Browser-local interactive workspace
Interactive workspace initializes in your browser. Engineering method, assumptions, validation, and references are available below.

Engineering reference

Statics & Strength: theory, method, and sources

This mechanical workspace publishes 10 governing equations, 6 stated assumptions, 9 documented boundaries, 1 worked example, and 1 source so the numbers it returns can be checked rather than taken on trust.

Calculations run locallyContent reviewed August 11, 2026Calculation & source methodology

How this tool works

The modules follow the order a first solid-mechanics course does. Forces are resolved and a free body is checked for equilibrium, that equilibrium gives the support reactions, the internal shear and moment are traced along the member, and only then are those internal actions turned into stresses and checked against a failure criterion.

Section properties and the chosen material feed the later modules directly, so changing the flange thickness or switching from steel to aluminium propagates through the bending stress, the deflection estimate, and the buckling capacity without re-entering anything.

Each module opens with the question it answers, the governing relationship, and the specific thing worth noticing in the numbers, rather than presenting a bare calculator.

Calculators and topics covered

  • statics
  • strength of materials
  • beams
  • stress
  • buckling
  • Mohr circle
  • pressure vessel
  • shear force and bending moment diagram
  • beam reaction calculator
  • method of joints truss
  • centroid and moment of inertia
  • bending stress My/I
  • transverse shear VQ/It
  • torsion of circular shafts

Core equations

R=Fx2+Fy2,  θ=atan2(Fy,  Fx)R = \sqrt{\sum Fx^{2} + \sum Fy^{2}},\; \theta = \operatorname{atan2} \left(\sum Fy,\; \sum Fx\right)yˉ=AyˉA,  I=(Iˉ+Ad2)\bar{y} = \frac{\sum A \bar{y}}{\sum A},\; I = \sum \left(\bar{I} + Ad^{2}\right)dMdx=V,  so  the  moment  peaks  where  the  shear  crosses  zero\frac{dM}{dx} = V,\; so\; the\; \text{moment}\; \text{peaks}\; \text{where}\; the\; \text{shear}\; \text{crosses}\; \text{zero}σ=MyI,  τ=VQIt,  δ=PLAE\sigma = \frac{My}{I},\; \tau = \frac{VQ}{It},\; \delta = \frac{PL}{AE}τ=TrJ,  φ=TLJG,  J=π(do4di4)32\tau = \frac{Tr}{J},\; \varphi = \frac{TL}{JG},\; J = \frac{\pi \left(do^{4} - di^{4}\right)}{32}σ=EαΔT  for  a  fully  restrained  member\sigma = - E\alpha \Delta T\; for\; a\; \text{fully}\; \text{restrained}\; \text{member}σ1,  2=σx+σy2±(σxσy2)2+τxy2\sigma _{1},\; 2 = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^{2} + \tau xy^{2}}Pcr=π2EI(KL)2,  σcr=π2E(KLr)2Pcr = \frac{\pi ^{2} EI}{\left(KL\right)^{2}},\; \sigma cr = \frac{\pi ^{2} E}{\left(\frac{KL}{r}\right)^{2}}σhoop=prt  and  σlong=pr2t  for  a  thinwalled  cylinder\sigma hoop = \frac{pr}{t}\; and\; \sigma long = \frac{pr}{2 t}\; for\; a\; \text{thin} - \text{walled}\; \text{cylinder}FoS=strengthapplied  stress,  margin=FoS1FoS = \frac{\text{strength}}{\text{applied}}\; \text{stress},\; \text{margin} = FoS - 1

Worked examples

Simply supported floor beam under a uniform load

A 6 m simply supported beam carries a uniformly distributed load of 12 kN/m over its full span, which is a typical factored floor loading. The beam is a 200 mm × 400 mm rectangular section bending about its strong axis. The task is to find the design shear and moment, then check the bending stress against the section.

Inputs
  • Span L = 6 m, simply supported
  • Uniform load w = 12 kN/m over the full span
  • Rectangular section 200 mm wide × 400 mm deep
Method
  1. Total load is wL = 72 kN. By symmetry each reaction carries half: R_A = R_B = 36 kN.
  2. Shear is maximum at the supports and equals the reaction there: V_max = 36 kN at x = 0.
  3. Shear passes through zero at midspan, x = 3.000 m, which is where the moment peaks.
  4. The workbench reports M_max = 54.000 kN·m at x = 3.000 m.
  5. For the section, I = bh³/12 = 0.2 × 0.4³/12 = 1.0667 × 10⁻³ m⁴ and the section modulus is S = I/c = 5.3333 × 10⁻³ m³.
  6. Bending stress is then σ = M/S = 54.0 × 10³ / 5.3333 × 10⁻³ = 10.125 MPa.

Result: Design actions are V_max = 36 kN at the supports and M_max = 54 kN·m at midspan, producing an extreme-fibre bending stress of 10.125 MPa in the 200 × 400 mm section.

Interpretation: Both actions match the standard closed-form results exactly: wL/2 = 36 kN and wL²/8 = 54 kN·m. That agreement is the point of running the case — it confirms the numerical shear and moment integration reproduces the textbook result before the tool is trusted on a load case with no closed form, such as several point loads combined with a partial-length distributed load. Note the stress figure assumes the full rectangular section is effective and the beam is restrained against lateral-torsional buckling; an unrestrained deep section will fail well below its bending capacity.

Method and assumptions

Assumptions

  • Linear-elastic, homogeneous, isotropic material behaviour below the yield point.
  • Small deflections, so equilibrium is written on the undeformed geometry.
  • Beams are prismatic and slender enough for Euler–Bernoulli theory, meaning plane sections stay plane and shear deformation is neglected.
  • Bending is about a principal axis of the section, with no unsymmetric bending, torsion coupling, or lateral-torsional instability.
  • Torsion applies to circular shafts only, where cross-sections do not warp.
  • Loads are static. Fatigue, impact, and creep are outside the scope of these modules.

Limitations and design boundaries

  • Beam analysis covers statically determinate members only: a simple span or a cantilever. Continuous and propped beams are statically indeterminate and need compatibility conditions this tool does not solve.
  • Deflection uses the standard closed-form cases. For a load set that matches none of them the tool says so rather than returning a plausible wrong number.
  • The truss module solves one pin joint with two unknown members, which is the teaching case. It does not assemble or solve a whole truss.
  • Transverse shear stress is given for a solid rectangle. Flanged sections concentrate shear in the web and need the section-specific Q.
  • Column capacity is the theoretical Euler and Johnson result with no imperfection, eccentricity, or code-based resistance factor. Real design values are lower.
  • Pressure vessel results use the thin-wall formulas, with the Lamé thick-wall values shown alongside. Below r/t of about 10 the tool says the thin-wall figure is unreliable rather than presenting it as the answer.
  • The ultimate strength offered in the factor-of-safety module is estimated at 1.5 times yield. Substitute the certified value before relying on it.
  • Combined loading superposes axial, bending, and torsional stress at a surface point. It does not account for stress concentrations at holes, fillets, or keyways, which is often what actually governs.
  • Nothing here is a substitute for a design code or for review by a licensed engineer.

Sources and references

Primary sources are preferred for ratings, standards, manufacturer data, and externally defined constants.

Source policy
  • Standard mechanics-of-materials formulationsThe relationships used are the conventional ones from an introductory solid-mechanics sequence: equilibrium, the parallel-axis theorem, Euler–Bernoulli beam theory, the elastic torsion formula, plane-stress transformation, and Euler buckling with the Johnson parabola for short columns.