = 11 GPa), 50 mm wide, spans 3.6 m and carries 1.5 kN/m. How deep must it be so that its deflection is below span/360?
The limit is 3600/360 = 10 mm. δ = 5wL⁴/(384EI) ≤ 0.010 m gives I ≥ 5 × 1500 × 3.6⁴ / (384 × 11×10⁹ × 0.010) ≈ 2.97×10⁻⁵ m⁴. With b = 0.05 m, h³ ≥ 12I/b ≈ 7.1×10⁻³, so h ≥ 0.193 m: about a 200 mm deep joist. Check it in the simulator.