Choose the model before entering numbers
A simply supported beam has an ideal pin and roller; both ends can rotate. A cantilever has one fully fixed end and one free end. The calculator supports a central point load for the first case and a free-end point load for the second. These load locations are part of the model.
The small-deflection relations below follow Euler–Bernoulli bending theory, as developed in MIT OpenCourseWare, Deflections due to Bending (PDF). Here P is load, L is span, E is Young’s modulus and I is second moment of area about the bending axis.
| Case | Deflection δ | Moment M |
|---|---|---|
| Simply supported, central P | PL³/(48EI) | PL/4 |
| Cantilever, free-end P | PL³/(3EI) | PL |
Worked example: a 1,000 mm span
Use P = 1,000 N, L = 1,000 mm, E = 200 GPa and I = 1,000,000 mm⁴. These are illustrative inputs. Convert E to 200,000 N/mm² before substituting into a formula using millimetres.
δ = (1,000 × 1,000³)/(48 × 200,000 × 1,000,000)
= 0.10416667 mm at midspan.
Each support reaction = P/2 = 500 N upward.
Change only the support/load case to a cantilever: δ = 1.66666667 mm at the free end. The fixed support carries a 1,000 N upward reaction and a balancing moment of 1,000,000 N·mm. The sixteenfold deflection difference comes from 48/3; it is not a change in material stiffness.
Deflection and stress answer different questions
If the elastic section modulus is Z = 20,000 mm³, nominal maximum bending stress is M/Z. The simply supported example gives 250,000/20,000 = 12.5 MPa; the cantilever gives 50 MPa. The relation follows from Z = I/c and the elastic bending equation in MIT’s Stresses: Beams in Bending (PDF).
Use I and elastic Z about the same centroidal principal axis. If the extreme-fibre distances differ, the smaller Z governs the largest stress magnitude. A plastic section modulus does not belong in this elastic calculation.
Common questions
What happens if the span doubles?
With the same point load and EI, deflection increases eightfold because it scales with L³. Maximum moment doubles. Changing the section or load at the same time changes that comparison.
Can I substitute total distributed load for P?
No. A uniformly distributed load has a different moment and deflection shape. Self-weight is also excluded from these two point-load cases.
Does a small result mean the beam is adequate?
No. This model assumes a straight, slender, prismatic, linear-elastic beam with constant EI and small rotations. Shear deformation, flexible supports, torsion, instability and local contact effects are omitted. Material strength and permissible deflection require separate checks.
Related: stress-unit conversion or all engineering guides.