INGENIA

MEC-35

Cantilever bending energy

U = P² L³ /(6 EI) for a tip-loaded cantilever. Castigliano δ = ∂U/∂P.

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EnergyCastigliano

Governing equation

U=P2L36EI,δ=PL33EIU=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI}

P∈[0.2,40]P \in \left[0.2, 40\right]

where

P
Tip load (kN)
L
Length (m)
E
Modulus (GPa)
I
Inertia (cm^4)
U
Stored energy (J)
\delta
Tip deflection (mm)

The figure, part by part

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MEC-35 · beam
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①

Governing law — red

U=P2L36EI,δ=PL33EIU=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI}

P∈[0.2,40]P \in \left[0.2, 40\right]

P=0.2P = 0.2U=0.072U = 0.072

②

Domain of the variable — blue

Pmin⁡=0.2,Pmax⁡=40P_{\min} = 0.2,\quad P_{\max} = 40

P∈[0.2,40]P \in \left[0.2, 40\right]

P=0.2P = 0.2P=0.2P = 0.2

③

Response curve — green

U=P2L36EI,δ=PL33EIU=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI}

P∈[0.2,40]P \in \left[0.2, 40\right]

P=0.2P = 0.2U=0.072U = 0.072

④

Rate of change — amber

dUdP\frac{\mathrm{d}U}{\mathrm{d}P}

P∈[0.2,40]P \in \left[0.2, 40\right]

P=0.45P = 0.45U′=0.36U^{\prime} = 0.36

Lecture brief

Historical brief

Machine design grew from Coulomb torsion and Hertz contact (1881) through Soderberg fatigue and the heat-engine cycle. The lab writes shaft, bearing, contact and thermodynamic limits in SI. This sheet (MEC-35 — Cantilever bending energy) is the form associated with Castigliano. Working symbols: PP, LL, EE, II →\rightarrow UU, δ\delta. Bending energy is ∫ M² dx / 2EI. For M = P(L−x) the integral is P²L³/6EI, so δ = PL³/3EI.

Purpose

Purpose: compute UU, δ\delta from PP, LL, EE, II in Mechanical via U=P2L36EI,δ=PL33EIU=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI} U = P² L³ /(6 EI) for a tip-loaded cantilever. Castigliano δ = ∂U/∂P. Use it when a real mechanical question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given P=5.000 kNP = 5.000\,\mathrm{kN}, L=1.200 mL = 1.200\,\mathrm{m}, E=200.000 GPaE = 200.000\,\mathrm{GPa}, I=800.000 cm4I = 800.000\,\mathrm{cm^4}, the governing relation U=P2L36EI,δ=PL33EIU=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI} yields U=4.50 JU = 4.50\,\mathrm{J}, δ=1.800 mm\delta = 1.800\,\mathrm{mm}. A cantilever, a tip load, a stored-energy bar. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • Stored energy U4.50 J
  • Tip deflection \delta1.800 mm
Reading speed

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MEC-35 · beam
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Narration of this film

A cantilever, a tip load, a stored-energy bar.

Bending energy is ∫ M² dx / 2EI. For M = P(L−x) the integral is P²L³/6EI, so δ = PL³/3EI.

Reading speed

Watch on YouTube