Beam Bending (Euler-Bernoulli)

The Euler-Bernoulli equation

For a slender beam bending under a distributed load w(x), the deflection y(x) satisfies:

EI · d⁴y/dx⁴ = w(x)

where:
  E = Young's modulus  (Pa)
  I = second moment of area  (m⁴)
  w(x) = transverse load per unit length  (N/m)
  y(x) = deflection (m)

Integrating twice gives the bending moment, twice more gives the deflection, with four constants of integration set by boundary conditions. The bending stress at a point is then σ = M·c / I, where c is the distance from the neutral axis.

The theory rests on three assumptions: cross-sections that were plane before bending stay plane after bending, those sections stay perpendicular to the deformed neutral axis (so shear deformation is ignored), and the material is linearly elastic. Together these collapse the beam from a 3D solid into a 1D problem in y(x).

Boundary conditions

	Cantilever	Simply supported	Fixed-fixed	Propped cantilever	Overhanging	Continuous (multi-span)
End A	Fixed (clamped)	Pin (no moment)	Fixed	Fixed	Pin	Pin
End B	Free	Roller	Fixed	Pin / roller	Roller w/ overhang	Pin (intermediate)
Static determinacy	Determinate	Determinate	Indeterminate (×2)	Indeterminate (×1)	Determinate	Indeterminate
Tip deflection (point P at midspan)	PL³/(3EI) at tip	PL³/(48EI)	PL³/(192EI)	0.0098·PL³/EI	varies by overhang	varies by spans
UDL midspan deflection	wL⁴/(8EI) at tip	5wL⁴/(384EI)	wL⁴/(384EI)	wL⁴/(185EI)	varies	varies
Max moment	wL²/2 at fixed end	wL²/8 at midspan	wL²/12 at ends	wL²/8 (fixed end)	varies	varies
Use case	Diving boards, balconies	Most simple beams	Slabs in moment frames	Continuous girder over a pier	Cantilever overhangs	Multi-span bridges

Worked example: cantilever tip deflection

A steel cantilever beam of length 2 m supports a 10 kN point load at its tip. The cross-section has second moment of area I = 4 × 10⁻⁵ m⁴. How much does the tip deflect?

δ = FL³ / (3EI)

  F = 10 kN     = 10,000 N
  L = 2 m       → L³ = 8 m³
  E = 200 GPa  = 2 × 10¹¹ Pa
  I = 4 × 10⁻⁵ m⁴

EI = 2 × 10¹¹ × 4 × 10⁻⁵
   = 8 × 10⁶ N·m²

δ = (10,000 × 8) / (3 × 8 × 10⁶)
  = 80,000 / 2.4 × 10⁷
  = 3.33 × 10⁻³ m
  = 3.3 mm

The tip drops 3.3 mm under load. To halve that deflection, doubling I (e.g. choosing the next size up in I-beams) is the cheapest path, because doubling E means switching from steel to a stiffer material — there are very few materials stiffer than steel that aren't exotic.

Worked example: bending stress in the same beam

What's the maximum bending stress, and is the beam safe? Take the cross-section depth as 0.2 m, so c = 0.1 m (distance from neutral axis to extreme fibre):

Maximum moment occurs at the fixed end:
  M_max = F × L = 10,000 × 2 = 20,000 N·m

Maximum bending stress:
  σ_max = M·c / I
        = 20,000 × 0.1 / (4 × 10⁻⁵)
        = 2,000 / 4 × 10⁻⁵
        = 5.0 × 10⁷ Pa
        = 50 MPa

Steel yield (typical S275): 275 MPa
Safety factor: 275 / 50 = 5.5

The beam is well below yield. In practice, code-prescribed safety factors and load combinations would take the design moment to perhaps 1.4 × 20 kN·m = 28 kN·m, still leaving stresses around 70 MPa — comfortably elastic.

Calculating I for common shapes

• Rectangle (b × h): I = b·h³ / 12. Doubling h gives 8× the I.
• Solid round (radius r): I = π·r⁴ / 4.
• Hollow round tube: I = π(r₀⁴ − r_i⁴) / 4. Most material near the outer fibre, so very efficient in bending — bicycle frames, scaffold poles.
• I-beam (idealised): I ≈ A_f · d² / 2 + I_web, dominated by the flange contribution if flanges are thin.
• Channel / angle: I depends on orientation; can be 5–10× different about the two axes.

Load types and their consequences

• Point load (P): sharp moment peak at the load location; deflection scales as PL³/EI.
• Uniformly distributed load (UDL, w): smooth parabolic moment diagram; deflection scales as wL⁴/EI.
• Triangular / varying distributed: integrate w(x) numerically or use influence-line tables.
• Pure moment at end (M): constant moment along the beam; deflection scales as ML²/EI.
• Multiple loads: superpose individual cases (linearity holds in elastic theory).
• Self-weight: always treat as a UDL of the beam's mass per unit length × g.

When Euler-Bernoulli breaks down

• Short, deep beams (L/d < 10). Shear deformation is no longer negligible. Switch to Timoshenko beam theory, which adds a shear term GA·κ to the stiffness matrix.
• Large deflections (δ/L > ~0.1). Geometry changes during loading invalidate the small-deflection assumption. Use elastica theory or nonlinear finite element analysis.
• Plastic hinges form. Past first yield, parts of the cross-section yield and the beam softens; use plastic limit analysis with M_p = σ_y × Z (plastic section modulus).
• Composite or layered beams. If shear lag between layers matters (sandwich panels), partial-interaction theory replaces the single-EI assumption.
• Buckling under axial load. A bending beam also under axial compression can buckle laterally before reaching its bending capacity — covered by separate column buckling theory.

Common failure modes

• Yield at extreme fibre. The most basic case — bending stress σ = M·c/I exceeds yield strength σ_y at the most highly loaded section. Design check: σ_max ≤ σ_y / safety factor.
• Lateral-torsional buckling (LTB). Compression flange of an I-beam buckles sideways out of plane; happens before yield in long unrestrained beams. Brace the compression flange or use a closed section.
• Shear yield in the web. For short, heavily-loaded beams, the web can yield in shear before the flanges yield in bending. Add web stiffeners or thicker web plates.
• Fatigue at stress concentrations. Welded connections, holes for bolts, and notches concentrate stress. Cyclic loading drives crack growth from these initiation sites; design for stress range, not just peak stress.
• Flange local buckling. Thin compression flanges can wrinkle locally before global LTB or yield. Code limits on flange b/t ratios prevent this.
• Excessive deflection at service. Even an unyielded beam can fail serviceability if it sags more than allowed (typically L/250 or L/360 for floors). This often controls design over strength.