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Cantilever Beams

Reference data and engineering information about cantilever beams for material properties applications.

cantileverbeams

Overview

Engineering reference data for Cantilever Beams in material science and properties.

Key Formulas

Stress

σ=FA\sigma = \frac{F}{A}

Force per unit area.

Strain

ε=ΔLL0\varepsilon = \frac{\Delta L}{L_0}

Change in length per original length.

Hooke's Law

σ=Eε\sigma = E \varepsilon

Stress proportional to strain in elastic region.

Thermal Expansion

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

Length change due to temperature.

Variables

SymbolDescriptionUnit
σ\sigmaStressPa
ε\varepsilonStrain—
EEYoung's modulusPa
α\alphaThermal expansion coefficient1/°C
ΔT\Delta TTemperature change°C

Cantilever Beam - Single Load at an Intermediate Point

This case describes a cantilever beam with a single point load (F) applied at a distance a from the fixed end.

Maximum Reaction Force at the fixed end (A):

RA=FR_A = F

Maximum Bending Moment at the fixed end (A):

Mmax=MA=−F⋅aM_{max} = M_A = -F \cdot a

Maximum Deflection at the free end (C):

δC=Fa33EI(1+3b2a)\delta_C = \frac{F a^3}{3 E I} \left(1 + \frac{3b}{2a}\right)

where b is the length of the overhang from the load point to the free end.

Deflection under the load point (B):

δB=Fa33EI\delta_B = \frac{F a^3}{3 E I}

Maximum Bending Stress:

σmax=ymax⋅F⋅aI\sigma_{max} = \frac{y_{max} \cdot F \cdot a}{I}

Cantilever Beam - Uniform Distributed Load

This case describes a cantilever beam with a constant load intensity q distributed along its entire length L.

Maximum Reaction Force at the fixed end (A):

RA=q⋅LR_A = q \cdot L

Maximum Bending Moment at the fixed end (A):

MA=−qL22M_A = -\frac{q L^2}{2}

Maximum Deflection at the free end (B):

δB=qL48EI\delta_B = \frac{q L^4}{8 E I}

Cantilever Beam - Declining Distributed Load (Triangular Load)

This case describes a cantilever beam with a load that linearly decreases from a maximum value q at the fixed end to zero at the free end.

Maximum Reaction Force at the fixed end (A):

RA=qL2R_A = \frac{q L}{2}

Maximum Bending Moment at the fixed end (A):

Mmax=MA=−qL26M_{max} = M_A = -\frac{q L^2}{6}

Maximum Deflection at the free end (B):

δB=qL430EI\delta_B = \frac{q L^4}{30 E I}

Principle of Superposition

For a cantilever beam subjected to multiple point loads and/or distributed loads simultaneously, the resulting internal forces (like bending moment) and deflections can be found by summing the effects of each individual load case acting separately. This is valid under the assumption of small deformations and linear elastic material behavior.

Cantilever Beam - Single Load at the End

The most fundamental cantilever beam configuration features a point load applied at the free end. This case serves as the basis for deriving many other cantilever beam solutions.

Maximum Reaction Force

The reaction force at the fixed support equals the applied load:

R_A = F \tag{1a}

where:

  • RAR_A = reaction force at support A (N, lb)
  • FF = applied point load at the free end B (N, lb)

Maximum Bending Moment

The maximum moment occurs at the fixed end:

M_{max} = M_A = -F \cdot L \tag{1b}

where:

  • MAM_A = maximum moment at support A (Nm, Nmm, lb·in)
  • LL = length of the cantilever beam (m, mm, in)

Maximum Deflection

The maximum deflection occurs at the free end:

\delta_B = \frac{F \cdot L^3}{3 \cdot E \cdot I} \tag{1c}

where:

  • δB\delta_B = maximum deflection at point B (m, mm, in)
  • EE = modulus of elasticity (Pa, N/mm², psi)
  • II = moment of inertia (m⁴, mm⁴, in⁴)

Bending Stress

The bending stress at any point in the cross-section is:

\sigma = \frac{y \cdot M}{I} \tag{1d}

where:

  • σ\sigma = bending stress (Pa, N/mm², psi)
  • yy = distance from the neutral axis to the point of interest (m, mm, in)
  • MM = bending moment at that section (Nm, lb·in)
  • II = moment of inertia (m⁴, mm⁴, in⁴)

The maximum stress occurs at the fixed end where the moment is maximum, and at the extreme fiber where y=ymaxy = y_{max}:

\sigma_{max} = \frac{y_{max} \cdot F \cdot L}{I} \tag{1e}

Example — Cantilever Beam with Single Load at the End (Metric Units)

Given:

  • Beam: UB 305 × 127 × 42 steel section
  • Length: L=5000 mmL = 5000\ \text{mm}
  • Moment of inertia: I=8196 cm4=8.196×107 mm4I = 8196\ \text{cm}^4 = 8.196 \times 10^7\ \text{mm}^4
  • Modulus of elasticity: E=200 GPa=2×105 N/mm2E = 200\ \text{GPa} = 2 \times 10^5\ \text{N/mm}^2
  • Applied load: F=3000 NF = 3000\ \text{N}

Maximum bending moment at the fixed end:

Mmax=3000 N×5000 mm=1.5×107 N⋅mm=1.5×104 N⋅mM_{max} = 3000\ \text{N} \times 5000\ \text{mm} = 1.5 \times 10^7\ \text{N·mm} = 1.5 \times 10^4\ \text{N·m}

Maximum deflection at the free end:

δB=3000×500033×2×105×8.196×107=7.6 mm\delta_B = \frac{3000 \times 5000^3}{3 \times 2 \times 10^5 \times 8.196 \times 10^7} = 7.6\ \text{mm}

Maximum bending stress:

For a UB 305 section with beam height 300 mm, ymax=150 mmy_{max} = 150\ \text{mm}:

σmax=150×3000×50008.196×107=27.4 N/mm2=27.4 MPa\sigma_{max} = \frac{150 \times 3000 \times 5000}{8.196 \times 10^7} = 27.4\ \text{N/mm}^2 = 27.4\ \text{MPa}

Note: The calculated maximum stress of 27.4 MPa is well below the ultimate tensile strength for most structural steels (typically 400–550 MPa), indicating significant safety margin in this application.

Example - Cantilever Beam with Single Load at the End (Metric Units)

Consider a UB 305 x 127 x 42 steel flange cantilever beam with the following properties:

  • Length (LL): 5000 mm
  • Moment of Inertia (II): 8196 cm⁴ = 81,960,000 mm⁴
  • Modulus of Elasticity (EE): 200 GPa = 200,000 N/mm²
  • Applied Load (FF): 3000 N at the free end
  • Beam Height: 300 mm (so distance to extreme fiber ymaxy_{max} = 150 mm)

Maximum Bending Moment at Fixed End: Mmax=−F⋅L=−(3000 N)(5000 mm)=−1.5×107 N⋅mm=−15,000 N⋅mM_{max} = -F \cdot L = -(3000 \text{ N})(5000 \text{ mm}) = -1.5 \times 10^7 \text{ N}\cdot\text{mm} = -15,000 \text{ N}\cdot\text{m}

Maximum Deflection at Free End:

## Bending Stress in Cantilever Beams The stress in a bending beam can be expressed as: $$\sigma = \frac{y \cdot M}{I}$$ where: - $\sigma$ = stress (Pa, N/mm^{2}, psi) - $y$ = distance to point from neutral axis (m, mm, in) - $M$ = bending moment (Nm, lb·in) - $I$ = moment of inertia (m^{4}, mm^{4}, in^{4}) The maximum moment in a cantilever beam occurs at the fixed support. By combining the stress formula with the moment equations, maximum stress can be calculated directly. ### Maximum Stress - Single Load at the End $$\sigma_{max} = \frac{y_{max} \cdot F \cdot L}{I}$$ ### Maximum Stress - Single Load at Intermediate Point $$\sigma_{max} = \frac{y_{max} \cdot F \cdot a}{I}$$ where $a$ = distance from fixed support to load location (m, mm, in). ## Important Properties - The **maximum bending moment** in a cantilever beam always occurs at the **fixed support**, regardless of load position - The **maximum stress** occurs at the extreme fiber (outermost point) where $y = y_{max}$ - For symmetric cross-sections, $y_{max}$ equals half the beam height ($h/2$) ## Design Considerations For the example UB 305 × 127 × 42 steel beam: - Beam height: 300 mm → $y_{max}$ = 150 mm - The calculated maximum stress (27.4 MPa) is well below the ultimate tensile strength of structural steel (typically 400–550 MPa), indicating adequate safety margin ## References - [Original Source](https://www.engineeringtoolbox.com/cantilever-beams-d_1848.html)
Source: engineeringtoolbox.com